Questions tagged [nested-radicals]
In algebra, a nested radical is a radical expression (one containing a square root sign, cube root sign, etc.) that contains (nests) another radical expression.
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Calculate $\frac{\pi}{2}-\sin(\frac{\pi}{2})+\sin(\sin(\frac{\pi}{2}))-\sin(\sin(\sin(\frac{\pi}{2})))\cdots$
I would like to calculate following sum:
$$\frac{\pi}{2}-\sin\left(\frac{\pi}{2}\right)+\sin\left(\sin\left(\frac{\pi}{2}\right)\right)-\sin\left(\sin\left(\sin\left(\frac{\pi}{2}\right)\right)\right)+...
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Prove that $\frac{1}{\sqrt[3]2}=\sqrt{\frac 5{\sqrt[3]4}-1}-\sqrt{(3-\sqrt[3]2)(\sqrt[3]2-1)}$
Playing around with denesting radicals, I arrived at the following formula which appears to be correct.
$$\frac 1{\sqrt[3]2}=\sqrt{\frac 5{\sqrt[3]4}-1}-\sqrt{(3-\sqrt[3]2)(\sqrt[3]2-1)}$$
If one ...
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2answers
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Find the value of $\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\cdots}}}}}$ .
We have:
$\sqrt{(1+0)+\sqrt{(4+1)+\sqrt{(9+2)+\sqrt{(16+3)+\sqrt{(25+4)+\cdots}}}}}.$
Basically I'm not getting any clue at the moment for reducing the infinite nested radicals. Any hint would be ...
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1answer
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Proving The Nested Radical
For $a>b^2$, prove that $\sqrt{a-b\sqrt{a+b\sqrt{a-b\sqrt{a+\cdots}}}} = \sqrt{a-\dfrac34b^2}-\dfrac12b$.
Attempt:
After assuming the value of the nested radicals to be $S$, I got $$S = \dfrac{\...
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4answers
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Find the two complex numbers whose sum is $5 - i$ and product is $8+i$
I have tried to solve this problem, but when I write the equations I end up with:
$$a + c = 5$$
$$b + d = -1$$
$$ab - cd = 8$$
$$ad + cb = 1$$
But substituting $a$ and $b$ get 2 equations I can't be ...
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2answers
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What is value of $\sqrt[n]{x\sqrt[n]{x\sqrt[n]{x\sqrt[n]{x\ldots}}}}=$?
My question: how to find nested radical having $n$th roots $$\large\sqrt[n]{x\sqrt[n]{x\sqrt[n]{x\sqrt[n]{x\ldots}}}}=?$$
My try:
$$\large\sqrt[n]{x\sqrt[n]{x\sqrt[n]{x\sqrt[n]{x\ldots}}}}=y$$
$$\...
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1answer
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Product $\left[\sin(x)\cos\left(\frac{x}2\right)\right]^{1/2}\cdot\left[\sin\left(\frac{x}{2}\right) \cos \left(\frac{x}4\right)\right]^{1/4}\ \cdots$
I came across this question in the following form:
Compute the following infinite product
$$\left[\sin (x)\cos \left(\frac{x}{2}\right)\right]^{1/2}\cdot \left[\sin \left(\frac{x}{2}\right) \cos \...
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On the nested square roots $\sqrt{1^2+\sqrt{2^2+\sqrt{3^2 …+\sqrt{(n-1)^2+\sqrt{n^2}}}}}$
I saw this on quora:
Use induction to show that
$\sqrt{1^2+\sqrt{2^2+\sqrt{3^2
...+\sqrt{(n-1)^2+\sqrt{n^2}}}}}
\le 2
$.
My question is simpler.
If
$x_n
=\sqrt{1^2+\sqrt{2^2+\sqrt{3^2
...+\sqrt{(n-...
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Conditions to be followed while Infinite Nesting so that the nested expression gives the intended initial expression?
I'd like to start from the problem which lead me to the whole question and then get into the broader question.
Given $ \sum _{k=1}^\infty \lfloor \frac n{2^k} \rfloor$
I thought the sum evaluates to ...
