- Sorry the zoom didn't work better!
- Please hand in your second homework assignment, for section 1.2 (Taylor series).
- You also have some 1.3 homework, which will be due Friday. Please scan it and email, or drop it off at my office.
- You also have some new homework problems for section 2.2, which will be due next Wednesday, 1/29.
- Addressed some questions about the Taylor Series stuff
- More on speed: more review, and a few more points
- Then we hit bisection, and used it to create an arcsin function (fake), based off our sinner fake of sine.
- Bisection is an iterative process, meaning that we start with the interval \([x_0,x_1]\), and cut the problem (to an interval of half the width of the original), meaning that we get closer to the solution by a factor of 2 ("Bi"section) with each iteration.
- Let's talk about Section 2.2: fixed-point iteration.
The author starts with an example, which illustrates that one can effortlessly solve the equation \[ \cos(x)=x \] by simply hitting your calculator's "cos" button over and over, starting from a reasonable first guess.
And starting from every point in the vicinity of the solution leads us steadily to the solution (or arbitrarily close, rather -- there's an error!). Here, let's try it (make sure that your calculator is in radian mode....).
Then he mentions that the same idea fails miserably if we try to arrive at the root \(x=1\) of \[ x^2=x \] Starting from every point in the vicinity of the solution leads us steadily away from the solution!
What distinguishes the two cases? Slopes....
- Proposition 2, p. 52: If $h(x)$ is differentiable on $(a,b)$ with
$|h'(x)|<1$ for all $x\in(a,b)$, then whenever
$x_{1},x_{2}\in(a,b)$, $|h(x_{2})-h(x_{1})|<|x_{2}-x_{1}|$.
"That is, points move toward one another under application of the function." -- when the slope is less than 1 over an interval.
- Proposition 3: Suppose $h(x)$ is continuous on $[a,b]$,
differentiable on $(a,b)$ with $|h'(x)|<1$ for all $x\in(a,b)$, and
$h([a,b])\subseteq[a,b]$. Then $h$ has a unique fixed point in
$[a,b]$.
- Let's use FPI to fake arcsine (rather than bisection). We'll need
a function that meets our requirements on the interval
\([0,\pi/2]\): how about
\[
f(x)=x-\sin(x)+x_0
\]
(remember, we're not root-finding here: we're looking for a
fixed point). So we re-write our root function,
\[
\sin(x)-x_0=0
\]
as
\[
x-\sin(x)+x_0=x
\]
et voila!
Does it meet the slope requirements? Where will we have trouble?
- Note that there are generally lots of candidates for the FPI
function, but only some of them will converge on a fixed point.
- A final really fun example illustrated in this section is that of
chaos, illustrated in this figure:
The example discussed illustrates a breakdown in convergence leading to cyclical behavior, and finally chaos....
- Our author's cobwebbing animation.
- fixed point in Mathematica
- Check out how my bisection works, and how it is applied to arcsine....
- Various sins....