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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.

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    $\begingroup$ 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. $\endgroup$
    – Calvin Lin
    Jun 16 '21 at 4:40
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    $\begingroup$ " 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. $\endgroup$
    – fleablood
    Jun 16 '21 at 4:54
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    $\begingroup$ 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. $\endgroup$
    – fleablood
    Jun 16 '21 at 4:57
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    $\begingroup$ "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.) $\endgroup$
    – fleablood
    Jun 16 '21 at 5:08
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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 $1/x = 0$.

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