Solve for $theta$: $x = theta - sintheta$ [duplicate]
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
This question already has an answer here:
How to solve Kepler's equation $M=E-varepsilon sin E$ for $E$?
2 answers
Solve for $theta$:
$$x = theta - sintheta$$
Is this type of isolation a matter of identities? If so, which one(s)?
trigonometry
edited Sep 7 at 6:50
Chase Ryan Taylor
4,24021530
asked Sep 7 at 5:54
user101434
244
marked as duplicate by Hans Lundmark, Jendrik Stelzner, Jack D'Aurizio♦ trigonometry
Users with the  trigonometry badge can single-handedly close trigonometry questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
Sep 7 at 13:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
 |Â
show 3 more comments
up vote
4
down vote
favorite
This question already has an answer here:
How to solve Kepler's equation $M=E-varepsilon sin E$ for $E$?
2 answers
Solve for $theta$:
$$x = theta - sintheta$$
Is this type of isolation a matter of identities? If so, which one(s)?
trigonometry
edited Sep 7 at 6:50
Chase Ryan Taylor
4,24021530
asked Sep 7 at 5:54
user101434
244
marked as duplicate by Hans Lundmark, Jendrik Stelzner, Jack D'Aurizio♦ trigonometry
Users with the  trigonometry badge can single-handedly close trigonometry questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
Sep 7 at 13:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
5
That’s transcendental
– InertialObserver
Sep 7 at 5:56
A closed solution for $theta$ doesn't exist. There are numerical approaches one can use to solve such a problem, such as Newton's Method
– WaveX
Sep 7 at 5:57
Are you serious? But I can create a function substituting y for θ and flip that function along y = x (a 45° diagonal through Quadrant I) and the function/curve passes the vertical line test. Doesn't that mean there is surely a way to arrive at a "θ = ..." equation?
– user101434
Sep 7 at 6:01
2
I don't quite see what you are doing there, but regardless of this you should ask yourself if that procedure applied to $y=x^3$ results in introducing a weird V-shaped symbol with a small three inside and then writing $x=sqrt[3]y$.
– Saucy O'Path
Sep 7 at 6:09
1
@zahbaz And the answer is no. You can define $theta(x)$ to be a function, but to compute values you will have to resort to other means: numerical methods, series, or whatever.
– Jean-Claude Arbaut
Sep 7 at 6:23
 |Â
show 3 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
This question already has an answer here:
How to solve Kepler's equation $M=E-varepsilon sin E$ for $E$?
2 answers
Solve for $theta$:
$$x = theta - sintheta$$
Is this type of isolation a matter of identities? If so, which one(s)?
trigonometry
edited Sep 7 at 6:50
Chase Ryan Taylor
4,24021530
asked Sep 7 at 5:54
user101434
244
This question already has an answer here:
How to solve Kepler's equation $M=E-varepsilon sin E$ for $E$?
2 answers
Solve for $theta$:
$$x = theta - sintheta$$
Is this type of isolation a matter of identities? If so, which one(s)?
This question already has an answer here:
How to solve Kepler's equation $M=E-varepsilon sin E$ for $E$?
2 answers
trigonometry
trigonometry
edited Sep 7 at 6:50
Chase Ryan Taylor
4,24021530
asked Sep 7 at 5:54
user101434
244
edited Sep 7 at 6:50
Chase Ryan Taylor
4,24021530
asked Sep 7 at 5:54
user101434
244
edited Sep 7 at 6:50
Chase Ryan Taylor
4,24021530
edited Sep 7 at 6:50
Chase Ryan Taylor
4,24021530
edited Sep 7 at 6:50
Chase Ryan Taylor
4,24021530
4,24021530
asked Sep 7 at 5:54
user101434
244
asked Sep 7 at 5:54
user101434
244
asked Sep 7 at 5:54
user101434
244
244
marked as duplicate by Hans Lundmark, Jendrik Stelzner, Jack D'Aurizio♦ trigonometry
Users with the  trigonometry badge can single-handedly close trigonometry questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
Sep 7 at 13:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Hans Lundmark, Jendrik Stelzner, Jack D'Aurizio♦ trigonometry
Users with the  trigonometry badge can single-handedly close trigonometry questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
Sep 7 at 13:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
5
That’s transcendental
– InertialObserver
Sep 7 at 5:56
A closed solution for $theta$ doesn't exist. There are numerical approaches one can use to solve such a problem, such as Newton's Method
– WaveX
Sep 7 at 5:57
Are you serious? But I can create a function substituting y for θ and flip that function along y = x (a 45° diagonal through Quadrant I) and the function/curve passes the vertical line test. Doesn't that mean there is surely a way to arrive at a "θ = ..." equation?
