Mixing
Find the sum $sqrt{5+sqrt{11+sqrt{19+sqrt{29+sqrt{41+cdots}}}}}$
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Okay so this can be written as $$sqrt{5+sqrt{(5+6)+sqrt{(5+6+8)+sqrt{(5+6+8+10)+sqrt{(5+6+8+10+12)cdots}}}}}$$
Putting it as $y$ and squaring both sides doesn't seem to help, and I don't know what else can be done.
sequences-and-series analysis
asked Jan 20 at 3:13
Goal123Goal123
507212
$endgroup$
add a comment |
$begingroup$
Okay so this can be written as $$sqrt{5+sqrt{(5+6)+sqrt{(5+6+8)+sqrt{(5+6+8+10)+sqrt{(5+6+8+10+12)cdots}}}}}$$
Putting it as $y$ and squaring both sides doesn't seem to help, and I don't know what else can be done.
sequences-and-series analysis
asked Jan 20 at 3:13
Goal123Goal123
507212
$endgroup$
$begingroup$
Numerically, it seems to converge to 3.
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– greelious
Jan 20 at 3:16
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Thanks, but I actually need to compute it theoretically.
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– Goal123
Jan 20 at 3:17
2
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Please give the rule for the coefficients $5, 11, 19, ldots$ in your formula (which, by the way, is not a sum).
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– Rob Arthan
Jan 20 at 3:31
2
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The terms seem to be given by $n^2 + 3n + 1$ for $n geq 1$.
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– rwbogl
Jan 20 at 3:40
1
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Just a nitpick on the title: this is not actually a sum.
$endgroup$
– Cheerful Parsnip
Jan 20 at 5:49
add a comment |
$begingroup$
Okay so this can be written as $$sqrt{5+sqrt{(5+6)+sqrt{(5+6+8)+sqrt{(5+6+8+10)+sqrt{(5+6+8+10+12)cdots}}}}}$$
Putting it as $y$ and squaring both sides doesn't seem to help, and I don't know what else can be done.
sequences-and-series analysis
asked Jan 20 at 3:13
Goal123Goal123
507212
$endgroup$
Okay so this can be written as $$sqrt{5+sqrt{(5+6)+sqrt{(5+6+8)+sqrt{(5+6+8+10)+sqrt{(5+6+8+10+12)cdots}}}}}$$
Putting it as $y$ and squaring both sides doesn't seem to help, and I don't know what else can be done.
sequences-and-series analysis
sequences-and-series analysis
asked Jan 20 at 3:13
Goal123Goal123
507212
asked Jan 20 at 3:13
Goal123Goal123
507212
asked Jan 20 at 3:13
Goal123Goal123
507212
asked Jan 20 at 3:13
Goal123Goal123
507212
asked Jan 20 at 3:13
Goal123Goal123
507212
507212
$begingroup$
Numerically, it seems to converge to 3.
$endgroup$
– greelious
Jan 20 at 3:16
$begingroup$
Thanks, but I actually need to compute it theoretically.
$endgroup$
– Goal123
Jan 20 at 3:17
2
$begingroup$
Please give the rule for the coefficients $5, 11, 19, ldots$ in your formula (which, by the way, is not a sum).
$endgroup$
– Rob Arthan
Jan 20 at 3:31
2
$begingroup$
The terms seem to be given by $n^2 + 3n + 1$ for $n geq 1$.
$endgroup$
– rwbogl
Jan 20 at 3:40
1
$begingroup$
Just a nitpick on the title: this is not actually a sum.
$endgroup$
– Cheerful Parsnip
Jan 20 at 5:49
add a comment |
$begingroup$
Numerically, it seems to converge to 3.
$endgroup$
– greelious
Jan 20 at 3:16
$begingroup$
Thanks, but I actually need to compute it theoretically.
$endgroup$
– Goal123
Jan 20 at 3:17
2
$begingroup$
Please give the rule for the coefficients $5, 11, 19, ldots$ in your formula (which, by the way, is not a sum).
$endgroup$
– Rob Arthan
Jan 20 at 3:31
2
$begingroup$
The terms seem to be given by $n^2 + 3n + 1$ for $n geq 1$.
$endgroup$
– rwbogl
Jan 20 at 3:40
1
$begingroup$
Just a nitpick on the title: this is not actually a sum.
$endgroup$
– Cheerful Parsnip
Jan 20 at 5:49
$begingroup$
Numerically, it seems to converge to 3.
$endgroup$
– greelious
Jan 20 at 3:16
$begingroup$
Numerically, it seems to converge to 3.
$endgroup$
– greelious
Jan 20 at 3:16
$begingroup$
Thanks, but I actually need to compute it theoretically.
$endgroup$
– Goal123
Jan 20 at 3:17
$begingroup$
Thanks, but I actually need to compute it theoretically.
