Mixing
How can I find the real and imaginary parts of $sqrt{i+sqrt{i+sqrt{i+dots}}}$? [closed]
$begingroup$
By using $x = sqrt{i+sqrt{i+sqrt{i+dots}}}$ , I can find a quadratic equation, but how can I separate the real and imaginary parts?
Edit:
My question was how to find the real and imaginary parts of the complex number $sqrt{i+sqrt{i+sqrt{i+dots}}}$.
After solving the quadratic equation we'll get the solution - $x_{1,2} = frac{1}{2}pm frac{sqrt{1+4i}}{2}$. Now the prblem is to find the real and imaginary parts of $sqrt{1+4i}$.
Sorry for the confusion.
complex-analysis complex-numbers
asked Jan 24 at 8:18
AruAru
375
$endgroup$
closed as off-topic by Eevee Trainer, Andrei, Did, ΘΣΦGenSan, José Carlos Santos Jan 24 at 13:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Did, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
By using $x = sqrt{i+sqrt{i+sqrt{i+dots}}}$ , I can find a quadratic equation, but how can I separate the real and imaginary parts?
Edit:
My question was how to find the real and imaginary parts of the complex number $sqrt{i+sqrt{i+sqrt{i+dots}}}$.
After solving the quadratic equation we'll get the solution - $x_{1,2} = frac{1}{2}pm frac{sqrt{1+4i}}{2}$. Now the prblem is to find the real and imaginary parts of $sqrt{1+4i}$.
Sorry for the confusion.
complex-analysis complex-numbers
asked Jan 24 at 8:18
AruAru
375
$endgroup$
closed as off-topic by Eevee Trainer, Andrei, Did, ΘΣΦGenSan, José Carlos Santos Jan 24 at 13:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Did, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
First off, complex square roots aren't well-defined. So you have a problem there already: does that expression even make sense?
$endgroup$
– Arthur
Jan 24 at 8:24
$begingroup$
Previously discussed here: math.stackexchange.com/questions/447227/…
$endgroup$
– Chris Culter
Jan 24 at 8:33
3
$begingroup$
Possible duplicate of Find the value of $sqrt{i+sqrt{i+sqrt{i+dots}}}$ .
$endgroup$
– Andrei
Jan 24 at 8:35
$begingroup$
So your real question is about solving a quadratic equation, not about $sqrt{i+sqrt{i+sqrt{i+dots}}}$ ?
$endgroup$
– Martin R
Jan 24 at 8:56
$begingroup$
@MartinR No. The problem was how to solve $sqrt{1+4i}$ which will appear after solving the quadratic equation.
$endgroup$
– Aru
Jan 24 at 9:02
add a comment |
$begingroup$
By using $x = sqrt{i+sqrt{i+sqrt{i+dots}}}$ , I can find a quadratic equation, but how can I separate the real and imaginary parts?
Edit:
My question was how to find the real and imaginary parts of the complex number $sqrt{i+sqrt{i+sqrt{i+dots}}}$.
After solving the quadratic equation we'll get the solution - $x_{1,2} = frac{1}{2}pm frac{sqrt{1+4i}}{2}$. Now the prblem is to find the real and imaginary parts of $sqrt{1+4i}$.
Sorry for the confusion.
complex-analysis complex-numbers
asked Jan 24 at 8:18
AruAru
375
$endgroup$
By using $x = sqrt{i+sqrt{i+sqrt{i+dots}}}$ , I can find a quadratic equation, but how can I separate the real and imaginary parts?
Edit:
My question was how to find the real and imaginary parts of the complex number $sqrt{i+sqrt{i+sqrt{i+dots}}}$.
After solving the quadratic equation we'll get the solution - $x_{1,2} = frac{1}{2}pm frac{sqrt{1+4i}}{2}$. Now the prblem is to find the real and imaginary parts of $sqrt{1+4i}$.
Sorry for the confusion.
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Jan 24 at 8:18
AruAru
375
asked Jan 24 at 8:18
AruAru
375
edited Jan 24 at 9:10
Aru
asked Jan 24 at 8:18
AruAru
375
asked Jan 24 at 8:18
AruAru
375
asked Jan 24 at 8:18
AruAru
375
375
closed as off-topic by Eevee Trainer, Andrei, Did, ΘΣΦGenSan, José Carlos Santos Jan 24 at 13:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Did, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Eevee Trainer, Andrei, Did, ΘΣΦGenSan, José Carlos Santos Jan 24 at 13:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Did, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
First off, complex square roots aren't well-defined. So you have a problem there already: does that expression even make sense?
$endgroup$
– Arthur
Jan 24 at 8:24
$begingroup$
Previously discussed here: math.stackexchange.com/questions/447227/…
$endgroup$
– Chris Culter
Jan 24 at 8:33
3
$begingroup$
Possible duplicate of Find the value of $sqrt{i+sqrt{i+sqrt{i+dots}}}$ .
$endgroup$
– Andrei
Jan 24 at 8:35
$begingroup$
So your real question is about solving a quadratic equation, not about $sqrt{i+sqrt{i+sqrt{i+dots}}}$ ?
$endgroup$
– Martin R
Jan 24 at 8:56
$begingroup$
@MartinR No. The problem was how to solve $sqrt{1+4i}$ which will appear after solving the quadratic equation.
$endgroup$
– Aru
Jan 24 at 9:02
add a comment |
4
$begingroup$
First off, complex square roots aren't well-defined. So you have a problem there already: does that expression even make sense?
$endgroup$
– Arthur
Jan 24 at 8:24
$begingroup$
Previously discussed here: math.stackexchange.com/questions/447227/…
$endgroup$
– Chris Culter
Jan 24 at 8:33
3
$begingroup$
Possible duplicate of Find the value of $sqrt{i+sqrt{i+sqrt{i+dots}}}$ .
$endgroup$
– Andrei
Jan 24 at 8:35
$begingroup$
So your real question is about solving a quadratic equation, not about $sqrt{i+sqrt{i+sqrt{i+dots}}}$ ?
$endgroup$
– Martin R
Jan 24 at 8:56
$begingroup$
@MartinR No. The problem was how to solve $sqrt{1+4i}$ which will appear after solving the quadratic equation.
$endgroup$
– Aru
Jan 24 at 9:02
4
4
$begingroup$
First off, complex square roots aren't well-defined. So you have a problem there already: does that expression even make sense?
$endgroup$
– Arthur
Jan 24 at 8:24
$begingroup$
First off, complex square roots aren't well-defined. So you have a problem there already: does that expression even make sense?
$endgroup$
– Arthur
Jan 24 at 8:24
$begingroup$
Previously discussed here: math.stackexchange.com/questions/447227/…
$endgroup$
– Chris Culter
Jan 24 at 8:33
$begingroup$
Previously discussed here: math.stackexchange.com/questions/447227/…
$endgroup$
– Chris Culter
Jan 24 at 8:33
3
3
$begingroup$
Possible duplicate of Find the value of $sqrt{i+sqrt{i+sqrt{i+dots}}}$ .
$endgroup$
– Andrei
Jan 24 at 8:35
$begingroup$
Possible duplicate of Find the value of $sqrt{i+sqrt{i+sqrt{i+dots}}}$ .
$endgroup$
– Andrei
Jan 24 at 8:35
$begingroup$
So your real question is about solving a quadratic equation, not about $sqrt{i+sqrt{i+sqrt{i+dots}}}$ ?
$endgroup$
– Martin R
Jan 24 at 8:56
$begingroup$
So your real question is about solving a quadratic equation, not about $sqrt{i+sqrt{i+sqrt{i+dots}}}$ ?
$endgroup$
– Martin R
Jan 24 at 8:56
$begingroup$
@MartinR No. The problem was how to solve $sqrt{1+4i}$ which will appear after solving the quadratic equation.
