# Another Euler number proof

OK, back to the issue of Euler’s number and the proof that it is $\lim _{x \to \infty}(1 + \frac{1}{x})^x$. This is a proof based on, though expanded from, what I picked up from a rather excellent website here.

Once again we start from the proposition that there exists a number $e$ such that $\frac{d}{dx}e^x = e^x$ and hence $\frac{d}{dx}log_e(x) = \frac{1}{x}$ (see here for why the second follows from the first).

Here we go…

$y = \lim_{x \to \infty}(1 + \frac{1}{x})^x$

$log_e(y) = log_e[\lim_{x \to \infty}(1 + \frac{1}{x})^x]$

$= \lim_{x \to \infty}[x log_e(1 +\frac{1}{x})]$

At this point we still have $\infty \times 0$, an indeterminate number, so we need to look for a determinate form.

So we take this as:

$= \lim_{x \to \infty}[\frac{log_e(1 + \frac{1}{x})}{\frac{1}{x}}]$

Now we have $\frac{0}{0}$, another indeterminate, but we can also apply L’Hospital’s rule.

This states that, $\lim_{x \to \pm\infty}[\frac{\frac{d}{dx}f(x)}{\frac{d}{dx}g(x)} = \frac{f(x)}{g(x)}]$ (a proof to follow sometime).

Here $f(x) = log_e(1+\frac{1}{x})$, $g(x) = \frac{1}{x}$ and, of course, $\lim_{x \to \infty}\frac{f(x)}{g(x)}= log_e[\lim_{x \to \infty}(1+\frac{1}{x})^x]$.

So, using the chain rule, $\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}$, for the numerator, $y = log_e(u)$, $u=1+\frac{1}{x}$: so $\frac{d}{dx}log_e(1 + \frac{1}{x}) = \frac{1}{1 + \frac{1}{x}}\times-\frac{1}{x^2}$

And for the denominator $\frac{d}{dx}\frac{1}{x} = -\frac{1}{x^2}$, giving for the whole thing, $\lim_{x \to \infty}[\frac{1}{1+\frac{1}{x}}] =1$

Hence $log_e[\lim_{x \to \infty}(1 + \frac{1}{x})^x] = 1$, so $e=\lim_{x \to \infty}(1 +\frac{1}{x})^x$.

# Good maths or bad maths?

Not sure, so maybe someone who knows can tell me.

Following on from the last blog, can we show $e = \lim_{n \to 0}(1 +n)^{\frac{1}{n}}$?

Assume a constant, $e$ exists such that $\frac{d}{dx}e^x=e^x$, could $\lim_{n \to 0}(1 + n)^{\frac{1}{n}}$ give us this $e$?

$\frac{d}{dx}e^x = \lim_{\delta \to 0} \frac{e^{x+\delta} - e^x}{\delta} =^{\lim_{\delta \to 0}}_{\lim_{n \to 0}}\frac{(1 +n)^{\frac{x+\delta}{n}} - (1 +n)^{\frac{x}{n}}}{\delta}$

$= (1+n)^{\frac{x}{n}}\frac{(1+n)^{\frac{\delta}{n}}-1}{\delta}$

Edit: Professor Rubin (see comments) tells me that, as I feared, what follows is not supportable:

Now, and this is the bit I have most doubts over, as $\delta \to 0$ and $n \to 0$ then $\frac{\delta}{n} \to 1$ and $1 + n - 1 \to 0$ and assuming $\frac {0}{0} = 1$ then we are left with $(1 + n)^{\frac{x}{n}}$.

Hence $\lim_{n \to 0} \frac{d}{dy} (1 + n)^{\frac{x}{n}} = (1 + n)^{\frac{x}{n}}$ and therefore $\lim_{n \to 0} (1+ n)^{\frac{1}{n}} = e$, the constant we are seeking.

But there are a lot of assumptions in there: anybody able to tell me how valid they are?

# Differential calculus reminder

This is just an online note to myself about differential calculus. A level maths again…

Calculating $\frac{d}{dx}2^x$

$2^x = e^u$ where $u=\ln(2^x)$

Using the chain rule: $\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}$

$\frac{dy}{du} = e^u$ (as $\frac{d}{dx}e^x = e^x$)

$\ln(2^x) = x\ln(2)$ hence $\frac{d}{dx}\ln(2^x) = \ln(2)$.

So $\frac{d}{dx}2^x = e^{\ln(2^x)}\ln(2) = 2^x\ln(2)$

And $e$, Euler’s number, really is magical.