1
$\begingroup$

I am trying to find a function such that

  1. $x$ reaches 0 at a set point $x^*$ such that $x* > 0$
  2. as $x \rightarrow 0 \implies y \rightarrow 1.$
  3. The curvature of the function between $x = 0$ and $x = x*$ can be changed such that it looks like the red lines in the figure.

I'm thinking the function has to be piecewise but I'm unsure of what to use for the part between $x=0$ and $x = x*$.

I've attached a picture of the kind of function I want, where the red lines represent the curve.

Picture of function I want

I've tried a negative exponential however I can't seem to get a curve such as the top red line of the figure.

Sorry this feels like a daft question. Any ideas? Thanks!!!

$\endgroup$
1
$\begingroup$

What about: $$ f_\alpha\colon x\in[0,x_\ast]\mapsto{x_\ast}^{-\alpha}(x_\ast-x)^\alpha $$ for $\alpha > 0$ ($\alpha < 1$ will be concave, $\alpha > 1$ convex, $\alpha=1$ linear)?

See e.g. this plot:

Plot http://www3.wolframalpha.com/Calculate/MSP/MSP4543212ded60fcgi25g900001i55c9egfdg979h7?MSPStoreType=image/gif&s=40&w=390.&h=193.&cdf=RangeControl

$\endgroup$
  • $\begingroup$ This may also be a good idea too! thank you! $\endgroup$ – Babbage Feb 18 '14 at 18:35
4
$\begingroup$

How about $y=(1-\frac x{x^*})^k$ The below shows $k=0.2,0.5.1.0,1.5,2$ The horizontal axis is $\frac x{x^*}$

enter image description here

$\endgroup$
  • $\begingroup$ Hmmm that might work! Thanks! $\endgroup$ – Babbage Feb 18 '14 at 18:35
  • $\begingroup$ Clement C. and I have the same function, expressed a bit differently. $\endgroup$ – Ross Millikan Feb 18 '14 at 18:52
0
$\begingroup$

For the linear piece, the line must pass through $(0,1)$ and $(m,0)$ has the equation $y = 1 - x/m$.

EDIT If you are looking for all 3 pieces, the second piece could be just a dropped down translated hyperbola $y = A + B/(x+1)$ so passing through $(0,1)$ forces $A+B=1$ and passing through $(m,0)$ forces $A + B/(m+1) = 0$ so solving two equations for $A,B$ will give you what you are seeking.

Similarly, for the inverse curve, try something like $y = -x^2 + Ax + B$ and two points will force the values of $A,B$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.