– user101434
Sep 7 at 6:01
2
I don't quite see what you are doing there, but regardless of this you should ask yourself if that procedure applied to $y=x^3$ results in introducing a weird V-shaped symbol with a small three inside and then writing $x=sqrt[3]y$.
– Saucy O'Path
Sep 7 at 6:09
1
@zahbaz And the answer is no. You can define $theta(x)$ to be a function, but to compute values you will have to resort to other means: numerical methods, series, or whatever.
– Jean-Claude Arbaut
Sep 7 at 6:23
 |Â
show 3 more comments
5
That’s transcendental
– InertialObserver
Sep 7 at 5:56
A closed solution for $theta$ doesn't exist. There are numerical approaches one can use to solve such a problem, such as Newton's Method
– WaveX
Sep 7 at 5:57
Are you serious? But I can create a function substituting y for θ and flip that function along y = x (a 45° diagonal through Quadrant I) and the function/curve passes the vertical line test. Doesn't that mean there is surely a way to arrive at a "θ = ..." equation?
– user101434
Sep 7 at 6:01
2
I don't quite see what you are doing there, but regardless of this you should ask yourself if that procedure applied to $y=x^3$ results in introducing a weird V-shaped symbol with a small three inside and then writing $x=sqrt[3]y$.
– Saucy O'Path
Sep 7 at 6:09
1
@zahbaz And the answer is no. You can define $theta(x)$ to be a function, but to compute values you will have to resort to other means: numerical methods, series, or whatever.
– Jean-Claude Arbaut
Sep 7 at 6:23
5
5
That’s transcendental
– InertialObserver
Sep 7 at 5:56
That’s transcendental
– InertialObserver
Sep 7 at 5:56
A closed solution for $theta$ doesn't exist. There are numerical approaches one can use to solve such a problem, such as Newton's Method
– WaveX
Sep 7 at 5:57
A closed solution for $theta$ doesn't exist. There are numerical approaches one can use to solve such a problem, such as Newton's Method
– WaveX
Sep 7 at 5:57
Are you serious? But I can create a function substituting y for θ and flip that function along y = x (a 45° diagonal through Quadrant I) and the function/curve passes the vertical line test. Doesn't that mean there is surely a way to arrive at a "θ = ..." equation?
– user101434
Sep 7 at 6:01
Are you serious? But I can create a function substituting y for θ and flip that function along y = x (a 45° diagonal through Quadrant I) and the function/curve passes the vertical line test. Doesn't that mean there is surely a way to arrive at a "θ = ..." equation?
– user101434
Sep 7 at 6:01
2
2
I don't quite see what you are doing there, but regardless of this you should ask yourself if that procedure applied to $y=x^3$ results in introducing a weird V-shaped symbol with a small three inside and then writing $x=sqrt[3]y$.
– Saucy O'Path
Sep 7 at 6:09
I don't quite see what you are doing there, but regardless of this you should ask yourself if that procedure applied to $y=x^3$ results in introducing a weird V-shaped symbol with a small three inside and then writing $x=sqrt[3]y$.
– Saucy O'Path
Sep 7 at 6:09
1
1
@zahbaz And the answer is no. You can define $theta(x)$ to be a function, but to compute values you will have to resort to other means: numerical methods, series, or whatever.
– Jean-Claude Arbaut
Sep 7 at 6:23
@zahbaz And the answer is no. You can define $theta(x)$ to be a function, but to compute values you will have to resort to other means: numerical methods, series, or whatever.
