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How do you solve the trigonometric equation $sin(x)+x=9$?
How do you solve the following trigonometric equation?Solution Sets of Trigonometric EquationsCalculate $cos(x / 3)$ from $sin(x)$ and $cos(x)$Substituting a value of sine function in a trigonometric equationSolve the trigonometric equation: $sin 3x = 4 sin^2 x$How do you solve for $alpha$ in this trigonometric equation?Solve the trigonometric equation with $sin$, $cos$How to properly choose a solution to a system of equations of trigonometric functions?How can I transform one equation about cycloidal cams to another via trigonometry? Does the author makes any assumptions on this?Can't resolve a simple equation
$begingroup$
How do you solve the trigonometric equation $sin(x)+x=9$? More generally, how do you solve equations with both trigs and 'x's without graphing? And maybe I only want real number answers.
Suggestions and edit: If you use the Taylor Series, you will basically end up with at least $6^th$ degree equations, if you want to make it close. Equation Solvers Are NOT Allowed. That will kind of limit you to this:(assuming you make careless mistakes, just like me, so you don't want cubic equations) $$x+x=9.$$ This yields $x=4.5$, which is DEFINITELY a bad estimation.
Thanks for helping me! It's just that I don't know what Newton-Ralphson and iterations are. Can you explain it to me or give me a link?
trigonometry
asked Mar 22 at 14:09
Math LoverMath Lover
17410
$endgroup$
add a comment |
$begingroup$
How do you solve the trigonometric equation $sin(x)+x=9$? More generally, how do you solve equations with both trigs and 'x's without graphing? And maybe I only want real number answers.
Suggestions and edit: If you use the Taylor Series, you will basically end up with at least $6^th$ degree equations, if you want to make it close. Equation Solvers Are NOT Allowed. That will kind of limit you to this:(assuming you make careless mistakes, just like me, so you don't want cubic equations) $$x+x=9.$$ This yields $x=4.5$, which is DEFINITELY a bad estimation.
Thanks for helping me! It's just that I don't know what Newton-Ralphson and iterations are. Can you explain it to me or give me a link?
trigonometry
asked Mar 22 at 14:09
Math LoverMath Lover
17410
$endgroup$
$begingroup$
I'd use Python or WolframAlpha.
$endgroup$
– Botond
Mar 22 at 14:13
$begingroup$
Two suggestions : (1) You could use iteration if you are just after a numerical solution. (2) Approximate sin(x) with, say a Taylor approximation with a suitable number of terms.
$endgroup$
– Martin Hansen
Mar 22 at 14:14
$begingroup$
You also know that $x$ must be at least $8$ because $sinx$ cannot be greater than $1$.
$endgroup$
– John Douma
Mar 22 at 14:17
$begingroup$
Likewise, $x leq 10$ because $sin x geq -1$.
$endgroup$
– Ertxiem
Mar 22 at 15:01
add a comment |
$begingroup$
How do you solve the trigonometric equation $sin(x)+x=9$? More generally, how do you solve equations with both trigs and 'x's without graphing? And maybe I only want real number answers.
Suggestions and edit: If you use the Taylor Series, you will basically end up with at least $6^th$ degree equations, if you want to make it close. Equation Solvers Are NOT Allowed. That will kind of limit you to this:(assuming you make careless mistakes, just like me, so you don't want cubic equations) $$x+x=9.$$ This yields $x=4.5$, which is DEFINITELY a bad estimation.
Thanks for helping me! It's just that I don't know what Newton-Ralphson and iterations are. Can you explain it to me or give me a link?
trigonometry
asked Mar 22 at 14:09
Math LoverMath Lover
17410
$endgroup$
How do you solve the trigonometric equation $sin(x)+x=9$? More generally, how do you solve equations with both trigs and 'x's without graphing? And maybe I only want real number answers.
Suggestions and edit: If you use the Taylor Series, you will basically end up with at least $6^th$ degree equations, if you want to make it close. Equation Solvers Are NOT Allowed. That will kind of limit you to this:(assuming you make careless mistakes, just like me, so you don't want cubic equations) $$x+x=9.$$ This yields $x=4.5$, which is DEFINITELY a bad estimation.
Thanks for helping me! It's just that I don't know what Newton-Ralphson and iterations are. Can you explain it to me or give me a link?
trigonometry
trigonometry
asked Mar 22 at 14:09
Math LoverMath Lover
17410
asked Mar 22 at 14:09
Math LoverMath Lover
17410
edited Mar 26 at 15:58
Math Lover
asked Mar 22 at 14:09
Math LoverMath Lover
17410
asked Mar 22 at 14:09
Math LoverMath Lover
17410
asked Mar 22 at 14:09
Math LoverMath Lover
17410
17410
$begingroup$
I'd use Python or WolframAlpha.
$endgroup$
– Botond
Mar 22 at 14:13
$begingroup$
Two suggestions : (1) You could use iteration if you are just after a numerical solution. (2) Approximate sin(x) with, say a Taylor approximation with a suitable number of terms.
$endgroup$
– Martin Hansen
Mar 22 at 14:14
$begingroup$
You also know that $x$ must be at least $8$ because $sinx$ cannot be greater than $1$.
$endgroup$
– John Douma
Mar 22 at 14:17
$begingroup$
Likewise, $x leq 10$ because $sin x geq -1$.
$endgroup$
– Ertxiem
Mar 22 at 15:01
add a comment |
$begingroup$
I'd use Python or WolframAlpha.
$endgroup$
– Botond
Mar 22 at 14:13
$begingroup$
Two suggestions : (1) You could use iteration if you are just after a numerical solution. (2) Approximate sin(x) with, say a Taylor approximation with a suitable number of terms.
$endgroup$
– Martin Hansen
Mar 22 at 14:14
$begingroup$
You also know that $x$ must be at least $8$ because $sinx$ cannot be greater than $1$.
$endgroup$
– John Douma
Mar 22 at 14:17
$begingroup$
Likewise, $x leq 10$ because $sin x geq -1$.
$endgroup$
– Ertxiem
Mar 22 at 15:01
$begingroup$
I'd use Python or WolframAlpha.
$endgroup$
– Botond
Mar 22 at 14:13
$begingroup$
I'd use Python or WolframAlpha.
$endgroup$
– Botond
Mar 22 at 14:13
$begingroup$
Two suggestions : (1) You could use iteration if you are just after a numerical solution. (2) Approximate sin(x) with, say a Taylor approximation with a suitable number of terms.
$endgroup$
– Martin Hansen
Mar 22 at 14:14
$begingroup$
Two suggestions : (1) You could use iteration if you are just after a numerical solution. (2) Approximate sin(x) with, say a Taylor approximation with a suitable number of terms.
$endgroup$
– Martin Hansen
Mar 22 at 14:14
$begingroup$
You also know that $x$ must be at least $8$ because $sinx$ cannot be greater than $1$.
$endgroup$
– John Douma
Mar 22 at 14:17
$begingroup$
You also know that $x$ must be at least $8$ because $sinx$ cannot be greater than $1$.
$endgroup$
– John Douma
Mar 22 at 14:17
$begingroup$
Likewise, $x leq 10$ because $sin x geq -1$.
$endgroup$
– Ertxiem
Mar 22 at 15:01
$begingroup$
Likewise, $x leq 10$ because $sin x geq -1$.
$endgroup$
– Ertxiem
Mar 22 at 15:01
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The function $sin x +x-9$ has the strictly non-negative derivative $cos x+1$, so the solution in $Bbb R$ is unique, confined to $[8,,10]$. The Newton-Raphson method $xmapsto x-fracsin x +x-9cos x +1$ iteratively improves the estimate $x_0=8$ to $8.0125448260685$ (according to my Python), if that's accurate enough.
answered Mar 22 at 14:44
J.G.J.G.
33.1k23251
$endgroup$
$begingroup$
Can you explain the Newton-Raphson method to me? Thanks!
$endgroup$
– Math Lover
Mar 26 at 15:45
1
$begingroup$
@MathLover This will explain it better than any comment of mine.
$endgroup$
– J.G.
Mar 26 at 16:07
add a comment |
$begingroup$
If you equate a numerical solution with graphing, generally you can't. You can use any of the one dimensional root finders, but I think of a graphing solution as equivalent. Either Alpha or the solver in a spreadsheet will make the numerics easy.
I am a fan of fixed point iteration. For your example, I would write it as $x=9-sin(x)$, start with $x_0=9$, plug $x_i$ into the right and compute $x_i+1$ on the left. It converges (for $x$ in radians) to about $8.012545$ in about $10$ iterations. The trick is to make the derivative of the right side less than $1$.
answered Mar 22 at 14:15
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Um, why does this work?
$endgroup$
– Math Lover
Mar 26 at 15:46
1
$begingroup$
See Wikipedia. If you can write your equation as $x=f(x)$ and at the fixed point $x_0$ imagine expanding $f$ as a Taylor series. To first order you have $f(x)=f(x_0)+(x-x_0)f'(x_0)$. If $|f'(x_0)|lt 1$ and you start close enough that the higher order terms don't hurt, you will get closer by a factor $f'(x_0)$ each time. There are other one dimensional root finding algorithms. I find this one easy to implement in a spreadsheet using copy down and you don't need to take derivatives like Newton-Raphson.
