Question

Use Lagrange’s form for the remainder to prove that the

Maclaurin series for the cosine,

$\cos x=\sum _{k=0}^{\mathrm{\infty }} \frac{(-1{)}^k}{\left(2k\right)!}{x}^{2k}$

converges for every real number.

Short answer

The Maclaurin series for the $\cos ine$for the given series converges for every number is proved.

Step-by-step solution

Given Information

Consider the function$f\left(x\right)=\cos x$

Proof

Lagrange form for the remainder is:

$R_p\left(x\right)=\frac{f^{a+1}\left(c\right)}{(n+1)!}{\left(x-x_0\right)}^{n+1}$

Here, $c$is between $x_0$and $x$

Now,

$f^{n+1}\left(c\right)=\pm \sin c$or $\pm \cos c$for every $n\ge 0$

In case of Maclaurin series: $x_0=0$

Thus,

$\left|R_n\left(x\right)\right|=\left|\frac{\pm \sin c\text{or}\pm \cos c}{(n+1)!}(x-0{)}^{n-1}\right|$

Take the limit.

$\begin{array}{rcl}\underset{k\to =}{\lim }\left|R_n\left(x\right)\right|& =& \lim \left|\frac{\pm \sin c\text{or}\pm \cos c}{(n+1)!}(x{)}^{n=1}\right|\\ & =& 0\end{array}$

This implies that the limit is zero because the quotient $\frac{x^{n+1}}{(n+1)!}\to 0$as$n\to 0$