– Jean-Claude Arbaut
Sep 7 at 6:23
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
5
down vote
As said in comments, you need some numerical methods for this kind of transcendental equations.
How,we can approximate solutions to get an estimate for, say, Newton method. For example, if the solution is in the range $0 leq theta leq pi$, the maginficent approximation
$$sin(theta) simeq frac16 (pi -theta) theta5 pi ^2-4 (pi -theta) thetaqquad textfor qquad 0leq thetaleqpi$$
proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (that is to say more than $1400$ years ago !) would lead to the cubic equation
$$4, theta ^3-4 (x+pi -4),theta ^2+ pileft(4 x+5 pi -16 right),theta-5
pi ^2 x=0$$ which can be solved.
In the table below, I produced some values for the solution.
$$left(
beginarrayccc
x & textapproximation & textexact \
0.0 & 0.00000 & 0.00000 \
0.1 & 0.84987 & 0.85375 \
0.2 & 1.08166 & 1.08369 \
0.3 & 1.24765 & 1.24852 \
0.4 & 1.38201 & 1.38228 \
0.5 & 1.49726 & 1.49730 \
0.6 & 1.59958 & 1.59959 \
0.7 & 1.69251 & 1.69259 \
0.8 & 1.77828 & 1.77851 \
0.9 & 1.85843 & 1.85881 \
1.0 & 1.93404 & 1.93456 \
1.1 & 2.00590 & 2.00655 \
1.2 & 2.07463 & 2.07538 \
1.3 & 2.14070 & 2.14151 \
1.4 & 2.20451 & 2.20534 \
1.5 & 2.26636 & 2.26717 \
1.6 & 2.32651 & 2.32726 \
1.7 & 2.38517 & 2.38584 \
1.8 & 2.44254 & 2.44308 \
1.9 & 2.49876 & 2.49915 \
2.0 & 2.55398 & 2.55420
endarray
right)$$
Another possible way would be Taylor expansion and series reversion to get
$$theta=t+fract^360+fract^51400+fract^725200+frac43
t^917248000+frac1213 t^117207200000+frac151439
t^1312713500800000$$ where $t=sqrt[3]6x$
answered Sep 7 at 6:41
Claude Leibovici
113k1155127
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
As said in comments, you need some numerical methods for this kind of transcendental equations.
How,we can approximate solutions to get an estimate for, say, Newton method. For example, if the solution is in the range $0 leq theta leq pi$, the maginficent approximation
$$sin(theta) simeq frac16 (pi -theta) theta5 pi ^2-4 (pi -theta) thetaqquad textfor qquad 0leq thetaleqpi$$
proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (that is to say more than $1400$ years ago !) would lead to the cubic equation
$$4, theta ^3-4 (x+pi -4),theta ^2+ pileft(4 x+5 pi -16 right),theta-5
pi ^2 x=0$$ which can be solved.
In the table below, I produced some values for the solution.
$$left(
beginarrayccc
x & textapproximation & textexact \
0.0 & 0.00000 & 0.00000 \
0.1 & 0.84987 & 0.85375 \
0.2 & 1.08166 & 1.08369 \
0.3 & 1.24765 & 1.24852 \
0.4 & 1.38201 & 1.38228 \
0.5 & 1.49726 & 1.49730 \
0.6 & 1.59958 & 1.59959 \
0.7 & 1.69251 & 1.69259 \
0.8 & 1.77828 & 1.77851 \
0.9 & 1.85843 & 1.85881 \
1.0 & 1.93404 & 1.93456 \
1.1 & 2.00590 & 2.00655 \
1.2 & 2.07463 & 2.07538 \
1.3 & 2.14070 & 2.14151 \
1.4 & 2.20451 & 2.20534 \
1.5 & 2.26636 & 2.26717 \
1.6 & 2.32651 & 2.32726 \
1.7 & 2.38517 & 2.38584 \
1.8 & 2.44254 & 2.44308 \
1.9 & 2.49876 & 2.49915 \
2.0 & 2.55398 & 2.55420
endarray
right)$$
Another possible way would be Taylor expansion and series reversion to get
$$theta=t+fract^360+fract^51400+fract^725200+frac43
t^917248000+frac1213 t^117207200000+frac151439
t^1312713500800000$$ where $t=sqrt[3]6x$
answered Sep 7 at 6:41
Claude Leibovici
113k1155127
add a comment |Â
up vote
5
down vote
As said in comments, you need some numerical methods for this kind of transcendental equations.
