Quotient

**ns1954** · January 24th 2010, 12:16 PM

This shouldn't be to heavy Challenge. Calculate

a) $ \frac{1+\frac{{\pi}^4}{5!}+\frac{{\pi}^8}{9!}+\fra c{{\pi}^{12}}{13!}+\cdots}{\frac 1{3!}+\frac{{\pi}^4}{7!}+\frac{{\pi}^8}{11!}+\frac {{\pi}^{12}}{15!}+\cdots} $ (very easy)

b) $ \frac{1+\frac{{\pi}^4}{2^4\cdot 4!}+\frac{{\pi}^8}{2^8\cdot 8!}+\frac{{\pi}^{12}}{2^{12}\cdot 12!}+\cdots}{\frac 1{2!}+\frac{{\pi}^4}{2^4\cdot 6!}+\frac{{\pi}^8}{2^8\cdot 10!}+\frac{{\pi}^{12}}{2^{12}\cdot 14!}+\cdots} $ (not so easy)

**simplependulum** · January 25th 2010, 02:29 AM

Originally Posted by ns1954

This shouldn't be to heavy Challenge. Calculate

a) $ \frac{1+\frac{{\pi}^4}{5!}+\frac{{\pi}^8}{9!}+\fra c{{\pi}^{12}}{13!}+\cdots}{\cdots} $ (very easy)

b) $ \frac{1+\frac{{\pi}^4}{2^4\cdot 4!}+\frac{{\pi}^8}{2^8\cdot 8!}+\frac{{\pi}^{12}}{2^{12}\cdot 12!}+\cdots}{\frac 1{2!}+\frac{{\pi}^4}{2^4\cdot 6!}+\frac{{\pi}^8}{2^8\cdot 10!}+\frac{{\pi}^{12}}{2^{12}\cdot 14!}+\cdots} $ (not so easy)

$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ...$

Sub $x = \pi$ we obtain

$\sin(\pi ) = \pi - \frac{\pi ^3}{3!} + \frac{\pi ^5}{5!} - \frac{\pi ^7}{7!} + ... = 0$

rearrange the series ,

$\pi ( 1+\frac{{\pi}^4}{5!}+\frac{{\pi}^8}{9!}+\frac{{\pi }^{12}}{13!}+ ... ) = \pi^3 (\frac 1{3!}+\frac{{\pi}^4}{7!}+\frac{{\pi}^8}{11!}+\frac {{\pi}^{12}}{15!}+ ...) $

so the answer to the first problem is $\pi^2$

**ns1954** · January 25th 2010, 02:39 AM

You absolytly right about a). Try b) by using power series. Define apropriate function and use standard power series for some functions...

**Unbeatable0** · January 25th 2010, 08:08 AM

The second one is exactly like the first one, but using the series expansion of $\cos{x}$ instead of $\sin{x}$ :

We have $\cos{x} = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} =\sum_{n=0}^\infty \frac{x^{4n}}{(4n)!} - \sum_{n=0}^\infty \frac{x^{4n+2}}{(4n+2)!}$

Put $x=\frac{\pi}{2}$ to get $\sum_{n=0}^\infty \frac{\left(\frac{\pi}{2}\right)^{4n}}{(4n)!} - \sum_{n=0}^\infty \frac{\left(\frac{\pi}{2}\right)^{4n+2}}{(4n+2)!} = 0$ or equivalently $\sum_{n=0}^\infty \frac{\pi^{4n}}{2^{4n}(4n)!} = \frac{\pi^2}{4} \sum_{n=0}^\infty \frac{\pi^{4n}}{2^{4n}(4n+2)!}$ .
Therefore:

$ \frac{\pi^2}{4} = \frac{\sum_{n=0}^\infty \frac{\pi^{4n}}{2^{4n}(4n)!}}{\sum_{n=0}^\infty \frac{\pi^{4n}}{2^{4n}(4n+2)!}} = \frac{1+\frac{{\pi}^4}{2^4\cdot 4!}+\frac{{\pi}^8}{2^8\cdot 8!}+\frac{{\pi}^{12}}{2^{12}\cdot 12!}+\cdots}{\frac 1{2!}+\frac{{\pi}^4}{2^4\cdot 6!}+\frac{{\pi}^8}{2^8\cdot 10!}+\frac{{\pi}^{12}}{2^{12}\cdot 14!}+\cdots} $

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