Can an equation of the form $sin(omega t) + at^2+bt=N$, with $aneq 0$, be solved analytically?












1












$begingroup$



Can the equation of the form
$$sin(omega t) + at^2+bt=N$$
where $a,b,N in mathbb{R}$ be solved analytically for $a ne 0$?




It, for example, arises when the rotating object falls down subject to gravity. Solving such equation can allow to compute the time of the object touchdown.










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$endgroup$












  • $begingroup$
    Not that I know of. Are you sure you got your equation right? The $sin omega t$ term implies an oscillation due to elastic force, not rotation.
    $endgroup$
    – Andrei
    Dec 17 '18 at 5:53










  • $begingroup$
    $ sin omega t$ is a harmonic oscillation due to the object rotation. It will be moving according to the quadratic law, but will also be spinning.
    $endgroup$
    – Jennifer M.
    Dec 17 '18 at 5:54












  • $begingroup$
    I am just curious how you got that equation.
    $endgroup$
    – Andrei
    Dec 17 '18 at 5:57










  • $begingroup$
    Imagine a rotating object in a freefall. Its Z coordinate is decreasing as $-frac{1}{2} g t^2$ and its rotation would add $sin omega t$ for each surface point.
    $endgroup$
    – Jennifer M.
    Dec 17 '18 at 6:04










  • $begingroup$
    If the domain of $sin omega t$ be $[0,π]$ then it is possible to get approximate solutions.
    $endgroup$
    – Awe Kumar Jha
    Dec 17 '18 at 6:05
















1












$begingroup$



Can the equation of the form
$$sin(omega t) + at^2+bt=N$$
where $a,b,N in mathbb{R}$ be solved analytically for $a ne 0$?




It, for example, arises when the rotating object falls down subject to gravity. Solving such equation can allow to compute the time of the object touchdown.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Not that I know of. Are you sure you got your equation right? The $sin omega t$ term implies an oscillation due to elastic force, not rotation.
    $endgroup$
    – Andrei
    Dec 17 '18 at 5:53










  • $begingroup$
    $ sin omega t$ is a harmonic oscillation due to the object rotation. It will be moving according to the quadratic law, but will also be spinning.
    $endgroup$
    – Jennifer M.
    Dec 17 '18 at 5:54












  • $begingroup$
    I am just curious how you got that equation.
    $endgroup$
    – Andrei
    Dec 17 '18 at 5:57










  • $begingroup$
    Imagine a rotating object in a freefall. Its Z coordinate is decreasing as $-frac{1}{2} g t^2$ and its rotation would add $sin omega t$ for each surface point.
    $endgroup$
    – Jennifer M.
    Dec 17 '18 at 6:04










  • $begingroup$
    If the domain of $sin omega t$ be $[0,π]$ then it is possible to get approximate solutions.
    $endgroup$
    – Awe Kumar Jha
    Dec 17 '18 at 6:05














1












1








1





$begingroup$



Can the equation of the form
$$sin(omega t) + at^2+bt=N$$
where $a,b,N in mathbb{R}$ be solved analytically for $a ne 0$?




It, for example, arises when the rotating object falls down subject to gravity. Solving such equation can allow to compute the time of the object touchdown.










share|cite|improve this question











$endgroup$





Can the equation of the form
$$sin(omega t) + at^2+bt=N$$
where $a,b,N in mathbb{R}$ be solved analytically for $a ne 0$?




It, for example, arises when the rotating object falls down subject to gravity. Solving such equation can allow to compute the time of the object touchdown.







trigonometry






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share|cite|improve this question













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share|cite|improve this question








edited Dec 18 '18 at 4:15









Blue

48.6k870154




48.6k870154










asked Dec 17 '18 at 5:47









Jennifer M.Jennifer M.

