The solution presented on the blog is my personal solutions for the exercises in the book ‘What is Mathematics: An Elementary Approach To Ideas And Methods’ by Herbert Robbins and Richard Courant, please leave a comment if you spot any mistakes or you have questions on the solution. Thanks in advance!
Since \(2^n \rightarrow \infty\), prove as a consequence that \(\sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots + \sqrt{2}}}}\) (with \(n\) square roots) converges to 2 as \(n \rightarrow \infty\).
\[\begin{align*} x &= \sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots}}} \\ x^2 &= 2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots}}} \\ x^2 &= 2 + x \end{align*}\]And we can get x = 2.
Show that every field contains all the rational numbers at least. (Hint: If \(a \neq 0\) is a number in the field \(F\) , then \(a/a = 1\) belongs to F , and from 1 we can obtain any rational number by rational operations.)
Ex: From \(p=1+\sqrt{2}, q=2-\sqrt{2}, r=-3+\sqrt{2}\) obtain the numbers \(\frac{p}{q}, \quad p^2, \quad (p-p^2)\frac{q}{r}, \quad \frac{p+qr}{p-r}, \quad \frac{p+r}{q+pr}\) in the form \(a+b\sqrt{2}\).
\[\begin{aligned} \frac{p}{q} &= \frac{1+\sqrt{2}}{2-\sqrt{2}} \\ &= \frac{(1+\sqrt{2})(2+\sqrt{2})}{(2-\sqrt{2})(2+\sqrt{2})} \\ &= \frac{2+2\sqrt{2}+\sqrt{2}+2}{4-2} \\ &= \frac{4+3\sqrt{2}}{2} \\ &= 2 + \frac{3\sqrt{2}}{2} \end{aligned}\] \[\begin{aligned} p + p^2 &= (1+\sqrt{2}) + (1+\sqrt{2})^2 \\ &= (1+\sqrt{2} + 1 + 2\sqrt{2} + 2) \\ &= 4 + 3\sqrt{2} \end{aligned}\] \[\begin{aligned} (p-p^2)\frac{q}{r} &= \left[(1+\sqrt{2}) - (1+\sqrt{2})^2\right]\frac{2-\sqrt{2}}{-3+\sqrt{2}} \\ &= \left[1+\sqrt{2} - (1+2\sqrt{2} + 2)\right]\frac{(2-\sqrt{2})(\sqrt{2}+3)}{(-3+\sqrt{2})(\sqrt{2}+3)} \\ &= \left[-2-\sqrt{2}\right]\frac{2\sqrt{2}+6-2+3\sqrt{2}}{2-9} \\ &= \left[-2-\sqrt{2}\right]\frac{4-\sqrt{2}}{-7} \\ &= \frac{6}{7} + \frac{2\sqrt{2}}{7} \end{aligned}\] \[\begin{aligned} \frac{pqr}{1+r^2} &= \frac{(1+\sqrt{2})(2-\sqrt{2})(-3+\sqrt{2})}{1+(-3+\sqrt{2})^2} \\ &= \frac{\sqrt{2}(-3+\sqrt{2})}{1+9-6\sqrt{2}+2} \\ &= \frac{2-3\sqrt{2}}{12-6\sqrt{2}} \\ &= \frac{(2-3\sqrt{2})(12+6\sqrt{2})}{(12-6\sqrt{2})(12+6\sqrt{2})} \\ &= \frac{24+12\sqrt{2}-36\sqrt{2}-36}{144-72} \\ &= \frac{-12-24\sqrt{2}}{72} \\ &= -\frac{1}{6}-\frac{1}{3}\sqrt{2} \\ \end{aligned}\] \[\begin{aligned} \frac{p+qp}{q+pr^2} &= \frac{(1+\sqrt{2})+(2-\sqrt{2})(-3+\sqrt{2})}{(2-\sqrt{2})+(1+\sqrt{2})(-3+\sqrt{2})^2} \\ &= \frac{(1+\sqrt{2})+(-6+2\sqrt{2}+3\sqrt{2}-2)}{(2-\sqrt{2})+(1+\sqrt{2})(11-6\sqrt{2})} \\ &= \frac{-7+6\sqrt{2}}{(2-\sqrt{2})+(5\sqrt{2}-1)}\\ &= \frac{-7+6\sqrt{2}}{1+4\sqrt{2}} \\ &= \frac{(-7+6\sqrt{2})(1-4\sqrt{2})}{(1+4\sqrt{2})(1-4\sqrt{2})} \\ &= \frac{-7+28\sqrt{2}+6\sqrt{2}-48}{1-32} \\ &= \frac{-55+34\sqrt{2}}{-31} \\ &= \frac{55}{31}-\frac{34}{31}\sqrt{2}\\ \end{aligned}\]Represent \((\sqrt{k})^3 , \frac{1+(\sqrt{k})^2}{1+\sqrt{k}} , \frac{\sqrt{2}\sqrt{k}+\frac{1}{\sqrt{2}}}{(\sqrt{k})^3-3}, \frac{(1+\sqrt{k})(2-\sqrt{k})(\sqrt{2}+\frac{1}{\sqrt{k}})}{1+\sqrt{2}k}\) in the form \(p+q\sqrt{k}\).
\[(\sqrt{k})^3 = k\sqrt{k}\] \[\begin{aligned} \frac{1+(\sqrt{k})^2}{1+\sqrt{k}} &= \frac{(1+k)(1-\sqrt{k})}{(1+\sqrt{k})(1-\sqrt{k})} \\ &= \frac{1+\sqrt{k}+k-k\sqrt{k}}{1-k} \\ &= \frac{1+k}{1-k} - \frac{1+k}{1-k}\sqrt{k} \end{aligned}\] \[\begin{aligned} \frac{\sqrt{2}\sqrt{k}+\frac{1}{\sqrt{2}}}{(\sqrt{k})^3-3} &= \frac{(\sqrt{2}\sqrt{k} + \frac{\sqrt{2}}{2})(k \sqrt{k} + 3)}{(k \sqrt{k} - 3)(k \sqrt{k} + 3)} \\ &= \frac{k^2\sqrt{2} + 3\sqrt{2} \sqrt{k} + \frac{\sqrt{2}}{2} k \sqrt{k} + 2 \frac{\sqrt{2}}{2}}{k^3 - 9} \\ &= \frac{k^2\sqrt{2}}{k^3 - 9} + \frac{3\sqrt{2} + \frac{\sqrt{2}}{2} k}{k^3 - 9}\sqrt{k} \end{aligned}\] \[\begin{aligned} \frac{(1+\sqrt{k})(2-\sqrt{k})(\sqrt{2}+\frac{1}{\sqrt{k}})}{1+\sqrt{2}k} &= \frac{(2 - \sqrt{k}+ \sqrt{2k} - \sqrt{k})(\sqrt{2} + \frac{1}{\sqrt{k}})}{1 + \sqrt{2k}} \\ &= \frac{2 \sqrt{2} + \frac{2 \sqrt{k}}{k} + 2 \sqrt{k} - 1 + 2\sqrt{k} + \sqrt{2} - \sqrt{2}k - \sqrt{k}}{1 + \sqrt{2k}} \\ &= \frac{2 \sqrt{2} + \sqrt{k} - \sqrt{2}k - 1}{1 + \sqrt{2}k} + \frac{\frac{2}{k}-\sqrt{2}+1}{1+\sqrt{2}k}\sqrt{k} \end{aligned}\]If two segments of lengths $l$ and $a$ are given, give actual constructions for \(1 + a + a^2\), \(1 + a + a^2 + a^3\), \((1+a)^2\), and \((1-a)^2\), \(a^3\).
The solution presented on the blog is my personal solutions for the exercises in the book ‘What is Mathematics: An Elementary Approach To Ideas And Methods’ by Herbert Robbins and Richard Courant, please leave a comment if you spot any mistakes or you have questions on the solution. Thanks in advance!
Prove that \(\sqrt[3]{2}, \sqrt{3}, \sqrt{5}, \sqrt[3]{3}\) are not rational. (Hint: Use the lemma of p. 47).
(1). \(\sqrt[3]{2}\)
Assume \(\sqrt[3]{2}\) is rational:
\[\sqrt[3]{2} = \frac{p}{q} \quad \text{gcd}(p, q) = 1\] \[\begin{aligned} 2 = \frac{p^3}{q^3} \Rightarrow 2q^3 = p^3 \end{aligned}\]Based on the lemma, p divdes ab, therefore p divides a or b. since \(\text{gcd}(p, q) = 1\), Therefore
\[2 \left| p^3 \Rightarrow 2\right|p\]therefore p is some even number, and can be denoted as 2k
\[2q^3 = (2k)^3\] \[q^3 = 4k^3 \quad \text{and q} \quad \text{is even}, \quad \text{contradicting gcd}(p, q) = 1.\]Therefore \(\sqrt[3]{2}\) is irrational number.
(2). \(\sqrt{3}\)
Assume \(\sqrt{3}\) is rational:
\[\sqrt{3} = \frac{p}{q} \quad \text{gcd}(p, q) = 1\] \[\begin{aligned} 3 = \frac{p^2}{q^2} \Rightarrow 3q^2 = p^2 \end{aligned}\]Based on the lemma, p divdes ab, therefore p divides a or b. since \(\text{gcd}(p, q) = 1\), Therefore
\[3 \left| p^2 \Rightarrow 3\right|p\]therefore can be denoted as 3k
\[3q^2 = (3k)^2\] \[q^2 = 3k^2 \quad \Rightarrow 3\left|q^2 \Rightarrow 3\right|q, \quad \text{which contradicting gcd}(p, q) = 1.\]Therefore \(\sqrt{3}\) is irrational number.
(3). \(\sqrt{5}\)
Assume \(\sqrt{5}\) is rational:
\[\sqrt{5} = \frac{p}{q} \quad \text{gcd}(p, q) = 1\] \[\begin{aligned} 5 = \frac{p^2}{q^2} \Rightarrow 5q^2 = p^2 \end{aligned}\]Based on the lemma, p divdes ab, therefore p divides a or b. since \(\text{gcd}(p, q) = 1\), Therefore
\[5 \left| p^2 \Rightarrow 3\right|p\]therefore can be denoted as 3k
\[5q^2 = (5k)^2\] \[q^2 = 5k^2 \quad \Rightarrow 5\left|q^2 \Rightarrow 5\right|q, \quad \text{which contradicting gcd}(p, q) = 1.\]Therefore \(\sqrt{5}\) is irrational number.
(4). \(\sqrt[3]{3}\)
Assume \(\sqrt[3]{3}\) is rational:
\[\sqrt[3]{3} = \frac{p}{q} \quad \text{gcd}(p, q) = 1\] \[\begin{aligned} 3 = \frac{p^3}{q^3} \Rightarrow 3q^3 = p^3 \end{aligned}\]Based on the lemma, p divdes ab, therefore p divides a or b. since \(\text{gcd}(p, q) = 1\), Therefore
\[3 \left| p^3 \Rightarrow 2\right|p\]therefore p can be denoted as 3k
\[3q^3 = (3k)^3\] \[q^3 = 9k^3 \quad \Rightarrow 9\left|q^2 \Rightarrow 9\right|q\]if a number can divides 9, it can also divdes 3, which contradicting with \(\text{gcd}(p,d)=1\).
Therefore \(\sqrt[3]{3}\) is irrational number.
Prove that \(\sqrt{2} + \sqrt{3}\) and \(\sqrt{2} + \sqrt[3]{2}\) are not rational. (Hint: If e.g. the first of these numbers were equal to a rational number ( r ), then, writing \(\sqrt{3} = r - \sqrt{2}\) and squaring, \(\sqrt{2}\) would be rational.)
(1). \(\sqrt{2} + \sqrt{3}\) If \(r = \sqrt{2} + \sqrt{3}\) is rational:
\[\sqrt{3} = r - \sqrt{2}, \quad \sqrt{3} \quad \text{is irrational}.\]squaring both side, we have
\[\begin{aligned} 3 &= (r - \sqrt{2})^2 \\ 3 &= r^2 - 2r\sqrt{2} + 2 \\ 1 &= r^2 - r\sqrt{2} \\ 1 &= r(r - 2\sqrt{2}) \end{aligned}\]1 is rational number, and we assume r is rational number, by subtracting an irrational number, it is impossible to gain 1. Therefore it contradicts the assumption, r should be irrational.
(2). \(\sqrt{2} + \sqrt[3]{2}\)
If \(\sqrt{2} + \sqrt[3]{2}\) is rational:
Prove that \(\sqrt{2} + \sqrt{3} + \sqrt{5}\) is irrational. Try to make up similar and more general examples.
If \(\sqrt{2} + \sqrt{3} + \sqrt{5}\) is rational,
\[\begin{aligned} r = \sqrt{2} + \sqrt{3} + \sqrt{5} (r - \sqrt{5}) &= \sqrt{2} + \sqrt{3} \\ (r - \sqrt{5})^2 &= (\sqrt{2} + \sqrt{3})^2 \\ r^2 - 2r\sqrt{5} + 5 &= 2 + 2\sqrt{6} + 3 \\ r^2 &= 2\sqrt{6} + 2\sqrt{5}r \\ 2\sqrt{6} &= r^2 - 2\sqrt{5}r \\ 24 &= r^4 - 6\sqrt{5}r^3 + 20r^2 \\ \end{aligned}\]24 is rational, but it contains \(6(\sqrt{5})\) on the right side, this is absurdly impossible.
Therefore, \(\sqrt{2} + \sqrt{3} + \sqrt{5}\) is irrational.
Calculate \(\sqrt[3]{2}\) and \(\sqrt[3]{5}\) with an accuracy of at least \(10^{-2}\)
By applying Newton’s Method
\[\begin{aligned} x &= \sqrt[3]{2} \\ x^3 &= 2 \\ \end{aligned}\] \[f(x) = x^3 - 2\] \[x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}= 1 - \frac{1^3 - 2}{3\cdot 1^2} = 1.33\] \[x_2 = x_1 - \frac{f(x_1)}{f'(x_1)}= 1.33 - \frac{1.33 - 2}{3\cdot (1.33)^2} \approx 1.26\]By using Interval (binary search).
\[a < \sqrt[3]{5} < b, \quad a = 1, \quad b = 2\] \[c = \frac{a+b}{2} = 1.5\] \[(1.5)^3 < 5\]Therefore it should be inside interval \(1.5 \sim 2\)
\[\frac{1.5 + 2}{2} = 1.75\] \[(1.75)^3 = 5.3579 > 5\]Therefore it should be inside interval \(1.5 \sim 1.75\)
\[1.75 + 1.5 = 1.625\] \[(1.625)^3 \approx 4.29 < 5\]Therefore it should be inside interval \(1.625 \sim 1.75\)
\[(\frac{1.625 + 1.75}{2})^3 \approx 4.80 < 5\] \[\therefore \text{between } 1.625 \text{ and } 1.75\]Prove that \(1 - q + q^2 - q^3 + q^4 - \cdots = \frac{1}{1 + q}, \quad \text{if} \quad |q| < 1.\)
\[\begin{align*} S_n &= 1 - q + q^2 - q^3 + q^4 - \cdots \\ qS_n &= q - q^2 + q^3 - q^4 + \cdots \\ S_n + qS_n &= 1 + (-1)^{n}q^n \to \infty \\ \end{align*}\]\(q^n\) will be 0 if n goes to infinity.
\[\begin{align*} S_n(1+q) &= 1 \\ S_n &= \frac{1}{1+q} \end{align*}\]What is the limit of the sequence \(a_1, a_2, a_3, \cdots\), where \(a_n = n/(n + 1)\)? (Hint: Write the expression in the form \(\frac{n}{n + 1} = 1 - \frac{1}{n + 1}\) and observe that the second term tends to zero.)
\[\begin{align*} a_n &= \left(\frac{1}{n+1}\right) = 1 - \frac{n+1}{n} \text{, the second term tends to 0} \\ & = 0 \end{align*}\]What is the limit of \(\frac{n^2 + n + 1}{n^2 - n + 1} \quad \text{for} \quad n \to \infty \quad ?\) (Hint: Write the expression in the form \(\frac{1 + \frac{1}{n} + \frac{1}{n^2}}{1 - \frac{1}{n} + \frac{1}{n^2}}\) .)
\[\begin{align*} \text{Divide by } n^2 \text{ for all terms,} \\ \frac{1 + \frac{1}{n} + \frac{1}{n^2}}{1 - \frac{1}{n} + \frac{1}{n^2}} & \to 1 \end{align*}\]Prove, for ( |q| < 1 ), that \(1 + 2q + 3q^2 + 4q^3 + \cdots = \frac{1}{(1 - q)^2}.\) (Hint: Use the result of exercise 3 on p. 18.)
\[\begin{align} S&=1+2q+3q^2+\qquad\cdots\qquad \qquad+nq^{n-1}\\ qS&=\qquad q+2q^2+3q^3+\cdots +\quad(n-1)q^{n-1}+nq^n \\ \text{Subtracting,}&\\ (1-q)S&=1+\;\ q \ +\ q^2 +\ q^3+\cdots \qquad \qquad +q^{n-1}-nq^n\\ &=\frac {\;\ 1-q^n}{1-q}-nq^n\\ S&=\frac{1-q^n-nq^n(1-q)}{(1-q)^2}\\ &=\frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2}\qquad \end{align}\]By analyzing the sum, \(\frac{(n+1)q^n}{(1-q)^2}\) and \(\frac{nq^{n+1}}{(1-q)^2}\) tend to 0, and \(\frac{1}{(1-q)^2}\) is the term left.
