- #1

- 93

- 1

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.

- Thread starter Atran
- Start date

- #1

- 93

- 1

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.

- #2

- #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

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

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

- Last Post

- Replies
- 4

- Views
- 2K

- Replies
- 6

- Views
- 27K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 5

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 13K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 500