Hölder's theorem explained

In mathematics, Hölder's theorem states that the gamma function does not satisfy any algebraic differential equation whose coefficients are rational functions. This result was first proved by Otto Hölder in 1887; several alternative proofs have subsequently been found.^[1]

The theorem also generalizes to the

q -gamma function.

Statement of the theorem

For every

n \in \N _{0,} there is no non-zero polynomial P \in \Complex [X;Y _{0,Y} _{1,\ldots,Y} _{n]} such that$$\backslash forall\; z\; \backslash in\; \backslash Complex\; \backslash setminus\; \backslash Z\; \_:\; \backslash qquad\; P\; \backslash left(z;\backslash Gamma(z),\backslash Gamma\text{'}(z),\backslash ldots,(z)\; \backslash right)\; =\; 0,$$where \Gamma is the gamma function.

For example, define

P \in \Complex [X;Y _{0,Y} _{1,Y} _{2]} by $$P\; ~\; \backslash stackrel\; ~\; X^2\; Y\_2\; +\; X\; Y\_1\; +\; (X^2\; -\; \backslash nu^2)\; Y\_0.$$

Then the equation$$P\; \backslash left\; (z;f(z),f\text{'}(z),f$$(z) \right) = z^2 f(z) + z f'(z) + \left (z^2 - \nu^2 \right) f(z) \equiv 0is called an algebraic differential equation, which, in this case, has the solutions

f = J _{\nu} and f = Y _{\nu} — the Bessel functions of the first and second kind respectively. Hence, we say that J _{\nu} and Y _{\nu} 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 differentially algebraic. Furthermore, all compositions of differentially algebraic functions are differentially algebraic. Hölder’s Theorem simply states that the gamma function, \Gamma , is not differentially algebraic and is therefore transcendentally transcendental.

Proof

Let

n \in \N _{0,} and assume that a non-zero polynomial P \in \Complex [X;Y _{0,Y} _{1,\ldots,Y} _{n]} exists such that$$\backslash forall\; z\; \backslash in\; \backslash Complex\; \backslash setminus\; \backslash Z\; \_:\; \backslash qquad\; P\; \backslash left(z;\backslash Gamma(z),\backslash Gamma\text{'}(z),\backslash ldots,(z)\; \backslash right)\; =\; 0.$$

As a non-zero polynomial in

\Complex [X] can never give rise to the zero function on any non-empty open domain of \Complex (by the fundamental theorem of algebra), we may suppose, without loss of generality, that P contains a monomial term having a non-zero power of one of the indeterminates Y _{0,Y} _{1,\ldots,Y} _{n} .

Assume also that

P has the lowest possible overall degree with respect to the lexicographic ordering Y _{0} < Y _{1} < … < Y _{n} < X. For example,because the highest power of Y _{0} in any monomial term of the first polynomial is smaller than that of the second polynomial.

Next, observe that for all

z \in \Complex \smallsetminus \Z _{\leq} we have:(z + 1),\ldots, \Gamma^(z + 1) \right) \\[1ex]&= P \left(z + 1;z \Gamma(z), [z \Gamma(z)]',[z \Gamma(z)],\ldots,[z \Gamma(z)]^ \right) \\[1ex]&= P \left(z + 1;z \Gamma(z), z \Gamma'(z) + \Gamma(z),z \Gamma(z) + 2\Gamma'(z),\ldots, z (z) + n (z) \right).\end

If we define a second polynomial

Q \in \Complex [X;Y _{0,Y} _{1,\ldots,Y} _{n]} by the transformation$$Q\; ~\; \backslash stackrel\; ~\; P(X\; +\; 1;X\; Y\_0,X\; Y\_1\; +\; Y\_0,X\; Y\_2\; +\; 2\; Y\_1,\backslash ldots,X\; Y\_n\; +\; n\; Y\_),$$then we obtain the following algebraic differential equation for \Gamma :$$\backslash forall\; z\; \backslash in\; \backslash Complex\; \backslash setminus\; \backslash Z\; \_:\; \backslash qquad\; Q\; \backslash left(z;\backslash Gamma(z),\backslash Gamma\text{'}(z),\backslash ldots,(z)\; \backslash right)\; \backslash equiv\; 0.$$

Furthermore, if

X ^{h} Y ^{h _{0}} _{0} Y ^{h _{1}} _{1} … Y ^{h _{n}} _{n} is the highest-degree monomial term in P , then the highest-degree monomial term in Q is $$X^\; Y\_0^\; Y\_1^\; \backslash cdots\; Y\_n^.$$

Consequently, the polynomial$$Q\; -\; X^\; P$$has a smaller overall degree than

P , and as it clearly gives rise to an algebraic differential equation for \Gamma , it must be the zero polynomial by the minimality assumption on P . Hence, defining R \in \Complex[X] by $$R\; ~\; \backslash stackrel\; ~\; X^,$$ we get$$Q=\; P(X\; +\; 1;X\; Y\_0,X\; Y\_1\; +\; Y\_0,X\; Y\_2\; +\; 2Y\_1,\backslash ldots,X\; Y\_n\; +\; n\; Y\_)=\; R(X)\; \backslash cdot\; P(X;Y\_0,Y\_1,\backslash ldots,Y\_n).$$

Now, let

X = 0 in Q to obtain$$Q(0;Y\_0,Y\_1,\backslash ldots,Y\_n)\; =\; P(1;0,Y\_0,2\; Y\_1,\backslash ldots,n\; Y\_)\; =\; R(0)\; \backslash cdot\; P(0;Y\_0,Y\_1,\backslash ldots,Y\_n)\; =\; 0\_.$$

A change of variables then yields $$P(1;0,Y\_1,Y\_2,\backslash ldots,Y\_n)\; =\; 0\_,$$ and an application of mathematical induction (along with a change of variables at each induction step) to the earlier expression$$P(X\; +\; 1;X\; Y\_0,X\; Y\_1\; +\; Y\_0,X\; Y\_2\; +\; 2Y\_1,\backslash ldots,X\; Y\_n\; +\; nY\_)=\; R(X)\; \backslash cdot\; P(X;Y\_0,Y\_1,\backslash ldots,Y\_n)$$reveals that$$\backslash forall\; m\; \backslash in\; \backslash N:\; \backslash qquad\; P(m;0,Y\_1,Y\_2,\backslash ldots,Y\_n)\; =\; 0\_.$$

This is possible only if

P is divisible by Y _{0} , which contradicts the minimality assumption on P . Therefore, no such P exists, and so \Gamma is not differentially algebraic.^{[2] [3]} Q.E.D.

Notes and References

1. Bank, Steven B. & Kaufman, Robert. “A Note on Hölder’s Theorem Concerning the Gamma Function”, Mathematische Annalen, vol 232, 1978.
2. Rubel, Lee A. “A Survey of Transcendentally Transcendental Functions”, The American Mathematical Monthly 96: pp. 777–788 (November 1989).
3. Boros, George & Moll, Victor. Irresistible Integrals, Cambridge University Press, 2004, Cambridge Books Online, 30 December 2011.