Question

EXAMPLE 2 Prove that 9ex is equal to the sum of its Maclaurin series. SOLUTION If f(x) = 9ex, then f (n + 1)(x) = for all n. If d is any positive number and |x| ≤ d, then |f (n + 1)(x)| = ≤ 9ed. So Taylor's Inequality, with a = 0 and M = 9ed, says that |Rn(x)| ≤ (n + 1)! |x|n + 1 for |x| ≤ d. Notice that the same constant M = 9ed works for every value of n. But, from this equation, we have lim n → [infinity] 9ed (n + 1)! |x|n + 1 = 9ed lim n → [infinity] |x|n + 1 (n + 1)! = . It follows from the Squeeze Theorem that lim n → [infinity] |Rn(x)| = 0 and therefore lim n → [infinity] Rn(x) = for all values of x. By this theorem, 9ex is equal to the sum of its Maclaurin series, that is, 9ex = [infinity] 9xn n! n = 0 for all x.

Answer 1

To Prove:
`9e^x` is equal to the sum of its Maclaurin series.

Explanation:

If
`f(x) = 9e^x`, then
`f ^((n + 1)(x)) =9e^x` for all n. If d is any positive number and |x| ≤ d, then
`|f^((n + 1)(x))| = 9e^x\leq 9e^d.`

So Taylor's Inequality, with a = 0 and M =
`9e^d`, says that
`|R_n(x)| \leq (9e^d)/((n+1)!) |x|^(n + 1) \:for\: |x| \leq d.`

Notice that the same constant
`M = 9e^d` works for every value of n.

But, since
`lim_(n\to\infty)(x^n)/(n!) =0 $ for every real number x$`,

We have
`lim_(n\to\infty) (9e^d)/((n+1)!) |x|^(n + 1) =9e^d lim_(n\to\infty) (|x|^(n + 1))/((n+1)!) =0`

It follows from the Squeeze Theorem that
`lim_(n\to\infty) |R_n(x)|=0` and therefore
`lim_(n\to\infty) R_n(x)=0` for all values of x.

`THEOREM\\If f(x)=T_n(x)+R_n(x), $where $T_n $is the nth degree Taylor Polynomial of f at a and $ lim_(n\to\infty) R_n(x)=0 \: for \: |x-a|<R, $then f is equal to the sum of its Taylor series on $ |x-a|<R`

By this theorem above,
`9e^x` is equal to the sum of its Maclaurin series, that is,

`9e^x=\sum_(n=0)^(\infty)(9x^n)/(n!)` for all x.