Math Question and Answers


Question and Answers Forum

All Questions Topic List
Arithmetic Questions
Previous in All Question Next in All Question
Previous in Arithmetic Next in Arithmetic
Question Number 62570 by aliesam last updated on 23/Jun/19

Commented by mathmax by abdo last updated on 23/Jun/19
$we have S(x) =Σ_(n=1) ^∞ ((sin(nx))/2^(n−1) ) =2 Σ_(n=1) ^∞ ((sin(nx))/2^n ) =2 Im(Σ_(n=1) ^∞ (e^(inx) /2^n )) Σ_(n=1) ^∞ (e^(inx) /2^n ) =Σ_(n=1) ^∞ ((e^(ix) /2))^n =Σ_(n=0) ^∞ ((e^(ix) /2))^n −1 =(1/(1−(e^(ix) /2))) −1 (∣(e^(ix) /2)∣=(1/2)<1) =(2/(2−e^(ix) )) −1 =(2/(2−cosx −isinx)) −1 =((2(2−cosx +isinx))/((2−cosx)^2 +sin^2 x)) −1 =((2(2−cosx)−{ (2−cosx)^2 +sin^2 x} +2i sinx)/((2−cosx)^2 +sin^2 x)) ⇒ S(x) =((4sinx)/((2−cosx)^2 +sin^2 x)) =((4sinx)/(4−4cosx +1)) =((4sinx)/(5−4cosx)) x =(π/4) ⇒S((π/4)) =((4sin((π/4)))/(5−4cos((π/4)))) =((4/(√2))/(5−(4/(√2)))) =(4/(5(√2)−4)) .$
$w e h a v e S (x) = \sum_{n = 1}^{\infty} \frac{s i n (n x)}{2^{n - 1}} = 2 \sum_{n = 1}^{\infty} \frac{s i n (n x)}{2^{n}} = 2 I m (\sum_{n = 1}^{\infty} \frac{e^{i n x}}{2^{n}})$ $\sum_{n = 1}^{\infty} \frac{e^{i n x}}{2^{n}} = \sum_{n = 1}^{\infty} {(\frac{e^{i x}}{2})}^{n} = \sum_{n = 0}^{\infty} {(\frac{e^{i x}}{2})}^{n} - 1 = \frac{1}{1 - \frac{e^{i x}}{2}} - 1 (∣ \frac{e^{i x}}{2} ∣= \frac{1}{2} < 1)$ $= \frac{2}{2 - e^{i x}} - 1 = \frac{2}{2 - c o s x - i s i n x} - 1 = \frac{2 (2 - c o s x + i s i n x)}{{(2 - c o s x)}^{2} + {s i n}^{2} x} - 1$ $= \frac{2 (2 - c o s x) - {{(2 - c o s x)}^{2} + {s i n}^{2} x} + 2 i s i n x}{{(2 - c o s x)}^{2} + {s i n}^{2} x} \Rightarrow$ $S (x) = \frac{4 s i n x}{{(2 - c o s x)}^{2} + {s i n}^{2} x} = \frac{4 s i n x}{4 - 4 c o s x + 1} = \frac{4 s i n x}{5 - 4 c o s x}$ $x = \frac{π}{4} \Rightarrow S (\frac{π}{4}) = \frac{4 s i n (\frac{π}{4})}{5 - 4 c o s (\frac{π}{4})} = \frac{\frac{4}{\sqrt{2}}}{5 - \frac{4}{\sqrt{2}}} = \frac{4}{5 \sqrt{2} - 4} .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum