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Sunday, July 29, 2007

One golden glance

After working steadily all weekend, I'm effectively halfway through assignment 3. Questions 1 to 4 simply involved working with the derivative matrices of multi-dimension functions - I've got one thing to tidy up in question 1, but otherwise those are all done. Question 5 is a little more work with directional derivatives and the angle between them, and I'll probably have that done by the end of the evening. A most productive two days.

I can't think of any better way to find the ranks of the derivatives [required for questions 2 and 4] than by row-reducing them as matrices. This is something not everyone in the class necessarily knows how to do - strangely enough, linear algebra isn't a prerequisite to this course - so presumably Chris has another method in mind which will be revealed at the tutorial. I wish he'd been able to start the assignment 3 tutorial exercises on Thursday. Although I know how to do the questions [for this assignment at least], I'm rather uncertain as how elementary he's expecting our work to be. For example, when asked to differentiate sin x, is it now acceptable to simply write "cos x"? Or are we still expected to go through all the epsilon-delta business first? Again, the tutorial will reveal all.

As it's a weekend, I'd like to conclude by recording the most interesting thing I discovered last week. It's not course-related - it's simply a proof of the Pythagorean theorem, c2=a2+b2 in a right-angled triangle. Since finding a student of my very own I live in constant fear of being caught taking something for granted, so I'm busy proving things left, right and centre which I had previously assumed to be true.

Having finally discovered this proof of Pythagoras I am absolutely scandalised that no one ever showed me this at high school. It's not hard - an astute fourth former could grasp it, and a sixth former should certainly be able to reproduce it - and it is rather astoundingly beautiful. So beautiful that I can't bring myself to just lay it down and say "there", so instead I'm going to record the walkthrough.

Get a piece of paper, now, and draw yourself a square. It doesn't have to be very accurate. Make a mark somewhere along one side of the square. Then reproduce the mark in the same position on each of the other sides of the square, so that each side is divided into two sections, of lengths A and B.

Connect up the marks, in pairs, with four straight lines joining marks on adjacent sides. Now the square is divided into five figures. Four of these figures are triangles. Observe that they are all identical right-angled triangles, with short sides [catheti] of length A and B, and a long side [hypotenuse] of length C. Recall that the area of each triangle is (1/2)AB.

Call the angle between sides A and C "x". Call the angle between sides B and C "y". Observe, from your knowledge that the sum of the internal angles of a triangle is 180°, that x+y=90°.

Look again at the marks you made on each side of the square. At each mark there are two lines meeting to make three angles on the straight line that is the side of the square. Two of these angles are x and y. Recall that the sum of angles on a straight line is also 180°; thus the third angle on the line must be 90°.

The central figure inside your original square, then, is a quadrilateral with four sides of length C and four angles of 90° - in fact, it's a square. This smaller square has area C2.

Thus the area of the original, larger square is equal to the sum of the areas of its constituent parts - 4((1/2)AB) + C2. But the area of the square is also equal to the length of its side - (A+B)2.

Equating these expressions for the area of the larger square gives us:
2AB + C2 = A2 + B2 + 2AB
Cancelling 2AB from each side:
C2 = A2 + B2

Magic!

Here is a graphic showing what your drawing should look like when you're done:

Forgive my enthusiasm, but when I saw this proof, I felt that I touched divinity for a moment. This is why I do maths.
 

4 comments:

lorne said...

A Kind of Magic. From the Highlander soundtrack if I'm not mistaken.

Gael said...

Of course! I knew you'd get this one :)

I shall just fix those line breaks now. *winces*

gliderguider said...

It's a nice proof, but I think I like the one where you drop an altitude from the right angle and show that you have similar triangles and use the ratios of the side lengths.

One pretty thing about that proof is that it strikes me as being similar to the little triangular proof that sqrt(2) is irrational.

Gael said...

No, I don't like that one. It requires too much work to show that the triangles are indeed similar, and the result doesn't just fall out the way it should do.

the little triangular proof that sqrt(2) is irrational

Please to elaborate?

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