Q.E.D
What is the limit of the infinite series \(1 - 2q + 3q^2 - 4q^3 + \cdots \quad ?\)
\[\begin{aligned} S_n &= 1 - 2q + 3q^2 - 4q^3 + \cdots \\ qS_n &= q - 2q^2 + 3q^3 - 4q^4 + \cdots \\ (1+q)S_n &= 1 - q + q^2 - q^3 + q^4 - \cdots \\ (1+q)S_n &= 1 + (-1)^{n+1}q^{n+1} \\ (-1)^{n+1}q^{n+1} &\to 0 \quad \quad \text{Therefore,} \\ S_n &= \frac{1}{(1+q)^2} \end{aligned}\]What is the limit of \(\frac{1 + 2 + 3 + \cdots + n}{n^2}, \quad \text{of} \quad \frac{1^2 + 2^2 + \cdots + n^2}{n^2} \quad ?\) (Hint: Use the results of pp. 12, 14, 15.) \(\begin{aligned} S_1 &= \frac{1 + 2 + 3 + \cdots + n}{n^2} = \frac{n(n+1)}{2n^2} = \frac{n+1}{2n} = \frac{1}{2} + \frac{1}{2n} \to \frac{1}{2} \\ S_2 &= \frac{1^2 + 2^2 + \cdots + n^2}{n^3} = \frac{n(n+1)(2n+1)}{6n^3} = \frac{(n+1)(2n+1)}{6n^2} \\ &= \frac{2n^2+3n+1}{6n^2} = \frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^2} \to \frac{1}{3}\\ S_3 &= \frac{1^3 + 2^3 + \cdots + n^3}{n^4} = \left( \frac{n(n+1)}{2n} \right)^2 = \frac{(n+1)^2}{4n^2} = \frac{1}{4} + \frac{1}{2n} + \frac{1}{4n^2} \to \frac{1}{4} \end{aligned}\)
Expand the fractions \(\frac{1}{11}\), \(\frac{1}{13}\), \(\frac{2}{13}\), \(\frac{3}{13}\), \(\frac{1}{17}\), \(\frac{2}{17}\) into decimal fractions and determine the period.
\[\begin{align*} \frac{1}{11} &= 0.\overline{09}\ldots \quad (\text{Period: } 2) \\ \frac{1}{13} &= 0.\overline{076923}\ldots \quad (\text{Period: } 6) \\ \frac{2}{13} &= 0.\overline{153846}\ldots \quad (\text{Period: } 6) \\ \frac{3}{13} &= 0.\overline{230769}\ldots \quad (\text{Period: } 6) \\ \frac{1}{17} &= 0.\overline{0588235294117647}\ldots \quad (\text{Period: } 16) \\ \frac{2}{17} &= 0.\overline{0.1176470588235294}\ldots \quad (\text{Period: } 16) \end{align*}\]The number 142857 has the property that multiplication with any one of the numbers 2, 3, 4, 5, or 6 produces only a cyclic permutation of its digits. Explain this property, using the expression of \(\frac{1}{7}\) into a decimal fraction.
\[\begin{align*} \frac{1}{7} &= 0.\overline{142857} \\ 2 \times \frac{1}{7} &= 0.\overline{285714} \\ 3 \times \frac{1}{7} &= 0.\overline{428571} \\ 4 \times \frac{1}{7} &= 0.\overline{571428} \\ 5 \times \frac{1}{7} &= 0.\overline{714285} \\ 6 \times \frac{1}{7} &= 0.\overline{857142} \end{align*}\]Expand the rational numbers \(\frac{1}{11}\), \(\frac{1}{13}\), \(\frac{2}{13}\), \(\frac{3}{13}\), \(\frac{1}{17}\), \(\frac{2}{17}\) as decimals with base 5, 7, and 12.
\[\begin{align*} \frac{1}{11} &= 0.\overline{09}\ldots \quad (\text{Period: } 2) \\ \frac{1}{13} &= 0.\overline{076923}\ldots \quad (\text{Period: } 6) \\ \frac{2}{13} &= 0.\overline{153846}\ldots \quad (\text{Period: } 6) \\ \frac{3}{13} &= 0.\overline{230769}\ldots \quad (\text{Period: } 6) \\ \frac{1}{17} &= 0.\overline{0588235294117647}\ldots \quad (\text{Period: } 16) \\ \frac{2}{17} &= 0.\overline{0.1176470588235294}\ldots \quad (\text{Period: } 16) \end{align*}\]Expand one third as a dyadic number.
Write \(.11212121 \cdots\) as a fraction. Find the value of this symbol if it is meant in the systems with the bases 3 or 5. \(\begin{aligned} 0.11212121\cdots &= \frac{1}{10} + 10^{-3} \cdot 12(1 + 10^{-2} + 10^{-4}+\cdots)\\ &= \frac{1}{10} + \frac{12}{1000} \times \frac{1}{1-10^{-2}}\\ &= \frac{1}{10} + \frac{12}{1000} \times \frac{100}{99}\\ &= \frac{111}{990} \end{aligned}\)
Show that the set of all positive and negative integers is denumerable. Show that the set of all positive and negative rational numbers is denumerable.
The set of all natural numbers is denumerable, denoted as \(\mathbb{N}\), and we define the set of all positive and negative integers as \(\mathbb{Z}\), and this is an inifite set.
To prove the set \(\mathbb{Z}\) is denumerable, we need to show it is countable by finding the bijection: \(f: \mathbb{Z} \to \mathbb{N}\)
\[f(n) = \begin{cases} 3^n & n>0 \\ 7^{-n} & n \leq 0 \end{cases}\]For every integers \(\mathbb{Z}\) inside we could find a one to one mapping to \(\mathbb{N}\), therefore the set \(\mathbb{Z}\) is denumerable.
For the set of all positive & negative rational numbers, all can be represented as \(\frac{b}{a}\) where a and b are natural number. We can build this matrix by having a rows and b columns. the nagative rational numbers can be inserted as intermediate.
\[\begin{aligned} \frac{1}{1}, -\frac{1}{1}, \frac{1}{2}, -\frac{1}{2}, \ldots\\ \frac{2}{1}, -\frac{2}{1}, \frac{2}{2}, -\frac{2}{2}, \ldots\\ \vdots\\ \frac{n}{1}, -\frac{n}{1}, \frac{n}{2}, -\frac{n}{2}, \ldots\\ \vdots\\ \end{aligned}\]By using Cantor’s Zig-Zag Method, we can see this set is also countable, therefore it is denumerable.
Show that the set \(S + T\) (see p. 110) ia denumerable if S and T are denumerable sets. Show the same for the sum of three, four, or any number, n, of sets, and finally for a set composed of denumerably many denumerable sets.
There exist bijections \(f: \mathbb{N} \to S\) and \(g: \mathbb{N} \to T\). Define \(h: \mathbb{N} \to S + T\) by:
\[h(n) = \begin{cases} f\left(\frac{n}{2}\right) & \text{if } n \text{ is even} \\ g\left(\frac{n-1}{2}\right) & \text{if } n \text{ is odd} \end{cases}\]h is a bijection, Therefore \(S + T\) is denumerable.
Let \(S_1, S_2, \ldots, S_n\) be denumerable sets. To show \(S_1 + S_2 + \cdots + S_n\) is denumerable.
There exist bijections \(f_i: \mathbb{N} \to S_i\) for \(i = 1, 2, \ldots, n\).
Define \(h: \mathbb{N} \to S_1 + S_2 + \cdots + S_n\) by:
\[h(n) =\]h is a bijection, so the sum of n denumerable sets is denumerable.
Show that any interval [A, B] of the number axis is equivalent to any other interval \([C, Z]\).
Define a function \(f\) to find bijection for end point:
\[f(A) = C, f(B) = D\]The function can be defined as to map \([A,B]\) to \([C,D]\):
\[f(x) = C + \frac{x-A}{B-A}\cdot (D-C)\]From this we can find x to f(x) is 1 to 1 mapping (bijection).
Q.E.D.
Prove that the same result holds for a denumerable set of points in the plane, replacing lengths of intervals by areas of squares.
We can represent all of the points in a circle by \(\theta\). Measure in radians, \([0, 2\pi]\).
Assume the enumerable set of angles is given by,
\[\begin{align*} \theta_1 &= 0.a_1b_1a_2a_3a_4 \ldots \\ \theta_2 &= 0.b_1b_2b_3b_4 \ldots \\ & \vdots \\ \theta_n &= 0.n_1n_2n_3 \ldots \end{align*}\]Then we can construct a new angle \(\Theta_i\) by using digits along the diagonal, \(\Theta_i = A_1b_2C_3 \ldots\)
Which is also within \([0, 2\pi]\) and it contradicts the assumption.
If A contains n elements, where n is a positive integer, show that B, defined as above, contains \(2^n\) elements. If A consists of the set of all positive integers, show that B is equivalent to the continuum of real numbers from 0 to 1, (Hint: Symbolize a subset of A in the first case by a finite and in the second case by an infinite sequence of the symbols 0 and 1
Express \(\frac{(1+i)(2+i)(3+i)}{(1-i)}\) in the form \(a + bi\).
\[\begin{align*} \frac{(1+i)(2+i)(3+i)}{(1-i)} & = \frac{(1+2i+i^2)(2+i)(3+i)}{(1-i)(1+i)} \\ & = \frac{(1+2i+i^2)(2+i)(3+i)}{1 - i^2} \\ & = i(2+i)(3+i) \\ & = (2i+i^2)(3+i) \\ & = (-1+2i)(3+i) \\ & = -3-i+6i+2i^2 \\ & = -5+5i \end{align*}\]Express \(\left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right)^3\) in the form \(a + bi\).
\[\begin{align*} \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right)^3 & = \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right)\left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right)^2 \\ & = \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) \left(\frac{1}{4} + 2\cdot(-\frac{1}{2})\cdot(\frac{\sqrt{3}}{2})i + \frac{3}{4}i^2\right) \\ & = \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) \\ & = \frac{1}{4} + \frac{\sqrt{3}}{4}i - \frac{\sqrt{3}}{4}i - \frac{3}{4}i^2\\ & = 1 \\ \end{align*}\]Express in the form \(a + bi\): \(\frac{1+i}{1-i}, \quad \frac{1+i}{2-i}, \quad \frac{1}{i^5}, \quad \frac{1}{(-2+i)(1-3i)}, \quad \frac{(4-5i)^2}{(2-3i)^2}\)
\[\begin{align*} \frac{1+i}{1-i} & = \frac{(1+i)^2}{1^2-i^2} \\ & = \frac{1+2i+i^2}{2} \\ & = 1 \end{align*}\] \[\begin{align*} \frac{1+i}{2-i} & = \frac{(1+i)(2+i)}{(2-i)(2+i)} \\ & = \frac{2+i+2i+2^2}{4-i^2}\\ & = \frac{1+3i}{5} \\ & = \frac{1}{5} + \frac{3}{5}i \end{align*}\] \[\begin{align*} \frac{1}{i^5} & = \frac{1}{i^2\cdot i^2\cdot i} \\ & = -i \end{align*}\] \[\begin{align*} \frac{1}{(-2+i)(1-3i)} & = \frac{1}{-2+6i+i-3i^2} \\ & = \frac{1-7i}{(1+7i)(1-7i)} \\ & = \frac{1-7i}{1-49i^2} \\ & = \frac{1}{50} - \frac{7}{50}i\\ \end{align*}\] \[\begin{align*} \frac{(4-5i)^2}{(2-3i)^2} & = \frac{16-40i+25i^2}{4-6i-6i+9i^2} \\ & = \frac{-9-40i}{-5-12i} \\ & = \frac{(-9-40i)(-5+12i)}{(-5-12i)(-5+12i)} \\ & = \frac{45-108i+200i-480i^2}{144+25} \\ & = \frac{525+92i}{169} \end{align*}\]Calculate \(\sqrt{5 + 12i}\). (Hint: Write \(\sqrt{5 + 12i} = x + yi\), square, and equate real and imaginary parts.)
\[\begin{align*} \sqrt{5 + 12i} &= x + yi \\ 5 + 12i &= x^2 + y^2 + 2xyi \\ 5 &= x^2 - y^2 \\ 12 &= 2xy \\ x^2 - \left(\frac{6}{x}\right)^2 &= 5 \\ x^2 - \frac{36}{x^2} &= 5 \implies x^4 - 36 = 5x^2 \\ \text{substitute} x^2 = t \\ t^2 - 5t - 36 &= 0 \\ t &= -4 \quad \text{or} \quad t = 9 \\ x &= \pm 3 \quad y = \pm 2 \end{align*}\]Therefore, \(\sqrt{5 + 12i} = 3 + 2i \quad \text{or} \quad \sqrt{5 + 12i} = -3 - 2i\)
Prove this theorem that directly from the definition of multiplication of two complex numbers, \(z_1=x_1+y_1i\) and \(z_2=x_2+y_2i\).
Theorem: \(\begin{aligned} |z_1 \cdot z_2| = |z_1| \cdot |z_2| \end{aligned}\).
\[\begin{aligned} |z_1 \cdot z_2| &= |(x_1 + y_1i) \cdot (x_2 + y_2i)| \\ z_1 \cdot z_2 &= x_1 x_2 - y_1 y_2 + (x_1 y_2 + x_2 y_1)i \\ |z_1 z_2| &= \sqrt{(x_1 x_2 - y_1 y_2)^2 + (x_1 y_2 + x_2 y_1)^2} \\ &= \sqrt{x_1^2 x_2^2 + x_1^2 y_2^2 + y_1^2 x_2^2 + y_1^2 y_2^2} \\ |z_1| &= \sqrt{x_1^2 + y_1^2} \\ |z_2| &= \sqrt{x_2^2 + y_2^2} \\ |z_1| \cdot |z_2| &= \sqrt{x_1^2 + y_1^2} \cdot \sqrt{x_2^2 + y_2^2} \\ &= \sqrt{(x_1^2 + y_1^2)(x_2^2 + y_2^2)} \\ &= \sqrt{x_1^2 x_2^2 + x_1^2 y_2^2 + y_1^2 x_2^2 + y_1^2 y_2^2} \\ \therefore |z_1 \cdot z_2| &= |z_1| \cdot |z_2| \end{aligned}\]From the fact that the product of two real numbers is 0 only if one of the factors is 0, prove the corresponding theorem for complex numbers. (Hint: Use the two theorems just stated.)
\[\begin{aligned} z_1 \cdot z_2 &= (x_1 + y_1 i) (x_2 + y_2 i)\\ &= (x_1 x_2 - y_1 y_2) + (x_1 y_2 + x_2 y_1)i &= 0 \end{aligned}\] \[\begin{aligned} = \begin{cases} x_1 x_2 - y_1 y_2 = 0 & \quad (1) \\ x_1 y_2 + x_2 y_1 = 0 & \quad (2) \end{cases} \end{aligned}\](1) implies that \(x_1 x_2 = y_1 y_2\). Either \(x_1 = 0\) or \(x_2 = 0\). and for \(y_1 = 0\) or \(y_2 = 0\).
(2) implies that \(x_1 y_2 = - x_2 y_1\). Then \(x_1 = 0\) or \(y_2 = 0\). and for \(x_2 = 0\) or \(y_1 = 0\).
Combine all of the situations together:
Either \(x_1 = 0\) and \(y_1 = 0\) or \(x_2 = 0\) and \(y_2 = 0\)
When does the equality \(|z_1 + z_2| = |z_1| + |z_2|\) hold?
\[\begin{aligned} |z_1 + z_2| &= |(x_1 + y_1i) + (x_2 + y_2i)| \\ &= |(x_1 + x_2) + (y_1 + y_2)i| \\ &= \sqrt{(x_1 + x_2)^2 + (y_1 + y_2)^2} \end{aligned}\] \[\begin{aligned} |z_1| + |z_2| &= \sqrt{(x_1 + y_1)^2} + \sqrt{(x_2 + y_2)^2} \end{aligned}\]In order to have these 2 equality hold, it requires \(x_1=x_2=y_1=y_2\), the intuition is that the sum of the other two sides of a triangle is equal to the hypotenuse if and only if it is an equilateral triangle.
Find the corresponding formulas for \(\sin 4\phi\) and \(\cos 4\phi\).
\[(a + b i)^4 = a^4 + 4a^3b i + 6a^2b^2 +4ab^3 + b^4\]Obtaining:
\[\begin{aligned} (\cos\phi + \sin\phi i)^4 &= \cos^4 \phi + 4 \cos^3 \phi \sin \phi i + 6\cos^2\phi\sin^2\phi i^2 + 4\cos \phi\sin^3\phi i^3 + \sin^4\phi i^4 \\ &= \cos^4 \phi + \sin^4\phi - 6\cos^2\phi\sin^2\phi + (4 \cos^3 \phi \sin \phi - 4\cos \phi\sin^3\phi) i \end{aligned}\]Therefore,
\[\begin{aligned} \cos 4\phi &= \cos^4\phi + (1-\cos^2\phi)^2 - 6\cos^2\phi(1-\cos^2\phi)\\ \sin 4\phi &= 4(1-\sin^2\phi)^{\frac{3}{2}}\sin\phi - 4\sqrt{1-\sin^2\phi}\sin^3\phi\\ \end{aligned}\]Prove that for a point, \(z = \cos \phi + i \sin \phi\), on the unit circle, \(\frac{1}{z} = \cos \phi - i \sin \phi\).
\[\begin{aligned} \frac{1}{z} &= \frac{1}{\cos \phi + i \sin \phi} \\ &= \cos \phi - i \sin \phi \end{aligned}\] \[\begin{aligned} 1 &= (\cos \phi - \sin \phi)(\cos \phi + \sin\phi) \\ \end{aligned}\]Prove without calculation that \(\frac{(a + bi)}{(a - bi)}\) always has the absolute value 1.
The expression we are interested in is
\[\frac{a + bi}{a - bi},\]which can be rewritten in terms of \(z\) and conjugate \(\bar{z}\):
\[\frac{a + bi}{a - bi} = \frac{z}{\bar{z}}.\]To find the magnitude of this expression, we use the property of magnitudes for complex numbers:
\[\left|\frac{z}{\bar{z}}\right| = \frac{|z|}{|\bar{z}|}.\]Since the modulo is the same, therefore it always has the solute value of 1 by calculating \(\frac{(a + bi)}{(a - bi)}\).
If \(z_1\) and \(z_2\) are two complex numbers, prove that the angle of \(z_1 - z_2\) is equal to the angle between the real axis and the vector pointing from \(z_2\) to \(z_1\).
Let \(z_1 = x_1 + iy_1\) and \(z_2 = x_2 + iy_2\) be two complex number. In the complex plane, \(z_1\) and \(z_2\) can be represented as points \((x_1, y_1)\) and \((x_2, y_2)\), respectively.
The difference \(z_1 - z_2\) is given by:
\[z_1 - z_2 = (x_1 - x_2) + i(y_1 - y_2).\]This difference represents the vector from \(z_2\) to \(z_1\) in the complex plane.
