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Question Number 73340 by Raxreedoroid last updated on 10/Nov/19
Isthereanypi/productnotationrulesIdiscoveredsomesuchas:∏bk=a[k]=b!(a−1)!∏bk=a[c]=cb−a+1∏bk=a[c⋅k]=∏bk=a[c]⋅∏bk=a[k]∏bk=a[k+c]=∏b+ck=a+c[k]Butwhataboutthis∏bk=a[c⋅k+d]
Commented by mathmax by abdo last updated on 10/Nov/19
whatmeans[...]?
Commented by MJS last updated on 10/Nov/19
∏bk=a[c×k+d]letd=c×e∏bk=a[c×(k+e)]=(b+e)!(a+e−1)!c1−a+b==∏bk=a[c]×∏bk=a[k+e]ifdc=e∈Q∖Z∏bk=a[k+dc]=(a+dc)Γ(b+dc+1)Γ(a+dc+1)
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