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Today:
You can just as easily count up, of course; but you want to give the user the ability to control the number.
Honestly though:
Let's go through these and see if there are any particular remarks you'd like to make. I'll point out a few things.
Danny considered a constrained problem: how to make a single Bezier cubic form a circle. It's not easy ("might fool a toddler...").
Danny also put ChatGPT to use in trying to solve this problem.
but struggled to get the "weird little wiggles" just right!
An important highlight is a description of the tedium involved. But the final results, wow!
Note the handwork (and hard work, and tedium) of transcribing the points from the signature.
We end up with an implicit formula for \(y\) -- buried in that integrand -- which allowed us to approximate points along the curve: \[ y(t_{1})=y(t_{0})+\int_{t_{0}}^{t_{1}}f(t,y)\,dt \] Each integration method produces a different estimate for that integral, and hence provides a different method for estimating the solution of the ODE.
Some methods leave us in a quandary as to how to replace a value (e.g. the derivation of Simpson's).
We know \(f\), and we know \(y_0\) (we study so called "initial value problems", IVPs, where we know the starting value) -- putting those together we can estimate \(y_1\); now do it again!
\[ y_{i+1}=y_{i}+\frac{h}{2}\left[f(t_{i},y_{i})+f(t_{i+1}, y_{i}+h \cdot f(t_{i},y_{i}))\right]. \]
Our author takes a crack at it, but admits that the first crack is not the best crack. Then, in a crumpet in the next section (which covers the error analysis of these methods), illustrates how RK4 is obtained from the typical "undetermined coefficents" approach by clever matching of Taylor series.
\[ \begin{eqnarray*} k_{1} & = & f(t_{i},y_{i})\\ k_{2} & = & f\left(t_{i}+\frac{h}{2},y_{i}+\frac{h}{2}k_{1}\right)\\ k_{3} & = & f\left(t_{i}+\frac{h}{2},y_{i}+\frac{h}{2}k_{2}\right)\\ k_{4} & = & f(t_{i+1},y_{i}+hk_{3})\\ y_{i+1} & = & y_{i}+\frac{h}{6}\left[k_{1}+2k_{2}+2k_{3}+k_{4}\right]. \end{eqnarray*} \]
We have split the middle term into two pieces, and essentially use two different slopes to approximate its value (creating a new estimate from two others).
The take-home part will be due on Wednesday, May 1st, at Midnight (I can't wait much further, because I'll be doing a lot of grading, and can't get backed up). I'm planning on grading those on Thursday.
You might also look over the homeworks for each section.
Danny has some notes that he's worked up (and perhaps continues to work up!), which he has indicated he is willing to share.
There are those who think that mathematics is great because "there's always a right answer." Well, phooey!:)
It ain't lyin', but it's the best we can do!
We saw that on this homework 4.4.
It's not quite what you're asked to do in this algorithm, however, since you're to keep track of the errors, and keep poking at the largest:
But the idea is really swell, and I am delighted by this method!
Here's another picture of how we might build one of these.