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

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

User Kodaloid
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6 votes

Answer:


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.

User Jason Warner
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