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

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Discussion Overview

The discussion revolves around finding a mathematical function that satisfies specific conditions: f(0)=0, f'(0)=+inf, and f(inf)<=1. Participants explore various functions and their properties, focusing on the behavior of derivatives and the monotonicity of the functions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that f(x) should represent a probability, hence f(inf)<=1, and expresses a desire for a function that is monotonically increasing with a decreasing rate.
  • Another participant proposes experimenting with the function x ln(x), but later clarifies that they meant to suggest a modified version that approaches 1 as x approaches infinity.
  • There is a discussion about the Heaviside and Dirac delta functions, with some participants expressing skepticism about their applicability to the problem.
  • A participant mentions a function involving log(x-1) and later proposes x^(1/3) - (x-1)^(1/3) + 1 as a potential solution, although they later question its monotonicity.
  • Another participant introduces f(x) = sqrt(x) / (1+sqrt(x)) and a more general form f(x) = (x^a)/(1+(x^a)) with 0
  • There is a correction regarding a typo about the desired derivative at zero, clarifying that f'(0) should be +inf, not f(0).

Areas of Agreement / Disagreement

Participants express various viewpoints and suggestions, but there is no clear consensus on a single function that meets all specified criteria. The discussion remains unresolved with multiple competing ideas presented.

Contextual Notes

Some functions proposed may not satisfy all conditions, such as monotonicity or the behavior of derivatives. Participants also reference distributions, which may not align with the original intent of the problem.

sargondjani1
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f(x) should give the chance of something happening, that's 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!
 
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Play around with xlnx.
 
When you said f( \infty )\leq 1 are you sure that's what you mean? So you could have f(\infty)=-\infty?
 
on first reply:
i don't 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.
 
sargondjani1 said:
f(x) should give the chance of something happening, that's 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.
 
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)
 
When I said play around with x lnx, I was thinking of something like (x lnx)/(1 + x2)
 
chiro said:
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.
 
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
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! :-) I am very happy with your quick response! thanks a lot!
 

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