
#1
Feb2810, 08:09 AM

P: 1,057

I was trying to expand a three and more parameter functions similarly to the twoparameter case f(x,y)=(f(x,y)+f(y,x))/2+(f(x,y)f(y,x))/2.
Anyway, to do the same for more parameters I need to solve [tex] \begin{pmatrix} 1 & 1 & 1 & \dotsb & 1\\ 1 & \omega & \omega^2 & \dotsb & \omega^{n1}\\ 1 & \omega^2 & \omega^4 & \dotsb & \omega^{2(n1)}\\ 1 & \omega^3 & \omega^6 & \dotsb & \omega^{3(n1)}\\ \vdots & &&& \vdots \\ 1 & \omega^{n1} & \omega^{2(n1)} & \dotsb & \omega^{(n1)(n1)} \end{pmatrix}\mathbf{x}= \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix} [/tex] with [itex]\omega=\exp(2\pi\mathrm{i}/n)[/itex] Is there a closed form expression for x? EDIT: Oh, silly me. I realized it's a discrete Fourier transform. So is this the correct way to expand then? In the 3 parameter case the solution would be [tex] f(x,y,z)=\frac13(f(x,y,z)+f(y,z,x)+f(z,x,y))+\frac13\left(f(x,y,z)+\ome ga f(y,z,x)+\omega^*f(z,x,y)\right)+\frac13\left(f(x,y,z)+\omega^*f(y,z,x) +\omega f(z,x,y)\right)[/tex] with [itex]\omega=\exp(2\pi\mathrm{i}/3)[/itex]? Now I'm just wondering why I get linear dependent terms when I consider the real part only? 



#2
Feb2810, 09:17 PM

Sci Advisor
PF Gold
P: 2,020

You're on the right track. It is convenient to write this equation in matrix notation as
[tex]Wx=b[/tex] where b is the vector you wrote on the RHS. Since W is fullrank and nonsingular, it is invertible and the solution to your problem is [tex]x=W^{1}b[/tex] . You need the inverse of W, but this is easy since we know from the properties of the discrete Fourier transform (DFT) that the individual vector columns of W are independent. Each is an orthonormal basis vector of the DFT. In words, this follows because the component of signal at one frequency (cosine or sine at w_m) is independent to that at every other frequency w_n. In fact, [tex]W^{1} = W^{\dagger} [/tex] that is, W is Hermitian (the dagger is the conjugate transpose operation) and unitary (you get the identity matrix in the next equation) [tex]WW^{\dagger}=I [/tex] . You can perform this multiplication explicitly it to see that this is so. Accordingly [tex]x=W^{\dagger}b[/tex] and you can write out the components of x explicitly. 



#3
Mar110, 06:00 AM

P: 1,057

I just wonder if that way to decompose a multiparameter function makes sense.
It's nicely symmetrical and generalizes to higher dimensions. However it introduces complex number where the initial function might actually be real only. And also I haven't included odd parity permutations of the function arguments... 



#4
Mar110, 09:50 AM

Sci Advisor
PF Gold
P: 2,020

Inversion of this Vandermonde matrix
Sorry, I'm not following your comments. I provided the closed solution to Wx=b, where W is complex, which was the question asked in your first post. Are you looking for something else?




#5
Mar110, 12:02 PM

P: 1,057

I did that exercise to find a way to extend the rule f(x,y)=(f(x,y)+f(y,x))/2+(f(x,y)f(y,x))/2 to higher dimensions (I didn't explain the connection; just mentioned it in the intro). I assumed some cyclic symmetry for the final form of f(x,y,z)=... and with the help of the discrete Fourier transform (which I didnt recognise at first), I can find some "decomposition".
Now I wasn't sure if the decomposition is useful this way. 


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