Re: Convolve: Different Looking Results. (2)
- To: mathgroup at smc.vnet.net
- Subject: [mg130642] Re: Convolve: Different Looking Results. (2)
- From: Bob Hanlon <hanlonr357 at gmail.com>
- Date: Sun, 28 Apr 2013 01:00:59 -0400 (EDT)
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- Delivered-to: l-mathgroup@wolfram.com
- Delivered-to: mathgroup-outx@smc.vnet.net
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- References: <20130427030837.E41B769EA@smc.vnet.net>
$Version
"9.0 for Mac OS X x86 (64-bit) (January 24, 2013)"
The canonical order of the variables can affect how algorithms are applied
and the resultant form; however, the form of any output may not be the
"simplest" unless you apply some degree of simplification.
cv1 = Convolve[Exp[-x] UnitStep[x], Sin[2 x] UnitStep[x],
x, y, Assumptions -> y > 0]
(-(2/5))*(Cos[2*y] - Cosh[y] -
Cos[y]*Sin[y] + Sinh[y])
cv2 = cv1 // TrigToExp // FullSimplify
(1/5)*(-2*Cos[2*y] + 2*Cosh[y] + Sin[2*y] -
2*Sinh[y])
cv3 = Convolve[Exp[-x] UnitStep[x], Sin[2 x] UnitStep[x],
x, t, Assumptions -> t > 0] /. t -> y
(1/5)*(2/E^y - 2*Cos[2*y] + Sin[2*y])
cv4 = cv3 // TrigToExp // FullSimplify
(1/5)*(-2*Cos[2*y] + 2*Cosh[y] + Sin[2*y] -
2*Sinh[y])
All of the forms are equivalent
cv1 == cv2 == cv3 == cv4 // Simplify
True
The two simplified results are identical
cv2 === cv4
True
Bob Hanlon
On Fri, Apr 26, 2013 at 11:08 PM, Polar Coords <pcoords29 at gmail.com> wrote:
>
> Dear Mathgroup,
>
> Sorry about my last incomplete mail.
>
> Consider the following in Mathematica 8 ( It's a convolution as in Laplace
> transforms)
>
> In[15]:= Convolve[Exp[-x] UnitStep[x], Sin[2 x] UnitStep[x], x, y,
> Assumptions -> y > 0]
>
> Out[15]= -(2/5) (Cos[2 y] - Cosh[y] - Cos[y] Sin[y] + Sinh[y])
>
> Now , use t instead of y
>
> In[16]:= Convolve[Exp[-x] UnitStep[x], Sin[2 x] UnitStep[x], x, t,
> Assumptions -> t > 0]
>
> Out[16]= 1/5 (2 E^-t - 2 Cos[2 t] + Sin[2 t])
>
>
> They're equivalent, but why the different looking results? We have the
> Out[16] results if we use symbols a to w, but as in Out[15] if using y
> or z !
>
> Do members have the same results in Mathematica 9?
>
> Finally, on a lighter note, one usually convolves 2 functions of x to get
> a third function of x. I can't suss out why Mathematica gives the answer
> in terms of
> a new variable.
>
> So, if I want to convolve 2 output functions of t, then I need to change to
> another variable and tell it to give the answer in terms of t. A hassle.
>
> Thanks for your help.
> Cheers.
>
> Sid
>
>
>
- References:
- Convolve: Different Looking Results. (2)
- From: Polar Coords <pcoords29@gmail.com>
- Convolve: Different Looking Results. (2)