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Suppose that I am given ${\tilde U_1,\ldots,\tilde U_N}$ as a sequence of numbers, and in addition, $U_1,\ldots,U_N$ is unknown, and $q$ is unknown and constant for all ${\tilde U_1,\ldots,\tilde U_N}$.

$$\tilde U_1 = U_1 \exp (f(q, 1))$$

$$\tilde U_2 = U_2 \exp (f(q,2))$$

$$\tilde U_3 = U_3 \exp (f(q,3))$$

The sequence continues up to:

$$\tilde U_N = U_N \exp (f(q,N))$$

Is there any numerical method or way to check and see if the exponential function has "disappeared", without knowing $f(q,p)$, but knowing that q is constant?

Suppose that for $ p = 1,...,N$:

${{\tilde U}_p} = {U_p}\exp ({k_{A,p}} + {ik_{B,p}})\exp \left[ { - \frac{{{\omega _p}}}{{2q}}\left[ {{{\left( {\frac{{{\omega _p}}}{{{\omega _h}}}} \right)}^{ - 1/\pi q}}} \right] - \left[ {\left[ {{{\left( {\frac{{{\omega _p}}}{{{\omega _h}}}} \right)}^{1/\pi q}} - 1} \right]{\omega _p}} \right]i} \right]$

The goal is to find a $q$ that will make $f(q,p) = 0$ without knowing $k_{A,p}$ or $k_{B,p}$, but knowing that $k_{A,p}$ and $k_{B,p}$ are positive.

${{\tilde U}_1} = {U_1}\exp ({k_{A,1}} + {ik_{B,1}})\exp \left[ { - \frac{{{\omega _1}}}{{2q}}\left[ {{{\left( {\frac{{{\omega _1}}}{{{\omega _h}}}} \right)}^{ - 1/\pi q}}} \right] - \left[ {\left[ {{{\left( {\frac{{{\omega _1}}}{{{\omega _h}}}} \right)}^{1/\pi q}} - 1} \right]{\omega _1}} \right]i} \right]$

${{\tilde U}_2} = {U_2}\exp ({k_{A,2}} + {ik_{B,2}})\exp \left[ { - \frac{{{\omega _2}}}{{2q}}\left[ {{{\left( {\frac{{{\omega _2}}}{{{\omega _h}}}} \right)}^{ - 1/\pi q}}} \right] - \left[ {\left[ {{{\left( {\frac{{{\omega _2}}}{{{\omega _h}}}} \right)}^{1/\pi q}} - 1} \right]{\omega _2}} \right]i} \right]$

${{\tilde U}_3} = {U_3}\exp ({k_{A,3}} + {ik_{B,3}})\exp \left[ { - \frac{{{\omega _3}}}{{2q}}\left[ {{{\left( {\frac{{{\omega _3}}}{{{\omega _h}}}} \right)}^{ - 1/\pi q}}} \right] - \left[ {\left[ {{{\left( {\frac{{{\omega _3}}}{{{\omega _h}}}} \right)}^{1/\pi q}} - 1} \right]{\omega _3}} \right]i} \right]$

This means that I am searching for a q such that the sequence above becomes:

${{\tilde U}_1} = {U_1}$

${{\tilde U}_2} = {U_2}$

${{\tilde U}_3} = {U_3}$

All that I know is the LHS, and I know that $q$ is constant on the RHS. I don't know $k_{A,N}, k_{B,N}$, $q$ and $U_1,\ldots,U_N$, but I do know $\omega_h$ as a constant and $\omega_p$ that changes for each element of the sequence. In the above, $i$ represents a complex number. In addition, $k_{A,p}, k_{B,p}$ are positive numbers.

Is there a way to check for the presence or absence of the exponential function in the sequence, and in doing so, determine $q$, which is constant for the entire sequence? Is there anything that I can do or change to get an approximation of $q$? Why or why not?

I suppose that the exponential function is still present in the expression, but I would like to make $\exp(f(q,p)) = 1$ for $p = 1,...,N$.

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1  
What is $f$? What do you mean by the "absence of the exponential function"? It's right there, in $\exp(f(q,1))$! –  Robert Israel Jul 31 '12 at 16:24
    
@RobertIsrael: Thanks for your comment. Perhaps you can help me with the notation here. Essentially what I know is $\tilde U_1,...,\tilde U_N$, and I know the form of the expression, but I don't know the RHS. I've updated my question above. –  Nicholas Kinar Jul 31 '12 at 17:15
    
At least to me your update doesn't clarify anything. Is $f$ a known or an unknown function? If known, what properties does it have? –  celtschk Jul 31 '12 at 17:42
    
If you don't tell us anything about $f$, $\exp(f(q,1)), \ldots, \exp(f(q,N))$ are just $N$ arbitrary nonzero numbers, and there's no way we can say anything useful. –  Robert Israel Jul 31 '12 at 17:53
    
@RobertIsrael: Okay, I hope that I'm heading in the right direction. I've updated the question above. Can anything useful still be said so that I can determine $q$? –  Nicholas Kinar Jul 31 '12 at 18:36

1 Answer 1

up vote 1 down vote accepted

Since you know neither $U_p$ nor $k_{A,p}$ you can't determine it, because if there's some solution with $U_p$ and $k_{A,p}$ then the same numbers $\tilde U_p$ will be achieved for $U'_p:=U_p\exp{-c}$ and $k'_{A,p}:=k_{A,p}+c$ with arbitrary $c>0$. If one of then fulfils your condition $U_p=\tilde U_p$, the other doesn't.

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Thanks, celtschk. Does this imply that there are many $q$, two different $q$ or a range of $q$ that will fulfill the condition? If I could restrict the range of $q$ or make some sort of assumption about $q$, can I still determine $q$? –  Nicholas Kinar Jul 31 '12 at 19:16
    
@NicholasKinar: No, it means that without further information about $k_{A,p}$ you cannot know whether such a $q$ exists at all (well, there might be special cases where you can explicitly exclude the existence, but there are definitely no cases where you can prove that it exists). I'm assuming that the $k_{A,p}$, while not known, are nevertheless fixed (i.e. you cannot just choose them to your liking). –  celtschk Jul 31 '12 at 19:45
    
@NicholasKinar: I slightly correct myself: There's one case where you can know for sure such a $q$ exists: If all $\tilde U_p=0$. Because that's only possible if also all $U_p=0$, and then the value of $q$ doesn't matter. –  celtschk Jul 31 '12 at 19:48
    
Okay, I understand now. Thanks for helping me think clearly. –  Nicholas Kinar Jul 31 '12 at 20:05

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