if-1-cosx-a-0-2-n-1-a-n-cos-nx-calculate-a-0-and-a-n-

Question Number 38204 by prof Abdo imad last updated on 22/Jun/18

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

c a l c u l a t e a_{0} a n d a_{n}

Commented by math khazana by abdo last updated on 26/Jun/18

l e t ? d e v e l o p p a t f o u r i e r ? s e r i e t b e ? f u n c t i o n

f (x) = \frac{1}{c o s x} (2 π p e r i o d i c e v e n)

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

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 (n x)}{c o s x} d x \Rightarrow

π a_{n} = \int_{- π}^{π} \frac{c o s (n x)}{c o s x} d x = R e (\int_{- π}^{π} \frac{e^{i n 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 n x}}{c o s x} d x = \int_{∣ z ∣= 1} 2 \frac{z^{n}}{z + z^{- 1}} \frac{d z}{i z}

= \int_{∣ z ∣= 1} \frac{- 2 {i z}^{n}}{z^{2} + 1} d z l e t φ (z) = \frac{- 2 {i z}^{n}}{z^{2} + 1} t h e p o l e s o f

φ a r e i a n d - i, r e s i d u s t h e o r e m ? g i v e

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

= 2 i π \frac{- 2 i . i^{n}}{2 i} = - 2 i π i^{n} = - 2 π i^{n + 1} \Rightarrow

π a_{2 n} = R e (- 2 π i^{2 n + 1}) = R e (- 2 π i {(- 1)}^{n}) = 0

π a_{2 n + 1} = R e (- 2 π i^{2 n + 2}) = R e (2 π {(- 1)}^{n})

= 2 π {(- 1)}^{n} \Rightarrow a_{2 n + 1} = 2 {(- 1)}^{n} \Rightarrow a_{0} = 0 a n d

\frac{1}{c o s x} = 2 \sum_{n = 0}^{\infty} {(- 1)}^{n} c o s (2 n + 1) x