cartesian product … stuff about computing mostly

An alternative second half?

Going back to the previous blog and looking for an alternative second half…

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

For this to be true $\frac{(1+n)^{\frac{\delta}{n}}-1}{\delta}=1$

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

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

Raise both sides to the $n$ th power:

$(\delta + 1)^n = (1+n)^\delta$

At $\lim_{\delta \to 0}$ and $\lim_{n \to 0}$ it can be seen that both sides equal 1.

Feels like quite a weak proof to me. But I am not in a position to claim expertise. Thoughts, anyone?

Adrian McMenamin

@Our_Frank

December 22, 2011 at 2:16 pm

Like your thinking and approach in looking at this from a different perspective (i.e. just using the assumption that there exists a number e such that the derivative of e^x = e^x). Good stuff. Certainly not in a position to claim any expertise at all either but thoughts are that, in general, lim(x->0) f(x).g(x) is only equal to the product of lim(x->0) f(x) and lim(x->0) g(x) if both of these limits are defined already – I think. e.g. (looking at it the other way round) lim(x->0) x/sin(x) = 1 but lim(x->0) x (=0) multiplied by lim(x->0) 1/sin(x) (undefined) is clearly not 1 (I’m not sure whether, technically, it’s 0 or undefined but anyway – it’s certainly not 1). So, even if lim(d,n->0) ((1+n)^(d/n)-1)/d is defined (and equal to 1) I’m not sure it necessarily follows that lim(n->0) of (1+n)^(x/n) is defined and therefore equal to e^x.