$endgroup$
– Aru
Jan 24 at 9:02
$begingroup$
@MartinR No. The problem was how to solve $sqrt{1+4i}$ which will appear after solving the quadratic equation.
$endgroup$
– Aru
Jan 24 at 9:02
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
If we assume that the following transformation makes sense (which makes a few important assumptions to be discussed later):
$$ z = sqrt{i + sqrt{i + sqrt{ i + ldots}}} = sqrt{i + z}$$
Let's assume that square roots are OK to use in the context of complex numbers, and ignore some complicating ideas regarding the roots of unity. We are playing a game with numbers, and having a bit of fun with it. So we can square both sides then complete the square.
$$begin{split}
z^2 &= z + i \
z^2 - z - i &= 0 \
4z^2 - 4z - 4i &= 0 \
4z^2 - 4z + 1 - 1 - 4i &= 0 \
(2z - 1)^2 - (1 + 4i) &= 0 \
2z - 1 &= sqrt{1 + 4i} \
z &= frac1{2}(1 + sqrt{1 + 4i}) \
end{split}$$
You've probably already made it this far. (No $pm$ because we're only considering the positive root for simplicity.)
So let's talk about what it means to take the square root of an imaginary number.
Any complex number can be represented in two forms, a rectangular form and a polar form. The rectangular form is your standard $a + bi$ that you ought to be familiar with. Whereas the polar form can be represented as $r angle theta$, where $theta$ is the angle from the positive real axis.
We can convert between the two much in the same manner as standard conversion between rectangular and polar coordinates. In fact, there's a handy formula that helps with this that's rather famous, it's called Euler's formula.
$$e^{itheta} = cos(theta) + i sin(theta)$$
Which the general case of the more famous $e^{pi i} + 1 = 0$ equation, that you might have heard about. I'm not going to prove this formula, there are other resources that do this much better than I can using only text.
The bottom line is you can use this formula to represent complex numbers without hiding behind a magic $angle$ symbol. Now everything is numbers. $rangle theta = r e^{i theta}$.
So. What happens when two numbers are multiplied together? Well!
$$r_1 e^{i theta_1} cdot r_2 e^{i theta_2} = r_1 r_2 e^{i (theta_1 + theta_2)}$$
When two of them are multiplied together, there's two components of the multiplication. There's a stretching and a rotation. The stretching occurs in the same manner as real numbers, but the rotation is the part that corresponds to the complex component of the multiplication. The angles are added together.
Now you may have noticed something interesting. Because both sine and cosine are periodic functions, more than one representation is possible of the same number. So bearing that in mind, let's talk about the roots of unity.
What does $sqrt{i}$ mean? Well, one interpretation of it is to find a number that when multiplied by itself gives you $i$. But, we know from real numbers that this isn't necessarily a unique number. For instance, both -8 and 8 multiply themselves to give 64. But $sqrt{64}$ is 8. There's only one value. That's because we've decided by convention to take only the positive value of this square root. But for complex numbers, the only "positive" numbers, are those boring numbers with imaginary part 0 and real part > 0. So this definition doesn't hold as well. But let's ignore that for now! Let's just explore what happens when we try to find numbers.
We want to find $z$ such that $z^2 = i$. So let's start by expressing $i$ in polar coordinates.
$$i = e^{i frac{pi}{2}}$$
This value can be obtained by using Euler's formula. So we know that whatever number z is, it needs to produce this when multiplied by itself. Remember that:
$$r_1 e^{i theta_1} cdot r_2 e^{i theta_2} = r_1 r_2 e^{i (theta_1 + theta_2)}$$
We can see that $r_1 = r_2 = 1$, and we have $theta_1 + theta_2 = frac{pi}{2}$, which seems to imply (because $theta_1 = theta_2$) that $theta_1 = frac{pi}{4}$. However, if you remember about that catch where this complex number can have more than one representation. What if we decide instead to represent $i$ slightly differently.
$$i = e^{i(frac{pi}{2} + 2pi)}$$
Now, this is exactly the same number. Just a different way of representing it. However, now when we do the same procedure we get $theta = frac{5pi}{4}$ which is a very different number. This is why taking the square root is somewhat questionable when dealing with complex numbers.
But hey, we're here to have fun and do some math. Let's ignore that and say that the "correct" square root is the one performed when $-pi < theta le pi$. (This is known as the principle root.) We can do this because we're strong independent people, and to hell with the consequences.
So now let's use this understanding to tackle your question. How do we find the real part?
$$begin{split}
z &= frac1{2}(1 + sqrt{1 + 4i}) \
&= frac1{2}left(1 + sqrt{sqrt{17}e^{i tan^{-1}(4)}}right) \
&= frac1{2}left(1 + sqrt[4]{17}e^{frac{i}{2} tan^{-1}(4)}right) \
end{split}$$
Now we can put this back into rectangular form.
$$begin{split}
z &= frac1{2}(1 + sqrt[4]{17}e^{frac{i}{2} tan^{-1}(4)}) \
&= frac1{2}left(1 + sqrt[4]{17}left(cosleft(frac{tan^{-1}(4)}{2}right) + i sinleft(frac{tan^{-1}(4)}{2}right)right)right)
end{split}$$
Which finally gives us our real part that we wanted.
$$Re(z) = frac{1 + sqrt[4]{17}cosleft(frac{tan^{-1}(4)}{2}right)}{2}$$
(It should be remarked that this number is the same as the one you found below. $frac{1 + sqrt[4]{17}cosleft(frac{tan^{-1}(4)}{2}right)}{2} = frac1{2} + frac{sqrt{sqrt{17}+1}}{2sqrt{2}}$)
Now as a footnotes, let's talk about what assumptions were made to get to the first bit. What does it actually mean for an infinite nested radical to exist? Well, one way of looking at it is that it's a value such that the following operation converges on a fixed point. $z_{n+1} = sqrt{z_n + i}$. But that begs the question, what is the starting seed value? (What is $z_0$?) And from there we can learn a bit more. For instance, are there any seed values that converge on a different point (for instance, we only considered the principle value of the square root, what occurs when $z_0$ starts at one of the alternate values? What happens if we chose alternate values of the square root during iteration? Are there any seed values that fail to converge?
Of course, this is only one way of considering the problem, there are others, which is precisely why other users remarked that the question does not seem well posed. Which is true, but it is less helpful than it could have been.
answered Jan 24 at 9:08
OmnipotentEntityOmnipotentEntity
477318
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add a comment |
$begingroup$
Consider the recursion
$$x_0=0,qquad x_{n+1}:=sqrt{i+x_n}quad(ngeq0) ,$$
where $sqrt{cdot}$ denotes the principal value of the square root. It is then easy to see that all $x_n$ $(ngeq1)$ lie in the first quadrant of the complex plane. Solving the equation $x=sqrt{i+x}$, i.e., $x^2=i+x$, we obtain a single point in this quadrant, namely
$$xi={1+sqrt{1+4i}over2}=1.30024 + 0.624811 i .$$
We now have to prove that the $x_n$ converge to this $xi$. To this end we consider
$$x_{n+1}-xi=sqrt{i+x_n}-xi={i+x_n-xi^2oversqrt{i+x_n}+xi}={x_n-xiover x_{n+1}+xi} .$$
As both $x_{n+1}$ and $xi$ lie in the first quadrant we can conclude that $$|x_{n+1}+xi|>|xi|>1.3>1 .$$
This shows that the differences $x_n-xi$ converge geometrically to $0$, hence $lim_{ntoinfty}x_n=xi$.
answered Jan 24 at 8:54
Christian BlatterChristian Blatter
175k8115327
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add a comment |
$begingroup$
x = $sqrt{(i+sqrt{(i+dots)})}$ = $sqrt{(i+x)}$
$x^2 - x - i = 0$
Solving for x we get,
$x_{1,2} = frac{1}{2}pm frac{sqrt{1+4i}}{2}$
Now,
$a + ib =sqrt{1+4i}$
$implies a^2 - b^2 + 2iab = 1+4i$
So we have, $a^2-b^2 = 1$ and 2ab = 4.
