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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76 views

Finding the value of $\sqrt{z \sqrt{z \sqrt{z}}}…$

I was working on the following nested square root problem: Let $a \in \mathbb R ^+$, what is the value of: $$\sqrt{a \sqrt{a \sqrt{a}}}...$$ I concluded that the answer is $a$ and then I thought ...
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$\sqrt{n+\sqrt{n+\sqrt{n+…+\sqrt{n+f(n)}}}}$

Let $\sqrt{n+\sqrt{n+\sqrt{n+...+\sqrt{n+f(n)}}}}=x$. Therefore, $$n+\sqrt{n+\sqrt{n+...+\sqrt{n+f(n)}}}=x^2$$ $$x^2-x-n=0$$ $$x=\sqrt{n+\sqrt{n+\sqrt{n+...+\sqrt{n+f(n)}}}}=\frac{1 \pm \sqrt{1+4n}}{2}...
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How to find the partial derivatives of the following nested expression?

I want to find the partial derivatives of the expression for $v_3(\boldsymbol{u})$ with respect to $u_1$, $u_2$ and $u_3$ from the expressions below. Here $\Phi$ denotes the cumulative distribution ...
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62 views

$\pi$, from Pentadecagon - infinitely expanding Balloon nested Radical

In this post, I would like to share the findings on derivation of $\pi$ with Pentadecagon inscribed in unit circle. Here the side of each chord is $2\sin12^\circ$ (Bisecting the chord by segment which ...
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First few Prime numbers and nested radicals in association with $\pi$

There is interesting association between $2\cos(\frac{\pi}{60})$ which is $\frac{1}{2}\sqrt{8+\sqrt{15}+\sqrt3+\sqrt{10-2\sqrt5}}$ and first few Prime numbers with infinite expansion of balloon nested ...
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1answer
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Simplifying $\sqrt{34+15\sqrt2}$ [closed]

$$\sqrt{34+15\sqrt2}$$ If we want $34+15\sqrt2$ to be a nice square $(a+b)^2=a^2+2ab+b^2$, most likely it is the case that $15\sqrt2$ corresponds to $2ab$. I don't know what to do from here. Is there ...
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1answer
70 views

Continued radical of powers of 4 equals 3 [closed]

Can someone explain to me why $$\sqrt{4 + \sqrt{4^2 + \sqrt{4^3 + \sqrt{4^4 + \dots}}}} = 3???$$ I need an answer
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2answers
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$\pi$ & $\phi$ (Golden ratio), Pentagon inscribed in unit circle

Everyone is aware that square inscribed in unit circle and infinite product giving rise to $\pi$. One of the simplest way to represent $\pi$ with the help of nested radical as follows $$\pi = \lim_{n\...
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Nested radicals, Golden ratio, number series

Finite and infinite expansion of nested Radical involving Golden ratio as follows $2\cos\frac{2\pi}{5} = \frac{1}{\phi} $ where $\phi$ is Golden ratio $(2\cos72°)$ Let us expand this with increasing ...
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1answer
127 views

Interesting cyclic infinite nested square roots of 2 and cosine values

It is interesting to note that any angle between 45° to 90° satisfying $1\over4$ < $p \over q$ <$1\over2$ where $ p \over q$ is of form $p = 2^n $ and $q$ is an odd number satisfying $2^{n+1} &...
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Infinite Nested Radical

Ramanujan discovered that $$x+n+a=\sqrt{ax+(n+a)^2+x\sqrt{a(x+n)+(n+a)^2+...}}$$ (see equation (27) here). I didn't understand how we can use this (basically what to put in place of $x, n, a$) to ...
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1answer
100 views

How to calculate $x_{n}= \sqrt{n+ \sqrt{n- 1+ \sqrt{n- 2+ \sqrt{\cdots + \sqrt{2+ \sqrt{1}}}}}}$ without using surd signs

How to calculate: $$x_{n}= \sqrt{n+ \sqrt{n- 1+ \sqrt{n- 2+ \sqrt{\cdots + \sqrt{2+ \sqrt{1}}}}}}$$ without using surd signs. My attempt: I saw that $$x^{2}_{n}= n+ x_{n- 1}$$ Therefore $$x= 1+ \frac{...
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Is $\sqrt[3]{7+5 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ a rational number?

