Question

Question: Show that the Maclaurin series for sin x converges to sin x . Hint: If f (x)= sin$f^{(n+1)}\left(X\right)=\pm \sin xor\pm \cos x$, and so $\left|f^{(n+1)}\left(x\right)\right|\le 1$

for all x and all n. Let$n\to \infty$in (14.2).

Short answer

Hence prove, Maclaurin Series for sin x converges to sin x .

Step-by-step solution

Define Maclaurin Series

Maclaurin series is basically a type of power series expansion of the function about the origin, with all the terms having positive values expanded as:$f\left(x\right)=f\left(0\right)+xf\text{'}\left(0\right)+\frac{x^2}{2}f"\left(0\right)+.......+\frac{x^2}{n}{f}^{\left(n\right)}\left(0\right)+..........$

Determine the proof for the Macular-in series

The given function is f(x)=sin x

Now, the formula for the remainder is given by:

$R_n\left(x\right)=\frac{x^{n+1}{f}^{(n+1)}\left(c\right)}{(n+1)}$

Consider the equation $f^{(n+1)}\left(x\right)=\pm \sin xor\pm \cos x$with $\left|f^{(n+1)}\left(x\right)\right|\le 1$

So, we taking limit of remainder as follow:

$\underset{n\to \infty }{\lim }\left|R_n\left(x\right)\right|=\underset{n\to \infty }{\lim }\left(\begin{array}{c}{x}^{n+1}\\ (n+1)\end{array}\right)$ = 0

Clearly, the remainder is zero, this implies that the series will converges to itself.

Hence prove, Maclaurin Series for sin x converges to sin x .