radius of convergence of function defined by alternated addition and multiplication
$begingroup$
Starting from $zinmathbb{C}$, define $r(z)$ starting from the initial value $v_{0}=0.0$ and repeat the following iteration from $1,2,ldots$
$$v_{n+1} = begin{cases} v_{n} + frac{z^{n}}{n!} &text{if } 2 notmid n \
v_{n} cdotleft(1-frac{z^{n}}{sinh{n}}right) & text{if } 2 mid n. end{cases} $$
until it converges (I used 40 iterations below, anything much larger and I run into overflow problems)
Is there any good analytic way of getting the radius of convergence for this process? If I had to eyeball it, I'd say $e$:
$[-3,3]times[-3,3]$ on $mathbb{C}$
The phase portrait was generated with mpmath
>>> def r(z):
... v = 0.0
... for n in range(1,40):
... v = v + (z**(2*n-1))/fp.fac(2*n-1)
... v = v * (1 - (z**(2*n))/fp.sinh(2*n))
... return v/abs(v)
fp.cplot(lambda z: r(z), [-3,3], [-3,3], points=100000, file="iter.png", verbose=True)
complex-analysis
edited Jan 14 at 18:18
Glorfindel
3,41581830
asked May 2 '13 at 3:35
graveolensagraveolensa
3,09711738
$endgroup$
add a comment |
$begingroup$
Starting from $zinmathbb{C}$, define $r(z)$ starting from the initial value $v_{0}=0.0$ and repeat the following iteration from $1,2,ldots$
$$v_{n+1} = begin{cases} v_{n} + frac{z^{n}}{n!} &text{if } 2 notmid n \
v_{n} cdotleft(1-frac{z^{n}}{sinh{n}}right) & text{if } 2 mid n. end{cases} $$
until it converges (I used 40 iterations below, anything much larger and I run into overflow problems)
Is there any good analytic way of getting the radius of convergence for this process? If I had to eyeball it, I'd say $e$:
$[-3,3]times[-3,3]$ on $mathbb{C}$
The phase portrait was generated with mpmath
>>> def r(z):
... v = 0.0
... for n in range(1,40):
... v = v + (z**(2*n-1))/fp.fac(2*n-1)
... v = v * (1 - (z**(2*n))/fp.sinh(2*n))
... return v/abs(v)
fp.cplot(lambda z: r(z), [-3,3], [-3,3], points=100000, file="iter.png", verbose=True)
complex-analysis
edited Jan 14 at 18:18
Glorfindel
3,41581830
asked May 2 '13 at 3:35
graveolensagraveolensa
3,09711738
$endgroup$
$begingroup$
@RossMillikan: thanks, I am not a number theorist.
$endgroup$
– graveolensa
May 2 '13 at 3:42
add a comment |
$begingroup$
Starting from $zinmathbb{C}$, define $r(z)$ starting from the initial value $v_{0}=0.0$ and repeat the following iteration from $1,2,ldots$
$$v_{n+1} = begin{cases} v_{n} + frac{z^{n}}{n!} &text{if } 2 notmid n \
v_{n} cdotleft(1-frac{z^{n}}{sinh{n}}right) & text{if } 2 mid n. end{cases} $$
until it converges (I used 40 iterations below, anything much larger and I run into overflow problems)
Is there any good analytic way of getting the radius of convergence for this process? If I had to eyeball it, I'd say $e$:
$[-3,3]times[-3,3]$ on $mathbb{C}$
The phase portrait was generated with mpmath
>>> def r(z):
... v = 0.0
... for n in range(1,40):
... v = v + (z**(2*n-1))/fp.fac(2*n-1)
... v = v * (1 - (z**(2*n))/fp.sinh(2*n))
... return v/abs(v)
fp.cplot(lambda z: r(z), [-3,3], [-3,3], points=100000, file="iter.png", verbose=True)
complex-analysis
edited Jan 14 at 18:18
Glorfindel
3,41581830
asked May 2 '13 at 3:35
graveolensagraveolensa
3,09711738
$endgroup$
Starting from $zinmathbb{C}$, define $r(z)$ starting from the initial value $v_{0}=0.0$ and repeat the following iteration from $1,2,ldots$
$$v_{n+1} = begin{cases} v_{n} + frac{z^{n}}{n!} &text{if } 2 notmid n \
v_{n} cdotleft(1-frac{z^{n}}{sinh{n}}right) & text{if } 2 mid n. end{cases} $$
until it converges (I used 40 iterations below, anything much larger and I run into overflow problems)
Is there any good analytic way of getting the radius of convergence for this process? If I had to eyeball it, I'd say $e$:
$[-3,3]times[-3,3]$ on $mathbb{C}$
The phase portrait was generated with mpmath
>>> def r(z):
... v = 0.0
... for n in range(1,40):
... v = v + (z**(2*n-1))/fp.fac(2*n-1)
... v = v * (1 - (z**(2*n))/fp.sinh(2*n))
... return v/abs(v)
fp.cplot(lambda z: r(z), [-3,3], [-3,3], points=100000, file="iter.png", verbose=True)
complex-analysis
complex-analysis
edited Jan 14 at 18:18
Glorfindel
3,41581830
asked May 2 '13 at 3:35
graveolensagraveolensa
3,09711738
edited Jan 14 at 18:18
Glorfindel
3,41581830
asked May 2 '13 at 3:35
graveolensagraveolensa
3,09711738
edited Jan 14 at 18:18
Glorfindel
3,41581830
edited Jan 14 at 18:18
Glorfindel
3,41581830
edited Jan 14 at 18:18
Glorfindel
3,41581830
3,41581830
asked May 2 '13 at 3:35
graveolensagraveolensa
3,09711738
asked May 2 '13 at 3:35
graveolensagraveolensa
3,09711738
asked May 2 '13 at 3:35
graveolensagraveolensa
3,09711738
3,09711738
$begingroup$
@RossMillikan: thanks, I am not a number theorist.