$endgroup$
– Ross Millikan
Mar 26 at 15:51
add a comment |
$begingroup$
Assuming you have a calculator in hand. Define $f(x)=sin x+x-9$. By Newton-Raphson method we have the following iteration formula: $$x_n+1=x_n-dfracsin x_n+x_n-9cos x_n +1$$
I would make the first guess to be $9$ because $9approx 3pi$ and sort of satisfies the equation with this bad approximation. Anyways the solution juggles around a bit before coming close to its actual solution that is approximately $8.013$.
$$beginarrayp3cmhline n & x_n \ hline 0 & 9 \ 1 & 4.36266794545 \ 2& 12.846245078\ 3& 10.7441080942\ 4 &9.71164729732\ 5 &−0.778750255328\6&5.34415560195 \ 7&8.14995865106\8&7.9996375311\9&8.01244847807\hline endarray$$
Motivation for choosing $x_0=8$. Another logic one may use to think of this first guess apart from that provided by @J.G. could be thinking in terms of bad approximations as I did in my previous approach, notice that $8approx 2pi +pi/2$ which about satisfies the equation in the following manner: $sin (2pi +pi/2)+2pi+pi/2-9approx 1+8-9=0$. Iterations in this case converge more quickly to the actual solution than in the first approach.
answered Mar 22 at 14:58
Paras KhoslaParas Khosla
2,883523
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$begingroup$
Since the root must be at least $8$, the first iteration here reveals $9$ is a bad starting point. That motivates trying either $8$ (which only needs three steps for high accuracy) or $8.5$.
$endgroup$
– J.G.
Mar 22 at 15:04
$begingroup$
Yes @J.G. that was what I realized after iterating upto $5$ steps with my first guess of $9$. Cheers :))
$endgroup$
– Paras Khosla
Mar 22 at 15:24
1
$begingroup$
This is a very good illustration of Darboux theorem : if you start with $x_0=9$, you have $f(9)times f''(9)=-sin^2(9) <0$ then overshoot of the solution while $f(8)times f''(8)=(1-sin (8)) sin (8)>0$ and no overshoot at all.
$endgroup$
– Claude Leibovici
Mar 24 at 9:47
$begingroup$
How do you iterate?
$endgroup$
– Math Lover
Mar 26 at 15:47
$begingroup$
Using the Iteration Formula: $x_n+1=x_n=f(x_n)/f'(x_n)$ with $f(x)$ as specified. Notice this is when we wish to solve $f(x)=0$. See Wikipedia-Newton's Method
$endgroup$
– Paras Khosla
Mar 27 at 9:01
add a comment |
$begingroup$
Consider that you look for the zero of function
$$f(x)=sin(x)+x-9$$
As said in comments and answers, by inspection, the solution is close to $frac 5 pi2$. So, let $x=t+frac 5 pi2$ to get
$$g(t)=t+cos (t)+frac5 pi 2-9$$ Now, using Taylor series around $t=0$ gives
$$g(t)=left(frac5 pi 2-8right)+t-fract^22+Oleft(t^4right)$$ Igoring the higher order terms, you need to solve a quadratic and the solution which is the closest to $0$ is given by
$$t=1-frac12 sqrt20 pi -60implies x=frac 5 pi2+1-frac12 sqrt20 pi -60approx 8.01258$$ which is quite close to the exact solution.
answered Mar 23 at 10:21
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The function $sin x +x-9$ has the strictly non-negative derivative $cos x+1$, so the solution in $Bbb R$ is unique, confined to $[8,,10]$. The Newton-Raphson method $xmapsto x-fracsin x +x-9cos x +1$ iteratively improves the estimate $x_0=8$ to $8.0125448260685$ (according to my Python), if that's accurate enough.
answered Mar 22 at 14:44
J.G.J.G.
33.1k23251
$endgroup$
$begingroup$
Can you explain the Newton-Raphson method to me? Thanks!
$endgroup$
– Math Lover
Mar 26 at 15:45
1
$begingroup$
@MathLover This will explain it better than any comment of mine.
$endgroup$
– J.G.
Mar 26 at 16:07
add a comment |
$begingroup$
The function $sin x +x-9$ has the strictly non-negative derivative $cos x+1$, so the solution in $Bbb R$ is unique, confined to $[8,,10]$. The Newton-Raphson method $xmapsto x-fracsin x +x-9cos x +1$ iteratively improves the estimate $x_0=8$ to $8.0125448260685$ (according to my Python), if that's accurate enough.
answered Mar 22 at 14:44
J.G.J.G.
33.1k23251
$endgroup$
$begingroup$
Can you explain the Newton-Raphson method to me? Thanks!
$endgroup$
– Math Lover
Mar 26 at 15:45
1
$begingroup$
@MathLover This will explain it better than any comment of mine.
$endgroup$
– J.G.
Mar 26 at 16:07
add a comment |
$begingroup$
The function $sin x +x-9$ has the strictly non-negative derivative $cos x+1$, so the solution in $Bbb R$ is unique, confined to $[8,,10]$. The Newton-Raphson method $xmapsto x-fracsin x +x-9cos x +1$ iteratively improves the estimate $x_0=8$ to $8.0125448260685$ (according to my Python), if that's accurate enough.
answered Mar 22 at 14:44
J.G.J.G.
33.1k23251
$endgroup$
The function $sin x +x-9$ has the strictly non-negative derivative $cos x+1$, so the solution in $Bbb R$ is unique, confined to $[8,,10]$. The Newton-Raphson method $xmapsto x-fracsin x +x-9cos x +1$ iteratively improves the estimate $x_0=8$ to $8.0125448260685$ (according to my Python), if that's accurate enough.
answered Mar 22 at 14:44
J.G.J.G.
