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A while ago one of my professors gave the class a problem "to think about when lying on the beach."

Well, I've been on the beach several times since then to no avail and my curiosity has finally outweighed my desire to solve this personally. The problem is this:

Let $a_1, \dotsc, a_n\in\mathbb{C}$ be distinct. Prove that:

\begin{equation} \sum_{i=1}^n\prod_{j\neq i}\frac{1}{a_i - a_j} = 0 \end{equation}

It's pretty easy, if tedious, to show this for a given $n$ but I'm unsure about how to generalise the result.

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  • $\begingroup$ No, the product is over all $1\leq j\leq n, j\neq i$ $\endgroup$ Commented Jul 23, 2014 at 1:01
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    $\begingroup$ If I had no idea on how to attack this, I'd look at the proof of the lagrange polynomial and try to adapt. $\endgroup$
    – Git Gud
    Commented Jul 23, 2014 at 1:04
  • $\begingroup$ Oh! Thanks, that does look promising $\endgroup$ Commented Jul 23, 2014 at 1:09
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    $\begingroup$ Look at the Lagrange polynomial of $P(a_i) = a_i$ $\endgroup$
    – Aryabhata
    Commented Jul 23, 2014 at 1:17

4 Answers 4

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First, assume $\prod a_i \ne 0$

Now consider the $(n-1)^{th}$ degree polynomial

$$P(z) = \sum_{i=1}^{n} a_i \prod_{j \neq i} \frac{z-a_j}{a_i - a_j }$$

We see that $P(a_i) = a_i$ for each $i$

Thus $P(z) - z $ has at least $n$ roots, and thus must be identically $0$.

$$ z \equiv \sum_{i=1}^{n} a_i \prod_{j \neq i} \frac{z-a_j}{a_i - a_j }$$

Now put $z= 0$ in the above divide by $(-1)^n\prod a_i$ to get your identity.

If $\prod a_i = 0$, then wlog, assume $a_1 = 0$.

Now take a sequence of complex numbers $c_n \to a_1$, $c_n \neq 0$, and use $c_n$ instead of $a_1$ and take limits.

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  • $\begingroup$ This is great, thanks! $\endgroup$ Commented Jul 23, 2014 at 1:47
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    $\begingroup$ A way to avoid the problem if $a_1 = 0$ without taking limits would be to take some $A\neq a_1\dotsc a_n$ and let $P(z) = \sum_{i=1}^n (a_i - A)\prod_{j\neq i} \frac{z-a_j}{a_i - a_j}$ so that $P(z) = z-A$. Then at the end we set $z = A$ and divide through by $(-1)^n\prod (a_i-A)$, which is always nonzero. $\endgroup$ Commented Jul 23, 2014 at 2:03
  • $\begingroup$ @DamianPavlyshyn: Yes, or more simply, replace $a_1$ by $z$, do some algebra to get a polynomial in $z$ which has infinite roots (any $z \ne a_i$), allowing setting $z=0$. $\endgroup$
    – Aryabhata
    Commented Jul 24, 2014 at 0:13
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Let $$f(z)=\frac{1}{(z-a_1)(z-a_2)\cdots(z-a_n)}$$ and consider $$\lim_{R\to\infty}\frac{1}{2\pi i}\int_{|z|=R}f(z)\,dz\ .$$ By residues (or the Cauchy Integral Formula), the limit is your expression. By estimating $|f(z)|$, the limit is zero. The conclusion follows.

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Another kind of answer (note that $n\geq 2$).

Put $\displaystyle R(z)=\frac{1}{\prod_{i=1}^n (z-a_i)}$. Then we have $\displaystyle R(z)=\sum_{j=1}^n \frac{c_j}{z-a_j}$. We immediately get that $\displaystyle c_j=\frac{1}{\prod_{i\not =j}^n (a_j-a_i)} $. We have $\displaystyle zR(z)=\sum_{j=1}^n \frac{zc_j}{z-a_j}\to \sum_{j=1}^n c_j$ as $z\to \infty$, and as $n\geq 2$, $zR(z)\to 0$ as $z\to \infty$ and we are done.

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Something related to Lagrange interpolation:

The polynomial

$$\omega_i(x) = \frac{\prod_{j\ne i} (x - a_j)}{\prod_{j\ne i} (a_i-a_j)}$$

has the properties

$$\omega_i(a_i) = 1 \\ \omega_i(a_j) = 0 \textrm{ for all } j \ne i$$

Therefore, the polynomial $$\sum_{i} \omega_i(x)$$ is of degree at most $n-1$ and takes value $1$ at $n$ points. We conclude that is is the constant polynomial $1$. Now identify the coefficient of $x^{n-1}$.

Note: In a similar way we check that for all $0 \le k \le n-2$ we have

$$\sum_i \frac{a_i^k}{\prod_{j\ne i} (a_i - a_j)} = 0$$

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