GammaRegularized

The regularized upper incomplete gamma function $Q(a, z) = \Gamma(a, z)/\Gamma(a)$, the survival function of the Gamma(a, 1) distribution.

✓ checked against that system · ! checked, with disagreements (hover a chip)Wolfram LanguageGammaRegularized!↗

GammaRegularized(a, z)the regularized upper incomplete gamma $Q(a, z) = \Gamma(a, z)/\Gamma(a)$.

Juliaenumeratio_ball(gamma_regularized, a, z)Rustgamma_regularized(a, z)

GammaRegularized(a, z0, z1)the regularized difference $(\Gamma(a, z_0) - \Gamma(a, z_1))/\Gamma(a)$; at $z_0 = 0$ the LOWER regularized $P(a, z_1) = 1 - Q(a, z_1)$.

Juliaenumeratio_ball((a, z0, z1) -> gamma_regularized(a, z0) - gamma_regularized(a, z1), a, z0, z1)

Typed signatures

Type	Package	Applies
(complex | infinity, complex | infinity) -> number	compute-engine	always
(complex | infinity, complex | infinity, (complex | infinity)?) -> number	analyticreplaces compute-engine	always

Domain: Special functions

Details

• Complementary to the lower regularized $P(a, z) = 1 - Q(a, z)$, which is the three-argument $Q(a, 0, z)$ (@enumeratio/analytic) -- the Gamma-distribution CDF, and with $a = k/2$, $z = x/2$ the $\chi^2_k$ CDF.
• $Q(a, 0) = 1$ for any $a > 0$: all the mass sits to the right of 0.
• For a positive integer n it has a closed polynomial form, $Q(n, z) = e^{-z}\sum_{k=0}^{n-1} z^k/k!$, which compute-engine uses to evaluate exactly, even for negative z.
• Gives the Poisson CDF tail and the Gamma-distribution survival function, so it shows up throughout reliability and queueing models. See Gamma.
• The three-argument form divides the difference by $\Gamma(a)$ rather than subtracting two $Q$ values, so $Q(a, 0, z) = 1 - Q(a, z)$ holds for negative $a$ too, where $\Gamma(a, 0)$ and $\Gamma(a)$ are separately infinite but their ratio is 1. That is Wolfram's convention.
• A genuinely complex $a$ or $z$, which the built-in handler declines, is evaluated as $\Gamma(a, z)/\Gamma(a)$ (@enumeratio/analytic) -- Gamma itself takes complex arguments. So is a non-integer real $a$ with $z < 0$, whose value is complex (principal branch, $\ln z = \ln|z| + i\pi$): $Q(5/2, -1) = 1 - 0.628\,i$. Computed in doubles from $Q = 1 - R\,e^{i\pi a}$ with a positive series for $R$; it stays unevaluated past what doubles hold, and where $\operatorname{Re} Q$ would lose digits to cancellation (an order near an even integer or near 0).

Plot

the complex plot cannot lower GammaRegularized

Examples

8 more kept as data, checked against the oracles

Basic #

$Q(a, 0) = 1$ for any $a$

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In$\operatorname{GammaRegularized}(1, 0)$Out$1$

$Q(1, z) = e^{-z}$, here at $z = 1$

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In$\operatorname{GammaRegularized}(1, 1)$Out$\frac{1}{e}$

Floating-point arguments evaluate directly to a high-precision decimal

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In$\operatorname{GammaRegularized}(2.5, 1.5)$Out$0.699\,985\,835\,878\,627\,509\,1$

$Q(1, z) = e^{-z}$ at $z = 1.5$

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In$\operatorname{GammaRegularized}(1, 1.5)$Out$0.223\,130\,160\,148\,429\,828\,933$

$Q(2, 3.3) = 4.3\,e^{-3.3}$

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In$\operatorname{GammaRegularized}(2, 3.3)$Out$0.158\,597\,619\,825\,332\,023\,416\,08$

Threads element-wise over a list, as Wolfram's Listable heads do

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In$\operatorname{GammaRegularized}(\bigl\lbrack1, 2\bigr\rbrack, 1)$Out$\bigl\lbrack\frac{1}{e}, \frac{2}{e}\bigr\rbrack$

Complex order, under N()

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In$N(\operatorname{GammaRegularized}(1+i, 2.5))$Out$0.007\,947\,060\,620\,119\,186\,222\,373\,870\,025\,622\,117\,84+0.152\,536\,778\,541\,018\,44i$

Three arguments: $Q(a, z_0) - Q(a, z_1)$, the Gamma(5/2) mass between $1/2$ and $3/2$

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In$\operatorname{GammaRegularized}(2.5, 0.5, 1.5)$Out$0.262\,579\,937\,368\,668\,97$

...which at an integer order equally evaluates, to $\frac32e^{-1/2} - \frac52e^{-3/2}$

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In$\operatorname{GammaRegularized}(2, 0.5, 1.5)$Out$0.351\,970\,589\,197\,875\,55$

Listable over its limits: $Q(1, -\infty) = \infty$, $Q(1, \infty) = 0$

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In$\operatorname{GammaRegularized}(1, \bigl\lbrack-∞, ∞\bigr\rbrack)$Out$\bigl\lbrack∞, 0\bigr\rbrack$

