Polynomial equivalent of $\frac{x+1}{x-1}$ Is there a way to express $\frac{x+1}{x-1}$ as a polynomial? I have tried:$$\frac{x+1}{x-1} = Ax+B$$ $$(Ax+B)\cdot(x-1)=x+1$$
But such a polynomial seems to contradict its own existence.
Is there any way to express this as a polynomial or is it impossible? Is there maybe a different method for it? A polynomial with negative terms? Maybe a taylor polynomial? Can I try substitute $x$ with $x+1$?
I would really appreciate any help with this.
 A: If you allow infinite expression like Taylor series and Laurent series into your definition of polynomial, then you can express the function in such a way.

Notice that
$$
\frac{x+1}{x-1} = \frac{2 +(x-1)}{x-1} = \frac{2}{x-1} +1 = \boxed{2(x-1)^{-1} +1 }\tag{1}
$$
The above equation corresponds to the Laurent series at $x=1$. You can see it as a polynomial in $x-1$ with a negative exponent. You can also see that the above expression is well defined for all $x \neq 1$.
Recalling the formula of a geometric series we know that $\frac{1}{x-1} = -\sum_{n = 0}^{\infty} x^n$ which converges for $|x| <1$. With this in mind, another way to express your function is
$$
\frac{x+1}{x-1} =  1+2\frac{1}{x-1}=1-2\sum_{n = 0}^{\infty} x^n =\boxed{ -1-2\sum_{n = 1}^{\infty} x^n, \quad \forall |x|<1} \tag{2}
$$
The above equation is the Taylor series centered at $x =0$ of your function.
Lastly, if you want another infinite polynomial that converges when $|x|>1$ we can do the following
$$
\frac{x+1}{x-1} = 1-\frac{2}{x}\frac{1}{\left(\frac{1}{x}-1\right)} =  1+\frac{2}{x}\sum_{n = 0}^{\infty} \left( \frac{1}{x}\right)^n = \boxed{1+2\sum_{n = 1}^{\infty} x^{-n},\quad \forall |x|>1} \tag{3}
$$
Since we applied the geometric series formula to $1/x$, by saying that $|1/x| <1$ this then implies that $|x| >1$, obtaining the desired convergence. This last equation corresponds to the Laurent expansion of your function at $x = \infty$, or at $1/x = 0$.
