My math notes

Gauss-Lucas Theorem

Published by Angel on Sunday, February 28, 2010 - 22:29:55 - Filed under Convex Sets

Im back with a very known theorem, I like it a lot, even if the proof is already on the Wikipedia Id like to write it here with an example because its a direct consequence of the Convex Hull post I wrote before.
———————-
Gauss-Lucas Theorem:
The roots of the derivative of a non constant complex polynomial belong to the convex hull generated by the roots of the polynomial.

Proof:
Let
Since we know that where z1,…,zn are the roots of the polynomial .

We can also write
Now:

If z is a root of the derivative, then:
This means that:

Then :

Which means:

Then:

But with this we know that z is a convex combination of all the roots, this is because :

Then by this we know that z is in the convex hull of

QED.

Example:
Proof that the diameter of the set of roots of the polynomial :
cannot be less than

Proof:
We know that :
Now we just use that:
and
Now its clear that:
.
Since the roots of the derivative are inside the convex hull generated by the roots of the original polynomial, then the diameter between them cannot be less than the diameter between the set of the roots of the second derivative of the polynomial.

—————-
The example was one of the problems on my exam. Good luck I liked this theorem and I didnt forget about it :).

Add a comment (791 views) - Permalink

Number theory - Möbius function

Published by Angel on Friday, February 19, 2010 - 02:20:52 - Filed under Notes, Number theory

I like this function a lot, proposition is not THAT interesting but whatevs
———-
Lets define the Möbius function as:

Proposition:

is multiplicative and

Proof:

Let gcd(m,n)=1

If or
Then

If , , then

If n or m = 1 proof is clear.

Now, let , since is multiplicative then F(n) is multiplicative.

It is clear that

Also:

.

( Möbius function is 0 on )

Now, if we use prime factorization on n, if n>1:

we can do this because F(n) is multiplicative.

——–
I <3 Möbius function

Add a comment (428 views) - Permalink

Convex sets - Convex Hull

Published by Angel on Wednesday, February 17, 2010 - 07:08:41 - Filed under Notes, Convex Sets

First post is on basic convex set theory, I liked this because it was simple, demonstration was not very hard either.

The convex hull for a set A ( Im using real A a subset of ) is the minimal convex set that contains A.
We’ll now define:

Lemma:

Let with A a convex set and then

Proof:

We’ll use induction over k.
If k=1 then x*1 is in A ( we choose x that way ).
Now we’ll asume that it works for every k.
Now, let where and

We wanna see that x is in A.

Since then at least there is one , lets say its .

Let
and
.

Then we know by our induction hypothesis that y is in A .
Now, since A is a convex set and then .

This shows that x is in A.

Definition:

A point its a convex combination of points if there is such that

Theorem:

Let then c(A) ( The convex hull of the set A ) is the set of all convex combinations of all points in A.

Proof:

Let B the set of all convex combinations of points in A.
We now have to show that B=c(A) (convex hull of A )

1)

Let x be an element of B, then with and , since A is in c(A) then , then because c(A) is convex.
This shows that .

2)

Note that if B is a convex set then A is a subset of B,this is because if x is in A then there is a such that px will be on B. Then c(A) is a subset of B

We have to show that B is convex.

Let x and y be in B
Then and where and and .
Let with such that
Then , it is clear now that
This shows that , which means B is convex.

Then B=c(A).

——-
‘Till next time :)

13 comments (979 views) - Permalink

New blog

Published by Angel on Saturday, February 13, 2010 - 02:47:00

The old blog has been down for about 3 weeks and I decided to give it a fresh start.

Old site was about giving application to pure mathematics, but I decided I’ll just use this for notes and results and demonstrations I find interesting.

Ill try to update as much as I can.

Angel.

Add a comment (424 views) - Permalink