Looking for a function with specific properties

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Discussion Overview

The discussion revolves around finding a single-variable function f(x) that meets specific mathematical properties, including symmetry, differentiability, and boundedness. Participants explore various function forms and their adherence to the stated conditions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes a function f(x) = 1 - ce^(-x^2) with 0 ≤ c ≤ 1, suggesting it has a minimum of 1 - c.
  • Another participant considers modifying this function to f(x) = 1 - ce^(-(bx)^2) to adjust the rate of approach to the upper bound by varying b.
  • A different participant suggests that the constant function f(x) = c (where 0 ≤ c ≤ 1) fulfills all requirements, emphasizing its simplicity.
  • There is a discussion about the phrasing of the upper and lower bounds, with one participant suggesting that "upper bound" should refer to "least upper bound" or "supremum," and "lower bound" to "greatest lower bound" or "infimum."

Areas of Agreement / Disagreement

Participants express differing views on the simplicity and suitability of various function forms. While some agree on the properties of the proposed functions, there is no consensus on a single function that meets all requirements without exceptions.

Contextual Notes

There are discussions about the implications of differentiability at certain points and the precise definitions of bounds, which may affect the validity of the proposed functions.

zeroseven
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Hi everyone,
I'm trying to find a function of single variable f(x) with the following properties:

-It is symmetric around zero
-It is differentiable everywhere
-f'(x)≥0 for all x>0
-f'(x)=0 when x=0
-f'(x)≤0 for all x<0
(I think these last two actually follow from the first three?)
-It has an upper bound of 1
-It has a lower bound between 0 and 1, which I can set using a parameter c

For example, the function
f(x)=(c/10)*abs(x)+1-c when abs(x)≤10
f(x)=1 when abs(x)>10
(0≤c≤1)

fulfills all the conditions, except that it is not differentiable at x=-10, x=0 and x=10.

But I'm hoping to find a relatively simple function that would fulfill all those requirements.

If anyone has any ideas, I'd be very thankful!
 
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How about [itex]f(x)=1-ce^{-x^2}[/itex] with [itex]0 \leq c \leq 1[/itex] and a minimum of [itex]1-c[/itex]?

Edit: And yes, the last two do follow from the first three. If [itex]f(-x)=f(x)[/itex], then [tex]f'(-x)=\lim_{h \to 0} \frac{f(-x+h)-f(-x)}{h}=\lim_{h \to 0} \frac{f(x-h)-f(x)}{h}= -\lim_{h \to 0} \frac{f(x)-f(x-h)}{h}=-f'(x)[/tex]

Or, if you prefer, [itex]f'(x)=\frac{d}{dx}f(x)= \frac{d}{dx}f(-x)=-f'(-x)[/itex]
 
Last edited:
Thanks HS! Impressively quick response.
That looks very promising actually. I might modify it with an extra parameter:
f(x)=1-ce^(-(bx)^2)

Then, by changing the value of b I can change the rate at which it approaches the upper bound.

Anyway, looks very good and helpful!PS Sorry about using text for the maths. Still trying to get the hang of this...
 
zeroseven said:
But I'm hoping to find a relatively simple function that would fulfill all those requirements.
[itex]f(x)=c[/itex] [itex](0 \leq c \leq 1)[/itex] surely fulfils all your requirements - and that's a pretty simple function!
 
That's true oay, thanks!

Maybe instead of
"-It has an upper bound of 1"
I should have written
-it's limit at +-Infinity = 1

Not sure if even that makes it watertight. But anyway, HS-Scientist's suggestion is more suitable for what I need ;)
 
zeroseven said:
That's true oay, thanks!

Maybe instead of
"-It has an upper bound of 1"
I should have written
-it's limit at +-Infinity = 1

Not sure if even that makes it watertight. But anyway, HS-Scientist's suggestion is more suitable for what I need ;)
I think when you say "upper bound", you mean to say "least upper bound" or "supremum". Similarly, "lower bound" should be "greatest lower bound" or "infimum".
 
Yes, that's sounds like what I was thinking of, and what I should have written!
It's been a while since I had to use this terminology...
 

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