Summation on HP 42S

[Thomas Klemm]

(09-24-2018 11:56 AM)Albert Chan Wrote:  This almost look like integration !
Any reference to show it is always true ?

Just have a look at the definition of Pascal's triangle:

\({n \choose k}={n-1 \choose k-1}+{n-1 \choose k}\)

Or then just check the diagonals.
E.g. for \(k=2\) the partial sum of the first elements is just below on the next line:

1 = 1
1 + 3 = 4
1 + 3 + 6 = 10
1 + 3 + 6 + 10 = 20
1 + 3 + 6 + 10 + 15 = 35
…

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:

\(f(x+a)=\sum_{n=0}^\infty{a \choose n}\Delta^nf(x)\)

Quote:the formula looks suspiciously like a finite analog of a Taylor series expansion

So for the given example we get:

x: 0 1 2 3 …
f: 0 -2 -4 -4 …
∆f: -2 -2 0 …
∆²f: 0 2 …
∆³f: 2 …

And so with \(x=0\) we end up with: \(f(a)=2{a \choose 3} - 2{a \choose 1}\)

This leads to an even shorter program:

Code:

00 { 11-Byte Prgm }
01 RCL ST X
02 3
03 COMB
04 X<>Y
05 -
06 2
07 ×
08 END

Conclusion: We don't really have to solve the linear system of equations.

Kind regards
Thomas

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We don't even have to calculate the sum f but can only calculate ∆f:

x: 0 1 2 3 …
f: 0 …
∆f: -2 -2 0 …
∆²f: 0 2 …
∆³f: 2 …