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

$$Findthefirstfourtermsofthepowerseries$$$$expansionof\frac{sinx}{1-x}$$

Answered by sandy_suhendra last updated on 23/Sep/16

$$fromthepowerseries:$$$$sinx=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{x^{2n+1}}{(2n+1)!}$$$$=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}\left(thefirstfourterm\right)$$$$\frac{1}{1-x}=\underset{n=0}{\sum ^{\mathrm{\infty }}}{x}^n$$$$=1+x+x^2+x^3\left(thefirstfourterm\right)$$$$so\frac{sinx}{1-x}=1\left(x\right)+x(-\frac{x^3}{3!})+x^2\left(\frac{x^5}{5!}\right)+x^3(-\frac{x^7}{7!})$$$$=x-\frac{x^4}{3!}+\frac{x^7}{5!}-\frac{x^{10}}{7!}$$

Commented by tawakalitu last updated on 23/Sep/16

$$Thankyousomuch.$$

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

$$(x-\frac{x^3}{6}+...)(1+x+x^2+x^3+..)$$$$=x+x^2+x^3+x^4-\frac{x^3}{6}-\frac{x^4}{6}+\left(\mathrm{higher}\mathrm{order}\right)$$$$=x+x^2-\frac{5x^3}{6}-\frac{5x^4}{6}$$

Commented by sandy_suhendra last updated on 28/Sep/16

$$Ithought$$$$\frac{sinx}{1-x}=sinx\left(\frac{1}{1-x}\right)$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{x^{2n+1}}{(2n+1)!}\left[\underset{n=0}{\sum ^{\mathrm{\infty }}}{x}^n\right]$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{x^{2n+1}}{(2n+1)!}{x}^n$$$$but{I}^{\prime }mwrong,{thank}^{\prime }sforyourcorrection$$

Commented by sandy_suhendra last updated on 28/Sep/16

$${thank}^{\prime }sforyourcorrection$$

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