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MathematicsIs this Batman equation for real?
[+465] [10] a_hardin
[2011-07-29 21:15:06]
[ geometry algebra-precalculus graphing-functions plane-curves ]
[ https://math.stackexchange.com/questions/54506/is-this-batman-equation-for-real ]

HardOCP [1] has an image with an equation which apparently draws the Batman logo. Is this for real?

Batman logo

Batman Equation in text form: \begin{align} &\left(\left(\frac x7\right)^2\sqrt{\frac{||x|-3|}{|x|-3}}+\left(\frac y3\right)^2\sqrt{\frac{\left|y+\frac{3\sqrt{33}}7\right|}{y+\frac{3\sqrt{33}}7}}-1 \right) \\ &\qquad \qquad \left(\left|\frac x2\right|-\left(\frac{3\sqrt{33}-7}{112}\right)x^2-3+\sqrt{1-(||x|-2|-1)^2}-y \right) \\ &\qquad \qquad \left(3\sqrt{\frac{|(|x|-1)(|x|-.75)|}{(1-|x|)(|x|-.75)}}-8|x|-y\right)\left(3|x|+.75\sqrt{\frac{|(|x|-.75)(|x|-.5)|}{(.75-|x|)(|x|-.5)}}-y \right) \\ &\qquad \qquad \left(2.25\sqrt{\frac{(x-.5)(x+.5)}{(.5-x)(.5+x)}}-y \right) \\ &\qquad \qquad \left(\frac{6\sqrt{10}}7+(1.5-.5|x|)\sqrt{\frac{||x|-1|}{|x|-1}} -\frac{6\sqrt{10}}{14}\sqrt{4-(|x|-1)^2}-y\right)=0 \end{align}

(149) Why don't you just try it? - user1120
(150) @Jim: If you mouse over the downvote button, you see: "This question does not show any research effort; it is unclear or not useful." I downvoted because the OP was too lazy to type in the equation himself to any plotting program or calculator, which would have immediately shown that the equation is "for real". If the OP were asking for an explanation of how such an equation might be derived, as ShreevatsaR has done, that would be an appropriate question. - Zev Chonoles
(3) @Zev: thanks for the tip. I didn't know about that feature. Now the post has an unbelievable 51 upvotes, which is clearly way out of proportion. - Cheerful Parsnip
There are test functions on a TI-84 (<,>,≤,≥) which return 1 when true, which bounds the pieces nicely. I've spent a few minutes converting them to GC friendly equations. Y1 = (.75+3|X|)(|X|≤.75)(|X|≥.5)+(9-8|X|)(|X|≤1)(|X|≥.75)+2.25(|X‌​|≤.5) Y2 = (7-3√33)X²/112+|X/2|-3+√(1-(||X|-2|-1)²) Y3 = ((6√10)(1-.5√(4-(|X|-1)²))/7+(3-|X|)/2)(|X|≥1)(|X|≤3) Y4 = {-(|X|≥4),(|X|≥3)}3√(1-x²/49) - user13953
OK - so here's the full Batman equation in Mathematica format. Proofread a couple of times in "traditional" format. I tried to get this to work with ContourPlot[theEquation,{x,-7,7},{y,-3,3}] and got a blank plot. Then I had the idea of forcing only the real part of the equation by plotting ContourPlot[Re[theEquationBeforeEqualsSign]==0,{x,-7,7},{y,-‌​3,3}]. Doing this shows that the equations work, but Mathematica's butchering it. Any suggestions to get Mathematica to make it look nice? - user13959
(11) @Zev Chonoles: i do not know of any web-site that can plot that equation. i wouldn't know where to begin. Also i don't understand how any solver could plot such a diagram. (Hence the question). - Ian Boyd
(6) The question is not (for me) "does this work?" but rather "how does it work?", for which ShreevatsaR provided an excellent reply. Trying it out in a program won't answer that question. - bart
(236) I don't understand why this question has so many upvotes. - user1120
(4) @Jonas Teuwen: You can contribute to the discussion at meta.math.stackexchange.com/questions/2707/… if you like. - Jonas Meyer
(58) @Jacob wolframalpha.com/input/?i=batman+equation - splattne
Corollary: Can you make google's graphing calculator plot it? - keflavich
(2) We've just had this one on TeX-SX. Here's the batman logo in TikZ: tex.stackexchange.com/q/47388/86 - Andrew Stacey
(6) The fact that this questio is the most up voted means that up votes are not so important in this site. - checkmath
(3) math is funny. to see 'the Pamela Anderson profile' google: exp(-((x-4)^2+(y-4)^2)^2/1000) + exp(-((x+4)^2+(y+4)^2)^2/1000) + 0.1exp(-((x+4)^2+(y+4)^2)^2)+0.1exp(-((x-4)^2+(y-4)^2)^2) - Helder Velez
(2) This question does not only seem to be the one with the most views, but also the one with the most downvotes (currently 46). - Dominik
(1) There's also a mathematical expression for the superman insignia, Superman - Rustyn
(1) @JonasTeuwen Because Batman. - temporary_user_name
(4) More logos equations on Alpha: wolframalpha.com/input/?i=logo+laminae - Jean-Claude Arbaut
(4) You should try it out yourself (-1). - Abhimanyu Pallavi Sudhir
(5) No efforts and still one of the most up-voted questions on M.S.E.. :( - Apurv
(1) Wolfram alpha has a bunch of "Pokemon Curves" made with parametric equations: Pikachu Curve on wolfram alpha - Shaurya Gupta
(3) To those who criticize the OP for not plotting the curve himself: I currently only know how to use Mathematica, and Mathematica 7 has no default way of plotting curves defined by implicit equations. This might be possible using 3rd party modules, but this is clearly out of reach for most casual users. - Alex M.
[+1075] [2011-07-30 05:15:47] ShreevatsaR [ACCEPTED]

