calculate-S-x-n-0-sin-nx-n-

Question Number 36741 by prof Abdo imad last updated on 04/Jun/18

c a l c u l a t e S (x) = \sum_{n = 0}^{\infty} \frac{s i n (n x)}{n!}

Commented by abdo.msup.com last updated on 05/Jun/18

S (x) = I m (\sum_{n = 0}^{\infty} \frac{e^{i n x}}{n!}) b u t

\sum_{n = 0}^{\infty} \frac{e^{i n x}}{n!} = \sum_{n = 0}^{\infty} \frac{{(e^{i x})}^{n}}{n!} = e^{e^{i x}}

= e^{c o s x + i s i n x} = e^{c o s x} {c o s (s i n x) + i s i n (s i n x)}

\Rightarrow S (x) = e^{c o s x} s i n (s i n x) .

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Jun/18

p = \frac{c o s n x}{n!}

q = \frac{s i n n x}{n!}

p + i q = \frac{e^{i n x}}{n!} = \frac{{(e^{i x})}^{n}}{n!}

T_{n} = \frac{{(e^{i x})}^{n}}{n!} =

T_{1} = \frac{{(e^{i x})}^{1}}{1!}

T_{2} = \frac{{(e^{i x})}^{2}}{2!}

T_{3} = \frac{{(e^{i x})}^{3}}{3!}

\dots

\dots

T_{n} = \frac{{(e^{i x})}^{n}}{n!}

\dots

\dots

T_{1} + T_{2} + \dots + T_{n} + \dots

w e k n o w t h a t e^{x} = \frac{x}{1!} + \frac{x^{2}}{2!} + \dots .

{s o T}_{1} + T_{2} + T_{3} + \dots

\frac{e^{(e^{i x})}}{1} = e^{c o s x + i s i n x} = e^{c o s x} . e^{i s i n x}

= e^{c o s x} {c o s (s i n x) + i s i n (s i n x)}

e^{c o s x} c o s (s i n x) + {i e}^{c o s x} s i n (s i n x)

s o r e q u i r e d a n s i s

e^{c o s x} s i n (s i n x)