Does equation $y = sin(x - y)$ imply an analytical form $f$ which satisfies $y = f(x)$?
$begingroup$
Or is there a way to calculate an approximate value of $y$ of any given $x$?
The graph looks like this:
y = sin(x - y) by WolframAlpha
Sorry for my poor English. I'm learning WebGL now, then I want to draw something and then run into this equation. So what I really want is a way to calculate $y$ when a certain $x$ is given.
Background: how did I get this equation
I want to draw my previous work using WebGL. It is a group of circles implying an equiangular spiral . When using canvas2d, I calculate some circles then draw them by z-index order (a.k.a. rasterization). But in a fragment shader, I have to compute the color directly of any given coordinates (a.k.a. sampling).
After mapping polar coordinates to Descartes coordinates, the image would look like this, the sine curves are circles and the implied line $y = x$ is the implied equiangular spiral in polar coordinates.
The sine curves are just isolines, what I have to do is: for given coordinates, find out which sine curve it is on. So here the equation appears.
Sorry for my poor English again.
functions
asked Dec 25 '18 at 4:16
ZzQZzQ
11
$endgroup$
add a comment |
$begingroup$
Or is there a way to calculate an approximate value of $y$ of any given $x$?
The graph looks like this:
y = sin(x - y) by WolframAlpha
Sorry for my poor English. I'm learning WebGL now, then I want to draw something and then run into this equation. So what I really want is a way to calculate $y$ when a certain $x$ is given.
Background: how did I get this equation
I want to draw my previous work using WebGL. It is a group of circles implying an equiangular spiral . When using canvas2d, I calculate some circles then draw them by z-index order (a.k.a. rasterization). But in a fragment shader, I have to compute the color directly of any given coordinates (a.k.a. sampling).
After mapping polar coordinates to Descartes coordinates, the image would look like this, the sine curves are circles and the implied line $y = x$ is the implied equiangular spiral in polar coordinates.
The sine curves are just isolines, what I have to do is: for given coordinates, find out which sine curve it is on. So here the equation appears.
Sorry for my poor English again.
functions
asked Dec 25 '18 at 4:16
ZzQZzQ
11
$endgroup$
1
$begingroup$
If $x approx y$, then $sin(x-y) approx x-y$ which yields $x = 2y$ for small $y$...
$endgroup$
– gt6989b
Dec 25 '18 at 4:21
1
$begingroup$
Welcome to Math SE. It would help us more easily help you if you add some context such as where this question came from, why you are asking it, what you have tried so far to solve it, etc.
$endgroup$
– John Omielan
Dec 25 '18 at 4:49
$begingroup$
There's a solution with $f$ having period $2pi$. You might be able to calculate a Fourier series for this function, though I don't really know how you'd go about it.
$endgroup$
– Milo Brandt
Dec 25 '18 at 5:44
$begingroup$
I did continue using up to $a_{19}$; it really improves the results up to $k=38$.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 5:54
add a comment |
$begingroup$
Or is there a way to calculate an approximate value of $y$ of any given $x$?
The graph looks like this:
y = sin(x - y) by WolframAlpha
Sorry for my poor English. I'm learning WebGL now, then I want to draw something and then run into this equation. So what I really want is a way to calculate $y$ when a certain $x$ is given.
Background: how did I get this equation
I want to draw my previous work using WebGL. It is a group of circles implying an equiangular spiral . When using canvas2d, I calculate some circles then draw them by z-index order (a.k.a. rasterization). But in a fragment shader, I have to compute the color directly of any given coordinates (a.k.a. sampling).
After mapping polar coordinates to Descartes coordinates, the image would look like this, the sine curves are circles and the implied line $y = x$ is the implied equiangular spiral in polar coordinates.
The sine curves are just isolines, what I have to do is: for given coordinates, find out which sine curve it is on. So here the equation appears.
Sorry for my poor English again.
functions
asked Dec 25 '18 at 4:16
ZzQZzQ
11
$endgroup$
Or is there a way to calculate an approximate value of $y$ of any given $x$?
The graph looks like this:
y = sin(x - y) by WolframAlpha
Sorry for my poor English. I'm learning WebGL now, then I want to draw something and then run into this equation. So what I really want is a way to calculate $y$ when a certain $x$ is given.
