Find the exact value of $x$ if $frac{sin x}x =sqrt x$












2












$begingroup$


It is quite easy to find that x is approximately .802. Is there any way in which we can solve this equation to find the exact value of x?



$frac{sin x}{x} = x^{1/2}$






algebra-precalculus graphing-functions







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Why not simplify and ask for $sin x = x^{3/2}$? And isn't $x=0$ enough?
    $endgroup$
    – David G. Stork
    Dec 31 '18 at 20:20








  • 2




    $begingroup$
    You might want to add some context, like how you came up with this question, and also what do you mean by "exact value"?
    $endgroup$
    – jgon
    Dec 31 '18 at 20:22






  • 1




    $begingroup$
    "quite easy" I presume not by hand?
    $endgroup$
    – TheSimpliFire
    Dec 31 '18 at 20:22






  • 6




    $begingroup$
    It seems like it's not solvable algebraically, so there is no exact answer. Unless someone manages to turn the sine into an exponential, and with subsequent series of steps isolate the x using the Lambert W function. If it's possible then more likely than not it is hard, tedious, and requires some complex analysis. I don't exactly know if I'm right, but with my naked eye, this is not solvable
    $endgroup$
    – KKZiomek
    Dec 31 '18 at 20:23








  • 9




    $begingroup$
    @DavidG.Stork I don't think $x=0$ is a solution, since to transform $sin x=x^{3/2}$ into $sin x/x=x^{1/2}$ one needs to assume $xneq 0$. The equation presented in the answer has $1$ at the LHS and $0$ in the RHS for $x=0$.
    $endgroup$
    – rafa11111
    Dec 31 '18 at 20:25


















2












$begingroup$


It is quite easy to find that x is approximately .802. Is there any way in which we can solve this equation to find the exact value of x?



$frac{sin x}{x} = x^{1/2}$






algebra-precalculus graphing-functions







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Why not simplify and ask for $sin x = x^{3/2}$? And isn't $x=0$ enough?
    $endgroup$
    – David G. Stork
    Dec 31 '18 at 20:20








  • 2




    $begingroup$
    You might want to add some context, like how you came up with this question, and also what do you mean by "exact value"?
    $endgroup$
    – jgon
    Dec 31 '18 at 20:22






  • 1




    $begingroup$
    "quite easy" I presume not by hand?
    $endgroup$
    – TheSimpliFire
    Dec 31 '18 at 20:22






  • 6




    $begingroup$
    It seems like it's not solvable algebraically, so there is no exact answer. Unless someone manages to turn the sine into an exponential, and with subsequent series of steps isolate the x using the Lambert W function. If it's possible then more likely than not it is hard, tedious, and requires some complex analysis. I don't exactly know if I'm right, but with my naked eye, this is not solvable
    $endgroup$
    – KKZiomek
    Dec 31 '18 at 20:23








  • 9




    $begingroup$
    @DavidG.Stork I don't think $x=0$ is a solution, since to transform $sin x=x^{3/2}$ into $sin x/x=x^{1/2}$ one needs to assume $xneq 0$. The equation presented in the answer has $1$ at the LHS and $0$ in the RHS for $x=0$.
    $endgroup$
    – rafa11111
    Dec 31 '18 at 20:25
















2












2








2


1



$begingroup$


It is quite easy to find that x is approximately .802. Is there any way in which we can solve this equation to find the exact value of x?



$frac{sin x}{x} = x^{1/2}$






algebra-precalculus graphing-functions







share|cite|improve this question











$endgroup$




It is quite easy to find that x is approximately .802. Is there any way in which we can solve this equation to find the exact value of x?



