Example 01:

Solve the following equations by Jacobi’s Method, performing three iterations only.

20x + y – 2z = 17, 3x + 20 y – z + 18 = 0, 2x – 3y + 20 z = 25.

Solution:

Given equations are

20x + y – 2z = 17, 3x + 20 y – z + 18 = 0, 2x – 3y + 20 z = 25.

Rewriting above equations we get

x = (1/20)(17 – y + 2z)    ………….. (1)

y = (1/20)(-18 -3x + z)    ………….. (2)

z = (1/20)(25 – 2x + 3y)    ………….. (3)

First Iteration:

x₁ = (1/20)(17 – y₀ + 2z₀)    ………….. (4)

y₁ = (1/20)(-18 -3x₀ + z₀)    ………….. (5)

z₁ = (1/20)(25 – 2x₀ + 3y₀)    ………….. (6)

Substituting x₀ = 0, y₀ = 0, z₀ = 0 in equations (4), (5), and (6)

x₁ = (1/20)(17 – (0) + 2(0)) = 17/20 = 0.85

y₁ = (1/20)(-18 -3(0) + (0)) = -18/20 = – 0.9

z₁ = (1/20)(25 – 2(0) + 3(0)) = 25/20 = 1.25

Second Iteration:

x₂ = (1/20)(17 – y₁ + 2z₁)    ………….. (7)

y₂ = (1/20)(-18 -3x₁ + z₁)    ………….. (8)

z₂ = (1/20)(25 – 2x₁ + 3y₁)    ………….. (9)

Substituting x₁ = 0.85, y₁ = – 0.9, z₁ = 1.25 in equations (7), (8), and (9)

x₂ = (1/20)(17 – (- 0.9) + 2(1.25)) = 1.02

y₂ = (1/20)(-18 -3(0.85) + (1.25)) = – 0.965

z₂ = (1/20)(25 – 2(0.85) + 3(- 0.9)) = 1.03

Third Iteration:

x₃ = (1/20)(17 – y₂ + 2z₂)    ………….. (10)

y₃ = (1/20)(-18 -3x₂ + z₂)    ………….. (11)

z₃ = (1/20)(25 – 2x₂ + 3y₂)    ………….. (12)

Substituting x₂ = 1.02, y₂ = – 0.965, z₂ = 1.03 in equations (10), (11), and (12)

x₃ = (1/20)(17 – (- 0.965) + 2(1.03)) = 1.0013 approx. x₃ = 1

y₃ = (1/20)(-18 -3(1.02) + (1.03)) = – 1.0015 approx. y₃ = – 1

z₃ = (1/20)(25 – 2(1.02) + 3(- 0.965)) = 1.0033 approx. z₃ = 1

After three iterations x = 1, y = -1, and z = 1 (approx.)

Example 02:

Solve the following equations by Jacobi’s Method, performing three iterations only.

10x + y + 2z = 13, 3x + 10y + z = 14, 2x + 3y + 10z = 15.

Solution:

Given equations are

10x + y + 2z = 13, 3x + 10y + z = 14, 2x + 3y + 10z = 15.

Rewriting above equations we get

x = (1/10)(13 – y – 2z)    ………….. (1)

y = (1/10)(14 – 3x – z)    ………….. (2)

z = (1/10)(15 – 2x – 3y)    ………….. (3)

First Iteration:

x₁ = (1/10)(13 – y₀ – 2z₀)    ………….. (4)

y₁ = (1/10)(14 – 3x₀ – z₀)    ………….. (5)

z₁ = (1/10)(15 – 2x₀ – 3y₀)    ………….. (6)

Substituting x₀ = 0, y₀ = 0, z₀ = 0 in equations (4), (5), and (6)

x₁ = (1/10)(13 – (0) – 2(0)) = 13/10 = 1.3

y₁ = (1/10)(14 – 3(0) – (0)) = 14/10 = 1.4

z₁ = (1/10)(15 – 2(0) – 3(0)) = 15/10 = 1.5

Second Iteration:

x₂ = (1/10)(13 – y₁ – 2z₁)    ………….. (7)

y₂ = (1/10)(14 – 3x₁ – z₁)    ………….. (8)

z₂ = (1/10)(15 – 2x₁ – 3y₁)    ………….. (9)