Solving for a and b we can get
$a = sqrt(frac{sqrt{17}+1}{2})$ and $b = sqrt(frac{sqrt{17}-1}{2})$.
So the final answer is-
$x = frac{1}{2} + (frac{sqrt{sqrt{17}+1}}{2sqrt{2}}) + i (frac{sqrt{sqrt{17}-1}}{2sqrt{2}})$.
answered Jan 24 at 8:51
AruAru
375
$endgroup$
$begingroup$
This is a fixed point. What about the main part of the question, which is to show the convergence to this point?
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– Did
Jan 24 at 9:02
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@Did the question is not about showing the convergence of the expression.
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– Aru
Jan 24 at 9:13
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Interesting that you think so, since this (that is, establishing convergence) is the only mathematically relevant part of the problem.
$endgroup$
– Did
Jan 24 at 10:11
add a comment |
$begingroup$
Assuming that we use the principal value of the roots, we find, as you noted, the equation:
$$
sqrt{i+x}=x Rightarrow x^2-x-i=0
$$
that has the solution
$$
x=frac{1pmsqrt{1+4i}}{2}
$$
so your problem reduce to find the square root of $1+4i$. Do you know how to do this?
answered Jan 24 at 8:36
Emilio NovatiEmilio Novati
52.2k43474
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$begingroup$
Yes. Now I know how to do it.
$endgroup$
– Aru
Jan 24 at 8:55
$begingroup$
This is a fixed point. What about the main part of the question, which is to show the convergence to this point?
$endgroup$
– Did
Jan 24 at 9:02
add a comment |
$begingroup$
You get $x=sqrt{x+i}implies x^2-x-i=0implies x=frac{-(-1)pmsqrt{1^2-4(1)(-i)}}{2(1)}=frac{1pmsqrt{1+4i}}2$. Now, $1+4i=sqrt{17}e^{(tan^{-1}4 )cdot i}$. So $sqrt{1+4i}=sqrt[4]{17}e^{frac{(tan^{-1}4)cdot i}2}$.
Finally, $x=frac12 pmfrac{sqrt[4]{17}e^{frac{(tan^{-1}4)cdot i}2}}2$, from which it follows the real part is $frac12 pmfrac{sqrt[4]{17}}2cosfrac {tan^{-1}4}2$.
As to the convergence, see @Christian Blatter's answer.
answered Jan 24 at 9:25
Chris CusterChris Custer
14.2k3827
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add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If we assume that the following transformation makes sense (which makes a few important assumptions to be discussed later):
$$ z = sqrt{i + sqrt{i + sqrt{ i + ldots}}} = sqrt{i + z}$$
Let's assume that square roots are OK to use in the context of complex numbers, and ignore some complicating ideas regarding the roots of unity. We are playing a game with numbers, and having a bit of fun with it. So we can square both sides then complete the square.
$$begin{split}
z^2 &= z + i \
z^2 - z - i &= 0 \
4z^2 - 4z - 4i &= 0 \
4z^2 - 4z + 1 - 1 - 4i &= 0 \
(2z - 1)^2 - (1 + 4i) &= 0 \
2z - 1 &= sqrt{1 + 4i} \
z &= frac1{2}(1 + sqrt{1 + 4i}) \
end{split}$$
You've probably already made it this far. (No $pm$ because we're only considering the positive root for simplicity.)
So let's talk about what it means to take the square root of an imaginary number.
Any complex number can be represented in two forms, a rectangular form and a polar form. The rectangular form is your standard $a + bi$ that you ought to be familiar with. Whereas the polar form can be represented as $r angle theta$, where $theta$ is the angle from the positive real axis.
We can convert between the two much in the same manner as standard conversion between rectangular and polar coordinates. In fact, there's a handy formula that helps with this that's rather famous, it's called Euler's formula.
$$e^{itheta} = cos(theta) + i sin(theta)$$
Which the general case of the more famous $e^{pi i} + 1 = 0$ equation, that you might have heard about. I'm not going to prove this formula, there are other resources that do this much better than I can using only text.
The bottom line is you can use this formula to represent complex numbers without hiding behind a magic $angle$ symbol. Now everything is numbers. $rangle theta = r e^{i theta}$.
So. What happens when two numbers are multiplied together? Well!
$$r_1 e^{i theta_1} cdot r_2 e^{i theta_2} = r_1 r_2 e^{i (theta_1 + theta_2)}$$
When two of them are multiplied together, there's two components of the multiplication. There's a stretching and a rotation. The stretching occurs in the same manner as real numbers, but the rotation is the part that corresponds to the complex component of the multiplication. The angles are added together.
Now you may have noticed something interesting. Because both sine and cosine are periodic functions, more than one representation is possible of the same number. So bearing that in mind, let's talk about the roots of unity.
What does $sqrt{i}$ mean? Well, one interpretation of it is to find a number that when multiplied by itself gives you $i$. But, we know from real numbers that this isn't necessarily a unique number. For instance, both -8 and 8 multiply themselves to give 64. But $sqrt{64}$ is 8. There's only one value. That's because we've decided by convention to take only the positive value of this square root. But for complex numbers, the only "positive" numbers, are those boring numbers with imaginary part 0 and real part > 0. So this definition doesn't hold as well. But let's ignore that for now! Let's just explore what happens when we try to find numbers.
We want to find $z$ such that $z^2 = i$. So let's start by expressing $i$ in polar coordinates.
$$i = e^{i frac{pi}{2}}$$
This value can be obtained by using Euler's formula. So we know that whatever number z is, it needs to produce this when multiplied by itself. Remember that:
$$r_1 e^{i theta_1} cdot r_2 e^{i theta_2} = r_1 r_2 e^{i (theta_1 + theta_2)}$$
We can see that $r_1 = r_2 = 1$, and we have $theta_1 + theta_2 = frac{pi}{2}$, which seems to imply (because $theta_1 = theta_2$) that $theta_1 = frac{pi}{4}$. However, if you remember about that catch where this complex number can have more than one representation. What if we decide instead to represent $i$ slightly differently.
$$i = e^{i(frac{pi}{2} + 2pi)}$$
Now, this is exactly the same number. Just a different way of representing it. However, now when we do the same procedure we get $theta = frac{5pi}{4}$ which is a very different number. This is why taking the square root is somewhat questionable when dealing with complex numbers.
But hey, we're here to have fun and do some math. Let's ignore that and say that the "correct" square root is the one performed when $-pi < theta le pi$. (This is known as the principle root.) We can do this because we're strong independent people, and to hell with the consequences.
So now let's use this understanding to tackle your question. How do we find the real part?
$$begin{split}
z &= frac1{2}(1 + sqrt{1 + 4i}) \
&= frac1{2}left(1 + sqrt{sqrt{17}e^{i tan^{-1}(4)}}right) \
&= frac1{2}left(1 + sqrt[4]{17}e^{frac{i}{2} tan^{-1}(4)}right) \
end{split}$$
Now we can put this back into rectangular form.
$$begin{split}
z &= frac1{2}(1 + sqrt[4]{17}e^{frac{i}{2} tan^{-1}(4)}) \
&= frac1{2}left(1 + sqrt[4]{17}left(cosleft(frac{tan^{-1}(4)}{2}right) + i sinleft(frac{tan^{-1}(4)}{2}right)right)right)
end{split}$$
Which finally gives us our real part that we wanted.
$$Re(z) = frac{1 + sqrt[4]{17}cosleft(frac{tan^{-1}(4)}{2}right)}{2}$$
(It should be remarked that this number is the same as the one you found below. $frac{1 + sqrt[4]{17}cosleft(frac{tan^{-1}(4)}{2}right)}{2} = frac1{2} + frac{sqrt{sqrt{17}+1}}{2sqrt{2}}$)
Now as a footnotes, let's talk about what assumptions were made to get to the first bit. What does it actually mean for an infinite nested radical to exist? Well, one way of looking at it is that it's a value such that the following operation converges on a fixed point. $z_{n+1} = sqrt{z_n + i}$. But that begs the question, what is the starting seed value? (What is $z_0$?) And from there we can learn a bit more. For instance, are there any seed values that converge on a different point (for instance, we only considered the principle value of the square root, what occurs when $z_0$ starts at one of the alternate values? What happens if we chose alternate values of the square root during iteration? Are there any seed values that fail to converge?
