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Math Help - Quotient

  1. #1
    Junior Member
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    Dec 2009
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    Quotient

    This shouldn't be to heavy Challenge. Calculate

    a) <br />
\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}<br />
(very easy)

    b) <br />
\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}<br />
(not so easy)
    Last edited by ns1954; January 24th 2010 at 01:17 PM.
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  2. #2
    Super Member
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    Quote Originally Posted by ns1954 View Post
    This shouldn't be to heavy Challenge. Calculate

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

    b) <br />
\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}<br />
(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!}+ ...)<br />

    so the answer to the first problem is  \pi^2
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  3. #3
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    You absolytly right about a). Try b) by using power series. Define apropriate function and use standard power series for some functions...
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  4. #4
    Member
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    Nov 2009
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    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:

    <br />
\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}<br />
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