Binary expansion of $frac{1}{pi}tan^{-1}left(frac{5}{12}right)$
$begingroup$
Prove that the binary expansion of $dfrac{1}{pi}tan^{-1}left(dfrac{5}{12}right)$ has strings of $0$s or $1$s of arbitrary length.
I didn't see how we can calculate the binary expansion of $tan^{-1}(x)$ or $pi$. Is there some other way of solving this question?
number-theory trigonometry decimal-expansion
edited Dec 28 '18 at 16:47
Martin Sleziak
44.9k10121273
asked Nov 6 '16 at 14:05
user19405892user19405892
7,63131055
$endgroup$
|
show 1 more comment
$begingroup$
Prove that the binary expansion of $dfrac{1}{pi}tan^{-1}left(dfrac{5}{12}right)$ has strings of $0$s or $1$s of arbitrary length.
I didn't see how we can calculate the binary expansion of $tan^{-1}(x)$ or $pi$. Is there some other way of solving this question?
number-theory trigonometry decimal-expansion
edited Dec 28 '18 at 16:47
Martin Sleziak
44.9k10121273
asked Nov 6 '16 at 14:05
user19405892user19405892
7,63131055
$endgroup$
1
$begingroup$
I do not know if this is really relevant, but since $arctan(z)=text{Im}log(1+iz)$ and $5+12i=(3+2i)^2$, the question is more or less equivalent to showing that $(3+2i)^{2^k}$ is often close to a real number (i.e.has a small imaginary part, compared to the real part).
$endgroup$
– Jack D'Aurizio
Nov 6 '16 at 15:30
3
$begingroup$
where did you find this question ?
$endgroup$
– mercio
Nov 6 '16 at 17:59
1
$begingroup$
@mercio I was solving a problem and needed to prove this as a step in the solution.
$endgroup$
– user19405892
Nov 6 '16 at 22:39
1
$begingroup$
this could very well be the sort of thing that is "obviously true" (because a number taken randomly in a unit interval will have this property with probability $1$) but completely hopeless to prove.
$endgroup$
– mercio
Nov 6 '16 at 22:46
1
$begingroup$
I knew this was a hidden duplicate : math.stackexchange.com/questions/1615708/…
$endgroup$
– mercio
Dec 2 '16 at 14:10
|
show 1 more comment
$begingroup$
Prove that the binary expansion of $dfrac{1}{pi}tan^{-1}left(dfrac{5}{12}right)$ has strings of $0$s or $1$s of arbitrary length.
I didn't see how we can calculate the binary expansion of $tan^{-1}(x)$ or $pi$. Is there some other way of solving this question?
number-theory trigonometry decimal-expansion
edited Dec 28 '18 at 16:47
Martin Sleziak
44.9k10121273
asked Nov 6 '16 at 14:05
user19405892user19405892
7,63131055
$endgroup$
Prove that the binary expansion of $dfrac{1}{pi}tan^{-1}left(dfrac{5}{12}right)$ has strings of $0$s or $1$s of arbitrary length.
I didn't see how we can calculate the binary expansion of $tan^{-1}(x)$ or $pi$. Is there some other way of solving this question?
number-theory trigonometry decimal-expansion
number-theory trigonometry decimal-expansion
edited Dec 28 '18 at 16:47
Martin Sleziak
44.9k10121273
asked Nov 6 '16 at 14:05
user19405892user19405892
7,63131055
edited Dec 28 '18 at 16:47
Martin Sleziak
44.9k10121273
asked Nov 6 '16 at 14:05
user19405892user19405892
7,63131055
edited Dec 28 '18 at 16:47
Martin Sleziak
44.9k10121273
edited Dec 28 '18 at 16:47
Martin Sleziak
44.9k10121273
edited Dec 28 '18 at 16:47
Martin Sleziak
44.9k10121273
44.9k10121273
asked Nov 6 '16 at 14:05
user19405892user19405892
7,63131055
asked Nov 6 '16 at 14:05
user19405892user19405892
7,63131055
asked Nov 6 '16 at 14:05
user19405892user19405892
7,63131055
7,63131055
1
$begingroup$
I do not know if this is really relevant, but since $arctan(z)=text{Im}log(1+iz)$ and $5+12i=(3+2i)^2$, the question is more or less equivalent to showing that $(3+2i)^{2^k}$ is often close to a real number (i.e.has a small imaginary part, compared to the real part).
