What are better approximations to $\pi$ as algebraic though irrational number? I know that $\pi \approx \sqrt{10}$, but that only gives one decimal place correct. I also found an algebraic number approximation that gives ten places but it's so cumbersome it's just much easier to just memorize those ten places.
What's a good approximation to $\pi$ as an irrational algebraic number (or algebraic integer if possible) that is easier to memorize than the number of places it gives correct?
EDIT: Algebraic number preferably of low degree, such as $2$ or $3$ (quadratic or cubic).
 A: How about,
$$ \sqrt[3] {31}=3.14138...$$
Where, $31$ is the length of a month.
If you want memorable, you could always use,
$$\pi \sim \sqrt{{{69} \over {7}}}=3.139...$$
Do I really need to explain this one?
You could also use,
$$\sqrt{{69 \cdot 1001} \over {7 \cdot 1000}}=3.14117...$$
Where, $1001$ refers to the book 1001 Arabian Nights
A: $\root 10 \of {93648}$ is marginally better than $\sqrt{10}$.
But one of the comments has a much better answer, with degree of just $4$.
A: I'm hardly the first to think of this, but I might be the first to say it in this thread: $\sqrt{10} \approx \pi$ suggests that we look at the powers of $\pi$ and see which come closest to integers. Then do floor or ceiling on $\pi^n$ and that gives you an approximation as an irrational algebraic integer  of degree $n$.
Hence $\root 3 \of 31$ (already mentioned by Zach), $\root 5 \of 306$, etc.
A: An interesting simple one is
$$
\pi\approx \frac{3(3+\sqrt{5})}{5} = \frac{6\varphi^2}{5} \approx 3.1416407
$$
where $\varphi$ is the golden ratio.
If you don't mind the algebraic numbers expressed in terms of their polynomials, here are some polynomials that have a root very close to $\pi$:

$$
x^3+120x^2+164x-\frac{86529}{50}=0
$$
for which all solutions are real, and one solution is $x\approx 3.14159265359006$

$$
x^4-48x^3-12x^2-33x+1613=0
$$
for which one of the two real solutions is $x\approx 3.14159265358842$

$$
x^5+2x^4-4x^3+76x^2+149x-1595=0
$$
which has exactly one real solution, $x\approx 3.14159265358998831$

