Image courtesy of Cormullion, code here.

This post is available as a Jupyter notebook here.

by Simon Byrne

Like most technical languages, Julia provides a variable constant for π. However Julia's handling is a bit special.

```
pi
π = 3.1415926535897...
```

It can also be accessed via the unicode symbol (you can get it at the REPL or in a notebook via the TeX completion `\pi`

followed by a tab)

```
π
π = 3.1415926535897...
```

You'll notice that it doesn't print like an ordinary floating point number: that's because it isn't one.

```
typeof(pi)
Irrational{:π}
```

π and a few other irrational constants are instead stored as special `Irrational`

values, rather than being rounded to `Float64`

. These act like ordinary numeric values, except that they can are converted automatically to any floating point type without any intermediate rounding:

```
1 + pi # integers are promoted to Float64 by default
4.141592653589793
```

```
Float32(1) + pi # Float32
4.141593f0
```

This is particularly useful for use with arbitrary-precision `BigFloat`

s, as π can be evaluated to full precision (rather than be truncated to `Float64`

and converted back).

```
BigFloat(1) + pi # 256 bits by default
4.141592653589793238462643383279502884197169399375105820974944592307816406286198
```

If π were stored as a `Float64`

, we would instead get

```
BigFloat(1) + Float64(pi)
4.141592653589793115997963468544185161590576171875000000000000000000000000000000
```

In fact `BigFloat`

(which uses the MPFR library) will compute π on demand to the current precision, which is set via `setprecision`

. This provides an easy way to get its digits:

```
# to 1024 bits
setprecision(BigFloat, 1024) do
BigFloat(pi)
end
3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724586997
```

The last few printed digits may be incorrect due to the conversion from the internal binary format of `BigFloat`

to the decimal representation used for printing. This is just a presentation issue, however – the internal binary representation is correctly rounded to the last bit.

Another neat property of `Irrational`

s is that inequalities are correct:

```
Float64(pi) < pi < nextfloat(Float64(pi))
true
```

by Simon Byrne

Julia provides a very low-level `llvmcall`

interface, which allows the user to directly write LLVM intermediate representation, including the use of inline assembly. The following snippet calls the `fldpi`

instruction ("**f**loating point **l**oa**d** **pi**") which loads the constant π onto the floating point register stack (this works only on x86 and x86_64 architectures)

```
function asm_pi()
Base.llvmcall(
""" %pi = call double asm "fldpi", "={st}"()
ret double %pi""",
Float64, Tuple{})
end
asm_pi (generic function with 1 method)
```

```
asm_pi()
3.141592653589793
```

We can look at the actual resulting code that is generated:

```
@code_native asm_pi()
.section __TEXT,__text,regular,pure_instructions
Filename: In[10]
pushq %rbp
movq %rsp, %rbp
Source line: 2
fldpi
fstpl -8(%rbp)
movsd -8(%rbp), %xmm0 ## xmm0 = mem[0],zero
popq %rbp
retq
```

If you're wondering what the rest of these instructions are doing:

the

`pushq`

and`movq`

adds to the call stack frame.`fldpi`

pushes π to the x87 floating point register stack

x87 is the older legacy floating point instruction set dating back to the original Intel 8087 coprocessor.

`fstpl`

and`movsd`

moves the value to the SSE floating point register`xmm0`

Julia, like most modern software, uses the newer SSE instruction set for its floating point operations. This also allows us to take advantage of things like SIMD operations.

`popq`

and`retq`

pops the call stack frame.

by Luis Benet, Instituto de Ciencias Físicas, Universidad Nacional Autónoma de México (UNAM))

This will demonstrate how to evaluate π using various Taylor series expansions via the TaylorSeries.jl package.

`using TaylorSeries`

One of the standard trigonmetric identities is

$\tan\left( \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}}.$Therefore, by taking the Taylor expansion of $6 \arctan(x)$ around 0 we may obtain the value of $\pi$, by evaluating it at $1/\sqrt{3}$, a value which is within the radius of convergence.

