I know that a polynomial in an indeterminate X ( say f), in order to be equal to another polynomial (for whatever value of X) has to share the same coefficients with it ( thus being itself). But what if we would like to try to write down the same polynomial into another form (other than the polynomial f itself) which doesn't have to be also a polynomial and can be used with the former one interchangeably. Would it be possible? If not, how can I prove it? P.s.: I'm 10th grade.

  • $\begingroup$ Well, sure. If you know logs and exponents, you could write $p(x)=\ln \,e^{p(x)}$. And if you are clever you could disguise this in all sorts of ways. Does that answer your question or did you have something more precise in mind? $\endgroup$ – lulu Sep 5 '17 at 19:15
  • $\begingroup$ Can you give an example, say, with $f(x)=x^2+2x+1$, of what you want to ask? Can I answer that $f(x)=\frac{x^3(x+1)^3}{x^3(x+1)}$? $\endgroup$ – Dietrich Burde Sep 5 '17 at 19:18
  • $\begingroup$ It would us to answer your question if you gave us some idea of what you wanted. For example, a silly thing to do would be to write $p(x)$ as $p(x)+\sin(x)-\sin(x)$. That is certainly a valid equality but it seems pointless. What sort of thing were you hoping for? $\endgroup$ – lulu Sep 5 '17 at 19:26
  • $\begingroup$ I see I wasn't quite on the point. I was actually referring to an example that is not constructed by cancellation ( like the ones above ). And if there's no one, I asked for a proof ( at least a starting point would be enough ). $\endgroup$ – Anonymus Sep 5 '17 at 21:11
  • $\begingroup$ If you know trigonometry, then you know that $\sin^2{x} = 1=\cos^2{x}$ so any expression involving $\sin^2{x}$ can be written in a form that does not involve $\sin^2{x}$. Are you asking if the same kind of thing is possible for polynomials? So instead of $3x^2-4x+1$ we have a different expression that has the same values for every $x$? $\endgroup$ – Jim H Sep 5 '17 at 21:40

A. Strictly speaking a polynomial is simply an ordered finite sequence of "coefficients" that can be operated on by rational operations. Two polynomials are equal if and only if they concides as sequences. Then the answer is negative, by definition.

B. If you mean by polynomial a polynomial function in one variable, that is, a map that linearly combines some (finitely many) positive powers of its variable by some coefficients then you are asking whether there exists an expression that can be used to define such a function other than the classical polynomial expression. (A polynomial expression is an expression where only finitely many positive powers of the indeterminate appear combined linearly). You can prove by yourself that there can be no two different polymonial expressions that define the same polynomial function. You can instead use espressions that involve other kind of elements in addition to those allowed by polynomial espressions: roots of variable or variable raised to the power of rational number (algebraic expressions), elementary functions of the variable like powers with irrational exponent, trigonometric functions, logarithms, ... (closed-form expressions), special functions, series or continued fractions (analytic expressions), derivatives, integrals, or limits (mathematical expressions). In this case the answer is affermative.

For instance: $$\frac{d}{dx}(x^3+ax^2+bx+c)$$ is a mathematical expression that is not a polynomial expression even though it defines a polynomial function (whose (unique) polynomial expression is $3x^2+2ax+b$)

As a much more complicated example next I show a simple polynomial function having the unique polynomial expression $x^2$ assigned with an analytic expression involving the product of two continued fractions (by Brouncker, 1655): $$\left\{(x-1)+\frac{1^2}{2(x-1)+\frac{3^2}{2(x-1)+\frac{5^2}{2(x-1)+...}}}\right\}\times\left\{(x+1)+\frac{1^2}{2(x+1)+\frac{3^2}{2(x+1)+\frac{5^2}{2(x+1)+...}}}\right\}$$

There are lots of other examples.

C. There are other way to think of a polynomial but I think that it is enough for now.

  • $\begingroup$ I see what are you coming from. But the asymptotical interpretation isn't accurate from my standpoint because the function will never be the same as x^2 no matter how close. But I appreciate the attempt though. $\endgroup$ – Anonymus Sep 6 '17 at 7:18

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