Show that $k^a=\sum_{m=1}^b\left ( c_m^a\prod_{n\neq m} \frac{k-c_n}{c_m-c_n} \right ).$

I used the following result in another post without providing proof (because I couldn't prove it):

$$k^a=\sum_{m=1}^b\left ( c_m^a\prod_{n\neq m} \frac{k-c_n}{c_m-c_n} \right ),$$

where $a$ and $b$ are non-negative integers, $b>a$, each $c_j\in\mathbb{C}$, each $|c_j|>0$, and each $c_j$ is unique. The product runs from $n=1$ to $b$ but skips $m$.

How can we prove this?

• The "each $|c_j|>0$" condition is clearly spurious. – darij grinberg Oct 8 '14 at 0:33

Given $\{c_n\}_{n=1}^b$ (where $c_n \not= c_m$ for $n\not = m$) we define

$$h(z) = \prod_{n=1}^b (z-c_n)$$ Then

$$h'(c_m) = \prod_{n\not = m}^b (c_m-c_n)$$

Using this the right hand side minus the left hand side of your equality can be written

$$g(z)\equiv \sum_{m=1}^bc_m^a\frac{h(z)}{h'(c_m)(z-c_m)} - z^a$$

which is a polynomial of degree $b-1$ satisfying (since $\lim_{z\to c_m} \frac{h(z)}{z-c_m} = h'(c_m)$)

$$g(c_n) = 0,~~~~n=1,2,\ldots,b$$

thus $g(z) \equiv 0$ as no non-zero polynomial of degree $b-1$ can have more than $b-1$ roots. Note that we do not need the restriction $|c_i|>0$.

For the case $b=a+1$ there is an even simpler proof. See this related question.

• An elegant answer. – SDiv Oct 8 '14 at 8:34

Start by encoding the sum call it $S_b$ using residues. We have by inspection that $$S_b = \sum_{m=1}^b \mathrm{Res} \left(f(z); z=c_m\right)$$ where $$f(z) = \frac{z^a}{k-z} \prod_{n=1}^b \frac{k-c_n}{z-c_n}$$ and $c_m\ne k$ and $b>a.$

We can therefore collect $S_b$ by integrating f(z) around a contour that encloses the $b+1$ poles. We will then use the residue at infinity to evaluate the sum of the residues inside the contour.

Now the residue at infinity of $f(z)$ is given by $$\mathrm{Res} \left(-\frac{1}{z^2} f\left(\frac{1}{z}\right); z=0\right).$$ The functional term becomes $$-\frac{1}{z^2} \frac{1}{z^a}\frac{1}{k-1/z} \prod_{n=1}^b \frac{k-c_n}{1/z-c_n} = -\frac{1}{z^2} \frac{1}{z^a}\frac{z}{zk-1} \prod_{n=1}^b \frac{z(k-c_n)}{1-zc_n} \\ = -\frac{1}{z^2} \frac{z^{b+1}}{z^a}\frac{1}{zk-1} \prod_{n=1}^b \frac{k-c_n}{1-zc_n} = \frac{1}{z^{a-b+1}}\frac{1}{1-zk} \prod_{n=1}^b \frac{k-c_n}{1-zc_n}.$$

But we have $b>a$ and hence $b-a-1 > -1$ or $b-a-1\ge 0$ so the term is in fact $$z^{b-a-1} \frac{1}{1-zk} \prod_{n=1}^b \frac{k-c_n}{1-zc_n}.$$ and the residue at zero of the substituted function is zero.

This means that $$S_b = -\mathrm{Res} \left(f(z); z=k\right)$$ which gives $$k^a \prod_{n=1}^b \frac{k-c_n}{k-c_n} = k^a,$$ done.

Addendum I. As an alternative to using the residue at infinity we could have used a circular contour enclosing all $b+1$ poles and observed that the integral along this contour goes to zero since on this circle we have

$$f(z) \in \Theta\left(R^{a-1} \times R^{-b}\right) = \Theta\left(R^{a-b-1}\right)$$ and $$\lim_{R\to\infty} 2\pi R \times R^{a-b-1} = 2\pi \lim_{R\to\infty} R^{a-b} = 0$$ because $b>a.$ This has the advantage of working for $a$ a positive real parameter as opposed to a positive integer only.

Addendum II. The case $k=c_m$ for some $m$ gives $$\sum_{m=1}^b\left(c_m^a\prod_{n\neq m} \frac{k-c_n}{c_m-c_n} \right) = k^a$$ because all the products vanish except the one for $c_m = k.$

A similar calculation can be found at this MSE link.

• Thanks for the solution to a more general problem. It seems from your Addendum I that the expression in the question should hold for non-integer $a$. I played with some numbers and it only seems to hold for integers. Perhaps I have interpreted your answer incorrectly? – SDiv Oct 8 '14 at 8:38
• Thanks for this observation. I also checked non-integer values but I get a match of the original formula and my residue formula for rational values, which seems intuitive by continuity. But the matching values are not equal to $k^a$ and this is because my proof does not go through in this case since $z^a = \exp(a\log(z))$ for non-integer $a$ introduces the branch cut of the logarithm, which crosses the contour and invalidates the residue calculation. – Marko Riedel Oct 8 '14 at 11:26