Find the exact value of $x$ if $frac{sin x}x =sqrt x$
$begingroup$
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
asked Dec 31 '18 at 20:14
Matthew ChristopherMatthew Christopher
1316
$endgroup$
|
show 1 more comment
$begingroup$
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
asked Dec 31 '18 at 20:14
Matthew ChristopherMatthew Christopher
1316
$endgroup$
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
|
show 1 more comment
$begingroup$
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
asked Dec 31 '18 at 20:14
Matthew ChristopherMatthew Christopher
1316
$endgroup$
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
asked Dec 31 '18 at 20:14
Matthew ChristopherMatthew Christopher
1316
asked Dec 31 '18 at 20:14
Matthew ChristopherMatthew Christopher
1316
edited Dec 31 '18 at 20:18
Matthew Christopher
asked Dec 31 '18 at 20:14
Matthew ChristopherMatthew Christopher
1316
asked Dec 31 '18 at 20:14
Matthew ChristopherMatthew Christopher
1316
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
|
show 1 more comment
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
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
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
$$left(
begin{array}{cc}
n & x_n \
0 & 0.785398163397448309615660845819875721049292350 \
1 & 0.802788359292669370099497626707352560682643311 \
2 & 0.802803731726750434568551610417466777052560631 \
3 & 0.802803731737889315511829476566222402429256954 \
4 & 0.802803731737889315511835324604000441222668911
end{array}
right)$$
answered Jan 1 at 7:18
Claude LeiboviciClaude Leibovici
126k1158134
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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
$$left(
begin{array}{cc}
n & x_n \
0 & 0.785398163397448309615660845819875721049292350 \
1 & 0.802788359292669370099497626707352560682643311 \
2 & 0.802803731726750434568551610417466777052560631 \
3 & 0.802803731737889315511829476566222402429256954 \
4 & 0.802803731737889315511835324604000441222668911
end{array}
right)$$
answered Jan 1 at 7:18
Claude LeiboviciClaude Leibovici
126k1158134
$endgroup$
add a comment |
$begingroup$
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
$$left(
begin{array}{cc}
n & x_n \
0 & 0.785398163397448309615660845819875721049292350 \
1 & 0.802788359292669370099497626707352560682643311 \
2 & 0.802803731726750434568551610417466777052560631 \
3 & 0.802803731737889315511829476566222402429256954 \
4 & 0.802803731737889315511835324604000441222668911
end{array}
right)$$
answered Jan 1 at 7:18
Claude LeiboviciClaude Leibovici
126k1158134
$endgroup$
add a comment |
$begingroup$
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
$$left(
begin{array}{cc}
n & x_n \
0 & 0.785398163397448309615660845819875721049292350 \
1 & 0.802788359292669370099497626707352560682643311 \
2 & 0.802803731726750434568551610417466777052560631 \
3 & 0.802803731737889315511829476566222402429256954 \
4 & 0.802803731737889315511835324604000441222668911
end{array}
right)$$
answered Jan 1 at 7:18
Claude LeiboviciClaude Leibovici
126k1158134
$endgroup$
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
$$left(
begin{array}{cc}
n & x_n \
0 & 0.785398163397448309615660845819875721049292350 \
1 & 0.802788359292669370099497626707352560682643311 \
2 & 0.802803731726750434568551610417466777052560631 \
3 & 0.802803731737889315511829476566222402429256954 \
4 & 0.802803731737889315511835324604000441222668911
end{array}
right)$$
answered Jan 1 at 7:18
Claude LeiboviciClaude Leibovici
126k1158134
answered Jan 1 at 7:18
Claude LeiboviciClaude Leibovici
126k1158134
answered Jan 1 at 7:18
Claude LeiboviciClaude Leibovici
126k1158134
answered Jan 1 at 7:18
Claude LeiboviciClaude Leibovici
126k1158134
126k1158134
add a comment |
add a comment |
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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