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How to approximate the solution of $y=a arctan(x/a)-arctan (x)$
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After this question, related to the Prandtl–Meyer function
$$nu(M)
= int fracsqrtM^2-11+fracgamma -12M^2frac,dMM=
sqrtfracgamma + 1gamma -1 tan^-1left( sqrtfracgamma -1gamma +1 (M^2 -1)right) - tan^-1 left(sqrtM^2 -1right)$$ where $gamma=frac CpCv$is the ratio of the specific heat capacities, I wondered if we could, at least, get an approximation of $x$, solution of the equation
$$colorbluey=a tan^-1 left(frac x aright)-tan^-1 (x)tag 1$$ when $y$ is close to the asymptote $(y_infty=frac pi2 (a-1))$ for the range of "physical" values of $a$ (from a numerical point of view, solving $(1)$ does not present any difficulty).
For the ideal gas state,
$$left(
beginarrayccc
colorbluetextmolecules &colorblue gamma & colorbluea \
textmonoatomic & frac53 & 2 \
textdiatomic & frac75 & sqrt6 \
texttriatomic & frac97 & sqrt8 \
textpolyatomic & frac43 & sqrt7
endarray
right)$$
Fortunately, for $a=2$, we can recombine the arctangents making the equation to be
$$y=tan^-1left(fracx^34+3x^2right)$$ the solution of which being
$$x=k left(1+2 cosh left(frac13 cosh
^-1left(1+frac2k^2right)right)right)qquad textwhereqquad k=tan(y)$$ which is valid for any $y$.
What is interesting to notice is that $x(k)$ is almost a linear function which is far away to be the case for $x(y)$ as one could expect. For large values of $k$, corresponding to $y to frac pi 2$, Taylor expansion gives
$$x=3 k+frac49 k-frac32243 k^3+Oleft(frac1k^5right)$$
My problem is that I do not see how this could be used to generate an approximation for other values of $a$.
Any idea will be more than welcome.
Edit
After @Mariusz Iwaniuk's comments, we could expand $y$ as a Taylor series for infinitely large values of $x$ and use series reversion.
Using $Delta=y_infty-y$, we should end with
$$x=fraca^2-1Delta -fracleft(a^2+1right) 3(a^2-1)Delta -fracleft(a^4+11
a^2+1right) 45 left(a^2-1right)^3Delta ^3+Oleft(Delta ^5right)$$ which, numerically, seems to be quite good.
Update
Combining
$$y_infty=frac pi2 (a-1) tag 2$$
$$y=a tan^-1 left(frac x aright)-tan^-1 (x)tag 3$$
$$Delta=y_infty-y=a left(frac pi2-tan^-1 left(frac x aright) right) -left( frac pi2-tan^-1 (x)right)=a tan^-1 left(frac a x right)-tan ^-1left(frac1xright)tag 4$$ seems to be an easier approach for large $x$.
Expanding as Taylor series
$$Delta=sum_k=1^infty (-1)^k+1frac left(a^2 k-1right) (2 k-1),x^2k-1$$
Using series reversion
$$x=fraca^2-1Delta -fracleft(a^2+1right) 3( a^2-1)Delta -fracleft(a^4+11
a^2+1right)45 left(a^2-1right)^3 Delta ^3-frac2
left(a^2+1right) left(a^4+56 a^2+1right) 945
left(a^2-1right)^5Delta ^5+Oleft(Delta ^7right)$$ which makes
$$Delta-left(a tan^-1 left(frac a x right)-tan ^-1left(frac1xright) right)=fracleft(a^8+247 a^6+723 a^4+247 a^2+1right) 4725
left(a^2-1right)^8Delta ^9+Oleft(Delta ^11right)$$ For diatomic molecules, such as air, $(a=sqrt 6)$, the coefficient is $frac11737263671875approx 4.5 times 10^-5$.