Substituting x₁ = 1.3, y₁ = 1.4, z₁ = 1. 5 in equations (7), (8), and (9)

x₂ = (1/10)(13 – (1.4) – 2(1.5)) = 0.86

y₂ = (1/10) (14 – 3(1.3) – (1.5))    = 0.86

z₂ = (1/10) (15 – 2(1.3) – 3(1.4))    = 0.82

Third Iteration:

x₃ = (1/10)(13 – y₂ – 2z₂)    ………….. (10)

y₃ = (1/10)(14 – 3x₂ – z₂)    ………….. (11)

z₃ = (1/10)(15 – 2x₂ – 3y₂)    ………….. (12)

Substituting x₂ = 0.86, y₂ = 0.86, z₂ = 0.82 in equations (10), (11), and (12)

x₃ = (1/10)(13 – (0.86) – 2(0.82)) = 1.05 approx. x₃ = 1

y₃ = (1/10) (14 – 3(0.86) – (0.82))    = 1.06 approx. y₃ = 1

z₃ = (1/10) (15 – 2(0.86) – 3(0.86))    = 1.07 approx. z₃ = 1

After three iterations x = y = z = 1 (approx.)

Example 03:

Solve the following equations by Jacobi’s Method, performing three iterations only.

5x – y + z = 10, 2x + 4y = 12, x + y +5z = -1.

Solution:

Given equations are

5x – y + z = 10, 2x + 4y = 12, x + y +5z = -1.

Rewriting above equations we get

x = (1/5)(10 + y – z)    ………….. (1)

y = (1/4)(12 – 2x)    ………….. (2)

z = (1/5)(- 1 – x – y)    ………….. (3)

First Iteration:

x₁ = (1/5)(10 + y₀ – z₀)    ………….. (4)

y₁ = (1/4)(12 – 2x₀)    ………….. (5)

z₁ = (1/5)(- 1 – x₀ – y₀)    ………….. (6)

Substituting x₀ = 0, y₀ = 0, z₀ = 0 in equations (4), (5), and (6)

x₁ = (1/5)(10 + (0) – (0)) = 2

y₁ = (1/4)(12 – 2(0)) = 3

z₁ = (1/5)(- 1 – (0) – (0)) = – 0.2

Second Iteration:

x₂ = (1/5)(10 + y₁ – z₁)    ………….. (7)

y₂ = (1/4)(12 – 2x₁)    ………….. (8)

z₂ = (1/5)(- 1 – x₁ – y₁)    ………….. (9)

Substituting x₁ = 2, y₁ = 3, z₁ = – 0.2 in equations (7), (8), and (9)

x₂ = (1/5)(10 + (3) – (- 0.2)) = 2.64

y₂= (1/4)(12 – 2(2)) = 2

z₂ = (1/5)(- 1 – (2) – (3)) = – 1.2

Third Iteration:

x₃ = (1/5)(10 + y₂ – z₂)    ………….. (10)

y₃ = (1/4)(12 – 2x₂)    ………….. (11)

z₃ = (1/5)(- 1 – x₂ – y₂)    ………….. (12)

Substituting x₀ = 2.64, y₀ = 2, z₀ = – 1.2 in equations (10), (11), and (12)

x₃ = (1/5)(10 + (2) – (- 1.2)) = 2.64

y₃ = (1/4)(12 – 2(2.64)) = 1.68

z₃ = (1/5)(- 1 – (2.64) – (2)) = – 1.128

After three iterations x = 2.64, y = 1.68, and z = – 1.128 approx.

Example 04:

Solve the following equations by Jacobi’s Method, performing three iterations only.

10x – 2y – 2z = 6, – x – y + 10z = 8, – x + 10y – 2z = 7.

Solution:

Given equations are

10x – 2y – 2z = 6, – x – y + 10z = 8, – x + 10y – 2z = 7.

Rewriting above equations we get

x = (1/10)(6 + 2y + 2z)    ………….. (1)

y = (1/10)(7 + x + 2z)    ………….. (2)

z = (1/10)(8 + x + y)    ………….. (3)

First Iteration:

x₁ = (1/10)(6 + 2y₀ + 2z₀)    ………….. (4)

y₁ = (1/10)(7 + x₀ + 2z₀)    ………….. (5)

z₁ = (1/10)(8 + x₀ + y₀)    ………….. (6)

Substituting x₀ = 0, y₀ = 0, z₀ = 0 in equations (4), (5), and (6)

x₁ = (1/10)(6 + 2(0) + 2(0)) = 0.6

y₁ = (1/10)(7 + (0) + 2(0)) = 0.7

z₁ = (1/10)(8 + (0) + (0)) = 0.8

Second Iteration:

x₂ = (1/10)(6 + 2y₁ + 2z₁)    ………….. (7)

y₂ = (1/10)(7 + x₁ + 2z₁)    ………….. (8)

z₂ = (1/10)(8 + x₁ + y₁)    ………….. (9)

