How do you calculate the decimal expansion of an irrational number?
Solution 1:
$\pi$
For computing $\pi$, many very convergent methods are known. Historically, popular methods include estimating $\arctan$ with its Taylor's series expansion and calculating $\pi/4$ using a Machin-like formula. A basic one would be
$$\frac{\pi}{4} = 4 \arctan\frac{1}{5} - \arctan\frac{1}{239}$$
The reason these formulas are used over estimating $\arctan 1 =\frac{\pi}{4}$ is because the series for $\arctan x$ is move convergent for $x \approx0$. Thus, small values of $x$ are better for estimating $\pi/4$, even if one is required to compute $\arctan$ more times. A good example of this is Hwang Chien-Lih's formula:
$$ \begin{align} \frac{\pi}{4} =& 183\arctan\frac{1}{239} + 32\arctan\frac{1}{1023} - 68\arctan\frac{1}{5832} + 12\arctan\frac{1}{113021}\\ & - 100\arctan\frac{1}{6826318} - 12\arctan\frac{1}{33366019650} + 12\arctan\frac{1}{43599522992503626068}\\ \end{align} $$ Though $\arctan$ needs to be computed 7 times to a desired accuracy, computing this formula interestingly requires less computational effort then computing $\arctan 1$ to the same accuracy.
Iterative algorithms, such as Borwein's algorithm or Gauss–Legendre algorithm can converge to $\pi$ extremely fast (Gauss–Legendre algorithm find 45 million correct digits in 25 iterations), but require much computational effort. Because of this, the linear convergence of Ramanujan's algorithm or the Chudnovsky algorithm is often preferred (these methods are mentioned in other posts here as well). These methods produce 6-8 digits and 14 digits respectively term added. It is interesting to mention that the Bailey–Borwein–Plouffe formula can calculate the $n^{th}$ binary digit of $\pi$ without needing to know the $n-1^{th}$ digit (these algorithms are known as "spigot algorithms"). Bellard's formula is similar but 43% faster.
The first few terms from the Chudnovsky algorithm are (note the accuracy increases by about 14 decimal places):
n Approx. sum Approx. error (pi-sum)
0 3.141592653 5.90 x 10^-14
1 3.141592653 -3.07 x 10^-28
2 3.141592653 1.72 x 10^-42
3 3.141592653 1.00 x 10^-56
See these two questions as well.
$e$
The most popular method for computing $e$ is its Taylor's series expansion, because it requires little computational effort and converges very quickly (and continues to speed up). $$e=\sum_{n=0}^\infty \frac{1}{n!}$$ The first sums created in this series are as follows:
n Approx. sum Approx. error (e-sum)
0 1 1.718281828.
1 2 0.718281828
2 2.5 0.218281828
3 2.666666666 0.051615161
...
10 2.718281801 2.73 x 10^-8
...
20 2.718281828 2.05 x 10^-20
One should also note that the limit definition of $e$ and the series may be used in conjunction. The canonical limit for $e$ is
$$e=\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n$$
Noting that this is the first two terms of the Taylor's series expansion for $\exp(\frac{1}{n})$ to the exponent of $n$ for $n$ large, it is clear that $\exp(\frac{1}{n})$ can be computed to a higher accuracy in fewer terms then $e^1$ in the series, because in two terms give a better and better estimate as $n \to \infty$. This means that if we add another few terms of the expansion of $\exp(\frac{1}{n})$, we can find the $n^{th}$ root of $e$ to high accuracy (higher then the limit and the series) and then we just multiply the answer $n$ times with itself (easy, if $n$ is an integer).
As a formula, we have, if $m$ and $a$ are large:
$$e \approx \left(\sum_{n=0}^m \frac{1}{n!a^n}\right)^a$$
If we use the series to find the $100^{th}$ root (i.e. using the above formula, $a=100$) of $e$, this is what results (note the fast rate of convergence):
n Approx. sum Approx. sum^100 Approx. error (e-sum)
0 1 1 1.718281828.
1 1.01 2.704813829 0.013467999
2 1.01005 2.718236862 0.000044965
3 1.010050166 2.718281716 1.12 x 10^-7
...
10 1.010050167 2.7182818284 6.74 x 10^-28
...
