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find the periodicity of a modular function

Is there an “exists” function for jQuery?How do you find out the caller function in JavaScript?var functionName = function() vs function functionName() How do I find out which DOM element has the focus?Designing function f(f(n)) == -nSet a default parameter value for a JavaScript functionWhere can I find documentation on formatting a date in JavaScript?Easy interview question got harder: given numbers 1..100, find the missing number(s)Is there a standard function to check for null, undefined, or blank variables in JavaScript?Find object by id in an array of JavaScript objects

I created my own version of the fractal (which in fact takes the same principle as the Pythagorean tree).

If you want to know what this gives you can do a little tour here.

 var myCanvas = document.getElementById('cnv');
 var ctx = myCanvas.getContext('2d');
 myCanvas.style.backgroundColor = "rgba(0, 0, 0, 0.1)";

My draw function like this.

 function draw(x, y, len, angle) 
 ctx.beginPath();
 ctx.save();
 ctx.translate(x, y);
 ctx.rotate((angle * Math.sin(len)) );
 ctx.moveTo(0, 0);
 ctx.lineTo(0, -len);
 ctx.stroke(); 
 ctx.strokeStyle = "grey";

 if(len < 1) 
 ctx.restore();
 return;
 
 var slider = document.getElementById("myRangeAngle");
 slider.oninput = function() 
 var p = document.getElementById('p');
 p.innerHTML = "L'angle : " + slider.value;
 ctx.clearRect(0, 0, 1366, 900); 
 draw(550,578,120,0);
 
 draw(0, -len, len*0.7, - slider.value);
 draw(0, -len, len*0.7, slider.value);
 ctx.restore();
 

 draw(550,578,200,0);

https://codepen.io/m-metore/pen/Mxvqdq

Here is my problem I have an iterative function, U0 = 200; A + 1 = A * 0.7

or Un = U0 * 0.7 ^ n.

Then I multiply my angle x in (rad) by this formula:

f (x) = x * sin (Un).

My question is is there an integer x for which (fx) gives an integer.

In other words, we must solve x * sin (U0 * 0.7 ^ n) = 2 * PI; (where x and n are integers) and we look for x (which is an integer)
I specify that U0 = 200;

I do not think there is an x such that this equation is true. Do you have any idea of how to prove it?

thank you

edited Mar 21 at 22:31

ahmet

14212

asked Mar 21 at 20:30

meteor meteor

This is more appropriate for Mathematics. But if both U0 and n are known, you can compute the sine and you are left with x * s = k * 2 pi. Further x = k * 2 pi / s. If we substitute 2 * pi / s = c, we want to find k, such that k * c is an integer. The easiest way to do is to try a few k (you probably don't want large k anyway).

– Nico Schertler
Mar 22 at 3:04

add a comment |

I created my own version of the fractal (which in fact takes the same principle as the Pythagorean tree).

If you want to know what this gives you can do a little tour here.

 var myCanvas = document.getElementById('cnv');
 var ctx = myCanvas.getContext('2d');
 myCanvas.style.backgroundColor = "rgba(0, 0, 0, 0.1)";

My draw function like this.

 function draw(x, y, len, angle) 
 ctx.beginPath();
 ctx.save();
 ctx.translate(x, y);
 ctx.rotate((angle * Math.sin(len)) );
 ctx.moveTo(0, 0);
 ctx.lineTo(0, -len);
 ctx.stroke(); 
 ctx.strokeStyle = "grey";

 if(len < 1) 
 ctx.restore();
 return;
 
 var slider = document.getElementById("myRangeAngle");
 slider.oninput = function() 
 var p = document.getElementById('p');
 p.innerHTML = "L'angle : " + slider.value;
 ctx.clearRect(0, 0, 1366, 900); 
 draw(550,578,120,0);
 
 draw(0, -len, len*0.7, - slider.value);
 draw(0, -len, len*0.7, slider.value);
 ctx.restore();
 

 draw(550,578,200,0);

https://codepen.io/m-metore/pen/Mxvqdq

Here is my problem I have an iterative function, U0 = 200; A + 1 = A * 0.7

or Un = U0 * 0.7 ^ n.

