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The Taylor series of function f ( z ) = 1 z for around z 0 = 2 i . The Taylor expansion of function f ( z ) = ∑ n = 0 ∞ ( − 1 ) n ( z − 2 i ) n ( 2 i ) n + 1 _ and region of convergence is R = 2 _ . 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 ) = 1 z Calculate the derivative of the function with respect to z and substitute z 0 = 2 i in differential function. f ′ ( z ) = [ d d z ( 1 z ) ] = − 1 z 2 f ′ ( 2 i ) = − 1 ( 2 i ) 2 = 1 2 2 Again, differentiate the function f ′ ( z ) = − 1 z 2 with respect to z and substitute z 0 = 2 i in differential function. f ″ ( z ) = [ d d z ( − 1 z 2 ) ] = 2 z 3 f ″ ( 2 i ) = 2 ( 2 i ) 3 = − 2 2 3 i Again, differentiate the function f ″ ( z ) = 2 z 3 with respect to z and substitute z 0 = 2 i in differential function. f ″ ′ ( z ) = [ d d z ( 2 z 3 ) ] = − 3 ! z 4 f ″ ′ ( 2 i ) = − 3 ! ( 2 i ) 4 = − 3 ! 2 4 Again, differentiate the function f ″ ′ ( z ) = − 3 ! z 4 with respect to z and substitute z 0 = 2 i in differential function. f ″ ″ ( z ) = [ d d z ( − 3 ! z 4 ) ] = 4 ! ( 2 i ) 5 f ″ ″ ( 2 i ) = 4 ! ( 2 i ) 5 = − i ( 4 ! ) ( 2 ) 5 Substitute, z 0 = 2 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 ( 2 i ) + f ′ ( 2 i ) 1 ! ( z − 2 i ) + f ″ ( 2 i ) 2 ! ( z − 2 i ) 2 + f ″ ′ ( 2 i ) 3 ! ( z − 2 i ) 3 ⋅ ⋅ ⋅ Substitute, f ( 2 i ) , f ′ ( 2 i ) , f ″ ( 2 i ) , f ″ ′ ( 2 i ) , and f ″ ″ ( 2 i ) in f ( z ) = f ( 2 i ) + f ′ ( 2 i ) 1 ! ( z − 2 i ) + f ″ ( 2 i ) 2 ! ( z − 2 i ) 2 + f ″ ′ ( 2 i ) 3 ! ( z − 2 i ) 3 ⋅ ⋅ ⋅ . f ( z ) = 1 2 i + ( 1 2 2 ) 1 ! ( z − 2 i ) + ( 2 ! i 2 3 ) 2 ! ( z − 2 i ) 2 + ( − 3 ! 2 4 ) 3 ! ( z − 2 i ) 3 + ( − 4 ! i 2 5 ) 4 ! ( z − 2 i ) 4 ⋅ ⋅ ⋅ = ∑ n = 0 ∞ ( − 1 ) n n ! ( z − 2 i ) n ( 2 i ) n + 1 ( n ! ) Hence, the Taylor series expansion of f ( z ) = 1 z is f ( z ) = ∑ n = 0 ∞ ( − 1 ) n n ! ( z − 2 i ) n ( 2 i ) n + 1 ( n ! ) . Now calculate the radius of convergence by using Cauchy-Hadamard formula. The Cauchy-Hadamard formula is R = lim n → ∞ | a n a n + 1 | . From the above calculation, it can be observed that a n = ( − 1 ) n ( 2 i ) n + 1 , a n + 1 = ( − 1 ) n + 1 ( 2 i ) n + 2 and z 0 = 2 i . Substitute, a n = ( − 1 ) n ( 2 i ) n + 1 , a n + 1 = ( − 1 ) n + 1 ( 2 i ) n + 2 in the Cauchy-Hadamard formula R = lim n → ∞ | a n a n + 1 | . R = lim n → ∞ | ( − 1 ) n ( 2 i ) n + 1 ( − 1 ) n ( 2 i ) n + 2 | = lim n → ∞ | ( − 1 ) n ( 2 i ) n + 1 × ( 2 i ) n + 2 ( − 1 ) n | = lim n → ∞ | 2 i | = 2 Therefore, the Taylor expansion of function f ( z ) = ∑ n = 0 ∞ ( − 1 ) n ( z − 2 i ) n ( 2 i ) n + 1 _ and region of convergence is R = 2 _ .