20 1.010050167 2.7182818284 4.08 x 10^-51
$\varphi$
The golden ratio is $$\varphi=\frac{\sqrt{5}+1}{2}$$ so once $\sqrt{5}$ is computed to a sufficient accuracy, so can $\varphi$. To estimate $\sqrt{5}$, many methods can be used, perhaps most simply through the Babylonian method. Newton's root-finding method may also be used to find $\varphi$ because it and its reciprocal, $\Phi$, are roots of $$0=x^2-x-1$$
If $\xi$ is a root of $f(x)$, Newtons method finds $\xi$:
$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ $$\xi=\lim_{n \to \infty}x_n$$
We thus assign $f(x)=x^2-x-1$ and $f'(x)=2x-1$. Then $$x_{n+1}=x_n-\frac{x_n^2-x_n-1}{2x_n-1}=\frac{x_n^2+1}{2x_n-1}$$
If $x_0=1$, the first few iterations yield:
n value of x_n Approx. error (phi-x_n)
1 2 -0.381966011
2 1.666666666 -0.048632677
3 1.619047619 -0.001013630
4 1.618034447 -4.59 x 10^-7
...
7 1.618033988 -7.05 x 10^-54
The quadratic convergence of this method is very clear in this example.
$\gamma$
Unfortunately, no quadratically convergent methods are known to compute $\gamma$.
As mentioned above, some methods are discussed here: What is the fastest/most efficient algorithm for estimating Euler's Constant $\gamma$?
The algorithm from here is
$$ \gamma= 1-\log k \sum_{r=1}^{12k+1} \frac{ (-1)^{r-1} k^{r+1}}{(r-1)!(r+1)} + \sum_{r=1}^{12k+1} \frac{ (-1)^{r-1} k^{r+1} }{(r-1)! (r+1)^2}+\mbox{O}(2^{-k}) $$
and this method gives the following approximation:
k Approx. sum Approx. error (gamma-sum)
1 0.7965995992978246 0.21938393439629178
5 0.5892082678451087 0.011992602943575847
10 0.5773243590712589 1.086941697260313 x 10^-4
15 0.5772165124955206 8.47593987773898 x 10^-7
This answer has even faster convergence.
Some other methods are also reviewed here: http://www.ams.org/journals/mcom/1980-34-149/S0025-5718-1980-0551307-4/S0025-5718-1980-0551307-4.pdf
$\zeta(3)$
A method for estimating $\zeta(3)$ is the Amdeberhan-Zeilberger formula ($O(n \log n^3)$):
$$\zeta(3)=\frac{1}{64}\sum_{k=0}^{\infty}\frac{(-1)^k(205k^2+250k+77)(k!)^{10}}{((2k+1)!)^5}$$
$G$ (Catalan's constant)
Fee, in his article, presents a method for computing Catalan's constant based on a formula of Ramanujan:
$$G=\sum_{k=0}^\infty \frac{2^{k-1}}{(2k+1)\binom{2k}{k}}\sum_{j=0}^k \frac1{2j+1}$$
Another rapidly-converging series from Ramanujan has also been used for computing Catalan's constant:
$$G=\frac{\pi}{8}\log(2+\sqrt 3)+\frac38\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!(2n+1)^2}$$
$\log 2$
The Taylor's series for $\log$ has disappointingly poor convergence and for that alternate methods are needed to efficiently compute $\log 2$. Common ways to compute $\log 2$ include "Machin-like formulae" using the $\operatorname{arcoth}$ function, similar to the ones used to compute $\pi$ with the $\arctan$ function mentioned above:
$$\log 2=144\operatorname {arcoth}(251)+54\operatorname {arcoth}(449)-38\operatorname {arcoth}(4801)+62\operatorname {arcoth}(8749)$$
$A$ (Glaisher-Kinkelin constant)
One usual method for computing the Glaisher-Kinkelin constant rests on the identity
$$A=\exp\left(\frac1{12}(\gamma+\log(2\pi))-\frac{\zeta'(2)}{2\pi ^2}\right)$$
where $\zeta'(s)$ is the derivative of the Riemann zeta function. Now,
$$\zeta'(2)=2\sum_{k=1}^\infty \frac{(-1)^k \log(2k)}{k^2}$$
and any number of convergence acceleration methods can be applied to sum this alternating series. Two of the more popular choices are the Euler transformation, and the CRVZ algorithm.
Another interesting website that has many fast algorithms for common constants is here.
Solution 2:
Different irrationals yield to different techniques. $\phi=(1+\sqrt5)/2$ just involves calculating $\sqrt5$, which can be done easily by Newton's method from introductory calculus. The infinite series $$e=1+1+1/2+1/6+1/24+\cdots$$ where the denominators are the factorials, can be used to calculate $e$. For pi, this article on Gauss-Legendre algorithm will give you some ideas.