Is there a way to show that $$\alpha=\sqrt[3]{7+5 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$$ is a rational number? I found $\alpha=3$ from doing simplifications. But, I would like to known a different ...
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Differents ways to evaluate the sum $\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$

Evaluate $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$$ My approach: Let $$x=\sqrt{12+ \sqrt{12+\sqrt{12+\cdots}}}$$ so, we have that $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}\iff \sqrt{12+...
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Peculiar Nested radicals, cosine values, Jacobsthal-Lucas numbers

Let us consider half angle formula for equilateral triangle in unit circle with initial angle as $\frac{\pi}{3}$ $2\cos\frac{\pi}{6} = \sqrt{2+2\cos\frac{\pi}{3}} = \sqrt{2+1}$ Further half angle when ...
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Golden ratio, cyclic infinite nested square roots of 2 and and Geometric progression and cosine value

In connection to previous posts here and herenow we know $2\cos\frac{2\pi}{5}$ can be represented as $\sqrt{2-\sqrt{2+...}}$ simply represented as $cin\sqrt2[1-1+]$ and this converges to inverse of ...
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For a recursion based on $x+y+z+2\sqrt{xy+yz+zx},$ does what happens in $\mathbb Z[\sqrt n]$ stay in $\mathbb Z[\sqrt n]$?

This problem has a geometric origin which I'll outline below, but I believe the concepts and explanation are algebraic. Given a function on triples $$K((x,y,z))=x+y+z+2\sqrt{xy+yz+zx}$$ we build a ...
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51 views

Cosine angles and finite nested square roots of 2

Let us consider interesting factors about cosine angles $2\cos(\frac{\pi}{2^3}) = \sqrt{2+\sqrt2}$ $2\cos(\frac{\pi}{2^4}) = \sqrt{2+\sqrt{2+\sqrt2}}$ To generalize  $2\cos(\frac{\pi}{2^n}) = \sqrt{2+\...
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1answer
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Interesting solutions for cyclic infinite nested square roots of 2

I have derived cosine values for following cyclic infinite nested square roots of 2 ( Hereafter simply referred as $cin\sqrt2$) $cin\sqrt2[1-]$ represents $\sqrt{2-\sqrt{2-...}}$ $cin\sqrt2[1-1+]$ ...
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Solving finite and infinite nested square roots of 2 - yet another interesting approach

Consider the following consecutive equalities: $\sqrt2=2\cos(\frac{1}{4})\pi$ $\sqrt{2-\sqrt2}=2\cos(\frac{3}{8})\pi$ $\sqrt{2-\sqrt{2-\sqrt{2}}}=2\cos(\frac{5}{16})\pi$ $\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{...
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1answer
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Interesting infinite nested square roots of 2 for $2\cos1°$ and $2\sin1°$

It is interesting to note that any angle between 45° to 90° satisfying $1\over4$ < $p \over q$ <$1\over2$ where $ p \over q$ is of form $p = 2^n $ and $q$ is an odd number satisfying $2^{n+1} &...
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3answers
135 views

Solving $\sqrt{2+\sqrt{2-\sqrt{2+x}}} = x$.

ETA: since commenters noted I missed a sign I am correcting that. This is one of those Mathsolutionzz instagram problems. I was curious if I did this correctly, because I looked at the comments, and ...
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1answer
115 views

Infinite ways to represent $\pi$ as product of nested square roots of $2$ and $2^n$ and odd numbers

Polygon inscribed in a circle leads to famous infinite product by Viete's formula.here One way to represent that as infinite nested radical of 2 as follows $$\pi = \lim_{n\to\infty}2^n \times \sqrt{2-\...
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1answer
69 views

Solve the inequality $\sqrt[3]{1+(3+x)\sqrt{x}+3x} - \sqrt[3]{1-(3+x)\sqrt{x}+3x} > x + a$

So am trying to solve this inequality, $\sqrt[3]{1+(3+x)\sqrt{x}+3x} - \sqrt[3]{1-(3+x)\sqrt{x}+3x} > x + a$, the problem is of course for values $0<a<1$, I tried working through it multiple ...
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An arithmetic problem that the sum of two irrationals involving cube roots makes an integer

Consider the following expression: $$(20-\sqrt{392})^{1/3}+(20+\sqrt{392})^{1/3}$$ This equals $4$, but how can I show this? Note that I do not want to make use of the following line of reasoning: 4 ...
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1answer
64 views

Find derivative of $f(x)=\tan^2x \sqrt{\tan x\sqrt[3]{\tan x\sqrt[4]{\tan x…}}}$

Find $f(x)$ for $f(x)=\tan^2x \sqrt{\tan x\sqrt[3]{\tan x\sqrt[4]{\tan x...}}}$ I found that the right expression is an infinite series of $\sum^{\infty}_{n=2}{\frac{1}{n!}}$. I know the series ...
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2answers
90 views

How to prove $f(x)=\sqrt{x+\sqrt{x+\sqrt{x}}}$ is differentiable?

