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Finding rational approximations for ln2 using integration Watch

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    Problem solved.
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    So, have you managed to find the integral or not?
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    I haven't found an expression for the integral because the nature of 'integration by parts' was leading me into expressions with increasingly larger sections to then re-integrate... So I gave up on that, plus it was also apparent that it wasn't going to give me an answer that related to ln2...

    I have used my computer and my graphical calculator which both tell me the numerical value of the integral is 6.64x10^-5 and I can't see how this relates to either ln2 or the fraction given...
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    Try long division.

    EDIT: This works
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    From what I can tell it's trying to get you manipulate the integral until you're integrating 2x/(1+x^2) with a bunch of stuff tagged on. Doing this between 0 and 1 gets ln(2) and then you can use the fact that the integral is larger than 0 to get an inequality out.

    However, I can't quite see how you're going to get there with the integral, and I'm pretty poor at integration. I feel like the fact that it's symmetrical around 0.5 should be useful, but I don't know why.
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    (Original post by Daniel Freedman)
    Try long division.

    EDIT: This works
    Using long division on what exactly...?
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    (Original post by TimWestwood)
    Using long division on what exactly...?
    For your function y, expand the numerator, perform long (polynomial) division using the denominator, and then integrate the whole lot and you'll get a term in ln(1+x^2), which with your limits will give you the ln2.

    A bit tedious, but can't see a slicker way.
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    I see... Well I'll give it a go and see what happens.

    Thanks for the response.
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    I just picked up some malware while viewing OP's link.
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    I've solved it; thanks for everyone's help.

    And I'll remove the link seeing as I'm finished.
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    (Original post by Get me off the £\?%!^@ computer)
    I just picked up some malware while viewing OP's link.
    Sorry to hear that, maybe it's because the website I used to host the image was free? Maybe they're not trustworthy...
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    (Original post by TimWestwood)
    Sorry to hear that, maybe it's because the website I used to host the image was free? Maybe they're not trustworthy...
    Well, my laptop is now ****ed.

    I suggest you edit your post to advise people not to click on the link.
 
 
 
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