Maclaurin Series

  • Thread starter Punkyc7
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  • #1
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F(x)= (cos(2x)/(1+x^2))





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

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  • #2
SammyS
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So, I take it that you need to find the Maclaurin Series for the function: f(x) = (cos(2x))/(1+x2) .

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+x2) or 1/(1+x), then No, you don't need to take a of derivatives.
 
  • #3
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The first 3 non zero terms,

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 (a1+a2+a3+...)·(b1+b2+b3+...) ?

It's: a1b1+a1b2+a2b1+a1b3+a2b2+a3b1...
 
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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 (x1) term? -Right, there is none.

What term(s) do you multiply, then sum, to get the squared (x2) term?
 
  • #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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