Calculate the first nonzero term in a Taylor Series expansion for x^3 about x = 0. What can you say about other nonzero terms in the expansion?

Question

anonymous · Answer

$$\large\begin{align*}
f(x)&=x^3\
f'(x)&=3x^2\
f''(x)&=6x\
f'''(x)&=6\
f^{(n)}(x)&=0&\text{for all }n\ge4
\end{align*}$$
Evaluated at $x=0$,
$$\large\begin{align*}
f(0)&=0\
f'(0)&=0\
f''(0)&=0\
f'''(0)&=6\
f^{(n)}(x)&=0&\text{for all }n\ge4
\end{align*}$$
The first (and only) non-zero term in the expansion is the function itself, since
$$f(x)\sim \frac{6}{3!}(x-0)^3=x^3$$
It makes sense that the expansion is the same as the given function because the given function is itself a polynomial.