Solution of a cubic equation (Part 2)

In the post Solution of a cubic equation Part 1 we found the solution of a cubic equation for a particular condition. Then, in the post Whether real or complex (Cubic Equation) we studied when the roots of a cubic equation will be real and when it will be complex. This post deals with the solution of cubic equation. The algebraic method of solving cubic equations is supposed to be due to the Italian, del Ferro (1465-1526). But it is called Cardano's method because it became known to people after the Italian, Girolamo Cardano, published it in 1545 in his 'Ars Magna'.

Omar Khayyam gave a great deal of thought to the cubic equations. Before him, Greek mathematicians obtained solutions for third degree equations by considering geometric methods that involved the intersection of conics. Although the solution is present but I am searching for a solution of cubic equation with almost a new perspective. I will continue my search but I am giving this method so that until I find my kind of solution the traditional solution is present on my blog.

Let us consider the equation (ax³ + bx² + cx + d = 0 ; a≠0)

Step 1: Express ax³ + bx² + cx + d = 0
as x³ + px² + qx + r = 0

Step 2: Shift the middle point of the curve on the axis x=−p/3.
Read post Solution of a cubic equation. A special solution and Why -b/3a? for detail.
x³ + 3(p/3)x² + 3(p²/9)x + p³/27 − 3(p²/9)x − p³/27 +qx+ r = 0
(x + p/3)³ − [3(p²/9) − q]x − p³/27 + r = 0
(x + p/3)³ − [p²/3 − q]x − p³/27 + r = 0


Step 3:  Let x + p/3 = y then x = y − p/3.
Substitute y and (y − p/3) for (x + p/3) and x in the above equation.
y³ +[q − p²/3](y − p/3) + r − p³/27 = 0
y³ +[q − p²/3]y − [qp/3 − p³/9 − r + p³/27] = 0
y³ +[q − p²/3]y + [r − qp/3 + 2p³/27] = 0
y³ +[q − p²/3]y + [2p³/27 − qp/3 + r] = 0

Step 4: Now the equation is of the form y³ + Ay + B = 0
where A = q − p²/3 and B = 2p³/27 − qp/3 + r
Let y = (s − A/3s)
then s³ − A³/(3s)³ − As + A²/3s + As − A²/3s + B = 0
s³ − A³/(3s)³ + B = 0
Multiplying throughout by s³, we get
s⁶ + Bs³ − A³/27 = 0
Let, s³ = z
z² + Bz − A³/27 = 0
z = [−B ±√(B² + 4A³/27)]/2

s₁³ = [−B +√(B² + 4A³/27)]/2
As we know that any equation has three cube roots. Let its cube root be s₁. then the roots are s₁, ωs₁ and ω²s₁.
Similarly,
s₂³ = [−B −√(B² + 4A³/27)]/2
Let its cube root be s₂. then the roots are s₂, ωs₂ and ω²s₂.
1/s₁³ = −(3s₂/A)³
⇒ 1/s₁³ = 1/[−B +√(B² + 4A³/27)]/2
= {[−B − √(B² + 4A³/27)]/2}/{[−B −√(B² + 4A³/27)][−B +√(B² + 4A³/27)]/4}
= {[−B − √(B² + 4A³/27)]/2}/{−4A³/(27×4)}
= {[−B − √(B² + 4A³/27)]/2}/{A³/27)}
= −(3s₂/A)³
s₁s₂ = −A/3
If we consider s₁ then −A/3s₁ = s₂.
The product of s₁ and s₂ is real so the possibilities of the roots for y (= s − A/3s) are
s₁ + s₂, (ωs₁ + ω²s₂) and (ω²s₁ + ωs₂).

Step 5: Shift the graph to the original position for the roots.
The three roots are
s₁ + s₂ − p/3, (ωs₁ + ω²s₂ − p/3) and (ω²s₁ + ωs₂ − p/3).

The solution of the equation
x³ + px² + qx + r = 0 is
s₁ + s₂ − p/3, (ωs₁ + ω²s₂ − p/3) and (ω²s₁ + ωs₂ − p/3).

