Answer:
[tex]|R_n (x)| \leq \dfrac{|x - \dfrac{\pi}{2}|^{n+1}}{(n+1)!}[/tex]
Step-by-step explanation:
From the given question; the objective is to show that :
[tex]\lim_{n \to \infty} R_n (x) = 0[/tex] 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 [tex]\lim_{n \to \infty} R_n (x) = 0[/tex]
where;
[tex]\mathtt{R_n (x) = \dfrac{f^{(n+1)} (c)}{n+1!}(x-a)^{n+1}}[/tex]
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:
[tex]\mathtt{R_n (x) = \dfrac{f^{(n+1)} (c)}{n+1!}(x-\dfrac{\pi}{2})^{n+1}}[/tex]
so c now happens to be the points between π/2 and x
If we recall; we know that:
[tex]f^{(n+1)}(c) = \pm \ sin \ c \ or \ cos \ c[/tex] (as a result of the value of n)
However, it is true that for all cases that [tex]|f ^{(n+1)} \ (c) | \leq 1[/tex]
Hence, the remainder terms is :
[tex]|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)!}[/tex]
If [tex]\lim_{n \to \infty} R_n (x) = 0[/tex] for all x and x is fixed, Then
[tex]|R_n (x)| \leq \dfrac{|x - \dfrac{\pi}{2}|^{n+1}}{(n+1)!}[/tex]
