Matrix and Determinant
PROPERTIES OF DETERMINANT
The determinant has many properties. Some basic properties of determinants are
1. Det (In)= 1 where In is the n × n identity matrix.
2. Det (AT)= Det (A)
3. Det( A-1)= 1/DetA= Det(A)-1
4. For square matrices A and B of equal size,
Det (AB)=Det(A)Det(B)
5. Det (cA)=cndet(A) for an n × n matrix.
6. If A is a triangular matrix, i.e. ai,j = 0 whenever i > j or, alternatively, whenever i < j, then its determinant equals the product of the diagonal entries:
푛
det 퐴 = 푎11 푎22 … 푎푛푛 = 푎푖푖 푖=1
A number of additional properties relate to the effects on the determinant of changing particular rows or columns:
1. Viewing an n × n matrix as being composed of n columns, the determinant is an n-linear function. This means that if one column of a matrix A is written as a sum v + w of two column vectors, and all other columns are left unchanged, then the determinant of A is the sum of the determinants of the matrices obtained from A by replacing the column by v and then by w (and a similar relation holds when writing a column as a scalar multiple of a column vector).
2. If in a matrix, any row or column is 0, then the determinant of that particular matrix is 0.
3. This n-linear function is an alternating form. This means that whenever two columns of a matrix are identical, or more generally some column can be expressed as a linear combination of the other columns (i.e. the columns of the matrix form a linearly dependent set), its determinant is 0.
PROBLEMS
0 푎 푏 1.Without expanding prove that −푎 0 푐 =0 WBUT 2007,2012 −푏 −푐 0 1 푎 푎2 푎3 + 푏푐푑 2 3 2. Without expanding prove that 2 푏 푏 푏 + 푐푑푎 =0 3 푐 푐2 푐3 + 푑푎푏 4 푑 푑2 푑3 + 푎푏푐 1 + 푎 1 1 1 1 1 + 푏 1 1 1 1 1 1 3. Show that = abcd 1 + + + + 1 1 1 + 푐 1 푎 푏 푐 푑 1 1 1 1 + 푑 MINOR AND COFACTOR
If A is a square matrix, then the minor of the entry in the i-th row and j-th column (also called the (i,j) minor, or a first minor is the determinant of the submatrix formed by deleting the i-th row and j-th column. This number is often denoted Mi,j. The (i,j) cofactor is obtained by multiplying the minor by .
To illustrate these definitions, consider the following 3 by 3 matrix,
To compute the minor M23 and the cofactor C23, we find the determinant of the above matrix with row 2 and column 3 removed. So the cofactor of the (2,3) entry is
LAPLACE EXPANSION OF A DETERMINANT BY COMPLEMENTARY MINORS
Laplace’s cofactor expansion can be generalised as follows.
EXAMPLE
Consider the matrix
The determinant of this matrix can be computed by using the Laplace's cofactor expansion along the first two rows as follows. Firstly note that there are 6 sets of two distinct numbers in {1, 2, 3, 4}, namely let be the aforementioned set.
By defining the complementary cofactors to be
,
, and the sign of their permutation to be
.
The determinant of A can be written out as
where is the complementary set to .
PROBLEMS
푎 −푏 −푎 푏 푏 푎 −푏 −푎 1. Prove that by laplace method =4 푎2 + 푏2 푐2 + 푑2 푐 −푑 푐 −푑 푑 푐 푑 푐 2. Expanding the determinant by laplace method in terms of minors of 2nd order formed from 0 푎 푏 푐 −푎 0 푑 푒 the first two rows prove that = 2 −푏 −푑 0 푓 푎푓 − 푏푒 + 푐푑 −푐 −푒 −푓 0
ADJOINT OR ADJUGATE DETERMINANT
푒 푓 푑 푓 푑 푒 − 푕 푖 푔 푖 푔 푕 푎 푏 푐 푏 푐 푎 푐 푎 푏 Adjof D = 푑 푒 푓 = − − 푔 푖 푔 푕 푔 푕 푖 푕 푖 푏 푐 푎 푐 푎 푏 − 푒 푓 푑 푒 푑 푒
JACOBI’S THEOREM
1. If D be 3rd order non zero determinant then adj D= 퐷2
2. If D be a non- zero determinant of order n then adj D =D(n-1)
PROBLEM
푏푐 − 푎2 푐푎 − 푏2 푎푏 − 푐2 Prove that 푐푎 − 푏2 푎푏 − 푐2 푏푐 − 푎2 = 푎3 + 푏3 + 푐3 − 3푎푏푐 2 푎푏 − 푐2 푏푐 − 푎2 푐푎 − 푏2
Definition of Singular Matrix
Singular matrix is a square matrix whose determinant is equal to zero .
Definition of Non - Singular Matrix
Non - Singular matrix is also square matrix whose determinant is not equal to zero.
ADJOINT OF MATRIX
푇 푎11 푎12 푎13 퐴11 퐴12 퐴13 A= 푎21 푎22 푎23 adj A= 퐴21 퐴22 퐴23 푎31 푎32 푎33 퐴31 퐴32 퐴33
INVERSE OF MATRIX
퐴−1 = 1 adj A 퐴
A nonsingular square matrix A is invertible.
If A & B are invertible matrices
퐴−1 −1 = 퐴
퐴퐵 −1 = 퐵−1퐴−1
퐴푇 −1 = 퐴−1 푇
ORTHOGONAL MATRIX
A square matrix A is said to be orthogonal if 퐴푇퐴 = 퐼 = 퐴퐴푇
Every orthogonal matrix A is nonsingular and det A=±1
If A & B are two orthogonal matrices then their product AB is also orthogonal.
If A is orthogonal matrix then 퐴푇&퐴−1 are also orthogonal.
TRACE OF MATRIX
Trace of a square matrix A is the sum of the diagonal elements of A. It is denoted by trA.
PROBLEMS
1. A & B are orthogonal. Hence and 퐴 + 퐵 = 0. 푃푟표푣푒 푡푕푎푡 퐴 + 퐵 is singular.WBUT 2012
2. If A be orthogonal matrix then 퐴−1 is orthogonal.
ECHELON MATRIX
A matrix A is said to be echelon matrix if
All zero rows of A follow all non zero rows of A. The number of zeros preceeding the first non zero element of a row increases as we pass from row to row downwards.
Restriction
To convert a matrix into an echelon matrix only elementary row operation is granted.
RANK OF MATRIX
Let A be a matrix , a natural number r is said to be rank of A if
There is atleast one r-th order nonsingular square submatrix of A
Every square submatrix of A of order greater than r is singular.
PROPERTIES OF RANK OF MATRIX
For an nth order square matrix A if detA ≠ 0 then rank of A=n.
The rank of a null matrix is 0.
Rank of nth order identity matrix is n.
PROBLEM
1 2 3 4 1 0 0 2 2 0 Find the rank of matrix 2 6 2 6 2 3 9 1 10 6
MATRIX INVERSION METHOD
Consider the system 2x+5y+3z=9, x+2y-z=6, 3x+y+2z=3. Solve it by matrix inversion method.
TRIVIAL SOLUTION OF A HOMOGENEOUS SYSTEM
In a homogeneous system of m equations in n unknowns x1,x2,……..xnwe see x1=0,x2=0,…….xn=0 must satisfy all the equations of the system. This solution (0,0,0,……,0) is called trivial solution of the system.
CONSISTENCY & INCONSISTENCY OF THE SYSTEM OF LINEAR EQUATIONS
A system of equations is said to be consistent if the system has atleast one solution.
A system of equations is said to be inconsistent if the system has no solution. A homogeneous system of equation must be consistent because it has trivial solution(0,0,0…0).
ON CONSISTENCY OF NON HOMOGENEOUS SYSTEM
Let a11x1+a12x2+…….+a1nxn=b1,…….,am1x1+…..+amnxn=bm be a system of m number of linear equation in n number of unknowns x1,x2,...... ,xn.
푎11 푎12 . . 푎1푛
푎21 푎22 . . . 푎11 . 푎1푛 푏1 푎 푎 . . . A= 31 32 B= ...... 푎푚1 . 푎푚푛 푏푛 푎푚1 . . . 푎푚푛
This system is consistent if and only if rank of A=rank of B.
It has unique solution if and only if rank of A=number of unknowns.
