Find-the-first-four-terms-of-the-power-series-expansion-of-sinx-1-x-

Question Number 7887 by tawakalitu last updated on 23/Sep/16

F i n d t h e f i r s t f o u r t e r m s o f t h e p o w e r s e r i e s

e x p a n s i o n o f \frac{s i n x}{1 - x}

Answered by sandy_suhendra last updated on 23/Sep/16

f r o m t h e p o w e r s e r i e s :

s i n x = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{x^{2 n + 1}}{(2 n + 1)!}

= x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} (t h e f i r s t f o u r t e r m)

\frac{1}{1 - x} = \underset{n = 0}{\sum^{\infty}} x^{n}

= 1 + x + x^{2} + x^{3} (t h e f i r s t f o u r t e r m)

s o \frac{s i n x}{1 - x} = 1 (x) + x (- \frac{x^{3}}{3!}) + x^{2} (\frac{x^{5}}{5!}) + x^{3} (- \frac{x^{7}}{7!})

= x - \frac{x^{4}}{3!} + \frac{x^{7}}{5!} - \frac{x^{10}}{7!}

Commented by tawakalitu last updated on 23/Sep/16

T h a n k y o u s o m u c h .

Commented by Rasheed Soomro last updated on 24/Sep/16

(x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!}) (1 + x + x^{2} + x^{3}) = x - \frac{x^{4}}{3!} + \frac{x^{7}}{5!} - \frac{x^{10}}{7!} ? ? ?

Commented by prakash jain last updated on 24/Sep/16

(x - \frac{x^{3}}{6} + \dots) (1 + x + x^{2} + x^{3} + . .)

= x + x^{2} + x^{3} + x^{4} - \frac{x^{3}}{6} - \frac{x^{4}}{6} + (higher order)

= x + x^{2} - \frac{5 x^{3}}{6} - \frac{5 x^{4}}{6}

Commented by sandy_suhendra last updated on 28/Sep/16

I t h o u g h t

\frac{s i n x}{1 - x} = s i n x (\frac{1}{1 - x})

= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{x^{2 n + 1}}{(2 n + 1)!} [\underset{n = 0}{\sum^{\infty}} x^{n}]

= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{x^{2 n + 1}}{(2 n + 1)!} x^{n}

b u t I^{'} m w r o n g, {t h a n k}^{'} s f o r y o u r c o r r e c t i o n

Commented by sandy_suhendra last updated on 28/Sep/16

{t h a n k}^{'} s f o r y o u r c o r r e c t i o n