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

  1. Mar 28, 2012 #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!
  2. jcsd
  3. Mar 28, 2012 #2


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    Play around with xlnx.
  4. Mar 28, 2012 #3


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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]?
  5. Mar 29, 2012 #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.
  6. Mar 29, 2012 #5


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    Look up the heaviside and dirac delta functions and use these to create your function.
  7. Mar 29, 2012 #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)
  8. Mar 29, 2012 #7


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    When I said play around with x lnx, I was thinking of something like (x lnx)/(1 + x2)
  9. Mar 29, 2012 #8
    Uuuh, these are distributions. I doubt the OP wants those.
  10. Mar 29, 2012 #9

    Char. Limit

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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: Mar 29, 2012
  11. Jul 7, 2012 #10
    a bit late, but thanks all for your efforts!!

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

  12. Jul 7, 2012 #11
    f(x) = sqrt(x) / (1+sqrt(x))
    more general : f(x) = (x^a)/(1+(x^a)) with 0<a<1
  13. Jul 7, 2012 #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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