$frac{sin x}{x} = x^{1/2}$






algebra-precalculus graphing-functions




algebra-precalculus graphing-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 20:18







Matthew Christopher

















asked Dec 31 '18 at 20:14









Matthew ChristopherMatthew Christopher

1316




1316








  • 2




    $begingroup$
    Why not simplify and ask for $sin x = x^{3/2}$? And isn't $x=0$ enough?
    $endgroup$
    – David G. Stork
    Dec 31 '18 at 20:20








  • 2




    $begingroup$
    You might want to add some context, like how you came up with this question, and also what do you mean by "exact value"?
    $endgroup$
    – jgon
    Dec 31 '18 at 20:22






  • 1




    $begingroup$
    "quite easy" I presume not by hand?
    $endgroup$
    – TheSimpliFire
    Dec 31 '18 at 20:22






  • 6




    $begingroup$
    It seems like it's not solvable algebraically, so there is no exact answer. Unless someone manages to turn the sine into an exponential, and with subsequent series of steps isolate the x using the Lambert W function. If it's possible then more likely than not it is hard, tedious, and requires some complex analysis. I don't exactly know if I'm right, but with my naked eye, this is not solvable
    $endgroup$
    – KKZiomek
    Dec 31 '18 at 20:23








  • 9




    $begingroup$
    @DavidG.Stork I don't think $x=0$ is a solution, since to transform $sin x=x^{3/2}$ into $sin x/x=x^{1/2}$ one needs to assume $xneq 0$. The equation presented in the answer has $1$ at the LHS and $0$ in the RHS for $x=0$.
    $endgroup$
    – rafa11111
    Dec 31 '18 at 20:25
















  • 2




    $begingroup$
    Why not simplify and ask for $sin x = x^{3/2}$? And isn't $x=0$ enough?
    $endgroup$
    – David G. Stork
    Dec 31 '18 at 20:20








  • 2




    $begingroup$
    You might want to add some context, like how you came up with this question, and also what do you mean by "exact value"?
    $endgroup$
    – jgon
    Dec 31 '18 at 20:22






  • 1




    $begingroup$
    "quite easy" I presume not by hand?
    $endgroup$
    – TheSimpliFire
    Dec 31 '18 at 20:22






  • 6




    $begingroup$
    It seems like it's not solvable algebraically, so there is no exact answer. Unless someone manages to turn the sine into an exponential, and with subsequent series of steps isolate the x using the Lambert W function. If it's possible then more likely than not it is hard, tedious, and requires some complex analysis. I don't exactly know if I'm right, but with my naked eye, this is not solvable
    $endgroup$
    – KKZiomek
    Dec 31 '18 at 20:23








  • 9




    $begingroup$
    @DavidG.Stork I don't think $x=0$ is a solution, since to transform $sin x=x^{3/2}$ into $sin x/x=x^{1/2}$ one needs to assume $xneq 0$. The equation presented in the answer has $1$ at the LHS and $0$ in the RHS for $x=0$.
    $endgroup$
    – rafa11111
    Dec 31 '18 at 20:25










2




2




$begingroup$
Why not simplify and ask for $sin x = x^{3/2}$? And isn't $x=0$ enough?
$endgroup$
– David G. Stork
Dec 31 '18 at 20:20






$begingroup$
Why not simplify and ask for $sin x = x^{3/2}$? And isn't $x=0$ enough?
$endgroup$
– David G. Stork
Dec 31 '18 at 20:20






2




2




$begingroup$
You might want to add some context, like how you came up with this question, and also what do you mean by "exact value"?
$endgroup$
– jgon
Dec 31 '18 at 20:22




$begingroup$
You might want to add some context, like how you came up with this question, and also what do you mean by "exact value"?
$endgroup$
– jgon
Dec 31 '18 at 20:22




1




1




$begingroup$
"quite easy" I presume not by hand?
$endgroup$
– TheSimpliFire
Dec 31 '18 at 20:22




$begingroup$
"quite easy" I presume not by hand?
$endgroup$
– TheSimpliFire
Dec 31 '18 at 20:22




6




6




$begingroup$
It seems like it's not solvable algebraically, so there is no exact answer. Unless someone manages to turn the sine into an exponential, and with subsequent series of steps isolate the x using the Lambert W function. If it's possible then more likely than not it is hard, tedious, and requires some complex analysis. I don't exactly know if I'm right, but with my naked eye, this is not solvable
$endgroup$
– KKZiomek
Dec 31 '18 at 20:23






