Approximations of $\pi$ using radicals

Question

There are many approximations of $\pi$ using trigonometric and rational numbers. But I created this one: $$\pi \approx \sqrt[11]{294204}$$ Which is correct to almost $8$ decimal places. Are there any other approximations of $\pi$ using radicals? I know of $\sqrt{10}$, $\sqrt[3]{31}$, $\sqrt[4]{97}$, and so on. But are there more beautiful ways (like using $\varphi$ since it is composed of radicals)?

Answer 1

There are lots of radical approximations to $\pi$. A classical example is given by Viete's formula, in which a series of increasingly complicated nested radicals may be used to render $2/\pi$:

$2/\pi=\Pi_{n=1}^\infty(a_n/2)$

$a_1=\sqrt2, a_n=\sqrt{2+a_{n-1}}.$

Many readers are aware that this series is derived by considering the perimeters of inscribed regular polygons having $2^k$ sides, these approaching the circumference of a circle as $k\to\infty$.

Just for fun, let's explore a method based on the Taylor series for the cosine function. Pick a value of $n$ for which $\cos(\pi/n)$ has a convenient radical expression and plug in the Taylor series approximation

$\cos(\pi/n)\approx 1-[(\pi/n)^2/2]+[(\pi/n)^4/24].$

Then substitute the radical expression for $\cos(\pi/n)$ on the left and solve the biquadratic polynomial by standard methods for "$\pi/n$".

With $n=5,\cos(\pi/5)=(\sqrt5+1)/4$ this gives

$\pi\approx5\sqrt{6-\sqrt3×(\sqrt5+1)}=\color{blue}{3.142}3...$

(correct to the significant digits indicated in blue.)

Answer 2

COMMENT: By way of simple pertinent information.

There are many remarkable approximations of $\pi$. For example $$\pi\approx\dfrac{22}{17}+\dfrac{37}{47}+\dfrac{88}{43}\\\pi\approx\sqrt[4]{\frac{2143}{22}}\\\pi\approx\sqrt[9]{\frac{34041350274878}{1141978491}}\\\pi\approx\frac{\ln(640320^3+744)}{\sqrt{163}} $$ The first and second each give $9$ exact decimal places, the third gives $15$ and the fourth gives $30$.

Buffon's experiment.- On any flat surface covered by parallel straight bands of equal width, a rod of length equal to the width of the bands is pulled a given number $T$ of times. In each shot the rod either stays within a band or cuts one of the parallel delimiting lines. Let $C$ be the number of cuts obtained; the quotient $\dfrac{2T}{c}$ it will be in principle the closer to $\pi$ the higher the number $T$.

In 1901, the Italian M. Lazzarini using a rod with a length equal to $\dfrac56$ of the width of the bands (in which case the theoretical probability of cutting is $\dfrac{5\pi}{3}$, he got in 3408 shots the rational approximation $\dfrac{355}{113}\approx3.141592$ .

Answer 3

The well-known $\pi \approx \frac{\ln\left(5280^3\,+\,744\right)}{\sqrt{67}}$ and $\pi \approx \frac{\ln\left(640320^3\,+\,744\right)}{\sqrt{163}}$ involves the j-function and integers. But we can also use the Dedekind eta function and radicals, $$\begin{align} \pi &\approx \frac{\ln\left(2^6\phi^6-24\right)}{\sqrt{5}}\\ \pi &\approx \frac{\ln\left(2^6\phi^{12}+24\right)}{\sqrt{10}}\\ \pi &\approx \frac{\ln\left(2^{12}\phi^8-24\right)}{\sqrt{15}}\\ \pi &\approx \frac{\ln\left(2^6\phi^{24}-24\right)}{\sqrt{25}} \end{align}$$ with golden ratio $\phi =\frac{1+\sqrt5}2$. The last two are correct to $9$ and $12$ decimal places, respectively.

P.S. The consistent appearance of $24$, the powers that are factors of $24$, and the radicands that are multiples of $5$ are clues the above are not coincidences and in fact have a mathematical reason behind them. Lovely, aren't they?

Answer 4

Playing your game, let me define $$A_n=\text{Round}\left[\pi ^{p_n}\right]$$ and compute $$\Delta_n=\log_{10}\Bigg[\left| A^{\frac{1}{p_n}}-\pi \right| \Bigg]$$ where $p_n$ is the $n^{\text{th}}$ prime number $$\left( \begin{array}{ccc} n & A_n & \Delta_n\\ 5 & 294204 & -7.758 \\ 6 & 2903677 & -7.647 \\ 7 & 282844564 & -9.568 \\ 8 & 2791563950 & -10.62 \\ 9 & 271923706894 & -12.71 \\ 10 & 261424513284461 & -16.27 \\ 11 & 2580156526864959 & -16.72 \\ 12 & 2480534602660760780 & -19.83 \\ 13 & 241626620923575111130 & -22.04 \\ 14 & 2384759161287667022585 & -23.76 \\ 15 & 232297222236041657886386 & -25.24 \\ 16 & 223328039155291367651964249 & -28.00 \\ 17 & 214705163466259581466085919504 & -32.68 \\ 18 & 2119055026283205737110078195749 & -31.92 \\ 19 & 2037236602860106711985144771884927 & -35.04 \\ 20 & 198445365705802004702001456166116206 & -37.30 \\ \end{array} \right)$$ showing that, more or less, $\Delta_n=6.156-2.190\,n$

Answer 5

Expanding on the Taylor series method given by https://math.stackexchange.com/a/4572788/945129

$\cos(x)+\cos(xe^{2\pi i/3})+\cos(xe^{4\pi i/3})=3 - x^6/240 + x^{12}/159667200 ...$

this is also equal to: $\cos(x) + 2 \cos(x/2) \cosh((x\sqrt 3)/2)$

setting $x= \pi$ removes the hyperbolic parts

we have $\pi/2 \approx\sqrt[6]{15}$ which differs from the claim by $3.7 \times 10^{-4}$

Related ones:

and $\pi/2 \approx\sqrt[12]{\frac{77}{691}} \sqrt[6]{45} $ which differs from the claim by $2.47 \times 10^{-7}$

and $\pi/2 \approx\sqrt[8]{\frac{70}{17}} \sqrt[4]{3} $ which differs from the claim by $3 \times 10^{-5}$