1
$\begingroup$

The following is something one of my teachers and I discussed but did not make any progress on, along with my own generalizations.

Let $p(x)=x^\alpha$ with $\alpha \in \mathbb{R}^+$. If we let $q(x) = (p(x))^\frac{a}{\alpha}$ with $a \in \mathbb{N}$, we have a polynomial of degree $a$.

Now let $p(x)=x^{\alpha_1} + x^{\alpha_0}$, $\alpha_1 > \alpha_0 > 0$. Does there exist a function $f(a,b,c)$ such that $q(x)=(p(x))^{f(a,\alpha_0,\alpha_1)}$ is a polynomial of degree $a$ with only integer powers? I have been able to solve cases for rational exponents which leads me to believe that such a function exists for rational $\alpha_0,\alpha_1$ (and possibly, that it depends only on $a$ and $\alpha_1$) but have been unable to find real power $\beta > 0$ such that $q(x) = (p(x))^{\beta}$ is a polynomial for non-trivial irrational $\alpha_1$ or $\alpha_0$, leading me to believe no such function exists when either power is irrational.

My question: Does there exist a function $f:\mathbb{R}^3 \to \mathbb{R}$ such that $(x^{\alpha_1} + x^{\alpha_0})^{f(a,\alpha_0,\alpha_1)}$ is a polynomial of degree $a$? How might I go about finding it, or disproving its existence? If no function exists for all powers, what are the (preferably non-trivial, because I can think of many) conditions the powers need to satisfy for one to exist? Further, might a similar function exist for $p(x) = x^{\alpha_k} + x^{\alpha_{k-1}} \dots x^{\alpha_{0}}$, with similar restrictions on $\alpha_k, \alpha_{k-1} \dots \alpha_0$?

Edit 1: quick afterthought, can this be proven via induction, where one inducts on the number of terms? I don't think so, but never a bad idea to try.

Edit 2: Perhaps one can forget about the function and simply seek to prove: given $p(x)=x^{\alpha_{k}} + x^{\alpha_{k-1}} \dots + x^{\alpha_{0}}$ with $\alpha_{k} > \alpha_{k-1} \dots > \alpha_{0} > 0$, for any $a \in \mathbb{N^+}$ there exists a $\beta \in \mathbb{R}$ ($\mathbb{C}$?) such that $(p(x))^\beta = q(x)$ where $q(x)$ is a polynomial of degree $a$. One can possibly induct over $k$ assuming that for all such $p(x)$ with more than $0$ and no more than $k$ terms that our proposition holds. The case for the monomial is trivial, and some clever algebra might help to prove the case for the 'polynomial' with $k+1$ terms using our assumption.

$\endgroup$
1
$\begingroup$

No, there is no such function for irrational $\alpha$ that will work for non-monomial $p$. One way to prove it is to find the Taylor series. The outline of this method is:

Consider $q(x) = (x^a + x^b)^c$, where $a$ and the ratio $b/a$ are irrational (pardon my simplification of the notation to avoid having to do a lot of greek letters and subscripts). Since we want $q$ to be well-defined where we are taking the Taylor series, we'll do it at $x = 1$.

  • $q(1) = 2$
  • $q'(x) = c(x^a + x^b)^{c-1}(ax^{a-1} + bx^{b-1})$, so $q'(1) = c2^{c-1}(a+b)$
  • $q''(x) = c(c-1)(x^a + x^b)^{c-2}(ax^{a-1} + bx^{b-1}) + c(x^a + x^b)^{c-1}(a(a-1)x^{a-2} + b(b-1)x^{b-2}$, so $q''(1) = c(c-1)2^{c-2}(a+b) + c2^{c-1}(a(a-1) + b(b-1))$
  • ...

This should be enough to see what happens: If $a$ is not an integer, then the progression $a(a-1)...$ will never be $0$, and while it takes some number theory to fully demonstrate it, if $a$ is irrational, the remaining terms will never cancel it out.

The Taylor series of a polynomial is the polynomial itself, which terminates (and thus is guaranteed to converge to the polynomial). Since $q$ has a non-terminating Taylor series, it cannot be a polynomial.

Of course, where the going got tough here, I just waved my hands. But this is the idea.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.