LINEAR ALGEBRA Matrices Linear Algebra Definitions m n matri with m rows and n columns is called an arra of mn real numbers The entr a a an A = a a an = ( a ij ) am am amn a ij denotes the element in the ith row and jth column If m = n then the arra is square, and A is then called square matri of order n In a square matri of order n, the diagonal containing elements a, a,, a nn is called principal or leading diagonal Diagonal matri D is a square matri that has its onl non-zero elements along the leading diagonal D = a 0 0 0 a 0 0 a nn 0 0 0 0 E = 0 A ver important special case of diagonal matri is the unit matri or identit matri E, for which a = a = = a = nn The zero or null matri is the matri with ever element zero The transposed matri interchanged T A = a a am a a am an an amn If the matri has one row or one column u = (,,, ) u u u n or v = T A of matri A is just the matri with rows and columns v v vn = ( v, v,, v ) then it is called a row vector or a column vector Basic Properties of Matrices Equalit A = B Two matrices A and B are said to be equal if and onl if all their corresponding elements are the same ( a = b for i, j) and the are of the same order m n ij ij n T - -
Linear Algebra Addition C = A + B We can onl add a m n matri to another m n matri, and an element of the sum is the sum of the corresponding elements C = ( c ij ) = ( aij + b ij ) for i, j Multiplication b a scalar k The matri ka has elements D = ka = ( d ij ) = ( ka ij ) for i, j Matri multiplication ka ij, ie, we just multipl each element b the scalar k If A is m p matri with elements a ij and B a p n matri with elements b ij then we define the product C = AB as the m n matri with elements C = ( c ij ) = ( ai b j + aibj + + aipbpj ) = aikb kj, for i=,,, mand j=,,, n p k = The ith row of A is multiplied term b term with the jth column of B to form the ijth component of C In order for multiplication to be possible, A must have p columns and B must have p rows otherwise their product AB is not defined Matri multiplication is not commutative in general: AB BA Properties of the transpose: From the definition, the transpose of matri is such that T T T ( A+ B) = A + B, T T T ( AB) = B A, T T ( A ) = A 4 Eample : Given the matrices A = and B = 0 Find the matrices A+B, A-B, B-A, A, 4B, A+4B, Solution: A+B = T T A + B 0 7 7, A-B =, B-A =, A = T A 6 9 4 0 0 9 0, 4B =, A+4B =, 6 0 8 4 4 5 4 =, 4 0 T B = 0, A T + B T = Eample: Given the matrices C = and D = Find the matrices K = CD and M = DC 0 0-4 -
Linear Algebra Solution: 9 9 5 9 K = CD =, M = DC =, 5 6 6 k = ++ =, m = + = 9, m = + = 9, k = +0+0 = 9, m = + = 5, m = +0 =, k = ++ = 5, m = +0 =, m = +0 =, k = +0+0 = m = +0 = 6, m = +0 = 6, m = +0 = CD DC Determinants Definition and Basic Properties The determinant of order n > a square matri A = a a an a a an an an ann is defined as a number det A = ( ) + n a D a D + + and n= = a an a a an a a, n + n a a + + ( ) an an ann an an ann an an, n We write det A = A = a a an a a an an an ann The commonl useful properties are as follows: Thus two rows (or columns) are the same, the determinant is zero We multipl a determinant b a number c if we multipl b this number c all the elements of a row or of a column Interchanging two rows (or columns) changes the sign of the determinant The addition rule: a a an a a an an an ann + Similarl for the columns b b b n a a an an an ann = a + b a + b an + b n a a an an an ann - 5 -
Linear Algebra Adding multiples of rows (or columns): a a an a a an an an ann a a a a + c a a a a a n n n n nn = a + ca a + ca an + can a a an an an ann Evaluation of Determinants For n =, is det A = a, a for n =, is det A = a a a = aa aa Eample : Evaluate the determinant det B = 4 5 Solution: det B = 5 4 = 5 - = -7 Sarrus s rule for a determinant of the third order: det A = a a a a a a a a a a a a a a = a a a a a a a a a a + = = aaa + aaa + aaa aaa aaa aaa Eample : Evaluate the determinant det C = 6 0 4 Solution: det C= 6 + 0 + 4(-) [4 + (-)6 + 0] = = 8 + 0-4 - ( 4 - + 0 ) = 4 - = In general for the determinant of order n > the Laplace s epansion according to the ith row holds: det A = n i ( ) + j j= a A = ij ij i+ i+ i+ n ( ) aiai + ( ) aiai + + ( ) ain A in, or the Laplace s epansion according to the jth column holds: det A = n i ( ) + j aij A ij = i= + j + j n+ j j j j j nj nj ( ) a A + ( ) a A + + ( ) a A Note that - 6 -