Of course, this is only one way of considering the problem, there are others, which is precisely why other users remarked that the question does not seem well posed. Which is true, but it is less helpful than it could have been.
answered Jan 24 at 9:08
OmnipotentEntityOmnipotentEntity
477318
$endgroup$
add a comment |
$begingroup$
If we assume that the following transformation makes sense (which makes a few important assumptions to be discussed later):
$$ z = sqrt{i + sqrt{i + sqrt{ i + ldots}}} = sqrt{i + z}$$
Let's assume that square roots are OK to use in the context of complex numbers, and ignore some complicating ideas regarding the roots of unity. We are playing a game with numbers, and having a bit of fun with it. So we can square both sides then complete the square.
$$begin{split}
z^2 &= z + i \
z^2 - z - i &= 0 \
4z^2 - 4z - 4i &= 0 \
4z^2 - 4z + 1 - 1 - 4i &= 0 \
(2z - 1)^2 - (1 + 4i) &= 0 \
2z - 1 &= sqrt{1 + 4i} \
z &= frac1{2}(1 + sqrt{1 + 4i}) \
end{split}$$
You've probably already made it this far. (No $pm$ because we're only considering the positive root for simplicity.)
So let's talk about what it means to take the square root of an imaginary number.
Any complex number can be represented in two forms, a rectangular form and a polar form. The rectangular form is your standard $a + bi$ that you ought to be familiar with. Whereas the polar form can be represented as $r angle theta$, where $theta$ is the angle from the positive real axis.
We can convert between the two much in the same manner as standard conversion between rectangular and polar coordinates. In fact, there's a handy formula that helps with this that's rather famous, it's called Euler's formula.
$$e^{itheta} = cos(theta) + i sin(theta)$$
Which the general case of the more famous $e^{pi i} + 1 = 0$ equation, that you might have heard about. I'm not going to prove this formula, there are other resources that do this much better than I can using only text.
The bottom line is you can use this formula to represent complex numbers without hiding behind a magic $angle$ symbol. Now everything is numbers. $rangle theta = r e^{i theta}$.
So. What happens when two numbers are multiplied together? Well!
$$r_1 e^{i theta_1} cdot r_2 e^{i theta_2} = r_1 r_2 e^{i (theta_1 + theta_2)}$$
When two of them are multiplied together, there's two components of the multiplication. There's a stretching and a rotation. The stretching occurs in the same manner as real numbers, but the rotation is the part that corresponds to the complex component of the multiplication. The angles are added together.
Now you may have noticed something interesting. Because both sine and cosine are periodic functions, more than one representation is possible of the same number. So bearing that in mind, let's talk about the roots of unity.
What does $sqrt{i}$ mean? Well, one interpretation of it is to find a number that when multiplied by itself gives you $i$. But, we know from real numbers that this isn't necessarily a unique number. For instance, both -8 and 8 multiply themselves to give 64. But $sqrt{64}$ is 8. There's only one value. That's because we've decided by convention to take only the positive value of this square root. But for complex numbers, the only "positive" numbers, are those boring numbers with imaginary part 0 and real part > 0. So this definition doesn't hold as well. But let's ignore that for now! Let's just explore what happens when we try to find numbers.
We want to find $z$ such that $z^2 = i$. So let's start by expressing $i$ in polar coordinates.
$$i = e^{i frac{pi}{2}}$$
This value can be obtained by using Euler's formula. So we know that whatever number z is, it needs to produce this when multiplied by itself. Remember that:
$$r_1 e^{i theta_1} cdot r_2 e^{i theta_2} = r_1 r_2 e^{i (theta_1 + theta_2)}$$
We can see that $r_1 = r_2 = 1$, and we have $theta_1 + theta_2 = frac{pi}{2}$, which seems to imply (because $theta_1 = theta_2$) that $theta_1 = frac{pi}{4}$. However, if you remember about that catch where this complex number can have more than one representation. What if we decide instead to represent $i$ slightly differently.
$$i = e^{i(frac{pi}{2} + 2pi)}$$
Now, this is exactly the same number. Just a different way of representing it. However, now when we do the same procedure we get $theta = frac{5pi}{4}$ which is a very different number. This is why taking the square root is somewhat questionable when dealing with complex numbers.
But hey, we're here to have fun and do some math. Let's ignore that and say that the "correct" square root is the one performed when $-pi < theta le pi$. (This is known as the principle root.) We can do this because we're strong independent people, and to hell with the consequences.
So now let's use this understanding to tackle your question. How do we find the real part?
$$begin{split}
z &= frac1{2}(1 + sqrt{1 + 4i}) \
&= frac1{2}left(1 + sqrt{sqrt{17}e^{i tan^{-1}(4)}}right) \
&= frac1{2}left(1 + sqrt[4]{17}e^{frac{i}{2} tan^{-1}(4)}right) \
end{split}$$
Now we can put this back into rectangular form.
$$begin{split}
z &= frac1{2}(1 + sqrt[4]{17}e^{frac{i}{2} tan^{-1}(4)}) \
&= frac1{2}left(1 + sqrt[4]{17}left(cosleft(frac{tan^{-1}(4)}{2}right) + i sinleft(frac{tan^{-1}(4)}{2}right)right)right)
end{split}$$
Which finally gives us our real part that we wanted.
$$Re(z) = frac{1 + sqrt[4]{17}cosleft(frac{tan^{-1}(4)}{2}right)}{2}$$
(It should be remarked that this number is the same as the one you found below. $frac{1 + sqrt[4]{17}cosleft(frac{tan^{-1}(4)}{2}right)}{2} = frac1{2} + frac{sqrt{sqrt{17}+1}}{2sqrt{2}}$)
Now as a footnotes, let's talk about what assumptions were made to get to the first bit. What does it actually mean for an infinite nested radical to exist? Well, one way of looking at it is that it's a value such that the following operation converges on a fixed point. $z_{n+1} = sqrt{z_n + i}$. But that begs the question, what is the starting seed value? (What is $z_0$?) And from there we can learn a bit more. For instance, are there any seed values that converge on a different point (for instance, we only considered the principle value of the square root, what occurs when $z_0$ starts at one of the alternate values? What happens if we chose alternate values of the square root during iteration? Are there any seed values that fail to converge?
Of course, this is only one way of considering the problem, there are others, which is precisely why other users remarked that the question does not seem well posed. Which is true, but it is less helpful than it could have been.
answered Jan 24 at 9:08
OmnipotentEntityOmnipotentEntity
477318
$endgroup$
add a comment |
$begingroup$
If we assume that the following transformation makes sense (which makes a few important assumptions to be discussed later):
$$ z = sqrt{i + sqrt{i + sqrt{ i + ldots}}} = sqrt{i + z}$$
Let's assume that square roots are OK to use in the context of complex numbers, and ignore some complicating ideas regarding the roots of unity. We are playing a game with numbers, and having a bit of fun with it. So we can square both sides then complete the square.
$$begin{split}
z^2 &= z + i \
z^2 - z - i &= 0 \
4z^2 - 4z - 4i &= 0 \
4z^2 - 4z + 1 - 1 - 4i &= 0 \
(2z - 1)^2 - (1 + 4i) &= 0 \
2z - 1 &= sqrt{1 + 4i} \
z &= frac1{2}(1 + sqrt{1 + 4i}) \
end{split}$$
You've probably already made it this far. (No $pm$ because we're only considering the positive root for simplicity.)