$endgroup$
– Goal123
Jan 20 at 3:17
2
2
$begingroup$
Please give the rule for the coefficients $5, 11, 19, ldots$ in your formula (which, by the way, is not a sum).
$endgroup$
– Rob Arthan
Jan 20 at 3:31
$begingroup$
Please give the rule for the coefficients $5, 11, 19, ldots$ in your formula (which, by the way, is not a sum).
$endgroup$
– Rob Arthan
Jan 20 at 3:31
2
2
$begingroup$
The terms seem to be given by $n^2 + 3n + 1$ for $n geq 1$.
$endgroup$
– rwbogl
Jan 20 at 3:40
$begingroup$
The terms seem to be given by $n^2 + 3n + 1$ for $n geq 1$.
$endgroup$
– rwbogl
Jan 20 at 3:40
1
1
$begingroup$
Just a nitpick on the title: this is not actually a sum.
$endgroup$
– Cheerful Parsnip
Jan 20 at 5:49
$begingroup$
Just a nitpick on the title: this is not actually a sum.
$endgroup$
– Cheerful Parsnip
Jan 20 at 5:49
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
We may adopt the technique for Ramanujan's infinite radical. Let $p(x) = x^2 + 3x + 1$ and define $F : [0, infty) to [0, infty)$ by
$$ F(x) = sqrt{p(x) + sqrt{p(x+1) + sqrt{p(x+2) + cdots }}} $$
Then $F$ solves
$$ F(x)^2 = p(x) + F(x+1). $$
Now we make an ansatz that $F(x)$ takes the form $F(x) = ax + b$. Plugging this and comparing coefficients shows that
$$ F(x) = x + 2 $$
solves the functional equation. Finally, since $(p(1), p(2), p(3), cdots) = (5, 11, 19, cdots) $, the infinite radical in question corresponds to the case $x = 1$, giving
$$ sqrt{5 + sqrt{11 + sqrt{19 + cdots}}} = F(1) = 3. $$
Rigorous justification. Let $mathcal{C}$ be the set of all continuous functions $f : [0, infty) to mathbb{R}$ such that
$$ | f| := sup_{xtoinfty} left( 2^{-x/2} |f(x)| right) $$
is finite. Notice that $mathcal{C}$ is a complete normed space with respect to $|cdot|$. Write $p(x) = x^2 + 3x + 1$ and define
$$mathcal{A} = { f in mathcal{C} : f(x) geq 0 text{ for all } x geq 0 }. $$
This is a closed subset of $mathcal{C}$. Now define $Phi : mathcal{A} to mathcal{A}$ by
$$ Phi[f](x) = sqrt{p(x) + f(x+1)}. $$
If $f in mathcal{A}$, then $ 2^{-x/2}|Phi[f](x)| leq 2^{-x/2}sqrt{p(x) + 2^{(x+1)/2}|f|} $ shows that $|Phi[f]| < infty$, hence $Phi$ is well-defined. Moreover, if $f, g in mathcal{A}$, then
begin{align*}
2^{-x/2} left| Phi[f](x) - Phi[g](x) right|
&= 2^{-x/2} cdot frac{left| f(x+1) - g(x+1) right|}{sqrt{p(x) + f(x+1)} + sqrt{p(x) + g(x+1)}} \
&leq 2^{-x/2} cdot frac{2^{(x+1)/2} | f - g |}{2} \
&= frac{1}{sqrt{2}} | f - g |.
end{align*}
So $Phi$ is a contraction mapping on $mathcal{A}$, and hence, by the contraction mapping theorem,
- There exists a unique $F in mathcal{A}$ for which $Phi[F] = F$, and
- Such $F$ is realized as the limit $Phi^{circ n}[f]$ as $ntoinfty$ for arbitrary initial choice $f in mathcal{A}$.
Finally, we already know that $F(x) = x+2$ is an element of $mathcal{A}$ that solves $Phi[F] = F$, and therefore,
$$ forall f in mathcal{A} : quad lim_{ntoinfty} Phi^{circ n}[f](x) = x+2 $$
answered Jan 20 at 3:42
Sangchul LeeSangchul Lee
95.4k12170278
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add a comment |
$begingroup$
Maybe works, $$3=sqrt{3^{2}}=sqrt{5+4}=sqrt{5+sqrt{16}}=sqrt{5+sqrt{11+5}}$$
$$=sqrt{5+sqrt{11+sqrt{25}}}=sqrt{5+sqrt{11+sqrt{19+6}}}=sqrt{5+sqrt{11+sqrt{19+sqrt{36}}}}$$
$$=sqrt{5+sqrt{11+sqrt{19+sqrt{29+7}}}}=sqrt{5+sqrt{11+sqrt{19+sqrt{29+sqrt{49}}}}}=sqrt{5+sqrt{11+sqrt{19+sqrt{29+sqrt{41+8}}}}}=ldots$$
This is just a beautiful way of writing 3.
answered Jan 20 at 4:20
Pablo_Pablo_
1946
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1
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Last line should have $sqrt{41+8}$ not $sqrt {41+7}.$
$endgroup$
– Mohammad Zuhair Khan
Jan 20 at 4:32
add a comment |
$begingroup$
This is a slightly more rigorous form of @Pablo_'s excellent insight. @Sangchul Lee covers the full, analytic answer.