Linear Algebra ( ) i+ j is called a sign of element a ij, determinant A ij originating from det A omitting the ith row and jth column, is called minor of order n- of det A belonging to the element a ij, cofactor * A of the element a ij ij is defined as the minor A ij multiplied b the appropriate sign ( ) i+ j : * ij A = ( ) i + j A ij Eample : Evaluate the determinant det D = 0 0 4 4 5 Solution: The second column contains onl even numbers therefore we can put it in form of product *det D a We do the Laplace s epansion according to the second column in the second step 0 4 0 4 4 det D = = = (-)(-) + 6 = (Sarrus s rule) = 0 5 6 0 4 4 0 4 4 = - [88 + 48 + 8 - ( 44 + 8 + 96 )] = 8 (Sarrus s rule) or Laplace s epansion according to the third row: 4 4 det D = - 6 = - 6 = -(-) + 4 6 = 8 4 4 0 0 Matri Inversion The determinant of order k formed b the elements in the intersections of arbitrar k rows and a a an k columns of matri A = a a an am am a mn is called a minor of order k of the matri A ( k min (m, n)) A matri A is of rank h if and onl if there eists a minor of A of order h different from zero, an minor of A of order higher than h being equal to zero The square matri A is called regular, if det A 0, singular, if det A = 0 The inverse of the square matri A of order n is a square matri - 7 - - A of order n such that
Linear Algebra - - AA = A A = E, where E is unit matri The inverse matri A Then holds A = adj A, det A Note that - - of the square matri A eist if and onl if A is regular adjoint matri adj A is defined as the transpose of matri of cofactors adj A = * * * T * * * A A A n A A A n * * * * * * A A A n = A A An * * * * * * An An A nn An An A nn Eample 4: Find the inverse matri Solution: Determinant of matri A: det A = the matri of cofactors * A : ij - A of the matri A = 0 * A, that is ij 0 = + 6+ 0 + 0 =, + 0 + 0 + ( ) ( ) ( ) * A = + + + ( ) ( ) ( ) 4 5 6 ij =, 4 + + + ( ) ( ) ( ) 0 0 4 the adjoint matri adj A = 4 5 6 = 5, 4 6 4 - A = det adj A A - Test: A A = = 4 4 5 = 5 6 4 6 4 4 0 0 0 5 = 0 0 6 4 0 0 Sstems of Linear Equations Definition B a sstem of m linear equations in n unknowns we understand the sstem T - 8 -
Linear Algebra a + a + + a = b a + a + + a = b a + a + + a = b n n n n m m mn n m where,,, n are called unknowns, real numbers a ij (i =,,, m, j =,,, n) are called coefficients,, real numbers b i are called right-hand side, if real numbers b i = 0 (i =,,, m) sstem of linear equations is called homogeneouse, if eists b i 0, sstem of linear equations is called nonhomogeneouse matri A = matri X = b b matri B = b m a a an a a an am am amn n is called solution vector, is called right-hand side vector, is called matri of the sstem, a a an b a a an b matri A B = is called augmented matri of the sstem a m am amn b m Then a sstem of equations can be rewritten in the matri form AX = B Theorem of Frobenius: The sstem of m linear equations in n unknowns is solvable if and onl if rank of matri of the sstem is equal to rank of the augmented matri of the sstem: h(a) = h(a B) = h If h = n, then sstem has onl one solution, If h < n, then sstem has n-h linearl independent solutions and ever solution of this sstem is a linear combination of this n-h solutions The homogeneouse sstem of m linear equations in n unknowns AX = O 0 0 a) has alwas the trivial (zero) solution X =, 0-9 -
b) has a non-trivial solution if and onl if det A = 0 Linear Algebra Gaussian Elimination Given sstem of n linear equations in n unknowns,,, n we solve in a serie of steps: The essence of the Gaussian elimination consists in transforming the sstem of m linear equations in n unknowns into equivalent sstem (possesses the same solution) whose a a an b 0 a augmented matri is upper triangular, ie the matri an b 0 0 amn b m Elimination procedures rel on the manipulation of equations or, equivalentl, the rows of the augmented matri of the sstem There are various elementar row operations used which do not alter the solution of the equations: multipl a row b a constant, interchange an two row, add or subtract one row from another We solve the last equation for n We then solve the penultimate equation and eliminate n in terms of We repeat the process in turn n, n,, which we can then solve Eample : Solve the sstem of equations: + + 5 = 9 + = 6 = 5 Solution: m = n = b Σ operation 5-9 - - 5 r -r -6-5 r -r 5-9 - 0 - - 6 0 - -6 5 4 r -4r 5-9 - 0 - - 6 0 0-8 8 0 h(a) = h(a B) = n =, equivalent upper triangular augmented matri of the sstem: n - 0 -
