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Let $z_j$ ($j=1,\dots, k$) be $k$ points on the complex plane none of which lies on the real line. Is it always true that the function $$ F(x)=\sum_{j=1}^k \frac{1}{|x-z_j|^2} $$ has at most $k$ local maxima on the real line?

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2 Answers

$F(x)=c$ has for any value of $c>0$ at most $2k$ solutions, which in a weak way supports the conjecture that the statement is always true.

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I must be blind, but why is it that $F(x)=c$ has at most $2k-1$ solutions ? –  Ewan Delanoy Nov 6 '12 at 9:33
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@EwanDelanoy: $F(x)=c$ is the quotient of two polynomials of degree $2k$. So my original $2k-1$ was one off. I corrected the mistake. –  Arnold Neumaier Nov 6 '12 at 11:52
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If $z_{j}=a_{j}+i\cdot b_{j}$ then $\left | x-z_{j} \right |^{2}=(x-a_{j})^{2}+b_{j}^{2}$ and $$F(x)=\sum_{j=1}^{k}\frac{1}{(x-a_{j})^{2}+b_{j}^{2}}$$ Function $F(x)$ is continuous and differentiable. $${F}'(x)=\sum_{j=1}^{k}\frac{-2(x-a_{j})}{[(x-a_{j})^{2}+b_{j}^{2}]^{2}}=-2\frac{P(x)}{\prod_{l=1}^{k} [(x-a_{l})^{2}+b_{l}^{2}]^{2}}$$ Where $$P(x)=\sum_{j=1}^{k}(x-a_{j})\prod_{l=1, l\neq j}^{k}[(x-a_{l})^{2}+b_{l}^{2}]^{2}$$ and $grad P(x)=4k-3$ which means there could be maximum $4k-3$ zeros of $P(x)$ or $F(x)$ has $4k-3$ extrema.

On top of this, if we assume the following ordering $$a_{1} \leqslant a_{2} \leqslant ... \leqslant a_{k}$$ $$x-a_{1} \geqslant x-a_{2} \geqslant ... \geqslant x-a_{k}$$ and $x = a_{1}$ $$0 \geqslant a_{1}-a_{2} \geqslant ... \geqslant a_{1}-a_{k}$$ if $x = a_{k}$ $$a_{k}-a_{1} \geqslant a_{k}-a_{2} \geqslant ... \geqslant a_{k}-a_{k-1} \geqslant 0$$ or basically ${F}'(a_{1}) \geqslant 0$ and ${F}'(a_{k}) \leqslant 0$ (*)

Further, ${F}'(x)$ is positive on $(-\infty ,a_{1}]$, looking at $P(x)$ and considering $x \leqslant a_{1} \leqslant a_{2} \leqslant ... \leqslant a_{k}$, thus ${F}(x)$ is ascending on this interval. ${F}'(x)$ is negative on $[a_{k},\infty )$, thus ${F}(x)$ is descending on this interval.

As a result, all the "fun" happens inside $(a_{1},a_{k})$ and according to (*) there is at least one zero for ${F}'(x)$ on this interval (or at least one extrema for $F(x)$).

The remaining part is to sort out $grad P(x)=4k-3$. Assuming the "oscillating" nature (max, min, max, min ...) within $(a_{1},a_{k})$ for $F(x)$ and ascending/descending nature outside that interval, the number of maxima should be greater than the number of minima by 1. Or $n+1$ - maxima, $n$ - minima and $2n + 1 = 4k - 3$ or $2k - 1$ maxima so far ...

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I'm not sure how to reconcile this with the intuitive (true?) fact that when $b_j$ are very small, we should have $k$ maxima.... –  leonbloy Jun 21 '12 at 14:16
    
That last argument assumes that all the maxima and minima are on the real line, which obviously can not be true. Everything up to the last paragraph here is correct, but it only gives a bound of $2k-1$, not $k$. –  Lukas Geyer Oct 18 '12 at 21:44
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In the special case where $k=2$, we can rescale without loss of generality to take $a_1=0, a_2=1$. In that case rtybase's polynomial $P(x)$ expands out as $$P(x)=2x^5-5x^4+(2b_1^2+2b_2^2+6)x^3+...$$ In this case, it's possible to show directly that there's at most $3$ roots (e.g. by Rolle's theorem any interval containing $4$ roots would also contain a root of $$P'''(x)=120x^2-120x+(12b_1^2+12b_2^2+36),$$ but that polynomial has negative discriminant). I don't see any obvious way of extending this to $k>2$ though. –  Kevin Costello Oct 18 '12 at 23:42
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For a specific case of $k=2$ where we can do things without Rolles: $F(x)=1/((x-1)^2+1)+1/((x+1)^2+1)$. Here the numerator is fifth degree and has a factor of $x$, with the other zeros of the numerator being the two real roots $+-\sqrt{-2+2\sqrt{2}}$ and the pair of complex roots $+-\sqrt{-2-2\sqrt{2}}$, if I copied the maple output right. So we see that two of the roots of rtybase's numerator term are complex, in this case. –  coffeemath Oct 21 '12 at 22:23
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