Stepwise regression

In mathematics, Hölder's theorem states that the gamma function does not satisfy any algebraic differential equation whose coefficients are rational functions. The result was first proved by Otto Hölder in 1887; several alternative proofs have subsequently been found.

The theorem also generalizes to the q-gamma function.

Statement of the Theorem

There is no non-constant polynomial $P(x;y_0,y_1,\dots ,y_n)$ such that

$P(x;\mathrm{\Gamma }(x),{\mathrm{\Gamma }}^{\prime }(x),\dots ,{\mathrm{\Gamma }}^{(n)}(x))\equiv 0.$

where $y_0,y_1,\dots ,y_n,$ are functions of x, Γ(x) is the gamma function, and P is a polynomial in $y_0,y_1,\dots ,y_n,$ with coefficients drawn from the ring of polynomials in x. That is,

$P(x;y_0,y_1,\dots ,y_n)=\sum _{(a_0,a_1,\dots ,a_n)}{A}_{(a_0,a_1,\dots ,a_n)}(x)\cdot (y_0{)}^{a_0}\cdot (y_1{)}^{a_1}\cdot \dots \cdot (y_n{)}^{a_n}$

where $(a_0,a_1,\dots ,a_n)$ indexes all possible terms of the polynomial and $A_{(a_0,a_1,\dots ,a_n)}(x)$ are polynomials in x acting as coefficients of the polynomial P. The $A_{(a_0,a_1,\dots ,a_n)}(x)$ may be constants or zero.

For example, if $P(x;y_0,y_1,y_2)=x^2{y}_2+x{y}_1+(x^2-{\nu }^2)y_0$ then $A_{(0,0,1)}(x)=x^2$, $A_{(0,1,0)}(x)=x$ and $A_{(1,0,0)}(x)=(x^2-{\nu }^2)$ where ν is a constant. All the other coefficients in the summation are zero. Then

$P(z;f,f^{\prime },f^{″})=x^2{f}^{″}+x{f}^{\prime }+(x^2-{\nu }^2)f=0$

is an algebraic differential equation which, in this example, has solutions $f=J_{\nu }(x)$ and $f=Y_{\nu }(x)$, the Bessel functions of either the first or second kind. So

$P(x;J_{\nu }(x),{J_{\nu }}^{\prime }(x),{J_{\nu }}^{″}(x))\equiv 0.$

and therefore both $J_{\nu }(x)$ and $Y_{\nu }(x)$ are differentially algebraic (also algebraically transcendental). Most of the familiar special functions of mathematical physics are differentially algebraic. All algebraic combinations of differentially algebraic functions are also differentially algebraic. Also, all compositions of differentially algebraic functions are differentially algebraic. Hölder's Theorem simply states that the gamma function, Γ(x) is not differentially algebraic and is, therefore, transcendentally transcendental.

Proof

Assume the existence of P as described in the statement of the theorem, that is

$P(x;\mathrm{\Gamma }(x),{\mathrm{\Gamma }}^{\prime }(x),\dots ,{\mathrm{\Gamma }}^{(n)}(x))\equiv 0.$

with

$P(x;y_0,y_1,\dots ,y_n)=\sum _{(a_0,a_1,\dots ,a_n)}{A}_{(a_0,a_1,\dots ,a_n)}(x)\cdot (y_0{)}^{a_0}\cdot (y_1{)}^{a_1}\cdot \dots \cdot (y_n{)}^{a_n}$

Also, assume that P is of lowest possible order/degree. This means that all the coefficients $A_{(a_0,a_1,\dots ,a_n)}$ have no common factor of the form (x − γ) and so P is not divisible by any factor of (x − γ). It also means that P is not the product of any two polynomials of lower order/degree.

