prove the sequence $x_k+1=frac 1 2 left (x_k+ frac 2 x_k right)$ doesn't converge The 2019 Stack Overflow Developer Survey Results Are InProve $leftfracn2n+3right$ and $leftfracn2n-3right$ converge?How to prove a sequence does not converge?Let be f a Isometry, show that the sequence doesn't convergeShow that $lim_ntoinfty frac1nsum_k=0^n x_k=x $ if $x_n to x$Check that that the sequence $(-1)^nleft ( 1-frac1n right )$ converge or not.Prove sequence $frac2sqrtn $ convergeWhy Newton's method work ? i.e. why $lim_kto infty fleft(x_k-fracf(x_k)f'(x_k)right)=0$?If $(x_n) to 2$ then $left(frac1x_nright) to frac12$Proving that the sequence $left fracx_ny_n right rightarrow fracxy$prove that $x_k+1=frac 1 2 left (x_k+ frac 2 x_k right)$ is well defined.
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prove the sequence $x_k+1=frac 1 2 left (x_k+ frac 2 x_k right)$ doesn't converge
The 2019 Stack Overflow Developer Survey Results Are InProve $leftfracn2n+3right$ and $leftfracn2n-3right$ converge?How to prove a sequence does not converge?Let be f a Isometry, show that the sequence doesn't convergeShow that $lim_ntoinfty frac1nsum_k=0^n x_k=x $ if $x_n to x$Check that that the sequence $(-1)^nleft ( 1-frac1n right )$ converge or not.Prove sequence $frac2sqrtn $ convergeWhy Newton's method work ? i.e. why $lim_kto infty fleft(x_k-fracf(x_k)f'(x_k)right)=0$?If $(x_n) to 2$ then $left(frac1x_nright) to frac12$Proving that the sequence $left fracx_ny_n right rightarrow fracxy$prove that $x_k+1=frac 1 2 left (x_k+ frac 2 x_k right)$ is well defined.
$begingroup$
given $x_1$ is positive real number, prove the sequence $x_k+1=frac 1 2 left (x_k+ frac 2 x_k right)$ doesn't converge in Q
I started with $abs(x_n-x)<epsilon$ and then by triangle of inequality
$left| x_n-xright| \left| frac 1 2x_n-1+frac 1 x_r-1-xright| leq left| frac 1 2x_n-1right| +left| frac 1 x_n-1-xright|$
I don't know how to proceed from here. Thanks!
real-analysis proof-writing
asked Apr 8 at 1:13
analysis1analysis1
274
New contributor
analysis1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
|
show 1 more comment
$begingroup$
given $x_1$ is positive real number, prove the sequence $x_k+1=frac 1 2 left (x_k+ frac 2 x_k right)$ doesn't converge in Q
I started with $abs(x_n-x)<epsilon$ and then by triangle of inequality
$left| x_n-xright| \left| frac 1 2x_n-1+frac 1 x_r-1-xright| leq left| frac 1 2x_n-1right| +left| frac 1 x_n-1-xright|$
I don't know how to proceed from here. Thanks!
real-analysis proof-writing
asked Apr 8 at 1:13
analysis1analysis1
274
New contributor
analysis1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
3
$begingroup$
What is the initial value for your sequence? Or are you claiming that it doesn't converge for any values.
$endgroup$
– Gareth Ma
Apr 8 at 1:16
1
$begingroup$
This is Newton's Method for the square root of $2$.
$endgroup$
– ZeroXLR
Apr 8 at 1:19
1
$begingroup$
Also notice that $x=sqrt2$ is a fixed point, so it obviously converges
$endgroup$
– Gareth Ma
Apr 8 at 1:19
$begingroup$
Also for any negative $x_0$.
$endgroup$
– Robert Israel
Apr 8 at 1:20
1
$begingroup$
Oh sorry, it was a new edit
$endgroup$
– Gareth Ma
Apr 8 at 1:26
|
show 1 more comment
$begingroup$
given $x_1$ is positive real number, prove the sequence $x_k+1=frac 1 2 left (x_k+ frac 2 x_k right)$ doesn't converge in Q
I started with $abs(x_n-x)<epsilon$ and then by triangle of inequality
$left| x_n-xright| \left| frac 1 2x_n-1+frac 1 x_r-1-xright| leq left| frac 1 2x_n-1right| +left| frac 1 x_n-1-xright|$
I don't know how to proceed from here. Thanks!
real-analysis proof-writing
asked Apr 8 at 1:13
analysis1analysis1
274
New contributor
analysis1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
given $x_1$ is positive real number, prove the sequence $x_k+1=frac 1 2 left (x_k+ frac 2 x_k right)$ doesn't converge in Q
I started with $abs(x_n-x)<epsilon$ and then by triangle of inequality
$left| x_n-xright| \left| frac 1 2x_n-1+frac 1 x_r-1-xright| leq left| frac 1 2x_n-1right| +left| frac 1 x_n-1-xright|$
I don't know how to proceed from here. Thanks!
real-analysis proof-writing
real-analysis proof-writing
asked Apr 8 at 1:13
analysis1analysis1
274
New contributor
analysis1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 1:13
analysis1analysis1
274
New contributor
analysis1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Apr 8 at 1:20
analysis1
asked Apr 8 at 1:13
analysis1analysis1
274
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analysis1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Apr 8 at 1:13
analysis1analysis1
274
asked Apr 8 at 1:13
analysis1analysis1
274
274
New contributor
analysis1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
analysis1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
analysis1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
$begingroup$
What is the initial value for your sequence? Or are you claiming that it doesn't converge for any values.
$endgroup$
– Gareth Ma
Apr 8 at 1:16
1
$begingroup$
This is Newton's Method for the square root of $2$.
