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Re: Assuming non-integer values in Mathematica simplifications
*To*: mathgroup at smc.vnet.net
*Subject*: [mg71094] Re: Assuming non-integer values in Mathematica simplifications
*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
*Date*: Wed, 8 Nov 2006 06:14:40 -0500 (EST)
*Organization*: The Open University, Milton Keynes, UK
*References*: <eihm5b$ou6$1@smc.vnet.net>
vladimir wrote:
> I just started using Mathematica. I need to simplify the following expressions assuming that w/Pi is not integer (see below). I used the command Element(w/Pi,Rationals] and Element[w/Pi,Reals], but I still get the answer containing If(w/Pi is Integers ...) in many places, making it difficult to extract the answer for non-integer w/Pi. It seems that the simplification commands in Mathematica do not listen to the assumption statements even when such a statement is given within the simplification command. Does anybody know how to tell Mathematice to stop evaluating the integer cases? Thanks in advance.
>
> Here is my expression:
Surely, what Mathematic cannot do is reading your mind! FullSimplify
*does* take in account the assumptions you gave; however, contrary to
what you seem to believe, these assumptions does *not* -- and have *no*
reason to -- exclude integer solutions.
In standard mathematics, an integer *is* a rational. The set of rational
numbers Q is equal to {x = p/q | p and q are integers and q != 0}.
Therefore, every natural number {1, 2, 3, ... } is a rational, zero is a
rational, every negative integer is a rational (and also a real, and
also a complex).
> FullSimplify[(Sum[1, {k, 0, n - 1}]*Sum[
> Cos[w*k]*Sin[w*k], {k, 0, n - 1}]*Sum[Sin[w*k]*x[k], {k,
> 0, n - 1}] - Sum[1, {k, 0, n - 1}]*Sum[Cos[w*k]*x[
> k], {k, 0, n - 1}]*Sum[Sin[w*k]^2, {k,
> 0, n - 1}] - Sum[Cos[w*k], {k, 0,
> n - 1}]*Sum[Sin[w*k], {k, 0, n - 1}]*Sum[Sin[w*k]*x[
> k], {k, 0, n - 1}] - Sum[Cos[w*k]*Sin[w*k], {k, 0, n - 1}]*
> Sum[Sin[w*k], {k, 0, n - 1}]*Sum[x[k], {k, 0, n - 1}] + Sum[Cos[w*
> k], {k, 0, n - 1}]*Sum[x[k], {k, 0, n - 1}]*Sum[Sin[w*k]^2, {k, 0,
> n - 1}] + Sum[Cos[w*
> k]*x[k], {k,
> 0, n - 1}]*Sum[Sin[w*k], {k, 0, n - 1}]^2)/(-2*Sum[Cos[
> w*k], {k, 0, n - 1}]*Sum[Sin[w*
> k], {k, 0, n - 1}]*Sum[Cos[w*k]*Sin[w*k], {k, 0,
> n - 1}] + Sum[Sin[w*k], {k, 0,
> n - 1}]^2*Sum[Cos[w*k]^2, {k, 0, n - 1}] + Sum[Cos[w*k]*
> Sin[w*k], {k, 0, n - 1}]^2*Sum[1, {k, 0, n - 1}] + Sum[
> Cos[w*k], {k, 0, n - 1}]^2*
> Sum[Sin[w*k]^2, {k, 0, n - 1}] - Sum[1, {k, 0, n -
> 1}]*Sum[Cos[w*k]^2, {k, 0, n - 1}]*Sum[Sin[w*k]^2, {k, 0, n -
> 1}]), w/Ï? â?? Rationals]
Consequently, the correct assumption is
Not[w/Pi \[Element] Integers]
Now, compare
FullSimplify[Sum[1, {k, 0, n - 1}]*
Sum[Cos[w*k]*Sin[w*k], {k, 0, n - 1}]*
Sum[Sin[w*k]*x[k], {k, 0, n - 1}],
w/Pi \[Element] Rationals]
that yields
(1/2)*n*If[w/Pi \[Element] Integers,
(I^((2*(-1 + n)*w)/Pi)*(-1 + n) + Cos[(-1 + n)*w])*
Sin[(-1 + n)*w], (-Csc[w])*Sin[n*w]*Sin[w - n*w]]*
Sum[Sin[k*w]*x[k], {k, 0, -1 + n}]
against
FullSimplify[Sum[1, {k, 0, n - 1}]*
Sum[Cos[w*k]*Sin[w*k], {k, 0, n - 1}]*
Sum[Sin[w*k]*x[k], {k, 0, n - 1}],
!w/Pi \[Element] Integers]
that yields
(-(1/2))*n*Csc[w]*Sin[n*w]*Sin[w - n*w]*
Sum[Sin[k*w]*x[k], {k, 0, -1 + n}]
Regards,
Jean-Marc
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