Mixing
Can I generalise $x_{n+1}$ in this case?
$begingroup$
I have the following algorithm for producing rational approximations of $sqrt x$. We take an initial $x_0$ and apply the following steps:
$$x_n=sqrt{mu_n}+sqrt{x+mu_n}tag1$$
$$x_{n+1}=sqrt{x+mu_n}tag2$$
$$x_{n+1}=sqrt{mu_{n+1}}+sqrt{x+mu_{n+1}}tag 3$$
etc. Since $mu_{t+1}<mu_t$, this converges on $sqrt x$
This formula $(1)$ can be algebraically manipulated:
$$2mu_n+x+2sqrt{mu_n}sqrt{x+mu_n}=x_n^2$$
$$tomu_n+sqrt{mu_n}sqrt{x+mu_n}=frac{x_n^2-x}{2}$$
$$to x_nsqrt{mu_n}=frac{x_n^2-x}{2}$$
$$mu_n=bigg(frac{x_n^2-x}{2x_n}bigg)^2$$
$$to x_{n+1}=sqrt{x+mu_n}=frac{x_n^2+x}{2x_n}$$
Which is the well known Babylonian Method.
We know, as the algorithm I present proves, that this is an overestimate of $sqrt x$. I tried switching up my algorithm to:
$$x_n=sqrt{x-nu_n}-sqrt{nu_n}tag 1$$
$$x_{n+1}=sqrt{x-nu_n}tag2$$
$$x_{n+1}=sqrt{x-nu_{n+1}}-sqrt{nu_{n+1}}tag3$$
etc in order to find an underestimate with roughly the same accuracy scaling, but couldn't find a similar generalisation. I squared $(1)$ as before and got to:
$$x_n=x-2sqrt{nu_n}sqrt{x-nu_n}$$ Then applied $sqrt{x-nu_n}=x_n+nu_n$ which led to:
$$-2x_nsqrt{nu_n}-2nu_n=x_n-x$$
but this would require solving a quadratic (taking $y=sqrt{nu_n}$ gives us $y^2+x_ny-frac12(x-x_n)=0$), leaving irrationals all over the place.
Is it possible to manipulate the second system to $x_{n+1}=f(x_n)$ where $f:Bbb Q mapsto Bbb Q$, as I have done with the first system, and if so, which steps have I overlooked towards achieving this?
Thanks for any assistance.
approximation
asked Feb 1 at 18:54
Rhys HughesRhys Hughes
7,0501630
$endgroup$
add a comment |
$begingroup$
I have the following algorithm for producing rational approximations of $sqrt x$. We take an initial $x_0$ and apply the following steps:
$$x_n=sqrt{mu_n}+sqrt{x+mu_n}tag1$$
$$x_{n+1}=sqrt{x+mu_n}tag2$$
$$x_{n+1}=sqrt{mu_{n+1}}+sqrt{x+mu_{n+1}}tag 3$$
etc. Since $mu_{t+1}<mu_t$, this converges on $sqrt x$
This formula $(1)$ can be algebraically manipulated:
$$2mu_n+x+2sqrt{mu_n}sqrt{x+mu_n}=x_n^2$$
$$tomu_n+sqrt{mu_n}sqrt{x+mu_n}=frac{x_n^2-x}{2}$$
$$to x_nsqrt{mu_n}=frac{x_n^2-x}{2}$$
$$mu_n=bigg(frac{x_n^2-x}{2x_n}bigg)^2$$
$$to x_{n+1}=sqrt{x+mu_n}=frac{x_n^2+x}{2x_n}$$
Which is the well known Babylonian Method.
We know, as the algorithm I present proves, that this is an overestimate of $sqrt x$. I tried switching up my algorithm to:
$$x_n=sqrt{x-nu_n}-sqrt{nu_n}tag 1$$
$$x_{n+1}=sqrt{x-nu_n}tag2$$
$$x_{n+1}=sqrt{x-nu_{n+1}}-sqrt{nu_{n+1}}tag3$$
etc in order to find an underestimate with roughly the same accuracy scaling, but couldn't find a similar generalisation. I squared $(1)$ as before and got to:
$$x_n=x-2sqrt{nu_n}sqrt{x-nu_n}$$ Then applied $sqrt{x-nu_n}=x_n+nu_n$ which led to:
$$-2x_nsqrt{nu_n}-2nu_n=x_n-x$$
but this would require solving a quadratic (taking $y=sqrt{nu_n}$ gives us $y^2+x_ny-frac12(x-x_n)=0$), leaving irrationals all over the place.
Is it possible to manipulate the second system to $x_{n+1}=f(x_n)$ where $f:Bbb Q mapsto Bbb Q$, as I have done with the first system, and if so, which steps have I overlooked towards achieving this?
