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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













0















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










share|improve this question
























  • 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















0















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










share|improve this question
























  • 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













0












0








0


1






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










share|improve this question
















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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 21 at 22:31









ahmet

14212




14212










asked Mar 21 at 20:30









meteor meteormeteor meteor

11




11












  • 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
















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












1 Answer
1






active

oldest

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0














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 !






share|improve this answer























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    1 Answer
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    active

    oldest

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    active

    oldest

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    0














    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 !






    share|improve this answer



























      0














      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 !






      share|improve this answer

























        0












        0








        0







        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 !






        share|improve this answer













        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 !







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Mar 22 at 5:21









        meteor meteormeteor meteor

        11




        11





























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