It has many solutions if rank of A Number of independent solutions= number of unknowns - rank of A ON EXIXTENCE OF NON TRIVIAL SOLUTION OF HOGENEOUS SYSTEM OF EQUATION The system has non trivial solution if rank of A Number of independent solution = number of unknowns - rank of A PROBLEMS 1. Determine the conditions under which the system of equations 푥 + 푦 + 푧 = 1, 푥 + 2푦 – 푧 = 푏, 5푥 + 7푦 + 푎푧 = 푏2admits of 1. Only one solution 2. No solution 3. Many solution. 2. Solve if possible 푥 + 푦 + 푧 = 1, 2푥 + 푦 + 2푧 = 2, 3푥 + 2푦 + 3푧 = 5 CHARACTERISTIC POLYNOMIAL We consider an n×n matrix A. The characteristic polynomial of A, denoted by pA(t), is the polynomial defined by where I denotes the n-by-n identity matrix. Suppose we want to compute the characteristic polynomial of the matrix We now compute the determinant of which is the characteristic polynomial of A. EIGEN VALUE AND EIGEN VECTOR Let A be a square matrix. A non-zero vector C is called an eigenvector of A if and only if there exists a number (real or complex) such that AC= C If such a number exists, it is called an eigenvalue of A. The vector C is called eigenvector associated to the eigenvalue . PROBLEMS 5 4 1. Find the eigen values &eigen vectors of the matrix 1 2 1 1 3 2. Find the eigen values &eigen vectors of the matrix 1 5 1 3 1 1 CAYLEY HAMILTON THEOREM A square matrix satisfies its own characteristic equation. PROBLEM 1 0 0 1. If A= 0 −1 1 , then verify that A satisfies its own characteristic equation. 0 1 0 Successive Differentiation The process of differentiating a given function again and again is called as Successive Differentiation and the results of such differentiation are called successive derivatives. The higher order differential coefficients will occur more frequently in spreading a function all fields of scientific and engineering applications. Solved Examples : 1. If y = sin(sin x), prove that 푦2 + 푦1푡푎푛푥 + 푦푐표푠2푥 = 0. 2. Find the nth derivative of y = 푥+2 + 푙표푔 푥+2 푥+1 푥+1 Some Standard Results: 푚 푛 푚 −푛 (i) 퐷푛 (푎푥 + 푏) = 푎 푚 푚 − 1 푚 − 2 … (푚 − 푛 + 1)(푎푥 + 푏) (−1)푛 푎푛 푛! (ii) 퐷 ( 1 ) = . 푛 푎푥+푏 (푎푥 +푏)푛 +1 푛−1 푛 (iii) 퐷 log 푎푥 + 푏 = (−1) 푎 (푛−1)! 푛 (푎푥+푏)푛 푚푥 푛 푚푥 푛 (iv) 퐷푛 푎 = 푚 푎 푙표푔푎 . 푚푥 푚푥 (v) 퐷푛 ( 푒 ) = 푚푛 푒 . (vi) 퐷 푠푖푛(푎푥 + 푏) = 푎푛 푠푖푛 (푛휋 + 푎푥 + 푏). 푛 2 (vii) 퐷 푐표푠(푎푥 + 푏) = 푎푛 푐표푠(푛휋 + 푎푥 + 푏). 푛 2 푛 푏 (viii) 퐷 [ 푒푎푥 푠푖푛( 푏푥 + 푐) ] = 푎2 + 푏2 2 푒푎푥 푠푖푛 { 푛 tan−1 + 푏푥 + 푐} 푛 푎 푛 푏 (ix) 퐷 [ 푒푎푥 푐표푠( 푏푥 + 푐) ] = 푎2 + 푏2 2 푒푎푥 푐표푠 { 푛 tan−1 + 푏푥 + 푐} 푛 푎 Leibnitz’s Theorem Leibnitz’s theorem is useful in the calculation of nth derivatives of the product of two functions. Statement of the theorem: If u and v are functions of x, then 푛 푛 n 푛−1 n 푛−2 2 n 푛−푟 푟 퐷 푢푣 = 퐷 푢푣 + C1퐷 푢퐷푣 + C2퐷 푢퐷 푣 + . . . . + Cr퐷 푢퐷 푣 + ...... + 푢 퐷푛 푣. Examples: 1. Find the nth derivative of the function 푥2 푙표푔 3푥. 2 Solution.We take u = log 3x and v = x , then v1 = 2x, v2 = 2 and v3 , v4 etc.. are all zero. By Leibnitz’s theorem, 2 2 n n ( x log 3x.) n = ( log 3x) n x + C1( log 3x) n-1 2x + C2 ( log 3x) n-2 2 = (-1) n-1 푛−1 ! 푥2 + (−1)n−2 푛. 2푥. 푛−2 ! + (−1)n−3 푛(푛−1) 2 푛−3 ! 푥 푛 푥 푛−1 2! 푥 푛−2 푛−3 = −1 푛−3 ![ (n - 1)(n - 2) x2 - n(n - 2) 2 x2 + n(n - 1) x2 ] 푥 푛 푛−3 = −1 푛−3 ![ (n - 1)(n - 2) x2 - n(n - 3) x2 ] 푥 푛 푚푐표푠 −1푥 2 2 2. If 푦 = 푒 , 푝푟표푣푒 푡푕푎푡 푖 1 − 푥 푦2 − 푥 푦1 = 푚 푦 2 2 2 (푖푖) 1 − 푥 푦푛+2 − (2푛 + 1) x 푦푛 +1 − (푛 +푚 )푦푛 = 0 . 푚푐표푠 −1푥 −푚 푚푦 Solution.y1 = 푒 = − . . . (i) 1−푥 2 1−푥 2 2 2 2 2 or, (1-x ) y1 = m y . Differentiating again, 2 2 2 (1-x ) 2 y1 y2 – 2x y1 – 2m y y1= 0 2 2 or, (1-x ) y2 – x y1 – m y= 0 . . . . (ii) Differentiating n- times following Leibnitz’s theorem, 2 푛(푛−1) 2 (1-x ) yn+2 + n yn+1 (-2x) + 푦 −2 − 푥 푦 − 푛 푦 −푚 푦 = 0 2 푛 푛 +1 푛 푛 2 2 2 i.e, (1-x ) yn+2− (2푛 + 1) 푥 푦푛+1 − (푛 + 푚 )푦푛 = 0 ...... (iii) 2 2 3. If 푥 = 푠푖푛 푡, 푦 = 푐표푠 푝푡,show that (1 – 푥 ) 푦2 – 푥 푦1 + 푝 푦 = 0. Hence deduce that 2 2 2 (1 – 푥 ) 푦푛+2 – ( 2푛 + 1) 푥 푦푛+1 – ( 푛 − 푝 ) 푦푛 = 0. 1 1 − 2 2 2 4. 퐼푓 푦푚 + 푦 푚 = 2푥 , Prove that( 푥 – 1) 푦푛+2 + (2푛 + 1) 푥 푦푛+1 + ( 푛 – 푚 ) 푦푛 = 0. Mean Value Theorems & Expansion of Functions Mean-Value Theorems In this topic we shall discuss a few important theorems of differential calculus- Rolle’s Theorem, Lagrange’s & Cauchy’s Mean-Value Theorem. Expansions of functions by Taylor’s and Maclaurin’s theorem. Rolle’s theorem and its application: Statement: Let a function f be defined on a closed interval [a,b]. Suppose, further that (i) 푓 is continuous on [푎, 푏] (ii) 푓 is differentiable in the open interval (푎, 푏) and (iii) 푓(푎) = 푓(푏) then, there exists at least one point 푥 = 푐 lying within 푎 < 푐 < 푏, such that 푓΄(푐) = 0. Geometrical Interpretation:The geometric interpretation of Rolle's Theorem is that if f is a continuous function whose domain is a closed interval and f has tangent lines at every point of its graph except possibly the endpoints, then at least one of those tangent lines is horizontal. Corollary: If 푎 < 푏 are two roots of the equation 푓(푥) = 0, then the equation 푓΄(푥) = 0 will have at least one root between a and b, provided (i) 푓(푥) is continuous on [푎, 푏] and (ii) 푓(푥) is differentiable in (푎, 푏). If 푓(푥) be a polynomial, the conditions (i) and (ii) are satisfied. Hencebetween any two roots of a polynomial 푓(푥) lies at least one zero of the polynomial ΄(푥). 2 Example :Verify Rolle’s Theorem for f(x) = 푥 −4푥 on [0,4]. 푥+2 Solution. Here f(x) is continuous on [0,4] and 2푥−4 푥+2 − 푥 2−4푥 .1 f΄(x) = (푥+2)2 2 2 = 2푥 −4푥+4푥−8−푥 +4푥 (푥+2)2 2 = 푥 +4푥−8 (푥+2)2 ⇒f(x) is differentiable on [0,4]. So by Rolle’s theorem 푓΄(푥) should have at least one zero within 2 (0,4), i.e, 푥 +4푥−8 = 0 for some 푥 ∈ (0,4). (푥+2)2 2 Equating 푥 +4푥−8 = 0, we find 푥 = −2 ± 2 3. Here 푥 = −2 + 2 3 lies within (0,4). (푥+2)2 Lagrange’s Mean-Value Theorem Statement. If a function 푓(푥) is a) continuous on [푎, 푏] b) differentiable in (푎, 푏) then there exists at least one value of x, say c, such that 푓 푏 − 푓 푎 = 푓΄(푐), for 푎 < 푐 < 푏. 푏−푎 Geometrical Interpretation: Geometrically, this is equivalent to stating that the tangent line to the graph of f at 푐 parallel to the chord joining the points (푎, 푓(푎)) and (푏, 푓(푏) ). Example : Verify Lagrange’s Mean-Value Theorem for the function 푓(푥) = 2푥2 – 7푥 – 10 over ( 2 , 5 ) and find ‘푐’ of the Lagrange’s Mean-Value Theorem. Solution; Here a=2 and b=5, hence by Lagrange’s Mean-Value Theorem 푓 푏 − 푓(푎) = f΄(c) or, f΄(c) = 5−(−16) = 21 = 7 (i) 푏−푎 5−2 3 Again f΄(x) = 4x – 7, so f΄(c) = 4c – 7 (ii) Combining (i) and (ii) we get 4c – 7 = 7 so, c = 7 and 7 lies within (2,5). 