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Deriving the value of e

  1. Sep 8, 2012 #1
    Hey, I am having some difficulties. So it's my understanding that the function ex comes around out of a desire to have a function whose derivative is equal to itself. Well we can show that if f(x)=ax, a>0, then f'(x) is equal to a multiple of itself using the limit definition of the derivative


    f'(x) = lim h-->0 (f(x+h)-f(x))/h = lim h-->0 (ax+h - ax)/h = lim h-->0 ax(ah-1)/h = ax ( lim h-->0 (ah-1)/h )


    So the goal is, if we can find a value a that makes (lim h--> 0 (ah-1)/h ) = 1, then f'(x) = f(x).


    My only issue is that when I actually take this limit, I don't understand how it can be anything other than 0.


    lim h-->0 (ah-1)/h ) = 0/0 so if we apply L'hopitals, we get


    lim h-->0 (h*ah-1)/(1) = [lim h--> 0 (h) * lim h-->0 (ah-1)]/lim h-->0 (1) = 0


    Right? What am I doing wrong in evaluating this limit? I mean I know I"m doing something wrong I just can't figure out what it is
     
    Last edited: Sep 8, 2012
  2. jcsd
  3. Sep 9, 2012 #2

    If you use L'Hopital's rule, you'd have to differentiate with respect to h and not a (you always differentiate with respect to the variable under lim with l'Hopital's rule):[tex]\lim_{h\to 0}\frac{a^h-1}{h} = \lim_{h\to 0}\frac{\frac{d}{dh}a^h-1}{\frac{d}{dh}h} = \lim_{h\to 0}\frac{d}{dh}a^h[/tex]which basically brings you back to where you started in finding the derivative of ax...

    Now, there is a way to rewrite the limit expression lim h-->0 (ah-1)/h ) to get another expression that you can use to approximate the value of a such that the limit above equals 1, which turns out to be a ≈ 2.71828, if that's what you wanted.
     
  4. Sep 9, 2012 #3
    To be honest, Taylor series can easily express the solution to the differential equation
    [tex]\frac{dy}{dx}=y[/tex]
    subject to the condition [itex]y(0)=1[/itex], which is the exponential function. e is its value at one.

    Your method will work. Try plugging in the limit formula for e after making the variable of the limit go to 0 via a substitution.
     
  5. Sep 9, 2012 #4

    HallsofIvy

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    You talk about "deriving" the value of e, but, in fact, in most texts the value of e is defined just as you give it:
    From the derivative calcluation, we can show that the derivative of [itex]a^x[/itex] for any positive a is
    [tex]a^x \lim_{h\to 0}\frac{a^h- 1}{h}[/tex]

    We define "e" to be the value of a that makes that coefficient 1:
    [tex]\lim_{h\to 0}\frac{e^h- 1}{h}= 1[/tex]

    We can roughly (and it can be made precise by "[itex]\delta- \epsilon[/itex] arguents") say that, for h close to 0, we have, approximately,
    [tex]\frac{e^h- 1}{h}~ 1[/tex]
    [tex]e^h- 1~ h[/tex]
    [tex]e^n~ 1+ h[/tex]
    [tex]e~ (1+ h)^h[/tex]
    and so arrive at the definition [itex]e= \lim_{h\to 0} (1+ h)^h[/itex] which is equivalent to
    [tex]e= \lim_{n\to\infty} (1+ \frac{1}{n})^{n}[/tex]

    You will also see that definition of "e" in situations that have nothing to do with derivatives. If you have a loan, or investment, or bank account, where the interest rate is r and the interest is compounded n times a year, after one year the return (or amount owed if a loan) will be the original amount times [itex](1+ r/n)^n[/itex]. The limit, as n goes to infinity, of that is [itex]\lim_{n\to\infty}(1+ r/n)^n= e^r[/itex].

    As for specific values of e, we can see that if a= 2, taking h= .001 gives [itex](a^h- 1)/h=[/itex] .6939< 1 while taking a= 3, h= .001, [itex](a^h- 1)/h=[/itex] 1.099> 1 so the value of a= e, giving a limit of 1 must be between 2 and 3. Taking a= 2.5, h= .001, [itex](a^h- 1)/h=[/itex]0.917< 1 so e must be between 2.5 and 3. Taking a= 2.75, h= .001, [itex](a^h- 1)/h=[/itex]1.012 so e must be between 2.5 and 2.75. Continuing in that way we can get e as accurately as we wish.

    I will also point out that many new Calculus texts first define [itex]ln(x)[/itex] by
    [tex]ln(x)= \int_1^x \frac{1}{t} dt[/tex]
    All of the properties of ln(x) can be shown from that, including that it is invertible. We can then define exp(x) to be the inverse function. It can then be shown that, for all x, [itex]exp(x)= (exp(1))^x[/itex] so, defining e= exp(1), we have exp(x)= e^x. That is the same as saying that ln(e)= 1 so that e can be evauated numerically by repeated numerical integrations of 1/x from 1 to different values of a, each time getting the integral closer to 1.
     
    Last edited by a moderator: Sep 9, 2012
  6. Sep 9, 2012 #5
    Thank you all so much for the responses! This has given me many different angles to think about this issue from. Also I see the stupid mistake I was making now (was differentiating in terms of a instead of h)!
     
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