As Willie Wong observed, including an expression of the form $\displaystyle \frac{|\alpha|}{\alpha}$ is a way of ensuring that $\alpha > 0$. (As $\sqrt{|\alpha|/\alpha}$ is $1$ if $\alpha > 0$ and non-real if $\alpha < 0$.)

The ellipse $\displaystyle \left( \frac{x}{7} \right)^{2} + \left( \frac{y}{3} \right)^{2} - 1 = 0$ looks like this:

ellipse

So the curve $\left( \frac{x}{7} \right)^{2}\sqrt{\frac{\left| \left| x \right|-3 \right|}{\left| x \right|-3}} + \left( \frac{y}{3} \right)^{2}\sqrt{\frac{\left| y+3\frac{\sqrt{33}}{7} \right|}{y+3\frac{\sqrt{33}}{7}}} - 1 = 0$ is the above ellipse, in the region where $|x|>3$ and $y > -3\sqrt{33}/7$:

ellipse cut

That's the first factor.

The second factor is quite ingeniously done. The curve $\left| \frac{x}{2} \right|\; -\; \frac{\left( 3\sqrt{33}-7 \right)}{112}x^{2}\; -\; 3\; +\; \sqrt{1-\left( \left| \left| x \right|-2 \right|-1 \right)^{2}}-y=0$ looks like:

second factor

This is got by adding $y = \left| \frac{x}{2} \right| - \frac{\left( 3\sqrt{33}-7 \right)}{112}x^{2} - 3$, a parabola on the positive-x side, reflected:

second factor first term

and $y = \sqrt{1-\left( \left| \left| x \right|-2 \right|-1 \right)^{2}}$, the upper halves of the four circles $\left( \left| \left| x \right|-2 \right|-1 \right)^2 + y^2 = 1$:

second factor second term

The third factor $9\sqrt{\frac{\left( \left| \left( 1-\left| x \right| \right)\left( \left| x \right|-.75 \right) \right| \right)}{\left( 1-\left| x \right| \right)\left( \left| x \right|-.75 \right)}}\; -\; 8\left| x \right|\; -\; y\; =\; 0$ is just the pair of lines y = 9 - 8|x|:

Third factor without cut

truncated to the region $0.75 < |x| < 1$.

Similarly, the fourth factor $3\left| x \right|\; +\; .75\sqrt{\left( \frac{\left| \left( .75-\left| x \right| \right)\left( \left| x \right|-.5 \right) \right|}{\left( .75-\left| x \right| \right)\left( \left| x \right|-.5 \right)} \right)}\; -\; y\; =\; 0$ is the pair of lines $y = 3|x| + 0.75$:

fourth factor without cut

truncated to the region $0.5 < |x| < 0.75$.

