This is an exercise in Stewart's Calculus (Exercise 19, Section 11.10 Taylor and Maclaurin Series):
Find the Taylor series for $f(x)$ centered at the given value of a. [Assume that f has a power series expansion. Do not show that $R_n(x) \to 0.$ Also find the associated radius of convergence.
Here $f(x)=x^6 – x^4 + 2$ and $a=-2$.
I'm having trouble finding a general formula of this Taylor series and therefore, also having problems finding the radius of convergence since I can't perform the ratio test.
Here is what I know:
\begin{align}
f'(x) = 6x^5 – 4x^3,\quad &f''(x) = 30x^4 – 12x^2,\\
f'''(x) = 120x^3 – 24x,\quad &f^{(4)}(x) = 360x^2 – 24,\\
f^{(5)}(x) = 720x,\quad &f^{(6)}(x) = 720.
\end{align}
And at $a=-2$,
\begin{align}
f(-2) = 50,\quad &f'(-2) = -160,\\
f''(-2) = 432,\quad &f'''(-2) = -912,\\
f^{(4)}(-2) = 1416,\quad &f^{(5)}(-2) = -1440,\\
f^{(6)}(-2) = 720.\quad &
\end{align}
I'm having trouble finding the general formula for each term. Without it, how am I supposed to find the radius of convergence?
Added:
So the general term I have for the n-th derivative of $f$ is:
$$f^{(n)}(x) = \frac{6!x^{6-n}}{(6-n)!}$$
So far the general term I have for the Taylor Series is:
$$\sum_{n=0}^{\infty} \frac{6! 2^{6-n}}{(6-n)!n!}(x+2)^n$$
I can see why the radius of convergence is $\infty$: because for any $x$the series converges.
But how do I show this formally? Can I use the ratio test?
Best Answer
Your general form of the derivative is wrong. It is valid only up to $n=6$. After that it is $0$. So your sum consist only of terms up to $(x+2)^6$. You therefore have a finite sum, not an infinite number of terms, neither of which diverges.