Substituting x₁ = 0.6, y₁ = 0.7, z₁ = 0.8 in equations (7), (8), and (9)

x₂ = (1/10)(6 + 2(0.7) + 2(0.8)) = 0.9

y₂ = (1/10)(7 + (0.6) + 2(0.8)) = 0.92

z₂ = (1/10)(8 + (0.6) + (0.7)) = 0.93

Third Iteration:

x₃ = (1/10)(6 + 2y₂ + 2z₂)    ………….. (10)

y₃ = (1/10)(7 + x₂ + 2z₂)    ………….. (11)

z₃ = (1/10)(8 + x₂ + y₂)    ………….. (12)

Substituting x₂ = 0.9, y₂ = 0.92, z₂ = 0.93 in equations (10), (11), and (12)

x₃ = (1/10)(6 + 2(0.92) + 2(0.93)) = 0.97, approx. x₃ = 1

y₃ = (1/10)(7 + (0.9) + 2(0.93)) = 0.976, approx. y₃ = 1

z₃ = (1/10)(8 + (0.9) + (0.92)) = 0.982, approx. z₃ = 1

After three iterations x = y = z = 1 approx.

Example 05:

Solve the following equations by Jacobi’s Method, performing three iterations only.

5x – y + z = 10, 2x + 4y = 12, x + y + 5z = 1 with initial solution (2, 3, 0).

Solution:

Given equations are

5x – y + z = 10, 2x + 4y = 12, x + y + 5z = 1.

Rewriting above equations we get

x = (1/5)(10 + y – z)    ………….. (1)

y = (1/4)(12 – 2x)    ………….. (2)

z = (1/5)(1 – x – y)    ………….. (3)

First Iteration:

x₁ = (1/5)(10 + y₀ – z₀)   ………….. (4)

y₁ = (1/4)(12 – 2x₀)    ………….. (5)

z₁ = (1/5)(1 – x₀ – y₀)    ………….. (6)

Substituting x₀ = 2, y₀ = 3, z₀ = 0 in equations (4), (5), and (6)

x₁ = (1/5)(10 + (3) – (0)) = 2.6

y₁ = (1/4)(12 – 2(2)) = 2

z₁ = (1/5)(1 – (2) – (3)) = – 0.8

Second Iteration:

x₂ = (1/5)(10 + y₁ – z₁)   ………….. (7)

y₂ = (1/4)(12 – 2x₁)    ………….. (8)

z₂ = (1/5)(1 – x₁ – y₁)    ………….. (9)

Substituting x₁ = 2.6, y₁ = 2, z₁ = – 0.8 in equations (7), (8), and (9)

x₂ = (1/5)(10 + (2) – (- 0.8)) = 2.56

y₂ = (1/4)(12 – 2(2.6)) = 1.7

z₂ = (1/5)(1 – (2.6) – (2)) = – 0.72

Third Iteration:

X₃ = (1/5)(10 + y₂ – z₂)   ………….. (10)

y₃ = (1/4)(12 – 2x₂)    ………….. (11)

z₃ = (1/5)(1 – x₂ – y₂)    ………….. (12)

Substituting x₂ = 2.56, y₂ = 1.7, z₂ = – 0.72 in equations (10), (11), and (12)

x₂ = (1/5)(10 + (1.7) – (- 0.72)) = 2.484

y₂ = (1/4)(12 – 2(2.56)) = 1.72

z₂ = (1/5)(1 – (2.56) – (1.7)) = – 0.652

After three iterations x = 2.484, y = 1.72, and z = – 0.652 approx.

The post Jacobi’s Method to Find Solution of Simultaneous Equations appeared first on The Fact Factor.

Intermediate Value Theorem :

• If a function f is continuous on closed interval [a, b] and f (a). f (b) < 0, i.e. ƒ (a) and ƒ (b) are opposite signs, then there exists at least one c ∈ (a, b) such that ƒ(c) = 0.
• This theorem is to be used without proof. We know that a polynomial function is continuous and so the above theorem can be applied to a polynomial function. Hence there exists at least one root of polynomial equation ƒ(x) = 0, in the open interval (a, b).
• We will be studying three methods of approximation of roots of equation ƒ (x) = 0.