$endgroup$
– ZeroXLR
Apr 8 at 1:19
1
$begingroup$
Also notice that $x=sqrt2$ is a fixed point, so it obviously converges
$endgroup$
– Gareth Ma
Apr 8 at 1:19
$begingroup$
Also for any negative $x_0$.
$endgroup$
– Robert Israel
Apr 8 at 1:20
1
$begingroup$
Oh sorry, it was a new edit
$endgroup$
– Gareth Ma
Apr 8 at 1:26
|
show 1 more comment
3
$begingroup$
What is the initial value for your sequence? Or are you claiming that it doesn't converge for any values.
$endgroup$
– Gareth Ma
Apr 8 at 1:16
1
$begingroup$
This is Newton's Method for the square root of $2$.
$endgroup$
– ZeroXLR
Apr 8 at 1:19
1
$begingroup$
Also notice that $x=sqrt2$ is a fixed point, so it obviously converges
$endgroup$
– Gareth Ma
Apr 8 at 1:19
$begingroup$
Also for any negative $x_0$.
$endgroup$
– Robert Israel
Apr 8 at 1:20
1
$begingroup$
Oh sorry, it was a new edit
$endgroup$
– Gareth Ma
Apr 8 at 1:26
3
3
$begingroup$
What is the initial value for your sequence? Or are you claiming that it doesn't converge for any values.
$endgroup$
– Gareth Ma
Apr 8 at 1:16
$begingroup$
What is the initial value for your sequence? Or are you claiming that it doesn't converge for any values.
$endgroup$
– Gareth Ma
Apr 8 at 1:16
1
1
$begingroup$
This is Newton's Method for the square root of $2$.
$endgroup$
– ZeroXLR
Apr 8 at 1:19
$begingroup$
This is Newton's Method for the square root of $2$.
$endgroup$
– ZeroXLR
Apr 8 at 1:19
1
1
$begingroup$
Also notice that $x=sqrt2$ is a fixed point, so it obviously converges
$endgroup$
– Gareth Ma
Apr 8 at 1:19
$begingroup$
Also notice that $x=sqrt2$ is a fixed point, so it obviously converges
$endgroup$
– Gareth Ma
Apr 8 at 1:19
$begingroup$
Also for any negative $x_0$.
$endgroup$
– Robert Israel
Apr 8 at 1:20
$begingroup$
Also for any negative $x_0$.
$endgroup$
– Robert Israel
Apr 8 at 1:20
1
1
$begingroup$
Oh sorry, it was a new edit
$endgroup$
– Gareth Ma
Apr 8 at 1:26
$begingroup$
Oh sorry, it was a new edit
$endgroup$
– Gareth Ma
Apr 8 at 1:26
|
show 1 more comment
1 Answer
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$begingroup$
If the limit exist, say it converges to $L$, then by taking $kto infty$ you get $L=frac12(L+frac2L)$, solving the equation you get $L=sqrt2$. So if the limit exist, it has to be $sqrt2$ which is not rational.
answered Apr 8 at 1:35
Julian MejiaJulian Mejia
54229
$endgroup$
add a comment |
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1 Answer
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$begingroup$
If the limit exist, say it converges to $L$, then by taking $kto infty$ you get $L=frac12(L+frac2L)$, solving the equation you get $L=sqrt2$. So if the limit exist, it has to be $sqrt2$ which is not rational.
answered Apr 8 at 1:35
Julian MejiaJulian Mejia
54229
$endgroup$
add a comment |
$begingroup$
If the limit exist, say it converges to $L$, then by taking $kto infty$ you get $L=frac12(L+frac2L)$, solving the equation you get $L=sqrt2$. So if the limit exist, it has to be $sqrt2$ which is not rational.
answered Apr 8 at 1:35
Julian MejiaJulian Mejia
54229
$endgroup$
add a comment |
$begingroup$
If the limit exist, say it converges to $L$, then by taking $kto infty$ you get $L=frac12(L+frac2L)$, solving the equation you get $L=sqrt2$. So if the limit exist, it has to be $sqrt2$ which is not rational.
answered Apr 8 at 1:35
Julian MejiaJulian Mejia
54229
$endgroup$
If the limit exist, say it converges to $L$, then by taking $kto infty$ you get $L=frac12(L+frac2L)$, solving the equation you get $L=sqrt2$. So if the limit exist, it has to be $sqrt2$ which is not rational.
answered Apr 8 at 1:35
Julian MejiaJulian Mejia
54229
answered Apr 8 at 1:35
Julian MejiaJulian Mejia
54229
answered Apr 8 at 1:35
Julian MejiaJulian Mejia
54229
answered Apr 8 at 1:35
Julian MejiaJulian Mejia
54229
54229
add a comment |
add a comment |
analysis1 is a new contributor. Be nice, and check out our Code of Conduct.
analysis1 is a new contributor. Be nice, and check out our Code of Conduct.
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3
$begingroup$
What is the initial value for your sequence? Or are you claiming that it doesn't converge for any values.
$endgroup$
– Gareth Ma
Apr 8 at 1:16
1
$begingroup$
This is Newton's Method for the square root of $2$.
$endgroup$
– ZeroXLR
Apr 8 at 1:19
1
$begingroup$
Also notice that $x=sqrt2$ is a fixed point, so it obviously converges
$endgroup$
– Gareth Ma
Apr 8 at 1:19
$begingroup$
Also for any negative $x_0$.
$endgroup$
– Robert Israel
Apr 8 at 1:20
1
$begingroup$
Oh sorry, it was a new edit
$endgroup$
– Gareth Ma
Apr 8 at 1:26