Several inequalities

Let $m,n\in \Bbb N, x\in \Bbb R, x\geq 0, x\neq 1$. Prove the followings: if $m>n$, then $\dfrac{x^m-1}{m}>\dfrac{x^n-1}{n}$; if $m<n$, then $\dfrac{x^m-1}{m}<\dfrac{x^n-1}{n}$.

Proof:
Consider the case when $m>n$. The other case is similar. It suffices to show that $n(x^m-1)-m(x^n-1)> 0$ for $x>1$. $$\begin{align}n(x^m-1)-m(x^n-1)&=n(x-1)(x^{m-1}+x^{m-2}+\cdots+1)-m(x-1)(x^{n-1}+x^{n-2}+\cdots+1)\\&=n(x-1)(x^{m-1}+x^{m-2}+\cdots+x^n)+n(x-1)(x^{n-1}+x^{n-2}+\cdots+1)\\&\quad-m(x-1)(x^{n-1}+x^{n-2}+\cdots+1)\\&=n(x-1)(x^{m-1}+x^{m-2}+\cdots+x^n)\\&\quad-(m-n)(x-1)(x^{n-1}+x^{n-2}+\cdots+1)\\&> (x-1)[n\underbrace{(x^n+\cdots+x^n)}_{m-n\; terms}-(m-n)\underbrace{(x^n+\cdots+x^n)}_{n\; terms}]\qquad(*)\\&=(x-1)x^n[n(m-n)-(m-n)n]\\&=0\qquad \Box\end{align}$$ In $(*)$, we have used the facts that $x^{m-i}\geq x^n$ when $i=1,\cdots,m-n$ and $-x^{n-j}\geq -x^n$ where $j=1,\cdots,n$. The case for $0\leq x<1$ also holds because we have $-x^{m-i}\geq -x^n$ when $i=1,\cdots,m-n$ and $-x^{n-j}\leq -x^n$ where $j=1,\cdots,n$.

Remark: This result implies that the sequence of functions $f_n(x)=\dfrac{x^n-1}{n}$ is increasing when $x$ is fixed and $x>1$. The term $-1$ there is crucial. Recall the taylor series of $e^x$ and exponential functions: $$e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots\\ 2^x=e^{(\ln 2)x}=1+(\ln 2)x+\dfrac{((\ln 2)x)^2}{2!}+\cdots.$$ Then $\dfrac{e^x-1}{x}=1+\dfrac{x}{2}+\dfrac{x^2}{6}+\dfrac{x^3}{24}+\cdots$, whereas $\dfrac{e^x}{x}$ or $\dfrac{e^x+5}{x}$ contains the expression $\dfrac{1}{x}$, which means there will be asymptotes.

We can use the proved inequalities to prove some other inequalities.

Let $a,b\in \Bbb R, a>0, b>0, a\neq b, r\in \Bbb Q$ Then $$rb^{r-1}(a-b)<a^r-b^r<ra^{r-1}(a-b)\quad r>1\\ ra^{r-1}(a-b)<a^r-b^r<rb^{r-1}(a-b)\quad 0<r<1.$$ Proof:
When we divide the inequalities by $b^r$, we will see that they resemble the inequalities that we have proved. We just prove the case when $r>1$. Let $r=\dfrac{m}{n}$, where $m,n\in \Bbb N$. $r>1\Rightarrow m>n$. Substituting $x=(\dfrac{a}{b})^{1/n}$ gives $$\dfrac{(\dfrac{a}{b})^{m/n}-1}{m}>\dfrac{(\dfrac{a}{b})^{m/n}-1}{n}\\ \dfrac{a^r-b^r}{mb^r}>\dfrac{a-b}{bn}\\ a^r-b^r>rb^{r-1}(a-b).$$ Similarly, substituting $x=(\dfrac{b}{a})^{1/n}$ yields $$\dfrac{(\dfrac{b}{a})^{m/n}-1}{m}>\dfrac{(\dfrac{a}{b})^{n/n}-1}{n}\\ \dfrac{b^r-a^r}{ma^r}>\dfrac{b-a}{an}\\ a^r-b^r<ra^{r-1}(a-b).\qquad \Box$$

