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sum of powers

Defines: 
Faulhaber's formula
Type of Math Object: 
Definition
Major Section: 
Reference
Groups audience: 

Mathematics Subject Classification

40A05 no label found

Comments

Here, Mr. Chi Woo, is what I think is an equivalent formulation to the "triangle" method you so elegantly present in your article, "sum of powers".

The key is to begin by reformulating s(n,k) -- directly -- as a recursion on sum(s(j,k-1)), j=1,...,n-1.

Express column A as column B. (And B translates to C.)

Your method begins with (something close to the) complementary upper triangle -- which is blank in the formulation sketched below.

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Express column A as column B. (And B translates to C.)

Your method begins with (something close to the) complementary upper triangle (left blank in the sketched below).

A B C

1**k 1 * 1**(k-1) 1**(k-1)
2**k 2 * 2**(k-1) 2**(k-1) + 2**(k-1)
. .
. .
. .
n**k n * n**(k-1) n**(k-1) + n**(k-1) + ... + n**(k-1)
----- -------------- -------- + -------- + ... + --------
s(n,k) s(n,k) s(n,k-1) + s(n,k-1) + ... + s(n,k-1)
- s(1,k-1) - ... - s(n-1,k-1)

or, in summary,

n-1
s(n,k) = (n * s(n,k-1)) - sum (s(r,k-1))
r=1

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If you see fit to reply, I'm hope I will be able to find it, being quite new to PlanetMath.

Tom Lewinson

Sorry for the mess I made of my original reply. Would love to delete it if I knew how. Am new to PlanetMath and don't know how to preserve white space.

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I offer what is perhaps a trivial though I think not uninteresting equivalent alternative point of departure to the recursive "triangle" method you describe.

We have the definition of sum of k-th powers,

A

1**k
2**k
.
j**k
.
.
n**k
-----
s(n,k)

We then reformulate A as B, below,

B

(1) * (1**(k-1))
(2) * (2**(k-1))
.
(j) * (j**(k-1))
.
.
(n) * (n**(k-1))
------------
s(n,k)

The key idea is rewriting the j-th line in B as a sum instead of a product, that is, as a sum of j terms, j**(k-1).

We then get a triangular array which is close to a sort of complementary array to the one you work with.

This approach has the possibly slight advantage in that it expresses the sum of k-th powers -- directly -- as a recursion.

s(n,k) = (n * s(n,k-1)) -

n-1
sum (s(r,k-1))
r=1

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