Linear Algebra () + + 5 = 9 () = () 8 = 8 we solve =, + we solve then (): + ( ) = = =, and we solve finall (): = 9 5 = 9+ 6+ 5= Eample : Solve the sstem of equations Solution: m = 4, n = b Σ operation 6 - - - r -r - r -r - r 4 -r 6 0 0-4 -4-8 r r 0-4 - -5 0 - -4-5 -0 6 0-4 - -5 0 0-4 -4-8 0 - -4-5 -0 r 4 +r 6 0-4 - -5 0 0-4 -4-8 0 0-8 -7-5 r 4 -r 6 0-4 - -5 0 0-4 -4-8 0 0 0 + + = + = + = + = h(a) =, h(a B) = 4 h(a) h(a B) sstem has no solution (last row: 0 + 0 + 0 = is not true) Eample : Solve the sstem of equations: Solution: homogeneous sstém with m =, n = 4 4 + 4 = 0 + + 54 = 0 + = 0 4 - -
Linear Algebra 4 b Σ operation - - 0 - -5 0 r -r 4 - - 0 r -4r - - 0-0 7 5-9 0 r 9 0 9 5-9 0 5 r (-7) - - 0-0 6 45-8 0 7 r :9 0-6 -5 6 0-5 r +r - - 0-0 7 5-9 0 0 0 0-8 0-8 h(a) = h(a B) =, n = 4 n-h = 4 = linearl independent solution 4 equivalent upper triangular augmented matri of the sstem: () + 4 = 0 () 7 + 5 94 = 0 () 0 8 = 0 4 9 p () 0 8p= 0 =, 5 9 p -5 () 7 + 5 9p = 0 9 p -9p = = = 0, 7 7 () + p = 0 8 p 8 p = p+ + = p + +0 = 5 5 For eample: p= 0 : = 0, = 0, = 0, 4 = 0 (trivial solution), p= 5 : = 8, = 0, = 9, = 5, Cramer s Rule 4 p = 5 : = 8, = 0, = 9, = 5 4 The sstem of n linear equations in n unknowns,,,n a + a + + a = b a + a + + a = b a + a + + a = b n n n n n n nn n n with a regular matri of the sstem has a unique solution,,, n, = p where det D i i =, i =,,, n, det D - -
Linear Algebra here det D is the determinant of the matri of the sstem and det D i is the determinant obtaining b replacing the ith column of det D b the column of elements forming the righthand side of equations + + = 5 Eample4: Solve the sstem of equations: 5 + = 7 7 + = 7 Solution: m = n = det D = 5 = - 76 ( 0) 0 7 5 5 5 det D = 7 5 = - 5, det D = 7 = 76, det D = 5 7 = 0, 7 7 0 7 0 7 7 det D 5 det D = = =, 76 det D = = =, 0 = = = 0 det D 76 det D 76 det D 76 - -
DIFFERENTIAL CALCULUS FUNCTIONS OF ONE REAL VARIABLE Functions of One Real Variable Definitions and Basic Properties A real-value function f relates each element of a set D( f ), with eactl one element of another set H( f ) We epress the relationship b the equation = f( ) or f: D( f ) is the domain of f and H( f ) is the range of f or codomain Smbol is an independent variable or argument of the function and the smbol is the dependent variable The graph of function is the set of ordered pairs [, ( )] f for D( f) Bounded functions We sa that the function f is bounded above on a set M D( f) if there eists a number h such that f( ) h M (Fig a), we sa that the function f is bounded below on a set M D( f ) if there eists a number d such that f( ) d M (Fig b), the function f is bounded provided it is bounded both from below and from above, it is if there eists a number k for which f( ) k M (Fig c) = h = - = - = sin = d Fig a Fig b Fig c = -k Monotonic functions We sa that the function f is increasing on an interval I D( f ) provided that for, I : < f( ) < f( ) (Fig a), we sa that the function f is decreasing on an interval I D( f ) provided that for, I : < f( ) > f( ) (Fig b), we sa that the function f is nondecreasing on an interval I D( f ) provided that for, I : < f( ) f( ) (Fig c), We sa that the function f is nonincreasing on an interval I D( f ) provided that for, I : < f( ) f( ) (Fig d) - 4 -
= - = - Fig a Fig b Fig c Fig d Periodic function We sa that the function f is periodic with period T on D( f ) provided that for D( f) : f( + T) = f( ) (Fig ) One-to-one function We sa that the function f is one-to-one on a set M D( f ) provided that for, M : f( ) f( ) (Fig 4) T = - = sin Fig Fig 4ˇ Composite function Let two functions = f( u) and u= g ( ) be given such that the domain of f( u ) intersects with the range of g ( ) Then the composite function = h ( ) is defined to be the function h, for which domain D consists of all Dg ( )) such that g ( ) lie in D( f ) and h ( ) = f( g ( )) for ever Dh ( ), (Fig 5) u = g() = f(u) D D = f(g()) Fig 5 Fig 6 H = e = = ln Inverse functions Let be an one-to-one function f with domain D( f ) and range H( f ), ie for ever H( f) there eists D( f) I for which = f( ) Then we define inverse function of f with domain H( f ) and range D( f ) b f( ) = Properties of inverse function: = f ( ) if and onl if - 5 -
