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

Finite and Infinite Sequences

If a set of numbers are such that they follow a specific pattern is called a sequence. If all the terms are added in order then it is called a series. The main types of series and sequences which are taught till intermediate are Arithmetic series and sequences, Geometric Series and Sequences and Harmonic Sequence.

Look at the following sequences:
1. 2, 4, 6, 8, 10, 12,...
2. 3, 9, 27, 81, 243,...
3. 1/2, 1/4, 1/6, 1/8, 1/10, 1/12,...

• Arithmetic sequence: The first sequence (2, 4, 6, 8, 10, 12,...) among the three sequences is an arithmetic sequence. 2+4+6+8+10+12+.... is an arithmetic series. A general arithmetic sequence looks like this.
  a, a+d, a+2d, a+3d,... and its corresponding series is a + a+d + a+2d + a+3d +....
  a is called the first term and d is the common difference. The first term of the example sequence is 2 and common difference is 2.
  
  The n^th term of an A.S. is
  t_n = a + (n-1)d
  and sum of first n terms is
  S_n = (n/2)(2a + (n-1)d)
  n is the number of terms.
  
  How to check whether a sequence is arithmetic sequence.
  
  Difference is found by subtracting any term with its successor term. When all the differences are the same then the sequence is arithmetic sequence and the difference is called common difference.
  


• Geometric sequence: The second series (3, 9, 27, 81, 243,...) among the three sequences is a geometric sequence. 3+9+27+81+243+.... is a geometric series. A general geometric sequence looks like this.
  a, ar, ar², ar³,... and its corresponding series is a + ar + ar² + ar³ + ....
  a is called the first term and r is the common ratio. The first term in the example is 3 and common ratio is 3.
  
  The n^th term of an G.S. is
  t_n = ar^(n-1)
  and sum of first n terms is
  S_n = a(1-rⁿ)/(1-r) if r<1
  else
  S_n = a(rⁿ-1)/(r-1) if r>1
  n is the number of terms.
  
  How to check whether a sequence is geometric sequence.
  
  Ratio is found by dividing any term with its previous term. When all the ratios are the same then the sequence is geometric sequence and ratio is called common ratio.
  


• Harmonic sequence: The third series (1/2, 1/4, 1/6, 1/8, 1/10, 1/12,...) among the three sequences is a harmonic sequence. 1/2+1/4+1/6+1/8+1/10+1/12+.... is a series. A general Harmonic sequence looks like this.
  1/a, 1/(a+d), 1/(a+2d), 1/(a+3d),... and its corresponding series is 1/a + 1/(a+d) + 1/(a+2d) + 1/(a+3d) +....
  
  (1/a) is called the first term. There is no formula to find the sum of n terms of the series.
  
  How to check whether a sequence is harmonic sequence.
  
  Find the reciprocal of all the terms. If the sequence formed is an arithmetic sequence then the sequence is harmonic sequence.

Quadratic equation

A quadratic expression when equated to zero is called a quadratic equation. A quadratic equation looks like this ax² + bx + c = 0. In this post I will show you a method which is very good to find the solution of a quadratic equation. I found this method.Let us look at the method.

ax² + bx + c = 0

We can write it as,
(x + b/a)x + c/a = 0
or (x + b/a)x = -c/a

Let A = x + b/a and B = x
Then,A-B = b/a, AB = -c/a and A+B = 2x + b/a
Applying the identity (A+B)² = (A-B)² + 4AB
(2x + b/a)² = (b/a)² - 4c/a = (b² - 4ac)/a²
2x + b/a = ±(1/a)√(b² - 4ac)
x = (-b ± √(b² - 4ac))/2a

As we can see there are two values which satisfy the equation hence the number of solutions is two and there are two roots. As the solutions of a quadratic equation are called roots.

Let us analyze the roots i.e. when they are real. The value under the square root is positive if b² - 4ac is positive. When such condition arises then the roots are real. The value b² - 4ac is called the discriminant. If the discriminant is equal to zero then both the roots are equal. If the discriminant is negative then both the roots are imaginary and they occur in conjugate pairs. If the roots are real and distinct then the graph cuts the x-axis at two different points. If the roots are real and equal then the graph cuts the x-axis at one point. If the roots are imaginary then the graph does not cut the x-axis.

The graph below shows two real roots.
x² + 5x - 2 = 0

The graph below shows two real roots
x² + 5x + 6.25 = 0

The graph below represents when roots are imaginary.

x² + 5x + 8 = 0

Linear equations in two variables

Let us first look at the form of linear equations in two variables.

ax + by = c

Such equations arise when we have two things changing at the same time. The simplest is to watch it on a graph. The x-axis forms one variable and the y-axis forms the other variable. A line has infinite number of solutions but we have to get a unique solution. And this is possible only when the number of lines is two and they intersect or in other words we have two linear equations in two variable. Hence to solve a system of two variable linear equation the required equations are

ax+ by = c
dx + ey = fA system of two variable linear equation is solved either by Substitution method or by Elimination method.

