How do I formally show the radius of convergence of the Taylor series of $f(x)=x^6 - x^4 + 2$ at $a=-2$?

Question

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 $a=-2$.

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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?

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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?

Answer 1

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.

Answer 2

The Taylor series around $a$ is simply given by $$\begin{align}f(a+x)&=(a+x)^6-(a+x)^4+2\\&=(a^6-a^4+2)+(6a^5-4a^3)x+(15a^4-6a^2)x^2+(20a^3-4a)x^3+(15a^2-1)x^4+6ax^5+x^6\end{align}$$ or $$f(x)= (a^6-a^4+2)+(6a^5-4a^3)(x-2)+(15a^4-6a^2)(x-a)^2+(20a^3-4a)(x-a)^3+(15a^2-1)(x-a)^4+6a(x-a)^5+(x-a)^6$$

and, as almost all coefficients are $=0$, converges for all $x$.

Answer 3

Sticking to the definition helps. Recall that the Taylor series of $f$ centered at a given value of $a$ is by definition

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

Now, given the expression of $f$, namely, $f(x)=x^6-x^4+2$, and value of $a=-2$, all you need to do is finding ${f^{(n)}(-2)}$ for each $n=0,1,2,\cdots$. You have already found the values of $f^{(n)}(-2)$ for $n=0,1,\cdots,6$, which gives you the first seven terms in the Taylor series: $$ f(-2)+\frac{f'(-2)}{1!}(x+2)+\frac{f''(-2)}{2!}(x+2)^2+\cdots+\frac{f^{(6)}(-2)}{3!}(x+2)^2.\tag{2} $$ On the other hand, $f^{(n)}(a)=0$ for any integer $n>6$. Hence, by (1), the Taylor series for $f$ at $a=-2$ is $$ f(-2)+\frac{f'(-2)}{1!}(x+2)+\frac{f''(-2)}{2!}(x+2)^2+\cdots+\frac{f^{(6)}(-2)}{3!}(x+2)^2+0+0+0+\cdots\tag{3} $$ which is a finite sum, and of course it is convergent for every $x$.

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Note that (3) can be written formally as a power series, which seems to be what you were looking for, as follows: $$ \sum_{n=0}^\infty c_n(x+2)^n\quad\textrm{where } c_0=f(-2),\ c_1=\frac{f'(-2)}{1!},\cdots, c_6=\frac{f^{(6)}(-2)}{6!},\ c_7=c_8=\cdots =0.\tag{4} $$ You do not need Root Test or Ratio Test to get the radius of convergence but only need to look at the definition of the "radius of convergence". See the definition below Theorem 4 of Section 11.8 in Stewart's Calculus (your textbook), in particular case (ii):