The Taylor series of function f ( z ) = 1 z for around z 0 = 2 i . The Taylor expansion of function f ( z ) = ∑ n = 0 ∞ ( − 1 ) n ( z − 2 i ) n ( 2 i ) n + 1 _ and region of convergence is R = 2 _ . 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 ) = 1 z Calculate the derivative of the function with respect to z and substitute z 0 = 2 i in differential function. f ′ ( z ) = [ d d z ( 1 z ) ] = − 1 z 2 f ′ ( 2 i ) = − 1 ( 2 i ) 2 = 1 2 2 Again, differentiate the function f ′ ( z ) = − 1 z 2 with respect to z and substitute z 0 = 2 i in differential function. f ″ ( z ) = [ d d z ( − 1 z 2 ) ] = 2 z 3 f ″ ( 2 i ) = 2 ( 2 i ) 3 = − 2 2 3 i Again, differentiate the function f ″ ( z ) = 2 z 3 with respect to z and substitute z 0 = 2 i in differential function. f ″ ′ ( z ) = [ d d z ( 2 z 3 ) ] = − 3 ! z 4 f ″ ′ ( 2 i ) = − 3 ! ( 2 i ) 4 = − 3 ! 2 4 Again, differentiate the function f ″ ′ ( z ) = − 3 ! z 4 with respect to z and substitute z 0 = 2 i in differential function. f ″ ″ ( z ) = [ d d z ( − 3 ! z 4 ) ] = 4 ! ( 2 i ) 5 f ″ ″ ( 2 i ) = 4 ! ( 2 i ) 5 = − i ( 4 ! ) ( 2 ) 5 Substitute, z 0 = 2 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 ( 2 i ) + f ′ ( 2 i ) 1 ! ( z − 2 i ) + f ″ ( 2 i ) 2 ! ( z − 2 i ) 2 + f ″ ′ ( 2 i ) 3 ! ( z − 2 i ) 3 ⋅ ⋅ ⋅ Substitute, f ( 2 i ) , f ′ ( 2 i ) , f ″ ( 2 i ) , f ″ ′ ( 2 i ) , and f ″ ″ ( 2 i ) in f ( z ) = f ( 2 i ) + f ′ ( 2 i ) 1 ! ( z − 2 i ) + f ″ ( 2 i ) 2 ! ( z − 2 i ) 2 + f ″ ′ ( 2 i ) 3 ! ( z − 2 i ) 3 ⋅ ⋅ ⋅ . f ( z ) = 1 2 i + ( 1 2 2 ) 1 ! ( z − 2 i ) + ( 2 ! i 2 3 ) 2 ! ( z − 2 i ) 2 + ( − 3 ! 2 4 ) 3 ! ( z − 2 i ) 3 + ( − 4 ! i 2 5 ) 4 ! ( z − 2 i ) 4 ⋅ ⋅ ⋅ = ∑ n = 0 ∞ ( − 1 ) n n ! ( z − 2 i ) n ( 2 i ) n + 1 ( n ! ) Hence, the Taylor series expansion of f ( z ) = 1 z is f ( z ) = ∑ n = 0 ∞ ( − 1 ) n n ! ( z − 2 i ) n ( 2 i ) n + 1 ( n ! ) . Now calculate the radius of convergence by using Cauchy-Hadamard formula. The Cauchy-Hadamard formula is R = lim n → ∞ | a n a n + 1 | . From the above calculation, it can be observed that a n = ( − 1 ) n ( 2 i ) n + 1 , a n + 1 = ( − 1 ) n + 1 ( 2 i ) n + 2 and z 0 = 2 i . Substitute, a n = ( − 1 ) n ( 2 i ) n + 1 , a n + 1 = ( − 1 ) n + 1 ( 2 i ) n + 2 in the Cauchy-Hadamard formula R = lim n → ∞ | a n a n + 1 | . R = lim n → ∞ | ( − 1 ) n ( 2 i ) n + 1 ( − 1 ) n ( 2 i ) n + 2 | = lim n → ∞ | ( − 1 ) n ( 2 i ) n + 1 × ( 2 i ) n + 2 ( − 1 ) n | = lim n → ∞ | 2 i | = 2 Therefore, the Taylor expansion of function f ( z ) = ∑ n = 0 ∞ ( − 1 ) n ( z − 2 i ) n ( 2 i ) n + 1 _ and region of convergence is R = 2 _ .

ADVANCED ENGINEERING MATH W/ACCESS

ADVANCED ENGINEERING MATH W/ACCESS

10th Edition

ISBN: 9781119096023

Author: Kreyszig

Publisher: WILEY

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ADVANCED ENGINEERING MATH W/ACCESS

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