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Can you derive e^x by first principles?

  1. Jul 15, 2014 #1
    Differentiation by first principles is as followed:
    $$y'=\lim_{h\rightarrow 0}\dfrac {f\left( x+h\right) -f\left( x\right) }{h}$$
    So, assuming that ##y= e^{x},## can we prove, using first principle, that:
    $$\dfrac{dy}{dx}\left( e^{x}\right) =e^x$$

    Or is there other methods that are primarily used to do so? Just curious, because my working lead me
    to the final end of:
    $$y'=\lim_{h\rightarrow 0}\dfrac {e^{x}\left( e^{h}-1\right) }{h}$$
     
  2. jcsd
  3. Jul 15, 2014 #2
    I think you need to pick a starting definition for "e" at least.
     
  4. Jul 15, 2014 #3

    I am referring to e=2.71828...


    ~| FilupSmith |~
     
  5. Jul 15, 2014 #4

    Char. Limit

    User Avatar
    Gold Member

    You pretty much have it right. Going any farther would require evaluating the limit
    [tex]\lim_{h \rightarrow 0} \frac{e^h - 1}{h}[/tex]

    Now, there's various ways to do this, including by estimation (take extremely small values of h and see what they tend to), delta-epsilon stuff (which I never learned), or by taking the Taylor series of e^h. Every method should lead you to the same answer, though.
     
  6. Jul 15, 2014 #5
    Oh, ok. Thanks :)


    ~| FilupSmith |~
     
  7. Jul 15, 2014 #6

    pwsnafu

    User Avatar
    Science Advisor

    ##e^h## is the limit of ##(1+\frac{h}{n})^n## as ##n\to\infty##. So
    [tex]e^h-1 = (h/n)^n + n(h/n)^{n-1}+\ldots+n \frac{h}{n}[/tex]
    Divide everything by h, means you have a bunch of terms depending on h, and a 1. As ##h\to0##, you are left with 1.
     
  8. Jul 15, 2014 #7
    How you go about showing that ##\frac{d}{dx}e^x=e^x## depends on your definition(s) of the number ##e## and/or the function ##f(x)=e^x##. See https://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function for more details.

    If we want to use the definition of the derivative to demonstrate that ##\frac{d}{dx}e^x=e^x##, then the final step would involve showing that ##\lim\limits_{h\rightarrow 0}\frac{e^h-1}{h}=1##. Again, how you go about this depends on your definition(s) of the number ##e## and/or the function ##f(x)=e^x##. The calculus text that my school uses defines ##e## as the unique real number satisfying ##\lim\limits_{h\rightarrow 0}\frac{e^h-1}{h}=1##. That may seem like cheating (it did to me at first), but ...

    I recall an exercise in a more advanced course in which I was to show that all of the "standard" definitions for the exponential function were equivalent; that is that all of the characterizations actually described the same function. What might be disappointing to you is that at no time in that proof did I need to spend any effort showing that ##\lim\limits_{h\rightarrow 0}\frac{e^h-1}{h}=1## using anything like an ##\epsilon-\delta## argument. That fact was either "free" based on assumptions about the function, or it was irrelevant in showing that ##\frac{d}{dx}e^x=e^x##.

    Also, and this is just a pet peeve of mine, the verb is "differentiate", not "derive".
     
  9. Jul 15, 2014 #8
    Thank you very much!


    And I will be sure to remember that ;)


    ~| FilupSmith |~
     
  10. Aug 16, 2014 #9
    Basically i had posted a similar question some time ago, but i finally got what i was doing wrong. Anyway, there are 2 ways of proving it, then one is the method you say, and the other is the chain rule.
    A known fact is that (a^x)' = a^x * lna so if you replace "a" with "e" you get what you want.
    The other way is this:
    [tex] \frac{d}{dx} (e^x) = \lim_{\substack{h\rightarrow 0}} \frac{e^{x+h} - e^x}{h} [/tex]

    [tex] \frac{d}{dx} (e^x) = \lim_{\substack{h\rightarrow 0}} e^x \frac{e^h - 1}{h} [/tex]

    Now here comes a more comfortable solution of setting. Remember, h is NOT 0 it approaches 0.
    You can set that [tex] e^h - 1 = t [/tex] so you get for "h" that:
    [tex] e^h = t + 1 [/tex]

    [tex] h = \ln {(t + 1)} [/tex]

    This also means, [tex] e^h - 1 [/tex] will also tend to 0.

