求解并化简表达式到“标准”的形式

Nested radicals

Chapter 1: Introduction

Nested radicals are something in the form of $$\sqrt[n_1]{x_1\sqrt[n_2]{x_2\sqrt[n_3]{x_3...}}}$$ where $n_i, x_i \in \mathbb{R}$. ($n_i$ is not written when $n_i=2$, and you should know that earlier!)

These nested radicals may repeated finitely many times like $\sqrt{1\sqrt{2\sqrt{3}}}$, and this type of nested radicals is known as Finite Nested Radicals. The example shows a finite nested radical with 3 nests.

Obviously if we have "finite", then we must have "infinite"! For nested radicals like $\sqrt{1\sqrt{2\sqrt{3...}}}$, we name it as infinite nested radicals.

In the next chapter, we are going to talk about denesting a nested radical. In fact most of the chapters will focus on denesting nested square roots. Stay tuned!

Credits to Elementary Analysis on Ramanujan’s Nested Radicals.

Nested Radicals

Chapter 2: Denesting 2-Nested and 3-Nested Radicals

Part 1: Denest a 2-Nested Radical.

Before I start, let's have some revision.

Now, let's start denesting! Denesting 2-Nested Radicals is just a piece of cake if we use algebraic methods, as shown below. Let

$$\sqrt{a+b\sqrt c}=\sqrt{d}+\sqrt{e}$$

Where RHS is made up of 2 one-nested radicals.

Squaring both sides, we have $$a+b\sqrt c= d+e+2\sqrt {de}$$ Now, we compare both sides.

So, by comparing both sides, or equating the integer part and radical part to each other, we get $$\begin{align*}a&=d+e-(1)\\b\sqrt c&=2\sqrt{de}-(2)\end{align*}$$

Solving (2) by squaring both sides yields $b^2c=4de$.

Substituting $d=a-e$ into (2) gives $4e^2-4ae+b^2c=0$ after some rearranging.

At this point, you may substitute the values of $a,b,c$ to solve for $e$ and thus denest the radical. But warning, if $e$ is not real, just ignore it for a real-valued radical.

Now, try to denest $\sqrt{3+2\sqrt2}$ and $\sqrt{9+4\sqrt5}$ using the steps above.

Now what? Let's denest a three-nested radical! Let $$\sqrt{a+b\sqrt{c+d\sqrt e}}=\sqrt f+\sqrt{g+\sqrt h}$$

Squaring both sides, $$a+b\sqrt {c+d\sqrt e}=f+g+\sqrt h+2\sqrt{fg+f\sqrt h}$$

Therefore $$a=f+g-(3)$$ and $$b\sqrt{c+d\sqrt e}=\sqrt h+2\sqrt{fg+f\sqrt h}-(4)$$

Squaring (4) yields $$b^2c+b^2d\sqrt e=h+4fg+4f\sqrt h+4\sqrt{fgh+fh\sqrt{h}}$$ giving $$b^2c=h+4fg$$ $$b^2d\sqrt e=4f\sqrt h+4\sqrt{fgh+fh\sqrt h}$$ Things are getting more complicated, so the rest of the steps to denesting it into one-nested radicals are left as an exercise to the reader.

In the next chapter, we are going to talk about the Ramanujan's Nested Radicals, which is even complicated. Stay tuned!

Credits to Elementary Analysis on Ramanujan’s Nested Radicals

Nested Radicals

Chapter 3: Ramanujan's Radicals

Ramanujan's Radicals are actually are radicals in the form of $$x+n+a=\sqrt{ax+(n+a)^2+x\sqrt{a(x+n)+(n+a)^2+(x+n)\sqrt{...\sqrt{a[x+(k-1)n]+(n+a)^2+x[x+(n+1)+a]}}}}$$

Why does the both sides of the equality holds? This is actually derived from a simple expansion that is $$(x+a)^2=x^2+2ax+a^2$$

