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

$${Find}\:{the}\:{first}\:{four}\:{terms}\:{of}\:{the}\:{power}\:{series}\: \\ $$$${expansion}\:{of}\:\:\:\:\:\frac{{sinx}}{\mathrm{1}\:−\:{x}}\:\:\: \\ $$

Answered by sandy_suhendra last updated on 23/Sep/16

from the power series : sin x = Σ_(n=0) ^∞ (−1)^n (x^(2n+1) /((2n+1)!)) = x−(x^3 /(3!))+(x^5 /(5!))−(x^7 /(7!)) (the first four term) (1/(1−x)) = Σ_(n=0) ^∞ x^n = 1+x+x^2 +x^3 (the first four term) so ((sin x)/(1−x)) = 1(x)+x(−(x^3 /(3!)))+x^2 ((x^5 /(5!)))+x^3 (−(x^7 /(7!))) = x − (x^4 /(3!)) + (x^7 /(5!)) − (x^(10) /(7!))

$${from}\:{the}\:{power}\:{series}\:: \\ $$$${sin}\:{x}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \:\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}!}\:\:\left({the}\:{first}\:{four}\:{term}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{x}}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:{x}^{{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} \:\left({the}\:{first}\:{four}\:{term}\right) \\ $$$${so}\:\:\:\:\:\frac{{sin}\:{x}}{\mathrm{1}−{x}}\:=\:\mathrm{1}\left({x}\right)+{x}\left(−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\right)+{x}^{\mathrm{2}} \left(\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}\right)+{x}^{\mathrm{3}} \left(−\frac{{x}^{\mathrm{7}} }{\mathrm{7}!}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{x}\:−\:\frac{{x}^{\mathrm{4}} }{\mathrm{3}!}\:+\:\frac{{x}^{\mathrm{7}} }{\mathrm{5}!}\:−\:\frac{{x}^{\mathrm{10}} }{\mathrm{7}!} \\ $$

Commented by tawakalitu last updated on 23/Sep/16

$${Thank}\:{you}\:{so}\:{much}. \\ $$

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

(x−(x^3 /(3!))+(x^5 /(5!))−(x^7 /(7!)))(1+x+x^2 +x^3 )=x − (x^4 /(3!)) + (x^7 /(5!)) − (x^(10) /(7!))???

$$\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}!}\right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} \right)={x}\:−\:\frac{{x}^{\mathrm{4}} }{\mathrm{3}!}\:+\:\frac{{x}^{\mathrm{7}} }{\mathrm{5}!}\:−\:\frac{{x}^{\mathrm{10}} }{\mathrm{7}!}??? \\ $$

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

(x−(x^3 /6)+...)(1+x+x^2 +x^3 +..) =x+x^2 +x^3 +x^4 −(x^3 /6)−(x^4 /6) +(higher order) =x+x^2 −((5x^3 )/6)−((5x^4 )/6)

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

Commented by sandy_suhendra last updated on 28/Sep/16

I thought ((sin x)/(1−x)) = sinx((1/(1−x))) =Σ_(n=0) ^∞ (−1)^n (x^(2n+1) /((2n+1)!)) [Σ_(n=0) ^∞ x^n ] =Σ_(n=0) ^∞ (−1)^n (x^(2n+1) /((2n+1)!)) x^n but I′m wrong, thank′s for your correction

$${I}\:{thought} \\ $$$$\frac{{sin}\:{x}}{\mathrm{1}−{x}}\:=\:{sinx}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right) \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \:\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\left[\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:{x}^{{n}} \right] \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \:\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\:{x}^{{n}} \\ $$$${but}\:{I}'{m}\:{wrong},\:{thank}'{s}\:{for}\:{your}\:{correction} \\ $$

Commented by sandy_suhendra last updated on 28/Sep/16

$${thank}'{s}\:{for}\:{your}\:{correction} \\ $$

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