8
$\begingroup$

Does an equation exist that can produce the curves shown in the attached image, by varying a single variable?

enter image description here

$\endgroup$
  • 1
    $\begingroup$ Do you mean in principle, or are you asking for specific examples? $\endgroup$ – PyRulez Jul 5 '16 at 3:17
10
$\begingroup$

Consider the equilateral hyperbola $xy=1$ and map the points $(1,0)$ and $(0,1)$ to two symmetrical points on the hyperbola $(t,t^{-1})$ and $(t^{-1},t)$ by translation/scaling (as if you were zooming in).

$$(x(t-t^{-1})+t^{-1})(y(t-t^{-1})+t^{-1})=1.$$

enter image description here

The straight line corresponds to two infinitely close points, $t=t^{-1}=1$, which is a degenerate case of the equation.


Previous answer:

You can think of a pencil of parabolas, $y=\frac1{\sqrt2}+\frac\lambda{\sqrt2}(2x^2-1)$, which you rotate by $45°$ right, giving

$$x+y=1+\lambda\left((x-y)^2-1\right).$$

You can solve the quadratic equation for $y$.

$\endgroup$
  • 1
    $\begingroup$ If I take $(x,y)$ to be $(1,0)$ or $(0,1)$, it seems to me that your equation is not satisfied. $\endgroup$ – Jeppe Stig Nielsen Jul 4 '16 at 21:51
  • 1
    $\begingroup$ @JeppeStigNielsen: right, I forgot a $\sqrt2$ factor. Now fixed (hopefully). $\endgroup$ – Yves Daoust Jul 4 '16 at 22:11
  • $\begingroup$ This has become promising, since the it goes through the two points as required, and is has the symmetry that swapping $x$ and $y$ leaves the equation unchanged. However for some values of $\lambda$ it may look different than expected. Like when $\lambda$ is close to $1$, the vertex of the parabola is close to the origin, and the parabola (after rotation) crosses the coordinate axes in the "opposite" direction, I think. $\endgroup$ – Jeppe Stig Nielsen Jul 4 '16 at 22:34
  • $\begingroup$ @JeppeStigNielsen: the exponent $2$ can be lessened, but at some point we are lacking curve specifications. $\endgroup$ – Yves Daoust Jul 4 '16 at 22:36
  • $\begingroup$ At what value of $\lambda$ will the two axes be tangents at the two "fixed" points? $\endgroup$ – Jeppe Stig Nielsen Jul 4 '16 at 22:40
2
$\begingroup$

Try a superellipse: $|x|^a + |y|^a =1$ with $0 <a \le 1$.

$\endgroup$
  • 3
    $\begingroup$ Yes, but that curve will not intersect (pass through) the coordinate axes and continue into the 2nd/4th quadrant as shown in the image. But maybe that was not a requirement. $\endgroup$ – Jeppe Stig Nielsen Jul 4 '16 at 22:02
  • $\begingroup$ @JeppeStigNielsen You could try analytic continuation, or simply remove the $|\cdot|$ $\endgroup$ – Tobias Kienzler Jul 5 '16 at 6:01
  • $\begingroup$ @TobiasKienzler The superellipse with $a<1$ has a cusp at the point in question. It is not meeting the axis at a non-zero angle like in the drawing of the question. $\endgroup$ – Jeppe Stig Nielsen Jul 5 '16 at 9:11
  • $\begingroup$ @JeppeStigNielsen Right you are, I spoke to soon... $\endgroup$ – Tobias Kienzler Jul 5 '16 at 9:50
2
$\begingroup$

Try also a quadratic Bézier curve $$ \gamma(t) = (1 - t)^{2} P_0 + 2(1 - t)t P_1 + t^{2} P_2 $$ where $P_0=(1,0), P_1=(a,a), P_2=(0,1)$, with $a \in [0,0.5]$.

$\endgroup$
2
$\begingroup$

You can first of all use a tool such as Engauge Digitizer to obtain tables of data points for each curve. Then either try a Log-log plot to determine a polynomial approximation or use a fitting tool of your choice to obtain fits of the same kind for each curve. Finally, find a parameter describing each curve such that all the fit-parameters previously obtained can also be fitted by simple functions, ideally linear or a low-order polynomial.

$\endgroup$
2
$\begingroup$

Assuming what you drew to be a parabola, and $p$ a constant to push each parabola along the line of symmetry,

$$ y = p (1-x^2) $$

Now rotate them clockwise by $\pi/4 \, , (x,y) \rightarrow ( x+y, x-y)/\sqrt2. $

EDIT1:

As you left it free as regards shape within constraints, we could use a cosh curve, sec curve etc.

In fact, we can generalize this using any open even function E(x) , passing through $(\pm 1/\sqrt{2},1/\sqrt{2} ). $

Letting

$$y(x,p) = (1 + p *(1/2 - x^2) E(x)) /\sqrt{2} \, $$

and building a differentiable function, rotating the curve by $-\pi/4$, we can make the family pass through singular points $(0,1),(1,0) $as shown above and illustrated below in a particular case.

In graph below $E(x) = \cos^2(x) $is chosen with $ p= (\pm 2,0). $ Set of such Curves

For the earlier special case you got parabolas using $E(x)=1 $ perturbed by arbitrary $p$ values.

$\endgroup$
0
$\begingroup$

Use Lagrange Interpolation formula: For a variable point $(t,t)$ on the line $x-y=0$ with $0<t\leq \sqrt2/2$, fit a degree 2 polynomial curve that passes through the three points $(0,1), (t,t)$ and $ (1,0)$.

For $t=\sqrt2/2$ the three points will be collinear and it will automatically give a straight line. This must produce the same result as Yves Daoust's answer above, but using different interpretation.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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