Taylor polynomial of $\frac{1}{2-x}$ Can someone show how to find the Taylor polynomial of $\frac{1}{2-x}$?
I tried this: $\frac{1}{2-x}=\frac{1}{1-(x-1)}$ and then use that $ \ T_n(\frac{1}{1-x})=1+x+\dots +x^n.$ But this gives $1+(x-1)+\dots +(x-1)^n$ which is the wrong answer. Why?
 A: Here is how you find it about the point $x=0$.
$$ \frac{1}{2-x} = \frac{1}{2(1-x/2)} = \frac{1}{2}( 1+x/2+x^2/2^2+\dots\,. ). $$
Added: If you want to derive the Taylor series at $x=1$ then you can advance like this
$$  \frac{1}{2-x} = \frac{1}{1-(x-1)}. $$
Now let $t=x-1$ which gives
$$ T_n(\frac{1}{1-t}) = 1+t+\dots+t^n. $$
Substitute back $t=x-1$ gives what you have

$$T_n(\frac{1}{1-t}) = = 1+(x-1)+\dots+(x-1)^n. $$

A: Full detail was not given, but I expect that from $\frac{1}{1-(x-1)}$ you intended to produce the degree $n$ polynomial
$$1+(x-1)+(x-1)^2+\cdots +(x-1)^n.$$
This is indeed a Taylor polynomial for our function. In general, there is no such thing as the Taylor polynomial. We must specify how many terms the polynomial should have, and what $a$ we are expanding about. 
The polynomials of shape $\frac{1}{2}+\frac{x}{4}+\frac{x^2}{8}+\frac{x^3}{16}+\cdots$ are a special case of a Taylor polynomial for our function, often called a Maclaurin polynomial.
That said, expansion about $a=0$ is by quite a bit the most common, and it may be the intended interpretation of "Taylor polynomial" in this problem. One could even call it the default interpretation. However, it definitely is not the only possible interpretation. 
A: Hint:
Try this:
$$\frac{1}{2-x} = \frac{\frac{1}{2}}{1-\frac{x}{2}} = \frac{1}{2}\sum_{n=0}^\infty \left(\frac{x}{2}\right)^n = \sum_{n=0}^\infty 2^{-(n+1)}x^n$$
Your method gives the Taylor expansion around $x=1$ with radius of convergence of $|x-1| < 1$ so for $x = 0$ you have $|x-1| = |-1| = 1 \not < 1$ you are on the boundary of the radius of convergence and you could end up with a delicate case.
$$\frac{1}{2-x} = \frac{1}{1-(x-1)} = \sum_{n=0}^\infty (x-1)^n = \sum_{n=0}^\infty\sum_{k=0}^n \binom{n}{k}(-1)^{n-k}x^k \\ \stackrel{*}{=} \lim_{m\to\infty} \sum_{n=0}^m \binom{n}{0}(-1)^n + \sum_{n=1}^m \binom{n}{1}(-1)^{n-1}x + \ldots + \sum_{n=k}^m \binom{n}{k}(-1)^{n-k}x^k + \ldots$$
$(*)$ is alowed when we have convergence (wich we don't  at $x=0$ as we will see). Note that at $x=0$  the first sum is divergent and the other ones are $0$. So you can see that we don't have convergence for $x=0$ in the Taylor expansion around $x=1$.
Note that the Taylor expansion at $x=0$ is often called Maclaurin series. If that's what you wanted then the method at the top is the one to go, otherwise, your method is a correct Taylor expansion, but centered at $x=1$.
A: We want to find $a_0, a_1, a_2,\ldots$ so that  $$\frac1{2-x} = a_0 + a_1x + a_2x^2 + \ldots.$$
Multiply both sides by $2-x$: $$
\begin{align}
1 & = (2-x)a_0 + (2-x)a_1x + (2-x)a_2x^2 + \cdots\\
& = (2a_0-a_0x)+(2a_1x-a_1x^2) + (2a_2x^2 - a_2x^3) + \cdots \\
\end{align}$$
Combine like terms:
$$ 1 = 2a_0 + (2a_1-a_0)x + (2a_2 -a_1)x^2 + \cdots$$
Equate coefficients on both sides:
$$\begin{align}
2a_0 & = 1 \\
2a_1 - a_0 & = 0 \\
2a_2 - a_1 & = 0 \\
& \vdots
\end{align}$$
Now just solve the equations starting with the top one:
$$\begin{align}
a_0 &= \frac12 \\
a_1 &= \frac14\\
a_2 &= \frac18 \\
& \vdots
\end{align}$$