The fifth factor $2.25\sqrt{\frac{\left| \left( .5-x \right)\left( x+.5 \right) \right|}{\left( .5-x \right)\left( x+.5 \right)}}\; -\; y\; =\; 0$ is the line $y = 2.25$ truncated to $-0.5 < x < 0.5$.

Finally, $\frac{6\sqrt{10}}{7}\; +\; \left( 1.5\; -\; .5\left| x \right| \right)\; -\; \frac{\left( 6\sqrt{10} \right)}{14}\sqrt{4-\left( \left| x \right|-1 \right)^{2}}\; -\; y\; =\; 0$ looks like:

sixth factor without cut

so the sixth factor $\frac{6\sqrt{10}}{7}\; +\; \left( 1.5\; -\; .5\left| x \right| \right)\sqrt{\frac{\left| \left| x \right|-1 \right|}{\left| x \right|-1}}\; -\; \frac{\left( 6\sqrt{10} \right)}{14}\sqrt{4-\left( \left| x \right|-1 \right)^{2}}\; -\; y\; =\; 0$ looks like

sixth factor

As a product of factors is $0$ iff any one of them is $0$, multiplying these six factors puts the curves together, giving: (the software, Grapher.app, chokes a bit on the third factor, and entirely on the fourth)

Wholly Batman


(1) Indeed, with your analysis, we see that "Batman's curve" is a true chimera. On that note, $|x|=x\,\mathrm{sign}\,x$, so if your environment supports a sign function, you can simplify the equations a bit. Now, I wonder if a less-contrived algebraic equation might exhibit that very shape... BTW, what did GrafEq have to say about this curve? ;) - J. M. ain't a mathematician
@J. M.: Heh, I did try it on GrafEq first. :-) But it was under Wine on Mac OS X so it kept crashing, also I couldn't get it to show the "doubtful" regions in a different colour from the "confirmed" ones, so I gave up. Good idea about replacing the terms $\sqrt{|x|/x}$ everywhere by $\sqrt{\mathrm{sgn}(x)}$. It does make the equations shorter, and doesn't seem to make, in Grapher.app, any difference to the plots. - ShreevatsaR
(187) I tip my hat to you for this comprehensive dissection. - Willie Wong
+∞ :) I'd love to see this changed to .... = z, and plotted with shading based on the z values. - nealmcb
@neal: I don't expect the surface to exist; the "curve" is already very contrived as it is. - J. M. ain't a mathematician
(4) If there were only no rep-cap, Shree would be swimming in rep now... ;P - J. M. ain't a mathematician
Since, there's a YComb post as you said, @Willie, it seems likely. :) - J. M. ain't a mathematician
(40) "I don’t know how ShreevatsaR did it but he sure brought a large luggage when intelligence showered the Earth." yangkidudel.wordpress.com/2011/08/02/love-and-mathematics - Jonas Meyer
(47) @Jonas Meyer: LOL, that's embarrassing! :P But then again, if Batman is what it takes for someone to appreciate mathematics a little, well good for Batman. :-) - ShreevatsaR
@Shr: What software did you use to draw the graph? Mathematica? - user9464
(4) @Jack: Grapher, which comes by default on Mac OS X. I mentioned it in the answer actually, just before the last figure. - ShreevatsaR
(4) Okay, now do this one: PSY curve. - Douglas S. Stones
(1) I can see a lot of people learning a lot just from this dissection, bravo ShreevatsaR. - F.Webber
(3) I just found out that the curve was devised by J. Mathew Register (then "teaching at a few art schools throughout the greater Sacramento area", now "a full time professor over at American River College"): quora.com/Who-wrote-the-Batman-equation/answer/… - ShreevatsaR
1
[+219] [2011-08-03 17:55:19] user9464

You may be able to see more easily the correspondences between the equations and the graph through the following picture which is from the link [1] I got after a curious search on Google(link broken now):

enter image description here

[1] http://everythingnew.net/wp-content/uploads/2011/07/Batman-Equation-solved.png