To find the angle of \(z_1 - z_2\) is given by \(\arg(z) = \theta\), where:
\[\arg(z_1 - z_2) = \tan^{-1} \left( \frac{y_1 - y_2}{x_1 - x_2} \right).\]Now, let’s consider the vector pointing from \(z_2\) to \((z_1\). In Cartesian coordinates, this vector is also:
\[(x_1 - x_2, y_1 - y_2).\]The angle between this vector and the positive real axis (the x-axis) is given by the same expression: \(\theta = \tan^{-1} \left( \frac{y_1 - y_2}{x_1 - x_2} \right).\)
Thus, the angle of the complex number \(z_1 - z_2\) is exactly the angle between the real axis and the vector pointing from \(z_2\) to \(z_1\).
Interpret the angle of the complex number \(\frac{(z_1 - z_2)}{(z_1 - z_3)}\) in the triangle formed by the points \(z_1\), \(z_2\), and \(z_3\).
\(z_1 - z_2\) represents the complex number corresponding to the vector from \(z_2\) to \(z_1\) and \(z_1 - z_3\) represents the complex number corresponding to the vector from \(z_3\) to \(z_1\).
Dividing two complex numbers \(\frac{z_1 - z_2}{z_1 - z_3}\) gives a new complex number whose magnitude is the ratio of the magnitudes of the two complex numbers and whose argument (angle) is the difference of their arguments. Let \(\theta_{12} = \arg(z_1 - z_2)\), which is the angle that the vector from \(z_2\) to \(z_1\) makes with the positive real axis. Let \(\theta_{13} = \arg(z_1 - z_3)\), which is the angle that the vector from \(z_3\) to \(z_1\) makes with the positive real axis.
The argument of the quotient \(\frac{(z_1 - z_2)}{(z_1 - z_3)}\) is \(\theta_{12} - \theta_{13}\).
In geometric terms:
If you start from the vector \(\overline{z_1z_3}\) and rotate counterclockwise by an angle \(\theta_{12} - \theta_{13}\), you will align with the vector \(\overline{z_1z_2}\).
Thus, the complex number \(\frac{(z_1 - z_2)}{(z_1 - z_3)}\) gives a complex number whose argument represents the oriented angle between the directed line segments from \(z_1\) to \(z_3\) and from \(z_1\) to \(z_2\).
Prove that the quotient of two complex numbers with the same angle is real.
If two complex numbers \(z_1\) and \(z_2\) have the same angle, then the argument (angle) of \(z_1\) is equal to the argument of \(z_2\). We can express these complex numbers in polar form:
\[z_1 = \rho _1 e^{i\theta}\] \[z_2 = \rho _2 e^{i\theta}\]where \(\rho _1\) and \(\rho _2\) are the magnitudes of \(z_1\) and \(z_2\), respectively, and \(\theta\) is their common angle.
The quotient of \(z_1\) and \(z_2\) is given by:
\[\begin{aligned} \frac{z_1}{z_2} &= \frac{\rho _1 e^{i\theta}}{\rho _2 e^{i\theta}} \\ &= \frac{\rho _1}{\rho _2} e^0 \\ &= \frac{\rho _1}{\rho _2} \end{aligned}\]Q.E.D.
Prove that if for four complex numbers \(z_1, z_2, z_3, z_4\) the angles of \(\frac{z_3 - z_1}{z_3 - z_2}\) and \(\frac{z_4 - z_1}{z_4 - z_2}\) are the same, then the four numbers lie on a circle or on a straight line, and conversely.
The solution is adapted from: math exchange
First of all, based on the prove on exercise 12,
\[\arg\left(\frac{z_4-z_1}{z_4-z_2}\right)\]represents the angle \(\angle Z_2 Z_4 Z_1\). Similarly
\[\arg\left(\frac{z_3-z_1}{z_3-z_2}\right)\]represents the angle \(\angle Z_2 Z_3 Z_1\).
This can be seen as 4 points to be concyclic
The \(A\) and \(B\) in the wikipedia graphic are your \(z_1\) and \(z_2\).
If the 4 lines lies in a straight line then the angle is 0, which is a special case.
Q.E.D.
Prove that four points \(z_1, z_2, z_3, z_4\) lie on a circle or on a straight line if and only if \(\frac{z_3 - z_1}{z_3 - z_2} = \frac{z_4 - z_1}{z_4 - z_2}\)
Proved above.
Find the 6th roots of 1.
\[x^6 = 1\]Based on De Moivre’s theorem, for unit circle:
\[z^n = (\cos \phi + i \sin \phi)^n = \cos(n \phi) + i \sin(n \phi)\] \[z^6 = 1 = \cos 0 + i \sin 0\] \[\begin{cases} \cos n \phi = 1 \\ \sin n \phi = 0 \end{cases} \Rightarrow \phi = \frac{2k\pi}{n} = \frac{k\pi}{3}, k \in \{1, 2, 3, 4, 5, 6\}\] \[z = \begin{cases} \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = \frac{1}{2} + i \frac{\sqrt{3}}{2} & k=1 \\ \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} = -\frac{1}{2} + i \frac{\sqrt{3}}{2} & k=2 \\ \cos \pi + i \sin \pi = -1 & k=3 \\ \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} = -\frac{1}{2} - i \frac{\sqrt{3}}{2} & k=4 \\ \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} = \frac{1}{2} - i \frac{\sqrt{3}}{2} & k=5 \\ \cos 2\pi + i \sin 2\pi = 1 & k=6 \\ \end{cases}\]Find \((1 + i)^{11}\).
\[1+i = \sqrt{2} (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})\] \[\begin{aligned} (1+i)^{11} &= (\sqrt{2})^{11} \left[ \cos \frac{11\pi}{4} + i \sin \frac{11\pi}{4} \right] \\ &= 32 \sqrt{2} \left[ \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right] \\ &= 32 \sqrt{2} \left[ -\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \right] \\ &= -32 + 32i \end{aligned}\]Find all the different values of \(\sqrt[3]{1 + i}\), \(\sqrt[3]{7 - 4i}\), \(\sqrt[3]{i}\), \(\sqrt[3]{-i}\).
(1).
\[1+i = \sqrt{2} (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})\] \[\sqrt{1+i} = \sqrt[4]{2} (\cos\frac{\pi}{8} + i \sin \frac{\pi}{8})\](2).
\[\sqrt[3]{7-4i} = \sqrt[6]{65} \{\cos [\tan^{-1}(\frac{-4}{7})] + i \sin [\tan^{-1}(\frac{-4}{7})]\}\](3).
\[i = \cos \phi + i \sin \phi \quad \phi = \frac{\pi}{2} + 2k\pi\] \[i^{1/3} = \cos \left( \frac{\pi}{6} + \frac{2k\pi}{3} \right) + i \sin \left( \frac{\pi}{6} + \frac{2k\pi}{3} \right), \quad k = 0, 1, 2\] \[\sqrt[3]{i} = \frac{i}{2} + \frac{\sqrt{3}}{2} \quad \text{or} \quad -i \quad \text{or} \quad \frac{i}{2} - \frac{\sqrt{3}}{2}\](4).
\[-i = \cos \phi + i \sin \phi \quad \phi = -\frac{\pi}{2} + 2k\pi, \quad k = 0, 1, 2, 3, 4\] \[-i^{1/5} = \cos \left( -\frac{\pi}{10} + \frac{2k\pi}{5} \right) + i \sin \left( -\frac{\pi}{10} + \frac{2k\pi}{5} \right)\]Calculate \(\frac{1}{2i} (i^7 - i^{-7})\).
\[i^7 = i^4 \cdot i^3 = (i^2)^2 \cdot i^3 = i^3 = -i\] \[i^{-7} = (-i)^7 = (-i)^4\cdot(-i)^3 = (-i)^3 = i\]Therefore
\[\frac{1}{2i} (-i - i) = \frac{1}{2i}(-2i)=-1\]The solution presented on the blog is my personal solutions for the exercises in the book ‘What is Mathematics: An Elementary Approach To Ideas And Methods’ by Herbert Robbins and Richard Courant, please leave a comment if you spot any mistakes or you have questions on the solution. Thanks in advance!
Set up the addition and multiplication tables in the duodecimal system and work some examples of the same sort.
The duodecimal system (also known as base 12 or dozenal) is the number system with a base of twelve.
\[\begin{aligned} & \text {Table 1.1. Addition table of Duodecimal }\\ & \begin{array}{c|cccccccccccc} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & B \\ \hline 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & B & 10 \\ 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & B & 10 & 11 \\ 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & B & 10 & 11 & 12 \\ 4 & 5 & 6 & 7 & 8 & 9 & A & B & 10 & 11 & 12 & 13 \\ 5 & 6 & 7 & 8 & 9 & A & B & 10 & 11 & 12 & 13 & 14 \\ 6 & 7 & 8 & 9 & A & B & 10 & 11 & 12 & 13 & 14 & 15 \\ 7 & 8 & 9 & A & B & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ 8 & 9 & A & B & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \\ 9 & A & B & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 \\ A & B & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 \\ B & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 1A \\ \end{array} \end{aligned}\] \[\begin{aligned} & \text{Table 1.2. Multiplication Table} \\\ & \begin{array}{c|cccccccccccc} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & B \\ \hline 1 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & B \\ 2 & 2 & 4 & 6 & 8 & A & 10 & 12 & 14 & 16 & 18 & 1A \\ 3 & 3 & 6 & 9 & 10 & 13 & 16 & 19 & 20 & 23 & 26 & 29 \\ 4 & 4 & 8 & 10 & 14 & 18 & 20 & 24 & 28 & 30 & 34 & 38 \\ 5 & 5 & A & 15 & 18 & 21 & 26 & 2B & 34 & 39 & 42 & 47 \\ 6 & 6 & 10 & 16 & 20 & 26 & 30 & 36 & 40 & 46 & 50 & 56 \\ 7 & 7 & 12 & 19 & 24 & 2B & 36 & 41 & 48 & 53 & 5A & 65 \\ 8 & 8 & 14 & 20 & 28 & 34 & 40 & 48 & 54 & 60 & 68 & 74 \\ 9 & 9 & 16 & 23 & 30 & 39 & 46 & 53 & 60 & 69 & 76 & 83 \\ A & A & 18 & 26 & 34 & 42 & 50 & 5A & 68 & 76 & 84 & 92 \\ B & B & 1A & 29 & 38 & 47 & 56 & 65 & 74 & 83 & 92 & A1 \\ \end{array} \end{aligned}\]Express ‘thirty’ and ‘one hundred and thirty-three’ in the systems with the bases 5, 7, 11, 12.
\[\\ \begin{aligned} & \text{base 5:} & \quad 110, \quad 1013 \\ & \text{base 7:} & \quad 42, \quad 245 \\ & \text{base 11:} & \quad 28, \quad 10A \\ & \text{base 12:} & \quad 26, \quad B1 \\ \end{aligned} \\\]What do the symbols 11111 and 21212 mean in these systems.
\[\\ \begin{aligned} &\text{Converting } 11111 \text{ from various bases to decimal:} \\ &\text{Base 5:} \\ &11111_5 = 1 \cdot 5^4 + 1 \cdot 5^3 + 1 \cdot 5^2 + 1 \cdot 5^1 + 1 \cdot 5^0 \\ &11111_5 = 1 \cdot 625 + 1 \cdot 125 + 1 \cdot 25 + 1 \cdot 5 + 1 \cdot 1 = 625 + 125 + 25 + 5 + 1 = 781 \\[10pt] &\text{Base 7:} \\ &11111_7 = 1 \cdot 7^4 + 1 \cdot 7^3 + 1 \cdot 7^2 + 1 \cdot 7^1 + 1 \cdot 7^0 \\ &11111_7 = 1 \cdot 2401 + 1 \cdot 343 + 1 \cdot 49 + 1 \cdot 7 + 1 \cdot 1 = 2401 + 343 + 49 + 7 + 1 = 2801 \\[10pt] &\text{Base 11:} \\ &11111_{11} = 1 \cdot 11^4 + 1 \cdot 11^3 + 1 \cdot 11^2 + 1 \cdot 11^1 + 1 \cdot 11^0 \\ &11111_{11} = 1 \cdot 14641 + 1 \cdot 1331 + 1 \cdot 121 + 1 \cdot 11 + 1 \cdot 1 = 14641 + 1331 + 121 + 11 + 1 = 16105 \\[10pt] &\text{Base 12:} \\ &11111_{12} = 1 \cdot 12^4 + 1 \cdot 12^3 + 1 \cdot 12^2 + 1 \cdot 12^1 + 1 \cdot 12^0 \\ &11111_{12} = 1 \cdot 20736 + 1 \cdot 1728 + 1 \cdot 144 + 1 \cdot 12 + 1 \cdot 1 = 20736 + 1728 + 144 + 12 + 1 = 22621 \\[20pt] &\text{Converting } 21212 \text{ from various bases to decimal:} \\ &\text{Base 5:} \\ &21212_5 = 2 \cdot 5^4 + 1 \cdot 5^3 + 2 \cdot 5^2 + 1 \cdot 5^1 + 2 \cdot 5^0 \\ &21212_5 = 2 \cdot 625 + 1 \cdot 125 + 2 \cdot 25 + 1 \cdot 5 + 2 \cdot 1 = 1250 + 125 + 50 + 5 + 2 = 1432 \\[10pt] &\text{Base 7:} \\ &21212_7 = 2 \cdot 7^4 + 1 \cdot 7^3 + 2 \cdot 7^2 + 1 \cdot 7^1 + 2 \cdot 7^0 \\ &21212_7 = 2 \cdot 2401 + 1 \cdot 343 + 2 \cdot 49 + 1 \cdot 7 + 2 \cdot 1 = 4802 + 343 + 98 + 7 + 2 = 5252 \\[10pt] &\text{Base 11:} \\ &21212_{11} = 2 \cdot 11^4 + 1 \cdot 11^3 + 2 \cdot 11^2 + 1 \cdot 11^1 + 2 \cdot 11^0 \\ &21212_{11} = 2 \cdot 14641 + 1 \cdot 1331 + 2 \cdot 121 + 1 \cdot 11 + 2 \cdot 1 = 29282 + 1331 + 242 + 11 + 2 = 30868 \\[10pt] &\text{Base 12:} \\ &21212_{12} = 2 \cdot 12^4 + 1 \cdot 12^3 + 2 \cdot 12^2 + 1 \cdot 12^1 + 2 \cdot 12^0 \\ &21212_{12} = 2 \cdot 20736 + 1 \cdot 1728 + 2 \cdot 144 + 1 \cdot 12 + 2 \cdot 1 = 41472 + 1728 + 288 + 12 + 2 = 43502 \\[20pt] \end{aligned} \\\] \[\\ \begin{aligned} &\text{Summary of conversions:} \\[10pt] &\begin{array}{|c|c|c|} \hline \text{Number} & \text{Base} & \text{Decimal Equivalent} \\ \hline 11111 & 5 & 781 \\ 11111 & 7 & 2801 \\ 11111 & 11 & 16105 \\ 11111 & 12 & 22621 \\ \hline 21212 & 5 & 1432 \\ 21212 & 7 & 5252 \\ 21212 & 11 & 30868 \\ 21212 & 12 & 43502 \\ \hline \end{array} \end{aligned}\]Form the addition and multiplication tables for the bases 5, 11, 13.
