There is an equation:
$p(x) = (\alpha \cdot x + 1)^{3/2}$
Are there ways to convert this equation into a polynomial of integer order with an arbitrary highest degree (with even or odd choice), for example:
$p(x) = (\alpha \cdot x + 1)^{3/2} \xrightarrow{Transform} p(x) = c_0 + c_1 \cdot x^1+ ... + c_m \cdot x^m$
where $m$ - even or odd arbitrary degree of polynomial, and $c_i$ - polynomial coefficients.
Here we are not talking about approximating this equation by polynomials and not about expanding it into a Taylor series.
In general, this is a bad way for this equation, because with an increase in the $\alpha$ coefficient, the Runge phenomenon manifests itself more and more.
This is a fractional order system, the structure is very similar to the polynomial equation, so I thought that there should be ways to convert the fractional order system into a similar integer order system that describes the same curve on any interval as the original fractional order system.
Is there such a transformation?
EDIT:
I would like to specify the problem:
There is the following ratio:
$f(x) = \frac{(x+1)^4}{(\alpha \cdot x + 1)^{3/2}}$
It is necessary to eliminate the fractional degree from this ratio and at the same time observe the condition under which the order of the numerator should not exceed the order of the denominator.
I will try to show why squaring is not a very good solution in this case:
$f(x) = \frac{((x+1)^4)^2}{((\alpha \cdot x + 1)^{3/2})^2} = \frac{(x+1)^8}{(\alpha \cdot x + 1)^{3}}$
In this case, the degree of the numerator is greater than the order of the denominator.
EDIT №2:
Squaring helps only if the order of the denominator is greater than the order of the numerator from the very beginning. Therefore, my comment should not be taken as a complete rejection of this method.
It seems to me that there should be a more elegant way.
Unfortunately, if I understand your question correctly, there is no polynomial that suits your needs.
You are looking for some polynomial $P(x)$ so that $$P(x)=(\alpha x+1)^{3/2}$$ for all real $x$. Squaring this, we must certainly have that $$P(x)^2=(\alpha x+1)^3$$ for all $x$. However, now both sides are actually polynomials, so they must in fact be the same polynomial. However, the degree of the polynomial on the left side is $2\deg P$, while the degree of the right side is $3$ (unless $\alpha=0$). Since $3$ is not even, $2\deg P\neq 3$, and thus we have reached a contradiction.
If you are satisfied with an approximation on an interval, take a look at the Stone--Weierstrass theorem. It tells you that, given any continuous function on any interval, you can always approximate it as well as you desire by a polynomial.