Evaluating Vandermonde Determinants

Suppose we have four points $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4)$. If we want to find a cubic polynomial $a+bx+cx^2+dx^3$ passing through these points, we have to solve $$\underbrace{\begin{pmatrix}1&x_1&x_1^2&x_1^3\\1&x_2&x_2^2&x_2^3\\1&x_3&x_3^2&x_3^3\\1&x_4&x_4^2&x_4^3\end{pmatrix}}_{=A}\begin{pmatrix}a\\b\\c\\d\end{pmatrix}=\begin{pmatrix}y_1\\y_2\\y_3\\y_4\end{pmatrix}.$$ There is always a unique solution when $A$ is invertible. $A$ is actually a Vandermonde matrix, which is invertible if all the $x_i$'s are distinct. How do we evaluate a Vandermonde determinant? As an example, let's consider $$A=\begin{pmatrix}1&3&9&27\\1&2&4&8\\1&4&16&64\\1&5&25&125\end{pmatrix}.$$ Replace the last row with the variable $x$. Let $$f(x)=\begin{pmatrix}1&3&9&27\\1&2&4&8\\1&4&16&64\\1&x&x^2&x^3\end{pmatrix}.$$ We see that $f(2)=f(3)=f(4)=0$ and $\det(A)=f(5)$. Since $f(x)$ is a cubic polynomial, we must have $f(x)=k(x-2)(x-3)(x-4)$, where $k$ is a constant. Now we perform cofactor expansion along the last row: $$f(x)=1\cdot c_{41}+x\cdot c_{42}+x^2\cdot c_{43}+x^3\cdot c_{44}.$$ Comparing the coefficients of $x^3$ of the two expressions, we have $$k=c_{44}=\begin{vmatrix}1&3&9\\1&2&4\\1&4&16\end{vmatrix}.$$Do you notice something? This problem is related to recursion! Using the same trick, we can argue that $c_{44}=(4-3)(4-2)(2-3)=-2$. Finally, $f(5)=-2(5-2)(5-3)(5-4)=-12$.

Reference:
vandermonde

Cosets and Decimal Expansions

Let us take a deeper look at fractions and their decimal expansions. $$\dfrac{1}{2}=0.5,\quad \dfrac{1}{3}=0.3333\cdots=0.\bar{3},\quad \dfrac{1}{4}=0.25,\quad \dfrac{1}{5}=0.2,\quad \dfrac{1}{6}=0.1666\cdots=0.1\bar{6},\\ \dfrac{1}{7}=0.142857142857\cdots=0.\overline{142857},\quad \dfrac{1}{8}=0.125,\quad \dfrac{1}{9}=0.1111\cdots=0.\bar{1},\quad \dfrac{1}{10}=0.1$$ Observe that the decimal expansions of $\dfrac{1}{3},\dfrac{1}{5},\dfrac{1}{6},\dfrac{1}{7},\dfrac{1}{9}$ are periodic. We claim that if the denominator $b$ of a fraction $a/b$ is relatively prime to $10$, then the fraction has a decimal expansion that is purely periodic. Note that the converse of this claim is not true as seen above: $(6,10)=2\neq 1$ but $1/6$ has a periodic decimal expansion. To prove this, recall a useful tool from high school: geometric series. Here's a special case for illustration: $$\dfrac{1}{7}=0.\overline{142857}=0.142857\cdot (1+10^{-6}+10^{-12}+\cdots)=\dfrac{0.142857}{1-10^{-6}}\\ 10^6-1=7(142857)\\ 10(10^5)-7(142857)=1\\ (7,10)=1$$ Let us now modify the arguments to prove our claim. We only prove the case when $a=1$. When $a$ is not $1$, $a/b$ is just $\underbrace{1/b+1/b+1/b+\cdots}_{a\; times}$ and it has a periodic decimal expansion when $1/b$ does. $$(b,10)=1 \Rightarrow \; \exists\; x,y\in \Bbb Z\; s.t.\; bx+10y=1\\ 10y-1=b(-x)\\ \dfrac{1}{b}=\dfrac{-x}{10y-1}=\dfrac{x}{1-10y}=x(1+10y+10y^2+\cdots)\; \Box$$ Consider $\dfrac{1}{7}$ and some of its multiples: $$\color{blue}{\dfrac{2}{7}=0.\overline{285714}},\quad \dfrac{3}{7}=0.\overline{428571},\quad \color{forestgreen}{\dfrac{4}{7}=0.\overline{571428}},\quad \color{red}{\dfrac{5}{7}=0.\overline{714285}},\quad \color{gray}{\dfrac{6}{7}=0.\overline{857142}}\\ \color{gray}{\dfrac{20}{7}=2.\overline{857142}},\quad \color{blue}{\dfrac{30}{7}=4.\overline{285714}},\quad \color{red}{\dfrac{40}{7}=5.\overline{714285}},\quad \dfrac{50}{7}=7.\overline{142857},\color{forestgreen}{\quad \dfrac{60}{7}=8.\overline{571428}}$$ Observe that if we cyclically shift the digits in the repeating part of the decimal expansions of a reduced fraction, we get the repeating part of the decimal expansion of another reduced fraction with the same denominator. It is by no means a coincidence. This phenomenon is actually related to cosets. We have $$\dfrac{30}{7}=\dfrac{2}{7}+4\\ \dfrac{60}{7}=\dfrac{4}{7}+8\\ \dfrac{20}{7}=\dfrac{6}{7}+2\\ \cdots$$ Rephrase these in modular arithmetic: $$30\equiv_7 2\\ 60\equiv_7 4\\ 20\equiv_7 6.$$ Recall that two integers being equal in modulo $n$ means they have the same remainder when divided by $n$. It makes sense now that $30/7$ and $2/7$ differ by an integer and have the same repeating part (remainder). Another observation is that when we shift the digits of $3/7=0.\overline{428571}$ to the left once, we get $2/7=0.\overline{285714}$. Note that $30\equiv_7 2$ is actually $3\cdot 10\equiv_7 2$. When we multiply a number by $10$, the digits are shifted to the left by a place. What if we shift two or three places to the left? Just multiply by $100$ or $1000$. In general, we have $$3\cdot 10^n \equiv_7 r,$$ where $0\leq r<7$, then $r/7$ will have a decimal expansion whose repeating part is the digits for $3/7$ shifted to the left $n$ times. We can display this cyclic behavior by placing the six digits on a 'circle diagram'.

