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Connection between Theodorus spiral and Archimedean spirals using an analytic expression
Consider figure below, made of right triangles with sides $1,sqrt{n}$ and $sqrt{n+1}$ spiraling around a common vertex (opposite side 1) beginning at $n=1$. This spiral is classical (Theodorus spiral) ; see for example https://mathematica.stackexchange.com/q/66969 .
I would like to establish that the limit curve is an Archimedean spiral.
Let us recall that such is spiral has a polar equation of the form :
$$dfrac{theta}{r}=a text{ for some constant} a (1)$$
$bf{My} bf{attempts}$ : My idea is to bring back this issue to the fact that the following limit exists :
$$a=lim_{N to infty} dfrac{1}{sqrt{N}} sum_{n=1}^N mathrm{atan} left(dfrac{1}{sqrt{n}} right) (2)$$
where the summation gives the general polar angle $theta$ and $sqrt{N}$ corresponds to radius $r$ in (1). Numerical computations give evidence for that with a value of $a$ looking slightly below $2$.
I have taken two ways :
- I have transformed the summation in (2) using relationship
$$mathrm{atan}(a)+mathrm{atan}(b)=mathrm{atan}dfrac{a+b}{1-ab} (3)$$
in a recurrent way, giving a single expression $mathrm{atan}dfrac{u_N}{v_N}$ through the double sequence :
$$begin{cases}u_{n}&=&sqrt{n} u_{n-1} & + & v_{n-1}\
v_{n}&=&-u_{n-1}&+&sqrt{n} v_{n-1}
end{cases}, u_1=v_1=1 (4)$$
but I am blocked there...
2) I have attempted a different way by using Euler-Maclaurin summation formula (see Appendix below) using the fact that an antiderivative of function defined by $f(x)=mathrm{atan}(tfrac{1}{sqrt{x}})$ is given by $$F(x)=sqrt{x}+x mathrm{atan}(tfrac{1}{sqrt{x}})-mathrm{atan}(sqrt{x}).$$
together with the development:
$$mathrm{atan}(tfrac{1}{sqrt{x}})=tfrac{1}{sqrt{x}}-tfrac{1}{3 x sqrt{x}}+tfrac{1}{5x^2sqrt{x}}- ...$$
which is a more rewarding direction...
Maybe, there other ways to prove that we have asymptotically an Archimedean spiral...
I would be grateful for any hint...
In fact, I discovered using the "Theodorus" keyword (that I hadn't when writing a first draft of this question) that the article
https://en.wikipedia.org/wiki/Spiral_of_Theodorus provides a lot of information about this (not so trivial) issue...
Appendix on Euler-Maclaurin formula : for any integers $a$ and $b$ with $0<a<b$ :
begin{eqnarray}
sum_{k = a}^{b} f(k) = int_{a}^{b} f(t) , dt + tfrac12 (f(a) + f(b)) + sum_{n = 1}^{N} frac{B_{2n}}{(2n)!} ( f^{(2n-1)}(b) - f^{(2n-1)}(a) ) + R_{N},
end{eqnarray}
where $B_{n}$ is the $n$th Bernoulli number with $B_{2} = tfrac{1}{6}$, $B_{4} = -tfrac{1}{30}$, $B_{6} = tfrac{1}{42}$, etc... and
begin{align}
|R_{N}| leq frac{|B_{2N} |}{(2n)!} int_{a}^{b} | f^{(2N)}(t) | , dt.
end{align}
for any arbitrary positive integer $N$.
real-analysis geometry
asked Nov 20 '18 at 10:34
Jean Marie
28.8k41949
add a comment |
Consider figure below, made of right triangles with sides $1,sqrt{n}$ and $sqrt{n+1}$ spiraling around a common vertex (opposite side 1) beginning at $n=1$. This spiral is classical (Theodorus spiral) ; see for example https://mathematica.stackexchange.com/q/66969 .
I would like to establish that the limit curve is an Archimedean spiral.
