Mixing
Does a given infinite nested radical have infinitely many solutions?
$begingroup$
Given a standard infinitely nested radical such as:
$$x = sqrt{1 + sqrt{1 + sqrt{ 1 + ...}}}$$
depending on where you choose to first substitute $x$ in the nest, aren't there infinitely many solutions when solving for $x$?
For example, you could substitute $x$ as follows:
$$x = sqrt{1 + x}$$
in which case, you get the equation $x^2 - x - 1 = 0$.
However, couldn't I just as well substitute $x$ in the following way:
$$x = sqrt{ 1 +sqrt{1 + x}}$$
in which case, you get the equation $x^4 - 2x^2 - x = 0$
Obviously, you could keep this up and generate infinitely different equations depending on where in the infinite nested loop you decided to substitute $x$. So, my questions are:
Is this an actual phenomenon or did I violate some sort of rule?
What are the implications of this?
nested-radicals
edited Jan 1 at 19:21
Shubham Johri
4,666717
asked Jan 1 at 19:15
S.CramerS.Cramer
13118
$endgroup$
|
show 9 more comments
$begingroup$
Given a standard infinitely nested radical such as:
$$x = sqrt{1 + sqrt{1 + sqrt{ 1 + ...}}}$$
depending on where you choose to first substitute $x$ in the nest, aren't there infinitely many solutions when solving for $x$?
For example, you could substitute $x$ as follows:
$$x = sqrt{1 + x}$$
in which case, you get the equation $x^2 - x - 1 = 0$.
However, couldn't I just as well substitute $x$ in the following way:
$$x = sqrt{ 1 +sqrt{1 + x}}$$
in which case, you get the equation $x^4 - 2x^2 - x = 0$
Obviously, you could keep this up and generate infinitely different equations depending on where in the infinite nested loop you decided to substitute $x$. So, my questions are:
Is this an actual phenomenon or did I violate some sort of rule?
What are the implications of this?
nested-radicals
edited Jan 1 at 19:21
Shubham Johri
4,666717
asked Jan 1 at 19:15
S.CramerS.Cramer
13118
$endgroup$
2
$begingroup$
Why do you think these give different answers? Both seem to give $frac {1+sqrt 5}2$. Note that, if the value exists at all, it is clearly $>0$ so we can discard non-positive roots.
$endgroup$
– lulu
Jan 1 at 19:18
1
$begingroup$
But those solutions are extraneous and therefore discarded, regardless of how many there are. The only correct solution is the golden ratio, as pointed out.
$endgroup$
– KM101
Jan 1 at 19:25
1
$begingroup$
@S.Cramer True but each time you square you pick up an "extra" solution which must be discarded. Here the fact that any value for this expression would have to exceed $1$ lets us discard the irrelevant values.
$endgroup$
– lulu
Jan 1 at 19:25
1
$begingroup$
Because $sqrt{1+{sqrt{1+…}}}$ is clearly positive. Any non-positive solution obtained is the solution to the new polynomial constructed, not the original nested radicals.
$endgroup$
– KM101
Jan 1 at 19:29
1
$begingroup$
@S.Cramer The problem is that squaring "loses" signs. Study my example of $sqrt x = x-1$. That only has a single real solution (easy to see) but squaring also picks up the real solution to $sqrt x = 1-x$.
$endgroup$
– lulu
Jan 1 at 19:30
|
show 9 more comments
$begingroup$
Given a standard infinitely nested radical such as:
$$x = sqrt{1 + sqrt{1 + sqrt{ 1 + ...}}}$$
depending on where you choose to first substitute $x$ in the nest, aren't there infinitely many solutions when solving for $x$?
For example, you could substitute $x$ as follows:
$$x = sqrt{1 + x}$$
in which case, you get the equation $x^2 - x - 1 = 0$.
However, couldn't I just as well substitute $x$ in the following way:
$$x = sqrt{ 1 +sqrt{1 + x}}$$
in which case, you get the equation $x^4 - 2x^2 - x = 0$
Obviously, you could keep this up and generate infinitely different equations depending on where in the infinite nested loop you decided to substitute $x$. So, my questions are:
Is this an actual phenomenon or did I violate some sort of rule?
What are the implications of this?
nested-radicals
edited Jan 1 at 19:21
Shubham Johri
4,666717
asked Jan 1 at 19:15
S.CramerS.Cramer
13118
$endgroup$
Given a standard infinitely nested radical such as:
$$x = sqrt{1 + sqrt{1 + sqrt{ 1 + ...}}}$$
depending on where you choose to first substitute $x$ in the nest, aren't there infinitely many solutions when solving for $x$?
For example, you could substitute $x$ as follows:
$$x = sqrt{1 + x}$$
in which case, you get the equation $x^2 - x - 1 = 0$.
However, couldn't I just as well substitute $x$ in the following way:
$$x = sqrt{ 1 +sqrt{1 + x}}$$
in which case, you get the equation $x^4 - 2x^2 - x = 0$
Obviously, you could keep this up and generate infinitely different equations depending on where in the infinite nested loop you decided to substitute $x$. So, my questions are:
Is this an actual phenomenon or did I violate some sort of rule?
