Answer:
[tex]14x cos(\frac{1}{15}x^{2})=14 \sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k+1}}{(2k)!15^{2k}}[/tex]
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:
[tex]cos(x)=\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{2k}}{(2k)!}[/tex]
So all we need to do is include the additional modifications to the series, for example, the angle of our current function is: [tex]\frac{1}{15}x^{2}[/tex] so for
[tex]cos(\frac{1}{15}x^{2})[/tex]
the modified series will look like this:
[tex]cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}(\frac{1}{15}x^{2})^{2k}}{(2k)!}[/tex]
So we can use some algebra to simplify the series:
[tex]cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}(\frac{1}{15^{2k}}x^{4k})}{(2k)!}[/tex]
which can be rewritten like this:
[tex]cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k}}{(2k)!15^{2k}}[/tex]
So finally, we can multiply a 14x to the series so we get:
[tex]14xcos(\frac{1}{15}x^{2})=14x\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k}}{(2k)!15^{2k}}[/tex]
We can input the x into the series by using power rules so we get:
[tex]14xcos(\frac{1}{15}x^{2})=14\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k+1}}{(2k)!15^{2k}}[/tex]
And that will be our answer.
