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Integration serie

  1. Aug 5, 2007 #1

    disregardthat

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    Hey, I am not trying to prove anything here, just merely asking a question to something I tryed:

    I found a relation to the Taylor series and how the integration by parts expand.

    for example:

    [tex]\int e^x dx = x e^x - \int xe^x dx[/tex]

    [tex]\int xe^x dx = \frac{x^2}{2}e^x - \int \frac{x^2}{2}e^x dx[/tex]

    [tex]\int \frac{x^2}{2}e^x dx = \frac{x^3}{6}e^x - \int \frac{x^3}{6}e^x dx[/tex]

    We see that this goes on and on to this:

    [tex]\int e^x dx = e^x[/tex]

    [tex]e^x = xe^x - \frac{x^2}{2}e^x + \frac{x^3}{6}e^x - \frac{x^4}{24}e^x...\frac{x^n}{n!}e^x[/tex]

    [tex]e^x=e^x \cdot \sum^{\infty}_{k=1}\frac{x^k}{k!}(-1)^{k-1}[/tex]

    [tex]\sum^{\infty}_{k=1}\frac{x^k}{k!}(-1)^{k-1} = 1[/tex]

    And that's definately NOT true! What is wrong?
     
  2. jcsd
  3. Aug 5, 2007 #2

    Mute

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    You forgot about the arbitrary constant that comes out of integrating. It should fix the problem.

    For example, instead of indefinite integration, choose limits of x and 0:

    [tex]\int_0^x dt e^{t} = e^{x} - 1[/tex]

    The series you get from the int. by parts can be written

    [tex] e^{x}\left[\sinh x - \cosh x + 1\right] = e^{x} - 1[/tex], so the results do indeed agree.

    (You have to be somewhat more careful in the case [itex]\int_{-\infty}^x dt \exp{t}[/itex], since the terms [itex]t^n \exp{t}[/itex] individually vanish as [itex]t \rightarrow -\infty[/itex], but the whole series actually converges to -1, which comes in as a +1 to cancel the -1 from the [itex]\exp{x}-1[/itex] result)
     
    Last edited: Aug 5, 2007
  4. Aug 6, 2007 #3

    disregardthat

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    All right, thanks mute
     
  5. Aug 6, 2007 #4

    disregardthat

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    Hi again, I think I found an expression for [tex]e^{-x}[/tex], but I am not sure if it is correct:

    [tex]e^{-x} = \sum_{n=0}^{\infty}\frac{x^n(-1)^n}{n!}[/tex]

    You probably know if it's wrong or not.
     
  6. Aug 6, 2007 #5

    quasar987

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  7. Aug 6, 2007 #6

    disregardthat

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    Yeah, I realized that now, and was going to change it... like this:
    [tex]e^{x} = \sum_{n=0}^{\infty}\frac{(-x)^n(-1)^n}{n!} = \sum_{n=0}^{\infty}\frac{x^n}{n!}[/tex]

    How is the taylor series of e^x made? Is it by integration?
     
  8. Aug 7, 2007 #7

    quasar987

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    suppose a power series solution to the diff. equ. y=y'.
     
  9. Aug 7, 2007 #8

    disregardthat

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    Oh I don't know much about differential equations..

    But I saw a flash "movie" about it. may it go something like this?

    y=y'
    y=dy/dx
    ydx=dy
    1dx=1/y dy
    x=lny
    y=e^x

    I don't know what a power series is.
     
    Last edited: Aug 7, 2007
  10. Aug 7, 2007 #9

    Gib Z

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    Well simply then, what quasar987 said asked was "find some function, where its derivative is equal to itself". Without having to know how to solve differential equations, we can see that the series posted is the solutions, just by finding the derivative of the series. We DEFINE e to be the solution to the differential equation,(there is a theorem that ensures uniquenesses of solutions) and we see the power series also satisfies it. Hence, they are equal.

    But there are many ways to define the exponential function, many!!
    One other one is: The Inverse function of the function [tex]\int^x_1 \frac{1}{t} dt[/tex]. What makes that obvious is that the integral is just the natural log function. You could define e in terms of many limits, products, integrals and series, countless definitions but some are more common than others, since it makes more sense to define it in a certain way. Eg It is quite easy to prove the limit [tex]\lim_{h\to 0} \frac{e^h-1}{h}=1[/tex] when we define e in terms of its power series, but not so easy otherwise!
     
    Last edited: Aug 7, 2007
  11. Aug 7, 2007 #10

    disregardthat

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    Yeah, the series is equal to itself when derivated because everything is "shifted".

    1+x+x^2/2+x^3/6+...+x^n/n!

    the derivative:

    (1)' +(x)' + (x^2/2)' + (x^3/6)'+...+(x^n/n!)'= 0 + 1 + x +x^2/2 + ... + x^(n-1)/(n-1)!

    Was that the point?

    What do you mean by this excactly?
     
  12. Aug 7, 2007 #11

    Gib Z

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    "Yeah, the series is equal to itself when derivated because everything is "shifted"." Correct, thats the point.

    "What do you mean by this excactly?" Well some problems have several solutions yes? polynomials, integrals with different constants of integration etc etc.

    For Differential Equations we are ensured 1 solution (with sufficient conditions) so we know that when we do find the power series solution to "find a function which is its own derivative", that it is equal to the ONLY solution, e^x, instead of perhaps some other solution that would be applicable if this theorem was not there.
     
  13. Aug 7, 2007 #12

    disregardthat

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    Can you perhaps give me a link to this theroem?
     
  14. Aug 7, 2007 #13
    Certainly, the 0 function is its own derivative, so there's something missing from this statement of the uniqueness theorem. I'm sure you know what it is. ;)
     
  15. Aug 8, 2007 #14

    Gib Z

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    "With Sufficient conditions" meaning setting the constant C in the solution [tex]y= Ce^x[/tex] to 1. :P

    http://hyperphysics.phy-astr.gsu.edu/hbase/diff.html has good info, but no proof. Different proof's exist for different types of differential equations.
     
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