1112




1112












  • $begingroup$
    Not that I know of. Are you sure you got your equation right? The $sin omega t$ term implies an oscillation due to elastic force, not rotation.
    $endgroup$
    – Andrei
    Dec 17 '18 at 5:53










  • $begingroup$
    $ sin omega t$ is a harmonic oscillation due to the object rotation. It will be moving according to the quadratic law, but will also be spinning.
    $endgroup$
    – Jennifer M.
    Dec 17 '18 at 5:54












  • $begingroup$
    I am just curious how you got that equation.
    $endgroup$
    – Andrei
    Dec 17 '18 at 5:57










  • $begingroup$
    Imagine a rotating object in a freefall. Its Z coordinate is decreasing as $-frac{1}{2} g t^2$ and its rotation would add $sin omega t$ for each surface point.
    $endgroup$
    – Jennifer M.
    Dec 17 '18 at 6:04










  • $begingroup$
    If the domain of $sin omega t$ be $[0,π]$ then it is possible to get approximate solutions.
    $endgroup$
    – Awe Kumar Jha
    Dec 17 '18 at 6:05


















  • $begingroup$
    Not that I know of. Are you sure you got your equation right? The $sin omega t$ term implies an oscillation due to elastic force, not rotation.
    $endgroup$
    – Andrei
    Dec 17 '18 at 5:53










  • $begingroup$
    $ sin omega t$ is a harmonic oscillation due to the object rotation. It will be moving according to the quadratic law, but will also be spinning.
    $endgroup$
    – Jennifer M.
    Dec 17 '18 at 5:54












  • $begingroup$
    I am just curious how you got that equation.
    $endgroup$
    – Andrei
    Dec 17 '18 at 5:57










  • $begingroup$
    Imagine a rotating object in a freefall. Its Z coordinate is decreasing as $-frac{1}{2} g t^2$ and its rotation would add $sin omega t$ for each surface point.
    $endgroup$
    – Jennifer M.
    Dec 17 '18 at 6:04










  • $begingroup$
    If the domain of $sin omega t$ be $[0,π]$ then it is possible to get approximate solutions.
    $endgroup$
    – Awe Kumar Jha
    Dec 17 '18 at 6:05
















$begingroup$
Not that I know of. Are you sure you got your equation right? The $sin omega t$ term implies an oscillation due to elastic force, not rotation.
$endgroup$
– Andrei
Dec 17 '18 at 5:53




$begingroup$
Not that I know of. Are you sure you got your equation right? The $sin omega t$ term implies an oscillation due to elastic force, not rotation.
$endgroup$
– Andrei
Dec 17 '18 at 5:53












$begingroup$
$ sin omega t$ is a harmonic oscillation due to the object rotation. It will be moving according to the quadratic law, but will also be spinning.
$endgroup$
– Jennifer M.
Dec 17 '18 at 5:54






$begingroup$
$ sin omega t$ is a harmonic oscillation due to the object rotation. It will be moving according to the quadratic law, but will also be spinning.
$endgroup$
– Jennifer M.
Dec 17 '18 at 5:54














$begingroup$
I am just curious how you got that equation.
$endgroup$
– Andrei
Dec 17 '18 at 5:57




$begingroup$
I am just curious how you got that equation.
$endgroup$
– Andrei
Dec 17 '18 at 5:57












$begingroup$
Imagine a rotating object in a freefall. Its Z coordinate is decreasing as $-frac{1}{2} g t^2$ and its rotation would add $sin omega t$ for each surface point.
$endgroup$
– Jennifer M.
Dec 17 '18 at 6:04




$begingroup$
Imagine a rotating object in a freefall. Its Z coordinate is decreasing as $-frac{1}{2} g t^2$ and its rotation would add $sin omega t$ for each surface point.
$endgroup$
– Jennifer M.
Dec 17 '18 at 6:04












$begingroup$
If the domain of $sin omega t$ be $[0,π]$ then it is possible to get approximate solutions.
$endgroup$
– Awe Kumar Jha
Dec 17 '18 at 6:05




$begingroup$
If the domain of $sin omega t$ be $[0,π]$ then it is possible to get approximate solutions.
$endgroup$
– Awe Kumar Jha
Dec 17 '18 at 6:05










2 Answers
2






active

oldest

votes


















2












$begingroup$

Your equation is a transcendental equation. It can not be exactly solved but you can find the roots using numerical methods like Newton's method. However if you are still interested in having a closed form solution, you may use Bhaskara's approximation for domain $[-frac {π}{2},frac {π}{2}]$. For example, consider another related equation,
$$cos omega t + t^2 + bt - N = 0$$
From Bhaskara's approximation formula,
$$cos omega t ≈ frac {π^2 - 4omega^2 t^2}{π^2 + omega^2 t^2}$$
The equation is reduced to a general quartic,
$$t^4+bt^3+(T^2 - (4+N))t^2+T^2bt+T^2(1-N)=0$$
Where $T=frac {π}{omega}$This can be solved using Ferrari's method (https://proofwiki.org/wiki/Ferrari%27s_Method). Although the computational cost is high and the solution may turn really ugly, yet it will give you much required insight into the dependence of the time of flight of the object on various parameters.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I totally agree with your approach but $$cos(x) simeqfrac{pi ^2-4x^2}{pi ^2+x^2}qquad (-frac pi 2 leq xleqfrac pi 2)$$
    $endgroup$
    – Claude Leibovici
    Dec 17 '18 at 9:02










