- Homework #1 returned; nice work, generally: if you have questions, let me know! Some skipped a problem, and I don't know if it was because it was too hard, or.... Here's a little Octave, but I did most of it in Mathematica.
- Again, we're going to skip Section 1.4 -- it doesn't really seem critical, but you might check out the Octave parts, in particular. It gives some good advice for programming using Octave.
- You also owe me your second homework assignment, for section 1.2 (Taylor series). If you've got it and want to turn it in, fine; otherwise, I'll take it on Wednesday.
- You also have some new homework problems for section 2.1, which will be due Monday, 1/29.
- We'll hopefully get started on section 2.2 next time.
- Reviewed Taylor Series, in particular the sins;
- Noted that, if we have sins, we have cosins, tansins, etc.
- Observed that, while we're close to having arcsins, we would need
to solve an equation, and for that we need to create a solver
(that's what we're doing today -- or at least starting....).
On the interval \(y \in [0,\frac{\pi}{2}]\), \[ \arcsin(x)=y \iff \sin(y)=x \] or \[ \arcsin(x)=y \iff \sin(y)-x=0 \] So we want to find \(y\) such that \(\sin(y)-x=0\), and if we can that, then we've got a fake arcsine.
- Then we discussed issues of speed (rate of convergence), and looked at some examples.
- Comments on #4
- #12a
- Any others?
- Suppose the sequence $\langle p_{n}\rangle$ converges to $p$: then
the order of convergence \(\alpha\) satisfies:
\[
\lim_{n\to\infty}\frac{\left|p_{n+1}-p\right|}{\left|p_{n}-p\right|^{\alpha}}=\lambda
\]
for some real number $\lambda>0$.
\(\alpha\) represents the rate at which a sequence converges, and bigger is better -- convergence is faster. And \(\alpha > 1\) -- or there is no convergence (see Crumpet 6).
Check out Table 1.4, on page 20: four sequences that converge to \(e\):
We can estimate \(\alpha\) by checking a few of the iterates: \[ \alpha\approx\frac{\ln\left|\frac{p_{n+2}-p}{p_{n+1}-p}\right|}{\ln\left|\frac{p_{n+1}-p}{p_{n}-p}\right|} \]
Last time we tried to determine the order of convergence of these sequences, and got strange results: my bad! I asked you to push the interates a little too far!
The problem was truncation error: if you used a value of the sequence that matched \(e\) to the given number of digits, then the calculation of \(2.71828182845904-e\) is nothing but garbage, and has nothing to do with the particular sequence that generated it.
- Let's show that, if $\alpha>1$,
\[
d(p_{n_{0}+k})\approx(d_{n_{0}}-C)\alpha^{k}+C
\]
where ${\displaystyle C=\frac{\log\left(\lambda|p|^{\alpha-1}\right)}{\alpha-1}}$
as our author asserts. This says that, if $\alpha>1$, the increase in significant digits is exponential; so it might better to express it this way: \[ d(p_{n_{0}+k})-C \approx(d_{n_{0}}-C)\alpha^{k} \] or \[ b_k \approx b_0 \alpha^{k} \] (This is the theme of Crumpet 7, but I'll do it using the "expand, guess, verify" method)
- Finally let's have a look at the "Big O" notation on page 24,
under the heading "rate of convergence of a sequence":
- So, for example,
\[
\left|\frac{2n+1}{4n}-\frac{1}{2}\right|=\frac{1}{4n}\leq\frac{1}{4}\cdot\frac{1}{n},
\]
means that we can write
\[
\frac{2n+1}{4n}=\frac{1}{2}+O(\frac{1}{n})
\]
"Normally, when we find a rate of convergence, we try to find
the fastest converging sequence from our stock of simple
examples that satisfies the definition. In this case, there
is none faster."
- "Basically all the sequences studied in any depth in calculus converge with linear order."
#1a, p. 29: Find the order of convergence for ${\displaystyle \left\langle \frac{n!}{n^{n}}\right\rangle \to0}$ (Danny's question: how do we know to expect rate of convergence of 1? Estimated using Mathematica, but also per our author's remark above...)
- "Essentially, it takes an exponentially growing exponent
to converge with an order greater than 1."
#3, p. 29: Show that the sequence $p_{n}=2^{1-2^{n}}$ is quadratically convergent.
#8, p. 30: Use a Taylor polynomial to find the rate of convergence of \[ \lim_{h\to0}(2-e^{h})=1. \]
- So, for example,
\[
\left|\frac{2n+1}{4n}-\frac{1}{2}\right|=\frac{1}{4n}\leq\frac{1}{4}\cdot\frac{1}{n},
\]
means that we can write
\[
\frac{2n+1}{4n}=\frac{1}{2}+O(\frac{1}{n})
\]
"Normally, when we find a rate of convergence, we try to find
the fastest converging sequence from our stock of simple
examples that satisfies the definition. In this case, there
is none faster."
- Finally, let's look at Exercise #20: Prove that the order of convergence of a sequence is unique.
- This is an equation solver, usually written as a "root-finder": find root \(x\) such that \[ f(x)=0 \]
- This method is very simple: it relies on the following:
- The function \(f\) is continuous on the interval \([x_0,x_1]\).
- The signs of \(f(x_0)\) and \(f(x_1)\) differ: that is, \(f(x_0)*f(x_1)<0\).
- Then, by the Intermediate Value Theorem (IVT), \[ \exists x\in(x_0, x_1) / f(x)=0 \]
This image shows that the IVT is more general than root-finding: it asserts that we can find a value of \(x\) such that \(f(x)=K\) for any "intermediate value" between the two values \(f(x_0)\) and \(f(x_1)\).
But of course we can re-write \(f(x)=K\) as \(f(x)-K=0\), and turn it into an exercise in finding roots of the function \(g(x)=f(x)-K\).
- 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.
We
- compute the midpoint \(c = \frac{x1-x0}{2}\)
- compute \(f(c)\); if \(f(c)=0\) then we're done, and an answer is \(c\) (there may be more than one!);
- Otherwise, check the sign of \(f(c)\), and replace the endpoint of matching sign with \((c,f(c))\);
- then do it again, until satisfied.
- So, how do we know when to stop -- when are we satisfied? We want
to find an approximate
solution \(x^*\) within a tolerance of \(\tau\) of the true solution
\(x\); that is, when the width of the interval \(x1-x0<\tau\).
And we may want to demand, in addition, that \(|f(c)| < \eps\).
- Let's check out how my bisection works, and we'll apply it to arcsine....
- Various sins....