🚀 go-pugleaf

In article <350C8ECD.2236BBD8@bellatlantic.net>, you write:
> > I need to expand some summations
> 
> A TI-92 expanded the sum of the values of k^2 from k=1 to k and gave me
> (k*(k+1)(2k+1))/6
> 
> I have unfortunaaately losst access to the TI-92 and need the following
> expanded:
> The sum of N^2 from N=0 to (N-1).
> 


Shouldn't you be able to get this from the one that you have?  Just let
k = N-1 and you get

(N-1)*(N)*(2N-1)/6  = 1/3 (N-1)^3 + 1/2 (N-1)^2 + 1/6 (N-1)


> if it is not much work could someone also show me how this expanding is done
> or point me to some texts I could read.
> 
> thanks
> the beemer
> 

How is it done?  Well, this may (or may not) be complicated.  Depends on what
you know.

Suppose I want to calculate the value of the sum

      Sum {i=0 to i=N} i^2

(the first N 2^nd powers).

Now, if you believe that the answer should be a polynomial in N (this is a big
IF), then you can write the polynomial in the form

a + bN + cN^2 + dN^3

Why did I stop at N^3?  Well, this is also a bit hard to argue.  But you can
see from your formula above that it is true.

So, now I just need to figure out what the constants a,b,c,d are.

Well, I can calculate the sum for small values of N:

N = 0, sum = 0

N = 1, sum = 1

N = 2, sum = 5

N = 3, sum = 14

Therefore,

a + b(0) + c(0) + d(0) = 0      ----- N = 0

a + b  + c + d         = 1      ----- N = 1

a + 2b + 4c + 8d       = 5      ----- N = 2

a + 3b + 9c + 27d      = 14     ----- N = 3

Solve these (linear!!) equations for a,b,c,d and you have your summation
formula:

a = 0

b = 1/6

c = 1/2

d = 1/3

and this is the same as your first formula (the one with the k's).


Suppose that you wish to compute the sum

Sum {i=15 to 100}  i^2

well, you can see that

Sum {i=15 to 100} i^2  = Sum {i=0 to 100} i^2  - sum {i = 0 to 14} i^2
                       = 1/3(100)^3 + 1/2(100)^2 + 1/6(100) 
                       - 1/3(14)^3   - 1/2(14)^2 - 1/6(14)
                       = 338350 - 1240 = 337110

So, you can compute many sums with this one formula.


Ok, what about sums of other powers?  Well, it turns out that the sum

Sum { i = 0 to i = N}  i^k

is a polynomial of degree k+1, so you can use exactly the same technique to
figure out what the polynomial should be.


For example,

Sum{i=0 to i = N} i^3 = 1/4 N^4 + 1/2 N^3 + 1/4 N^2

Sum{ i=0 to i = N} i^4 = 1/5 N^5 + 1/2 N^4 + 1/3 N^3 - 1/30 N

Sum{i=0 to i = N}  i^5 = 1/6 N^6 + 1/2 N^5 + 5/12 N^4 - 1/12 N^2


(For those of you that know some integration, it is no accident that
the first term is 1/(k+1) N^(k+1) ).
Hope this helps (and doesn't confuse you too much!).

Cheers,

Franklin Mendivil