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

Solution is a solution

Let us start with two equations

1. ax + b = 0
2. ax² + bx + c = 0

Among the above two equations one is linear and the other is quadratic. The solution of the linear equation is x = -b/a and the solution of the quadratic is (1/2a)(-b±√(b² - 4ac)). The point which I want to state is that the solution of a quadratic equation contains the solution of a linear equation. When we substitute c for zero in the quadratic equation then we get ax² + bx = 0. On factorizing we get (ax + b) = 0 and x = 0. Whose solutions are x = -b/a and 0. -b/a is the solution of ax + b = 0. And x = 0 comes because x is multiplied throughout to get the equation ax² + bx = 0.

Now according to me if solution of cubic exist then it must have the solution of quadratic contained in it. Similar is the case for the others. Let us compute the solution of a cubic equation when one root is known

The equation is ax³ + bx² + cx + d =0. If the roots are α, β and γ and γ is known then
α + β + γ = -b/a
αβ + βγ + γα = c/a
αβγ = -d/a
Expressing sum of two roots in terms of third
α + β = -b/a - γ
Expressing product of two roots in terms of third
αβ = c/a - (βγ + γα)
αβ = c/a - γ(β + α)
αβ = c/a - γ(-γ - b/a)
αβ = c/a + γ² + bγ/a
(α - β)² = (α + β)² - 4αβ
(α - β)² = (-b/a - γ)² - 4(c/a + γ² + bγ/a)
(α - β)² = (-b/a)² + (γ)² + 2bγ/a - 4c/a - 4γ² - 4bγ/a
(α - β)² = b²/a² - 4c/a - 3(γ)² - 2bγ/a
α - β = ±√[(b² - 4ac)/a² - 3(γ)² - 2bγ/a]
adding α - β and α + β
2α or 2β = -b/a - γ ± √[(b² - 4ac)/a² - 3(γ)² - 2bγ/a]
α or β = -b/2a - γ/2 ± (1/2a)√[(b² - 4ac) - 3a²(γ)² - 2baγ]

As we can see when one root γ = 0 and the equation is independent of d then the solution reduces to the solution of a quadratic equation.

Now it can be the solution of 5^th degree equation exist and it contains the solution of 4^th degree in it.

Operations in algebra (Multiplication and Division)

Multiplication and division in algebra rests on exponents and ordinary division. The ordinary division is applied to coefficients and the exponents are applied to symbolic constants and variables. Exponents are result of successive multiplication. Some rules of exponents that are useful in multiplication and division of algebra are:

1. a^maⁿ = a^m+n
2. a^m/aⁿ = a^m−n
3. 1/a^m = a^−m

Here the first rule is useful in multiplication and the second rule is useful in division. One more important point is to be noted here. The law of multiplication for exponents is applied on similar symbolic constants and variables. Similar is the case for division. All the symbolic constants and variables are grouped and the law of exponents are applied. Then the different symbolic constants are multiplied or divided as the case may be.

Procedure for division and multiplication in algebra

1. Group the symbolic constants and variables.
2. Perform the division or multiplication on coefficients.
3. Apply the rule of exponents for similar symbolic constants and variables.
4. Solve.

Example 1:
Multiply 2a²b³ and 5a³b²
2a²b³×5a³b²
= 2×5a²a³b³b²
= 2×5a⁽²⁺³⁾b⁽³⁺²⁾
= 10a⁵b⁵

Divide 2a²b³ by 5a³b²
2a²b³/(5a³b²)
= (2/5)a²b³×a⁻³b⁻²
= (2/5)a²a⁻³b³b⁻²
= (2/5)a⁽²⁻³⁾b⁽³⁻²⁾
= (.4)a⁻¹b¹

Operations in algebra (Addition and Subtraction)

There are four operations which can be performed on variables. These are addition, subtraction, multiplication and division. Repeated multiplication and division are parts of exponents.

Let us consider three arithmetic expressions.
1×3+2×3+3×3+4×3 = 3 + 6 + 9 + 12 = 30
1×5+2×5+3×5+4×5 = 5 + 10 + 15 + 20 = 50
1×8+2×8+3×8+4×8 = 8 + 16 + 24 + 32 = 80
When we look at the above expressions we find that 3 changes to 5 and 5 changes to 8.

As the value changes we can put a symbolic constant in place of it. Let it be a. Then the expression turns to
1×a+2×a+3×a+4×a
We can write
1a+2a+3a+4a
as the symbol of product in algebra can be replaced by nothing.

The numbers occurring in front of the symbolic constants are called the coefficients. They are also called coefficients if the symbolic constants are replaced by variables.

To perform the operation of addition simply add the coefficients of the symbolic constants.

