Chapter 10.3, Problem 16E

To determine

To prove: The Maclaurin series for the given function converges for all real x using remainder estimation theorem

Explanation of Solution

Given:

The given function is $\sin 5x$ .

Proof:

As,

  $\mathrm{Sin}5x={\displaystyle \sum _{k=0}^{\infty }{\left(-1\right)}^{k+1}\frac{{\left(5x\right)}^{2k+1}}{\left(2k+1\right)!}\text{for}x\in \left(-\infty ,\infty \right).}$

So, to show that the Maclaurin series for $\sin 5x$ converges to $\sin 5x$ for all x , show

  $R_n\left(x\right)\to 0$ for all x as $n\to +\infty$

Now,

Using Remainder theorem, it states that

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

Here M and r are constants,

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

Here, a = 0 as we are dealing with Maclaurin series

Since $f\left(x\right)=\mathrm{Sin}5x$ , the derivative of cosine is sine, so

Either $\left|f^{\left(n+1\right)}\left(t\right)\right|=\left|\pm 5^{n+1}\cos \left(5t\right)\right|\le 5^{n+1}or\left|f^{\left(n+1\right)}\left(t\right)\right|=\left|\pm 5^{n+1}\sin \left(5t\right)\right|\le 5^{n+1}$

Which means that

  $\left|f^{\left(n+1\right)}\left(t\right)\right|\le M{r}^{n+1}\le 5^{n+1}$

Then, $R_n\left(x\right)$

  $\to 0$ for all x as $n\to +\infty$

Since,

  $\underset{n\to \infty }{\lim }\left|R_n\left(x\right)\right|\le \underset{n\to \infty }{\lim }M\frac{r^{n+1}{\left|x\right|}^{n+1}}{\left(n+1\right)!}\le \underset{n\to \infty }{\lim }\frac{5^{n+1}{\left|x\right|}^{n+1}}{\left(n+1\right)!}=\underset{n\to \infty }{\lim }\frac{{\left|5x\right|}^{n+1}}{\left(n+1\right)!}$

Also $\underset{n\to \infty }{\lim }\frac{{\left|5x\right|}^{n+1}}{\left(n+1\right)!}=0$

Since factorials increase faster than exponentials. This can also be done using the Ratio test.

So, conditions are satisfied $R_n\left(x\right)\to 0$ for all x as $n\to +\infty$ ,

Hence proved that

  $\mathrm{Sin}5x={\displaystyle \sum _{k=0}^{\infty }{\left(-1\right)}^{k+1}\frac{{\left(5x\right)}^{2k+1}}{\left(2k+1\right)!}forx\in \left(-\infty ,\infty \right)}$