  • $begingroup$
    Now I have corrected it. Well , thanks for pointing my error.
    $endgroup$
    – Awe Kumar Jha
    Dec 17 '18 at 10:53










  • $begingroup$
    You are welcome ! Interesting problem from a numerical point of veiw. Cheers.
    $endgroup$
    – Claude Leibovici
    Dec 17 '18 at 10:57



















1












$begingroup$

In the same spirit as Awe Kumar Jha, considering $$sin (omega t) + at^2+bt=N$$ let
$$x=omega tqquad alpha=frac a{omega^2}qquadbeta=frac b{omega}$$ to make the equation
$$sin(x)+alpha x^2+beta x=N$$ and use Bhaskara I's sine approximation formula
$$sin(x) simeq frac{16 (pi -x) x}{5 pi ^2-4 (pi -x) x}qquad (0leq xleqpi)$$
$$4 alpha x^4+ 4( beta - pi alpha )x^3+ left(5 pi ^2 alpha -4 pi beta -4 N-16right)x^2+pi (5 pi beta +4
N+16)x-5 pi ^2 N=0tag 1$$
which, as said,can be solved using Ferrari's method (this would be quite tedious).



You can use another approximation (from far, not as good asBhaskara I's one ) minimizing with respect to $k$
$$int_0^pi left[sin(x)-k (pi-x)xright]^2,dx=frac{pi ^5 }{30}k^2-8 k+frac{pi }{2}$$ to get $color{blue}{k=frac{120}{pi ^5}}$ and then $sin(x)simeq k(pi-x)x$. This leads to
$$ (alpha -k)x^2+ (beta +pi k)x-N=0tag 2$$ which is simpler for sure. Using its solution as $x_0$ you can start Newton iterations to polish the root accorging to
$$x_{n+1}=x_n-frac{sin (x_n)+alpha x_n^2+beta x_n-N}{ cos (x_n)+2 alpha x_n+beta}$$



Let us try with $alpha=0.4$, $beta=0.6$, $N=2.345$. Solving $(1)$ gives $x^{(1)}_0 =1.26126$ while solving $(2)$ gives $x^{(2)}_0=1.27312$. Now, the iterates of Newton method starting with $x_0=x^{(2)}_0$
$$left(
begin{array}{cc}
n & x_n \
0 & 1.273117520 \
1 & 1.260970996 \
2 & 1.260976964
end{array}
right)$$
which is the solution for ten significant figures.



Edit



To stay with a quadratic equation, we could be even more empirical (not respecting the values at end points) and write
$$sin(x) simeq a_0+a_1x+a_2x^2$$ The same minimization method would lead to
$$a_0=frac{12 left(pi ^2-10right)}{pi ^3}qquad a_1=-frac{60 left(pi ^2-12right)}{pi ^4}qquad a_2=frac{60 left(pi ^2-12right)}{pi ^5}$$ which would lead to $x^{(3)}_0=1.26757$.






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    2 Answers
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    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Your equation is a transcendental equation. It can not be exactly solved but you can find the roots using numerical methods like Newton's method. However if you are still interested in having a closed form solution, you may use Bhaskara's approximation for domain $[-frac {π}{2},frac {π}{2}]$. For example, consider another related equation,
    $$cos omega t + t^2 + bt - N = 0$$
    From Bhaskara's approximation formula,
    $$cos omega t ≈ frac {π^2 - 4omega^2 t^2}{π^2 + omega^2 t^2}$$
    The equation is reduced to a general quartic,
    $$t^4+bt^3+(T^2 - (4+N))t^2+T^2bt+T^2(1-N)=0$$
    Where $T=frac {π}{omega}$This can be solved using Ferrari's method (https://proofwiki.org/wiki/Ferrari%27s_Method). Although the computational cost is high and the solution may turn really ugly, yet it will give you much required insight into the dependence of the time of flight of the object on various parameters.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      I totally agree with your approach but $$cos(x) simeqfrac{pi ^2-4x^2}{pi ^2+x^2}qquad (-frac pi 2 leq xleqfrac pi 2)$$
      $endgroup$
      – Claude Leibovici
      Dec 17 '18 at 9:02










