Confusion about Complex Continued Fraction
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I faced some problems when I tried to evaluate the following continued fraction
$$cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cdots}}}}}} $$
The common trick as usual is to set
$$x= cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cdots}}}}}} $$
Then, I have
$$x=frac{1}{i+x}$$
Using the quadratic formula, I get
$$x=frac{-ipmsqrt{3}}{2}$$
There are two solutions for $x$, but I don't know which one is correct.
For any real continued fraction, for example
$$cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cdots}}}}}} $$
it is clear to see that the value is $frac{-1+sqrt{5}}{2}$ instead of $frac{-1-sqrt{5}}{2}$.
But for the complex continued fraction, I have difficulty distinguishing the correct solution.
complex-numbers continued-fractions
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add a comment |
$begingroup$
I faced some problems when I tried to evaluate the following continued fraction
$$cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cdots}}}}}} $$
The common trick as usual is to set
$$x= cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cdots}}}}}} $$
Then, I have
$$x=frac{1}{i+x}$$
Using the quadratic formula, I get
$$x=frac{-ipmsqrt{3}}{2}$$
There are two solutions for $x$, but I don't know which one is correct.
For any real continued fraction, for example
$$cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cdots}}}}}} $$
it is clear to see that the value is $frac{-1+sqrt{5}}{2}$ instead of $frac{-1-sqrt{5}}{2}$.
But for the complex continued fraction, I have difficulty distinguishing the correct solution.
complex-numbers continued-fractions
$endgroup$
add a comment |
$begingroup$
I faced some problems when I tried to evaluate the following continued fraction
$$cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cdots}}}}}} $$
The common trick as usual is to set
$$x= cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cdots}}}}}} $$
Then, I have
$$x=frac{1}{i+x}$$
Using the quadratic formula, I get
$$x=frac{-ipmsqrt{3}}{2}$$
There are two solutions for $x$, but I don't know which one is correct.
For any real continued fraction, for example
$$cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cdots}}}}}} $$
it is clear to see that the value is $frac{-1+sqrt{5}}{2}$ instead of $frac{-1-sqrt{5}}{2}$.
But for the complex continued fraction, I have difficulty distinguishing the correct solution.
complex-numbers continued-fractions
$endgroup$
I faced some problems when I tried to evaluate the following continued fraction
$$cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cdots}}}}}} $$
The common trick as usual is to set
$$x= cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cdots}}}}}} $$
Then, I have
$$x=frac{1}{i+x}$$
Using the quadratic formula, I get
$$x=frac{-ipmsqrt{3}}{2}$$
There are two solutions for $x$, but I don't know which one is correct.
For any real continued fraction, for example
$$cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cdots}}}}}} $$
it is clear to see that the value is $frac{-1+sqrt{5}}{2}$ instead of $frac{-1-sqrt{5}}{2}$.
But for the complex continued fraction, I have difficulty distinguishing the correct solution.
complex-numbers continued-fractions
complex-numbers continued-fractions
asked 16 hours ago
LarryLarry
2,52331131
2,52331131
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add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Well, Vasily has put his finger on one problem, but I would like to point out a much more serious one.
We write a continued fraction to get a number that is the limit of the convergents, that is, of the expressions that you get when you cut your continued fraction off, to be a finite c.f.
The first convergent is $frac1i$, no problem, but the second is
$$frac1{i+frac1i}=frac10,,$$ a most unfortunate development. My recommendation would be to go on to other complex continued fractions, where the partial denominators are rather larger than $i$.
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4
$begingroup$
[+1] very thorough remark.
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– Jean Marie
13 hours ago
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I'm not sure what you mean by your last sentence. What about $[x;x,x,cdots]$ where $x = sqrt{2}i$?
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– user21820
7 mins ago
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I didn’t mean much. But $sqrt2i$ is not what I had in mind by “rather larger”. Indeed, even $x=2i$ seems to give an uninteresting limiting value.