    [tex] \frac{d}{dx} (e^x) = \lim_{\substack{t\rightarrow 0}} e^x \frac{t}{\ln {(t + 1)}} [/tex]

    Exchanging [tex] e^x [/tex] with [tex] t [/tex]

    [tex] \frac{d}{dx} (e^x) = \lim_{\substack{t\rightarrow 0}} t \frac{e^x}{\ln {(t + 1)}} [/tex]

    Now, knowing that [tex] a = \frac{1}{\frac{1}{a}} [/tex] , we can invert the number "t" to multiply it with the denumerator.

    [tex] \frac{d}{dx} (e^x) = \lim_{\substack{t\rightarrow 0}} \frac{e^x}{\ln {(1 + t)} * \frac{1}{t}} [/tex]

    [tex] \frac{d}{dx} (e^x) = \lim_{\substack{t\rightarrow 0}} \frac{e^x}{\ln {(1 + t)}^\frac{1}{t}} [/tex]

    Since "t" tends to 0, we get 1/t tends to infinity. Now what that reminds us of is that:

    [tex] \lim_{\substack{n\rightarrow ∞}} (1 + \frac{1}{n})^n = e [/tex]

    We have here respectively that 1/n tends to 0 and the exponent outside tends to infinity. Much like our case here.

    [tex] \frac{d}{dx} (e^x) = \lim_{\substack{t\rightarrow 0}} \frac{e^x}{\ln e} [/tex]

    [tex] \frac{d}{dx} (e^x) = e^x [/tex]
     
    Last edited: Aug 16, 2014
  11. Aug 16, 2014 #10
    If we define the [itex]e[/itex] as

    [tex]
    e = \lim_{N\to\infty}\Big(1 + \frac{1}{N}\Big)^N
    [/tex]

    then by definition we have

    [tex]
    e^x = \Big(\lim_{N\to\infty}\Big(1 + \frac{1}{N}\Big)^N\Big)^x
    [/tex]

    It is not obvious at all how to get [itex]D_xe^x=e^x[/itex] from that.

    Do you know how to prove

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

    When I was a high school kid, I thought I knew how to prove this. Then I completed some studies in university, and understood that in fact I had not known how to prove this. The problem is that the proof involves a change of order of a limit and an infinite series. Such change of order is never trivial, and the standard tool to deal with it would be the Lebesgue dominated convergence theorem. However, I have never bothered to go through the effort of finding what the suitable dominating function would be. So in the end, I still don't know how to prove this! :tongue: Did I just help some people here to understand the same? That you have never really proven this?

    Another question: How do you define [itex]a^x[/itex] for arbitrary [itex]a,x\in\mathbb{R}[/itex] with [itex]a>0[/itex]? Perhaps we assume that we have this defined for [itex]a\in\mathbb{Q}[/itex], and then we define it as

    [tex]
    a^x = \lim_{N\to\infty} a_n^x
    [/tex]

    where [itex]a_n\to a[/itex] and [itex]a_n\in\mathbb{Q}[/itex]? Again, I have not bothered to go through the effort of proving that this produces a well defined [itex]a^x[/itex], but I believe it can be done. Perhaps somebody here could dig something out of his or her notes, if you are familiar with this?

    If we define [itex]a^x[/itex] like that, and if we know that the definition is proper, then we get

    [tex]
    e^x = \lim_{N\to\infty} \Big(1 + \frac{1}{N}\Big)^{Nx}
    [/tex]

    By using the definition of the limit it can be proven that this is the same as

    [tex]
    e^x = \lim_{N\to\infty} \Big(1 + \frac{x}{N}\Big)^N
    [/tex]

    Next, do you know how to prove that this is

    [tex]
    1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots
    [/tex]

    It is not trivial, because an order of a limit and an infinite series must be changed again. Once that is accomplished, the worst is behind.
     
  12. Aug 19, 2014 #11
    I remembered some things incorrectly. The most reasonable way to define [itex]a^x[/itex] would be to define as a limit of [itex]a^{q_n}[/itex] where [itex]q_n\in\mathbb{Q}[/itex] such that [itex]q_n\to x[/itex].

    Anyway, there is another problem. Suppose we have managed to prove that

    [tex]
    \underset{N\in\mathbb{N}}{\lim_{N\to\infty}} \Big(1 +\frac{1}{N}\Big)^N
    [/tex]

    converges. How do you prove that

    [tex]
    \underset{\alpha\in\mathbb{R}}{\lim_{\alpha\to\infty}} \Big(1 + \frac{1}{\alpha}\Big)^{\alpha}
    [/tex]

    converges too? It is obvious that if the second limit converges, it must be the same as the first one, but how do prove that the second limit does converge in the first place? Since the result is elementary, the use of exponential function or the logarithm is dangerous. Their use is allowed only if you can prove that you are not guilty of circular reasoning.
     
