Question

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?

Answer 1

\[\large\begin{align*}
f(x)&=x^3\\
f'(x)&=3x^2\\
f''(x)&=6x\\
f'''(x)&=6\\
f^{(n)}(x)&=0&\text{for all }n\ge4
\end{align*}\]
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.