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Is there anyway to do this without taking a lot of derivatives and looking for a pattern?

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- Thread starter Punkyc7
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- #1

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Is there anyway to do this without taking a lot of derivatives and looking for a pattern?

- #2

SammyS

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How many terms do you need, or do you need to write the result using ∑ notation.

If you know the Maclaurin Series for cos(x) and 1/(1+x

- #3

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I know the Maclaurin series for both of them but when I try doing it like that I don't get the right answer. I have only been able to solve it by taking a bunch of derivatives

- #4

SammyS

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What is the series for for each?

Can you multiply (a_{1}+a_{2}+a_{3}+...)·(b_{1}+b_{2}+b_{3}+...) ?

It's: a_{1}b_{1}+a_{1}b_{2}+a_{2}b_{1}+a_{1}b_{3}+a_{2}b_{2}+a_{3}b_{1}...

Can you multiply (a

It's: a

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- #5

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((-1)^n *((2x)^2n)/(2n!) and (-1^n)((x^2)^n) since our x is -x^2 in the 1/(1+x^2)

- #6

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Ok so how would you know when to stop the a1 and then move onto the a2 and so on... and yes your allowed to multiply them

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- #7

SammyS

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((-1)^n *((2x)^2n)/(2n!) and (-1^n)((x^2)^n) since our x is -x^2 in the 1/(1+x^2)

I suppose you mean that:

[tex]\cos(2x)=\sum_{n=0}^\infty \ (-1)^n\,\frac{(2x)^{2n}}{(2n)!}[/tex]

and

[tex]\frac{1}{1+x^2}=\sum_{n=0}^\infty \ (-1)^{n}\,x^{2n}[/tex]

Write out the first three terms for each and multiply. Of course group all the terms having the same power of x together.

[tex]\cos(2x)=1-2x^2+\frac{2x^4}{3}-\dots[/tex]

[tex]\frac{1}{1+x^2}=1-\frac{1}{x^2}+\frac{1}{x^4}-\dots[/tex]

Multiply them together.

What term(s) do you multiply, then sum, to get the constant term?

What term(s) do you multiply, then sum, to get the linear (x

What term(s) do you multiply, then sum, to get the squared (x

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- #8

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Thank you for your help it makes sense now

- #9

SammyS

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Cool! Thanks for the THANKS!!! We often don't get that!

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