We obtain the Taylor series of order 37th, using `BigFloat`

s:

```
series1 = 6atan( Taylor1(BigFloat, 37) )
convert(Taylor1{Rational{BigInt}},series1)
6//1 t - 2//1 t³ + 6//5 t⁵ - 6//7 t⁷ + 2//3 t⁹ - 6//11 t¹¹ + 6//13 t¹³ - 2//5 t¹⁵ + 6//17 t¹⁷ - 6//19 t¹⁹ + 2//7 t²¹ - 6//23 t²³ + 6//25 t²⁵ - 2//9 t²⁷ + 6//29 t²⁹ - 6//31 t³¹ + 2//11 t³³ - 6//35 t³⁵ + 6//37 t³⁷ + 𝒪(t³⁸)
```

Note that the series above has only odd powers, so we will be using in this case 18 coefficients.

Evaluating that expression in $1/\sqrt{3}$ we get

```
pi_approx1 = evaluate(series1, 1/sqrt(big(3)))
3.141592653647826046431202390582141253830948237428790668441592864548346569098516
```

Then, the 37th order Taylor expansion yields a value which differs from $\pi$ in:

```
abs(pi - pi_approx1)
5.803280796855900730263836963377883805368484746664827224053016281231814650118929e-11
```

To obtain more accurate results, we may simply increase the order of the expansion:

```
series2 = 6atan( Taylor1(BigFloat,99) ) # 49 coefficients of the series
pi_approx2 = evaluate(series2, 1/sqrt(BigInt(3)))
3.141592653589793238462643347272152237127662423839333289949470742535834074912581
```

```
abs(pi - pi_approx2)
3.600735064706950697553577253102547384977198233137361734413175534929622111373249e-26
```

This formulation is one of the *Madhava* or *Gregory–Leibniz series*:

Following the same idea, John Machin derived an algorithm which converges much faster, using the identity

$\frac{\pi}{4} = 4 \arctan\left(\frac{1}{5}\right) - \arctan\left(\frac{1}{239}\right).$Following what we did above, using again a 37th Taylor expansion:

```
ser = atan( Taylor1(BigFloat, 37) )
pi_approx3 = 4*( 4*evaluate(ser, 1/big(5)) - evaluate(ser, 1/big(239)) )
3.141592653589793238462643383496777424642594661632063407072684671069773618535135
```

```
abs(pi - pi_approx3)
2.17274540445425262256957586097740078761957212248936631045983596428448951876822e-28
```

by David P. Sanders, Department of Physics, Faculty of Sciences, National University of Mexico (UNAM)

We will calculate *guaranteed* (i.e., *validated*, or mathematically rigorous) bounds on $\pi$ using just floating-point arithmetic. This requires "directed rounding", i.e. the ability to control in which direction floating-point operations are rounded.

This is based on the book *Validated Numerics* (Princeton, 2011) by Warwick Tucker.

Consider the infinite series

$S := \sum_{n=1}^\infty \frac{1}{n^2},$whose exact value is known to be $S = \frac{\pi^2}{6}$. Thus, if finding guaranteed bounds on $S$ will give guaranteed bounds on $\pi$.

The idea is to split $S$ up into two parts, $S = S_N + T_N$, where $S_N := \sum_{n=1}^N \frac{1}{n^2}$ contains the first $N$ terms, and $T_N := S - S_N = \sum_{n=N+1}^\infty \frac{1}{n^2}$ contains the rest (an infinite number of terms).

We will evalute $S_N$ numerically, and use the following analytical bound for $T_N$:

$\frac{1}{N+1} \le T_N \le \frac{1}{N}$.

