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.
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 $1/x = 0$.