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1answer
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Roots of a quadratic equation that is obtained from nested radicals
Take any $ 0 \lt k \in \mathbb R$, and "algebraically massage" it into a nested radical:
$$k=\sqrt{k^2}=\sqrt{k^2+k-k}=\sqrt{k+(k-1)k}=\sqrt{k+(k-1)\sqrt{k^2}}= \ .\ .\ .\ = \sqrt{k+(k-1)\sqrt{k+(k-1)\...
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Can someone help me evaluate this infinite square root expression? I canāt seem to find a way to simplify it.
I tried to separate the 5 into 1 + 4 and the 11 into 8 + 3 and so on for the next number. This isnāt like any other problems with nested radicals that iāve solved. I hope i can learn something from ...
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2answers
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Find a closed form to the solution of $\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-x}}}}}=x$
Hi I try to solve the following nested radical :
$$\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-x}}}}}=x$$
Miraculously the related polynomials is a quintic .More precisely :
$$ x^5 - x^4 - 4 x^3 + 3 x^...
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3answers
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Why $\sqrt{23-\sqrt{17}}-2\sqrt{7-\sqrt{17}}=\sqrt{71-17\sqrt{17}}$ is true?
Easy to show this identity after squaring twice of the both sides.
But why it turned out true?
For example, if we want to prove that
$$\sqrt{23-3\sqrt{5}}-2\sqrt{3-\sqrt{5}}=\sqrt{3+\sqrt{5}},$$
we ...
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votes
1answer
158 views
On the formula, $\pi = \frac 5\varphi\cdot\frac 2{\sqrt{2+\sqrt{2+\varphi}}}\cdot\frac 2{\sqrt{2+\sqrt{2+\sqrt{2+\varphi}}}}\cdots$
I found a formula on google images when I was looking at some formulas for $\pi$ just for the fun of it, and I came across one that really startled me, and was quite reminiscent of ViĆØte's product.
...
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Proof $\frac{-\sqrt{2 - \sqrt{2 - \sqrt{2}}}}{\sqrt2} + \frac{\sqrt{2 + \sqrt2}}{\sqrt2 \sqrt{2 - \sqrt{2 - \sqrt2}}} =\sqrt{2-\sqrt{2+\sqrt2}}$
I want to prove the following equation
$$\frac{-\sqrt{2 - \sqrt{2 - \sqrt{2}}}}{\sqrt{2}} + \frac{\sqrt{2 + \sqrt{2}}}{\sqrt{2} \sqrt{2 - \sqrt{2 - \sqrt{2}}}}$$
I know this must be equal to $\sqrt{2-...
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1answer
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Two nested radicals with similar result
Sorry for my last post it's my bad .So I ask to this (the true ^^)nested radical :
$$\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{\cdots }}}}}}}}=\frac{1+\sqrt{5}+\sqrt{30-6\sqrt{5}}}...
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1answer
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A false nested radical of Ramanujan and a true nested radical
On the wikipedia page I can see a nested radical by Ramanujan :
$$\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{\cdots}}}}}}}=\frac{2+\sqrt{5}+\sqrt{15-6\sqrt{5}}}{2}$$
Wolfram alpha says it'...
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1answer
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Finding the value of x, $x=\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+…} } } } $
I want to find the value of x where x is given by the following: $$x=\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+....\infty} } } } $$
Noting that the above can be written in the following form: $x=\sqrt{0+x} $, ...
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1answer
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Does this infinite nested radical show that $\pi$ is irrational?
A $\sqrt{2}$ is at the end (?!) of this infinite nested radical expression. Is that enough to show that $\pi$ is irrational? Also, I'm curious if there is a better way to write this? Thanks!
$$\lim_{n\...
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7answers
578 views
When does $\sqrt{x+\sqrt{x+1+\sqrt{x+2+…}}}=0$?
Consider the function $f$ defined as the limit of the functions
$$f_0(x)=\sqrt{x}$$
$$f_1(x)=\sqrt{x+\sqrt{x+1}}$$
$$f_2(x)=\sqrt{x+\sqrt{x+1+\sqrt{x+2}}}$$
$$...$$
so that $f(x)$ is defined iff $f_n(...
