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Anonymous Wed 24 Nov 13:12:18 2021 No.13886261 View Reply Original Report

Quoted By: >>13886310 >>13886933

Can /sci/ solve it?

Anonymous

Anonymous Wed 24 Nov 2021 13:30:06 No.13886300 Report

Quoted By: >>13886773

It's trivial, but I'm not doing your homework retard.

Anonymous

Anonymous Wed 24 Nov 2021 13:33:40 No.13886310 Report

Quoted By: >>13886315 >>13886773 >>13886962

>>13886261
$f^{(n)}(x) = \frac{P_n(x)}{(x^k - 1)^{n+1}}$
Differentiate it with respect to x
$f^{(n+1)}(x) = \frac{(x^k - 1) P'_n(x)}{(x^k - 1)^{n+2}} - \frac{(n+1) k x^{k-1} P_n(x)}{(x^k - 1)^{n+2}}$

So
$P_{n+1} (x) = (x^k - 1) P'_n(x) - (n+1) k x^{k-1} P_n(x)$
Plug in x = 1 to get
$P_{n+1} (1) = - (n+1) k P_n(1)$
With the initial value
$P_0(1) = 1$
we get
$P_n(1) = (-k)^n n!$

Anonymous

Anonymous Wed 24 Nov 2021 13:36:12 No.13886315 Report

Quoted By:

>>13886310
Nice gg

Anonymous

Anonymous Wed 24 Nov 2021 17:44:24 No.13886773 Report

Quoted By: >>13886783

>>13886310
thanks vro
>>13886300
not homework just a problem I was doing for fun retard

Anonymous

Anonymous Wed 24 Nov 2021 17:46:33 No.13886783 Report

Quoted By: >>13886796

>>13886773
Show the full context of where it comes from then. Otherwise I see no reason why jannies shouldn't assume it's homework.

Anonymous

Anonymous Wed 24 Nov 2021 17:51:04 No.13886796 Report

Quoted By:

>>13886783
https://kskedlaya.org/putnam-archive/2005.pdf

Anonymous

Anonymous Wed 24 Nov 2021 18:40:08 No.13886933 Report

Quoted By:

>>13886261
We only need the leading term for $x \approx 1$ .
$\frac{1}{x^k - 1} \approx \frac{1}{k}\frac{1}{x - 1}$
$\left(\frac{\partial}{\partial x}\right)^{n} \frac{1}{k}\frac{1}{x - 1} = \frac{1}{k} \frac{(-1)^n \, n!}{(x -1)^{n+1}} = \frac{(-1)^n k^n n!}{k^{n+1}(x-1)^{n+1}} \approx \frac{P_n(1)}{(x^k-1)^{n+1}}$
So $P_n(1) = (-1)^n k^n n!$ .

Anonymous

Anonymous Wed 24 Nov 2021 18:48:28 No.13886962 Report

Quoted By:

>>13886310
kys

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