What are the implications of this?
nested-radicals
nested-radicals
edited Jan 1 at 19:21
Shubham Johri
4,666717
asked Jan 1 at 19:15
S.CramerS.Cramer
13118
edited Jan 1 at 19:21
Shubham Johri
4,666717
asked Jan 1 at 19:15
S.CramerS.Cramer
13118
edited Jan 1 at 19:21
Shubham Johri
4,666717
edited Jan 1 at 19:21
Shubham Johri
4,666717
edited Jan 1 at 19:21
Shubham Johri
4,666717
4,666717
asked Jan 1 at 19:15
S.CramerS.Cramer
13118
asked Jan 1 at 19:15
S.CramerS.Cramer
13118
asked Jan 1 at 19:15
S.CramerS.Cramer
13118
13118
2
$begingroup$
Why do you think these give different answers? Both seem to give $frac {1+sqrt 5}2$. Note that, if the value exists at all, it is clearly $>0$ so we can discard non-positive roots.
$endgroup$
– lulu
Jan 1 at 19:18
1
$begingroup$
But those solutions are extraneous and therefore discarded, regardless of how many there are. The only correct solution is the golden ratio, as pointed out.
$endgroup$
– KM101
Jan 1 at 19:25
1
$begingroup$
@S.Cramer True but each time you square you pick up an "extra" solution which must be discarded. Here the fact that any value for this expression would have to exceed $1$ lets us discard the irrelevant values.
$endgroup$
– lulu
Jan 1 at 19:25
1
$begingroup$
Because $sqrt{1+{sqrt{1+…}}}$ is clearly positive. Any non-positive solution obtained is the solution to the new polynomial constructed, not the original nested radicals.
$endgroup$
– KM101
Jan 1 at 19:29
1
$begingroup$
@S.Cramer The problem is that squaring "loses" signs. Study my example of $sqrt x = x-1$. That only has a single real solution (easy to see) but squaring also picks up the real solution to $sqrt x = 1-x$.
$endgroup$
– lulu
Jan 1 at 19:30
|
show 9 more comments
2
$begingroup$
Why do you think these give different answers? Both seem to give $frac {1+sqrt 5}2$. Note that, if the value exists at all, it is clearly $>0$ so we can discard non-positive roots.
$endgroup$
– lulu
Jan 1 at 19:18
1
$begingroup$
But those solutions are extraneous and therefore discarded, regardless of how many there are. The only correct solution is the golden ratio, as pointed out.
$endgroup$
– KM101
Jan 1 at 19:25
1
$begingroup$
@S.Cramer True but each time you square you pick up an "extra" solution which must be discarded. Here the fact that any value for this expression would have to exceed $1$ lets us discard the irrelevant values.
$endgroup$
– lulu
Jan 1 at 19:25
1
$begingroup$
Because $sqrt{1+{sqrt{1+…}}}$ is clearly positive. Any non-positive solution obtained is the solution to the new polynomial constructed, not the original nested radicals.
$endgroup$
– KM101
Jan 1 at 19:29
1
$begingroup$
@S.Cramer The problem is that squaring "loses" signs. Study my example of $sqrt x = x-1$. That only has a single real solution (easy to see) but squaring also picks up the real solution to $sqrt x = 1-x$.
$endgroup$
– lulu
Jan 1 at 19:30
2
2
$begingroup$
Why do you think these give different answers? Both seem to give $frac {1+sqrt 5}2$. Note that, if the value exists at all, it is clearly $>0$ so we can discard non-positive roots.
$endgroup$
– lulu
Jan 1 at 19:18
$begingroup$
Why do you think these give different answers? Both seem to give $frac {1+sqrt 5}2$. Note that, if the value exists at all, it is clearly $>0$ so we can discard non-positive roots.
$endgroup$
– lulu
Jan 1 at 19:18
1
1
$begingroup$
But those solutions are extraneous and therefore discarded, regardless of how many there are. The only correct solution is the golden ratio, as pointed out.
$endgroup$
– KM101
Jan 1 at 19:25
$begingroup$
But those solutions are extraneous and therefore discarded, regardless of how many there are. The only correct solution is the golden ratio, as pointed out.
$endgroup$
– KM101
Jan 1 at 19:25
1
1
$begingroup$
@S.Cramer True but each time you square you pick up an "extra" solution which must be discarded. Here the fact that any value for this expression would have to exceed $1$ lets us discard the irrelevant values.
$endgroup$
– lulu
Jan 1 at 19:25
$begingroup$
@S.Cramer True but each time you square you pick up an "extra" solution which must be discarded. Here the fact that any value for this expression would have to exceed $1$ lets us discard the irrelevant values.
$endgroup$
– lulu
Jan 1 at 19:25
1
1
$begingroup$
Because $sqrt{1+{sqrt{1+…}}}$ is clearly positive. Any non-positive solution obtained is the solution to the new polynomial constructed, not the original nested radicals.
