To show that $$z_n=\left(1+\dfrac{z}{n} \right)^{n} \underset{n \longmapsto +\infty}{\longrightarrow \exp(z)}$$ the author of textbook use the following method but there is some steps that i'm not sure if i got it right so would someone elaborate it
Let $x = \Re(z)$ and $y=\Im(z)$
- ${\displaystyle r_n=\sqrt{\left(1+\dfrac{x}{n} \right)^{2}+\dfrac{y^{2}}{n^2}}=\sqrt{1+\dfrac{2x}{n}+o\left(\dfrac{1}{n}\right)}=1+\dfrac{x}{n}+o\left(\dfrac{x}{n}\right)}$
- why if $r_n=1+\dfrac{x}{n}+o\left(\dfrac{x}{n} \right)$ then $\ln(r_n)\sim \dfrac{x}{n}$
- how we can get the expression of $\tan(\alpha_n)$
- why if $\tan\left( \alpha_n \right)\sim \dfrac{y}{n}$ then $\alpha_{n}\sim \dfrac{y}{n}$
My thoughts:
first i think there is typo in $z_n=r_n^{n}e^{n\alpha_n}$ should we write $z_n=r_n^{n}e^{i n\alpha_n}$ instead.
- ${\displaystyle r_n=\sqrt{\left(1+\dfrac{x}{n} \right)^{2}+\dfrac{y^{2}}{n^2}}=\sqrt{1+\dfrac{2x}{n}+o\left(\dfrac{1}{n}\right)}=1+\dfrac{x}{n}+o\left(\dfrac{x}{n}\right)}$
note that
$$\left(1+x \right)^{\alpha}\underset{x\to 0}{=}1+\alpha x +o\left( x\right) $$
$$\begin{cases} \left(1+\dfrac{x}{n} \right)^{2}=1+2\dfrac{x}{n}+o\left( \dfrac{1}{n}\right) \\ \dfrac{y^{2}}{n^{2}}=o\left(\dfrac{1}{n}\right)\end{cases}\implies 1+\dfrac{2x}{n}+o\left(\dfrac{1}{n}\right)$$
on the other hand \begin{aligned}\sqrt{1+\dfrac{2x}{n}+o\left(\dfrac{1}{n}\right)}=\left(1+\dfrac{2x}{n}+o\left(\dfrac{1}{n}\right) \right)^{\frac{1}{2}}&=1+\dfrac{1}{2}\left( \dfrac{2x}{n}+o\left( \dfrac{1}{n}\right)\right)+o\left(o\left(\dfrac{1}{n}\right)\right) \\ &=1+\dfrac{x}{n}+o\left( \dfrac{1}{n}\right)+o\left(\dfrac{1}{n}\right)\\ &=1+\dfrac{x}{n}+o\left( \dfrac{1}{n}\right) \end{aligned} then $$\fbox{$r_{n}=1+\dfrac{x}{n}+o\left( \dfrac{1}{n}\right)$}$$
- why if $r_n=1+\dfrac{x}{n}+o\left(\dfrac{x}{n} \right)$ then $\ln(r_n)\sim \dfrac{x}{n}$
note that :
if $u_n\sim v_n $ and $v_n \sim w_n $ then $u_n \sim w_n$ $u_n\sim v_n \iff u_n=v_n+o(v_n)$
i can't show this
- how we can get the expression of $\tan(\alpha_n)$
If $a+ib=\rho e^{i\theta}$ with $a>0$ then $\tan(\theta)=\frac b a$
since $\left( 1+\dfrac{z}{n}\right)=\left( 1+\dfrac{x}{n}\right)+i\dfrac{y}{n}=r_{n}e^{i\alpha_n} $ then $$\tan(\alpha_n)=\dfrac{\dfrac{y}{n}}{1+\dfrac{x}{n}}=\dfrac{y}{x+n}$$ $$\fbox{$\tan(\alpha_n)=\dfrac{y}{x+n} $} $$
- why if $\tan\left( \alpha_n \right)\sim \dfrac{y}{n}$ then $\alpha_{n}\sim \dfrac{y}{n}$
if $u_n\sim v_n $ and $v_n \sim w_n $ then $u_n \sim w_n$
So should show that $\alpha_n \underset{n \to +\infty}{\overset{}{\longrightarrow}}0$ to be able to say that $\tan(\alpha_n)\sim r_n$
we've $\tan(\alpha_n)=\dfrac{y}{x+n}$ then $\alpha_n=\arctan\left(\dfrac{y}{x+n}\right) $ $$\lim_{n\to +\infty}\alpha_n=\lim_{n\to +\infty} \arctan\left(\dfrac{y}{x+n}\right)=\arctan\left(\dfrac{y}{x+\lim_{n\to +\infty} n}\right)=\arctan(0)=0 $$ then $$\begin{cases}\tan(\alpha_n) \sim \alpha_n \\ \tan(\alpha_n)\sim \dfrac{y}{n} \end{cases} \implies \alpha_n\sim \dfrac{y}{n}$$
- If my proof wrong would you elaborate the steps
This is a very strange proof, apart from the handwaving colloquial style in printed matter. The usual setup is the following: $$\eqalign{e^z-\left(1+{z\over n}\right)^n&=\bigl((e^{z/n}\bigr)^n-\left(1+{z\over n}\right)^n\cr &=\left(e^{z/n}-\bigl(1+{z\over n}\bigr)\right)\cdot\sum_{k=0}^{n-1}\bigl(e^{z/n}\bigr)^{n-1-k}\left(1+{z\over n}\right)^k\cr &=R_n\cdot \Sigma_n\ .\cr}$$ Here $$R_n={z^2\over n^2}g\left({z\over n}\right)$$ with $\lim_{\zeta\to0}g(\zeta)={1\over2}$ by Taylor's theorem, and $$\eqalign{|\Sigma_n|&\leq \sum_{k=0}^{n-1}\bigl(e^{|z|/n}\bigr)^{n-1-k}\left(1+{|z|\over n}\right)^k\leq \sum_{k=0}^{n-1}\bigl(e^{|z|/n}\bigr)^{n-1-k}\left(1+{|z|\over n}\right)^k\cr &\leq \sum_{k=0}^{n-1}\bigl(e^{|z|/n}\bigr)^{n-1-k}\bigl(e^{|z|/n}\bigr)^k\leq n\>e^{|z|}\ .\cr}$$