Need Help?

  • Notes
  • Comments & Questions

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 \(x\)

No comments yet

Join the conversation

Join Notatee Today!