(12) Geometer's Sketchpad? - Isaac
@Isaac: Probably right. - Tim Seguine
(2) to see the graph just google the equation: 2*sqrt(-abs(abs(x)-1)*abs(3-abs(x))/((abs(x)-1)*(3-abs(x))))‌​(1+abs(abs(x)-3)/(ab‌​s(x)-3))sqrt(1-(x/7)‌​^2)+(5+0.97(abs(x-.5‌​)+abs(x+.5))-3(abs(x‌​-.75)+abs(x+.75)))(1‌​+abs(1-abs(x))/(1-ab‌​s(x))),-3sqrt(1-(x/7‌​)^2)sqrt(abs(abs(x)-‌​4)/(abs(x)-4)),abs(x‌​/2)-0.0913722(x^2)-3‌​+sqrt(1-(abs(abs(x)-‌​2)-1)^2),(2.71052+(1‌​.5-.5abs(x))-1.35526‌​sqrt(4-(abs(x)-1)^2)‌​)sqrt(abs(abs(x)-1)/‌​(abs(x)-1))+0.9 - Helder Velez
what program is that? - Joao
(4) @HelderVelez 2014: "'abs' (and any subsequent words) was ignored because we limit queries to 32 words." - Google - Baby
@Jonjongot check the Patel answer (equal to my comment), and past the equation to the google search text window. The batman graph will appear then. (in my eq the symbol '–' following the last 0.9 do not belong to the eq). My google chrome browser accepts much more than 32 words. - Helder Velez
@HelderVelez Ah! now I can see it :D - Baby
(1) x1(y), x2(y) should be 7 * sqrt( 1 - y^2 / 9) and -7 * sqrt( 1 - y^2 / 9), respectively - moorepants
2
[+86] [2011-07-29 21:59:12] GEdgar

Here's what I got from the equation using Maple...

enter image description here


(105) What if Commissioner Gordon uses Mathematica ???? - The Chaz 2.0
(22) @The Chaz: Then Commissioner Gordon should support the Mathematica SE Site Proposal on Area 51. - Isaac
(9) touché ....$$$$ - The Chaz 2.0
(1) Here's Mathematica code. See Heike's post. I tried it on M8 and it works fine. groups.google.com/group/comp.soft-sys.math.mathematica/… - Sol
@Sol: You can do better; see my answer. - J. M. ain't a mathematician
3
[+67] [2011-07-29 22:22:44] Willie Wong

Looking at the equation, it looks like it contains terms of the form $$ \sqrt{\frac{| |x| - 1 |}{|x| - 1}} $$ which evaluates to $$\begin{cases} 1 & |x| > 1\\ i & |x| < 1\end{cases} $$

Since any non-zero real number $y$ cannot be equal to a purely imaginary non-zero number, the presence of that term is a way of writing a piece-wise defined function as a single expression. My guess is that if you try to plot this in $\mathbb{C}^2$ instead of $\mathbb{R}^2$ you will get all kinds of awful.


(5) Yeah, the equation looks too contrived to me. :) A parametric form (it's just quadratic and linear arcs sewn together, it looks) would still be messy, but not as messy. (Probably a good job for splines...) - J. M. ain't a mathematician
+1 i was wondering how they split it up into sections. - Ian Boyd
(29) " My guess is that if you try to plot this in C2 instead of R2 you will get all kinds of awful." What did you expect? The analytic continuation of the Batman symbol?? - jwg
4
[+55] [2011-08-07 17:01:16] J. M. ain't a mathematician

Since people (not from this site, but still...) keep bugging me, and I am unable to edit my previous answer, here's Mathematica code for plotting this monster:

Plot[{With[{w = 3 Sqrt[1 - (x/7)^2], 
            l = 6/7 Sqrt[10] + (3 + x)/2 - 3/7 Sqrt[10] Sqrt[4 - (x + 1)^2], 
            h = (3 (Abs[x - 1/2] + Abs[x + 1/2] + 6) -
                 11 (Abs[x - 3/4] + Abs[x + 3/4]))/2, 
            r = 6/7 Sqrt[10] + (3 - x)/2 - 3/7 Sqrt[10] Sqrt[4 - (x - 1)^2]}, 
           w + (l - w) UnitStep[x + 3] + (h - l) UnitStep[x + 1] +
           (r - h) UnitStep[x - 1] + (w - r) UnitStep[x - 3]],
      1/2 (3 Sqrt[1 - (x/7)^2] + Sqrt[1 - (Abs[Abs[x] - 2] - 1)^2] + Abs[x/2] -
      ((3 Sqrt[33] - 7)/112) x^2 - 3) (Sign[x + 4] - Sign[x - 4]) - 3*Sqrt[1 - (x/7)^2]},
     {x, -7, 7}, AspectRatio -> Automatic,  Axes -> None, Frame -> True,
     PlotStyle -> Black]

