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.