Where ω is a cube root of 1 i.e (−1 + i√3)/2
Where s₁ is a cube root of [−B +√(B² + 4A³/27)]/2 and
s₂ is a cube root of [−B − √(B² + 4A³/27)]/2.
and
where A = q − p²/3 and B = 2p³/27 − qp/3 + r

How many Solutions of x + y + z = k

In this post we will consider all the positive integer solutions of the equation x + y + z = k. At the end of the post we will generalize the method.

Let us first solve the equation. There is a x, a y and a z. We can consider x 1's, y 1's and z 1's. So there are a total of (x + y + z) 1's. We can find its integer solutions by separating  1's at different positions. Consider the equation x + y + z = 9.

Some of the solutions of the equation are

11|111|1111 (2+3+4)
111|111|111 (3+3+3)
1111|111|11 (4+3+2)
1111|11|111 (4+2+3)
11|11|11111 (2+2+5)

Add all the ones in one group separated by |. The number of solutions of this type is the solution of the equation x + y + z = 9. The number of solutions of the equation is (9+2)!/(9!2!) = (11×10)/2 = 55. Hence there are 55 solutions. If we have the total as k and the number of plus sign is n. Then the total number of solutions is (k+n)!/(k!n!). This can be written in the combination symbol as C(n+k,k).

This equation is similar as taking the combination of r things in n containers of similar kind. The number of combinations is (n+r−1)!/(n−1)!r!. This can be written in the combination symbol as C(n+r−1,r).

Preliminaries in plane geometry (Part 1)

Many shapes in a plane are well represented by equations. Circles, parabolas and hyperbolas are some of the shapes which can be represented by equations.
But to deal with them we need to know the basics of geometry. The posts related to basics of coordinate geometry is given in the related posts "Preliminaries in plane geometry". The posts will be given in many parts. each part will cover two to three topics. In this post we will discuss about Distance formula and Section formula.

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Distance formula

The Cartesian coordinates are used to represent points in a plane. Every point in a place has a one to one relationship to a coordinate point. The distance (d)  between two points A(x₁, y₁) and B(x₂, y₂) is given by the formula
d = √[(x₂ − x₁)²+(y₂ − y₁)²]

The above formula can be found with the help of Pythagoras theorem. Draw a right angled triangle as shown in the graph. When we draw a right angled triangle we draw it such that the two edges including the right angle are parallel to the two axis. This helps us to find the coordinate of the vertex. Name the right angled vertex as C. The coordinate of the point C is (x₂, y₁). a line parallel to the coordinate axis has its other coordinate same. Suppose the line is parallel to x-axis then all the points on the line has its y-coordinate equal.  The distance between the points A and B is d. The distance between the points A and C is AC =(x₂ − x₁) and distance between the points B and C is BC =(y₂ − y₁). By Pythagoras Theorem we have (AB)² = (AC)² + (BC)². Substituting the value of AC and BC we get

d² = (x₂ − x₁)²+(y₂ − y₁)²Taking square root and as distance cannot be negative, we get
d = √[(x₂ − x₁)²+(y₂ − y₁)²]

Example:
Distance between A(3,5) and B(1,2) is
AB = d
d = √[(1 − 3)²+(2 − 5)²]
= √(4+9)
= √13

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Section Formula

A point C(x,y) is present between A(x₁,y₁) and B(x₂,y₂). C divides AB internally in the ratio m:n. Then the coordinate of C is given by

C≡( [mx₂+nx₁)/(m+n)] , [(my₂+ny₁)/(m+n)] )

Example:
C divides AB internally in the ratio 1:2. A(3,5) and B(1,2). Find C.
Let the coordinate of C be (x,y). Then
Here m=1, n=2, x₁= 3, x₂ = 1, y₁ = 5, y₂ = 2.
x = (1×1 + 2×3)/(2+1)
= 7/3
  y= (1×2 + 2×5)/(2+1)
= 12/3
= 4. Hence the coordinate of C is (7/3,4).

