## Some help with these equations?

For the discussion of math. Duh.

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olliver
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### Some help with these equations?

I just kind of stumbled onto these equations when i was bored a few years ago and found them saved on my computer. Could someone relate them to more general equations or tell me if they have a name? I remember feeling proud about them and i was gonna show my math teacher (like 10th grade or something) but never did. thanks for the help

CodeLabMaster
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Location: Just sum the possible locations times thier probabilities to find where you should expect me...

### Re: Some help with these equations?

Well, the coefficients of the k's in the sigmas are the binomal coefficients corresponding with the exponent of the n on the left hand side. I could write out the more general case for nm, but I can't do LaTeX, so I'll let the next person do that. It looks to me like an identity you might be able to find on the wikipedia or wolfram page for binomial coefficients. As for a name, sorry, I've no idea.

t0rajir0u
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### Re: Some help with these equations?

All you have to do is compute [imath](n+1)^k - n^k[/imath] by the binomial theorem. For example, [imath](n+1)^2 - n^2 = 2n+1[/imath].

jaap
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### Re: Some help with these equations?

olliver wrote:
I just kind of stumbled onto these equations when i was bored a few years ago and found them saved on my computer. Could someone relate them to more general equations or tell me if they have a name? I remember feeling proud about them and i was gonna show my math teacher (like 10th grade or something) but never did. thanks for the help

Or more simply:
$n^p = \sum_{k=1}^{n} k^p-(k-1)^p$

By expanding the (k-1)p, you indeed get the binomial coefficients, with alternating signs.

[edited to change n's into k's]
Last edited by jaap on Mon Jul 06, 2009 3:00 pm UTC, edited 1 time in total.

TheWaterBear
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### Re: Some help with these equations?

t0rajir0u wrote:All you have to do is compute [imath](n+1)^k - n^k[/imath] by the binomial theorem. For example, [imath](n+1)^2 - n^2 = 2n+1[/imath].

I'm not sure that your formula is correct

olliver
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### Re: Some help with these equations?

Well I see that you can reduce the right side to something more general taking the binomial expansion in reverse k^m - (k - 1)^m but that isn't really the heart of the problem is it? Or am I just not seeing it right? Perhaps its a boring question.

Charlie!
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### Re: Some help with these equations?

Sums are generally just mean, not very conducive to general solutions.

But to get things flowing, I'll try calculating a few and looking for patterns. n1 is 1. n2 is 1+3=4. n3 is 1+7+19. n4 is 1+15+67+185=268.

Hm, seems nasty. It's possible to note that the components there are all odd, leading to alternating even/odd, but aside from that, I got nothin'. Just looks like a series.

If you ever want to approximate it, though, the related integrals are nicer solution-wise. And since I'm going to go sleep now I'll leave it at that.
Some people tell me I laugh too much. To them I say, "ha ha ha!"

transcendency
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### Re: Some help with these equations?

the equations are basically $n^p = \sum_{k=1}^{n} k^p-(k-1)^p$

it's doing [imath]n^3 = (n^3 - (n-1)^3) + ((n-1)^3 - (n-2)^3) + ... + (1^3 - 0)[/imath], which is .. [imath]n^3[/imath]

t0rajir0u
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### Re: Some help with these equations?

olliver wrote:Well I see that you can reduce the right side to something more general taking the binomial expansion in reverse k^m - (k - 1)^m but that isn't really the heart of the problem is it?

That depends on what you mean by the heart of the problem. In some sense, this is just an application of the telescoping series method, i.e. [imath]a_n - a_0 = \sum_{k=1}^{n} a_k - a_{k-1}[/imath].

But the interesting things about these formulas is that taking linear combinations of them tells you how to compute sums like [imath]\sum_{k=1}^{n} k^p[/imath] - is that what you mean? This is a highly nontrivial problem, since it's not clear what linear combinations to take. The answer is given by Faulhaber's formula, which is nontrivial to understand.

Another way to understand this problem is combinatorially; that's where the hockey-stick identity and Stirling numbers come into play. this is also nontrivial to understand.