Explanation The given series is, J v ( x ) = ∑ n = 0 ∞ ( − 1 ) n n ! Γ ( 1 + v + n ) ( x 2 ) 2 n + v The function to be proved as real is given as follows: I v ( x ) = i − v J v ( i x ) = i − v ∑ n = 0 ∞ ( − 1 ) n n ! Γ ( 1 + v + n ) ( i x 2 ) 2 n + v = ∑ n = 0 ∞ ( − 1 ) n n ! Γ ( 1 + v + n ) i 2 n + v − v ( x 2 ) 2 n + v = ∑ n = 0 ∞ ( − 1 ) n n ! Γ ( 1 + v + n ) ( i 2 ) n ( x 2 ) 2 n + v Further solve the above equation

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ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 5.3, Problem 21E

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To verify: The result Iv(x)=i−vJv(ix) is a real function, using the series Jv(x)=∑n=0∞(−1)nn!Γ(1+v+n)(x2)2n+v.

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Chapter 5.3, Problem 20E

Chapter 5.3, Problem 22E

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To verify: The result I v ( x ) = i − v J v ( i x ) is a real function, using the series J v ( x ) = ∑ n = 0 ∞ ( − 1 ) n n ! Γ ( 1 + v + n ) ( x 2 ) 2 n + v .

Solution Summary: The author explains that the function to be proved as real is I_v(x)=i-v

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To verify: The result Iv(x)=i−vJv(ix) is a real function, using the series Jv(x)=∑n=0∞(−1)nn!Γ(1+v+n)(x2)2n+v.

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Chapter 5 Solutions