Set $a_n = n^2 + 5n + 5$ for $n geq 0$. This sequence gives the coefficients of the "infinite radical." Rather than consider the full infinite radical, consider the "partial radicals," defined as $$r_n = sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_n + (4 + n)}}}.$$
As Pablo_Lee notes, $r_n = 3$ for all $n$. To see this, observe that $a_n + (n + 4) = (n + 3)^2$. This allows us to "unroll" the radical back to $a_0$. For example, $$a_{n - 1} + sqrt{a_n + (n + 4)} = a_{n - 1} + n + 3 = a_{n - 1} + ((n - 1) + 4) = ((n - 1) + 3)^2.$$ Therefore,
$$
begin{align*}
r_n &= sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_{n - 1} + sqrt{a_n + (n + 4)}}}} \
&= sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_{n - 1} + ((n - 1) + 4)}}} \
&= sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_{n - 2} + ((n - 2) + 4)}}} \
&vdots \
&= sqrt{a_0 + sqrt{a_1 + 5}} \
&= sqrt{a_0 + 4} \
&= sqrt{(0 + 3)^2} \
&= 3.
end{align*}
$$
(There is likely a snappy way to do this by induction, but I don't see it yet.)
If we are willing to define the full radical as $lim_{n to infty} r_n$, then this should also be an acceptable answer.
Edit: For any integer $r geq 2$, setting $p_n = n^2 + (2r - 1)n + r^2 - r - 1$ and $q_n = n + r + 1$ should yield, through the same arguments, $$r = sqrt{p_0 + sqrt{p_1 + cdots + sqrt{p_n + q_n}}}$$ for all $n geq 0$. Note that $p_n$ is merely a shifted form of the Fibonacci polynomial $n^2 - n - 1$ at integer values.
For example, $$4 = sqrt{11 + sqrt{19 + sqrt{29 + sqrt{41 + 8}}}}.$$
answered Jan 20 at 5:05
rwboglrwbogl
1,027617
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add a comment |
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3 Answers
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3 Answers
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$begingroup$
We may adopt the technique for Ramanujan's infinite radical. Let $p(x) = x^2 + 3x + 1$ and define $F : [0, infty) to [0, infty)$ by
$$ F(x) = sqrt{p(x) + sqrt{p(x+1) + sqrt{p(x+2) + cdots }}} $$
Then $F$ solves
$$ F(x)^2 = p(x) + F(x+1). $$
Now we make an ansatz that $F(x)$ takes the form $F(x) = ax + b$. Plugging this and comparing coefficients shows that
$$ F(x) = x + 2 $$
solves the functional equation. Finally, since $(p(1), p(2), p(3), cdots) = (5, 11, 19, cdots) $, the infinite radical in question corresponds to the case $x = 1$, giving
$$ sqrt{5 + sqrt{11 + sqrt{19 + cdots}}} = F(1) = 3. $$
Rigorous justification. Let $mathcal{C}$ be the set of all continuous functions $f : [0, infty) to mathbb{R}$ such that
$$ | f| := sup_{xtoinfty} left( 2^{-x/2} |f(x)| right) $$
is finite. Notice that $mathcal{C}$ is a complete normed space with respect to $|cdot|$. Write $p(x) = x^2 + 3x + 1$ and define
$$mathcal{A} = { f in mathcal{C} : f(x) geq 0 text{ for all } x geq 0 }. $$
This is a closed subset of $mathcal{C}$. Now define $Phi : mathcal{A} to mathcal{A}$ by
$$ Phi[f](x) = sqrt{p(x) + f(x+1)}. $$
If $f in mathcal{A}$, then $ 2^{-x/2}|Phi[f](x)| leq 2^{-x/2}sqrt{p(x) + 2^{(x+1)/2}|f|} $ shows that $|Phi[f]| < infty$, hence $Phi$ is well-defined. Moreover, if $f, g in mathcal{A}$, then
begin{align*}
2^{-x/2} left| Phi[f](x) - Phi[g](x) right|
&= 2^{-x/2} cdot frac{left| f(x+1) - g(x+1) right|}{sqrt{p(x) + f(x+1)} + sqrt{p(x) + g(x+1)}} \
&leq 2^{-x/2} cdot frac{2^{(x+1)/2} | f - g |}{2} \
&= frac{1}{sqrt{2}} | f - g |.
end{align*}
So $Phi$ is a contraction mapping on $mathcal{A}$, and hence, by the contraction mapping theorem,
- There exists a unique $F in mathcal{A}$ for which $Phi[F] = F$, and
- Such $F$ is realized as the limit $Phi^{circ n}[f]$ as $ntoinfty$ for arbitrary initial choice $f in mathcal{A}$.