Background: how did I get this equation
I want to draw my previous work using WebGL. It is a group of circles implying an equiangular spiral . When using canvas2d, I calculate some circles then draw them by z-index order (a.k.a. rasterization). But in a fragment shader, I have to compute the color directly of any given coordinates (a.k.a. sampling).
After mapping polar coordinates to Descartes coordinates, the image would look like this, the sine curves are circles and the implied line $y = x$ is the implied equiangular spiral in polar coordinates.
The sine curves are just isolines, what I have to do is: for given coordinates, find out which sine curve it is on. So here the equation appears.
Sorry for my poor English again.
functions
functions
asked Dec 25 '18 at 4:16
ZzQZzQ
11
asked Dec 25 '18 at 4:16
ZzQZzQ
11
edited Dec 25 '18 at 7:28
ZzQ
asked Dec 25 '18 at 4:16
ZzQZzQ
11
asked Dec 25 '18 at 4:16
ZzQZzQ
11
asked Dec 25 '18 at 4:16
ZzQZzQ
11
11
1
$begingroup$
If $x approx y$, then $sin(x-y) approx x-y$ which yields $x = 2y$ for small $y$...
$endgroup$
– gt6989b
Dec 25 '18 at 4:21
1
$begingroup$
Welcome to Math SE. It would help us more easily help you if you add some context such as where this question came from, why you are asking it, what you have tried so far to solve it, etc.
$endgroup$
– John Omielan
Dec 25 '18 at 4:49
$begingroup$
There's a solution with $f$ having period $2pi$. You might be able to calculate a Fourier series for this function, though I don't really know how you'd go about it.
$endgroup$
– Milo Brandt
Dec 25 '18 at 5:44
$begingroup$
I did continue using up to $a_{19}$; it really improves the results up to $k=38$.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 5:54
add a comment |
1
$begingroup$
If $x approx y$, then $sin(x-y) approx x-y$ which yields $x = 2y$ for small $y$...
$endgroup$
– gt6989b
Dec 25 '18 at 4:21
1
$begingroup$
Welcome to Math SE. It would help us more easily help you if you add some context such as where this question came from, why you are asking it, what you have tried so far to solve it, etc.
$endgroup$
– John Omielan
Dec 25 '18 at 4:49
$begingroup$
There's a solution with $f$ having period $2pi$. You might be able to calculate a Fourier series for this function, though I don't really know how you'd go about it.
$endgroup$
– Milo Brandt
Dec 25 '18 at 5:44
$begingroup$
I did continue using up to $a_{19}$; it really improves the results up to $k=38$.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 5:54
1
1
$begingroup$
If $x approx y$, then $sin(x-y) approx x-y$ which yields $x = 2y$ for small $y$...
$endgroup$
– gt6989b
Dec 25 '18 at 4:21
$begingroup$
If $x approx y$, then $sin(x-y) approx x-y$ which yields $x = 2y$ for small $y$...
$endgroup$
– gt6989b
Dec 25 '18 at 4:21
1
1
$begingroup$
Welcome to Math SE. It would help us more easily help you if you add some context such as where this question came from, why you are asking it, what you have tried so far to solve it, etc.
$endgroup$
– John Omielan
Dec 25 '18 at 4:49
$begingroup$
Welcome to Math SE. It would help us more easily help you if you add some context such as where this question came from, why you are asking it, what you have tried so far to solve it, etc.
$endgroup$
– John Omielan
Dec 25 '18 at 4:49
$begingroup$
There's a solution with $f$ having period $2pi$. You might be able to calculate a Fourier series for this function, though I don't really know how you'd go about it.
$endgroup$
– Milo Brandt
Dec 25 '18 at 5:44
$begingroup$
There's a solution with $f$ having period $2pi$. You might be able to calculate a Fourier series for this function, though I don't really know how you'd go about it.
$endgroup$
– Milo Brandt
Dec 25 '18 at 5:44
$begingroup$
I did continue using up to $a_{19}$; it really improves the results up to $k=38$.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 5:54
$begingroup$
I did continue using up to $a_{19}$; it really improves the results up to $k=38$.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 5:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I do not know how good or bad this will be.
