Some fun with π in Julia

14 Mar 2017 |  Simon Byrne, Luis Benet and David Sanders

This post is available as a Jupyter notebook here

π in Julia

(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 BigFloats, 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 bitssetprecision(BigFloat,1024)doBigFloat(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 Irrationals is that inequalities are correct:

Float64(pi)<pi<nextfloat(Float64(pi))

true

π via inline assembly instructions

(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 (“floating point loadpi”) 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_nativeasm_pi()

    .section        __TEXT,__text,regular,pure_instructionsFilename: In[10]        pushq   %rbp        movq    %rsp, %rbpSource 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:

1. the pushq and movq adds to the call stack frame.
2. 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.
3. 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.
4. popq and retq pops the call stack frame.

π using a Taylor series expansions

(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.

usingTaylorSeries

Madhava’s formula

One of the standard trigonmetric identities is

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 BigFloats:

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 + ?(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 seriespi_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:

\begin{equation} \pi = 6 \sum_{n=0}^{\infty} (-1)^n \frac{(1/\sqrt{3})^{2n+1}}{2n+1}. \end{equation}

Machin’s approach

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

\begin{equation} \frac{\pi}{4} = 4 \arctan\left(\frac{1}{5}\right) - \arctan\left(\frac{1}{239}\right). \end{equation}

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

Finding guaranteed bounds on π

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

Using standard floating-point arithmetic

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

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 $T_N$:

.

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

$S_N$ may be calculated easily by summing either forwards or backwards:

function forward_sum(N,T=Float64)total=zero(T)foriin1:Ntotal+=one(T)/(i^2)endtotalendfunction reverse_sum(N,T=Float64)total=zero(T)foriinN:-1:1total+=one(T)/(i^2)endtotalend

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=10^6lowerbound_S_N=setrounding(Float64,RoundDown)doforward_sum(N)endupperbound_S_N=setrounding(Float64,RoundUp)doforward_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=10^6lower_π=setrounding(Float64,RoundDown)dolower_bound=forward_sum(N)+1/(N+1)sqrt(6*lower_bound)endupper_π=setrounding(Float64,RoundUp)doupper_bound=forward_sum(N)+1/Nsqrt(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=10^6lower_π=setrounding(Float64,RoundDown)dolower_bound=reverse_sum(N)+1/(N+1)sqrt(6*lower_bound)endupper_π=setrounding(Float64,RoundUp)doupper_bound=reverse_sum(N)+1/Nsqrt(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 principal, we could attain arbitrarily good precision with higher-precision BigFloats, but the result is hampered by the slow convergence of the series.

Summing a series using interval arithmetic

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

usingValidatedNumerics

setdisplay(:standard)# abbreviated display of intervals

6

N=10000S=forward_sum(N,Interval)S+=1/(N+1)..1/N# interval bound on the remainder of the seriesπ_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)

It’s 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 displayInterval(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

Calculating an area

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 $\pi$:

usingPlots;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

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]foriin1:length(xs)-1]endintervals=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)forXinintervalsY=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)endplot!(f,0,2)p

Now we just sum up the areas:

N=20intervals=make_intervals(N)width=2/Nwidth*sum(√(4-X^2)forXinintervals)

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")forNin50:50:1000intervals=make_intervals(N)area=(2/N)*sum(√(4-X^2)forXinintervals)println("$N \t$area \t$(diam(area))")end

N    area interval           diameter50       [3.0982, 3.1783]        0.0800000000000165100      [3.1204, 3.1605]        0.040000000000032454150      [3.1276, 3.1543]        0.02666666666670814200      [3.1311, 3.1512]        0.02000000000006308250      [3.1332, 3.1493]        0.016000000000075065300      [3.1346, 3.1481]        0.013333333333415354350      [3.1356, 3.1472]        0.011428571428676815400      [3.1364, 3.1465]        0.010000000000123688450      [3.137, 3.146]          0.008888888889027502500      [3.1374, 3.1455]        0.008000000000148333550      [3.1378, 3.1452]        0.007272727272884527600      [3.1381, 3.1449]        0.006666666666829357650      [3.1384, 3.1446]        0.006153846154013376700      [3.1386, 3.1444]        0.0057142857144931725750      [3.1388, 3.1443]        0.005333333333562784800      [3.139, 3.1441]         0.005000000000246363850      [3.1391, 3.1439]        0.004705882353203794900      [3.1393, 3.1438]        0.004444444444719142950      [3.1394, 3.1437]        0.0042105263160761021000     [3.1395, 3.1436]        0.004000000000294435