Finally, we already know that $F(x) = x+2$ is an element of $mathcal{A}$ that solves $Phi[F] = F$, and therefore,
$$ forall f in mathcal{A} : quad lim_{ntoinfty} Phi^{circ n}[f](x) = x+2 $$
answered Jan 20 at 3:42
Sangchul LeeSangchul Lee
95.4k12170278
$endgroup$
add a comment |
$begingroup$
We may adopt the technique for Ramanujan's infinite radical. Let $p(x) = x^2 + 3x + 1$ and define $F : [0, infty) to [0, infty)$ by
$$ F(x) = sqrt{p(x) + sqrt{p(x+1) + sqrt{p(x+2) + cdots }}} $$
Then $F$ solves
$$ F(x)^2 = p(x) + F(x+1). $$
Now we make an ansatz that $F(x)$ takes the form $F(x) = ax + b$. Plugging this and comparing coefficients shows that
$$ F(x) = x + 2 $$
solves the functional equation. Finally, since $(p(1), p(2), p(3), cdots) = (5, 11, 19, cdots) $, the infinite radical in question corresponds to the case $x = 1$, giving
$$ sqrt{5 + sqrt{11 + sqrt{19 + cdots}}} = F(1) = 3. $$
Rigorous justification. Let $mathcal{C}$ be the set of all continuous functions $f : [0, infty) to mathbb{R}$ such that
$$ | f| := sup_{xtoinfty} left( 2^{-x/2} |f(x)| right) $$
is finite. Notice that $mathcal{C}$ is a complete normed space with respect to $|cdot|$. Write $p(x) = x^2 + 3x + 1$ and define
$$mathcal{A} = { f in mathcal{C} : f(x) geq 0 text{ for all } x geq 0 }. $$
This is a closed subset of $mathcal{C}$. Now define $Phi : mathcal{A} to mathcal{A}$ by
$$ Phi[f](x) = sqrt{p(x) + f(x+1)}. $$
If $f in mathcal{A}$, then $ 2^{-x/2}|Phi[f](x)| leq 2^{-x/2}sqrt{p(x) + 2^{(x+1)/2}|f|} $ shows that $|Phi[f]| < infty$, hence $Phi$ is well-defined. Moreover, if $f, g in mathcal{A}$, then
begin{align*}
2^{-x/2} left| Phi[f](x) - Phi[g](x) right|
&= 2^{-x/2} cdot frac{left| f(x+1) - g(x+1) right|}{sqrt{p(x) + f(x+1)} + sqrt{p(x) + g(x+1)}} \
&leq 2^{-x/2} cdot frac{2^{(x+1)/2} | f - g |}{2} \
&= frac{1}{sqrt{2}} | f - g |.
end{align*}
So $Phi$ is a contraction mapping on $mathcal{A}$, and hence, by the contraction mapping theorem,
- There exists a unique $F in mathcal{A}$ for which $Phi[F] = F$, and
- Such $F$ is realized as the limit $Phi^{circ n}[f]$ as $ntoinfty$ for arbitrary initial choice $f in mathcal{A}$.
Finally, we already know that $F(x) = x+2$ is an element of $mathcal{A}$ that solves $Phi[F] = F$, and therefore,
$$ forall f in mathcal{A} : quad lim_{ntoinfty} Phi^{circ n}[f](x) = x+2 $$
answered Jan 20 at 3:42
Sangchul LeeSangchul Lee
95.4k12170278
$endgroup$
add a comment |
$begingroup$
We may adopt the technique for Ramanujan's infinite radical. Let $p(x) = x^2 + 3x + 1$ and define $F : [0, infty) to [0, infty)$ by
$$ F(x) = sqrt{p(x) + sqrt{p(x+1) + sqrt{p(x+2) + cdots }}} $$
Then $F$ solves
$$ F(x)^2 = p(x) + F(x+1). $$
Now we make an ansatz that $F(x)$ takes the form $F(x) = ax + b$. Plugging this and comparing coefficients shows that
$$ F(x) = x + 2 $$
solves the functional equation. Finally, since $(p(1), p(2), p(3), cdots) = (5, 11, 19, cdots) $, the infinite radical in question corresponds to the case $x = 1$, giving
$$ sqrt{5 + sqrt{11 + sqrt{19 + cdots}}} = F(1) = 3. $$
Rigorous justification. Let $mathcal{C}$ be the set of all continuous functions $f : [0, infty) to mathbb{R}$ such that
$$ | f| := sup_{xtoinfty} left( 2^{-x/2} |f(x)| right) $$
is finite. Notice that $mathcal{C}$ is a complete normed space with respect to $|cdot|$. Write $p(x) = x^2 + 3x + 1$ and define
$$mathcal{A} = { f in mathcal{C} : f(x) geq 0 text{ for all } x geq 0 }. $$
This is a closed subset of $mathcal{C}$. Now define $Phi : mathcal{A} to mathcal{A}$ by
$$ Phi[f](x) = sqrt{p(x) + f(x+1)}. $$
If $f in mathcal{A}$, then $ 2^{-x/2}|Phi[f](x)| leq 2^{-x/2}sqrt{p(x) + 2^{(x+1)/2}|f|} $ shows that $|Phi[f]| < infty$, hence $Phi$ is well-defined. Moreover, if $f, g in mathcal{A}$, then
begin{align*}
2^{-x/2} left| Phi[f](x) - Phi[g](x) right|
&= 2^{-x/2} cdot frac{left| f(x+1) - g(x+1) right|}{sqrt{p(x) + f(x+1)} + sqrt{p(x) + g(x+1)}} \
&leq 2^{-x/2} cdot frac{2^{(x+1)/2} | f - g |}{2} \
&= frac{1}{sqrt{2}} | f - g |.
