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    Problem 506

    This is a very nice one indeed, evaluate:

    \displaystyle \int_0^1 \frac{\ln (x+1)}{x^2 + 1} \, \mathrm{d}x
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    (Original post by Zacken)
    It's called Latex, and here's how you use it on TSR.
    I thank you, good sir
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    (Original post by Alexion)
    I thank you, good sir
    You're welcome, have a stab at my previous integral, you'll find it a valiant challenge.
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    (Original post by Zacken)
    You're welcome, have a stab at my previous integral, you'll find it a valiant challenge.
    Haha, I'll pass it's late, and I'm not the best at these sorta challenges anyway :lol:
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    (Original post by Alexion)
    Haha, I'll pass it's late, and I'm not the best at these sorta challenges anyway :lol:
    Fair enough. :rofl:
    I'm heading to bed myself!

    Anyways, what's your story, A-Levels, degree maths, prospective mathmo?
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    (Original post by Zacken)
    Fair enough. :rofl:
    I'm heading to bed myself!

    Anyways, what's your story, A-Levels, degree maths, prospective mathmo?
    A-level Further Mathematician prospective Electronic Engineer actually...
    Doesn't stop me from being that nerd who aced C3 though
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    (Original post by Alexion)
    A-level Further Mathematician prospective Electronic Engineer actually...
    Doesn't stop me from being that nerd who aced C3 though
    Filthy engineer.

    Kidding, kidding. Y12 or 13?
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    (Original post by Zacken)
    Filthy engineer.

    Kidding, kidding. Y12 or 13?
    Y13... only 31/2 people in our FM class, it's great

    All the easier to brag about my higher test and exam scores :lol:
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    (Original post by Alexion)
    Y13... only 31/2 people in our FM class, it's great
    1/2? Do I want to ask? xD
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    (Original post by Zacken)
    1/2? Do I want to ask? xD
    Yeah, we sliced him up and threw the other half out the window :mwuaha:

    Nah, it's just that he dropped the subject at A2, so he's only in half of our lessons now (hence the .5)
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    (Original post by Alexion)
    Yeah, we sliced him up and threw the other half out the window :mwuaha:

    Nah, it's just that he dropped the subject at A2, so he's only in half of our lessons now (hence the .5)
    What's the M/F ratio in that class? :rofl:
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    (Original post by Zacken)
    What't the M/F ratio in that class? :rofl:
    #DIV/0!

    All guys
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    (Original post by Zacken)
    Problem 506

    This is a very nice one indeed, evaluate:

    \displaystyle \int_0^1 \frac{\ln (x+1)}{x^2 + 1} \, \mathrm{d}x
    Set \displaystyle x = \frac{1+t}{1-t} then \begin{aligned} I = \int_0^1 \frac{\ln (x+1)}{x^2 + 1} \, \mathrm{d}x = \int_0^1 \frac{\ln (\frac{2}{1+t})}{t^2 + 1} \, \mathrm{d}t = \int_0^1 \frac{\ln (2)}{t^2 + 1} \, \mathrm{d}t - \int_0^1 \frac{\ln (1+t)}{t^2 + 1} \, \mathrm{d}t.\end{aligned}

    So \displaystyle 2I = \frac{\pi}{4} \log(2)} therefore \displaystyle I = \frac{\pi}{8}\log(2).
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    (Original post by Kummer)
    Set \displaystyle x = \frac{1+t}{1-t} then \begin{aligned} I = \int_0^1 \frac{\ln (x+1)}{x^2 + 1} \, \mathrm{d}x = \int_0^1 \frac{\ln (\frac{2}{1+t})}{t^2 + 1} \, \mathrm{d}t = \int_0^1 \frac{\ln (2)}{t^2 + 1} \, \mathrm{d}t - \int_0^1 \frac{\ln (1+t)}{t^2 + 1} \, \mathrm{d}t.\end{aligned}

    So \displaystyle 2I = \frac{\pi}{4} \log(2)} therefore \displaystyle I = \frac{\pi}{8}\log(2).
    That's certainly one way to do it, bravo!
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    Problem 507

    Does the series

      (\ln x) + \frac{(\ln x)^2}{2!} + \frac{(\ln x)^3}{3!} +... + \frac{(\ln x)^n}{n!}

    Converge as  n \longrightarrow + \infty? If so, to what value.
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    Problem 508

     M is an N x N matrix with N distinct eigenvalues all in the range  - 1 < \lambda < 1 . Prove that:

     I + \displaystyle \sum_{r = 1}^{\infty} M^r

    Converges. {NB: I denotes the N x N identity matrix}
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    Problem 509

    Let  a,b,c be the sides of a triangle. Prove that
     \dfrac{a}{b+c}+\dfrac{b}{c+a}+ \dfrac{c}{a+b}+\dfrac{3abc}{(a+b  )(b+c)(c+a)}<2
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    (Original post by 16Characters....)
    Problem 507
    Solution 507

    You like your convergence questions, don't you?

    By the ratio test, we have

    \displaystyle \frac{(\ln x)^{n+1}}{(n+1)!} \cdot \frac{n!}{(\ln x)^n} = \lim_{n \to \infty} \frac{\ln x}{n+1} = 0 < 1.

    So our series converges. To determine what is converges to is slightly easier to see if we define y = \ln x so that our series becomes

    \displaystyle y + \frac{y^2}{2!} + \frac{y^3}{3!} + \cdots + \frac{y^n}{n!} + \cdots = \sum_{n=1}^{\infty} \frac{y^n}{n!}

    This looks suspiciously like a certain well-known McLaurin series, so lo and behold:

    \displaystyle  \sum_{n=1}^{\infty} \frac{y^n}{n!} = e^{y} - e^0 = x - 1.

    This was fun!
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    (Original post by Zacken)
    Solution 507

    You like your convergence questions, don't you?
    This was fun!
    Haha maybe... :-)
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    (Original post by 16Characters....)
    Haha maybe... :-)
    I don't know much about eigenvalues to attempt your other question, sadly!
 
 
 
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