    • $begingroup$
      Now I have corrected it. Well , thanks for pointing my error.
      $endgroup$
      – Awe Kumar Jha
      Dec 17 '18 at 10:53










    • $begingroup$
      You are welcome ! Interesting problem from a numerical point of veiw. Cheers.
      $endgroup$
      – Claude Leibovici
      Dec 17 '18 at 10:57
















    2












    $begingroup$

    Your equation is a transcendental equation. It can not be exactly solved but you can find the roots using numerical methods like Newton's method. However if you are still interested in having a closed form solution, you may use Bhaskara's approximation for domain $[-frac {π}{2},frac {π}{2}]$. For example, consider another related equation,
    $$cos omega t + t^2 + bt - N = 0$$
    From Bhaskara's approximation formula,
    $$cos omega t ≈ frac {π^2 - 4omega^2 t^2}{π^2 + omega^2 t^2}$$
    The equation is reduced to a general quartic,
    $$t^4+bt^3+(T^2 - (4+N))t^2+T^2bt+T^2(1-N)=0$$
    Where $T=frac {π}{omega}$This can be solved using Ferrari's method (https://proofwiki.org/wiki/Ferrari%27s_Method). Although the computational cost is high and the solution may turn really ugly, yet it will give you much required insight into the dependence of the time of flight of the object on various parameters.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      I totally agree with your approach but $$cos(x) simeqfrac{pi ^2-4x^2}{pi ^2+x^2}qquad (-frac pi 2 leq xleqfrac pi 2)$$
      $endgroup$
      – Claude Leibovici
      Dec 17 '18 at 9:02










    • $begingroup$
      Now I have corrected it. Well , thanks for pointing my error.
      $endgroup$
      – Awe Kumar Jha
      Dec 17 '18 at 10:53










    • $begingroup$
      You are welcome ! Interesting problem from a numerical point of veiw. Cheers.
      $endgroup$
      – Claude Leibovici
      Dec 17 '18 at 10:57














    2












    2








    2





    $begingroup$

    Your equation is a transcendental equation. It can not be exactly solved but you can find the roots using numerical methods like Newton's method. However if you are still interested in having a closed form solution, you may use Bhaskara's approximation for domain $[-frac {π}{2},frac {π}{2}]$. For example, consider another related equation,
    $$cos omega t + t^2 + bt - N = 0$$
    From Bhaskara's approximation formula,
    $$cos omega t ≈ frac {π^2 - 4omega^2 t^2}{π^2 + omega^2 t^2}$$
    The equation is reduced to a general quartic,
    $$t^4+bt^3+(T^2 - (4+N))t^2+T^2bt+T^2(1-N)=0$$
    Where $T=frac {π}{omega}$This can be solved using Ferrari's method (https://proofwiki.org/wiki/Ferrari%27s_Method). Although the computational cost is high and the solution may turn really ugly, yet it will give you much required insight into the dependence of the time of flight of the object on various parameters.






    share|cite|improve this answer











    $endgroup$



    Your equation is a transcendental equation. It can not be exactly solved but you can find the roots using numerical methods like Newton's method. However if you are still interested in having a closed form solution, you may use Bhaskara's approximation for domain $[-frac {π}{2},frac {π}{2}]$. For example, consider another related equation,
    $$cos omega t + t^2 + bt - N = 0$$
    From Bhaskara's approximation formula,
    $$cos omega t ≈ frac {π^2 - 4omega^2 t^2}{π^2 + omega^2 t^2}$$
    The equation is reduced to a general quartic,
    $$t^4+bt^3+(T^2 - (4+N))t^2+T^2bt+T^2(1-N)=0$$
    Where $T=frac {π}{omega}$This can be solved using Ferrari's method (https://proofwiki.org/wiki/Ferrari%27s_Method). Although the computational cost is high and the solution may turn really ugly, yet it will give you much required insight into the dependence of the time of flight of the object on various parameters.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 17 '18 at 15:56

























    answered Dec 17 '18 at 6:38









    Awe Kumar JhaAwe Kumar Jha

    43813




    43813








    • 1




      $begingroup$
      I totally agree with your approach but $$cos(x) simeqfrac{pi ^2-4x^2}{pi ^2+x^2}qquad (-frac pi 2 leq xleqfrac pi 2)$$
      $endgroup$
      – Claude Leibovici
      Dec 17 '18 at 9:02










