More Dimensions Less freedom

You may wonder how can a person suffer from less freedom if he has more and more dimensions. In mathematics the more dimensions you have the more freedom you have. If there is one dimension then the life of a person is confined to a straight line. If he has two dimensions then he can move in a plane. If he has three dimensions then he can move in space. These three dimensions in physics are called of space. The next dimension which comes is fourth dimension and in physics it is called a dimension of time. These four dimensions clearly define the life of a human being. The first three define the position and the last defines the time he is present in. But according to me if someone lives in many dimensions and the number of dimensions increase then his life becomes more and more specific. Let us see how.

When a person lives in one dimension and its other dimensions are sleeping then  his sleeping dimensions can take any value at any instant of time as they are not defined. Our life will become simple if we consider a person living in three dimensions and the fourth dimension is sleeping. As no specific value is assigned to the fourth dimension so the fourth co-ordinate can be anything. If a person is at (2,2,3) then his fourth coordinate if it comes into being can be 1 or 2 or 3 or 4 or anything among infinite values of numbers. Now if we call the fourth dimension a dimension of time then the person can reach any time if he is able to interact with the fourth dimension. And this thing forms the basis of time travel. Now as we know that we cannot be present at many places at the same time so there can be no more freedom for the fourth dimension and our time dimension exist and is united with the three dimensions we have.  If there would have been only three dimensions then if we are able to interact with the fourth dimension or create a dimension which runs through the three dimensions then we could travel time. Now you would have understood that the more dimensions are defined the more freedom we loose.

The more dimensions are defined the more freedom we loose.

Binomial theorem

We are interested in finding relation between many things. When (a + b) is raised to different powers then relation between previous expansion with the next expansion has a very interesting result. This post is about this and binomial theorem. It is called binomial as it has two variables in it. In this post I will show you how we can arrive at the coefficients of the Binomial Expansion. The foundation of binomial theorem lies in Pascals Triangle. Pascals triangle is an interesting topic and I will cover it in detail. A Pascals Triangle looks like this.

                              1
                            1   1
                          1   2   1
                        1   3   3   1
                      1   4   6   4   1
                    1   5   10  10  5   1
                      so on.  

A pascals triangle is generated from coefficients of the expansion of different powers raised to (a + b). Let us expand some of the powers.

(a + b)1 = (a + b)

(a + b)2 = (a2 + 2ab + b2)

(a + b)3 = (a3 + 3a2b + 3ab2 + b3)

(a + b)4 = (a4 + 4a3b + 6a2b2 + 4ab3 + b4)

(a + b)5 = (a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5)

Let us express the coefficients in other form:

1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5   10  10  5   1
1   6   15  20  15  6  1
so on

Some points to note

1. The first column is a constant.
2. The second column is an arithmetic sequence whose common difference is 1. i.e. the terms are represented by n/1.
3. The third column needs some explanation
   The sequence is 1 3 6 10 15...
   The difference between successive terms is 2 3 4 5....
   The difference of elements again is 1 1 1.
   Don't worry about the method which I am going to follow. Learn the method. If on finding successive differences you reach at a constant in 2 steps. Suppose the sequence is formed by the second power. Here, the equation is an² + bn + c. Where n is the position of term. Let us take the first three terms of the sequence. Then we get a system of simultaneous equations.
   
   a + b + c = 1
   4a + 2b + c = 3
   9a + 3b + c = 6
   solving we get, a = 1/2, b = 1/2 and c = 0
   The equation is (1/2)(n²+n). In the table the sequence start from row 2 so we will replace n by (n-1).
   Then the equation becomes (1/2)((n-1)² + (n-1)).
   Reducing into factors we get n(n-1)/2 = n(n-1)/(1x2)
   The above thing in combinations is represented as ⁿC₂.
4. Following the above process and assuming the sequence depends on third power and solving and replacing n by (n-2) we get ⁿC₃.
5. Following the above process and assuming the sequence depends on fourth power and solving and replacing n by (n-3) we get ⁿC₄.
6. Following the above process we get ⁿC₅.
7. Following the above process we get ⁿC₆.

Hence the expansion of (a+b) raised to n^th power is

The binomial theorem
(a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁a^n-1b + ⁿC₂a^n-2b² + ... + ⁿC_n-1ab^n-1 + ⁿC_nbⁿ