$endgroup$
– KM101
Jan 1 at 19:29
$begingroup$
Because $sqrt{1+{sqrt{1+…}}}$ is clearly positive. Any non-positive solution obtained is the solution to the new polynomial constructed, not the original nested radicals.
$endgroup$
– KM101
Jan 1 at 19:29
1
1
$begingroup$
@S.Cramer The problem is that squaring "loses" signs. Study my example of $sqrt x = x-1$. That only has a single real solution (easy to see) but squaring also picks up the real solution to $sqrt x = 1-x$.
$endgroup$
– lulu
Jan 1 at 19:30
$begingroup$
@S.Cramer The problem is that squaring "loses" signs. Study my example of $sqrt x = x-1$. That only has a single real solution (easy to see) but squaring also picks up the real solution to $sqrt x = 1-x$.
$endgroup$
– lulu
Jan 1 at 19:30
|
show 9 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Every time you square this you add extraneous solutions which arise from taking the negative value of some of the square roots. The use of $x$ implies that the pattern is recurrent, but the quartic arising from squaring twice corresponds to the four possible choices for the square roots you have removed.
answered Jan 1 at 19:19
Mark BennetMark Bennet
80.7k981179
$endgroup$
$begingroup$
Does this logic apply if, instead of square roots, I used cube roots to construct this nesting? Then, I would have odd powered solutions half of the time.
$endgroup$
– S.Cramer
Jan 1 at 19:37
$begingroup$
@S.Cramer There are three (complex) cube roots of any non-zero real number - so you have three solutions of the resulting cubic - but quite often only one of these is a real number, corresponding to the original expression.
$endgroup$
– Mark Bennet
Jan 1 at 19:40
add a comment |
$begingroup$
The issue is the convergence of the nested radical.
Let $(x_n)$ be the sequence of iterated radicals. We'll have
$$x_n = left{begin{matrix}
1 & n = 1\
sqrt{1+x_{n-1}} & n > 1
end{matrix}right.$$
A sequence can logically only converge to one unique value, i.e. $(x_n) to x$. Take $n to infty$ and we have the infinite nested radical described,
$$x = sqrt{1 + sqrt{1+sqrt{...}}}$$
Suppose we have some number $alpha$ we want to express as an infinite root. A video by blackpenredpen on YouTube goes over generating such expressions, and touches on why some of these roots may seem to be equal to multiple values or have multiple solutions - namely, the implicit introduction of extraneous solutions.
We can start by saying the following: let $x = alpha$ be a solution to our hypothetical infinite root. Then, throwing another arbitrary constant $beta$ in, where $beta neq alpha$,
$$begin{align}
x = alpha &implies x - alpha = 0 \
&implies (x - alpha)(x - beta) =0 \
&implies x^2 -(beta+alpha)x + betaalpha = 0 \
&implies x^2 = (beta+alpha)x - beta alpha \
&implies x = sqrt{(beta+alpha)x - beta alpha} \
end{align}$$
And from here, we can use this sort of recursive definition to generate our infinitely nested radical:
$$begin{align}
x &= sqrt{ -beta alpha + ( beta + alpha ) x } \
&= sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) x } } \
&= sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) x } } } \
&= sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) ... } } } \
end{align}$$
Notice how through all this, the other root $beta$ still was introduced as an extraneous solution. And of course our chose to multiply by $(x - beta)$ was arbitrary: we could've gone
$$begin{align}
x = alpha &implies x - alpha &= 0 \
&implies (x - alpha)(x - beta_1) &=0 \
&implies (x - alpha)(x - beta_1)(x - beta_2) &=0 \
&implies (x - alpha)(x - beta_1)(x - beta_2)(x - beta_3) &=0 \
&implies (x - alpha)(x - beta_1)(x - beta_2)(x - beta_3)...(x - beta_n) &=0 \
end{align}$$
and thus introduced $n$ arbitrary constants $beta_1 ... beta_n$, all not equal to $alpha$, as these "fake solutions." (Fake in that we begin with the presumption that $alpha$ is the only solution, so to say $beta_k$ for any $k$ is a solution when $beta_k neq alpha$ is an immediate contradiction.)
The algebra would be a pain in the butt in constructing these radicals for $n>1$, but it's doable. Our situation when presented with solving such a root, instead of constructing it, is essentially the same, but in reverse: we have no inherent idea of how many arbitrary solutions might have been generated, and indeed in a sense infinitely many were depending on your interpretation of the root and its recursion. So we always have to hearken back to what would make the root a proper solution:
That being, convergence of the iterated root to that value.