Taking positive square roots on both sides, we have $$x+a+n=\sqrt{ax+(n+a)^2+x(x+a+2n)}-(1)$$ Replacing $x$ with $x+n$, $$(x+n)+a+n=\sqrt{a(x+n)+(n+a)^2+(x+n)((x+n)+a+2n)}$$ $$x+2n+a=\sqrt{a(x+n)+(n+a)^2+(x+n)(x+3n+a)}-(2)$$ Similarly, $$x+3n+a=\sqrt{a(x+2n)+(n+a)^2+(x+n)(x+4n+a)}-(3)$$ And $$x+4n+a=\sqrt{a(x+3n)+(n+a)^2+(x+n)(x+5n+a)}-(4)$$

Substituting (2) into (1) implies $$x+a+n=\sqrt{ax+(n+a)^2+x\sqrt{a(x+n)+(n+a)^2+(x+n)(x+3n+a)}}-(5)$$ Again, substituting (3) into (5) gives $$x+a+n=\sqrt{ax+(n+a)^2+x\sqrt{a(x+n)+(n+a)^2+(x+n)\sqrt{a(x+2n)+(n+a)^2+(x+n)(x+4n+a)}}}$$

Great! We had derived the Ramanujan's radical! In fact $x=\sqrt{x\sqrt{x\sqrt{x...}}}$ Why?

Challenge

1. i) Write $3$ in the form of infinite nested radicals. Start your solution by setting $x=a=n=1$.

ii) Then write $3u$ in the form of infinite nested radicals. Set $x=a=n=u$.

iii) What can you say about the solutions in (i) and (ii)?

Chapter 4 Sneak Peak: Nested Radicals in the form of $\sqrt{a+ b{\sqrt{a+ b{\sqrt{a+ b{\sqrt{...}}}}}}}$

Credits to Elementary Analysis on Ramanujan’s Nested Radicals

Nested Radicals

Chapter 4: Nested Radicals of the type $\sqrt{a+b\sqrt{a+b\sqrt{a+b\sqrt{...}}}}$

In this chapter we want to show that this type of radical can converge to certain forms where $a,b\in\mathbb{N}_0$.

Let $k$ be the value that the radical converges to, or simply, $$k=\sqrt{a+b\sqrt{a+b\sqrt{a+b\sqrt{...}}}}$$ $$\implies k=\sqrt{a+bk}-(*)$$

In $(*)$, write $k$ in terms of $a,b$ gives $$\begin{align*}k^2-bk-a&=0\\\implies k&=\frac{b+\sqrt{b^2+4a}}{2}\end{align*}$$

And, we are done! In fact, you can also use the same approach when $b=-c$ or $a=-d$. What do you get in the end? What if $a=0$ or $b=0$?

Chapter 5 Sneak Peek: Relationship between Nested Radicals and Continued Fractions. Credits to Elementary Analysis on Ramanujan Nested Radicals.

Nested Radicals

Chapter 5: Relationship between Nested Radicals and Continued Fractions.

Part 1: Introduction of continued fractions

Continued fractions are those fractions in the type of $$a_1+\frac{b_1}{a_2+\frac{b_2}{a_3+\frac{b_3}{a_4+...}}}$$ where $a_i, b_i\in\mathbb{C}$. But we will only focus in $\mathbb{Z}$.

Most numbers can be written in the form of continued fractions as an example, $$3\frac{6}{17}=3+\frac{1}{2+\frac{1}{1+\frac{1}{5}}}$$ Sometimes, for the sake of laziness, we may write continued fractions as $$a_1+\frac{b_1}{a_2+\frac{b_2}{a_3+\frac{b_3}{a_4+...}}}=[a_1;a_2,a_3,...]$$ Iff $b_i=1$, aka a simple continued fraction. Like nested radicals, continued fractions can be infinite.

In Part 2, we are going to study the relationship between the nested radicals and the continued fractions.

Are you ready? Click to continue.

Part 2: Relationship between the nested radicals and the continued fractions.

In the last chapter, we studied the nested radicals of the type $\sqrt{a+b\sqrt{a+b\sqrt{a+b\sqrt{...}}}}$. So the continued fraction we want to study now is of course a little bit special, which is $$a+\frac{b}{a+\frac{b}{a+\frac{b}{a+\frac{b}{a+...}}}}$$ And it goes infinite~

Now, we want to show that this type of continued fraction converges where $a,b\in\mathbb{N}_0$.Let $$\begin{align*}k&=a+\frac{b}{a+\frac{b}{a+\frac{b}{a+\frac{b}{a+...}}}}\\\implies k&=a+\frac{b}{k}-(1)\end{align*}$$ Solving $k$ in terms of $a,b$, we get $$k=\frac{a+\sqrt{a^2+4b}}{2}-(2)$$ And the step from $(1)$ to $(2)$ is left as an exercise to the reader.