Graph of the function Composition of inverse functions: f ( f( )) = for H( f), f is the reflection of the graph of f in the line = (Fig 6) For domain and range of the inverse functions is valid: D( f) H( f = ), H( f) = D( f ) Even and odd functions f ( f( )) = for D( f) We sa that the function f is even on an interval ( a, a) D( f) provided that for ( a, a) : f( ) = f( ) Graph of the even functions is smmetrical about vertical ais (Fig 7a) We sa that the function f is odd on an interval ( a, a) D( f ) provided that for ( a, a) : f( ) = f( ) Graph of the odd functions is smmetrical about the origin (Fig7b) = = - Fig 7a Fig 7b Elementar Functions Polnomial functions A polnomial function has the general form n n f( ) = an + an + + a + a0, where n is a positive integer and coefficients a i, i = 0,,,, n, a n 0 are real numbers The inde n is called the degree of the polnomial Domain D( f) n = 0: constant function (polnomial of degree 0): = a0 = k = R= (-, + ) Graph is a line parallel to -ais, constant function is not one-to-one function (Fig 8), = = + = - + Fig 8 Fig 9a Fig 9b n = : linear function (polnomial of degree ): = a+ a0 = k+ q Graph is a line with slope k = tgα which intersects -ais at [0, q] - 6 -
For k > 0 is linear function increasing (Fig 9a), for k < 0 is decreasing (Fig 9b), linear function is one-to-one n = : quadratic function (polnomial of degree ): = a + a + a0 = a + b + c Graph is a parabola, it is not one-to-one function (Fig 0a, b) For a > 0 it is bounded below (Fig 0a), for a < 0 it is bounded above (Fig 0b) = - + = - + + Fig 0a Fig 0b Rational function A rational function has the general form Pm ( ) = Qn ( ), where Pm ( ) and Qn ( ) are polnomials of degree m a n Domain D( f) R { R: Q ( ) 0} = = If m < n it is said to be a proper rational function, if m n it is improper rational function An improper rational function can alwas be epressed as a polnomial function plus a proper rational function b algebraic division Eponential function An eponential function has the general form where real constant a is called base, a > 0, a Domain D( f) = a, = R= (-, + ), range H( f) = R + = (0, + ) Graph of the eponential function is called eponential curve (Fig a, b, c) = = ( _ ) = e Fig a Fig b Fig c It is bounded below and it is increasing for a > (Fig a) and it is decreasing for 0< a < (Fig b), it is one-to-one function and it intersects -ais at [0, ] - 7 -
The standard eponential function is Let us recall some basic properties: r s r s a a = a +, Logarithmic functions a a r s r s = e, (Fig c), Euler s number e =,788 r s = a, ( a ) A logarithmic function has the general form where real constant a is called base, a > 0, a = log a, It is defined as inverse function of eponential function rs = a, = a, that is: 0 a = Domain D( f) = R + = (0, + ), range H( f) = R= (-, + ) Graph of the logarithmic function is called logarithmic curve (Fig a, b, c) = log = log _ 0 = ln Fig a Fig b Fig c For a > it is increasing (Fig a), for 0 < a < it is decreasing (Fig b), it is one-to-one function and it intersects -ais at [, 0] The inverse function of standard eponential function = = e is called the natural logarithmic function and is written = ln (Fig c) Let us recall some basic properties: for u > 0, v > 0 is valid: log auv = logau+ logav, loga a = ( a = a), u loga logau logav v =, log a = 0 ( 0 v a = ), logau = vlogau Trigonometric (circular) functions All trigonometric functions are periodic, hence trigonometric functions are not one-to-one Functions Domain D( f) = sin and = cos have basic properties: = R= (-, + ), range H( f ) = <-,>, hence the are bounded (Fig a, b) - 8 -
= sin = cos Fig a Fig b Functions are periodic with period T = π: sin( + kπ ) = sin, cos( + kπ ) = cos, k is integer number Function = sin is odd function: sin( ) = sin, Function = cos is even function: cos( ) = cos Functions sin = tg = and cos cos = cotg = have basic properties: sin Domain D (tg ) = R {(k+) π }, D(cotg ) = R {kπ }, k is integer number, range H(tg ) = H(cotg ) = R = (-, + ), hence the are not bounded (Fig 4a, b) Functions are periodic with period T = π: tg( + kπ) = tg, cotg( + kπ) = cotg, k is integer number Functions = tg and = cotg are odd functions: tg ( ) = tg, cotg ( ) = cotg Function Function π π = tg is increasing on the intervals ( k ), ( k + ) ( ) = cotg is decreasing on the intervals kπ, ( k+ ) π = tg = cotg Fig 4a Fig 4b Let us recall some basic properties: sin + cos =, tg cotg =, sin = sin cos, cos = cos - sin, sin = ( - cos ), cos = ( + cos ) - 9 -