For example, for the sequence $(x_n)$ introduced initially, you can show that:
- It is monotone increasing
- It is bounded above by $phi = (1 + sqrt 5) / 2$
- As a consequence of the previous two, $(x_n) to phi$
Thus, $phi$ is the only actual solution to this nested radical. Sure, you can have manipulations and that supply other solutions - but take $x_1, x_2, x_3, x_4, ...$, take the limit as $n to infty$ of $(x_n)$: it's not going to approach any of them but $phi$.
answered Jan 1 at 19:56
Eevee TrainerEevee Trainer
5,3881836
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$begingroup$
No, there aren't infinite solutions, because
$$ f(x)=sqrt{x+1} $$
is a contraction of $[1,2]$ ($frac{1}{2sqrt{3}}leq f'(x)leq frac{1}{2sqrt{2}}$ for any $xinleft[1,2right]=I$), hence the Banach fixed point theorem ensures that the iteration
$$ x_{n+1} = sqrt{1+x_n} $$
with starting point $x_0=0$ (such that both $x_1$ and $x_2$ belong to $I$) converges to the unique solution of $f(x)=x$ in $I$, i.e. $frac{1+sqrt{5}}{2}$. In fancy terms,
$$ 1+frac{1}{1+frac{1}{1+frac{1}{1+frac{1}{ldots}}}}=sqrt{1+sqrt{1+sqrt{1+sqrt{1+ldots}}}} $$
In general, when squaring both sides of an equation involving one or many square roots you may introduce spurious solutions.
answered Jan 2 at 2:01
Jack D'AurizioJack D'Aurizio
288k33280659
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add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
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active
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oldest
votes
$begingroup$
Every time you square this you add extraneous solutions which arise from taking the negative value of some of the square roots. The use of $x$ implies that the pattern is recurrent, but the quartic arising from squaring twice corresponds to the four possible choices for the square roots you have removed.
answered Jan 1 at 19:19
Mark BennetMark Bennet
80.7k981179
$endgroup$
$begingroup$
Does this logic apply if, instead of square roots, I used cube roots to construct this nesting? Then, I would have odd powered solutions half of the time.
$endgroup$
– S.Cramer
Jan 1 at 19:37
$begingroup$
@S.Cramer There are three (complex) cube roots of any non-zero real number - so you have three solutions of the resulting cubic - but quite often only one of these is a real number, corresponding to the original expression.
$endgroup$
– Mark Bennet
Jan 1 at 19:40
add a comment |
$begingroup$
Every time you square this you add extraneous solutions which arise from taking the negative value of some of the square roots. The use of $x$ implies that the pattern is recurrent, but the quartic arising from squaring twice corresponds to the four possible choices for the square roots you have removed.
answered Jan 1 at 19:19
Mark BennetMark Bennet
80.7k981179
$endgroup$
$begingroup$
Does this logic apply if, instead of square roots, I used cube roots to construct this nesting? Then, I would have odd powered solutions half of the time.
$endgroup$
– S.Cramer
Jan 1 at 19:37
$begingroup$
@S.Cramer There are three (complex) cube roots of any non-zero real number - so you have three solutions of the resulting cubic - but quite often only one of these is a real number, corresponding to the original expression.
$endgroup$
– Mark Bennet
Jan 1 at 19:40
add a comment |
$begingroup$
Every time you square this you add extraneous solutions which arise from taking the negative value of some of the square roots. The use of $x$ implies that the pattern is recurrent, but the quartic arising from squaring twice corresponds to the four possible choices for the square roots you have removed.
answered Jan 1 at 19:19
Mark BennetMark Bennet
80.7k981179
$endgroup$
Every time you square this you add extraneous solutions which arise from taking the negative value of some of the square roots. The use of $x$ implies that the pattern is recurrent, but the quartic arising from squaring twice corresponds to the four possible choices for the square roots you have removed.
answered Jan 1 at 19:19
Mark BennetMark Bennet
80.7k981179
edited Jan 1 at 19:25
answered Jan 1 at 19:19
Mark BennetMark Bennet
80.7k981179
answered Jan 1 at 19:19
Mark BennetMark Bennet
80.7k981179
answered Jan 1 at 19:19
Mark BennetMark Bennet
80.7k981179
80.7k981179
$begingroup$
Does this logic apply if, instead of square roots, I used cube roots to construct this nesting? Then, I would have odd powered solutions half of the time.
$endgroup$
– S.Cramer
Jan 1 at 19:37
$begingroup$
@S.Cramer There are three (complex) cube roots of any non-zero real number - so you have three solutions of the resulting cubic - but quite often only one of these is a real number, corresponding to the original expression.
$endgroup$
– Mark Bennet
Jan 1 at 19:40
add a comment |
$begingroup$
Does this logic apply if, instead of square roots, I used cube roots to construct this nesting? Then, I would have odd powered solutions half of the time.
$endgroup$
– S.Cramer
Jan 1 at 19:37
$begingroup$
@S.Cramer There are three (complex) cube roots of any non-zero real number - so you have three solutions of the resulting cubic - but quite often only one of these is a real number, corresponding to the original expression.
$endgroup$
– Mark Bennet
Jan 1 at 19:40
$begingroup$
Does this logic apply if, instead of square roots, I used cube roots to construct this nesting? Then, I would have odd powered solutions half of the time.
$endgroup$
– S.Cramer
Jan 1 at 19:37
$begingroup$
Does this logic apply if, instead of square roots, I used cube roots to construct this nesting? Then, I would have odd powered solutions half of the time.