On the other side, for small $x$, we can build the simplest Padé approximant and get
$$k=frac5 left(a^2-1right) x^315a^2+9 left(a^2+1right) x^2$$ leading to the cubic equation
$$5left( a^2-1right) x^3-9 left(a^2 +1right)k,x^2-15 a^2 k=0 $$ which can be solved using the hyperbolic method for one real root. The expression is too messy to be reported here.
trigonometry taylor-expansion approximation
edited Aug 9 at 3:48
Michael Hardy
204k23186463
asked Aug 6 at 7:42
Claude Leibovici
112k1055126
 |Â
show 1 more comment
up vote
2
down vote
favorite
After this question, related to the Prandtl–Meyer function
$$nu(M)
= int fracsqrtM^2-11+fracgamma -12M^2frac,dMM=
sqrtfracgamma + 1gamma -1 tan^-1left( sqrtfracgamma -1gamma +1 (M^2 -1)right) - tan^-1 left(sqrtM^2 -1right)$$ where $gamma=frac CpCv$is the ratio of the specific heat capacities, I wondered if we could, at least, get an approximation of $x$, solution of the equation
$$colorbluey=a tan^-1 left(frac x aright)-tan^-1 (x)tag 1$$ when $y$ is close to the asymptote $(y_infty=frac pi2 (a-1))$ for the range of "physical" values of $a$ (from a numerical point of view, solving $(1)$ does not present any difficulty).
For the ideal gas state,
$$left(
beginarrayccc
colorbluetextmolecules &colorblue gamma & colorbluea \
textmonoatomic & frac53 & 2 \
textdiatomic & frac75 & sqrt6 \
texttriatomic & frac97 & sqrt8 \
textpolyatomic & frac43 & sqrt7
endarray
right)$$
Fortunately, for $a=2$, we can recombine the arctangents making the equation to be
$$y=tan^-1left(fracx^34+3x^2right)$$ the solution of which being
$$x=k left(1+2 cosh left(frac13 cosh
^-1left(1+frac2k^2right)right)right)qquad textwhereqquad k=tan(y)$$ which is valid for any $y$.
What is interesting to notice is that $x(k)$ is almost a linear function which is far away to be the case for $x(y)$ as one could expect. For large values of $k$, corresponding to $y to frac pi 2$, Taylor expansion gives
$$x=3 k+frac49 k-frac32243 k^3+Oleft(frac1k^5right)$$
My problem is that I do not see how this could be used to generate an approximation for other values of $a$.
Any idea will be more than welcome.
Edit
After @Mariusz Iwaniuk's comments, we could expand $y$ as a Taylor series for infinitely large values of $x$ and use series reversion.
Using $Delta=y_infty-y$, we should end with
$$x=fraca^2-1Delta -fracleft(a^2+1right) 3(a^2-1)Delta -fracleft(a^4+11
a^2+1right) 45 left(a^2-1right)^3Delta ^3+Oleft(Delta ^5right)$$ which, numerically, seems to be quite good.
Update
Combining
$$y_infty=frac pi2 (a-1) tag 2$$
$$y=a tan^-1 left(frac x aright)-tan^-1 (x)tag 3$$
$$Delta=y_infty-y=a left(frac pi2-tan^-1 left(frac x aright) right) -left( frac pi2-tan^-1 (x)right)=a tan^-1 left(frac a x right)-tan ^-1left(frac1xright)tag 4$$ seems to be an easier approach for large $x$.
Expanding as Taylor series
$$Delta=sum_k=1^infty (-1)^k+1frac left(a^2 k-1right) (2 k-1),x^2k-1$$
Using series reversion
$$x=fraca^2-1Delta -fracleft(a^2+1right) 3( a^2-1)Delta -fracleft(a^4+11
a^2+1right)45 left(a^2-1right)^3 Delta ^3-frac2
left(a^2+1right) left(a^4+56 a^2+1right) 945
left(a^2-1right)^5Delta ^5+Oleft(Delta ^7right)$$ which makes
$$Delta-left(a tan^-1 left(frac a x right)-tan ^-1left(frac1xright) right)=fracleft(a^8+247 a^6+723 a^4+247 a^2+1right) 4725
left(a^2-1right)^8Delta ^9+Oleft(Delta ^11right)$$ For diatomic molecules, such as air, $(a=sqrt 6)$, the coefficient is $frac11737263671875approx 4.5 times 10^-5$.
On the other side, for small $x$, we can build the simplest Padé approximant and get
$$k=frac5 left(a^2-1right) x^315a^2+9 left(a^2+1right) x^2$$ leading to the cubic equation
$$5left( a^2-1right) x^3-9 left(a^2 +1right)k,x^2-15 a^2 k=0 $$ which can be solved using the hyperbolic method for one real root. The expression is too messy to be reported here.
trigonometry taylor-expansion approximation
edited Aug 9 at 3:48
Michael Hardy
204k23186463
asked Aug 6 at 7:42
Claude Leibovici
112k1055126
With MMA you can expand with series:X = (InverseSeries[Series[a ArcTan[x/a] - ArcTan[x], x, 0, 10]] // Normal) /. x -> y /. a -> 2(*for a = 2*), Plot[(1 + 2 Cosh[1/3 ArcCosh[1 + 2 Cot[y]^2]]) Tan[y], X, y, 0, 1]
– Mariusz Iwaniuk
Aug 6 at 8:31
@MariuszIwaniuk. This is fine but I am looking for $x$ when $y$ is large. For $a=2$ I have been happy yo be able to solve the cubic and the solution is valid for any $y$. If $y$ is small, I do not have any problem inversing the Taylor series. I suspect that this linearity of $x(k)$ is hidding something.
