Digamma function

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Digamma function ψ(s) in the complex plane. The color of a point s encodes the value of ψ(s). Strong colors denote values close to zero and hue encodes the value's argument.

In mathematics, the digamma function is defined as the logarithmic derivative of the gamma function:[1][2]

\psi(x) =\frac{d}{dx} \ln{\Gamma(x)}= \frac{\Gamma'(x)}{\Gamma(x)}.

It is the first of the polygamma functions.

Contents

[edit] Relation to harmonic numbers

The digamma function, often denoted also as ψ0(x), ψ0(x) or \digamma (after the shape of the archaic Greek letter Ϝ digamma), is related to the harmonic numbers in that

\psi(n) = H_{n-1}-\gamma\!

where Hn is the n 'th harmonic number, and γ is the Euler-Mascheroni constant. For half-integer values, it may be expressed as

\psi\left(n+{\frac{1}{2}}\right) = -\gamma - 2\ln 2 + 
\sum_{k=1}^n \frac{2}{2k-1}

[edit] Integral representations

It has the integral representation

\psi(x) = \int_0^{\infty}\left(\frac{e^{-t}}{t} - \frac{e^{-xt}}{1 - e^{-t}}\right)\,dt

valid if the real part of x is positive. This may be written as

\psi(s+1)= -\gamma + \int_0^1 \frac {1-x^s}{1-x} dx

which follows from Euler's integral formula for the harmonic numbers.

[edit] Series formula

Digamma can be computed in the complex plane outside negative integers (Abramowitz and Stegun 6.3.16)[1], using

\psi(z+1)= -\gamma +\sum_{n=1}^\infty \frac{z}{n(n+z)} \qquad z \neq -1, -2, -3, \ldots

or

\psi(z)=-\gamma+\sum_{n=0}^{\infty}\frac{z-1}{(n+1)(n+z)}=-\gamma+\sum_{n=0}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+z}\right)\qquad z\neq0,-1,-2,-3,\ldots

This can be utilized to evaluate infinite sums of rational functions, i.e., \sum_{n=0}^{\infty}u_{n}=\sum_{n=0}^{\infty}\frac{p(n)}{q(n)}, where p(n) and q(n) are polynomials of n.

Performing partial fraction on un in the complex field, in the case when all roots of q(n) are simple roots,

u_{n} =\frac{p(n)}{q(n)}=\sum_{k=1}^{m}\frac{a_{k}}{n+b_{k}}.

For the series to converge,

\lim_{n\to\infty}nu_{n}=0,

or otherwise the series will be greater than harmonic series and thus diverges.

Hence

\sum_{k=1}^{m}a_{k}=0,

and

\sum_{n=0}^{\infty}u_{n}=\sum_{n=0}^{\infty}\sum_{k=1}^{m}\frac{a_{k}}{n+b_{k}}=\sum_{n=0}^{\infty}\sum_{k=1}^{m}a_{k}\left(\frac{1}{n+b_{k}}-\frac{1}{n+1}\right)=
=\sum_{k=1}^{m}\left(a_{k}\sum_{n=0}^{\infty}\left(\frac{1}{n+b_{k}}-\frac{1}{n+1}\right)\right)=-\sum_{k=1}^{m}a_{k}\left(\psi(b_{k})+\gamma\right)=-\sum_{k=1}^{m}a_{k}\psi(b_{k}).

With the series expansion of higher rank polygamma function a generalized formula can be given as

\sum_{n=0}^{\infty}u_{n}=\sum_{n=0}^{\infty}\sum_{k=1}^{m}\frac{a_{k}}{(n+b_{k})^{r_{k}}}=\sum_{k=1}^{m}\frac{(-1)^{r_{k}}}{(r_{k}-1)!}a_{k}\psi^{(r_{k}-1)}(b_{k}),

provided the series on the left converges.

[edit] Taylor series

The digamma has a rational zeta series, given by the Taylor series at z=1. This is

\psi(z+1)= -\gamma -\sum_{k=1}^\infty \zeta (k+1)\;(-z)^k,

which converges for |z|<1. Here, ζ(n) is the Riemann zeta function. This series is easily derived from the corresponding Taylor's series for the Hurwitz zeta function.

[edit] Newton series

The Newton series for the digamma follows from Euler's integral formula:

\psi(s+1)=-\gamma-\sum_{k=1}^\infty \frac{(-1)^k}{k} {s \choose k}

where \textstyle{s \choose k} is the binomial coefficient.

[edit] Reflection formula

The digamma function satisfies a reflection formula similar to that of the Gamma function,

\psi(1 - x) - \psi(x) = \pi\,\!\cot{ \left ( \pi x \right ) }

[edit] Recurrence formula and characterization

The digamma function satisfies the recurrence relation

\psi(x + 1) = \psi(x) + \frac{1}{x}. (see proof)


Thus, it can be said to "telescope" 1/x, for one has

\Delta [\psi] (x) = \frac{1}{x}

where Δ is the forward difference operator. This satisfies the recurrence relation of a partial sum of the harmonic series, thus implying the formula

 \psi(n)\ =\ H_{n-1} - \gamma

where \gamma\, is the Euler-Mascheroni constant.

More generally, one has

\psi(x+1) = -\gamma + \sum_{k=1}^\infty 
\left( \frac{1}{k}-\frac{1}{x+k} \right).

