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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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closed as off topic by Scott Morrison Oct 31 '09 at 2:53

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

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! – Scott Morrison Oct 31 '09 at 2:55
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 – Yemon Choi Oct 31 '09 at 3:54
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! – vonjd Oct 31 '09 at 6:33
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! – vonjd Oct 31 '09 at 7:24
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. – Yemon Choi Oct 31 '09 at 9:41
up vote 5 down vote accepted

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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If you lack the patience to read this, see Qioachu's answer. – Michael Lugo Oct 31 '09 at 1:48
"Then since exp(0) = 1, we must have a0 = 0." Should that be a0 = 1, or am I having one of those days? – Vectornaut Dec 15 '09 at 2:20
That should be a0 = 1, and indeed if you read the next line you'll see that I proceeded with a0 = 1. – Michael Lugo Dec 15 '09 at 3:02

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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Oops; I see I've been out-sketched! – Qiaochu Yuan Oct 31 '09 at 1:56

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