How,we can approximate solutions to get an estimate for, say, Newton method. For example, if the solution is in the range $0 leq theta leq pi$, the maginficent approximation
$$sin(theta) simeq frac16 (pi -theta) theta5 pi ^2-4 (pi -theta) thetaqquad textfor qquad 0leq thetaleqpi$$
proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (that is to say more than $1400$ years ago !) would lead to the cubic equation
$$4, theta ^3-4 (x+pi -4),theta ^2+ pileft(4 x+5 pi -16 right),theta-5
pi ^2 x=0$$ which can be solved.
In the table below, I produced some values for the solution.
$$left(
beginarrayccc
x & textapproximation & textexact \
0.0 & 0.00000 & 0.00000 \
0.1 & 0.84987 & 0.85375 \
0.2 & 1.08166 & 1.08369 \
0.3 & 1.24765 & 1.24852 \
0.4 & 1.38201 & 1.38228 \
0.5 & 1.49726 & 1.49730 \
0.6 & 1.59958 & 1.59959 \
0.7 & 1.69251 & 1.69259 \
0.8 & 1.77828 & 1.77851 \
0.9 & 1.85843 & 1.85881 \
1.0 & 1.93404 & 1.93456 \
1.1 & 2.00590 & 2.00655 \
1.2 & 2.07463 & 2.07538 \
1.3 & 2.14070 & 2.14151 \
1.4 & 2.20451 & 2.20534 \
1.5 & 2.26636 & 2.26717 \
1.6 & 2.32651 & 2.32726 \
1.7 & 2.38517 & 2.38584 \
1.8 & 2.44254 & 2.44308 \
1.9 & 2.49876 & 2.49915 \
2.0 & 2.55398 & 2.55420
endarray
right)$$
Another possible way would be Taylor expansion and series reversion to get
$$theta=t+fract^360+fract^51400+fract^725200+frac43
t^917248000+frac1213 t^117207200000+frac151439
t^1312713500800000$$ where $t=sqrt[3]6x$
answered Sep 7 at 6:41
Claude Leibovici
113k1155127
add a comment |Â
up vote
5
down vote
up vote
5
down vote
As said in comments, you need some numerical methods for this kind of transcendental equations.
How,we can approximate solutions to get an estimate for, say, Newton method. For example, if the solution is in the range $0 leq theta leq pi$, the maginficent approximation
$$sin(theta) simeq frac16 (pi -theta) theta5 pi ^2-4 (pi -theta) thetaqquad textfor qquad 0leq thetaleqpi$$
proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (that is to say more than $1400$ years ago !) would lead to the cubic equation
$$4, theta ^3-4 (x+pi -4),theta ^2+ pileft(4 x+5 pi -16 right),theta-5
pi ^2 x=0$$ which can be solved.
In the table below, I produced some values for the solution.
$$left(
beginarrayccc
x & textapproximation & textexact \
0.0 & 0.00000 & 0.00000 \
0.1 & 0.84987 & 0.85375 \
0.2 & 1.08166 & 1.08369 \
0.3 & 1.24765 & 1.24852 \
0.4 & 1.38201 & 1.38228 \
0.5 & 1.49726 & 1.49730 \
0.6 & 1.59958 & 1.59959 \
0.7 & 1.69251 & 1.69259 \
0.8 & 1.77828 & 1.77851 \
0.9 & 1.85843 & 1.85881 \
1.0 & 1.93404 & 1.93456 \
1.1 & 2.00590 & 2.00655 \
1.2 & 2.07463 & 2.07538 \
1.3 & 2.14070 & 2.14151 \
1.4 & 2.20451 & 2.20534 \
1.5 & 2.26636 & 2.26717 \
1.6 & 2.32651 & 2.32726 \
1.7 & 2.38517 & 2.38584 \
1.8 & 2.44254 & 2.44308 \
1.9 & 2.49876 & 2.49915 \
2.0 & 2.55398 & 2.55420
endarray
right)$$
Another possible way would be Taylor expansion and series reversion to get
$$theta=t+fract^360+fract^51400+fract^725200+frac43
t^917248000+frac1213 t^117207200000+frac151439
t^1312713500800000$$ where $t=sqrt[3]6x$
answered Sep 7 at 6:41
Claude Leibovici
113k1155127
As said in comments, you need some numerical methods for this kind of transcendental equations.