$endgroup$
– S.Cramer
Jan 1 at 19:37
$begingroup$
@S.Cramer There are three (complex) cube roots of any non-zero real number - so you have three solutions of the resulting cubic - but quite often only one of these is a real number, corresponding to the original expression.
$endgroup$
– Mark Bennet
Jan 1 at 19:40
$begingroup$
@S.Cramer There are three (complex) cube roots of any non-zero real number - so you have three solutions of the resulting cubic - but quite often only one of these is a real number, corresponding to the original expression.
$endgroup$
– Mark Bennet
Jan 1 at 19:40
add a comment |
$begingroup$
The issue is the convergence of the nested radical.
Let $(x_n)$ be the sequence of iterated radicals. We'll have
$$x_n = left{begin{matrix}
1 & n = 1\
sqrt{1+x_{n-1}} & n > 1
end{matrix}right.$$
A sequence can logically only converge to one unique value, i.e. $(x_n) to x$. Take $n to infty$ and we have the infinite nested radical described,
$$x = sqrt{1 + sqrt{1+sqrt{...}}}$$
Suppose we have some number $alpha$ we want to express as an infinite root. A video by blackpenredpen on YouTube goes over generating such expressions, and touches on why some of these roots may seem to be equal to multiple values or have multiple solutions - namely, the implicit introduction of extraneous solutions.
We can start by saying the following: let $x = alpha$ be a solution to our hypothetical infinite root. Then, throwing another arbitrary constant $beta$ in, where $beta neq alpha$,
$$begin{align}
x = alpha &implies x - alpha = 0 \
&implies (x - alpha)(x - beta) =0 \
&implies x^2 -(beta+alpha)x + betaalpha = 0 \
&implies x^2 = (beta+alpha)x - beta alpha \
&implies x = sqrt{(beta+alpha)x - beta alpha} \
end{align}$$
And from here, we can use this sort of recursive definition to generate our infinitely nested radical:
$$begin{align}
x &= sqrt{ -beta alpha + ( beta + alpha ) x } \
&= sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) x } } \
&= sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) x } } } \
&= sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) ... } } } \
end{align}$$
Notice how through all this, the other root $beta$ still was introduced as an extraneous solution. And of course our chose to multiply by $(x - beta)$ was arbitrary: we could've gone
$$begin{align}
x = alpha &implies x - alpha &= 0 \
&implies (x - alpha)(x - beta_1) &=0 \
&implies (x - alpha)(x - beta_1)(x - beta_2) &=0 \
&implies (x - alpha)(x - beta_1)(x - beta_2)(x - beta_3) &=0 \
&implies (x - alpha)(x - beta_1)(x - beta_2)(x - beta_3)...(x - beta_n) &=0 \
end{align}$$
and thus introduced $n$ arbitrary constants $beta_1 ... beta_n$, all not equal to $alpha$, as these "fake solutions." (Fake in that we begin with the presumption that $alpha$ is the only solution, so to say $beta_k$ for any $k$ is a solution when $beta_k neq alpha$ is an immediate contradiction.)
The algebra would be a pain in the butt in constructing these radicals for $n>1$, but it's doable. Our situation when presented with solving such a root, instead of constructing it, is essentially the same, but in reverse: we have no inherent idea of how many arbitrary solutions might have been generated, and indeed in a sense infinitely many were depending on your interpretation of the root and its recursion. So we always have to hearken back to what would make the root a proper solution:
That being, convergence of the iterated root to that value.
For example, for the sequence $(x_n)$ introduced initially, you can show that:
- It is monotone increasing
- It is bounded above by $phi = (1 + sqrt 5) / 2$
- As a consequence of the previous two, $(x_n) to phi$
Thus, $phi$ is the only actual solution to this nested radical. Sure, you can have manipulations and that supply other solutions - but take $x_1, x_2, x_3, x_4, ...$, take the limit as $n to infty$ of $(x_n)$: it's not going to approach any of them but $phi$.
answered Jan 1 at 19:56
Eevee TrainerEevee Trainer
5,3881836
$endgroup$
add a comment |
$begingroup$
The issue is the convergence of the nested radical.
Let $(x_n)$ be the sequence of iterated radicals. We'll have
$$x_n = left{begin{matrix}
1 & n = 1\
sqrt{1+x_{n-1}} & n > 1
end{matrix}right.$$
A sequence can logically only converge to one unique value, i.e. $(x_n) to x$. Take $n to infty$ and we have the infinite nested radical described,
$$x = sqrt{1 + sqrt{1+sqrt{...}}}$$
Suppose we have some number $alpha$ we want to express as an infinite root. A video by blackpenredpen on YouTube goes over generating such expressions, and touches on why some of these roots may seem to be equal to multiple values or have multiple solutions - namely, the implicit introduction of extraneous solutions.
