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Calculus Level 3

$\large \prod_{p \text{ prime}} \dfrac{p^2}{p^2 + 1 }$

If the product above is equal to $\frac{\pi^A}{B}$ for positive integers $A$ and $B$ , find $2A+B$ .

The answer is 19.

1 solution

Aditya Kumar
Jan 11, 2016

First we use the formula: $\zeta \left( s \right) =\prod _{ p= prime }^{ }{ \frac { 1 }{ 1-{ p }^{ -s } } }$

See the proof here

Now,

$\zeta \left( 4 \right) =\prod _{ p=prime }^{ }{ \frac { 1 }{ 1-{ p }^{ -4 } } } \\ \zeta \left( 4 \right) =\prod _{ p=prime }^{ }{ \frac { 1 }{ 1-{ p }^{ -2 } } } \prod _{ p=prime }^{ }{ \frac { 1 }{ 1+{ p }^{ -2 } } } \\ \zeta \left( 4 \right) =\zeta \left( 2 \right) \prod _{ p=prime }^{ }{ \frac { 1 }{ 1+{ p }^{ -2 } } } \\ \therefore \quad \prod _{ p=prime }^{ }{ \frac { 1 }{ 1+{ p }^{ -2 } } } =\frac { \zeta \left( 4 \right) }{ \zeta \left( 2 \right) } =\frac { { \pi }^{ 2 } }{ 15 } \$

Moderator note:

Good recognition of how to use the zeta function.

$\zeta \left( s \right) =\prod _{ p\quad prime }^{ }{ \frac { 1 }{ 1-{ p }^{ -s } } }$ is not a definition. If it is so, why would it have a proof ! It is just a formula by Euler. Do edit the wording. By the way, nice solution :) +1.

Venkata Karthik Bandaru - 5 years, 5 months ago

Thanks edited

Aditya Kumar - 5 years, 5 months ago

Could you pls explain how u obtained zeta(4)/zeta(2) ?

Milind Blaze - 5 years, 2 months ago

This looks familiar?

The answer is 19.

1 solution

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