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Let $x_0,x_1,x_2, ...,x_{99}$ be 100 distinct real numbers. Show that $$\sum_{j=0}^{99}x_j^{99}\prod_{0 \le k \le 99}^{k \neq j} \frac{x-x_k}{x_j-x_k}=x^{99}$$

I found that the left side of the equation has the form of Lagrange Interpolation with points $(x_i,x_i^{99})(i=0,1,2,...,99)$, and I proved the equation above by the discrete points $x_i$. But I think it is not rigorous proof. So how can this equation be proved by a more rigorous way?

Gerry Myerson
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Bowen
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    You're right, and almost done. If you wirte the LHS as $P(X)$, you have indeed an interpolating Lagrange polynomial, so $P(x_i)=x_i^{99}$ for $i \in {0,\ldots,99}$. The RHS is also a polynomial $Q(X)=X^{99}$, such that $Q(x_i)=x_i^{99}$ too. Now observe that the degree of $P$ and $Q$ is $99$. If $A$ and $B$ are two polynomials of degree $n$ take the same value at least at $n+1$ distinct points, then they are equals, because $A-B$ is a polynomial of degree at most $n$ with $n+1$ distinct roots, so it can only be $0$. So here, $P=Q$ – yago May 06 '14 at 11:16
  • This is the Lagrange interpolation formula. – Marko Riedel May 06 '14 at 21:58

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