What is the trick to simplifying the Taylor series of 1/(1 + x^2)?

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SUMMARY

The discussion focuses on simplifying the Taylor series of the function 1/(1 + x^2), specifically centered around x = 0. Participants suggest calculating a few derivatives to identify a pattern, noting that the McLaurin series yields f(0) = 1, f'(0) = 0, f''(0) = -2, and f'''(0) = 0. A hypothesis is proposed that the nth derivative at 0 is 0 for odd n and (-1)^n n! for even n. Additionally, methods such as expanding the function using the geometric series or applying the Leibniz product rule are recommended for further simplification.

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animboy
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The equation starts at B and this is my attempt. As you can see it soon complicates and doesn't look like what t should since I already know what the Taylor series of his function should look like. Is there some clever trick to it that I am missing? PS the series is centred around x = 0.

[PLAIN]http://img823.imageshack.us/img823/3459/phys.png
 
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A Taylor's series about what central point? Rather than work out a large number of derivatives, I would calculate a few derivatives at the given point and try to find a pattern.

For example, the Taylor's series about x= 0 (the McLaurin series), has f(0)= 1, f'(0)= 0, f''(0)= -2, f'''(0)= 0, f''''(0)= 4!, etc. so I would hypothesize that the nth derivative, at 0, is 0 for odd n, (-1)^n n! for n odd. Then I would try to prove that is true by induction.
 
HallsofIvy said:
A Taylor's series about what central point? Rather than work out a large number of derivatives, I would calculate a few derivatives at the given point and try to find a pattern.

For example, the Taylor's series about x= 0 (the McLaurin series), has f(0)= 1, f'(0)= 0, f''(0)= -2, f'''(0)= 0, f''''(0)= 4!, etc. so I would hypothesize that the nth derivative, at 0, is 0 for odd n, (-1)^n n! for n odd. Then I would try to prove that is true by induction.

it's at x = 0, I will try your method and post a pic.
 
Either write
1/(1+x^2)=1/(1-(-x^2))
and expand in geometric series
or apply Leibniz product rule to
[(1+x^2)/(1+x^2)]
and note
(1+x^2)'''=0
 

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