– Claude Leibovici
Aug 6 at 8:36
@MariuszIwaniuk. To be more precise, what I was working (in vain !) is $$x_(a)=x_(2)+f(x_(2),k)times (a-2)+g(x_(2),k)times (a-2)^2+cdots$$ Please, understand that for a given $y$, the solution $x_(a)$ varies extrely fast.
– Claude Leibovici
Aug 6 at 8:45
@MariuszIwaniuk. INspired by your answer, I stopped looking for something rigorous and used series reversion. This leads to something quite accurate. Thanks again. Cheers.
– Claude Leibovici
Aug 7 at 5:44
You're welcome :)
– Mariusz Iwaniuk
Aug 7 at 7:22
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
After this question, related to the Prandtl–Meyer function
$$nu(M)
= int fracsqrtM^2-11+fracgamma -12M^2frac,dMM=
sqrtfracgamma + 1gamma -1 tan^-1left( sqrtfracgamma -1gamma +1 (M^2 -1)right) - tan^-1 left(sqrtM^2 -1right)$$ where $gamma=frac CpCv$is the ratio of the specific heat capacities, I wondered if we could, at least, get an approximation of $x$, solution of the equation
$$colorbluey=a tan^-1 left(frac x aright)-tan^-1 (x)tag 1$$ when $y$ is close to the asymptote $(y_infty=frac pi2 (a-1))$ for the range of "physical" values of $a$ (from a numerical point of view, solving $(1)$ does not present any difficulty).
For the ideal gas state,
$$left(
beginarrayccc
colorbluetextmolecules &colorblue gamma & colorbluea \
textmonoatomic & frac53 & 2 \
textdiatomic & frac75 & sqrt6 \
texttriatomic & frac97 & sqrt8 \
textpolyatomic & frac43 & sqrt7
endarray
right)$$
Fortunately, for $a=2$, we can recombine the arctangents making the equation to be
$$y=tan^-1left(fracx^34+3x^2right)$$ the solution of which being
$$x=k left(1+2 cosh left(frac13 cosh
^-1left(1+frac2k^2right)right)right)qquad textwhereqquad k=tan(y)$$ which is valid for any $y$.
What is interesting to notice is that $x(k)$ is almost a linear function which is far away to be the case for $x(y)$ as one could expect. For large values of $k$, corresponding to $y to frac pi 2$, Taylor expansion gives
$$x=3 k+frac49 k-frac32243 k^3+Oleft(frac1k^5right)$$
My problem is that I do not see how this could be used to generate an approximation for other values of $a$.
Any idea will be more than welcome.
Edit
After @Mariusz Iwaniuk's comments, we could expand $y$ as a Taylor series for infinitely large values of $x$ and use series reversion.
Using $Delta=y_infty-y$, we should end with
$$x=fraca^2-1Delta -fracleft(a^2+1right) 3(a^2-1)Delta -fracleft(a^4+11
a^2+1right) 45 left(a^2-1right)^3Delta ^3+Oleft(Delta ^5right)$$ which, numerically, seems to be quite good.
Update
Combining
$$y_infty=frac pi2 (a-1) tag 2$$
$$y=a tan^-1 left(frac x aright)-tan^-1 (x)tag 3$$
$$Delta=y_infty-y=a left(frac pi2-tan^-1 left(frac x aright) right) -left( frac pi2-tan^-1 (x)right)=a tan^-1 left(frac a x right)-tan ^-1left(frac1xright)tag 4$$ seems to be an easier approach for large $x$.