$endgroup$
– Jack D'Aurizio
Nov 6 '16 at 15:30
3
$begingroup$
where did you find this question ?
$endgroup$
– mercio
Nov 6 '16 at 17:59
1
$begingroup$
@mercio I was solving a problem and needed to prove this as a step in the solution.
$endgroup$
– user19405892
Nov 6 '16 at 22:39
1
$begingroup$
this could very well be the sort of thing that is "obviously true" (because a number taken randomly in a unit interval will have this property with probability $1$) but completely hopeless to prove.
$endgroup$
– mercio
Nov 6 '16 at 22:46
1
$begingroup$
I knew this was a hidden duplicate : math.stackexchange.com/questions/1615708/…
$endgroup$
– mercio
Dec 2 '16 at 14:10
|
show 1 more comment
1
$begingroup$
I do not know if this is really relevant, but since $arctan(z)=text{Im}log(1+iz)$ and $5+12i=(3+2i)^2$, the question is more or less equivalent to showing that $(3+2i)^{2^k}$ is often close to a real number (i.e.has a small imaginary part, compared to the real part).
$endgroup$
– Jack D'Aurizio
Nov 6 '16 at 15:30
3
$begingroup$
where did you find this question ?
$endgroup$
– mercio
Nov 6 '16 at 17:59
1
$begingroup$
@mercio I was solving a problem and needed to prove this as a step in the solution.
$endgroup$
– user19405892
Nov 6 '16 at 22:39
1
$begingroup$
this could very well be the sort of thing that is "obviously true" (because a number taken randomly in a unit interval will have this property with probability $1$) but completely hopeless to prove.
$endgroup$
– mercio
Nov 6 '16 at 22:46
1
$begingroup$
I knew this was a hidden duplicate : math.stackexchange.com/questions/1615708/…
$endgroup$
– mercio
Dec 2 '16 at 14:10
1
1
$begingroup$
I do not know if this is really relevant, but since $arctan(z)=text{Im}log(1+iz)$ and $5+12i=(3+2i)^2$, the question is more or less equivalent to showing that $(3+2i)^{2^k}$ is often close to a real number (i.e.has a small imaginary part, compared to the real part).
$endgroup$
– Jack D'Aurizio
Nov 6 '16 at 15:30
$begingroup$
I do not know if this is really relevant, but since $arctan(z)=text{Im}log(1+iz)$ and $5+12i=(3+2i)^2$, the question is more or less equivalent to showing that $(3+2i)^{2^k}$ is often close to a real number (i.e.has a small imaginary part, compared to the real part).
$endgroup$
– Jack D'Aurizio
Nov 6 '16 at 15:30
3
3
$begingroup$
where did you find this question ?
$endgroup$
– mercio
Nov 6 '16 at 17:59
$begingroup$
where did you find this question ?
$endgroup$
– mercio
Nov 6 '16 at 17:59
1
1
$begingroup$
@mercio I was solving a problem and needed to prove this as a step in the solution.
$endgroup$
– user19405892
Nov 6 '16 at 22:39
$begingroup$
@mercio I was solving a problem and needed to prove this as a step in the solution.
$endgroup$
– user19405892
Nov 6 '16 at 22:39
1
1
$begingroup$
this could very well be the sort of thing that is "obviously true" (because a number taken randomly in a unit interval will have this property with probability $1$) but completely hopeless to prove.
$endgroup$
– mercio
Nov 6 '16 at 22:46
$begingroup$
this could very well be the sort of thing that is "obviously true" (because a number taken randomly in a unit interval will have this property with probability $1$) but completely hopeless to prove.
$endgroup$
– mercio
Nov 6 '16 at 22:46
1
1
$begingroup$
I knew this was a hidden duplicate : math.stackexchange.com/questions/1615708/…
$endgroup$
– mercio
Dec 2 '16 at 14:10
$begingroup$
I knew this was a hidden duplicate : math.stackexchange.com/questions/1615708/…
$endgroup$
– mercio
Dec 2 '16 at 14:10
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
NOT A SOLUTION:
Something that may be of use
begin{equation}
arctanleft(frac{x + y}{1 - xy}right) = arctan(x) + arctan(y)
end{equation}
Here,
begin{equation}
frac{x + y}{1 - xy} = frac{5}{12}
end{equation}
Which has the integer solutions $x,y = -5$
and so,
begin{equation}
frac{1}{pi}left[arctanleft(frac{5}{12}right)right] = frac{1}{pi}left[arctan(-5) + arctan(-5)right] = -frac{2}{pi}arctan(5)
end{equation}
As before, unsure if this will be of help.
