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Proof for exponential number

  1. Nov 1, 2011 #1
    If anybody knows could they please tell me how it is known that there exists some number 'e' with the property that the derivative of ex is itself.

    I have a feeling this might require some definition of ab.

    Surely this isn't a complicated problem, but for some reason I can't work it out or even find a good answer online.

    Thanks for the help.
     
  2. jcsd
  3. Nov 1, 2011 #2

    I like Serena

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    Welcome to PF, DylanG! :smile:

    Actually, the property that the derivative of ex is itself, is the definition of ex (or at least 1 of the definitions).

    From wikipedia:
    and also from wikipedia:
     
  4. Nov 1, 2011 #3
    Look at this series definition/expansion of the exponential.

    [tex]{e^p} = 1 + p + \frac{{{p^2}}}{{2!}} + \frac{{{p^3}}}{{3!}} + \frac{{{p^4}}}{{4!}} + ...[/tex]

    differentiate with respect to p

    [tex]\frac{{d\left( {{e^p}} \right)}}{{dp}} = 0 + 1 + \frac{{2p}}{{2*1}} + \frac{{3{p^2}}}{{3*2*1}} + \frac{{4{p^3}}}{{4*3*2*1}} + ...[/tex]

    =[tex]1 + p + \frac{{{p^2}}}{{2!}} + \frac{{{p^3}}}{{3!}} + ... = {e^p}[/tex]

    does this help?
     
  5. Nov 1, 2011 #4

    lurflurf

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    Yes you would need to define ax first.
    Such as f(x)=ax has two properties
    f(x+y)=f(x)f(y)
    and
    f'(0)=1/f-1(e)
    then we have
    f(1)=a
    f'(x)=f'(0)f(x)
    so we take a=e<->f'(0)=1 as the standard case of such functions
     
  6. Nov 1, 2011 #5
    The power series is a little like pulling a rabbit out of a hat. Doesn't show you how it's done. Just shows you it can be done. What you can do to rectify that is you can assume that the function is equal to 1 when x equals zero, and then the rest of the Taylor series is determined by that. Which gives you the thing that Studiot posted.

    That just tells you there is such a function. Doesn't tell you it's a^x for some a. But you can can argue that it satisfies the property exp(x + y) = exp(x) exp (y), which implies what you want.

    There's also a geometric construction of the exponential function, but I can't draw pictures on here, so it's hard to explain. Basically, you start out at the point (0,1) and you do a discrete approximation and take the limit. It leads to the definition of e as a limit. I came up with it in high school when I was studying calculus, and later I found the same argument appears in Needham's Visual Complex Analysis, somewhere in the book, as a warm-up for the complex number version.
     
  7. Nov 1, 2011 #6
    As ILS has said, there are many ways to approach the exponential function.
    It all depends upon the order you define things when you build up your theory of analysis.

    Another way is to state

    y = exp(x) if and only if x = log(y)

    @ homeomorphic

    The following properties may be proved directly from the series definition:

    exp(0) = 1
    exp(-x) = 1/exp(x)
    exp(x) > 0 ie it never vanishes

    We may also use series comparison techniques to establish the derivative from first principles by rearranging and expanding

    [tex]\frac{{\exp (x + h) - \exp (x)}}{h}[/tex]

    But my original offering is simpler.

    go well

    exp(x+h)
     
  8. Nov 1, 2011 #7

    AlephZero

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    For a "simple" introduction to this I prefer to start by defining ln(x) as an integral:
    When x > 0, [tex]\ln x = \int_1^x \frac{dt}{t}[/tex]

    It is then easy to see that[tex]\frac{d}{dx}\ln x = \frac 1 x[/tex] and ln(x) is a continuous and monotonic function.

    From the definition ln(1) = 0, and by approximating the "area under the graph" by three rectangles, ln(4) > 1/2 + 1/3 + 1/4 > 1.

    So there is a number e such that 1 < e < 4 and ln(e) = 1.

    It is straightforward to show properties like ln(xy) = ln(x) + ln(y) direct from the definition, using techniqes from elementary calculus.

    And finally, define [itex]e^x[/itex] as the inverse function of ln(x).

    The advantage of this, IMO, is that it doesn't require anything more complcated than "calculus", which is much more accessible to "beginners" than a proper justification of operations on an infinite series that was "pulled out of a hat" as the definition of [itex]e^x[/itex].

    And a few years later, if and when you have taken a course on analytic functions, you can use the power series to define [itex]e^z[/itex] when z is complex, which is much more useful than restricting yourself to real variables.
     
  9. Nov 2, 2011 #8
    I definitely prefer the geometric approach (and you'll find I'm usually biased towards geometric approaches). Essentially, you're solving an ODE with an initial condition, so you can use Euler's method, which is pretty geometric. You have a slope field whose slope is equal to its height and you just follow the slopes up successively, until you get to the x-value you want, dividing it up into smaller and smaller time intervals. I think that's what my geometric construction amounts to if you unravel it. It's been a while since I've thought it through. Of course, Euler's method might be difficult to demonstrate rigorously (though it definitely can be done).

    And in this context (flow along a vector field on the line or slope field), I might add that the property exp(x+y) = exp(x)exp(y) has the very elegant interpretation that time evolution by x + y is the same thing as time evolution by x followed by time evolution by y.
     
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