so I'll just have to say it in hoots!

They say they'll let me out of my cage for half an hour so I can swing around the trees and be delirious for half an hour.

Whooo..

HooH hoo-hoo-hoo

Yeeeee Hah!

Yippetty yippitti yip yip yip!!

No I'm not going to tell you the mark. Anything north of 50 is cause for celebration.

In fact if you've made it into the 40's don't feel bad. I've been there...

I know I didn't feel too terrible about my fail last year S2. Was actually quite

Ja. Don't feel bad if you tried and stumbled a bit. Look at me and say "Hey he's also only human; if he can make it I can". And then Do It Again. Just work harder this time. And come and try to get this forum alive.

sheepapple, if you're still lurking there, mate, a lot of this is thanks to you keeping us forum contributors properly focused. Thanks. You can measure how much in whoops up there.

Make your forum work for you. Don't wait around for someone else. Do it yourself. It works. It improves your understanding.]]>

16e

Even in a 12-hour paper my blood would run cold if I came across something like that.

Failing an error on the 4th derivative, may I hope to have made a mistake about the method we use? (ie. maximise the xi term to give a "worst case error")]]>

It's for any function that can be represented as a power series. None of us are really meant to have a clue what that means, yet.

Unfortunately in my current haste I can't find the nice graphical resource I had on this. Add terms and you'll see the graph of the TS starts to accommodate itself to the graph of f better and better.

The main formula is for f(x). However we use f(a+h) (or a-h) more. You can use x

Try substituting a-h or a+h for a few of the x's in the "standard formula" of p 569.

First term has no x. Next one .. (x - a) = (a + h - a) = h. Actually let's call that

No need to do more calcs. They all use that (x - a) term, basically.

OK. Now imagine doing x = a - h .. Your h now has a negative sign.

And I think maybe I'll just leave this part all hanging like that, since it's that sign of h that turns out to be what makes terms cancel when you step out your central difference approximation of eg. f'

=========================

OK and I might as well mention those formulas while I'm here.

All you do is take this "abbreviated TS", and solve for the derivative you want.

f(x) = f(a) + x f'(a) + k f''(xi) ... (truncation at the last)

Solve for f', using just the non-error terms before the truncation.

f'(a) = - (f(a))/ x. (The details are wrong, of course; I'm just trying to show that the principle is fairly simple).

(Or I hope it is, otherwise I'm in trouble).]]>

I hope things don't get to technical.

But we worked hard, and I hope it shows in our marks.

:)-D]]>

Also, has anyone got the solution for 5c?]]>

A = Newton

B = Secant

C = Regular Falsi]]>

31a)

Firstly to calculate the analytical result if you integrate x

Then use x = 0.5 to get 0.015625. The book says the result SHOULD be 0.015

Already alarm bells are ringing.

f''(x) = 6x

To perform the trapezoidal rule, you need some value for h or n. This is not supplied in the question anywhere!

How does one complete this question?]]>

Step 1. Follow text book's approach to solving for f'(x) Keep that result for later.

Step 2. Produce the right Taylor Series (which might be difficult to see straight off) to solve for f''.

Step 3. Do that. Get f'' out of that TS and on its own side of the equation.

I think before the manipulations is the time to plug in the f' expression you got in Step 1. Whatever. Plug it in at an appropriate time so that the entire other side expression is in terms of f and h.

Somehow or other that first error term falls away? Or do we have to keep an additional f'' term in the error? We certainly can't just simply wish it away.

There ought to be an f''' term for error.

Now one of the problems is that unless you simply know the formula you must solve for (and they're not all in the text book) how would you know whether you're working with forward, backward, or central approximations?

I don't intend investing too vast an amount of the little time left on this, but how does the general scheme of it sound?]]>

While this is enjoyable in a twisted kind of way, if you look at that question, you'll see it's not necessary. In fact the x column of that table could be restated as ...

x

0.6

0.7

0.8

red

herring

kipper

toby

This is because your point needs to be centred inside the steps of the table. Here 0.63 only fits between 0.6 and 0.7. This means the polynomial can only go up to first function difference (even spacing means no need to divide).

Had it been 0.73 you could've brought in point 0.8 and 0.9, even...

That's good news isn't it? No need to calculate the whole lot unless explicitly required to do so.

So the answer would be f

And the error is from the next term rule. Do one more row of differences and get the second divided difference.]]>

f(x, y) = cos

g(x, y) = x

Q2 has me a little stumped. I've calculated all the partial derivative equations df/dx, df/dy, dg/dx and dg/dy.

Once more I have no idea how to determine what the starting x and y values should be now. I had this very same problem some weeks ago and ended up assuming they will give them to us.

How the heck are we supposed to know what f(x, y) and g(x, y) look like? I can see f is a squashed cos graph which never goes below zero. I can assume g(x, y) is some kind of oval with a y radius of 2, but how can we figure out where to start?]]>

I know you just take the error term of your estimating polynomial and integrate that. It's the actual nitty gritty of that "Calculus 1 integration" that's clobbering me. Is it integration by parts or something like that?]]>

[Edit]

Are we looking to show that f(x) = f1(x) for the boundary value of 2, and also show that the first and second derivatives are equal?

If that is the case then this isn't a spline since the function values aren't equal though the derivatives are?

[Edited2] can't post a new message.

Thanks slow_eddy. The question is:

f(x) = 11 ? 24x + 18x^2 ? 4x^3, for x [1;2]

f(x) = ?54 + 72x ? 30x^2 + 4x^3, for x [2;3]

It just seems to me that for 6 marks question showing the value for f and the derivatives is very little work]]>

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.]]>

My calculation is:

P

P

Simplify from there but you can already see its not the same polynomial as whats given.]]>

The one section I'm battling with a little (not battling just going slow) is solving nonlinear equations using Newton. Thoughts/pointers to resources?]]>

please help.]]>

Does anyone have some past papers in the meantime?]]>

For the sake of convenience, the questions asks for the integral between 0 and 3 of e

Thoughts?]]>

Anyways, I'm using the first formula at the top of page 272 to do Question 2 of assignment 3. How do you find O(h

b) 0.9666

c) 1.5666

Its all very disturbing because only a seems actually close to the real definite integral. Busy double-checking so long...

Edit: Found the issue with b - had a sign wrong. New result = 3.16115. Happier. Trying to figure out c now...]]>

Does anyone have a full example with step by step workings for me to check my logic against. I need some basis of correctness to compare against.]]>

I am reaaally struggling with this subject... What am I missing? I feel like I don't know anything...

I'm ok with Chapter 0,1, and half of chapter 2, but from there it is all a blur.

I looked at Assignment 2 and can honestly say I have no clue where to start with some of those questions.

I have never been this lost

Regards]]>

I think all three options 1, 3, 4, are all correct...

Let me know if what you guys think.]]>

Does anybody know what textbook we need to get for this module?]]>