Re: Generalized functions and integral transforms
- To: mathgroup at smc.vnet.net
- Subject: [mg34445] Re: Generalized functions and integral transforms
- From: Name <mee at home.com.redline.ru>
- Date: Tue, 21 May 2002 03:36:22 -0400 (EDT)
- References: <abvtj7$m3a$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
I think I finally figured it out -- this is just an error in
InverseLaplaceTransform:
Tanh[y] == (E^y-E^-y) / (E^y+E^-y) == 1+2Sum[(-1)^k E^(-2k y),
{k,1,Infinity}],
so InverseLaplaceTransform is
DiracDelta[x]+2Sum[(-1)^k DiracDelta[2k-x],{k,1,Infinity}],
while Mathematica gives
2Sum[(-1)^k DiracDelta[2k-x],{k,0,Infinity}].
Maxim
(correct e-mail address is m.r at prontomail.com; mailer acting up...)
DrBob wrote:
> My previous experience with the La Place transform functions is that the
> examples in Help are the ONLY ones that work properly. (Good choice.)
>
> Bobby
>
> -----Original Message-----
> From: Name [mailto:mee at home.com.redline.ru]
To: mathgroup at smc.vnet.net
> Subject: [mg34445] Generalized functions and integral transforms
>
> Tanh[y]-
> LaplaceTransform[InverseLaplaceTransform[Tanh[y],y,x],x,y]//
> Simplify
>
> Out[1]=
> -1
>
> Where does it go wrong?
> Doing it step by step:
>
> InverseLaplaceTransform[Tanh[y], y, x]
>
> Out[2]=
> 2*Sum[(-1)^K$144*DiracDelta[2*K$144 - x], {K$144, 0, Infinity}]
>
> LaplaceTransform[(-1)^K$144*DiracDelta[2*K$144 - x], x, y]
>
> Out[3]=
> ((-1)^K$144*UnitStep[2*K$144])/E^(2*K$144*y)
>
> Why everything works as expected if we set UnitStep[0]==1/2?