circle digram for 1/7

From the diagram, the decimal expansion of any proper fraction with denominator $7$ can be read off by going around the wheel. As an example, to find the decimal expansion of $5/7$, start at digit $7$ and go around clockwise to obtain $0.\overline{714285}$.

Let's consider one more example: $1/13$ and its multiples. $$\dfrac{1}{13}=0.\overline{076923},\quad \dfrac{3}{13}=0.\overline{230769},\quad \dfrac{4}{13}=0.\overline{307692},\quad \dfrac{9}{13}=0.\overline{692307},\quad \dfrac{10}{13}=0.\overline{769230},\quad \dfrac{12}{13}=0.\overline{923076}\\ \dfrac{2}{13}=0.\overline{153846},\quad \dfrac{5}{13}=0.\overline{384615},\quad \dfrac{6}{13}=0.\overline{461538},\quad \dfrac{7}{13}=0.\overline{538461},\quad \dfrac{8}{13}=0.\overline{615384},\quad \dfrac{11}{13}=0.\overline{846153}$$This time there are two distinct classes of decimal digits. Six of them contain the repeating digits $076923$; the other six $153846$. Why is this? Indeed, the fractions on the first row are congruent to $10^k \;(\text{mod}\; 13)$; those on the second row are congruent to $2\cdot 10^k\;(\text{mod}\; 13)$.

circle diagrams for 1/13

We see that $$(10)=\{10^k\;(\text{mod}\; 13)\mid k\in \Bbb Z^+\}=\{1,10,9,12,3,4\}\\ 2(10)=\{2\cdot 10^k\;(\text{mod}\; 13)\mid k\in \Bbb Z^+\}=\{2,7,5,11,6,8\}$$ are two disjoint sets whose union is $Z_{13}$. $(10)$ is a cyclic subgroup $H$ generated by the congruence class of $10$ in $\Bbb Z_{13}$, whereas $2(10)$ represent the coset $2H$.

We conclude that if $a/b$ is a reduced fraction where $(10,b)=1$, then the other reduced fractions whose decimal has a repeating block that is a shift of the repeating block of the decimal for $a/b$ are precisely those fractions with denominator $b$ and numerator in the coset $a\cdot 10\;(\text{mod}\; b)$ of $\Bbb Z_b$.

Questions to be answered:
In the circle diagrams for $13$, why does any pair of opposite decimal digits adds up to $9$? Why does any pair of opposite fractions adds up to $1$? These two questions can be answered by Midy's theorem.
How do we know how many cosets there should be?

Reference:
notes by Keith Conrad
7th decimal wheel

More to know:
cyclic number
primitive root modulo n