2 2 Example: Prove that if b−a < tan−1 b − tan−1 a < b−a , 0 < 푎 < 푏. 1+b2 1+a2 Hence show that 휋 + 3 Example: Estimate 3 28. Solution. We take (푥) = 3 푥 , 푥 ∈ [27,28]. 2 1 − So 푡푕푎푡 푓΄(푥) = 푥 3 . Then using Lagrange’s Mean-Value Theorem 3 푓 28 − 푓 27 = 푓΄(푥 ), 푥 ∈ (27,28) 28 − 27 0 0 i.e, f(28) = f(27) + f΄(x0) 2 3 1 − = 27 + 푥 3 3 0 1 1 2 = 3 + ( )3, 27 < 푥0 < 28 3 푥0 1 1 2 1 < 3 + ( )3 = 3 + . 3 27 27 1 Thus, 3 28< 3 . 27 Cauchy’s Mean-Value Theorem. Statement: If 푓(푥) 푎푛푑 푔(푥) are continuous in [푎, 푏] and differentiable in (푎, 푏) and 푔΄(푥) ≠ 0 for any 푥 in (푎, 푏), then there exists at least one point 푥 = 푐 in (푎, 푏) such that 푓 푏 − 푓 푎 = 푓΄(푐). 푔 푏 − 푔(푎) 푔΄(푐) Example: If in the Cauchy’s Mean-Value Theorem we take 푥 = 푒푥 푎푛푑 푔 푥 = 푒−푥 , then prove that c is the arithmetic mean between 푎 푎푛푑 푏. Solution:푓(푥) and 푔(푥)satisfies the conditions of Cauchy’s Mean-Value Theorem, so there exists at least one point 푥 = 푐 in (푎, 푏) such that 푓 푏 − 푓 푎 = 푓΄(푐) , 푎 < 푐 < 푏 푔 푏 − 푔(푎) 푔΄(푐) 푏 푎 푐 표푟, 푒 −푒 = 푒 푒 −푏 −푒 −푎 −푒 −푐 표푟, 푒푎+푏 = 푒2푐 i.e, c = 푎+푏. 2 2 Example:Using Cauchy Mean Value Theorem, show that 1 - 푥 < 푐표푠 푥 for 푥 ≠ 0. 2! Example:For each of the following, verify that the hypotheses of Rolle's Theorem are satisfied on the given interval. Then find all value(s) of c in that interval that satisfy the conclusion of the theorem. (푖) 푓(푥) = −푥2 − 4푥 – 11 on [0,4], (푖푖) 푓(푥) = − 푠푖푛 푥 on [0,2휋]. Example:Use the mean value theorem (MVT) to establish the following inequalities. (i) 푒푥 >1 + 푥for푥 ∈ 푅 and (ii) 푥−1<푙표푔 푥 < 푥 – 1 for 푥 > 1. 푥 Generalized Mean-Value Theorem 1. Taylor’s Theorem with Lagrange’s Form of Remainder: Statement: Let 푓(푥) be a function defined in the closed interval [a,a+h] such that (i) (n-1)th derivative 푓푛 −1is continuous on [푎, 푎 + 푕] and (ii) nth derivative 푓푛 exists in (푎, 푎 + 푕). Then there exists at least one number 휃, where 0 <휃< 1 such that 2 푛 −1 푓 푎 + 푕 = 푓 푎 + 푕 푓 ΄ 푎 + 푕 푓 ΄΄ 푎 + ...... + 푕 푓푛−1(푎) + 푅푛….(1) 2! 푛−1 ! 푛 where푅푛 = 푕 푓푛 (푎 + 휃푕) is called the Lagrange’s Form of Remainder after n terms. 푛! 2. Taylor’s Theorem with Cauchy’s Form of Remainder: Statement: Let 푓(푥) be a function defined in the closed interval [푎, 푎 + 푕] such that (i) (푛 − 1)푡푕derivative푓푛−1is continuous on [a,a+h] and (ii) nth derivative 푓푛 exists in (푎, 푎 + 푕). Then there exists at least one number 휃, where 0 < 휃 < 1 such that 푕2 푕푛−1 푓(푎 + 푕) = 푓(푎) + 푕 푓 ΄(푎) + 푓 ΄΄(푎) + ...... + 푓푛−1(푎) + 푅푛 2! 푛 − 1 ! 푛 푛−1 where 푅푛 = 푕 (1−휃) 푓푛 (푎 + 휃푕) is called the Cauchy’s Form of Remainder after n terms. (푛−1)! Note1: If we take n=1 in (1), Taylor’s theorem reduces to Lagrange’s Mean-Value Theorem. 3. Maclaurin’s Theorem Statement: Let 푓(푥) be a function defined in the closed interval [0, 푥] such that (i) 푓푛 −1is continuous on [0 , 푥] and (ii) 푓푛 exists in (0 , 푥) then there exists at least one number 휃, where 0 < 휃 < 1 such that 푥2 푥푛−1 푓(푥) = 푓(0) + 푥 푓 ΄(0) + 푓 ΄΄(0) + ...... + 푓푛−1(0) + 푅푛 2! 푛 − 1 ! 푛 where푅푛 = 푥 푓푛 (휃푥) , 0 < 휃 < 1 [Lagrange’s Form] 푛! 푛 푛−1 and푅푛 = 푥 (1−휃) 푓푛 (휃푥) , 0 < 휃 < 1 [ Cauchy’s Form] (푛−1)! Example: Find the Maclaurin’s theorem with Lagrange’s form of remainder for 푓(푥) = 푠푖푛 푥. 5 Example: Verify Maclaurin’s theorem for f(푥) = (1 − 푥)2 with Lagrange’s form of remainder upto three terms when 푥 = 1. Solution: Here we use the formula, 푥2 푥푛 푓(푥) = 푓(0) + 푥 푓 ΄(0) + 푓 ΄΄(0) + 푓푛 (휃푥) , 0 < 휃 < 1 2! 푛! At 푥 = 1, 1 푥푛 푓(1) = 푓(0) + 푓 ΄(0) + 푓 ΄΄(0) + 푓푛 (휃) 2! 푛! 1 5 1 5 3 1 5 3 1 − 표푟, 0 = 1 + (−1) + (−1) (−1) + (−1) (−1) (−1) (1 − 휃) 2 2 2! 2 2 3! 2 2 2 Simplifying, = 0.25. Here, 휃 = 0.25 lies within (0,1), hence Maclaurin’s theorem with Lagrange’s form of remainder is verified. Taylor’s Series Statement:Let 푓(푥), 푓΄(푥), 푓΄΄(푥), ...... , 푓푛 (푥) exist finitely however large 푛 may be in any interval (푥 − 훿 , 푥 + 훿) enclosing the point 푥 and let 푅푛 → 0 푎푠 푛 → ∞ . Then Taylor’s series of finite form can be extended to an infinite series of the form 푕2 푕푛 푓(푥 + 푕) = 푓(푥) + 푕 푓΄(푥) + 푓΄΄(푥) + ...... + 푓푛 (푥) + . . .. , 푕 < 훿 2! 푛! Maclaurin’s Series Statement: Let 푓(푥), 푓΄(푥), 푓΄΄(푥), ...... , 푓푛 (푥) exist finitely however large 푛 may be in any interval (−훿 , 훿) and 푅푛 → 0 푎푠 푛 → ∞. Then Maclaurin’s series of finite form can be extended to an infinite series of the form 2 푛 푓(푥) = 푓(0) + 푥 푓΄(0) + 푥 푓΄΄(0) + ...... + 푥 푓푛 (0) + . . . . ., 푥 < 훿. 2! 푛! Example: Expand the function 푓 푥 = 푒푥 in the form of Maclaurin’s series in the neighbour- hood of the point 푥 = 0. 푛 Solution: Here Lagrange’s remainder after n terms 푅푛 = 푥 푒휃푥 , ( 0 < 휃 < 1) → 0 푎푠 푛 → 푛! ∞ ∀ 푥. Then Maclaurin’s series for 푒푥 is given by 푥2 푥푛 푓(푥) = 푓(0) + 푥 푓΄(0) + 푓΄΄(0) + ...... + 푓(푛) (0) + . . . . . 2! 푛! 2 푛 i.e, 푒푥 = 1 + 푥 + 푥 + ...... + 푥 + . . . . . 2! 푛! Example: Maclaurin’s series for 푠푖푛 푥 and 푐표푠 푥. Example: Maclaurin’s series for 푙표푔 (1 + 푥). Reduction Formulae 풏 1. Let In = 풔풊풏 풙 풅풙 = 푠푖푛푛−1푥 sin 푥 푑푥 = 푠푖푛푛−1푥 (− cos 푥) − 푛 − 1 푠푖푛푛−2푥푐표푠 푥(− cos 푥) 푑푥 = - 푠푖푛푛−1푥 cos 푥 + 푛 − 1 푠푖푛푛−2푥 푐표푠2푥 푑푥 = - 푠푖푛푛−1푥 cos 푥 + (푛 − 1) 푠푖푛푛−2푥 (1 − 푠푖푛2푥) 푑푥 = -푠푖푛푛−1푥 cos 푥 + (푛 − 1) 푠푖푛푛−2푥 푑푥 - (n-1) 푠푖푛푛 푥 푑푥 푛−1 = -푠푖푛 푥 cos 푥 + 푛 − 1 In-2 - (n - 1) In 푛−1 Or, { 1 + (n-1)}In = - 푠푖푛 푥 cos 푥 + 푛 − 1 In-2 ퟏ 풏−ퟏ 풏−ퟏ ⇒In = − 풔풊풏 풙 퐜퐨퐬 풙 + In—2 풏 풏 휋/2 푛 Jn = 0 푠푖푛 푥 푑푥 1 풏−ퟏ 휋/2 = [ − 푠푖푛푛−1푥 cos 푥 ]흅/ퟐ + 푠푖푛푛−2푥 푑푥 푛 ퟎ 풏 0 풏−ퟏ = 0 + Jn-2 풏 푛−1 = Jn-2 푛 풏 2. In = 풄풐풔 풙 풅풙 = 푐표푠푛−1푥 푐표푠 푥 푑푥 = 푐표푠푛−1푥 sin 푥 − 푛 − 1 푐표푠푛−2푥 푠푖푛 푥(− sin 푥) 푑푥 = 푐표푠푛−1푥 sin 푥 + (푛 − 1) 푐표푠푛−2푥 푠푖푛2푥 푑푥 푛−1 = 푐표푠 푥 sin 푥 + (푛 − 1) In-2 – ( n – 1 ) In ퟏ 풏−ퟏ 풏−ퟏ ⇒In = 풄풐풔 풙 퐬퐢퐧 풙 + In—2 풏 풏 휋/2 푛 IfJn = 0 푐표푠 푥 푑푥 1 푛−1 휋/2 = [ 푐표푠푛−1푥 sin 푥 ]휋/2 + 푐표푠푛−2푥 푑푥 푛 0 푛 0 푛−1 = Jn-2 푛 휋/2 7 Example:Using reduction formula, evaluate 0 푐표푠 푥 푑푥. Ans: 16. 