We can start by saying the following: let $x = alpha$ be a solution to our hypothetical infinite root. Then, throwing another arbitrary constant $beta$ in, where $beta neq alpha$,
$$begin{align}
x = alpha &implies x - alpha = 0 \
&implies (x - alpha)(x - beta) =0 \
&implies x^2 -(beta+alpha)x + betaalpha = 0 \
&implies x^2 = (beta+alpha)x - beta alpha \
&implies x = sqrt{(beta+alpha)x - beta alpha} \
end{align}$$
And from here, we can use this sort of recursive definition to generate our infinitely nested radical:
$$begin{align}
x &= sqrt{ -beta alpha + ( beta + alpha ) x } \
&= sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) x } } \
&= sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) x } } } \
&= sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) ... } } } \
end{align}$$
Notice how through all this, the other root $beta$ still was introduced as an extraneous solution. And of course our chose to multiply by $(x - beta)$ was arbitrary: we could've gone
$$begin{align}
x = alpha &implies x - alpha &= 0 \
&implies (x - alpha)(x - beta_1) &=0 \
&implies (x - alpha)(x - beta_1)(x - beta_2) &=0 \
&implies (x - alpha)(x - beta_1)(x - beta_2)(x - beta_3) &=0 \
&implies (x - alpha)(x - beta_1)(x - beta_2)(x - beta_3)...(x - beta_n) &=0 \
end{align}$$
and thus introduced $n$ arbitrary constants $beta_1 ... beta_n$, all not equal to $alpha$, as these "fake solutions." (Fake in that we begin with the presumption that $alpha$ is the only solution, so to say $beta_k$ for any $k$ is a solution when $beta_k neq alpha$ is an immediate contradiction.)
The algebra would be a pain in the butt in constructing these radicals for $n>1$, but it's doable. Our situation when presented with solving such a root, instead of constructing it, is essentially the same, but in reverse: we have no inherent idea of how many arbitrary solutions might have been generated, and indeed in a sense infinitely many were depending on your interpretation of the root and its recursion. So we always have to hearken back to what would make the root a proper solution:
That being, convergence of the iterated root to that value.
For example, for the sequence $(x_n)$ introduced initially, you can show that:
- It is monotone increasing
- It is bounded above by $phi = (1 + sqrt 5) / 2$
- As a consequence of the previous two, $(x_n) to phi$
Thus, $phi$ is the only actual solution to this nested radical. Sure, you can have manipulations and that supply other solutions - but take $x_1, x_2, x_3, x_4, ...$, take the limit as $n to infty$ of $(x_n)$: it's not going to approach any of them but $phi$.
answered Jan 1 at 19:56
Eevee TrainerEevee Trainer
5,3881836
$endgroup$
add a comment |
$begingroup$
The issue is the convergence of the nested radical.
Let $(x_n)$ be the sequence of iterated radicals. We'll have
$$x_n = left{begin{matrix}
1 & n = 1\
sqrt{1+x_{n-1}} & n > 1
end{matrix}right.$$
A sequence can logically only converge to one unique value, i.e. $(x_n) to x$. Take $n to infty$ and we have the infinite nested radical described,
$$x = sqrt{1 + sqrt{1+sqrt{...}}}$$
Suppose we have some number $alpha$ we want to express as an infinite root. A video by blackpenredpen on YouTube goes over generating such expressions, and touches on why some of these roots may seem to be equal to multiple values or have multiple solutions - namely, the implicit introduction of extraneous solutions.
We can start by saying the following: let $x = alpha$ be a solution to our hypothetical infinite root. Then, throwing another arbitrary constant $beta$ in, where $beta neq alpha$,
$$begin{align}
x = alpha &implies x - alpha = 0 \
&implies (x - alpha)(x - beta) =0 \
&implies x^2 -(beta+alpha)x + betaalpha = 0 \
&implies x^2 = (beta+alpha)x - beta alpha \
&implies x = sqrt{(beta+alpha)x - beta alpha} \
end{align}$$
And from here, we can use this sort of recursive definition to generate our infinitely nested radical:
$$begin{align}
x &= sqrt{ -beta alpha + ( beta + alpha ) x } \
&= sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) x } } \
&= sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) x } } } \
&= sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) ... } } } \
end{align}$$
Notice how through all this, the other root $beta$ still was introduced as an extraneous solution. And of course our chose to multiply by $(x - beta)$ was arbitrary: we could've gone
$$begin{align}
x = alpha &implies x - alpha &= 0 \
&implies (x - alpha)(x - beta_1) &=0 \
&implies (x - alpha)(x - beta_1)(x - beta_2) &=0 \
&implies (x - alpha)(x - beta_1)(x - beta_2)(x - beta_3) &=0 \
&implies (x - alpha)(x - beta_1)(x - beta_2)(x - beta_3)...(x - beta_n) &=0 \
end{align}$$
and thus introduced $n$ arbitrary constants $beta_1 ... beta_n$, all not equal to $alpha$, as these "fake solutions." (Fake in that we begin with the presumption that $alpha$ is the only solution, so to say $beta_k$ for any $k$ is a solution when $beta_k neq alpha$ is an immediate contradiction.)
The algebra would be a pain in the butt in constructing these radicals for $n>1$, but it's doable. Our situation when presented with solving such a root, instead of constructing it, is essentially the same, but in reverse: we have no inherent idea of how many arbitrary solutions might have been generated, and indeed in a sense infinitely many were depending on your interpretation of the root and its recursion. So we always have to hearken back to what would make the root a proper solution:
That being, convergence of the iterated root to that value.
