cartesian product … stuff about computing mostly

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?

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Adrian McMenamin

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2 responses to “Good maths or bad maths?”

  1. Paul A. Rubin Avatar
    Paul A. Rubin

    The bit about delta/n going to 1 is not supportable. Replace delta with n^2 (which approaches zero when n does) and your argument proves that n approaches 1 as n approaches 0 (immediately curing the Greek debt crisis, among other things).

    Interchanging the order of limits is also tricky business. For any positive y, the limit of x/y as x approaches 0 from above is 0; so the limit of that limit as y approaches 0 from above is 0. Swap the order of the limits, treat the limit as y approaches 0 of x/y as infinity for positive x, and you get infinity for the outer limit.

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