answered Dec 25 '18 at 4:09
DavidGDavidG
1
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
NOT A SOLUTION:
Something that may be of use
begin{equation}
arctanleft(frac{x + y}{1 - xy}right) = arctan(x) + arctan(y)
end{equation}
Here,
begin{equation}
frac{x + y}{1 - xy} = frac{5}{12}
end{equation}
Which has the integer solutions $x,y = -5$
and so,
begin{equation}
frac{1}{pi}left[arctanleft(frac{5}{12}right)right] = frac{1}{pi}left[arctan(-5) + arctan(-5)right] = -frac{2}{pi}arctan(5)
end{equation}
As before, unsure if this will be of help.
answered Dec 25 '18 at 4:09
DavidGDavidG
1
$endgroup$
add a comment |
$begingroup$
NOT A SOLUTION:
Something that may be of use
begin{equation}
arctanleft(frac{x + y}{1 - xy}right) = arctan(x) + arctan(y)
end{equation}
Here,
begin{equation}
frac{x + y}{1 - xy} = frac{5}{12}
end{equation}
Which has the integer solutions $x,y = -5$
and so,
begin{equation}
frac{1}{pi}left[arctanleft(frac{5}{12}right)right] = frac{1}{pi}left[arctan(-5) + arctan(-5)right] = -frac{2}{pi}arctan(5)
end{equation}
As before, unsure if this will be of help.
answered Dec 25 '18 at 4:09
DavidGDavidG
1
$endgroup$
add a comment |
$begingroup$
NOT A SOLUTION:
Something that may be of use
begin{equation}
arctanleft(frac{x + y}{1 - xy}right) = arctan(x) + arctan(y)
end{equation}
Here,
begin{equation}
frac{x + y}{1 - xy} = frac{5}{12}
end{equation}
Which has the integer solutions $x,y = -5$
and so,
begin{equation}
frac{1}{pi}left[arctanleft(frac{5}{12}right)right] = frac{1}{pi}left[arctan(-5) + arctan(-5)right] = -frac{2}{pi}arctan(5)
end{equation}
As before, unsure if this will be of help.
answered Dec 25 '18 at 4:09
DavidGDavidG
1
$endgroup$
NOT A SOLUTION:
Something that may be of use
begin{equation}
arctanleft(frac{x + y}{1 - xy}right) = arctan(x) + arctan(y)
end{equation}
Here,
begin{equation}
frac{x + y}{1 - xy} = frac{5}{12}
end{equation}
Which has the integer solutions $x,y = -5$
and so,
begin{equation}
frac{1}{pi}left[arctanleft(frac{5}{12}right)right] = frac{1}{pi}left[arctan(-5) + arctan(-5)right] = -frac{2}{pi}arctan(5)
end{equation}
As before, unsure if this will be of help.
answered Dec 25 '18 at 4:09
DavidGDavidG
1
answered Dec 25 '18 at 4:09
DavidGDavidG
1
answered Dec 25 '18 at 4:09
DavidGDavidG
1
answered Dec 25 '18 at 4:09
DavidGDavidG
1
1
add a comment |
add a comment |
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1
$begingroup$
I do not know if this is really relevant, but since $arctan(z)=text{Im}log(1+iz)$ and $5+12i=(3+2i)^2$, the question is more or less equivalent to showing that $(3+2i)^{2^k}$ is often close to a real number (i.e.has a small imaginary part, compared to the real part).
$endgroup$
– Jack D'Aurizio
Nov 6 '16 at 15:30
3
$begingroup$
where did you find this question ?
$endgroup$
– mercio
Nov 6 '16 at 17:59
1
$begingroup$
@mercio I was solving a problem and needed to prove this as a step in the solution.
$endgroup$
– user19405892
Nov 6 '16 at 22:39
1
$begingroup$
this could very well be the sort of thing that is "obviously true" (because a number taken randomly in a unit interval will have this property with probability $1$) but completely hopeless to prove.
$endgroup$
– mercio
Nov 6 '16 at 22:46
1
$begingroup$
I knew this was a hidden duplicate : math.stackexchange.com/questions/1615708/…
$endgroup$
– mercio
Dec 2 '16 at 14:10