end{align*}
So $Phi$ is a contraction mapping on $mathcal{A}$, and hence, by the contraction mapping theorem,
- There exists a unique $F in mathcal{A}$ for which $Phi[F] = F$, and
- Such $F$ is realized as the limit $Phi^{circ n}[f]$ as $ntoinfty$ for arbitrary initial choice $f in mathcal{A}$.
Finally, we already know that $F(x) = x+2$ is an element of $mathcal{A}$ that solves $Phi[F] = F$, and therefore,
$$ forall f in mathcal{A} : quad lim_{ntoinfty} Phi^{circ n}[f](x) = x+2 $$
answered Jan 20 at 3:42
Sangchul LeeSangchul Lee
95.4k12170278
$endgroup$
We may adopt the technique for Ramanujan's infinite radical. Let $p(x) = x^2 + 3x + 1$ and define $F : [0, infty) to [0, infty)$ by
$$ F(x) = sqrt{p(x) + sqrt{p(x+1) + sqrt{p(x+2) + cdots }}} $$
Then $F$ solves
$$ F(x)^2 = p(x) + F(x+1). $$
Now we make an ansatz that $F(x)$ takes the form $F(x) = ax + b$. Plugging this and comparing coefficients shows that
$$ F(x) = x + 2 $$
solves the functional equation. Finally, since $(p(1), p(2), p(3), cdots) = (5, 11, 19, cdots) $, the infinite radical in question corresponds to the case $x = 1$, giving
$$ sqrt{5 + sqrt{11 + sqrt{19 + cdots}}} = F(1) = 3. $$
Rigorous justification. Let $mathcal{C}$ be the set of all continuous functions $f : [0, infty) to mathbb{R}$ such that
$$ | f| := sup_{xtoinfty} left( 2^{-x/2} |f(x)| right) $$
is finite. Notice that $mathcal{C}$ is a complete normed space with respect to $|cdot|$. Write $p(x) = x^2 + 3x + 1$ and define
$$mathcal{A} = { f in mathcal{C} : f(x) geq 0 text{ for all } x geq 0 }. $$
This is a closed subset of $mathcal{C}$. Now define $Phi : mathcal{A} to mathcal{A}$ by
$$ Phi[f](x) = sqrt{p(x) + f(x+1)}. $$
If $f in mathcal{A}$, then $ 2^{-x/2}|Phi[f](x)| leq 2^{-x/2}sqrt{p(x) + 2^{(x+1)/2}|f|} $ shows that $|Phi[f]| < infty$, hence $Phi$ is well-defined. Moreover, if $f, g in mathcal{A}$, then
begin{align*}
2^{-x/2} left| Phi[f](x) - Phi[g](x) right|
&= 2^{-x/2} cdot frac{left| f(x+1) - g(x+1) right|}{sqrt{p(x) + f(x+1)} + sqrt{p(x) + g(x+1)}} \
&leq 2^{-x/2} cdot frac{2^{(x+1)/2} | f - g |}{2} \
&= frac{1}{sqrt{2}} | f - g |.
end{align*}
So $Phi$ is a contraction mapping on $mathcal{A}$, and hence, by the contraction mapping theorem,
- There exists a unique $F in mathcal{A}$ for which $Phi[F] = F$, and
- Such $F$ is realized as the limit $Phi^{circ n}[f]$ as $ntoinfty$ for arbitrary initial choice $f in mathcal{A}$.
