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.

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.

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.

Calculus

Calculus is an important branch of Mathematics. Calculus is studied in three parts. First part is single variable calculus, second part is multi-variable calculus and third part is differential calculus. In our daily life we come across problems which belong to single variable calculus or can be solved by methods applied by single variable calculus. But if we have to deal with complex problems we have to come across multi-variable calculus and differential equations. A little bit of differential equations is taught in single variable calculus because we come across simple differential equations in physics. For example to describe motion of springs and simple pendulum which is simple harmonic.

In single variable calculus topics covered are differentiation, integration, application of differentiation and application of integration. Application of differentiation is completely dependent upon differentiation. Similarly application of integration is dependent upon integration. Integration is reverse process of differentiation. But integration of all functions don't exist. So we have to study differentiation first to have a view of what can be integrated. Also integration requires practice. When we integrate a lot of problems we develop skills to integrate.

Differentiation is related to rate. Rate is found of something which changes. Now the change can be constant or variable. We come to know about change by finding the rate. Here differentiation comes in handy. We can find the rate if we can express a method by which we can check change in one variable w.r.t other variable. Relationship between two variables is known as a map. A class of mapping is known as functions. Calculus is full of functions.

Integration is related to sum. Integration can be used to find sum, area, etc.