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Proof of Differentiation

  1. Jan 16, 2005 #1
    hi

    I'm in a bit of a hurry but I'm looking for the proof or derivation of the formula for differentiation :
    nx^n-1

    I don't mean to show it works for all (most) functions but where it comes from
    Thanks

    Roger
     
  2. jcsd
  3. Jan 16, 2005 #2

    Curious3141

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    You can prove it for positive integral n using the Binomial Theorem.

    Hint : consider [tex](x + \delta x)^n - x^n[/tex]
     
  4. Jan 16, 2005 #3
    I am looking for the rigorous proof, not the binomial theory.

    Perhaps as a series ?

    Roger
     
  5. Jan 16, 2005 #4
    Use the natural logarithm to turn exponents into products.
     
  6. Jan 16, 2005 #5

    Curious3141

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    This would mean assuming a less "obvious" result. I suppose the way this could be done would go something like this :

    [tex]f(x) = x^n = e^{n\ln x}[/tex]

    [tex]f'(x) = (e^{n\ln x})(\frac{n}{x}) = nx^{n-1}[/tex]

    To me this is using a more advanced result to prove a less advanced one.

    Instead, consider using the general Binomial expansion (which holds for all complex exponents) to prove the result. Please see the expansion in my post below to work out the details.
     
    Last edited: Jan 16, 2005
  7. Jan 16, 2005 #6

    cepheid

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    So, as I understand it, you'd do this expansion inside the limit (from the definition of a derivative) in order to be as rigorous as possible, right?
     
  8. Jan 16, 2005 #7

    Curious3141

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    I just edited my post, please read it. It took me a while to work out the LaTex, that's really harder than it should be. :rofl:
     
  9. Jan 16, 2005 #8

    dextercioby

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    And what would that be????Is it the one which includes the Gamma Euler function??Please,enlighten us...

    Would u say Gamma Euler is simpler than natural logarithms and exponentials?? :surprised

    Daniel.
     
  10. Jan 16, 2005 #9

    mathwonk

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    the derivative of x^n means the coefficient of the linear part of the change in y, compared to the change in x. so we write x, as it changes, as x+dx, and then raise it to the nth power, getting (x+dx)^n = x^n + n x^(n-1) dx
    + (n)(n-1)/2 x^(n-2)(dx)^2 +.....

    so we see that the linear part of this change in x, i.e. linear in dx, is n x^(n-1)dx. Thus the coefficient is n x^(n-1).

    much of the discussion here so far has seemed ludicrous to me. so maybe this will seem so to you.
     
  11. Jan 16, 2005 #10

    Curious3141

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    Actually, it's not as difficult as you make out.

    The binomial expansion of (1 + x)^n for any complex n and |x| < 1 is :

    [tex](1 + x)^n = 1 + x + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ...[/tex]

    is it not ?

    Doesn't involve gamma functions now does it ? Where's the difficulty ?

    It's fairly simple to work the rest out.
     
  12. Jan 16, 2005 #11

    dextercioby

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    You're right.My formula is general.Yours does not cover real and complex exponents.

    [tex] (1+x)^{n}= \sum_{k=0}^{+\infty} \frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}x^{k} [/tex]

    ,for all complex "n",with:
    [tex] |x|<1 [/tex]

    Daniel.
     
    Last edited: Jan 16, 2005
  13. Jan 16, 2005 #12

    Curious3141

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    Actually it does. When you simplify the gamma functions, it comes out to the exact same thing. Try it. :tongue2:

    Hint : Let n = N + a, where N is an integer and a is the nonintegral portion between 0 and 1. Can you see where this is going ?

    That would cover all real exponents. Complex exponents, I haven't tried yet.
     
    Last edited: Jan 16, 2005
  14. Jan 16, 2005 #13

    dextercioby

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    Compare my formula with yours for "k=3" and "n=i+5".

    I edited my previous post. :tongue2:

    Daniel.
     
  15. Jan 16, 2005 #14

    Curious3141

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    I know for a fact that my formula covers all complex exponents. I can quite easily prove it for all reals at least, I just scribbled it out, do you want the proof ?
     
  16. Jan 16, 2005 #15

    dextercioby

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    I asked you to see whether my formula and yours give the same values for "k=3" and "n=i+5".I didn't ask you for any proof...

    Daniel.

    P.S.There's no point in doing the calculation,because they don't... :tongue2:
     
  17. Jan 17, 2005 #16

    Curious3141

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    You are quite, quite wrong.

    My way : coefficient of term with k = 3 is

    [tex]\frac{(i + 5)(i + 4)(i + 3)}{3!}[/tex]

    Your way :

    [tex]\frac{\Gamma (i + 6)}{(\Gamma(4))(\Gamma(i + 3))} = \frac{(i + 5)(i + 4)(i + 3)(i + 2)(i + 1)(i)\Gamma(i)}{3!(i + 2)(i + 1)(i)(\Gamma(i))} = \frac{(i + 5)(i + 4)(i + 3)}{3!}[/tex]

    Correct ?
     
  18. Jan 17, 2005 #17

    dextercioby

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    Yeah,you're right,just because those gamma-s get symplified.But "area of a circle"??? :yuck: :tongue2:

    Daniel.
     
  19. Jan 17, 2005 #18

    Curious3141

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    You can use the same principle of cancellation to prove it for all reals (with the method I suggested). I honestly don't know enough about the gamma functions of complex arguments to try and tackle the proof for all complex exponents.
     
  20. Jan 17, 2005 #19

    dextercioby

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    U don't have to know anything about the complex variable Gamma Euler.
    [tex] \Gamma(z+1)=z\Gamma(z) [/tex]

    would be enough to prove that the Gamma ratio can be written:
    [tex] \frac{\Gamma(z+1)}{\Gamma[(z-k)+1]}=\frac{\Gamma(z+1)}{\Gamma[(z+1)-k]}=(z+1-k)_{k}=(z+1-k)[(z+1-k)+1][(z+1-k)+2]...[(z+1-k)+(k-1)] [/tex]

    ,where i made use of a substitution and the definiton of the Pochhammer symbol.

    Daniel.

    P.S.It's better if u didn't put the Pochhammer symbol in the binomial expansion.Just leave it with the 3 gammas Euler.
     
  21. Jan 17, 2005 #20

    Curious3141

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    Yes, that would work. :smile:
     
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