Understanding a proof that a given recursive complex sequence converges
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Currently I am struggling to understand a solution to the following exercise:
Let $z_0 = x_0+iy_0 ne 0$ be a complex number and let the sequence $(z_n)_n$ be recursively defined as
$$z_{n+1} = frac{1}{2} left( z_n+frac{1}{z_n} right)$$
for $n ge 0$. Show that if $x_{0} > 0$ then $lim_{n to infty} z_n = 1$.
$\$
We notice that the only possible values for a limit would be $pm 1$. We further observe that since $z_0$ is in the right half plane all $z_n$ are in there too. Next we observe the sequence $w_n$ defined by
$$ w_n := frac{z_n-1}{z_n+1}$$
It holds that $w_{n+1} = w_n^2$. And since $|w_n| < 1 $ we see that $w_n$ converges to $0$. From $|z_n+1| ge 1$ we deduce that $z_n$ converges to $1$.
I do not understand the part
From $|z_n+1| ge 1$ we deduce that $z_n$ converges to $1$.
Could you explain that to me ?
sequences-and-series complex-analysis convergence
asked Dec 17 '18 at 17:48
3nondatur3nondatur
404111
$endgroup$
|
show 3 more comments
$begingroup$
Currently I am struggling to understand a solution to the following exercise:
Let $z_0 = x_0+iy_0 ne 0$ be a complex number and let the sequence $(z_n)_n$ be recursively defined as
$$z_{n+1} = frac{1}{2} left( z_n+frac{1}{z_n} right)$$
for $n ge 0$. Show that if $x_{0} > 0$ then $lim_{n to infty} z_n = 1$.
$\$
We notice that the only possible values for a limit would be $pm 1$. We further observe that since $z_0$ is in the right half plane all $z_n$ are in there too. Next we observe the sequence $w_n$ defined by
$$ w_n := frac{z_n-1}{z_n+1}$$
It holds that $w_{n+1} = w_n^2$. And since $|w_n| < 1 $ we see that $w_n$ converges to $0$. From $|z_n+1| ge 1$ we deduce that $z_n$ converges to $1$.
I do not understand the part
From $|z_n+1| ge 1$ we deduce that $z_n$ converges to $1$.
Could you explain that to me ?
sequences-and-series complex-analysis convergence
asked Dec 17 '18 at 17:48
3nondatur3nondatur
404111
$endgroup$
3
$begingroup$
FWIW, I do not understand "From $|z_n+1| ge 1$ we deduce that $z_n$ converges to $1$" either. From $w_nto0$ and $$z_n=frac{1+w_n}{1-w_n}$$ it is direct that $z_nto1$ and the argument that $|z_n+1| ge 1$ seems quite unrelated and not needed at all to get that $z_nto1$.
$endgroup$
– Did
Dec 17 '18 at 17:59
1
$begingroup$
@Did It seems... circumvoluted. I understood it as "the denominator is bounded away from $0$, so we don't have to worry about anything at all anyway."
$endgroup$
– Clement C.
Dec 17 '18 at 18:00
2
$begingroup$
@ClementC. It seems... wrong, actually. If one knows that $w_n=u_n/v_nto0$, what is needed to deduce that $u_nto0$ is that $(v_n)$ stays bounded, not that $(v_n)$ stays bounded away from $0$ (since, well... obviously, $u_n=w_nv_n$).
$endgroup$
– Did
Dec 17 '18 at 18:03
2
$begingroup$
@Did You're right. (Again, I don't know where this proof is from, but my best-case assumption is that the person who wrote the argument wanted to say that that $w_n$ was well-defined since no division by zero occurred, and put it in the wrong place, in the most confusing manner)
$endgroup$
– Clement C.
Dec 17 '18 at 18:06
1
$begingroup$
Probably the intention was to say that $g:zmapstofrac{z-1}{z+1}$ is holomorphic in $B_1(-1)^c$ and the only root is $1$, so $g(z_n)to 0$ iff $z_nto1$. Very poorly phrased
$endgroup$
– Federico
Dec 17 '18 at 18:26
|
show 3 more comments
$begingroup$
Currently I am struggling to understand a solution to the following exercise:
Let $z_0 = x_0+iy_0 ne 0$ be a complex number and let the sequence $(z_n)_n$ be recursively defined as
$$z_{n+1} = frac{1}{2} left( z_n+frac{1}{z_n} right)$$
for $n ge 0$. Show that if $x_{0} > 0$ then $lim_{n to infty} z_n = 1$.
$\$
We notice that the only possible values for a limit would be $pm 1$. We further observe that since $z_0$ is in the right half plane all $z_n$ are in there too. Next we observe the sequence $w_n$ defined by
$$ w_n := frac{z_n-1}{z_n+1}$$
It holds that $w_{n+1} = w_n^2$. And since $|w_n| < 1 $ we see that $w_n$ converges to $0$. From $|z_n+1| ge 1$ we deduce that $z_n$ converges to $1$.
I do not understand the part
From $|z_n+1| ge 1$ we deduce that $z_n$ converges to $1$.
