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Orthogonalize[{E^x, E^(w^1*x), E^(w^2*x), E^(w^3*x)}, 
  Integrate[#1 *Conjugate[#2], {x, -Pi, Pi}] &]

I am trying to orthogonalize the first 4 expression using my own custom inner product which is the integral of F(x)[G(x)]* from -pi to pi. I want the second argument of the product (G(x)) to be the complex conjugate but I cannot figure out how to do that. Any tips?

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  • 1
    $\begingroup$ At a minimum try E^x instead of e^x. $\endgroup$
    – Somos
    Feb 3 '19 at 2:47
  • $\begingroup$ Thanks for the heads up. I changed it but it still seems to error out due to the integration. $\endgroup$ Feb 3 '19 at 2:58
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    $\begingroup$ Next step. What is the value of w? Is it a complex number? Also, what is the error message? $\endgroup$
    – Somos
    Feb 3 '19 at 8:27
  • $\begingroup$ I changed some things and it is no longer erroring but instead giving me a long answer. Should Mathematica be displaying Conjugate[i] instead of -i? $\endgroup$ Feb 3 '19 at 17:40
  • $\begingroup$ Try replacing i with I if you want $\sqrt{-1}$. $\endgroup$
    – Somos
    Feb 3 '19 at 17:42
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I think the inner-product function needs to be linear in the second argument, meaning you need to reverse the order of the parameters in your inner-product function. If you set w=a+I*b and reverse your inner-product function:

F = Integrate[ComplexExpand[Conjugate[#1]]*#2, {x, -Pi, Pi}] &

then this simplified example works:

S = Orthogonalize[{E^x, E^(w^1*x)}, F]

Test:

F[S[[1]], S[[1]]]
(* 1 *)
F[S[[1]], S[[2]]]
(* 0 *)

With your four terms it should work too, but will take a long time.

Keep in mind that the manual on Orthogonalize says that "The $e_i$ can be any expressions for which $f$ always yields real results." Your example does not satisfy this condition (some of the overlap integrals are complex-valued); what I wrote above about linearity in the second argument may or may not work properly.

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  • $\begingroup$ There is an example u = Orthogonalize[RandomComplex[1 + I, {4, 4}]] in the help. $\endgroup$
    – user64494
    Feb 3 '19 at 18:49
  • $\begingroup$ Look in en.wikipedia.org/wiki/Inner_product_space concerning linearity in the second argument. This property is not required. $\endgroup$
    – user64494
    Feb 3 '19 at 18:56
  • $\begingroup$ Seem to work, thanks! $\endgroup$ Feb 3 '19 at 19:03
  • $\begingroup$ @user64494 you're correct about the mathematical definitions (wiki link); however this is not how it's implemented in Mathematica. When I define F = Integrate[#1*ComplexExpand[Conjugate[#2]], {x, -Pi, Pi}] & to make the inner-product function linear in the first argument, as initially proposed, then the result is incorrect in the sense that F[S[[1]], S[[2]]] does not give zero. $\endgroup$
    – Roman
    Feb 3 '19 at 19:36
  • $\begingroup$ Upgrade your math. BTW, Maple produces answer to the original question with $w=(-1)^{\frac 1 4}$ in approximately 400 seconds. $\endgroup$
    – user64494
    Feb 3 '19 at 19:47

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