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    (Original post by Bhaal85)
    **** that. Looks hard, either that or you mean secx (which I doubt), is it a composite thing? Or is it a trig identity that I don't know?
    sech = 1/cosh, a hyperbolic function done in P4-6.
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    (Original post by Jonny W)
    sech = 1/cosh, a hyperbolic function done in P4-6.
    I thought so, though I've only gone up to P3.
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    (Original post by Jonny W)
    (Step 3ish.) Integrate sech(x), sech^2(x) and sech^3(x).
    Interesting. I can integrate sec x by partial fractions, but because of the subtle change in the trig identity I don't think the same way will work. I'll have a think about it. Cheers.
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    (Original post by rahaydenuk)
    Integrate wrt. t:
    sin( sin( t ) ) / cos( sin( t ) )
    Nasty. Even Mathematica can't do it.
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    (Original post by Jonny W)
    Nasty. Even Mathematica can't do it.
    Yes that would be because I typed out the wrong thing *looks embarassed*.
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    OK,

    Given the Gamma function defined as:

    Gamma(x) = Integral(0 to +infin.) { t^(x-1) * e^(-t) dt }

    Show (for real x) that Gamma(x) satisfies the functional equation:

    Gamma(x) = (x - 1)Gamma(x - 1)

    Hence show that for any integer n >= 1, Gamma(n) reduces to (n - 1)!.

    I'm proud of that question, I could write exam papers!
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    this is prettttty solid. The thing you're trying to integrate has quite a nice derivative, but my attempts to integrate as 1 x f(t) are getting a bit silly. Tried a few other methods but none of them are obviously leading towards a neat integral. I'll have a think about the sech x integration again.
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    (Original post by mik1a)
    Lol, oh a third year mathematic methods at oxford prelim. Not hard at all.

    differentiate x^x
    funnily enough, that was in the paper too.

    here: http://users.ox.ac.uk/~chri1957/P03_cp3j.pdf
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    Is the answer 1 + lnx?
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    (Original post by imasillynarb)
    Is the answer 1 + lnx?
    to differentiating x^x? no, not quite.
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    (Original post by elpaw)
    to differentiating x^x? no, not quite.
    oh wait, ill get it

    y = x^x
    lny = xlnx
    1/ydy/dx = 1+ lnx

    dy/dx = y(1+lnx)

    y = x^x

    dy/dx = x^x(1+lnx) ??
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    (Original post by fishpaste)
    I got the following answer, I will eat my underpants live on webcam if it's right because it's very silly looking.
    So, when do we get to see this?
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    (Original post by imasillynarb)
    Is the answer 1 + lnx?
    Think about what x^x means for irrational reals or complex numbers. Given its definition (hint: in terms of the exponential function), differentiating it is easy. In fact, you can't really expect to do anything with it if you don't know what it means, i.e. its definition.
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    (Original post by rahaydenuk)
    Think about what x^x means for irrational reals or complex numbers. Given its definition (hint: in terms of the exponential function), differentiating it is easy. In fact, you can't really expect to do anything with it if you don't know what it means, i.e. its definition.
    irrational reals...complex numbers...mean nothing to me
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    (Original post by imasillynarb)
    dy/dx = x^x(1+lnx) ??
    yep, correct.
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    (Original post by imasillynarb)
    irrational reals...complex numbers...mean nothing to me
    Not surprising, especially when you think my girlfriend looks 30 :eek:
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    (Original post by elpaw)
    yep, correct.
    Woo, hoo. First time I forgot the y Schoolboy error.
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    (Original post by Jonny W)
    (Step 3ish.) Integrate sech(x), sech^2(x) and sech^3(x).
    can you do this one by invoking the exponential defenitions?
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    (Original post by Joey_Johns)
    Not surprising, especially when you think my girlfriend looks 30 :eek:
    So did everyone else who saw the picture
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    (Original post by imasillynarb)
    irrational reals...complex numbers...mean nothing to me
    OK... if I asked you what 2^2 was, you'd answer 4 yes?

    What if I asked you what 2^(pi) was to 3 decimal places? Without using a calculator. Do you know what I mean when I say "2 to the power of pi"? Is this 2 times 2 pi times? What does that mean?

    In fact, x^a is defined to be e^(a * ln(x)), and this definition helpfully reduces to the 'accepted' results for x^(p/q) where p and q are integers. e^(pi * ln(2)) is what you mean when you say 2^(pi), and this could be expanded using the series expansion for e^(x) (its definition) and ln(x) without using a calculator.
 
 
 
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