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Suppose that

$$y_i=\big(\frac{2x_i}{\sum^n(x_i)}\big)^{\frac{1}{b-1}}$$

Where $b$ is some constant >1, then what is $$\sum^n y_i$$

I want to say simply $2^\frac{1}{b-1}$, but that doesn't seem right, does it?


Just a simple question I don't know how to solve, please don't downvote without commenting what's wrong with the question

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    $\begingroup$ This is not abstract-algebra - it is just algebra. $\endgroup$ – marty cohen Aug 20 '17 at 1:24
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Let $c = \frac{1}{b-1}$. Then

$\begin{array}\\ \sum_{j=1}^n y_j &=\sum_{j=1}^n\big(\frac{2x_j}{\sum_{i=1}^n(x_i)}\big)^{c}\\ &=\sum_{j=1}^n\frac{(2x_j)^c}{(\sum_{i=1}^n(x_i))^{c}}\\ &=\frac1{(\sum_{i=1}^n(x_i))^c}\sum_{j=1}^n(2x_j)^c\\ &=\frac{2^c}{(\sum_{i=1}^n(x_i))^c}\sum_{j=1}^n(x_j)^c\\ &=2^c\\ \end{array} $

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  • $\begingroup$ Why does $\sum x_j^c = (\sum x_j)^c$? $\endgroup$ – Olivier Aug 20 '17 at 2:21
  • $\begingroup$ It does not seem to be right. $\endgroup$ – Claude Leibovici Aug 20 '17 at 3:23
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I do not think that there is a solution (even using special functions).

Consider the simple case $x_i=i$, $n=3$ and $b=3$. This would give $$y_i=\frac{\sqrt{i}}{\sqrt{3}}\implies \sum_{i=1}^3 y_i=\frac{1+\sqrt 2+\sqrt 3}{\sqrt{3}}$$ which does not look like ${\sqrt{2}}$.

For a general value of $n$, we should have $$y_i=\frac{\sqrt{2}\sqrt{i}}{\sqrt{n(n+1)}}\implies S_n=\sum_{i=1}^n y_i=\frac{\sqrt{2}}{\sqrt{n(n+1)}}H_n^{\left(-\frac{1}{2}\right)}$$ where appear the generalized harmonic numbers.

Making $n$ infinitely large and using asymptotics, we should have $$S_n=\frac{2 \sqrt{2} }{3}\sqrt{n}+\frac{1}{3 \sqrt{2n}}+O\left(\frac{1}{n}\right)$$

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