1

I have 3 points ($x_0$, $y_0$), ($x_1$, $y_1$) and ($x_2$, $y_2$) that lie on a monotonically-increasing (asymptotic) curve (whose function is not known a priori).

The only unknown value is $x_2$, which corresponds to reaching an asymptote (in the limit sense) of $y_2$.

My question is: Is there a straightforward numeric way to determine $x_2$? My eventual goal is to proceed iteratively until convergence is reached.

I am familiar with numerical root-finding algorithms, of which Muller's method crossed my mind, since it relies on three known points to construct a parabola that interpolates the points. Apart from this, my problem is different in that I wish to know the value of $x_2$ that gives $f(x_2) = y_2$.

Any thoughts on a sound approach?

Parcly Taxel
  • 103,344
compbiostats
  • 115
  • 6
  • Are you sure? The curve will never reach the asymptote by definition. An explicit $x_2$ doesn't exist. – Parcly Taxel Mar 11 '18 at 00:10
  • @ParclyTaxel You're absolutely right! But, for all intents and purposes, suppose we could reach an asymptote (kind of a thought experiment). Is there an existing approach? – compbiostats Mar 11 '18 at 00:15
  • Well you definitely can find a curve that goes through two points and reaches an asymptote $y=y_2$... Would you like this as an answer? – Parcly Taxel Mar 11 '18 at 00:16
  • If I understand you correctly, that's just a straight line, since the function passes through two points. – compbiostats Mar 11 '18 at 00:20
  • But it also has to reach an asymptote. A straight line has no asymptote. – Parcly Taxel Mar 11 '18 at 00:21
  • Yes, an asymptote must be reached. My last comment was simply stating that a straight line is still monotone and concave (with constant slope) – compbiostats Mar 11 '18 at 00:23
  • I presume you want to find an $x_2$ such that its corresponding $y_2$ is "close enough" to the asymptote, right? – Parcly Taxel Mar 11 '18 at 00:36
  • Yes. "Close enough" is sufficient, say in terms of absolute or relative approximation error. – compbiostats Mar 11 '18 at 00:54

1 Answers1

0

There is a very simple function that can be made to fit the data exactly. $$y=a+\frac b{x+c}$$ $a=y_2$ to reach the asymptote. To find $b$ and $c$, rearrange the equation to be linear in them and solve: $$(y_0-y_2)(x_0+c)=(y_1-y_2)(x_1+c)=b\tag1$$ $$c=\frac{x_0(y_2-y_0)+x_1(y_1-y_2)}{y_0-y_1}$$ where the value obtained for $c$ is back-substituted into $(1)$ to find $b$. Note that $b$ will always be negative, since the function is monotonically increasing.

Now suppose $x_2$ is to be found which makes the difference between the function value there and $y_2$ less than some constant $\varepsilon>0$. This entails solving $$y_2-\left(y_2+\frac b{x_2+c}\right)<\varepsilon$$ This is also easy: $$-\frac b{x_2+c}<\varepsilon$$ $$x_2>-\frac b\varepsilon-c$$

Parcly Taxel
  • 103,344