How,we can approximate solutions to get an estimate for, say, Newton method. For example, if the solution is in the range $0 leq theta leq pi$, the maginficent approximation
$$sin(theta) simeq frac16 (pi -theta) theta5 pi ^2-4 (pi -theta) thetaqquad textfor qquad 0leq thetaleqpi$$
proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (that is to say more than $1400$ years ago !) would lead to the cubic equation
$$4, theta ^3-4 (x+pi -4),theta ^2+ pileft(4 x+5 pi -16 right),theta-5
pi ^2 x=0$$ which can be solved.
In the table below, I produced some values for the solution.
$$left(
beginarrayccc
x & textapproximation & textexact \
0.0 & 0.00000 & 0.00000 \
0.1 & 0.84987 & 0.85375 \
0.2 & 1.08166 & 1.08369 \
0.3 & 1.24765 & 1.24852 \
0.4 & 1.38201 & 1.38228 \
0.5 & 1.49726 & 1.49730 \
0.6 & 1.59958 & 1.59959 \
0.7 & 1.69251 & 1.69259 \
0.8 & 1.77828 & 1.77851 \
0.9 & 1.85843 & 1.85881 \
1.0 & 1.93404 & 1.93456 \
1.1 & 2.00590 & 2.00655 \
1.2 & 2.07463 & 2.07538 \
1.3 & 2.14070 & 2.14151 \
1.4 & 2.20451 & 2.20534 \
1.5 & 2.26636 & 2.26717 \
1.6 & 2.32651 & 2.32726 \
1.7 & 2.38517 & 2.38584 \
1.8 & 2.44254 & 2.44308 \
1.9 & 2.49876 & 2.49915 \
2.0 & 2.55398 & 2.55420
endarray
right)$$
Another possible way would be Taylor expansion and series reversion to get
$$theta=t+fract^360+fract^51400+fract^725200+frac43
t^917248000+frac1213 t^117207200000+frac151439
t^1312713500800000$$ where $t=sqrt[3]6x$
answered Sep 7 at 6:41
Claude Leibovici
113k1155127
edited Sep 7 at 6:59
answered Sep 7 at 6:41
Claude Leibovici
113k1155127
answered Sep 7 at 6:41
Claude Leibovici
113k1155127
answered Sep 7 at 6:41
Claude Leibovici
113k1155127
113k1155127
add a comment |Â
add a comment |Â
5
That’s transcendental
– InertialObserver
Sep 7 at 5:56
A closed solution for $theta$ doesn't exist. There are numerical approaches one can use to solve such a problem, such as Newton's Method
– WaveX
Sep 7 at 5:57
Are you serious? But I can create a function substituting y for θ and flip that function along y = x (a 45° diagonal through Quadrant I) and the function/curve passes the vertical line test. Doesn't that mean there is surely a way to arrive at a "θ = ..." equation?
– user101434
Sep 7 at 6:01
2
I don't quite see what you are doing there, but regardless of this you should ask yourself if that procedure applied to $y=x^3$ results in introducing a weird V-shaped symbol with a small three inside and then writing $x=sqrt[3]y$.
– Saucy O'Path
Sep 7 at 6:09
1
@zahbaz And the answer is no. You can define $theta(x)$ to be a function, but to compute values you will have to resort to other means: numerical methods, series, or whatever.
– Jean-Claude Arbaut
Sep 7 at 6:23