- 30 Oct '02 22:06Gilbert, do you have a proof? Can you prove your formula to be true

for when n = 100 say?

Mark

The Squirrel Lover

PS. There is more than one way of doing it. There is a particularly

attractive way of doing it however that does in fact kind of tie in with

the other thread on arithmetic progressions!!!! - 30 Oct '02 23:21Oh Mark, that has been years ago since I tried things like this. I

cannot do that just like that. At most try to 'plan' a proof on intuitive

basis. The formal proof would have to follow. Let me see:

If I can proof that when it is true for in-1, then it holds also for n, AND

knowing it is true for say n=1, then it is ok isn't it? That is another

way of expressing the progression, I think.

I think the point is to prove that with n-1 non-parallel lines, the n-th

line creates n NEW regions, because at any (of the n-1) intersection

(s) the line splitses a region in two, but except from one extreme, the

others are counted twice (keft and right).

That's as far as I get right now. Saved by the bedtime-bell. lol.

Gil. - 07 Nov '02 12:30

f(n) denotes the number of regions in the circle with n lines in it.

As you say, by drawing one more line there will be n points on that

line. In other words, of the f(n) regions the line will traverse n+1 of

them. Each region gets split into two so n+1 regions are added as a

result of this extra line.

So before we had f(n), now we have f(n+1) and this extra line has

given us n+1 extra regions.

Therefore, f(n+1) = f(n) + n+1

Now the cunning bit...

Rearranging the above line will give f(n+1) - f(n) = n+1

Consider what this is saying (sorry for all the dots, they are necessary

to line up the columns, if I leave spaces the forum software scrunches

everything up).

f(1) - f(0) = 1 (since here n = 0 so n+1 = 0+1 = 1)

f(2) - f(1) = 2 (since here n = 1 so n+1 = 1+1 = 2)

f(3) - f(2) = 3 (since...etc)

f(4) - f(5) = 4

and so on up to:-

f(n-1) - f(n-2) = n-1

f(n) - f(n-1) = n

Add up the columns (note how the top left to bottom right diagonal on

the left hand side of the equals sign cancels, this is what Acolyte

mentioned in one of his posts where all the terms cancel leaving just

two, (usually) just the first one and the last one).

We get f(n) - f(0) = 1+2+3+4+...+(n-1)+n

The right hand side is simply the sum of the first n integers!!! This is

where it ties in with the other thread involving Gauss and arithmetic

progressions!! I love maths! We can use the formula Hotpawn and

Gilbert gave in that thread (the proof of which follows the same

pattern as the explanation I gave Jon with n replacing 100).

So, we have:-

f(n) - f(0) = n(n+1)/2

We know f(0) to be 1 so...

f(n) - 1 = n(n+1)/2

Which, adding one to both sides, is the formula that Gilbert arrived at

by intuition and testing the first few cases.

f(n) = n(n+1)/2 + 1 - 30 Oct '02 20:53How are we drawing the lines? Through a central point? If so, then the

figure would have:

1 lines = 2 regions

2 lines = 4 regions

3 lines = 6 regions

4 lines = 8 regions

Which would wake the pattern regions = 2n

Or more simply, just noticing that every line divides 2 regions into 4

regions (double), we get the same result.

However, if you can make the like anywhere, the circle with three lines

in it can have 7 regions, the circle with 4 can have 10, the circle with 5

can have 14 regions (am I correct) and I fail to see a pattern.

Rein