Review for test 2 in MTH 229

Test 2 will cover the 4th and 5th packages: zeros of functions and limits. Some features we used are only available once the MTH229 package is loaded:

using MTH229
using SymPy

Zeros

The MTH229 package loads the Roots package which provides functions to return zeros of a function.

Polynomials

For polynomials functions we have the following:

roots(f) to find all real and (possibly) complex roots of a polynomial function.

fzeros(f) to find the real roots of a polynomial function.

For an $n$th degree polynomial, there are exactly $n$ roots when multiplicities are accounted for and complex numbers are used. There are $n-2k$ real roots, for some integer $k$, as complex roots come in pairs.

For example:

f(x) = x^5 - x - 1
roots(f)

5-element Array{Complex{Float64},1}:
    1.1673+0.0im     
  0.181232+1.08395im 
  0.181232-1.08395im 
 -0.764884+0.352472im
 -0.764884-0.352472im

(There are 5 complex roots to match the degree of the polynomial). Yet, only one is real, which you can check with fzeros(f):

fzeros(f)

1-element Array{Real,1}:
 1.1673

The following facts about polynomial zeros (roots in this case) are mentioned in the project:

The fundamental theorem of algebra: For a real polynomial of degree $n$ there are $n$ roots over the complex numbers if multiplicities are accounted for. The number of real roots is $n-2k$ for some integer $k$, possibly $k=0$.

Descarte's rule of signs: The number of positive real roots is $N-2k$ for some integer $k$, possible $k=0$ where $N$ counts the number of sign changes in the coefficients.

General functions

Polynomials are special enough, that one doesn't need to specify where to look for zeros. This is not so in general. We used two functions to find zeros:

fzero(f, a, b) will find a zero in $[a,b]$ when $[a,b]$ is a bracketing interval and $f$ is continuous. (That means $f(a)$ and $f(b)$ have different signs, so the intermediate value theorem guarantees some $c$ with $a \leq c \leq b$ and $f(c) = 0$.) This uses the bisection method.

fzeros(f, a, b) – not the extra "s" – will try to use bisection to find all the roots within the interval $[a,b]$. This $[a,b]$ need not be a bracket.

For example. Find a solution to $cos(x) = x/2$. We reframe this to be find a zero of $f(x) = \cos(x) - x/2$ and not that $f(0)=1$ and $f(pi/2) = -pi/4$, so $[0, \pi/2]$ is a bracket. Using this we have:

f(x) = cos(x) - x/2
fzero(f, 0, pi/2)

1.0298665293222589

Are there others? We can check. Since $|x/2| > 1$ when $|x| > 2$ and $\cos(x)$ is between $[-1,1]$, any solution must lie in $[-2,2]$. So we ask:

fzeros(f, -2, 2)

1-element Array{Float64,1}:
 1.02987

And we see that is it. Not so for a "flatter" line, say $x/10$:

f(x) = cos(x) - x/10
fzeros(f, -10, 10)

7-element Array{Float64,1}:
 -9.67888
 -8.96602
 -4.2711 
 -1.74633
  1.42755
  5.26712
  7.06889

These functions only solve for answers to $f(x) = 0$. You may have to rearrange your problem to fit that. For example, to find out when $\sinh(x) = 1/2$, we can solve for when $\sinh(x) - 1/2 = 0$:

f(x) = sinh(x) - 1/2
fzeros(f, -2, 2)  # one answer here

1-element Array{Float64,1}:
 0.481212

Limits

We discussed three ways to investigate the answer (if it exists) to

$$~ \lim_{x \rightarrow c} f(x) ~$$

graphical approach: make a graph and look (in the right place). For example, the limit of

$$~ f(t) = \frac{t - \sqrt{3t+4}}{4-t}, \text{ as } t \rightarrow 4 ~$$

Can be investigated by creating a graph around $c=4$:

f(t) = (t - sqrt(3t+4)) / (4-t)
plot(f, 3, 5)

The answer looks to be about $-0.625$.

tables. We had the helper function lim in the MTH229 package that makes a table

With this, making a table to investigate a limit is as easy as:

lim(f, 4, dir="+") ## going to 4 from right

6x2 Array{Float64,2}:
 4.1      -0.626742
 4.01     -0.625176
 4.001    -0.625018
 4.0001   -0.625002
 4.00001  -0.625   
 4.0      -0.625

symbolic: using the SymPy package, the command limit(f, c) will often find the limit. Here we compute the limit above symbolically:

limit(f, 4)

-5/8

In this case, all three give the same answer – the limit is $-5/8 = -0.625$.

Other questions may include finding the distinction between left and right limits, limits of functions with jumps, or whether a limit can be found just by "plugging in".

For example, to find the left limit at $0$ of $\sin(|x|)/x$ we have:

f(x) = sin(abs(x))/x
limit(f, 0, dir="-")

-1