Could you explain that to me ?
sequences-and-series complex-analysis convergence
asked Dec 17 '18 at 17:48
3nondatur3nondatur
404111
$endgroup$
Currently I am struggling to understand a solution to the following exercise:
Let $z_0 = x_0+iy_0 ne 0$ be a complex number and let the sequence $(z_n)_n$ be recursively defined as
$$z_{n+1} = frac{1}{2} left( z_n+frac{1}{z_n} right)$$
for $n ge 0$. Show that if $x_{0} > 0$ then $lim_{n to infty} z_n = 1$.
$\$
We notice that the only possible values for a limit would be $pm 1$. We further observe that since $z_0$ is in the right half plane all $z_n$ are in there too. Next we observe the sequence $w_n$ defined by
$$ w_n := frac{z_n-1}{z_n+1}$$
It holds that $w_{n+1} = w_n^2$. And since $|w_n| < 1 $ we see that $w_n$ converges to $0$. From $|z_n+1| ge 1$ we deduce that $z_n$ converges to $1$.
I do not understand the part
From $|z_n+1| ge 1$ we deduce that $z_n$ converges to $1$.
Could you explain that to me ?
sequences-and-series complex-analysis convergence
sequences-and-series complex-analysis convergence
asked Dec 17 '18 at 17:48
3nondatur3nondatur
404111
asked Dec 17 '18 at 17:48
3nondatur3nondatur
404111
asked Dec 17 '18 at 17:48
3nondatur3nondatur
404111
asked Dec 17 '18 at 17:48
3nondatur3nondatur
404111
asked Dec 17 '18 at 17:48
3nondatur3nondatur
404111
404111
3
$begingroup$
FWIW, I do not understand "From $|z_n+1| ge 1$ we deduce that $z_n$ converges to $1$" either. From $w_nto0$ and $$z_n=frac{1+w_n}{1-w_n}$$ it is direct that $z_nto1$ and the argument that $|z_n+1| ge 1$ seems quite unrelated and not needed at all to get that $z_nto1$.
$endgroup$
– Did
Dec 17 '18 at 17:59
1
$begingroup$
@Did It seems... circumvoluted. I understood it as "the denominator is bounded away from $0$, so we don't have to worry about anything at all anyway."
$endgroup$
– Clement C.
Dec 17 '18 at 18:00
2
$begingroup$
@ClementC. It seems... wrong, actually. If one knows that $w_n=u_n/v_nto0$, what is needed to deduce that $u_nto0$ is that $(v_n)$ stays bounded, not that $(v_n)$ stays bounded away from $0$ (since, well... obviously, $u_n=w_nv_n$).
$endgroup$
– Did
Dec 17 '18 at 18:03
2
$begingroup$
@Did You're right. (Again, I don't know where this proof is from, but my best-case assumption is that the person who wrote the argument wanted to say that that $w_n$ was well-defined since no division by zero occurred, and put it in the wrong place, in the most confusing manner)
$endgroup$
– Clement C.
Dec 17 '18 at 18:06
1
$begingroup$
Probably the intention was to say that $g:zmapstofrac{z-1}{z+1}$ is holomorphic in $B_1(-1)^c$ and the only root is $1$, so $g(z_n)to 0$ iff $z_nto1$. Very poorly phrased
$endgroup$
– Federico
Dec 17 '18 at 18:26
|
show 3 more comments
3
$begingroup$
FWIW, I do not understand "From $|z_n+1| ge 1$ we deduce that $z_n$ converges to $1$" either. From $w_nto0$ and $$z_n=frac{1+w_n}{1-w_n}$$ it is direct that $z_nto1$ and the argument that $|z_n+1| ge 1$ seems quite unrelated and not needed at all to get that $z_nto1$.
$endgroup$
– Did
Dec 17 '18 at 17:59
1
$begingroup$
@Did It seems... circumvoluted. I understood it as "the denominator is bounded away from $0$, so we don't have to worry about anything at all anyway."
$endgroup$
– Clement C.
Dec 17 '18 at 18:00
2
$begingroup$
@ClementC. It seems... wrong, actually. If one knows that $w_n=u_n/v_nto0$, what is needed to deduce that $u_nto0$ is that $(v_n)$ stays bounded, not that $(v_n)$ stays bounded away from $0$ (since, well... obviously, $u_n=w_nv_n$).
$endgroup$
– Did
Dec 17 '18 at 18:03
2
$begingroup$
@Did You're right. (Again, I don't know where this proof is from, but my best-case assumption is that the person who wrote the argument wanted to say that that $w_n$ was well-defined since no division by zero occurred, and put it in the wrong place, in the most confusing manner)
$endgroup$
– Clement C.
Dec 17 '18 at 18:06
1
$begingroup$
Probably the intention was to say that $g:zmapstofrac{z-1}{z+1}$ is holomorphic in $B_1(-1)^c$ and the only root is $1$, so $g(z_n)to 0$ iff $z_nto1$. Very poorly phrased
$endgroup$
– Federico
Dec 17 '18 at 18:26
3
3
$begingroup$
FWIW, I do not understand "From $|z_n+1| ge 1$ we deduce that $z_n$ converges to $1$" either. From $w_nto0$ and $$z_n=frac{1+w_n}{1-w_n}$$ it is direct that $z_nto1$ and the argument that $|z_n+1| ge 1$ seems quite unrelated and not needed at all to get that $z_nto1$.
