Question

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

f(x)= 14x cos(1/15x^2)

Answer 1

`14x cos((1)/(15)x^(2))=14 \sum _(k=0) ^(\infty) ((-1)^(k)x^(4k+1))/((2k)!15^(2k))`

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) ((-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:
`(1)/(15)x^(2)` so for

`cos((1)/(15)x^(2))`

the modified series will look like this:

`cos((1)/(15)x^(2))=\sum _(k=0) ^(\infty) ((-1)^(k)((1)/(15)x^(2))^(2k))/((2k)!)`

So we can use some algebra to simplify the series:

`cos((1)/(15)x^(2))=\sum _(k=0) ^(\infty) ((-1)^(k)((1)/(15^(2k))x^(4k)))/((2k)!)`

which can be rewritten like this:

`cos((1)/(15)x^(2))=\sum _(k=0) ^(\infty) ((-1)^(k)x^(4k))/((2k)!15^(2k))`

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

`14xcos((1)/(15)x^(2))=14x\sum _(k=0) ^(\infty) ((-1)^(k)x^(4k))/((2k)!15^(2k))`

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

`14xcos((1)/(15)x^(2))=14\sum _(k=0) ^(\infty) ((-1)^(k)x^(4k+1))/((2k)!15^(2k))`

And that will be our answer.