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anonymous

  • 5 years ago

Prove that x-x^3/(3*3!)+x^5/(5*5!)-x^7/(7*7!)... converges to pi/2

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  1. watchmath
    • 5 years ago
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    This can't be true. Try x=0 :).

  2. anonymous
    • 5 years ago
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    haha

  3. anonymous
    • 5 years ago
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    hey are u still online?

  4. anonymous
    • 5 years ago
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    i need help with an answer you have given me

  5. anonymous
    • 5 years ago
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    which one?

  6. anonymous
    • 5 years ago
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    Watchmath, x = infinity :D

  7. anonymous
    • 5 years ago
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    Where do you guys come up with this stuff? Your question doesn't make sense as stated. Are you looking for a value of \(x\) so that this series will converge to \(\pi/2\)? The series that you've written is a little tricky to deal with. Start with the series for \(\sin(x)\). \[ \sin(x) = x - \frac{x^3}{3!}+\frac{x^5}{5!}-\cdots \] The series you have comes from dividing by \(x\) and integrating, \[ \int \frac{\sin (x)}{x} dx = x-\frac{x^3}{3\cdot 3!}+\frac{x^5}{5\cdot 5!}-\frac{x^7}{7\cdot 7!}+\cdots \] but the thing on the left is a "sine integral" and is not an elementary function. You can read a little more about it here: http://en.wikipedia.org/wiki/Trigonometric_integral#Sine_integral Can you clarify your question?

  8. watchmath
    • 5 years ago
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    Hi commutant. He was working on my question when he asked that: http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4dd0a7769fe58b0b4ec339f7

  9. anonymous
    • 5 years ago
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    I see! That's where the \(\pi/2\) business comes from, the limit as \(x\rightarrow \infty \). By the way, I think you can compute that integral (the one you asked about) by contour integration in the complex plane, but you may have to use the "fractional" residue theorem. Is that the way you did it? We seriously need a sub-forum for more advanced math; it's selfish of me but I don't like sifting through all the "do my algebra homework" questions to find more interesting things. Perhaps we should suggest this to the site developers?

  10. watchmath
    • 5 years ago
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    I solve the problem by differentiating under the integral tric :). You were right commutant. But I think it won't be of the interest of the developer. Actually we can discuss here in my site if you want a more advance discussion. Right now the site only has few users. http://www.ask.watchmath.com Let me know what do you think about it.

  11. anonymous
    • 5 years ago
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    Okay, I'll have to try that (diff under the integral). I'll take a look at your site.

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