35 풎 풏 3. Im,n = 풔풊풏 풙 풄풐풔 풙 풅풙 = 푐표푠푛−1푥 푠푖푛푚 푥 cos 푥 푑푥 푠푖푛푚+1푥 푠푖푛푚+1푥 = 푐표푠푛−1푥 – (푛 − 1)푐표푠푛−2푥 (− sin 푥) dx 푚+1 푚+1 푚+1 푠푖푛 푥 푛−1 = 푐표푠푛−1푥 + 푐표푠푛−2푥 푠푖푛푚 +2푥 푑푥 푚+1 푚 +1 푚+1 푠푖푛 푥 푛−1 = 푐표푠푛−1푥 + 푐표푠푛−2푥 푠푖푛푚 푥 푠푖푛2푥 푑푥 푚+1 푚+1 푚+1 푠푖푛 푥 푛−1 = 푐표푠푛−1푥 + 푐표푠푛−2푥 푠푖푛푚 푥 (1 − 푐표푠2푥) 푑푥 푚+1 푚+1 푚+1 푛−1 푠푖푛 푥 푛−1 푛−1 = 푐표푠 푥 + Im,n-2 - Im,n 푚+1 푚 +1 푚+1 푚+1 푛−1 푛−1 푠푖푛 푥 푛−1 Or, (1 + + ) Im,n = 푐표푠 푥 + Im,n-2 푚 +1 푚+1 푚 +1 푚+1 푚 +1 푛−1 푠푖푛 푥 푚 +1 푛−1 Or, Im,n = 푐표푠 푥 + . Im,n-2 푚+푛 푚+1 푚 +푛 푚 +1 ퟏ 풏−ퟏ 풎+ퟏ 풏−ퟏ = 풄풐풔 풙풔풊풏 풙 + Im,n-2 풎+풏 풎+풏 Alternately, m−1 n Writing, Im,n = sin x (cos x sin 푥) 푑푥 and proceeding as earlier, we can show that 1 푚 −1 푛+1 푚 −1 Im,n = - 푠푖푛 푥푐표푠 푥 + Im-2,n 푚 +푛 푚 +푛 휋 2 푚 푛 If Jm,n = 0 sin 푥 cos 푥 푑푥 푚+1 푛−1 푠푖푛 푥 휋/2 푛−1 = [ 푐표푠 푥 ] + Jm,n-2 , using (8) 푚+1 0 푚+푛 푛−1 = Jm,n-2 푚 +푛 INFINITE SERIES To learn the infinite series we have to learn first, “Sequence” 1. Def. of Sequence: A sequence of real number is a mapping from set of natural numbers ℕ to the set of real numbers ℝ , which is denoted by xn where n ∈ ℕ. Note: If the number of elements in the sequence is finite then the sequence is called FINITE SEQUENCE Otherwise it is INFINITE SEQUENCE When nothing is mentioned, then only sequence ⟹ INFINITE SEQUENCE Ex1. Let xn = 2n − 1, n ∈ ℕ. Then this is a sequence of odd +ve integer. i.e. 1, 3, 5, 7, … … … … n−1 Ex2. Let xn = −1 , n ∈ ℕ. Then xn = 1, −1, 1, −1, … … … . This type of sequence is known as OCCILATERY SEQUENCE. A sequence can develop in 4 ways Divergent The terms keep growing The terms converge on a single Convergent value, in this case 0. The sequence repeats itself after Periodic a set number of terms. The sequence oscillates between Oscillating 2 values. If you add the terms of a sequence together, you get a series A. 3 + 6 + 9 +12 + 15 B. 2 + 8 + 18 + 32 Arithmetic Sequence Sequences of numbers that follow a pattern of adding a fixed number from one term to the next are called arithmetic sequences. Definition A sequence with general term an+1 = an + d is called an arithmetic sequence. an = nth term and d = common difference Examples Find the general (nth) term for the following arithmetic sequences: A. 2,6,10,14,18,22, ... B. -5,-3,-1,1,3,... C. 1,4,7,10,13,16,... Solution All of these have one thing in common. To get to the next term we add a fixed number st th (d). Let a1 = 1 term and an = n term a1 = 2 a2 = 2 + 4 = 6 a3 = 2 + 2(4) = 10 a4 = 2 + 3(4) = 14 and so on . an = a1 + (n – 1)4 If we let d = 4 this becomes an = a1 + (n – 1)d Geometric Sequence Sequences of numbers that follow a pattern of multiplying a fixed number from one term to the next are called geometric sequences. Definition A sequence with general term an+1 = an r is called an geometric sequence. an = nth term and r = common ratio Examples Find the general (nth) term for the following geometric sequences: A. 2,6,18,54, ... B. 27,9,3,1,….. C. 16,-8,4,-2,1,... Solution All of these have one thing in common. To get to the next term we multiply a fixed number (r). st th Let a1 = 1 term and an = n term a1 = 2 a2 = 2(3) = 6 2 a3 = 2(3) = 18 3 a4 = 2(3) = 54 and so on . n-1 an = a1 (3) n-1 If we let r = 3 this becomes an = a1 r The Arithmetic Series The following theorem provides us with an easy way to calculate the arithmetic series. Theorem If an = a1 + (n - 1)d is an arithmetic sequence then the sum of the sequence is n n S (a a ) [2a (n 1)d] n 2 1 n 2 This can be proven but let’s just convince ourselves that it works. It is easy to determine the sum of the following arithmetic sequence: 3 + 5 + 7 = 15 This is an arithmetic series with common difference 2 Now let’s use the formula We have a1 = 3, an = 7, d = 2 3 S (3 7) 15 n 2 Examples: Find the following sums A. 3 + 7 + 11 + 15 + ... + 35 B. -2 + 1 + 4 + 7 + …+ 25 Solutions: We have a1 = 3, an = 35, d = 4 To find n we note that 35 = 3 + (n - 1)4 so that 32 = (n - 1)4 and n = 9 Now we are ready to use the formula 9 S (3 35) 171 n 2 A. We have a1 = -2, an = 25 and n = 10 10 So that S (2 25) 5(23) 115 10 2 Exercises: Find the sums 1. 5 + 10 + 15 +.... + 500 2. 3 + 6 + 9 + .... + 99 3. -5 + -15 + -25 + -35 + ... + -95 4. What is the sum of the numbers 1 to 100? 5. Find the sum of the first 27 terms of the series that starts 7 + 3 – 1 – 5 The Finite Geometric Series The following theorem provides us with an easy way to calculate the arithmetic series. Theorem If n-1 an = a r is a geometric sequence then the sum of the sequence is n 1 r n Sn an a i1 1 r This can be proven but let’s just convince ourselves that it works. It is easy to determine the sum of the following geometric sequence: 3 + 6 + 12 = 21 This is a geometric series with common ratio 2 Now let’s use the formula We have a = 3, an = 12, r = 2 1 23 7 Sn 3 3 3(7) 21 1 2 1 Examples: Find the following sums 1. 1st 5 terms of -6 + 18 – 54 + …. 2. 5 + 10 + 20 + 40 + ... + 2560 Solutions: 1. We have a = -6, r = -3 and n = 5 5 5 1 (3) 244 So that 2(3)n 6 6 366 r1 1 (3) 4 The Infinite Geometric Series Theorem If n-1 an = a r is a geometric sequence and r 1 then the sum of the infinite sequence is a Sn an i1 1 r Examples: Find the following sums A. 2 4 8 16 ..... B. 24 + 12 + 6 + 3 + 3/2 + ¾ + ……. Solutions: A. We have a = -2, r =-2 This infinite series diverges because r = -2 and 2 is not less than 1. There is no sum. B. We have a = 24, r = 1/2 24 So that S 48 n 1 1 2 An infinite series is the limit of a sequence of finite sums. For example, we can define the finite (or partial) sum of the terms ai, for i = 1,2,3,…,N as N SN ai a1 a2 aN i1 which must be finite. We call SN a partial sum and S1, S2, …, SN taken together is referred to as the sequence of partial sums. The infinite series, if it exists, is defined as S ai lim SN N i1 This infinite series should not be viewed as actually adding together infinitely many things; since one could never finish such a task. Rather it is a limit, often never reached, of the partial sums.1These partial sums siddle-up arbitrarily near the limit of the partial sums. Many people wrongly think that an infinite series is a formalization of the process of adding infinitely many things together. It is instead a limit of partial sums, where each term in the sequence is the adding together of finitely many things. At no time are infinitely many things being added. The point of the limit is to avoid having to say that infinitely many additions are being done. But, while such infinity is never reached, many people nevertheless find it useful to think of infinite series as an infinite summation. When an infinite series exists we say that the series converges. What this means is that the sequence (of partial sums) limits to something which is finite and unique. In fact, it is the sequence