For example, for the sequence $(x_n)$ introduced initially, you can show that:
- It is monotone increasing
- It is bounded above by $phi = (1 + sqrt 5) / 2$
- As a consequence of the previous two, $(x_n) to phi$
Thus, $phi$ is the only actual solution to this nested radical. Sure, you can have manipulations and that supply other solutions - but take $x_1, x_2, x_3, x_4, ...$, take the limit as $n to infty$ of $(x_n)$: it's not going to approach any of them but $phi$.
answered Jan 1 at 19:56
Eevee TrainerEevee Trainer
5,3881836
$endgroup$
The issue is the convergence of the nested radical.
Let $(x_n)$ be the sequence of iterated radicals. We'll have
$$x_n = left{begin{matrix}
1 & n = 1\
sqrt{1+x_{n-1}} & n > 1
end{matrix}right.$$
A sequence can logically only converge to one unique value, i.e. $(x_n) to x$. Take $n to infty$ and we have the infinite nested radical described,
$$x = sqrt{1 + sqrt{1+sqrt{...}}}$$
Suppose we have some number $alpha$ we want to express as an infinite root. A video by blackpenredpen on YouTube goes over generating such expressions, and touches on why some of these roots may seem to be equal to multiple values or have multiple solutions - namely, the implicit introduction of extraneous solutions.
We can start by saying the following: let $x = alpha$ be a solution to our hypothetical infinite root. Then, throwing another arbitrary constant $beta$ in, where $beta neq alpha$,
$$begin{align}
x = alpha &implies x - alpha = 0 \
&implies (x - alpha)(x - beta) =0 \
&implies x^2 -(beta+alpha)x + betaalpha = 0 \
&implies x^2 = (beta+alpha)x - beta alpha \
&implies x = sqrt{(beta+alpha)x - beta alpha} \
end{align}$$
And from here, we can use this sort of recursive definition to generate our infinitely nested radical:
$$begin{align}
x &= sqrt{ -beta alpha + ( beta + alpha ) x } \
&= sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) x } } \
&= sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) x } } } \
&= sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) sqrt{ -beta alpha + ( beta + alpha ) ... } } } \
end{align}$$
Notice how through all this, the other root $beta$ still was introduced as an extraneous solution. And of course our chose to multiply by $(x - beta)$ was arbitrary: we could've gone
$$begin{align}
x = alpha &implies x - alpha &= 0 \
&implies (x - alpha)(x - beta_1) &=0 \
&implies (x - alpha)(x - beta_1)(x - beta_2) &=0 \
&implies (x - alpha)(x - beta_1)(x - beta_2)(x - beta_3) &=0 \
&implies (x - alpha)(x - beta_1)(x - beta_2)(x - beta_3)...(x - beta_n) &=0 \
end{align}$$
and thus introduced $n$ arbitrary constants $beta_1 ... beta_n$, all not equal to $alpha$, as these "fake solutions." (Fake in that we begin with the presumption that $alpha$ is the only solution, so to say $beta_k$ for any $k$ is a solution when $beta_k neq alpha$ is an immediate contradiction.)
The algebra would be a pain in the butt in constructing these radicals for $n>1$, but it's doable. Our situation when presented with solving such a root, instead of constructing it, is essentially the same, but in reverse: we have no inherent idea of how many arbitrary solutions might have been generated, and indeed in a sense infinitely many were depending on your interpretation of the root and its recursion. So we always have to hearken back to what would make the root a proper solution:
That being, convergence of the iterated root to that value.
For example, for the sequence $(x_n)$ introduced initially, you can show that:
- It is monotone increasing
- It is bounded above by $phi = (1 + sqrt 5) / 2$
- As a consequence of the previous two, $(x_n) to phi$
Thus, $phi$ is the only actual solution to this nested radical. Sure, you can have manipulations and that supply other solutions - but take $x_1, x_2, x_3, x_4, ...$, take the limit as $n to infty$ of $(x_n)$: it's not going to approach any of them but $phi$.
answered Jan 1 at 19:56
Eevee TrainerEevee Trainer
5,3881836
answered Jan 1 at 19:56
Eevee TrainerEevee Trainer
5,3881836
answered Jan 1 at 19:56
Eevee TrainerEevee Trainer
5,3881836
answered Jan 1 at 19:56
Eevee TrainerEevee Trainer
5,3881836
5,3881836
add a comment |
add a comment |
$begingroup$
No, there aren't infinite solutions, because
$$ f(x)=sqrt{x+1} $$
is a contraction of $[1,2]$ ($frac{1}{2sqrt{3}}leq f'(x)leq frac{1}{2sqrt{2}}$ for any $xinleft[1,2right]=I$), hence the Banach fixed point theorem ensures that the iteration
$$ x_{n+1} = sqrt{1+x_n} $$
with starting point $x_0=0$ (such that both $x_1$ and $x_2$ belong to $I$) converges to the unique solution of $f(x)=x$ in $I$, i.e. $frac{1+sqrt{5}}{2}$. In fancy terms,
$$ 1+frac{1}{1+frac{1}{1+frac{1}{1+frac{1}{ldots}}}}=sqrt{1+sqrt{1+sqrt{1+sqrt{1+ldots}}}} $$
In general, when squaring both sides of an equation involving one or many square roots you may introduce spurious solutions.
answered Jan 2 at 2:01
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
add a comment |
$begingroup$
No, there aren't infinite solutions, because
$$ f(x)=sqrt{x+1} $$
is a contraction of $[1,2]$ ($frac{1}{2sqrt{3}}leq f'(x)leq frac{1}{2sqrt{2}}$ for any $xinleft[1,2right]=I$), hence the Banach fixed point theorem ensures that the iteration
$$ x_{n+1} = sqrt{1+x_n} $$
with starting point $x_0=0$ (such that both $x_1$ and $x_2$ belong to $I$) converges to the unique solution of $f(x)=x$ in $I$, i.e. $frac{1+sqrt{5}}{2}$. In fancy terms,
$$ 1+frac{1}{1+frac{1}{1+frac{1}{1+frac{1}{ldots}}}}=sqrt{1+sqrt{1+sqrt{1+sqrt{1+ldots}}}} $$
In general, when squaring both sides of an equation involving one or many square roots you may introduce spurious solutions.
answered Jan 2 at 2:01
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
add a comment |
$begingroup$
No, there aren't infinite solutions, because
$$ f(x)=sqrt{x+1} $$
is a contraction of $[1,2]$ ($frac{1}{2sqrt{3}}leq f'(x)leq frac{1}{2sqrt{2}}$ for any $xinleft[1,2right]=I$), hence the Banach fixed point theorem ensures that the iteration
$$ x_{n+1} = sqrt{1+x_n} $$
with starting point $x_0=0$ (such that both $x_1$ and $x_2$ belong to $I$) converges to the unique solution of $f(x)=x$ in $I$, i.e. $frac{1+sqrt{5}}{2}$. In fancy terms,
$$ 1+frac{1}{1+frac{1}{1+frac{1}{1+frac{1}{ldots}}}}=sqrt{1+sqrt{1+sqrt{1+sqrt{1+ldots}}}} $$
In general, when squaring both sides of an equation involving one or many square roots you may introduce spurious solutions.
answered Jan 2 at 2:01
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
No, there aren't infinite solutions, because
$$ f(x)=sqrt{x+1} $$
is a contraction of $[1,2]$ ($frac{1}{2sqrt{3}}leq f'(x)leq frac{1}{2sqrt{2}}$ for any $xinleft[1,2right]=I$), hence the Banach fixed point theorem ensures that the iteration
$$ x_{n+1} = sqrt{1+x_n} $$
with starting point $x_0=0$ (such that both $x_1$ and $x_2$ belong to $I$) converges to the unique solution of $f(x)=x$ in $I$, i.e. $frac{1+sqrt{5}}{2}$. In fancy terms,
$$ 1+frac{1}{1+frac{1}{1+frac{1}{1+frac{1}{ldots}}}}=sqrt{1+sqrt{1+sqrt{1+sqrt{1+ldots}}}} $$
In general, when squaring both sides of an equation involving one or many square roots you may introduce spurious solutions.
answered Jan 2 at 2:01
Jack D'AurizioJack D'Aurizio
288k33280659
answered Jan 2 at 2:01
Jack D'AurizioJack D'Aurizio
288k33280659
answered Jan 2 at 2:01
Jack D'AurizioJack D'Aurizio
288k33280659
answered Jan 2 at 2:01
Jack D'AurizioJack D'Aurizio
288k33280659
288k33280659
add a comment |
add a comment |
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$begingroup$
Why do you think these give different answers? Both seem to give $frac {1+sqrt 5}2$. Note that, if the value exists at all, it is clearly $>0$ so we can discard non-positive roots.
$endgroup$
– lulu
Jan 1 at 19:18
1
$begingroup$
But those solutions are extraneous and therefore discarded, regardless of how many there are. The only correct solution is the golden ratio, as pointed out.
$endgroup$
– KM101
Jan 1 at 19:25
1
$begingroup$
@S.Cramer True but each time you square you pick up an "extra" solution which must be discarded. Here the fact that any value for this expression would have to exceed $1$ lets us discard the irrelevant values.
$endgroup$
– lulu
Jan 1 at 19:25
1
$begingroup$
Because $sqrt{1+{sqrt{1+…}}}$ is clearly positive. Any non-positive solution obtained is the solution to the new polynomial constructed, not the original nested radicals.
$endgroup$
– KM101
Jan 1 at 19:29
1
$begingroup$
@S.Cramer The problem is that squaring "loses" signs. Study my example of $sqrt x = x-1$. That only has a single real solution (easy to see) but squaring also picks up the real solution to $sqrt x = 1-x$.
$endgroup$
– lulu
Jan 1 at 19:30