Mathematica graphics

This should work even for versions that do not have the Piecewise[] construct. Enjoy. :P


5
[+46] [2011-07-30 18:46:41] J. M. ain't a mathematician

In fact, the five linear pieces that consist the "head" (corresponding to the third, fourth, and fifth pieces in Shreevatsa's answer) can be expressed in a less complicated manner, like so:

$$y=\frac{\sqrt{\mathrm{sign}(1-|x|)}}{2}\left(3\left(\left|x-\frac12\right|+\left|x+\frac12\right|+6\right)-11\left(\left|x-\frac34\right|+\left|x+\frac34\right|\right)\right)$$

This can be derived by noting that the functions

$$\begin{cases}f(x)&\text{if }x<c\\g(x)&\text{if }c<x\end{cases}$$

and $f(x)+(g(x)-f(x))U(x-c)$ (where $U(x)$ is the unit step function) are equivalent, and using the "relation"

$$U(x)=\frac{x+|x|}{2x}$$

Note that the elliptic sections (both ends of the "wings", corresponding to the first piece in Shreevatsa's answer) were cut along the lines $y=-\frac37\left((2\sqrt{10}+\sqrt{33})|x|-8\sqrt{10}-3\sqrt{33}\right)$, so the elliptic potion can alternatively be expressed as

$$\left(\left(\frac{x}{7}\right)^2+\left(\frac{y}{3}\right)^2-1\right)\sqrt{\mathrm{sign}\left(y+\frac37\left((2\sqrt{10}+\sqrt{33})|x|-8\sqrt{10}-3\sqrt{33}\right)\right)}=0$$

Theoretically, since all you have are arcs of linear and quadratic curves, the chimera can be expressed parametrically using rational B-splines, but I'll leave that for someone else to explore...


(10) Okay, make that an hour. Sheesh. *facepalm* - J. M. ain't a mathematician
(5) Great, this would be much clearer. BTW, I think it's safer to leave it at "can be expressed using B-splines" than to actually try it out: who knows how many hours that will waste, right? :-) - ShreevatsaR
6
[+45] [2011-08-03 18:58:54] user9464

The following is what I got from the equations using MATLAB: enter image description here

Here is the M-File (thanks to this link [1]):

clf; clc; clear all; 
syms x y

eq1 = ((x/7)^2*sqrt(abs(abs(x)-3)/(abs(x)-3))+(y/3)^2*sqrt(abs(y+3/7*sqrt(33))/(y+3/7*sqrt(33)))-1);
eq2 = (abs(x/2)-((3*sqrt(33)-7)/112)*x^2-3+sqrt(1-(abs(abs(x)-2)-1)^2)-y);
eq3 = (9*sqrt(abs((abs(x)-1)*(abs(x)-.75))/((1-abs(x))*(abs(x)-.75)))-8*abs(x)-y);
eq4 = (3*abs(x)+.75*sqrt(abs((abs(x)-.75)*(abs(x)-.5))/((.75-abs(x))*(abs(x)-.5)))-y);
eq5 = (2.25*sqrt(abs((x-.5)*(x+.5))/((.5-x)*(.5+x)))-y);
eq6 = (6*sqrt(10)/7+(1.5-.5*abs(x))*sqrt(abs(abs(x)-1)/(abs(x)-1))-(6*sqrt(10)/14)*sqrt(4-(abs(x)-1)^2)-y);


axes('Xlim', [-7.25 7.25], 'Ylim', [-5 5]);
hold on

ezplot(eq1,[-8 8 -3*sqrt(33)/7 6-4*sqrt(33)/7]);
ezplot(eq2,[-4 4]);
ezplot(eq3,[-1 -0.75 -5 5]);
ezplot(eq3,[0.75 1 -5 5]);
ezplot(eq4,[-0.75 0.75 2.25 5]);
ezplot(eq5,[-0.5 0.5 -5 5]);
ezplot(eq6,[-3 -1 -5 5]);
ezplot(eq6,[1 3 -5 5]);
colormap([0 0 1])

title('Batman');
xlabel('');
ylabel('');
hold off
[1] https://gist.github.com/1119139

7
[+23] [2012-06-02 08:38:35] copper.hat

The 'Batman equation' above relies on an artifact of the plotting software used which blithely ignores the fact that the value $\sqrt{\frac{|x|}{x}}$ is undefined when $x=0$. Indeed, since we’re dealing with real numbers, this value is really only defined when $x>0$. It seems a little ‘sneaky’ to rely on the solver to ignore complex values and also to conveniently ignore undefined values.

