developp-at-fourier-serie-f-x-3-1-2cosx-by-use-of-two-methods-

Question Number 143380 by Mathspace last updated on 13/Jun/21

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

f (x) = \frac{3}{1 + 2 c o s x}

b y u s e o f t w o m e t h o d s

Answered by mathmax by abdo last updated on 14/Jun/21

method 1 f (x) = \frac{3}{1 + 2 cosx} = \frac{3}{1 + 2 \frac{e^{ix} + e^{- ix}}{2}} = \frac{3}{1 + e^{ix} + e^{- ix}}

=_{e^{ix} = z} \frac{3}{1 + z + z^{- 1}} = \frac{3 z}{z + z^{2} + 1} = \frac{3 z}{z^{2} + z + 1} = Ψ (z)

z^{2} + z + 1 = 0 \to Δ = - 3 \Rightarrow z_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{2 i π}{3}} and z_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{2 i π}{3}}

\Rightarrow Ψ (z) = \frac{3 z}{(z - z_{1}) (z - z_{2})} = \frac{3 z}{z_{1} - z_{2}} (\frac{1}{z - z_{1}} - \frac{1}{z - z_{2}})

= \frac{3}{i \sqrt{3}} (\frac{z}{z - z_{1}} - \frac{z}{z - z_{2}}) = - i \sqrt{3} (\frac{z - z_{1} + z_{1}}{z - z_{1}} - \frac{z - z_{2} + z_{2}}{z - z_{2}})

= - i \sqrt{3} (\frac{z_{1}}{z - z_{1}} - \frac{z_{2}}{z - z_{2}}) = i \sqrt{3} (\frac{z_{1}}{z_{1} - z} - \frac{z_{2}}{z_{2} - z})

= i \sqrt{3} {\frac{1}{1 - \frac{z}{z_{1}}} - \frac{1}{1 - \frac{z}{z_{2}}}} we have ∣ \frac{z}{z_{1}} ∣=∣ \frac{z}{z_{2}} ∣= 1 \Rightarrow

Ψ (z) = i \sqrt{3} (\sum_{n = 0}^{\infty} \frac{z^{n}}{z_{1}^{n}} - \sum_{n = 0}^{\infty} \frac{z^{n}}{z_{2}^{n}})

= i \sqrt{3} \sum_{n = 0}^{\infty} (e^{- \frac{2 in π}{3}} + e^{\frac{2 in π}{3}}) z^{n}

= i \sqrt{3} \sum_{n = 0}^{\infty} 2 \cos (\frac{2 n π}{3}) z^{n} = 2 i \sqrt{3} \sum_{n = 0}^{\infty} \cos (\frac{2 n π}{3}) e^{inx}

= 2 i \sqrt{3} \sum_{n = 0}^{\infty} \cos (\frac{2 n π}{3}) (\cos (nx) + isin (nx))

= 2 i \sqrt{3} \sum_{n = 0}^{\infty} \cos (\frac{2 n π}{3}) \cos (nx) - 2 \sqrt{3} \sum_{n = 0}^{\infty} \cos (\frac{2 n π}{3}) \sin (nx)

but Ψ (z) = f (x) real \Rightarrow f (x) = - 2 \sqrt{3} \sum_{n = 0}^{\infty} \cos (\frac{2 n π}{3}) \sin (nx)