$endgroup$
– Lubin
34 secs ago
add a comment |
$begingroup$
The problem with this expression is that function $f(x)=(i+x)^{-1}$ is a tricky one:
$$
f(f(x)) = frac{1}{i+frac{1}{i+x}}=frac{i+x}{-1+ix+1}=-i+frac1x,\
f(f(f(x))) = frac{1}{i+(-i+1/x)}=x.
$$
So every number generates the orbit of length 3, except for two numbers you have found, which are fixed points. Thus, the series of nesting functions $f$ does not converge if you haven't already started with a fixed point. So the question as it is stated has no sense: you cannot assign any number to this expression.
$endgroup$
add a comment |
$begingroup$
For future complex continued fractions...
For a continued fraction to converge, the sequence of convergents (values finite initial segments of the partial fraction expression) must converge to a particular complex number, that is, to a specific argument and magnitude. For the real continued fraction you mention, the algebraic expression allows the argument to be either $0$ (containing "$+sqrt{5}$") or $pi$ (containing "$-sqrt{5}$"). The same thing can be done with the complex version. Your two expressions have arguments $-pi/6$ and $7pi/6$. If your continued fraction had any hope of converging (which, as others have shown, it does not) it converges to something with a specific argument. You generically hope that one of the algebraic solutions you have has an argument matching the limit argument of the convergents. Finding that limit argument usually requires another computation, typically solving another recurrence.
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Well, Vasily has put his finger on one problem, but I would like to point out a much more serious one.
We write a continued fraction to get a number that is the limit of the convergents, that is, of the expressions that you get when you cut your continued fraction off, to be a finite c.f.
The first convergent is $frac1i$, no problem, but the second is
$$frac1{i+frac1i}=frac10,,$$ a most unfortunate development. My recommendation would be to go on to other complex continued fractions, where the partial denominators are rather larger than $i$.
$endgroup$
4
$begingroup$
[+1] very thorough remark.
$endgroup$
– Jean Marie
13 hours ago
$begingroup$
I'm not sure what you mean by your last sentence. What about $[x;x,x,cdots]$ where $x = sqrt{2}i$?
$endgroup$
– user21820
7 mins ago
$begingroup$
I didn’t mean much. But $sqrt2i$ is not what I had in mind by “rather larger”. Indeed, even $x=2i$ seems to give an uninteresting limiting value.
$endgroup$
– Lubin
34 secs ago
add a comment |
$begingroup$
Well, Vasily has put his finger on one problem, but I would like to point out a much more serious one.
We write a continued fraction to get a number that is the limit of the convergents, that is, of the expressions that you get when you cut your continued fraction off, to be a finite c.f.
The first convergent is $frac1i$, no problem, but the second is
$$frac1{i+frac1i}=frac10,,$$ a most unfortunate development. My recommendation would be to go on to other complex continued fractions, where the partial denominators are rather larger than $i$.
$endgroup$
4
$begingroup$
[+1] very thorough remark.
$endgroup$
– Jean Marie
13 hours ago
$begingroup$
I'm not sure what you mean by your last sentence. What about $[x;x,x,cdots]$ where $x = sqrt{2}i$?
$endgroup$
– user21820
7 mins ago
$begingroup$
I didn’t mean much. But $sqrt2i$ is not what I had in mind by “rather larger”. Indeed, even $x=2i$ seems to give an uninteresting limiting value.
$endgroup$
– Lubin
34 secs ago
add a comment |
$begingroup$
Well, Vasily has put his finger on one problem, but I would like to point out a much more serious one.
We write a continued fraction to get a number that is the limit of the convergents, that is, of the expressions that you get when you cut your continued fraction off, to be a finite c.f.
The first convergent is $frac1i$, no problem, but the second is
$$frac1{i+frac1i}=frac10,,$$ a most unfortunate development. My recommendation would be to go on to other complex continued fractions, where the partial denominators are rather larger than $i$.
$endgroup$
Well, Vasily has put his finger on one problem, but I would like to point out a much more serious one.
We write a continued fraction to get a number that is the limit of the convergents, that is, of the expressions that you get when you cut your continued fraction off, to be a finite c.f.