Then I multiply my angle x in (rad) by this formula:

f (x) = x * sin (Un).

My question is is there an integer x for which (fx) gives an integer.

In other words, we must solve x * sin (U0 * 0.7 ^ n) = 2 * PI; (where x and n are integers) and we look for x (which is an integer)
I specify that U0 = 200;

I do not think there is an x such that this equation is true. Do you have any idea of how to prove it?

thank you

edited Mar 21 at 22:31

ahmet

14212

asked Mar 21 at 20:30

meteor meteor

This is more appropriate for Mathematics. But if both U0 and n are known, you can compute the sine and you are left with x * s = k * 2 pi. Further x = k * 2 pi / s. If we substitute 2 * pi / s = c, we want to find k, such that k * c is an integer. The easiest way to do is to try a few k (you probably don't want large k anyway).

– Nico Schertler
Mar 22 at 3:04

add a comment |

I created my own version of the fractal (which in fact takes the same principle as the Pythagorean tree).

If you want to know what this gives you can do a little tour here.

 var myCanvas = document.getElementById('cnv');
 var ctx = myCanvas.getContext('2d');
 myCanvas.style.backgroundColor = "rgba(0, 0, 0, 0.1)";

My draw function like this.

 function draw(x, y, len, angle) 
 ctx.beginPath();
 ctx.save();
 ctx.translate(x, y);
 ctx.rotate((angle * Math.sin(len)) );
 ctx.moveTo(0, 0);
 ctx.lineTo(0, -len);
 ctx.stroke(); 
 ctx.strokeStyle = "grey";

 if(len < 1) 
 ctx.restore();
 return;
 
 var slider = document.getElementById("myRangeAngle");
 slider.oninput = function() 
 var p = document.getElementById('p');
 p.innerHTML = "L'angle : " + slider.value;
 ctx.clearRect(0, 0, 1366, 900); 
 draw(550,578,120,0);
 
 draw(0, -len, len*0.7, - slider.value);
 draw(0, -len, len*0.7, slider.value);
 ctx.restore();
 

 draw(550,578,200,0);

https://codepen.io/m-metore/pen/Mxvqdq

Here is my problem I have an iterative function, U0 = 200; A + 1 = A * 0.7

or Un = U0 * 0.7 ^ n.

Then I multiply my angle x in (rad) by this formula:

f (x) = x * sin (Un).

My question is is there an integer x for which (fx) gives an integer.

In other words, we must solve x * sin (U0 * 0.7 ^ n) = 2 * PI; (where x and n are integers) and we look for x (which is an integer)
I specify that U0 = 200;

I do not think there is an x such that this equation is true. Do you have any idea of how to prove it?

thank you

edited Mar 21 at 22:31

ahmet

14212

asked Mar 21 at 20:30

meteor meteor

I created my own version of the fractal (which in fact takes the same principle as the Pythagorean tree).

If you want to know what this gives you can do a little tour here.

 var myCanvas = document.getElementById('cnv');
 var ctx = myCanvas.getContext('2d');
 myCanvas.style.backgroundColor = "rgba(0, 0, 0, 0.1)";

My draw function like this.

 function draw(x, y, len, angle) 
 ctx.beginPath();
 ctx.save();
 ctx.translate(x, y);
 ctx.rotate((angle * Math.sin(len)) );
 ctx.moveTo(0, 0);
 ctx.lineTo(0, -len);
 ctx.stroke(); 
 ctx.strokeStyle = "grey";

 if(len < 1) 
 ctx.restore();
 return;
 
 var slider = document.getElementById("myRangeAngle");
 slider.oninput = function() 
 var p = document.getElementById('p');
 p.innerHTML = "L'angle : " + slider.value;
 ctx.clearRect(0, 0, 1366, 900); 
 draw(550,578,120,0);
 
 draw(0, -len, len*0.7, - slider.value);
 draw(0, -len, len*0.7, slider.value);
 ctx.restore();
 

 draw(550,578,200,0);

https://codepen.io/m-metore/pen/Mxvqdq

Here is my problem I have an iterative function, U0 = 200; A + 1 = A * 0.7

or Un = U0 * 0.7 ^ n.

