a glimmer of hope

Have you guys (slowedd and sheapple) handed in the example paper for marking?

Would you be interested divding the paper into tree parts to hand it in and get some feed back?

One concern of mine is the descution questions, Don't sound smart enough when I try to explain things.

Lemmie know.

I'm on a supp. Would be grateful if you posted the questions here, and I'll give my version of the answer to all of them if you like. (I'm presuming year mark is now out of the way by quite a long way.)

This is one of those where whether you pass or fail in the end, you gain something. Just by putting your shoulder to this boulder you improve your brain a bit. Don't be too het up about the prospective outcome, just get the benefit of all the pain while you can.

(Really. It's good practice at keeping several things in your head at once, often enough. It's also interesting getting a look into the guts of numerical work - has a practical aroma to it.)

www.smaatiboy.com

I've already done another question 7b of Oct/Nov 2008 and sent on to the lecturer. Waiting for feedback. Its almost identical to an example done by patrickJMT on Youtube.

I'm keen to break up a paper. Shall we start with Jan/Feb 2010? I can do questions 1-2 and submit.

OK I got carried away and finished most of that paper. I skipped straight over Q3 as I havent started revision of least squares and bezier yet.

I am a bit stuck on 4a though:

f⁽⁴⁾ = sin(x)

so: -((pi - 0)/180) * h⁴ * f⁽⁴⁾ (E) < 0.0001

E should be pi/2 according to my logic for sin to be maximized, but that doesn't mean anything at this stage since it looks like I'm going to have to find the 4th root of a negative number. What did I stuff up?

Only paper I have is Oct/Nov 2009. I'm sure I had many others at some stage, but maybe that's just my imagination playing tricks on me.

Your "q4" means as much to me as a reference to the Rhind papyrus would, unfortunately. Maybe I must have another look at myUnisa, but last time I looked they only had last years' stuff there for my kind.

Your f⁽⁴⁾ looks like a 4th derivative to me? Which means you got there oscillating between various forms of sin and cos? +cos, -sin, -cos, +sin? I don't see there which negative number you'd be taking the 4th root of.

PM me your email address.

-((pi - 0)/180) * h⁴ * f⁽⁴⁾ (E) < 0.0001
h⁴ * f⁽⁴⁾ (E) < 0.0001 / ( -((pi - 0)/180) )
h⁴ < ( 0.0001 / ( -((pi - 0)/180) ) ) / f⁽⁴⁾ (E)


Can you see I will have to find the 4th root of h to solve for h?

OK. So you want to solve for h. Yes then that makes sense. 4th root and RHS is negative.

But if MINUS {some number} < {Some Positive Number} don't you have cases in which {some number} is positive? In fact most of your values for {some number} would have to be positive ... Brain burning lots of oil here, so I'll guess this observation is about as useful as reference to the Rhind papyrus then.

Are you restricting the range of your sin^-1 (if any?) That's the only thing I can think of offhand.

Yeow! OK, 2010 paper integration. I'll have to go and mull that over for a few weeks before I have any ideas, I think. At least I can stop guessing now...

Um.. first thing is that you know h is positive. Before you got to the set of symbols you have here, you had defined h as an increment - a small positive value. That is an implicit fact of your inequality. It "overrides" any general cases. In other words it's up to the term on the other side to arrange itself so as to be positive overall, so as to agree with h's positivity. Agreed?

«Edit» Deleted edit. Waffle is no help to any of us. I'm right back at the basics again, so first thing is let's say this is the interval with 3 points (for a quadratic polynomial to integrate to get the 1/3 law):

|left-of-centre|right-of-centre| ..... We have one single quadratic approximating the entire interval there. That's one error term and two panels. Double the number of panels and we have:

|LHS1|RHS1.|.LHS2|RHS2.| ...... 2 pairs of panels. 2 quadratics (a bit like splines?). 2 error terms. Total error is the sum of these two.

|L|R|L|R|......................|L|R| ..... n pairs of panels. n quadratics to integrate in a straightforward manner. Each quadratic has some error.

And I'll leave it all hanging in mid air like that for now. If I'm fundamentally getting it all back to front, please correct me, folks.

Question 1 and 2 of that(Jan/Feb 2010) paper is to easy. answers not gonna be informative enough.

I would like us to tri the example paper she gave us. (Oct/Nov 2009 exam)

I busy working on the Question 2 stuff. so gonna tri Q2.

Don't think we're gonna get any feedback from Dr Rapoo. since I have not recieved any mark or my marked answer paper for Ass 2.

Would have loved to get some feed back on that.