33.1k23251
answered Mar 22 at 14:44
J.G.J.G.
33.1k23251
answered Mar 22 at 14:44
J.G.J.G.
33.1k23251
answered Mar 22 at 14:44
J.G.J.G.
33.1k23251
33.1k23251
$begingroup$
Can you explain the Newton-Raphson method to me? Thanks!
$endgroup$
– Math Lover
Mar 26 at 15:45
1
$begingroup$
@MathLover This will explain it better than any comment of mine.
$endgroup$
– J.G.
Mar 26 at 16:07
add a comment |
$begingroup$
Can you explain the Newton-Raphson method to me? Thanks!
$endgroup$
– Math Lover
Mar 26 at 15:45
1
$begingroup$
@MathLover This will explain it better than any comment of mine.
$endgroup$
– J.G.
Mar 26 at 16:07
$begingroup$
Can you explain the Newton-Raphson method to me? Thanks!
$endgroup$
– Math Lover
Mar 26 at 15:45
$begingroup$
Can you explain the Newton-Raphson method to me? Thanks!
$endgroup$
– Math Lover
Mar 26 at 15:45
1
1
$begingroup$
@MathLover This will explain it better than any comment of mine.
$endgroup$
– J.G.
Mar 26 at 16:07
$begingroup$
@MathLover This will explain it better than any comment of mine.
$endgroup$
– J.G.
Mar 26 at 16:07
add a comment |
$begingroup$
If you equate a numerical solution with graphing, generally you can't. You can use any of the one dimensional root finders, but I think of a graphing solution as equivalent. Either Alpha or the solver in a spreadsheet will make the numerics easy.
I am a fan of fixed point iteration. For your example, I would write it as $x=9-sin(x)$, start with $x_0=9$, plug $x_i$ into the right and compute $x_i+1$ on the left. It converges (for $x$ in radians) to about $8.012545$ in about $10$ iterations. The trick is to make the derivative of the right side less than $1$.
answered Mar 22 at 14:15
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Um, why does this work?
$endgroup$
– Math Lover
Mar 26 at 15:46
1
$begingroup$
See Wikipedia. If you can write your equation as $x=f(x)$ and at the fixed point $x_0$ imagine expanding $f$ as a Taylor series. To first order you have $f(x)=f(x_0)+(x-x_0)f'(x_0)$. If $|f'(x_0)|lt 1$ and you start close enough that the higher order terms don't hurt, you will get closer by a factor $f'(x_0)$ each time. There are other one dimensional root finding algorithms. I find this one easy to implement in a spreadsheet using copy down and you don't need to take derivatives like Newton-Raphson.
$endgroup$
– Ross Millikan
Mar 26 at 15:51
add a comment |
$begingroup$
If you equate a numerical solution with graphing, generally you can't. You can use any of the one dimensional root finders, but I think of a graphing solution as equivalent. Either Alpha or the solver in a spreadsheet will make the numerics easy.
I am a fan of fixed point iteration. For your example, I would write it as $x=9-sin(x)$, start with $x_0=9$, plug $x_i$ into the right and compute $x_i+1$ on the left. It converges (for $x$ in radians) to about $8.012545$ in about $10$ iterations. The trick is to make the derivative of the right side less than $1$.
answered Mar 22 at 14:15
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Um, why does this work?
$endgroup$
– Math Lover
Mar 26 at 15:46
1
$begingroup$
See Wikipedia. If you can write your equation as $x=f(x)$ and at the fixed point $x_0$ imagine expanding $f$ as a Taylor series. To first order you have $f(x)=f(x_0)+(x-x_0)f'(x_0)$. If $|f'(x_0)|lt 1$ and you start close enough that the higher order terms don't hurt, you will get closer by a factor $f'(x_0)$ each time. There are other one dimensional root finding algorithms. I find this one easy to implement in a spreadsheet using copy down and you don't need to take derivatives like Newton-Raphson.
$endgroup$
– Ross Millikan
Mar 26 at 15:51
add a comment |
$begingroup$
If you equate a numerical solution with graphing, generally you can't. You can use any of the one dimensional root finders, but I think of a graphing solution as equivalent. Either Alpha or the solver in a spreadsheet will make the numerics easy.
I am a fan of fixed point iteration. For your example, I would write it as $x=9-sin(x)$, start with $x_0=9$, plug $x_i$ into the right and compute $x_i+1$ on the left. It converges (for $x$ in radians) to about $8.012545$ in about $10$ iterations. The trick is to make the derivative of the right side less than $1$.
answered Mar 22 at 14:15
Ross MillikanRoss Millikan
301k24200375
$endgroup$
If you equate a numerical solution with graphing, generally you can't. You can use any of the one dimensional root finders, but I think of a graphing solution as equivalent. Either Alpha or the solver in a spreadsheet will make the numerics easy.