Expanding as Taylor series
$$Delta=sum_k=1^infty (-1)^k+1frac left(a^2 k-1right) (2 k-1),x^2k-1$$
Using series reversion
$$x=fraca^2-1Delta -fracleft(a^2+1right) 3( a^2-1)Delta -fracleft(a^4+11
a^2+1right)45 left(a^2-1right)^3 Delta ^3-frac2
left(a^2+1right) left(a^4+56 a^2+1right) 945
left(a^2-1right)^5Delta ^5+Oleft(Delta ^7right)$$ which makes
$$Delta-left(a tan^-1 left(frac a x right)-tan ^-1left(frac1xright) right)=fracleft(a^8+247 a^6+723 a^4+247 a^2+1right) 4725
left(a^2-1right)^8Delta ^9+Oleft(Delta ^11right)$$ For diatomic molecules, such as air, $(a=sqrt 6)$, the coefficient is $frac11737263671875approx 4.5 times 10^-5$.
On the other side, for small $x$, we can build the simplest Padé approximant and get
$$k=frac5 left(a^2-1right) x^315a^2+9 left(a^2+1right) x^2$$ leading to the cubic equation
$$5left( a^2-1right) x^3-9 left(a^2 +1right)k,x^2-15 a^2 k=0 $$ which can be solved using the hyperbolic method for one real root. The expression is too messy to be reported here.
trigonometry taylor-expansion approximation
edited Aug 9 at 3:48
Michael Hardy
204k23186463
asked Aug 6 at 7:42
Claude Leibovici
112k1055126
After this question, related to the Prandtl–Meyer function
$$nu(M)
= int fracsqrtM^2-11+fracgamma -12M^2frac,dMM=
sqrtfracgamma + 1gamma -1 tan^-1left( sqrtfracgamma -1gamma +1 (M^2 -1)right) - tan^-1 left(sqrtM^2 -1right)$$ where $gamma=frac CpCv$is the ratio of the specific heat capacities, I wondered if we could, at least, get an approximation of $x$, solution of the equation
$$colorbluey=a tan^-1 left(frac x aright)-tan^-1 (x)tag 1$$ when $y$ is close to the asymptote $(y_infty=frac pi2 (a-1))$ for the range of "physical" values of $a$ (from a numerical point of view, solving $(1)$ does not present any difficulty).
For the ideal gas state,
$$left(
beginarrayccc
colorbluetextmolecules &colorblue gamma & colorbluea \
textmonoatomic & frac53 & 2 \
textdiatomic & frac75 & sqrt6 \
texttriatomic & frac97 & sqrt8 \
textpolyatomic & frac43 & sqrt7
endarray
right)$$
Fortunately, for $a=2$, we can recombine the arctangents making the equation to be
$$y=tan^-1left(fracx^34+3x^2right)$$ the solution of which being
$$x=k left(1+2 cosh left(frac13 cosh
^-1left(1+frac2k^2right)right)right)qquad textwhereqquad k=tan(y)$$ which is valid for any $y$.
What is interesting to notice is that $x(k)$ is almost a linear function which is far away to be the case for $x(y)$ as one could expect. For large values of $k$, corresponding to $y to frac pi 2$, Taylor expansion gives
$$x=3 k+frac49 k-frac32243 k^3+Oleft(frac1k^5right)$$
My problem is that I do not see how this could be used to generate an approximation for other values of $a$.
Any idea will be more than welcome.
Edit
After @Mariusz Iwaniuk's comments, we could expand $y$ as a Taylor series for infinitely large values of $x$ and use series reversion.
Using $Delta=y_infty-y$, we should end with
$$x=fraca^2-1Delta -fracleft(a^2+1right) 3(a^2-1)Delta -fracleft(a^4+11
a^2+1right) 45 left(a^2-1right)^3Delta ^3+Oleft(Delta ^5right)$$ which, numerically, seems to be quite good.
Update
Combining
$$y_infty=frac pi2 (a-1) tag 2$$
$$y=a tan^-1 left(frac x aright)-tan^-1 (x)tag 3$$
$$Delta=y_infty-y=a left(frac pi2-tan^-1 left(frac x aright) right) -left( frac pi2-tan^-1 (x)right)=a tan^-1 left(frac a x right)-tan ^-1left(frac1xright)tag 4$$ seems to be an easier approach for large $x$.
