And by substituting (n+1) for n in 1.03,
k(n+1)=k(k-1)...(k-n+1)(k-n) [1.04]
Therefore:
k(n+1)=k(n)(k-n)
Values
for 0 and 1
Substituting 0 for n in 1.02 we find: [1.05]
And substituring 1 for n: [1.06]
Expression
for Negative Factorials
We have shown that k(0) is 1, and we make the
following claim: [1.07]
By induction, we claim that if 1.07 is true then so is (substituting
(-n+1 for -n) [1.08]
We recall that [1.09]
And [1.10]
Hence, by substituting these in 1.07 and 1.08 [1.11]
[1.12]
From 1.11 and 1.12 we note:
Which is true by definition (1.02). Hence because 1.07 is true for n=0,
it is true for n=-1, and so true for all n.
Examples
for n=-1, -2, -3
If n=(-1), using
k(-1)=1/(k+1)1
k(-1)=1/(k+1)
For n=(-2),
k(-2)=1/(k+2)2
k(-2)=1/(k+2)(k+1)
For n=(-3),
k(-3)=1/(k+3)3
k(-3)=1/(k+3)(k+2)(k+1)
[1.02]
[1.05]
[1.06]
[1.07]
[1.08]
[1.09]
[1.10]
[1.11]
[1.12]