QBinomial
QBinomial[n,m,q]
gives the 
-binomial coefficient 
.
Details
- Mathematical function, suitable for both symbolic and numerical manipulation.
.- QBinomial automatically threads over lists.
Examples
open all close allBasic Examples (4)
Exact evaluation with numbers:
QBinomial[4, 2, 1 / 2]Plot over a subset of the reals:
Plot[QBinomial[1, 1 / 3, x], {x, 0, 2}, Exclusions -> None]Series expansion at the origin:
Series[QBinomial[1 / 2, 1, x], {x, 0, 2}]Use FunctionExpand to obtain Gaussian polynomials:
QBinomial[10, 5, q]//FunctionExpandScope (22)
Numerical Evaluation (5)
QBinomial[5, -3.1, E]QBinomial[0, 12, 5]N[QBinomial[15 / 17, 5, 1], 50]The precision of the output tracks the precision of the input:
QBinomial[0.211111111111111111, 5, 1]N[QBinomial[23 / 47, 5 - I, 2]]Evaluate efficiently at high precision:
QBinomial[23 / 47, 5, 1`100]//TimingQBinomial[15 / 71, 5, 1`1000];//TimingCompute the elementwise values of an array:
QBinomial[1, {{1 / 2, 1}, {0, -1 / 2}}, .2]Or compute the matrix QBinomial function using MatrixFunction:
MatrixFunction[QBinomial[1, #, .2]&, {{1 / 2, -1}, {0, 1 / 2}}]//FullSimplifySpecific Values (5)
QBinomial for symbolic parameters:
QBinomial[n, 4, 2]//FunctionExpandQBinomial[1, m, 2]//FunctionExpandQBinomial[4, 2, q]//FunctionExpandQBinomial[0, 0, 0]Find the minimum of QBinomial[3,2,q]:
qmin = q /. FindRoot[D[QBinomial[3, 2, q], q] == 0, {q, 0}]Plot[QBinomial[3, 2, q], {q, -3, 2}, Epilog -> Style[Point[{qmin, QBinomial[3, 2, qmin]}], PointSize[Large], Red]]QBinomial threads elementwise over lists:
QBinomial[5, {0, 1, 2, 3, 4, 5}, 1 / 3]TraditionalForm formatting:
QBinomial[n, m, q]//TraditionalFormVisualization (3)
Plot the QBinomial function for various parameters:
Plot[{QBinomial[2, 1, x], QBinomial[3, 1, x], QBinomial[4, 1, x]}, {x, -3, 2}]Plot the QBinomial function over a subset of the complexes:
ComplexPlot3D[QBinomial[1 / 2, 1, z], {z, -1 - I, 1 + I}, PlotLegends -> Automatic]![TemplateBox[{3, 2, z}, QBinomial]](Files/QBinomial.en/4.png "TemplateBox[{3, 2, z}, QBinomial]"){kind=small}
ComplexContourPlot[Re[QBinomial[3, 2, z]], {z, -4 - 4 I, 4 + 4 I}, Contours -> 20]![TemplateBox[{3, 2, z}, QBinomial]](Files/QBinomial.en/5.png "TemplateBox[{3, 2, z}, QBinomial]"){kind=small}
ComplexContourPlot[Im[QBinomial[3, 2, z]], {z, -4 - 4 I, 4 + 4 I}, Contours -> 20]Function Properties (4)
![TemplateBox[{1, {1, /, 3}, x}, QBinomial]](Files/QBinomial.en/6.png "TemplateBox[{1, {1, /, 3}, x}, QBinomial]"){kind=small}
has both singularities and discontinuities for 
and for 
:
FunctionSingularities[QBinomial[1, 1 / 3, x], x]//QuietFunctionDiscontinuities[QBinomial[1, 1 / 3, x], x]//Quiet![TemplateBox[{1, {1, /, 3}, x}, QBinomial]](Files/QBinomial.en/9.png "TemplateBox[{1, {1, /, 3}, x}, QBinomial]"){kind=small}
is neither non-negative nor non-positive:
FunctionSign[QBinomial[1, 1 / 3, x], x]QBinomial is neither convex nor concave:
