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I am trying to figure out how to solve the following expression for $x$ and I'm surprised that I don't know what to do.

$$\frac{2n}{x} = \sum_{i=1}^{n} \frac{1}{x-y_{i}}$$

We have that $n$ and $x$ are constants, any ideas?

I've noticed that this expression is equivalent to solving

$$\frac{2n}{x} \prod_{i=1}^{n} (x-y_{i}) = \sum_{i=1}^{n}\left(\prod_{j=1, i\neq j}^{n} (x - y_{j})\right)$$

which looks even harder.

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  • $\begingroup$ Unless there is a typo and/or some data is missing (what are the $\,y_i'$s , for example?), I think this is not an easy exercise. In fact, it looks awfully messy and boring, though not that hard. But definitely not trivial at all: $\,x\,$ appears as summand of denominators of a sum...yucks! $\endgroup$ – DonAntonio Mar 18 '13 at 5:15
  • $\begingroup$ @DonAntonio: $\{y_{1},...,y{n}\}$ is an arbitrary set of non-negative integers with $x \neq y_{i}, \forall i$. It seems really hard to me, any ideas about how it could be not that hard? $\endgroup$ – Samuel Reid Mar 18 '13 at 5:18
  • $\begingroup$ I really is very hard, not only messy, even for low positive integral values of $\,n\,$... $\endgroup$ – DonAntonio Mar 18 '13 at 5:26
  • $\begingroup$ Not only hard: it looks hopeless...are you sure this is the exact identity you need to solve for $\,x\,$ ? Where does it come from? $\endgroup$ – DonAntonio Mar 18 '13 at 5:28
  • $\begingroup$ By clearing denominators you get a polynomial equation of degree $n$ on the variable $x$. If $n\geq5$, well, you know, life becomes hard... $\endgroup$ – Matemáticos Chibchas Mar 18 '13 at 5:30

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