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Suppose that I have a function:

$f(x,U) = x + \log(U)$

where $x$ and $U$ can be complex numbers. I take $f(x,U)$ and evaluate it a number of times for different $x$ and $U$.

Since $U$ can be complex, I take $\log(U)$ as the complex logarithm (Wikipedia Link). For example, $\log(-1) = 3.1416i$.

Now suppose that for all of my function evaluations, $x = 0$ so:

$f(x=0,U) =\log(U)$

But given only $f(x,U)$, and not knowing $x$ and $U$, is there a way to use curve-fitting or some other method to determine that $f(x,U) = \log(U)$ and $x = 0$?

Can numerical methods be used to distinguish between $f(x,U) = x + \log(U)$ and $f(x=0,U) =\log(U)$, and how would I deal with the real and complex parts?

I assume that $\log(U) \neq 0$ for all function evaluations, and that $x \neq \log(P)$, where $P$ is some complex number.

I've found a similar question on Stack Overflow (link), but I am uncertain as to whether something from that question can be used to solve this problem.

Is this problem less of an issue when considered with the complex set of numbers in lieu of the reals?

2 Answers 2

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If you have a function evaluation where $f(x,U)=0$, then $x \neq 0$ since $log(U) \neq 0$ for all function evaluations.

Otherwise, you'll have to rely on statistical methods, and assume $U$ is randomly distributed. You can represent $U$ as $a+bi$, where $a$ and $b$ are normally distributed, if you have some idea what the means and variances of $a$ and $b$ could be.

But since that involves four parameters you might have to fudge (I'm assuming you know as little as possible about the distribution of $U$), representing $U$ in polar form ($r e^{i\theta}$) could be easier: assume $\theta$ is uniformly distributed between $0$ and $2\pi$, and $r$ follows an exponential distribution. Then you'll just need to come up with a guess for what the mean of $r$ could be (the $\lambda$ parameter).

Regardless of how you generate your $U$s, evaluate $f$ on them assuming $x=0$. Create a histogram of the results, and compare this to a histogram of the function evaluations for the unknown $U$s and $x$s.

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    Thank you so much for all of your comments; I have learned a lot from you. And even more, thank you very much for your patience, and for the wonderful solution.2012-08-02
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Let's first deal with the reals. The range of $\log U$ as $U$ goes through the positive reals is the entire real line. The range of $x+\log U$ for real $x$ and positive real $U$ is the entire real line. It follows that if you are given the outputs and no information about the inputs you can't tell $\log U$ from $x+\log U$.

Now if $U$ can be a complex number, then first you have to tell me what you mean by $\log U$.

EDIT: I take it from your update that you are using the principal branch of the complex logarithm, so the imaginary part of $\log U$ lies between $-\pi$ (exclusive) and $\pi$ (inclusive). That means that if you get an output with an imaginary part outside those bounds, then you know you're getting $x+\log U$ for some $x\ne0$, and you even have some bounds on the imaginary part of $x$. The more outputs you get (with different imaginary parts), the tighter the bounds you get on the imaginary part of $x$.

On the other hand, if every output has imaginary part consistent with being $\log U$ for some $U$ and the principal branch, then you can make no conclusion as to whether or not $x=0$. All you can do is bound the imaginary part of any possible $x$, and you can't say anything about the real part.

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    It's interesting to look at the function as bounded on the imaginary part of $x$. Thanks for the edit, Gerry.2012-08-03