1
$\begingroup$

Prove the identity

$$1 + \sum_{i=0}^n \left(\frac1{x_i}\prod_{j\neq i} \left(1+\frac1{x_j-x_i} \right) \right)=\prod_{i=0}^n \left(1+\frac1{x_i} \right)$$

and hence deduce the inequality in Problem 2:

http://www.imc-math.org.uk/imc2010/imc2010-day2-solutions.pdf

Of course here we need all the $x_i$s distinct. (By the way, this somehow resembles the Lagrange Interpolation formula.)

$\endgroup$
  • 2
    $\begingroup$ The question would be clearer if you wrote the inequality you mention. $\endgroup$ – Seirios Sep 21 '13 at 7:36
  • $\begingroup$ Make several variable transforms, and proceed by induction. $\endgroup$ – Ethan Sep 21 '13 at 7:59
2
$\begingroup$

Method 1

Let $f(z) = \prod_i (z - x_i)$, we have

$$\prod_{j\ne i}(1 + \frac{1}{x_j-x_i}) = \prod_{j\ne i}\frac{(x_i - 1) - x_j}{x_i - x_j} = - \frac{f(x_i-1)}{f'(x_i)}$$

We can rewrite LHS of the equality as

$$1 - \sum_i \frac{f(x_i-1)}{x_if'(x_i)} = 1 - \sum_i \frac{1}{2\pi i}\int_{C_i} \frac{f(z-1)}{zf'(x_i)(z-x_i)} dz = 1 - \frac{1}{2\pi i}\int_{\sum_i C_i} \frac{f(z-1)}{zf(z)} dz $$ where $C_i$ is a bunch of small circular contours surrounding $x_i$ counterclockwisely. By deforming the contours, it is easy to see above contour integral is equal to the difference of two contour integral, one with a big circle $C_{\infty}$ near infinity and another small circle $C_o$ at origin:

$$\frac{1}{2\pi i}\int_{\sum_i C_i} \frac{f(z-1)}{zf(z)} dz = \frac{1}{2\pi i} \left( \int_{C_\infty} - \int_{C_o} \right) \frac{f(z-1)}{zf(z)} dz$$

Since $\displaystyle\;\;\frac{f(z-1)}{z f(z)} = \frac{1}{z} + O(\frac{1}{z^2})\;\;$ for large $z$, we have

$$\frac{1}{2\pi i} \int_{C_\infty} \frac{f(z-1)}{zf(z)} dz = \frac{1}{2\pi i} \int_{C_\infty} ( \frac{1}{z} + O(\frac{1}{z^2}) ) dz = 1$$

This cancels out the $1$ in LHS of inequality and we get

$$\text{LHS} = \frac{1}{2\pi i}\int_{C_o}\frac{f(z-1)}{zf(z)}dz = \frac{f(-1)}{f(0)} =\frac{\prod_i(-1 - x_i)}{\prod_i ( -x_i )} = \prod_i (1+\frac{1}{x_i}) =\text{RHS} $$

Method 2

If one don't want to use complex analysis, an alternate way is apply Lagrange interpolation formula to polynomial $f(z-1) - f(z)$ whose degree is one less than the number of $x_i$, we have:

$$f(z-1) - f(z) = \sum_i \frac{f(z)(f(x_i-1) - f(x_i))}{(z-x_i)f'(x_i)} = f(z) \sum_{i}\frac{f(x_i-1)}{(z-x_i)f'(x_i)} $$ This implies $$1 + \sum_i \frac{f(x_i-1)}{(z-x_i)f'(x_i)} = \frac{f(z-1)}{f(z)}$$ Set $z = 0$, we obtain $$\text{LHS} = 1 - \sum_i \frac{f(x_i-1)}{x_if'(x_i)} = \frac{f(-1)}{f(0)} = \text{RHS}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.