Review for test 1 in MTH 229

[Get this a an ipynb file.]

Test 1 will cover the first 4 projects. This review workbook serves two purposes:

• to go over the major concepts we've learned

• to give you a workbook to start your exam from. The exam itself will be distributed on paper and the answers written on paper.

To begin, you should "run all" so that these packages are loaded into your Julia session.

using Plots
using Roots

Calculator skills:

The basic ideas of a calculator easily correspond to ideas of julia:

• basic computations involving +, -, *, / and ^ utilize the same order of operations (PEMDAS) that they do on most calculators, save for expressions like 5x where that multiplication happens before the "P".

• The buttons on a calculator for functions are replaced by names. So the "sin" button becomes sin(). Buttons like "sin$^{-1}$" are not spelled that way as functions (asin in this case). In fact the biggest gotcha is translating the shorthand of math related to powers to valid commands in julia.

• The = in julia is assignment. In math it usually implies solving an equation. This is completely different. The left hand side of the = sign in julia is just a name or function name with arguments, not an expression.

Functions

Basic functions in julia and in math look almost identical: $f(x) = \sin(x) + \cos(2x)$ becomes

f(x) = sin(x) + cos(2x)

f (generic function with 1 method)

Julia has some alternative forms though, for example when parameters are involved these can be added as additional arguments to the function with default values.

Functions in math that involve cases are often easily handled with the ternary operator: predicate ? iftrue : iffalse. An example was a model for cost when the first 100 cost 5 dollars per unit and afterwards 4 dollars per unit:

cost(x) = (x <= 100) ? 5*x : 5*100 + 4*(x - 100)

cost (generic function with 1 method)

Here are some examples:

• A function describing a secant line between 0 and $\pi/2$ for the sine function is given using the point-slope form of a line $(y=y0 + m \cdot (x-x0))$:

f(x) = sin(0) + (sin(pi/2) - sin(0)) / (pi/2 - 0) * (x - 0)

f (generic function with 1 method)

• A function modeling exponential growth with $r=5$% per year and initial amount of 1,000 would be given by the following. It's value at 2 is given:

f(x) = 1000 * exp(0.05 * x)
f(2)

1105.1709180756477

Plotting

Plotting is done in julia through external packages. We discussed the Plots package that makes it easy to plot functions by name: plot(f, a, b). We can also plot two or more functions over $[a,b]$ – just place the functions into a [] container:

f(x) = x^2 - 2x + 3
plot(f, -3, 4)

f(x) = cos(x)
g(x) = 1 + f(2*x)
plot([f,g], 0, 2pi)

Making a plot is easy, where to plot is the possibly difficult question.

Zeros

Finding zeros of a function is aided by some functions in the Roots package:

• If f is a polynomial function, then fzeros(f) will return all the real zeros and roots(f) all the zeros (which are called "roots" when speaking of polynomials.

f(x) = (x-1) * (x-2) * (x^2 + 2x + 2)
roots(f)

4-element Array{Complex{Float64},1}:
 -1.0+1.0im
 -1.0-1.0im
  1.0+0.0im
  2.0+0.0im

(Note, this uses complex values as some of the values are complex.) Whereas this finds the real roots:

fzeros(f)

2-element Array{Real,1}:
 1
 2

• if f is not a polynomial, there is nothing general, but: fzero(f, [a,b]) is guaranteed to find a zero in the interval $[a,b]$ provided there is obviously one there, whichi is when $f$ is continuous and $f(a)$ and $f(b)$ have different signs. Further, fzeros(f, [a,b]) will try to find all simple zeros in the interval $[a,b]$. This is not guaranteed to work, but usually does.

f(x) = exp(-3x) - 1/2
plot(f, 0, 5)			# zero between 0 and 1
fzero(f, [0, 1])

0.23104906018664842

Okay, start your engines

Use these cells or add some more to answer the questions on the exam.