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Re: InverseFourierTransform strange behaviour
*To*: mathgroup at smc.vnet.net
*Subject*: [mg82335] Re: InverseFourierTransform strange behaviour
*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
*Date*: Thu, 18 Oct 2007 04:44:53 -0400 (EDT)
*Organization*: The Open University, Milton Keynes, UK
*References*: <fesjid$ks8$1@smc.vnet.net><ff1qf0$9ao$1@smc.vnet.net> <ff4hrl$k1c$1@smc.vnet.net>
roby.nowak at gmail.com wrote:
> Hi Frank,
>
> sorry but you are not correct.
>
> 2 Sqrt[2/Pi] Exp[(w/2)^2] Sin[w]/w
> is pefectly even (as it is without the negative sign as you suggested)
>
>
> Exp[(w/2)^2] is even
> Sin[w] is odd
> 1/w is odd
> Sin[w]/w is even
>
> a simple numerical test too shows that the function is even ( f[x] ==
> f[-x] )
And one can even (no pun intended) use Mathematica's symbolic
capabilities to check that f(x) is even, indeed.
In[1]:= f[x_] := 2 Sqrt[2/Pi] Exp[(w/2)^2] Sin[w]/w
In[2]:= f[x] == f[-x]
Out[2]= True
Just my two cents,
Jean-Marc
> @everybody
>
> I am in doubt now if the IFT exists for my function. any suggestions
> to this ?
>
> regards robert
>
>
> On 16 Okt., 09:49, Frank Iannarilli <frank... at cox.net> wrote:
>> On Oct 14, 4:21 am, roby.no... at gmail.com wrote:
>>
>>> In[1]:= 2 Sqrt[2/Pi] Exp[(w/2)^2] Sin[w]/w;
>>> InverseFourierTransform[%, w, x]
>> Your function is actually not even, but I believe you intended it to
>> be:
>> 2 Sqrt[2/Pi] Exp[-(w/2)^2] Sin[w]/w
>> ^^^^
>> (note inserted negative sign)
>>
>> In this case, everything will work as you expected.
>>
>> Regards,
>
>
>
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