$begingroup$
It seems like it's not solvable algebraically, so there is no exact answer. Unless someone manages to turn the sine into an exponential, and with subsequent series of steps isolate the x using the Lambert W function. If it's possible then more likely than not it is hard, tedious, and requires some complex analysis. I don't exactly know if I'm right, but with my naked eye, this is not solvable
$endgroup$
– KKZiomek
Dec 31 '18 at 20:23






9




9




$begingroup$
@DavidG.Stork I don't think $x=0$ is a solution, since to transform $sin x=x^{3/2}$ into $sin x/x=x^{1/2}$ one needs to assume $xneq 0$. The equation presented in the answer has $1$ at the LHS and $0$ in the RHS for $x=0$.
$endgroup$
– rafa11111
Dec 31 '18 at 20:25






$begingroup$
@DavidG.Stork I don't think $x=0$ is a solution, since to transform $sin x=x^{3/2}$ into $sin x/x=x^{1/2}$ one needs to assume $xneq 0$. The equation presented in the answer has $1$ at the LHS and $0$ in the RHS for $x=0$.
$endgroup$
– rafa11111
Dec 31 '18 at 20:25












1 Answer
1






active

oldest

votes


















4












$begingroup$

The exact value, no way but good approximations (remember that $x=cos(x)$ does not show explicit solutions).



Plotting the function, we can see that the root is close to $frac pi 4$. So, to make expressions "simple", let us use $[1,n]$ Padé approximants for $frac{sin x}{x} - x^{1/2}$. They will write
$$frac{sin x}{x} - x^{1/2}=frac {frac{4 sqrt{2}-pi ^{3/2}}{2 pi }+a_1^{(n)} left(x-frac pi 4 right)}{1+sum_{k=1}^n b_k left(x-frac pi 4 right)^k }$$ from which the approximate solution
$$x_{(n)}=frac pi 4 +frac{pi ^{3/2}-4 sqrt{2}}{2 pi a_1^{(n)}}$$ This would give quite nasty formulae since
$$a_1^{(0)}=frac{-8 sqrt{2}+2 sqrt{2} pi -pi ^{3/2}}{pi ^2}$$
$$a_1^{(1)}=frac{64-24 pi +16 sqrt{2} pi ^{3/2}-3 pi ^2+sqrt{2} pi ^{5/2}-60 sqrt{2 pi
}}{2 pi left(8 sqrt{2}-2 sqrt{2} pi +pi ^{3/2}right)}$$
$$left(
begin{array}{cc}
n & x_{(n)} approx \
0 & 0.8027883593 \
1 & 0.8028042932 \
2 & 0.8028037219 \
3 & 0.8028037319 \
4 & 0.8028037317
end{array}
right)$$ which is the solution for ten significant figures.



If you use Newton method, it would be much less tedious and you could get as many decimal places as you wish
$$left(
begin{array}{cc}
n & x_n \
0 & 0.785398163397448309615660845819875721049292350 \
1 & 0.802788359292669370099497626707352560682643311 \
2 & 0.802803731726750434568551610417466777052560631 \
3 & 0.802803731737889315511829476566222402429256954 \
4 & 0.802803731737889315511835324604000441222668911
end{array}
right)$$






share|cite|improve this answer









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    1 Answer
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    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    The exact value, no way but good approximations (remember that $x=cos(x)$ does not show explicit solutions).