\[\begin{aligned} & \text{Table 1.3. Addition Table of Base 5} \\ & \begin{array}{c|ccccc} & 0 & 1 & 2 & 3 & 4 \\ \hline 0 & 0 & 1 & 2 & 3 & 4 \\ 1 & 1 & 2 & 3 & 4 & 10 \\ 2 & 2 & 3 & 4 & 10 & 11 \\ 3 & 3 & 4 & 10 & 11 & 12 \\ 4 & 4 & 10 & 11 & 12 & 13 \\ \end{array} \end{aligned} \begin{aligned} & \text{Table 1.4. Multiplication Table of Base 5} \\ & \begin{array}{c|ccccc} & 0 & 1 & 2 & 3 & 4 \\ \hline 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 & 3 & 4 \\ 2 & 0 & 2 & 4 & 11 & 13 \\ 3 & 0 & 3 & 11 & 14 & 22 \\ 4 & 0 & 4 & 13 & 22 & 31 \\ \end{array} \end{aligned}\] \[\\ \begin{aligned} & \text{Table 1.5. Addition Table: Base 11} \\ & \begin{array}{c|ccccccccccc} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A \\ \hline 0 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A \\ 1 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & 10 \\ 2 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & 10 & 11 \\ 3 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & 10 & 11 & 12 \\ 4 & 4 & 5 & 6 & 7 & 8 & 9 & A & 10 & 11 & 12 & 13 \\ 5 & 5 & 6 & 7 & 8 & 9 & A & 10 & 11 & 12 & 13 & 14 \\ 6 & 6 & 7 & 8 & 9 & A & 10 & 11 & 12 & 13 & 14 & 15 \\ 7 & 7 & 8 & 9 & A & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ 8 & 8 & 9 & A & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \\ 9 & 9 & A & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 \\ A & A & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 \\ \end{array} \end{aligned} \\\] \[\\ \begin{aligned} & \text{Table 1.6. Multiplication Table: Base 11} \\ & \begin{array}{c|ccccccccccc} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A \\ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A \\ 2 & 0 & 2 & 4 & 6 & 8 & A & 11 & 13 & 15 & 17 & 19 \\ 3 & 0 & 3 & 6 & 9 & 11 & 14 & 17 & 1A & 22 & 25 & 28 \\ 4 & 0 & 4 & 8 & 11 & 15 & 19 & 22 & 26 & 2A & 33 & 37 \\ 5 & 0 & 5 & A & 14 & 19 & 23 & 28 & 32 & 37 & 41 & 46 \\ 6 & 0 & 6 & 11 & 17 & 22 & 28 & 33 & 39 & 44 & 4A & 55 \\ 7 & 0 & 7 & 13 & 1A & 26 & 32 & 39 & 45 & 51 & 58 & 64 \\ 8 & 0 & 8 & 15 & 22 & 2A & 37 & 44 & 51 & 59 & 66 & 73 \\ 9 & 0 & 9 & 17 & 25 & 33 & 41 & 4A & 58 & 66 & 74 & 82 \\ A & 0 & A & 19 & 28 & 37 & 46 & 55 & 64 & 73 & 82 & 91 \\ \end{array} \end{aligned} \\\] \[\\ \begin{aligned} & \text{Table 1.7. Addition Table: Base 13} \\ & \begin{array}{c|ccccccccccccc} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & B & C \\ \hline 0 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & B & C \\ 1 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & B & C & 10 \\ 2 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & B & C & 10 & 11 \\ 3 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & B & C & 10 & 11 & 12 \\ 4 & 4 & 5 & 6 & 7 & 8 & 9 & A & B & C & 10 & 11 & 12 & 13 \\ 5 & 5 & 6 & 7 & 8 & 9 & A & B & C & 10 & 11 & 12 & 13 & 14 \\ 6 & 6 & 7 & 8 & 9 & A & B & C & 10 & 11 & 12 & 13 & 14 & 15 \\ 7 & 7 & 8 & 9 & A & B & C & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ 8 & 8 & 9 & A & B & C & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \\ 9 & 9 & A & B & C & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 \\ A & A & B & C & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 \\ B & B & C & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 1A \\ C & C & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 1A & 1B \\ \end{array} \end{aligned} \\\] \[\\ \begin{aligned} & \text{Table 1.8. Multiplication Table: Base 13} \\ & \begin{array}{c|ccccccccccccc} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & B & C \\ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & B & C \\ 2 & 0 & 2 & 4 & 6 & 8 & A & C & 11 & 13 & 15 & 17 & 19 & 1B \\ 3 & 0 & 3 & 6 & 9 & C & 12 & 15 & 18 & 1B & 21 & 24 & 27 & 2A \\ 4 & 0 & 4 & 8 & C & 13 & 17 & 1B & 22 & 26 & 2A & 31 & 35 & 39 \\ 5 & 0 & 5 & A & 12 & 17 & 1C & 24 & 29 & 31 & 36 & 3B & 43 & 48 \\ 6 & 0 & 6 & C & 15 & 1B & 24 & 2A & 33 & 39 & 42 & 48 & 51 & 57 \\ 7 & 0 & 7 & 11 & 18 & 22 & 29 & 33 & 3A & 44 & 4B & 55 & 5C & 66 \\ 8 & 0 & 8 & 13 & 1B & 26 & 31 & 39 & 44 & 4C & 57 & 62 & 6A & 75 \\ 9 & 0 & 9 & 15 & 21 & 2A & 36 & 42 & 4B & 57 & 63 & 6C & 78 & 84 \\ A & 0 & A & 17 & 24 & 31 & 3B & 48 & 55 & 62 & 6C & 79 & 86 & 93 \\ B & 0 & B & 19 & 27 & 35 & 43 & 51 & 5C & 6A & 78 & 86 & 94 & A2 \\ C & 0 & C & 1B & 2A & 39 & 48 & 57 & 66 & 75 & 84 & 93 & A2 & B1 \\ \end{array} \end{aligned} \\\]Exercise: Consider the question of representing integers with the base $a$. In order to name the integers in this system we need words for the digits \(0, 1, \ldots, a - 1\) and for the various powers of \(a\): \(a, a^1, a^2, \ldots\). How many different number words are needed to name all numbers from zero to one thousand, for \(a = 2, 3, 4, 5, \ldots, 15\)? Which base requires the fewest? (Examples: If \(a = 10\), we need ten words for the digits, plus words for \(10, 100,\) and \(1000\), making a total of 13. For \(a = 20\), we need twenty words for the digits, plus words for \(20\) and $400$, making a total of 22. If \(a = 100\), we need 100 plus 1.)
From the description we can easily get the following count by listsing them.
\[\begin{aligned} a = 10 & : \quad 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10^1, 10^2, 10^3\\ a = 11 & : \quad 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, 11^1, 11^2 \\ a = 20 & : \quad 0, \ldots, R,S, 20^1, 20^2 \end{aligned}\]It can be eaily derived that the number of words x needed to name all numbers from 0 to 1000 is number number of a + the maximum power of a that is less or equal to 1000. for 10 that is 3 in \(10^3\) because \(10^3\leq1000\).Therefore we can have the formula:
\[x = a + \lfloor\frac{\log 1000}{\log a}\rfloor\]This shows that 4 uses the fewest words.
Exercise: Prove by mathematical induction
\(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\cdots+\frac{1}{n(n+1)}=\frac{n}{n+1}\) .
Base Case: For \(n = 1\),
\[\frac{1}{1 \cdot 2} = \frac{1}{2}\]So, the base case holds.
Inductive Step: Assume the statement is true for some arbitrary positive integer \(k\), i.e.,
\[\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} = \frac{k}{k+1}.\]Now, we prove it for \(k+1\):
\[\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} + \frac{1}{(k+1)((k+1)+1)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}.\]To simplify the right-hand side:
\[\begin{aligned}\frac{k}{k+1} + \frac{1}{(k+1)(k+2)} &= \frac{k(k+2) + 1}{(k+1)(k+2)}\\ &= \frac{k^2 + 2k + 1}{(k+1)(k+2)}\\ &= \frac{(k+1)^2}{(k+1)(k+2)}\\ &= \frac{k+1}{k+2}.\end{aligned}\]Q.E.D.
\(\frac{1}{2}+\frac{2}{2^n}+\frac{3}{2^n}+\cdots+\frac{n}{2^n}=2-\frac{n+2}{2^n}\).
Base Case: For \(n = 1\),
\[\frac{1}{2} = 2 - \frac{1+2}{2^1} = 2 - \frac{3}{2} = \frac{1}{2}.\]So, the base case holds.
Inductive Step:Assume the statement is true for some arbitrary positive integer \(k\) sum denoted as $A_k$
\[A_k = \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^2} + \cdots + \frac{k}{2^2} = 2 - \frac{k+2}{2^k}.\]Then for sum of \(A_{k+1}\) is:
\[\begin{aligned} A_{k+1} &= 2 - \frac{k+2}{2^k} + \frac{k+1}{2^{k+1}} \\ &= 2 - \frac{2k+4}{2^{k+1}} + \frac{k+1}{2^{k+1}} \\ &= 2 - \frac{k+3}{2^{k+1}} \end{aligned}\]Q.E.D.
\(1+2 q+3 q^2+\cdots+n q^{n-1}=\frac{1-(n+1) q^n+n q^{n+1}}{(1-q)^2}\).
Base Case: \(n = 1\)
\[\frac{1 - 2q + q^2}{(1-q)^2} = \frac{(1-q)^2}{(1-q)^2} = 1.\]Inductive Step:
Assume the identity holds for some arbitrary \(n = k\):
\[\\ 1 + 2q + 3q^2 + \cdots + kq^{k-1} = \frac{1 - (k+1)q^k + kq^{k+1}}{(1-q)^2}. \\\]Now, we need to show it holds for \(n = k + 1\):
\[\\ 1 + 2q + 3q^2 + \cdots + (k+1)q^k = \frac{1 - ((k+1)+1)q^{k+1} + (k+1)q^{k+2}}{(1-q)^2}. \\\]To prove this, consider the sum up to \(k+1\):
\[\\ 1 + 2q + 3q^2 + \cdots + (k+1)q^k = \left( 1 + 2q + 3q^2 + \cdots + kq^{k-1} \right) + (k+1)q^k. \\\]Using the induction hypothesis:
\[\\ 1 + 2q + 3q^2 + \cdots + kq^{k-1} = \frac{1 - (k+1)q^k + kq^{k+1}}{(1-q)^2}. \\\]Therefore,
\[\\ \begin{aligned} 1 + 2q + 3q^2 + \cdots + (k+1)q^k &= \frac{1 - (k+1)q^k + kq^{k+1}}{(1-q)^2} + (k+1)q^k \\ &= \frac{1 - (k+1)q^k + kq^{k+1} + (k+1)q^k (1-q)^2}{(1-q)^2} \\ &= \frac{1 - (k+2) q^{k+1} + (k+1) q^{k+2}}{(1 - q)^2} \end{aligned} \\\]Q.E.D.
\((1+q)\left(1+q^2\right)\left(1+q^4\right) \cdots\left(1+q^{2^n}\right)=\frac{1-q^{2^{n+1}}}{1-q}\).
Base Case: For \(n=1\).
\[\begin{aligned} (1+q)(1+q^2) &= \frac{1 - q^4}{1 - q} \\ 1 + q^2 + q + q^3 &= \frac{1 - q^4}{1 - q} \\ (1 + q^2 + q + q^3)(1-q) &= 1-q^4 \end{aligned}\]Simpily the left term we have \(1-q^4 = 1-q^4\). therefore \(n=1\) is true.
Assume for n case is correct, for n+1 case: \(\begin{aligned} (1+q)\left(1+q^2\right)\left(1+q^4\right) \cdots\left(1+q^{2^n}\right)(1+q^{2{n+1}}) &= \frac{1-q^{2^{n+1}}}{1-q}(1+q^{2^{n+1}})\\ &= \frac{1-q^{2^{n+2}}}{1-q} \end{aligned}\)
Q.E.D.
Find sum of the following:
\(\frac{1}{1+x^2}+\frac{1}{\left(1+x^2\right)^2}+\cdots+\frac{1}{\left(1+x^2\right)^n}\).
Denote \(S_n\) be the sum
\begin{equation} \label{sum_5_1} S_n = \frac{1}{1+x^2}+\frac{1}{\left(1+x^2\right)^2}+\cdots+\frac{1}{\left(1+x^2\right)^n} \end{equation}
Construct another sum \begin{equation} \label{sum_5_2} \frac{1}{1+x^2}S_n = \frac{1}{\left(1+x^2\right)^2}+\cdots+\frac{1}{\left(1+x^2\right)^n}+\frac{1}{\left(1+x^2\right)^{n+1}} \end{equation}
By using \eqref{sum_5_1} - \eqref{sum_5_2} we can have
\[\begin{aligned} S_n - \frac{1}{1+x^2}S_n &= \frac{1}{\left(1+x^2\right)}-\frac{1}{\left(1+x^2\right)^{n+1}} \\ \frac{(1+x^2)S_n-S_n}{1+x^2} &=\frac{1}{\left(1+x^2\right)}-\frac{1}{\left(1+x^2\right)^{n+1}} \\ \frac{x^2}{1+x^2}S_n &=\frac{1}{\left(1+x^2\right)}-\frac{1}{\left(1+x^2\right)^{n+1}} \\ Sn &=\frac{1}{x^2} - \frac{1}{\left(1+x^2\right)^{n}x^2} \end{aligned}\]Thus the sum is:
\[S_n = \frac{1}{x^2} \left(1 - \frac{1}{(1+x^2)^n}\right)\]\(1+\frac{x}{1+x^2}+\frac{x^2}{\left(1+x^2\right)^2}+\cdots+\frac{x^n}{\left(1+x^2\right)^n}\).
we denote the sum by \(S_{n+1}\). \begin{equation} \label{sum_6_1} 1+\frac{x}{1+x^2}+\frac{x^2}{\left(1+x^2\right)^2}+\cdots+\frac{x^n}{\left(1+x^2\right)^n} \end{equation}
Construct another sum \(\frac{x}{1+x^2}S_{n+1}\). \begin{equation} \label{sum_6_2} \begin{aligned} 1+\frac{x}{1+x^2}+\frac{x^2}{\left(1+x^2\right)^2}+\cdots+\frac{x^n}{\left(1+x^2\right)^n} \end{aligned} \end{equation}
By using \eqref{sum_6_1} - \eqref{sum_6_2} we can have
\[\begin{aligned} S_{n+1} - \frac{x}{1+x^2}S_{n+1} &= 1 -\frac{x^n}{\left(1+x^2\right)^{n+1}} \\ \frac{1+x^2-x}{1+x^2}S_{n+1} &= 1 -\frac{x^n}{\left(1+x^2\right)^{n+1}} \\ S_{n+1} &= \frac{1+x^2}{1+x^2-x} - \frac{x^n}{\left(1+x^2\right)^{n+1}}\frac{1+x^2}{1+x^2-x} \\ S_{n+1} &= \frac{\left(1+x^2\right)^{n+1}}{(1+x^2-x)(\left(1+x^2\right)^{n})} - \frac{x^{n+1}}{(1+x^2-x)(\left(1+x^2\right)^{n})} \end{aligned}\]Thus, the sum of the series is:
\[S_{n+1} = \frac{\left(1+x^2\right)^{n+1}-x^{n+1}}{(1+x^2-x)(\left(1+x^2\right)^{n})}\]\(\frac{x^2-y^2}{x^2+y^2}+\left(\frac{x^2-y^2}{x^2+y^2}\right)^2+\cdots+\left(\frac{x^2-y^2}{x^2+y^2}\right)^n\).
The common ratio of the geometric progression is \(q = \frac{x^2-y^2}{x^2+y^2}\)
Based on the formula of summation for geometric sequence:
\[S_n = a_1\frac{1-q^n}{1-q}\]By substituting \(q = \frac{x^2-y^2}{x^2+y^2}\), we can have
\[S_n = \frac{x^2-y^2}{x^2+y^2}\frac{1-(\frac{x^2-y^2}{x^2+y^2})^n}{1-\frac{x^2-y^2}{x^2+y^2}} \\\]Using formulas (4) and (5) in the book to prove:
\(1^2+3^2+\cdots+(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}\).
What we’ve already known: \(1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}\).
If \(S_n\) is the original progression, notice that each base is odd, we can construct it by using \(2n+1\) terms and subtract it with the progress where each base is even.
\[\begin{aligned} Sn &= [1^2+2^2+\cdots+(2n+1)^2=] - (2^2+4^2+6^2+\cdots+(2n)^2) \\ &= \frac{(2n+1)(2n+2)(4n+3)}{6} - 2^2[1^2+2^2+\cdots+(n)^2] \\ &= \frac{(2n+1)(2n+2)(4n+3)}{6} - \frac{4n(n+1)(2n+1)}{3} \\ &= \frac{(2n+1)(2n+2)(4n+3)}{6} - \frac{2n(2n+2)(2n+1)}{6} \\ &= \frac{(2n+1)(2n+2)[(4n+3)-2n]}{6}\\ &= \frac{(n+1)(2n+1)(2n+3)}{3} \end{aligned}\]\(1^3+3^3+\cdots+(2 n+1)^2=(n+1)^2\left(2 n^2+4 n+1\right)\).
Using the same trick as above.
Known: \((1^3+2^3+3^3+\cdots+n^3)=[\frac{n(n+1)}{2}]^2\)
If \(S_n\) is the original progression:
\[\begin{aligned} S_n &= \left[ 1^3 + 2^3 + \ldots + (2n+1)^3 \right] - \left[ 2^3 + 4^3 + \ldots + (2n)^3 \right]\\ &= \left[ \frac{(2n+1)(2n+2)}{2} \right]^2 - 2^3 \left[ \frac{n(n+1)}{2} \right]^2\\ &= \frac{(2n+1)^2 (2n+2)^2}{4} - 2^3 \cdot \frac{n^2 (n+1)^2}{4}\\ &= (4n^4 + 12n^3 + 13n^2 + 6n + 1) - (2n^4 + 4n^3 + 2n^2)\\ &= 2n^4 + 8n^3 + 11n^2 + 6n + 1 \\ &= (n+1)^2\left(2 n^2+4 n+1\right) \end{aligned}\]Prove the same results directly by mathematical induction.
To prove \(1^3+3^3+\cdots+(2n+1)^3=\frac{(n+1)(2n+1)(2n+3)}{3}\).
Base Case:
\[S_1 = \frac{2 \times 3 \times 5}{3} = 10 \quad \text{True.}\]Assume \(S_n\) holds
\[S_n = \frac{(n+1)(2n+1)(2n+3)}{3}\]For \(S_{n+1}\)
\[\begin{aligned} S_{n+1} &= \frac{(n+1)(2n+1)(2n+3)}{3} + (2n+3)^2 \\ &= \frac{(2n+3)\left[(n+1)(2n+1)+3(2n+3)\right]}{3} \\ &= \frac{(2n+3)\left[2n^2 + n + 2n + 1 + 6n + 9\right]}{3} \\ &= \frac{(2n+3)(n+2)(2n+5)}{3} \end{aligned}\]Q.E.D.
To prove \(1^3+3^3+\cdots+(2 n+1)^3=(n+1)^2\left(2 n^2+4 n+1\right)\).
Base Case:
\[S_1 = 2^2 (2 + 4 + 1) = 28 \quad \text{True}\]Assume \(S_n\) holds
We need to show that the statement holds for n+1 That is, we need to prove: \(1^3 + 3^3 + \cdots + (2n+1)^3 + (2(n+1)+1)^3 = ((n+1)+1)^2 (2(n+1)^2 + 4(n+1) + 1)\)
Simplify the left hand we have
\[(n+1)^2 (2n^2 + 4n + 1) + (2n + 3)^3\]So we need to show: \((n+1)^2 (2n^2 + 4k + 1) + (2n + 3)^3 = (n+2)^2 (2(n+1)^2 + 4(n+1) + 1)\)
Expand it and we can get:
\[\begin{aligned} 2(n+1)^2 + 4(n+1) + 1 &= 2(n^2 + 2n + 1) + 4n + 4 + 1 \\ &= 2n^2 + 4n + 2 + 4n + 4 + 1 = 2n^2 + 8n + 7 \\ \end{aligned}\]Then,
\[\begin{aligned} S_{n+1} &=(n+2)^2 (2n^2 + 8n + 7) \\ &= (n^2 + 4n + 4)(2n^2 + 8n + 7) \\ &= n^2(2n^2 + 8n + 7) + 4n(2n^2 + 8n + 7) + 4(2n^2 + 8n + 7) \\ &= 2n^4 + 8n^3 + 7n^2 + 8n^3 + 32n^2 + 28n + 8n^2 + 32n + 28 \\ &= 2n^4 + 16n^3 + 47n^2 + 60n + 28 \\ \end{aligned}\]Now, let’s expand the left-hand side: \(\begin{aligned} S_{n+1} &=(n+1)^2(2n^2 + 4n + 1) + (2n+3)^3 \\ &= (n^2 + 2n + 1)(2n^2 + 4n + 1) + (2n + 3)^3 \\ &= 2n^4 + 4n^3 + n^2 + 4n^3 + 8n^2 + 2n + 2n^2 + 4n + 1 + (2n+3)^3 \\ &= 2n^4 + 8n^3 + 11n^2 + 6n + 1 + (8n^3 + 36n^2 + 54n + 27) \\ &= 2n^4 + 16n^3 + 47n^2 + 60n + 28 \\ \end{aligned}\)
Both sides are equal, so the inductive step holds.
