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The Taylor series of function f ( z ) = z 4 for around z 0 = i . The Taylor expansion of function is f ( z ) = 1 − 4 i ( z − i ) − 6 ( z − i ) 2 + 4 i ( z − i ) 3 + ( z − i ) 4 _ and region of convergence is R = 1 _ . Formula used: Taylor series : The expansion of function f ( z ) = f ( z 0 ) + f ′ ( z 0 ) 1 ! ( z − z 0 ) + f ″ ( z 0 ) 2 ! ( z − z 0 ) 2 + ⋅ ⋅ ⋅ and the radius of convergence of a power series f ( z ) at center z 0 is the distance from z 0 to the nearest singular point of f ( z ) . Calculation: The given equation is: f ( z ) = z 4 Calculate the derivative of the function with respect to z and substitute z 0 = i in differential function. f ′ ( z ) = [ d d z ( z 4 ) ] = [ 4 z 3 ] f ′ ( i ) = 4 ( i ) 3 = − 4 i Again, differentiate the function f ′ ( z ) = 4 z 3 with respect to z and substitute z 0 = i in differential function. f ″ ( z ) = [ d d z ( 4 z 3 ) ] = 12 z 2 f ″ ( i ) = 12 ( i ) 2 = − 12 Again, differentiate the function f ″ ( z ) = 12 z 2 with respect to z and substitute z 0 = i in differential function. f ″ ′ ( z ) = [ d d z ( 12 z 2 ) ] = 24 z f ″ ′ ( π 2 ) = 24 ( i ) = 24 i Again, differentiate the function f ″ ′ ( z ) = 24 z with respect to z and substitute z 0 = i in differential function. f ″ ″ ( z ) = [ d d z ( 24 z ) ] = 24 f ″ ″ ( i ) = 24 Substitute, z 0 = i in the Taylor series function f ( z ) = f ( z 0 ) + f ′ ( z 0 ) 1 ! ( z − z 0 ) + f ″ ( z 0 ) 2 ! ( z − z 0 ) 2 + ⋅ ⋅ ⋅ . f ( z ) = f ( i ) + f ′ ( i ) 1 ! ( z − i ) + f ″ ( i ) 2 ! ( z − i ) 2 + f ″ ′ ( i ) 3 ! ( z − i ) 3 ⋅ ⋅ ⋅ Substitute, f ( i ) , f ′ ( i ) , f ″ ( i ) , and f ″ ′ ( i ) in f ( z ) = f ( i ) + f ′ ( i ) 1 ! ( z − i ) + f ″ ( i ) 2 ! ( z − i ) 2 + f ″ ′ ( i ) 3 ! ( z − i ) 3 ⋅ ⋅ ⋅ . f ( z ) = i 4 + ( − 4 i ) ( z − i ) + ( − 12 ) 2 ! ( z − i ) 2 + ( 24 i ) 3 ! ( z − i ) 3 + ( 24 ) 4 ! ( z − i ) 4 = 1 − 4 i ( z − i ) − 6 ( z − i ) 2 + 4 i ( z − i ) 3 + ( z − i ) 4 Hence, the Taylor series expansion of f ( z ) = z 4 is f ( z ) = 1 − 4 i ( z − i ) − 6 ( z − i ) 2 + 4 i ( z − i ) 3 + ( z − i ) 4 _ . Now calculate the radius of convergence of the series. The Taylor series of the function f ( z ) = z 4 is f ( z ) = 1 − 4 i ( z − i ) − 6 ( z − i ) 2 + 4 i ( z − i ) 3 + ( z − i ) 4 . The general term of the series is ( i + ( z − i ) ) 4 . Let f ( z ) = ( i + ( z − i ) ) 4 . Then, | z n + 1 z n | = | ( i + ( z − i ) ) 5 ( i + ( z − i ) ) 4 | = | ( i + ( z − i ) ) | = | z | The series is convergent if lim n → ∞ | z n + 1 z n | < 1 , that is | z | < 1 . Therefore, the radius of convergence of the series is R = 1 _ Hence, the Taylor expansion of function is f ( z ) = 1 − 4 i ( z − i ) − 6 ( z − i ) 2 + 4 i ( z − i ) 3 + ( z − i ) 4 _ and region of convergence is R = 1 _ .

