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I have this function, un complex numbers

[itex]\frac{1}{(1+z^2)}[/itex]

I know that the Taylor serie of that function is

[itex]\frac{1}{(1+z^2)}[/itex] = [itex]\sum (-1)^k.z^(2.k) [/itex]

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- Thread starter nicolas.ard
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- #1

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I have this function, un complex numbers

[itex]\frac{1}{(1+z^2)}[/itex]

I know that the Taylor serie of that function is

[itex]\frac{1}{(1+z^2)}[/itex] = [itex]\sum (-1)^k.z^(2.k) [/itex]

- #2

HallsofIvy

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If you are asking about how to get that series, you could, of course, find the derivatives of that function, and apply the usual formula for Taylor's series.

However, for this particular function, the simpler way to handle it is to think of it as the sum of a

Here, "[itex]a/(1- r)[/itex]" is [itex]1/(1+ z^2)[/itex] so that a= 1 and [itex]r= -z^2[/itex]. [itex]\sum ar^n[/itex] becomes [itex]\sum (1)(-z^2)^n= \sum (-1)^nz^{2n}[/itex].

By the way, to get all of "(2k)" in the exponent, use "z^{(2k)}" rather than just "z^(2k)".

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[1] https://www.physicsforums.com/showthread.php?t=297842

- #4

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However, for this particular function, the simpler way to handle it is to think of it as the sum of a geometric series. The sum of a geometric series, ∑arn is a/(1−r).

Here, "a/(1−r)" is 1/(1+z2) so that a= 1 and r=−z2. ∑arn becomes ∑(1)(−z2)n=∑(−1)nz2n.

By the way, to get all of "(2k)" in the exponent, use "z^{(2k)}" rather than just "z^(2k)".

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