Substitution method
In substitution method we find the value of one variable in terms of the other and substitute it in the other equation.

ax + by = c ----(i)
dx + ey = f ----(ii)From (i) x = (c - by)/a

Substituting in second we get
d(c - by)/a + ey = f
dc/a - dby/a + ey= f
dc/a - (db - ea)y/a = f
dc - (db - ea)y = fa
(db - ea)y = (dc - fa)
y = (dc - fa)/(db - ea)

Substituting in (i) we get
ax + b(dc - fa)/(db - ea) = c
ax = c - b(dc - fa)/(db - ea)
ax = [c(db - ea) - b(dc - fa)]/(db - ea)
x = [cdb - cea - bdc + bfa]/a(db - ea)
x = (bf - ce)a/a(db - ea)
x = (bf - ce)/(db - ea)

x = (bf - ce)/(db - ea) = (ce - bf)/(ea - db)
y = (dc - fa)/(db - ea) = (fa - dc)/(ea - db)

Elimination method
In elimination method we eliminate one variable by equaling the other variable in both the equations. Then we substitute the value of first variable to get the value of other variable.

ax + by = c ----(i)
dx + ey = f ----(ii)
Multiplying (i) by e and (ii) by b, we get
aex + bey = ce ----(iii)
dbx + bey = fb ----(iv)

Subtracting (iv) from (iii) we get
(ae - db)x = (ce - fb)
x = (ce - fb)/(ae - db)
As substituted above in substitution method we get,
y = (fa - dc)/(ea - db)
x = (ce - bf)/(ea - db)
y = (fa - dc)/(ea - db)

Linear Equations in one variable

Linear Equations are the starting point of two branches of mathematics. One deals with roots and other deals with solutions. Linear equation can be single variable or multi variable. The complexity of the solutions increases as we move to higher number of variables. As we will see every degree equation can be represented as multi variable linear equations. The point here is that the equations are related and a quadratic and cubic has 3 and 4 unknowns in it. Let us talk about the single variable linear equation. A single variable linear equation is very simple and is taught from the very basic grades or classes. A single variable linear equation looks like this:

ax + b = 0.

There can be many variant of it. ax + b = c or ax + c = bx + d or ax + c = bx, etc.

All we need to know is basic arithmetic. All the laws of basic arithmetic also works in algebra. To solve a linear equation in one variable we perform operations like this:

Let us solve different types of linear equations:

1. ax + b = 0
   Transpose b to right hand side and divide by a.
   x = -b/a
2. ax + b = c
   Transpose b to RHS and divide by a.
   x = (c - b)/a
3. ax + b = cx
   Transpose cx to LHS and b to RHS
   ax - cx = -b
   (a-c)x = -b
   x = -b/(a-c)

Solution

The simplest way to solve a linear equation is to transpose all the constants to RHS and terms containing variable to LHS and divide RHS by the coefficient of resulting LHS.

Factors, LCM and HCF or GCD

Factors
2a²bc has 2,a,b and c as factors. Factors are used to find the L.C.M. and H.C.F. of expressions. L.C.M. and H.C.F. also play an important role in mathematics. Suppose you are given some sticks of different sizes and you are asked to find a length whose length will be divisible by all the sticks of different sizes. You need the knowledge of L.C.M. Again consider you are given many sticks and you are asked to find a length which will measure all of them in integer terms. You will need the knowledge of H.C.F.

L.C.M.

Rules to find L.C.M. (Lowest common multiple)

1. Find the L.C.M. of the constant terms.
2. Find the power of all the expressions for each variable or symbolic constant.
3. Find the maximum power for each variable or symbolic constant among each expression.
4. List the variables with the maximum power of each.
5. Multiply it with the L.C.M. of the constant term.

Evaluate L.C.M. of 3a²b⁶c and 8a³b²cd

1. L.C.M. of the constant terms is 24.
2. Power of all the expressions for each variable or symbolic constant.
   a:2 and 3
   b:6 and 2
   c:1 and 1
   d:0 and 1
3. Maximum power for each variable or symbolic constant among each expression.
   a:2 and 3 max = 3
   b:6 and 2 max = 6
   c:1 and 1 max = 1
   d:0 and 1 max = 1
4. List the variables with the maximum power of each.
   a³b⁶cd
5. Multiply it with the L.C.M. of the constant term.
   24a³b⁶cd

L.C.M. is 24a³b⁶cd

H.C.F. or G.C.D.

Rules to find G.C.D. or H.C.F. (Greatest common Divisor or Highest Common Factor)

1. Find the G.C.D. of the constant terms.
2. Find the power of all the expressions for each variable or symbolic constant.
3. Find the minimum power for each variable or symbolic constant among each expression.
4. List the variables with the minimum power of each.
5. Multiply it with the G.C.D. of the constant term.

Evaluate G.C.D. of 3a²b⁶c and 8a³b²cd

1. G.C.D. of the constant terms is 1.
2. Power of all the expressions for each variable or symbolic constant.
   a:2 and 3
   b:6 and 2
   c:1 and 1
   d:0 and 1
3. Minimum power for each variable or symbolic constant among each expression.
   a:2 and 3 max = 2
   b:6 and 2 max = 2
   c:1 and 1 max = 1
   d:0 and 1 max = 0
4. List the variables with the maximum power of each.
   a²b²cd⁰
5. Multiply it with the G.C.D. of the constant term.
   a²b²c

G.C.D. is a²b²c