Expanding as Taylor series
$$Delta=sum_k=1^infty (-1)^k+1frac left(a^2 k-1right) (2 k-1),x^2k-1$$
Using series reversion
$$x=fraca^2-1Delta -fracleft(a^2+1right) 3( a^2-1)Delta -fracleft(a^4+11
a^2+1right)45 left(a^2-1right)^3 Delta ^3-frac2
left(a^2+1right) left(a^4+56 a^2+1right) 945
left(a^2-1right)^5Delta ^5+Oleft(Delta ^7right)$$ which makes
$$Delta-left(a tan^-1 left(frac a x right)-tan ^-1left(frac1xright) right)=fracleft(a^8+247 a^6+723 a^4+247 a^2+1right) 4725
left(a^2-1right)^8Delta ^9+Oleft(Delta ^11right)$$ For diatomic molecules, such as air, $(a=sqrt 6)$, the coefficient is $frac11737263671875approx 4.5 times 10^-5$.
On the other side, for small $x$, we can build the simplest Padé approximant and get
$$k=frac5 left(a^2-1right) x^315a^2+9 left(a^2+1right) x^2$$ leading to the cubic equation
$$5left( a^2-1right) x^3-9 left(a^2 +1right)k,x^2-15 a^2 k=0 $$ which can be solved using the hyperbolic method for one real root. The expression is too messy to be reported here.
trigonometry taylor-expansion approximation
edited Aug 9 at 3:48
Michael Hardy
204k23186463
asked Aug 6 at 7:42
Claude Leibovici
112k1055126
edited Aug 9 at 3:48
Michael Hardy
204k23186463
edited Aug 9 at 3:48
Michael Hardy
204k23186463
edited Aug 9 at 3:48
Michael Hardy
204k23186463
204k23186463
asked Aug 6 at 7:42
Claude Leibovici
112k1055126
asked Aug 6 at 7:42
Claude Leibovici
112k1055126
asked Aug 6 at 7:42
Claude Leibovici
112k1055126
112k1055126
With MMA you can expand with series:X = (InverseSeries[Series[a ArcTan[x/a] - ArcTan[x], x, 0, 10]] // Normal) /. x -> y /. a -> 2(*for a = 2*), Plot[(1 + 2 Cosh[1/3 ArcCosh[1 + 2 Cot[y]^2]]) Tan[y], X, y, 0, 1]
– Mariusz Iwaniuk
Aug 6 at 8:31
@MariuszIwaniuk. This is fine but I am looking for $x$ when $y$ is large. For $a=2$ I have been happy yo be able to solve the cubic and the solution is valid for any $y$. If $y$ is small, I do not have any problem inversing the Taylor series. I suspect that this linearity of $x(k)$ is hidding something.
– Claude Leibovici
Aug 6 at 8:36
@MariuszIwaniuk. To be more precise, what I was working (in vain !) is $$x_(a)=x_(2)+f(x_(2),k)times (a-2)+g(x_(2),k)times (a-2)^2+cdots$$ Please, understand that for a given $y$, the solution $x_(a)$ varies extrely fast.
– Claude Leibovici
Aug 6 at 8:45
@MariuszIwaniuk. INspired by your answer, I stopped looking for something rigorous and used series reversion. This leads to something quite accurate. Thanks again. Cheers.
– Claude Leibovici
Aug 7 at 5:44
You're welcome :)
– Mariusz Iwaniuk
Aug 7 at 7:22
 |Â
show 1 more comment
With MMA you can expand with series:X = (InverseSeries[Series[a ArcTan[x/a] - ArcTan[x], x, 0, 10]] // Normal) /. x -> y /. a -> 2(*for a = 2*), Plot[(1 + 2 Cosh[1/3 ArcCosh[1 + 2 Cot[y]^2]]) Tan[y], X, y, 0, 1]
– Mariusz Iwaniuk
Aug 6 at 8:31
@MariuszIwaniuk. This is fine but I am looking for $x$ when $y$ is large. For $a=2$ I have been happy yo be able to solve the cubic and the solution is valid for any $y$. If $y$ is small, I do not have any problem inversing the Taylor series. I suspect that this linearity of $x(k)$ is hidding something.
– Claude Leibovici
Aug 6 at 8:36
@MariuszIwaniuk. To be more precise, what I was working (in vain !) is $$x_(a)=x_(2)+f(x_(2),k)times (a-2)+g(x_(2),k)times (a-2)^2+cdots$$ Please, understand that for a given $y$, the solution $x_(a)$ varies extrely fast.
– Claude Leibovici
Aug 6 at 8:45
@MariuszIwaniuk. INspired by your answer, I stopped looking for something rigorous and used series reversion. This leads to something quite accurate. Thanks again. Cheers.
