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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
javascript math
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
javascript math
This is more appropriate for Mathematics. But if bothU0
andn
are known, you can compute the sine and you are left withx * s = k * 2 pi
. Furtherx = k * 2 pi / s
. If we substitute2 * pi / s = c
, we want to findk
, such thatk * c
is an integer. The easiest way to do is to try a fewk
(you probably don't want largek
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
javascript math
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
javascript math
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 bothU0
andn
are known, you can compute the sine and you are left withx * s = k * 2 pi
. Furtherx = k * 2 pi / s
. If we substitute2 * pi / s = c
, we want to findk
, such thatk * c
is an integer. The easiest way to do is to try a fewk
(you probably don't want largek
anyway).
– Nico Schertler
Mar 22 at 3:04
add a comment |
This is more appropriate for Mathematics. But if bothU0
andn
are known, you can compute the sine and you are left withx * s = k * 2 pi
. Furtherx = k * 2 pi / s
. If we substitute2 * pi / s = c
, we want to findk
, such thatk * c
is an integer. The easiest way to do is to try a fewk
(you probably don't want largek
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
add a comment |
1 Answer
1
active
oldest
votes
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 !
add a comment |
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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 !
add a comment |
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 !
add a comment |
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 !
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 meteormeteor meteor
11
11
add a comment |
add a comment |
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This is more appropriate for Mathematics. But if both
U0
andn
are known, you can compute the sine and you are left withx * s = k * 2 pi
. Furtherx = k * 2 pi / s
. If we substitute2 * pi / s = c
, we want to findk
, such thatk * c
is an integer. The easiest way to do is to try a fewk
(you probably don't want largek
anyway).– Nico Schertler
Mar 22 at 3:04