MathAction SandBoxCombfuncDiscussion

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Nicer display from binomials:

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--)set output tex off
--)set output algebra on
binomial(n,m)

$\label{eq1}\hbox{\axiomType{BINOMIAL}\ } \left({n , \: m}\right)$

(1)

Type: Expression(Integer)

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opbinom := operator(operator 'binomial)$CombinatorialFunction(INT,EXPR INT)

$\label{eq2}binomial$

(2)

Type: BasicOperator?

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setProperty(opbinom, '%specialDisp::Symbol,
  ((l:List EXPR INT):OutputForm +-> paren vconcat(first(l)::OutputForm,second(l)::OutputForm)) pretend None)

$\label{eq3}binomial$

(3)

Type: BasicOperator?

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binomial(n,m)

$\label{eq4}\left( \begin{array}{c} n \ m \$

(4)

Type: Expression(Integer)

Tests from GouldBk?.pdf --rrogers, Fri, 07 Apr 2017 16:22:25 +0000 reply

http://www.dsi.unifi.it/~resp/GouldBK.pdf

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binomial(5,2)-binomial(5,5-2)

$\label{eq5}0$

(5)

Type: NonNegativeInteger?

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binomial(a,b)-binomial(a,a-b)

$\label{eq6}\begin{array}{@{}l} \displaystyle {\left( \begin{array}{c} a \ b \$

(6)

Type: Expression(Integer)

Ah well no joy in Mudville I guess I need a special function of some sort, any comments?

simplification of binomials --Bill Page, Fri, 07 Apr 2017 21:38:48 +0000 reply

The simplifications provided by Combfunc apply only to specific integer values of the 2nd argument of a single kernel. What you want to do usually involves combinations of more than one kernel. For this you might expect something like simplify or expand to work but these commands are not aware of binomial. The next best thing might be to use some custom rules. E.g.

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BS := rule
  binomial(a,a-b) == binomial(a,b)

$\label{eq7}= = \left({{\left( \begin{array}{c} a \ {- b + a} \$

(7)

Type: RewriteRule?(Integer,Integer,Expression(Integer))

then we have

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Is(binomial(3,3-x),lhs BS)

$\label{eq8}\left[{b = x}, \:{a = 3}\right]$

(8)

Type: List(Equation(Expression(Integer)))

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BS( binomial(3,x)-binomial(3,3-x) )

$\label{eq9}0$

(9)

Type: Expression(Integer)

This rule is already built-in

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BN := rule
  binomial(-n+k-1,k | even? k) == binomial(n,k)
  binomial(-n+k-1,k | odd? k) == -binomial(n,k)

$\label{eq10}\begin{array}{@{}l} \displaystyle \left\{{= = \left({{\left( \begin{array}{c} {- n + k - 1} \ k \$

(10)

Type: Ruleset(Integer,Integer,Expression(Integer))

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binomial(-n+1,2)

$\label{eq11}{{{n}^{2}}- n}\over 2$

(11)

Type: Expression(Integer)

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binomial(-n+2,3)

$\label{eq12}{-{{n}^{3}}+{3 \ {{n}^{2}}}-{2 \ n}}\over 6$

(12)

Type: Expression(Integer)

Cross product. Note that because pattern matching is syntactic we need to take care about the lexical ordering of the variables.

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BX := rule
  binomial(k,n)*binomial(n,j) == binomial(k,j)*binomial(k-j,n-j)
  binomial(k,j)*binomial(n,k) == binomial(n,j)*binomial(n-j,k-j)

$\label{eq13}\begin{array}{@{}l} \displaystyle \left\{{= = \left({{\%C \ {\left( \begin{array}{c} k \ n \$

(13)

Type: Ruleset(Integer,Integer,Expression(Integer))

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ex1 := binomial(x+y,y)*binomial(y,y-z)

$\label{eq14}{\left({ \begin{array}{c} y \ {- z + y} \$

(14)

Type: Expression(Integer)

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ex2 := BX ex1

$\label{eq15}{\left({ \begin{array}{c} {y + x} \ {- z + y} \$

(15)

Type: Expression(Integer)

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eval(ex1-ex2,[x=3,y=5,z=7])

$\label{eq16}0$

(16)

Type: Expression(Integer)

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