complex function ?

Forum for the GRE subject test in mathematics.
hadimotamedi
Posts: 34
Joined: Thu Dec 30, 2010 4:36 am

complex function ?

Postby hadimotamedi » Tue Mar 12, 2013 6:16 am

Dear All
The z is the complex variable and the function f(z) is defined as (exp(z^2)-1)/(z^2) if z not equal zero and defined as 1 if z=0 . It is asked for its differentiation up to 2k steps evaluated at z=0 i.e. f(2k)(0) . The answers are as :
1) 2k(2k-1) .... (k+2)
2) (2k)!/(k)!
3) 1
4) not computable , since not analytic at z=0
Which one is correct in your opinion ?

math_applicant
Posts: 157
Joined: Sun Oct 14, 2012 12:15 pm

Re: complex function ?

Postby math_applicant » Tue Mar 12, 2013 9:10 am

its not analytic at 0 since the cauchy-riemann equations are not satisfied (unless i made a calculation error :D )

randomposter
Posts: 7
Joined: Tue Mar 05, 2013 10:22 am

Re: complex function ?

Postby randomposter » Tue Mar 12, 2013 10:18 am

I think you've made a mistake math applicant. The pole of e^(z^2)-1 at z=0 is order 2. Notice that the first derivative is 2ze^(z^2) which has a zero at z=0 and taking another derivative we find that the second derivative is not zero, so the zero is second order. Ergo e^(z^2)-1/z^2 is entire. Anyway I'd suggest taking the Taylor series for e^(z^2)-1 and then dividing by z^2 to get the correct Taylor series which should give you all the derivatives you could possibly need.




Return to “Mathematics GRE Forum: The GRE Subject Test in Mathematics”



Who is online

Users browsing this forum: No registered users and 2 guests

cron