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

by nicolas.ard
Tags: 1 or 1, function, serie, taylor
 P: 4 Hello folks, I have this function, un complex numbers $\frac{1}{(1+z^2)}$ I know that the Taylor serie of that function is $\frac{1}{(1+z^2)}$ = $\sum (-1)^k.z^(2.k)$
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,705 Okay, do you have a [b]question[/b}? What do you want to know about that function? 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 geometric series. The sum of a geometric series, $\sum ar^n$ is $a/(1- r)$. Here, "$a/(1- r)$" is $1/(1+ z^2)$ so that a= 1 and $r= -z^2$. $\sum ar^n$ becomes $\sum (1)(-z^2)^n= \sum (-1)^nz^{2n}$. By the way, to get all of "(2k)" in the exponent, use "z^{(2k)}" rather than just "z^(2k)".
 P: 4 I'm sorry, i wrote the question and i posted it by mistake, i found the solution over here [1]. I did't found the option to delete my post, this is phpBB? [1] http://www.physicsforums.com/showthread.php?t=297842
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)".

 Related Discussions Calculus & Beyond Homework 7 Math & Science Software 1 Calculus & Beyond Homework 1 Calculus 1 Introductory Physics Homework 1