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The Steepest Descent steps of the given function for 3 steps. The Steepest Descent steps of the given function for 3 steps are x 1 = [ 4 3 , − 4 3 ] _ , x 2 = [ 8 9 , − 16 9 ] _ and x 3 = [ 28 27 , − 52 27 ] _ . Formula used: The gradient of f ( x ) is ∇ f ( x ) = [ ∂ f ∂ x , ∂ f ∂ y ] and next approximation is z ( t ) = x − t ∇ f ( x ) with function g ( t ) = f ( z ( t ) ) . Calculation: The function is given as f ( x ) = 2 x 1 2 + x 2 2 − 4 x 1 + 4 x 2 . The boundary condition for the function is given as x 0 = 0 . Differentiate the equation f ( x ) = 2 x 1 2 + x 2 2 − 4 x 1 + 4 x 2 with respect to x 1 . d d x 1 ( f ( x 1 ) ) = d d x 1 ( 2 x 1 2 + x 2 2 − 4 x 1 + 4 x 2 ) = 4 x 1 − 4 Again, differentiate the equation f ( x ) = 2 x 1 2 + x 2 2 − 4 x 1 + 4 x 2 with respect to x 2 . d d x 2 ( f ( x 2 ) ) = d d x 2 ( 2 x 1 2 + x 2 2 − 4 x 1 + 4 x 2 ) = 2 x 2 + 4 The gradient of f ( x ) is ∇ f ( x ) = [ ∂ f ∂ x , ∂ f ∂ y ] . Substitute the value of ∂ f ∂ x and ∂ f ∂ y in ∇ f ( x ) = [ ∂ f ∂ x , ∂ f ∂ y ] . ∇ f ( x ) = [ 4 x 1 − 4 , 2 x 2 + 4 ] Calculate the Steepest Descent first step. The first approximation is given as x 0 = 0 . Substitute x 0 = 0 in gradient of ∇ f ( x ) = [ 4 x 1 − 4 , 2 x 2 + 4 ] . ∇ f ( 0 , 0 ) = [ 4 × 0 − 4 , 2 × 0 + 4 ] = [ − 4 , 4 ] The next approximation is given as z 1 ( t ) = x 0 − t ∇ f ( x 0 ) . Substitute the value of x 0 and ∇ f ( 0 , 0 ) in z 1 ( t ) = x 0 − t ∇ f ( x 0 ) . z 1 ( t ) = [ 0 , 0 ] − t [ − 4 , 4 ] = [ 4 t , − 4 t ] The function of next approximation is given as g 1 ( t ) = f ( z 1 ( t ) ) . Substitute the value of z 1 ( t ) in g 1 ( t ) = f ( z 1 ( t ) ) . g 1 ( t ) = [ 2 ( 4 t ) 2 + ( − 4 t ) 2 − 4 ( 4 t ) + 4 ( − 4 t ) ] = 32 t 2 + 16 t 2 − 16 t − 16 t = 48 t 2 − 32 t Differentiate g 1 ( t ) = 48 t 2 − 32 t with respect to t. d d t ( g 1 ( t ) ) = d d t ( 48 t 2 − 32 t ) g 1 ′ ( t ) = 96 t − 32 Again differentiate g 1 ′ ( t ) = 96 t − 32 with respect to t . d d t ( g 1 ′ ( t ) ) = d d t ( 96 t − 32 ) g 1 ′ ′ ( t ) = 96 Therefore, g 1 ′ ′ ( t ) = 96 > 0 , substitute the value of g 1 ′ ( t ) = 0 in g 1 ′ ( t ) = 96 t − 32 . 96 t − 32 = 0 96 t = 32 t = 32 96 t = 1 3 Hence, g 1 ( t ) has a minimum at t = 1 3 . Substitute the value of t = 1 3 in z 1 ( t ) = [ 4 t , − 4 t ] for next approximation x 1 = z 1 ( t ) . x 1 = [ 4 × 1 3 , − 4 × 1 3 ] x 1 = [ 4 3 , − 4 3 ] Hence, the Steepest Descent first step x 1 = [ 4 3 , − 4 3 ] _ . The gradient of f ( x ) is ∇ f ( x ) = [ ∂ f ∂ x , ∂ f ∂ y ] . Substitute the value of ∂ f ∂ x and ∂ f ∂ y in ∇ f ( x ) = [ ∂ f ∂ x , ∂ f ∂ y ] . ∇ f ( x ) = [ 4 x 1 − 4 , 2 x 2 + 4 ] Calculate the Steepest Descent second step. The next approximation is given as x 1 = 4 3 , − 4 3 . Substitute x 1 = 4 3 , − 4 3 in gradient of ∇ f ( x ) = [ 4 x 1 − 4 , 2 x 2 + 4 ] . ∇ f ( 4 3 , − 4 3 ) = [ 4 × 4 3 − 4 , 2 × ( − 4 3 ) + 4 ] = [ 4 3 , 4 3 ] The next approximation is given as z 2 ( t ) = x 1 − t ∇ f ( x 1 ) . Substitute the value of x 0 and ∇ f ( 4 3 , − 4 3 ) in z 2 ( t ) = x 1 − t ∇ f ( x 1 ) . z 2 ( t ) = [ 4 3 , − 4 3 ] − t [ 4 3 , 4 3 ] = [ 4 3 − 4 3 t , − 4 3 − 4 3 t ] The function of next approximation is given as g 2 ( t ) = f ( z 2 ( t ) ) . Substitute the value of z 2 ( t ) in g 2 ( t ) = f ( z 2 ( t ) ) . g 2 ( t ) = [ 2 ( 4 3 − 4 3 t ) 2 + ( − 4 3 − 4 3 t ) 2 − 4 ( 4 3 − 4 3 t ) + 4 ( − 4 3 − 4 3 t ) ] = 16 3 t 2 − 32 9 t − 16 3 Differentiate g 2 ( t ) = 16 3 t 2 − 32 9 t − 16 3 with respect to t. d d t ( g 2 ( t ) ) = d d t ( 16 3 t 2 − 32 9 t − 16 3 ) g 2 ′ ( t ) = 32 3 t − 32 9 Again differentiate g 2 ′ ( t ) = 32 3 t − 32 9 with respect to t . d d t ( g 2 ′ ( t ) ) = d d t ( 32 3 t − 32 9 ) g 2 ′ ′ ( t ) = 32 3 Therefore, g 2 ′ ′ ( t ) = 32 3 > 0 , substitute the value of g 2 ′ ( t ) = 0 in g 2 ′ ( t ) = 32 3 t − 32 9 . 32 3 t − 32 9 = 0 32 3 t = 32 9 t = 32 9 × 3 32 t = 1 3 Hence, g 2 ( t ) has a minimum at t = 1 3 . Substitute the value of t = 1 3 in z 2 ( t ) = [ 4 3 − 4 3 t , − 4 3 − 4 3 t ] for next approximation x 2 = z 2 ( t ) . x 2 = [ 4 3 − 4 3 × 1 3 , − 4 3 − 4 3 × 1 3 ] x 2 = [ 4 3 − 4 9 , − 4 3 − 4 9 ] x 2 = [ 12 − 4 9 , − 12 − 4 9 ] x 2 = [ 8 9 , − 16 9 ] Hence, the Steepest Descent first step x 2 = [ 8 9 , − 16 9 ] _ . The gradient of f ( x ) is ∇ f ( x ) = [ ∂ f ∂ x , ∂ f ∂ y ] . Substitute the value of ∂ f ∂ x and ∂ f ∂ y in ∇ f ( x ) = [ ∂ f ∂ x , ∂ f ∂ y ] . ∇ f ( x ) = [ 4 x 1 − 4 , 2 x 2 + 4 ] Calculate the Steepest Descent second step. The next approximation is given as x 2 = ( 8 9 , − 16 9 ) . Substitute x 2 = ( 8 9 , − 16 9 ) in gradient of ∇ f ( x ) = [ 4 x 1 − 4 , 2 x 2 + 4 ] . ∇ f ( 8 9 , − 16 9 ) = [ 4 × 8 9 − 4 , 2 × − 16 9 + 4 ] = [ − 4 9 , 4 9 ] The next approximation is given as z 3 ( t ) = x 2 − t ∇ f ( x 2 ) . Substitute the value of x 2 and ∇ f ( 4 3 , − 4 3 ) in z 3 ( t ) = x 2 − t ∇ f ( x 2 ) . z 3 ( t ) = [ 8 9 , − 16 9 ] − t [ − 4 9 , 4 9 ] = [ 8 9 + 4 9 t , − 16 9 − 4 9 t ] The function of next approximation is given as g 3 ( t ) = f ( z 3 ( t ) ) . Substitute the value of z 3 ( t ) in g 3 ( t ) = f ( z 3 ( t ) ) . g 3 ( t ) = [ 2 ( 8 9 + 4 9 t ) 2 + ( − 16 9 − 4 9 t ) 2 − 4 ( 8 9 + 4 9 t ) + 4 ( − 16 9 − 4 9 t ) ] = 16 27 t 2 − 32 81 t − 160 27 Differentiate g 3 ( t ) = 16 27 t 2 − 32 81 t − 160 27 with respect to t. d d t ( g 3 ( t ) ) = d d t ( 16 27 t 2 − 32 81 t − 160 27 ) g 3 ′ ( t ) = 32 27 t − 32 81 Again differentiate g 3 ′ ( t ) = 32 27 t − 32 81 with respect to t . d d t ( g 3 ′ ( t ) ) = d d t ( 32 27 t − 32 81 ) g 3 ′ ′ ( t ) = 32 27 Therefore, g 3 ′ ′ ( t ) = 32 27 > 0 , substitute the value of g 3 ′ ( t ) = 0 in g 3 ′ ( t ) = 32 27 t − 32 81 . 32 27 t − 32 81 = 0 32 27 t = 32 81 t = 32 81 × 27 32 t = 1 3 Hence, g 3 ( t ) has a minimum at t = 1 3 . Substitute the value of t = 1 3 in z 3 ( t ) = [ 8 9 + 4 9 t , − 16 9 − 4 9 t ] for next approximation x 3 = z 3 ( t ) . x 3 = ( 8 9 + 4 9 × 1 3 , − 16 9 − 4 9 × 1 3 ) x 3 = ( 8 9 + 4 27 , − 16 9 − 4 27 ) x 3 = ( 24 + 4 27 , − 48 − 4 27 ) x 3 = ( 28 27 , − 52 27 ) Hence, the Steepest Descent first step x 3 = ( 28 27 , − 52 27 ) _ .

