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hello community.

I need some help

given $k = \sum A_q e^{ia_q s}$ where $k$, and $s$ is known. Can $A_q$ s now be expressed in terms of $a_q$ .

any help in that direction will be appreciated.

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closed as not a real question by Felipe Voloch, fedja, Andrés E. Caicedo, Gerry Myerson, Ryan Budney Dec 5 '11 at 4:23

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Is the sum finite or infinite? Is it just one equation or a system? – fedja Dec 4 '11 at 12:47
finite. one equation. so we have one equation with $n$ unknowns. that is, $a_q$ (or $A_q$) for $ q \in \{1,...,n\}$ – Sean Dec 4 '11 at 14:21
You should indicate if the $A_q$ and $a_q$ are assumed real (which they likely are). What about $k$ and $s$ ? – Jacques Carette Dec 4 '11 at 14:25
in the general case $k$ is complex, $s$ is real. – Sean Dec 4 '11 at 14:39
Is there something I'm missing? How can you hope to solve one equation for N variables? – Antoine Levitt Dec 4 '11 at 14:57
up vote 1 down vote accepted

No, you cannot, at least not at this level of generality. Consider for example $a_q = \frac{\pi}{q}$, $s=1$ and $k=0$. Basically a roots-of-unity situation; then each $A_q$ can be a multiple of the proper coefficient of a cyclotomic polynomial. That does 'solve' the problem, in that you get the $A_q$'s in terms of the $a_q$, up to a single free variable. But of course you can also express a finite linear combination of many roots of unity in the same way, thus getting as many free variables as you wish.

If the $a_q$ have some structure, then you might be able to say something -- thought the problem often reduces to some nasty diophantine equations OR to requiring Schanuel's conjecture to separate things out.

Your question is likely to get closed soon unless you make it a lot more precise and motivated.

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thanks, i got some headstart – Sean Dec 4 '11 at 16:02

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