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Currently I am trying to solve the following equality for $|p|<1$:

$$p\left(\frac{-\ln(a)-\ln(1+\gamma p+p^2)+\ln(1-p+ap+\gamma p+p^2)}{1-p+\gamma p+p^2}\right)=a-1,$$

where $0<a<1$ and $\gamma > 0$.

I am doing this because I have a probability generating function which is given as a quotient, where the numerator is a difficult integral. In order to say something about the numerator, I am trying to find the root of the denominator. This gave me the equality shown above.

Honestly, I have no idea how to handle this. Does anybody have any idea?

Thank you in advance.

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If you make the substitution $$ \zeta := 1-p+\gamma p + p^2$$ your equation becomes $$ \ln{\dfrac{\zeta+ap}{a(\zeta-p)}} = \dfrac{a-1}{p}\zeta. $$ using $\ln(a b) = \ln(a)\ln(b),\,\, \ln{\dfrac{a}{b}} = \ln(a)-\ln(b)$. Now let us substitute further $$ \eta:= \dfrac{\zeta+ap}{a(\zeta-p)} \Leftrightarrow \zeta = \dfrac{ap(1+\eta)}{a\eta-1} $$ for $\zeta \neq p, 1 \neq a\eta$. If you substitute this in the second equation you get independent (explicitly) of $p$ equation: $$ \ln{\eta} = a(a-1)\dfrac{1+\eta}{a\eta-1}\cdot $$ Now this equation can be solved approximately and you can get idea about the solutions graphically for different values of $a$ very easily. In the picture you see the graph for $a = 0.12$

If $\eta = 1$ then $\zeta = 0$ and you can calculate $p$ from the quadratic equation $$p^2 + p(\gamma-1) + 1 =\zeta$$ depending on $\gamma$.

Simulating over $a \in (0,1)$ seems to give that after $a \simeq 0.18$ there are no roots to this equation. Have also in mind that the case $\zeta = p$ should be considered separately.

P.S. The Wolfram Mathematica code that produced the manipulation is

Manipulate[ Plot[{Log[b], a (a - 1) (b + 1)/(a b - 1)}, {b, -10, 20}], {a, 0.01, 0.99, 0.01}]

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  • $\begingroup$ Thanks for the response. I redid your steps, but I think there is something wrong. When computing $\zeta$ I find $\frac{ap(\eta +1)}{1-a \eta}$. Would there still be an 'easy' root then? $\endgroup$ – Zdenek Rovnez May 17 '17 at 21:13
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    $\begingroup$ @ZdenekRovnez Yes, you are right. It is even $\dfrac{ap(\eta+1)}{a\eta-1}$ :) In this way I don't see a nice root. Moreover, as you can see from my answer after a certain point there is no solution to the equation.. $\endgroup$ – Veliko May 17 '17 at 21:37

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