How to find $L$ if $L=\frac{c}{(1-L)^a}$
I was trying to apply log but $\ln L +a\ln (1-L)=\ln c$. How can continued please?
Thank you
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$\begingroup$ This equation does not have a closed form (nice) solution for $L$. If $a$ is a positive integer then $L$ will be the root of a polynomial of degree $a$. That might be useful too you in some cases. $\endgroup$– Ethan BolkerCommented Sep 23, 2022 at 19:11
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$\begingroup$ Thank you, i forgot, $a>0$. How can continued to arrive that root please? $\endgroup$– weymar andresCommented Sep 23, 2022 at 19:14
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$\begingroup$ If $a>0$ is an integer you expand the binomial in the denominator and then multiply both sides by the denominator. That will give you a polynomial of degree $a$ on the left. Subtract $c$ and find the roots. For $a > 4$ there won't be a formula. $\endgroup$– Ethan BolkerCommented Sep 23, 2022 at 19:26
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$\begingroup$ thank you, an d when $a$ is not integer what happen? $\endgroup$– weymar andresCommented Sep 23, 2022 at 19:42
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1$\begingroup$ When $a$ is not an integer you will need numerical methods. $\endgroup$– Ethan BolkerCommented Sep 23, 2022 at 19:47
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1 Answer
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As Ethan Bolker notes, numerical methods are needed in general. But the details work out rather nicely nevertheless. We assume $a$ and $c$ are positive and define $$f(L) \doteq \ln L +a\ln (1-L) - \ln c.$$ The goal is to find all solutions to $f(L) = 0$ for $0 < L < 1$. The function $f$ is smooth and concave, and $f(L) \rightarrow -\infty$ as $L \rightarrow 0$ and as $L \rightarrow 1$. Therefore $f$ has a unique maximum on $(0,1)$ and it occurs at the solution to $$f'(L) = \frac{1}{L} - \frac{a}{1-L} = 0.$$ This solution is $L^* \doteq 1/(a+1)$, and the maximum that $f$ achieves there is $$f^* \doteq f(L^*) = a \ln a - (a+1) \ln(a+1) - \ln c.$$ We define $c^* \doteq a^a/(a+1)^{a+1}$ to be the value of $c$ for which this maximum is $0$.
For $c > c^*$ there are no solutions to $f(L) = 0$.
For $c = c^*$ there is one solution, namely $L = L^*$.
And for $c < c^*$ there are exactly two solutions, one with $0 < L < L^*$ and one with $L^* < L < 1$. To find these two solutions numerically, it is helpful to bound these roots a little better than with locations where $f(L) = -\infty$. We can do this by solving $f(L) = 0$ but with the second or the first term of $f$ omitted. This yields values of $L$ (on opposite sides of $L^*$) where $f(L) < 0$, so we obtain the intervals $(c,L^*)$ and $(L^*,1-c^{1/a})$ that contain each root in the $c < c^*$ case. You may now confidently call a numerical root finder, telling it to find the unique root that exists in each of the given intervals.
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$\begingroup$ i liked,thank you so much!! $\endgroup$ Commented Sep 24, 2022 at 11:46