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• Notes

Taylor and Maclaurin series:

$f(x)=\sum_{n=0}^{\infty} \frac{f^{n}{(c)}}{n !}(x-c)^{n}$

$T_{n}(x) =\sum_{i=0}^{n} \frac{f^{(i)}{(c)}}{i !}(x-c)^{i} =f(c)+\frac{f^{\prime}{(c)}}{1 !}(x-c) +\frac{f^{\prime \prime}{(c)}}{2 !} (x-c)^{2}+\dots$

$\cdots \frac {f^{(n)}(c)}{n!}(x-c)^{n}$

$f(x)=\sum_{n=0}^{\infty} \frac{f^{n}{(c)}}{n !}(x-c)^{n} \longrightarrow \lim _{n \rightarrow \infty} R_{n}=0$

$R_{n}(x)=f(x)-{T_n}(x)$

$\lim _{n \rightarrow \infty} \frac{x^{n}}{n !}=0$ for all $x$

Taylor Inequality: If $\left|f^{(n+1)}{(n)}\right| \leq M$ for $|x-c| \leq d$

$\rightarrow$ The Remainder $R_{n}(x)$ satisfies the inequality

$|R_{n}(x)| \leq \frac{M}{(n+1) !}| x-\left.c\right|^{n+1}$          $|x-a| \leq d$

Prove that $e^{x}$ is equal to the sum of its Maclaurin series.

$f(x)=e^{x} \longrightarrow f^{(n+1)}(x)=e^{x}$ for all $n.$

If dis any + ve number & $|x| \leq d$

$\Rightarrow then\ \left|f^{n+1}(x)\right| =e^{x} \leq e^{d}$

Using Taylor In equality with $C=0, \ M=e^{d}$

$|{R_n}(x)| \leq \frac{e^{d}}{(n+1) !}|x|^{n+1}$ for $|x| \leq d$

because $\lim _{n \rightarrow \infty} \frac{x^{n}}{n !}=0$ for every $x$.

then $\longrightarrow\lim _{n \rightarrow \infty}|{R_n}(x)| \leq \lim _{n \rightarrow \infty} \frac{e^{d}}{(n+1) !}|x|^{n+1}$

$=e^{d} \lim _{n \rightarrow \infty} \frac{|x|^{n+1}}{(n+1) !}=0$

$∴ \lim _{n \rightarrow \infty}\left|R_{n}(x)\right|=0$ by sqneeze theory $∴ \lim _{n \rightarrow \infty} R_{n}(x)=0$

$∴ e^{x}$ Is equal to the sum of its Maclaurin series

$e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n !}$ for all $x$