Bisection Method:

Example 01:

• Using the bisection method solve the equation x² + 2x – 8 = 0 in the interval [1, 4]. Use two iterations.

Solution:

Let ƒ(x) = x² + 2x – 8
∴ ƒ(1) = (1)² + 2(1) – 8 = 1 + 2 – 8 = -5 < 0 (negative)
∴ ƒ(4) = (4)² + 2(4) – 8 = 16 + 8 – 8 = 16 > 0 (positive)
Thus we see that ƒ(1) < 0 and ƒ(4) > 0
∴ By intermediate value theorem, the root of equation ƒ(x) = 0 lies in interval (1, 4) = (a, b)

• Initial approximation : (First approximation)

Let x_o be the initial (first) approximation to the root.
By the bisection formula

x_o = (a + b)/2 = (1 + 4)/2 = 5/2 = 2.5

• First iteration (Second approximation):

Now, ƒ(2.5) = (2.5)² + 2(2.5) – 8
ƒ(2.5) = 6.25 + 5 – 8
ƒ(2.5) = 3.25 > 0 (positive)

Now, ƒ(1) < 0 and ƒ(2.5) > 0, hence by intermediate value theorem the root of equation ƒ(x) = 0 lies in interval (1, 2.5) = (a, b).
Let x₁ be the second approximation to the root.
By the bisection formula
x₁ = (a + b)/2 = (1 + 2.5)/2 = 3.5/2 = 1.75

• Second iteration (Third approximation):

Now, ƒ(1.75) = (1.75)² + 2(1.75) – 8
= 3.0635 + 3.5 – 8
= -1.4375 < 0 (negative)

Now, ƒ(1.75) < 0 and ƒ(2.5) > 0 and , hence by intermediate value theorem the root of equation ƒ(x) = 0 lies in interval (1.75, 2.5) = (a, b)
Let x₂ be the third approximation to the root.
By the bisection formula
x₁ = (a + b)/2 = (1.75 + 2.5)/2 = 4.25/2 = 2.125
After two iteration by bisection method solution of equation x² + 2x – 8 = 0 is x = 2.125. (approx.)

Example 02:

• Show that the root of the equation x2 + 3x – 5 = 0 lies in (1,2). Find the first three approximations to the roots of this equation using the bisection method.

Solution :

Let ƒ(x) = x² + 3x – 5
∴ ƒ(1) = (1)² + 3(1) – 5 = 1 + 3 – 5 = -1 < 0 (negative)
∴ ƒ(2) = (2)² + 3(2) – 5 = 4 + 6 – 5 = 5 > 0 (positive)
Thus we see that ƒ(1) < 0 and ƒ(2) > 0
∴ By intermediate value theorem, the root of equation ƒ(x) = 0 lies in interval (1, 2) = (a, b)

• Initial approximation : (First approximation)

Let x₀ be the initial approximation (first approximation) to the root.
By the bisection formula
x₀ = (a + b)/2 = (1 + 2)/2 = 3/2 = 1.5

• First iteration (Second approximation):

Now, ƒ(1.5) = (1.5)² + 3(1.5) – 5
= 2.25 + 4.5 – 5
= 1.75 > 0 (positive)

Now, ƒ(1) < 0 and ƒ1.5) > 0, hence by intermediate value theorem the root of equation ƒ(x) = 0 lies in interval (1, 1.5) = (a, b)
Let x₁ be the second approximation to the root.
By the bisection formula

x₁ = (a + b)/2 = (1 + 1.5)/2 = 2.5/2 = 1.25

• Second iteration (Third approximation):

Now, ƒ(1.25) = (1.25)² + 3(1.25) – 5
= 1.5625 + 3.75 – 5
= 0.3125 > 0 (positive)

Now, ƒ(1) < 0 and ƒ(1.25) > 0 and , hence by intermediate value theorem the root of equation ƒ(x) = 0 lies in interval (1, 1.25) = (a, b)
Let x₂ be the third approximation to the root.
By the bisection formula
x2 = (a + b)/2 = (1 + 1.25)/2 = 2.25/2 = 1.125
Thus the first three approximations to the root of equation x² + 3x – 5 = 0 by bisection method are 1.5, 1.25 and 1.125

Example 03:

• Show that the root of the equation x³ – x – 1 = 0 lies in (1,2). Find the first three approximations to the roots of this equation using the bisection method.

