......advanced calculus...... when ∣z∣<1 and:: Ω:=((sin(x))/(z^2 +2z cos(x)+1)) =Σ_(n=0) ^∞ a_n z^n are satisfied , then solve , a


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Question Number 140141 by mnjuly1970 last updated on 04/May/21

$. . . . . . a d v a n c e d c a l c u l u s . . . . . .$ $w h e n ∣ z ∣< 1 a n d ::$ $Ω := \frac{s i n (x)}{z^{2} + 2 z c o s (x) + 1} = \underset{n = 0}{\sum^{\infty}} a_{n} z^{n}$ $a r e s a t i s f i e d, t h e n s o l v e, a_{n} . . .$ $. . . . . . . . . . . . . . . . . .$
	Answered by qaz last updated on 04/May/21
	$Ω=((e^(ix) −e^(−ix) )/(2i[z^2 +z(e^(ix) +e^(−ix) )+1])) =((e^(ix) −e^(−ix) )/(2i(ze^(ix) +1)(ze^(−ix) +1))) =(1/(2iz))[(1/(ze^(−ix) +1))−(1/(ze^(ix) +1))] =(1/(2iz)){Σ_(n=0) ^∞ (−1)^n z^n (e^(−inx) −e^(inx) )} =Σ_(n=0) ^∞ (−1)^(n+1) z^(n−1) sin (nx) ⇒a_n =(((−1)^(n+1) )/z)sin (nx)$
	$Ω = \frac{e^{i x} - e^{- i x}}{2 i [z^{2} + z (e^{i x} + e^{- i x}) + 1]}$ $= \frac{e^{i x} - e^{- i x}}{2 i ({z e}^{i x} + 1) ({z e}^{- i x} + 1)}$ $= \frac{1}{2 i z} [\frac{1}{{z e}^{- i x} + 1} - \frac{1}{{z e}^{i x} + 1}]$ $= \frac{1}{2 i z} {\underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} z^{n} (e^{- i n x} - e^{i n x})}$ $= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n + 1} z^{n - 1} \sin (n x)$ $\Rightarrow a_{n} = \frac{{(- 1)}^{n + 1}}{z} \sin (n x)$
	Commented bymnjuly1970 last updated on 04/May/21

	$n i c e v e r y n i c e . . .$
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