Let us recall that such is spiral has a polar equation of the form :
$$dfrac{theta}{r}=a text{ for some constant} a (1)$$
$bf{My} bf{attempts}$ : My idea is to bring back this issue to the fact that the following limit exists :
$$a=lim_{N to infty} dfrac{1}{sqrt{N}} sum_{n=1}^N mathrm{atan} left(dfrac{1}{sqrt{n}} right) (2)$$
where the summation gives the general polar angle $theta$ and $sqrt{N}$ corresponds to radius $r$ in (1). Numerical computations give evidence for that with a value of $a$ looking slightly below $2$.
I have taken two ways :
- I have transformed the summation in (2) using relationship
$$mathrm{atan}(a)+mathrm{atan}(b)=mathrm{atan}dfrac{a+b}{1-ab} (3)$$
in a recurrent way, giving a single expression $mathrm{atan}dfrac{u_N}{v_N}$ through the double sequence :
$$begin{cases}u_{n}&=&sqrt{n} u_{n-1} & + & v_{n-1}\
v_{n}&=&-u_{n-1}&+&sqrt{n} v_{n-1}
end{cases}, u_1=v_1=1 (4)$$
but I am blocked there...
2) I have attempted a different way by using Euler-Maclaurin summation formula (see Appendix below) using the fact that an antiderivative of function defined by $f(x)=mathrm{atan}(tfrac{1}{sqrt{x}})$ is given by $$F(x)=sqrt{x}+x mathrm{atan}(tfrac{1}{sqrt{x}})-mathrm{atan}(sqrt{x}).$$
together with the development:
$$mathrm{atan}(tfrac{1}{sqrt{x}})=tfrac{1}{sqrt{x}}-tfrac{1}{3 x sqrt{x}}+tfrac{1}{5x^2sqrt{x}}- ...$$
which is a more rewarding direction...
Maybe, there other ways to prove that we have asymptotically an Archimedean spiral...
I would be grateful for any hint...
In fact, I discovered using the "Theodorus" keyword (that I hadn't when writing a first draft of this question) that the article
https://en.wikipedia.org/wiki/Spiral_of_Theodorus provides a lot of information about this (not so trivial) issue...
Appendix on Euler-Maclaurin formula : for any integers $a$ and $b$ with $0<a<b$ :
begin{eqnarray}
sum_{k = a}^{b} f(k) = int_{a}^{b} f(t) , dt + tfrac12 (f(a) + f(b)) + sum_{n = 1}^{N} frac{B_{2n}}{(2n)!} ( f^{(2n-1)}(b) - f^{(2n-1)}(a) ) + R_{N},
end{eqnarray}
where $B_{n}$ is the $n$th Bernoulli number with $B_{2} = tfrac{1}{6}$, $B_{4} = -tfrac{1}{30}$, $B_{6} = tfrac{1}{42}$, etc... and
begin{align}
|R_{N}| leq frac{|B_{2N} |}{(2n)!} int_{a}^{b} | f^{(2N)}(t) | , dt.
end{align}
for any arbitrary positive integer $N$.
real-analysis geometry
asked Nov 20 '18 at 10:34
Jean Marie
28.8k41949
It is $arctan$ in these circles, or do you have something else in mind?
– Christian Blatter
Nov 20 '18 at 11:02
@Christian Blatter Is your question about notation "atan" vs. notation "arctan" ? If yes, they are completely synonymous for me.
– Jean Marie
Nov 20 '18 at 11:14
add a comment |
Consider figure below, made of right triangles with sides $1,sqrt{n}$ and $sqrt{n+1}$ spiraling around a common vertex (opposite side 1) beginning at $n=1$. This spiral is classical (Theodorus spiral) ; see for example https://mathematica.stackexchange.com/q/66969 .
I would like to establish that the limit curve is an Archimedean spiral.
Let us recall that such is spiral has a polar equation of the form :
$$dfrac{theta}{r}=a text{ for some constant} a (1)$$
$bf{My} bf{attempts}$ : My idea is to bring back this issue to the fact that the following limit exists :
$$a=lim_{N to infty} dfrac{1}{sqrt{N}} sum_{n=1}^N mathrm{atan} left(dfrac{1}{sqrt{n}} right) (2)$$
where the summation gives the general polar angle $theta$ and $sqrt{N}$ corresponds to radius $r$ in (1). Numerical computations give evidence for that with a value of $a$ looking slightly below $2$.