Thanks for any assistance.
approximation
asked Feb 1 at 18:54
Rhys HughesRhys Hughes
7,0501630
$endgroup$
$begingroup$
Did you consider applying the method to $frac{1}{x}$?
$endgroup$
– Keith McClary
Feb 2 at 3:20
$begingroup$
Could you clarify this? Because applying $frac 1x$ to the first method would result in something very close to $0$, not $sqrt x$ as intended.
$endgroup$
– Rhys Hughes
Feb 2 at 3:54
$begingroup$
I mean, the reciprocals of the approximants to $sqrt frac{1}{x}$ converge to $sqrt x$ from below.
$endgroup$
– Keith McClary
Feb 2 at 4:00
$begingroup$
Interesting. Thank you, I'll give that a try.
$endgroup$
– Rhys Hughes
Feb 2 at 4:02
add a comment |
$begingroup$
I have the following algorithm for producing rational approximations of $sqrt x$. We take an initial $x_0$ and apply the following steps:
$$x_n=sqrt{mu_n}+sqrt{x+mu_n}tag1$$
$$x_{n+1}=sqrt{x+mu_n}tag2$$
$$x_{n+1}=sqrt{mu_{n+1}}+sqrt{x+mu_{n+1}}tag 3$$
etc. Since $mu_{t+1}<mu_t$, this converges on $sqrt x$
This formula $(1)$ can be algebraically manipulated:
$$2mu_n+x+2sqrt{mu_n}sqrt{x+mu_n}=x_n^2$$
$$tomu_n+sqrt{mu_n}sqrt{x+mu_n}=frac{x_n^2-x}{2}$$
$$to x_nsqrt{mu_n}=frac{x_n^2-x}{2}$$
$$mu_n=bigg(frac{x_n^2-x}{2x_n}bigg)^2$$
$$to x_{n+1}=sqrt{x+mu_n}=frac{x_n^2+x}{2x_n}$$
Which is the well known Babylonian Method.
We know, as the algorithm I present proves, that this is an overestimate of $sqrt x$. I tried switching up my algorithm to:
$$x_n=sqrt{x-nu_n}-sqrt{nu_n}tag 1$$
$$x_{n+1}=sqrt{x-nu_n}tag2$$
$$x_{n+1}=sqrt{x-nu_{n+1}}-sqrt{nu_{n+1}}tag3$$
etc in order to find an underestimate with roughly the same accuracy scaling, but couldn't find a similar generalisation. I squared $(1)$ as before and got to:
$$x_n=x-2sqrt{nu_n}sqrt{x-nu_n}$$ Then applied $sqrt{x-nu_n}=x_n+nu_n$ which led to:
$$-2x_nsqrt{nu_n}-2nu_n=x_n-x$$
but this would require solving a quadratic (taking $y=sqrt{nu_n}$ gives us $y^2+x_ny-frac12(x-x_n)=0$), leaving irrationals all over the place.
Is it possible to manipulate the second system to $x_{n+1}=f(x_n)$ where $f:Bbb Q mapsto Bbb Q$, as I have done with the first system, and if so, which steps have I overlooked towards achieving this?
Thanks for any assistance.
approximation
asked Feb 1 at 18:54
Rhys HughesRhys Hughes
7,0501630
$endgroup$
I have the following algorithm for producing rational approximations of $sqrt x$. We take an initial $x_0$ and apply the following steps:
$$x_n=sqrt{mu_n}+sqrt{x+mu_n}tag1$$
$$x_{n+1}=sqrt{x+mu_n}tag2$$
$$x_{n+1}=sqrt{mu_{n+1}}+sqrt{x+mu_{n+1}}tag 3$$
etc. Since $mu_{t+1}<mu_t$, this converges on $sqrt x$
This formula $(1)$ can be algebraically manipulated:
$$2mu_n+x+2sqrt{mu_n}sqrt{x+mu_n}=x_n^2$$
$$tomu_n+sqrt{mu_n}sqrt{x+mu_n}=frac{x_n^2-x}{2}$$
$$to x_nsqrt{mu_n}=frac{x_n^2-x}{2}$$
$$mu_n=bigg(frac{x_n^2-x}{2x_n}bigg)^2$$
$$to x_{n+1}=sqrt{x+mu_n}=frac{x_n^2+x}{2x_n}$$
Which is the well known Babylonian Method.
