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I found a really weird function

  1. May 2, 2015 #1
    Hey all, so I don't even know how I managed to find this, So I'll walk you all through how I've spent my last couple hours.

    I've been studying intro calculus, and I recently came across the chain rule. I was screwing around with finding the derivative of various functions like the square-root of 4x^2 + 4. I then took the derivative of square-root 4x^2 + 4 (I think that's the function I took, I screwed up somewhere, but oh well, it doesn't change the function that I found), and found it's derivative using the quotient rule, and after simplifying I was left with

    (4/(sqrt(4x^2+4)) - (16x^2/sqrt(4x^2+4)^3/2)

    You can find it on at this link: (It's a graphing calculator website), https://www.desmos.com/calculator/l68uv67zay

    It looks remarkably similar to a normal distribution, and seems to exhibit some properties of a normal distribution, is it doing this for any particular reason?
     
  2. jcsd
  3. May 2, 2015 #2
    Why don't you reduce it to a common denominator and also factor out √4 from within the radical? What are you left with then?

    Chet
     
  4. May 2, 2015 #3
    Give me a bit, i'll work on it. Thanks for your comment :D
     
  5. May 4, 2015 #4

    Svein

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  6. May 9, 2015 #5

    mfb

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    Staff: Mentor

    Your function can be simplified to$$(x^2+1)^{-\frac{3}{2}}$$
    Functions like this don't look completely different from a normal distribution on a linear plot, but they have a completely different shape in the tails.
     
  7. May 15, 2015 #6
    Your function has second derivative ##f''(x)=\frac{-3x}{(x^2+1)^{5/2}}##, which is approximately equal to ##-3xf(x)## when ##x## is small.
    The function ##g(x)=e^{-\frac{3x^2}{2}}## has derivative ##-3xg(x)##, and furthermore ##f'(0)=g(0)=1##; so ##g## can be said to approximate ##f'## near ##0##. Coincidentally ##g## describes a normal PDF, so that explains why your function looks like a normal distribution. I would link a graph comparing the two (they are very close!) but I happen to be posting from my phone presently.
     
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