What is the function that describes this Asymptotic behaviour?

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  • #1
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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
fresh_42
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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.
This is far too vague to give a reasonable answer. How about
$$
a(x):=\begin{cases} f(x)&\text{ if }x\leq x_c\\1&\text{ if }x> x_c \end{cases}
$$
 
  • #4
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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
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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.
 
  • #6
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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
FactChecker
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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
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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
mathman
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Try ##a(x)=\frac{2.arctan(x)}{\pi}##
 
  • #10
fresh_42
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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
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Well, currently we dont 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
fresh_42
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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
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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
robphy
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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
 

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