Write $$y=sum_{i=0}^n a_ix^i$$ Replace in the equation and expand as a Taylor series built at $x=0$. Identify the coefficients to get $a_{2n}=0$ and, for the first odd coefficients,
$$a_1=frac{1}{2} qquad a_3=-frac{1}{96} qquad a_5=-frac{1}{1920} qquad a_7=-frac{43}{1290240}qquad a_9=-frac{223}{92897280}$$ We could continue for more terms.
Using $x=frac{k pi}{48}$, we get the following results
$$left(
begin{array}{ccc}
k & text{approximation} & text{solution} \
1 & 0.03272 & 0.03272 \
2 & 0.06543 & 0.06543 \
3 & 0.09810 & 0.09810 \
4 & 0.13071 & 0.13071 \
5 & 0.16326 & 0.16326 \
6 & 0.19571 & 0.19571 \
7 & 0.22806 & 0.22806 \
8 & 0.26028 & 0.26028 \
9 & 0.29236 & 0.29236 \
10 & 0.32426 & 0.32426 \
11 & 0.35598 & 0.35598 \
12 & 0.38749 & 0.38749 \
13 & 0.41876 & 0.41876 \
14 & 0.44978 & 0.44978 \
15 & 0.48051 & 0.48051 \
16 & 0.51093 & 0.51093 \
17 & 0.54101 & 0.54101 \
18 & 0.57072 & 0.57072 \
19 & 0.60002 & 0.60002 \
20 & 0.62889 & 0.62888 \
21 & 0.65727 & 0.65726 \
22 & 0.68514 & 0.68512 \
23 & 0.71244 & 0.71241 \
24 & 0.73912 & 0.73909 \
25 & 0.76513 & 0.76508 \
26 & 0.79042 & 0.79034 \
27 & 0.81492 & 0.81479 \
28 & 0.83855 & 0.83835 \
29 & 0.86124 & 0.86094 \
30 & 0.88290 & 0.88245 \
31 & 0.90344 & 0.90277 \
32 & 0.92276 & 0.92177 \
33 & 0.94073 & 0.93929 \
34 & 0.95722 & 0.95515 \
35 & 0.97210 & 0.96912 \
36 & 0.98521 & 0.98094 \
37 & 0.99636 & 0.99029 \
38 & 1.00537 & 0.99676 \
39 & 1.01202 & 0.99984 \
40 & 1.01607 & 0.99883 \
41 & 1.01726 & 0.99283 \
42 & 1.01528 & 0.98055 \
43 & 1.00982 & 0.96010 \
44 & 1.00052 & 0.92847 \
45 & 0.98697 & 0.88044 \
46 & 0.96873 & 0.80533 \
47 & 0.94533 & 0.67355
end{array}
right)$$
For a quite large range, this seems to work quite properly.
answered Dec 25 '18 at 5:32
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
$begingroup$
Thank you for your answer, I'll have a try.
$endgroup$
– ZzQ
Dec 25 '18 at 7:52
add a comment |
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$begingroup$
I do not know how good or bad this will be.
Write $$y=sum_{i=0}^n a_ix^i$$ Replace in the equation and expand as a Taylor series built at $x=0$. Identify the coefficients to get $a_{2n}=0$ and, for the first odd coefficients,
$$a_1=frac{1}{2} qquad a_3=-frac{1}{96} qquad a_5=-frac{1}{1920} qquad a_7=-frac{43}{1290240}qquad a_9=-frac{223}{92897280}$$ We could continue for more terms.