of finite sums which converge. An infinite series which does not converge is said to be a divergent series. A series which converges when each term of the series is replaced by its absolute value is said to be absolutely connvergent. A series which is not absolutely convergent, but which is nevertheless convergent, is said to be conditionally convergent. Loosely one can say that absolutely convergent series are ones that behave much like finite sums. By contrast, conditionally convergent series are totally non- intuitive. For example, by simply rearranging the terms of a conditionally convergent series, one can make the series limit to any real number. Despite these peculiarities, infinite series are immensely useful and sometimes indispensible in mathematics. In addition to infinite series defined in terms of constant terms, mathematicians are interested in the series representation of functions. The most common of these is the power series representation of functions. A power series representation of a function can be written as k f (x) ak x k0 1The sequence 1,1/2,1/3,1/4,… limits to zero, but never reaches zero. However, the sequence 1,0,1/2,0,1/3,… reaches its limit infinitely many times, but still has infinitely many terms that differ from the limit. The sequence 1,2,0,0,0… limits to zero and reaches the limit after the second term. This series is naturally defined as the limit of partial sums of the power series. The power series does not always converge for all values of x. In addition, not all functions have a power series representation. 1 Example Determine whether the series 2 converges or diverges. If it converges, n 3 n 2n then find its sum. This series was one of our examples given above. 1 We will rewrite the fraction n 2 2n using partial fraction decomposition. 1 A B = + 1 A(n 2) Bn n(n 2) n n 2 1 To solve for A, choose n 0 : 1 2 A A 2 1 To solve for B, choose n 2: 1 2B B 2 1 1 2 2 1 1 1 Thus, = = + = . n n 2 2 n 2 n 1 1 1 1 1 1 Thus, = = n 3 2 n 2 n 2 n 3 n 2 n 1 1 We will find the sequence { S n } of partial sums for the series , where n 3 n 2 n 1 1 a n n 2 n . Thus, 1 S a 1 1 3 3 1 1 1 S a a S a 1 2 3 4 1 4 3 2 4 1 1 1 1 1 1 1 1 S a a a S a 1 1 3 3 4 5 2 5 3 2 4 3 5 2 4 5 1 1 1 1 1 1 1 1 S a a a a S a 1 1 4 3 4 5 6 3 6 2 4 5 4 6 2 5 6 1 1 1 1 1 1 1 1 S a a a a a S a 1 1 5 3 4 5 6 7 4 7 2 5 6 5 7 2 6 7 1 1 1 3 1 1 S a a a a 1 n 3 4 5 n 2 2 n 1 n 2 2 n 1 n 2 3 1 1 3 3 lim Then lim S n = = 0 0 = . n n 2 n 1 n 2 2 2 1 1 3 1 1 1 1 Thus, . Thus, 2 = = n 3 n 2 n 2 n 3 n 2n 2 n 3 n 2 n 1 3 3 = 2 2 4 3 Answer: Converges; 4 Theorem1. If a series a n is convergent, then lim a n 0 . n n N Test for Divergence:If lim a n 0 , then the series is divergent. n COMMENT: The Test for Divergence is the second most misused statement by Calculus students. Students want to apply the converse of the previous theorem, which is the statement if , then the series is convergent. However, this is NOT true. 1 1 An easy example to keep in mind is the series . We have that lim 0. However, n n 1 n n we will show in a later lesson that this series is DIVERGENT. The series is called the (divergent) harmonic series. Theorem2. If a n and bn are convergent series with sums A and B, respectively, n N n N then 1. (a n bn ) is a convergent series and has of sum of A B . n N 2. if c is a constant, then ca n is a convergent series and has of sum of c A. n N 3. (a n bn ) is a convergent series and has of sum of A B . n N Examples Determine whether the following series converge or diverge. If the series converges, then give its sum. n n 1 1. 3 4 n 1 n 1 n 1 n 1 n n 1 4 1 4 1 4 3 4 = n = n 1 = n 1 n 1 3 n 1 3 3 n 1 3 3 1 4 4 This is a geometric series where a and r . Since r 1, then the geometric 3 3 3 series diverges. Answer: Diverges n 1 5 2. 2 n 1 8 5 5 This is a geometric series where a 2 and r . Since r 1, then the 8 8 geometric series converges and has a sum of 2 16 16 a S = 5 . 1 r 1 8 5 3 8 16 Answer: Converges; 3 The Absolute convergence: Ratio and root test 1. A series an is called absolutely convergent if the series of absolute values an is n1 n1 convergent. 2. A series is called conditionally convergent if the series is convergent but not absolutely convergent 3. Theorem If a series is absolutely convergent then the series is convergent. The ratio test: a i) If lim n1 L 1, then the series is absolutely convergent and n an therefore convergent. a a ii) If lim n1 L 1, or lim n1 then the series is divergent n n an an a iii) If lim n1 L 1, then the ratio test is inconclusive. n an iv) The root test: n i) If lim an L 1, then the series is absolutely convergent and n therefore convergent. n ii) If lim an L 1, or then the series is divergent n n iii) If lim an L 1, then the root test is inconclusive. n Examples 1. Test the convergence of the series (1) n n3 a a) . We have lim n1 1/ 3 1. So the given series is absolutely convergent by n n n1 3 an the ratio test. (3n 2)n n n b) (1) . We have lim an 3/ 4 1. So the given series is absolutely n n n1 (4n 3) convergent by the root test. Show that following series are conditionally convergent: (1) n1 a a) . We have lim n1 1, by ratio test the it is inconclusive. But by absolute 4 n n1 n an 1 convergence test lim an lim is a p - series with p =1/4 < 1, divergent, on the other n n n1/ 4 1 hand by alternating series test it is convergent, since lim bn lim 0, bn bn1 . So the n n n1/ 4 given series is conditionally convergent. (1) n1 n b) 2 . We have , by ratio test the it is inconclusive. But by limit n1 n 1 2 an n /(n 1) 1 comparision test (Section 11.4) lim lim 1 0 is divergent since bn is a n n bn 1/ n n divergent p - series with p =1. On the other hand by alternating series test it is convergent, n since lim bn lim 0, bn bn1 . So the given series is conditionally convergent. n n n2 1 Strategy for Testing Series We have learnt the following: 1 1. p is a divergent p – series, converges when p > 1 and diverges when p 1. n1 n 2. ar n1 orar n is a geometric series, converges when r 1, diverges when r 1. n1 n1 3. If lim bn 0 the series diverges (Divergent test) n 4. Series with factorials use ratio test. Vector Algebra and Calculus: Vector Algebra: Definition: A vector is a physical quantity (notation: 푎 ) which has a direction, and a length 푎 and follows the laws of vector addition. Geometrically, a