– Claude Leibovici
Aug 7 at 5:44
You're welcome :)
– Mariusz Iwaniuk
Aug 7 at 7:22
With MMA you can expand with series:
X = (InverseSeries[Series[a ArcTan[x/a] - ArcTan[x], x, 0, 10]] // Normal) /. x -> y /. a -> 2(*for a = 2*), Plot[(1 + 2 Cosh[1/3 ArcCosh[1 + 2 Cot[y]^2]]) Tan[y], X, y, 0, 1]– Mariusz Iwaniuk
Aug 6 at 8:31
With MMA you can expand with series:
X = (InverseSeries[Series[a ArcTan[x/a] - ArcTan[x], x, 0, 10]] // Normal) /. x -> y /. a -> 2(*for a = 2*), Plot[(1 + 2 Cosh[1/3 ArcCosh[1 + 2 Cot[y]^2]]) Tan[y], X, y, 0, 1]– Mariusz Iwaniuk
Aug 6 at 8:31
@MariuszIwaniuk. This is fine but I am looking for $x$ when $y$ is large. For $a=2$ I have been happy yo be able to solve the cubic and the solution is valid for any $y$. If $y$ is small, I do not have any problem inversing the Taylor series. I suspect that this linearity of $x(k)$ is hidding something.
– Claude Leibovici
Aug 6 at 8:36
@MariuszIwaniuk. This is fine but I am looking for $x$ when $y$ is large. For $a=2$ I have been happy yo be able to solve the cubic and the solution is valid for any $y$. If $y$ is small, I do not have any problem inversing the Taylor series. I suspect that this linearity of $x(k)$ is hidding something.
– Claude Leibovici
Aug 6 at 8:36
@MariuszIwaniuk. To be more precise, what I was working (in vain !) is $$x_(a)=x_(2)+f(x_(2),k)times (a-2)+g(x_(2),k)times (a-2)^2+cdots$$ Please, understand that for a given $y$, the solution $x_(a)$ varies extrely fast.
– Claude Leibovici
Aug 6 at 8:45
@MariuszIwaniuk. To be more precise, what I was working (in vain !) is $$x_(a)=x_(2)+f(x_(2),k)times (a-2)+g(x_(2),k)times (a-2)^2+cdots$$ Please, understand that for a given $y$, the solution $x_(a)$ varies extrely fast.
– Claude Leibovici
Aug 6 at 8:45
@MariuszIwaniuk. INspired by your answer, I stopped looking for something rigorous and used series reversion. This leads to something quite accurate. Thanks again. Cheers.
– Claude Leibovici
Aug 7 at 5:44
@MariuszIwaniuk. INspired by your answer, I stopped looking for something rigorous and used series reversion. This leads to something quite accurate. Thanks again. Cheers.
– Claude Leibovici
Aug 7 at 5:44
You're welcome :)
– Mariusz Iwaniuk
Aug 7 at 7:22
You're welcome :)
– Mariusz Iwaniuk
Aug 7 at 7:22
 |Â
show 1 more comment
1 Answer
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oldest
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up vote
2
down vote
accepted
An analytic approximation for large $k$ and $a ge 2$ is
$$ x=frac-k-acos(pi/a)sin(pi/a) + a^2,k,sin^2(pi/a)
sin^2(pi/a) + a,k,cos(pi/a),sin(pi/a) .$$
What's nice about it is that it reduces to $x=3k$ as given in Claude's initial exposition for $a=2.$ This is a large $k$ expansion. For a numerical check I let $a$ run from 2 to 3 by 0.05, and $k=5+1.5^m$ for $m$ = 0 to 20. The largest relative error of about 1.2% was obtained on the 'small k' side of $k=6.$
The technique was to take the tangent of eq. (1) above and write it, using the tangent addition formula, in the equivalent form
$$x=tan(a,tan^-1(x/a)) - kbig(1+xtan(a,tan^-1(x/a))big). $$
At first I tried expanding $tan(a,tan^-1(x/a))$ in a Laurent series around its singular point to get an equation in $x$ that could be solved analytically. Unfortunately it takes too many terms to give an accurate representation of the RHS of the equation. They will have to be included to go beyond the given approximation. In making the plots I noticed that a ballpark approximation to the solution was $x=a tan(pi/a).$
Furthermore the curve is decently approximated by a line over a fairly large region. Finally, a $k to infty$ analysis suggests that the point $x=a tan(pi/a)$ is the best. Thus the RHS of the previous equation is expanded around this point to first order in $(x-a tan(pi/a))$ and, solve for $x$ to get the equation
at the top of this answer.
answered Aug 7 at 19:46
skbmoore
1,12026
Very nice work, for sure. Thanks for posting it. Cheers.