    • $begingroup$
      Now I have corrected it. Well , thanks for pointing my error.
      $endgroup$
      – Awe Kumar Jha
      Dec 17 '18 at 10:53










    • $begingroup$
      You are welcome ! Interesting problem from a numerical point of veiw. Cheers.
      $endgroup$
      – Claude Leibovici
      Dec 17 '18 at 10:57














    • 1




      $begingroup$
      I totally agree with your approach but $$cos(x) simeqfrac{pi ^2-4x^2}{pi ^2+x^2}qquad (-frac pi 2 leq xleqfrac pi 2)$$
      $endgroup$
      – Claude Leibovici
      Dec 17 '18 at 9:02










    • $begingroup$
      Now I have corrected it. Well , thanks for pointing my error.
      $endgroup$
      – Awe Kumar Jha
      Dec 17 '18 at 10:53










    • $begingroup$
      You are welcome ! Interesting problem from a numerical point of veiw. Cheers.
      $endgroup$
      – Claude Leibovici
      Dec 17 '18 at 10:57








    1




    1




    $begingroup$
    I totally agree with your approach but $$cos(x) simeqfrac{pi ^2-4x^2}{pi ^2+x^2}qquad (-frac pi 2 leq xleqfrac pi 2)$$
    $endgroup$
    – Claude Leibovici
    Dec 17 '18 at 9:02




    $begingroup$
    I totally agree with your approach but $$cos(x) simeqfrac{pi ^2-4x^2}{pi ^2+x^2}qquad (-frac pi 2 leq xleqfrac pi 2)$$
    $endgroup$
    – Claude Leibovici
    Dec 17 '18 at 9:02












    $begingroup$
    Now I have corrected it. Well , thanks for pointing my error.
    $endgroup$
    – Awe Kumar Jha
    Dec 17 '18 at 10:53




    $begingroup$
    Now I have corrected it. Well , thanks for pointing my error.
    $endgroup$
    – Awe Kumar Jha
    Dec 17 '18 at 10:53












    $begingroup$
    You are welcome ! Interesting problem from a numerical point of veiw. Cheers.
    $endgroup$
    – Claude Leibovici
    Dec 17 '18 at 10:57




    $begingroup$
    You are welcome ! Interesting problem from a numerical point of veiw. Cheers.
    $endgroup$
    – Claude Leibovici
    Dec 17 '18 at 10:57











    1












    $begingroup$

    In the same spirit as Awe Kumar Jha, considering $$sin (omega t) + at^2+bt=N$$ let
    $$x=omega tqquad alpha=frac a{omega^2}qquadbeta=frac b{omega}$$ to make the equation
    $$sin(x)+alpha x^2+beta x=N$$ and use Bhaskara I's sine approximation formula
    $$sin(x) simeq frac{16 (pi -x) x}{5 pi ^2-4 (pi -x) x}qquad (0leq xleqpi)$$
    $$4 alpha x^4+ 4( beta - pi alpha )x^3+ left(5 pi ^2 alpha -4 pi beta -4 N-16right)x^2+pi (5 pi beta +4
    N+16)x-5 pi ^2 N=0tag 1$$
    which, as said,can be solved using Ferrari's method (this would be quite tedious).



    You can use another approximation (from far, not as good asBhaskara I's one ) minimizing with respect to $k$
    $$int_0^pi left[sin(x)-k (pi-x)xright]^2,dx=frac{pi ^5 }{30}k^2-8 k+frac{pi }{2}$$ to get $color{blue}{k=frac{120}{pi ^5}}$ and then $sin(x)simeq k(pi-x)x$. This leads to
    $$ (alpha -k)x^2+ (beta +pi k)x-N=0tag 2$$ which is simpler for sure. Using its solution as $x_0$ you can start Newton iterations to polish the root accorging to
    $$x_{n+1}=x_n-frac{sin (x_n)+alpha x_n^2+beta x_n-N}{ cos (x_n)+2 alpha x_n+beta}$$



    Let us try with $alpha=0.4$, $beta=0.6$, $N=2.345$. Solving $(1)$ gives $x^{(1)}_0 =1.26126$ while solving $(2)$ gives $x^{(2)}_0=1.27312$. Now, the iterates of Newton method starting with $x_0=x^{(2)}_0$
    $$left(
    begin{array}{cc}
    n & x_n \
    0 & 1.273117520 \
    1 & 1.260970996 \
    2 & 1.260976964
    end{array}
    right)$$
    which is the solution for ten significant figures.



