Math Is Fun Forum

Hi,benice;
I review the sites and infer that the earliest one is before 2011-07-29 09:35:26 (+8). But the site you gave just show "9 months ago", when the equation was flooding, so I don't know whether that is the original one.
We tried to plot it, only to found it abused "sqrt" so there is no plot. I tried to correct but gave up later because of the complex numbers (complex, not "i").

Try this: HTTP://www.xamuel.com/inverse-graphing-calculator.php?phrase=anonimnystefy&clientAction=15.click

It couldn't be nicer for you to do that. But what you're considering is right so I hope that benice can put out them himself. Thank you all very much.

Aaa...Where's the dragonfly ?

Oh...still not work...Otherwise you're in China. The goverment do such things for no reason.
But I got this(may be rude but just a joke)

May I ask for some help? What's the inequalities of the graphs? The website is blocked.

Hi,MathsIsFun;
You've got a wonderful idea! But I have to say that someone did have the same idea and he made an wonderful grapher called GrafEq. But unfortunatly, the function of graphing has been out of time, so it's only for win6 or winXP.
He is Jeff Tupper, and he found the super inequality bellow. Why super? Its graph contains itself where 0<=x<=105 and n<=y<=n+16 ! n=96093937991895888497167296212785275471500433966012930665150551927170280239526642
46896428421743507181212671537827706233559932372808741443078913259639413377234878
57735749823926629715517173716995165232890538221612403238855866184013235585136048
82869333790249145422928866708109618449609170518345406782773155170540538162738096
76025656250169814820834187831638491155902256100036523513703438744618483787372381
98224849863465033159410054974700593138339226497249461751545728366702369745461014
655997933798537483143786841806593422227898388722980000748404719
As expected, this grapher has bugs, too.It allows x/0=0, x^(1/3)¡Ý0. It is the same problem of f(x,y)≈0 that you faced with.
Why not share your commands? Let's do this wonderful project together ! Many friends of mine and I want to improve it after our Chinese University Enterance Exam.
I've got a Jeff Tupper's paper about this, would you like to read ?

A leg of turkey exactly!

(x+13+sqrt((x+4)^2+4)-2sqrt((x+8)^2+4))^2*(12x*y^2+x^3-11y^2+x^2+7x-4)^3+(8y^2+6x^2-16x+4)^3*y^4*(x+13)^2=0 where x=-15 to 5, guess what's in it?
Tip:Thanksgiving Days

ρ=1-cosθ , an old equation shows heart cruve (bottom instead I think).But I found "(x²+y²-1)³=x²y³" on WolframAlpha.
"Who found this?",I search it on Google and finally I learned it was Siehe Beutel.But how he found this?
We chosed Archimedes spiral: x²+y²=arc tan²(y/x)
we find that "tan x" is simillar to "x³",so we got
(x²+y²)³=(y/x)²
we find heart isn't a symmetrical shape,so we change "y²" by "y³"
As x becomes larger,the difference between "tan x" and "x³" becomes larger,so we plus "x⁴"
Aha~just a joke ! Maybe following can be turth.
He used a simple ellipse "x²+y²-xy=1".
We transform it to "x²+y²-1=|x|y".
NO ABS ! We get "(x²+y²-1)²=x²y²".
But now we lose the relation of that "x²+y²-1" is the same to "y" in "±" (I can't convey in proper way)
So we try the equation "(a²x²+b²y²-1)^(2k+m)=(by)^m*(abxy)^2k"
when a、b→1,there is the equation "
(x²+y²-1)^m=y^m"
So we have "|x|^m=((x²+y²-1)/y)^m=1"
Compare plots of "x²+y²-xy=1" and "x²+y²-y=1",we find only when x∈[-1/2,1/2] their difference become obvious,so m→0.
We make m=1，so k→+∞.
By trying,due to the equation's interval,when k=1 we get such a beautiful plot : (x²+y²-1)³=x²y³

Oh,I got it ! Different softwares define "x^(2/3)" differently.It should be (x^2)^(1/3)

Hi anonimnystefy;
Aaaa......they are created in different ways.Some are using the transformation in which we replace x by x±f(x).