    Plotting the function, we can see that the root is close to $frac pi 4$. So, to make expressions "simple", let us use $[1,n]$ Padé approximants for $frac{sin x}{x} - x^{1/2}$. They will write
    $$frac{sin x}{x} - x^{1/2}=frac {frac{4 sqrt{2}-pi ^{3/2}}{2 pi }+a_1^{(n)} left(x-frac pi 4 right)}{1+sum_{k=1}^n b_k left(x-frac pi 4 right)^k }$$ from which the approximate solution
    $$x_{(n)}=frac pi 4 +frac{pi ^{3/2}-4 sqrt{2}}{2 pi a_1^{(n)}}$$ This would give quite nasty formulae since
    $$a_1^{(0)}=frac{-8 sqrt{2}+2 sqrt{2} pi -pi ^{3/2}}{pi ^2}$$
    $$a_1^{(1)}=frac{64-24 pi +16 sqrt{2} pi ^{3/2}-3 pi ^2+sqrt{2} pi ^{5/2}-60 sqrt{2 pi
    }}{2 pi left(8 sqrt{2}-2 sqrt{2} pi +pi ^{3/2}right)}$$
    $$left(
    begin{array}{cc}
    n & x_{(n)} approx \
    0 & 0.8027883593 \
    1 & 0.8028042932 \
    2 & 0.8028037219 \
    3 & 0.8028037319 \
    4 & 0.8028037317
    end{array}
    right)$$     which is the solution for ten significant figures.



    If you use Newton method, it would be much less tedious and you could get as many decimal places as you wish
    $$left(
    begin{array}{cc}
    n & x_n \
    0 & 0.785398163397448309615660845819875721049292350 \
    1 & 0.802788359292669370099497626707352560682643311 \
    2 & 0.802803731726750434568551610417466777052560631 \
    3 & 0.802803731737889315511829476566222402429256954 \
    4 & 0.802803731737889315511835324604000441222668911
    end{array}
    right)$$






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      The exact value, no way but good approximations (remember that $x=cos(x)$ does not show explicit solutions).



      Plotting the function, we can see that the root is close to $frac pi 4$. So, to make expressions "simple", let us use $[1,n]$ Padé approximants for $frac{sin x}{x} - x^{1/2}$. They will write
      $$frac{sin x}{x} - x^{1/2}=frac {frac{4 sqrt{2}-pi ^{3/2}}{2 pi }+a_1^{(n)} left(x-frac pi 4 right)}{1+sum_{k=1}^n b_k left(x-frac pi 4 right)^k }$$ from which the approximate solution
      $$x_{(n)}=frac pi 4 +frac{pi ^{3/2}-4 sqrt{2}}{2 pi a_1^{(n)}}$$ This would give quite nasty formulae since
      $$a_1^{(0)}=frac{-8 sqrt{2}+2 sqrt{2} pi -pi ^{3/2}}{pi ^2}$$
      $$a_1^{(1)}=frac{64-24 pi +16 sqrt{2} pi ^{3/2}-3 pi ^2+sqrt{2} pi ^{5/2}-60 sqrt{2 pi
      }}{2 pi left(8 sqrt{2}-2 sqrt{2} pi +pi ^{3/2}right)}$$
      $$left(
      begin{array}{cc}
      n & x_{(n)} approx \
      0 & 0.8027883593 \
      1 & 0.8028042932 \
      2 & 0.8028037219 \
      3 & 0.8028037319 \
      4 & 0.8028037317
      end{array}
      right)$$       which is the solution for ten significant figures.



      If you use Newton method, it would be much less tedious and you could get as many decimal places as you wish
      $$left(
      begin{array}{cc}
      n & x_n \
      0 & 0.785398163397448309615660845819875721049292350 \
      1 & 0.802788359292669370099497626707352560682643311 \
      2 & 0.802803731726750434568551610417466777052560631 \
      3 & 0.802803731737889315511829476566222402429256954 \
      4 & 0.802803731737889315511835324604000441222668911
      end{array}
      right)$$






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        The exact value, no way but good approximations (remember that $x=cos(x)$ does not show explicit solutions).