So let's talk about what it means to take the square root of an imaginary number.
Any complex number can be represented in two forms, a rectangular form and a polar form. The rectangular form is your standard $a + bi$ that you ought to be familiar with. Whereas the polar form can be represented as $r angle theta$, where $theta$ is the angle from the positive real axis.
We can convert between the two much in the same manner as standard conversion between rectangular and polar coordinates. In fact, there's a handy formula that helps with this that's rather famous, it's called Euler's formula.
$$e^{itheta} = cos(theta) + i sin(theta)$$
Which the general case of the more famous $e^{pi i} + 1 = 0$ equation, that you might have heard about. I'm not going to prove this formula, there are other resources that do this much better than I can using only text.
The bottom line is you can use this formula to represent complex numbers without hiding behind a magic $angle$ symbol. Now everything is numbers. $rangle theta = r e^{i theta}$.
So. What happens when two numbers are multiplied together? Well!
$$r_1 e^{i theta_1} cdot r_2 e^{i theta_2} = r_1 r_2 e^{i (theta_1 + theta_2)}$$
When two of them are multiplied together, there's two components of the multiplication. There's a stretching and a rotation. The stretching occurs in the same manner as real numbers, but the rotation is the part that corresponds to the complex component of the multiplication. The angles are added together.
Now you may have noticed something interesting. Because both sine and cosine are periodic functions, more than one representation is possible of the same number. So bearing that in mind, let's talk about the roots of unity.
What does $sqrt{i}$ mean? Well, one interpretation of it is to find a number that when multiplied by itself gives you $i$. But, we know from real numbers that this isn't necessarily a unique number. For instance, both -8 and 8 multiply themselves to give 64. But $sqrt{64}$ is 8. There's only one value. That's because we've decided by convention to take only the positive value of this square root. But for complex numbers, the only "positive" numbers, are those boring numbers with imaginary part 0 and real part > 0. So this definition doesn't hold as well. But let's ignore that for now! Let's just explore what happens when we try to find numbers.
We want to find $z$ such that $z^2 = i$. So let's start by expressing $i$ in polar coordinates.
$$i = e^{i frac{pi}{2}}$$
This value can be obtained by using Euler's formula. So we know that whatever number z is, it needs to produce this when multiplied by itself. Remember that:
$$r_1 e^{i theta_1} cdot r_2 e^{i theta_2} = r_1 r_2 e^{i (theta_1 + theta_2)}$$
We can see that $r_1 = r_2 = 1$, and we have $theta_1 + theta_2 = frac{pi}{2}$, which seems to imply (because $theta_1 = theta_2$) that $theta_1 = frac{pi}{4}$. However, if you remember about that catch where this complex number can have more than one representation. What if we decide instead to represent $i$ slightly differently.
$$i = e^{i(frac{pi}{2} + 2pi)}$$
Now, this is exactly the same number. Just a different way of representing it. However, now when we do the same procedure we get $theta = frac{5pi}{4}$ which is a very different number. This is why taking the square root is somewhat questionable when dealing with complex numbers.
But hey, we're here to have fun and do some math. Let's ignore that and say that the "correct" square root is the one performed when $-pi < theta le pi$. (This is known as the principle root.) We can do this because we're strong independent people, and to hell with the consequences.
So now let's use this understanding to tackle your question. How do we find the real part?
$$begin{split}
z &= frac1{2}(1 + sqrt{1 + 4i}) \
&= frac1{2}left(1 + sqrt{sqrt{17}e^{i tan^{-1}(4)}}right) \
&= frac1{2}left(1 + sqrt[4]{17}e^{frac{i}{2} tan^{-1}(4)}right) \
end{split}$$
Now we can put this back into rectangular form.
$$begin{split}
z &= frac1{2}(1 + sqrt[4]{17}e^{frac{i}{2} tan^{-1}(4)}) \
&= frac1{2}left(1 + sqrt[4]{17}left(cosleft(frac{tan^{-1}(4)}{2}right) + i sinleft(frac{tan^{-1}(4)}{2}right)right)right)
end{split}$$
Which finally gives us our real part that we wanted.
$$Re(z) = frac{1 + sqrt[4]{17}cosleft(frac{tan^{-1}(4)}{2}right)}{2}$$
(It should be remarked that this number is the same as the one you found below. $frac{1 + sqrt[4]{17}cosleft(frac{tan^{-1}(4)}{2}right)}{2} = frac1{2} + frac{sqrt{sqrt{17}+1}}{2sqrt{2}}$)
Now as a footnotes, let's talk about what assumptions were made to get to the first bit. What does it actually mean for an infinite nested radical to exist? Well, one way of looking at it is that it's a value such that the following operation converges on a fixed point. $z_{n+1} = sqrt{z_n + i}$. But that begs the question, what is the starting seed value? (What is $z_0$?) And from there we can learn a bit more. For instance, are there any seed values that converge on a different point (for instance, we only considered the principle value of the square root, what occurs when $z_0$ starts at one of the alternate values? What happens if we chose alternate values of the square root during iteration? Are there any seed values that fail to converge?
Of course, this is only one way of considering the problem, there are others, which is precisely why other users remarked that the question does not seem well posed. Which is true, but it is less helpful than it could have been.
answered Jan 24 at 9:08
OmnipotentEntityOmnipotentEntity
477318
$endgroup$
If we assume that the following transformation makes sense (which makes a few important assumptions to be discussed later):
$$ z = sqrt{i + sqrt{i + sqrt{ i + ldots}}} = sqrt{i + z}$$
Let's assume that square roots are OK to use in the context of complex numbers, and ignore some complicating ideas regarding the roots of unity. We are playing a game with numbers, and having a bit of fun with it. So we can square both sides then complete the square.
$$begin{split}
z^2 &= z + i \
z^2 - z - i &= 0 \
4z^2 - 4z - 4i &= 0 \
4z^2 - 4z + 1 - 1 - 4i &= 0 \
(2z - 1)^2 - (1 + 4i) &= 0 \
2z - 1 &= sqrt{1 + 4i} \
z &= frac1{2}(1 + sqrt{1 + 4i}) \
end{split}$$
You've probably already made it this far. (No $pm$ because we're only considering the positive root for simplicity.)
So let's talk about what it means to take the square root of an imaginary number.
Any complex number can be represented in two forms, a rectangular form and a polar form. The rectangular form is your standard $a + bi$ that you ought to be familiar with. Whereas the polar form can be represented as $r angle theta$, where $theta$ is the angle from the positive real axis.
We can convert between the two much in the same manner as standard conversion between rectangular and polar coordinates. In fact, there's a handy formula that helps with this that's rather famous, it's called Euler's formula.
$$e^{itheta} = cos(theta) + i sin(theta)$$
Which the general case of the more famous $e^{pi i} + 1 = 0$ equation, that you might have heard about. I'm not going to prove this formula, there are other resources that do this much better than I can using only text.
The bottom line is you can use this formula to represent complex numbers without hiding behind a magic $angle$ symbol. Now everything is numbers. $rangle theta = r e^{i theta}$.
So. What happens when two numbers are multiplied together? Well!
$$r_1 e^{i theta_1} cdot r_2 e^{i theta_2} = r_1 r_2 e^{i (theta_1 + theta_2)}$$
When two of them are multiplied together, there's two components of the multiplication. There's a stretching and a rotation. The stretching occurs in the same manner as real numbers, but the rotation is the part that corresponds to the complex component of the multiplication. The angles are added together.