Listable: threads elementwise over a matrix

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In$\operatorname{GammaRegularized}(2, \bigl\lbrack\bigl\lbrack3.5, 0\bigr\rbrack, \bigl\lbrack0, 6.5\bigr\rbrack\bigr\rbrack)$Out$\bigl\lbrack\bigl\lbrack0.135\,888\,225\,400\,433\,24, 1\bigr\rbrack, \bigl\lbrack1, 0.011\,275\,793\,947\,331\,794\bigr\rbrack\bigr\rbrack$

Interval arithmetic: $Q$ is decreasing in $z$

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In$\operatorname{GammaRegularized}(\frac{2}{5}, \mathrm{Interval}(0.21, 0.22))$Out$\mathrm{Interval}(0.421\,244\,895\,791\,953\,23, 0.430\,390\,666\,940\,599\,5)$

To 30 significant digits, the last one correctly rounded

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In$N(\operatorname{GammaRegularized}(2, 1), 30)$Out$0.735\,758\,882\,342\,884\,643\,191\,047\,540\,323$

$Q(s, z)$ at a complex order $s$ — beyond what the native two-argument handler covers

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In$N(\operatorname{GammaRegularized}(2+i, 1.5))$Out$0.617\,652\,231\,045\,064\,133\,665\,033\,103\,829\,741\,524\,6+0.333\,481\,540\,658\,552\,3i$

The CDF of a Poisson($\lambda=2$) random variable at $k=2$ is $Q(3, 2) = 5/e^2$

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In$\operatorname{GammaRegularized}(3, 2)$Out$\frac{5}{e^2}$

Three arguments: $P(5/2, 3/2) = Q(5/2, 0, 3/2)$, the Gamma(5/2, 1) distribution's CDF at $3/2$

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In$\operatorname{GammaRegularized}(2.5, 0, 1.5)$Out$0.300\,014\,164\,121\,371\,94$

Known value · mpmath 1.3.0 gammainc

The $\chi^2_1$ CDF, $P(1/2, x/2)$, at the familiar critical value $x = 3.8415$: 95%

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In$1-\operatorname{GammaRegularized}(0.5, 1.920\,729\,410\,347\,042\,5)$Out$0.949\,999\,999\,999\,998\,778\,643\,7$

The polynomial closed form for an integer order extends validly to negative $z$ too, following the analytic continuation

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In$\operatorname{GammaRegularized}(5, -1)$Out$\frac{3e}{8}$

$Q(a, 0, z) = 1 - Q(a, z)$ even for negative $a$, where $\Gamma(a, 0)$ and $\Gamma(a)$ are each infinite but their ratio is 1

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In$\operatorname{GammaRegularized}(-1.5, 0, 1.5)$Out$0.985\,260\,074\,360\,347\,2$

Known value · Wolfram 15.0.0 GammaRegularized

$Q(a, 0) = 1$ for symbolic $a$

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In$\operatorname{GammaRegularized}(a, 0)$Out$1$

$Q(a, \infty) = 0$

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In$\operatorname{GammaRegularized}(a, ∞)$Out$0$

$Q(1, x) = e^{-x}$ symbolically

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In$\operatorname{GammaRegularized}(1, x)$Out$e^{-x}$

$Q(2, x) = (1 + x)e^{-x}$, through $Q(s,x) = \Gamma(s,x)/\Gamma(s)$ now that $\Gamma(2,x)$ has a closed form

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In$\operatorname{GammaRegularized}(2, x)$Out$(x+1)e^{-x}$

$Q(1/2, x) = \operatorname{erfc}(\sqrt x)$. See Erfc

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In$\operatorname{GammaRegularized}(\frac{1}{2}, x)$Out$\operatorname{erfc}(\sqrt{x})$

A non-integer order combined with negative z is complex, on the principal branch of the logarithm

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In$\operatorname{GammaRegularized}(2.5, -1)$Out$1-0.628\,008\,230\,288\,381\,4i$

Possible issues #

$Q(0, 0)$: both $\Gamma(0, 0)$ and $\Gamma(0)$ diverge at the same rate — Indeterminate, not NaN

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In$\operatorname{GammaRegularized}(0, 0)$Out$\operatorname{Indeterminate}$

About

Examples	34
Checked against	Wolfram Language 31 agree3 differSageMath 19 agree1 unevaluated1 errorSymPy 19 agree1 differ2 unevaluatedJulia 25 agree3 errorRust 8 agree5 error
Crosswalk	Wolfram LanguageJulia ×2Rust

Implementation

mappedsympypackages/oracle/src/mappings.ts

SymPy has no regularized incomplete gamma; it is written out as its definition, Gamma(a, z) / Gamma(a).

mappedsagepackages/oracle/src/mappings.ts

mappedjuliapackages/oracle/src/mappings.ts

mappedjuliapackages/oracle/src/mappings.ts

mappedrustpackages/oracle/src/mappings.ts

a, z > 0 only.

See also: Gamma, BetaRegularized