# sigma reccurence

• October 12th 2010, 11:20 PM
dunsta
sigma reccurence
f:N-->N
f(0) = 1

$1+\sum_{j=0}^{n-1} jf(j)$, n>=1

i.e.
1 + ((0*f*0) + (1*f*1) + (2 * f *2) ....(n-1)*f*(n-1)) <-- I'm sure this isn't right

then f(4) =

1+ ((0*4*0) + (1*4*1) + (2 * 4 *2) + (3 * 4 *3))
= 1 + 0 + 4 + 16 + 36
= 55

Have I worked this correctly?
• October 13th 2010, 12:02 AM
aman_cc
No doesn't appear correct
f(n) = n.f(n-1)
• October 13th 2010, 04:49 AM
dunsta
Am I correct in assuming f stands for function, f is not a numeric variable. only the n inside the function(n-1) is substituted for a number.

f(n) = n.f(n-1) < -- ok, I see, that makes sense. The the final value goes up to n(n-1)