Chapter 10.3, Problem 18E

To determine

To show that the Maclaurin series for ${\mathrm{e}}^{5\mathrm{x}}$ converges to ${\mathrm{e}}^{5\mathrm{x}}$ for all $\mathrm{x}$ using remainder estimation theorem.

Explanation of Solution

Given:

The given Maclaurin series for ${\mathrm{e}}^{5\mathrm{x}}$ .

Calculation:

To show that the Maclaurin series for ${\mathrm{e}}^{5\mathrm{x}}$ converges to ${\mathrm{e}}^{5\mathrm{x}}$ for all x, you must show:

  ${\mathrm{e}}^{5\mathrm{x}}={\displaystyle \sum _{\mathrm{k}=0}^{{(5\mathrm{x})}^{\mathrm{k}}}\frac{{(5\mathrm{x})}^{\mathrm{k}}}{(2\mathrm{k})!}}$ For $\mathrm{x}\in (-\mathrm{\infty },\mathrm{\infty })$ .

This proof can be completed using the Remainder Estimation Theorem, which states:

  $\left|{\mathrm{R}}_{\mathrm{n}}(\mathrm{x})\right|\le \mathrm{M}\frac{{\mathrm{r}}^{\mathrm{n}+1}{\left|\mathrm{x}-\mathrm{a}\right|}^{\mathrm{n}+1}}{(\mathrm{n}+1)!}$

where $\mathrm{M}$ and $\mathrm{r}$ are constants such that

  $\left|{\mathrm{f}}^{(\mathrm{n}+1)}(\mathrm{t})\right|\le \mathrm{M}{\mathrm{r}}^{\mathrm{n}+1}$ for all t in between a and x.

Note that $\mathrm{a}=0$ since you are dealing with a Maclaurin series.

Since the $\mathrm{f}(\mathrm{x})={\mathrm{e}}^{5\mathrm{x}}$ , the derivative is the same exponential multiplied by a factor of 5, so

  ${\mathrm{f}}^{(\mathrm{n}+1)}(\mathrm{t})=5^{\mathrm{n}+1}{\mathrm{e}}^{5\mathrm{t}}$

which means that for negative values of t, you let $\mathrm{M}=1$ and $\mathrm{r}=5$ so that

  ${\mathrm{f}}^{(\mathrm{n}+1)}(\mathrm{t})=5^{\mathrm{n}+1}{\mathrm{e}}^{5\mathrm{t}}\le 5^{\mathrm{n}+1}(1)\le {\mathrm{r}}^{\mathrm{n}+1}\mathrm{M}$

and for positive values of t you let $\mathrm{M}={\mathrm{e}}^{5\mathrm{t}}$ and $\mathrm{r}=5$ so that

  ${\mathrm{f}}^{(\mathrm{n}+1)}(\mathrm{t})=5^{\mathrm{n}+1}{\mathrm{e}}^{5\mathrm{t}}\le {\mathrm{r}}^{\mathrm{n}+1}\mathrm{M}$

Now, you can show that

  ${\mathrm{R}}_{\mathrm{n}}(\mathrm{x})\to 0$ for all x as $\mathrm{n}\to +\mathrm{\infty }$

since, by the Remainder Estimation Theorem you know that

  $\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left|{\mathrm{R}}_{\mathrm{n}}(\mathrm{x})\right|\le \underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\mathrm{M}\frac{{\mathrm{r}}^{\mathrm{n}+1}{\left|\mathrm{x}-\mathrm{a}\right|}^{\mathrm{n}+1}}{(\mathrm{n}+1)!}$

and you know that

both $\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left|{\mathrm{R}}_{\mathrm{n}}(\mathrm{x})\right|=0$ and $\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\mathrm{M}\frac{{\mathrm{r}}^{\mathrm{n}+1}{\left|\mathrm{x}-\mathrm{a}\right|}^{\mathrm{n}+1}}{(\mathrm{n}+1)!}=0$

since factorials grow so much faster than exponentials.

So it is proved that

  ${\mathrm{e}}^{5\mathrm{x}}={\displaystyle \sum _{\mathrm{k}=0}^{{(5\mathrm{x})}^{\mathrm{k}}}\frac{{(5\mathrm{x})}^{\mathrm{k}}}{(2\mathrm{k})!}}$ For $\mathrm{x}\in (-\mathrm{\infty },\mathrm{\infty })$ .