This is obtained by approximating the sum in $T_N$ using integrals from below and above:

$\int_{x=N+1}^\infty \frac{1}{x^2} dx \le T_N \le \int_{x=N}^\infty \frac{1}{x^2} dx.$$S_N$ may be calculated easily by summing either forwards or backwards:

```
function forward_sum(N, T=Float64)
total = zero(T)
for i in 1:N
total += one(T) / (i^2)
end
total
end
function reverse_sum(N, T=Float64)
total = zero(T)
for i in N:-1:1
total += one(T) / (i^2)
end
total
end
reverse_sum (generic function with 2 methods)
```

To find *rigorous* bounds for $S_N$, we use "directed rounding", that is, we round downwards for the lower bound and upwards for the upper bound:

```
N = 10^6
lowerbound_S_N =
setrounding(Float64, RoundDown) do
forward_sum(N)
end
upperbound_S_N =
setrounding(Float64, RoundUp) do
forward_sum(N)
end
(lowerbound_S_N, upperbound_S_N)
(1.6449330667377557,1.644933066959796)
```

We incorporate the respective bound on $T_N$ to obtain the bounds on $S$, and hence on $\pi$:

```
N = 10^6
lower_π =
setrounding(Float64, RoundDown) do
lower_bound = forward_sum(N) + 1/(N+1)
sqrt(6 * lower_bound)
end
upper_π =
setrounding(Float64, RoundUp) do
upper_bound = forward_sum(N) + 1/N
sqrt(6 * upper_bound)
end
(lower_π, upper_π, lowerbound_S_N)
(3.1415926534833463,3.1415926536963346,1.6449330667377557)
```

```
upper_π - lower_π
2.1298829366855898e-10
```

We may check that the true value of $\pi$ is indeed contained in the interval:

```
lower_π < pi < upper_π
true
```

Summing in the opposite direction turns out to give a more accurate answer:

```
N = 10^6
lower_π =
setrounding(Float64, RoundDown) do
lower_bound = reverse_sum(N) + 1/(N+1)
sqrt(6 * lower_bound)
end
upper_π =
setrounding(Float64, RoundUp) do
upper_bound = reverse_sum(N) + 1/N
sqrt(6 * upper_bound)
end
(lower_π, upper_π)
(3.1415926535893144,3.141592653590272)
```

```
upper_π - lower_π
9.57456336436735e-13
lower_π < pi < upper_π
true
```

In principle, we could attain arbitrarily good precision with higher-precision `BigFloat`

s, but the result is hampered by the slow convergence of the series.

We repeat the calculation using *interval arithmetic*, provided by the ValidatedNumerics.jl package.

`using ValidatedNumerics`

```
setdisplay(:standard) # abbreviated display of intervals
6
```

```
N = 10000
S = forward_sum(N, Interval)
S += 1/(N+1) .. 1/N # interval bound on the remainder of the series
π_interval = √(6S)
[3.14159, 3.1416]
```

Here we used an abbreviated display for the interval. Let's see the whole thing:

```
setdisplay(:full)
π_interval
Interval(3.1415926488148807, 3.141592658365341)
```

Its diameter (width) is

```
diam(π_interval)
9.550460422502738e-9
```

Thus, the result is correct to approximately 8 decimals.

In this calculation, we used the fact that arithmetic operations of intervals with numbers automatically promote the numbers to an interval:

```
setdisplay(:full) # full interval display
Interval(0) + 1/3^2
Interval(0.1111111111111111, 0.11111111111111112)
```

This is an interval containing the true real number $1/9$ (written `1//9`

in Julia):

```
1//9 ∈ convert(Interval{Float64}, 1/3^2)
true
```

Finally, we can check that the true value of $\pi$ is indeed inside our interval:

```
pi ∈ π_interval
true
```

Although the calculation above is simple, the derivation of the series itself is not. In this section, we will use a more natural way to calculate $\pi$, namely that the area of a circle of radius $r$ is $A(r) = \pi r^2$. We will calculate the area of one quadrant of a circle of radius $r=2$, which is equal to $\pi$:

`using Plots; gr();`

```
f(x) = √(4 - x^2)
f (generic function with 1 method)
```

`plot(f, 0, 2, aspect_ratio=:equal, fill=(0, :orange), alpha=0.2, label="")`

The circle of radius $r=2$ is given by $x^2 + y^2 = 2^2 = 4$, so

$\pi = \frac{1}{4} A(2) = \int_{x=0}^2 y(x) \, dx = \int_{x=0}^2 \sqrt{4 - x^2}.$In calculus, we learn that we can approximate integrals using **Riemann sums**. Interval arithmetic allows us to make these Riemann sums **rigorous** in a very simple way, as follows.

