julia-first-steps.ipynb

Kernel: Julia 1.5

Julia – a modern approach to scientific computing

https://www.julialang.org

Evaluate the following cell by selecting it and then hit Shift+Return. Watch the top right status indicator of the Jupyter Notebook kernel to have started up Julia. The result will appear below.

1 + 1

2

Check the version of Julia you're currently running:

VERSION

v"1.5.0"

Vectors and matrices have a simple syntax. Notice, that the ' means to transpose this vector.

b = [1 4.4 -9]'

3×1 LinearAlgebra.Adjoint{Float64,Array{Float64,2}}:
  1.0
  4.4
 -9.0

A = [  1   9   1
       0   1   0
    -1.1   0  -1]

3×3 Array{Float64,2}:
  1.0  9.0   1.0
  0.0  1.0   0.0
 -1.1  0.0  -1.0

… and here we solve this as a linear system of equations: $A \cdot x = b$

x = A \ b

3×1 Array{Float64,2}:
  475.9999999999999
    4.400000000000001
 -514.5999999999999

A * x

3×1 Array{Float64,2}:
  1.0
  4.400000000000001
 -9.0

Batman Curve

Transcribed from Julia: A Fast Language for Numerical Computing, by Alan Edelman, SIAM News, Volume 49 | Number 02 | March 2016

# Note: the first time in a project, it can take over a minute to precompile PyPlot
using PyPlot
function batman()

    # bat functions: semicircle, ellipse, shoulders, bottom, cowl
    σ(x) = sqrt.(1 .- x.^2)
    e(x) = 3σ(x/7)
    s(x) = 4.2 .- .5x - 2.8σ(.5x .- .5)
    b(x) = σ(abs.(2 .- x) .- 1) - x.^2/11 .+ .5x .- 3
    c(x) = [1.7, 1.7, 2.6, .9]

    # plot symmetrically across y-axis
    p(x,f) = plot(-x,f(x), color="black") , plot(x,f(x), color="black")
    p((3:.1:7),e);p(4:.1:7,t->-e(t))   # ellipse
    p(1:.1:3,s);p(0:.1:4,b)            # shoulders and bottom
    p([0,.5,.8,1],c)
end
batman();

┌ Info: Precompiling PyPlot [d330b81b-6aea-500a-939a-2ce795aea3ee]
└ @ Base loading.jl:1278

Matplotlib is building the font cache; this may take a moment.