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Proving e^x is equals its own derivative?

  1. Feb 24, 2014 #1
    Hi, I need this for part of a math project I'm working on. So as most of us know, e^x, when differentiated, is equal to e^x, hence it is its own derivative.

    Now, I could prove this quite easily with the Chain Rule, but my teacher said that this is predicated on too many assumptions (not sure what he meant in particular, but i get it), thus I need to prove it by first principles. Could anyone help me prove this? Thanks
     
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  3. Feb 24, 2014 #2

    tiny-tim

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    Hi Cookies?! :smile:

    How are you defining ex ? :wink:
     
  4. Feb 24, 2014 #3

    mathman

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    You could do it by using the series for ex. In any case you need to start from a definition.
     
  5. Feb 24, 2014 #4
    Try using the limit definition of e or ex, of the form limn→∞(???)
     
  6. Feb 24, 2014 #5

    Redbelly98

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    If you are referring to the Taylor Series, don't you need to know the derivative before you can use that series?
     
  7. Feb 24, 2014 #6

    D H

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    Not necessarily. The powers series is one way of many ways to define the exponential function.

    That's why this question of how the function is defined is central. One of the definitions makes this question absolutely trivial. It's not quite as trivial with other definitions.
     
  8. Feb 24, 2014 #7
    Some people define [itex]e^x[/itex] by its power series. Some define it as a continuous function with [itex]e^0=1[/itex] and [itex]e^{s+t}=e^se^t[/itex]. When I was in undergrad, we defined [itex]\log(y)[/itex] as [itex]\int_1^y\dfrac{1}{t}\text{d}t[/itex] and [itex]e^x[/itex] as [itex]\log^{-1}(x)[/itex]. Some even define it as a differentiable function with [itex]e^0=1[/itex] which is equal to its own derivative.

    What constitutes a valid proof depends on which you consider "the" definition.
     
  9. Feb 24, 2014 #8
    That function isn't unique. You also have to give another value of the function, so you can't get out of saying what "e" is.
     
  10. Feb 24, 2014 #9
    Mathmans suggestion is probably best. Who doesn't love binomial theorem with a little bit of calculus?!
     
  11. Feb 24, 2014 #10
    [tex]e^x=Lim_{n→∞}{\left( 1+\frac{x}{n}\right)^n}[/tex]
     
  12. Feb 24, 2014 #11
    Whoops! Nice catch. Same goes for the derivative-based definition.
     
  13. Feb 25, 2014 #12

    D H

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    The derivative-based definition is that exp(x) is the function whose derivative is equal to itself and such that exp(0)=1. There is only one such function. That certainly isn't the definition being used as the problem is far too trivial with this definition.
     
  14. Feb 25, 2014 #13
    [I should shut up. Sorry for the serial errors.]
     
  15. Feb 26, 2014 #14

    mathwonk

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    if you assume you have a continuous function f satisfying f(x+y) = f(x).f(y) for all x,y, and f(1) = a>0, then it does equal a^x, and has derivative k.a^x, for some constant k, but just showing it is differentiable is not at all easy, in my opinion.

    proving functions given by convergent power series are differentiable is not trivial either. the easiest way of dealing with exponentials, is the usual one of using the fundamental theorem of calculus to prove the function defined by the integral of 1/x is differentiable, and then proving it is also invertible, and using the chain rule (and inverse function theorem) to get the derivative of the inverse to equal itself, and then proving also that the inverse satisfies the properties listed earlier that characterize an exponential function.

    Maybe the series approach would be cleaner after all. but you have to deal with uniform convergence which is usually considered more advanced than inverse functions and the FTC. It would be worth learning about though.

    maybe the differential equations approach is worth trying too. In that approach you look at the de y'=y, y(0) = 1, and you prove that such a function must also satisfy y(a+b) = y(a).y(b). Then you deduce that it is an exponential function of form a^x for some a, and you just define e to be that value of a. Of course in this approach you have to prove the d.e. has a solution.

    there is a nice proof of existence of solutions of first order d.e.'s using sequences of approximations, which should generate something like the series expansion of e^x.

    so this approach works backwards: first you prove there is some function which equals its own derivative, and then afterwards you prove that function is an exponential function. then the method of proof may give
    you a series expansion and you can compute the base e via that expansion.

    i think that proof method picard's method. here is a link that may help you.

    http://www.math.msu.edu/~seal/teaching/f09/picard_iteration.pdf
     
    Last edited: Feb 26, 2014
  16. Feb 27, 2014 #15
  17. Feb 28, 2014 #16

    mathwonk

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    this seems to assume that d/dx(ln(x)) = 1/x and then uses the chain rule to conclude the required fact (i got bored and did not watch it all). this is the easy part of one of the approaches given above in much more detail.
     
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