The Steepest Descent steps of the given function for 3 steps. The Steepest Descent steps of the given function for 3 steps are x 1 = [ 4 3 , − 4 3 ] _ , x 2 = [ 8 9 , − 16 9 ] _ and x 3 = [ 28 27 , − 52 27 ] _ . Formula used: The gradient of f ( x ) is ∇ f ( x ) = [ ∂ f ∂ x , ∂ f ∂ y ] and next approximation is z ( t ) = x − t ∇ f ( x ) with function g ( t ) = f ( z ( t ) ) . Calculation: The function is given as f ( x ) = 2 x 1 2 + x 2 2 − 4 x 1 + 4 x 2 . The boundary condition for the function is given as x 0 = 0 . Differentiate the equation f ( x ) = 2 x 1 2 + x 2 2 − 4 x 1 + 4 x 2 with respect to x 1 . d d x 1 ( f ( x 1 ) ) = d d x 1 ( 2 x 1 2 + x 2 2 − 4 x 1 + 4 x 2 ) = 4 x 1 − 4 Again, differentiate the equation f ( x ) = 2 x 1 2 + x 2 2 − 4 x 1 + 4 x 2 with respect to x 2 . d d x 2 ( f ( x 2 ) ) = d d x 2 ( 2 x 1 2 + x 2 2 − 4 x 1 + 4 x 2 ) = 2 x 2 + 4 The gradient of f ( x ) is ∇ f ( x ) = [ ∂ f ∂ x , ∂ f ∂ y ] . Substitute the value of ∂ f ∂ x and ∂ f ∂ y in ∇ f ( x ) = [ ∂ f ∂ x , ∂ f ∂ y ] . ∇ f ( x ) = [ 4 x 1 − 4 , 2 x 2 + 4 ] Calculate the Steepest Descent first step. The first approximation is given as x 0 = 0 . Substitute x 0 = 0 in gradient of ∇ f ( x ) = [ 4 x 1 − 4 , 2 x 2 + 4 ] . ∇ f ( 0 , 0 ) = [ 4 × 0 − 4 , 2 × 0 + 4 ] = [ − 4 , 4 ] The next approximation is given as z 1 ( t ) = x 0 − t ∇ f ( x 0 ) . Substitute the value of x 0 and ∇ f ( 0 , 0 ) in z 1 ( t ) = x 0 − t ∇ f ( x 0 ) . z 1 ( t ) = [ 0 , 0 ] − t [ − 4 , 4 ] = [ 4 t , − 4 t ] The function of next approximation is given as g 1 ( t ) = f ( z 1 ( t ) ) . Substitute the value of z 1 ( t ) in g 1 ( t ) = f ( z 1 ( t ) ) . g 1 ( t ) = [ 2 ( 4 t ) 2 + ( − 4 t ) 2 − 4 ( 4 t ) + 4 ( − 4 t ) ] = 32 t 2 + 16 t 2 − 16 t − 16 t = 48 t 2 − 32 t Differentiate g 1 ( t ) = 48 t 2 − 32 t with respect to t. d d t ( g 1 ( t ) ) = d d t ( 48 t 2 − 32 t ) g 1 ′ ( t ) = 96 t − 32 Again differentiate g 1 ′ ( t ) = 96 t − 32 with respect to t . d d t ( g 1 ′ ( t ) ) = d d t ( 96 t − 32 ) g 1 ′ ′ ( t ) = 96 Therefore, g 1 ′ ′ ( t ) = 96 > 0 , substitute the value of g 1 ′ ( t ) = 0 in g 1 ′ ( t ) = 96 t − 32 . 