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Motivation:

I have a few polynomials and am trying to find a representation for them in terms of special functions. I'm more interested in the techniques here, so I won't give any too particular example, only let's say I find that they look a bit like

$$p_1(x)=a_0+a_1x,~p_2(x)=b_0+b_1x,~p_2(x)=c_0+c_1x+c_2x^2,\dots$$

i.e. they stay the same degree for two steps of $n,$ before their degree increases by $1$. So $p_2(x)$ and $p_3(x)$ are quadratics, then $p_4(x)$ and $p_5(x)$ are cubics, and so on.

A natural first guess is something like

$$p_n(x)=\frac{Q_n(x)}{x^n},$$

where $Q_n(x)$ is something like a Hermite or Chebyshev or Legendre polynomial. Basically, if $Q_n(x)$ is something like a hypergeometric series that alternates between an odd and even function of $x,$ it at least forms that pattern even if the coefficients don't exactly match.

But, when I look at the differential equation polynomials of this type should satisfy, it still has three regular singular points, which could be shifted so that they are at $x=0,1,\infty,$ and they generally satisfy some differential equation which looks like

$$x(1+A_nx)\frac{d^2p}{dx^2}+(B_n+C_nx)\frac{dp}{dx}+D_np=0,$$

$A_n,\dots,D_n$ being some sequences depending on $n,$ and exactly what their values are depending on the special polynomial $Q_n(x)$ is.

So after some playing around, I expect that polynomials that look like

$$\frac{{_2}F_1(a,b,c;x)}{x^n}$$

should probably be expressible themselves as hypergeometric functions of some type.

Question:

Is there a natural kind of transformation that would fit the bill here, so that I can write

$${_2}F_1(a',b';c';x')x^n={_2}F_1(a,b;c;x)?$$

Or, is there some 'special form' of the arguments of ${_2}F_1(a,b;c;x)$ which tends to give polynomials which keep the same degree for two steps?

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The differential equation $$x(1+a x)\frac{d^2p}{dx^2}+(b+cx)\frac{dp}{dx}+d\,p=0$$ admits as general solution $$p(x)=C_1 a^{1-b} x^{1-b} \, _2F_1\left(\alpha_1,\alpha_2;2-b;-a x\right)+C_2 \, _2F_1\left(\alpha_3,\alpha_4;b;-a x\right)$$ where $$\alpha_1=\frac{a+c-2ab-\sqrt{(a-c)^2-4 a d}}{2a} \qquad \text{and} \qquad \alpha_2=\frac{a+c-2ab+\sqrt{(a-c)^2-4 a d}}{2 a}$$ $$\alpha_3=\frac{c-a-\sqrt{(a-c)^2-4 a d}}{2 a}\qquad \text{and} \qquad \alpha_4=\frac{c-a+\sqrt{(a-c)^2-4 a d}}{2 a}$$

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  • $\begingroup$ Is this something that is a well-known result, or is there some method to be aware of here? $\endgroup$ – Hobbyist Mar 3 at 22:23
  • $\begingroup$ It is well-known to Claude, as are many things that are not to me and you. $\endgroup$ – marty cohen Mar 3 at 22:53
  • $\begingroup$ @Hobbyist. One of my former PhD students worked this differential equation. $\endgroup$ – Claude Leibovici Mar 4 at 5:41

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