Re: simplifying inside sum, Mathematica 5.1
- To: mathgroup at smc.vnet.net
- Subject: [mg53762] Re: simplifying inside sum, Mathematica 5.1
- From: Maxim <ab_def at prontomail.com>
- Date: Wed, 26 Jan 2005 04:37:34 -0500 (EST)
- Organization: MTU-Intel ISP
- References: <ct4h70$av2$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Mathematica does use some implicit assumptions about the iteration bounds:
In[1]:=
Sum[1, {k, 1/2, n}]
Out[1]=
1/2 + n
This assumes that the sum is taken over the points k = 1/2, 3/2, ..., n;
that is, when evaluating Sum[f[k], {k, a, b}] symbolically Mathematica
assumes that (b-a) is integer. For a value of n such that (n-1/2) is not
integer (say n=5) Out[1] would be incorrect in the sense that Sum[1, {k,
1/2, 5}] does not equal 1/2 + 5.
So if we take
In[2]:=
Sum[Sin[Pi*k]^2, {k, n}]
Out[2]=
(1 + 2*n - 2*I^(2*n)*n*Cos[n*Pi] - Cos[2*n*Pi])/4
the result again doesn't hold for an arbitrary non-integer n (Out[2] can
be complex), and since n is assumed to be integer, Mathematica indeed
could have simplified the summand by doing Refine[Sin[Pi*k]^2, Element[k,
Integers]] to obtain 0 immediately.
What seems to be the real problem is that Mathematica doesn't handle those
implicit assumptions in a consistent way:
In[3]:=
Sum[Cos[Pi*k], {k, 1/2, 2*n}]
Out[3]=
1
In[4]:=
Sum[UnitStep[1 - k], {k, 1/2, Infinity}]
Out[4]=
3/2
In[5]:=
Sum[Sqrt[k], {k, 1/2, n + 1/2}]
Out[5]=
HarmonicNumber[1/2 + n, -1/2]
It is easy to verify that Out[3] is incorrect for any value of n (again,
in the sense that if we substitute any numerical value for n before
evaluating the sum, we will obtain 0, not 1), Out[4] cannot be correct
because a sum of UnitStep terms cannot be half-integer, and Out[5] may
happen to be correct for some values of n, but not for integer n.
Next, what if we take a sum from a to b with a>b?
In[6]:=
Sum[1, {k, Infinity, 1}]
Out[6]=
-Infinity
In[7]:=
Sum[1/k^2, {k, Infinity, 1}]
Out[7]=
Pi^2/6
In[8]:=
Sum[1/(k^2 + 1), {k, Infinity, -Infinity}]
Out[8]=
0
Out[6] can be justified if we assume
Sum[f[k], {k, a, b}] == -Sum[f[k], {k, b, a}].
Out[7] is fine under the assumption
Sum[f[k], {k, a, b}] == Sum[f[k], {k, b, a}].
Out[8] is valid if we agree that
Sum[f[k], {k, a, b}] == 0 provided that a>b.
Each of these three conventions makes sense and can be useful. However, it
is impossible to reconcile all three answers with one another (I prefer
the third case, again because it agrees with numerical summation). We
should choose one and only one convention and stick to it, otherwise it's
just as good as returning a random result.
Finally, the general form of the iterator is {k, a, b, d}:
In[9]:=
Sum[1/k^2, {k, Infinity, 1, -1}]
<Infinity::indet warning>
Out[9]=
Indeterminate
Even in these trivial examples we can see that Mathematica simply doesn't
have a clear definition of how a symbolic sum is understood in such cases,
and consequently, doesn't have a consistent implementation either.
Maxim Rytin
m.r at inbox.ru
- Follow-Ups:
- Re: Re: simplifying inside sum, Mathematica 5.1
- From: DrBob <drbob@bigfoot.com>
- Re: Re: simplifying inside sum, Mathematica 5.1