Actually, ψ is the only solution of the functional equation \psi(x + 1) = \psi(x) + \frac{1}{x} that is monotone on (0,\infty) and such that ψ(1) = − γ.

[edit] Gaussian sum

The digamma has a Gaussian sum of the form

\frac{-1}{\pi k} \sum_{n=1}^k 
\sin \left( \frac{2\pi nm}{k}\right) \psi \left(\frac{n}{k}\right) =
\zeta\left(0,\frac{m}{k}\right) = -B_1 \left(\frac{m}{k}\right) = 
\frac{1}{2} - \frac{m}{k}

for integers 0 < m < k. Here, ζ(s,q) is the Hurwitz zeta function and Bn(x) is a Bernoulli polynomial. A special case of the multiplication theorem is

\sum_{n=1}^k \psi \left(\frac{n}{k}\right)
 =-k(\gamma+\log k),

and a neat generalization of this is

\sum_{p=0}^{q-1}\psi(a+p/q)=q(\psi(qa)-\log(q)),

in which it is assumed that q is a natural number, and that 1-qa is not.

[edit] Gauss's digamma theorem

For positive integers m and k (with m < k), the digamma function may be expressed in terms of elementary functions as:

\psi\left(\frac{m}{k}\right) = -\gamma -\ln(2k) 
-\frac{\pi}{2}\cot\left(\frac{m\pi}{k}\right)
+2\sum_{n=1}^{\lfloor (k-1)/2\rfloor}
\cos\left(\frac{2\pi nm}{k} \right)
\ln\left(\sin\left(\frac{n\pi}{k}\right)\right)

[edit] Computation & approximation

According to J.M. Bernardo AS 103 algorithm[3] the digamma function for x, a real number, can be approximated by

 \psi(x) = \ln(x) - \frac{1}{2x} - \frac{1}{12x^2} + \frac{1}{120x^4} - \frac{1}{252x^6} + O\left(\frac{1}{x^8}\right)

or the asymptotic series

 \psi(x) = \ln(x) - \frac{1}{2x} + \sum_{n=1}^\infty \frac{\zeta(1-2n)}{x^{2n}}
 \psi(x) = \ln(x) - \frac{1}{2x} - \sum_{n=1}^\infty \frac{B_{2n}}{2n\cdot x^{2n}}

n as integer, where Bk is the kth Bernoulli number and ζ is the Riemann zeta function.

 \psi(x) \in [\ln(x-1), \ln x]
(see proof)
 \exp(\psi(x)) \approx \begin{cases} \frac{x^2}{2} &: x\in[0,1] \\ x - \frac{1}{2} &: x>1 \end{cases}

From the above asymptotic series for ψ you can derive asymptotic series for \exp \circ\, \psi that contain only rational functions and constants. The first series matches the overall behaviour of \exp \circ\, \psi well, that is, it behaves asymptotically identically for large arguments and has a zero of unbounded multiplicity at the origin, too. It can be considered a Taylor expansion of exp( − ψ(1 / y)) at y = 0.

 \exp(\psi(x)) = \frac{1}{\frac{1}{x}+\frac{1}{2\cdot x^2}+\frac{5}{4\cdot3!\cdot x^3}+\frac{3}{2\cdot4!\cdot x^4}+\frac{47}{48\cdot5!\cdot x^5} - \frac{5}{16\cdot6!\cdot x^6} + \dots}

The other expansion is more precise for large arguments and saves computing terms of even order.

 \exp(\psi(x+\tfrac{1}{2})) = x + \frac{1}{4!\cdot x} - \frac{37}{8\cdot6!\cdot x^3} + \frac{10313}{72\cdot8!\cdot x^5} - \frac{5509121}{384\cdot10!\cdot x^7} + O\left(\frac{1}{x^9}\right)\quad\mbox{for } x>1

(See derivation of all coefficients.)

[edit] Special values

The digamma function has values in closed form for rational numbers, as a result of Gauss's digamma theorem. Some are listed below:

 \psi(1) = -\gamma\,\!
 \psi\left(\frac{1}{2}\right) = -2\ln{2} - \gamma
 \psi\left(\frac{1}{3}\right) = -\frac{\pi}{2\sqrt{3}} -\frac{3}{2}\ln{3} - \gamma
 \psi\left(\frac{1}{4}\right) = -\frac{\pi}{2} - 3\ln{2} - \gamma
 \psi\left(\frac{1}{6}\right) = -\frac{\pi}{2}\sqrt{3} -2\ln{2} -\frac{3}{2}\ln(3) - \gamma
 \psi\left(\frac{1}{8}\right) = -\frac{\pi}{2} - 4\ln{2} - \frac{1}{\sqrt{2}} \left\{\pi + \ln(2 + \sqrt{2}) - \ln(2 - \sqrt{2})\right\} - \gamma

[edit] See also

[edit] References

  1. ^ a b Abramowitz, M.; Stegun, I. A., eds. (1972). "6.3 psi (Digamma) Function.". Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. New York: Dover. pp. 258-259. http://www.math.sfu.ca/~cbm/aands/page_258.htm. 
  2. ^ Weisstein, Eric W., "Digamma function" from MathWorld.
  3. ^ Bernardo, José M. (1976). "Algorithm AS 103 psi(digamma function) computation". Applied Statistics 25: 315-317. http://www.uv.es/~bernardo/1976AppStatist.pdf. 

[edit] External links

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