Now you may have noticed something interesting. Because both sine and cosine are periodic functions, more than one representation is possible of the same number. So bearing that in mind, let's talk about the roots of unity.
What does $sqrt{i}$ mean? Well, one interpretation of it is to find a number that when multiplied by itself gives you $i$. But, we know from real numbers that this isn't necessarily a unique number. For instance, both -8 and 8 multiply themselves to give 64. But $sqrt{64}$ is 8. There's only one value. That's because we've decided by convention to take only the positive value of this square root. But for complex numbers, the only "positive" numbers, are those boring numbers with imaginary part 0 and real part > 0. So this definition doesn't hold as well. But let's ignore that for now! Let's just explore what happens when we try to find numbers.
We want to find $z$ such that $z^2 = i$. So let's start by expressing $i$ in polar coordinates.
$$i = e^{i frac{pi}{2}}$$
This value can be obtained by using Euler's formula. So we know that whatever number z is, it needs to produce this when multiplied by itself. Remember that:
$$r_1 e^{i theta_1} cdot r_2 e^{i theta_2} = r_1 r_2 e^{i (theta_1 + theta_2)}$$
We can see that $r_1 = r_2 = 1$, and we have $theta_1 + theta_2 = frac{pi}{2}$, which seems to imply (because $theta_1 = theta_2$) that $theta_1 = frac{pi}{4}$. However, if you remember about that catch where this complex number can have more than one representation. What if we decide instead to represent $i$ slightly differently.
$$i = e^{i(frac{pi}{2} + 2pi)}$$
Now, this is exactly the same number. Just a different way of representing it. However, now when we do the same procedure we get $theta = frac{5pi}{4}$ which is a very different number. This is why taking the square root is somewhat questionable when dealing with complex numbers.
But hey, we're here to have fun and do some math. Let's ignore that and say that the "correct" square root is the one performed when $-pi < theta le pi$. (This is known as the principle root.) We can do this because we're strong independent people, and to hell with the consequences.
So now let's use this understanding to tackle your question. How do we find the real part?
$$begin{split}
z &= frac1{2}(1 + sqrt{1 + 4i}) \
&= frac1{2}left(1 + sqrt{sqrt{17}e^{i tan^{-1}(4)}}right) \
&= frac1{2}left(1 + sqrt[4]{17}e^{frac{i}{2} tan^{-1}(4)}right) \
end{split}$$
Now we can put this back into rectangular form.
$$begin{split}
z &= frac1{2}(1 + sqrt[4]{17}e^{frac{i}{2} tan^{-1}(4)}) \
&= frac1{2}left(1 + sqrt[4]{17}left(cosleft(frac{tan^{-1}(4)}{2}right) + i sinleft(frac{tan^{-1}(4)}{2}right)right)right)
end{split}$$
Which finally gives us our real part that we wanted.
$$Re(z) = frac{1 + sqrt[4]{17}cosleft(frac{tan^{-1}(4)}{2}right)}{2}$$
(It should be remarked that this number is the same as the one you found below. $frac{1 + sqrt[4]{17}cosleft(frac{tan^{-1}(4)}{2}right)}{2} = frac1{2} + frac{sqrt{sqrt{17}+1}}{2sqrt{2}}$)
Now as a footnotes, let's talk about what assumptions were made to get to the first bit. What does it actually mean for an infinite nested radical to exist? Well, one way of looking at it is that it's a value such that the following operation converges on a fixed point. $z_{n+1} = sqrt{z_n + i}$. But that begs the question, what is the starting seed value? (What is $z_0$?) And from there we can learn a bit more. For instance, are there any seed values that converge on a different point (for instance, we only considered the principle value of the square root, what occurs when $z_0$ starts at one of the alternate values? What happens if we chose alternate values of the square root during iteration? Are there any seed values that fail to converge?
Of course, this is only one way of considering the problem, there are others, which is precisely why other users remarked that the question does not seem well posed. Which is true, but it is less helpful than it could have been.
answered Jan 24 at 9:08
OmnipotentEntityOmnipotentEntity
477318
edited Jan 31 at 17:23
answered Jan 24 at 9:08
OmnipotentEntityOmnipotentEntity
477318
answered Jan 24 at 9:08
OmnipotentEntityOmnipotentEntity
477318
answered Jan 24 at 9:08
OmnipotentEntityOmnipotentEntity
477318
477318
add a comment |
add a comment |
$begingroup$
Consider the recursion
$$x_0=0,qquad x_{n+1}:=sqrt{i+x_n}quad(ngeq0) ,$$
where $sqrt{cdot}$ denotes the principal value of the square root. It is then easy to see that all $x_n$ $(ngeq1)$ lie in the first quadrant of the complex plane. Solving the equation $x=sqrt{i+x}$, i.e., $x^2=i+x$, we obtain a single point in this quadrant, namely
$$xi={1+sqrt{1+4i}over2}=1.30024 + 0.624811 i .$$
We now have to prove that the $x_n$ converge to this $xi$. To this end we consider
$$x_{n+1}-xi=sqrt{i+x_n}-xi={i+x_n-xi^2oversqrt{i+x_n}+xi}={x_n-xiover x_{n+1}+xi} .$$
As both $x_{n+1}$ and $xi$ lie in the first quadrant we can conclude that $$|x_{n+1}+xi|>|xi|>1.3>1 .$$
This shows that the differences $x_n-xi$ converge geometrically to $0$, hence $lim_{ntoinfty}x_n=xi$.
answered Jan 24 at 8:54
Christian BlatterChristian Blatter
175k8115327
$endgroup$
add a comment |
$begingroup$
Consider the recursion
$$x_0=0,qquad x_{n+1}:=sqrt{i+x_n}quad(ngeq0) ,$$
where $sqrt{cdot}$ denotes the principal value of the square root. It is then easy to see that all $x_n$ $(ngeq1)$ lie in the first quadrant of the complex plane. Solving the equation $x=sqrt{i+x}$, i.e., $x^2=i+x$, we obtain a single point in this quadrant, namely
$$xi={1+sqrt{1+4i}over2}=1.30024 + 0.624811 i .$$
We now have to prove that the $x_n$ converge to this $xi$. To this end we consider
$$x_{n+1}-xi=sqrt{i+x_n}-xi={i+x_n-xi^2oversqrt{i+x_n}+xi}={x_n-xiover x_{n+1}+xi} .$$
As both $x_{n+1}$ and $xi$ lie in the first quadrant we can conclude that $$|x_{n+1}+xi|>|xi|>1.3>1 .$$
This shows that the differences $x_n-xi$ converge geometrically to $0$, hence $lim_{ntoinfty}x_n=xi$.
answered Jan 24 at 8:54
Christian BlatterChristian Blatter
175k8115327
$endgroup$
add a comment |
$begingroup$
Consider the recursion
$$x_0=0,qquad x_{n+1}:=sqrt{i+x_n}quad(ngeq0) ,$$
where $sqrt{cdot}$ denotes the principal value of the square root. It is then easy to see that all $x_n$ $(ngeq1)$ lie in the first quadrant of the complex plane. Solving the equation $x=sqrt{i+x}$, i.e., $x^2=i+x$, we obtain a single point in this quadrant, namely
$$xi={1+sqrt{1+4i}over2}=1.30024 + 0.624811 i .$$
We now have to prove that the $x_n$ converge to this $xi$. To this end we consider
$$x_{n+1}-xi=sqrt{i+x_n}-xi={i+x_n-xi^2oversqrt{i+x_n}+xi}={x_n-xiover x_{n+1}+xi} .$$
As both $x_{n+1}$ and $xi$ lie in the first quadrant we can conclude that $$|x_{n+1}+xi|>|xi|>1.3>1 .$$
This shows that the differences $x_n-xi$ converge geometrically to $0$, hence $lim_{ntoinfty}x_n=xi$.
