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 Question from Sourik, a student: Dear Expert, In my Amithabha Mitra and Shambhunath Ganguly's "A Text Book of Mathematics" I found the formula of log (1+x) where the base is e and x lies in between -1 and +1.As I want to learn Mathematics,I am not satisfied with the mere statement of the formula. Please help giving me the full proof. Thanking you, Sourik

I presume that the formula to which you refer is the Taylor series

$x - \frac{x^2}{2} + \frac{x^3}{3} -...+ \; (-1)^{n+1} \frac{x^n}{n} +...$

This can be derived from the power series for $\large \frac{1}{1+x}$

$1 - x + x^2 - x^3 ...$

by integrating. There is a theorem that power series can be integrated termwise, so that the convergence of the second series to $\large \frac{1}{1+x}$ on the specified interval implies the convergence of the first one to $log(1+x).$ This is nontrivial but can be found in any good elementary analysis text, or rigorous calculus text.

The second convergence can be justified formally as follows. By almost any convergence test you choose, $1 - x + \frac{x^2}{2} -...$ converges pointwise to something on (-1,1). Multiply it by (1+x) and all terms cancel except for the first: so what it converges to must be $\large \frac{1}{1+x}$.

Alternatively, you can invoke Taylor's Theorem directly and compute the successive iterated derivatives of $log(1+x);$ after the first, they are all just negative integer powers of x, so it's easy!

Good Hunting!
RD

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