Summation and Factorial

  • Thread starter Atran
  • Start date
  • #1
93
1
Hi, I've been trying to make another expression for,

m
[tex]\sum[/tex](k!) = f(m)
k=0

I did one expression that is,

f(m) = 2(3(4(5...((m-1)(m+1)+1)...+1)+1)+1)+1

For instance, f(4) = 2(3(4+1)+1)+1

Can you hint me on finding a finite expression for the above expression?
Thanks for help.
 

Answers and Replies

  • #3
70
0
FYI your function f(x) can be extended to non-integer argument x as follows:

Since

s! = integral(0,infinity)[ x^s e^-x dx]

it follows that

f(s) = integral(0,infinity)[ (1+x+x^2+...+x^s) e^-x dx]
= integral(0,infinity)[ (1-x^(s+1))/(1-x) *e^-x dx]

Numerical integration (not sure how accurate) yields
f(-1.5)=-1.21027
f(-0.5) = 0.562187
f(0.5) = 1.44841
f(1.5) = 2.77775
f(2.5) = 6.1011
f(i) = 0.963332 + 0.751815i
 
Last edited:
  • #4
129
0
f0(n)= sum(i=0, n-1, i!)
f1(n)= my(r=1,s=1); for(i=1,n-1,s*=i;r+=s);r
f2(n)= my(r=n);forstep(i=n-2,1,-1,r=r*i+1);r
 

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