F Nasim's Weblog

CSE 141: Modular Arithmatic and Congruences

fnasim — Fri, 27 Feb 2009 09:54:12 +0000

Definition:

1. If then .

Where and are integers and is a positive integer.

2. If

and

then .

Where and are integers and is a positive integer.

3. If

then, .

Where , and are integers and is a positive integer.

Example:

Algebra of Congruence:

and

then

also .

The Princeton Companion to Mathematics

fnasim — Fri, 31 Oct 2008 11:31:47 +0000

Princeton University Press has recently published a book called The Princeton Companion to Mathematics. The chief editor of the book is Professor Timothy Gowers, who is a Fields Medal winner. It is a reference for every mathematician or anyone interested in Mathematics. Although world leaders in Mathematics have contributed to the book but, it will attract a wide range of audience.

Princeton Companion to Mathematics

It discusses almost all of the branches of present day mathematics, fundamental concepts of mathematics, biographical essays on famous mathematicians,influence of mathematics on other branches of knowledge etc. I believe it will provide a good introduction the subject to anyone.

I have tried some sample chapters from the book. Among them I found the Advice to a Young Mathematician very useful and stimulating. This chapter is contributed by world leading mathematicians like Michael Atiyah (Fields Medalist) and others.

Anyone interested may try some free chapters here.

WordPress posts on Princeton Companion may found here.

Solution to STEP problem-I

fnasim — Sun, 06 Jul 2008 11:34:34 +0000

I have solved a STEP problem from Stephen Siklos’ book Advanced Problems in Core Mathematics. The problem and the solution follows:

Problem: Use the substitution to evaluate

Hence or otherwise show that for

Where and

Solution:

At the first look it seems to me that solving the first part may give an idea of the second part. So lets begin with the first part of the question.

Lets substitute with

Now lets deal with the limits:

for

becomes

Similarly for is

Now the integral becomes:

Now, we notice that this is a special case of the equation:

Where and

Now we turn to the second part of the problem. I think if we could express the fraction as then we would find the solution.In order to do so we should find a number for which

And

So we should replace with

Now lets evaluate the limits for

For is

Similarly for

Now the second integral becomes:

And we solve the problem!

Galilean and Lorentz transformation

fnasim — Wed, 02 Jul 2008 06:27:09 +0000

Suppose two inertial reference frames S and S’ are moving with velocity v with respect to each other. Also suppose that an observer at reference frame S measures an event as and another observer at frame S’ measures the same event as . Now if we want to relate the measurements of S frame with S’ frame we have to use transformation equations.

If we take Newtonian mechanics granted then we use the Galilean transformation. Galilean or Classical transformation equations are:

Notice that the time co-ordinates are measured the same by both observer—that is .

But if we take Special Relativity instead of Newtonian Mechanics than we use Lorentz transformation. Lorentz transformation equations are:

The fraction is also called relativistic gamma .

Hello world!

fnasim — Tue, 01 Jul 2008 05:10:28 +0000

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