7
$\begingroup$

I posted this question on Math StackExchange but did not get any answer. I am trying my luck here.

Let $a_{1},a_{2},\dotsc,a_{n+1}$ be a sequence of distinct non-zero real numbers with $$\sum_{j=1}^{n+1}a^2_{j}=1,\qquad\sum_{j=1}^{n+1}a_{j}=0.$$ Show \begin{equation} \tag{1} \label{1} 0<\sum_{k=1}^{n+1}\dfrac{1}{\lvert a_{k}\rvert}\prod_{\substack{j=1 \\ j\neq k}}^{n+1}\dfrac{a_{k}}{a_{k}-a_{j}}\le\sqrt{2}. \end{equation}

I found the equality on the right-hand side when $n=1$. But I can't prove this inequality \eqref{1}. First of all, this inequality is a bit like Lagrange's interpolation formula https://math.stackexchange.com/questions/500139/prove-1-sum-i-0n-frac1x-i-prod-j-neq-i1-frac1x-j-x-i-prod-i. I tried to prove it using Lagrange's interpolation formula but can't.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ What does "Webb" refer to here? $\endgroup$
    – Sam Hopkins
    Commented Jan 14, 2021 at 4:59
  • $\begingroup$ what is the equal condition on the right hand? seems equal only hold when two of $x_i$ is $({\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}})$, other is zeroes. $\endgroup$
    – katago
    Commented Jan 14, 2021 at 5:31
  • $\begingroup$ a quick remark for the upper bound is, as said in the answer of Fedor Petrov we can understand the sum $\sum_{k=1}^{n+1} \frac{1}{\left|a_{k}\right|} \prod_{j=1 \atop j \neq k}^{n+1} \frac{a_{k}}{a_{k}-a_{j}}$ as the coefficient of the degree n term of the interpolation polynomial to $f(x)=x^{n-2}|x|=x^{n} /|x|$ at $n+1$ points, and a dual view point is we can in fact view this as the coefficient $a$ of $ax^n$ which can intersect with linear combination of a polynomial with degree $n-1$ and $f(x)=x^{n-2}|x|=x^{n} /|x|$ at $n+1$ different points, (continued) $\endgroup$
    – katago
    Commented Jan 14, 2021 at 15:49
  • $\begingroup$ , this almost get a finite upper bound(maybe can not get $\sqrt{2}$ but not related to n), if we carefully look at the singularity 0, at this singularity point up to $n-2$th derivative of $x^{n} /|x|$ vanish. $\endgroup$
    – katago
    Commented Jan 14, 2021 at 15:51
  • 1
    $\begingroup$ The result was proved by Simon Webb in his PhD thesis discovery.ucl.ac.uk/id/eprint/10102127/1/out.pdf (If you were aware of that you should have stated it.) $\endgroup$
    – esg
    Commented Jan 20, 2021 at 19:27

1 Answer 1

Reset to default
4
$\begingroup$

Here is the proof that the sum is non-negative.

Denote $f(x)=x^{n-2}|x|=x^n/|x|$. Then $$ A:=\sum_{k=1}^{n+1}\dfrac{1}{|a_{k}|}\prod_{j=1,j\neq k}^{n+1}\dfrac{a_{k}}{a_{k}-a_{j}}= [x^n]\sum_{k=1}^{n+1}f(a_k)\prod_{j\ne k}\frac{x-a_j}{a_k-a_j}=:[x^n]h(x), $$ where the polynomial $h(x)=Ax^n+\ldots$, $\deg h\leqslant n$, interpolates $f$ in points $a_1,\ldots,a_{n+1}$.

The function $h(x)-f(x)$ is $(n-2)$ times continuously differentiable and has $n+1$ roots at $a_i$'s, thus by Rolle theorem its $(n-2)$-st derivative $\frac{n!}2Ax^2+Bx+C-(n-1)!|x|$ has three distinct roots. But if $A<0$, this function is strictly concave and can not have three roots. So $A\geqslant 0$.

Also this sum appears in the theory of splines and in the algebraic combinatorics / integrable probability. Probably everything is done and written somewhere.

Improve this answer
$\endgroup$
6
  • $\begingroup$ It's nice proof it.+1 $\endgroup$
    – math110
    Commented Jan 12, 2021 at 4:43
  • $\begingroup$ I hope someone can help prove this right hand inequality Thanks $\endgroup$
    – math110
    Commented Jan 14, 2021 at 4:21
  • 1
    $\begingroup$ It would follow from $\int_0^\infty \frac{dx}{\sqrt{\prod (1+a_j^2x^2)}}\leqslant \pi/\sqrt{2}=\int_0^\infty dx/(1+x^2/2)$. If we had $\sum a_j^4\leqslant 1/2$, this would hold pointwise, but in general it does not. $\endgroup$
    – Fedor Petrov
    Commented Jan 14, 2021 at 11:11
  • 1
    $\begingroup$ @katago compute the integral of $1/\prod(1+i a_k x)$ against the real line using residues $\endgroup$
    – Fedor Petrov
    Commented Jan 14, 2021 at 13:10
  • 2
    $\begingroup$ Alas, this integral inequality is not true at all: if $a_1^2=\ldots=a_n^2=1/(n^2+n)$, $a_{n+1}^2=n/(n+1)$ (the signs are appropriate), and $n$ is large, the integral tends to infinity. Bounding the real part by teh absolute value was far too brave. $\endgroup$
    – Fedor Petrov
    Commented Jan 14, 2021 at 16:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .