Editing Digamma function (section)

==Computation and approximation==
The asymptotic expansion gives an easy way to compute {{math|''ψ''(''x'')}} when the real part of {{mvar|''x''}} is large.  To compute {{math|''ψ''(''x'')}} for small {{mvar|x}}, the recurrence relation

:<math> \psi(x+1) = \frac{1}{x} + \psi(x)</math>

can be used to shift the value of {{mvar|x}} to a higher value. Beal<ref>{{cite thesis |first1=Matthew J. |last1=Beal |title=Variational Algorithms for Approximate Bayesian Inference|year= 2003 |type=PhD thesis |publisher= The Gatsby Computational Neuroscience Unit, University College London |pages=265–266 |url=http://www.cse.buffalo.edu/faculty/mbeal/papers/beal03.pdf}}</ref> suggests using the above recurrence to shift {{mvar|x}} to a value greater than 6 and then applying the above expansion with terms above {{math|''x''<sup>14</sup>}} cut off, which yields "more than enough precision" (at least 12 digits except near the zeroes).

As {{mvar|x}} goes to infinity, {{math|''ψ''(''x'')}} gets arbitrarily close to both {{math|ln(''x'' − {{sfrac|1|2}})}} and {{math|ln ''x''}}. Going down from {{math|''x'' + 1}} to {{mvar|x}}, {{mvar|ψ}} decreases by {{math|{{sfrac|1|''x''}}}}, {{math|ln(''x'' − {{sfrac|1|2}})}} decreases by {{math|ln(''x'' + {{sfrac|1|2}}) / (''x'' − {{sfrac|1|2}})}}, which is more than {{math|{{sfrac|1|''x''}}}}, and {{math|ln ''x''}} decreases by {{math|ln(1 + {{sfrac|1|''x''}})}}, which is less than {{math|{{sfrac|1|''x''}}}}. From this we see that for any positive {{mvar|x}} greater than {{math|{{sfrac|1|2}}}},

:<math>\psi(x)\in \left(\ln\left(x-\tfrac12\right), \ln x\right)</math>

or, for any positive {{mvar|x}},

:<math>\exp \psi(x)\in\left(x-\tfrac12,x\right).</math>

The exponential {{math|exp ''ψ''(''x'')}} is approximately {{math|''x'' − {{sfrac|1|2}}}} for large {{mvar|x}}, but gets closer to {{mvar|x}} at small {{mvar|x}}, approaching 0 at {{math|''x'' {{=}} 0}}.

For {{math|''x'' < 1}}, we can calculate limits based on the fact that between 1 and 2, {{math|''ψ''(''x'') ∈ [−''γ'', 1 − ''γ'']}}, so

:<math>\psi(x)\in\left(-\frac{1}{x}-\gamma, 1-\frac{1}{x}-\gamma\right),\quad x\in(0, 1)</math>
or
:<math>\exp \psi(x)\in\left(\exp\left(-\frac{1}{x}-\gamma\right),e\exp\left(-\frac{1}{x}-\gamma\right)\right).</math>

From the above asymptotic series for {{mvar|ψ}}, one can derive an asymptotic series for {{math|exp(−''ψ''(''x''))}}. The series matches the overall behaviour well, that is, it behaves asymptotically as it should for large arguments, and has a zero of unbounded multiplicity at the origin too.

:<math> \frac{1}{\exp \psi(x)} \sim \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} + \cdots</math>

This is similar to a Taylor expansion of {{math|exp(−''ψ''(1 / ''y''))}} at {{math|''y'' {{=}} 0}}, but it does not converge.<ref>If it converged to a function {{math|''f''(''y'')}} then {{math|ln(''f''(''y'') / ''y'')}} would have the same [[Maclaurin series]] as {{math|ln(1 / ''y'') − ''φ''(1 / ''y'')}}. But this does not converge because the series given earlier for {{math|''φ''(''x'')}} does not converge.</ref> (The function is not [[analytic function|analytic]] at infinity.) A similar series exists for {{math|exp(''ψ''(''x''))}} which starts with <math>\exp \psi(x) \sim x- \frac 12.</math>

If one calculates the asymptotic series for {{math|''ψ''(''x''+1/2)}} it turns out that there are no odd powers of {{mvar|x}} (there is no {{mvar|x}}<sup>−1</sup> term). This leads to the following asymptotic expansion, which saves computing terms of even order.

:<math> \exp \psi\left(x+\tfrac{1}{2}\right) \sim 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} + \cdots</math>

Similar in spirit to the [[Lanczos approximation]] of the <math>\Gamma</math>-function is [[Spouge's approximation]].

Another alternative is to use the recurrence relation or the multiplication formula to shift the argument of <math>\psi(x)</math> into the range <math>1\le x\le 3</math> and to evaluate the Chebyshev series there.<ref>{{cite journal|first1=Jet|last1=Wimp | title=Polynomial approximations to integral transforms|journal=Math. Comp. |year=1961|volume=15|issue=74 |pages=174–178| doi=10.1090/S0025-5718-61-99221-3|jstor=2004225}}</ref><ref>{{cite journal|title=Chebyshev series expansion of inverse polynomials|first1=R. J.|last1=Mathar|journal=Journal of Computational and Applied Mathematics |year=2004|volume=196 |issue=2 |pages=596–607 |doi=10.1016/j.cam.2005.10.013 |arxiv=math/0403344}} App. E</ref>