I am a fan of fixed point iteration. For your example, I would write it as $x=9-sin(x)$, start with $x_0=9$, plug $x_i$ into the right and compute $x_i+1$ on the left. It converges (for $x$ in radians) to about $8.012545$ in about $10$ iterations. The trick is to make the derivative of the right side less than $1$.
answered Mar 22 at 14:15
Ross MillikanRoss Millikan
301k24200375
answered Mar 22 at 14:15
Ross MillikanRoss Millikan
301k24200375
answered Mar 22 at 14:15
Ross MillikanRoss Millikan
301k24200375
answered Mar 22 at 14:15
Ross MillikanRoss Millikan
301k24200375
301k24200375
$begingroup$
Um, why does this work?
$endgroup$
– Math Lover
Mar 26 at 15:46
1
$begingroup$
See Wikipedia. If you can write your equation as $x=f(x)$ and at the fixed point $x_0$ imagine expanding $f$ as a Taylor series. To first order you have $f(x)=f(x_0)+(x-x_0)f'(x_0)$. If $|f'(x_0)|lt 1$ and you start close enough that the higher order terms don't hurt, you will get closer by a factor $f'(x_0)$ each time. There are other one dimensional root finding algorithms. I find this one easy to implement in a spreadsheet using copy down and you don't need to take derivatives like Newton-Raphson.
$endgroup$
– Ross Millikan
Mar 26 at 15:51
add a comment |
$begingroup$
Um, why does this work?
$endgroup$
– Math Lover
Mar 26 at 15:46
1
$begingroup$
See Wikipedia. If you can write your equation as $x=f(x)$ and at the fixed point $x_0$ imagine expanding $f$ as a Taylor series. To first order you have $f(x)=f(x_0)+(x-x_0)f'(x_0)$. If $|f'(x_0)|lt 1$ and you start close enough that the higher order terms don't hurt, you will get closer by a factor $f'(x_0)$ each time. There are other one dimensional root finding algorithms. I find this one easy to implement in a spreadsheet using copy down and you don't need to take derivatives like Newton-Raphson.
$endgroup$
– Ross Millikan
Mar 26 at 15:51
$begingroup$
Um, why does this work?
$endgroup$
– Math Lover
Mar 26 at 15:46
$begingroup$
Um, why does this work?
$endgroup$
– Math Lover
Mar 26 at 15:46
1
1
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See Wikipedia. If you can write your equation as $x=f(x)$ and at the fixed point $x_0$ imagine expanding $f$ as a Taylor series. To first order you have $f(x)=f(x_0)+(x-x_0)f'(x_0)$. If $|f'(x_0)|lt 1$ and you start close enough that the higher order terms don't hurt, you will get closer by a factor $f'(x_0)$ each time. There are other one dimensional root finding algorithms. I find this one easy to implement in a spreadsheet using copy down and you don't need to take derivatives like Newton-Raphson.
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– Ross Millikan
Mar 26 at 15:51
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See Wikipedia. If you can write your equation as $x=f(x)$ and at the fixed point $x_0$ imagine expanding $f$ as a Taylor series. To first order you have $f(x)=f(x_0)+(x-x_0)f'(x_0)$. If $|f'(x_0)|lt 1$ and you start close enough that the higher order terms don't hurt, you will get closer by a factor $f'(x_0)$ each time. There are other one dimensional root finding algorithms. I find this one easy to implement in a spreadsheet using copy down and you don't need to take derivatives like Newton-Raphson.
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– Ross Millikan
Mar 26 at 15:51
add a comment |
$begingroup$
Assuming you have a calculator in hand. Define $f(x)=sin x+x-9$. By Newton-Raphson method we have the following iteration formula: $$x_n+1=x_n-dfracsin x_n+x_n-9cos x_n +1$$
I would make the first guess to be $9$ because $9approx 3pi$ and sort of satisfies the equation with this bad approximation. Anyways the solution juggles around a bit before coming close to its actual solution that is approximately $8.013$.
$$beginarrayp3cmhline n & x_n \ hline 0 & 9 \ 1 & 4.36266794545 \ 2& 12.846245078\ 3& 10.7441080942\ 4 &9.71164729732\ 5 &−0.778750255328\6&5.34415560195 \ 7&8.14995865106\8&7.9996375311\9&8.01244847807\hline endarray$$
Motivation for choosing $x_0=8$. Another logic one may use to think of this first guess apart from that provided by @J.G. could be thinking in terms of bad approximations as I did in my previous approach, notice that $8approx 2pi +pi/2$ which about satisfies the equation in the following manner: $sin (2pi +pi/2)+2pi+pi/2-9approx 1+8-9=0$. Iterations in this case converge more quickly to the actual solution than in the first approach.
answered Mar 22 at 14:58
Paras KhoslaParas Khosla
2,883523
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$begingroup$
Since the root must be at least $8$, the first iteration here reveals $9$ is a bad starting point. That motivates trying either $8$ (which only needs three steps for high accuracy) or $8.5$.
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– J.G.
