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The definition of Euler's constant to the power $x$, $e^x$, is

$$e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + {...}$$

And of course, we have the number $e$ defined as

$$e = \sum_{n=0}^\infty \frac{1}{n!} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + {...}$$ or $$e = \lim_{n\to \infty} (1+\frac{1}{n})^{n}$$

$e$ and $e^x$ here are expressions of a sum of infinite series. When one calculate $e^x$, he doesn't go by the definition of $e^x$, but instead calculates the numerical value of $e$, and takes the power of that numerical value directly.

How can one simply take the power of the numerical value of $e$ directly, and be sure the answer is $e^x$? And what about in the context of arbitrary powers of $e$?

p.s There are also different definitions of $e$, like: $$\int_1^{e^x}{\frac{1}{t}dt}=x$$ $$\frac{d}{dx}e^x = e^x$$ $$\frac{d}{dx}log_e{x}=\frac{1}{x}$$ But they do not explain the concern too.

user158163
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  • The answer to your question depends on how you are defining the power function. You can define $b^x$ as a sequence of rational powers $b^{r_n}$, where $r_n\rightarrow x$. However, a more common definition is in terms of logarithms: $b^x=e^{x\cdot\log{b}}$. – David H Jul 14 '14 at 08:09
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    https://i.imgflip.com/aas9z.jpg – Asaf Karagila Jul 14 '14 at 08:13
  • What does do you mean by "take the power of the numerical value of $e$ directly"? If you try to spell out what that means, you will see your confusion. You seem to be thinking of all exponentiation along the lines of exponentiating an integer to an integer power. – symplectomorphic Jul 14 '14 at 11:47

3 Answers3

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I assume that the question is (with the definition of the exponential function $\exp(x):=\sum\limits_{k=0}^{\infty} \frac{x^k}{k!}$):

Why do we have $\exp(1)^x = \exp(x)$?

There are (at least) two definitions of powers $a^x$ of real numbers. The first one uses $\exp$ and makes the claim trivial. The second (and probably more natural) one first defines $a^x$ when $x$ is an integer, then when $x$ is a rational number and finally when $x$ is a real number. I won't explain the details, because these are contained in every book on analysis.

So let us verify $\exp(1)^x = \exp(x)$ with this definition. We will only need the formula $$\exp(x+y)=\exp(x) \cdot \exp(y).$$ It immediately implies by induction $\exp(1)^x = \exp(x)$ when $x$ is an integer. If $x=p/q$ is rational, it follows $$(\exp(1)^x)^q = \exp(1)^p = \exp(p)=\exp(x)^q$$ and hence $\exp(1)^x = \exp(x)$. Finally, if $x$ is a real number, there is a sequence of rationals $x_i$ convering to $x$. Hence, $$\exp(1)^x = \lim_i \, \exp(1)^{x_i} = \lim_i \, \exp(x_i) = \exp(x).$$

Martin Brandenburg
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  • Your answer actually addressed my issue by not defining powers by exponentials. Thanks! By the way, is it true that exp(x+y)=exp(x)⋅exp(y) can be proven true no matter which definition of exp(x) is used? – user158163 Jul 14 '14 at 17:28
  • Sure, since all def's are equivalent. But it is in particular simple for $\exp(x)=\sum_k x^k/k!$ using the Cauchy product formula. – Martin Brandenburg Jul 14 '14 at 17:39
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When you say calculate the value of $e$ and then take the power $e^x$, what does taking a power $a^x$ mean? By definition, we let $$ a^x = \exp(x\log(a)), $$ where $\exp$ is defined as the power series you mentioned and $\log$ is its left inverse. Thus, $$ e^x = \exp(x\log(e)) = \exp(x\log(\exp(1)) = \exp(x), $$ since $e$ is defined as $\exp(1)$.

Christoph
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  • But then, how can one ensure this result matches with the simple "multiply the base that number of times" calculation? While redefining all the operations are perfectly fine, I can't see why those two results match... – user158163 Jul 14 '14 at 10:17
  • @user158163 You need to verify $\exp(x+y)=\exp(x)\exp(y)$. – Christoph Jul 14 '14 at 12:29
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Here is a proof of the equivalence of the two definitions. I take $$e=\lim\limits_{n \to\infty} \left(1+\frac{1}{n}\right)^n.$$ as the definition of $e$ and denote by $e^x$ the ordinary exponent function ($a^b$ can be defined directly for all $a>0$ and $b$) now we show $$e^x=\lim\limits_{n \to\infty} \left(1+\frac{x}{n}\right)^n.$$ This follows by writing

$$\left(1+\frac{x}{n}\right)^n= \left[ \left(1+\frac{1}{n/x}\right)^{n/x}\right]^x$$

And $$\left(1+\frac{1}{n/x}\right)^{n/x}\to e$$ so taking exponents we get the result.

Rene Schipperus
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