Using $x=frac{k pi}{48}$, we get the following results
$$left(
begin{array}{ccc}
k & text{approximation} & text{solution} \
1 & 0.03272 & 0.03272 \
2 & 0.06543 & 0.06543 \
3 & 0.09810 & 0.09810 \
4 & 0.13071 & 0.13071 \
5 & 0.16326 & 0.16326 \
6 & 0.19571 & 0.19571 \
7 & 0.22806 & 0.22806 \
8 & 0.26028 & 0.26028 \
9 & 0.29236 & 0.29236 \
10 & 0.32426 & 0.32426 \
11 & 0.35598 & 0.35598 \
12 & 0.38749 & 0.38749 \
13 & 0.41876 & 0.41876 \
14 & 0.44978 & 0.44978 \
15 & 0.48051 & 0.48051 \
16 & 0.51093 & 0.51093 \
17 & 0.54101 & 0.54101 \
18 & 0.57072 & 0.57072 \
19 & 0.60002 & 0.60002 \
20 & 0.62889 & 0.62888 \
21 & 0.65727 & 0.65726 \
22 & 0.68514 & 0.68512 \
23 & 0.71244 & 0.71241 \
24 & 0.73912 & 0.73909 \
25 & 0.76513 & 0.76508 \
26 & 0.79042 & 0.79034 \
27 & 0.81492 & 0.81479 \
28 & 0.83855 & 0.83835 \
29 & 0.86124 & 0.86094 \
30 & 0.88290 & 0.88245 \
31 & 0.90344 & 0.90277 \
32 & 0.92276 & 0.92177 \
33 & 0.94073 & 0.93929 \
34 & 0.95722 & 0.95515 \
35 & 0.97210 & 0.96912 \
36 & 0.98521 & 0.98094 \
37 & 0.99636 & 0.99029 \
38 & 1.00537 & 0.99676 \
39 & 1.01202 & 0.99984 \
40 & 1.01607 & 0.99883 \
41 & 1.01726 & 0.99283 \
42 & 1.01528 & 0.98055 \
43 & 1.00982 & 0.96010 \
44 & 1.00052 & 0.92847 \
45 & 0.98697 & 0.88044 \
46 & 0.96873 & 0.80533 \
47 & 0.94533 & 0.67355
end{array}
right)$$
For a quite large range, this seems to work quite properly.
answered Dec 25 '18 at 5:32
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
$begingroup$
Thank you for your answer, I'll have a try.
$endgroup$
– ZzQ
Dec 25 '18 at 7:52
add a comment |
$begingroup$
I do not know how good or bad this will be.
Write $$y=sum_{i=0}^n a_ix^i$$ Replace in the equation and expand as a Taylor series built at $x=0$. Identify the coefficients to get $a_{2n}=0$ and, for the first odd coefficients,
$$a_1=frac{1}{2} qquad a_3=-frac{1}{96} qquad a_5=-frac{1}{1920} qquad a_7=-frac{43}{1290240}qquad a_9=-frac{223}{92897280}$$ We could continue for more terms.
Using $x=frac{k pi}{48}$, we get the following results
$$left(
begin{array}{ccc}
k & text{approximation} & text{solution} \
1 & 0.03272 & 0.03272 \
2 & 0.06543 & 0.06543 \
3 & 0.09810 & 0.09810 \
4 & 0.13071 & 0.13071 \
5 & 0.16326 & 0.16326 \
6 & 0.19571 & 0.19571 \
7 & 0.22806 & 0.22806 \
8 & 0.26028 & 0.26028 \
9 & 0.29236 & 0.29236 \
10 & 0.32426 & 0.32426 \
11 & 0.35598 & 0.35598 \
12 & 0.38749 & 0.38749 \
13 & 0.41876 & 0.41876 \
14 & 0.44978 & 0.44978 \
15 & 0.48051 & 0.48051 \
16 & 0.51093 & 0.51093 \
17 & 0.54101 & 0.54101 \
18 & 0.57072 & 0.57072 \
19 & 0.60002 & 0.60002 \
20 & 0.62889 & 0.62888 \
21 & 0.65727 & 0.65726 \
22 & 0.68514 & 0.68512 \
23 & 0.71244 & 0.71241 \
24 & 0.73912 & 0.73909 \
25 & 0.76513 & 0.76508 \
26 & 0.79042 & 0.79034 \
27 & 0.81492 & 0.81479 \
28 & 0.83855 & 0.83835 \
29 & 0.86124 & 0.86094 \
30 & 0.88290 & 0.88245 \
31 & 0.90344 & 0.90277 \
32 & 0.92276 & 0.92177 \
33 & 0.94073 & 0.93929 \
34 & 0.95722 & 0.95515 \
35 & 0.97210 & 0.96912 \
36 & 0.98521 & 0.98094 \
37 & 0.99636 & 0.99029 \
38 & 1.00537 & 0.99676 \
39 & 1.01202 & 0.99984 \
40 & 1.01607 & 0.99883 \
41 & 1.01726 & 0.99283 \
42 & 1.01528 & 0.98055 \
43 & 1.00982 & 0.96010 \
44 & 1.00052 & 0.92847 \
45 & 0.98697 & 0.88044 \
46 & 0.96873 & 0.80533 \
47 & 0.94533 & 0.67355
end{array}
right)$$
For a quite large range, this seems to work quite properly.
