# Better methods to approximate $2^{2\over 3}$

Recently while solving a problem on thermodynamics I ended up with $$2^{2\over 3}$$ .

Now the problem was on a test where no calculators were allowed and answer was required upto $$2$$ decimal digits.

I then resorted to binomial theorem for help (for $$x\lt 1$$) $$\left. \begin{array} { l } { ( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) x ^ { 2 } } { 2 ! } + \frac { n ( n - 1 ) ( n - 2 ) } { 3 ! } x ^ { 3 } + \ldots \ldots + \frac { n ( n - 1 ) \ldots \ldots ( n - r + 1 ) } { r ! } x ^ { r } \ldots } \\ \end{array} \right.\text{upto}\, \, \infty$$ So the original problem can be written as :

$$2^{2\over 3}=4^{1\over 3}=(8-4)^{1\over 3}=2\left(1-\frac{1}{2}\right)^{1\over 3}$$

Now after evaluation first $$3$$ terms I ended up with $$\left(2-\frac{1}{3}-\frac{1}{18}\right) \approx 1.61$$ but the correct answer was $$1.59$$.

Also the average time you got per question was around $$2$$ minutes and I had already used more than half of it so I did not think of calculating more terms.

Now I am looking for a method which can help me evaluate $$2^{2\over 3}$$ faster and more precisely.

• If you can evaluate $log_e(x)$ easily, then you can choose to rewrite the thing as $2^{ \frac {2}{3}} = 2^{log_e(e^{\frac {2}{3}})}$ and solve, which may result in another cumbersome calculation. Commented Aug 3, 2020 at 7:18
• No idea why I went to such great depths but that's what I am - a person who twists his arm around his face to touch his nose while one can directly touch it (at times only !). Plus, I am 15, so what I did may not be of use, but please do tell me. Commented Aug 3, 2020 at 7:21
• @SamRubenAbraham Do you mean to use $x=e^{ln(x)}$? Doesn't that require expansion of $e^x$ and value of $ln(2)$? I guess that would be even more complicated? I want a faster to do by hand way. Commented Aug 3, 2020 at 7:22
• Maybe Newton Raphson :en.m.wikipedia.org/wiki/Newton%27s_method Commented Aug 3, 2020 at 7:24
• Great, @GregMartin ! You've corrected my folly ! Commented Aug 3, 2020 at 7:24

Consider that you look for the zero of function $$f(x)=x^3-4$$ We have $$f(1)=-3$$ and $$f(2)=+4$$. Using $$x_0=\frac 32$$ (the midpoint), the first iterate of Newton method is $$\frac{43}{27} \approx 1.59259$$ while the exact value is $$1.58740$$.

The answer is then $$1.59$$.

Edit

You do not need to know about Newton method since what I did is just linearization (or Taylor series) $$f(x)=f(x_0)+f'(x_0) (x-x_0)+ O((x-x_0)^2$$

so for you case $$0=-\frac 58+\frac {27}4 (x-\frac 32)=\frac{27 }{4}x-\frac{43}{4}\implies x=\frac{43}{27}$$

• Unfortunately I do not know Newton Raphson. But thanks for the answer! Commented Aug 3, 2020 at 7:34
• There's a link in my comment to your question. Commented Aug 3, 2020 at 9:11

If you are around powers of 2 a lot (as computer scientists are, for example), then you can recognize $$4.096=2^{12}/10^3=(2^4/10)^3=1.6^3$$, and then $$4^{1/3} = 1.6(1-\frac{0.096}{4.096})^{1/3}$$ is much better approximated by its power series. Indeed, the approximation $$1.6(1-\frac13\frac{0.096}{4.096}) \approx 1.6(1-\frac13\frac{0.1}4) = 1.6 - \frac1{75}$$ is already within $$0.05\%$$ of the true answer $$\approx 1.5874$$.