I've been working on this question for over 2 hours now and I tried to use the limit definition of a derivative to show that it's differentiable but that got me nowhere because I was completely ...
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1answer
210 views

How to denest $\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}$ from scratch?

I have seen several questions asking for the proof of $$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}$$ However, I want to simplify $\sqrt[3]{\sqrt[3]{2}-1}$ into the ...
4
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2answers
94 views

Nested radicals like Ramanujan's infinite radicals

$$\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+...}}}}}=?$$ This like the My question is about "how to start this problem ?".I've been thinking for over two hours but get stuck at the end. I ...
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1answer
125 views

Taylor series expansion and value of cosine angle

This is a related question in deriving cosine values by 2 different means Deriving values of Trigonometric angles will be easier with Taylor series expansion for first few terms for some of the ...
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4answers
79 views

Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$?

Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$. Here's my progress. Let $x = \sqrt[4]{2}$. Then our expression can be written as $x^4/(x^4 - x)$, which simplifies to $x^3/(x^3 - 1)$. ...
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1answer
67 views

Does the sequence $t_n=\sqrt{a_1+\sqrt{a_2+\cdots+\sqrt{a_n}}} (a_k>0,k=1,2,\cdots)$ converge? [duplicate]

Does the sequence $t_n=\sqrt{a_1+\sqrt{a_2+\cdots+\sqrt{a_n}}} (a_k>0,k=1,2,\cdots)$ converge? I need to prove that if $$ \limsup\limits_{n\to\infty} \dfrac{\ln\ln a_n}{n}<\ln 2, $$ the sequence ...
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2answers
212 views

Prove that $\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}+\frac{c}{1+a^2+b^2+c^2}+\frac{d}{1+a^2+b^2+c^2+d^2}\leq\frac{3}{2}$

For any reals $a$, $b$, $c$ and $d$ prove that: $$\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}+\frac{c}{1+a^2+b^2+c^2}+\frac{d}{1+a^2+b^2+c^2+d^2}\leq\frac{3}{2}$$ C-S in the IMO 2001 stile does not help here:...
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2answers
42 views

Infinite Nested Square Root, Prove or disprove that there is at least one real number 𝑥 satisfy

$$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}} = 4$$ If I let $X_1=\sqrt x$, $X_2= \sqrt {x+X_1}$, then $X_{n+1}= \sqrt{x+X_n}$ It is a increasing sequence since $X_n<X_{n+1}$ for all $n$. However, is ...
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1answer
81 views

Representing $\cos(\frac{π}{11})$ as cyclic infinite nested square roots of $2$

How can we represent $2\cos(\frac{\pi}{11})$ and $2\cos(\frac{\pi}{13})$ as cyclic infinite nested square roots of 2 I have partially answered below for $2\cos(\frac{\pi}{11})$ which I derived it ...
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3answers
625 views

$\sqrt{a^2+5b^2}+\sqrt{b^2+5c^2}+\sqrt{c^2+5a^2}\geq\sqrt{10(a^2+b^2+c^2)+8(ab+ac+bc)}$ for any real numbers.

I think that this inequality is strong, though I do not have knowledge of many techniques. There goes my work: Positive variables only make the inequality stronger, hence suppose $a,b,c\geqslant0$ $$ \...
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4answers
147 views

Simplify $(1+\sqrt{3}) \cdot \sqrt{2-\sqrt{3}}$

Can someone help me simplify $(1+\sqrt{3})\times\sqrt{2-\sqrt{3}}$? The end result is $\sqrt{2}$, however, I honestly do not know how to get there using my current skills. I asked a teacher/tutor and ...
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1answer
99 views

Solving cyclic infinite nested square roots of 2 as cosine functions

Common infinite nested square roots of 2 are well known from school grade. We used to solve $$\sqrt{2+\sqrt{2+\sqrt{2+...}}}$$ as $x=\sqrt{2+x}$ which becomes $x^2 = x+2$ ==> $x^2-x-2=0$ The ...
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1answer
24 views

