Looking for a function with specific properties

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SUMMARY

The discussion centers on finding a single-variable function f(x) that is symmetric around zero, differentiable everywhere, and meets specific derivative conditions. The proposed function f(x) = 1 - c e^(-b x^2) with 0 ≤ c ≤ 1 is highlighted as a promising candidate, allowing for adjustment of the rate at which it approaches its upper limit of 1. Participants clarify the terminology regarding upper and lower bounds, emphasizing the need for precise mathematical language in defining these properties.

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  • Familiarity with the concepts of symmetry in functions
  • Knowledge of differentiability and its implications
  • Basic grasp of limits and bounds in mathematical analysis
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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 f(x)=1-ce^{-x^2} with 0 \leq c \leq 1 and a minimum of 1-c?

Edit: And yes, the last two do follow from the first three. If f(-x)=f(x), then f&#039;(-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&#039;(x)

Or, if you prefer, f&#039;(x)=\frac{d}{dx}f(x)= \frac{d}{dx}f(-x)=-f&#039;(-x)
 
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.
f(x)=c (0 \leq c \leq 1) 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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