Inverse trigonometric functions None of the trigonometric functions are one-to-one since the are periodic In order to define inverses, it is customar to restrict the domains in which the trigonometric functions are oneto-one as follows Function π π = sin is increasing on the interval <, >, hence it is one-to-one on this interval and it covers the range <-,> So its inverse function eists and is denoted b = arc sin π π We define = arcsin, <,>, if and onl if, = sin, <, > π π Domain D( arc sin ) = <,>, range H( arc sin ) = <, > Function = arc sin is increasing on its domain (Fig 5a) Function = cos is decreasing on the interval < 0,π >, hence it is one-to-one on this interval and covers the range <-, > So its inverse function eists and is denoted b = arccos We define arccos = arccos, <,>, if and onl if, = cos, < 0,π > Domain D( arccos ) = <,>, range H( arccos ) = < 0,π > Function =arccos is decreasing on its domain (Fig 5b) Fig 5a Fig 5b - 0 -
π π Function = tg is increasing on the interval,, hence it is one-to-one on this interval and it covers the range (-, + ) So its inverse function eists and is denoted b = arctan We define π π = arctan, (-, + ), if and onl if, = tg,, π π Domain D( arctan ) = (-, + ), range H( arctan ) =, Function Function = arctan is increasing on its domain (Fig 6a) = cotg is decreasing on the interval ( 0,π ), hence it is one-to-one on this interval and covers the range (-, + ) So its inverse function eists and is denoted b = arc cot We define = arccot, (-, + ) if and onl if, cotg =, ( ) 0,π Domain D( arccot ) = (-, + ) range H( arccot ) = ( 0,π ) Function = arc cot is decreasing on its domain (Fig 6b) Fig 6a Fig 6b Limits of Functions Definition Intuitive idea of limit: The notation lim f( ) = A a means that as gets close to a (but does not equal a), f( ) gets close to A (Fig 7) - -
Definition of limit: A function = f( ) is said to approach a limit A as approaches the value a if, given small quantit ε, i tis possible to find a positive number δ such that f( ) A < ε for all a < δ, a, (Fig 7) Notation: lim f( ) = A a Smbolic notation of limit: lim f ( ) = A for ε > 0 δ > 0, that for a D( f) is valid: a < δ f( ) A < ε A ε ε δ a - δ a δ δ a + = A + = f() = A - ε ε Two facts regarding limits must be kept in mind: Fig 7 a) The limit of a function as approaches a is independent of the value of the function at a Even though lim f ( ) eists, the value of the function at a ma be undefined or ma be the a same as the limit or ma be defined but different from the limit (Fig 8a, b, c) Fig 8a Fig 8b Fig 8c b) The limit is said to eist onl if the following condition is satisfied: The limit as approaches a from left, written lim f( ), equals the limit as approaches a a from right, written lim f( ) : lim f( ) = lim f( ) + + a a a Fig 9a, b show two cases where lim f () a does not eist - -
= _ -5 4 8 Fig 9a Fig9b Basic Rules Suppose that lim f( ) = A, lim g ( ) = B, than hold: a a There eists at most one A R such that lim f ( ) = A, a lim kf ( ) = k lim f ( ) = ka, k R, a a lim [ f( ) ] n = lim f ( ) A a = a lim [ ( ) ( )] n n, n N, f ± g = lim f ( ) ± lim g ( ) = A ± B, a a a lim [ ( ) ( )] f g a lim f( ) f( ) lim = a = A a g ( ) lim g ( ) B a = lim f ( ) lim g ( ) = AB, a a Limits of Selected Functions B 0 lim c = c, c R, lim = 0, a lim n = a n k, a R, n N, lim = 0, n > 0, a lim + lim 0 n = +, n N, =, ± n lim 0 + = +, (Fig 9a), lim e = 0, lim e = +, (Fig c), + lim ln =, lim ln = +, (Fig c), + 0 + sin lim =, 0 lim + = e, - -
a lim + = e a Eample Find the limits: a) d) lim 6 + D = + Solution: a) A = = b) B = A = lim, B = lim +, e) + 6 E = lim + +, ( ) lim = ( )( + ) lim + = =, + 4 4, c) lim C =, c) C = lim = lim = lim = 0 0 = 0, 6 + 6 + 6 + 6 d) D = lim = lim = lim = = + + + + + + e) + + 6 + + 4 4 lim lim lim = + + + + + + E = = == + + 4 4 4 = lim + = lim + + = e 4 = e 4 + + + + +, Continuit of Functions A function = f( ) is said to be continuous at a point a if lim f ( ) = f ( a ) a Note that the definition requires three conditions to be satisfied: a) f( ) must be defined at the point a, b) f( ) must have a limit at point a, c) this limit must be equal to the value f( a ) We sa that a function f( ) is continuous on open interval (a, b), if it is continuous at each point of the interval (a, b) - 4 -
