How do I prove that $\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}} \approx 1$

Question

How do I prove that

$$\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}} \approx 1$$

without using the calculator?

Answer 1

If we consider an infinite chain.

Suppose $x = \sqrt{20 +\sqrt{20+\sqrt{20+\sqrt{20+\sqrt{\cdots}}}}}$

$x = \sqrt{20 +x}\\ x^2 = 20 + x\\ x^2 - x - 20 = 0\\ (x-5)(x+4) = 0$

$x$ must be greater than $0, x = 5$

and $y = \sqrt{20 -\sqrt{20-\sqrt{20-\sqrt{20-\sqrt{\cdots}}}}}$

$y = \sqrt{20 - y}\\ y^2 + y - 20=0\\ y = 4$

$x-y = 1$

As we add more terms under those square roots $\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}}$ converges toward 1.

Answer 2

Even without a calculator, you can do the numerics fairly easily. 20 is about halfway between 16 and 25, so $\sqrt{20} \approx 4.5$. So $20\pm\sqrt{20}$ is about 15.5 and 24.5, respectively. These in turn have square roots of about (a little less than) 4 and 5. This leaves $\sqrt{25}−\sqrt{16}\approx 1$. The "little less than" might contribute significant error on its own, but since it's similar sizes we can expect it to largely cancel out: $\sqrt{25-\epsilon} - \sqrt{16-\epsilon} \approx \sqrt{25} - \sqrt{16}$, since we can expect it shift each root over by a similar amount. Here $\epsilon = 1/2$.

Answer 3

In general, it holds that $$\sqrt{n(n-1)+\sqrt{n(n-1)+\sqrt{n(n-1)}}}=n-\frac{1}{8n^2}+O\left(\frac1{n^3}\right)$$ and that $$\sqrt{n(n-1)-\sqrt{n(n-1)-\sqrt{n(n-1)}}}=(n-1)+\frac{1}{8n^2}+O\left(\frac1{n^3}\right)\,$$ for all $n\geq 1$. Hence, their difference is $$1-\frac1{4n^2}+O\left(\frac1{n^3}\right)\,.$$ In particular, for $n=5$, the difference should be about $1-\dfrac1{100}=0.99$. This is quite close to the actual value of $0.9872649...$.

From $\sqrt{1+x}=1+\frac{1}{2}x+O\left(x^2\right)$, we have $\sqrt{1-\frac1n}=1-\frac1{2n}+O\left(\frac{1}{n^2}\right)$. This means $$\sqrt{n(n-1)}=n\,\sqrt{1-\frac{1}{n}}=n\,\Biggl(1-\frac1{2n}+O\left(\frac{1}{n^2}\right)\Biggr)=n-\frac{1}{2}+O\left(\frac{1}{n}\right)\,,$$ which can also be written as $$\sqrt{n(n-1)}=(n-1)+\frac{1}{2}+O\left(\frac{1}{n}\right)\,.$$ Ergo, $$\begin{align}\sqrt{n(n-1)+\sqrt{n(n-1)}}&=\sqrt{n^2-\frac{1}{2}+O\left(\frac{1}{n}\right)}&=&n\,\sqrt{1-\frac{1}{2n^2}+O\left(\frac{1}{n^3}\right)}\\&=n\,\Biggl(1-\frac{1}{4n^2}+O\left(\frac{1}{n^3}\right)\Biggr)&=&n-\frac{1}{4n}+O\left(\frac{1}{n^2}\right)\,.\end{align}$$ Similarly, $$\sqrt{n(n-1)-\sqrt{n(n-1)}}=(n-1)-\frac{1}{4n}+O\left(\frac{1}{n^2}\right)\,.$$ Hence, $$\begin{align}\sqrt{n(n-1)+\sqrt{n(n-1)+\sqrt{n(n-1)}}}&=\sqrt{n^2-\frac{1}{4n}+O\left(\frac{1}{n^2}\right)}\\&=n\,\sqrt{1-\frac{1}{4n^3}+O\left(\frac{1}{n^4}\right)}\\&=n\,\Biggl(1-\frac{1}{8n^3}+O\left(\frac{1}{n^4}\right)\Biggr)\\&=n-\frac{1}{8n^3}+O\left(\frac{1}{n^3}\right)\,.\end{align}$$ Likewise, $$\sqrt{n(n-1)-\sqrt{n(n-1)-\sqrt{n(n-1)}}}=(n-1)+\frac{1}{8n^2}+O\left(\frac{1}{n^3}\right)\,.$$ In fact, we can prove by induction on $k$ that $$f^+_k\big(n(n-1)\big)=n-\frac{1}{2^kn^{k-1}}+O\left(\frac{1}{n^k}\right)$$ and $$f^-_k\big(n(n-1)\big)=(n-1)-\frac{(-1)^k}{2^kn^{k-1}}+O\left(\frac{1}{n^k}\right)\,,$$ where $f^+_k(x):=\sqrt{x+f^+_{k-1}(x)}$ with $f^+_0(x):=0$ and $f^-_k(x):=\sqrt{x-f^-_{k-1}(x)}$ with $f^-_0(x):=0$ for all $x\geq 0$ and for each $k=1,2,3,\ldots$.