How about using some Calculus? We got to compute $$\sqrt[3]{4}$$, so let $$y=f(x) =\sqrt[3]{x}$$. Now, we know that $$f(8)= 2$$ and $$\frac{dy}{dx}= \frac{1}{3} \frac{1}{x^{2/3}}$$

$$\Delta x = 4 - 8 = -4 \\ \Delta y = 1/3 \frac{1}{x^{2/3}} \Delta x \\ \Delta y = \frac{1}{3 \times 4} (-4) \\ \Delta y = -0.3333$$

So, $$f(4) = 2 - 0.3333= 1.667$$.

I know my answer is very far from original one, and to be honest it looks to me that they have used the calculator for the correct answer, I added this method because you've mentioned the time limit and wanted something quick.

Perhaps more elaboration on how to use Newton's method, since I personally find it very easy to use to find the first few digits of $$n$$th roots using only basic calculations.

From the binomial expansion, we know that

$$(x+\Delta x)^3=x^3+3x^2\Delta x+\mathcal O((\Delta x)^2)$$

Our goal is essentially finding the $$\Delta x$$ that fixes $$x^3$$ so that $$(x+\Delta x)^3$$ is closer to $$4$$. In this way, we can find $$\sqrt[3]4$$.

Setting it equal and solving, we end up with

$$x^3+3x^2\Delta x=4\implies\Delta x=\frac{4-x^3}{3x^2}=\dfrac{\dfrac4{x^2}-x}3$$

Adding $$x$$ to both sides to find our new $$x$$ gives us

$$x+\Delta x=\frac{4+2x^3}{3x^2}=\dfrac{\dfrac4{x^2}+2x}3$$

Starting with values of $$x$$ closer to $$\sqrt[3]4$$ such as the $$1.61$$ you found will lead to faster correction, but for example purposes we start far away at $$x=1$$, which makes the next estimate become

$$x+\Delta x=\frac{4+2}3=2$$

Substituting $$x=2$$ in now gives us

$$x+\Delta x=\frac{4+16}{12}=\frac53\simeq1.66$$

The next estimate is a tad messier but still manageable:

$$x+\Delta x=\frac{4+2\times\frac{125}{27}}{3\times\frac{25}9}=\frac{4\times27+2\times125}{3\times3\times25}=\frac{108+250}{225}=\frac{358}{225}\simeq1.591$$

In general you will find that

$$x+\Delta x=\frac{a+(n-1)x^n}{nx^{n-1}}$$

for computing $$\sqrt[n]a$$, which works out to give the first few digits fairly fast and easily provided a good initial estimate (as you had done) and that $$n$$ is not too large.

$$\displaystyle\,\sqrt[\large 3]{\, 4\, } = \sqrt[\large 3]{512 - 12 \over 125} = {8 \over 5}\left(1 - {3 \over 128}\right)^{1/3} \approx 1.6 \times \left(1 - {1 \over 128}\right) = 1.587\color{red}{5}$$

In general ( expanding the above binomial ), with $$\displaystyle{8 \over 5}\sum_{k = 0}^{n}{1/3 \choose k} \left(\,{-\,{1 \over 128}}\,\right)^{k}$$: $${\Large\,\sqrt[\large 3]{\, 4\, }\,}\ \approx\ \left\{\begin{array}{lll} {\displaystyle 1.\color{red}{6}} & {\displaystyle n = } & {\displaystyle 0} \\ {\displaystyle 1.587\color{red}{5\ldots}} & {\displaystyle n = } & {\displaystyle 1} \\ {\displaystyle 1.58740\color{red}{2\ldots}} & {\displaystyle n = } & {\displaystyle 2} \\ {\displaystyle 1.5874010\color{red}{7\ldots}} & {\displaystyle n = } & {\displaystyle 3} \\ {\displaystyle 1.58740105\color{red}{2\ldots}} & {\displaystyle n = } & {\displaystyle 4} \\ {\displaystyle 1.5874010519\color{red}{7\ldots}} & {\displaystyle n = } & {\displaystyle 5} \\ {\displaystyle 1.587401051968\color{red}{3\ldots}} & {\displaystyle n = } & {\displaystyle 6} \end{array}\right.$$
Result for some machine software: $$1.587401051968199474751705639272308260391493327899853009808\ldots$$