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nicolas.ard is offline
Nov28-11, 12:38 PM
P: 4

Taylor serie of a function 1/(1+Z^2)

Thanks!, the use of geometric series it's a way to do it easier. :)

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)".