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Monday, July 30, 2007

Computationally freaking intractable

Never again will I predict in this blog that a particular question or assignment will be "easy", "straightforward", be "done soon" or similar. It's obviously a hex.

I'm still working on question 5 because I can't work out how to find a directional derivative. All that the internet will give me is the definition, and I can't work from the definition because I can't find the limit. All that my 200-level math notes will give me is a method for scalar-valued functions, and my attempts to generalise this to higher dimensions have so far failed to generate a plausible answer.

Not even Maple can solve this one; I spent the whole morning trying to make it do so, but it spat an error message at me and then sulked. If you, the reader, know how to find the directional derivative of a vector-valued function [from R2 -> R2, in this case], please tell me. I'm in despair.

I'm also on the floor below Chris's office, and I'm about to go up there and question him. Furiously.

In the lecture room today there was a URL written on the blackboard inside a DNE. The URL was http://xkcd.com/295/, which kinda made my day. Chris, not knowing the joke, lost no time in erasing it. The beautiful drawing of Tux in coloured chalk survived a little longer though :)

4 comments:

gliderguider said...

::whoosh:: over my head .. but ...

How is the "direction" specified? Can you turn it into a unit vector? In that case I would expect that for each dimension of the result you just take the dot product of the unit vector with a vector consisting of the partial derivative of that dimension of the result with respect to the axes.

Um.

suppose your function is R^2 -> R^2 and is (cos x + sin y, x^2 + y^3) and your direction is (0.5, 0.866). The result would be (-0.5*sin x + 0.866*cos y, x + 2.598*y^2)

I'm just making this up on the spot though...

Gael said...

What do you mean "over your head", boy? You just gave me the damn definition. Again.

Well, it's called a "definition" in second year, anyways.

gliderguider said...

Um. So does that mean it was a help, or not?

I'm happy (and a little surprised) if my ramblings somehow matched the proper definition, but sad if that didn't help you due to just being what you already didn't "get".

Gael said...

Heh. I apologise for being a little short. The problem was getting me down.

What you gave was effectively the same method I'd already found in a number of places, but hadn't yet untangled sufficiently from my [lopsided, erroneous] working to see whether it was what I was already doing. It's not actually the definition of the directional derivative; that's something misleadingly easy-looking, involving a limit as a parameter tends to zero.

My problem was that working from this definition was failing to yield any limit at all, while working with various other methods failed to give what I expected as an answer. I still don't have an answer [nor have I gone back to my earliest workings to see what was wrong with my attempts at taking the limit - I've now confirmed with Maple that it does indeed exist], but I'm close to finding it by effectively the method you've outlined.

All I have to do now is prove that 1+r^(-4)*2*r^(-2)*a is equal to r^2+r^(-2)+2*r^(-1)*a , or else find the error in my working that leads to them not being equal.

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