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

Answer:

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

Step-by-step 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 $|R_n(x)| \leq \dfrac{9e^d}{(n+1)!} |x|^{n + 1} \:for\: |x| \leq d.$

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

But, since $lim_{n\to\infty}\dfrac{x^n}{n!} =0 for every real number x$,

We have $lim_{n\to\infty} \dfrac{9e^d}{(n+1)!} |x|^{n + 1} =9e^d lim_{n\to\infty} \dfrac{|x|^{n + 1}}{(n+1)!} =0$

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.

$THEOREM\\If f(x)=T_n(x)+R_n(x), where T_n is the nth degree Taylor Polynomial of f at a and lim_{n\to\infty} R_n(x)=0 \: for \: |x-a|$

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

$9e^x=\sum_{n=0}^{\infty}\frac{9x^n}{n!}$  for all x.