96 t − 32 = 0 96 t = 32 t = 32 96 t = 1 3 Hence, g 1 ( t ) has a minimum at t = 1 3 . Substitute the value of t = 1 3 in z 1 ( t ) = [ 4 t , − 4 t ] for next approximation x 1 = z 1 ( t ) . x 1 = [ 4 × 1 3 , − 4 × 1 3 ] x 1 = [ 4 3 , − 4 3 ] Hence, the Steepest Descent first step x 1 = [ 4 3 , − 4 3 ] _ . The gradient of f ( x ) is ∇ f ( x ) = [ ∂ f ∂ x , ∂ f ∂ y ] . Substitute the value of ∂ f ∂ x and ∂ f ∂ y in ∇ f ( x ) = [ ∂ f ∂ x , ∂ f ∂ y ] . ∇ f ( x ) = [ 4 x 1 − 4 , 2 x 2 + 4 ] Calculate the Steepest Descent second step. The next approximation is given as x 1 = 4 3 , − 4 3 . Substitute x 1 = 4 3 , − 4 3 in gradient of ∇ f ( x ) = [ 4 x 1 − 4 , 2 x 2 + 4 ] . ∇ f ( 4 3 , − 4 3 ) = [ 4 × 4 3 − 4 , 2 × ( − 4 3 ) + 4 ] = [ 4 3 , 4 3 ] The next approximation is given as z 2 ( t ) = x 1 − t ∇ f ( x 1 ) . Substitute the value of x 0 and ∇ f ( 4 3 , − 4 3 ) in z 2 ( t ) = x 1 − t ∇ f ( x 1 ) . z 2 ( t ) = [ 4 3 , − 4 3 ] − t [ 4 3 , 4 3 ] = [ 4 3 − 4 3 t , − 4 3 − 4 3 t ] The function of next approximation is given as g 2 ( t ) = f ( z 2 ( t ) ) . Substitute the value of z 2 ( t ) in g 2 ( t ) = f ( z 2 ( t ) ) . g 2 ( t ) = [ 2 ( 4 3 − 4 3 t ) 2 + ( − 4 3 − 4 3 t ) 2 − 4 ( 4 3 − 4 3 t ) + 4 ( − 4 3 − 4 3 t ) ] = 16 3 t 2 − 32 9 t − 16 3 Differentiate g 2 ( t ) = 16 3 t 2 − 32 9 t − 16 3 with respect to t. d d t ( g 2 ( t ) ) = d d t ( 16 3 t 2 − 32 9 t − 16 3 ) g 2 ′ ( t ) = 32 3 t − 32 9 Again differentiate g 2 ′ ( t ) = 32 3 t − 32 9 with respect to t . d d t ( g 2 ′ ( t ) ) = d d t ( 32 3 t − 32 9 ) g 2 ′ ′ ( t ) = 32 3 Therefore, g 2 ′ ′ ( t ) = 32 3 > 0 , substitute the value of g 2 ′ ( t ) = 0 in g 2 ′ ( t ) = 32 3 t − 32 9 . 32 3 t − 32 9 = 0 32 3 t = 32 9 t = 32 9 × 3 32 t = 1 3 Hence, g 2 ( t ) has a minimum at t = 1 3 . Substitute the value of t = 1 3 in z 2 ( t ) = [ 4 3 − 4 3 t , − 4 3 − 4 3 t ] for next approximation x 2 = z 2 ( t ) . x 2 = [ 4 3 − 4 3 × 1 3 , − 4 3 − 4 3 × 1 3 ] x 2 = [ 4 3 − 4 9 , − 4 3 − 4 9 ] x 2 = [ 12 − 4 9 , − 12 − 4 9 ] x 2 = [ 8 9 , − 16 9 ] Hence, the Steepest Descent first step x 2 = [ 8 9 , − 16 9 ] _ . The gradient of f ( x ) is ∇ f ( x ) = [ ∂ f ∂ x , ∂ f ∂ y ] . Substitute the value of ∂ f ∂ x and ∂ f ∂ y in ∇ f ( x ) = [ ∂ f ∂ x , ∂ f ∂ y ] . ∇ f ( x ) = [ 4 x 1 − 4 , 2 x 2 + 4 ] Calculate the Steepest Descent second step. The next approximation is given as x 2 = ( 8 9 , − 16 9 ) . Substitute x 2 = ( 8 9 , − 16 9 ) in gradient of ∇ f ( x ) = [ 4 x 1 − 4 , 2 x 2 + 4 ] . ∇ f ( 8 9 , − 16 9 ) = [ 4 × 8 9 − 4 , 2 × − 16 9 + 4 ] = [ − 4 9 , 4 9 ] The next approximation is given as z 3 ( t ) = x 2 − t ∇ f ( x 2 ) . Substitute the value of x 2 and ∇ f ( 4 3 , − 4 3 ) in z 3 ( t ) = x 2 − t ∇ f ( x 2 ) . z 3 ( t ) = [ 8 9 , − 16 9 ] − t [ − 4 9 , 4 9 ] = [ 8 9 + 4 9 t , − 16 9 − 4 9 t ] The function of next approximation is given as g 3 ( t ) = f ( z 3 ( t ) ) . Substitute the value of z 3 ( t ) in g 3 ( t ) = f ( z 3 ( t ) ) . g 3 ( t ) = [ 2 ( 8 9 + 4 9 t ) 2 + ( − 16 9 − 4 9 t ) 2 − 4 ( 8 9 + 4 9 t ) + 4 ( − 16 9 − 4 9 t ) ] = 16 27 t 2 − 32 81 t − 160 27 Differentiate g 3 ( t ) = 16 27 t 2 − 32 81 t − 160 27 with respect to t. d d t ( g 3 ( t ) ) = d d t ( 16 27 t 2 − 32 81 t − 160 27 ) g 3 ′ ( t ) = 32 27 t − 32 81 Again differentiate g 3 ′ ( t ) = 32 27 t − 32 81 with respect to t . d d t ( g 3 ′ ( t ) ) = d d t ( 32 27 t − 32 81 ) g 3 ′ ′ ( t ) = 32 27 Therefore, g 3 ′ ′ ( t ) = 32 27 > 0 , substitute the value of g 3 ′ ( t ) = 0 in g 3 ′ ( t ) = 32 27 t − 32 81 . 32 27 t − 32 81 = 0 32 27 t = 32 81 t = 32 81 × 27 32 t = 1 3 Hence, g 3 ( t ) has a minimum at t = 1 3 . Substitute the value of t = 1 3 in z 3 ( t ) = [ 8 9 + 4 9 t , − 16 9 − 4 9 t ] for next approximation x 3 = z 3 ( t ) . x 3 = ( 8 9 + 4 9 × 1 3 , − 16 9 − 4 9 × 1 3 ) x 3 = ( 8 9 + 4 27 , − 16 9 − 4 27 ) x 3 = ( 24 + 4 27 , − 48 − 4 27 ) x 3 = ( 28 27 , − 52 27 ) Hence, the Steepest Descent first step x 3 = ( 28 27 , − 52 27 ) _ .

ADVANCED ENGINEERING MATH W/ACCESS

ADVANCED ENGINEERING MATH W/ACCESS

10th Edition

ISBN: 9781119096023

Author: Kreyszig

Publisher: WILEY

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Ch. 22.1 - Prob. 1P Ch. 22.1 - Prob. 2P Ch. 22.1 - Prob. 3P Ch. 22.1 - Prob. 4P Ch. 22.1 - Prob. 5P Ch. 22.1 - Prob. 6P Ch. 22.1 - Prob. 7P Ch. 22.1 - Prob. 8P Ch. 22.1 - Prob. 9P Ch. 22.1 - CAS EXPERIMENT. Steepest Descent. (a) Write a...

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