Summation on HP 42S

[Albert Chan]

(09-24-2018 01:52 PM)Thomas Klemm Wrote:  

Quote:I normally just fit the data to search for the formula (assume I don't cheat by googling)
For sum(x^2 - 3*x), formula should be a cubic, with no constant term (sum=0 when n=0)

Try:
n = 1, sum = (1^2 - 3) = -2
n = 2, sum = -2 + (2^2 - 3*2) = -4
n = 3, sum = -4 + (3^2 - 3*3) = -4

3 equations, 3 unknowns (cubic coefficients), we get sum = n^3/3 - n^2 - 4/3*n

So, for n = 100, sum = n/3 * (n^2 - 3*n - 4) = 100/3 * 9696 = 323200

You might be interested in Newton's Forward Difference Formula:

I could never remember Forward Difference formula, without looking up.
Instead, I use the Lagrange formula, which work for uneven intervals too.

The formula look complicated, but it is very mechanical, easy to remember.
Fitting a cubic sum = n * quadratic,

sum / n
= (-2/1) \((n-2)(n-3)\over(1-2)(1-3)\) + (-4/2) \((n-1)(n-3)\over(2-1)(2-3)\) + (-4/3) \((n-1)(n-2)\over(3-1)(3-2)\)
= -(n-2)(n-3) + 2(n-1)(n-3) - (2/3)(n-1)(n-2)
= (-n^2 + 5 n - 6) + (2 n^2 - 8 n + 6) + (-2/3 n^2 + 2 n - 4/3)
= n^2/3 - n - 4/3

sum = n * (n^2/3 - n - 4/3) = n/3 * (n^2 - 3 n - 4) = n(n+1)(n-4) / 3

Edit: Forward Difference Formula is very neat, without even evaluate sums.