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An example of nested radical and power tower .$e^{-1}=(e^{-1})^{\sqrt{e^{-1}+(e^1-1)(e^{-1})^{\sqrt{e^{-1}+(e^1-1)(e^{-1})^{\sqrt{\cdots}}}}}}$
I want to share with you some of my last work:
$$e^{-1}=(e^{-1})^{\sqrt{e^{-1}+(e^1-1)(e^{-1})^{\sqrt{e^{-1}+(e^1-1)(e^{-1})^{\sqrt{\cdots}}}}}}$$
It's easy to solve using logarithm but I would like ...
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1answer
60 views
Conjecture on a functional equality and nested radical of Ramanujan :$3=\sqrt{1+f(2)\sqrt{1+f(3)\sqrt{1+\cdots}}}$
Hi i was wondering something about the great Ramanujan :
I think moreover I'm not the only one who propose this kind of problem (so if you have a link related to this subject).
We have :
$$3=\sqrt{1+...
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Almost integer with nested radicals and power tower .
playing with power tower and nested radicals I get :
Prove that
Let $a_1=\sqrt{2}$ ,$a_2=\sqrt{2}^{\sqrt{2}}$,$a_3=\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}$,$a_4=\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}...
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How to evaluate the integral $\int_{-1}^{0}\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{\cdots\sqrt{1+x}}}}}dx=?$
We have :
$$\int_{-1}^{0}\sqrt{1+\sqrt{1+\sqrt{1+x}}}dx=\frac{8}{315}\sqrt{2}\Big(16+\sqrt{233+317\sqrt{2}}\Big)$$
We are lucky because this integral have an anti-derivative like here.
More ...
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1answer
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On proving an infinite-nested radical
I was playing around with square roots and I noticed that the number $1$ can be seemingly expressed as an infinite nested radical with an easy pattern. I then noticed that if this is true, this would ...
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How to solve $10\sqrt{10\sqrt[3]{10\sqrt[4]{10…}}}$? [closed]
How to solve $10\sqrt{10\sqrt[3]{10\sqrt[4]{10...}}}$?
I tried to solve this problem by letting $x=10\sqrt{10\sqrt[3]{10\sqrt[4]{10...}}}$ to observe the pattern.
Based on the pattern, the result is
...
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AbelāRuffini theorem, connection to nested radicals
I hope someone can help me with understanding the AbelāRuffini theorem, especially the connection to nested radicals. I do know the proof based on Galois theory, but I wondered if this text which is ...
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1answer
37 views
Complex nested radicals ${\Re}\Big(\sqrt{1+\frac{i}{2}\sqrt{1+\frac{i}{2^2}\sqrt{1+\frac{i}{2^3}\sqrt{1+\frac{i}{2^4}\sqrt{\cdots}}}}}\Big)=1$
A Last question on nested radicals but this time with complex value :
$${\Re}\Big(\sqrt{1+\frac{i}{2}\sqrt{1+\frac{i}{2^2}\sqrt{1+\frac{i}{2^3}\sqrt{1+\frac{i}{2^4}\sqrt{\cdots}}}}}\Big)=1$$
I have ...
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Algebra point of view on Tribonacci constant with nested radical
Trying to find an expression in term of nested radical for the tribonacci constant we get a messy result starting with :
$$S=\sqrt{1+S+\frac{1}{S}}$$
I have tried another expression like :
$$S=\...
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Special power tower $x^{\sqrt{x+x^{\sqrt{x+x^{\sqrt{\cdots}}}}}}$ and generalized Lambert's function
I'm interested by the following "nested radical-power tower" we have :
$$x^{\sqrt{x+x^{\sqrt{x+x^{\sqrt{\cdots}}}}}}=S$$
My try :
We have taking logarithm on both side :
$$\ln(S)=\sqrt{x+S}\ln(x)$$
...
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Inequality of the April month $a^{\sqrt{b+b^{\sqrt{2b}}}}+b^{\sqrt{a+a^{\sqrt{2a}}}}\leq 1$
the main idea was to create an inequality based on the well know inequality of Vasile Cirtoaje and add some nested radicals it gives :
Let $a,b>0$ such that $a+b=1$ then we have :
$$a^{...