The first convergent is $frac1i$, no problem, but the second is
$$frac1{i+frac1i}=frac10,,$$ a most unfortunate development. My recommendation would be to go on to other complex continued fractions, where the partial denominators are rather larger than $i$.
answered 14 hours ago
LubinLubin
45k44687
45k44687
4
$begingroup$
[+1] very thorough remark.
$endgroup$
– Jean Marie
13 hours ago
$begingroup$
I'm not sure what you mean by your last sentence. What about $[x;x,x,cdots]$ where $x = sqrt{2}i$?
$endgroup$
– user21820
7 mins ago
$begingroup$
I didn’t mean much. But $sqrt2i$ is not what I had in mind by “rather larger”. Indeed, even $x=2i$ seems to give an uninteresting limiting value.
$endgroup$
– Lubin
34 secs ago
add a comment |
4
$begingroup$
[+1] very thorough remark.
$endgroup$
– Jean Marie
13 hours ago
$begingroup$
I'm not sure what you mean by your last sentence. What about $[x;x,x,cdots]$ where $x = sqrt{2}i$?
$endgroup$
– user21820
7 mins ago
$begingroup$
I didn’t mean much. But $sqrt2i$ is not what I had in mind by “rather larger”. Indeed, even $x=2i$ seems to give an uninteresting limiting value.
$endgroup$
– Lubin
34 secs ago
4
4
$begingroup$
[+1] very thorough remark.
$endgroup$
– Jean Marie
13 hours ago
$begingroup$
[+1] very thorough remark.
$endgroup$
– Jean Marie
13 hours ago
$begingroup$
I'm not sure what you mean by your last sentence. What about $[x;x,x,cdots]$ where $x = sqrt{2}i$?
$endgroup$
– user21820
7 mins ago
$begingroup$
I'm not sure what you mean by your last sentence. What about $[x;x,x,cdots]$ where $x = sqrt{2}i$?
$endgroup$
– user21820
7 mins ago
$begingroup$
I didn’t mean much. But $sqrt2i$ is not what I had in mind by “rather larger”. Indeed, even $x=2i$ seems to give an uninteresting limiting value.
$endgroup$
– Lubin
34 secs ago
$begingroup$
I didn’t mean much. But $sqrt2i$ is not what I had in mind by “rather larger”. Indeed, even $x=2i$ seems to give an uninteresting limiting value.
$endgroup$
– Lubin
34 secs ago
add a comment |
$begingroup$
The problem with this expression is that function $f(x)=(i+x)^{-1}$ is a tricky one:
$$
f(f(x)) = frac{1}{i+frac{1}{i+x}}=frac{i+x}{-1+ix+1}=-i+frac1x,\
f(f(f(x))) = frac{1}{i+(-i+1/x)}=x.
$$
So every number generates the orbit of length 3, except for two numbers you have found, which are fixed points. Thus, the series of nesting functions $f$ does not converge if you haven't already started with a fixed point. So the question as it is stated has no sense: you cannot assign any number to this expression.
$endgroup$
add a comment |
$begingroup$
The problem with this expression is that function $f(x)=(i+x)^{-1}$ is a tricky one:
$$
f(f(x)) = frac{1}{i+frac{1}{i+x}}=frac{i+x}{-1+ix+1}=-i+frac1x,\
f(f(f(x))) = frac{1}{i+(-i+1/x)}=x.
$$
So every number generates the orbit of length 3, except for two numbers you have found, which are fixed points. Thus, the series of nesting functions $f$ does not converge if you haven't already started with a fixed point. So the question as it is stated has no sense: you cannot assign any number to this expression.
$endgroup$
add a comment |
$begingroup$
The problem with this expression is that function $f(x)=(i+x)^{-1}$ is a tricky one:
$$
f(f(x)) = frac{1}{i+frac{1}{i+x}}=frac{i+x}{-1+ix+1}=-i+frac1x,\
f(f(f(x))) = frac{1}{i+(-i+1/x)}=x.
$$
So every number generates the orbit of length 3, except for two numbers you have found, which are fixed points. Thus, the series of nesting functions $f$ does not converge if you haven't already started with a fixed point. So the question as it is stated has no sense: you cannot assign any number to this expression.