A nicer solution would be one that is unequivocally defined everywhere (in the real, as opposed to complex, world). Furthermore, a nice solution would be ‘robust’ in that small variations (such as those arising from, say, roundoff) would perturb the solution slightly (as opposed to eliminating large chunks).

Try the following in Maxima (actually wxmaxima) which is free. The resulting plot is not quite as nice as the plot above (the lines around the head don’t have that nice ‘straight line’ look), but seems more ‘legitimate’ to me (in that any reasonable solver should plot a similar shape). Please excuse the code mess.

/* [wxMaxima batch file version 1] [ DO NOT EDIT BY HAND! ]*/
/* [ Created with wxMaxima version 0.8.5 ] */

/* [wxMaxima: input   start ] */
load(draw);
/* [wxMaxima: input   end   ] */

/* [wxMaxima: input   start ] */
f(a,b,x,y):=a*x^2+b*y^2;
/* [wxMaxima: input   end   ] */

/* [wxMaxima: input   start ] */
c1:sqrt(26);
/* [wxMaxima: input   end   ] */

/* [wxMaxima: input   start ] */
draw2d(implicit(
f(1/36,1/9,x,y)
+max(0,2-f(1.5,1,x+3,y+2.7))
+max(0,2-f(1.5,1,x-3,y+2.7))
+max(0,2-f(1.9,1/1.7,(5*(x+1)+(y+3.5))/c1,(-(x+1)+5*(y+3.5))/c1))
+max(0,2-f(1.9,1/1.7,(5*(x-1)-(y+3.5))/c1,((x-1)+5*(y+3.5))/c1))
+max(0,2-((1.1*(x-2))^4-(y-2.1)))
+max(0,2-((1.1*(x+2))^4-(y-2.1)))
+max(0,2-((1.5*x)^8-(y-3.5)))
-1,
x,-6,6,y,-4,4));
/* [wxMaxima: input   end   ] */

/* Maxima can't load/batch files which end with a comment! */
"Created with wxMaxima"$

The resulting plot is: enter image description here

(Note that this is, more or less, a copy of the entry I made on http://blog.makezine.com.)


I really think that an indeterminate value multiplied by zero equals zero, so it seems to be legit. Is there any reason 0 * 0/0 should not be defined to be zero? - dbanet
@dbanet: What are you referring to? The issue above is that the original equations rely on the plotting software ignoring undefined values, which is peculiar, to say the least. The expression $\sqrt{\frac{|x|}{x}}$ (with $x$ being replaced by some expression) is what I referred to and it appears without being multiplied by $x$. - copper.hat
@copper-hat: The function $f(x)=\sqrt{\frac{|x|}{x}}$ appears only in boolean expressions $F:x\to \{\text{True},\text{False}\}$ of form $f(x)g(x)=0$, so if $g(x)$ is defined as $g:x\in\mathbb{C}\to{0}$, I would rather evaluate $F(0)$ to $\text{True}$ than to $\text{False}$, as $\forall{x}:f(x)g(x)=0\Longleftrightarrow \Big(f(x)=0\lor g(x)=0\Big)$. - dbanet
@dbanet: I'm really not sure what you are getting at. Look at the expressions in the question. They rely on the expression $\sqrt{\frac{|x|}{x}}$ returning zero for $x \le 0$, which is strange (look at Willie's answer math.stackexchange.com/a/54521/27978). My answer plots level sets, which avoids this whole issue. - copper.hat
@copper-hat: Why is it strange? For $x<0:\operatorname{Im}\left(\sqrt{\frac{|x|}{x}}\right)\neq{0}$. For $x=0$ it is indeterminate but that does not matter as long as any other factor evaluates to zero, so that the whole boolean expression holds. - dbanet
(1) @dbanet: I don't really get your point. In the formula in the question there are lots of expressions of the above form that are multiplied by quantities that do not evaluate to zero when $x<0$. If they did, there would be no need to have the strange expression in the first place. - copper.hat
@copper-hat: In your answer you state that "the solver ignores complex and indeterminate values". That is not true. Can you please give an example point $(x;y)$ which belongs to the plot constructed by a (bad) plotting program, but only due to the fact that "the solver ignores complex and indeterminate values"? - dbanet
@dbanet: Look at this answer to see what is happening: math.stackexchange.com/a/54568/27978. The plot is composed of sections that are, in my opinion, artificially stitched together using the 'ignore if complex' plotting approach. There are lots of example points in the answer at the start of this comment. It would help me answer your question if you could tell me what you trying to get at. - copper.hat
@copper-hate: Can you please give an example point which belongs to the plot constructed by a (bad) plotting program, but only due to the fact that "the solver ignores complex and indeterminate values"? There is nothing that is "ignored" by the plotter. The complex or indeterminate value is getting multiplied by zero. - dbanet
@dbanet: No need to be rude. I never wrote that the plotting program was bad. If you take $x=3$ in the first ellipse, then the quantity $\left( \frac{x}{7} \right)^{2}\sqrt{\frac{\left| \left| x \right|-3 \right|}{\left| x \right|-3}}$ is indeterminate and is not multiplied by zero. - copper.hat
@copper-hat: I'm sorry, never meant to be rude to you. Indeed, no multiple at point e. g. $(3;1)$ evaluates to zero, so the whole LHS evaluates to indeterminate due to the first multiple, so one can't determine whether to include the $(3;1)$ point to the plot or not. Thank you. - dbanet
@dbanet: I was kidding; you wrote at-copper-hate above... - copper.hat
@copper-hat: Oops a typo. :P - dbanet
8
[+17] [2011-07-30 02:18:16] stoicfury

