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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- 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)