Solution:

Let f (x) = x³ – x – 1
∴ ƒ(1) = (1)³ – (1) – 1 = 1 –1 -1 = -1 < 0 (negative)
∴ ƒ(2) = (2)³ – (2) – 1 = 8 – 2 – 1 = 5 > 0 (positive)
Thus, we see that ƒ(1) < 0 and ƒ(2) > 0
∴ By intermediate value theorem, the root of equation ƒ(x) = 0 lies in interval (1, 2) = (a, b)

• Initial approximation : (First approximation)

Let x_o be the initial approximation (first approximation) to the root.
By the bisection formula
x_o = (a + b)/2 = (1 + 2)/2 = 3/2 = 1.5

• Second approximation :(First iteration)

Now, ƒ(1.5) = (1.5)³ – 1.5 – 1
= 3.375 – 1.5 –1
= 0.875 > 0 (positive)

Now, ƒ(1) < 0 and ƒ(1.5) > 0, hence by intermediate value theorem the root of equation ƒ(x) = 0 lies in interval (1, 1.5) = (a, b)
Let x₁ be the second approximation to the root.

By the bisection formula

x1 = (a + b)/2 = (1 + 1.5)/2 = 2.5/2 = 1.25

• Third approximation :(Second iteration)

Now, ƒ(1.25) = (1.25)³ – 1.25 – 1
= 1.953125 – 1.25 – 1
= – 0.296875 < 0 (negative)

Now, ƒ(1.25) < 0 and ƒ(1.5) > 0 and , hence by intermediate value theorem the root of equation ƒ(x) = 0 lies in interval (1.25, 1.5) = (a, b).
Let x₂ be the third approximation to the root.
By bisection formula,

x₂ = (a + b)/2 = (1.25 + 1.5)/2 = 2.75/2 = 1.375
Thus the first three approximations to the root of equation x³ – x – 1 = 0 by bisection method are 1.5, 1.25 and 1.375.

Example 04:

• Using the bisection method find the approximate value of square root of 3 in the interval (1, 2) by performing two iterations.

Solution:

Let x = √3,

Squaring both the sides we have,
x² = 3

∴ x² – 3 = 0
The positive root of this equation is √3.
Let f(x) = X2 – 3
∴ ƒ(1) = (1)2 – 3 = 1 – 3 = -2 < 0 (negative)
∴ ƒ(2) = (2)2 – 3 = 4 – 3 = 1 > 0 (positive)
∴ By intermediate value theorem, the root of equation ƒ(x) = 0 lies in interval (1, 2) = (a, b)

• Initial approximation : (First approximation)

Let x₀ be the initial approximation (first approximation) to the root.
By the bisection formula
x₀ = (a + b)/2 = (1 + 2)/2 = 3/2 = 1.5

• First Iteration (Second approximation) :

Now, ƒ(1.5) = (1.5)² – 3
= 2.25 – 3
= – 0.75 < 0 (negative)

Now, ƒ(1.5) < 0 and ƒ(2) > 0, hence by intermediate value theorem the root of equation ƒ(x) = 0 lies in interval (1.5, 2) = (a, b)
Let x₁ be the second approximation to the root.

By bisection formula

x₁ = (a + b)/2 = (1.5 + 2)/2 = 3.5/2 = 1.75

• Second Iteration (Third approximation) :

Now, ƒ(1.75) = (1.75)² – 3
= 3.0625 – 3
= 0.0625 > 0 (positive)

Now, ƒ(1.5) < 0 and ƒ(1.75) > 0 and , hence by intermediate value theorem the root of equation ƒ(x) = 0 lies in interval (1.5, 1.75) = (a, b)
Let, x₂ be the third approximation to the root.
By bisection formula,

x₂ = (a + b)/2 = (1.5 + 1.75)/2 = 3.25/2 = 1.625
After two iterations by the Bisection method √3 = 1.625

Example 05:

• Find the square root of 6 using the bisection method and two iterations, (Ans. 2.375)

Example 06:

• Find the fourth root of 27 using the bisection method and two iterations, (Ans. 2.375)

Example 07:

• Show that the root of equation x3 – 2×2 + 2 = 0 in the interval (-1, 0) by using bisection method three times (Ans. – 0.875)

The post The solution of an Equation: Bisection Method appeared first on The Fact Factor.

Logic helps in the development of systematic and logical reasoning skills. It helps in understanding the precise meaning of statements of theorems, the converse of theorems, and corollaries of the theorem. It is the basis of circuit designing, artificial intelligence, and computer programming.

