Inequalities (1 Viewer)

[Hotdog1]

There are a couple of things in Bill Pender's Harder 3U portfolio, concerning Inequalities that i didnt get:

First What is Combinatorics? ans What does this mean?

 

[Hotdog1]

Secondly, I didn't quite understand the proof using Newton's Method for the 1988 Q8:

The part I don't understand is circled... or rectangled rather... in red. Can some one please explain this?

 

[Hotdog1]

I'm really lost so please help

 

[underthesun]

Oh, so this thing is Bill Pender's?

Combinatorics

Combinatorics is the branch of mathematics that deals with determining the number of ways to arrange things, the number of subsets of a set that have a certain property, etc.
The two fundamental concepts in combinatorics are permutations and combinations.

A permutation is an arrangement of a set of objects, or specifically an arrangement of a subset of a certain size of some set of objects.

A combination is a subset of a certain size of some set of objects, without regard to their order.

These concepts are inherent to some kinds of probability problems.
source : everything2.com

 

[Hotdog1]

ok i understand what that means now, but i still don't understand the solution to that question

 

[ND]

Originally posted by Hotdog
There are a couple of things in Bill Pender's Harder 3U portfolio, concerning Inequalities that i didnt get:

First What is Combinatorics? ans What does this mean?

Ok i'm not sure what part you don't understand but i'm guessing it's how he got from the 2nd last line to the last?

I think you can see where he go the n from, now for the n(n-1)/2:

If you look at the first line of the solution, you can see that it is a product of a sum of n terms and another sum of n terms. Therefore when it's expanded there would be n*n terms, but, because n of the terms cancel to 1's, there are n(n-1) terms + the 1's (1*n). Now, there would be n(n-1) terms in the form x_i/x_j, but because these are grouped together in forms of (x_i/x_j+x_j/x_i), there are n(n-1)/2 'grouped' terms. Now it's >= n(n-1)/2*2 cos (x_i/x_j+x_j/x_i) >= 2 (this must be given somewhere above).

 

[ND]

Originally posted by Hotdog
Secondly, I didn't quite understand the proof using Newton's Method for the 1988 Q8:

The part I don't understand is circled... or rectangled rather... in red. Can some one please explain this?

All that's being used here is the identity x_(n+1)-2^(3/2)<(x_n-2^(3/2))^2 from (iii).
i.e.
x_(n+1)-2^(3/2)<(x_n-2^(3/2))^2
let n=11:
x_12-2^(3/2)<(x_11-2^(3/2))^2
then use the same thing, only letting n=10 and squaring both sides. etc. etc. until you get down to n=1.

 

[maniacguy]

I'm also going to make the observation that Bill Pender has stuck to the simple 2-case AM-GM inequality here, although I believe you can assume the more general case (confirmation, anyone?).

If so, then a simpler method would be:

(x_1 + x_2 + ... + x_n)/n >= (x_1 x_2 x_3 ... x_n)^(1/n) > 0
(1/x_1 + 1/x_2 + ... + 1/x_n)/n >= (x_1 x_2 ... x_n)^(-1/n) > 0

(note the negative sign in the exponent)

Multiplying:

(x_1+...+x_n)(1/x_1+...+1/x_n)/n^2 >= 1

The desired result falls out.

In point of fact, the result:
(x_1 + x_2 + ... + x_n)/n >= n/(1/x_1 + ... + 1/x_n)

is called the Arithmetic-Harmonic mean inequality, and has some uses in other forms of maths. (You don't need to know that last part for the HSC though)

 

[maniacguy]

To elaborate a little on ND's answer to the second one:

x_12 - 2^(1/3) < (x_11 - 2^(1/3))^2 using the identity from (iii)

x_11 - 2^(1/3) < (x_10 - 2^(1/3))^2 using the same identity

(x_11 - 2^(1/3))^2 < (x_10 - 2^(1/3))^4 squaring both sides

x_12 - 2^(1/3) < (x_10 - 2^(1/3))^4 since (x_11 - 2^(1/3))^2 is between them

Then repeat this idea over and over again
(so (x_10 - 2^(1/3))^4 < ((x_9 - 2^(1/3))^2)^4 = (x_9 - 2^(1/3))^2)^8

etc.