I'm having difficulty calculating the Taylor series of $\frac{1}{1-x}$ about $a=3$, and was wondering if anyone on here could help me out.
Here's what I've tried so far:
Attempt 1 - Take $y=x-3$ and rearrange as $x=y+3$.
Then do $\frac{1}{1-x}=\frac{1}{1-(y+3)}=\frac{1}{-y-2}=\frac{-1}{y+2}=\frac{-1}{1+(y+1)}$.
Now, knowing $\frac{-1}{1+x}=\sum_{n=0}^{\infty} (-1)^{n+1}x^n$, I write $\frac{-1}{1+(y+1)}=\sum_{n=0}^{\infty} (-1)^{n+1}(y+1)^n$.
We change now and get $\sum_{n=0}^{\infty} (-1)^{n+1}(y+1)^n=\sum_{n=0}^{\infty} (-1)^{n+1}((x-3)+1)^n$.
And then simplify and get our final result which is $\sum_{n=0}^{\infty} (-1)^{n+1}(x-2)^n$.
Attempt 2 - Recall first the formal definition of a Taylor series centered about $x=a$;
the formal definition is as I remember $f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$.
Take $f(x)=\frac{1}{1-x}$.
Now, I see that $f'(x)=\frac{1}{(1-x)^2}$, $f''(x)=\frac{2}{(1-x)^3}$, ..., $f^{(n)}(x)=\frac{n!}{(1-x)^{n+1}}$.
Replacing $x$ as $a$ I get now $f^{(n)}(a)=\frac{n!}{(1-a)^{n+1}}$.
Taking $a=3$, we write $f^{(n)}(a)=\frac{n!}{(1-a)^{n+1}}=f^{(n)}(3)=\frac{n!}{(1-3)^{n+1}}=f^{(n)}(3)=\frac{n!}{(-2)^{n+1}}$.
By the formal definition we now write $f(x)=\sum_{n=0}^{\infty} \frac{\frac{n!}{(-2)^{n+1}}}{n!}(x-3)^n$.
Simplifying the above we get the end result $f(x)=\sum_{n=0}^{\infty} \frac{(x-3)^n}{(-2)^{n+1}}$.
My attempts gave me two completely different answers. What gives? Is there something I'm doing wrong in either attempt? Is there something I'm doing wrong in both?
Help regarding this issue would be greatly appreciated. Thanks in advance.
Your first attempt is wrong, because after you do the substitution $y=x-3$, you should aim at a power series about $0$, not about $-1$.
Your second attempt is corect (but I would put $(-1)^n$ in the numerator and $2^n$ in the denominator).