Wanted ... a function

krassi_holmz · 2006-07-22 03:57:08

Hi.
I need a function, which has the property:

for every integer k>n, where n is some integer.

Thank you.

Ricky · 2006-07-22 08:38:50

Krassi, it seems to be that:

f(0) = f(-2)
f(1) = f(-3)
f(2) = f(-4)

and so on. At least that's the only constaint that I could come up with. So it seems that:

f(k) = k for k >= 0
f(k) = f(-k + 2) for k < 0

Works. But I have yet to really test it.

krassi_holmz · 2006-07-22 10:31:36

f(k)=k for k>=0?
But then f(3)=3f(2)+1=7...

George,Y · 2006-07-22 18:03:42

f(n+1)= nf(n)+f(n-1)

f(n+2)= (n+1) f(n+1)+ f(n)
= ((n+1)n+1) f(n)+ (n+1) f(n-1)

f(n+3)= (n+2) f(n+2) +f(n+1)
= ... I don't want to write it out

krassi_holmz · 2006-07-22 19:52:45

George,Y wrote:

f(n+1)= nf(n)+f(n-1)
f(n+2)= (n+1) f(n+1)+ f(n)
= ((n+1)n+1) f(n)+ (n+1) f(n-1)
f(n+3)= (n+2) f(n+2) +f(n+1)
= ... I don't want to write it out

It will go bigger and bigger, besause in the expansion of f(x) there are 2 f-s.
You will get something as a Fibbonacci tree.

I'm starting to think that there won't be a simple function with this property. What about an integral?

John E. Franklin · 2006-07-31 12:50:09

I guess it's a couple ways bigger than exponential.
Because exponential multiplies by a constant over and over, but
this multiplies by an increasing number, plus the small addition of the second term looking back a couple.
How do you think up these things, anyway?

krassi_holmz · 2006-08-02 01:23:04

It must involve the factorials.
I can find limits for this functions (m(x)<=f(x)<=M(x) for every x), and thus some asymptotic relations.
But I want the exact function.

Math Is Fun Forum

#1 2006-07-22 03:57:08

Wanted ... a function

#2 2006-07-22 08:38:50

Re: Wanted ... a function

#3 2006-07-22 10:31:36

Re: Wanted ... a function

#4 2006-07-22 18:03:42

Re: Wanted ... a function

#5 2006-07-22 19:52:45

Re: Wanted ... a function

#6 2006-07-31 12:50:09

Re: Wanted ... a function

#7 2006-08-02 01:23:04

Re: Wanted ... a function

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