$\begin{array}{rl}& P(x+1;\mathrm{\Gamma }(x+1),{\mathrm{\Gamma }}^{(1)}(x+1),\dots ,{\mathrm{\Gamma }}^{(n)}(x+1))=\\ & =P(x+1;x\mathrm{\Gamma }(x),{[x\mathrm{\Gamma }(x)]}^{(1)},{[x\mathrm{\Gamma }(x)]}^{(2)},\dots ,{[x\mathrm{\Gamma }(x)]}^{(n)})\\ & =P(x+1;x\mathrm{\Gamma }(x),x{\mathrm{\Gamma }}^{(1)}(x)+\mathrm{\Gamma }(x),x{\mathrm{\Gamma }}^{(2)}(x)+2{\mathrm{\Gamma }}^{(1)}(x),\dots ,x{\mathrm{\Gamma }}^{(n)}(x)+n{\mathrm{\Gamma }}^{(n-1)}(x))\end{array}$

and so we can define a second polynomial, Q, defined by the transformation

$\begin{array}{r}Q(x;y_0,y_1,\dots ,y_n)=P(x+1;x{y}_0,x{y}_1+y_0,x{y}_2+2y_1,x{y}_3+3y_2,\dots ,x{y}_n+n{y}_{(n-1)})\end{array}$

and $Q(x;\mathrm{\Gamma }(x),{\mathrm{\Gamma }}^{\prime }(x),\dots ,{\mathrm{\Gamma }}^{(n)}(x))=0$ is also an algebraic differential equation for Γ(x). This substitution forces the highest order/degree term of Q to be

$x^{a_0+a_1+\dots +a_n}{A}_{(h_0,h_1,\dots ,h_n)}(x+1)\cdot (y_0{)}^{h_0}\cdot (y_1{)}^{h_1}\cdot \dots \cdot (y_n{)}^{h_n}$

where $(h_0,h_1,\dots ,h_n)$ are the exponents of the term of P with highest order/degree. This indicates that Q and P both have the same order/degree and an application of the Euclidean Algorithm to Q and P shows that P must divide Q. If not, there would be a remainder and that would mean P was not of minimal order/degree. Call R(x) the ratio between P and Q:

$\begin{array}{rl}Q(x;y_0,y_1,\dots ,y_n)& =P(x+1;x{y}_0,x{y}_1+y_0,x{y}_2+2y_1,x{y}_3(x)+3y_2,\dots ,x{y}_n+n{y}_{(n-1)})\\ & =R(x)P(x;y_0,y_1,\dots ,y_n)\end{array}$

and consider the two leading terms, which must be equal:

$\begin{array}{rl}R(x)A_{(h_0,\dots ,h_n)}(x)\cdot (y_0{)}^{h_0}\cdot \dots (y_n{)}^{h_n}& =x^{h_0+\dots +h_n}{A}_{(h_0,\dots ,h_n)}(x+1)\cdot (y_0{)}^{h_0}\cdot \dots (y_n{)}^{h_n}\\ R(x)A_{(h_0,\dots ,h_n)}(x)& =x^{h_0+\dots +h_n}{A}_{(h_0,\dots ,h_n)}(x+1)\end{array}$

Consider γ to be a zero of R(x) and $\gamma \ne 0$. Then substituting γ into

$P(\gamma +1;\gamma y_0,\gamma y_1+y_0,\gamma y_2+2y_1,\gamma y_3+3y_2,\dots ,\gamma y_n+n{y}_{n-1})=0$

This last equality indicates that $(z-(\gamma +1))$ is a factor of P, contradicting the assumption that P was of minimal order/degree. Therefore the only root of R(x) is 0 and we can take $R(x)=x^n$, although we will not need to for this version of the proof. Therefore, with $\gamma =0$

$\begin{array}{rl}P(\gamma +1;\gamma y_0,\gamma y_1+y,\gamma y_2+2y_1,\gamma y_3+3y_2,\dots ,\gamma y_n+n{y}_{n-1})& =P(1;0,y_0,2y_1,3y_2,\dots ,n{y}_{n-1})\\ & =P(1;0,z_1,z_2,z_3,\dots ,z_{n-1})\\ & =0.\end{array}$

But if $P(1;0,z_1,z_2,z_3,\dots ,z_{n-1})=0$ then our earlier expression

$\begin{array}{rl}P(x+1;0,x{y}_1+y_0,x{y}_2+2y_1,x{y}_3(x)+3y_2,\dots ,x{y}_n+n{y}_{(n-1)})& =R(x)P(x;0,z_1,\dots ,z_n)\\ \end{array}$

tells us

$P(m;0,z_1,z_2,z_3,\dots ,z_{n-1})=0$

for any natural number m. The only way this is possible is if P is divisible by $y_0$ contradicting the assumption that P was of minimal order/degree. Therefore, no such P exists and Γ(x) is not differentially algebraic.

References

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