answered Dec 25 '18 at 5:32
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
$begingroup$
Thank you for your answer, I'll have a try.
$endgroup$
– ZzQ
Dec 25 '18 at 7:52
add a comment |
$begingroup$
I do not know how good or bad this will be.
Write $$y=sum_{i=0}^n a_ix^i$$ Replace in the equation and expand as a Taylor series built at $x=0$. Identify the coefficients to get $a_{2n}=0$ and, for the first odd coefficients,
$$a_1=frac{1}{2} qquad a_3=-frac{1}{96} qquad a_5=-frac{1}{1920} qquad a_7=-frac{43}{1290240}qquad a_9=-frac{223}{92897280}$$ We could continue for more terms.
Using $x=frac{k pi}{48}$, we get the following results
$$left(
begin{array}{ccc}
k & text{approximation} & text{solution} \
1 & 0.03272 & 0.03272 \
2 & 0.06543 & 0.06543 \
3 & 0.09810 & 0.09810 \
4 & 0.13071 & 0.13071 \
5 & 0.16326 & 0.16326 \
6 & 0.19571 & 0.19571 \
7 & 0.22806 & 0.22806 \
8 & 0.26028 & 0.26028 \
9 & 0.29236 & 0.29236 \
10 & 0.32426 & 0.32426 \
11 & 0.35598 & 0.35598 \
12 & 0.38749 & 0.38749 \
13 & 0.41876 & 0.41876 \
14 & 0.44978 & 0.44978 \
15 & 0.48051 & 0.48051 \
16 & 0.51093 & 0.51093 \
17 & 0.54101 & 0.54101 \
18 & 0.57072 & 0.57072 \
19 & 0.60002 & 0.60002 \
20 & 0.62889 & 0.62888 \
21 & 0.65727 & 0.65726 \
22 & 0.68514 & 0.68512 \
23 & 0.71244 & 0.71241 \
24 & 0.73912 & 0.73909 \
25 & 0.76513 & 0.76508 \
26 & 0.79042 & 0.79034 \
27 & 0.81492 & 0.81479 \
28 & 0.83855 & 0.83835 \
29 & 0.86124 & 0.86094 \
30 & 0.88290 & 0.88245 \
31 & 0.90344 & 0.90277 \
32 & 0.92276 & 0.92177 \
33 & 0.94073 & 0.93929 \
34 & 0.95722 & 0.95515 \
35 & 0.97210 & 0.96912 \
36 & 0.98521 & 0.98094 \
37 & 0.99636 & 0.99029 \
38 & 1.00537 & 0.99676 \
39 & 1.01202 & 0.99984 \
40 & 1.01607 & 0.99883 \
41 & 1.01726 & 0.99283 \
42 & 1.01528 & 0.98055 \
43 & 1.00982 & 0.96010 \
44 & 1.00052 & 0.92847 \
45 & 0.98697 & 0.88044 \
46 & 0.96873 & 0.80533 \
47 & 0.94533 & 0.67355
end{array}
right)$$
For a quite large range, this seems to work quite properly.
answered Dec 25 '18 at 5:32
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
I do not know how good or bad this will be.
Write $$y=sum_{i=0}^n a_ix^i$$ Replace in the equation and expand as a Taylor series built at $x=0$. Identify the coefficients to get $a_{2n}=0$ and, for the first odd coefficients,
$$a_1=frac{1}{2} qquad a_3=-frac{1}{96} qquad a_5=-frac{1}{1920} qquad a_7=-frac{43}{1290240}qquad a_9=-frac{223}{92897280}$$ We could continue for more terms.