Statement in Logic:

Language is the medium of communication of thoughts. We do so by using sentences simple or complex. They may be assertive or imperative or exclamatory or interrogative or suggestive or wishes. A declarative sentence, which is either true or false, but not both simultaneously, is called a statement in logic.

Sentences which are incomplete or imperative or interrogative or exclamatory or suggestive or wishes or perceptions are not taken as statements in logic. An open sentence is a sentence whose truth value can vary according to some conditions which are not stated in the sentence. e.g. It is white in colour.

In logic, the statements are denoted by small case letters particularly p, q, r, ….

Law of the Excluded Middle:

A statement is either true or false. It can not be both true and false and also neither true nor false. This fact is known as the law of the excluded middle.

Truth Value of a Statement:

• The truth or falsity of a statement is called the truth value of the statement.
• If the statement is true then its truth value is denoted by the letter ‘T’.  If the statement is false then its truth value is denoted by the letter ‘F’. In boolean algebra 1 is used for T and 0 is used for F.

To find the Truth Value of a Statement:

Determine which of the following sentences are statements in logic. If not give a reason. If a statement, then find its truth value.

1. Do your homework today: It is a command or suggestion. Hence it is not a statement.
2. x² – 5x + 6 = 0, when x = 2: It is a statement. Its truth value is ‘T’
3. x² – 5x + 6 = 0, x ∈ R: For x = 2 or x = 3 it is true for x ≠ 2 and x ≠ 3, it is false. Hence it is open sentence. It is not a statement.
4. It is white in colour.: We cannot decide whether the statement is true or false. It is not a statement.
5. x + 3 = 5: If x = 2 it is true and if x ≠ 2 it is false. Hence it is open sentence. It is not a statement.
6. Oh! What a beautiful scene!: It is an exclamation. Hence it is not a statement.
7. Let us go for a walk: It is a suggestion. Hence it is not a statement.
8. I wish the man had wings: It is a wish. Hence it is not a statement.
9. Please give me a glass of water: It is a request i.e. imperative sentence. Hence it is not a statement.
10. Get out of the class immediately: It is a Command i.e. imperative sentence. Hence it is not a statement.
11. When is your examination going to start?: It is an interrogative sentence. Hence it is not a statement.
12. The sum of two odd integers is always odd: It is a statement. Its truth value is ‘F’
13. The product of two odd integers is always odd: It is a statement. Its truth value is ‘T’
14. Please come here: It is request i.e. imperative sentence. Hence it is not a statement.
15. The quadratic equation x² – 3x + 2 = 0 has two real roots: It is a statement. Its truth value is ‘T’
16. The square of any real number is always positive: It is a statement. Its truth value is ‘F’. Because the square of 0 is 0 which is neither positive nor negative.
17. 5 + 4 = 11: It is a statement. Its truth value is ‘F’
18. Every square of an odd number is always even: It is a statement. Its truth value is ‘F’
19. 1 is a prime number: It is a statement. Its truth value is ‘F’
20. Every natural number is a whole number: It is a statement. Its truth value is ‘T’
21. Are you ready for a picnic?: It is an interrogative sentence. Hence it is not a statement.
22. Sin 2θ = 2 sin θ cos θ, for all θ: It is a statement. Its truth value is ‘T’
23. 829 is divisible by 9: It is a statement. Its truth value is ‘F’
24. What a great fall of Humpty Dumpty!: It is an exclamatory sentence. Hence it is not a statement.
25. New Delhi is the capital of India: It is a statement. Its truth value is ‘T’
26. Shut the door: It is an imperative sentence. Hence it is not a statement.
27. Please give me a pen: It is an imperative sentence. Hence it is not a statement.
28. Do you like fruits?: It is an interrogative sentence. Hence it is not a statement.
29. What a heavy downpour!: It is an exclamatory sentence. Hence it is not a statement.
30. Have you ever seen a rainbow?: It is an interrogative sentence. Hence it is not a statement.
31. x + 2 = 7: It is true when x = 5 and false when x ≠ 5. It is an open sentence. Hence it is not a statement.
32. He is a musician: Its truth cannot be determined. Hence it is not a statement.
33. Statistics is an easy subject: Its truth cannot be determined because it is a perception which may change from person to person. Hence it is not a statement.
34. The sun is a star: It is a statement. Its truth value is ‘T’
35. May God bless you!: It is a wish. Hence it is not a statement.
36. The sum of the interior angles of a triangle is 180°: It is a statement. Its truth value is ‘T’
37. Every real number is a complex number: It is a statement. Its truth value is ‘T’. because every real number can be written in the form a + bi
38. Why are you upset?: It is an interrogative sentence. Hence it is not a statement.
39. The square root of – 9 is a rational number: It is a statement. Its truth value is ‘F’
40. The sum of cube roots of unity is 1: It is a statement. Its truth value is ‘F’. Because the sum is zero.
41. x² – 3x + 2 implies that x = -1 or x = -2: It is a statement. Its truth value is ‘T’
42. He is a good person: Its truth value cannot be determined because it is a perception which may change from person to person. Hence it is not a statement.
43. Two is the only even prime number: It is a statement. Its truth value is ‘T’
44. Do not disturb: It is an imperative sentence. Hence it is not a statement.
45. x² – 3x – 4 = 0 when x = -1: It is a statement. Its truth value is ‘F’
46. It is red in colour: We cannot decide whether the sentence is true or false. Hence it is not a statement.
47. Every parallelogram is a rhombus: It is a statement. Its truth value is ‘F’.
48. Every set is a finite set: It is a statement. Its truth value is ‘F’
49. Indians are intelligent: Its truth value cannot be determined because it is a perception which may change from person to person. Hence it is not a statement.
50. Do your work properly: It is a suggestion. Hence it is not a statement.
51. x + 3 = 10: It is true when x = 7 and false when x ≠ 7. It is an open sentence. Hence it is not a statement.
52. Will you help me?: It is an imperative sentence. Hence it is not a statement.
53. Square of an odd number is odd: It is a statement. Its truth value is ‘T’
54. Zero is a complex number: It is a statement. Its truth value is ‘T’. Because any real number is a complex number.
55. All real numbers are rational numbers: It is a statement. Its truth value is ‘F’
56. 2 + 3 < 6: It is a statement. Its truth value is ‘T’
57. x² – 5x + 6 = 0 when x = 2: It is a statement. Its truth value is ‘T’.
58. The door is open: Its truth cannot be determined because it is a perception (How much?) which may change from person to person. Hence it is not a statement.
59. I am lying: Its truth cannot be determined because it is dependent on the truthfulness of the person saying it. Hence it is not a statement. It is called the liar’s paradox.

