It is a major theorem that

Attachment 18320

Thanks!

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- Jul 24th 2010, 08:55 PMsoyeahiknowa number logic probem
It is a major theorem that

Attachment 18320

Thanks! - Jul 25th 2010, 09:28 AMSoroban
Hello, soyeahiknow!

Quote:

It is a major theorem that:

. . $\displaystyle \displaystyle{S \;=\; \sum^{\infty}_{n=1}\frac{1}{n^2} \;=\;\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \hdots \;=\;\frac{\pi^2}{6} }$

Use this to calculate: .$\displaystyle [1]\;\dfrac{1}{2^2} + \dfrac{1}{4^2} + \dfrac{1}{6^2} + \hdots$

We have: .$\displaystyle X \;=\;\dfrac{1}{2^2} + \dfrac{1}{4^2} + \dfrac{1}{6^2} + \hdots $

. . . . . . . .$\displaystyle X \;=\;\dfrac{1}{2^2}\underbrace{\bigg[\dfrac{1}{1^2} + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \hdots \bigg]}_{\text{This is }S} $

. . . . . . . .$\displaystyle X \;=\;\dfrac{1}{4}\cdot\dfrac{\pi^2}{6}$

. . . . . . . .$\displaystyle X \;=\;\dfrac{\pi^2}{24}$

Quote:

$\displaystyle [2]\;\dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{5^2} + \hdots$

Are the denominators primes? .$\displaystyle 2, 3, 5, 7, 11, 13,\:\hdots$

or a Fibonacci-type sequence? .$\displaystyle 2, 3, 5, 8, 13, 21,\:\hdots$

- Jul 25th 2010, 12:21 PMsoyeahiknow
I believe they are primes

- Jul 25th 2010, 12:34 PMsoyeahiknow
can u tell me why you took out the 1/(2^2) ?

- Jul 25th 2010, 01:04 PMtonio
- Jul 25th 2010, 01:15 PMundefined
Prime Zeta Function -- from Wolfram MathWorld

id:A085548 - OEIS Search Results

EDIT: I should have added some comment to be clear; it's likely that tonio is right that the problem is misstated; nevertheless, if it is correctly stated then the above links are relevant. - Jul 25th 2010, 01:20 PMtonio
- Jul 25th 2010, 01:21 PMundefined
- Jul 25th 2010, 01:22 PMsoyeahiknow
wait wait,

i think there may have been a mistake.

The 2nd problem is just simply:

1/(1^2) + 1(3^2) + 1/(5^2)

So it is simply the odd numbers. So would I just do 1/(3^2) * S= 1/9 * (Pi^2)/6 = (Pi^2)/ 54? Thanks - Jul 25th 2010, 01:26 PMundefined
In post #2, Soroban factored out (1/2)^2 because it evenly divides each term, and because doing so allows us to express the sum as a product of already known quantities.

For this problem, think about what happens when you add this series to the series that Soroban solved. - Jul 25th 2010, 01:33 PMsoyeahiknow
if I add this series to the one Soroban solves, it would just be back to the original series right?

OOOOO so all I have to do is ( original series - series with all the even numbers)= series for problem 2 ( the odd numbers). So (pi^2)/6-(pi^2)/24= (pi^2)/8

and what kind of problems are these called? Thanks so much and sorry for the typo in the beginning! I posted it at like 3 am. - Jul 25th 2010, 01:36 PMtonio