By mathematical induction, the statement is proven:
\[1^3 + 3^3 + \cdots + (2n+1)^3 = (n+1)^2 (2n^2 + 4n + 1)\]Exercise: Carry out this construction starting with \(p_1 = 2\), \(p_2\)= 3 and find 5 more primes The construction of proving that prime numbers are infinite is as followed:
\[A = p_1p_2\ldots p_n + 1\]Based on this constrction, the next 5 primes is
\[\begin{aligned} p_3 &= p_1p_2 + 1 = 7 \\ p_4 &= p_1p_2p_3 + 1 = 43 \\ p_5 &= p_1p_2p_3p_4 + 1 = 1807\\ p_6 &= p_1p_2p_3p_4p_5 + 1 = 3263443 \\ p_7 &= p_1p_2p_3p_4p_5p_6 + 1 = 10650056950807 \end{aligned}\]side note: This is the famous Sylvester’s sequence
Exercise: In order to find all the divisors of any number a we need only decompose a into a product
\[a = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_r^{\alpha_r},\]where the \(p\)’s are distinct primes, each raised to a certain power. All the divisors of $a$ are the numbers
\[b = p_1^{\beta_1} p_2^{\beta_2} \cdots p_r^{\beta_r},\]where the \(\beta\)’s are any integers satisfying the inequalities
\[0 \leq \beta_1 \leq \alpha_1, \quad 0 \leq \beta_2 \leq \alpha_2, \quad \cdots, \quad 0 \leq \beta_r \leq \alpha_r.\]Prove this statement. As a consequence, show that the number of different divisors of \(a\) (including the divisors \(a\) and 1) is given by the product
\[(\alpha_1 + 1)(\alpha_2 + 1) \cdots (\alpha_r + 1).\]For example,
\[144 = 2^4 \cdot 3^2\]has \(5 \cdot 3\) divisors. They are 1, 2, 4, 8, 16, 3, 6, 12, 24, 48, 9, 18, 36, 72, 144.
Proof:
For the form of b that is divisors of a, every prime \(p_i\) in a must appear in b with an exponent that is less than or equal to the exponent in a. For example: \(b_1 = p_1^1\) is a divisor where \(\beta_1\) is 1 and the rest are 0
Hence, \(0 \leq \beta_i \leq \alpha_i\).
To find the total number of distinct divisors, we need to count all possible combinations of the exponents \(\beta_i\) where \(0 \leq \beta_i \leq \alpha_i\).
For each prime \(p_i\), the exponent \(\beta_i\) can take any integer value from 0 to \(\alpha_i\). This gives \(\alpha_i + 1\) possible values (including 0).
Since the exponents are independent for different primes, the total number of distinct divisors is the product of the number of choices for each \(\beta_i\):
\[(\alpha_1 + 1)(\alpha_2 + 1) \cdots (\alpha_r + 1).\]Prove the corresponding theorem for the progression \(6n+5\) that there are infinitely number of primes inside.
Assume there are only infinit amount of prime numbers inside \(6n+5\)
Any prime greater than 2 is odd. And all of the odds greater than 2 can be described by
\[6n+3, 6n+5, 6n+7\]Furthermore, the product of 2 numbers of the form \(6n+3\) and \(6n+7\) are again of that form
\(\begin{aligned} (6a+3)(6b+3) = 36ab + 18a + 18b + 9 = 6(6ab+3a+3b+1) + 3 \end{aligned}\) \(\begin{aligned} (6a+7)(6b+7) = 36ab + 42a + 42b + 49 = 6(6ab+7a+7b+7) + 7 \end{aligned}\)
Construct N where: \(N = 6(p_1p_2\cdots p_n) - 1 = 6(p_1\cdots p_n - 1) + 5\)
None of the product from \(p_1\) to \(p_n\) could be N’s factor since it always remains with 1
And the factor can not be \(6n+3\) or \(6n+7\) because the product of them is still the form of themselves.
So it can only be 6n+5, which contractdicts with the assumption, therefore there are infinit number of primes inside it.s
Related proof: Dirichlet’s theorem. if gcd(a, b) = 1 in the progression \(an+b\), then there are infinite number of prime numbers inside.
Find a similar rule for divisibility by 13.
For congruences modulo 13 we have:
\[\begin{aligned} 10^0 &\equiv 1 \pmod{13} \\ 10^1 &\equiv 10 \pmod{13} \\ 10^2 &\equiv 100 \equiv 100 - 7 \cdot 13 \equiv 100 - 91 \equiv 9 \pmod{13}\\ 10^3 &\equiv 10 \cdot 10^2 \equiv 10 \cdot 9 \equiv 90 \equiv 90 - 7 \cdot 13 \equiv 90 - 91 \equiv -1 \equiv 12 \pmod{13}\\ 10^4 &\equiv 10 \cdot 10^3 \equiv 10 \cdot 12 \equiv 120 \equiv 120 - 9 \cdot 13 \equiv 120 - 117 \equiv 3 \pmod{13} \\ 10^5 &\equiv 10 \cdot 10^4 \equiv 10 \cdot 3 \equiv 30 \equiv 30 - 2 \cdot 13 \equiv 30 - 26 \equiv 4 \pmod{13} \\ 10^6 &\equiv 10 \cdot 10^5 \equiv 10 \cdot 4 \equiv 40 \equiv 40 - 3 \cdot 13 \equiv 40 - 39 \equiv 1 \pmod{13} \\ \end{aligned}\]the successive remainders then repeat, for the expression
\[z = a_0 + a_110 + a2_10^2 + \cdots + a_n10^n\]z is divisible by 13 if the following expression is divisible by 13.
\[r = a_0 + 10a_1 + 9a_2 + 12a_3 + 3a_4 + 4a_5 + a_6 + \cdots\]Show that the following law of cancellation holds for congruences with respect to a prime modulus: If \(ab \equiv ac\) and \(a\not\equiv0\), then \(b \equiv c\).
Using law of subtraction:
\[\begin{aligned} ab-ac \equiv 0 \pmod{p} \\ a(b-c) \equiv 0 \pmod{p} \end{aligned}\]By applying the law 7 stated in the book, since \(a\not\equiv0 \pmod{p}\), therefore
\[(b-c) \equiv 0 \pmod{p}\]which leads to
\(b \begin{aligned} \equiv c \pmod{p} \end{aligned}\).
Q.E.D.
To what number between 0 and 6 inclusive is the product \(11\times 18\times 2322\times 13\times 19\) congruent modulo 7? By computing the smaller congruences from the factors in the product.
\[\begin{aligned} 11 \equiv 4 \pmod{7} \\ 18 \equiv 4 \pmod{7} \\ 2322 \equiv 5 \pmod{7} \\ 13 \equiv 6 \pmod{7} \\ 19 \equiv 5 \pmod{7} \\ \end{aligned}\]Then we have \(11\times 18\times 2322\times 13\times 19 \equiv 4\times 4\times 5\times 6\times 5 \pmod{7}\)
\[\begin{aligned} 4\times 4 \equiv 2 \pmod{7} \\ 5\times 6 \equiv 2 \pmod{7} \\ \end{aligned}\]Therefore we have \(2\times 2 \times 5 \equiv 6\pmod{7}\), 6 the congruence of the product.
To what number between 0 and 12 inclusive is \(3\cdot7\cdot11\cdot17\cdot19\cdot23\cdot29\cdot113\) congruent modulo 13? By computing the smaller congruences from the factors in the product.
\[\begin{aligned} 17 \equiv 4 \pmod{13} \\ 19 \equiv 6 \pmod{13} \\ 23 \equiv 10 \pmod{13} \\ 29 \equiv 3 \pmod{13} \\ 113 \equiv 9 \pmod{13} \\ \end{aligned}\]To furthur make it smaller
\[\begin{aligned} 3\times 7 \equiv 8 \pmod{13} \\ 11\times 4 \equiv 5 \pmod{13} \\ 6\times 10 \equiv 8 \pmod{13} \\ 3\times 9 \equiv 1 \pmod{13} \\ \end{aligned}\]Therefore we have \(9\times 8 \times 5 \times 8 \equiv 8 \pmod{13}\), 8 is the congruence of the product.
To what number between 0 and 4 inclusive is the sum \(1 + 2 + 2^2 +\cdots +2^{19}\) congruent modulo 5?
We can observe from the following that the form 1,2,4,3 is repeated for the congruence in each term.
\[\begin{aligned} 1 \equiv 1 \pmod{5} \\ 2 \equiv 2 \pmod{7} \\ 2^2 \equiv 4 \pmod{5} \\ 2^3 \equiv 3 \pmod{5} \\ 2^4 \equiv 1 \pmod{5} \\ 2^5 \equiv 2 \pmod{5} \\ \cdots \end{aligned}\]Therefore the original sum can be rewritten as:
\[1 + 2 + 4 + 3 + \cdots\]There are 20 terms inside, for every 4 terms \(1+2+4+3 \equiv 0 \pod{5}\), so the congrence is 0.
Show by similar computation that
\[\begin{align*} 2^8 &\equiv 1 \pmod{17}; \\ 3^8 &\equiv -1 \pmod{17}; \\ 3^{14} &\equiv \pmod{29}; \\ 2^{14} &\equiv -1 \pmod{29}; \\ 4^{14} &\equiv 1 \pmod{29}; \\ 5^{14} &\equiv 1 \pmod{29}\\ \end{align*}\] \[\begin{aligned} 2^8 &\equiv 2^4 \cdot 2^4 \equiv (-1)\cdot(-1) \equiv 1 \pmod{17} \\ 3^8 &\equiv 3^4 \cdot 3^4 \cdot (-4)\cdot(-4) \equiv -1 \pmod{17} \\ 3^{14} &\equiv 3^4 \cdot 3^4 \cdot 3^4 \cdot 3^2 \equiv (-6) \cdot (-6) \cdot(-6) \cdot(9) \\ &\equiv 7 \cdot 9 \cdot (-6) \equiv -1 \pmod{29} \\ 2^{14} &\equiv 2^5 \cdot 2^5 \cdot 2^4 \equiv 3 \cdot 3 \cdot 16 \equiv 28 \equiv -1 \pmod{29} \\ 4^{14} &\equiv 4^3 \cdot 4^3 \cdot 4^3 \cdot 4^3 \cdot 4^2 \equiv 6 \cdot 6 \cdot 6 \cdot 6 \cdot 16 \equiv 1 \pmod{29} \\ 5^{14} &\equiv 5^2 \cdot 5^4 \cdot 5^8 \equiv 4 \cdot 16 \dot 24 \equiv 1 \pmod{29} \\ \end{aligned}\]Check Fermat’s theorem for p = 5, 7, 11, 17, and 23 with different values of a.
For a = 3
\[\begin{aligned} 3^4 &\equiv 3^2 \cdot 3^2 \equiv 9 \cdot 9 \equiv 4 \cdot 4 \equiv 1 \pmod{5} \\ 3^6 &\equiv 3^2 \cdot 3^2 \cdot 3^2 \equiv 2 \cdot 2\cdot 2 \equiv 1 \pmod{7} \\ 3^{10} &\equiv 3^2 \cdot 3^4 \cdot 3^4 \equiv 9 \cdot 4 \cdot 4 \equiv 1 \pmod{11} \\ 3^{16} &\equiv 3^4 \cdot 3^4 \cdot 3^4 \cdot 3^4 \equiv 13^4 \equiv 1 \pmod{17}\\ 3^{22} &\equiv 3^8 \cdot 3^8 \cdot 3^4 \cdot 3^2 \equiv (-6) \cdot (-6) \cdot 12 \cdot 9 \equiv 1 \pmod{23}\\ \end{aligned}\]Prove the general theorem: The smallest positive integer e for which \(a^e \equiv 1 \pmod{p}\) must be a divisor of p-1. (Hint: Divide p — 1 by e, obtaining:
\[p-1 = ke + r\]where \(0 < r < e\), and use the fact that \(a^{p-1} \equiv a^e \equiv 1 \pmod{p}\) )
Proof: Based on Fermat’s Little theorem, we have \(a^{p-1} \equiv 1 \pmod{p}\)
Dividing p-1 by e and obtain:
\[p-1 = ke + r\]where k is the quotient, r is the remainder, \(0 \leq r < e\).
Substitute it in the Fermat’s Little theorem
\[a^{p-1} = a^{ke + r} \equiv 1 \pmod{p}\]Simplify using the fact that \(a^e \equiv 1 \pmod{p}\) :
\[(a^e)^k \equiv 1^k \equiv 1 \pmod{p}\]Therefore,
\[a^{ke + r} \equiv 1 \cdot a^r \equiv a^r \pmod{p}\]And this implies
\[a^r \equiv 1 \pmod{p}\]r can only be 0 because e is defined as the smallest positive integer for which \(0 \leq r < e\)
Therefore, r must be 0, meaning:
\[p-1 = ke\]Q.E.D.
\(6^2 = 36 \equiv 13 \pmod{23}\). Is 23 a quadratic residue (mod 13)?
Based on the definition of quadratic resudule, 23 is a quadratic residule if that:
\[x^2 \equiv 23 \pmod{13}\]First, we can simplify \(23 \pmod{13}\): \(23 \equiv 10 \pmod{13}\)
The integers modulo 13 are \(0, 1, 2, \ldots, 12\). By calculating the squares of each of these integers modulo 13:
\[\begin{aligned} 1^2 &\equiv 1 \pmod{13} \\ 2^2 &\equiv 4 \pmod{13} \\ 3^2 &\equiv 9 \pmod{13} \\ 4^2 &\equiv 16 \equiv 3 \pmod{13} \\ 5^2 &\equiv 25 \equiv 12 \pmod{13} \\ 6^2 &\equiv 36 \equiv 10 \pmod{13} \\ 7^2 &\equiv 49 \equiv 10 \pmod{13} \\ 8^2 &\equiv 64 \equiv 12 \pmod{13} \\ 9^2 &\equiv 81 \equiv 3 \pmod{13} \\ 10^2 &\equiv 100 \equiv 9 \pmod{13} \\ 11^2 &\equiv 121 \equiv 4 \pmod{13} \\ 12^2 &\equiv 144 \equiv 1 \pmod{13} \end{aligned}\]we have
\[6^2 \equiv 10 \pmod{13} \quad \text{and} \quad 7^2 \equiv 10 \pmod{13}\]Thus, there exist integers \(x\) specifically \(x = 6\) and \(x = 7\) such that \(x^2 \equiv 10 \pmod{13}\). Therefore, 10 is a quadratic residue modulo 13.
Also we could use the law of Quadratic reciprocity
\[\frac{p-1}{2} \cdot \frac{q-1}{2} = \frac{13-1}{2} \cdot \frac{23-1}{2} = 66\]66 is even, therefore 23 is a quadratic residue modulo 13.
We have seen that \(x^2 \equiv (p - x)^2 \pmod{p}\). Show that these are the only congruences among the numbers \(1^2, 2^2, 3^2, \ldots, (p - 1)^2\).
Let’s construct another variable y, we have distinct x and y where \(x,y \in \{1,2,...;p-1\}\)
The goal of this proof is to claim if we can have \(x^2 \equiv y^2 \pmod{p}\) leads to \(y=p-x\)
rewrite the equation and we have
\[(x-y)(x+y) \equiv 0 \pmod{p}\]since p is a prime, it must devide either \(x-y \equiv 0\) where $$\left | x-y\right | < p-2 < p\(or\)x+y \equiv 0$$ where |
\(2 \leq x+y < 2p\).
x-y can only be divisible by p is \(x - y = 0\) which contradicts to the distinct consurtcion, therefore
\[x+y = p\]which leads to \(y = p-x\).
Q.E.D.
Prove the last statement. Page 41.
Prove the last statement: (u, v) have no common factor and are not both odd, yield all primitive Pythagorean number triples.
The statement has:
\[\gcd(u, v) = 1\]To cover all primitive Pythagorean triples:
Construct \(u = n\), \(v = n + 1\). where n is any positive integer.
It satisfis both are not odd, and \(\gcd(n, n+1) = 1\).
\[\begin{aligned} a &= (n+1)^2 - n^2 = 2n + 1 \\ b &= 2(n+1)n = 2n^2 + 2n \\ c &= (n+1)^2 + n^2 = 2n^2 + 2n + 1 \end{aligned}\] \[\begin{aligned} a^2 + b^2 &= (2n+1)^2 + (2n^2+2n)^2 \\ &= 4n^2 + 4n + 1 + 4n^4 + 8n^3 + 8n^2 + 4n^2 \\ c^2 &= (2n^2+2n+1)^2 = 4n^4 + 8n^3 + 8n^2 + 4n + 1 \\ \end{aligned}\]and it satisfied \(a^2 + b^2 = c^2\).
Carry out the Euclidean algorithm for finding the greatest common divisor of (a) 187, 77. (b) 105, 385. (c) 245, 193.
\[\gcd(187, 77) = 11\] \[\begin{array}{r|r} 187 & 77 \\ 77 & 33 \\ 33 & 11 \\ 11 & 0 \\ \end{array}\] \[\gcd(105, 385) = 35\] \[\begin{array}{r|r} 385 & 105 \\ 105 & 70 \\ 70 & 35 \\ 35 & 0 \\ \end{array}\] \[\gcd(245, 193) = 1\] \[\begin{array}{r|r} 245 & 193 \\ 193 & 52 \\ 52 & 37 \\ 37 & 15 \\ 15 & 7 \\ 7 & 1 \\ 1 & 0 \\ \end{array}\]The extension of this argument to products of any number n of integers requires the explicit or implicit use of the principle of mathematical induction. Supply the details of this argument. Page 47.
Assume that the statement is true for some integer \(n \geq 2\). That is, if $p$ divides the product \(a_1 a_2 \cdots a_n\), then \(p\) must divide at least one of the integers \(a_1, a_2, \ldots, a_n\).
We need to show that the statement is also true for \(n+1\). After adding a new term \(a_{n+1}\)
Consider the product \(a_1 a_2 \cdots a_n\). There are two cases to consider:
Case 1: \(p\) divides \(a_1 a_2 \cdots a_n\): By the inductive hypothesis, since \(p\) divides \(a_1 a_2 \cdots a_n\), $p$ must divide at least one of the integers \(a_1, a_2, \ldots, a_n\). Adding another term \(a_{n+1}\) is the same.
