Explanation Given: The given equation is, J 1 / 2 ( x ) = 2 π x sin x . Approach: The solution to the Bessel equation of first kind of order v is given as, J p ( x ) = ∑ k = 0 ∞ ( − 1 ) k Γ ( k + 1 ) Γ ( p + k + 1 ) ( x 2 ) 2 k + p . The formulas are, Γ ( 1 2 ) = π . sin x = ∑ k = 0 ∞ ( − 1 ) k ( x ) 2 k + 1 ( 2 k + 1 ) ! . Calculation: Consider the solution to the Bessel equation of first kind of order p is, J p ( x ) = ∑ k = 0 ∞ ( − 1 ) k Γ ( k + 1 ) Γ ( p + k + 1 ) ( x 2 ) 2 k + p . As, Γ ( k + 1 ) = k ! , then the solution is given as, J p ( x ) = ∑ k = 0 ∞ ( − 1 ) k k ! Γ ( p + k + 1 ) ( x 2 ) 2 k + p For the order 1 2 the solution becomes, J 1 / 2 ( x ) = ∑ k = 0 ∞ ( − 1 ) k k ! Γ ( 1 2 + k + 1 ) ( x 2 ) 2 k + 1 2 = ( x 2 ) − 1 2 ∑ k = 0 ∞ ( − 1 ) k k ! Γ ( 1 2 + k + 1 ) ( x 2 ) 2 k + 1 Consider, Γ ( 1 + 1 2 + k ) = ( k + 1 2 ) Γ ( k + 1 2 ) = ( k + 1 2 ) ( k + 1 2 − 1 ) Γ ( k + 1 2 − 2 ) = ( k + 1 2 ) ( k + 1 2 − 1 )

Differential Equations And Linear Algebra, Books A La Carte Edition (4th Edition)

4th Edition

ISBN: 9780321985811

Author: Stephen W. Goode, Scott A. Annin

Publisher: Pearson (edition 4)

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Chapter 11.6, Problem 9P

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The equation J1/2(x)=2πxsinx.

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************* ********************************* Q.1) Classify the following statements as a true or false statements: a. If M is a module, then every proper submodule of M is contained in a maximal submodule of M. b. The sum of a finite family of small submodules of a module M is small in M. c. Zz is directly indecomposable. d. An epimorphism a: M→ N is called solit iff Ker(a) is a direct summand in M. e. The Z-module has two composition series. Z 6Z f. Zz does not have a composition series. g. Any finitely generated module is a free module. h. If O→A MW→ 0 is short exact sequence then f is epimorphism. i. If f is a homomorphism then f-1 is also a homomorphism. Maximal C≤A if and only if is simple. Sup Q.4) Give an example and explain your claim in each case: Monomorphism not split. b) A finite free module. c) Semisimple module. d) A small submodule A of a module N and a homomorphism op: MN, but (A) is not small in M.

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Differential Equations And Linear Algebra, Books A La Carte Edition (4th Edition)

Ch. 11.1 - True-False Review For Questions a-j, decide if the...Ch. 11.1 - Prob. 2TFR Ch. 11.1 - Prob. 3TFR Ch. 11.1 - Prob. 4TFR Ch. 11.1 - Prob. 5TFR Ch. 11.1 - True-False Review For Questions a-j, decide if the...Ch. 11.1 - Prob. 7TFR Ch. 11.1 - Prob. 8TFR Ch. 11.1 - Prob. 9TFR Ch. 11.1 - Prob. 10TFR

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The equation J 1 / 2 ( x ) = 2 π x sin x .

Solution Summary: The author explains the solution to the Bessel equation of first kind of order v.

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The equation J1/2(x)=2πxsinx.

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The equation J 1 / 2 ( x ) = 2 π x sin x .

Solution Summary: The author explains the solution to the Bessel equation of first kind of order v.

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To show: The equation J1/2(x)=2πxsinx.

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Chapter 11 Solutions

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The equation J1/2(x)=2πxsinx.