Also, here's an interesting way to express one of the other famous ways to approximate $\pi$, particularly $\sqrt[4]{2143/22}$:
$$
\pi\approx \sqrt{\sqrt{\frac12+\frac{3!+4}{5+6}+7+89}}
$$
Note that the numbers 1 through 9 appear once each, in order.
A: Dalzell's integral is related to the rational approximation $\pi\approx \frac{22}{7}$.
$$\pi=\frac{22}{7}-\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx\approx\frac{22}{7}$$
Similar small integrals are related to simple irrational approximations using $\sqrt{2}$ and $\sqrt{3}$.
$$\pi=\frac{20\sqrt{2}}{9}-\frac{2\sqrt{2}}{3} \int_0^1 \frac{x^4(1-x)^4}{1+x^2+x^4+x^6}dx\approx \frac{20\sqrt{2}}{9}$$
$$\pi=\frac{9\sqrt{3}}{5}+\frac{6\sqrt{3}}{5}\int_0^1\frac{x^3(1-x)^2}{1+x^2+x^4}dx\approx \frac{9\sqrt{3}}{5}
$$
Fractions $\frac{20}{9}$ and $\frac{9}{5}$ are convergents of $\frac{\pi}{\sqrt{2}}$ and $\frac{\pi}{\sqrt{3}}$ respectively.
A: Some nice approximations can be produced by exploiting the ideas of Archimedes. The difference between a unit circle and an inscribed regular $n$-agon is made by $n$ circle segments. If we approximate them with parabolic segments and call
$$ A_n = \frac{n}{2}\sin\frac{2\pi}{n}=n\sin\frac{\pi}{n}\cos\frac{\pi}{n} $$
the area of the inscribed $n$-agon, we get that
$$ \pi \approx \frac{4 A_{2n}-A_n}{3} = \frac{n}{3}\sin\frac{\pi}{n}\left(4-\cos\frac{\pi}{n}\right)$$
where the absolute error behaves like $\frac{C}{n^5}$. Here some approximations derived through this geometric method:
$$\begin{array}{l|c|l}\hline n=12 & 4\sqrt{6}-4\sqrt{2}-1 & 3.141104722\\
\hline n=24 & \sqrt{2}-\sqrt{6}+8 \sqrt{8-4 \sqrt{2+\sqrt{3}}}&3.141561971\\ \hline\end{array}$$
This can be further improved. For instance, since the MacLaurin series of $\frac{x}{\sin x}$ and $\frac{1}{15}\left(68+11\cos(x)-64\cos(x/2)\right)$ agree up to the $x^6$ term (the same idea has been exploited here) we have
$$ \pi \approx \frac{n}{15}\sin\frac{\pi}{n}\left(68+11\cos\frac{\pi}{n}-15\cos\frac{\pi}{2n}\right) $$
and the following algebraic approximations:
$$\begin{array}{l|c|l}\hline n=6 & \frac{1}{10}\left(136+11\sqrt{3}-64\sqrt{2+\sqrt{3}}\right) & 3.141405312\\
\hline n=12 & \frac{\sqrt{3}-1}{5\sqrt{2}}\left(136+11\sqrt{2+\sqrt{3}}-64\sqrt{2+\sqrt{2+\sqrt{3}}}\right) &3.141589664\\ \hline \end{array}$$
Plenty of other approximations (both accurate and reasonably simple) can be derived by combining some version of the Shafer-Fink inequality and Machin formulas, for instance
$$\pi\approx \frac{180}{16 \sqrt{20+6 \sqrt{10}}+6 \sqrt{10}+21}+\frac{90}{8 \sqrt{10+4 \sqrt{5}}+3 \sqrt{5}+7}$$
whose error is $<10^{-6}$, or
$$ \pi \approx \frac{360}{7+7\sqrt{2}+6 \sqrt{2 \left(2+\sqrt{2}\right)}+16 \sqrt{2 \left(2+\sqrt{2}\right) \left(\sqrt{2+\sqrt{2}}+2\right)}}$$
whose error is $<4\cdot 10^{-7}$.
A: How about $$\pi \simeq \sqrt [3]{\cfrac{31}{1-\cfrac{12}{39^3-40}}}\tag{11 d.p.}$$ Easy to remember because $\pi^3\approx 31$ and $39$ is the number before $40$.
Also, a rather cool approximation is: $$\pi - e\simeq 1-\frac 1{\sqrt 3} + \frac{1}{\sqrt{11^6+13^5+19^4+24^3+33^2+\sqrt 5}}\tag{13 d.p.}$$

Here are a few approximations given by Ramanujan:
$$\pi\simeq \frac{12}{\sqrt {130}}\log_e \bigg\{\frac{(2+\sqrt 5)(3+\sqrt {13})}{\sqrt 2}\bigg\}\tag{15 d.p.}$$
$$\pi \simeq \frac{24}{\sqrt {142}}\log_e \Bigg\{\sqrt{\frac{10+7\sqrt 2}{4}}+\sqrt{\frac{10+11\sqrt 2}{4}}\Bigg\}\tag{16 d.p.}$$
$$\pi \simeq \frac{12}{\sqrt {190}}\log_e\big\{(2\sqrt 2+10)(3+\sqrt {10})\big\}\tag{18 d.p.}$$

A Chebyshev-type approximation: $$\pi\simeq \bigg(9^2+\frac{19^2}{22}\bigg)^{1/4}\tag{8 d.p.}$$
A: It is not a low degree polynomial, but easy to remember for sure. 
We know that $$\frac{1}{1+x}=1-x+x^2-x^3+\cdots$$
Substituting $x^2$, $$\frac1{1+x^2}=1-x^2+x^4-x^6+\cdots$$
But we also know that $\int\frac1{1+x^2}dx=\arctan x$. 

So let us integrate both sides (from $x=0$ to $x=y$), $$\arctan y=y-\frac{y^3}3+\frac{y^5}{5}-\frac{y^7}{7}+\cdots$$
Substitute $y=1$ and we get $$\pi=4(1-\frac13+\frac15-\frac17+\cdots)$$. 
A: How about $$\frac {3.1415926535}{1}$$  
It's fairly easy to memorize, and it's good to10 decimal places.