    Edit



    To stay with a quadratic equation, we could be even more empirical (not respecting the values at end points) and write
    $$sin(x) simeq a_0+a_1x+a_2x^2$$ The same minimization method would lead to
    $$a_0=frac{12 left(pi ^2-10right)}{pi ^3}qquad a_1=-frac{60 left(pi ^2-12right)}{pi ^4}qquad a_2=frac{60 left(pi ^2-12right)}{pi ^5}$$ which would lead to $x^{(3)}_0=1.26757$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      In the same spirit as Awe Kumar Jha, considering $$sin (omega t) + at^2+bt=N$$ let
      $$x=omega tqquad alpha=frac a{omega^2}qquadbeta=frac b{omega}$$ to make the equation
      $$sin(x)+alpha x^2+beta x=N$$ and use Bhaskara I's sine approximation formula
      $$sin(x) simeq frac{16 (pi -x) x}{5 pi ^2-4 (pi -x) x}qquad (0leq xleqpi)$$
      $$4 alpha x^4+ 4( beta - pi alpha )x^3+ left(5 pi ^2 alpha -4 pi beta -4 N-16right)x^2+pi (5 pi beta +4
      N+16)x-5 pi ^2 N=0tag 1$$
      which, as said,can be solved using Ferrari's method (this would be quite tedious).



      You can use another approximation (from far, not as good asBhaskara I's one ) minimizing with respect to $k$
      $$int_0^pi left[sin(x)-k (pi-x)xright]^2,dx=frac{pi ^5 }{30}k^2-8 k+frac{pi }{2}$$ to get $color{blue}{k=frac{120}{pi ^5}}$ and then $sin(x)simeq k(pi-x)x$. This leads to
      $$ (alpha -k)x^2+ (beta +pi k)x-N=0tag 2$$ which is simpler for sure. Using its solution as $x_0$ you can start Newton iterations to polish the root accorging to
      $$x_{n+1}=x_n-frac{sin (x_n)+alpha x_n^2+beta x_n-N}{ cos (x_n)+2 alpha x_n+beta}$$



      Let us try with $alpha=0.4$, $beta=0.6$, $N=2.345$. Solving $(1)$ gives $x^{(1)}_0 =1.26126$ while solving $(2)$ gives $x^{(2)}_0=1.27312$. Now, the iterates of Newton method starting with $x_0=x^{(2)}_0$
      $$left(
      begin{array}{cc}
      n & x_n \
      0 & 1.273117520 \
      1 & 1.260970996 \
      2 & 1.260976964
      end{array}
      right)$$
      which is the solution for ten significant figures.



      Edit



      To stay with a quadratic equation, we could be even more empirical (not respecting the values at end points) and write
      $$sin(x) simeq a_0+a_1x+a_2x^2$$ The same minimization method would lead to
      $$a_0=frac{12 left(pi ^2-10right)}{pi ^3}qquad a_1=-frac{60 left(pi ^2-12right)}{pi ^4}qquad a_2=frac{60 left(pi ^2-12right)}{pi ^5}$$ which would lead to $x^{(3)}_0=1.26757$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        In the same spirit as Awe Kumar Jha, considering $$sin (omega t) + at^2+bt=N$$ let
        $$x=omega tqquad alpha=frac a{omega^2}qquadbeta=frac b{omega}$$ to make the equation
        $$sin(x)+alpha x^2+beta x=N$$ and use Bhaskara I's sine approximation formula
        $$sin(x) simeq frac{16 (pi -x) x}{5 pi ^2-4 (pi -x) x}qquad (0leq xleqpi)$$
        $$4 alpha x^4+ 4( beta - pi alpha )x^3+ left(5 pi ^2 alpha -4 pi beta -4 N-16right)x^2+pi (5 pi beta +4
        N+16)x-5 pi ^2 N=0tag 1$$
        which, as said,can be solved using Ferrari's method (this would be quite tedious).