So I'm gonna be working on the Oct/Nov 2009 exam and posting it here or mailing it to you guys.

Maybe start a new thread for that.

If you guys feel starting the thread yourself, call it "Oct/Nov 2009 exam" or something.

I think the minus out front of the error formula just indicates that the value should be subtracted from the approximated integral.

This means everything should be fine:

0.0001 > (pi / 180) * h⁴ * sin(pi/2)

I get h = 0.275125 when I do the algebra. That means you have pi / h panels = 11.41878.

Since the panel count needs to be whole and even I would say M = 12

I agree after seeing Q1 and Q2 and have thus moved on straight after.

Dr Rapoo has in fact registered on this forum so if you mail her the link to your posting she will be able to respond. She mailed me saying she will look at our questions this weekend if we require. Please ask nicely, I'm sure she is willing to help.

OK I will work through Oct/Nov 2009 this morning.

Cool, I'll give it a go aswell

I've also seen some pages on the web that use absolute value for this. Somewhere on my disk I have a solved problem exactly like this that google took me to, but do think I can find it now??

Actually google taking you direct to pdfs like that is a bit of a nuisance, isn't it? Because then you don't get a link you can copy and paste to share.

OK! Either it's a case of us fools never differing or of these great minds thinking alike ...

I need to spell it out to myself in a bit more detail.
1. We want a "worst case scenario" for the error. (I got that on the web somewhere). This means we must maximise the f⁽⁴⁾(xi) that we're bounding. We want the highest possible value for this. The cycle is sin, cos, -sin, -cos, sin. At f⁽⁴⁾ we're back at sin again.
2. How to maximise it? The ordinary way. Find out where it takes on its greatest possible value in the interval of interest. Now somewhere else I saw someone recommending taking the derivative of f⁽⁴⁾ ie. f⁽⁵⁾ ie. cos, here; and then setting that derivative of the 4th derivative to zero. Either I'm not calculating it right, or it somehow doesn't work, even though it sounds a perfectly sensible suggestion. I took the easy way out. Just look at sin x. What's its maximum? 1. On this interval it hits 1 at pi/2 and then falls back to zero. So maybe the best way to work out WHAT the maximum is would be to just "look at the graph" in your mind.
3. Just so it doesn't get lost in all the chatter, we've maximised f⁽⁴⁾x to give ourselves a worst case for this term, and the x value for which this derivative is at a maximum is pi/2.
4. With that chosen the rest of what we have to solve for is just plugins. pi - 0 = pi = "b - a". And solve for the unknown h. It's the 4th root of 180/pi * 0.0001, then. 0.275 .... is the result.
5. We don't want h; we want n, so we need an expression for it. We can use the definition of h = (b-a)/n and restate that as n = (b-a)/h. b-a, we have (pi - 0). h we now have (or the "worst case h" we wanted), so plug those in and voila! .. h = 11.418. We want an even number of panels so choose 12.

And the reason we need an even number, of course is that we're estimating the curve over the top of each panel as a quadratic ie. n+1 = 2+1 = 3 points. In picture form: | | | (and just imagine a smooth curve over that ...

Ah! That minus sign... Would it not be to do with the fact that our parabola is always concave down? ...

Não, as they say in Brasil, I don't think so... oh well ...

The panel count needs to be even.

That means we need an extra fence post. An odd number of points.

M = 13.

eg ||| 2 panels , ||||3 panels...X..or.. :-( , ||||| 4 panels ...

Dr Rapoo's response to 7b of 2008 (mailed her my scanned workings):

You are correct in that you need to find the fourth derivative and find its largest (absolute) value. (Here the largest absolute value will be negative so the error expression will be positive, so you can solve for the h directly from the inequality; if the function is convex the 4th derivative would be positive and the total error would be negative in which case you would surely want the abs value of the error to be less than 0.0001 or whatever). Here the fourth derivative is (according to my Scientific Workplace) -120 exp(2x )*cos(3x) -119 exp(2x)*sin(3x). The largest negative value of this is reached at x=2, and the value is -33158, again according to my software. So we would need to select h such that

-(2-0)*(-33158)h^4/180<0.0001

which gives h<=2. 2825×10^-2; and since (b-a)/h with that value is 27.6, we should take M=28 subintervals. The final h you should use is then (b-a)/h.

OK so M is number of subintervals, not of points needed. So then yes, it must be even.

Good to get confirmation that it's absolute value that's required.

I hope that derivative is of some other example. (Meaning I hope we get something like sin, plain, and not something more complicated).

Right. Now all we need is some clues as to how that Taylor Series approximation of f⁽²⁾ works...