Case 2: \(p\) does not divide \(a_1 a_2 \cdots a_n\): Since \(p\) divides \(a_1 a_2 \cdots a_n a_{n+1}\) and does not divide \(a_1 a_2 \cdots a_n\), it must be the case that \(p\) divides \(a_{n+1}\).
In both cases, \(p\) divides at least one of the integers \(a_1, a_2, \ldots, a_n, a_{n+1}\).
By the principle of mathematical induction, the statement is true for all integers \(n \geq 2\).
Q.E.D.
Exercise: Prove the theorem: If an integer r divides a product ab and is relatively prime to a, then r must divide b.
r is relatively prime to a, we can have integers k and l such that
\[kr + la = 1\]Multiply both sides of this equation by b.
\[bkr + lab = b\]If r divides ab, we can have
\[rq = ab\]Therefore we have
\[\begin{aligned} bkr + lrq &= b \\ r(bk + lq) & = b \end{aligned}\]from which it is evident that r divides b.
Q.E.D.
Using Euler’s \(\phi\) function, generalize Fermat’s theorem of page 37. The general theorem states: If n is any integer, and a is relatively prime to n, then
\[a^{\phi(n)} \equiv 1 \ (\text{mod } n).\]To generalize Fermat’s theorem, first consider the multiplies (set A) the number of intergers \(\phi{n}\) with n, the multiplies (or the modulo n if it is larger than n) are again belongs to \(\phi(n)\).
\[\{m_1 = a \cdot A_1 , m_2 = a \cdot A_2 , \cdots m_{A_{\phi{(n)}}}\}\]Since a is coprime to n, \(gcd(n, a)=1\)
\[\{A_1 \equiv a \cdot A_1 \pmod{n}, A_2 \equiv a \cdot A_2 \pmod{n}, \cdots... A_i \equiv a \cdot A_i \cdot \phi{(n)}\pmod{n}\}\]Therefore, the products of the elements of the two sets are congruent modulo n, we have
\[\{A_1 \cdot A_2 \cdots A_n \equiv A_1 \pmod{n}, A_2 \equiv a \cdot A_2 \pmod{n}, \cdots... A_i \equiv a \cdot A_i \cdot \phi{(n)}\pmod{n}\}\]Using cancellation rule we can have
\[a^{\phi(n)} \equiv 1 \ (\text{mod } n).\]Q.E.D.
Exercise: Find the continued fraction developments of
\[\frac{2}{5}, \quad \frac{43}{30}, \quad \frac{169}{70}.\]By applying the same procedure in the book page 49, we can have
\[\frac{2}{5} = 0+\frac{1}{2+\frac{1}{2}}\] \[\frac{43}{30} = 1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}\] \[\frac{169}{70} =2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}}\]Solve the Diophantine equations \((a) 3x — 4y = 29. (b)3 - 4y = 58. (c) 11x - 12 y = 51\).
\[(a). 3x - 4y = 29\] \[\gcd(3, 4) = 1 \quad \text{it has integer solutions}\] \[\begin{aligned} 4 - 3 &= 1 \\ 3(-1) - 4(-1) &= 1 \\ \end{aligned}\]And we can have
\[3(-29) - 4(-29) = 29\]Based on tge general solution of linear diophantinue equation we can have
\[\begin{aligned} x &= -29 + 4k \\ y &= -29 + 3k \\ \end{aligned}\] \[(b). 11x + 12y = 58\] \[\gcd(11, 12) = 1\] \[11(-1) + 12(1) = 1\] \[11(-58) + 12(58) = 58\]Based on tge general solution of linear diophantinue equation we can have
\[\begin{aligned} x &= -58 + 12k \\ y &= 58 - 11k \\ \end{aligned}\] \[(c). 153x - 34y = 51\] \[\gcd(153, 34) = 17\] \[153 + 4 \times 34(1) = 17\] \[153(3) + 4 \times 34(3) = 51\]Based on tge general solution of linear diophantinue equation we can have
\[\begin{aligned} x &= 3 + 2k \\ y &= 12 + 9k \\ \end{aligned}\]The information presented on the blog is ONLY intended for knowledge sharing and understanding how supplements work, it is NOT intended as medical advice or as a substitute for medical treatment. You should always consult with your physician or other health care professional before taking any medication or nutritional, herbal or homeopathic supplement.
Last week, I had a thought-provoking discussion with my colleagues. The topic of our conversation? Supplements. What intrigued me was that, just like many of my peers, I too have ventured into the world of supplements because of David Sinclair and various of health-related podcasts like Huberman lab. While the reasons taking supplements is easy for me: looking to combat inflammation. For my colleague, this could be boost performance when working out.
A pill that promises health improvements, anti-aging effects, but to be honest I had very limited knowledge of those small pills I’m taking, I yearn for comprehension.What exactly are these supplements composed of at a molecular level? How do these minuscule pills exert their influence within our bodies biologically? Moreover, is there an optimal approach to taking these supplements to maximize their potential benefits for our health and well-being?
Over the past few days, I’ve immersed myself into articles, research papers, and scientific literature. My aim was clear: to understand these tiny capsules that promise health. In the sections to come, I included 10 different supplements and introduced them each from the very essence of these their chemical identities, their categorizations, their mechanisms of action with the academic evidence. In the end a suggest dose for those supplements was given.
Resveratrol (RV) is a potent polyphenolic compound found in a variety of fruits and vegetables, including grapes, skin of peanuts, mulberries, blueberries, bilberries, soybeans, pomegranates, cranberries, soy, dark chocolate, pistachios. It possesses a distinct chemical structure, primarily composed of two aromatic rings connected by a methylene bridge. Biologically when you take resveratorl, it is rapidly metabolized by phase II enzymes in the intestine and liver. The RV’s low solubility(< 0.05mg/ml) in water is caused by its enzymatic structure, which might affect its consumption.[1]
Resveratrol occurs naturally in two forms: trans and cis isomers, with the trans-isomer being the more commonly associated with health benefits. It is typically a white powder extracted using methanol, characterized by a melting point between 253°C and 255°C, and has a relatively low molecular weight of 228.25.[1] While trans-resveratrol is the form primarily studied for its potential health advantages, cis-resveratrol is less prevalent and is typically produced synthetically rather than being the predominant form found in natural sources. The variation in the spatial arrangement of atoms around the double bond results in differences in the chemical and biological properties of these two isomers. Trans-resveratrol is generally considered the more biologically active and bioavailable form, making it the focal point of most research on the potential health benefits of resveratrol.
It has been discovered that RV has many potential health benefits like antioxidant , cardioprotective, neurological, anti-inflammatory, antiplatelet, blood glucose-lowering, and anticancer activities.[1] In another famous experiments in nonhuman primates carried out by Laboratory of Cardiovascular Science, resveratrol has been found to prevent high fat/sucrose (HFS) diet-induced central arterial wall inflammation and stiffening.[2] Below is a simplified summary of its actions:
SIRT1/3 Activation: Resveratrol activates SIRT1 and SIRT3 enzymes (SIRT1 and SIRT3 is one of Sirtuin, which are types of proteins involved in metabolic regulation, inflammation), Sirt1 has been implicated in many vital processes, such as DNA repair, cell survival, gluconeogenesis, muscle cell differentiation, cell-cycle regulation, lipid metabolism, fat mobilization and insulin sensitivity. [3]
Neuroprotective Effects: It reduces endoplasmic reticulum (ER) stress, inhibits carbonyl protein activity, and supports factors associated with brain health.
Cardiometabolic Benefits: Resveratrol improves mitochondrial (The mitochondria are like little workers in a factory. They take the food you eat, use the oxygen you breathe, and create energy for your body) respiratory activity in the heart, increases adenosine monophosphate kinase (p-AMPK) levels (AMPK acts as a sensor of cellular energy status and helps regulate various metabolic processes to ensure the cell has enough energy to function), and boosts SOD2 (a family of enzymes that play a crucial role in protecting cells from damage caused by free radicals) levels, promoting cardiovascular health.
Inflammation Regulation: Resveratrol reduces inflammatory parameters, including interleukin-6 IL-6 (IL-6 involved in promoting inflammation, regulating immune cell activity) and Tumor Necrosis Factor-Alpha TNF-α, which plays a significant role in managing oxidative stress.
Anti-Cancer Properties: Resveratrol inhibits the production of NADPH oxidase NOX and myeloperoxidase, showing potential anti-cancer effects.
Neurological Health: Resveratrol inhibits Nitric oxide synthase iNOS production, protects against lipid peroxidation, and has potential therapeutic effects for Alzheimer’s disease (AD) by promoting amyloid precursor protein cleavage and overcoming oxidative stress.
Reproductive System Benefits: Resveratrol improves sperm mobility, increases zinc concentration, and counteracts prooxidative effects in the reproductive system.
Intestinal Health: Resveratrol reduces markers like IL-8, Vascular Endothelial Growth Factor VEGF (VEGF’s normal function is to create new blood vessels, but overexpression of VEGF can cause vascular disease), and Reactive oxygen species ROS (ROS are like energetic troublemakers in your body that, when there are too many of them, can create a bit of chaos and cause damage to your cells), indicating a role in managing intestinal disorders. [4]
The figure below shows the general benefits of resveratrol.
Nicotinamide Mononucleotide (NMN). In humans, several enzymes use NMN to generate nicotinamide adenine dinucleotide (NADH). (NAD+ (Nicotinamide Adenine Dinucleotide) is a crucial coenzyme in cells that plays a fundamental role in various biological processes, you can think of it as coal for energy production, NAD+ was found gradually declining during aging.[6]) NMN can be found in fruits and vegetables such as broccoli, cabbage, cucumber and avocado at a concentration of about 1 mg per 100g. In mammals most NMN is synthesized from vitamin B3 in the form of nicotinamide. [5]
NMN consists of three main components: a nicotinamide group, a ribose sugar, and a phosphate group. The nicotinamide group is connected to the ribose unit, and the phosphate group is attached to the ribose, forming the complete NMN molecule. In the human body, it can be found in placenta tissue and body fluids such as blood and urine. [7]
NAD+ Decline During Aging:
Aging is naturally associated with a decline in NAD+ levels. Enzymes like sirtuins,CD38/CD157, PARP, TNKS, and BST contribute to NAD+ consumption during aging. [7] Chronic inflammation and oxidative stress hinder NAD+ biosynthesis. Reduced NAD+ levels lead to interruptions in cellular communication and mitochondrial dysfunction, contributing to age-related complications.
Why NMN Could Work as a Solution:
NMN (Nicotinamide Mononucleotide) is an intermediate compound in the NAD+ biosynthesis (simple compounds are modified, converted into other compounds) pathway. Administering NMN can compensate for the decline in NAD+ levels during aging. NMN supplementation may activate SIRT1, a protein associated with anti-aging effects. Studies suggest NMN as a potential solution to restore NAD+ levels and counteract aging effects. The effect could be summarized by the figure [7] shown below
The term “ubiquinol” derives from “ubiquitous,” signifying its widespread presence. Ubiquinol is an electron-rich (reduced) form of coenzyme Q (ubiquinone), aka CoQ10. In 1957, Professor Frederick Crane and colleagues discovered CoQ10 from beef heart mitochondria at the University of Wisconsin-Madison Enzyme Institute.6 R.A. Morton, from the United Kingdom, isolated the compound in rat liver just after Dr. Crane’s discovery. It was Morton who named the compound ubiquinone, meaning ubiquitous quinone or one that “exists everywhere.”. In 1962, Peter D. Mitchell, PhD from University of Edinburgh determined how CoQ10 produces energy at the cellular level and in 1978 he was awarded the Nobel Prize for chemistry based on his discovery. Q10 normally could be found in Fish, Calf’s Liver (and other organ meats), Germ portion of whole grains in nature. [8]
CoQ10, also known as ubiquinone, is a necessary component of cell respiration and ATP production. Ubiquinone is a fat soluble substance used to form ubiquinol, the fully reduced form of ubiquinone and potent antioxidant. [8]
Chemical Structure of CoQ10: The most common form in humans is Coenzyme Q10 (CoQ10 or ubiquinone-10). CoQ10 is a 1,4-benzoquinone, where the “Q” refers to the quinone chemical group, and “10” signifies the number of isoprenyl chemical subunits in its tail. (Ubiquinone Tail Variability: In natural ubiquinones, there can be from six to ten subunits in the tail. The specific variant, ubiquinone-10, is predominantly found in humans.)
Role in Energy Production: CoQ10 is a family of fat-soluble substances, resembling vitamins, and is present in all respiring eukaryotic cells, primarily in the mitochondria.
Mitochondrial Function: As a vital component of the electron transport chain, CoQ10 participates in aerobic cellular respiration, a process that generates energy in the form of ATP.
Energy Generation: A substantial 95% of the human body’s energy is generated through aerobic cellular respiration, highlighting the significance of CoQ10.
Organ-Specific Concentrations: Organs with high energy demands, such as the heart, liver, and kidney, exhibit the highest concentrations of CoQ10, emphasizing its critical role in meeting the energy requirements of these vital organs.
The effect of Ubiquinol is much better than that of ordinary Q10 (ubiquinone), and currently the only supplier of Ubiquinol on the market is Kaneka from Japan. [8]
Moreover, CoQ10 can increase the production of key antioxidants such as superoxide dismutase, an enzyme capable of reducing vascular oxidative stress in hypertensive patients. [9] CoQ10 reduces levels of lipid peroxidation via the reduction of pro-oxidative compounds. [10] So peroxidation is a chain reaction commonly found in polyunsaturated fatty acids, and its products may be carcinogenic and mutagenic. Since lipids are components of cell membranes, lipid peroxidation may have disastrous consequences if it’s out of control inside body.
Astaxanthin (ASX) is a xanthophyll carotenoid, chemically known as 3,3′-dihydroxyβ,β-carotene-4,4′-dione, It is a lipid-soluble pigment with red coloring properties. It was originally isolated from lobster by Kuhn in 1938. [11] ASX is primarily biosynthesised by various algae, bacteria and fungi and is consumed by marine animals, such as salmon, trout, crab, lobster and shrimp. [12]
Haematococcus pluvialis, is reported to accumulate the highest levels of ASX in nature.
The skin comprises epidermis, dermis and hypodermis, which form the protectively outermost barrier against external environmental stresses such as repeated sun UV ray exposure. UV radiation comprises three types: UVA(Long-wave UV, Passes largely through glass and clothing), UVB(Medium-wave UV, mostly absorbed by the ozone layer, Does not pass through ‘normal’ glass. but does pass through quartz glass, Causes other forms of skin cancer.) and UVC(Short-wave UV, germicidal UV, ionizing radiation at shorter wavelengths, completely absorbed by the ozone layer and atmosphere: hard UV..s very destructive to skin cells). UVC is filtered out by atmospheric ozone for the most part, while both UVA and UVB can cause a biological change in the skin. The overview could be found from the Figure [11] below.
The mechanism of action of Astaxanthin (ASX) involves its multifaceted impact on cellular and molecular processes, particularly in mitigating skin aging.
Through different and various studies, it was proved that the proanthocyanidin rich grape seed extract provides benefits against many diseases i.e. inflammation, cardiovascular disease, hypertension, diabetes, cancer, peptic ulcer, microbial infections, etc.
The chemical composition of Grape Seed Extract (GSE) is rich and diverse, encompassing various components obtained through specific extraction processes. Grape seeds are manually separated from grapes, air-dried, and ground to a fine powder.The powdered grape seed is macerated in 70% ethanol, filtered, and the filtrate is dried to obtain powdered GSE. Dried grape seeds contain approximately 35% fiber, 29% extractable components, including phenolic compounds, proteins (11%), minerals (3%), and water (7%). [13]
The mechanisms of action of Grape Seed Extract (GSE) are diverse and involve a range of bioactive compounds present in the extract.
Serratiopeptidase is a zinc containing metalloprotease of molecular weight 45–60 kDa. ($1Da = 1.66053906660(50) \times 10^{-27}kg$). It is originally obtained from Serratia marcescens isolated from the intestine of the silkworm Bombyx mori. [14]
The graph below is Serratia marcescens:
While errapeptase and other proteolytic enzymes break down proteins by hydrolyzing them, breaking the peptide bonds between amino acids (the building blocks of proteins). It is well documented that oral administration of proteolytic enzymes, including serrapeptase, is fully absorbed into the bloodstream and exerts effects throughout the body.
Vitamin D is a lipid-soluble prohormone that is vital for the maintenance of bone and muscle health by promoting the absorption and metabolism of calcium and phosphate. [16] In addition to food sources such as fatty fish, eggs, fortified milk and cod liver oil, the human body uses ultraviolet B (UVB) radiation from sunlight to synthesise a significant portion of vitamin D requirements.
The figure below is from Catalyst University, which shows the process of biosynthesis of Vitamin D.
Starting from the up-left of the graph above, the synthesis of vitamin D starts with the oxidation of cholesterol to 7-dehydrocholesterol (7-DHC). (7-dehydrocholesterol is part of the metabolic pathway that controls the synthesis of cholesterol in human cells.) 7-DHC is then transported to the skin and is stored in the cell membranes of keratinocytes and fibroblasts in the epidermis of skin. [16] In the skin, 7-DHC is photolysed by UVB (290–315 nm) to previtamin $D_3$, which is converted to inactive vitamin D (Cholecalciferol) by photolysis-mediated thermo-isomerisation. (What is thermal isomerization?)
To become biologically active, the vitamin D originating from dermal production or dietary sources undergoes a series of enzymatic conversions in the liver and kidney. Vitamin D is transported to the liver by vitamin D binding protein (DBP), together they are transported to the liver where the cytochrome P450 enzyme 25-hydroxylase ([CYP2R1]) adds a hydroxylgroup on carbon 25 to produce a major circulating form of vitamin D (25-(OH)D), aka Calcifediol, this process on the chemical level can be seen from the graph below:
The Calcifediol then circulates in the bloodstream bound to DBP and it must be further hydroxylated at a different site in the kidney tubules to gain hormonal bioactivity. Hydroxylation at position $1\alpha$ by the mitochondrial cytochrome P450 enzyme 25-hydroxyvitamin D-$1\alpha$-hydroxylase (CYP27B1) of kidney converts Calcifediol to 1α,25-dihydroxyvitamin D (aka calcitriol, is the most active form of Vitamin D, ‘trio’ because it has 3 hydroxl groups), the active form of vitamin D that plays an essential role in mineral homeostasis.
