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Can somebody show me how to go about solving the following (really easy) equation for $\alpha$ please? $$\displaystyle 0 = \sum_{n=1}^{N}\left( \frac{y_n}{\alpha} + \frac{1-y_n}{1-\alpha}\right)$$

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Sorry, I made a change (forgot plus sign) in between times. It's ok though, turns out Wolfram Alpha can do these things! Thanks anyway – rbell May 18 '11 at 20:33
@rbell: look at picakhu's solution. Assuming that $y_i, i=1,..,n$ are known quantities, you can solve for $\alpha$ in terms of $\sum_{i=1}^n y_i$ and $n$. – Weltschmerz May 18 '11 at 20:40
Are there only finitely many values of $n$ that you are summing over? If the sum is an infinite sum, I'm struggling to think of an example where the sum converges. – Aaron May 18 '11 at 21:08
@Weltschmerz So I've worked this through for a while, but I keep getting stuck at $-N\alpha = \sum_n y_n(1 - 2\alpha)$, where N is the total number of $y$-terms. Can you give me a hint what I'm missing? – rbell May 18 '11 at 21:21
Please see proof, I am genuinely stuck working on a past paper, – rbell May 18 '11 at 21:46

Multiplying through by $\alpha(1-\alpha)$ you get $$\begin{align*} 0 &= \sum_n\left(\frac{y_n}{\alpha} + \frac{1-y_n}{1-\alpha}\right)\\ 0 &= \sum_{n}\left( (1-\alpha)y_n + \alpha(1-y_n)\right)\\ 0 &= \sum_n \left( y_n + (1-2y_n)\alpha\right)\\ 0&= \sum_n y_n + \sum_n \alpha(1-2y_n)\\ \sum_n \alpha(2y_n -1) &= \sum_n y_n\\ \alpha\sum_n(2y_n-1) &= \sum_n y_n\\ \alpha &= \frac{\displaystyle \sum_n y_n}{\displaystyle\sum_n (2y_n-1)}. \end{align*} $$

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$$\displaystyle 0 = \sum_n\left( \frac{y_n}{\alpha} + \frac{1-y_n}{1-\alpha}\right)$$ $$0 = \sum_n\left( \frac{y_n}{\alpha} \right) + \sum_n\left(\frac{1-y_n}{1-\alpha}\right)$$ $$0 = \frac{1}{\alpha} \sum_n y_n + \frac{1}{1-\alpha} \sum_n {(1-y_n)}$$

Then you can solve it.

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