If C divides AB externally in the ratio m:n. Then the coordinate of C is given by

C≡( [mx₂− nx₁)/(m − n)] , [(my₂− ny₁)/(m − n)] )

Example:
C divides AB externally in the ratio 2:1. A(3,5) and B(1,2). Find C.
Let the coordinate of C be (x,y). Then
Here m=2, n=1, x₁= 3, x₂ = 1, y₁ = 5, y₂ = 2.
x = (2×1 − 1×3)/(2 − 1)
= −1
  y = (2×2 − 1×5)/(2 − 1)
= −1
Hence the coordinate of C is (−1,−1).

Recurrence relation (Part 2) : Linear Equations

In the post Recurrence Relation (Part 1) we studied three examples of recurrence relations. In this post we will solve them. The recurrence relation (T_n = T_n-1 + T_n-2) of Fibonacci problem is linear and has order 2 and degree 1. It is also homogeneous. The recurrence relation (T_n = 2T_n-1 + 1) of Tower of Hanoi problem is also linear. It has order 1 and degree 1 but it is non homogeneous. Recurrence relations can exist in many forms as linear or non-linear.

A linear recurrence relation looks like this

T_n = f₁(n)T_n-1 + f₂(n)T_n-2 + f₃(n)T_n-3 + ...+ f_r(n)T_n-r+...+f_n-1(n)T₁ + f(n)
The order of the recurrence relation is r if f_i(n) = 0 for i > r.
It is called homogeneous if f(n) = 0 and non-homogeneous if f(n) ≠ 0.
We will consider linear homogeneous and non-homogeneous recurrence relation with constant coefficients. A linear recurrence relation with constant coefficients is given by

T_n = C₁T_n-1 + C₂T_n-2 + C₃T_n-3 + ...+ C_rT_n-r+...+C_n-1T₁ + f(n) where C_i is a constant, 0 < i < n.
The order of the recurrence relation is r if C_i = 0 for i > r.
It is called homogeneous if f(n) = 0 and non-homogeneous if f(n) ≠ 0.

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We will solve the linear homogeneous recurrence relation with constant coefficients first. The example of this type of equation which we are going to solve is Fibonacci Problem.
T_n = T_n−1 + T_n−2
Substitute the Terms with the r^{(the subscript)}
We get,
rⁿ = rⁿ⁻¹ + rⁿ⁻²
Divide throughout by the lowest power term which is r^n-2.
We get,
r² = r + 1
r² − r − 1 = 0
The above equation is called the characteristic polynomial of the recurrence relation and its roots are called characteristic roots.
The solution of the polynomial is r₁ = (1 + √5)/2 and r₂ = (1 − √5)/2

The solution of the recurrence relation can be written like this
T_n = A(r₁)ⁿ + B(r₂)ⁿ = A[(1 + √5)/2]ⁿ + B[(1 − √5)/2]ⁿ

T₁ = 1 = A[(1 + √5)/2] + B[(1 − √5)/2] = (A + B)/2 + (√5)/2(A − B)
T₂ = 1 = A[(1 + √5)/2]² + B[(1 − √5)/2]²
= A [1 + 5 + 2√5]/4 + B [1 + 5 − 2√5]/4
= A [6 + 2√5]/4 + B[6 − 2√5]/4

We get two relations
(A + B) + (√5)(A − B) = 2 ----(i)
and
3(A + B) + (√5)(A − B) = 2 ----(ii)

Solving the equations simultaneously, we get
3(i) - (ii)
2√5(A − B) = 4
A − B = 2/√5 and substituting in (i)  we get
A + B = 0

Again solving simultaneously,
2A = 2/√5
A = 1/√5
B = -1/√5
Hence the solution of the recurrence is
T_n = [1/√5][(1 + √5)/2]ⁿ - [1/√5][(1 − √5)/2]ⁿ

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Next we will solve the recurrence relation T_n = 2T_n−1 + 1 which belongs to the tower of Hanoi problem.