We know, as the algorithm I present proves, that this is an overestimate of $sqrt x$. I tried switching up my algorithm to:
$$x_n=sqrt{x-nu_n}-sqrt{nu_n}tag 1$$
$$x_{n+1}=sqrt{x-nu_n}tag2$$
$$x_{n+1}=sqrt{x-nu_{n+1}}-sqrt{nu_{n+1}}tag3$$
etc in order to find an underestimate with roughly the same accuracy scaling, but couldn't find a similar generalisation. I squared $(1)$ as before and got to:
$$x_n=x-2sqrt{nu_n}sqrt{x-nu_n}$$ Then applied $sqrt{x-nu_n}=x_n+nu_n$ which led to:
$$-2x_nsqrt{nu_n}-2nu_n=x_n-x$$
but this would require solving a quadratic (taking $y=sqrt{nu_n}$ gives us $y^2+x_ny-frac12(x-x_n)=0$), leaving irrationals all over the place.
Is it possible to manipulate the second system to $x_{n+1}=f(x_n)$ where $f:Bbb Q mapsto Bbb Q$, as I have done with the first system, and if so, which steps have I overlooked towards achieving this?
Thanks for any assistance.
approximation
approximation
asked Feb 1 at 18:54
Rhys HughesRhys Hughes
7,0501630
asked Feb 1 at 18:54
Rhys HughesRhys Hughes
7,0501630
edited Feb 1 at 23:18
Rhys Hughes
asked Feb 1 at 18:54
Rhys HughesRhys Hughes
7,0501630
asked Feb 1 at 18:54
Rhys HughesRhys Hughes
7,0501630
asked Feb 1 at 18:54
Rhys HughesRhys Hughes
7,0501630
7,0501630
$begingroup$
Did you consider applying the method to $frac{1}{x}$?
$endgroup$
– Keith McClary
Feb 2 at 3:20
$begingroup$
Could you clarify this? Because applying $frac 1x$ to the first method would result in something very close to $0$, not $sqrt x$ as intended.
$endgroup$
– Rhys Hughes
Feb 2 at 3:54
$begingroup$
I mean, the reciprocals of the approximants to $sqrt frac{1}{x}$ converge to $sqrt x$ from below.
$endgroup$
– Keith McClary
Feb 2 at 4:00
$begingroup$
Interesting. Thank you, I'll give that a try.
$endgroup$
– Rhys Hughes
Feb 2 at 4:02
add a comment |
$begingroup$
Did you consider applying the method to $frac{1}{x}$?
$endgroup$
– Keith McClary
Feb 2 at 3:20
$begingroup$
Could you clarify this? Because applying $frac 1x$ to the first method would result in something very close to $0$, not $sqrt x$ as intended.
$endgroup$
– Rhys Hughes
Feb 2 at 3:54
$begingroup$
I mean, the reciprocals of the approximants to $sqrt frac{1}{x}$ converge to $sqrt x$ from below.
$endgroup$
– Keith McClary
Feb 2 at 4:00
$begingroup$
Interesting. Thank you, I'll give that a try.
$endgroup$
– Rhys Hughes
Feb 2 at 4:02
$begingroup$
Did you consider applying the method to $frac{1}{x}$?
$endgroup$
– Keith McClary
Feb 2 at 3:20
$begingroup$
Did you consider applying the method to $frac{1}{x}$?
$endgroup$
– Keith McClary
Feb 2 at 3:20
$begingroup$
Could you clarify this? Because applying $frac 1x$ to the first method would result in something very close to $0$, not $sqrt x$ as intended.
$endgroup$
– Rhys Hughes
Feb 2 at 3:54
$begingroup$
Could you clarify this? Because applying $frac 1x$ to the first method would result in something very close to $0$, not $sqrt x$ as intended.
$endgroup$
– Rhys Hughes
Feb 2 at 3:54
$begingroup$
I mean, the reciprocals of the approximants to $sqrt frac{1}{x}$ converge to $sqrt x$ from below.
$endgroup$
– Keith McClary
Feb 2 at 4:00
$begingroup$
I mean, the reciprocals of the approximants to $sqrt frac{1}{x}$ converge to $sqrt x$ from below.
$endgroup$
– Keith McClary
Feb 2 at 4:00
$begingroup$
Interesting. Thank you, I'll give that a try.
$endgroup$
– Rhys Hughes
Feb 2 at 4:02
$begingroup$
Interesting. Thank you, I'll give that a try.
$endgroup$
– Rhys Hughes
Feb 2 at 4:02
add a comment |
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$begingroup$
Did you consider applying the method to $frac{1}{x}$?
$endgroup$
– Keith McClary
Feb 2 at 3:20
$begingroup$
Could you clarify this? Because applying $frac 1x$ to the first method would result in something very close to $0$, not $sqrt x$ as intended.
$endgroup$
– Rhys Hughes
Feb 2 at 3:54
$begingroup$
I mean, the reciprocals of the approximants to $sqrt frac{1}{x}$ converge to $sqrt x$ from below.
$endgroup$
– Keith McClary
Feb 2 at 4:00
$begingroup$
Interesting. Thank you, I'll give that a try.
$endgroup$
– Rhys Hughes
Feb 2 at 4:02