Can somebody show me how to go about solving the following (really easy) equation for $\alpha$ please? $$\displaystyle 0 = \sum_{n=1}^{N}\left( \frac{y_n}{\alpha} + \frac{1-y_n}{1-\alpha}\right)$$
Really easy algebra problem
5
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algebra-precalculus
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0Sorry, I made a change (forgot plus sign) in between times. It's ok though, turns out Wolfram Alpha can do these things! Thanks anyway – 2011-05-18
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0@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 $n$. – 2011-05-18
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0Are 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. – 2011-05-18
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0@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? – 2011-05-18
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0Please see proof, I am genuinely stuck working on a past paper, http://imgur.com/Dlk8c – 2011-05-18
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0@rbell: Several people (myself included) have posted derivations, preceding your comments above. Are you stuck even after seeing those solutions? – 2011-05-18
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0@Arturo Magidin I'm ashamed to say that's the case. I'm trying to show that $\alpha = \frac{1}{N}\sum_{n=1}^{N}y_n$. – 2011-05-19
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0Sorry, turns out I made a mistake elsewhere. – 2011-05-19