Fig 0a Fig 0b Consider the two functions (Fig 0a, b) Function f( ) in Fig 0a we can draw the whole curve without lifting the pencil from the paper, but this is not possible for the function g ( ) in Fig 0b Hence the function f( ) is continuous everwhere, while the function g ( ) is continuous on intervals (, 0) and (0, + ), and it has a discontinuit at = 0 Properties of continuous functions: Constant functions are continuous Sums, differences, and products of continuous functions are continuous Quotients and rational powers of continuous functions (where defined) are continuous The composite function f( g ( )) is continuous at a if g is continuous at a and f is continuous at ga ( ) 4 The Derivative of Function 4 Definition Let = f( ) be a function and point 0 D(f) The derivative of the function f() at the point 0 is defined b limits f( ) f( lim 0) f( = lim 0 + h) f( 0) f( = lim 0) 0 0 h 0 h 0 df ( The derivative is denoted b smbols: f ( 0), ( 0), 0) d( 0), d d If derivative f ( 0 ) is a finite number, we sa that function f( ) is differentiable at 0 Let f( ) be a function differentiable at an point from (a, b) Then we sa that the function is differentiable on interval (a, b) The function f ( ) is called the derivative function of f( ) The process is called differentiation - 5 -
f( + h) 0 f( ) 0 T = f() S s t Fig 0 + h 0 Eample 4: Find the derivative of the function = f( ) = cat point 0 Solution: f ( 0 ) = f( + ) f( ) = lim 0 0 0 c c lim 0 = lim 0 0 Eample 4: Find the derivative of the function = at point 0 = lim 0 = 0 0 Solution: f ( 0 ) = = lim 0 = f( lim 0 + ) f( 0) 0 = ( lim 0 + ) ( 0) 0 = lim 0 = Geometric sense of derivative The derivative f ( 0 ) is the slope k t of the tangent t to the graph of function = f( ) at f( ) f( 0, 0 = T[ 0, f( 0 )]: k = kt = f ( 0 ) = lim 0), (Fig ) point T[ ] The tangent t to the graph of function = f( ) at point = T[ 0, f( 0 )] is described b equation k n = or f( 0) = f ( 0)( 0), 0 kt ( 0) The normal line n to the graph of function = f( ) at point = T[ 0, f( 0 )] has slope = = (while f ( k f 0) 0) and it is described b equation ( ) t 0-0 = k n ( - 0 ) or f ( 0) = ( 0) f ( ) Eample 4: Find the equations of the tangent and normal line of the function point T[,?] Solution: 0 = = 9, that is T[, 9], slope k t of the tangent t: = k, = () = = 6, the equation of the tangent t: 9 = 6( ), slope k n and the equation of the normal line n: t 0 0 k n =, 6 0 9 = ( ) 6 = at - 6 -
Phsical sense of derivative The derivative f ( t0) is the instantaneous velocit v of phsical point T[ t 0, f( t 0 )] f( t at time t 0 : vt ( 0) = f ( t0) = 0 + h) f( t0) h lim h 0 4 Basic Rules for differentiation Let f( ) and g ( ) be functions differentiable at point, let c R be a constant Then the f( ) function cf (, ) f( ) ± g ( ), f( ) gand ( ) g ( ), ( g ( ) 0) are differentiable at point The following hold: ( ( f( ) ± g( )) = f ( ) ± g ( ), ( f( ) g( )) = f ( ) g( ) + f( ) g ( ) and hence for g ( ) f( ) = g ( ) f ( ) g( ) f( ) g ( ), g ( ) the derivative of a composite function (the chain rule): = c: ( cf ( ) ) = cf ( ) Let g ( ) be a function differentiable at point and f be a function differentiable at point g, ( ) then the composite function f( g ( )) is differentiable at point, and ( f( g ( ))) = f ( g ( )) g ( ) If a function f( ) is differentiable at point 0, then f( ) is continuous at point 0 L Hospital rule (for limits of fraction with infinite denominator) Let for a R be lim f ( ) = 0, lim g ( ) = 0, respectivel a a lim f( ) =±, lim g ( ) =±, and lim f ( ) eists a a a g ( ) f( ) f( ) f ( ) Then also eists lim and it holds lim = lim a g ( ) a g ( ) a g ( ) 4 Derivatives of Selected Functions (c) = c = 0, for R, n n ( ) = n for R, for n R, hence ( ) = a a ( ) = a for (0,+ ), for a R, ( e ) = e for R, ( a ) = a ln a R, for a R, a > 0, a, - 7 -
( ln ) = for R, 0, (log a ) ln a for R, 0, for a R, a > 0, a, (sin ) = cos for R, (cos ) = sin for R, (tg ) = (cotg ) = (arcsin ) = cos (arccos ) = for R, kπ, sin for R, (k+) π, for (-,), for (-,), (arctan ) = for R, + (arc cot ) = for R + Eample 44: Find the derivatives of the function at point D(f): a) 4 = + 4+ Solution: b) 4 = 4 + 4+ 0 = 8 + 6 4 4 = + 5 Solution: 4 = 4 + 5 4, 4 4 4 = 4( ) + 5 = + 0 4 4 c) = ln Solution: = ( ) ln + (ln ) = ln + = ln + d) = e cos 4 4 Solution: = ( e ) cos + e (cos ) = e cos + e sin ) = e (cos sin ) e) 7 = ( )sin - 8 -
