Question-196303

Question Number 196303 by sonukgindia last updated on 22/Aug/23

Answered by sniper237 last updated on 22/Aug/23

\underset{n = 1}{\sum^{\infty}} I m {(\frac{e^{i π / 4}}{2})}^{n} = I m (\underset{n = 1}{\sum^{\infty}} {(e^{i π / 4} / 2)}^{n})

= I m (\frac{e^{i π / 4} / 2}{1 - e^{i π / 4} / 2})

Commented by sonukgindia last updated on 22/Aug/23

e x p l a i n t h i s

Commented by JDamian last updated on 22/Aug/23

1 . \sin α = Im {e^{i α}}

2 . Σ (Im {z_{i}}) = Im {Σ z_{i}}

Answered by Mathspace last updated on 22/Aug/23

s (x) = \sum_{n = 1}^{\infty} \frac{s i n (n x)}{2^{n}} \Rightarrow

s (x) = I m (\sum_{n = 1}^{\infty} \frac{e^{i n x}}{2^{n}}) b u t

\sum_{n = 1}^{\infty} \frac{e^{i n x}}{2^{n}} = \sum_{n = 1}^{\infty} {(\frac{1}{2} e^{i x})}^{n}

\frac{1}{1 - \frac{1}{2} e^{i x}} (l o o k t h a t ∣ \frac{1}{2} e^{i x} ∣< 1)

= \frac{2}{2 - c o s x - i s i n x}

= \frac{2 (2 - c o s x + i s i n x)}{{(2 - c o s x)}^{2} + {s i n}^{2} x}

\Rightarrow s (x) = \frac{2 s i n x}{4 - 4 c o s x + 1}

= \frac{2 s i n x}{5 - 4 c o s x}

a n d \sum_{n = 1}^{\infty} \frac{s i n (\frac{n π}{4})}{2^{n}}

= s (\frac{π}{4}) = \frac{2 s i n (\frac{π}{4})}{5 - 4 c o s (\frac{π}{4})}

= \frac{2 \times \frac{1}{\sqrt{2}}}{5 - 4 . \frac{\sqrt{2}}{2}} = \frac{\sqrt{2}}{5 - 2 \sqrt{2}}