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 .

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Answer 1

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Chapter 6.4, Problem 21E

To determine

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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Answer 2

Question

Chapter 6.4, Problem 21E

To determine

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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Answer 3

Question

Chapter 6.4, Problem 21E

To determine

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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Answer 4

Question

Chapter 6.4, Problem 21E

To determine

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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Answer 5

Chapter 6.4, Problem 21E

To determine

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.

Answer 6

Chapter 6.4, Problem 21E

To determine

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.

Answer 7

Chapter 6.4, Problem 21E

To determine

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.

Answer 8

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Answer 9

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Answer 10

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Answer 11

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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