Answer:
![|R_n (x)| \leq \dfrac{|x - \dfrac{\pi}{2}|^{n+1}}{(n+1)!}](https://tex.z-dn.net/?f=%7CR_n%20%28x%29%7C%20%5Cleq%20%5Cdfrac%7B%7Cx%20-%20%5Cdfrac%7B%5Cpi%7D%7B2%7D%7C%5E%7Bn%2B1%7D%7D%7B%28n%2B1%29%21%7D)
Step-by-step explanation:
From the given question; the objective is to show that :
for all x in the interval of convergence f(x)=cos x, a= π/2
Assuming for the convergence f the taylor's series , f happens to be the derivative on an open interval I with a . Then the Taylor series for the convergence of f , for all x in I , if and only if
where;
![\mathtt{R_n (x) = \dfrac{f^{(n+1)} (c)}{n+1!}(x-a)^{n+1}}](https://tex.z-dn.net/?f=%5Cmathtt%7BR_n%20%28x%29%20%3D%20%5Cdfrac%7Bf%5E%7B%28n%2B1%29%7D%20%28c%29%7D%7Bn%2B1%21%7D%28x-a%29%5E%7Bn%2B1%7D%7D)
is a remainder at x and c happens to be between x and a.
Given that:
a= π/2
Then; the above equation can be written as:
![\mathtt{R_n (x) = \dfrac{f^{(n+1)} (c)}{n+1!}(x-\dfrac{\pi}{2})^{n+1}}](https://tex.z-dn.net/?f=%5Cmathtt%7BR_n%20%28x%29%20%3D%20%5Cdfrac%7Bf%5E%7B%28n%2B1%29%7D%20%28c%29%7D%7Bn%2B1%21%7D%28x-%5Cdfrac%7B%5Cpi%7D%7B2%7D%29%5E%7Bn%2B1%7D%7D)
so c now happens to be the points between π/2 and x
If we recall; we know that:
(as a result of the value of n)
However, it is true that for all cases that ![|f ^{(n+1)} \ (c) | \leq 1](https://tex.z-dn.net/?f=%7Cf%20%5E%7B%28n%2B1%29%7D%20%5C%20%28c%29%20%7C%20%5Cleq%201)
Hence, the remainder terms is :
![|R_n (x)| = | \dfrac{f^{(n+1)}(c)}{(n+1!)}(x-\dfrac{\pi}{2})^{n+1}| \leq \dfrac{|x - \dfrac{\pi}{2}|^{n+1}}{(n+1)!}](https://tex.z-dn.net/?f=%7CR_n%20%28x%29%7C%20%3D%20%7C%20%5Cdfrac%7Bf%5E%7B%28n%2B1%29%7D%28c%29%7D%7B%28n%2B1%21%29%7D%28x-%5Cdfrac%7B%5Cpi%7D%7B2%7D%29%5E%7Bn%2B1%7D%7C%20%5Cleq%20%5Cdfrac%7B%7Cx%20-%20%5Cdfrac%7B%5Cpi%7D%7B2%7D%7C%5E%7Bn%2B1%7D%7D%7B%28n%2B1%29%21%7D)
If
for all x and x is fixed, Then
![|R_n (x)| \leq \dfrac{|x - \dfrac{\pi}{2}|^{n+1}}{(n+1)!}](https://tex.z-dn.net/?f=%7CR_n%20%28x%29%7C%20%5Cleq%20%5Cdfrac%7B%7Cx%20-%20%5Cdfrac%7B%5Cpi%7D%7B2%7D%7C%5E%7Bn%2B1%7D%7D%7B%28n%2B1%29%21%7D)