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Re: Re: need help with integration
*To*: mathgroup at smc.vnet.net
*Subject*: [mg52060] Re: [mg52044] Re: need help with integration
*From*: DrBob <drbob at bigfoot.com>
*Date*: Tue, 9 Nov 2004 01:36:50 -0500 (EST)
*References*: <cmhv6l$pqv$1@smc.vnet.net> <cmklju$jk2$1@smc.vnet.net> <200411080813.DAA07936@smc.vnet.net>
*Reply-to*: drbob at bigfoot.com
*Sender*: owner-wri-mathgroup at wolfram.com
Symbio seems to be right, but a solution CAN be salvaged.
Here's Wolfgang's code from David's link, executed in Mathematica 5.0.1:
x[t_] = 3*Exp[-t/2]*UnitStep[t] +
DiracDelta[t + 3];
h[t_] = UnitStep[t] - UnitStep[t - 2];
eq1 = Integrate[x[a]*h[t - a],
{a, -Infinity, Infinity}];
eq2 = Integrate[x[a]*h[t - a],
{a, 0, Infinity}];
eq3 = Integrate[x[a]*h[t - a],
{a, -Infinity, 0}];
Simplify[eq1, t \[Element] Reals]
FullSimplify[eq2 + eq3, t \[Element] Reals]
(1/2)*(-((6*(-2 + t))/ Abs[-2 + t]) +
(6*t)/Abs[t] - (1 + t)/Abs[1 + t] + (3 + t)/Abs[3 + t])
-((1 + t)/(2*Abs[1 + t])) + (3 + t)/(2*Abs[3 + t]) +
6*((-1 + E^(1 - t/2))*UnitStep[-2 + t] + UnitStep[t] - UnitStep[t]/
E^(t/2))
Plot[{(1/2)*(-((6*(-2 + t))/Abs[-2 + t]) +
(6*t)/Abs[t] - (1 + t)/Abs[1 + t] +
(3 + t)/Abs[3 + t]),
(1/2)*(-Sign[1 + t] + Sign[3 + t] +
12*(-UnitStep[-2 + t] +
(E*UnitStep[-2 + t] - UnitStep[t])/
E^(t/2) + UnitStep[t]))},
{t, -10, 10}, PlotRange -> All]
The Plots are NOT the same, so what worked for Wolfgang in Mathematica 4.0 doesn't work in 5.0.1.
His Integrate with Assumptions gives yet a third expression (after a LOOOONG calculation):
eq1 = Integrate[x[a]*h[t - a],
{a, -Infinity, Infinity},
Assumptions -> t \[Element] Reals]
(1 + t + Abs[1 + t])/(-2 - 2*t) +
(3 + t + Abs[3 + t])/(6 + 2*t) -
3*((2 - 2*E^(1 - t/2))* UnitStep[-2 + t] +
2*(-1 + E^(-t/2))*UnitStep[t])
This one plots the same as eq2+eq3, so that solves the problem....
...IF we're convinced the original eq1 was wrong and the other two answers are right.
I can't muster much confidence in that.
Bobby
On Mon, 8 Nov 2004 03:13:30 -0500 (EST), symbio <symbio at has.com> wrote:
> actually this is different, i used both assumptions and simplify but won't
> work. Please try running the code, you'll see for yourself when you look at
> the plots, they are different even if you use assumptions.
>
> "David W. Cantrell" <DWCantrell at sigmaxi.org> wrote in message
> news:cmklju$jk2$1 at smc.vnet.net...
>> "symbio" <symbio at sha.com> wrote:
>>> Below I define two functions x[t] and h[t], then in eq1 I integrate
>>> the integrand ( x[tau] h[t-tau] ) from -inf to +inf and get one set of
>>> results and plots, then I integrate the same integrand as before but this
>>> time in two steps, once from -inf to 0 and once from 0 to +inf, but the
>>> results and plots from the integration performed in two steps are NOT the
>>> same as the results and plots from integration in one step. Is this a
>>> bug and if so what's the work around?
>>
>> This looks essentially like what you asked here about two weeks ago in
>> "Integration of UnitStep has bugs!? help!"
>>
>> <http://groups.google.com/groups?threadm=clneor%24obq%241%40smc.vnet.net>
>>
>> I would have answered your question in the previous thread had you not
>> already gotten two good answers. Look at them carefully. I suspect that
>> the
>> trouble you're having now is the same as you were having previously; if
>> so,
>> then your question has already been answered, twice.
>>
>> David
>>
>>> Try this in the input cell
>>>
>>> In[19]:=
>>> x[t_] = (3Exp[-.5t]UnitStep[t]) + DiracDelta[t + 3]
>>> h[t_] = UnitStep[t] - UnitStep[t - 2]
>>> eq1 = Integrate[x[\[Tau]]h[t - \[Tau]], {\[Tau], -\[Infinity],
>>> \[Infinity]}] Plot[eq1, {t, -10, 10}, PlotRange -> All]
>>> eq2 = Integrate[x[\[Tau]]h[t - \[Tau]], {\[Tau], 0, \[Infinity]}]
>>> Plot[eq2, {t, -10, 10}, PlotRange -> All]
>>> eq3 = Integrate[x[\[Tau]]h[t - \[Tau]], {\[Tau], -\[Infinity], 0}]
>>> Plot[eq3, {t, -10, 0}, PlotRange -> All]
>>>
>>> You will get the following output expressions (I couldn't paste the plots
>>> here, but if you run the above input cells you should get the plots too,
>>> then it will be more obvious what the problem is)
>>>
>>> Out[21]=
>>> \!\(1\/2\ \((1 + \(6.`\ \@\((\(-2\) + t)\)\^2\)\/\(\(\(2.`\)\(\
>>> \[InvisibleSpace]\)\) - 1.`\ t\) + \(6.`\ \@t\^2\)\/t + \@\((3 +
>>> t)\)\^2\/\(3 \
>>> + t\) - \(1 + t + \@\((1 + t)\)\^2\)\/\(1 + t\))\)\)
>>>
>>> Out[23]=
>>> \!\(\((\(-6.`\) +
>>> 16.30969097075427`\ \[ExponentialE]\^\(\(-0.5`\)\ t\))\)\ UnitStep[\
>>> \(-2\) + t] + \((\(\(6.`\)\(\[InvisibleSpace]\)\) -
>>> 6.`\ \[ExponentialE]\^\(\(-0.5`\)\ t\))\)\ UnitStep[t]\)
>>>
>>> Out[25]=
>>> \!\(1\/2\ \((\(-\(\(1 +
>>> t\)\/\@\((1 + t)\)\^2\)\) + \(3 + t\)\/\@\((3 + t)\)\^2)\)\)
>>
>
>
>
>
--
DrBob at bigfoot.com
www.eclecticdreams.net
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