answered Jan 24 at 8:54
Christian BlatterChristian Blatter
175k8115327
$endgroup$
Consider the recursion
$$x_0=0,qquad x_{n+1}:=sqrt{i+x_n}quad(ngeq0) ,$$
where $sqrt{cdot}$ denotes the principal value of the square root. It is then easy to see that all $x_n$ $(ngeq1)$ lie in the first quadrant of the complex plane. Solving the equation $x=sqrt{i+x}$, i.e., $x^2=i+x$, we obtain a single point in this quadrant, namely
$$xi={1+sqrt{1+4i}over2}=1.30024 + 0.624811 i .$$
We now have to prove that the $x_n$ converge to this $xi$. To this end we consider
$$x_{n+1}-xi=sqrt{i+x_n}-xi={i+x_n-xi^2oversqrt{i+x_n}+xi}={x_n-xiover x_{n+1}+xi} .$$
As both $x_{n+1}$ and $xi$ lie in the first quadrant we can conclude that $$|x_{n+1}+xi|>|xi|>1.3>1 .$$
This shows that the differences $x_n-xi$ converge geometrically to $0$, hence $lim_{ntoinfty}x_n=xi$.
answered Jan 24 at 8:54
Christian BlatterChristian Blatter
175k8115327
answered Jan 24 at 8:54
Christian BlatterChristian Blatter
175k8115327
answered Jan 24 at 8:54
Christian BlatterChristian Blatter
175k8115327
answered Jan 24 at 8:54
Christian BlatterChristian Blatter
175k8115327
175k8115327
add a comment |
add a comment |
$begingroup$
x = $sqrt{(i+sqrt{(i+dots)})}$ = $sqrt{(i+x)}$
$x^2 - x - i = 0$
Solving for x we get,
$x_{1,2} = frac{1}{2}pm frac{sqrt{1+4i}}{2}$
Now,
$a + ib =sqrt{1+4i}$
$implies a^2 - b^2 + 2iab = 1+4i$
So we have, $a^2-b^2 = 1$ and 2ab = 4.
Solving for a and b we can get
$a = sqrt(frac{sqrt{17}+1}{2})$ and $b = sqrt(frac{sqrt{17}-1}{2})$.
So the final answer is-
$x = frac{1}{2} + (frac{sqrt{sqrt{17}+1}}{2sqrt{2}}) + i (frac{sqrt{sqrt{17}-1}}{2sqrt{2}})$.
answered Jan 24 at 8:51
AruAru
375
$endgroup$
$begingroup$
This is a fixed point. What about the main part of the question, which is to show the convergence to this point?
$endgroup$
– Did
Jan 24 at 9:02
$begingroup$
@Did the question is not about showing the convergence of the expression.
$endgroup$
– Aru
Jan 24 at 9:13
$begingroup$
Interesting that you think so, since this (that is, establishing convergence) is the only mathematically relevant part of the problem.
$endgroup$
– Did
Jan 24 at 10:11
add a comment |
$begingroup$
x = $sqrt{(i+sqrt{(i+dots)})}$ = $sqrt{(i+x)}$
$x^2 - x - i = 0$
Solving for x we get,
$x_{1,2} = frac{1}{2}pm frac{sqrt{1+4i}}{2}$
Now,
$a + ib =sqrt{1+4i}$
$implies a^2 - b^2 + 2iab = 1+4i$
So we have, $a^2-b^2 = 1$ and 2ab = 4.
Solving for a and b we can get
$a = sqrt(frac{sqrt{17}+1}{2})$ and $b = sqrt(frac{sqrt{17}-1}{2})$.
So the final answer is-
$x = frac{1}{2} + (frac{sqrt{sqrt{17}+1}}{2sqrt{2}}) + i (frac{sqrt{sqrt{17}-1}}{2sqrt{2}})$.
answered Jan 24 at 8:51
AruAru
375
$endgroup$
$begingroup$
This is a fixed point. What about the main part of the question, which is to show the convergence to this point?
$endgroup$
– Did
Jan 24 at 9:02
$begingroup$
@Did the question is not about showing the convergence of the expression.
$endgroup$
– Aru
Jan 24 at 9:13
$begingroup$
Interesting that you think so, since this (that is, establishing convergence) is the only mathematically relevant part of the problem.
$endgroup$
– Did
Jan 24 at 10:11
add a comment |
$begingroup$
x = $sqrt{(i+sqrt{(i+dots)})}$ = $sqrt{(i+x)}$
$x^2 - x - i = 0$
Solving for x we get,
$x_{1,2} = frac{1}{2}pm frac{sqrt{1+4i}}{2}$
Now,
$a + ib =sqrt{1+4i}$
$implies a^2 - b^2 + 2iab = 1+4i$
So we have, $a^2-b^2 = 1$ and 2ab = 4.
Solving for a and b we can get
$a = sqrt(frac{sqrt{17}+1}{2})$ and $b = sqrt(frac{sqrt{17}-1}{2})$.
So the final answer is-
$x = frac{1}{2} + (frac{sqrt{sqrt{17}+1}}{2sqrt{2}}) + i (frac{sqrt{sqrt{17}-1}}{2sqrt{2}})$.
answered Jan 24 at 8:51
AruAru
375
$endgroup$
x = $sqrt{(i+sqrt{(i+dots)})}$ = $sqrt{(i+x)}$
$x^2 - x - i = 0$
Solving for x we get,
$x_{1,2} = frac{1}{2}pm frac{sqrt{1+4i}}{2}$
Now,
$a + ib =sqrt{1+4i}$
$implies a^2 - b^2 + 2iab = 1+4i$
So we have, $a^2-b^2 = 1$ and 2ab = 4.
Solving for a and b we can get
$a = sqrt(frac{sqrt{17}+1}{2})$ and $b = sqrt(frac{sqrt{17}-1}{2})$.
So the final answer is-
$x = frac{1}{2} + (frac{sqrt{sqrt{17}+1}}{2sqrt{2}}) + i (frac{sqrt{sqrt{17}-1}}{2sqrt{2}})$.
answered Jan 24 at 8:51
AruAru
375
answered Jan 24 at 8:51
AruAru
375
answered Jan 24 at 8:51
AruAru
375
answered Jan 24 at 8:51
AruAru
375
375
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This is a fixed point. What about the main part of the question, which is to show the convergence to this point?
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– Did
Jan 24 at 9:02
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@Did the question is not about showing the convergence of the expression.
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– Aru
Jan 24 at 9:13
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Interesting that you think so, since this (that is, establishing convergence) is the only mathematically relevant part of the problem.
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– Did
Jan 24 at 10:11
add a comment |
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This is a fixed point. What about the main part of the question, which is to show the convergence to this point?
$endgroup$
– Did
Jan 24 at 9:02
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@Did the question is not about showing the convergence of the expression.
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– Aru
Jan 24 at 9:13
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Interesting that you think so, since this (that is, establishing convergence) is the only mathematically relevant part of the problem.
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– Did
Jan 24 at 10:11
$begingroup$
This is a fixed point. What about the main part of the question, which is to show the convergence to this point?
$endgroup$
– Did
Jan 24 at 9:02
$begingroup$
This is a fixed point. What about the main part of the question, which is to show the convergence to this point?
$endgroup$
– Did
Jan 24 at 9:02
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@Did the question is not about showing the convergence of the expression.
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– Aru
Jan 24 at 9:13
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@Did the question is not about showing the convergence of the expression.
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– Aru
Jan 24 at 9:13
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Interesting that you think so, since this (that is, establishing convergence) is the only mathematically relevant part of the problem.
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– Did
Jan 24 at 10:11
$begingroup$
Interesting that you think so, since this (that is, establishing convergence) is the only mathematically relevant part of the problem.