Linear non homogeneous recurrences require us to solve for their homogeneous part then substitute a relation which satisfies the non homogeneous recurrence. There are many methods which are different from it and can be used to solve many recurrences. These methods are

1. Method of iteration
2. Method of substitution
3. Method of telescoping sums
4. Method of inspection

Here, we will use method of iteration to solve the recurrence.
In this method we substitute the earlier term in the recurrence and look at the pattern and solve the series.
T_n = 2T_n−1 + 1
T_n = 2(2T_n−2 + 1) + 1
T_n = 2(2(2T_n−3 + 1) + 1) + 1
As T₁ = 1, so we get
T_n = 2^n-1 + 1 + 2 + 2² +....+ 2^n-2
T_n = 2^n-1 + (2^n-1 − 1) = 2ⁿ − 1

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Now we will solve the recurrence T_n = T_n-1 + k. This problem is multiplication problem. This recurrence will be solved by inspection. In inspection we write down enough terms such that we can guess the solution.
T₁ = k. Hence,
T_n  = T_n−2 + k + k
T_n  = T_n−3 + k + k + k
T_n  = T_n−4 + k + k + k + k
We can guess the solution to be T_n = nk as the number of k's increases as we use the lower terms of the recurrence and it increases by a definite number..

Whether real or complex (Cubic Equation)

This post deal with the conditions to check whether the roots are real or complex of a cubic equation. A cubic equation always has a real root. The other two roots are real or complex according as the graph cuts the x axis in one or three positions. If the graph touches the x-axis then two roots are equal. If the graph cuts the x-axis in three positions then all the roots are real. If the graph cuts the x-axis in one position then other two roots are complex. When we look at the graph of a general cubic equation we see a cup like structure and a cap like structure. The base of the cup like structure gives local minima and the top of cap like structure gives local maxima. Let us find the positions where the curve bends and changes its direction. We call the cup like structure concave upward and a cap like structure convex upward. The quadratic equation is ax³ + bx² + cx + d = 0. Let it be equal to y or f(x). y = ax³ + bx² + cx + d. Now the rate of change of curve is f '(x) = 3ax² + 2bx + c. f(x) is differentiated to obtain f '(x). We again differentiate f '(x) to obtain f ''(x),

f ''(x) = 6ax + 2b
f ''(x) = 6a(x + b/3a)

Let us check, for what values of x it may give maxima and for what values it may give minima.
There are four cases for it and is given in the following table

f ''(x)	x	a	state
> 0	x < (−b/3a)	a < 0	minima
< 0	x > (−b/3a)	a < 0	maxima
< 0	x < (−b/3a)	a > 0	maxima
> 0	x > (−b/3a)	a > 0	minima

When the local maxima and and local minima are on opposite sides of x-axis then all the roots are real.

Let us find the x-coordinate of maxima or minima. It is when f '(x)=0.

Solving 3ax² + 2bx + c = 0 we get
x = [−2b±√(4b² − 12ac)]/6a
x = [−b±√(b² − 3ac)]/3a
or
x₁ = [−b+√(b² − 3ac)]/3a
and
x₂ = [−b−√(b² − 3ac)]/3a Find the value of y for x₁ and x₂, substituting them in place of x.

For x₁ we get y₁ = [4b³ − 15abc + 27a²d + (6ac − 2b²)√(b² −3ac)]/27a²

For x₂ we get y₂ = [4b³ − 15abc + 27a²d − (6ac − 2b²)√(b² −3ac)]/27a²

Now local maxima and local minima lie on the opposite sides of the x-axis when y₁ and y₂ lie on the opposite sides of x-axis. This will happen when they are of opposite sign.

Now four cases arise:

y1	y2	Remark
y1< 0	y2< 0	One root is real and other two are complex. Graph of  y = x3 + 4x2 + 4x − 7
y1< 0	y2> 0	All the three roots are real.  Graph of y = x3 + 6x2 + 4x − 7 If y1 = 0 or y2 = 0 then there are two equal roots and it may be   [−b+√(b2 − 3ac)]/3a or [−b-√(b2 − 3ac)]/3a
y1> 0	y2< 0	All the three roots are real. Graph of y = − x3 + 4x2 + 4x − 7 If y1 = 0 or y2 = 0 then there are two equal roots. and it may be   [−b+√(b2 − 3ac)]/3a or [−b-√(b2 − 3ac)]/3a
y1> 0	y2> 0	One root is real and other two are complex. Graph of y = x3 + 4x2 + 4x + 7

When (b² − 3ac) is less than zero then y₁ and y₂ are complex and there is no maxima and minima. Then the equation has only one real root and other two are complex. See the graph for 3x² + 2x² + 3x + 4 = 0.
Where (b² − 3ac) = (4 − 27) < 0.

