let g(x)=ln(cos(ax)) developp g at fourier serie (a real given)


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Question Number 116556 by Bird last updated on 04/Oct/20

$l e t g (x) = l n (c o s (a x))$ $d e v e l o p p g a t f o u r i e r s e r i e$ $(a r e a l g i v e n)$
	Answered by maths mind last updated on 05/Oct/20
	$let f(t)=ln(cos(t)) t∈]−(π/2),(π/2)[,cos(t)=((e^(it) +e^(−it) )/2) f(t)=ln(((e^(it) +e^(−it) )/2))=ln(1+e^(2it) )−ln(2e^(it) ) =Σ_(k≥0) (((−1)^k e^(2i(k+1)t) )/((k+1)))−ln(2)−it f(t)=f(−t)⇒f(t)=((f(t)+f(−t))/2) ⇒f(t)=(1/2)[Σ_(k≥0) (((−1)^k )/((k+1)))e^(2i(k+1)t) −ln(2)−it+Σ_(k≥0) (((−1)^k )/((k+1)))e^(−2i(k+1)t) −ln(2)+it] =−ln(2)+Σ_(k≥0) (((−1)^k )/((k+1))){((e^(2i(k+1)t) +e^(−2i(k+1)t) )/2)} =−ln(2)+Σ_(k≥0) (((−1)^k )/((k+1)))cos(2(k+1)t) g(x)=f(ax),∀x∈]−∣(π/(2a))∣,(π/(∣2a∣))[,∀a≠0 g(x)=−ln(2)+Σ_(k≥0) (((−1)^k )/(k+1))cos(2a(k+1)x)$
	$l e t$ $f (t) = l n (c o s (t))$ $t \in] - \frac{π}{2}, \frac{π}{2} [, c o s (t) = \frac{e^{i t} + e^{- i t}}{2}$ $f (t) = l n (\frac{e^{i t} + e^{- i t}}{2}) = l n (1 + e^{2 i t}) - l n (2 e^{i t})$ $= \sum_{k ⩾ 0} \frac{{(- 1)}^{k} e^{2 i (k + 1) t}}{(k + 1)} - l n (2) - i t$ $f (t) = f (- t) \Rightarrow f (t) = \frac{f (t) + f (- t)}{2}$ $\Rightarrow f (t) = \frac{1}{2} [\sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{(k + 1)} e^{2 i (k + 1) t} - l n (2) - i t + \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{(k + 1)} e^{- 2 i (k + 1) t} - l n (2) + i t]$ $= - l n (2) + \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{(k + 1)} {\frac{e^{2 i (k + 1) t} + e^{- 2 i (k + 1) t}}{2}}$ $= - l n (2) + \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{(k + 1)} c o s (2 (k + 1) t)$ $g (x) = f (a x), \forall x \in] - ∣ \frac{π}{2 a} ∣, \frac{π}{∣ 2 a ∣} [, \forall a \neq 0$ $g (x) = - l n (2) + \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{k + 1} c o s (2 a (k + 1) x)$
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