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Euler's number

Revision as of 20:51, 23 June 2006 by ComplexZeta (talk | contribs) (→‎Where <math>e</math> comes from)

The mathematical constant e is defined as the following limit: $e=\lim_{n\rightarrow \infty}{(1+\frac1n)}^n$. In calculus, the fact that $e^x = \sum{\frac{x^n}{n!}}$ is used often, based on the above definition and the Binomial Theorem.


Where $e$ comes from

Suppose $b$ is a positive real number, and $f(x) = b^x$ for all real numbers $x$. Let's try to figure out what $f'(x)$ is.


$f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} = \lim_{\Delta x \to 0} \frac{b^{x+\Delta x} - b^x}{\Delta x} = \lim_{\Delta x \to 0} b^x \left( \frac {b^{\Delta x} - 1}{\Delta x}\right) = b^x \left(\lim_{\Delta x \to 0} \frac{b^{\Delta x} - 1}{\Delta x}\right)$.


Now look what has happened. Something special has almost happened. We found the derivative of $b^x$, and what we got was almost $b^x$. In other words, the derivative of this function is almost the same function that we started with. However, there is that annoying and kind of messy limit on the right that is messing things up.


It seems like it would be at least cool, if not downright useful, to have a function whose derivative is equal to itself. (In fact, it turns out that such a function is very useful indeed, for example in finding solutions to certain differential equations.) So let's ask this question: Is it possible to cleverly pick a special value of $b$ in order to make that annoying limit on the right turn out to be equal to $1$? If this were possible, then for that special value of $b$, it would in fact be true that the derivative of $b^x$ is just $b^x$, exactly the same function we started with.


Well, the fact is that there is a special value of $b$ which accomplishes this goal. It is approximately $2.718$, and it is called $e$. ("$e$" stands for exponential and not Euler, despite the fact that Euler was one of the first mathematicians to use it and the one to name it. $e$ is sometimes (perhaps incorrectly) called "Euler's number.")


So, how do we choose $b$ so that $\lim_{\Delta x \to 0} \frac{b^{\Delta x} - 1}{\Delta x} = 1$? Could we pick $b$ in some clever way to make this expression simplify? As a first rough idea, imagine what would happen to the expression $\frac{b^{\Delta x}-1}{\Delta x}$ if $b$ were equal to $(1+\Delta x)^{\frac{1}{\Delta x} }$. Everything cancels out nicely, and we are left with just $1$. This suggests that we should select $b$ to be $\lim_{\Delta x \to 0} (1+\Delta x)^{\frac{1}{\Delta x}}$. So here we have our definition of $e$. This limit is approximately $2.718$.


To make this rigorous, it should be possible to prove that this limit actually exists. And once we define $e$ to be this limit, it should be possible to prove that $\lim_{\Delta x \to 0} \frac{e^{\Delta x}-1}{\Delta x} = 1$.


(This definition of $e$ is equivalent to the one given at the beginning of this article. Just let $M = \frac{1}{\Delta x}$. Then $\lim_{\Delta x \to 0} (1+\Delta x)^{\frac{1}{\Delta x}} = lim_{M \to \infty} \left(1 + \frac{1}{M}\right)^M$.)

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