Answer 4

Repeatedly using $\sqrt{1+x} \approx 1+x/2$,

$\begin{array}\\ d(a) &=\sqrt{a+\sqrt{a+\sqrt{a}}}-\sqrt{a-\sqrt{a-\sqrt{a}}}\\ &=(\sqrt{a+\sqrt{a+\sqrt{a}}}-\sqrt{a-\sqrt{a-\sqrt{a}}})\dfrac{\sqrt{a+\sqrt{a+\sqrt{a}}}+\sqrt{a-\sqrt{a-\sqrt{a}}}}{\sqrt{a+\sqrt{a+\sqrt{a}}}+\sqrt{a-\sqrt{a-\sqrt{a}}}}\\ &=\dfrac{(a+\sqrt{a+\sqrt{a}})-(a-\sqrt{a-\sqrt{a}})}{\sqrt{a+\sqrt{a+\sqrt{a}}}+\sqrt{a-\sqrt{a-\sqrt{a}}}}\\ &=\dfrac{\sqrt{a+\sqrt{a}}+\sqrt{a-\sqrt{a}}}{\sqrt{a+\sqrt{a+\sqrt{a}}}+\sqrt{a-\sqrt{a-\sqrt{a}}}}\\ &=\dfrac{\sqrt{a}(\sqrt{1+1/\sqrt{a}}+\sqrt{1-1/\sqrt{a}})}{\sqrt{a}\sqrt{1+(1/a)\sqrt{a+\sqrt{a}}}+\sqrt{a}\sqrt{1-(1/a)\sqrt{a-\sqrt{a}}}}\\ &=\dfrac{\sqrt{1+1/\sqrt{a}}+\sqrt{1-1/\sqrt{a}}}{\sqrt{1+\sqrt{1/a+1/a^{3/2}}}+\sqrt{1-\sqrt{1/a-1/a^{3/2}}}}\\ &\approx\dfrac{1+1/(2\sqrt{a})+1-1/(2\sqrt{a})} {1+(1/2)\sqrt{1/a+1/a^{3/2}}+1-(1/2)\sqrt{1/a-1/a^{3/2}}}\\ &=\dfrac{2} {2+(1/(2\sqrt{a}))\sqrt{1+1/a}-(1/(2\sqrt{a}))\sqrt{1-1/a}}\\ &\approx\dfrac{2} {2+(1/(2\sqrt{a}))(1+1/(2a)-(1/(2\sqrt{a}))(1-1/(2a)}\\ &\approx\dfrac{2} {2+(1/(2\sqrt{a}))(1/(2a))}\\ &=\dfrac{1} {1+(1/(8a^{3/2}))}\\ &\approx 1-(1/(8a^{3/2}))\\ \end{array} $

Answer 5

By binomial formulas and cancellation, you get that \begin{align} ...&=\frac{\sqrt{20+\sqrt{20}}+\sqrt{20-\sqrt{20}}}{\sqrt{20+\sqrt{20+\sqrt{20}}}+\sqrt{20-\sqrt{20-\sqrt{20}}}} \\&=\frac{ \sqrt{5+\sqrt{\frac54}}+\sqrt{5-\sqrt{\frac54}} }{ \sqrt{5+\sqrt{\frac54+\sqrt{\frac5{64}}}}+\sqrt{5-\sqrt{\frac54-\sqrt{\frac5{64}}}} } \end{align} As one can see, the denominator is a small perturbation of the numerator, so that the quotient will be close to $1$.