Prove the convergence of the infinite product involving nested radicals [duplicate]

How to show that $\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2+\sqrt{2}}}{2}\cdot \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot ... =\prod\limits_{n\in \mathbb{N}} \frac{1}{2} ~ \overbrace{\sqrt{2+\sqrt{2+...+\...
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2answers
81 views

Find value of $\sin x-\frac{1}{\cot x}$

If $\sin x+\frac{1}{\cot x}=3$, calculate the value of $\sin x-\frac{1}{\cot x}$ Please kindly help me Let $\sin x -\frac{1}{\cot x}=t$ Then, $$\sin x= \frac{3+t}{2}, \cot x= \frac{2}{3-t}$$ By ...
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1answer
73 views

Is $\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}$ even? prime? composite? an even prime? [duplicate]

I am not being able to solve this question. It is basically a high school level question and chapter I think is basic fundamentals of maths. The value of $$\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}...
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2answers
174 views

Solving infinite nested square roots of 2 converging to finite nested radical

Can anyone explain to solve the identity posted by my friend $$2\cos12°= \sqrt{2+{\sqrt{2+\sqrt{2-\sqrt{2-...}}} }}$$ which is an infinite nested square roots of 2. (Pattern $++--$ repeating ...
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1answer
105 views

Alternate method to solve $\sqrt{11\sqrt{11\sqrt{11…4\, \text{times}}}}$

Question : What is the value of $$\sqrt{11\sqrt{11\sqrt{11...4\,\text{times}}}}$$ I did it by solving square root one by one. $$\sqrt{11\sqrt{11\sqrt{11\times11^\frac{1}{2}}}}$$ $$\sqrt{11\sqrt{11\...
5
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3answers
167 views

Does a value for $\sqrt{x+\sqrt{x+\sqrt{x+…}}}$ actually exist? [duplicate]

I have seen questions of this type being solved as follows : $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$'s value does not change if we add an $x$ to the expression and square root it. Let the value of this ...
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2answers
25 views

Infinite nested radical with variables

Let $a \bowtie b = a+\sqrt{b+\sqrt{b+\sqrt{b+...}}}$. If $4\bowtie y = 10$, find the value of $y$. I know that $\sqrt{y+\sqrt{y+\sqrt{y+...}}}=6,$ but what do I do know? I know how to solve normal ...
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1answer
45 views

Simplify a radical inside a radical inside a radical

The question is this: Simplify $\sqrt{1+\sqrt{21+12\sqrt{3}}}$ I defined the value as x, then got $(x+1)^2(x-1)^2$=$21+12\sqrt{13}$. I don't know what to do from there. Any help?
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1answer
43 views

How to get rid of such radicals?

I would like to know if there is any way I can get rid of these cubic radicals bellow (1). I am allowing both complex and real values. $$ \sqrt[3]{ -\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27} }...
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2answers
86 views

An analogue to Ramanujan's identity $\left\{3\left(\left(a^{3}+b^{3}\right)^{1/3}-a\right)\left(\left(a^{3}+b^{3}\right)^{1/3}-b\right)\right\}^{1/3}$

In this paper discussing Ramanujan's submitted questions to the Mathematical journal, Question 785 states: Show that; $$\small{\left\{3\left(\left(a^{3}+b^{3}\right)^{1/3}-a\right)\left(\left(a^{3}+b^...
12
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1answer
209 views

How would I simplify this function $\rho(x)=x+\sqrt{x-\sqrt{x-\sqrt{x+\sqrt{\dots}}}}$

How do I simplify $\rho(x)$ into simple terms? $$\rho(x)=x+\sqrt{x-\sqrt{x-\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x+\sqrt{x-\sqrt{\dots}}}}}}}}$$ where the subtracting and the adding follows the Thue–Morse ...
12
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3answers
283 views

Simplify the radical $\sqrt{x-\sqrt{x+\sqrt{x-…}}}$

I need help simplifying the radical $$y=\sqrt{x-\sqrt{x+\sqrt{x-...}}}$$ The above expression can be rewritten as $$y=\sqrt{x-\sqrt{x+y}}$$ Squaring on both sides, I get $$y^2=x-\sqrt{x+y}$$ ...

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