When y₁ or y₂ is equal to zero then there are two equal roots.

Recurrence Relation (Part 1)

In mathematics we come across many series which can be expressed in terms of its some or every term. These terms can be distributed in the whole of the series. For example if the n^th term is termed as T_n and it depends on its previous two terms then T_n can be expressed as many recurrence relation with these two terms. One of it is T_n = T_n-1 + T_n-2. Recurrence relations are used in computer science to find the time complexity of many algorithms. Let us consider some examples of recurrence relations. We will solve them in some other post.

• Multiple by consecutive numbers: Look at the series 2,4,6,8,10,12,14,16...The series can be represented as 0+2, 2+2, 4+2, 6+2, 8+2, 10+2, 12+2, 14+2...Here every term can be represented as sum of the previous term and 2.Hence, multiplication of 2 and consecutive natural numbers can be represented like this. T_n = T_n−1 + 2 where the first term is T₁ = 2. Similarly, multiplication of can be represented like T_n = T_n−1 + k where T₁ = k.



• Fibonacci Series: Leonardo di Pisa also known as Fibonacci posed a problem about rabbits in his book Liber Abaci in 1202. The problem stated: One pair of rabbits, one male and one female, are left on an island. These rabbits begin breeding at the end of two months and produce a pair of rabbits of opposite sex at the end of each month thereafter. Can you determine the number of pairs of rabbits after n months assuming no rabbit dies on this island?
  
  Let us try to solve this problem. In first month there is one pair. So, T₁ = 1. In the second month there is one pair. So, T₂ = 1. In the third month there are two pairs. So, T₃ = 2. Two are the rabbits in the first month and additional two are due to birth. In fourth month the pair of children produced are equal to the pairs of rabbits in the second month which can produce, this is equal to 1. The additional number of pairs are those in third month which equals 2. So, T₄ = 3. So it can be expressed in the recurrence relation as T_n = T_n-1 + T_n-2. Where T_n, T_n-1 and T_n-2 is pair of rabbits in n^th, n-1^th and n-2^th month respectively.


• The tower of Hanoi: The problem is invented by Edouard Lucas in 1883. We are given a tower of eight discs, initially stacked in decreasing size on one of three pegs. The problem is to transfer the entire tower to one of three pegs, moving only one disc at a time without ever moving a larger disc onto a smaller one.
  
  Let us solve this problem.
  
  
  
  1. With one disk we can move it in 1 move. So T₁ = 1.
  2. With two disks we can move it in 3 moves. The procedure is like this. There are three pegs A, B and C. A has both the disks in increasing order from top arranged in the order 2,1. We move 1 to B. Then move 2 to C. Then move 1 to C. So the number of moves for two disks is three. So, T₂ = 3.
  3. For three disks, we can move it in 7 moves. The procedure is like this. There are three pegs A, B and C. A has all the three disks in increasing order from top arranged in the order 3,2,1. We, move 1 to C, move 2 to B, move 1 to B, move 3 to C, move 1 to A. Then move 2 to C. Then move 1 to C. So the number of moves for three disks is seven. So, T₃ = 7.
  We can generalize the procedure like this, move n−1 disks to one peg then move the n^th disk and bring back the n−1 disks above the n^th disk. We can solve this problem in T_n = 2T_n-1 + 1 moves.

Read solutions of the above recurrences

Whether this or this or both or none (Logical connectives)

This post deals with propositional calculus. In propositional calculus we deal with propositions. Propositions are those sentences which are either true of false. One and the most important thing which is required in propositional calculus is reasoning. Our ancestors were able to form civilizations as they were able to reason things. But a rigorous study of logical reasoning was not done for a long time. The first such study that has been found is by the Greek philosopher Aristotle (384-322 BC). Leibnitz (1646-1716) and George Boole (1815-1864) seriously studied this and came up with with a theory and called it symbolic logic.