Recalling back the equation we derived in the last chapter, $$\sqrt{a+b\sqrt{a+b\sqrt{a+b\sqrt{...}}}}=\frac{b+\sqrt{b^2+4a}}{2}$$ And compare what we got just now.

Chapter 6 Sneak Peek: Identities derived from nested radicals Credits to Wikipedia and Elementary Analysis on Ramanujan's Nested Radicals.

Nested Radicals

Chapter 6: Identities derived from nested radicals

Part 1: $\frac{b+b\sqrt{1+4\omega}}{2}=\sqrt{\omega b^2+\sqrt{\omega b^4+\sqrt{\omega b^8+\sqrt{...}}}}$

Recalling the formula derived in Chapter 4, $$\sqrt{a+b\sqrt{a+b\sqrt{a+b\sqrt{...}}}}=\frac{b+\sqrt{b^2+4a}}{2}$$ We can rewrite LHS as $$\sqrt{a+\sqrt{ab^2+\sqrt{ab^6+\sqrt{ab^{14}+\sqrt{...}}}}}$$

Therefore, $$\frac{b+\sqrt{b^2+4a}}{2}=\sqrt{(\frac{a}{b^2})(b^2)+\sqrt{(\frac{a}{b^2})(b^4)+\sqrt{(\frac{a}{b^2})(b^8)+\sqrt{...}}}}$$ Setting $\omega=\frac{a}{b^2}$ gives $$\frac{b+b\sqrt{1+4\omega}}{2}=\sqrt{\omega b^2+\sqrt{\omega b^4+\sqrt{\omega b^8+\sqrt{...}}}}$$ So, we are done!

Part 2: $x+a=\sqrt{a^2+x\sqrt{a^2+(x+a)\sqrt{a^2+(x+2a)\sqrt{...\sqrt{a^2+(x+a)[x+(n+1)a]}}}}}$

In chapter 4, we learned on how to write sum of any three numbers in the form of nested radicals. But how about two numbers? We are going to use the same concept here, which is $$(x+a)^2=x^2+a^2+2ax=x^2+a^2+ax+ax$$ Taking square root on both sides, $$x+a=\sqrt{x^2+a^2+ax+ax}=\sqrt{a^2+x(x+a+a)}-(0)$$ Replacing $x$ with $x+a$ and $x+2a$ gives $$x+a+a=\sqrt{a^2+(x+a)(x+2a+a)}-(1)$$ $$\text{and }x+2a+a=\sqrt{a^2+(x+2a)(x+3a+a)}-(2)$$ respectively. Substituting $(2)$ into $(1)$ yields $$x+a=\sqrt{a^2+x\sqrt{a^2+(x+a)(x+2a+a)}}-(3)$$

Wala! DoNe!

Chapter 7 Sneak Peek: Denesting a n-th degree nested radical Credits to Elementary Analysis on Ramanujan's Nested Radicals.

Lets say we want to find the sum and product of the roots of $5x^2+3x+1=0$. Unfortunately the roots are complex.

Well if $r_1$ and $r_2$ are the roots then we can write $5(x-r_1)(x-r_2) = 0$. Expanding this we obtain $5x^2-5(r_1+r_2)x+5r_1r_2$.

Comparing this to the coefficients of the original quadratic we get $-5(r_1+r_2) = 3$ and $r_1+r_2 = -\frac{3}{5}$ Also we get $5r_1r_2 = 1$ so $r_1r_2 = \frac{1}{5}$.

We found the sum and product of the roots without actually solving for them!

In general for a quadratic $ax^2+bx+c=0$ we can write $a(x-r_1)(x-r_2) = 0$ and using the same methods as above we get $ax^2-a(r_1+r_2)x+ar_1r_2 = 0$.

Comparing to the coefficients of the original quadratic we get $r_1+r_2 = -\frac{b}{a}$ and $r_1r_2 = \frac{c}{a}$. These are Vieta's Formulas for a Quadratic.

