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This question is probably very basic, but I've been away from school for a while and the answer eludes me.

I was tempted to prove that d/dx(e^x) = (e^x) for old times sake and that was easy enough. I just expressed e^x as a power series where n goes from 0 to infinity for ((x^n)/n!).

During the derivation I started to wonder, how did they know that a power series where n goes from 0 to infinity for A(subscript n)*X^n would converge to the form of (someconstant)^x.

Is there some theorem that proves this?

This question is pretty probably trivial for the hardcore math types but it's been bothering me, so I thought I'd ask :-) ....

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  • $\begingroup$ Err... this clearly fails "What kind of questions can I ask here?" from the FAQ. Please don't even answer questions like this -- comment pointing questions like this to Ask Dr. Math or similar. -1 to everyone! $\endgroup$
    – Kim Morrison
    Commented Oct 31, 2009 at 2:55
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    $\begingroup$ I didn't think this one was too bad - it wasn't H/W, and it (perhaps inadvertently) touches on quite a subtle point, namely the proper definition of exp(x) and the business of the proving it behaves like constant^x. There's something conceptually interesting going on, at the interface between school maths and university maths... But I suppose this could be taken somewhere like NRICH $\endgroup$
    – Yemon Choi
    Commented Oct 31, 2009 at 3:54
  • $\begingroup$ I think it is unfair to close a question like this because it even touches on deep connections between integral transforms (like e.g. Laplace transofrms) in the continous domain and power series in the discrete domain. I think by closing these kinds of questions you discourage people that are willing to learn! I would vote to reopen it if I could! $\endgroup$
    – vonjd
    Commented Oct 31, 2009 at 6:33
  • $\begingroup$ Scott: You said "-1 to everybody": What a kind of policy of a moderator is this supposed to be that you downvote sophisticated and elaborate answers from experts with high reputation just because they answered a question you didn't like. I think this is the best way to destroy the open culture of this forum! $\endgroup$
    – vonjd
    Commented Oct 31, 2009 at 7:24
  • $\begingroup$ Actually, vonjd, I don't think it does have anything to do with a connection between integral transforms and power series, except in a superficial sense. Trying to make connections between everything just devalues the connections. $\endgroup$
    – Yemon Choi
    Commented Oct 31, 2009 at 9:41

2 Answers 2

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Some people would say that your question is trivial because we define the power function by a^x = exp(x log a).

However, that's not a very satisfying answer.

Clearly one wants the power series for exp(x) to satisfy exp(z+w) = exp(z) exp(w), and exp(0) = 1, from what we know the power function should be if z and w are integers. (I'm writing exp(x), not e^x, because I'm assuming exp(x) hasn't yet been shown to have this property.)

So say exp(x) = a0 + a1 x + a2 x^2 + a3 x^3 ... is a formal power series satisfying exp(z+w) = exp(z) exp(w).

Then since exp(0) = 1, we must have a0 = 0.

So exp(x) = 1 + a1 x + a2 x^2 + a3 x^3 + ...; therefore

exp(2x) = exp(x) exp(x) = (1 + a1 x + a2 x^2 + a3 x^3 + ...) (1 + a1 x + a2 x^2 + a3 x^3 + ...)

and expanding the rightmost member of this equation as a formal power series,

exp(2x) = 1 + 2a1 x + (2a2 + a1^2) + (2a3 + 2 a2 a1) x^3 + ...

However, exp(2x) = 1 + 2a1 x + 4a2 x^3 + 8a3 x^3 + ... by substituting 2x into the formal power series for exp(x).

By equating the coefficients of x, x^2, and x^3, you get

2 a1 = 2 a1 2 a2 + a1^2 = 4 a2 2 a3 + 2 a2 a1 = 8 a3

and so on. The first equation tells you nothing. The second tells you a1^2 = 2a2, so a2 = a1^2/2. The third becomes

2 a3 + 2 (a1^2 / 2) a1 = 8 a3

from which you get a1^3 = 6 a3, and a3 = a1^3/6. The pattern here continues, with an = a1^n/n!, as can be proven by induction.

This gives the series

exp(x) = 1 + a1 x + a1^2/2! x^2 + a1^3/3! x^3 + ...

and now we just have to choose a1. We pick 1 just because it's simple to do so.

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  • $\begingroup$ If you lack the patience to read this, see Qioachu's answer. $\endgroup$
    – Michael Lugo
    Commented Oct 31, 2009 at 1:48
  • $\begingroup$ "Then since exp(0) = 1, we must have a0 = 0." Should that be a0 = 1, or am I having one of those days? $\endgroup$
    – Vectornaut
    Commented Dec 15, 2009 at 2:20
  • $\begingroup$ That should be a0 = 1, and indeed if you read the next line you'll see that I proceeded with a0 = 1. $\endgroup$
    – Michael Lugo
    Commented Dec 15, 2009 at 3:02
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One answer is that the power series sum x^n/n! is used to define exponentials, not the other way around. The way to understand this is in terms of the defining property of the exponential, which is that e^{x+y} = e^x e^y. It turns out that this property is more or less equivalent to the property that d/dx e^x = e^x, and this property in turn is equivalent to having a certain Taylor series expansion.

Edit: Let me sketch the arguments here. In one direction, if e^{x+y} = e^x e^y then d/dx e^x = lim (e^{x+h} - e^x)/h = e^x lim (e^h - 1)/h. Since lim (e^h - 1)/h is just the derivative of e^x evaluated at 1, we can pretend that this exists and normalize it to 1, and then d/dx e^x = e^x as desired.

In the other direction, if d/dx e^x = e^x then d/dx e^{x+y} = e^{x+y} for fixed y. On the other hand, the space of functions satisfying d/dx f(x) = f(x) has to be one-dimensional (just take this on faith; it's more trouble to justify than it's worth right now), so e^{x+y} has to be a constant multiple of e^x, say e^{x+y} = Ke^x, and setting x=0 gives K = e^y as expected.

As for how these both relate to Taylor series, if d/dx e^x = e^x this implies that every Taylor coefficient of e^x is equal.

Edit #2: Anyway, one lesson to take away from all this is that there are multiple equivalent definitions of e^x, and the way you go about proving any of its properties depends strongly on what definition you use. There are good technical reasons to start with the power series definition: 1) you are automatically guaranteed that the function is well-defined everywhere, and 2) the power series definition makes sense in a very general context, such as over the complex numbers or with matrix inputs.

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  • $\begingroup$ Oops; I see I've been out-sketched! $\endgroup$
    – Qiaochu Yuan
    Commented Oct 31, 2009 at 1:56

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