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Re: What assumptions to use to check for orthogonality

  • To: mathgroup at smc.vnet.net
  • Subject: [mg112915] Re: What assumptions to use to check for orthogonality
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Tue, 5 Oct 2010 05:38:09 -0400 (EDT)

Nasser M. Abbasi wrote:
> This is basic thing, and I remember doing this or reading about it before.
> 
> I am trying to show that Cos[m Pi x], Cos[n Pi x] are orthogonal 
> functions, m,n are integers, i.e. using the inner product definition
> 
> Integrate[Cos[n*Pi*x]*Cos[m*Pi*x], {x, 0, 1}];
> 
> So, the above is ZERO when n not equal to m and 1/2 when n=m. hence 
> orthogonal functions.
> 
> This is what I tried:
> 
> ------ case 1 -------------
> Clear[n, m, x]
> r = Integrate[Cos[n*Pi*x]*Cos[m*Pi*x], {x, 0, 1}];
> Assuming[Element[{n, m}, Integers], Simplify[r]]
> 
> Out[167]= 0
> ----------------
> 
> I was expecting to get a result with conditional on it using Piecewise 
> notation.
> 
> Then I tried
> 
> ---------case 2 ------------
> Clear[n, m, x]
> r = Integrate[Cos[n*Pi*x]*Cos[m*Pi*x], {x, 0, 1}];
> Assuming[Element[{n, m}, Integers] && n != m, Simplify[r]]
> 
> Out[140]= 0
> 
> Assuming[Element[{n, m}, Integers] && n == m, Simplify[r]]
> 
> Out[184]= Indeterminate
> ----------------
> 
> So, it looks like one has to do the limit by 'hand' to see that for n=m 
> we get non-zero?
> 
> -------------------
> Clear[n, m, x]
> r = Integrate[Cos[n*Pi*x]*Cos[m*Pi*x], {x, 0, 1}];
> Limit[Limit[r, n -> m], m -> 1]
> 
> Out[155]= 1/2
> 
> Limit[Limit[r, n -> 1], m -> 99]
> 
> Out[187]= 0
> ----------------------------
> 
> 
> So, is there a way to get Mathematica to tell me that the integral is 
> zero for m!=n and 1/2 when n=m? (tried Reduce, Refine). It seems the 
> problem is that the Integrate is not taking the limit automatically to 
> determine what happens when n=m? Should it at least in case have told me 
> that when n!=m it is zero, and when n=m it is  Indeterminate? It just 
> said zero which is not correct when n=m and I did say n,m are integers.
> 
> thanks
> --Nasser

Here is what I get.

In[3]:= r = Integrate[Cos[n*Pi*x]*Cos[m*Pi*x], {x, 0, 1}]

Out[3]//InputForm=
(m*Cos[n*Pi]*Sin[m*Pi] - n*Cos[m*Pi]*Sin[n*Pi])/(m^2*Pi - n^2*Pi)

I am hard pressed to imagine a better result from Integrate. A result 
littered with Piecewise to handle a non-generic (indeed, discrete) set 
of values would be a really bad idea. And one that would be quite hard 
to make work in any general manner.

Also observe that the result is (also) nonzero for m = -n and integral. 
One way to see this is to let m approach -n.

In[5]:= Limit[r, m->-n]
Out[5]//InputForm= (2 + Sin[2*n*Pi]/(n*Pi))/4

Regarding the result from Simplify, which is (in some sense) generically 
correct, see:

http://forums.wolfram.com/mathgroup/archive/2008/Sep/msg00492.html

Daniel Lichtblau
Wolfram Research


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