Ken Ward's Mathematics
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Negative Factorials First Difference
Series Contents
Page Contents
Preamble
Previously, we claimed that
Δk^{(n)} =n·k^{(n-1)}
[1.01]
And we proved it for integer n>0
Now we wish to show that 1.01, above, works for negative and zero n too.
Proof
We wish to prove 1.01. We have proved it for integer n >0.
If n=-1, then
Δk^{(0)} =k^{(0)} -k^{(0)}
=0
If n=-1, then
Δk^{(-1)} =(k+1)^{(-1)} -k^{(-1)}
Δk^{(-1)} =(k+1)^{(-1)} -k^{(-1)}
Δk^{(-1)} =1/(k+2)-1/(k+1)
= -1/ (k+2)(k+1)
=(-1)k^{(-2)}
So 1.01 works for n=-1
We wish to prove that
[1.01 repeated]
is true for negative integers.
Let n=-p, where p is a positive integer, so
[1.02]
So, the first difference is by definition:
[1.03]
[1.04]
Expanding the factorials, we have:
[1.05]
Extracting a factor:
[1.06]
Evaluating gives:
[1.07]
Rewriting 1.07 as a factorial:
[1.08]
substituting n=-p, we have:
■ [1.09]
Hence, 1.01 is true for negative integers too.
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