'The sun rises in the East.' is a proposition. It is a proposition because it can only take two values either true or false. The sentences which are either true or false are propositions or statements. In mathematics we come across many propositions and statements. x>2 is not a proposition as it can be true or false according as x>2 or x<2. So, 2<3 is a proposition but x<2 is not a proposition.

There are three main connectives in propositional calculus. It is conjunction(∧), disjunction(∨) and negation(¬).

Consider the two statements.
p: I play badminton.
q: I play football.

We can frame two statements with the combination of A and B. It is called compound statement.

I play badminton and I play football.

The above statement is formed by the combination of the two statements with the help of 'and'. This is called conjunction. It is denoted by the symbol '∧'. We can frame a table of all the possibilities. It is called truth table.
 

F is false. T is true.
p	q	p∧q
F	F	F
F	T	F
T	F	F
T	T	T

As we can see from the above that conjunction is true only when all the possibilities are true and false if anyone is false.

I play badminton or I play football.

The above statement is formed by the combination of the two statements with the help of 'or'. This is called disjunction. It is denoted by the symbol '∨' We can frame a truth table of all the possibilities.
 

F is false. T is true.
p	q	p∨q
F	F	F
F	T	T
T	F	T
T	T	T

As we can see from the above that conjunction is false only when all the possibilities are false and true if anyone is true.

I don't play football.

The statement 'I play football.' is transformed to 'I don't play football.' with the help of 'negation'. It is denoted by '¬'. The truth table for 'negation' is

F is false. T is true.
q	¬q
F	T
T	F

Negation of q is true when q is false and false when q is true.

Whether real or complex (Quadratic Equation)

This post deals with the conditions to check whether the roots are real or complex. The roots are real when the graph really cuts the x-axis. When we look at the graph of a quadratic equation we see a cup like structure or cap like structure. Let us check when it is cup like and cap like. We call the cup like structure concave upward and a cap like structure convex upward. The quadratic equation is ax² + bx + c = 0. Let it be equal to y or f(x). Now the rate of change of curve is f'(x) = 2ax + b. f(x) is differentiated to obtain f'(x). We again differentiate f'(x) to obtain f''(x), f''(x) = 2a. Now we find the x-coordinate of the point where the slope is zero i.e. f'(x) = 0.

2ax₁ + b = 0
⇒ x₁ = −b/2aNow the point on the graph is concave upward or convex upward according as the point at the x-coordinate x₁ = −b/2a is maxima or minima. This thing is determined by checking the value of f''(x). If it is positive then the curve has minima and if it is negative then the curve is maxima. We have f''(x) = 2a. It is positive if a is positive and negative when a is negative. Hence we can make following conclusions

1. concave : The graph is concave upward if a is positive.
   
   
   
   The graph given above is of the function f(x) = 2x² + 4x - 3.


2. convex : The graph is convex upward if a is negative.
   
   
   The graph given above is of the function f(x) = −2x² + 4x + 3

Now, we find the y coordinate of the point where it moves upward or downward traveling from x = −∞ at x = -b/2a. It changes its direction at −b/2a to opposite direction of y-axis.
y = ax² + bx + c
at x = x₁ = −b/2a
y₁ = a(−b/2a)² + b(−b/2a) + c
= b²/4a − b²/2a + c
= − b²/4a + c
= − (b² − 4ac)/4a

Now, four conditions arise as given in the following table

a	b2−4ac	Comment
−ve	−ve	The value of y1 is −ve and the curve is concave downward. The curve will be present below y1. Hence it will cut the x-axis at no points. So the roots are imaginary.
−ve	+ve	The value of y1 is +ve and the curve is concave downward. The curve will be present below y1. Hence it will cut the x-axis at one or two points. So the roots are real and may be equal if y1 = 0.
+ve	−ve	The value of y1 is +ve and the curve is concave upward. The curve will be present above y1. Hence it will cut the x-axis at no points. So the roots are imaginary.
+ve	+ve	The value of y1 is −ve and the curve is concave upward. The curve will be present above y1. Hence it will cut the x-axis at one or two points. So the roots are real and may be equal if  y1 = 0.

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

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