    Last edited: Aug 19, 2014
  13. Aug 19, 2014 #12
    Thank you guys. Your explanations were all very informative!
     
  14. Aug 20, 2014 #13

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The simplest way to find the derivative of [itex]e^x[/itex] from "first principles" is to start with completely different definitions!

    It is perfectly valid to define ln(x) by [itex]ln(x)= \int_1^x \frac{dt}{t}[/itex].

    From that it is easy to determine the important properties of the ln(x) function:
    1) Since 1/x is defined and continuous for all x except 0, and this integral always starts at 1, ln(x) is defined and continuously differentiable for all positive x.

    2) Since 1/x is positive for all positive x, ln(x) is an increasing function for all positive x.

    3) ln(1)= 0. If 0< x< 1, ln(x) is negative. If x> 1, ln(x) is positive.

    4) [itex]ln(1/x)= \int_1^{1/x} \frac{dt}{t}[/itex]. Let u= 1/t. Then [itex]du= -dt/t^2[/itex] so that [itex]dt= -t^2 du= -du/u^2[/itex]. When t= 1, u= 1 and when t= 1/x, u= x. so the integral becomes [tex]\int_1^x 1/(1/u)(-du/u^2)= -\int_1^x \frac{u}{u^2}du= -\int_1^x\frac{du}{u}= -ln(x)[/tex].
    That is, [itex]ln(1/x)= -ln(x)[/itex].

    5. [itex]ln(xy)= \int_1^{xy} \frac{dt}{t}[/itex]. Let u= t/y so that t= yu. Then dt= ydu. When t= 1, u= 1/y and when t= xy, u= x. The integral becomes [tex]y\int_{1/y}^x (u/y)(y/u^2)du= yint_{1/y}^x \frac{du}{u}= \int_1^x \frac{du}{u}+ \int_{1/y}^1 \frac{du}{u}= \int_1^x \frac{du}{u}- \int_1^{1/y}\frac{y}{dy}= ln(x)- ln(1/y)= ln(x)+ ln(y)[/tex]
    That is, [itex]ln(xy)= ln(x)+ ln(y)[/itex].

    6. [itex]ln(x^y)= \int_1^{x^y} \frac{dt}{t}[/itex]. If y is not 0, let [itex]u= x^{1/y}[/itex]. Then [itex]x= u^y[/itex] and [itex]dx= yu^{y-1}du[/itex]. When t= 1, u= 1 and when [itex]t= x^y[/itex], [itex]u= x[/itex]. The integral becomes [itex]\int_1^x \frac{yu^{y-1}du}{x^y}= y\int_1^x \frac{du}{u}= yln(x)[/itex].
    If y= 0, then [itex]x^y= x^0= 1[/itex] and ln(1)= 0 so that [itex]ln(x^0)= 0= 0ln(x)[/itex].
    That is, [itex]ln(x^y)= yln(x)[/itex].

    7. Since ln(x) is differentiable for all positive x we can apply the mean value theorem to any interval of positive real numbers and, in particular, to [1, X] for X any real number. That is, [itex](ln(X)- ln(1))/(X- 1)=\frac{1}{c}[/itex] for some number, c, between 1 and X. Taking X= 2, we have [itex]ln(2)= \frac{1}{c}[/itex] for some c between 1 and 2. The largest possible value of c is 2 so the smallest possible value of ln(2) is 1/2: [itex]ln(2)\ge 1/2[/itex].
    The reason that is important is that, for any positive X, we have [itex]ln(2^X)= Xln(2)\ge X/2[/itex]. Since X can be any positive number, so can X/2. That is, ln(x) is NOT bounded above. Since ln(x) is increasing, [itex]\lim_{x\to +\infty} ln(x)= +\infty[/itex] and, since [itex]ln(1/x)= -ln(x)[/itex], [itex]\lim_{x\to -\infty} ln(x)= 0[/itex].
    That is, since, also, ln(x) is an increasing function, ln(x) maps the set of all positive real numbers, one to one onto the set of all real numbers. From that it follows that it has an inverse function that maps the set of all real numbers onto the set of all positive real numbers.

    We define that inverse to be Exp(x). It then follows that Exp(x) is a differentiable function and that, writing y= Exp(x) so that ln(y)= x, [itex]dExp(x)/dx= dy/dx= 1/(dx/dy)= 1/(d ln(y)/dy)= 1/(1/y)= y= Exp(x)[/itex].