As the graph (from Catalyst University) shown above, the receptor of Calcifediol is not existed in the plasma membrane since it is hydrophobic, instead it is inside nucleus. Calcifediol is bound with Vitamin D receptor (VDR) VDR thens binds to DNA as VDR/VDR homodimers or VDR/RXR heterodimers in order to regulate gene expression. [16]. The dimers subsequently recognise and bind with transcription factor IIB (TFIIB) to a vitamin D response element (VDRE) located in the promoter region of target genes and leads to transcriptional suppression or activation of vitamin D response genes.
Vitamin K is a family of structurally similar, fat-soluble vitamers found in foods and marketed as dietary supplements. Vitamin K is important for the function of numerous proteins within the body, such as the coagulation factors. [18]
Vitamin K exists naturally as vitamin K1 (phylloquinone) and vitamin K2 (menaquinone(), MK-4 through MK-10). Vitamin K1 is mainly found in green leafy vegetables as well as olive oil and soyabean oil, whereas vitamin K2 (menaquinone) is found in small amounts in chicken, butter, egg yolks, cheese and fermented soyabeans (better known as natto).
Vitamin K1 is involved in the activation of other clotting factors, including Factors VII, IX, and X. These factors play roles in the coagulation process, preventing excessive bleeding. The absorption rate of vitamin K1 from kale and spinach – foods identified as having a high vitamin K content – are on the order of 4% to 17% regardless of whether raw or cooked. Vitamin K2 is synthesized by bacteria in the human gut. Some of the vitamin K1 from dietary sources is converted into vitamin K2 by these gut bacteria.
Electrolytes such as sodium, potassium, calcium, and magnesium are physiologically important due to their integral roles in metabolic and biologic processes. [21] Magnesium (${Mg}^{2+}$) possesses two hydration shells, making its hydrated radius larger than that of other cations (${Na}^+$, $K^+$, and ${Ca}^{2+}$).
The ‘hydrated radius’ is the effective size of the ion when surrounded by water molecules. In the case of magnesium, it has two layers or shells of water molecules around it. Because magnesium has two hydration shells, the presence of these water layers makes the effective size or radius of the hydrated magnesium ion larger compared to other cations (positively charged ions).
Magnesium is an essential mineral which is a cofactor in over 300 enzymatic reactions for protein synthesis such as muscle contraction, nerve function, blood pressure and hormone binding. [22]. Magnesium is required for energy production, oxidative phosphorylation, and glycolysis. It contributes to the structural development of bone and is required for the synthesis of DNA, RNA, and the antioxidant glutathione. Magnesium also plays a role in the active transport of calcium and potassium ions across cell membranes, a process that is important to nerve impulse conduction, muscle contraction, and normal heart rhythm. [23]
As we discussed in the Vitamin D section, vitamin D can’t work without first being converted into a form your body can absorb. The level to which a vitamin or mineral can be absorbed is known as its “bioavailability.” Vitamin D’s bioavailability depends on magnesium. The enzymes (in the liver and kidneys) that enable vitamin D metabolism—converting it into its active form, calcitriol — can’t work without sufficient amounts of magnesium to draw upon.
Early signs of magnesium deficiency include loss of appetite, nausea, vomiting, fatigue, and weakness. Therefore proper intake of magnesium from food or supplements is important. Magnesium supplements are available in a variety of forms. The following information comes from Dr. Jin W. Sung. ‘s YouTube Channel [23].
Fish Oil: Fish oil is a dietary supplement derived from the tissues of fatty fish, such as salmon, mackerel, and sardines. It is rich in omega-3 fatty acids, which are essential for various physiological functions in the body.
Omega-3 Fatty Acids: Omega-3 fatty acids are a type of polyunsaturated fat that includes three main types: eicosapentaenoic acid (EPA), docosahexaenoic acid (DHA), and alpha-linolenic acid (ALA). EPA and DHA are commonly found in fish oil, while ALA is present in plant-based sources like flaxseeds and walnuts.
Omega-6 Fatty Acids: Omega-6 fatty acids are another group of polyunsaturated fats that include linoleic acid (LA) and arachidonic acid (AA). These fats are found in various vegetable oils and nuts.
Supplements come in various forms, and understanding their safe levels is crucial. The following is a brief corresponding supplement suggested dosage based on Observed Safe Level (OSL) of human body.
Understanding the motivation behind exploring supplements is essential. However, it’s equally important to approach these substances with caution. Many supplements listed here are best absorbed when taken with a fatty meal. This enhances their bioavailability, ensuring your body can make the most of their benefits.
Remember, the decision to supplement should not be taken lightly. Always consult with a healthcare professional before introducing any new supplements into your meal. In such cases, seeking guidance from a doctor or checking a reliable drug interaction website (e.g. https://reference.medscape.com/drug-interactionchecker) is advisable to ensure compatibility and prevent any potential adverse effects.
[1] Zhang, Li-Xue, Chang-Xing Li, Mohib Ullah Kakar, Muhammad Sajjad Khan, Pei-Feng Wu, Rai Muhammad Amir, Dong-Fang Dai, et al. ‘Resveratrol (RV): A Pharmacological Review and Call for Further Research’. Biomedicine & Pharmacotherapy 143 (November 2021): 112164. https://doi.org/10.1016/j.biopha.2021.112164.
[2] Mattison, Julie A., Mingyi Wang, Michel Bernier, Jing Zhang, Sung-Soo Park, Stuart Maudsley, Steven S. An, et al. ‘Resveratrol Prevents High Fat/Sucrose Diet-Induced Central Arterial Wall Inflammation and Stiffening in Nonhuman Primates’. Cell Metabolism 20, no. 1 (July 2014): 183–90. https://doi.org/10.1016/j.cmet.2014.04.018.
[3] Marques, Francine Z., M. Andrea Markus, and Brian J. Morris. ‘Resveratrol: Cellular Actions of a Potent Natural Chemical That Confers a Diversity of Health Benefits’. The International Journal of Biochemistry & Cell Biology 41, no. 11 (November 2009): 2125–28. https://doi.org/10.1016/j.biocel.2009.06.003.
[4] Y. Bai, H. Yang, G. Zhang, L. Hu, Y. Lei, Y. Qin, Y. Yang, Q. Wang, R. Li, Q. Mao, Inhibitory effects of resveratrol on the adhesion, migration and invasion of human bladder cancer cells, Mol. Med. Rep. 15 (2) (2017) 885–889.
[5] Shade C. The Science Behind NMN-A Stable, Reliable NAD+Activator and Anti-Aging Molecule. Integr Med (Encinitas). 2020 Feb;19(1):12-14. PMID: 32549859; PMCID: PMC7238909.
[6] Schultz MB et al. Why NAD+ Declines during Aging: It’s Destroyed. Cell Metab. 2016 June 14; 23(6): 965–966
[7] Nadeeshani, H., Li, J., Ying, T., Zhang, B., & Lu, J. (2021). Nicotinamide mononucleotide (NMN) as an anti-aging health product – Promises and safety concerns. Journal of Advanced Research. doi:10.1016/j.jare.2021.08.003
[8] Dyer, Andrew R. ‘CoQ10: A Literature Review’ 40, no. 1 (n.d.).
[9] Kędziora-Kornatowska K, Czuczejko J, Motyl J, Szewczyk-Golec K, Kozakiewicz M, Pawluk H, Kędziora J, Błaszczak R, Banach M, Rysz J. Effects of coenzyme Q10 supplementation on activities of selected antioxidative enzymes and lipid peroxidation in hypertensive patients treated with indapamide. A pilot study. Arch Med Sci. 2010 Aug 30;6(4):513-8.
[10] Cordero MD, Cano-García FJ, Alcocer-Gómez E, De Miguel M, Sánchez-Alcázar JA. Oxidative stress correlates with headache symptoms in fibromyalgia: coenzyme Q₁₀ effect on clinical improvement. PLoS One. 2012;7(4):e35677.
[11] Zhou, Xiangyu, Qingming Cao, Caroline Orfila, Jian Zhao, and Lin Zhang. ‘Systematic Review and Meta-Analysis on the Effects of Astaxanthin on Human Skin Ageing’. Nutrients 13, no. 9 (24 August 2021): 2917. https://doi.org/10.3390/nu13092917.
[12] Higuera-Ciapara, I.; Felix-Valenzuela, L.; Goycoolea, F. Astaxanthin: A review of its chemistry and applications. Crit. Rev. Food Sci. Nutr. 2006, 46, 185–196.
[13] Gupta, Madhavi, Sanjay Dey, Daphisha Marbaniang, Paulami Pal, Subhabrata Ray, and Bhaskar Mazumder. ‘Grape Seed Extract: Having a Potential Health Benefits’. Journal of Food Science and Technology 57, no. 4 (April 2020): 1205–15. https://doi.org/10.1007/s13197-019-04113-w.
[14] Jadhav, Swati B., Neha Shah, Ankit Rathi, Vic Rathi, and Abhijit Rathi. ‘Serratiopeptidase: Insights into the Therapeutic Applications’. Biotechnology Reports 28 (December 2020): e00544. https://doi.org/10.1016/j.btre.2020.e00544.
[15] L. Desser, A. Rehberger, E. Kokron, W. Paukovits, Cytokine synthesis in human peripheral blood mononuclear cells after oral administration of polyenzyme preparations, Oncology. 50 (1993) 403–407.
[16] Basit, S. ‘Vitamin D in Health and Disease: A Literature Review’. British Journal of Biomedical Science 70, no. 4 (1 January 2013): 161–72. https://doi.org/10.1080/09674845.2013.11669951.
[17] Ginde AA, Mansbach JM, Camargo CA Jr. Association between serum 25-hydroxyvitamin D level and upper respiratory tract infection in the Third National Health and Nutrition Examination Survey. Arch Intern Med 2009; 169: 384–90.
[18] DiNicolantonio, James J, Jaikrit Bhutani, and James H O’Keefe. ‘The Health Benefits of Vitamin K’. Open Heart 2, no. 1 (October 2015): e000300. https://doi.org/10.1136/openhrt-2015-000300.
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[21] L.L. Yu, W.C. Davis, Y. Nuevo Ordonez, S.E. Long, Fast and accurate determination of K, Ca, and Mg in human serum by sector field ICP-MS, Anal. Bioanal. Chem. 405 (27) (2013) 8761–8768.
[22] Author Dr. Jin W. Sung. “Magnesium—-Comprehensive Guide. Forms of magnesium, benefits, dosages, impact on Vitamin D.” YouTube, uploaded by Dr. Jin W. Sung, June 2023,https://www.youtube.com/watch?v=Z4ohSzChgII&ab_channel=Dr.JinW.Sung
[23] Rude RK. Magnesium. In: Ross AC, Caballero B, Cousins RJ, Tucker KL, Ziegler TR, eds. Modern Nutrition in Health and Disease. 11th ed. Baltimore, Mass: Lippincott Williams & Wilkins; 2012:159-75.
[24] Simopoulos, Ap. ‘Omega-3 Fatty Acids in Health and Disease and in Growth and Development’. The American Journal of Clinical Nutrition 54, no. 3 (September 1991): 438–63. https://doi.org/10.1093/ajcn/54.3.438.
]]>人生的秩序性来自于安排好的一顿饭，一个和朋友的见面，一场旅行，我的2022开始于波罗的海三小国的一次旅行: 在拉脱维亚一家参观喝南瓜汤，吃了随便点的一道菜，随机性给人惊喜。在爱沙尼亚，我在火车上仔细端详一位在报纸上读读写写的老奶奶，拍着窗外徐徐向后行走的雪地。下了火车，在路边看蜷缩在河边的鸭子和天鹅，一父子在旁边逗白天鹅，还喊着白天鹅和他们回家。后来，搭上去芬兰的轮渡，再一次到赫尔辛基，已然是冰天雪地，寒风凛冽，想起上一次夏天来的时候，这种冷峻的气息悄然隐藏了，可能夏天是芬兰的保护色，行人还是一如既往得冷漠。最后在瑞典看完瓦萨博物馆和斯德哥尔摩的地铁，一天之内瑞典这个国家的美把我的审美上限提升了一个级别。
而人生的混乱来自于各种不确定性，可以是吃饭噎到，也可以是飞机误点，或者是一个离谱政策的落地，甚至是一场战争。劳东燕在一月写的《直面真实的世界》遭到封禁，我那一段时间每天晚上睡前都要拿出来读一下，那一句 “疫情之下，岁月静好的想象终于难以维续，因为兵荒马乱的日子，距离每个人都几乎只有一步之遥。”，而这兵荒马乱的来源，不是病毒，更多的是这个刻舟求剑的清零政策，一个不以人为中心的冷酷冰冷政策下，剩下的只有荒谬和无情，人性的丑恶，在这个时间里像弱者伸出援助之手的人，一定会被一直记住。世界的另外一端，普京宣布进攻乌克兰，我难以想象2021年我在乌克兰基辅的地铁下，还和同学开玩笑说这地铁这么深，打战了肯定全部逃进地铁。随着乌克兰东部逐渐沦陷，战争的影响远超蝴蝶效应的速度，很快欧洲陷入了供应链的危机，能源危机接踵而来，超市里的食用油供不应求，能源账单指数上升，物价上升，肉眼可见超市的所有产品涨价，为了抑制全球性的通货膨胀，美联储加息的步伐宛如车轮战，导致美股暴跌，比特币暴跌，硅谷科技公司开始裁员，全球金融萧条近在咫尺。到今天为止，这场战争还在继续，这个世界上从来没有所谓正义的战争，发起战争的人，比如普金，希特勒之流，成为众矢之的是第一步，而血债血偿是所有被殃及的乌克兰人的心声。
最让人无力的是，这些混乱和恐怖，还来自遥远的哭声，他们是丰县铁链女八孩妈妈，唐山烧烤店打人事件，四月的上海，东航坠机，贵州深夜转移放舱的大巴上的乘客，乌鲁木齐火灾，还有更多因为疫情而待业失业或者失去生命的普通人。如果我们对遥远的逝去的生命都没有一点难过，那我们如何面对自己的丧亲之痛？那一段时间，我每天都听不合时宜《在远方与日常的割裂中，人该如何自处》，这期podcast我今年估计听了100次，就在这无数次的回放中，好像我和主播一起在分担这份痛苦，不断轮回每一份沉甸甸的社会问题，像针扎一样让自己时刻保持清醒。年初的时候，官方总有一些媒体会去教育民众，让民众非必要不出行，但与此同时，这一片土地一直在上演具体的事情不重要，重要的事情不具体的诡异现象。没有什么事是非必要的，身边的一切都是必要的，一张白纸的内容，一次和朋友的聚会，一次旅程，一次呐喊，一场精彩的世界杯球赛，人的尊严人的价值，所有人都应该被当做人看待，勇敢，公平，正义，自由，这些都是必要的。
我是一个特别容易共情的人，每次在微博上看到一些类似唐山烧烤店打人事件的视频，我有时候会有难以呼吸的窒息感，之后转化为长达好几天的抑郁。或者听到一个白人朋友说去泰国能任意让亚洲人挥之即来，招之即去，好像在种族的优越感面前，屈从意味着常态，他们的存在是可以随时dispensable的，这一切不论是互联网的，还是现实世界的东西都让我反胃和恶心，可能是一个人待久了，今年这种状态尤为强烈。为了适当管理调整自己的多巴胺分泌，我做了许多医学的笔记，还买了几盒补剂，后来自己一个人背上书包，去了一趟葡萄牙，轻装上阵，我只是去放松心情，感受花怎么开，水怎么流。走到距离里斯本几公里外的Sintra, 石碑上写着“陆止于此”，这种尽头感给人一种重获新生的力量感，在海边看海浪拍打礁石，自然的一切都是如此随机但恒定，驻足不久，就坐上回城的巴士，仿佛从那之后人生开始了新的篇章。多巴胺释放的时刻当然还有很多，我去看了Ed Sheeran, Coldplay, AJR的演唱会。我去了一趟纽约，因为时间有限没有逗留很久，但是短暂的时间里，我去了百老汇看音乐剧，在时代广场休憩，在纽约排列整齐的高层建筑之间穿梭，猎奇地在博物馆里拍陈列的古董，在中央公园漫步，我十分享受这种青春感，我真的全情投入，热爱这一切。
几个月前，我辞职了，我很感谢我第一份工作的所有同事，最后一天的午餐，我尴尬于自己无法周全预定饭馆的慌乱和好像沉浸在刚毕业的小心翼翼里，但又感受到同事加倍的暖意，好像有点无罪释放的不安。这两年的工作我获得了很多技术上的成长，但内心里我也想着“打一枪换一个地方”，我需要换一碗水测试自己的能力。至此之后我个人的工作生活从布鲁塞尔转移到了阿姆斯特丹，感觉又一次站在人生的路口, 每一次做出选择，我都想起风雨哈佛路里面的一幕: 女主角在经历人生家庭各种混乱之后回到社区学校，在地铁上的独白“我的人生终于有了一些秩序感”。我一直在给我人生做一些维持秩序感的盘算，等我真的到了荷兰，发现一些更加拥有约束感的秩序性让我有点难以适应，比如更加高冷的邻居，条款更加细致和严肃的外籍事务办公室和市政厅，上下公交都需要刷卡，谈话更sharp的同事。有一天坐在公交车上，阴雨绵绵，我看着外面草原上的牛马，有一股想回到布鲁塞尔那种相对“混乱”社会的冲动和对当下的悔恨，我第一次如此迅速地后悔，虽然布鲁塞尔有不太干净的街道，尿骚味的地铁，夜里撒酒疯的醉汉，但这种少了条条框框的布鲁塞尔更加兜得住各种各样的人，它不缺乏温度，好像从某种程度上提供了适度的安全感。
尽管我有时在抱怨，但我还是成功得到了我想去的地方。我愈发觉得，只要你一直聚焦，聚焦所有的细节，往对的方向跑下去，一直跑，你能到达你想到达的任何地方，我一直很感激我一直有往前走的动力，因为我相信 Your strength will equal your days. 但是前几天我在微博上写道“享受当下的每一个时刻，看乾坤大，草木青，无穷无尽地往所谓更高更更纸醉金迷的世界前进未必是人生唯一的答案，人应该有更高的追求”。人生还有许多其他更精彩的事情，去电影院看《瞬息全宇宙》，在酒店里熬夜看《风骚律师》，写一个脚本去刷美签slot，写一篇技术博客，看《送你一颗子弹》，《叫魂》，写一封感谢信，第一次当舅舅，去感谢别人的善意，去和朋友吃一餐散伙饭，或者……去争取谈一场恋爱。
世界的分崩离析仍然在继续，生活还是要继续，就在几个小时前我坐在地上弯腰拧螺丝的时候，门铃突然响了，打开门一看，原来是隔壁邻居的老爷爷，年过60，戴着个精致的草帽，过来打招呼互相认识，握手后他就回去了。几分钟后，门铃又响了，还是这位老爷爷，他手上端着一碟荷兰人跨年会吃的两个油球，说送给我吃，还问我食物够不够，需不需要帮忙，走之前很真挚地和我说“you’re young and live alone, if you need anything, don’t hesitate and come to us”. 我内心感激无比，我会一直记得这份善意，这是我混沌或稳定的人生轨迹中的时刻，它与我漫长人生中的其他闪闪发光的琐碎一样，细碎却无比重要。
]]>Dopamine (DA, a contraction of 3,4-dihydroxyphenethylamine) is a neuromodulatory molecule that plays several important roles in cells. Dopamine is the substance that curculating in your system, it is responsible for motivation, craving thing, and also for time perception. The depletion of dopamine will lead to diseases, for example, Parkinson’s disease, a degenerative condition causing tremor and motor impairment, is caused by a loss of dopamine-secreting neurons in an area of the midbrain called the substantia nigra. Tonic and Phasic refers to the low and peaks level of your dopamin baseline.
