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Question Number 39025 by maxmathsup by imad last updated on 01/Jul/18

l e t f (x) = \frac{c o s (α x)}{c o s x} (2 π p e r i o d i c e v e n)

d e v e l o p p f a t f o u r i e r s e r i e .

Commented by math khazana by abdo last updated on 05/Jul/18

f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} c o s (n x) w i t h

a_{n} = \frac{2}{T} \int_{[T]} f (x) c o s (n x) d x

= \frac{2}{2 π} \int_{- π}^{π} \frac{c o s (α x) c o s (n x)}{c o s x} d x \Rightarrow

π a_{n} = \frac{1}{2} \int_{- π}^{π} \frac{c o s (n + α) x + c o s (n - α) x}{c o s x} d x

2 π a_{n} = \int_{- π}^{π} \frac{c o s (n + α) x}{c o s x} d x + \int_{- π}^{π} \frac{c o s (n - α) x}{c o s x} d x

l e t f i n d f i r s t I_{λ} = \int_{- π}^{π} \frac{c o s (λ x)}{c o s x} d x

I_{λ} = R e (\int_{- π}^{π} \frac{e^{i λ x}}{c o s x} d x) c h a n g e m e n t e^{i x} = z g i v e

\int_{- π}^{π} \frac{e^{i λ x}}{c o s x} d x = \int_{∣ z ∣= 1} \frac{2 z^{λ}}{z + z^{- 1}} \frac{d z}{i z}

= \int_{∣ z ∣= 1} \frac{- 2 i z^{λ}}{z^{2} + 1} d z l e t φ (z) = \frac{- 2 {i z}^{λ}}{z^{2} + 1}

\int_{∣ z ∣= 1} φ (z) d z = 2 i π {R e s (φ, i) + R e s (φ, - i)}

R e s (φ, i) = \frac{- 2 i i^{λ}}{2 i} = - i^{λ}

R e s (\bar{φ}, - i) = \frac{- 2 i {(- i)}^{λ}}{- 2 i} = {(- i)}^{λ} \Rightarrow

\int_{∣ z ∣= 1} φ (z) d z = 2 i π {- i^{λ} + {(- i)}^{λ}}

= - 2 i π {i^{λ} - {(- i)}^{λ}} = - 2 i π 2 i I m (i^{λ})

= 4 π I m (c o s (\frac{λ π}{2}) + i s i n (\frac{λ π}{2})} = 4 π s i n (\frac{λ π}{2}) \Rightarrow

I_{λ} = 4 π s i n (\frac{λ π}{2}) \Rightarrow

2 π a_{n} = 4 π s i n (\frac{(n + α) π}{2}) + 4 π s i n (\frac{(n - α) π}{2}) \Rightarrow

a_{n} = \frac{1}{2} {s i n (\frac{(n + α) π}{2}) + s i n (\frac{(n - α) π}{2})}

a_{0} = \frac{1}{π} \int_{- π}^{π} \frac{c o s (α x)}{c o s x} d x = \frac{1}{π} 4 π s i n (\frac{α π}{2})

= 4 s i n (\frac{α}{2}) \Rightarrow \frac{a_{0}}{2} = 2 s i n (\frac{α}{2}) \Rightarrow

f (x) = 2 s i n (\frac{α}{2}) + \frac{1}{2} \sum_{n = 1}^{\infty} {s i n (\frac{(n + α) π}{2}) + s i n (\frac{(n - α) π}{2})}