What is the function that describes this Asymptotic behaviour?

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In summary, the conversation discusses finding a function that satisfies the conditions of $$a(x) \rightarrow 1~\text{for}~(x \gg x_c)$$ and $$a(x) \rightarrow f(x)~\text{for}~(x \ll x_c)$$. Various suggestions are made, but the final answer proposed is $$a(x) = \frac{f(x)x_c - x}{x_c - x}$$ with the conditions that $$a(x) = 1~\text{for}~x \gg x_c$$ and $$a(x) = f(x)~\text{for}~x \ll x_c$$. Different ideas and assumptions are discussed, such
  • #1
Arman777
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I would like to find a function such that for

$$a(x) \rightarrow 1~\text{for}~(x \gg x_c)$$

$$a(x) \rightarrow f(x)~\text{for}~(x \ll x_c)$$

What could be the ##a(x)## ? I have tried some simple functions but could not figure it out. Maybe I am just blind to see the correct result.
 
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  • #2
Arman777 said:
I would like to find a function such that for

$$a(x) \rightarrow 1~\text{for}~(x \gg x_c)$$

$$a(x) \rightarrow f(x)~\text{for}~(x \ll x_c)$$

What could be the ##a(x)## ? I have tried some simple functions but could not figure it out. Maybe I am just blind to see the correct result.
This is far too vague to give a reasonable answer. How about
$$
a(x):=\begin{cases} f(x)&\text{ if }x\leq x_c\\1&\text{ if }x> x_c \end{cases}
$$
 
  • #3
fresh_42 said:
This is far too vague to give a reasonable answer.
How so ?
fresh_42 said:
How about
Hmm..that also works I guess
 
  • #4
What if I say, finding simplest possible $a(x)$. Is that makes sense ? Of course there could be infinetly many functions, but I am looking for least complicated/simplest one
 
  • #5
Arman777 said:
What if I say, finding simplest possible $a(x)$. Is that makes sense ? Of course there could be infinetly many functions, but I am looking for least complicated/simplest one
The one that @fresh_42 suggested is pretty simple. Without knowing what f(x) is, you probably won't find anything simpler than that.
 
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  • #6
No, I am looking for one single function, I mean. When I take the asymptotic behavior, it produces the results I have mentioned.

I am not looking for the same answer in a different form.
 
  • #7
Do you want it to be a continuous function of x?
Consider this:
calculate ##r = (\arctan(x)/\pi + 0.5)##
calculate ##h(x) = f(x)(1-r) + r##
 
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  • #8
Okay, I have found. It seems the answer is

For $$a(x) = \frac{f(x)x_c - x}{x_c - x}$$
$$a(x) = 1~\text{for}~x \gg x_c$$
$$a(x) = f(x)~\text{for}~x \ll x_c$$
 
  • #9
Try ##a(x)=\frac{2.arctan(x)}{\pi}##
 
  • #10
Arman777 said:
Okay, I have found. It seems the answer is

For $$a(x) = \frac{f(x)x_c - x}{x_c - x}$$
$$a(x) = 1~\text{for}~x \gg x_c$$
$$a(x) = f(x)~\text{for}~x \ll x_c$$
Still heavily depends on ##f(x)##. And what do you mean by ##x\ll x_c##? What if ##x\to -\infty ##?

This doesn't solve your problem unless you make several assumptions on ##f(x)## which you didn't tell us about!
 
  • #11
Well, currently we don't know the ##f(x)## but we know that it does not contain ##x_c##. BTW ##x_c## is just a constant, and ##x## ranges from ##[0, \infty)##. I am not interested the behaviour of the ##f(x)## as ##x \ll x_c## I am just interested in the behaviour of the ##a(x)## as a whole. ##f(x)## could also be a some constant. Or linearly dependent on ##x## but its not much complicated...

I guess you mean that if we open up the ##f(x)##, the behavior of the asymptotes will change?
 
  • #12
Arman777 said:
I guess you mean that if we open up the ##f(x)##, the behavior of the asymptotes will change?

Yes. I haven't checked in detail all possibilities, but it seemed you assumed that ##f(x)## is bounded in some sense and continuous. Otherwise, it could "overwrite" the rest of the formula.
 
  • #13
fresh_42 said:
Yes. I haven't checked in detail all possibilities, but it seemed you assumed that ##f(x)## is bounded in some sense and continuous. Otherwise, it could "overwrite" the rest of the formula.
Well yes its. Also it seems that the f(x) must be in the 0th order or 1st order. Otherwise the asymptotic relation fails. But of course this is true for this function. I am not sure we can find a more general one
 
  • #14
Arman777 said:
Okay, I have found. It seems the answer is

For $$a(x) = \frac{f(x)x_c - x}{x_c - x}$$
$$a(x) = 1~\text{for}~x \gg x_c$$
$$a(x) = f(x)~\text{for}~x \ll x_c$$
You might like to play around with
https://www.desmos.com/calculator/da9wzphuxc

[You can rescale the axes by shift-drag near each axis.]

1632813294811.png
 

FAQ: What is the function that describes this Asymptotic behaviour?

What is an asymptotic behavior?

An asymptotic behavior is a mathematical concept that describes the behavior of a function as its input approaches a certain value, typically infinity or zero. It is often used to analyze the long-term behavior of a function.

What is the purpose of describing asymptotic behavior?

The purpose of describing asymptotic behavior is to better understand the overall behavior of a function and to make predictions about its behavior as its input approaches a certain value. It can also help in analyzing the efficiency and complexity of algorithms in computer science.

What are some common types of asymptotic behavior?

Some common types of asymptotic behavior include linear, logarithmic, exponential, and polynomial. These types of behavior can be identified by looking at the rate of change of a function as its input approaches a certain value.

How is asymptotic behavior different from the actual behavior of a function?

Asymptotic behavior only describes the long-term behavior of a function as its input approaches a certain value, while the actual behavior of a function may vary for different input values. Asymptotic behavior does not take into account any sudden changes or fluctuations in the function's behavior.

Can asymptotic behavior change?

Yes, asymptotic behavior can change if the function itself changes. For example, if a function's growth rate increases or decreases, its asymptotic behavior will also change. Additionally, different types of asymptotic behavior can occur at different input values for a single function.

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