In conclusion, Vietas formulas for a quadratic state that for a general quadratic $ax^2+bx+c$ the sum of the roots is $-\frac{b}{a}$ and the product of the roots is $\frac{c}{a}$

Nested Radicals

Chapter 7: Convergence of a n-th degree nested radical

Recalling that in Chapter 4, we proved that a nested square root converges, but now, what if I ask you to prove that a n-th nested root converges, where $n>2$?

Let's consider a nested square root of the type $$\sqrt[n]{a+b\sqrt[n]{a+b\sqrt[n]{a+b\sqrt[n]{...}}}}$$ where it converges when $a,b\in\mathbb{N}_0$. Set $$\begin{align*}k&=\sqrt[n]{a+b\sqrt[n]{a+b\sqrt[n]{a+b\sqrt[n]{...}}}}\\k&=\sqrt[n]{a+bk}\\0&=k^n-bk-a\end{align*}$$

To solve this, let $a=\alpha k^n, b=\beta k^{n-1}$. We have $$\begin{align*}k^n-\beta k^n-\alpha k^n&=0\\\beta&=1-\alpha\end{align*}$$ Means that $$k=\sqrt[n]{\alpha k^n+\beta k^{n-1}\sqrt[n]{\alpha k^n+\beta k^{n-1}\sqrt[n]{...}}}$$ Proving that it converges. In fact, you can stop at $0=k^n-bk-a$ by substituting the values of $a, b$

Chapter 8 Sneak Peek: Existence of denesting in 2-nd degree of nested radicals of the form $\sqrt{\sqrt[3]\alpha-\sqrt[3]\beta}$ Credits to Elementary Analysis on Ramanujan's Nested Radicals.

Nested Radicals

Chapter 8: Existence of denesting in 2nd degree of nested radicals of the form $\sqrt{\sqrt[3]\alpha-\sqrt[3]\beta}$

In this chapter, we are going to determine if a nested radical in the form $\sqrt{\sqrt[3]\alpha-\sqrt[3]\beta}$. We claim that $\forall \alpha, \beta \in \mathbb{Q}\backslash\{0\}$, if in $$F_{\beta/\alpha}(t)=t^4+4t^3+8\frac{\beta}{\alpha}t-4\frac{\beta}{\alpha}$$ and $\exists t\in\mathbb{Q},f(t)=0$, then $\sqrt{\sqrt[3]\alpha-\sqrt[3]\beta}$ is denestable.

Sketch of proof: Realizing that $\sqrt{\sqrt[3]\alpha-\sqrt[3]\beta}$ is denestable iff $\sqrt{1+\sqrt[3]{\frac{\beta}{\alpha}}}$ is denestable. This happens iff $\exists f,x,y,z\in\mathbb{Q}$ such that $$1+\sqrt[3]{\frac{\beta}{\alpha}}=f(x(1)+y\sqrt[3]{\frac{\beta}{\alpha}}+z\sqrt[3]{\frac{\beta^2}{\alpha^2}})^2$$

Expanding RHS and compare the like powers of $\sqrt[3]{\frac{\beta^2}{\alpha^2}}$ with LHS to get

$$\begin{align*}\frac{1}{f}&=x^2+\frac{2yz\beta}{\alpha}\\0&=y^2+2xz\\\frac{1}{f}&=\frac{\beta z^2}{\alpha}+2xy\end{align*}$$

It is easy to see that $z\neq0, F_{\beta/\alpha}(\frac{y}{z})=0$. Conversely, if $F_{\beta/\alpha}(s)=0$, then the denesting is $$\sqrt{\sqrt[3]\alpha-\sqrt[3]\beta}=\pm\frac{1}{\sqrt f}\Big(\frac{-s^2(\sqrt[3]\alpha^2)}{2}+s\sqrt[3]{\alpha\beta}+\sqrt{\beta^2}\Big), f=\beta-s^3\alpha\blacksquare$$ Now try to denest $\sqrt{\sqrt[3]5-\sqrt[3]4}, \sqrt{\sqrt[3]{28}-\sqrt[3]{27}}$ using the steps above.

Credits to Ramanujan Day.

So that's it for this series. $THE$ $END$.