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I have the ratio of two functions $D(\rho)$ and $S(\rho)$ where $$D(\rho)= g(\rho)e^{ia\rho}-g(\rho)e^{-ia\rho} \ \ \text{and} \ \ S(\rho)= g(\rho)e^{ia\rho}+g(\rho)e^{-ia\rho}.$$ Obviously $\dfrac{D(\rho)}{S(\rho)}= i \tan(\rho)$ is independent of the function $g(\rho)$.

But I also have an asymmetric case where

$$D(\rho)= g'(\rho)e^{ia\rho}-g'(-\rho)e^{-ia\rho} \ \ \text{and} \ \ S(\rho)= g'(\rho)e^{ia\rho}+g'(-\rho)e^{-ia\rho}$$

and $g'(\rho)$ is not symmetrical.

Is the ratio $\dfrac{D(\rho)}{S(\rho)}$ still independent if $g'(\rho)\neq g'(-\rho)$?

If so, what is the ratio? I'm having trouble seeing how to approach this, one way or the other.

EDIT
I have worked through my problem in a bit more detail, and the ratio works out as follows:

$$\dfrac{D(\rho)}{S(\rho)}=i \dfrac{\sum_{n=1}^N {a_n sin((n-1/2)\rho)}}{\sum_{n=1}^N {a_n cos((n-1/2)\rho)}}$$

It is from this that I'm trying to determine whether it can be rewritten in a way that is either independent of the coefficients or in a form like this:

$$\dfrac{D(\rho)}{S(\rho)}= iK \tan(\rho)$$

where K is a function of the coefficients $a_n$ but not a function of $\rho$.

The reason I am trying to do this is that some computer modeling has shown this to be the case, but I'm having some trouble with the analysis to show why this is the case and how to determine the factor, $K$.

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No, the ratio is not independent. To see this, we have to consider one counter example.

Let $g(\rho)=\rho+c\rho^2$, where $c$ is a constant (in the end we will see that the ratio depends on $c$, and thus on $g$). Then, a small calculation yields $$ \frac{D(\rho)}{S(\rho)}=\frac{2c\rho\cos(a\rho)+i\sin(a\rho)}{\cos(a\rho)+2icr\sin(a\rho)}. $$ Now, different $c$ (read: different $g$) gives different (non-equal) expressions for the quotient. Thus, the quotient will, in general, depend on $g$.

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  • $\begingroup$ Thanks you. I have added a detail to my question that may affect the answer. $\endgroup$ – Jim Jan 18 '16 at 18:27

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