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 | | From: | NKProductionZ | | Subject: | factorials and powers -- the relationship | | Date: | 23 Jan 2005 20:31:51 GMT |
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 | The differences between successive integers to the first power is one
which is equal to 1!
The differences between successive integers to the second power are as
follows:
3 (4-1) 5 (9-4) 7 (16-9) ...
The differences between these differences is 2 which is equal to 2!
The differences between successive integers cubed are as follows:
7 (8-1) 19 (27-8) 37 (64-27) 51 (125-64) 91 (216-125) ...
The differences between these differences are:
12 18 24 30
The differences between these diffences are 6=3!
For fourth powers and depicted in a more compact format we have:
1 16 81 256 625 1296 2401 4096
6561
15 65 175 369 671 1105 1695
2465
50 110 194 302 434 590 770
60 84 108 132 156 180
24 24 24 24 24
and 24=4!
For fifth powers, after 5 sets of successive differences, we obtain
120=5!
how can we prove this in general?
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| | From: | Peter Webb | | Subject: | Re: factorials and powers -- the relationship | | Date: | Mon, 24 Jan 2005 17:41:04 +1100 |
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"NKProductionZ" wrote in message
news:20050123153151.13786.00000157@mb-m23.aol.com...
> The differences between successive integers to the first power is one
> which is equal to 1!
>
> The differences between successive integers to the second power are as
> follows:
>
> 3 (4-1) 5 (9-4) 7 (16-9) ...
>
> The differences between these differences is 2 which is equal to 2!
>
> The differences between successive integers cubed are as follows:
>
> 7 (8-1) 19 (27-8) 37 (64-27) 51 (125-64) 91 (216-125) ...
>
> The differences between these differences are:
>
> 12 18 24 30
>
> The differences between these diffences are 6=3!
>
> For fourth powers and depicted in a more compact format we have:
>
>
> 1 16 81 256 625 1296 2401 4096
> 6561
> 15 65 175 369 671 1105 1695
> 2465
> 50 110 194 302 434 590 770
> 60 84 108 132 156 180
> 24 24 24 24 24
>
> and 24=4!
>
> For fifth powers, after 5 sets of successive differences, we obtain
> 120=5!
>
>
> how can we prove this in general?
As nobody else has leapt in, here is my not-very-elegant proof
Its a proof by induction ...
Assume true for all values up to n, that is the finite differences of k^n go
to n! after n subtractions.
Lets have a look at the differences of 1^(n+1), 2^(n+1), 3^(n+1) ....
The first level differences are 2^(n+1) - 1^(n+1), 3^(n+1) - 2^(n+1), ...
Or
(k+1)^(n+1) - k^(n+1)
Now we know that the binomial expansion of (k+1)^(n+1) = k^(n+1) +
(n+1)k^(n) + (n+1)^2/2)k^(n-1) ....
Subtract k^(n+1)
The first row of differences will consist of terms of the form
(n+1)k^n + terms in k with exponents smaller than n.
By our assumption, all terms in k with exponent less than n will go to zero
in the continued difference. The difference of the (n+1)k^n terms will be
(n+1) times the finite differences of k^n. The finite difference for n=1 is
1.
So the finite difference for n=2 = 2*1 = 2
So the finite difference for n=3 = 3*2 = 6
So the finite difference for n is n!
Note also for comparison that the sequence of finite differences is the
discrete case of repeatedly differentiating.
(d^n/dx^n) x^n = n!
So its a plausible argument.
HTH
Peter Webb
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