Here we get (1+2+3+4)a = 10a

When we put different values of a, we get
a=3, 10a = 30
a=5, 10a = 50
a=8, 10a = 80.
Hence we have verified the operation.

Subtraction
To perform subtraction simply do subtraction for the coefficients.
23a − 7a = (23 − 7)a = 16a

Like and unlike terms

When the symbol for the symbolic constant or the variables are same for all the terms then the terms are called the like terms. Else they are unlike terms.

3a and 6a are like terms. Some examples of like terms are
3d and 7d, 23ad and 67ad, 3x and 7x, 4xy and 5xy, etc.
When the variables or symbolic constants are different for the considering terms then the terms are called unlike terms.Some examples of unlike terms are
3x and 6y, 56df and 78hu, etc.

Rules for addition and subtraction

1. Group the like terms.
2. Add or subtract the coefficients.
3. Only like terms can be added to or subtracted from each other.

Example 1:
Add 2x + 3y + 4z and 3x + 7x + 7y + 6z
2x + 3y + 4z + 3x + 7x + 7y + 6z
= 2x + 3x + 7x + 3y + 7y + 4z + 6z
= (2+3+7)x + (3+7)y + (4+6)z
= 12x + 10y + 10z

A cubic gives cubic

I ask you a question and the answer is very simple. But it has a very interesting conclusion in it. The question is "If we take a cubic equation and consider A as one number. Now If we find the difference of the roots from this point A and frame an equation which gives the values. What will we get."

The answer is very simple and is a cubic. The expression will be a cubic because we know that there are three roots and each value will be at a constant distance from the given number A. So there will be three numbers and a cubic equation only gives three values i.e. three numbers when equated to zero.

Now the question arises how can we find that equation. Suppose the equation is f(x) and the number is A. Then, the required equation is
f(x+δ) = f(x) + f'(x)δ /1!+ f''(x)(δ)² /2!+ f'''(x)(δ)³/3! +...

The above equation can be helpful for a cubic if we know one root.

Suppose the equation is f(x) = (x - 1)(x - 2)(x - 3)
= x³ - 6x² + 11x - 6

Now as 1 is a root then
f(1+δ) = f(1) + f'(1)δ/1! + f''(1)(δ)² /2!+ f'''(1)(δ)³/3!
f(1+δ) = f(1) + (3x² - 12x + 11)δ + (6x - 12)(δ)²/2 + 6(δ)³/6   x=1
f(1+δ) = 2δ - 3δ² + δ³
f(1+δ) = δ(2 - 3δ + δ²)
For the other roots f(1+δ) = 0 So,
Solving, δ(2 - 3δ + δ²) = 0
we get, δ = 0,1,2. Hence the roots are 1,2,3.

Method of differences

There are many occurrences when we have certain numbers and we have to find relation between them. The relation is usually in form of expressions. The expression can contain log function, exponential function and algebraic functions.The simplest among them is to find polynomial expressions.In this post I will describe how to find functions when we know a series which has its terms separated by certain constant. The method is similar to differentiation but a lot different from them.

Suppose we have the numbers (from a function) like given in the following table

1	6	15	28	45	66	91
We find the difference of successive term
5		9	13	17	21	25
We again find the difference of successive term
4			4	4	4	4
We find the differences till we get a constant term for all

As we got the constant term so we can create a table of possible polynomials for all. We start from the bottom.

c''	4			4	4	4	4
a'x+b'	5		9	13	17	21	25
ax2+bx+c	1	6	15	28	45	66	91

We got a quadratic because we had three rows in the table.
Now we have
f(x) =ax²+bx+c
f(0) = c = 1
f(1) = a + b + c = 6
f(2) = 4a + 2b + c = 15
from f(0): c =1
from f(1): a + b = 5
from f(2): 4a + 2b = 14
so solving the simultaneous equations a = 2 and b = 3 hence the function is f(x) = 2x² + 3x + 1.
Above it must be known from where the values start for x and what is their differences in each step.

Right Angled Triangle

When we add the squares of 3 and 4 we get 5² or in other words 9+16=25. This property was first found as these numbers. Later with the help of geometry and algebra it was proved that the sides of a right angled triangle follow the rule a² + b² = c², where a,b and c are the length of the sides of the triangle. This theorem was later called as Pythagoras Theorem.

Let us derive this theorem.

The triangles BCD and ABC are similar
we have, BD/AB = DC/BC = BC/AC             (i)
Also triangles ABC and ADB are similar
we have, AD/AB = AB/AC = BD/CB             (ii)
As triangles ABC is similar to ADB.
From above AD/AB = AB/AC ;
AB² = AD·AC = (AC − DC)AC
=AC² − DC·AC       from (i)
=AC² − BC²

Pythagoras Theorem is in a triangle ABC
right angled at B
AB² + BC² = AC²