        Plotting the function, we can see that the root is close to $frac pi 4$. So, to make expressions "simple", let us use $[1,n]$ Padé approximants for $frac{sin x}{x} - x^{1/2}$. They will write
        $$frac{sin x}{x} - x^{1/2}=frac {frac{4 sqrt{2}-pi ^{3/2}}{2 pi }+a_1^{(n)} left(x-frac pi 4 right)}{1+sum_{k=1}^n b_k left(x-frac pi 4 right)^k }$$ from which the approximate solution
        $$x_{(n)}=frac pi 4 +frac{pi ^{3/2}-4 sqrt{2}}{2 pi a_1^{(n)}}$$ This would give quite nasty formulae since
        $$a_1^{(0)}=frac{-8 sqrt{2}+2 sqrt{2} pi -pi ^{3/2}}{pi ^2}$$
        $$a_1^{(1)}=frac{64-24 pi +16 sqrt{2} pi ^{3/2}-3 pi ^2+sqrt{2} pi ^{5/2}-60 sqrt{2 pi
        }}{2 pi left(8 sqrt{2}-2 sqrt{2} pi +pi ^{3/2}right)}$$
        $$left(
        begin{array}{cc}
        n & x_{(n)} approx \
        0 & 0.8027883593 \
        1 & 0.8028042932 \
        2 & 0.8028037219 \
        3 & 0.8028037319 \
        4 & 0.8028037317
        end{array}
        right)$$         which is the solution for ten significant figures.



        If you use Newton method, it would be much less tedious and you could get as many decimal places as you wish
        $$left(
        begin{array}{cc}
        n & x_n \
        0 & 0.785398163397448309615660845819875721049292350 \
        1 & 0.802788359292669370099497626707352560682643311 \
        2 & 0.802803731726750434568551610417466777052560631 \
        3 & 0.802803731737889315511829476566222402429256954 \
        4 & 0.802803731737889315511835324604000441222668911
        end{array}
        right)$$






        share|cite|improve this answer









        $endgroup$



        The exact value, no way but good approximations (remember that $x=cos(x)$ does not show explicit solutions).



        Plotting the function, we can see that the root is close to $frac pi 4$. So, to make expressions "simple", let us use $[1,n]$ Padé approximants for $frac{sin x}{x} - x^{1/2}$. They will write
        $$frac{sin x}{x} - x^{1/2}=frac {frac{4 sqrt{2}-pi ^{3/2}}{2 pi }+a_1^{(n)} left(x-frac pi 4 right)}{1+sum_{k=1}^n b_k left(x-frac pi 4 right)^k }$$ from which the approximate solution
        $$x_{(n)}=frac pi 4 +frac{pi ^{3/2}-4 sqrt{2}}{2 pi a_1^{(n)}}$$ This would give quite nasty formulae since
        $$a_1^{(0)}=frac{-8 sqrt{2}+2 sqrt{2} pi -pi ^{3/2}}{pi ^2}$$
        $$a_1^{(1)}=frac{64-24 pi +16 sqrt{2} pi ^{3/2}-3 pi ^2+sqrt{2} pi ^{5/2}-60 sqrt{2 pi
        }}{2 pi left(8 sqrt{2}-2 sqrt{2} pi +pi ^{3/2}right)}$$
        $$left(
        begin{array}{cc}
        n & x_{(n)} approx \
        0 & 0.8027883593 \
        1 & 0.8028042932 \
        2 & 0.8028037219 \
        3 & 0.8028037319 \
        4 & 0.8028037317
        end{array}
        right)$$         which is the solution for ten significant figures.



        If you use Newton method, it would be much less tedious and you could get as many decimal places as you wish
        $$left(
        begin{array}{cc}
        n & x_n \
        0 & 0.785398163397448309615660845819875721049292350 \
        1 & 0.802788359292669370099497626707352560682643311 \
        2 & 0.802803731726750434568551610417466777052560631 \
        3 & 0.802803731737889315511829476566222402429256954 \
        4 & 0.802803731737889315511835324604000441222668911
        end{array}
        right)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 7:18









        Claude LeiboviciClaude Leibovici

        126k1158134




        126k1158134






























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