I'm slightly stuck on a homework question. I'm supposed to:
(a) Find a power series representation for the function $f(x) = \frac{1}{x} $ with the center of the series at $a = 1$ by realizing that $f$ is the sum of a convergent geometric series. Then, find the series interval and radius of convergence using the established fact that a geometric series, $ \sum_{n=1}^\infty ar^{n-1}$, converges when $|r| < 1$.
(b) Find the Taylor series expansion of $f(x) = \frac{1}{x} $ about $a = 1$, making sure to show steps involved in finding the coefficients $c_n = \frac{f^n(a)}{n!}$. Then, using the ratio test, find the series interval and radius of convergence.
So far, this is what I've figured out so far:
(a) Because it is given that $f$ is the sum of a geometric series, and we know that $\sum_{n=1}^\infty ar^{n-1} $ sums to $\frac{a}{1-r}$ when $|r| < 1$, we can put $f$ in the form $f(x) = \frac{a}{1-r} = \frac{1}{1-1+x} = \frac{1}{1-(1-x)} = \sum_{n=1}^\infty 1(1-x)^{n-1} = \sum_{n=0}^\infty 1(1-x)^n$ .
(b) I found the Taylor series for $f(x) = \frac {1}{x}$ about $a=1$ and ended up with $$\frac{1}{x} = 1+(-1)+\frac{2}{2!}(x-1)^2+\frac{-6}{3!}(x-1)^3+\frac{24}{4!}(x-1)^4+… $$$$= 1-(x-1)+(x-1)^2-(x-1)^3+(x-1)^4+…$$
I know I still have to figure out a very large chunk of these problems. Any help is greatly appreciated!
Best Answer
hint
To expand $ f(x) $ around $ x=a$, You define $g(x)=f(x+a)$ and you expand $ g(x) $ around $ x=0 $.
You get $$g(x)=P_n(x) + x^n\epsilon_1(x)$$ with $$\lim_0\epsilon_1(x)=0$$ then you will have
$$f(x)=g(x-a)=P_n(x-a) + (x-a)^n\epsilon_2(x)$$ with $$\lim_a\epsilon_2(x)=0$$
In your case, $$g(x)=\frac{1}{1+x}$$ $$=1-x+x^2-x^3+...+(-1)^nx^n+x^n\epsilon_1(x)$$