Analysis 2-Fourier Series The L2 Norm and Calculating Fourier Series, guaranteed and verified 100% Pass

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Fourier Series: The 𝐿2 Norm and Calculating Fourier Series


The Fourier series for a 2𝜋-periodic function, 𝑓, which is bounded and Riemann
integrable on [−𝜋, 𝜋] is given by:
𝑎0
+ ∑∞
𝑘=1(𝑎𝑘 cos 𝑘𝑥 + 𝑏𝑘 sin 𝑘𝑥)
2
where the Fourier coefficients are given by:
1 𝜋
𝑎𝑘 = ∫−𝜋 𝑓(𝑡 ) cos 𝑘𝑡 𝑑𝑡
𝜋
1 𝜋
𝑏𝑘 = 𝜋 ∫−𝜋 𝑓 (𝑡) sin 𝑘𝑡 𝑑𝑡 .
Note that:
1 𝜋 1 𝜋
|𝑎𝑘 | ≤ ∫−𝜋|𝑓(𝑡 ) cos 𝑘𝑡| 𝑑𝑡 ≤ ∫−𝜋|𝑓(𝑡)|𝑑𝑡
𝜋 𝜋
1 𝜋 1 𝜋
|𝑏𝑘 | ≤ ∫−𝜋|𝑓(𝑡) sin 𝑘𝑡 | 𝑑𝑡 ≤ ∫−𝜋|𝑓 (𝑡 )|𝑑𝑡.
𝜋 𝜋
Since 𝑓 is bounded:
1 𝜋 1 𝜋 1
∫ |𝑓 (𝑡 )|𝑑𝑡
𝜋 −𝜋
≤ 𝜋 ∫−𝜋‖𝑓‖∞ 𝑑𝑡 = 𝜋 (2𝜋)‖𝑓‖∞ = 2‖𝑓‖∞
Thus,
|𝑎𝑘 | ≤ 2‖𝑓‖∞
|𝑏𝑘 | ≤ 2‖𝑓‖∞ .


We will denote the partial sums of a Fourier series by:
𝑎
𝑆𝑛 (𝑓)(𝑥 ) = 20 + ∑𝑛𝑘=1(𝑎𝑘 cos 𝑘𝑥 + 𝑏𝑘 sin 𝑘𝑥).
Notice 𝑆𝑛 (𝑓) is a trig polynomial of degree at most 𝑛, or 𝑆𝑛 (𝑓) ∈ 𝑇𝑛 .
We will be interested in what sense 𝑆𝑛 (𝑓) converges to 𝑓 (Pointwise? Uniformly?
In 𝐿2 ?)

, 2


Recall that the functions: 1, cos 𝑥 , sin 𝑥 , cos 2𝑥 , sin 2𝑥 , … are orthogonal
with respect to the inner product:
𝜋
< 𝑓, 𝑔 > = ∫−𝜋 𝑓(𝑥 ) 𝑔(𝑥 ) 𝑑𝑥
𝜋 𝜋
since ∫−𝜋(cos 𝑚𝑥)(cos 𝑛𝑥) 𝑑𝑥 = ∫−𝜋 (sin 𝑚𝑥)(sin 𝑛𝑥) 𝑑𝑥
𝜋
= ∫−𝜋 cos 𝑚𝑥 sin 𝑛𝑥 𝑑𝑥 = 0
for 𝑚 ≠ 𝑛 (the last integral is 0 for 𝑚 = 𝑛 as well).

𝜋 𝜋
Also, ∫−𝜋 cos 2 𝑚𝑥 𝑑𝑥 = ∫−𝜋 sin2 𝑚𝑥 𝑑𝑥 = 𝜋 , for 𝑚 ≠ 0,
𝜋
∫−𝜋 1 𝑑𝑥 = 2𝜋.

There is nothing special about the interval [−𝜋, 𝜋]. If we have a periodic function
of period 2𝐿 instead of 2𝜋 then the Fourier series for 𝑓 becomes:
𝑎0 𝑘𝜋𝑥 𝑘𝜋𝑥
+ ∑∞
𝑘=1(𝑎𝑘 cos + 𝑏𝑘 sin )
2 𝐿 𝐿
where:
1 𝐿 𝑘𝜋𝑥
𝑎𝑘 = 𝐿 ∫−𝐿 𝑓(𝑥 ) cos 𝑑𝑥
𝐿
1 𝐿 𝑘𝜋𝑥
𝑏𝑘 = 𝐿 ∫−𝐿 𝑓(𝑥 ) sin 𝑑𝑥.
𝐿



Notice if 𝐿 = 𝜋 we get our original formulas. In fact, sometimes it’s easier
to express a function, 𝑓, of a period 2𝐿 by giving a formula for 𝑓 on an interval
[𝑐, 𝑐 + 2𝐿]. In that case, the formula for the series stays the same, but the
formulas for the coefficients become:
1 𝑐+2𝐿 𝑘𝜋𝑥
𝑎𝑘 = 𝐿 ∫𝑐 𝑓(𝑥 ) cos 𝑑𝑥
𝐿
1 𝑐+2𝐿 𝑘𝜋𝑥
𝑏𝑘 = 𝐿 ∫𝑐 𝑓(𝑥 ) sin 𝑑𝑥.
𝐿