October/November 2008

What did you get for 5b? 0.904025812

Also, has anyone got the solution for 5c?

I get 0.9087765 (correction below).

My four points were x = 0.4, 0.6, 0.8, 1.0

My s came out as 1.65. ---> correction of correction. I think that's right.

Did you use the same? Or are we using a different calculation base altogether?

Recalculation, remembering to divide by the factorials this time: 0.89322525

As a start to my solution to c I'd give a next term as: s(s-1)(s-2)(s-3) delta-f⁴/ 4!

Hmm ... I must just make sure I did all my divisions for (a). I think I may have skipped the whole lot, somehow.

Thanks for reminding me that I need to select the 4 points around the interpolating approximation. I selected the wrong x0. I used 0.6, 0.8, 1.0, 1.2 instead thats why I went wrong - you are right.

Dr Rapoo agrees with your answer point selection. This way you are also able to do the next term rule properly.

[Edit] Am redoing the work now. More to follow.

Either way it's good to get the next waypoint like that. No matter whose turn it is this time round to make the error, eh? As long as in the end it's all nice and clear we're all happy.

OK. Here's the error term calc.

1.65 X 0.65 X -0.35 X -1.35 X 0.211 / 24 = 0.004455232

(Nice to have operations that don't require brackets).

Yes, I get the same answer as you: 0.89322525

The next term rule for part c: gives me 0.00445523

What did you get?

OK good. The numbers say we're finally getting it just-so. (My error term is in the previous post, so I presume the postings overlapped).

I keep forgetting my factorials, though. Hope I don't go forgetting them tomorrow. I'm going to try a very sawn-off version of "The Taylor Series" here. It'll require reference to the back of the book, I think.

Yeah. Overlap.

Had a bit of an attempt at the exponential function least squares approx.

It starts off quite simple. Take ln on both sides and you get ln y = ax + ln b (for example). "So we'll just plot that on log paper ... " Not so fast Jose...

(E)² = (Y - y)² ... we're supposed to be minimising the square of the Error.

I stagger forward just this little bit before I collapse again: "Minimise" means find a spot where the derivative = zero. (Or the partial. And I think it better be a decreasing function, too...)

Hoo boy...

Had any joy on that, or have you consigned it to /dev/null ?

dev/null for now bud. I asked the lecturer whether it is a required item on a few occasions and she ignored me. Not sure if thats a good sign or a bad one....

Yes, I think for me that one goes to /dev/null and wing it if it crops up and the rest of the paper is such a disaster that I've got lots of time for reflection left over ...