vector is represented by a directed line segment 푃 푄 from one point 푃 to another point 푄. Here 푃 is called origin or initial point and 푄 is called terminal point, end or terminus of the vector. In a coordinate system it’s expressed by components: in space, 푎 = 푎1, 푎2, 푎3 = 푎1푖 + 푎2푗 + 푎3푘 (Recall in space 푥-axis points to the lower-left, y to the right, z up). Here 푎1, 푎2, 푎3 are scalars and are called component of the vector 푎 in 푥, 푦, 푧 direction respectively. Magnitude or modulus of a vector: A positive number which is used to measure of length of a vector is called magnitude. The magnitude of the vector 푎 = 푎1푖 + 푎2푗 + 푎3푘 is denoted by 푎 and calculated by 푎 = 2 2 2 푎1 + 푎2 + 푎3. Note: (i)A vector with unit magnitude is called unit vector. (ii) In three dimensional geometry unit vectors 푖 , 푗 , 푘 along 푥-axis,푦-axis and 푧-axis respectively are called fundamental unit vectors. (iii) Two vectors are equal if they have the same magnitude and direction regardless their initial point. (iv) A vector having opposite to that of a given vector 푎 but having the same magnitude is denoted by −푎 . And is called negative of 푎 . Algebra of vectors: Addition of vectors: Triangular law of addition: Let 푎 and 푏 be two vectors then the sum of resultant of 푎 and 푏 , is a vector 푐 formed by placing the initial point of 푏 on the terminal point of 푎 and then joining the initial point of 푎 and final point of 푏 . The sum 푐 is written as 푐 = 푎 + 푏 .This is known as triangular law of vector addition. Parallelogram law of addition: Let 푎 and 푏 be two vectors have same initial point. Then the sum or resultant of 푎 and 푏 , is a vector 푐 is the diagonal of the parallelogram formed by drawing parallel lines at the terminal points of 푎 and 푏 . Generally when we have two vectors having same initial point then we use parallelogram law of addition and when we have more than two vectors then we use triangular law of vector addition to form polygon and find the resultant or sum of the vectors. Note: (i) The difference of two vectors 푎 and 푏 denoted by 푎 − 푏 is that vector 푐 which can be obtained by sum of the vectors 푎 and −푏 . (ii) If 푎 = 푏 , then 푎 − 푏 is defined to as the null or zero vector and it is denoted by 푂 . That is it has zero magnitude but direction undefined. (iii) A vector that is not null vector is called proper vector. Scalar Multiplication: Multiplication of a vector 푎 by a scalar 푚 is vector 푚푎 with magnitude 푚 times the magnitude of 푎 and the direction of 푚푎 is in the same or opposite of 푎 according as 푚 is positive or negative. Note: If 푚 = 0then 푚푎 = 푂 , the null vector. Laws of vector Algebra: If 푎 , 푏 and 푐 are vectors and 푚 and 푛 are scalars then the following relations are satisfied: (i)Associative law for addition: 푎 + 푏 + 푐 = 푎 + 푏 + 푐 (ii)Existence of zero element: ∃ a zero vector 푂 such that for every vector 푎 , 푎 + 푂 = 푂 + 푎 = 푎 . (iii)Existence of negative: ∀vector 푎 , ∃a vector −푎 such that 푎 + −푎 = −푎 + 푎 = 푂 . (iv)Commutative law of addition.푎 + 푏 = 푏 + 푎 . (v)Distributive law of scalar multiplication over addition: 푚 푎 + 푏 = 푚푎 + 푚푏 . (vi)Distributive law of scalar multiplication over addition of scalars: 푚 + 푛 푎 = 푚푎 + 푛푎 . (vii)Associative law of scalar multiplication over multiplication of scalars: 푚 푛푎 = 푚푛 푎 . (viii) Unit multiplication: 1 푎 = 푎 . (ix) If 푎 = 푎1푖 + 푎2푗 + 푎3푘 and 푏 = 푏1푖 + 푏2푗 + 푏3푘then 푎 ± 푏 = 푎1 ± 푏1 푖 + 푎2 ± 푏2 푗 + 푎3 ± 푏3 푘. Collinear vectors: Two vectors 푎 and 푏 are said to be parallel or collinear if 푎 = 휆푏 where 휆 is a scalar. 푎1 푎2 Note: If two vectors 푎 = (푎1, 푎2, 푎3) and 푏 = (푏1, 푏2, 푏3) are parallel or collinear then = = 푏1 푏2 푎3. 푏3 Coplanar vectors: A system of vectors is said to be coplanar if they are parallel to the same plane. Linear . Linear combination of vectors: A vector 푏 is said to be linear combination of 푛 number of vectors 푎 1, 푎 2, 푎 3, ⋯ 푎 푛 if 푏 can be expressed as 푏 = 푐1푎 1 + 푐2푎 2 + 푐3푎 3 + ⋯ + 푐푛 푎 푛 where 푐1, 푐2, 푐3, ⋯ , 푐푛 are scalars. Linearly dependent vectors: Vectors 푎 1, 푎 2, 푎 3, ⋯ 푎 푛 are said to be linearly dependent if there exist scalars 푐1, 푐2, 푐3, ⋯ , 푐푛 not all zero, such that 푐1푎 1 + 푐2푎 2 + 푐3푎 3 + ⋯ + 푐푛 푎 푛 = 푂 . Linearly independent vectors: Vectors 푎 1, 푎 2, 푎 3, ⋯ 푎 푛 are said to be linearly independent if the vector equation 푐1푎 1 + 푐2푎 2 + 푐3푎 3 + ⋯ + 푐푛 푎 푛 = 푂 , has only solution 푐1 = 0, 푐2 = 0, 푐3 = 0, ⋯ , 푐푛 = 0 where 푐1, 푐2, 푐3, ⋯ , 푐푛 are unknown scalars. Note: (i) If there is a solution with some 푐푗 ≠ 0, then the above set of vectors are linearly dependent. (ii)If 푎 is a non null vector, then 푎 by itself, is linearly independent, as 푚푎 = 푂 , and 푎 ≠ 0 ⇒ 푚 = 0 . (iii) Two or more vectors are linearly dependent if and only if one of them is linear combination the others. Position vector of a point: Let 푃(푥, 푦, 푧) be a point in 3dimensional space. The vector 푟 from the origin 푂 to the point 푃 is called position vector (or radius vector of 푃. Thus 푟 may be written as 푟 = 푥푖 + 푦푗 + 푧푘 , and the magnitude 푟 = 푥2 + 푦2 + 푧2. Note: If two points 푃(푥1, 푦1, 푧1) and 푄(푥2, 푦2, 푧2) forms a vector 푃 푄 , then the position vector of the vector 푃 푄 is given by 푃 푄 = (푥2 − 푥1, 푦2 − 푦1, 푧2 − 푧1). Distance between two points: The distance between two points 푃(푥1, 푦1, 푧1) and 푄(푥2, 푦2, 푧2) is to be given by the modulus of the vector 푃 푄 , that is distance between 푃 and 푄 is 푃 푄 = 2 2 2 푥2 − 푥1 , 푦2 − 푦1 , 푧2 − 푧1 . Ratio formula: (i) If a point 푅 divides the line 푃푄 internally in the ratio 푚: 푛, then the coordinates of 푅 are given by 푚푥2+푛푥1 , 푚푦2 +푛푦1 , 푚푧2+푛푧1 . 푚+푛 푚 +푛 푚 +푛 (ii) If a point 푅 divides the line 푃푄 externally in the ratio 푚: 푛, then the coordinates of 푅 are given by 푚푥2−푛푥1 , 푚 푦2−푛푦1 , 푚 푧2−푛푧1 . 푚−푛 푚 −푛 푚 −푛 (iii) Coordinates of the middle point of 푃푄 are given by 푥1+푥2 , 푦1 +푦2 , 푧1+푧2 . 2 2 2 Vector Products: Dot or Scalar product: The dot or scalar product of two vectors 푎 and 푏 , denoted by 푎 ∙ 푏 (read 푎 dot 푏 ), is defined as 푎 ∙ 푏 = 푎 푏 cos 휃, where 휃 is the angle between the vectors 푎 and 푏 and 0 ≤ 휃 ≤ 휋 . Note: (i) Dot product of two vectors are scalar not vector. (ii) If 푎 = 푎1푖 + 푎2푗 + 푎3푘 and 푏 = 푏1푖 + 푏2푗 + 푏3푘then 푎 ∙ 푏 = 푎1푏1 + 푎2푏2 + 푎3푏3. 2 2 2 2 (iii) If 푎 = 푎1푖 + 푎2푗 + 푎3푘then 푎 ∙ 푎 = 푎 = 푎1 + 푎2 + 푎3. (iv) For any two non-null vectors 푎 and 푏 , if 푎 ∙ 푏 = 0, then the vectors are perpendicular to each other. (v) Angle 휃 between and two vectors 푎 and 푏 is given by 휃 = cos−1 푎 ∙푏 . 