Explanation Given: The given condition is: The fact that E * is defined as: E * = ∫ − π π f 2 ( x ) d x − π ( a 0 2 2 + a 1 2 + ⋯ + a N 2 + b 1 2 + ⋯ + b N 2 ) . Formula: The Fourier series formula is: f ( x ) = a 0 2 + ∑ n = 1 ∞ { a n cos n π x L + b n sin n π x L } Proof: The function f ( x ) is continuous on [ π , − π ] The associate Fourier series is; f ( x ) = a 0 2 + ∑ n = 1 ∞ { a n cos n π x L + b n sin n π x L } Let f ( x ) be a square-integral function on the interval [ π , − π ] . ∫ − π π [ f ( x ) ] 2 d x < ∞ Let, s w ( x ) = a 0 + ∑ n = 1 N { a n cos n x + b n sin n x } Thus, we get: [ f ( x ) − s w ( x ) ] 2 = f 2 ( x ) − 2 s w ( x ) f ( x ) + s w 2 ( x ) Integrate both sides, we get: ∫ − π π [ f ( x ) − s w ( x ) ] 2 d x = ∫ − π π f 2 ( x ) − 2 s w ( x ) f ( x ) + s w 2 ( x ) d x = ∫ − π π f 2 ( x ) d x − 2 ∫ − π π s w ( x ) f ( x ) d x + ∫ − π π s w 2 ( x ) d x Consider the integral ∫ − π π s w 2 ( x ) d x : ∫ − π π s w 2 ( x ) d x = ∫ − π π ( a 0 2 + ∑ n = 1 N { a n cos n x + b n sin n x } ) 2 d x = a 0 2 4 ∫ − π π d x + ∑ n = 1 N a n 2 ∫ − π π cos 2 n x d x + b n 2 ∫ − π π sin 2 n x d x = a 0 2 4 ∫ − π π d x + 1 2 ∑ n = 1 N a n 2 [ ( 1 + cos 2 n x ) ] − π π + 1 2 b n 2 [ ( 1 − cos 2 n x ) ] − π π = a 0 2 4 ( 2 x ) + 1 2 ∑ n = 1 N a n 2 ( 2 x ) + 1 2 b n 2 ( 2 x ) Further simplify as: a 0 2 4 ( 2 π ) + 1 2 ∑ n = 1 N a n 2 ( 2 π ) + 1 2 b n 2 ( 2 π ) = π [ a 0 2 2 + ∑ n = 1 N a n 2 + b n 2 ] Consider the integral ∫ − π π f ( x ) s w ( x ) d x :

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 10.3, Problem 38E

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To prove:

The Bessel’s inequality a022+∑n=1∞{an2+bn2}≤∫−ππf2(x)dx for the given condition.

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Chapter 10.3, Problem 37E

Chapter 10.3, Problem 39E

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To prove: The Bessel’s inequality a 0 2 2 + ∑ n = 1 ∞ { a n 2 + b n 2 } ≤ ∫ − π π f 2 ( x ) d x for the given condition.

Solution Summary: The author explains the Bessel's inequality a_02+displaystyle

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To prove:

The Bessel’s inequality a022+∑n=1∞{an2+bn2}≤∫−ππf2(x)dx for the given condition.

Want to see the full answer?

Chapter 10 Solutions

To prove: The Bessel’s inequality a 0 2 2 + ∑ n = 1 ∞ { a n 2 + b n 2 } ≤ ∫ − π π f 2 ( x ) d x for the given condition.

Solution Summary: The author explains the Bessel's inequality a_02+displaystyle

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To prove: The Bessel’s inequality a022+∑n=1∞{an2+bn2}≤∫−ππf2(x)dx for the given condition.

Want to see the full answer?

Chapter 10 Solutions

To prove:

The Bessel’s inequality a022+∑n=1∞{an2+bn2}≤∫−ππf2(x)dx for the given condition.