Elegance Comes Only Periodically

Calculus Level 5

P ( x ) = x 2 0 ( n = 1 n 2 + t n 2 ) d t P\left( x \right) =\int _{ { -x }^{ 2 } }^{ 0 }{ \left( \prod _{ n=1 }^{ \infty }{ \frac { { n }^{ 2 }+t }{ { n }^{ 2 } } } \right) dt }

When x Z x \in \mathbb{Z} and P ( x ) 0 P(x) \neq 0 , the value of P ( x ) P\left( x \right) can be expressed in the form α π β \frac { \alpha }{ { \pi }^{ \beta } } , where α { \alpha } is a perfect square and β { \beta } is a prime number. Find α β { \alpha }^{ \beta } .


This problem is original. The picture of the graph was produced by Wolfram .


The answer is 16.

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1 solution

Mark Hennings
Jan 22, 2016

The question needs re-expressing. The integral is P ( x ) = x 2 0 n = 1 ( 1 + t n 2 ) d t = 2 0 x u n = 1 ( 1 u 2 n 2 ) d u P(x) \; = \; \int_{-x^2}^0 \prod_{n=1}^\infty \left(1 + \tfrac{t}{n^2}\right)\,dt \; = \; 2 \int_0^x u \prod_{n=1}^\infty \left(1 - \tfrac{u^2}{n^2}\right)\,du and hence P ( x ) = 2 0 x u × sin π u π u d u = 2 π 0 x sin π u d u = 2 π 2 ( 1 cos π x ) . P(x) \; = \; 2\int_0^x u \times \frac{\sin \pi u}{\pi u}\,du \; = \; \tfrac{2}{\pi}\int_0^x \sin\pi u\,du \; = \; \tfrac{2}{\pi^2}(1 - \cos \pi x) \;. When x x is an integer, this integral is either 4 π 2 \frac{4}{\pi^2} or zero.

The way the question is written, it wants P ( x ) P(x) to take the nonzero value for all integers x x , which is not true. The question should either say "When x Z x \in \mathbb{Z} and P ( x ) 0 P(x) \neq 0 ...", or else "When x Z x \in \mathbb{Z} is odd ... " to be accurate, and obtain the intended answer of 4 2 = 16 4^2 = \boxed{16} .

Hello, Mark. I thought saying P ( x ) P(x) besides zero would suffice, but I could see why that would be ambiguous. I just fixed it!

What I have written now is what I really meant to say.

Jonas Katona - 5 years, 4 months ago

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Yeah, it's fixed now. Interesting problem.

Michael Mendrin - 5 years, 4 months ago

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Thank you!!

Jonas Katona - 5 years, 4 months ago

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