Show that,
\[\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots\]
Last edited by Sudharaka; January 20th, 2013 at 22:22. Reason: Used LaTeX to write equations.
Can we write this as a Taylor's series as f(x) = Log(1+x), then f'(x)=1\1+x so on.
Of course you can. But using that method you only obtain the Taylor series of $f(x)=\log (1+x)$, that is $\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{x^n}{n}$. To prove that $\log (1+x)=\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{x^n}{n}$ in $(-1,1)$ (also in $x=1$) you need to verify that the remainder of the Taylor series converges to $0$.
If you want to use the long method, remember that a Maclaurin series for a function is given by $\displaystyle \begin{align*} f(x) = \sum_{n = 0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n \end{align*}$
So evaluating the derivatives gives
$\displaystyle \begin{align*} f(x) &= \ln{(1 + x)} \\ f(0) &= 0 \\ f'(x) &= \frac{1}{1 + x} \\ f'(0) &= 1 \\ f''(x) &= -\frac{1}{(1 + x)^2} \\ f''(0) &= -1 \\ f'''(x) &= \frac{2}{(1 +x)^3} \\ f'''(0) &= 2 \\ f^{(4)}(x) &= -\frac{3!}{(1 + x)^4} \\ f^{(4)}(0) &= -3! \\ f^{(5)}(x) &= \frac{4!}{(1 + x)^5} \\ f^{(5)}(0) &= 4! \\ \vdots \end{align*}$
So substituting these in gives
$\displaystyle \begin{align*} \ln{(1 + x)} &= \frac{0}{0!}x^0 + \frac{1}{1!}x^1 + \frac{(-1)}{2!}x^2 + \frac{2}{3!}x^3 - \frac{3!}{4!}x^4 + \frac{4!}{5!}x^5 - \dots + \dots \\ &= x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \dots + \dots \end{align*}$
Now, we have to prove the Taylor's remainder $ \displaystyle R_n(x)$:
$$R_n(x)=\dfrac{f^{(n+1)}(\xi)}{(n+1)!}x^n=\dfrac{(-1)^n}{(1+\xi)^{n+1}}\dfrac{x^{n+1}}{(n+1)!}\quad (\xi \mbox{ between }0\mbox{ and }x)$$
has limit $ \displaystyle 0$ for $ \displaystyle x\in (-1,1)$ as $ \displaystyle n\to \infty$. For that reason is better to use the series expansion of $ \displaystyle 1/(1+x)$.
Taking the derivative of the MacLaurin series gives you
$1 -x +x^2 - x^3 + x^4 + \ldots$
Since this is a geometric series with ratio $-x$, it equals $\frac{1}{1 + x}$ when x is in $(-1, 1)$.
This shows the expression $\ln(1+x)$ and its MacLaurin expansion to have the same derivative over $(-1, 1)$, which means they are equal within a constant. And, since they are equal at $x=0$, this constant is zero.
If my reasoning is correct, this is simpler than proving the limit of the Taylor remainder.
The logic and context are not the same, as I was answering the question whether you can use the Taylor series. In general, you prove the validity of the Taylor expansion over a given interval by proving the Taylor reminder tends to zero as n goes to infinity. But here, because we can bring out a geometric series, as Fernando Revilla does in the initial answer, we have a simpler alternative. It is actually an exceptional case that is worth noting.