Answer 6

Use the classical approximation: $$\sqrt{a^2 + b} \approx a + \frac{b}{2a}$$ With $a = \sqrt{20}$ and $b = \sqrt{20 + \sqrt{20}}$ we have $$\sqrt{20 + \sqrt{20 + \sqrt{20}}} \approx \sqrt{20} + \frac{\sqrt{20 + \sqrt{20}}}{2\sqrt{20}} = \sqrt{20} + \frac{\sqrt{400 + 20\sqrt{20}}}{40} $$ Now use the same classical approximation again, this time working with the numerator of the second term. This time with $a=20$ and $b=20\sqrt{20}$ we get $$\sqrt{400 + 20\sqrt{20}} \approx 20 + \frac{20\sqrt{20}}{40} = 20 + \frac{\sqrt{20}}{2}$$ Combining these, we've got: $$\sqrt{20 + \sqrt{20 + \sqrt{20}}} \approx \sqrt{20} + \frac{20 + \frac{\sqrt{20}}{2}}{40} = \sqrt{20} + \frac{1}{2} + \frac{\sqrt{20}}{80}$$

Using similar methods, we get $$\sqrt{20 - \sqrt{20 - \sqrt{20}}} \approx \sqrt{20} - \frac{1}{2} + \frac{\sqrt{20}}{80}$$

Finally, subtracting one from the other we end up with $$\sqrt{20 + \sqrt{20 + \sqrt{20}}} -\sqrt{20 - \sqrt{20 - \sqrt{20}}} \approx \left( \sqrt{20} + \frac{1}{2} + \frac{\sqrt{20}}{80} \right) - \left( \sqrt{20} - \frac{1}{2} + \frac{\sqrt{20}}{80} \right)$$ and in this last expression everything cancels out except for $$\frac{1}{2} - \left(-\frac{1}{2} \right) = 1$$

Answer 7

Since Alex Meiburg didn't write an answer I'm going to steal his idea:

(Oh, I see he did after all... oh, well, I'm still going to steal his answer.)

$\sqrt{20 + \sqrt{20 + \sqrt{20}}} =$

$\sqrt{20 + \sqrt{20 + \sqrt{25*4/5}}}=$

$\sqrt{20 + \sqrt{20 + 5\sqrt{4/5}}}=$

$\sqrt{20 + \sqrt{20 + 5 +5(\sqrt{4/5}-1)}}=$

$\sqrt{20 + \sqrt{25 +5(\sqrt{4/5}-1)}}=$

$\sqrt{20 + 5\sqrt{1 +(\sqrt{4/5}-1)}}=$

$\sqrt{20 + 5\sqrt[4]{4/5}}=$

$\sqrt{20 + 5 + 5(\sqrt[4]{4/5}-1)}=$

$\sqrt{25 + 5(\sqrt[4]{4/5}-1)}=$

$5\sqrt{1 + (\sqrt[4]{4/5}-1)}=$

$5\sqrt[8]{4/5}\approx 5$ ($\sqrt[8]{4/5}$ = 0.97249247246607303150644442684673)

Likewise

$\sqrt{20 - \sqrt{20 - \sqrt{20}}} =$

$\sqrt{20 - \sqrt{20 - \sqrt{16*5/4}}}=$

$\sqrt{20 - \sqrt{20 - 4\sqrt{5/4}}}=$

$\sqrt{20 - \sqrt{20 - 4 +4(\sqrt{5/4}-1)}}=$

$\sqrt{20 - \sqrt{16 + 4(\sqrt{5/4}-1)}}=$

$\sqrt{20 - 4\sqrt{1 +(\sqrt{5/4}-1)}}=$

$\sqrt{20 - 4\sqrt[4]{5/4}}=$

$\sqrt{20 - 4 + 4(\sqrt[4]{5/4}-1)}=$

$\sqrt{16 + 4(\sqrt[4]{5/4}-1)}=$

$4\sqrt{1 + (\sqrt[4]{5/4}-1)}=$

$4\sqrt[8]{5/4}\approx 4$ (as $\sqrt[8]{5/4}$ = 1.0282855942978896558606746137882)

So $\sqrt{20 + \sqrt{20 + \sqrt{20}}}-\sqrt{20 - \sqrt{20 - \sqrt{20}}}\approx 1$

or precisely $5\sqrt[8]{4/5} - 4\sqrt[8]{5/4} = 1 -5(1- \sqrt[8]{4/5})- 4(\sqrt[8]{5/4}-1)$.