푎 푏 (vi)Projection: Component orProjection of the vector 푎 along 푏 = 푎 cos 휃 = 푎 ∙푏. 푏 Properties:If 푎 , 푏 and 푐 are three vectors and 푚 is a scalar, then the following laws hold: (i)Commutative law of dot products: 푎 ∙ 푏 = 푏 ∙ 푎 (ii)Distributive law over addition:푎 ∙ 푏 + 푐 = 푎 ∙ 푏 + 푎 ∙ 푐 . (iii)푚 푎 ∙ 푏 = 푚푎 ∙ 푏 = 푎 ∙ 푚푏 = 푎 ∙ 푏 푚. (iv)푖 ∙ 푖 = 푗 ∙ 푗 = 푘 ∙ 푘 = 1, 푖 ∙ 푗 = 푗 ∙ 푘 = 푘 ∙ 푖 = 0. Cross product: The cross product of the vectors 푎 and 푏 is a vector 푐 = 푎 × 푏 (read 푎 cross 푏 ), defined as 푎 × 푏 = 푎 푏 sin 휃 푛 , where 휃 is the angle between the vectors 푎 and 푏 and 0 ≤ 휃 ≤ 휋 and 푛 is the unit vector perpendicular to the plane of the vectors 푎 and 푏 indicating the direction of 푎 × 푏 . Note: (i)푎 , 푏 and 푛 form a right handed system. (ii) If two non null vectors 푎 and 푏 are such that, 푎 = 푏 or 푎 and 푏 are parallel or collinear then sin 휃 = 0 and we define 푎 × 푏 = 푂 . (iii) If 푎 = 푎1푖 + 푎2푗 + 푎3푘 and 푏 = 푏1푖 + 푏2푗 + 푏3푘 then 푖 푗 푘 푎 푎 푎 푎 푎 푎 2 3 1 3 1 2 푎 × 푏 = 푎1 푎2 푎3 = 푖 − 푗 + 푘. 푏2 푏3 푏1 푏3 푏1 푏2 푏1 푏2 푏3 Properties:If 푎 , 푏 and 푐 are three vectors and 푚 is a scalar, then the following laws hold: (i)Non commutative:푎 × 푏 = −( 푏 × 푎 ). (ii)Distributive law over addition: 푎 × 푏 + 푐 = 푎 × 푏 + 푎 × 푐 . (iii) 푚 푎 × 푏 = 푚푎 × 푏 = 푎 × 푚푏 = 푎 × 푏 푚. (iv)푖 × 푖 = 푗 × 푗 = 푘 × 푘 = 푂 , 푖 × 푗 = 푘 , 푗 × 푘 = 푖 , 푘 × 푖 = 푗, 푗 × 푖 = −푘 , 푘 × 푗 = −푖 , 푖 × 푘 = −푗 . (v) The magnitude of 푎 × 푏 is same as the area of a parallelogram with sides 푎 and 푏 . 푎 ×푏 (vi)Unit vector perpendicular to the plane of 푎 and 푏 is given by . 푎 ×푏 푎 ×푏 (vii) Angle between any two vectors 푎 and 푏 is given by sin−1 . 푎 푏 Triple scalar product The triple scalar product (also known as box product) of three vectors 푎 , 푏 and 푐 is to be denoted by 푎 , 푏 , 푐 or 푎 푏 푐 is defined by 푎 , 푏 , 푐 = 푎 ∙ 푏 × 푐 = 푎 × 푏 ∙ 푐 . Note: (i) If 푎 = 푎1푖 + 푎2푗 + 푎3푘 and 푏 = 푏1푖 + 푏2푗 + 푏3푘 and 푐 = 푐1푖 + 푐2푗 + 푐3푘 then 푎1 푎2 푎3 푎 , 푏 , 푐 = 푏1 푏2 푏3 . 푐1 푐2 푐3 (ii) 푎 , 푏 , 푐 is a scalar quantity. (iii)If three vectors 푎 , 푏 and 푐 are coplanar then 푎 , 푏 , 푐 = 0. (iv) 푎 , 푏 , 푐 = 푏 , 푐 , 푎 = 푐 , 푎 , 푏 = − 푎 , 푐 , 푏 = − 푏 , 푎 , 푐 = − 푐 , 푏 , 푎 (v) 푎 , 푏 , 푏 = 푎 , 푎 , 푐 = 푐 , 푏 , 푐 = 0. (vii) 휆푎 , 푏 , 푐 = 푎 , 휆푏 , 푐 = 푎 , 푏 , 휆푐 = 휆 푎 , 푏 , 푐 for any scalar 휆. (viii) 푎 + 푏 , 푐 , 푑 = 푎 , 푐 , 푑 + 푏 , 푐 , 푑 . (ix) 푖 , 푗 , 푘 = 푗 , 푘 , 푖 = 푘 , 푖 , 푗 = 1and 푖 , 푘 , 푗 = 푗 , 푖 , 푘 = 푘 , 푗 , 푖 = 0. Vector triple product: For any three vectors 푎 , 푏 and 푐 , vector triple product denoted by 푎 × 푏 × 푐 or 푎 × 푏 × 푐 defined by 푎 × 푏 × 푐 = 푎 ∙ 푐 푏 − 푎 ∙ 푏 푐 Or, 푎 × 푏 × 푐 = 푎 ∙ 푐 푏 − 푏 ∙ 푐 푎 . Note: (i) 푎 × 푏 × 푐 ≠ 푎 × 푏 × 푐 . (ii) 푎 × 푏 × 푐 = −푐 × 푎 × 푏 . Limit, Continuity Differentiability of vector function: Scalar field: Suppose that corresponding to each point 푥, 푦, 푧 of a region 퐷in three dimensional space, there corresponds a scalar 휑 푥, 푦, 푧 . Then 휑 is called a scalar function of position, and we say that a scalar field 휑 has been defined on 퐷. Note: (i) Height, temperature etc. at any point on the Earth’s surface at a certain place at a certain time define scalar field. (ii)A scalar field 휑 which is independent of time is called a stationary or steady state scalar field. Vector Field: Let to each point 푥, 푦, 푧 in a region 퐷 in three dimensional space there corresponds a vector 푣 푥, 푦, 푧 . Then 푣 is called a vector field of position, and we say that the vector field 푣 has been defined over 퐷. Note: (i) Velocity, acceleration etc. of any moving object a certain time define vector field. (ii)A vector field 푣 which is independent of time is called a stationary or steady state vector field. Vector Function: Let 푃 be a variable point on a curve in space and the position vector of 푃 relative to a fixed origin be 푟 . If there exists an independent scalar variable 푡 such that corresponding to each value of 푡 in a definite domain, we get definite position of 푃, that is, a unique vector 푟 , then 푟 is called a single valued vector function of the scalar variable 푡 in the domain. It is usually denoted by 푟 = 푓 푡 . Note: (i) 푓 푐 denotes the particular vector for some fixed value 푐of 푡. (ii) If 푖 , 푗 , 푘 denote a fixed triad of mutually orthogonal vectors, then the vector function 푓 (푡) of the scalar parameter 푡 can be decomposed to express as in the form 푟 = 푓 푡 = 푓1 푡 푖 + 푓2 푡 푗 + 푓3 푡 푘 in which 푓1 푡 , 푓2 푡 , 푓3 푡 are scalar functions of 푡. (iii)The point 푃, whose Cartesian co-ordinates are 푓1, 푓2 , 푓3 , describes a Cartesian curve as 푡 varies and hence 푓 푡 represents a curve. Limit of a vector function: A vector function 푓 푡 of a scalar parameter 푡 is said to tend to a limit 푙 as 푡 tend to 푡0, if corresponding to any pre assigned positive quantity 휀, however small, we can find out another positive quantity 훿, such that 푓 푡 − 푙 < 휀, when ever 0 < 푡 − 푡0 < 훿 . This is expressed as lim푡→푡0 푓 푡 = 푙. Continuity of a vector function: A vector function 푓 푡 is said to be continuous at 푡 = 푡0if lim푡→푡0 푓 푡 = 푓 푡0 . Note: If 푓 푡 is continuous at every value of 푡 in a domain, then it is said to be continuous in that domain. Derivative of a vector function: The derivative of a vector function 푓 (푡) is denoted by 푓 ′(푡) and defined by 푓 ′ 푡 = 푑푓 = lim 푓 푡+∆푡 −푓(푡) , if the limiting value exists finitely. 푑푡 ∆t→0 ∆푡 Note: (i) Derivative of a vector function is again a vector function. (ii) If a vector function 푓 푡 can be expressed as 푓 푡 = 푓1 푡 푖 + 푓2 푡 푗 + 푓3 푡 푘, where 푓1 푡 , 푓2 푡 , 푓3 푡 are scalar functions of 푡. Then 푓 ′ 푡 = 푑푓 = 푑푓1 푡 푖 + 푑푓2 푡 푗 + 푑푓3 푡 푘 = 푓′ 푡 + 푓′ 푡 + 푓 ′ 푡 . 푑푡 푑푡 푑푡 푑푡 1 2 3 Space curves Let is consider a variable point 푃(푥, 푦, 푧) of three dimensional co-ordinate system and 푂 be the origin. Let the position vector of the line joining 푂 and 푃 be 푟 푡 = 푥 푡 푖 + 푦 푡 푗 + 푧 푡 푘 , where 푥 = 푥 푡 , 푦 = 푦 푡 , 푧 = 푧 푡 are functions of 푡 and 푡 changes the terminal point 푟 describes a space curve having parametric equation 푥 = 푥 푡 , 푦 = 푦 푡 , 푧 = 푧 푡 . Note: (i)The derivative of a constant vector is zero vector. (ii) If 푡 denotes the time, 푑푟 represents the velocity 푣 with which the terminal point 푟 describes 푑푡 2 the curve. Similarly, 푑푣 = 푑 푟 represents its acceleration along the curve. 