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1answer
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Find $m^3$ if $m=\sqrt[3]{a + \frac{a+8}{3}\sqrt{\frac{a-1}{3}}} + \sqrt[3]{a - \frac{a+8}{3}\sqrt\frac{a-1}{3}}$
Please help me solve this question in a easy way:
$$
\sqrt[3]{a + \frac{a+8}{3}\sqrt{\frac{a-1}{3}}} + \sqrt[3]{a - \frac{a+8}{3}\sqrt\frac{a-1}{3}} = m
$$
Find $m^3$.
(The answer is $8$.)
I ...
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3answers
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Showing $\frac34=\sqrt{1-\frac{1}{2}\sqrt{1-\frac{1}{4}\sqrt{1-\frac{1}{8}\sqrt{\cdots}}}}$
$$\frac34=\sqrt{1-\frac{1}{2}\sqrt{1-\frac{1}{4}\sqrt{1-\frac{1}{8}\sqrt{\cdots}}}}$$
We cannot use the Herschfeld's Convergence Theorem because you can see that it's negative. I have tried some ...
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3answers
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Solutions of the equation $x=\sqrt{2 + \sqrt{2 +\sqrt 2+…}}$
I have seen the equation $x=\sqrt{2 + \sqrt{2 +\sqrt 2+.....}}$ in many places and the answer is $x=2$ which is obtained by making the substitution for $x$ inside the radical sign i.e $x=\sqrt{2+x}$ ...
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1answer
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Proving that there exists $n,a \in \mathbb{N}$ such that $n = \sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+…}}}}$ for all $n \in \mathbb{N}$
I suspect that there are infinite solutions of $n,a \in \mathbb{N}$ for all $n \in \mathbb{N}$, $n>1$ such that $$n = \sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+...}}}}$$
Some solutions include, for $(n,a):$ $...
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1answer
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How to simplify fractions involving surds? $\frac{(\sqrt{Īø^2+1})(2Īø)-(Īø^2-1)(\frac{Īø}{\sqrt{Īø^2+1}})}{Īø^2+1}$
I came across this example of the Quotient Rule and can't understand how it was simplified from step $2$ to $3$. How were the roots turned into the power of $3/2$?
Find $\frac{du}{d\theta}$ when $u = ...
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votes
2answers
74 views
Determine whether $\sqrt[n]{1+\sqrt[n]{2+\sqrt[n]{3+\cdots+\sqrt[n]{n}}}}$ diverges or not
Determine whether $\{a_n\}$ is convergent or not, where
$$a_n:=\sqrt[n]{1+\sqrt[n]{2+\sqrt[n]{3+\cdots+\sqrt[n]{n}}}}.$$
At least, we can obtain$$\sqrt[n]{1+\sqrt[n]{2+\sqrt[n]{3+\cdots+\sqrt[n]{n}}...
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0answers
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Does this have repeated operation have a name?
For some $k\in \mathbb{R}$ taking $\sqrt{\sqrt{\dotsb\sqrt{k}}}$ with $N$ radicals, then taking $N\to \infty$, will give $1$. Does this series of operations have a name?
6
votes
2answers
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Evaluating $\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-\cdots}}}}}$.
I was wondering if it was possible to evaluate
$$\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-\cdots}}}}}}}$$
I let the expression equal $x>0$ and wrote $$x=\sqrt{9-5\sqrt{3-x}}$$ ...
2
votes
2answers
140 views
$a_n=\sqrt{2-a_{n-1}},a_1=\sqrt{2}$
Given
$$a_n=\sqrt{2-a_{n-1}},a_1=\sqrt{2}$$
I calculated $a_1$ to $a_5$
$$\sqrt{2},
\sqrt{2-\sqrt{2}},
\sqrt{2-\sqrt{2-\sqrt{2}}}, \\
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}, \\
\sqrt{2-\sqrt{2-\sqrt{2-\...
6
votes
0answers
107 views
Value of $\sqrt{\sum_{k=1}^{1}k+\sqrt{\sum_{k=1}^{2}k+\sqrt{\sum_{k=1}^{3}k…}}}$
Note: I would like to add this is very similar to this problem posted a couple months ago, which is the inspiration for this question, was proven to converge in one of the answers, and I believe a ...