I have taken two ways :
- I have transformed the summation in (2) using relationship
$$mathrm{atan}(a)+mathrm{atan}(b)=mathrm{atan}dfrac{a+b}{1-ab} (3)$$
in a recurrent way, giving a single expression $mathrm{atan}dfrac{u_N}{v_N}$ through the double sequence :
$$begin{cases}u_{n}&=&sqrt{n} u_{n-1} & + & v_{n-1}\
v_{n}&=&-u_{n-1}&+&sqrt{n} v_{n-1}
end{cases}, u_1=v_1=1 (4)$$
but I am blocked there...
2) I have attempted a different way by using Euler-Maclaurin summation formula (see Appendix below) using the fact that an antiderivative of function defined by $f(x)=mathrm{atan}(tfrac{1}{sqrt{x}})$ is given by $$F(x)=sqrt{x}+x mathrm{atan}(tfrac{1}{sqrt{x}})-mathrm{atan}(sqrt{x}).$$
together with the development:
$$mathrm{atan}(tfrac{1}{sqrt{x}})=tfrac{1}{sqrt{x}}-tfrac{1}{3 x sqrt{x}}+tfrac{1}{5x^2sqrt{x}}- ...$$
which is a more rewarding direction...
Maybe, there other ways to prove that we have asymptotically an Archimedean spiral...
I would be grateful for any hint...
In fact, I discovered using the "Theodorus" keyword (that I hadn't when writing a first draft of this question) that the article
https://en.wikipedia.org/wiki/Spiral_of_Theodorus provides a lot of information about this (not so trivial) issue...
Appendix on Euler-Maclaurin formula : for any integers $a$ and $b$ with $0<a<b$ :
begin{eqnarray}
sum_{k = a}^{b} f(k) = int_{a}^{b} f(t) , dt + tfrac12 (f(a) + f(b)) + sum_{n = 1}^{N} frac{B_{2n}}{(2n)!} ( f^{(2n-1)}(b) - f^{(2n-1)}(a) ) + R_{N},
end{eqnarray}
where $B_{n}$ is the $n$th Bernoulli number with $B_{2} = tfrac{1}{6}$, $B_{4} = -tfrac{1}{30}$, $B_{6} = tfrac{1}{42}$, etc... and
begin{align}
|R_{N}| leq frac{|B_{2N} |}{(2n)!} int_{a}^{b} | f^{(2N)}(t) | , dt.
end{align}
for any arbitrary positive integer $N$.
real-analysis geometry
asked Nov 20 '18 at 10:34
Jean Marie
28.8k41949
Consider figure below, made of right triangles with sides $1,sqrt{n}$ and $sqrt{n+1}$ spiraling around a common vertex (opposite side 1) beginning at $n=1$. This spiral is classical (Theodorus spiral) ; see for example https://mathematica.stackexchange.com/q/66969 .
I would like to establish that the limit curve is an Archimedean spiral.
Let us recall that such is spiral has a polar equation of the form :
$$dfrac{theta}{r}=a text{ for some constant} a (1)$$
$bf{My} bf{attempts}$ : My idea is to bring back this issue to the fact that the following limit exists :
$$a=lim_{N to infty} dfrac{1}{sqrt{N}} sum_{n=1}^N mathrm{atan} left(dfrac{1}{sqrt{n}} right) (2)$$
where the summation gives the general polar angle $theta$ and $sqrt{N}$ corresponds to radius $r$ in (1). Numerical computations give evidence for that with a value of $a$ looking slightly below $2$.
I have taken two ways :
- I have transformed the summation in (2) using relationship
$$mathrm{atan}(a)+mathrm{atan}(b)=mathrm{atan}dfrac{a+b}{1-ab} (3)$$
in a recurrent way, giving a single expression $mathrm{atan}dfrac{u_N}{v_N}$ through the double sequence :
$$begin{cases}u_{n}&=&sqrt{n} u_{n-1} & + & v_{n-1}\
v_{n}&=&-u_{n-1}&+&sqrt{n} v_{n-1}
end{cases}, u_1=v_1=1 (4)$$
but I am blocked there...