7 Solution: 7 ( ) ( ) 6 7 ( ) ( ) = sin + (sin ) = 7 6 sin + cos ln e) = (ln ) ln ( ) ln Solution: = = = ln ( ln ) ln = = = ( ) 4 4 4 f) = tg = sin cos (sin ) cos sin (cos ) cos cos sin ( sin ) Solution: = = = (cos ) (cos ) (cos ) f) = e + 5 Solution: = e, u = + 5: u u u + 5 + 5 = ( e ) u = e = e = e g) = sin Solution: = sin u, u = : = (sin u) u = cos u = cos = cos h) = sin Solution: = sin u, u = : = (sin u) u = cos u = cos = cos i) = sin = (sin ) Solution: = u, u = sin : = ( u ) u = u cos = sin cos j) 5 4 = ( ) 4 Solution: = u, 5 u = : 4 4 = ( u ) u = 4 u (0 ) = 5 4 5 4 = 4( ) (0 ) = 8( ) (5 ) i) = arcsin(4 + ) Solution: = arcsin u, u = 4 + : 4 = 8 = (4 + ) (4 + ) = (arcsin u) u = (8 ) = u 44 Differential of the Function Let f( ) be a function differentiable at point 0 The differential of the function f( ) at point 0 is called the linear function d0 = d( 0) = df ( 0) = f ( 0) d = ( 0) d, where d = 0 The differential d( ) is used to describe a small change in the dependent variable, and d is a small change in the independent variable - 9 -
45 Highes-Order Derivative The derivative = f ( ) of the function f( ) is a function itself We can compute the derivative ( ) = ( f ( )) of the function f ( ) f (), which is called second derivative of the function f( ) and it is denoted b d d f, f ( ),, d d d d f Analogousl ( ) = ( f ( )) = = f ( ) = = d d is called third derivative of the function f( ), etc Generall the n-th derivative of the function f( ) is defined b ( ( n ) ( n ) ) ( f ( ) ) = and denoted b Eample 45: Find all derivatives of the function ( n) ( n), f ( ), 4 n = 5 + 8 n d f( ) d, n n d d Solution: = 0 +,, (4) = 0, (5) = = 60, = 0, (6) = = 0 Eample 46: Find first and second derivative of the function = sin at point 0 = 0 Solution: = sin + cos, (0) = sin 0 + 0cos 0 = 0 + 0 = 0, (0) = cos 0 0sin 0 = 00 = 46 Parametric Differentiation We sa that a function is defined parametricall b = f( ), where = t () and = t (), t <a, b> is parameter d d For a derivative of this function is hold: = f ( ) = = dt = d d dt Eample 47: Find a derivative of the function = at ( sin t), = a( cos t), a> 0, t R Solution: = a( cos t), = asin t and then asin t sin t = = a( cos t) ( cos t) 5 Applications of the Derivatives 6 Basic Theorems The Mean Value Theorem - 0 -
Let = f( ) be a function differentiable on ( ab, ) and continuous on < ab, > Then there is f( b) f( a) at least one point c ( ab, ) such that f () c =, Fig b a Fig Fig Rolle s Theorem Let = f( ) be a function differentiable on ( ab, ) and continuous on < ab, > If f( a) = f( b), then there eists at least one point c ( ab, ) such that f () c = 0, Fig Cauch s Theorem Let functions f( ) and g ( ) be continuous on < ab, >, function f( ) be differentiable on interval (a, b), function g ( ) has a finite derivative g ( ) 0 on ( ab, ) Then there eists f() b f() a f () c a point c (a, b) such that = gb () ga () g () c 56 Monotonic Functions and Optimal Value From the Mean Value Theorem there is valid: Suppose that two functions f( ) and g ( ) are differentiable on ( ab, ) and continuous and bounded on < ab, > Then the following statements are true: If f ( ) > 0 for ( ab, ), then f( ) is increasing on ( ab,, ) if f ( ) < 0 for ( ab, ), then f( ) is decreasing on ( ab,, ) if f ( ) 0 for ( ab, ), then f( ) is nondecreasing on ( ab,, ) if f ( ) 0 for ( ab, ), then f( ) is nonincreasing on ( ab,, ) if f ( ) = 0 for ( ab, ), then f( ) is constant on ( ab, ) Definition: A function f( ) with domain D( f ) is said to have an absolute maimum (respectivel absolute minimum) at point 0 D( f) if f( ) f( 0 ), (respectivel if f( ) f( 0 ) for D( f) The number f( 0) ) is called the absolute maimum (respectivel absolute minimum) of f( ) on D( f ) The function f( ) is said to have a local or relative maimum (respectivel local or relative minimum) at point 0 D( f) if there is some open interval ( ab, ) D( f) containing 0 and f( 0) is the absolute maimum (respectivel absolute minimum) of - -