$endgroup$
– graveolensa
May 2 '13 at 3:42
add a comment |
$begingroup$
@RossMillikan: thanks, I am not a number theorist.
$endgroup$
– graveolensa
May 2 '13 at 3:42
$begingroup$
@RossMillikan: thanks, I am not a number theorist.
$endgroup$
– graveolensa
May 2 '13 at 3:42
$begingroup$
@RossMillikan: thanks, I am not a number theorist.
$endgroup$
– graveolensa
May 2 '13 at 3:42
add a comment |
1 Answer
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$begingroup$
$sinh n$ is approximately $e^n/2$ when $n$ is large. So if $|z|>e$, then eventually you are multiplying by huge numbers every even step. That explains the divergence for $|z|>e$ (since the odd steps are adding a total contribution that is bounded for any given $z$).
Similarly, when $|z|<e$, eventally every even step is just multiplying by a number exponentially close to $1$. So that explains the convergence when $|z|<e$.
answered May 2 '13 at 5:58
Greg MartinGreg Martin
36.5k23565
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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active
oldest
votes
$begingroup$
$sinh n$ is approximately $e^n/2$ when $n$ is large. So if $|z|>e$, then eventually you are multiplying by huge numbers every even step. That explains the divergence for $|z|>e$ (since the odd steps are adding a total contribution that is bounded for any given $z$).
Similarly, when $|z|<e$, eventally every even step is just multiplying by a number exponentially close to $1$. So that explains the convergence when $|z|<e$.
answered May 2 '13 at 5:58
Greg MartinGreg Martin
36.5k23565
$endgroup$
add a comment |
$begingroup$
$sinh n$ is approximately $e^n/2$ when $n$ is large. So if $|z|>e$, then eventually you are multiplying by huge numbers every even step. That explains the divergence for $|z|>e$ (since the odd steps are adding a total contribution that is bounded for any given $z$).
Similarly, when $|z|<e$, eventally every even step is just multiplying by a number exponentially close to $1$. So that explains the convergence when $|z|<e$.
answered May 2 '13 at 5:58
Greg MartinGreg Martin
36.5k23565
$endgroup$
add a comment |
$begingroup$
$sinh n$ is approximately $e^n/2$ when $n$ is large. So if $|z|>e$, then eventually you are multiplying by huge numbers every even step. That explains the divergence for $|z|>e$ (since the odd steps are adding a total contribution that is bounded for any given $z$).
Similarly, when $|z|<e$, eventally every even step is just multiplying by a number exponentially close to $1$. So that explains the convergence when $|z|<e$.
answered May 2 '13 at 5:58
Greg MartinGreg Martin
36.5k23565
$endgroup$
$sinh n$ is approximately $e^n/2$ when $n$ is large. So if $|z|>e$, then eventually you are multiplying by huge numbers every even step. That explains the divergence for $|z|>e$ (since the odd steps are adding a total contribution that is bounded for any given $z$).
Similarly, when $|z|<e$, eventally every even step is just multiplying by a number exponentially close to $1$. So that explains the convergence when $|z|<e$.
answered May 2 '13 at 5:58
Greg MartinGreg Martin
36.5k23565
answered May 2 '13 at 5:58
Greg MartinGreg Martin
36.5k23565
answered May 2 '13 at 5:58
Greg MartinGreg Martin
36.5k23565
answered May 2 '13 at 5:58
Greg MartinGreg Martin
36.5k23565
36.5k23565
add a comment |
add a comment |
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$begingroup$
@RossMillikan: thanks, I am not a number theorist.
$endgroup$
– graveolensa
May 2 '13 at 3:42