– Claude Leibovici
Aug 8 at 2:33
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
An analytic approximation for large $k$ and $a ge 2$ is
$$ x=frac-k-acos(pi/a)sin(pi/a) + a^2,k,sin^2(pi/a)
sin^2(pi/a) + a,k,cos(pi/a),sin(pi/a) .$$
What's nice about it is that it reduces to $x=3k$ as given in Claude's initial exposition for $a=2.$ This is a large $k$ expansion. For a numerical check I let $a$ run from 2 to 3 by 0.05, and $k=5+1.5^m$ for $m$ = 0 to 20. The largest relative error of about 1.2% was obtained on the 'small k' side of $k=6.$
The technique was to take the tangent of eq. (1) above and write it, using the tangent addition formula, in the equivalent form
$$x=tan(a,tan^-1(x/a)) - kbig(1+xtan(a,tan^-1(x/a))big). $$
At first I tried expanding $tan(a,tan^-1(x/a))$ in a Laurent series around its singular point to get an equation in $x$ that could be solved analytically. Unfortunately it takes too many terms to give an accurate representation of the RHS of the equation. They will have to be included to go beyond the given approximation. In making the plots I noticed that a ballpark approximation to the solution was $x=a tan(pi/a).$
Furthermore the curve is decently approximated by a line over a fairly large region. Finally, a $k to infty$ analysis suggests that the point $x=a tan(pi/a)$ is the best. Thus the RHS of the previous equation is expanded around this point to first order in $(x-a tan(pi/a))$ and, solve for $x$ to get the equation
at the top of this answer.
answered Aug 7 at 19:46
skbmoore
1,12026
Very nice work, for sure. Thanks for posting it. Cheers.
– Claude Leibovici
Aug 8 at 2:33
add a comment |Â
up vote
2
down vote
accepted
An analytic approximation for large $k$ and $a ge 2$ is
$$ x=frac-k-acos(pi/a)sin(pi/a) + a^2,k,sin^2(pi/a)
sin^2(pi/a) + a,k,cos(pi/a),sin(pi/a) .$$
What's nice about it is that it reduces to $x=3k$ as given in Claude's initial exposition for $a=2.$ This is a large $k$ expansion. For a numerical check I let $a$ run from 2 to 3 by 0.05, and $k=5+1.5^m$ for $m$ = 0 to 20. The largest relative error of about 1.2% was obtained on the 'small k' side of $k=6.$
The technique was to take the tangent of eq. (1) above and write it, using the tangent addition formula, in the equivalent form
$$x=tan(a,tan^-1(x/a)) - kbig(1+xtan(a,tan^-1(x/a))big). $$
At first I tried expanding $tan(a,tan^-1(x/a))$ in a Laurent series around its singular point to get an equation in $x$ that could be solved analytically. Unfortunately it takes too many terms to give an accurate representation of the RHS of the equation. They will have to be included to go beyond the given approximation. In making the plots I noticed that a ballpark approximation to the solution was $x=a tan(pi/a).$
Furthermore the curve is decently approximated by a line over a fairly large region. Finally, a $k to infty$ analysis suggests that the point $x=a tan(pi/a)$ is the best. Thus the RHS of the previous equation is expanded around this point to first order in $(x-a tan(pi/a))$ and, solve for $x$ to get the equation
at the top of this answer.
answered Aug 7 at 19:46
skbmoore
1,12026
Very nice work, for sure. Thanks for posting it. Cheers.
– Claude Leibovici
Aug 8 at 2:33
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
An analytic approximation for large $k$ and $a ge 2$ is
$$ x=frac-k-acos(pi/a)sin(pi/a) + a^2,k,sin^2(pi/a)
sin^2(pi/a) + a,k,cos(pi/a),sin(pi/a) .$$
What's nice about it is that it reduces to $x=3k$ as given in Claude's initial exposition for $a=2.$ This is a large $k$ expansion. For a numerical check I let $a$ run from 2 to 3 by 0.05, and $k=5+1.5^m$ for $m$ = 0 to 20. The largest relative error of about 1.2% was obtained on the 'small k' side of $k=6.$
The technique was to take the tangent of eq. (1) above and write it, using the tangent addition formula, in the equivalent form
$$x=tan(a,tan^-1(x/a)) - kbig(1+xtan(a,tan^-1(x/a))big). $$
At first I tried expanding $tan(a,tan^-1(x/a))$ in a Laurent series around its singular point to get an equation in $x$ that could be solved analytically. Unfortunately it takes too many terms to give an accurate representation of the RHS of the equation. They will have to be included to go beyond the given approximation. In making the plots I noticed that a ballpark approximation to the solution was $x=a tan(pi/a).$
Furthermore the curve is decently approximated by a line over a fairly large region. Finally, a $k to infty$ analysis suggests that the point $x=a tan(pi/a)$ is the best. Thus the RHS of the previous equation is expanded around this point to first order in $(x-a tan(pi/a))$ and, solve for $x$ to get the equation
at the top of this answer.