We split up the $x$ axis into intervals, for example of equal width:

```
function make_intervals(N=10)
xs = linspace(0, 2, N+1)
return [xs[i]..xs[i+1] for i in 1:length(xs)-1]
end
intervals = make_intervals()
10-element Array{ValidatedNumerics.Interval{Float64},1}:
Interval(0.0, 0.2)
Interval(0.19999999999999998, 0.4)
Interval(0.39999999999999997, 0.6000000000000001)
Interval(0.6, 0.8)
Interval(0.7999999999999999, 1.0)
Interval(1.0, 1.2000000000000002)
Interval(1.2, 1.4000000000000001)
Interval(1.4, 1.6)
Interval(1.5999999999999999, 1.8)
Interval(1.7999999999999998, 2.0)
```

Given one of those intervals, we evaluate the function of interest:

```
II = intervals[1]
Interval(0.0, 0.2)
```

```
f(II)
Interval(1.9899748742132397, 2.0)
```

The result is an interval that is **guaranteed to contain** the true range of the function $f$ over that interval. So the lower and upper bounds of the intervals may be used as lower and upper bounds of the height of the box in a Riemann integral:

```
intervals = make_intervals(30)
p = plot(aspect_ratio=:equal)
for X in intervals
Y = f(X)
plot!(IntervalBox(X, Interval(0, Y.lo)), c=:blue, label="", alpha=0.1)
plot!(IntervalBox(X, Interval(Y.lo, Y.hi)), c=:red, label="", alpha=0.1)
end
plot!(f, 0, 2)
p
```

Now we just sum up the areas:

```
N = 20
intervals = make_intervals(N)
width = 2/N
width * sum(√(4 - X^2) for X in intervals)
Interval(3.0284648797549782, 3.2284648797549846)
```

As we increase the number of sub-intervals, the approximation gets better and better:

```
setdisplay(:standard, sigfigs=5)
println("N \t area interval \t \t diameter")
for N in 50:50:1000
intervals = make_intervals(N)
area = (2/N) * sum(√(4 - X^2) for X in intervals)
println("$N \t $area \t $(diam(area))")
end
N area interval diameter
50 [3.0982, 3.1783] 0.0800000000000165
100 [3.1204, 3.1605] 0.040000000000032454
150 [3.1276, 3.1543] 0.02666666666670814
200 [3.1311, 3.1512] 0.02000000000006308
250 [3.1332, 3.1493] 0.016000000000075065
300 [3.1346, 3.1481] 0.013333333333415354
350 [3.1356, 3.1472] 0.011428571428676815
400 [3.1364, 3.1465] 0.010000000000123688
450 [3.137, 3.146] 0.008888888889027502
500 [3.1374, 3.1455] 0.008000000000148333
550 [3.1378, 3.1452] 0.007272727272884527
600 [3.1381, 3.1449] 0.006666666666829357
650 [3.1384, 3.1446] 0.006153846154013376
700 [3.1386, 3.1444] 0.0057142857144931725
750 [3.1388, 3.1443] 0.005333333333562784
800 [3.139, 3.1441] 0.005000000000246363
850 [3.1391, 3.1439] 0.004705882353203794
900 [3.1393, 3.1438] 0.004444444444719142
950 [3.1394, 3.1437] 0.004210526316076102
1000 [3.1395, 3.1436] 0.004000000000294435
```