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Math Help - an integral from the 2010 MIT Integration Bee

  1. #1
    Super Member Random Variable's Avatar
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    an integral from the 2010 MIT Integration Bee

    The following integral was on the qualifying test:

     \int^{1}_{0} \sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}} \ dx

    Any ideas? I'm stumped.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    The following integral was on the qualifying test:

     \int^{1}_{0} \sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}} \ dx

    Any ideas? I'm stumped.
    Let x=\sec^2(z) or x=\sinh^2(z)?
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  3. #3
    Super Member Random Variable's Avatar
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    How about the following:

    let  y = \sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}}

    then  y^{2} = 1+x\sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}} = 1+xy

    solving for y

     y = \frac{x}{2} \pm \frac{\sqrt{x^{2}+4}}{2}

    then using the positive root since we're integrating over positive values of x

     \int^{1}_{0} \sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}} \ dx \ = \int^{1}_{0} \Big(\frac{x}{2} + \frac{\sqrt{x^{2}+4}}{2} \Big) \ dx ?
    Last edited by Random Variable; January 28th 2010 at 10:14 PM.
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  4. #4
    MHF Contributor Calculus26's Avatar
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    then you solve for y using quadratic formula ?

    y= x/2 + sqrt(x^2 +4)/2 ?

    or am I way off base?
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    What about the following:

    let  y = \sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}}

    then  y^{2} = 1+x\sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}} = 1+xy ?
    Quote Originally Posted by Calculus26 View Post
    then you solve for y using quadratic formula ?

    y= x/2 + sqrt(x^2 +4)/2 ?

    or am I way off base?
    No no! That is correct. I didn't realize it was supposed to be an infinite surd . Yes, that method works...if you can prove that the surd converges .

    P.S. You can choose the \pm according to the fact that y>0 for x\in[0,1]
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  6. #6
    MHF Contributor Calculus26's Avatar
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    Then since y > 0 we take y= x/2 + sqrt(x^2 +4)/2 ?
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Calculus26 View Post
    Then since y > 0 we take y= x/2 + sqrt(x^2 +4)/2 ?
    I would agree with that. the answer is \frac{1+\sqrt{5}}{4}+\text{arccsch}\left(2\right)
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