I know that Euler solved this somehow for a=1 and came to the value of e. I also found a series expansion that Euler developed. However, is there someone who could explain the derivation ending up in exp(a) using "high school math"? Maybe one needs to be genius or have high math skills to understand. But if this is not the case I would be happy for an understandable explanation, probably just framing the idea...
What do you think of using the ln(1 + x) Taylor expansion around 0 ?
(1+a/n)^n = exp( n * ln( 1 + a/n ) )
ln( 1 + x ) = x + o(x) => ln(1 + a/n) = (a/n) + o(a/n)
n*ln(1 + a/n) = a + o(a)
(1+a/n)^n = exp(n * ln(1 + a/n)) = exp(a + o(a))
Hence the requested limit...
Thank you! So I have to learn what the Taylor expansion is. What is o(a)? From your text I read o(x) = ln(1+x)-x, so o(a) = ln(1+a)-a. Therefore, exp(a+o(a)) = exp(a+ln(1+a)-a) = exp(a)*exp(ln(1+a))/exp(a) = 1+a. Doen't seem to be interpreted this way...
o(f(x)) means "something that tends towards 0 when x tends to 0 quicker than f(x)" --- hence, o(a) means something tending to 0 quicker than the a constant, which simply means it tends to 0 as x tends to 0. Here, 1/n tends to 0 as n tends to infinity, hence the usage.
Taylor expansion is a way to replace, in a vicinity of a point, a complex (but "regular") function by a polynomial expression, in such a way that the error tends to 0 quicker than the polynome itself. This leads to the expression
f(x) = f(x0) + f'(x0)(x - x0) + f"(x0) (x-x0)^2/2 + ... + f(n)(x0) (x-x0)^n/n! + o( (x-x0)^n )
provided of course different conditions, which here are met. It is a generalisation of the usual linearisation, using the tangent instead of the curve itself, which corresponds to n = 1 in the previous expression. the o(...) expression is in fact the most important, in a way, since it ensures that the error by using the polynomial part to compute the function is controlled, and the closer x is from x0, the smaller this error.
- Badges
- Science topic
- Similar topics
- Mathematical Sciences
- Algebra