    Finally, to show that Exp(x) really is an exponential, if y= Exp(x), the x= ln(y). If [itex]x\ne 0[/itex], we have [itex]1= (1/x)ln(y)= ln(y^{1/x})[/itex] and then, going back to Exp, [itex]y^{1/x}= Exp(1)[/itex] and [itex]Exp(x)= y= Exp(1)^x[/itex]. That is, "Exp(x)" really is some number to the x power. Defining e= Exp(1), we have [itex]Exp(x)= e^x[/itex] and that the derivative of [itex]y= e^x[/itex] is [itex]dy/dx= e^x[/itex].
     
  15. Aug 20, 2014 #14
    By using the derivative of the logarithm you can prove that

    [tex]
    \lim_{N\to\infty} N\log\Big(1 + \frac{1}{N}\Big) = 1
    [/tex]

    which will help to identify the Exp(1) with [itex]e[/itex], if hit has already been defined with the limit.
     
  16. Aug 20, 2014 #15
    I would like to show how to prove the formula

    [tex]
    \lim_{N\to\infty}\Big(1 + \frac{1}{N}\Big)^N = \sum_{n=0}^{\infty}\frac{1}{n!}
    [/tex]

    because I'm not convinced that many people know how to accomplish it, and I'm also interested to learn if the proof can be found in some elementary book.

    With the binomial formula we get

    [tex]
    \Big(1 + \frac{1}{N}\Big)^N = \sum_{n=0}^N \frac{N!}{n!(N-n)!}\frac{1}{N^n}
    [/tex]

    With fixed [itex]n[/itex] we can prove the following limit:

    [tex]
    \lim_{N\to\infty} \frac{N!}{n!(N-n)!}\frac{1}{N^n} = \frac{1}{n!}\lim_{N\to\infty} \frac{(N-n+1)\cdots (N-1)N}{N\cdot N\cdot\ldots \cdot N}
    [/tex]
    [tex]
    = \frac{1}{n!}\lim_{N\to\infty} \Big(1 - \frac{n-1}{N}\Big)\Big(1 - \frac{n-2}{N}\Big)\cdots \Big(1 - \frac{1}{N}\Big) = \frac{1}{n!}
    [/tex]

    At this point the uneducated would probably believe that the proof is complete, but the contrary becomes clearer with the following notations: Let's define

    [tex]
    a(n,N) = \left\{\begin{array}{ll}
    \frac{N!}{n!(N-n)!}\frac{1}{N^n},\quad &0\leq n\leq N \\
    0,\quad & 0\leq N < n \\
    \end{array}\right.
    [/tex]

    We know the following formulas:

    [tex]
    \lim_{N\to\infty} \Big(1 + \frac{1}{N}\Big)^N = \lim_{N\to\infty} \sum_{n=0}^{\infty} a(n,N)
    [/tex]

    and

    [tex]
    \sum_{n=0}^{\infty} \lim_{N\to\infty} a(n,N) = \sum_{n=0}^{\infty} \frac{1}{n!}
    [/tex]

    So the final question is that do we know the formula

    [tex]
    \lim_{N\to\infty} \sum_{n=0}^{\infty} a(n,N) = \sum_{n=0}^{\infty} \lim_{N\to\infty} a(n,N)
    [/tex]

    The answer is of course that if we haven't proven it, then we don't know it. How to prove the change of order of the limit and the series? There are at least two ways. We have already shown that the [itex]a(n,N)[/itex] can be written as

    [tex]
    a(n, N) = \frac{1}{n!}\Big(1 - \frac{n-1}{N}\Big)\Big(1 - \frac{n-2}{N}\Big)\cdots \Big(1 - \frac{1}{N}\Big)
    [/tex]

    for [itex]n\leq N[/itex], while [itex]a(n,N)=0[/itex] for [itex]0\leq N<n[/itex], and from this we see that the sequence [itex]a(n,0),a(n,1),a(n,2),\ldots[/itex] is monotously increasing with all [itex]n[/itex], and the change of order can be justified with the Lebesgue's monotone convergence theorem.

    Another possibility is to use the Lebesgue's dominated convergence theorem. It says that if there exists a function [itex]g(n)[/itex] such that

    [tex]
    \sum_{n=0}^{\infty} g(n)<\infty
    [/tex]

    and

    [tex]
    |a(n,N)|\leq g(n)\quad \forall N
    [/tex]

    then the change of order is justified. Setting [itex]g(n)=\frac{1}{n!}[/itex] works for this purpose.
     
  17. Aug 20, 2014 #16
    Awesome. Thanks! This is so interesting!
     
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