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I'm currently doing some physics and while trying to solve a problem I came across an equation that I can't rearrange for the variable (the question is pure mathematics so I'm asking it here).

The equation is: $$H = \frac{\sum\limits_n^N E_n \exp ( \lambda E_n)}{ \sum\limits_n^N \exp(\lambda E_n) }$$

I need to calculate $\lambda$. $H$ and all $E_n$ are known. Clearly it should be possible to calculate $\lambda$ but I have no idea how to rearrange for $\lambda$. I think it can't be done, but even if I had $H$ and $E_n$ as actual numbers I can't see how to calculate $\lambda$. Can it only be done numerically or am I missing something?

Thanks.

EDIT:

I also know: $\sum\limits_{n=1}^N a_n E_n = H$ with $\sum_{n=1}^N a_n = 1$, $a_n = \frac{\exp (\lambda E_n)}{\sum\limits_{n=1}^N \exp(\lambda E_n)}$ and $N>1$ ( I used those to build the equation)

user44789
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    Are there any conditions on $E_n$, $\lambda$ or $H$? I am not even sure there generally exists a solution for $N=1$. E.g if $N=1$ and $E_1 = 1$ I think there is no solution for $H \in [0,e)$ (I think $H \neq \exp(\lambda)/\lambda$) – flawr Nov 15 '14 at 23:22
  • No, they are completely general. And it's only one variable, one equation, so it should be enough. – user44789 Nov 15 '14 at 23:29
  • You are right with your counter example. I know that $H = \sum a_n E_n$ with $\sum a_n = 1$ and since it's an entropy problem we have more than 1 particle, so $N > 1$. But I used those conditions already to build the equation, so I don't think it helps. – user44789 Nov 15 '14 at 23:38
  • sorry guys, I made a typo in the denominator. – user44789 Nov 15 '14 at 23:54
  • Not quite sure I understand the notation but it looks like the derivative of the log of a characteristic function, i.e. if $G=\ln (\sum \exp(\lambda E_n))$ then $H=\frac{d G}{d \lambda}$. – user103828 Nov 16 '14 at 10:51

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This isn't yet an Answer to your question, but note that you can write

$$H\sum_{n=1}^N \mu^{E_n} = \sum_{n=1}^N E_n \mu^{E_n}$$ with $\mu = e^\lambda$.

So the problem is finding a positive positive (since $\mu = e^\lambda>0$) zero of

$$ 0 = \sum_{n=1}^N (E_n - H)\mu^{En}.$$

(Which is not necessarily possible, e.g. if all $E_n-H$ do have the same sign.)

flawr
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