Re: error with Sum and Infinity

*To*: mathgroup at smc.vnet.net*Subject*: [mg102265] Re: error with Sum and Infinity*From*: pfalloon <pfalloon at gmail.com>*Date*: Mon, 3 Aug 2009 05:47:09 -0400 (EDT)*References*: <h53o1e$1i8$1@smc.vnet.net>

On Aug 2, 7:59 pm, Llewlyn <tommaso.biancal... at gmail.com> wrote: > Hi experts, > > i'm very new to Mathematica and this is my first problem. I define a > function that is 0 everywhere: > t[i_] := 0 > expect for point i =3 that is: > t[3] = 1 > If now i evaluate the finite sum i obtain 1 as expected > Sum[ t[i], {i, 1, 10}] > Out = 1 > but if i try do the same with infinity i obtain > In[808]:= Sum[ t[i], {i, 1, Infinity}] > Out[808]= 0 > instead of 1 expected. Any advice? > > tb It doesn't work with Infinity as the upper limit because in that case the function works by first symbolically evaluating the argument. In this case, for symbolic i, t[i] happily returns 0. Using a more strict definitions will resolve the issue. One solution is to modify the first definition to only match integer arguments: In[202]:= Clear[t]; t[i_Integer] := 0 t[3] = 1; Sum[t[i], {i,1,Infinity}] Out[205]= \!\( \*UnderoverscriptBox[\(\[Sum]\), \(i = 1\), \(\[Infinity]\)]\(t[i]\)\) In[206]:= N[%] Out[206]= 1. Note that here you need to use N to get a numeric result; this is non- ideal, but at least it is no longer returning a spurious result. A better solution is a piecewise definition (in the Piecewise function the default value is 0, so you only need to specify the value at i=3): In[215]:= Clear[t]; t[i_] = Piecewise[{{1, i == 3}}]; Sum[t[i], {i,1,Infinity}] Out[217]= 1 But for the specific case you give, the neatest solution would be to define t as a Kronecker delta function: In[221]:= t[i_] := KroneckerDelta[i-3] Sum[t[i], {i,1,Infinity}] Out[222]= 1 There are many ways to solve this, but I guess the moral is to be careful when using multiple definitions relying on pattern matching of the argument -- since it can often have this kind of unexpected effect. Cheers, Peter.