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Suppose that we have a function:

$\tilde U(\tau ,\omega ) = U(\tau ,\omega )f(\tau ,\omega ,Q)$

By evaluating this function, I would like to be able to find $Q$. Suppose that I don't know the explicit form of $U(\tau ,\omega )$ but I do know the explicit form of function $f(\tau ,\omega ,Q)$.

The $U(\tau ,\omega )$ and the $f(\tau ,\omega ,Q)$ can be complex-valued numbers. The $Q$ is a real integer.

Suppose that I evaluate the function a number of times (perhaps more times than shown here):

$\tilde U(\tau ,{\omega _0}) = U(\tau ,{\omega _0})f(\tau ,{\omega _0},Q)$

$\tilde U(\tau ,{\omega _1}) = U(\tau ,{\omega _1})f(\tau ,{\omega _1},Q)$

$\tilde U(\tau ,{\omega _2}) = U(\tau ,{\omega _2})f(\tau ,{\omega _2},Q)$

and so on, for as many function evaluations as required. When evaluating the function, I've fixed $\tau$ and $Q$ and I know $\omega_0$, $\omega_1$, $\omega_2$. The $Q$ remains constant for each function evaluation.

Is there a way (using numerical methods or non-linear curve-fitting) to find $Q$, only given $\tilde U(\tau ,\omega_0 )$, $\tilde U(\tau ,\omega_1 )$, $\tilde U(\tau ,\omega_2 )$ and $\omega_0$, $\omega_1$, $\omega_2$ and $\tau$ variables?

What might I be able to do in order to find Q? Might there be a way to change the problem so that I can find $Q$?

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1 Answer 1

up vote 1 down vote accepted

If you don't know anything about $U$, the product $Uf$ carries very little information about $f$. Namely, all we can conclude is that $f\ne 0$ whenever $Uf\ne 0$. This little information might still be helpful in ruling out possible candidates for $Q$. If you suspect that $Q=Q_1$ and can find $\tau,\omega$ such that $f(\tau, \omega,Q_1)=0$, then the evaluation of $\tilde U(\tau, \omega )$ can refute the hypothesis.

To get more, you should carefully consider the nature of $U$ and extract as much analytic information about $U$ as possible.

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Thanks Leonid; that's really interesting. I will investigate. –  Nicholas Kinar Jul 23 '12 at 19:16
    
I agree that more analytic information is indeed required for $U$. –  Nicholas Kinar Jul 24 '12 at 17:54

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