# 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.

• There is no such polynomial (with finitely many terms). If one existed, substituting in $x = 1$ would lead to a contradiction. However, if you allow for a power series (polynomial with infinite degree), then yes you can use the taylor polynomial. Jun 16 '21 at 4:40
• " A polynomial with negative terms?" If that is acceptable terminology (it isn't to me) then $\frac {x+1}{x-1}=1+\frac 2{x-1}= 1+\frac {2 -\frac 2x + \frac 2x}{x-1} = 1+2x^{-1} + \frac {\frac 2x}{x-1}=1+2x^{-1} + \frac {\frac 2x -\frac 2{x^2}+\frac 2{x^2}}{x-1} = 1 + 2x^{-1} + 2x^{-2} + 2x^{-3} + .......$ Which is a polynomial with an infinite number of negative terms. In my opinion that's an oxymoron but ... If you are allowed to make up your terms why not. Jun 16 '21 at 4:54
• Shouldn't downvote. It's a legitimate question (where the answer is, no, it isn't a polynomial) and the OP thought about things and came up with legitimate observations. Someone ought to answer why "rational expressions" (which this is) will not be a polynomial. Jun 16 '21 at 4:57
• "But such a polynomial seems to contradict its own existence" It does indeed! If $\frac{x+1}{x-1}=\sum_{k=0}^n a_k x^k$ then $x+1= (x-1)(\sum_{k=0}^n a_k x^k)= a_nx^{n+1} + \sum_{k=1}^n (a_k-a_{k-1})x^k - a_0$. But we assume it is an $n$ degree polynomial so $a_n\ne 0$ and $a_nx^{n+1} = x$ so $n=0$ and .... it just won't work. So no such finite non-negative degree polynomial can exist. (Also as Calvin Lin pointed out, if it were $P(x) =\sum_{k=0}^n a_k x^k$ then $P(1) = \sum a_k \in \mathbb R$. But $\frac {1+1}{1-1}$ is undefined.) Jun 16 '21 at 5:08

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 $$1/x = 0$$.