        You can use another approximation (from far, not as good asBhaskara I's one ) minimizing with respect to $k$
        $$int_0^pi left[sin(x)-k (pi-x)xright]^2,dx=frac{pi ^5 }{30}k^2-8 k+frac{pi }{2}$$ to get $color{blue}{k=frac{120}{pi ^5}}$ and then $sin(x)simeq k(pi-x)x$. This leads to
        $$ (alpha -k)x^2+ (beta +pi k)x-N=0tag 2$$ which is simpler for sure. Using its solution as $x_0$ you can start Newton iterations to polish the root accorging to
        $$x_{n+1}=x_n-frac{sin (x_n)+alpha x_n^2+beta x_n-N}{ cos (x_n)+2 alpha x_n+beta}$$



        Let us try with $alpha=0.4$, $beta=0.6$, $N=2.345$. Solving $(1)$ gives $x^{(1)}_0 =1.26126$ while solving $(2)$ gives $x^{(2)}_0=1.27312$. Now, the iterates of Newton method starting with $x_0=x^{(2)}_0$
        $$left(
        begin{array}{cc}
        n & x_n \
        0 & 1.273117520 \
        1 & 1.260970996 \
        2 & 1.260976964
        end{array}
        right)$$
        which is the solution for ten significant figures.



        Edit



        To stay with a quadratic equation, we could be even more empirical (not respecting the values at end points) and write
        $$sin(x) simeq a_0+a_1x+a_2x^2$$ The same minimization method would lead to
        $$a_0=frac{12 left(pi ^2-10right)}{pi ^3}qquad a_1=-frac{60 left(pi ^2-12right)}{pi ^4}qquad a_2=frac{60 left(pi ^2-12right)}{pi ^5}$$ which would lead to $x^{(3)}_0=1.26757$.






        share|cite|improve this answer











        $endgroup$



        In the same spirit as Awe Kumar Jha, considering $$sin (omega t) + at^2+bt=N$$ let
        $$x=omega tqquad alpha=frac a{omega^2}qquadbeta=frac b{omega}$$ to make the equation
        $$sin(x)+alpha x^2+beta x=N$$ and use Bhaskara I's sine approximation formula
        $$sin(x) simeq frac{16 (pi -x) x}{5 pi ^2-4 (pi -x) x}qquad (0leq xleqpi)$$
        $$4 alpha x^4+ 4( beta - pi alpha )x^3+ left(5 pi ^2 alpha -4 pi beta -4 N-16right)x^2+pi (5 pi beta +4
        N+16)x-5 pi ^2 N=0tag 1$$
        which, as said,can be solved using Ferrari's method (this would be quite tedious).



        You can use another approximation (from far, not as good asBhaskara I's one ) minimizing with respect to $k$
        $$int_0^pi left[sin(x)-k (pi-x)xright]^2,dx=frac{pi ^5 }{30}k^2-8 k+frac{pi }{2}$$ to get $color{blue}{k=frac{120}{pi ^5}}$ and then $sin(x)simeq k(pi-x)x$. This leads to
        $$ (alpha -k)x^2+ (beta +pi k)x-N=0tag 2$$ which is simpler for sure. Using its solution as $x_0$ you can start Newton iterations to polish the root accorging to
        $$x_{n+1}=x_n-frac{sin (x_n)+alpha x_n^2+beta x_n-N}{ cos (x_n)+2 alpha x_n+beta}$$



        Let us try with $alpha=0.4$, $beta=0.6$, $N=2.345$. Solving $(1)$ gives $x^{(1)}_0 =1.26126$ while solving $(2)$ gives $x^{(2)}_0=1.27312$. Now, the iterates of Newton method starting with $x_0=x^{(2)}_0$
        $$left(
        begin{array}{cc}
        n & x_n \
        0 & 1.273117520 \
        1 & 1.260970996 \
        2 & 1.260976964
        end{array}
        right)$$
        which is the solution for ten significant figures.



        Edit



        To stay with a quadratic equation, we could be even more empirical (not respecting the values at end points) and write
        $$sin(x) simeq a_0+a_1x+a_2x^2$$ The same minimization method would lead to
        $$a_0=frac{12 left(pi ^2-10right)}{pi ^3}qquad a_1=-frac{60 left(pi ^2-12right)}{pi ^4}qquad a_2=frac{60 left(pi ^2-12right)}{pi ^5}$$ which would lead to $x^{(3)}_0=1.26757$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 18 '18 at 3:52

























        answered Dec 17 '18 at 10:42









        Claude LeiboviciClaude Leibovici

        122k1157134




        122k1157134






























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