The post Statement in Logic appeared first on The Fact Factor.

Example – 1:

If f(x) = x² + x + 1, construct a forward difference table by taking x = 0, 1, 2, 3, 4, 5 i.e. 0(1)5. Hence find Df(o), D²f(1) and D³f(2).

Solution:

Given f(x) = x² + x + 1 and x = 0, 1, 2, 3, 4, 5 i.e. 0(1)5.

f(0) = 0² + 0 + 1  = 0 + 0 + 1 = 1

f(1) = 1² + 1 + 1  = 1 + 1 + 1 = 3

f(2) = 2² + 2 + 1  = 4 + 2 + 1 = 7

f(3) = 3² + 3 + 1  = 9 + 3 + 1 = 13

f(4) = 4² + 4 + 1  = 16 + 4 + 1 = 21

f(5) = 5² + 5 + 1  = 25 + 5 + 1 = 31

The forward difference table is constructed as follows.

From table we can see that Δ²f(x) = 2 = constant and Δ³f(x) = 0

Note:

• Given function is f(x) = x² + x + 1. The highest power is 2, which obviously means Δ²f(x) = constant and Δ³f(x) = 0.
• No need to show steps of calculations as shown in Δf(x),  Δ²f(x) and Δ³f(x) columns. You can write the values directly. To explain the concept for some problems steps are shown.

Example – 2:

If f(x) = x² – 3x + 1, construct a forward difference table by taking x = 0, 1, 2, 3, 4, 5 i.e. 0(1)5. Verify that the second differences are constant.

Solution:

Given f(x) = x² – 3x + 1 and x = 0, 1, 2, 3, 4, 5 i.e. 0(1)5.

f(0) = 0² – 3(0) + 1 = 0 – 0 + 1 = 1

f(1) = 1² – 3(1) + 1 = 1 – 3 + 1 = -1

f(2) = 2² – 3(2) + 1 = 4 – 6 + 1 = -1

f(3) = 3² – 3(3) + 1 = 9 – 9 + 1 = 1

f(4) = 4² – 3(4) + 1 =16 – 12 + 1 = 5

f(5) = 5² – 3(5) + 1 = 25 – 15 + 1 = 11

The forward difference table is constructed as follows.

