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Re: Using FindRoot in complex funtion
*To*: mathgroup at smc.vnet.net
*Subject*: [mg71945] Re: Using FindRoot in complex funtion
*From*: Peter Pein <petsie at dordos.net>
*Date*: Tue, 5 Dec 2006 06:04:43 -0500 (EST)
*Organization*: 1&1 Internet AG
*References*: <ekucvg$830$1@smc.vnet.net>
tatsec schrieb:
> The program is :
>
> dotnum=40;
> sigma=Sqrt[2];
> For[L=2,L<=dotnum,L=L+2,
> FindRoot[Integrate[2^(1/2-L/2)*sigma^(-1-L)*(Abs[t])^(1/2*(-1-L))*BesselK[(1-L)/2,Abs[t]/sigma]/(Sqrt[pi]*Gamma[L/2]),{t,0,y}],{y,1}];]
>
> when L<30, FindRoot can find the root of function.but L=>30,FindRoot
> can't calculation.So How to solve this problem?
> Thanks in advance for any help you can give me.
>
> Raymond
>
Hi Raymond,
have a closer look at your integrand:
in the case L==2 you get
FullSimplify[#1, t >= 0]&/@ Collect[ExpandAll[Normal[Series[
(2^(1/2 - L/2)*sigma^(-1 - L)*Abs[t]^((1/2)*(-1 - L))* BesselK[(1 - L)/2,
Abs[t]/sigma])/(Sqrt[Pi]*Gamma[L/2]) /. {sigma -> Sqrt[2], L -> 2},
{t, 0, 10}]]], t]
--> 1/(E^(t/Sqrt[2])*(4*2^(1/4)*t^2))
This can't give a value < infinity, when integrated from 0 to somewhere (David
Park mentioned this).
For other L you will get similar "results".
P²
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