Example Show that the following MacLaurin expansion holds．

$$\log(1+x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \cdots + (-1)^{n-1}\frac{x^{n}}{n} + \cdots , \ (-1 < x \leq 1))$$

Answer Suppose that $f(x)$ is differentiable for infinitely many times．Now we consider expressing $f(x)$ using polynomials．In other words,

$$f(x) \approx a_{0} + a_{1}x + a_{2}x^2 + a_{3}x^3 + \cdots + a_{n}x^{n}$$

If $f(x) = a_{0} + a_{1}x + a_{2}x^2 + a_{3}x^3 + \cdots + a_{n}x^{n}$, then

$$f'(x) = a_{1} + 2a_{2}x + 3a_{3}x^2 + \cdots + na_{n}x^{n-1}$$

Thus, $f'(0) = a_{1}$ and similarly, $f''(0) = 2a_{2}$, $f'''(0) = 3! a_{3} ,\ldots f^{(n)}(0) = n!a_{n}$. From this, we have

$$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^{n}$$

But this can not be true. For if we take $f^{(n+1)}(0)$, then it does not have to be 0. But the right-hand side is 0.． Then to make the equality hold，we put the remainder term.

$$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots + \frac{f^{(n-1)}(0)}{(n-1)!}x^{n-1} + R_{n}$$

Here, $R_{n} = \frac{f^{(n)}(\theta)}{n!}x^{n} \rightarrow 0\ (n \to \infty)$, $0 < \theta < 1$. This is called MacLaurin's expansion.

Now let $f(x) = \log(1+x)$. Then $f'(x) = \frac{1}{1+x} = (1+x)^{-1}$, $f''(x) = -(1+x)^{-2}, f'''(x) = 2(1+x)^{-3}$. By the induction, we have $f^{(n)}(x) = (-1)^{n-1}(n-1)!(1+x)^{-n}$．

We finally show that $\lim_{n \to \infty} \vert R_{n}\vert = 0$．

$$\vert R_{n}\vert = \vert\frac{f^{(n)}(\theta)}{n!}x^{n}\vert = \vert\frac{(-1)^{n-1}(n-1)!(1+\theta)^{-n}}{n!}x^n\vert$$

$$\leq \vert\frac{x^n}{n(1+x)^{n}}\vert \rightarrow 0 \ (n \to \infty)$$