Here's the equations typed out if you want save time with writing it yourself.

(x/7)^2*SQRT(ABS(ABS(x)-3)/(ABS(x)-3))+(y/3)^2\*SQRT(ABS(y+3*SQRT(33)/7)/(y+3*SQRT(33)/7))-1=0
ABS(x/2)-((3*SQRT(33)-7)/112)*x^2-3+SQRT(1-(ABS(ABS(x)-2)-1)^2)-y=0
9*SQRT(ABS((ABS(x)-1)*(ABS(x)-0.75))/((1-ABS(x))*(ABS(x)-0.75)))-8*ABS(x)-y=0
3*ABS(x)+0.75*SQRT(ABS((ABS(x)-0.75)*(ABS(x)-0.5))/((0.75-ABS(x))*(ABS(x)-0.5)))-y=0
2.25*SQRT(ABS((x-0.5)*(x+0.5))/((0.5-x)*(0.5+x)))-y=0
(6*SQRT(10))/7+(1.5-0.5*ABS(x))*SQRT(ABS(ABS(x)-1)/(ABS(x)-1))-((6*SQRT(10))/14)*SQRT(4-(ABS(x)-1)^2)-y=0

Also: http://pastebin.com/x9T3DSDp


The multiple "=0" lines are different than the mulitiplications in the original, there are some backslashes in there that throw things off, and the formatting is hard to read. The pastebin is better. - nealmcb
(1) I've made a meta post about this answer. - Zev Chonoles
(2) I would have commented it but I don't have enough rep to comment :P - stoicfury
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[+5] [2014-02-19 16:37:05] Shivam Patel

Sorry but this is not the answer but too long for a comment: Probably the easiest verification is to type the equation on Google you'l be surprised : The easiest way is to Google :2 sqrt(-abs(abs(x)-1)abs(3-abs(x))/((abs(x)-1)(3-abs(x))))(1+abs(abs(x)-3)/(abs(x)-3))sqrt(1-(x/7)^2)+(5+0.97(abs(x-.5)+abs(x+.5))-3(abs(x-.75)+abs(x+.75)))(1+abs(1-abs(x))/(1-abs(x))),-3sqrt(1-(x/7)^2)sqrt(abs(abs(x)-4)/(abs(x)-4)),abs(x/2)-0.0913722(x^2)-3+sqrt(1-(abs(abs(x)-2)-1)^2),(2.71052+(1.5-.5abs(x))-1.35526sqrt(4-(abs(x)-1)^2))sqrt(abs(abs(x)-1)/(abs(x)-1))+0.9


@copper, it is just because of the algorithm that draw uses for drawing implicit functions. You need to setup the variables ip_grid and ip_grid_in, that are the sampling values in your region. For example draw2d(ip_grid=[60,60], ip_grid_in=[20,20], implicit(y^2=x^3-2*x+1, x, -4,4, y, -4,4) ); - nicoguaro
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