$endgroup$
– Did
Dec 17 '18 at 17:59
$begingroup$
FWIW, I do not understand "From $|z_n+1| ge 1$ we deduce that $z_n$ converges to $1$" either. From $w_nto0$ and $$z_n=frac{1+w_n}{1-w_n}$$ it is direct that $z_nto1$ and the argument that $|z_n+1| ge 1$ seems quite unrelated and not needed at all to get that $z_nto1$.
$endgroup$
– Did
Dec 17 '18 at 17:59
1
1
$begingroup$
@Did It seems... circumvoluted. I understood it as "the denominator is bounded away from $0$, so we don't have to worry about anything at all anyway."
$endgroup$
– Clement C.
Dec 17 '18 at 18:00
$begingroup$
@Did It seems... circumvoluted. I understood it as "the denominator is bounded away from $0$, so we don't have to worry about anything at all anyway."
$endgroup$
– Clement C.
Dec 17 '18 at 18:00
2
2
$begingroup$
@ClementC. It seems... wrong, actually. If one knows that $w_n=u_n/v_nto0$, what is needed to deduce that $u_nto0$ is that $(v_n)$ stays bounded, not that $(v_n)$ stays bounded away from $0$ (since, well... obviously, $u_n=w_nv_n$).
$endgroup$
– Did
Dec 17 '18 at 18:03
$begingroup$
@ClementC. It seems... wrong, actually. If one knows that $w_n=u_n/v_nto0$, what is needed to deduce that $u_nto0$ is that $(v_n)$ stays bounded, not that $(v_n)$ stays bounded away from $0$ (since, well... obviously, $u_n=w_nv_n$).
$endgroup$
– Did
Dec 17 '18 at 18:03
2
2
$begingroup$
@Did You're right. (Again, I don't know where this proof is from, but my best-case assumption is that the person who wrote the argument wanted to say that that $w_n$ was well-defined since no division by zero occurred, and put it in the wrong place, in the most confusing manner)
$endgroup$
– Clement C.
Dec 17 '18 at 18:06
$begingroup$
@Did You're right. (Again, I don't know where this proof is from, but my best-case assumption is that the person who wrote the argument wanted to say that that $w_n$ was well-defined since no division by zero occurred, and put it in the wrong place, in the most confusing manner)
$endgroup$
– Clement C.
Dec 17 '18 at 18:06
1
1
$begingroup$
Probably the intention was to say that $g:zmapstofrac{z-1}{z+1}$ is holomorphic in $B_1(-1)^c$ and the only root is $1$, so $g(z_n)to 0$ iff $z_nto1$. Very poorly phrased
$endgroup$
– Federico
Dec 17 '18 at 18:26
$begingroup$
Probably the intention was to say that $g:zmapstofrac{z-1}{z+1}$ is holomorphic in $B_1(-1)^c$ and the only root is $1$, so $g(z_n)to 0$ iff $z_nto1$. Very poorly phrased
$endgroup$
– Federico
Dec 17 '18 at 18:26
|
show 3 more comments
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3
$begingroup$
FWIW, I do not understand "From $|z_n+1| ge 1$ we deduce that $z_n$ converges to $1$" either. From $w_nto0$ and $$z_n=frac{1+w_n}{1-w_n}$$ it is direct that $z_nto1$ and the argument that $|z_n+1| ge 1$ seems quite unrelated and not needed at all to get that $z_nto1$.
$endgroup$
– Did
Dec 17 '18 at 17:59
1
$begingroup$
@Did It seems... circumvoluted. I understood it as "the denominator is bounded away from $0$, so we don't have to worry about anything at all anyway."
$endgroup$
– Clement C.
Dec 17 '18 at 18:00
2
$begingroup$
@ClementC. It seems... wrong, actually. If one knows that $w_n=u_n/v_nto0$, what is needed to deduce that $u_nto0$ is that $(v_n)$ stays bounded, not that $(v_n)$ stays bounded away from $0$ (since, well... obviously, $u_n=w_nv_n$).
$endgroup$
– Did
Dec 17 '18 at 18:03
2
$begingroup$
@Did You're right. (Again, I don't know where this proof is from, but my best-case assumption is that the person who wrote the argument wanted to say that that $w_n$ was well-defined since no division by zero occurred, and put it in the wrong place, in the most confusing manner)
$endgroup$
– Clement C.
Dec 17 '18 at 18:06
1
$begingroup$
Probably the intention was to say that $g:zmapstofrac{z-1}{z+1}$ is holomorphic in $B_1(-1)^c$ and the only root is $1$, so $g(z_n)to 0$ iff $z_nto1$. Very poorly phrased
$endgroup$
– Federico
Dec 17 '18 at 18:26