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 2)

In this post we will discuss about equations of a line in different form. Every equation is related to other equation and can be derived from one other. In the post I have derived the different forms of equations of a straight line. I have started from an equation which is called two-point form and end with the normal form. normal form is also called perpendicular form. I have given graphs of four equations.

The equation of a line can be found if we know anyone of

1. two points
2. one point and slope of the line
3. slope of the line and y intercept
4. angle made by the perpendicular and its length to the line from the origin

The corresponding four types of equations are as follows

1. two points form: (y−y₁)/(x−x₁) = (y₂−y₁)/(x₂−x₁)
2. point-slope form: (y−y₁) = m (x−x₁)
3. slope-intercept form: y = mx + c
4. normal form: x cos α + y sin α = p

The four equations in the graph are
y = 3x + 4, y = x + 7 , y = 2x− 5 and y = −x + 3.

Let A(x₁, y₁) and B(x₂, y₂) be two points. Then the four types of equations can be framed as follows:

1. As the slope of the line will be constant. So, if a variable point is (x,y) then
   (y−y₁)/(x−x₁) = (y₂−y₁)/(x₂−x₁)


2. If the value of the slope is m then we can substitute,
   (y₂−y₁)/(x₂−x₁) = m
   and get the equation of the line as
   (y−y₁)/(x−x₁) = m
   (y−y₁)= m (x−x₁)
   


3. Expanding (y−y₁)=m(x−x₁)
   y−y₁= mx − mx₁
   y = mx−(mx₁− y₁)
   As [−(mx₁ − y₁)] is a constant and can be substituted for c
   y = mx + c
   
   From the above equation we get y = c when x = 0. Hence, c is the y intercept.


4. Normal form is found by considering the angle which the perpendicular from the origin  to the line makes with the x-axis and its length.
   
   
   
   
   In the figure the angle is α and the length of the perpendicular is p. Equation of line in two point form is
   (x₁, y₁) ≡ (a,0) and (x₂, y₂) ≡ (0,b)
   
   Using the two point form
   (y−y₁)/(x−x₁) = (y₂−y₁)/(x₂−x₁)
   y/(x − a) = b/(− a)
   −ay = bx − ab
   bx + ay = ab
   x/a + y/b = 1
   a = p sec α
   b = p cosec α
   x/(p sec α) + y/(p cosec α) = 1
   x cos α + y sin α = p

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 (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.

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.

Linear equations

Linear equations are those equations which are linear in its variables i.e. the power of the variables are 1. There can be many kinds of linear equations depending upon the number of variables used. a single variable linear equation has one variable, a double variable linear equation has two variables, a triple variable linear equation has three variables and so on. The linear equations with two or more variables is called a multi-variable linear equation. A single variable linear equation looks like this: ax = b where x is the variable. A double variable linear equation looks like this: ax + by = c and a triple variable linear equation looks like this: ax + by + cz = d. The number of variables can extend to any number of variables.

The solution of a linear requires as much equations as there are variables in it. A single variable linear equation requires one equation. a double variable linear equation requires two variables and a three variable linear equation requires three equations. The number of variables is equal to the number of equations. When we find the solution of a set of linear equations then the solutions can be unique or dependent or no solution. The solution of a linear equation can be found with the help of substitution, elimination or with the help of determinants. The solution using determinants is very cumbersome so to simplify the steps we use Gauss algorithms which use matrix to solve such equations. Let us look at the solutions of some linear equations.

Single variable linear equation
A single variable linear equation ax = b has solution x = b/a. When we plot it on a graph then we get a straight line parallel to y-axis. As the value of y is not present in the equation therefore the solution is independent of y.

Double variable linear equation

A double variable linear equation can be solved either by substitution method or elimination method. The solution on a graph is the point of intersection of the lines represented by the two equations.

The equations are 2x − y = 3 and 4x + y = −6

Let us solve the above two equations by substitution method and elimination method.

Substitution method

In substitution method we express one variable in terms of the other and substitute it to get value of one variable. Then substitute it in any equation to get value of other variable.

Step 1:Express any one equation in terms of one variable.
2x−y = 3 ⇒ y = 2x − 3
Step 2:Substitute it in next equation.
4x + y = −6
⇒ 4x + 2x − 3 = −6
⇒ 6x = −3
⇒ x = −(1/2) = −.5
Step 3:Substitute value in the first equation.
y = 2x − 3 = 2(−.5) − 3 = −4
So solution is x = −.5 and y = −4

Elimination method

The elimination method is based on the fact that if one variable is removed from the equation then the reduced equation contains only one variable. After we can find the solution. Substitute one solution in one equation to get the other solution.

Step 1: Eliminate one variable by making coefficient of that variable equal in two equations.

2x − y = 3
4x + y = −6
Add the two equations
6x = −3
x = −(1/2) = −.5
Substitute in anyone
2(−.5) − y = 3
y = − 4

the linear equations of multi variables is found by Gauss elimination method. We will discuss about such methods in some other posts.