Answer 8

You can do some ad hoc evaluation:

$4=\sqrt{20-\sqrt{20-\sqrt{16}}} \leq\sqrt{20-\sqrt{20-\sqrt{20}}}\leq \sqrt{20-\sqrt{14-\sqrt{25}}}=\sqrt{17}$

$\sqrt{24}\leq\sqrt{20+\sqrt{24}}= \sqrt{20+\sqrt{20+\sqrt{16}}} \leq\sqrt{20+\sqrt{20+\sqrt{20}}}\leq \sqrt{20+\sqrt{20+\sqrt{25}}}=5$

And thus, $$1\geq\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}}\geq \sqrt{24}-\sqrt{17}\approx 0.776$$

This is not that good so let us approximate $\sqrt{20-\sqrt{20-\sqrt {25}}} = \sqrt{20-\sqrt{15}}$ instead by using Newton's method. In this particular case when we want to approximate $\sqrt a$, we want zero of polynomial $x^2-a$ which in turn gives us formula $$x_{n+1}=x_n-\frac{x_n^2-a}{2x_n}$$ For $\sqrt {15}$ we start with $x_0 = 4$ to get $\sqrt{15} \approx 4-\frac 1 8$ (which for geometric reasons is an overestimate, but actually really close).

So, $\sqrt{20-\sqrt{15}}\approx \sqrt{16+\frac 18}\approx 4- \frac{16-(16+\frac 18)}{8} = 4+ \frac 1{64}$

Similarly, $\sqrt{20+\sqrt {24}}\approx \sqrt{24.9}\approx 4.99$ and our bound becomes $$1\geq\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}}\geq 4.99-(4+\frac 1{64})\approx 0.9744$$

Note: The lower bound is no longer true lower bound, but only approximate lower bound, since we overestimated square roots every time we used Newton's method. So, let us have some comparisons:

$\sqrt {20+\sqrt{24}}\approx 4.99$ (our estimate), while $\sqrt{20+\sqrt{20+\sqrt{20}}}\approx 4.99469$ (computer estimate)

$\sqrt{20-\sqrt{15}}\approx 4.015625$ (our estimate), while $\sqrt{20-\sqrt{20-\sqrt{20}}}\approx 4.00743$ (computer estimate)

And finally, since we have that $0.9744 \leq x \leq 1$, we could take arithmetic mean to get estimate $x\approx 0.9872$, while computer estimates $x\approx 0.9872649$.

Answer 9

As in @mweiss 's answer we use repeatedly the approximation $$\sqrt{a^2+b}\approx a+{b\over 2a}\qquad(|b|\ll a^2)\ .$$In this way we obtain on the one hand $$\eqalign{ \sqrt{20}&=\sqrt{25-5}\approx 5-{1\over2},\quad 20+\sqrt{20}\approx25-{1\over2},\cr \sqrt{20+\sqrt{20}}&\approx5-{1\over20},\quad 20+\sqrt{20+\sqrt{20}}\approx 25-{1\over20},\cr \sqrt{20+\sqrt{20+\sqrt{20}}}&\approx5-{1\over200}, \cr}$$ and on the other hand $$\eqalign{ \sqrt{20}&=\sqrt{16+4}\approx 4+{1\over2},\quad 20-\sqrt{20}\approx16-{1\over2},\cr \sqrt{20-\sqrt{20}}&\approx4-{1\over16},\quad 20-\sqrt{20-\sqrt{20}}\approx 16+{1\over16},\cr \sqrt{20-\sqrt{20-\sqrt{20}}}&\approx4+{1\over128}. \cr}$$ It follos that the quantity $Q$ in question is approximatively given by $$Q\approx1-{1\over200}-{1\over128}\approx0.9872.$$