Using $x=frac{k pi}{48}$, we get the following results
$$left(
begin{array}{ccc}
k & text{approximation} & text{solution} \
1 & 0.03272 & 0.03272 \
2 & 0.06543 & 0.06543 \
3 & 0.09810 & 0.09810 \
4 & 0.13071 & 0.13071 \
5 & 0.16326 & 0.16326 \
6 & 0.19571 & 0.19571 \
7 & 0.22806 & 0.22806 \
8 & 0.26028 & 0.26028 \
9 & 0.29236 & 0.29236 \
10 & 0.32426 & 0.32426 \
11 & 0.35598 & 0.35598 \
12 & 0.38749 & 0.38749 \
13 & 0.41876 & 0.41876 \
14 & 0.44978 & 0.44978 \
15 & 0.48051 & 0.48051 \
16 & 0.51093 & 0.51093 \
17 & 0.54101 & 0.54101 \
18 & 0.57072 & 0.57072 \
19 & 0.60002 & 0.60002 \
20 & 0.62889 & 0.62888 \
21 & 0.65727 & 0.65726 \
22 & 0.68514 & 0.68512 \
23 & 0.71244 & 0.71241 \
24 & 0.73912 & 0.73909 \
25 & 0.76513 & 0.76508 \
26 & 0.79042 & 0.79034 \
27 & 0.81492 & 0.81479 \
28 & 0.83855 & 0.83835 \
29 & 0.86124 & 0.86094 \
30 & 0.88290 & 0.88245 \
31 & 0.90344 & 0.90277 \
32 & 0.92276 & 0.92177 \
33 & 0.94073 & 0.93929 \
34 & 0.95722 & 0.95515 \
35 & 0.97210 & 0.96912 \
36 & 0.98521 & 0.98094 \
37 & 0.99636 & 0.99029 \
38 & 1.00537 & 0.99676 \
39 & 1.01202 & 0.99984 \
40 & 1.01607 & 0.99883 \
41 & 1.01726 & 0.99283 \
42 & 1.01528 & 0.98055 \
43 & 1.00982 & 0.96010 \
44 & 1.00052 & 0.92847 \
45 & 0.98697 & 0.88044 \
46 & 0.96873 & 0.80533 \
47 & 0.94533 & 0.67355
end{array}
right)$$
For a quite large range, this seems to work quite properly.
answered Dec 25 '18 at 5:32
Claude LeiboviciClaude Leibovici
126k1158135
answered Dec 25 '18 at 5:32
Claude LeiboviciClaude Leibovici
126k1158135
answered Dec 25 '18 at 5:32
Claude LeiboviciClaude Leibovici
126k1158135
answered Dec 25 '18 at 5:32
Claude LeiboviciClaude Leibovici
126k1158135
126k1158135
$begingroup$
Thank you for your answer, I'll have a try.
$endgroup$
– ZzQ
Dec 25 '18 at 7:52
add a comment |
$begingroup$
Thank you for your answer, I'll have a try.
$endgroup$
– ZzQ
Dec 25 '18 at 7:52
$begingroup$
Thank you for your answer, I'll have a try.
$endgroup$
– ZzQ
Dec 25 '18 at 7:52
$begingroup$
Thank you for your answer, I'll have a try.
$endgroup$
– ZzQ
Dec 25 '18 at 7:52
add a comment |
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$begingroup$
If $x approx y$, then $sin(x-y) approx x-y$ which yields $x = 2y$ for small $y$...
$endgroup$
– gt6989b
Dec 25 '18 at 4:21
1
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Welcome to Math SE. It would help us more easily help you if you add some context such as where this question came from, why you are asking it, what you have tried so far to solve it, etc.
$endgroup$
– John Omielan
Dec 25 '18 at 4:49
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There's a solution with $f$ having period $2pi$. You might be able to calculate a Fourier series for this function, though I don't really know how you'd go about it.
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– Milo Brandt
Dec 25 '18 at 5:44
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I did continue using up to $a_{19}$; it really improves the results up to $k=38$.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 5:54