$endgroup$
The problem with this expression is that function $f(x)=(i+x)^{-1}$ is a tricky one:
$$
f(f(x)) = frac{1}{i+frac{1}{i+x}}=frac{i+x}{-1+ix+1}=-i+frac1x,\
f(f(f(x))) = frac{1}{i+(-i+1/x)}=x.
$$
So every number generates the orbit of length 3, except for two numbers you have found, which are fixed points. Thus, the series of nesting functions $f$ does not converge if you haven't already started with a fixed point. So the question as it is stated has no sense: you cannot assign any number to this expression.
answered 15 hours ago
Vasily MitchVasily Mitch
2,5941312
2,5941312
add a comment |
add a comment |
$begingroup$
For future complex continued fractions...
For a continued fraction to converge, the sequence of convergents (values finite initial segments of the partial fraction expression) must converge to a particular complex number, that is, to a specific argument and magnitude. For the real continued fraction you mention, the algebraic expression allows the argument to be either $0$ (containing "$+sqrt{5}$") or $pi$ (containing "$-sqrt{5}$"). The same thing can be done with the complex version. Your two expressions have arguments $-pi/6$ and $7pi/6$. If your continued fraction had any hope of converging (which, as others have shown, it does not) it converges to something with a specific argument. You generically hope that one of the algebraic solutions you have has an argument matching the limit argument of the convergents. Finding that limit argument usually requires another computation, typically solving another recurrence.
$endgroup$
add a comment |
$begingroup$
For future complex continued fractions...
For a continued fraction to converge, the sequence of convergents (values finite initial segments of the partial fraction expression) must converge to a particular complex number, that is, to a specific argument and magnitude. For the real continued fraction you mention, the algebraic expression allows the argument to be either $0$ (containing "$+sqrt{5}$") or $pi$ (containing "$-sqrt{5}$"). The same thing can be done with the complex version. Your two expressions have arguments $-pi/6$ and $7pi/6$. If your continued fraction had any hope of converging (which, as others have shown, it does not) it converges to something with a specific argument. You generically hope that one of the algebraic solutions you have has an argument matching the limit argument of the convergents. Finding that limit argument usually requires another computation, typically solving another recurrence.
$endgroup$
add a comment |
$begingroup$
For future complex continued fractions...
For a continued fraction to converge, the sequence of convergents (values finite initial segments of the partial fraction expression) must converge to a particular complex number, that is, to a specific argument and magnitude. For the real continued fraction you mention, the algebraic expression allows the argument to be either $0$ (containing "$+sqrt{5}$") or $pi$ (containing "$-sqrt{5}$"). The same thing can be done with the complex version. Your two expressions have arguments $-pi/6$ and $7pi/6$. If your continued fraction had any hope of converging (which, as others have shown, it does not) it converges to something with a specific argument. You generically hope that one of the algebraic solutions you have has an argument matching the limit argument of the convergents. Finding that limit argument usually requires another computation, typically solving another recurrence.
$endgroup$
For future complex continued fractions...
For a continued fraction to converge, the sequence of convergents (values finite initial segments of the partial fraction expression) must converge to a particular complex number, that is, to a specific argument and magnitude. For the real continued fraction you mention, the algebraic expression allows the argument to be either $0$ (containing "$+sqrt{5}$") or $pi$ (containing "$-sqrt{5}$"). The same thing can be done with the complex version. Your two expressions have arguments $-pi/6$ and $7pi/6$. If your continued fraction had any hope of converging (which, as others have shown, it does not) it converges to something with a specific argument. You generically hope that one of the algebraic solutions you have has an argument matching the limit argument of the convergents. Finding that limit argument usually requires another computation, typically solving another recurrence.
answered 10 hours ago
Eric TowersEric Towers
32.8k22370
32.8k22370
add a comment |
add a comment |
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