The Taylor series of function f ( z ) = z 4 for around z 0 = i . The Taylor expansion of function is f ( z ) = 1 − 4 i ( z − i ) − 6 ( z − i ) 2 + 4 i ( z − i ) 3 + ( z − i ) 4 _ and region of convergence is R = 1 _ . Formula used: Taylor series : The expansion of function f ( z ) = f ( z 0 ) + f ′ ( z 0 ) 1 ! ( z − z 0 ) + f ″ ( z 0 ) 2 ! ( z − z 0 ) 2 + ⋅ ⋅ ⋅ and the radius of convergence of a power series f ( z ) at center z 0 is the distance from z 0 to the nearest singular point of f ( z ) . Calculation: The given equation is: f ( z ) = z 4 Calculate the derivative of the function with respect to z and substitute z 0 = i in differential function. f ′ ( z ) = [ d d z ( z 4 ) ] = [ 4 z 3 ] f ′ ( i ) = 4 ( i ) 3 = − 4 i Again, differentiate the function f ′ ( z ) = 4 z 3 with respect to z and substitute z 0 = i in differential function. f ″ ( z ) = [ d d z ( 4 z 3 ) ] = 12 z 2 f ″ ( i ) = 12 ( i ) 2 = − 12 Again, differentiate the function f ″ ( z ) = 12 z 2 with respect to z and substitute z 0 = i in differential function. f ″ ′ ( z ) = [ d d z ( 12 z 2 ) ] = 24 z f ″ ′ ( π 2 ) = 24 ( i ) = 24 i Again, differentiate the function f ″ ′ ( z ) = 24 z with respect to z and substitute z 0 = i in differential function. f ″ ″ ( z ) = [ d d z ( 24 z ) ] = 24 f ″ ″ ( i ) = 24 Substitute, z 0 = i in the Taylor series function f ( z ) = f ( z 0 ) + f ′ ( z 0 ) 1 ! ( z − z 0 ) + f ″ ( z 0 ) 2 ! ( z − z 0 ) 2 + ⋅ ⋅ ⋅ . f ( z ) = f ( i ) + f ′ ( i ) 1 ! ( z − i ) + f ″ ( i ) 2 ! ( z − i ) 2 + f ″ ′ ( i ) 3 ! ( z − i ) 3 ⋅ ⋅ ⋅ Substitute, f ( i ) , f ′ ( i ) , f ″ ( i ) , and f ″ ′ ( i ) in f ( z ) = f ( i ) + f ′ ( i ) 1 ! ( z − i ) + f ″ ( i ) 2 ! ( z − i ) 2 + f ″ ′ ( i ) 3 ! ( z − i ) 3 ⋅ ⋅ ⋅ . f ( z ) = i 4 + ( − 4 i ) ( z − i ) + ( − 12 ) 2 ! ( z − i ) 2 + ( 24 i ) 3 ! ( z − i ) 3 + ( 24 ) 4 ! ( z − i ) 4 = 1 − 4 i ( z − i ) − 6 ( z − i ) 2 + 4 i ( z − i ) 3 + ( z − i ) 4 Hence, the Taylor series expansion of f ( z ) = z 4 is f ( z ) = 1 − 4 i ( z − i ) − 6 ( z − i ) 2 + 4 i ( z − i ) 3 + ( z − i ) 4 _ . Now calculate the radius of convergence of the series. The Taylor series of the function f ( z ) = z 4 is f ( z ) = 1 − 4 i ( z − i ) − 6 ( z − i ) 2 + 4 i ( z − i ) 3 + ( z − i ) 4 . The general term of the series is ( i + ( z − i ) ) 4 . Let f ( z ) = ( i + ( z − i ) ) 4 . Then, | z n + 1 z n | = | ( i + ( z − i ) ) 5 ( i + ( z − i ) ) 4 | = | ( i + ( z − i ) ) | = | z | The series is convergent if lim n → ∞ | z n + 1 z n | < 1 , that is | z | < 1 . Therefore, the radius of convergence of the series is R = 1 _ Hence, the Taylor expansion of function is f ( z ) = 1 − 4 i ( z − i ) − 6 ( z − i ) 2 + 4 i ( z − i ) 3 + ( z − i ) 4 _ and region of convergence is R = 1 _ .

Advanced Engineering Mathematics

Advanced Engineering Mathematics

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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Chapter 15 Solutions

Advanced Engineering Mathematics

Ch. 15.1 - Prob. 1P Ch. 15.1 - Prob. 2P Ch. 15.1 - Prob. 3P Ch. 15.1 - Prob. 4P Ch. 15.1 - Prob. 5P Ch. 15.1 - Prob. 6P Ch. 15.1 - Prob. 7P Ch. 15.1 - Prob. 8P Ch. 15.1 - Prob. 9P Ch. 15.1 - Prob. 10P