Mar 22 at 15:04
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Yes @J.G. that was what I realized after iterating upto $5$ steps with my first guess of $9$. Cheers :))
$endgroup$
– Paras Khosla
Mar 22 at 15:24
1
$begingroup$
This is a very good illustration of Darboux theorem : if you start with $x_0=9$, you have $f(9)times f''(9)=-sin^2(9) <0$ then overshoot of the solution while $f(8)times f''(8)=(1-sin (8)) sin (8)>0$ and no overshoot at all.
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– Claude Leibovici
Mar 24 at 9:47
$begingroup$
How do you iterate?
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– Math Lover
Mar 26 at 15:47
$begingroup$
Using the Iteration Formula: $x_n+1=x_n=f(x_n)/f'(x_n)$ with $f(x)$ as specified. Notice this is when we wish to solve $f(x)=0$. See Wikipedia-Newton's Method
$endgroup$
– Paras Khosla
Mar 27 at 9:01
add a comment |
$begingroup$
Assuming you have a calculator in hand. Define $f(x)=sin x+x-9$. By Newton-Raphson method we have the following iteration formula: $$x_n+1=x_n-dfracsin x_n+x_n-9cos x_n +1$$
I would make the first guess to be $9$ because $9approx 3pi$ and sort of satisfies the equation with this bad approximation. Anyways the solution juggles around a bit before coming close to its actual solution that is approximately $8.013$.
$$beginarrayp3cmhline n & x_n \ hline 0 & 9 \ 1 & 4.36266794545 \ 2& 12.846245078\ 3& 10.7441080942\ 4 &9.71164729732\ 5 &−0.778750255328\6&5.34415560195 \ 7&8.14995865106\8&7.9996375311\9&8.01244847807\hline endarray$$
Motivation for choosing $x_0=8$. Another logic one may use to think of this first guess apart from that provided by @J.G. could be thinking in terms of bad approximations as I did in my previous approach, notice that $8approx 2pi +pi/2$ which about satisfies the equation in the following manner: $sin (2pi +pi/2)+2pi+pi/2-9approx 1+8-9=0$. Iterations in this case converge more quickly to the actual solution than in the first approach.
answered Mar 22 at 14:58
Paras KhoslaParas Khosla
2,883523
$endgroup$
$begingroup$
Since the root must be at least $8$, the first iteration here reveals $9$ is a bad starting point. That motivates trying either $8$ (which only needs three steps for high accuracy) or $8.5$.
$endgroup$
– J.G.
Mar 22 at 15:04
$begingroup$
Yes @J.G. that was what I realized after iterating upto $5$ steps with my first guess of $9$. Cheers :))
$endgroup$
– Paras Khosla
Mar 22 at 15:24
1
$begingroup$
This is a very good illustration of Darboux theorem : if you start with $x_0=9$, you have $f(9)times f''(9)=-sin^2(9) <0$ then overshoot of the solution while $f(8)times f''(8)=(1-sin (8)) sin (8)>0$ and no overshoot at all.
$endgroup$
– Claude Leibovici
Mar 24 at 9:47
$begingroup$
How do you iterate?
$endgroup$
– Math Lover
Mar 26 at 15:47
$begingroup$
Using the Iteration Formula: $x_n+1=x_n=f(x_n)/f'(x_n)$ with $f(x)$ as specified. Notice this is when we wish to solve $f(x)=0$. See Wikipedia-Newton's Method
$endgroup$
– Paras Khosla
Mar 27 at 9:01
add a comment |
$begingroup$
Assuming you have a calculator in hand. Define $f(x)=sin x+x-9$. By Newton-Raphson method we have the following iteration formula: $$x_n+1=x_n-dfracsin x_n+x_n-9cos x_n +1$$
I would make the first guess to be $9$ because $9approx 3pi$ and sort of satisfies the equation with this bad approximation. Anyways the solution juggles around a bit before coming close to its actual solution that is approximately $8.013$.
$$beginarrayp3cmhline n & x_n \ hline 0 & 9 \ 1 & 4.36266794545 \ 2& 12.846245078\ 3& 10.7441080942\ 4 &9.71164729732\ 5 &−0.778750255328\6&5.34415560195 \ 7&8.14995865106\8&7.9996375311\9&8.01244847807\hline endarray$$
Motivation for choosing $x_0=8$. Another logic one may use to think of this first guess apart from that provided by @J.G. could be thinking in terms of bad approximations as I did in my previous approach, notice that $8approx 2pi +pi/2$ which about satisfies the equation in the following manner: $sin (2pi +pi/2)+2pi+pi/2-9approx 1+8-9=0$. Iterations in this case converge more quickly to the actual solution than in the first approach.
answered Mar 22 at 14:58
Paras KhoslaParas Khosla
2,883523
$endgroup$
Assuming you have a calculator in hand. Define $f(x)=sin x+x-9$. By Newton-Raphson method we have the following iteration formula: $$x_n+1=x_n-dfracsin x_n+x_n-9cos x_n +1$$
I would make the first guess to be $9$ because $9approx 3pi$ and sort of satisfies the equation with this bad approximation. Anyways the solution juggles around a bit before coming close to its actual solution that is approximately $8.013$.