Finally, we already know that $F(x) = x+2$ is an element of $mathcal{A}$ that solves $Phi[F] = F$, and therefore,
$$ forall f in mathcal{A} : quad lim_{ntoinfty} Phi^{circ n}[f](x) = x+2 $$
answered Jan 20 at 3:42
Sangchul LeeSangchul Lee
95.4k12170278
edited Jan 20 at 4:20
answered Jan 20 at 3:42
Sangchul LeeSangchul Lee
95.4k12170278
answered Jan 20 at 3:42
Sangchul LeeSangchul Lee
95.4k12170278
answered Jan 20 at 3:42
Sangchul LeeSangchul Lee
95.4k12170278
95.4k12170278
add a comment |
add a comment |
$begingroup$
Maybe works, $$3=sqrt{3^{2}}=sqrt{5+4}=sqrt{5+sqrt{16}}=sqrt{5+sqrt{11+5}}$$
$$=sqrt{5+sqrt{11+sqrt{25}}}=sqrt{5+sqrt{11+sqrt{19+6}}}=sqrt{5+sqrt{11+sqrt{19+sqrt{36}}}}$$
$$=sqrt{5+sqrt{11+sqrt{19+sqrt{29+7}}}}=sqrt{5+sqrt{11+sqrt{19+sqrt{29+sqrt{49}}}}}=sqrt{5+sqrt{11+sqrt{19+sqrt{29+sqrt{41+8}}}}}=ldots$$
This is just a beautiful way of writing 3.
answered Jan 20 at 4:20
Pablo_Pablo_
1946
$endgroup$
1
$begingroup$
Last line should have $sqrt{41+8}$ not $sqrt {41+7}.$
$endgroup$
– Mohammad Zuhair Khan
Jan 20 at 4:32
add a comment |
$begingroup$
Maybe works, $$3=sqrt{3^{2}}=sqrt{5+4}=sqrt{5+sqrt{16}}=sqrt{5+sqrt{11+5}}$$
$$=sqrt{5+sqrt{11+sqrt{25}}}=sqrt{5+sqrt{11+sqrt{19+6}}}=sqrt{5+sqrt{11+sqrt{19+sqrt{36}}}}$$
$$=sqrt{5+sqrt{11+sqrt{19+sqrt{29+7}}}}=sqrt{5+sqrt{11+sqrt{19+sqrt{29+sqrt{49}}}}}=sqrt{5+sqrt{11+sqrt{19+sqrt{29+sqrt{41+8}}}}}=ldots$$
This is just a beautiful way of writing 3.
answered Jan 20 at 4:20
Pablo_Pablo_
1946
$endgroup$
1
$begingroup$
Last line should have $sqrt{41+8}$ not $sqrt {41+7}.$
$endgroup$
– Mohammad Zuhair Khan
Jan 20 at 4:32
add a comment |
$begingroup$
Maybe works, $$3=sqrt{3^{2}}=sqrt{5+4}=sqrt{5+sqrt{16}}=sqrt{5+sqrt{11+5}}$$
$$=sqrt{5+sqrt{11+sqrt{25}}}=sqrt{5+sqrt{11+sqrt{19+6}}}=sqrt{5+sqrt{11+sqrt{19+sqrt{36}}}}$$
$$=sqrt{5+sqrt{11+sqrt{19+sqrt{29+7}}}}=sqrt{5+sqrt{11+sqrt{19+sqrt{29+sqrt{49}}}}}=sqrt{5+sqrt{11+sqrt{19+sqrt{29+sqrt{41+8}}}}}=ldots$$
This is just a beautiful way of writing 3.
answered Jan 20 at 4:20
Pablo_Pablo_
1946
$endgroup$
Maybe works, $$3=sqrt{3^{2}}=sqrt{5+4}=sqrt{5+sqrt{16}}=sqrt{5+sqrt{11+5}}$$
$$=sqrt{5+sqrt{11+sqrt{25}}}=sqrt{5+sqrt{11+sqrt{19+6}}}=sqrt{5+sqrt{11+sqrt{19+sqrt{36}}}}$$
$$=sqrt{5+sqrt{11+sqrt{19+sqrt{29+7}}}}=sqrt{5+sqrt{11+sqrt{19+sqrt{29+sqrt{49}}}}}=sqrt{5+sqrt{11+sqrt{19+sqrt{29+sqrt{41+8}}}}}=ldots$$
This is just a beautiful way of writing 3.
answered Jan 20 at 4:20
Pablo_Pablo_
1946
edited Jan 20 at 4:33
answered Jan 20 at 4:20
Pablo_Pablo_
1946
answered Jan 20 at 4:20
Pablo_Pablo_
1946
answered Jan 20 at 4:20
Pablo_Pablo_
1946
1946
1
$begingroup$
Last line should have $sqrt{41+8}$ not $sqrt {41+7}.$
$endgroup$
– Mohammad Zuhair Khan
Jan 20 at 4:32
add a comment |
1
$begingroup$
Last line should have $sqrt{41+8}$ not $sqrt {41+7}.$
$endgroup$
– Mohammad Zuhair Khan
Jan 20 at 4:32
1
1
$begingroup$
Last line should have $sqrt{41+8}$ not $sqrt {41+7}.$
$endgroup$
– Mohammad Zuhair Khan
Jan 20 at 4:32
$begingroup$
Last line should have $sqrt{41+8}$ not $sqrt {41+7}.$
$endgroup$
– Mohammad Zuhair Khan
Jan 20 at 4:32
add a comment |
$begingroup$
This is a slightly more rigorous form of @Pablo_'s excellent insight. @Sangchul Lee covers the full, analytic answer.
