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freckles

  • one year ago

I want to evaluate: \[\int\limits_{0}^{\pi} \pi \sin^2(\sin(x)) dx\] ... Wolfram tells me it involves Bessel function.. How can I get wolfram's answer of \[\frac{1}{2} \pi (\pi-\pi J_0(2))\]

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  1. Kainui
    • one year ago
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    Ohhhh can you link it to us?

  2. freckles
    • one year ago
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    http://www.wolframalpha.com/input/?i=integrate%28pi*sin%5E2%28sin%28x%29%29%2Cx%3D0..pi%29

  3. freckles
    • one year ago
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    there is this one lemma that says: \[\sin(x)=2 \sum_{i=0}^\infty (-1)^{n} J_{2n+1}(x)\]

  4. freckles
    • one year ago
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    but replacing x with sin(x) might involve sum inside the sum

  5. Kainui
    • one year ago
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    Hmmm well I haven't really played with these much, I just know they're like orthogonal or something. I feel like I could give you an answer but it wouldn't be anything other than steps... There wouldn't really be a strategy to it, it'd just be pattern matching to the definition of \(J_0\).

  6. freckles
    • one year ago
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    Also right now I know nothing about this function lol it came up in @Jdosio 's cal 1 problem which he was allowed to use a calculator for

  7. freckles
    • one year ago
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    Can I have the steps? Maybe I can try to follow it.

  8. Kainui
    • one year ago
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    Sure, actually I might have found something.

  9. Kainui
    • one year ago
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    Ok I'm going to try a slightly different approach than what I was going to use, I'll start with what I see here: https://en.wikipedia.org/wiki/Bessel_function#Bessel.27s_integrals \[J_n(x)=\frac{1}{\pi} \int_0^\pi \cos(n \tau - x \sin(\tau)) d \tau \] To match our question I plug in \(x=2\) and \(n=0\) and remove the negative sign because cosine is even. \[J_0(2)=\frac{1}{\pi} \int_0^\pi \cos(2 \sin(\tau)) d \tau \] Now we can apply the double angle formula on here: \[ 2 \sin^2(\sin x) = 1 - \cos(2 \sin x)\] I think the rest might be clear from here. If not I can write out the rest, but you can see how this isn't very illuminating. What I do know is that the Bessel functions describe hitting a drum head in the middle and the vibrations on it, which came up in solving the wave equation with the proper boundary conditions which is fun, but other than that I haven't really touched them since.

  10. anonymous
    • one year ago
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    wow, that escalated quickly

  11. Kainui
    • one year ago
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    Yeah other than confirming that this is right, I sure as hell wouldn't have been able to just like come here without wolfram alpha saying what the answer was haha.

  12. freckles
    • one year ago
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    \[\int\limits_0^\pi \pi \sin^2(\sin(x)) dx \\ =\int\limits_0^\pi \frac{\pi}{2}(1-\cos(2\sin(x)) dx \\ =\frac{\pi^2}{2}- \frac{\pi}{2}\int\limits_0^\pi \cos(2\sin(x)) dx \\ =\frac{\pi^2}{2}-\pi^2 \frac{1}{\pi} \int\limits_0^\pi \cos(2\sin(x)) dx \\ =\frac{\pi^2}{2}-\frac{\pi^2}{2} J_0(2)\] ok i get yeah i wish wolfram could tell us how it thought on that one like what made it think to use the bessel function

  13. Kainui
    • one year ago
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    Yeah probably some flow chart it just works through like a database of a thousand different things or something lol.

  14. freckles
    • one year ago
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    \[\int\limits_0^\pi \pi \sin^2(\sin(x)) dx \\ =\int\limits_0^\pi \frac{\pi}{2}(1-\cos(2\sin(x)) dx \\ =\frac{\pi^2}{2}- \frac{\pi}{2}\int\limits_0^\pi \cos(2\sin(x)) dx \\ =\frac{\pi^2}{2}-\frac{\pi^2}{2} \frac{1}{\pi} \int\limits_0^\pi \cos(2\sin(x)) dx \\ =\frac{\pi^2}{2}-\frac{\pi^2}{2} J_0(2)\] and even after I corrected my latex once I still left an error

  15. freckles
    • one year ago
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    if they invented the technology that would instantly give you a brain that could have these flow charts in it would you do it?

  16. freckles
    • one year ago
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    i mean would you let them use their invention on you

  17. thomas5267
    • one year ago
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    Maybe Wolfram saw a integral of trig in trig so it tried to make it fit into Bessel's function?

  18. thomas5267
    • one year ago
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    I believe that flowchart is so complex that even if you could memorise it it would simply be quicker to consult Wolfram Alpha than using the flowchart.

  19. freckles
    • one year ago
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    if it didn't take up too much space it would be nice to have a wolframalpha in my brain

  20. Kainui
    • one year ago
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    Hahaha well honestly I kinda like it. I think we should change what they teach in school and how they teach it because reality has fundamentally been changed by computers and the internet.

  21. Kainui
    • one year ago
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    I know almost nothing, I just am really good at quickly remembering what exists, looking it up, and figuring out how to put them together. I just ask mommy google and papa wikipedia... and uncle wolfram... lol

  22. freckles
    • one year ago
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    err why couldn't wolfram be a girl @Kainui

  23. thomas5267
    • one year ago
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    Consult papa Wikipedia! https://en.wikipedia.org/wiki/Symbolic_integration https://en.wikipedia.org/wiki/Risch_algorithm

  24. Kainui
    • one year ago
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    Because of this! https://en.wikipedia.org/wiki/Uncle_Tungsten

  25. Kainui
    • one year ago
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    Also I realize that says Tungsten but if you look on the periodic table it has a W there, why's that? It gives a cute little thing here: http://education.jlab.org/itselemental/ele074.html Anyways just some bit of trivia I learned about while taking chemistry.

  26. thomas5267
    • one year ago
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    I finally get your joke lol!

  27. Kainui
    • one year ago
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    Yeah I realized I didn't quite explain that very well hahaha.... XD

  28. freckles
    • one year ago
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    Oh so that is why wolfram is an uncle and not an aunt.

  29. Kainui
    • one year ago
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    Yeah, maybe... Auntie YouTube?

  30. Kainui
    • one year ago
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    Honestly mother google makes everyone look bad in comparison, nothing's better than her haha.

  31. freckles
    • one year ago
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    just as long as a girl is in charge

  32. freckles
    • one year ago
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    i'm happy

  33. thomas5267
    • one year ago
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    Actually calling google mommy is a better choice than calling her (?) daddy. Google is literally like a mom, she knows where you are, what you are doing and why you are doing it even without help from the alphabetical agencies. XD

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