$$beginarrayp3cmhline n & x_n \ hline 0 & 9 \ 1 & 4.36266794545 \ 2& 12.846245078\ 3& 10.7441080942\ 4 &9.71164729732\ 5 &−0.778750255328\6&5.34415560195 \ 7&8.14995865106\8&7.9996375311\9&8.01244847807\hline endarray$$
Motivation for choosing $x_0=8$. Another logic one may use to think of this first guess apart from that provided by @J.G. could be thinking in terms of bad approximations as I did in my previous approach, notice that $8approx 2pi +pi/2$ which about satisfies the equation in the following manner: $sin (2pi +pi/2)+2pi+pi/2-9approx 1+8-9=0$. Iterations in this case converge more quickly to the actual solution than in the first approach.
answered Mar 22 at 14:58
Paras KhoslaParas Khosla
2,883523
edited Mar 22 at 15:19
answered Mar 22 at 14:58
Paras KhoslaParas Khosla
2,883523
answered Mar 22 at 14:58
Paras KhoslaParas Khosla
2,883523
answered Mar 22 at 14:58
Paras KhoslaParas Khosla
2,883523
2,883523
$begingroup$
Since the root must be at least $8$, the first iteration here reveals $9$ is a bad starting point. That motivates trying either $8$ (which only needs three steps for high accuracy) or $8.5$.
$endgroup$
– J.G.
Mar 22 at 15:04
$begingroup$
Yes @J.G. that was what I realized after iterating upto $5$ steps with my first guess of $9$. Cheers :))
$endgroup$
– Paras Khosla
Mar 22 at 15:24
1
$begingroup$
This is a very good illustration of Darboux theorem : if you start with $x_0=9$, you have $f(9)times f''(9)=-sin^2(9) <0$ then overshoot of the solution while $f(8)times f''(8)=(1-sin (8)) sin (8)>0$ and no overshoot at all.
$endgroup$
– Claude Leibovici
Mar 24 at 9:47
$begingroup$
How do you iterate?
$endgroup$
– Math Lover
Mar 26 at 15:47
$begingroup$
Using the Iteration Formula: $x_n+1=x_n=f(x_n)/f'(x_n)$ with $f(x)$ as specified. Notice this is when we wish to solve $f(x)=0$. See Wikipedia-Newton's Method
$endgroup$
– Paras Khosla
Mar 27 at 9:01
add a comment |
$begingroup$
Since the root must be at least $8$, the first iteration here reveals $9$ is a bad starting point. That motivates trying either $8$ (which only needs three steps for high accuracy) or $8.5$.
$endgroup$
– J.G.
Mar 22 at 15:04
$begingroup$
Yes @J.G. that was what I realized after iterating upto $5$ steps with my first guess of $9$. Cheers :))
$endgroup$
– Paras Khosla
Mar 22 at 15:24
1
$begingroup$
This is a very good illustration of Darboux theorem : if you start with $x_0=9$, you have $f(9)times f''(9)=-sin^2(9) <0$ then overshoot of the solution while $f(8)times f''(8)=(1-sin (8)) sin (8)>0$ and no overshoot at all.
$endgroup$
– Claude Leibovici
Mar 24 at 9:47
$begingroup$
How do you iterate?
$endgroup$
– Math Lover
Mar 26 at 15:47
$begingroup$
Using the Iteration Formula: $x_n+1=x_n=f(x_n)/f'(x_n)$ with $f(x)$ as specified. Notice this is when we wish to solve $f(x)=0$. See Wikipedia-Newton's Method
$endgroup$
– Paras Khosla
Mar 27 at 9:01
$begingroup$
Since the root must be at least $8$, the first iteration here reveals $9$ is a bad starting point. That motivates trying either $8$ (which only needs three steps for high accuracy) or $8.5$.
$endgroup$
– J.G.
Mar 22 at 15:04
$begingroup$
Since the root must be at least $8$, the first iteration here reveals $9$ is a bad starting point. That motivates trying either $8$ (which only needs three steps for high accuracy) or $8.5$.
$endgroup$
– J.G.
Mar 22 at 15:04
$begingroup$
Yes @J.G. that was what I realized after iterating upto $5$ steps with my first guess of $9$. Cheers :))
$endgroup$
– Paras Khosla
Mar 22 at 15:24
$begingroup$
Yes @J.G. that was what I realized after iterating upto $5$ steps with my first guess of $9$. Cheers :))
$endgroup$
– Paras Khosla
Mar 22 at 15:24
1
1
$begingroup$
This is a very good illustration of Darboux theorem : if you start with $x_0=9$, you have $f(9)times f''(9)=-sin^2(9) <0$ then overshoot of the solution while $f(8)times f''(8)=(1-sin (8)) sin (8)>0$ and no overshoot at all.
