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Problem 3. In this problem, we will determine a closed formula for the the n-th partial sum of a Fourier series of a function ƒ : (−π, π) → R. You will need the following trigonometric identity (which you do not have to prove): 1 +cosu+cos 2u + ... + cos nu = sin(n+)u 2 sin(u/2) Suppose that f has period 2, with the Fourier series f(x) ~ +) (ak cos(kx) + bk sin(kx)). k=1 If sn(x) denotes the n-th partial sum of this series, show that n 8,(x) := 0 + (a* cos(kr) + b² sin(kr)) = k=1 П L f(t) sin[(n+)(t)]dt. 2 sin((t-x))

Problem 3. In this problem, we will determine a closed formula for the the n-th partial sum of a Fourier series of a function ƒ : (−π, π) → R. You will need the following trigonometric identity (which you do not have to prove): 1 +cosu+cos 2u + ... + cos nu = sin(n+)u 2 sin(u/2) Suppose that f has period 2, with the Fourier series f(x) ~ +) (ak cos(kx) + bk sin(kx)). k=1 If sn(x) denotes the n-th partial sum of this series, show that n 8,(x) := 0 + (a* cos(kr) + b² sin(kr)) = k=1 П L f(t) sin[(n+)(t)]dt. 2 sin((t-x))

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Problem 3. In this problem, we will determine a closed formula for the the n-th partial sum of a Fourier series of a function f : (-π, π) → R. You will need the following trigonometric identity (which you do not have to prove): 1 + cosu + cos 2u + ... + cos nu = 2 sin(n+1)u 2 sin(u/2) Suppose that has period 2, with the Fourier series f(x) ~ +Σ(a* cos(kx) + b* sin(kx)). k=1 If sn(x) denotes the n-th partial sum of this series, show that bk Sn(x): == + Σ (a* cos(kx) + b² sin(kx)) = = = ["_ ƒ(t)³ sin[(n+)(t)]dt. πT 2 sin((t - x)) k=1
Transcribed Image Text:Problem 3. In this problem, we will determine a closed formula for the the n-th partial sum of a Fourier series of a function f : (-π, π) → R. You will need the following trigonometric identity (which you do not have to prove): 1 + cosu + cos 2u + ... + cos nu = 2 sin(n+1)u 2 sin(u/2) Suppose that has period 2, with the Fourier series f(x) ~ +Σ(a* cos(kx) + b* sin(kx)). k=1 If sn(x) denotes the n-th partial sum of this series, show that bk Sn(x): == + Σ (a* cos(kx) + b² sin(kx)) = = = ["_ ƒ(t)³ sin[(n+)(t)]dt. πT 2 sin((t - x)) k=1
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