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Somebody slap me

  1. May 9, 2005 #1

    Hurkyl

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    Nothing enlightening in this post, or even informative. Just the overwhelming need to complain about my own stupidity. :frown:

    I've been staring at this integral all day:

    [tex]
    \int_{-\infty}^{\infty} f(x)^{n-1} f'(x) dx
    [/tex]

    *sigh*

    It doesn't dawn on me that this is a simple u-substitution until the drive home after a nice bridge session. Admittedly, f wasn't something with which I was eminently familiar -- it was [itex]1 + \mathop{erf} (x / \sqrt{2})[/itex], but still, there's no excuse!


    But it gets worse. While I was working out how to set up the integral, I had already deduced the key step that would allow me to set up an easily solved recursive formula for my integral. I enjoy solving things with recursion. But did this dawn on me? Noooo... not until my drive home tonight.

    *sigh*

    But wait, my stupidity doesn't even end there! Before I was even interested in setting up and solving this integral, I had already figured out how the value behaves as n grows! (And the behavior as I adjust n was my primary interest)

    But do I remember that? Nooo... not until the drive home today.

    *sigh*

    Sorry I took up so much space for this rant!
     
  2. jcsd
  3. May 9, 2005 #2

    Math Is Hard

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    I guess even geniuses like Hurkyl have off days. Makes me feel a little better!
     
  4. May 9, 2005 #3

    Hurkyl

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    The best part about it is this -- the constant out front of that integral had an n in it, and I actually thought to myself:

    "Neat -- n (1 + erf (x/√2))n-1 looks like part of a derivative."

    :grumpy:
     
  5. May 10, 2005 #4

    mathwonk

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    just to put things in perspewctive, heres a little one that stumped most of my class:

    differentiate x^e (wrt x).



    another puzzle for me was to decide whether to give half credit for

    saying the derivative of x^(ln(x)) is ln(x)[x^(ln(x)-1)]. (work out the answer).

    you sort of have to be a mind reader to know for sure.
     
  6. May 10, 2005 #5

    saltydog

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    So you have:


    [tex]f(x)=[1+\frac{2}{\sqrt{\pi}}\int_{0}^{\frac{x}{\sqrt{2}}} e^{-u^2}du][/tex]

    Then:

    [tex]\frac{df}{dx}=\frac{2}{\sqrt{2\pi}}e^{\frac{-x^2}{2}[/tex]

    So that:

    [tex]\int_{-\infty}^{\infty}[f(x)]^{n-1}f^{'}(x)dx=\int_{-\infty}^{\infty}[f(x)]^{n-1}e^{-\frac{x^2}{2}}dx[/tex]

    Ok, now what?
     
  7. May 10, 2005 #6

    krab

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    salty: The integral is [tex]\int^1_0 f^{n-1}df[/tex]
     
  8. May 10, 2005 #7

    honestrosewater

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    Must I know what you're talking about if I can't slap you?
     
  9. May 11, 2005 #8

    saltydog

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    Would have taken longer than a drive home for me to see the integrand in that form. I don't see how the limits change to 0 and 1 but I'll spend some time on it.

    Thanks.
     
  10. May 11, 2005 #9

    saltydog

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    No. It's not happening for me but that's all right. It's like that integral Daniel solved the other day. I wondered out of the 85 people who looked at it, who other than me couldn't solve it even after he gave the hint.
     
  11. May 11, 2005 #10

    saltydog

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    Alright, I let:

    [tex]u=f(x)[/tex]

    [tex]du=f^{'}(x)dx=df[/tex]

    So, switching to u and changing limits I get:

    [tex]f(\infty)=2[/tex]

    [tex]f(-\infty)=0[/tex]

    Thus I get:

    [tex]\int_0^2 u^{n-1}du[/tex]

    When I integrate directly the integral (for n=5):

    [tex]\int_{-\infty}^{\infty}[f(x)]^{n-1}f^{'}(x)dx[/tex]

    both numerically and analytically in Mathematica, I get 6.4. The above integral in u is also 6.4. Note the plot of the integrand. It's about 3 high by about 3 wide so say 9 or say somewhere between 5 and 10, I mean we're not building the shuttle or nothing. So I believe the upper limit of 2 is correct.

    Edit: And you knew this from the get-go too Hurkyl, no driving around or nothing.
     

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    Last edited: May 11, 2005
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