푑푡 푑푡2 (iii) If 푎 , 푏 , 푐 be three differentiable vector function of a scalar 푡 and 휑 be a differentiable scalar function of 푡, then (a) 푑 푎 + 푏 = 푑푎 + 푑푏 푑푡 푑푡 푑푡 (b) 푑 푎 ∙ 푏 = 푎 ∙ 푑푏 + 푑푎 ∙ 푏 푑푡 푑푡 푑푡 (c) 푑 푎 × 푏 = 푎 × 푑푏 + 푑푎 × 푏 푑푡 푑푡 푑푡 (d) 푑 휑푎 = 휑 푑푎 + 푑휑 푎 푑푡 푑푡 푑푡 Results: (i)The necessary and sufficient condition for a vector function 푓 푡 to be constant is that 푑 푓 푡 = 푂 . 푑푡 (ii)The necessary and sufficient condition for a vector function 푓 푡 to have constant magnitude is that 푓 푡 . 푑 푓 푡 = 0. 푑푡 (iii) The necessary and sufficient condition for a vector function 푓 푡 to have constant direction is that 푓 푡 × 푑 푓 푡 = 푂 . 푑푡 Gradient Divergence and Curl The vector differential operator ∇ (called vector del or nabla)is defined as follows: ∇ = 휕 푖 + 휕 푗 + 휕 푘 = 푖 휕 + 푗 휕 + 푘 휕 . 휕푥 휕푦 휕푧 휕푥 휕푦 휕푧 Gradient: Let 휑 푥, 푦, 푧 be a scalar function defined ad differentiable at each point 푥, 푦, 푧 in a certain region of three dimensional space. Then the gradient of 휑, written as ∇ 휑 or 푔푟푎푑 휑 is defined as ∇ 휑 = 휕 푖 + 휕 푗 + 휕 푘 휑 = 푖 휕 + 푗 휕 + 푘 휕 휑 = 휕휑 푖 + 휕휑 푗 + 휕휑 푘 휕푥 휕푦 휕푧 휕푥 휕푦 휕푧 휕푥 휕푦 휕푧 Note: (i) ∇ 휑 defines a vector field. (ii)∇ 휑 gives a vector normal (out ward)to the surface of 휑 푥, 푦, 푧 = 0. Directional Derivative: Consider a scalar function 휑 = 휑 푥, 푦, 푧 . Then the directional derivative of 휑 in the direction of a vector 푎 is denoted by 퐷 휑 . If 푎 = 푎 , the unit vector in the direction of 푎 , then 퐷 휑 = 푎 푎 푎 ∇ 휑 ∙ 푎 . Note: The directional derivative has its maximum magnitude in the direction of ∇ 휑. Divergence: Suppose 푣 푥, 푦, 푧 = 푣1푖 + 푣2푗 + 푣3푘 is defined and differentiable at each point 푥, 푦, 푧 in a region of three dimensional space. Then the divergence of 푣 written as ∇ ∙ 푣 or 푑푖푣 푣 , is defined as follows: ∇ ∙ 푣 = 휕 푖 + 휕 푗 + 휕 푘 ∙ 푣 푖 + 푣 푗 + 푣 푘 = 휕푣1 + 휕푣2 + 휕푣3 휕푥 휕푦 휕푧 1 2 3 휕푥 휕푦 휕푧 Note: (i)Though푣 is a vector field, ∇ ∙ 푣 is a scalar. (ii) the vector field 푣 is source vector, sink vector or solenoidal according as ∇ ∙ 푣 > 0, ∇ ∙ 푣 < 0 or ∇ ∙ 푣 = 0. 2 2 2 (iii)∇ ∙ ∇ ≡ 휕 푖 + 휕 푗 + 휕 푘 ∙ 휕 푖 + 휕 푗 + 휕 푘 = 휕 + 휕 + 휕 ≡ ∇2is the Laplacian 휕푥 휕푦 휕푧 휕푥 휕푦 휕푧 휕푥 2 휕푦 2 휕푧 2 operator. (iv) ∇2휑 = 0is called Laplace equation. Curl: Suppose 푣 푥, 푦, 푧 = 푣1푖 + 푣2푗 + 푣3푘 be a differentiable vector field. Then the rotation or curl of 푣 is written as ∇ × 푣 , 푐푢푟푙 푣 or 푟표푡 푣 , defined as follows: 푖 푗 푘 휕 휕 휕 휕 휕 휕 ∇ × 푣 = 푖 + 푗 + 푘 × 푣1푖 + 푣2푗 + 푣3푘 = . 휕푥 휕푦 휕푧 휕푥 휕푦 휕푧 푣1 푣2 푣3 Note: (i)Curl of a vector field gives the rotational properties of a vector field. (ii) If ∇ × 푣 = 푂 , then the vector field 푣 is called irrotational vector field or conservative vector field or equipotential vector field. (iii) If ∇ × 푣 = 푂 , that is if 푣 irrotational, then 푣 can be expressed as 푣 = ∇ 휑, where 휑 푥, 푦, 푧 is a scalar function, called scalar potential function. (iv) If∇ × 푣 ≠ 푂 then 푣 is not irrotational and is sometimes called vortex field. (v) ∇ × ∇ 휑 = 푂 , where 휑 푥, 푦, 푧 is a scalar function. (vii) ∇ ∙ ∇ × 푣 = 0, where 푣 is a vector field. Results Involving ∇ . If 푎 and 푏 be two differentiable vector functions and 휑 and 휓 are differentiable scalar function of position 푥, 푦, 푧 , then the following laws hold: (i) ∇ 휑 + 휓 = ∇ 휑 + ∇ 휓 (ii)∇ ∙ 푎 + 푏 = ∇ ∙ 푎 + ∇ ∙ 푏 (iii) ∇ × 푎 + 푏 = ∇ × 푎 + ∇ × 푏 (iv)∇ ∙ 휑푎 = ∇ 휑 ∙ 푎 + 휑 ∇ ∙ 푎 (v)∇ × 휑푎 = ∇ 휑 × 푎 + 휑 ∇ × 푎 (vi)∇ ∙ 푎 × 푏 = 푏 ∙ ∇ × 푎 − 푎 ∙ ∇ × 푏 (vii) ∇ × 푎 × 푏 = 푏 ∙ ∇ 푎 − 푏 ∇ ∙ 푎 − 푎 ∙ ∇ 푏 + 푎 ∇ ∙ 푏 (viii)∇ 푎 ∙ 푏 = 푏 ∙ ∇ 푎 + 푎 ∙ ∇ 푏 + 푏 × ∇ × 푎 + 푎 × ∇ × 푏 (ix)∇ × ∇ × 푎 = ∇ ∇ ∙ 푎 − ∇2푎 . Vector Integration: Ordinary integrals of vector valued functions: Let 푓 푡 = 푓1 푡 푖 + 푓2 푡 푗 + 푓3 푡 푘 be a vector on a scalar variable 푡, where 푓1 푡 , 푓2 푡 , 푓3 푡 are continuous functions in some specific interval. Then 푓 푡 푑푡 = 푖 푓1 푡 푑푡 + 푗 푓2 푡 푑푡 + 푘 푓3 푡 푑푡is called an indefinite integral of 푓 푡 . If there exists a vector 푠 푡 , such that 푓 푡 = 푑 푠 푡 , 푑푡 푑 then 푓 푡 푑푡 = 푠 푡 푑푡 = 푠 푡 + 푐 , 푑푡 where푐 is an arbitrary constant vector independent of 푡. The definite integral between the limits 푡 = 푡0 and 푡 = 푡1 can in such case be written 푡1 푡1 푑 푓 푡 푑푡 = 푠 푡 푑푡 = 푠 푡1 − 푠 푡0 . 푡0 푡0 푑푡 Line Integrals: Let 푟 푡 = 푥 푡 푖 + 푦 푡 푗 + 푧 푡 푘 is the position vector of points 푃 푥, 푦, 푧 and suppose 푟 푡 defines a curve 퐶 joining points 푃1 and 푃2 where 푡 = 푡1and 푡 = 푡2, respectively. Let us assume that 퐶 composed of a finite number of curves for each of which 푟 푡 has a continuous derivative. Let 푎 = 푎1푖 + 푎2푗 + 푎3푘be a vector function of position defined and continuous along 퐶. Then the integral of the tangential component of 푎 along 퐶 form 푃1 to 푃2 푃2 written as 푎 ∙ 푑푟 = 푎 ∙ 푑푟 = 푎1푑푥 + 푎2푑푦 + 푎3푑푧 푃1 퐶 퐶 is a line integral. Note: If 퐶 be a closed curve, the integral around 퐶 is often denoted by 퐶 푎 ∙ 푑푟 = 퐶 푎1푑푥 + 푎2푑푦 + 푎3푑푧 . Surface Integral: Let 푆 be a two sided surface. Let one side of 푆 be considered arbitrarily as the positive side (If S is closed surface, such as sphere, then the outer side considered as positive side). A unit normal 푛 to any point of the positive side of 푆 is called a positive or outward drawn unit normal. Associate with differential surface area 푑푆 whose magnitude is 푑푆 and whose direction is that of 푛 . Then 푑푆 = 푛 푑푆. The integral 푆 푎 ∙ 푑푆 = 푆 푎 ∙ 푛 푑푆 is an example of a surface integral called flux of 푎 over 푆. Volume Integrals: Consider a closed surface in space enclosing a volume 푉, Then the following denote volume integrals or space integrals: 푉 푎 푑푉 or 푉 휑 푑푉. Divergence Theorem, Stoke’s theorem, Green’s theorem: Divergence theorem of Gauss: Let 푉 is the volume bounded by a closed surface 푆 and 푎 is a vector function of position with continuous derivatives. Then ∇ ∙ 푎 푑푉 = 푎 ∙ 푛 푑푆 = 푎 ∙ 푑푆 푆 푉 푆 Where 푛 is the positive (outward drawn) normal to 푆. Stoke’s theorem: Let 푆 be an open, two sided surface bounded by a closed nonintersecting curve 퐶(simple closed curve), and let 푎 is a vector function of position with continuous derivatives. Then 푎 ∙ 푑푟 = ∇ × 푎 ∙ 푛 푑푆 = ∇ × 푎 ∙ 푑푆 푆 퐶 푆 where퐶 is traversed in the positive direction. Note: The direction of 퐶is called positive if an observer, walking on the boundary of 푆 in that direction, with his head pointing in the direction of the positive normal to 푆, has the surface on his left. Green’s theorem in the plane: Let 푅be a closed regioninthe 푥푦 −plane bounded by a simple closed curve 퐶, and suppose 푀 and 푁 are continuous functions of 푥 and 푦 having continuous derivatives in 푅. Then 휕푁 휕푀 푀 푑푥 + 푁푑푦 = − 푑푥 푑푦. 퐶 푅 휕푥 휕푦