answered Aug 7 at 19:46
skbmoore
1,12026
An analytic approximation for large $k$ and $a ge 2$ is
$$ x=frac-k-acos(pi/a)sin(pi/a) + a^2,k,sin^2(pi/a)
sin^2(pi/a) + a,k,cos(pi/a),sin(pi/a) .$$
What's nice about it is that it reduces to $x=3k$ as given in Claude's initial exposition for $a=2.$ This is a large $k$ expansion. For a numerical check I let $a$ run from 2 to 3 by 0.05, and $k=5+1.5^m$ for $m$ = 0 to 20. The largest relative error of about 1.2% was obtained on the 'small k' side of $k=6.$
The technique was to take the tangent of eq. (1) above and write it, using the tangent addition formula, in the equivalent form
$$x=tan(a,tan^-1(x/a)) - kbig(1+xtan(a,tan^-1(x/a))big). $$
At first I tried expanding $tan(a,tan^-1(x/a))$ in a Laurent series around its singular point to get an equation in $x$ that could be solved analytically. Unfortunately it takes too many terms to give an accurate representation of the RHS of the equation. They will have to be included to go beyond the given approximation. In making the plots I noticed that a ballpark approximation to the solution was $x=a tan(pi/a).$
Furthermore the curve is decently approximated by a line over a fairly large region. Finally, a $k to infty$ analysis suggests that the point $x=a tan(pi/a)$ is the best. Thus the RHS of the previous equation is expanded around this point to first order in $(x-a tan(pi/a))$ and, solve for $x$ to get the equation
at the top of this answer.
answered Aug 7 at 19:46
skbmoore
1,12026
answered Aug 7 at 19:46
skbmoore
1,12026
answered Aug 7 at 19:46
skbmoore
1,12026
answered Aug 7 at 19:46
skbmoore
1,12026
1,12026
Very nice work, for sure. Thanks for posting it. Cheers.
– Claude Leibovici
Aug 8 at 2:33
add a comment |Â
Very nice work, for sure. Thanks for posting it. Cheers.
– Claude Leibovici
Aug 8 at 2:33
Very nice work, for sure. Thanks for posting it. Cheers.
– Claude Leibovici
Aug 8 at 2:33
Very nice work, for sure. Thanks for posting it. Cheers.
– Claude Leibovici
Aug 8 at 2:33
add a comment |Â
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With MMA you can expand with series:
X = (InverseSeries[Series[a ArcTan[x/a] - ArcTan[x], x, 0, 10]] // Normal) /. x -> y /. a -> 2(*for a = 2*), Plot[(1 + 2 Cosh[1/3 ArcCosh[1 + 2 Cot[y]^2]]) Tan[y], X, y, 0, 1]– Mariusz Iwaniuk
Aug 6 at 8:31
@MariuszIwaniuk. This is fine but I am looking for $x$ when $y$ is large. For $a=2$ I have been happy yo be able to solve the cubic and the solution is valid for any $y$. If $y$ is small, I do not have any problem inversing the Taylor series. I suspect that this linearity of $x(k)$ is hidding something.
– Claude Leibovici
Aug 6 at 8:36
@MariuszIwaniuk. To be more precise, what I was working (in vain !) is $$x_(a)=x_(2)+f(x_(2),k)times (a-2)+g(x_(2),k)times (a-2)^2+cdots$$ Please, understand that for a given $y$, the solution $x_(a)$ varies extrely fast.
– Claude Leibovici
Aug 6 at 8:45
@MariuszIwaniuk. INspired by your answer, I stopped looking for something rigorous and used series reversion. This leads to something quite accurate. Thanks again. Cheers.
– Claude Leibovici
Aug 7 at 5:44
You're welcome :)
– Mariusz Iwaniuk
Aug 7 at 7:22