Ch. 15.1 - Prob. 12P Ch. 15.1 - Prob. 13P Ch. 15.1 - Prob. 14P Ch. 15.1 - Prob. 15P Ch. 15.1 - Prob. 16P Ch. 15.1 - Prob. 17P Ch. 15.1 - Prob. 18P Ch. 15.1 - Prob. 19P Ch. 15.1 - Prob. 20P Ch. 15.1 - Prob. 21P Ch. 15.1 - Prob. 22P Ch. 15.1 - Prob. 23P Ch. 15.1 - Prob. 24P Ch. 15.1 - Prob. 25P Ch. 15.1 - Prob. 26P Ch. 15.1 - Prob. 27P Ch. 15.1 - Prob. 29P Ch. 15.1 - Prob. 30P Ch. 15.2 - Prob. 1P Ch. 15.2 - Prob. 2P Ch. 15.2 - Prob. 3P Ch. 15.2 - Prob. 4P Ch. 15.2 - Prob. 5P Ch. 15.2 - Prob. 6P Ch. 15.2 - Prob. 7P Ch. 15.2 - Prob. 8P Ch. 15.2 - Prob. 9P Ch. 15.2 - Prob. 10P Ch. 15.2 - Prob. 11P Ch. 15.2 - Prob. 12P Ch. 15.2 - Prob. 13P Ch. 15.2 - Prob. 14P Ch. 15.2 - Prob. 15P Ch. 15.2 - Prob. 16P Ch. 15.2 - Prob. 17P Ch. 15.2 - Prob. 18P Ch. 15.3 - Prob. 1P Ch. 15.3 - Prob. 2P Ch. 15.3 - Prob. 3P Ch. 15.3 - Prob. 4P Ch. 15.3 - Prob. 5P Ch. 15.3 - Prob. 6P Ch. 15.3 - Prob. 7P Ch. 15.3 - Prob. 8P Ch. 15.3 - Prob. 9P Ch. 15.3 - Prob. 10P Ch. 15.3 - Prob. 11P Ch. 15.3 - Prob. 12P Ch. 15.3 - Prob. 13P Ch. 15.3 - Prob. 14P Ch. 15.3 - Prob. 15P Ch. 15.3 - Prob. 16P Ch. 15.3 - Prob. 17P Ch. 15.3 - Prob. 18P Ch. 15.3 - Prob. 19P Ch. 15.4 - Prob. 1P Ch. 15.4 - Prob. 2P Ch. 15.4 - Prob. 3P Ch. 15.4 - Prob. 4P Ch. 15.4 - Prob. 5P Ch. 15.4 - Prob. 6P Ch. 15.4 - Prob. 7P Ch. 15.4 - Prob. 8P Ch. 15.4 - Prob. 9P Ch. 15.4 - Prob. 10P Ch. 15.4 - Prob. 11P Ch. 15.4 - Prob. 12P Ch. 15.4 - Prob. 13P Ch. 15.4 - Prob. 14P Ch. 15.4 - Prob. 16P Ch. 15.4 - Prob. 18P Ch. 15.4 - Prob. 19P Ch. 15.4 - Prob. 20P Ch. 15.4 - Prob. 21P Ch. 15.4 - Prob. 22P Ch. 15.4 - Prob. 23P Ch. 15.4 - Prob. 24P Ch. 15.4 - Prob. 25P Ch. 15.5 - Prob. 2P Ch. 15.5 - Prob. 3P Ch. 15.5 - Prob. 4P Ch. 15.5 - Prob. 5P Ch. 15.5 - Prob. 6P Ch. 15.5 - Prob. 7P Ch. 15.5 - Prob. 8P Ch. 15.5 - Prob. 9P Ch. 15.5 - Prob. 10P Ch. 15.5 - Prob. 11P Ch. 15.5 - Prob. 12P Ch. 15.5 - Prob. 13P Ch. 15.5 - Prob. 14P Ch. 15.5 - Prob. 15P Ch. 15.5 - Prob. 16P Ch. 15.5 - Prob. 17P Ch. 15 - Prob. 1RQ Ch. 15 - Prob. 2RQ Ch. 15 - Prob. 3RQ Ch. 15 - Prob. 4RQ Ch. 15 - Prob. 5RQ Ch. 15 - Prob. 6RQ Ch. 15 - Prob. 7RQ Ch. 15 - Prob. 8RQ Ch. 15 - Prob. 9RQ Ch. 15 - Prob. 10RQ Ch. 15 - Prob. 11RQ Ch. 15 - Prob. 12RQ Ch. 15 - Prob. 13RQ Ch. 15 - Prob. 14RQ Ch. 15 - Prob. 15RQ Ch. 15 - Prob. 16RQ Ch. 15 - Prob. 17RQ Ch. 15 - Prob. 18RQ Ch. 15 - Prob. 19RQ Ch. 15 - Prob. 20RQ Ch. 15 - Prob. 21RQ Ch. 15 - Prob. 22RQ Ch. 15 - Prob. 23RQ Ch. 15 - Prob. 24RQ Ch. 15 - Prob. 25RQ Ch. 15 - Prob. 26RQ Ch. 15 - Prob. 27RQ Ch. 15 - Prob. 28RQ Ch. 15 - Prob. 29RQ Ch. 15 - Prob. 30RQ

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