테일러·매클로린 급수

• Solution 1: Since $f(x)$ is odd, $f(-x)=-f(x)$ for all $x$ in its domain. We plug this into our power series, then consider the even-indexed terms and the odd-indexed terms separately.

$$\begin{aligned} f(-x)&=-f(x)\\ \sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}(-x)^n&=-\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n\\ \sum_{n=0}^\infty\frac{f^{(2n+1)}(0)}{(2n+1)!}(-x)^{2n+1}+ \sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}(-x)^{2n}&= -\sum_{n=0}^\infty\frac{f^{(2n+1)}(0)}{(2n+1)!}x^{2n+1} -\sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}\\ -\sum_{n=0}^\infty\frac{f^{(2n+1)}(0)}{(2n+1)!}x^{2n+1}+ \sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}&= -\sum_{n=0}^\infty\frac{f^{(2n+1)}(0)}{(2n+1)!}x^{2n+1} -\sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}\\ \sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}&= -\sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}\\ 2 \sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}&=0\\ \sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}&=0 \end{aligned}$$

• Solution 2: Alternately, we could note the following:

• Since all derivative of $f(x)$ exist, all its derivatives are continuous.

• The derivative of an odd function is even, and the derivative of an even function is odd.

• So, the even-indexed derivatives of $f(x)$ are continuous, odd functions.

• Every continuous, odd function passes through the origin. That is, $f^{(2n)}(0)=0$.

• So, every term in the series is $0$.