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Re: If and Piecewise don't quite do what I need
*To*: mathgroup at smc.vnet.net
*Subject*: [mg113457] Re: If and Piecewise don't quite do what I need
*From*: Bob Hanlon <hanlonr at cox.net>
*Date*: Fri, 29 Oct 2010 06:29:35 -0400 (EDT)
You get a wrong result that way because the special case of (n = 0) is not handled by Sum
Clear[d]
d[0] = 1;
d[n_] = Integrate[a^2 Cos[n a], {a, -Pi, Pi}];
Sum[d[n], {n, 0, Infinity}]
Pi^3/3
% // N
10.3354
Handling the first term separately,
1 + Sum[d[n], {n, 1, Infinity}]
1 - Pi^3/3
% // N
-9.33543
Bob Hanlon
---- Bill Rowe <readnews at sbcglobal.net> wrote:
=============
On 10/27/10 at 5:15 AM, sam.takoy at yahoo.com (Sam Takoy) wrote:
>I have found that If doesn't do winside it like I would want it to
>do. Piecewise does do it, but then it doesn't work the way I would
>want it. Is there a solution that does both?
>Example:
>c[n_] := Piecewise[{{1, {n, 0}}},
>Integrate[a^2 Cos[n a], {a, -Pi, Pi}]]
>c[n] (* Calculates the Integral *)
>Sum[c[n], {n, 0, Infinity}] (* But fails to Sum *)
>d[n_] := If[n == 0, 1, Integrate[a^2 Cos[n a], {a, -Pi, Pi}]]
>(* Fails to calculate the integral *)
>d[n] (* But does sum OK *)
>Sum[d[n], {n, 0, Infinity}]
>I want it to evaluate the integral and to sum properly.
Instead of using either Piecewise or If why not do:
In[5]:= d[0] = 1;
d[n_] = Integrate[a^2 Cos[n a], {a, -Pi, Pi}];
In[7]:= Sum[d[n], {n, 0, \[Infinity]}]
Out[7]= Pi^3/3
Note, I use Set (=) not SetDelayed (:=) when defining d. This
avoids any need to re-compute the integral when doing the summation.
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