Then I multiply my angle x in (rad) by this formula:

f (x) = x * sin (Un).

My question is is there an integer x for which (fx) gives an integer.

In other words, we must solve x * sin (U0 * 0.7 ^ n) = 2 * PI; (where x and n are integers) and we look for x (which is an integer)
I specify that U0 = 200;

I do not think there is an x such that this equation is true. Do you have any idea of how to prove it?

thank you

javascript math

edited Mar 21 at 22:31

ahmet

14212

asked Mar 21 at 20:30

meteor meteor

edited Mar 21 at 22:31

ahmet

14212

asked Mar 21 at 20:30

meteor meteor

edited Mar 21 at 22:31

ahmet

14212

edited Mar 21 at 22:31

ahmet

14212

edited Mar 21 at 22:31

ahmet

14212

asked Mar 21 at 20:30

meteor meteor

asked Mar 21 at 20:30

meteor meteor

asked Mar 21 at 20:30

meteor meteor

This is more appropriate for Mathematics. But if both U0 and n are known, you can compute the sine and you are left with x * s = k * 2 pi. Further x = k * 2 pi / s. If we substitute 2 * pi / s = c, we want to find k, such that k * c is an integer. The easiest way to do is to try a few k (you probably don't want large k anyway).

– Nico Schertler
Mar 22 at 3:04

add a comment |

This is more appropriate for Mathematics. But if both U0 and n are known, you can compute the sine and you are left with x * s = k * 2 pi. Further x = k * 2 pi / s. If we substitute 2 * pi / s = c, we want to find k, such that k * c is an integer. The easiest way to do is to try a few k (you probably don't want large k anyway).

– Nico Schertler
Mar 22 at 3:04

This is more appropriate for Mathematics. But if both U0 and n are known, you can compute the sine and you are left with x * s = k * 2 pi. Further x = k * 2 pi / s. If we substitute 2 * pi / s = c, we want to find k, such that k * c is an integer. The easiest way to do is to try a few k (you probably don't want large k anyway).

– Nico Schertler
Mar 22 at 3:04

add a comment |

1 Answer
1

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according to the Lindemann-Weierstrass theorem sin (200 * O.7 ^ n) is a transcendent number https://planetmath.org/proofoflindemannweierstrasstheoremandthateandpiaretranscendental
so I have to show that arcsin (2 * PI / x) is an irrational number.

The best way to do this is to try some k (anyway, you do not want big
k).

in fact I want to demonstrate for all numbers k that's why.
Thank you for your help !

answered Mar 22 at 5:21

meteor meteor

add a comment |

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The best way to do this is to try some k (anyway, you do not want big
k).

in fact I want to demonstrate for all numbers k that's why.
Thank you for your help !

answered Mar 22 at 5:21

meteor meteor

add a comment |

The best way to do this is to try some k (anyway, you do not want big
k).

in fact I want to demonstrate for all numbers k that's why.
Thank you for your help !

answered Mar 22 at 5:21

meteor meteor

add a comment |

The best way to do this is to try some k (anyway, you do not want big
k).

in fact I want to demonstrate for all numbers k that's why.
Thank you for your help !

answered Mar 22 at 5:21

meteor meteor

The best way to do this is to try some k (anyway, you do not want big
k).

in fact I want to demonstrate for all numbers k that's why.
Thank you for your help !

answered Mar 22 at 5:21

meteor meteor

answered Mar 22 at 5:21

meteor meteor

answered Mar 22 at 5:21

meteor meteor

answered Mar 22 at 5:21

meteor meteor

add a comment |

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