If one assumes the function f(x) has a power series representation

$$\ f(x) = \sum_1^\infty C_n (x-5)^n$$

what would you say the formula for $\ C_n$ is here?


How does one find the Taylor series for the following function centered at 4 if

$$\ f^n(4) = \frac{(-1)^nn!}{3^n(n+1)} $$

Also.. how do you find the interval of convergence for this??


One last question about series.. What would the Maclaurin Series and interval of convergence be for these two functions?

$$ F(x) = \frac{x}{4+9x^2} $$


$$ F(x) = ln(4-x) $$

I've attempted these two, but I'd like to have something to check my work against.

Thanks for any assistance! I've been looking at these for several hours along with some other things so I've just gotten a little burnt out on them at the moment.


To general form of the Taylor expansion of a function $f$ around $x_0$ is $$ f(x) = \sum_{k=0}^\infty c_k(x-x_0)^k \quad\text{where}\quad c_k = \frac{f^{(k)}(x_0)}{k!} \text{.} $$ This is a power series, and as such it's radius of convergence $r$ obeys $$ \frac{1}{r} = \limsup_{k\to\infty } \sqrt[k]{c_k} \text{.} $$

Note, however, that the radius of convergence only tells you where the Taylor series converges, not where it faithfully represents the original function. For example, for $f(x)=e^{-\frac{1}{x^2}}$, you have $f'(x) = \frac{2}{x^3}e^{-\frac{1}{x^2}}$, $f''(x) = \frac{-4}{x^6}e^{-\frac{1}{x^2}} + \frac{-6}{x^3}e^{-\frac{1}{x^2}}$ and so on. In general, $f^{(n)}(x) = p_n\left(\frac{1}{x}\right)e^{-\frac{1}{x^2}}$ for some polynomial $p_n$, and since as $x \to 0$, $e^{-\frac{1}{x^2}} \to 0$ faster than any polynomial, you have $$ c_n = \frac{f^{(n)}(0)}{n!} = 0 \text{,} $$ i.e. the "Taylor expanson" of $e^{-\frac{1}{x^2}}$ around $0$ is identically zero! It's radius of convergence is $r = \infty$, yet the taylor expansion agrees with the original function only at $x=0$.

So, for your problems

1: Compare to the general formula above to find $x_0$, then the formula for the $c_k$ follows trivially

2: Just put this into the general formula above.

3: This is ambiguous - the Taylor series depends on with point $x_0$ you pick. So pick one, say $x_0 = 0$, and again use the general formula above. You'll need to find all the derivatives of $F$ - the best way to do that is usually to compute the first few, and see if a pattern emerges.

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