Quote (feanur @ 1 Nov 2019 04:50)
Quote (feanur @ 1 Nov 2019 15:55)
I missed the 'homogenous" part, I'm sorry !
With a degree 3 relationship :
We're looking for real numbers a, b, c, d such that, for every integer n :
a.t_{n+3} + b.t_{n+2} + c.t_{n+1} + d.t_n = 0
replace and expand (after multiplying by 2) :
a(n²+7n+12) + b(n²+5n+6) + c(n²+3n+2) + d(n²+n) = 0
rearrange the terms :
n² (a+b+c+d) + n(7a+5b+3c+d) + (12a+6b+2c) = 0
We just want a, b, c, d satisfying the linear homogenous system :
{ a+b+c+d = 0
{ 7a+5b+3c+d = 0
{ 12a+6b+2c = 0
Solving this leads to :
{ a = -d
{ b = 3d
{ c = -3d
We can choose d = 1, then a = -1, b = 3, and c = -3.
Finally, for every integer n :
-t_{n+3} + 3.t_{n+2} - 3.t_{n+1} + t_n = 0
?!
You were on the right track but this is needlessly complicated... just differentiate, just because it is discrete does not mean differentiation is not allowed.
Start with tn, the nth triangular number. Either using the formula tn=n(n+1)/2 or noticing that the nth triangular number is just 1+2+...+n, we get that t_{n)-t_{n-1}=n
Then we differentiate (I hope that's how you found the first relation): (t_{n)-t_{n-1}) - (t_{n-1)-t_{n-2}) = t_{n} - 2t_{n-1} + t_{n-2} = n - (n-1) = 1 (notice how the degree drops like in usual differentiation... though one needs to be careful, the formula for the discrete differentiation is not in general the same as the "analytic" one)
Then differentiate one last time to get the desired formula: (t_{n} - 2t_{n-1} + t_{n-2}) - (t_{n-1} - 2t_{n-2} + t_{n-3}) = t_{n} -3t_{n-1} + 3t_{n-2} -t_{n-3} = 1 - 1 = 0
This post was edited by Hanako on Nov 3 2019 02:18pm