Set $a_n = n^2 + 5n + 5$ for $n geq 0$. This sequence gives the coefficients of the "infinite radical." Rather than consider the full infinite radical, consider the "partial radicals," defined as $$r_n = sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_n + (4 + n)}}}.$$
As Pablo_Lee notes, $r_n = 3$ for all $n$. To see this, observe that $a_n + (n + 4) = (n + 3)^2$. This allows us to "unroll" the radical back to $a_0$. For example, $$a_{n - 1} + sqrt{a_n + (n + 4)} = a_{n - 1} + n + 3 = a_{n - 1} + ((n - 1) + 4) = ((n - 1) + 3)^2.$$ Therefore,
$$
begin{align*}
r_n &= sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_{n - 1} + sqrt{a_n + (n + 4)}}}} \
&= sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_{n - 1} + ((n - 1) + 4)}}} \
&= sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_{n - 2} + ((n - 2) + 4)}}} \
&vdots \
&= sqrt{a_0 + sqrt{a_1 + 5}} \
&= sqrt{a_0 + 4} \
&= sqrt{(0 + 3)^2} \
&= 3.
end{align*}
$$
(There is likely a snappy way to do this by induction, but I don't see it yet.)
If we are willing to define the full radical as $lim_{n to infty} r_n$, then this should also be an acceptable answer.
Edit: For any integer $r geq 2$, setting $p_n = n^2 + (2r - 1)n + r^2 - r - 1$ and $q_n = n + r + 1$ should yield, through the same arguments, $$r = sqrt{p_0 + sqrt{p_1 + cdots + sqrt{p_n + q_n}}}$$ for all $n geq 0$. Note that $p_n$ is merely a shifted form of the Fibonacci polynomial $n^2 - n - 1$ at integer values.
For example, $$4 = sqrt{11 + sqrt{19 + sqrt{29 + sqrt{41 + 8}}}}.$$
answered Jan 20 at 5:05
rwboglrwbogl
1,027617
$endgroup$
add a comment |
$begingroup$
This is a slightly more rigorous form of @Pablo_'s excellent insight. @Sangchul Lee covers the full, analytic answer.
Set $a_n = n^2 + 5n + 5$ for $n geq 0$. This sequence gives the coefficients of the "infinite radical." Rather than consider the full infinite radical, consider the "partial radicals," defined as $$r_n = sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_n + (4 + n)}}}.$$
As Pablo_Lee notes, $r_n = 3$ for all $n$. To see this, observe that $a_n + (n + 4) = (n + 3)^2$. This allows us to "unroll" the radical back to $a_0$. For example, $$a_{n - 1} + sqrt{a_n + (n + 4)} = a_{n - 1} + n + 3 = a_{n - 1} + ((n - 1) + 4) = ((n - 1) + 3)^2.$$ Therefore,
$$
begin{align*}
r_n &= sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_{n - 1} + sqrt{a_n + (n + 4)}}}} \
&= sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_{n - 1} + ((n - 1) + 4)}}} \
&= sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_{n - 2} + ((n - 2) + 4)}}} \
&vdots \
&= sqrt{a_0 + sqrt{a_1 + 5}} \
&= sqrt{a_0 + 4} \
&= sqrt{(0 + 3)^2} \
&= 3.
end{align*}
$$
(There is likely a snappy way to do this by induction, but I don't see it yet.)
If we are willing to define the full radical as $lim_{n to infty} r_n$, then this should also be an acceptable answer.
Edit: For any integer $r geq 2$, setting $p_n = n^2 + (2r - 1)n + r^2 - r - 1$ and $q_n = n + r + 1$ should yield, through the same arguments, $$r = sqrt{p_0 + sqrt{p_1 + cdots + sqrt{p_n + q_n}}}$$ for all $n geq 0$. Note that $p_n$ is merely a shifted form of the Fibonacci polynomial $n^2 - n - 1$ at integer values.
For example, $$4 = sqrt{11 + sqrt{19 + sqrt{29 + sqrt{41 + 8}}}}.$$
answered Jan 20 at 5:05
rwboglrwbogl
1,027617
$endgroup$
add a comment |
$begingroup$
This is a slightly more rigorous form of @Pablo_'s excellent insight. @Sangchul Lee covers the full, analytic answer.