7
votes
2answers
114 views
Minimal polynomial of $\sqrt{3+2\sqrt{3}}$
Let $a=\sqrt{3+2\sqrt{3}}$. Then
\begin{align*}
&a=\sqrt{3+2\sqrt{3}}\\
&\implies a^2=3+2\sqrt{3}\\
&\implies a^2-3 = 2\sqrt{3}\\
&\implies (a^2-3)^2 = 4\cdot 3=12\\
&\implies (a^2-...
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votes
1answer
87 views
Denesting $\sqrt{20+ \sqrt{96}+\sqrt{12}}$ into four possible radicals
Denesting $\sqrt{20+ \sqrt{96}+\sqrt{12}}$ into possible radicals. This is an answer to an obscure closed question, here on the site. While there is an answer posted, it isn't the complete solution.
2
votes
3answers
128 views
Prove that $\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}} = 3$ without using Cardano's formula. (Hint, what is $(3\pm \sqrt{13})^3$
Prove that $\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}} = 3$ without using Cardano's formula. (Hint, what is $(3\pm \sqrt{13})^3$
I have that $$(3 + \sqrt{13})^3 = 144 + 40 \sqrt{13} $$ ...
2
votes
1answer
60 views
Infinite complex nested radical and it's complex conjugate.
Today I want to play with $i$ the imaginary unit I have this :
$$\overline{\sqrt{1+i\sqrt{1+i^2\sqrt{1+i^3\sqrt{1+i^4\sqrt{\cdots}}}}}}=\sqrt{1+\frac{1}{i}\sqrt{1+\frac{1}{i^2}\sqrt{1+\frac{1}{i^3}\...
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vote
3answers
108 views
Show $\frac{\sqrt{\sqrt[4]8-\sqrt{\sqrt2+1}\;}}{\sqrt{\sqrt[4]8+\sqrt{\sqrt2-1}\;} -\sqrt{\sqrt[4]8-\sqrt{\sqrt2-1}\;}}=\frac1{\sqrt2}$
Days ago, I tried to demonstrate this equality, reducing radicals, multiplying by the conjugate of the denominator, etc. But, I did not reach anything similar to the right side.
$$
\frac{\sqrt{\sqrt[...
4
votes
1answer
108 views
How to prove that $\sqrt{1+\frac{1}{2}\sqrt{1+\frac{1}{3}\sqrt{1+\frac{1}{4}\sqrt{\cdots}}}}<\sqrt[3] 2$
It's an estimation that I find interesting :
$$\sqrt{1+\frac{1}{2}\sqrt{1+\frac{1}{3}\sqrt{1+\frac{1}{4}\sqrt{1+\frac{1}{5}\sqrt{\cdots}}}}}<\sqrt[\leftroot{-0}\uproot{0}3]{2}$$
I think to ...
8
votes
1answer
120 views
Nested radical and rationality
I did not found it on the forum so :
$$S=\sqrt{1+\frac{1}{2}\sqrt{1+\frac{1}{2^2}\sqrt{1+\frac{1}{2^3}\sqrt{1+\frac{1}{2^4}\sqrt{\cdots}}}}}=1.25$$
I try to denested the radical but I'm really stuck ...
2
votes
1answer
48 views
Nested radical of Ramanujan and cyclic inequality
Inspired by a nested radical of Ramanujan I propose this :
Let $a,b,c\in (2\sqrt[4\,]{5},6-2\sqrt[4\,]{5})$ such that $a+b+c=9$ then we have :
$$\sqrt[4\,]{\frac{a+2\sqrt[4\,]{5}}{a-2\sqrt[4\,]{5}...
1
vote
1answer
46 views
Denest $\sqrt{3+\sin\frac{\pi}{4}-4\cos\frac{\pi}{8}}$ in terms of possible trigonometric functions.
As the title suggests, denest;
$$\sqrt{3+\sin\frac{\pi}{4}-4\cos\frac{\pi}{8}}$$
in terms of possible trigonometric functions.