2) I have attempted a different way by using Euler-Maclaurin summation formula (see Appendix below) using the fact that an antiderivative of function defined by $f(x)=mathrm{atan}(tfrac{1}{sqrt{x}})$ is given by $$F(x)=sqrt{x}+x mathrm{atan}(tfrac{1}{sqrt{x}})-mathrm{atan}(sqrt{x}).$$
together with the development:
$$mathrm{atan}(tfrac{1}{sqrt{x}})=tfrac{1}{sqrt{x}}-tfrac{1}{3 x sqrt{x}}+tfrac{1}{5x^2sqrt{x}}- ...$$
which is a more rewarding direction...
Maybe, there other ways to prove that we have asymptotically an Archimedean spiral...
I would be grateful for any hint...
In fact, I discovered using the "Theodorus" keyword (that I hadn't when writing a first draft of this question) that the article
https://en.wikipedia.org/wiki/Spiral_of_Theodorus provides a lot of information about this (not so trivial) issue...
Appendix on Euler-Maclaurin formula : for any integers $a$ and $b$ with $0<a<b$ :
begin{eqnarray}
sum_{k = a}^{b} f(k) = int_{a}^{b} f(t) , dt + tfrac12 (f(a) + f(b)) + sum_{n = 1}^{N} frac{B_{2n}}{(2n)!} ( f^{(2n-1)}(b) - f^{(2n-1)}(a) ) + R_{N},
end{eqnarray}
where $B_{n}$ is the $n$th Bernoulli number with $B_{2} = tfrac{1}{6}$, $B_{4} = -tfrac{1}{30}$, $B_{6} = tfrac{1}{42}$, etc... and
begin{align}
|R_{N}| leq frac{|B_{2N} |}{(2n)!} int_{a}^{b} | f^{(2N)}(t) | , dt.
end{align}
for any arbitrary positive integer $N$.
real-analysis geometry
real-analysis geometry
asked Nov 20 '18 at 10:34
Jean Marie
28.8k41949
asked Nov 20 '18 at 10:34
Jean Marie
28.8k41949
edited Nov 25 '18 at 17:30
asked Nov 20 '18 at 10:34
Jean Marie
28.8k41949
asked Nov 20 '18 at 10:34
Jean Marie
28.8k41949
asked Nov 20 '18 at 10:34
Jean Marie
28.8k41949
28.8k41949
It is $arctan$ in these circles, or do you have something else in mind?
– Christian Blatter
Nov 20 '18 at 11:02
@Christian Blatter Is your question about notation "atan" vs. notation "arctan" ? If yes, they are completely synonymous for me.
– Jean Marie
Nov 20 '18 at 11:14
add a comment |
It is $arctan$ in these circles, or do you have something else in mind?
– Christian Blatter
Nov 20 '18 at 11:02
@Christian Blatter Is your question about notation "atan" vs. notation "arctan" ? If yes, they are completely synonymous for me.
– Jean Marie
Nov 20 '18 at 11:14
It is $arctan$ in these circles, or do you have something else in mind?
– Christian Blatter
Nov 20 '18 at 11:02
It is $arctan$ in these circles, or do you have something else in mind?
– Christian Blatter
Nov 20 '18 at 11:02
@Christian Blatter Is your question about notation "atan" vs. notation "arctan" ? If yes, they are completely synonymous for me.
– Jean Marie
Nov 20 '18 at 11:14
@Christian Blatter Is your question about notation "atan" vs. notation "arctan" ? If yes, they are completely synonymous for me.
– Jean Marie
Nov 20 '18 at 11:14
add a comment |
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It is $arctan$ in these circles, or do you have something else in mind?
– Christian Blatter
Nov 20 '18 at 11:02
@Christian Blatter Is your question about notation "atan" vs. notation "arctan" ? If yes, they are completely synonymous for me.
– Jean Marie
Nov 20 '18 at 11:14