Set $a_n = n^2 + 5n + 5$ for $n geq 0$. This sequence gives the coefficients of the "infinite radical." Rather than consider the full infinite radical, consider the "partial radicals," defined as $$r_n = sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_n + (4 + n)}}}.$$
As Pablo_Lee notes, $r_n = 3$ for all $n$. To see this, observe that $a_n + (n + 4) = (n + 3)^2$. This allows us to "unroll" the radical back to $a_0$. For example, $$a_{n - 1} + sqrt{a_n + (n + 4)} = a_{n - 1} + n + 3 = a_{n - 1} + ((n - 1) + 4) = ((n - 1) + 3)^2.$$ Therefore,
$$
begin{align*}
r_n &= sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_{n - 1} + sqrt{a_n + (n + 4)}}}} \
&= sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_{n - 1} + ((n - 1) + 4)}}} \
&= sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_{n - 2} + ((n - 2) + 4)}}} \
&vdots \
&= sqrt{a_0 + sqrt{a_1 + 5}} \
&= sqrt{a_0 + 4} \
&= sqrt{(0 + 3)^2} \
&= 3.
end{align*}
$$
(There is likely a snappy way to do this by induction, but I don't see it yet.)
If we are willing to define the full radical as $lim_{n to infty} r_n$, then this should also be an acceptable answer.
Edit: For any integer $r geq 2$, setting $p_n = n^2 + (2r - 1)n + r^2 - r - 1$ and $q_n = n + r + 1$ should yield, through the same arguments, $$r = sqrt{p_0 + sqrt{p_1 + cdots + sqrt{p_n + q_n}}}$$ for all $n geq 0$. Note that $p_n$ is merely a shifted form of the Fibonacci polynomial $n^2 - n - 1$ at integer values.
For example, $$4 = sqrt{11 + sqrt{19 + sqrt{29 + sqrt{41 + 8}}}}.$$
answered Jan 20 at 5:05
rwboglrwbogl
1,027617
$endgroup$
This is a slightly more rigorous form of @Pablo_'s excellent insight. @Sangchul Lee covers the full, analytic answer.
Set $a_n = n^2 + 5n + 5$ for $n geq 0$. This sequence gives the coefficients of the "infinite radical." Rather than consider the full infinite radical, consider the "partial radicals," defined as $$r_n = sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_n + (4 + n)}}}.$$
As Pablo_Lee notes, $r_n = 3$ for all $n$. To see this, observe that $a_n + (n + 4) = (n + 3)^2$. This allows us to "unroll" the radical back to $a_0$. For example, $$a_{n - 1} + sqrt{a_n + (n + 4)} = a_{n - 1} + n + 3 = a_{n - 1} + ((n - 1) + 4) = ((n - 1) + 3)^2.$$ Therefore,
$$
begin{align*}
r_n &= sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_{n - 1} + sqrt{a_n + (n + 4)}}}} \
&= sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_{n - 1} + ((n - 1) + 4)}}} \
&= sqrt{a_0 + sqrt{a_1 + cdots + sqrt{a_{n - 2} + ((n - 2) + 4)}}} \
&vdots \
&= sqrt{a_0 + sqrt{a_1 + 5}} \
&= sqrt{a_0 + 4} \
&= sqrt{(0 + 3)^2} \
&= 3.
end{align*}
$$
(There is likely a snappy way to do this by induction, but I don't see it yet.)
If we are willing to define the full radical as $lim_{n to infty} r_n$, then this should also be an acceptable answer.
Edit: For any integer $r geq 2$, setting $p_n = n^2 + (2r - 1)n + r^2 - r - 1$ and $q_n = n + r + 1$ should yield, through the same arguments, $$r = sqrt{p_0 + sqrt{p_1 + cdots + sqrt{p_n + q_n}}}$$ for all $n geq 0$. Note that $p_n$ is merely a shifted form of the Fibonacci polynomial $n^2 - n - 1$ at integer values.
For example, $$4 = sqrt{11 + sqrt{19 + sqrt{29 + sqrt{41 + 8}}}}.$$
answered Jan 20 at 5:05
rwboglrwbogl
1,027617
edited Jan 20 at 16:42
answered Jan 20 at 5:05
rwboglrwbogl
1,027617
answered Jan 20 at 5:05
rwboglrwbogl
1,027617
answered Jan 20 at 5:05
rwboglrwbogl
1,027617
1,027617
add a comment |
add a comment |
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$begingroup$
Numerically, it seems to converge to 3.
$endgroup$
– greelious
Jan 20 at 3:16
$begingroup$
Thanks, but I actually need to compute it theoretically.
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– Goal123
Jan 20 at 3:17
2
$begingroup$
Please give the rule for the coefficients $5, 11, 19, ldots$ in your formula (which, by the way, is not a sum).
$endgroup$
– Rob Arthan
Jan 20 at 3:31
2
$begingroup$
The terms seem to be given by $n^2 + 3n + 1$ for $n geq 1$.
$endgroup$
– rwbogl
Jan 20 at 3:40
1
$begingroup$
Just a nitpick on the title: this is not actually a sum.
$endgroup$
– Cheerful Parsnip
Jan 20 at 5:49