$endgroup$
– Claude Leibovici
Mar 24 at 9:47
$begingroup$
This is a very good illustration of Darboux theorem : if you start with $x_0=9$, you have $f(9)times f''(9)=-sin^2(9) <0$ then overshoot of the solution while $f(8)times f''(8)=(1-sin (8)) sin (8)>0$ and no overshoot at all.
$endgroup$
– Claude Leibovici
Mar 24 at 9:47
$begingroup$
How do you iterate?
$endgroup$
– Math Lover
Mar 26 at 15:47
$begingroup$
How do you iterate?
$endgroup$
– Math Lover
Mar 26 at 15:47
$begingroup$
Using the Iteration Formula: $x_n+1=x_n=f(x_n)/f'(x_n)$ with $f(x)$ as specified. Notice this is when we wish to solve $f(x)=0$. See Wikipedia-Newton's Method
$endgroup$
– Paras Khosla
Mar 27 at 9:01
$begingroup$
Using the Iteration Formula: $x_n+1=x_n=f(x_n)/f'(x_n)$ with $f(x)$ as specified. Notice this is when we wish to solve $f(x)=0$. See Wikipedia-Newton's Method
$endgroup$
– Paras Khosla
Mar 27 at 9:01
add a comment |
$begingroup$
Consider that you look for the zero of function
$$f(x)=sin(x)+x-9$$
As said in comments and answers, by inspection, the solution is close to $frac 5 pi2$. So, let $x=t+frac 5 pi2$ to get
$$g(t)=t+cos (t)+frac5 pi 2-9$$ Now, using Taylor series around $t=0$ gives
$$g(t)=left(frac5 pi 2-8right)+t-fract^22+Oleft(t^4right)$$ Igoring the higher order terms, you need to solve a quadratic and the solution which is the closest to $0$ is given by
$$t=1-frac12 sqrt20 pi -60implies x=frac 5 pi2+1-frac12 sqrt20 pi -60approx 8.01258$$ which is quite close to the exact solution.
answered Mar 23 at 10:21
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Consider that you look for the zero of function
$$f(x)=sin(x)+x-9$$
As said in comments and answers, by inspection, the solution is close to $frac 5 pi2$. So, let $x=t+frac 5 pi2$ to get
$$g(t)=t+cos (t)+frac5 pi 2-9$$ Now, using Taylor series around $t=0$ gives
$$g(t)=left(frac5 pi 2-8right)+t-fract^22+Oleft(t^4right)$$ Igoring the higher order terms, you need to solve a quadratic and the solution which is the closest to $0$ is given by
$$t=1-frac12 sqrt20 pi -60implies x=frac 5 pi2+1-frac12 sqrt20 pi -60approx 8.01258$$ which is quite close to the exact solution.
answered Mar 23 at 10:21
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Consider that you look for the zero of function
$$f(x)=sin(x)+x-9$$
As said in comments and answers, by inspection, the solution is close to $frac 5 pi2$. So, let $x=t+frac 5 pi2$ to get
$$g(t)=t+cos (t)+frac5 pi 2-9$$ Now, using Taylor series around $t=0$ gives
$$g(t)=left(frac5 pi 2-8right)+t-fract^22+Oleft(t^4right)$$ Igoring the higher order terms, you need to solve a quadratic and the solution which is the closest to $0$ is given by
$$t=1-frac12 sqrt20 pi -60implies x=frac 5 pi2+1-frac12 sqrt20 pi -60approx 8.01258$$ which is quite close to the exact solution.
answered Mar 23 at 10:21
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
Consider that you look for the zero of function
$$f(x)=sin(x)+x-9$$
As said in comments and answers, by inspection, the solution is close to $frac 5 pi2$. So, let $x=t+frac 5 pi2$ to get
$$g(t)=t+cos (t)+frac5 pi 2-9$$ Now, using Taylor series around $t=0$ gives
$$g(t)=left(frac5 pi 2-8right)+t-fract^22+Oleft(t^4right)$$ Igoring the higher order terms, you need to solve a quadratic and the solution which is the closest to $0$ is given by
$$t=1-frac12 sqrt20 pi -60implies x=frac 5 pi2+1-frac12 sqrt20 pi -60approx 8.01258$$ which is quite close to the exact solution.
answered Mar 23 at 10:21
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 23 at 10:21
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 23 at 10:21
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 23 at 10:21
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
add a comment |
add a comment |
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$begingroup$
I'd use Python or WolframAlpha.
$endgroup$
– Botond
Mar 22 at 14:13
$begingroup$
Two suggestions : (1) You could use iteration if you are just after a numerical solution. (2) Approximate sin(x) with, say a Taylor approximation with a suitable number of terms.
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– Martin Hansen
Mar 22 at 14:14
$begingroup$
You also know that $x$ must be at least $8$ because $sinx$ cannot be greater than $1$.
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– John Douma
Mar 22 at 14:17
$begingroup$
Likewise, $x leq 10$ because $sin x geq -1$.
$endgroup$
– Ertxiem
Mar 22 at 15:01