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– Did
Jan 24 at 10:11
add a comment |
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Assuming that we use the principal value of the roots, we find, as you noted, the equation:
$$
sqrt{i+x}=x Rightarrow x^2-x-i=0
$$
that has the solution
$$
x=frac{1pmsqrt{1+4i}}{2}
$$
so your problem reduce to find the square root of $1+4i$. Do you know how to do this?
answered Jan 24 at 8:36
Emilio NovatiEmilio Novati
52.2k43474
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Yes. Now I know how to do it.
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– Aru
Jan 24 at 8:55
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This is a fixed point. What about the main part of the question, which is to show the convergence to this point?
$endgroup$
– Did
Jan 24 at 9:02
add a comment |
$begingroup$
Assuming that we use the principal value of the roots, we find, as you noted, the equation:
$$
sqrt{i+x}=x Rightarrow x^2-x-i=0
$$
that has the solution
$$
x=frac{1pmsqrt{1+4i}}{2}
$$
so your problem reduce to find the square root of $1+4i$. Do you know how to do this?
answered Jan 24 at 8:36
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
$begingroup$
Yes. Now I know how to do it.
$endgroup$
– Aru
Jan 24 at 8:55
$begingroup$
This is a fixed point. What about the main part of the question, which is to show the convergence to this point?
$endgroup$
– Did
Jan 24 at 9:02
add a comment |
$begingroup$
Assuming that we use the principal value of the roots, we find, as you noted, the equation:
$$
sqrt{i+x}=x Rightarrow x^2-x-i=0
$$
that has the solution
$$
x=frac{1pmsqrt{1+4i}}{2}
$$
so your problem reduce to find the square root of $1+4i$. Do you know how to do this?
answered Jan 24 at 8:36
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
Assuming that we use the principal value of the roots, we find, as you noted, the equation:
$$
sqrt{i+x}=x Rightarrow x^2-x-i=0
$$
that has the solution
$$
x=frac{1pmsqrt{1+4i}}{2}
$$
so your problem reduce to find the square root of $1+4i$. Do you know how to do this?
answered Jan 24 at 8:36
Emilio NovatiEmilio Novati
52.2k43474
answered Jan 24 at 8:36
Emilio NovatiEmilio Novati
52.2k43474
answered Jan 24 at 8:36
Emilio NovatiEmilio Novati
52.2k43474
answered Jan 24 at 8:36
Emilio NovatiEmilio Novati
52.2k43474
52.2k43474
$begingroup$
Yes. Now I know how to do it.
$endgroup$
– Aru
Jan 24 at 8:55
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This is a fixed point. What about the main part of the question, which is to show the convergence to this point?
$endgroup$
– Did
Jan 24 at 9:02
add a comment |
$begingroup$
Yes. Now I know how to do it.
$endgroup$
– Aru
Jan 24 at 8:55
$begingroup$
This is a fixed point. What about the main part of the question, which is to show the convergence to this point?
$endgroup$
– Did
Jan 24 at 9:02
$begingroup$
Yes. Now I know how to do it.
$endgroup$
– Aru
Jan 24 at 8:55
$begingroup$
Yes. Now I know how to do it.
$endgroup$
– Aru
Jan 24 at 8:55
$begingroup$
This is a fixed point. What about the main part of the question, which is to show the convergence to this point?
$endgroup$
– Did
Jan 24 at 9:02
$begingroup$
This is a fixed point. What about the main part of the question, which is to show the convergence to this point?
$endgroup$
– Did
Jan 24 at 9:02
add a comment |
$begingroup$
You get $x=sqrt{x+i}implies x^2-x-i=0implies x=frac{-(-1)pmsqrt{1^2-4(1)(-i)}}{2(1)}=frac{1pmsqrt{1+4i}}2$. Now, $1+4i=sqrt{17}e^{(tan^{-1}4 )cdot i}$. So $sqrt{1+4i}=sqrt[4]{17}e^{frac{(tan^{-1}4)cdot i}2}$.
Finally, $x=frac12 pmfrac{sqrt[4]{17}e^{frac{(tan^{-1}4)cdot i}2}}2$, from which it follows the real part is $frac12 pmfrac{sqrt[4]{17}}2cosfrac {tan^{-1}4}2$.
As to the convergence, see @Christian Blatter's answer.
answered Jan 24 at 9:25
Chris CusterChris Custer
14.2k3827
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add a comment |
$begingroup$
You get $x=sqrt{x+i}implies x^2-x-i=0implies x=frac{-(-1)pmsqrt{1^2-4(1)(-i)}}{2(1)}=frac{1pmsqrt{1+4i}}2$. Now, $1+4i=sqrt{17}e^{(tan^{-1}4 )cdot i}$. So $sqrt{1+4i}=sqrt[4]{17}e^{frac{(tan^{-1}4)cdot i}2}$.
Finally, $x=frac12 pmfrac{sqrt[4]{17}e^{frac{(tan^{-1}4)cdot i}2}}2$, from which it follows the real part is $frac12 pmfrac{sqrt[4]{17}}2cosfrac {tan^{-1}4}2$.
As to the convergence, see @Christian Blatter's answer.
answered Jan 24 at 9:25
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
You get $x=sqrt{x+i}implies x^2-x-i=0implies x=frac{-(-1)pmsqrt{1^2-4(1)(-i)}}{2(1)}=frac{1pmsqrt{1+4i}}2$. Now, $1+4i=sqrt{17}e^{(tan^{-1}4 )cdot i}$. So $sqrt{1+4i}=sqrt[4]{17}e^{frac{(tan^{-1}4)cdot i}2}$.
Finally, $x=frac12 pmfrac{sqrt[4]{17}e^{frac{(tan^{-1}4)cdot i}2}}2$, from which it follows the real part is $frac12 pmfrac{sqrt[4]{17}}2cosfrac {tan^{-1}4}2$.
As to the convergence, see @Christian Blatter's answer.
answered Jan 24 at 9:25
Chris CusterChris Custer
14.2k3827
$endgroup$
You get $x=sqrt{x+i}implies x^2-x-i=0implies x=frac{-(-1)pmsqrt{1^2-4(1)(-i)}}{2(1)}=frac{1pmsqrt{1+4i}}2$. Now, $1+4i=sqrt{17}e^{(tan^{-1}4 )cdot i}$. So $sqrt{1+4i}=sqrt[4]{17}e^{frac{(tan^{-1}4)cdot i}2}$.
Finally, $x=frac12 pmfrac{sqrt[4]{17}e^{frac{(tan^{-1}4)cdot i}2}}2$, from which it follows the real part is $frac12 pmfrac{sqrt[4]{17}}2cosfrac {tan^{-1}4}2$.
As to the convergence, see @Christian Blatter's answer.
answered Jan 24 at 9:25
Chris CusterChris Custer
14.2k3827
answered Jan 24 at 9:25
Chris CusterChris Custer
14.2k3827
answered Jan 24 at 9:25
Chris CusterChris Custer
14.2k3827
answered Jan 24 at 9:25
Chris CusterChris Custer
14.2k3827
14.2k3827
add a comment |
add a comment |
4
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First off, complex square roots aren't well-defined. So you have a problem there already: does that expression even make sense?
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– Arthur
Jan 24 at 8:24
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Previously discussed here: math.stackexchange.com/questions/447227/…
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– Chris Culter
Jan 24 at 8:33
3
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Possible duplicate of Find the value of $sqrt{i+sqrt{i+sqrt{i+dots}}}$ .
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– Andrei
Jan 24 at 8:35
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So your real question is about solving a quadratic equation, not about $sqrt{i+sqrt{i+sqrt{i+dots}}}$ ?
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– Martin R
Jan 24 at 8:56
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@MartinR No. The problem was how to solve $sqrt{1+4i}$ which will appear after solving the quadratic equation.
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– Aru
Jan 24 at 9:02