Analysis 2-Fourier Series L2 Convergence and Parsevals, guaranteed and verified 100% Pass

1


Fourier Series: 𝐿2 Convergence and Parseval’s Identity



Suppose 𝑓 ∈ 𝑅[−𝜋, 𝜋], then 𝑓 2 ∈ 𝑅[−𝜋, 𝜋] as we saw earlier and:


1 𝜋 2 1 𝜋
∫−𝜋
(𝑓(𝑥)) 𝑑𝑥 ≤ ‖𝑓‖2∞ ∫−𝜋 1 𝑑𝑥 = 2‖𝑓‖2∞ < ∞ .
𝜋 𝜋


𝑐0
Let’s let 𝑇(𝑥 ) = + ∑𝑛𝑘=1(𝑐𝑘 cos 𝑘𝑥 + 𝑑𝑘 sin 𝑘𝑥).
2



For what values of 𝑐0 , 𝑐𝑘 and 𝑑𝑘 , 𝑘 = 1, … , 𝑛, is
1 𝜋
‖𝑓 − 𝑇‖22 = ∫−𝜋(𝑓 (𝑥 ) − 𝑇(𝑥))2 𝑑𝑥 minimized?
𝜋



Let’s show that for all 𝑛, 𝑇(𝑥 ) = 𝑆𝑛 (𝑓)(𝑥 ), where:


𝑎
𝑆𝑛 (𝑓)(𝑥 ) = 20 + ∑𝑛𝑘=1(𝑎𝑘 cos 𝑘𝑥 + 𝑏𝑘 sin 𝑘𝑥)

1 𝜋
𝑎𝑘 = 𝜋 ∫−𝜋 𝑓 (𝑥 ) cos 𝑘𝑥 𝑑𝑥
1 𝜋
𝑏𝑘 = ∫−𝜋 𝑓(𝑥 ) sin 𝑘𝑥 𝑑𝑥
𝜋


1 𝜋
minimizes
𝜋
∫−𝜋(𝑓(𝑥) − 𝑇(𝑥))2 𝑑𝑥 = ‖𝑓 − 𝑇‖22 .

, 2


Notice that:
𝜋 𝜋 𝜋 𝜋
∫−𝜋(𝑓 − 𝑇)2 𝑑𝑥 = ∫−𝜋 𝑓 2 𝑑𝑥 − 2 ∫−𝜋 𝑓 𝑇 𝑑𝑥 + ∫−𝜋 𝑇 2 𝑑𝑥


𝜋 𝜋 𝑐
∫−𝜋 𝑓 𝑇𝑑𝑥 = ∫−𝜋 𝑓 (𝑥 ) ( 20 + ∑𝑛𝑘=1(𝑐𝑘 cos 𝑘𝑥 + 𝑑𝑘 sin 𝑘𝑥)) 𝑑𝑥
𝑐 𝜋 𝜋
= 20 ∫−𝜋 𝑓(𝑥 ) 𝑑𝑥 + ∑𝑛𝑘=1(𝑐𝑘 ∫−𝜋 𝑓(𝑥 ) cos 𝑘𝑥 𝑑𝑥)
𝜋
+ ∑𝑛𝑘=1(𝑑𝑘 ∫−𝜋 𝑓 (𝑥 ) sin 𝑘𝑥 𝑑𝑥 )
𝑐0𝑎0
= 𝜋[ + ∑𝑛𝑘=1(𝑐𝑘 𝑎𝑘 + 𝑑𝑘 𝑏𝑘 )]
2




𝜋
∫−𝜋 𝑇 2 𝑑𝑥
𝜋 𝑐
= ∫−𝜋( 20 + ∑𝑛𝑘=1(𝑐𝑘 cos 𝑘𝑥 + 𝑑𝑘 sin 𝑘𝑥 ))
𝑐
( 20 + ∑𝑛𝑗=1(𝑐𝑗 cos 𝑗𝑥 + 𝑑𝑗 sin 𝑗𝑥)) 𝑑𝑥 .




Since 1, cos 𝑥 , sin 𝑥 , cos 2𝑥, sin 2𝑥 , … , cos 𝑛𝑥, sin 𝑛𝑥 are orthogonal and:
𝜋 𝜋 𝜋
∫−𝜋 cos 2 𝑘𝑥 𝑑𝑥 = ∫−𝜋 sin2 𝑘𝑥 𝑑𝑥 = 𝜋 , ∫−𝜋 1 𝑑𝑥 = 2𝜋




we get:
𝜋 𝑐2
∫−𝜋 𝑇 2 𝑑𝑥 = 𝜋 [ 20 + ∑𝑛𝑘=1(𝑐𝑘2 + 𝑑𝑘2 )] .