Function with f(0)=0,f'(0)=+inf,f(inf)=<1

  • #1
f(x) should give the chance of something happening, thats the reason for f(inf)=<1.

i used f(x)=1-exp(-a*x) until now, which is ok but f'(0) is not +inf. i would prefer a function with f(0)=+inf.

i want the function to be (monotonically) increasing at a decreasing rate (f' decreasing monotonically from +inf to 0)

does a function like this exist?

many thanks in advance!
 

Answers and Replies

  • #2
mathman
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Play around with xlnx.
 
  • #3
Mentallic
Homework Helper
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When you said [itex]f( \infty )\leq 1[/itex] are you sure that's what you mean? So you could have [itex]f(\infty)=-\infty[/itex]?
 
  • #4
on first reply:
i dont see how i can get what i want from x ln(x)... can you eloborate? i mean x ln (x) goes to infinity

on second reply:
i meant i want f(inf)=1 (monotonically increasing from 0 to inf.), so f'(inf)=0.
 
  • #5
chiro
Science Advisor
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f(x) should give the chance of something happening, thats the reason for f(inf)=<1.

i used f(x)=1-exp(-a*x) until now, which is ok but f'(0) is not +inf. i would prefer a function with f(0)=+inf.

i want the function to be (monotonically) increasing at a decreasing rate (f' decreasing monotonically from +inf to 0)

does a function like this exist?

many thanks in advance!
Look up the heaviside and dirac delta functions and use these to create your function.
 
  • #6
i was shocked at the functions you came up with... then is saw my typo:

i would prefer f'(0)=inf (and not f(0)=inf, since i want f(0)=0)
 
  • #7
mathman
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When I said play around with x lnx, I was thinking of something like (x lnx)/(1 + x2)
 
  • #8
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Look up the heaviside and dirac delta functions and use these to create your function.
Uuuh, these are distributions. I doubt the OP wants those.
 
  • #9
Char. Limit
Gold Member
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What about log(x-1)/(x-1) + i pi?

EDIT: Never mind this. It assumed f'(1)=+inf, not f'(0)=+inf.

x^(1/3) - (x-1)^(1/3) + 1 should work though.
 
Last edited:
  • #10
a bit late, but thanks all for your efforts!!

Char Limit's thing works perfectly!! :-) ... after changing the signs:

-x^(1/3)+(x-1)^(1/3)+1
 
  • #11
798
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f(x) = sqrt(x) / (1+sqrt(x))
more general : f(x) = (x^a)/(1+(x^a)) with 0<a<1
 
  • #12
hmmm, at first i thought -x^(1/3)+(x-1)^(1/3)+1 worked (i tested only for integers) but f is not monotonically increasing (and f' is not monotically decreasing)... and worse, f even gets negative...

but jacquelin your solutions does work perfectly! :-) im very happy with your quick response! thanks alot!
 

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