Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
    
With reference to the newly added homework tag: I assure you that I am not asking for help in connection to a homework assignment. It's too late for that! This is now exam preparation! –  rbell May 18 '11 at 21:40

2 Answers 2

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

share|improve this answer

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

Then you can solve it.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.