FunctionConvexity[QBinomial[m, n, x], {m, n, x}]TraditionalForm formatting:
QBinomial[n, m, q]//TraditionalFormDifferentiation (2)
The first derivative with respect to n when m=1/2:
D[QBinomial[n, 1 / 2, q], n]The first derivative with respect to m when n=1 and q=2:
D[QBinomial[1, m, 2], m]Higher derivatives with respect to m:
Table[D[QBinomial[n, m, q], {m, k}], {k, 1, 3}]//FullSimplifyPlot the higher derivatives with respect to m when n=3 and q=2:
Plot[Evaluate[% /. { n -> 3, q -> 2}], {m, -5, 5}, PlotLegends -> {"First Derivative", "Second Derivative", "Third Derivative"}]Series Expansions (3)
Find the Taylor expansion using Series:
Series[QBinomial[n, x, q], {x, 0, 2}]//Normal//FullSimplifyPlots of the first three approximations around 
:
terms = Normal@Table[Series[QBinomial[2, x, 2], {x, 0, m}], {m, 1, 5, 2}];
Plot[{QBinomial[2, x, 2], terms}, {x, -2, 2}]Series expansion at Infinity:
Series[QBinomial[1 / 2, 2, x], {x, ∞, 2}]//NormalThe Taylor expansion at a generic point:
Series[QBinomial[n, x, q], {x, x0, 2}]// Normal//FullSimplifyGeneralizations & Extensions (1)
QBinomial can be applied to a power series:
QBinomial[100, 50, q + O[q] ^ 10]Applications (4)
Explicit combinatorial construction of QBinomial:
n = 10;m = 3;Total[q ^ ((Total /@ Subsets[Range[n], {m}]) - m(m + 1) / 2)]QBinomial[n, m, q]//FunctionExpand//Expand
-binomial is a generating function for the sequence in a grid-shading problem:
n = 5;m = 3;
QBinomial[n + m, m, q]//FunctionExpandCoefficientList[%, q]Count[IntegerPartitions[#], x_ /; Length[x] ≤ n && Max[x] ≤ m]& /@ Range[0, n * m]Elements in the 
-Pascal triangle satisfy two recurrence relations:
Table[QBinomial[n, m, q] == QBinomial[n - 1, m - 1, q] + q^m QBinomial[n - 1, m, q], {n, 5}, {m, n - 1}]//FullSimplifyTable[QBinomial[n, m, q] == q^n - m QBinomial[n - 1, m - 1, q] + QBinomial[n - 1, m, q], {n, 5}, {m, n - 1}]//FullSimplifyThe number of subspaces in the 
-dimensional vector space over 
with prime-power 
:
GaloisNumber[n_, q_] := Sum[QBinomial[n, m, q], {m, 0, n}]Total number of subspaces in three-dimensional vector space over 
:
GaloisNumber[3, 2]Check using recurrence equation for Galois numbers:
RecurrenceTable[{g[n + 1] == 2g[n] + (q ^ n - 1)g[n - 1], g[0] == 1, g[1] == 2}, g, {n, 3}]//Last//ExpandGaloisNumber[3, q]//FunctionExpandProperties & Relations (2)
Use FunctionExpand and FullSimplify to manipulate expressions containing QBinomial:
QBinomial[n, m, q]//FunctionExpandSeries[QBinomial[5, 5 / 2, q], {q, 0, 4}]Series[QBinomial[5, 5 / 2, q], {q, Infinity, 4}]Related Links
MathWorldText
Wolfram Research (2008), QBinomial, Wolfram Language function, https://reference.wolfram.com/language/ref/QBinomial.html.
CMS
Wolfram Language. 2008. "QBinomial." Wolfram Language & System Documentation Center. Wolfram Research. https://reference.wolfram.com/language/ref/QBinomial.html.
APA
Wolfram Language. (2008). QBinomial. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/QBinomial.html