Last time | Next time |
Today:
Homework pp. 93--, #1-5, 6, 8, 11, 13, 14, 16, 22 (use your Muller code), 23
I could stare at that all day!
That one I've got to let him know about!:)
\[ a_{n}(\cdots a_{n}(a_{n}(a_{n}t+a_{n-1})+a_{n-2})+\cdots+a_{1}) \] becomes \[ a_0+t(a_{1} + t(a_{2}+t(a_{3}+t(a_{n}+\cdots+t(a_{n-1}+t(a_{n}))\cdots)) \]
Let me show you how that comes about....
We can even get the 2nd derivative at \(t\) as well, in case that's useful! Let's see....
(In the exercises you'll extend this idea to a cubic!)
I'll gripe a little about our author's choice for how he represents the algorithm on pp. 91-92 (he's taking some abuse in this section!:). My feeling is that this would be a good time to introduce divided differences. If you look at the formulas on p. 91, you'll see that we re-use some material -- and any time you re-use material, you should avoid re-calculating.
Divided differences are what we use to approximate derivatives, say: \[ DD(x_1,x_0) \equiv \frac{f(x_1)-f(x_0)}{x_1-x_0} \]
Of course, once you have divided differences, you can take divided differences of them (essentially higher-order derivatives):
\[ DD(x_2,x_1,x_0) \equiv \frac{DD(x_2,x_1)-DD(x_1,x_0)}{x_2-x_0} \]
This allows us to write \[ a=DD(p_1,p_2,p_0)=\frac{DD(p_1,p_2)-DD(p_2,p_0)}{p_0-p_1} \] and \[ b=DD(p_1,p_2)-(p_1-p_2)a \]
This is sort of a "2.7" thing, however....