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Let n e N. You are going to prove that 1.3-5.. (2n – 1) = 27 2 -4 ·6 . .· (2n) 27 In := | cos2" 0 d0 (a) Relate In to the following integral (++) 2n Jn := {lz|=1}

Let n e N. You are going to prove that 1.3-5.. (2n – 1) = 27 2 -4 ·6 . .· (2n) 27 In := | cos2" 0 d0 (a) Relate In to the following integral (++) 2n Jn := {lz|=1}

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.6: The Matrix Of A Linear Transformation
Problem 29EQ
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Let n e N. You are going to prove that 1.3-5.. (2n – 1) = 27 2 -4 ·6 . .· (2n) 27 In := | cos?" 0 do (a) Relate In to the following integral 2n := {l2|=1} 1 (b) For an appropriate zo € C and an appropriate analytic function f(z), write Jn in the form {(2) dz . (c) Using the Cauchy Integral Formula, show that f(2) 2in g(2n) (z0) . (2n)! dz = =|=1} I+uz(0z – 2) (d) Put together the above pieces to obtain the value of In.
Transcribed Image Text:Let n e N. You are going to prove that 1.3-5.. (2n – 1) = 27 2 -4 ·6 . .· (2n) 27 In := | cos?" 0 do (a) Relate In to the following integral 2n := {l2|=1} 1 (b) For an appropriate zo € C and an appropriate analytic function f(z), write Jn in the form {(2) dz . (c) Using the Cauchy Integral Formula, show that f(2) 2in g(2n) (z0) . (2n)! dz = =|=1} I+uz(0z – 2) (d) Put together the above pieces to obtain the value of In.
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