Question

Use a Maclaurin series to obtain the Maclaurin series for the given function.

Answer 1

Answer:

$14x cos(\frac{1}{15}x^{2})=14 \sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k+1}}{(2k)!15^{2k}}$

Step-by-step explanation:

In order to find this Maclaurin series, we can start by using a known Maclaurin series and modify it according to our function. A pretty regular Maclaurin series is the cos series, where:

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

So all we need to do is include the additional modifications to the series, for example, the angle of our current function is: $\frac{1}{15}x^{2}$ so for

$cos(\frac{1}{15}x^{2})$

the modified series will look like this:

$cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}(\frac{1}{15}x^{2})^{2k}}{(2k)!}$

So we can use some algebra to simplify the series:

$cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}(\frac{1}{15^{2k}}x^{4k})}{(2k)!}$

which can be rewritten like this:

$cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k}}{(2k)!15^{2k}}$

So finally, we can multiply a 14x to the series so we get:

$14xcos(\frac{1}{15}x^{2})=14x\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k}}{(2k)!15^{2k}}$

We can input the x into the series by using power rules so we get:

$14xcos(\frac{1}{15}x^{2})=14\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k+1}}{(2k)!15^{2k}}$

And that will be our answer.