From table we can see that second order differences i.e. Δ²f(x) = 2 = constant

Example – 3:

If f(x) = 2x² + 5, construct a forward difference table by taking x = 0, 2, 4, 6, 8 i.e. 0(2)8. Verify that the third differences are zero.

Solution:

Given f(x) = 2x² + 5 and x = 0, 2, 4, 6, 8 i.e. 0(2)8.

f(0) = 2(0)² + 5 = 0 + 5 = 5

f(2) = 2(2)² + 5 = 8 + 5 = 13

f(4) = 2(4)² + 5 = 32 + 5 = 37

f(6) = 2(6)² + 5 = 72 + 5 = 77

f(8) = 2(8)² + 5 = 128 + 5 = 133

The forward difference table is constructed as follows.

From table we can see that third order differences i.e.  Δ³f(x) = 0

Example – 4:

If f(x) = 2x³ – x² + 3x + 1, construct a forward difference table by taking x = 0, 1, 2, 3, 4, 5 i.e. 0(1)5. Verify that the third differences are constant.

Solution:

Given f(x) = 2x³ – x² + 3x + 1 and x = 0, 1, 2, 3, 4, 5 i.e. 0(1)5.

f(0) = 2(0)³ – (0)² + 3(0) + 1 = 0 – 0 + 0 + 1 = 1

f(1) = 2(1)³ – (1)² + 3(1) + 1 = 2 – 1 + 3 + 1 = 5

f(2) = 2(2)³ – (2)² + 3(2) + 1 = 16 – 4 + 6 + 1 = 19

f(3) = 2(3)³ – (3)² + 3(3) + 1 = 54 – 9 + 9 + 1 = 55

f(4) = 2(4)³ – (4)² + 3(4) + 1 = 128 – 16 + 12 + 1 = 125

f(5) = 2(5)³ – (5)² + 3(5) + 1 = 250 – 25 + 15 + 1 = 241

The forward difference table is constructed as follows.

From table we can see that third order differences i.e.  Δ³f(x) = 12 = constant

Example – 5:

If f(x) = x³ – 2x² + 1, construct a forward difference table by taking x = 0, 1, 2, 3, 4 i.e. 0(1)4. Verify that the fourth differences are zero.

Solution:

Given f(x) = x³ – 2x² + 1 and x = 0, 1, 2, 3, 4 i.e. 0(1)4.

f(0) = 0³ – 2(0)² + 1 = 0 – 0 + 1 = 1

f(1) = 1³ – 2(1)² + 1 = 1 – 2 + 1 = 0

f(2) = 2³ – 2(2)² + 1 = 8 – 8 + 1 = 1

f(3) = 3³ – 2(3)² + 1 = 27 – 18 + 1 = 10

f(4) = 4³ – 2(4)² + 1 = 64 – 32 + 1 = 33

The forward difference table is constructed as follows.

From the table, we can see that fourth-order differences i.e. Δ⁴f(x) are zero.

Example – 6:

By constructing a forward difference table find the 7^th and 8^th terms of a sequence 8, 14, 22, 32, 44, 58,….

Solution:

Let f(1) = 8, f(2) = 14, f(3) = 22, f(4) = 32, f(5) = 44, f(6) = 58

We have to find f(7) and f(8)

We prepare following forward difference table.

We can see that the second differences i.e. Δ²f(x) are constant.

• To find f(7), extra 2 (shown in red colour) is written in D²f(x) column. The entry in Df(x) is 2 + 14 = 16 (shown in red colour) is added. The entry in f(x) is 16 + 58 = 74 (shown in red colour) is added. Thus f(7) = 74.
• To find f(8), extra 2 (shown in green colour) is written in D²f(x) column. The entry in Df(x) is 2 + 16 = 18 (shown in green colour) is added. The entry in f(x) is 18 + 74 = 92 (shown in green colour) is added. Thus f(8) = 92.

Thus 7 th and 8 th terms of series are 74 and 92 respectively.

Example – 7:

By constructing a forward difference table find the 6^th and 7^th terms of a sequence 6, 11, 18, 27, 38,….

Solution:

Let f(1) = 6, f(2) = 11, f(3) = 18, f(4) = 27, f(5) = 38

We have to find f(6) and f(7)

We prepare following forward difference table.

We can see that the second differences are 2 i.e. constant

The post Forward Differences 01 appeared first on The Fact Factor.