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Explanation & Answer

Uff... I hope there are no mistakes. Please ask if something is unclear.
The Fourier series
∞
∞
𝑛=1
𝑛=1
𝑎0
+ ∑ 𝑎𝑛 cos 𝑛𝑡 + ∑ 𝑏𝑛 sin 𝑛𝑡
2
of a function 𝑔 defined on [−𝜋, 𝜋] has the coefficients
1 𝜋
1 𝜋
1 𝜋
𝑎0 = ∫ 𝑓(𝑡)𝑑𝑡 , 𝑎𝑛 = ∫ 𝑓(𝑡) cos 𝑛𝑡 𝑑𝑡 , 𝑏𝑛 = ∫ 𝑓(𝑡) sin 𝑛𝑡 𝑑𝑡 , 𝑛 ≥ 1.
𝜋 −𝜋
𝜋 −𝜋
𝜋 −𝜋
We are given the function 𝑓(𝑡) = 𝜋 − 𝑡, 0 < 𝑡 < 𝜋 defined on (0. 𝜋).
1. If we perform the even extension of 𝑓, the resulting function will be even and defined on (−𝜋, 𝜋).
When computing the coefficients for that even function, all 𝑏𝑛 become zeros while 𝑎𝑛 will be
𝑎0 =
2
𝜋
2 𝜋
2 𝜋
∫ 𝑓(𝑡)𝑑𝑡 , 𝑎𝑛 = ∫ 𝑓(𝑡) cos 𝑛𝑡 𝑑𝑡.
𝜋 0
𝜋 0
2
𝜋
1
Compute them: 𝑎0 = 𝜋 ∫0 (𝜋 − 𝑡)𝑑𝑡 = 𝜋 (𝜋𝑡 − 2 𝑡 2 )
2
𝑡=0
1
= 𝜋 (𝜋 2 − 2 𝜋 2 ) = 𝜋.
𝜋
𝜋
2
2 𝜋
(𝜋
𝑎𝑛 = ∫
− 𝑡) cos 𝑛𝑡 𝑑𝑡 = 2 ∫ cos 𝑛𝑡 𝑑𝑡 ...