f( ) on ( ab, ) The number f( 0) is called a local or relative maimum (respectivel local or relative minimum) of f( ) on ( ab, ) An absolute maimum or absolute minimum of f( ) is called an absolute etreme of f() A local maimum or local minimum of f( ) is called a local etreme of f( ), Fig 4a,b) = f() = f() 0 Fig 4a 0 JINý James 476 Fig 4b Proposition: If a function f( ) has a local etreme at point 0, then either f ( ) = 0, or f ( ) does not eist (When searching for local etremes of a function f( ), in view of this result, it suffices to test onl those points 0, for which f ( 0 ) = 0 or f ( 0 ) does not eist These points are called critical or stationar point) The first derivative test for etreme Let = f( ) be a function continuous on ( ab, ) and critical point 0 ( ab, ) If f ( ) > 0 on ( a, 0) and f ( ) < 0 on ( 0, b ), then at point 0 there is a local maimum f( 0) of the function f() on ( ab, ) If f () < 0 on (a, 0 ) and f ( ) > 0 on ( 0, b ), then at point 0 there is a local minimum f( 0) of the function f( ) on ( ab, ) The second derivative test for etreme Suppose that f( ), f ( ), f ( ) eist on ( ab, ) and 0 ( ab, ) Let f ( 0 ) = 0 Then the following statements are true: If f ( ) > 0, then at point 0 there is a local minimum f( 0) of the function f( ) on ( ab, ) If f ( ) < 0, then at point 0 there is a local maimum f( 0) of the function f( ) on ( ab, ) Determination of absolute maimum or absolute minimum of function f( ) on < ab, >: Ever function f( ) continuous on < ab, > attains both its absolute maimum and absolute minimum there Therefore, if we determine absolute maimum and absolute minimum of function f( ), we proceed as follows: a) Find critical points of f( ) - -
b) Compute the values of f( ) at all critical points and f( a), f( b ) c) The largest value among them is absolute maimum, the least value among them is absolute minimum of the function f( ) on < ab, > 5 Conveit and Concavit of a Function We sa, that a function f( ) is conve at point 0, if there eists an interval ( 0 δ, 0 + δ) such that the graph of the function f( ) restricted to ( 0 δ, 0 + δ) lies above the tangent drawn at the point [[ 0, f( 0) ], Fig 5a If f( ) is conve at ever point of ( ab,, ) we sa that f( ) is conve (or concave up or concave upward) on ( ab, ) We sa, that a function f( ) is concave at point 0, if there eists a interval ( 0 δ, 0 + δ) such that the graph of the function f( ) restricted to ( 0 δ, 0 + δ) lies below the tangent drawn at the point [ 0, f( 0) ], Fig 5b If f( ) is concave at ever point of ( ab,, ) we sa that f( ) is concave (or concave down or concave downward) on ( ab, ) We sa, that a point [, ( )] 0 f 0 is a point of inflection (inflection point) of a function f( ) if there eists some δ > 0 such that either the graph of f( ) is conve on ( 0 δ, 0) and concave down on ( 0, 0 + δ ), Fig 5c, or the graph of f( ) is concave down on ( 0 δ, 0) and conve on ( 0, 0 + δ ) = f() 0 t t = f() [, f( )] 0 0 0 f( ) 0 Fig 5a Fig 5b Fig 5c 0 t = f() Proposition: Test of conveit and concavit and inflection point Suppose that f ( ) of the function f( ) eists on (a, b): If f ( ) > 0 for ( ab, ), then f( ) is conve on ( ab,, ) if f ( ) < 0 for ( ab, ), then f( ) is concave on ( ab,, ) if f ( 0 ) = 0 and f ( 0 ) 0 then [, ( )] Eample 5: Draw a graph of the function 0 f 0, is an inflection point of f( ) + = Solution: a) 0, than domain D( f ) = (-, 0) (0, + ) = R - {0} - -
( ) + b) ( ) = = smmetrical about the origin) c) + =, therefore the function is odd (its graph is + 0 for R, therefore there is no point of intersection with -ais 0 D( f ), therefore there is no point of intersection with -ais too d) ( + ) = =, = = 0, = 0, critical point: The function is increasing for > 0: (, ) (, + ), The function is decreasing for < 0: =, = > 0, > 0, >, At point = there is a local minimum: () =, At point = there is a local maimum, (-) = - < 0, < 0, <, (, + ) -{0} ( ) e) = = =, 4 4 equation = = 0 has no solution, therefore a function has no inflection point The function is conve for > 0: The function is concave for < 0: f) The graph of the function > 0, > 0, > 0, (0, + ), < 0, < 0, < 0, (, 0) + = is drawn on Fig 6 - = + Fig 6-4 -
Literature: [] BUBENÍK, F: Mathematics for Engineers Vdavatelství ČVUT, Praha997 ISBN 80-0-0574- [] Demlová, M, HAMHALTER,J: Calculus I Vdavatelství ČVUT, Praha 998, ISBN 80-0-00-0 [] HARSHBARGER, RJ-REYNOLDS, JJ: Calculus with Applications DCHeath and Compan, Leington990, ISBN 0-669-45- [4] JAMES, G: Modern Engineering Mathematics Addison-Wesle, 99 ISBN 0-0-805456 [5] JAMES, G: Advanced Modern Engineering Mathematics Addison-Wesle, 99 ISBN 0-0-5659-6 - 5 -