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**Sumasoltin****Member**- Registered: 2012-04-28
- Posts: 30

I'm a Chinese high school student who love equation and function CRAZILY ! And I've found a lot about this mathematical game.Let's look into!

Absolution,I think it useful but unpleasant,can make any project practicable.As is known to all,the equation "ab=0" is equal to"a=0 or b=0".Similarly,"g(x,y)f(x,y)=0" is equal to "g(x,y)=0 or f(x,y)=0",so now you can use "plus" to combine different plots.And another equation "f²(x,y)+g²(x,y)=0" is equal to "f(x,y)=0 and g(x,y)=0",and in this way you can cut a part of plot.How? There comes the equation "|x-a|+|x-b|-|a-b|=0",whose graph is a part of the plane.Now,let's graph "(x-y)²+(|x-1|+|x+1|-2)²=0".It's a segment.We find |x-1|+|x+1|-2>0,so this equation can be simplified as (x-y)²+|x-1|+|x+1|=2.

This is easy,but not best.Keep patient and you're due to find an equation showing special meanings.Take your aim apart and find equations describe them,and combine them together.

But I dislike absolution,so I found the "fusion".We plus "x=1" and "(x+1/2)²+y²=1",and then we got "x((x+1/2)²+y²)=1".whose plot is the fusion of a circle and straight line.Similarly,many plots can be combine in this way.

There are lots of tips but I can only convey those with my poor English.To show more about this geeks' game I will show you some my original equations.

What's more,have you found some interesting graphs of equation?Share it if it's convenient to you!

*Last edited by Sumasoltin (2012-04-28 22:18:03)*

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**Sumasoltin****Member**- Registered: 2012-04-28
- Posts: 30

Aaaaa......the most beautiful one is larger than 100kb

The function is:

y=abs[x]-abs[x/2+3]-abs[x/2-3]+3(x-14)(x-16)/abs[2(x-14)(x-16)]+3(x-12)(x-18)/abs[(x-12)(x-18)]+(sqrt[4-(abs[x-14]-abs[x-16])^2]+sqrt[36-(abs[x-12]-abs[x-18])^2])/4-3+(18+abs[2x+24]+abs[2x-24]-abs[2x+12]-abs[2x-12]-6(x-14)(x-16)/abs[(x-14)(x-16)]-6(x-12)(x-18)/abs[(x-12)(x-18)]-18(x+14)(x+16)/abs[(x+14)(x+16)]+sqrt[4-(abs[x-14]-abs[x-16])^2]-sqrt[36-(abs[x-12]-abs[x-18])^2])sin[5pi*x]/4

Everyone knows that the heart cruve is a closed cruve so we can't find a functiom with a heart-shaped plot.Really?

y=x^(2/3)+sqrt(4-x^2)sin(5pi*x)

*Last edited by Sumasoltin (2012-05-01 04:41:19)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,042

Hi Sumasoltin;

Welcome to the forum.

For your heart curve, what is the interval for x?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**Sumasoltin****Member**- Registered: 2012-04-28
- Posts: 30

Hi bobbym;

Thank you.I'm sorry to have forgot to say that.The first one is from x=-25 to 20,and the second is in [-5,5]

I found something interesting I set down before:Nike equation

*Last edited by Sumasoltin (2012-04-29 23:19:01)*

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

These images are so cool! How did you derive the heart curve and Nike curve?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,042

Hi Sumasoltin;

What are you using to do the graphing?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**Sumasoltin****Member**- Registered: 2012-04-28
- Posts: 30

Hi bobbym;

There are lots of mathematical softwares,like matlb,mathematic,and to graph equations I prefer GrafEq(w w w .xamuel.com/graphs-of-implicit-equations)

*Last edited by Sumasoltin (2012-04-30 00:05:31)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,042

y=x^(2/3)+sqrt(4-x^2)sin(5pi*x)

Are you sure this is the right equation?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**Sumasoltin****Member**- Registered: 2012-04-28
- Posts: 30

Hi anonimnystefy;

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

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**Sumasoltin****Member**- Registered: 2012-04-28
- Posts: 30

To bobbym;

Aaaa...what's wrong?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,042

Nothing yet, I am rechecking, probably the program.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Sumasoltin****Member**- Registered: 2012-04-28
- Posts: 30

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

Still not a heart graph!

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**Sumasoltin****Member**- Registered: 2012-04-28
- Posts: 30

WolframAlpha shows as this.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Oh,yes! Got it now! Thanks for the cool graph! Got more?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,042

Yes, I got that, I did not see the heart at first.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

Here's another heart graph that you get by varying the constant with pi*x inside the sine function:

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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What is cbrt?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

Cube root function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Got MIF's graph to do it.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

Good! It is nice. Did you try using sin(k*pi*x) instead of sin(5*pi*x) for arbitrary values of k?

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**Sumasoltin****Member**- Registered: 2012-04-28
- Posts: 30

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

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**Sumasoltin****Member**- Registered: 2012-04-28
- Posts: 30

Oh, that's funny ! Without blanks "= (" was transformed into negetive face!

I did use k instead of 5,but a known number looks better.

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**Sumasoltin****Member**- Registered: 2012-04-28
- Posts: 30

(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

*Last edited by Sumasoltin (2012-04-30 01:26:19)*

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

Check this out:Some values of k

*Last edited by anonimnystefy (2012-04-30 01:24:49)*

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