Two main circuits for dopamine
How dopamine communicate?
There are two ways for dopamine to communicate, one is what we called local synapse release, basically it is a way where 2 neurons are communicating through the electric signal. And the other is a broader way, the volumetric release, this is like a broader way to release high density or much more neurons at the same time (Because dopamine is a neuron neuromodulator).
We all have our own baseline to feel happy and pleasure, if you are taking drug, the kind of pleasure will instantly leads you the a very peak of dopamine, of course higher than your baseline, but if you’re experiencing this type of peak pleasure in a long period of time, your baseline will be higher too. It is very intuitive because you won’t be happier like when you’re kid when you received candies. Therefore controlling relativity gap between your baseline and peak is important.
Dopamine hit
Your behaviour can somehow depict your dopamine state, if you have no motivation to do things, you’re in a low dopamine state, when you try your first skydive and you feel scared or you are about to do a public speaking, you’re in a high dopamine state. And every time you enter into the high dopamine state, or you have a dopamine difference compared to the last state, you’re experiencing a dopamine hit, this could be when you eat a chocolate, you’re scrolling Instagram or Tiktok. But when you repeated engaged in things you are doing, your dopamine threshold hold will be higher.
Parkingson and Drugs
Andrew talked about a story about some young people who were addicted to heroine trying to get some drug called MPPP but mistakenly bought MPTP somewhere in california, what ended up is that the they couldn’t speak and have no movement to do things. It turns out the MPTP kills the nigrostriatal pathway where it generates the movement.
Control dopamine peaks & baseline
All people have different levels of dopamine baseline because of several reasons. (e.g. Gene) Dopamine and epinephrine (aka adrenaline) are cousins. The release of epinephrine depends on the release of dopamine
The list of things that affect your baseline:
Dopamine Setpoint
Dopamine is the universal currency of all the mammals, this substance allows us seeking, motivating us to do things. But counter intuitively is that when you reach the peak of your dopamine, after a while, you might feel low, it is because your dopamine drop to the lower level than before. An example is that postpartum depression aka PPD, when you had a greater reward, after that your dopamine will be back to a lower level. Same as when you finished your graduation or a long trip, you might experience sometime of de-motivation or ‘depression’. It always takes some time for the dopamine to come back to stability from the ‘setpoint’.
And one thing to keep in mind is that you only have certain amount of dopamine what we called pool of dopamine and only those are in readily releasable pool that could be ‘deployed’ or say released in a certain amount of time.
Another thing is that for those ‘play hard and party hard’ type of person, they could do multi-tasking and feel energetic through the week, at some point, they will experience the low level of dopamine in there lives, which is not caused by aging or slow metabolism, it is because for all of the ‘hard’ mode things in their life, it all consumes dopamine and it might all reach to a peak to some extends, and gradually, the baseline of your dopamine will drop. Then you will feel much harder to feel pleasure of doing things as before.
Ensure the Best Evoking Dopamine Release
How do we maintain a good level of our dopamine baseline while reaching our peaks? The key thing is not to a expect or chase high levels of dopamine release every time we’re engaging in certain things. What you need is a random intermittent reinforcement schedules. Things you could do is to start by observing the things that gives you high pleasure or motivation, and randomly modified the ‘factor’ that enhancing this activity. For example, when you’re doing exercise, you’re also drinking your energetic drinks beforehand and listening to your favorite music when you’re running, what you can do is to cut one of those things, only focus on one thing that give your relatively high but not peak dopamine release activity. You could do a flip coin before to decide whether to take those dopamine supportive elements with you into the gym.
Smart phone (social media) is a very usual dopamine supportive element in our modern daily life and it alters the way our dopamine baseline changes. A good practice is to remove the smart phone element in our activity as much as possible, e.g. When working out or doing other things.
Activities that increase our dopamine baseline
Activities that increase our dopamine baseline: Cold exposure (shower with cold water). Another interesting fact is that when you’re getting reward as you finish something, you’re actually demotivating throughout the engagement the task. The key thing is having the growth mindset, which is only focusing the effort and the process you’re doing, and at the same time removing other things that gives you dopamine, only focus on the effort part and remind yourself that this thing could bring the release of dopamine and you will love it in the end. In summary, don’t spike your dopamine prior or after your activity, spike your dopamine in the effort when you’re doint it.
Intermittent fasting could also help because of the fact that when we’re eating, we’re releasing dopamine. Quitting sugar and palatable food and instead eat whole foods. Same as preventing constantly spending time and receive pleasure through watching the pornography.
Supplement that increase our dopamine baseline
Some supplement that could increase the dopamine baseline, one is Macuna Pruiens, this could also help manage stress, which also helps improve sexual health. L-tyrosine: Tyrosine is a nonessential amino acid the body makes from another amino acid called phenylalanine. It is an essential component for the production of several important brain chemicals called neurotransmitters, including epinephrine, norepinephrine, and dopamine. Those are supplement and not meant to use it regularly, could use it once or twice a week. Phenylethylamine, PEA is also good for releasing dopamine (the host take it from time to time for work focus, 500ml PEA and 300ml of Alpha-GPC). One compound called Huperzine A also gained some popularity recently. For Molatonin, there is a study in 2001 about reducing dopamine, so this is something we might need to avoid.
]]>contract ERC721 { event Transfer(address indexed _from, address indexed _to, uint256 indexed _tokenId); event Approval(address indexed _owner, address indexed _approved, uint256 indexed _tokenId);
function balanceOf(address _owner) external view returns (uint256); function ownerOf(uint256 _tokenId) external view returns (address); function transferFrom(address _from, address _to, uint256 _tokenId) external payable; function approve(address _approved, uint256 _tokenId) external payable; }
balanceOf
This function simply takes an address, and returns how many tokens that address owns.
Note: Remember, uint256 is equivalent to uint. We’ve been using uint in our code up until now, but we’re using uint256 here because we copy/pasted from the spec.
ownerOf
This function takes a token ID , and returns the address of the person who owns it.
transferFrom
Note that the ERC721 spec has 2 different ways to transfer tokens:
In supervised learning, concretely we’re learning from given training set \(\left(x_i, y_i\right)\) and formulate our hypotheses \(h_\theta(x)\). Here, the \(\theta_i\) ‘s are the parameters (also called weights) parameterizing the space of linear functions mapping from \(\mathcal{X}\) to \(\mathcal{Y}\). When there is no risk of confusion, we will drop the \(\theta\) subscript in \(h_\theta(x)\), and write it more simply as \(h(x)\).
The goal of our learning is to minimize the distance between hypotheses and the real y, and Loss function is a function to measure such distance.
The difference between Loss Function and Cost Function
Mean Squared Error (MSE) L2 Loss
Mean Squared Error is the most common cost function used in regression problems, also called L2 loss, the formula is as followed:
\[\left.J(\theta)=\frac{1}{n} \sum_{i=1}^n\left(y^{(i)}-\hat{y}^{(i)}\right)\right)^2\]We could see from the graph below, the square error is increased in a quadratic way, the lowest loss is 0 and and highest loss could be infinite.
The MSE is very useful in regression problem, from the perspective of bias and variance perspective, we could do a bias and variance decomposition from MSE function.
By using the decomposition trick in MSE equation shown in 2. We could get that MSE is actually the addition of variance and the square of bias.
\[\begin{aligned} M S E(\hat{\theta}) & =\mathbb{E}\left[(\hat{\theta}-\mathbb{E}(\hat{\theta})+\mathbb{E}(\hat{\theta})-\theta)^2\right] \\ & =\mathbb{E}\left[(\hat{\theta}-\mathbb{E}(\hat{\theta}))^2+2((\hat{\theta}-\mathbb{E}(\hat{\theta}))(\mathbb{E}(\hat{\theta})-\theta))+(\mathbb{E}(\hat{\theta})-\theta)^2\right] \\ & =\mathbb{E}\left[(\hat{\theta}-\mathbb{E}(\hat{\theta}))^2\right]+2 \mathbb{E}[(\hat{\theta}-\mathbb{E}(\hat{\theta}))(\mathbb{E}(\hat{\theta})-\theta)]+\mathbb{E}\left[(\mathbb{E}(\hat{\theta})-\theta)^2\right] \\ & =\mathbb{E}\left[(\hat{\theta}-\mathbb{E}(\hat{\theta}))^2\right]+2(\mathbb{E}(\hat{\theta})-\theta) \mathbb{E}(\hat{\theta}-\mathbb{E}(\hat{\theta}))^2+\mathbb{E}\left[(\mathbb{E}(\hat{\theta})-\theta)^2\right] \\ & =\mathbb{E}\left[(\hat{\theta}-\mathbb{E}(\hat{\theta}))^2\right]+\mathbb{E}\left[(\mathbb{E}(\hat{\theta})-\theta)^2\right] \\ & =\operatorname{Var}(\hat{\theta})+\operatorname{Bias}(\hat{\theta}, \theta)^2\end{aligned}\]Probabilistic interpretation of MSE
A better explanation for using the MSE could be derived from a probabilistic interpretation. The relationship of the target hypothesis and real value y could be formed as
\[y^{(i)}=\theta^T x^{(i)}+\epsilon^{(i)}\]where \(\epsilon^{(i)}\) is an error term that captures either unmodeled effects, or random noise. Let us further assume that the \(\epsilon^{(i)}\) are distributed IID (independently and identically distributed) according to a Gaussian distribution (also called a Normal distribution) with mean zero and some variance \(\sigma^2\). We can write this assumption as ‘ \(\epsilon{ }^{(i)} \sim \mathcal{N}\left(0, \sigma^2\right)\).’ l.e., the density of \(\epsilon^{(i)}\) is given by
\[p\left(\epsilon^{(i)}\right)=\frac{1}{\sqrt{2 \pi} \sigma} \exp \left(-\frac{\left(\epsilon^{(i)}\right)^2}{2 \sigma^2}\right)\]This implies that
\[p\left(y^{(i)} \mid x^{(i)} ; \theta\right)=\frac{1}{\sqrt{2 \pi} \sigma} \exp \left(-\frac{\left(y^{(i)}-\theta^T x^{(i)}\right)^2}{2 \sigma^2}\right)\]When we wish to explicitly view this as a function of \(\theta\), we will instead call it the likelihood function:
\[L(\theta)=L(\theta ; X, \vec{y})=p(\vec{y} \mid X ; \theta)\]The MLE (Maximization of Likelihood Estimation) is a method we used to do the parameter estimation, here we have the parameter theta to be estimated, in order to maximize the likelihood, normally we took the log of this likelihood function (because likelihood function involves tons of probability products, using log form to transform it to summation form), so we have:
\[\begin{aligned} L(\theta) & =\prod_{i=1}^n p\left(y^{(i)} \mid x^{(i)} ; \theta\right) \\ & =\prod_{i=1}^n \frac{1}{\sqrt{2 \pi} \sigma} \exp \left(-\frac{\left(y^{(i)}-\theta^T x^{(i)}\right)^2}{2 \sigma^2}\right)\end{aligned}\] \[\begin{aligned} \ell(\theta) & =\log L(\theta) \\ & =\log \prod_{i=1}^n \frac{1}{\sqrt{2 \pi} \sigma} \exp \left(-\frac{\left(y^{(i)}-\theta^T x^{(i)}\right)^2}{2 \sigma^2}\right) \\ & =\sum_{i=1}^n \log \frac{1}{\sqrt{2 \pi} \sigma} \exp \left(-\frac{\left(y^{(i)}-\theta^T x^{(i)}\right)^2}{2 \sigma^2}\right) \\ & =n \log \frac{1}{\sqrt{2 \pi} \sigma}-\frac{1}{\sigma^2} \cdot \frac{1}{2} \sum_{i=1}^n\left(y^{(i)}-\theta^T x^{(i)}\right)^2 .\end{aligned}\]Hence, maximizing \(\ell\left(\theta\right)\) gives the same answer as minimizing
\(\frac{1}{2} \sum_{i=1}^n\left(y^{(i)}-\theta^T x^{(i)}\right)^2\) The equation 9 is also known as ordinary least square. When linear regression model is built, you would usually use the least square error (LSE) method that is minimizing the total euclidean distance between a line and the data points.
Once the model is built, in order to evaluate its performances. A metric is introduced to evaluate ‘how far’ is your model to the actual real data points in average. The MSE is a good estimate function.
Therefore, LSE is a method that builds a model and MSE is a metric that evaluate your model’s performances, but this 2 have a lot in common in the probabilistic perspective, that is the reason I used hypothesis in the derivation, so you could see the same but in 2 different context.
Mean Absolute Error (MAE) L1 Loss
Mean Absolute Error (MAE) is another class of loss function used in regression problem, also known as L1 loss, the cost function is shown in equation 10.
\[J_{\theta}={\frac{1}{n}}\sum_{i=1}^{n}|y_{i}-{\hat{y}}_{i}|\]The loss of mae when assumed y real is 0 could be plotted below. We could tell from the graph that the biggest loss could be infinite and the lowest is 0, and the loss increased linearly.
Probabilistic interpretation of MAE
Same as the derivation of MSE, when we’re considering the loss of MAE, we assumed that the error is distributed as Laplace distribution \((\mu=0, b=1)\), the error \(\epsilon\) distribution of could be written as 11
\[p\left(y_{i}\mid x_{i}\right)=\frac{1}{2}\exp\left(-\left|y_{i}-\hat{y}_{i}\right|\right)\]Using the Maximum Likelihood Estimation (MLE) as in mean square error example, we could have the following derivation
\[L(x,y)=\prod_{i=1}^{n}\frac{1}{2}\exp\left(-\left|y_{i}-\hat{y}_{i}\right|\right)\] \[L L(x,y)=-n\log2-\sum_{i=1}^{n}|y_{i}-\hat{y_{i}}|\] \[N L L(x,y)=\sum_{i=1}^{n}|y_{i}-\hat{y}_{i}|\]As we can see after that we could get the form of MAE, by maximize the LL is the same as minimize NLL.
Difference between MSE and MAE
The MSE loss (L2) generally converges faster than the MAE loss (L1), but the MAE loss is more robust to outliers.
MSE generally converges faster than MAE. When using the gradient descent algorithm, and the gradient of MAE loss is \(-\hat{y_{i}}\), that is, the scale of the gradient of MSE will change with the size of the error, while the scale of the gradient of MAE will always remain 1 , Even when the absolute error is very small, the gradient scale of MAE is also 1, which is actually very unfavorable for model training. This is also the reason why MSE is more popular.
MAE is more robust to outliers. We can understand this from the 2 perspectives:
Firstly, the following figure shows the MAE and MSE losses drawn into the same picture. Since the MAE loss and the absolute error are linear, the MSE loss and the error have a quadratic relationship. When the error is very large, The MSE loss will be much larger than the MAE loss. Therefore, when there is an outlier with a very large error in the data, MSE will generate a very large loss, which will have a greater impact on the training of the model.
Secondly, when we look at the assumption of the two loss functions. MSE assumes that the error is distributed as a Gaussian distribution, and MAE assumes that the error is distributed as a Laplace distribution. The Laplace distribution by itself is more robust to outliers. when outliers appear on the right side of the right figure, the Laplace distribution is much less affected than the Gaussian distribution. Graph is from Machine Learning A Probabilistic Perspective
Code
Graph could be found in my github
Reference
]]>with pd.ExcelWriter(path, engine=’xlsxwriter’) as ew # Add a header format. workbook = ew.book header_format = workbook.add_format({ ‘bold’: True, ‘text_wrap’: True, ‘valign’: ‘top’, ‘fg_color’: ‘#D7E4BC’, ‘border’: 1}) # save dataframe to excel df.to_excel(ew, sheet_name=’Sheet1’, header=true, index=False) # apply fmt worksheet = ew.sheets[‘Sheet1’] worksheet.set_column(0, 0, width=15, cell_format=header_format)
General Function
pd.eval
:pd.eval(“double_age = df.age * 2”, target=df)
Dataframe
Sorting, reindexing, renaming
Subset Observation: df.length, df.drop_duplicates(subset=None), df_sample, df_nsmallest, df.head(n), df.tail(n), df.filter(), df.query()
Select columns whose name matches regular expression regex.
Summarize Data
Handling Missing Data & Sanity Check Empty Data
Group Data
Window Function
df.rolling(window, min_periods=None, center=False, win_type=None, on=None, axis=0, closed=None, method=’single’)
min_periods[int]: Minimum number of observations in window required to have a value
df.expanding(min_periods=1, center=None, axis=0, method=’single’)
Explode
Reference
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