Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 164418 by HongKing last updated on 16/Jan/22

Commented by Kamel last updated on 16/Jan/22

Commented by Kamel last updated on 16/Jan/22

With differential equation.

$${With}\:{differential}\:{equation}. \\ $$

Answered by Kamel last updated on 16/Jan/22

Commented by Kamel last updated on 16/Jan/22

With Laplas transform′s

$${With}\:{Laplas}\:{transform}'{s} \\ $$

Answered by mindispower last updated on 18/Jan/22

sin(x)=−ish(ix) =−i∫_0 ^π cot((x/2))sh(iasin(x))sh(acos(x))dx =−i∫_0 ^π cot((x/2))(ch(ae^(ix) )−ch(ae^(−ix) ) =2Im∫_0 ^π cot((x/2))ch(ae^(ix) ) =2Im∫_0 ^π cot((x/2))Σ_(n≥0) (((ae^(ix) )^(2n) )/((2n)!)) =2Σ_(n≥1) a^(2n) ∫_0 ^π ((sin(2nx))/((2n)!))cot((x/2))dx =Σ_(n≥1) (a^(2n) /((2n)!))∫_0 ^π sin(2nx)(((cos((x/2)))/(sin((x/2))))−((sin((x/2)))/(cos((x/2))))) =2Σ_(n≥1) (a^(2n) /((2n)!))∫_0 ^π ((sin(2nx)cos(x))/(sin(x)))dx=Ω ∫_0 ^π ((sin(2nx)cos(x))/(sin(x)))dx=∫_π ^(2π) ((sin(2nx)cos(x))/(sin(x)))dx Ω=Σ_(n≥1) (a^(2n) /((2n)!))∫_0 ^(2π) ((sin(2nx)cos(x))/(sin(x))) I_(2n+2) −I_(2n) =∫_0 ^(2π) ((sin(2n+2x)−sin(2nx))/(sin(x)))cos(x) sin(2n+2)x−sin(2nx)=2sin(x)coos(n+1)x =∫_0 ^(2π) 2cos(n+1)xcos(x) =∫_0 ^(2π) cos(nx)−cos(n+2)x dx=0 ∀n∈N^∗ I_(2n+2) =I_(2n) ⇔I_(2n) =I_2 ⇔I_(2n) =∫_0 ^(2π) ((sin(2nx)cos(x))/(sin(x)))dx=∫_0 ^(2π) ((sin(2x)cos(x))/(sin(x)))dx ∫_0 ^(2π) 2cos^2 (x)dx=∫_0 ^(2π) (1+cos(2x))dx=2π we get Σ_(n≥1) (a^(2n) /((2n)!)).π =π(Σ_(n≥0) (a^(2n) /((2n)!))−1)=π(ch(a)−1)

$${sin}\left({x}\right)=−{ish}\left({ix}\right) \\ $$$$=−{i}\int_{\mathrm{0}} ^{\pi} {cot}\left(\frac{{x}}{\mathrm{2}}\right){sh}\left({iasin}\left({x}\right)\right){sh}\left({acos}\left({x}\right)\right){dx} \\ $$$$=−{i}\int_{\mathrm{0}} ^{\pi} {cot}\left(\frac{{x}}{\mathrm{2}}\right)\left({ch}\left({ae}^{{ix}} \right)−{ch}\left({ae}^{−{ix}} \right)\right. \\ $$$$=\mathrm{2}{Im}\int_{\mathrm{0}} ^{\pi} {cot}\left(\frac{{x}}{\mathrm{2}}\right){ch}\left({ae}^{{ix}} \right) \\ $$$$=\mathrm{2}{Im}\int_{\mathrm{0}} ^{\pi} {cot}\left(\frac{{x}}{\mathrm{2}}\right)\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left({ae}^{{ix}} \right)^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!} \\ $$$$=\mathrm{2}\underset{{n}\geqslant\mathrm{1}} {\sum}{a}^{\mathrm{2}{n}} \int_{\mathrm{0}} ^{\pi} \frac{{sin}\left(\mathrm{2}{nx}\right)}{\left(\mathrm{2}{n}\right)!}{cot}\left(\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{a}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}\int_{\mathrm{0}} ^{\pi} {sin}\left(\mathrm{2}{nx}\right)\left(\frac{{cos}\left(\frac{{x}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}−\frac{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{{cos}\left(\frac{{x}}{\mathrm{2}}\right)}\right) \\ $$$$=\mathrm{2}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{a}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}\int_{\mathrm{0}} ^{\pi} \frac{{sin}\left(\mathrm{2}{nx}\right){cos}\left({x}\right)}{{sin}\left({x}\right)}{dx}=\Omega \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{sin}\left(\mathrm{2}{nx}\right){cos}\left({x}\right)}{{sin}\left({x}\right)}{dx}=\int_{\pi} ^{\mathrm{2}\pi} \frac{{sin}\left(\mathrm{2}{nx}\right){cos}\left({x}\right)}{{sin}\left({x}\right)}{dx} \\ $$$$\Omega=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{a}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{sin}\left(\mathrm{2}{nx}\right){cos}\left({x}\right)}{{sin}\left({x}\right)} \\ $$$${I}_{\mathrm{2}{n}+\mathrm{2}} −{I}_{\mathrm{2}{n}} =\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{sin}\left(\mathrm{2}{n}+\mathrm{2}{x}\right)−{sin}\left(\mathrm{2}{nx}\right)}{{sin}\left({x}\right)}{cos}\left({x}\right) \\ $$$${sin}\left(\mathrm{2}{n}+\mathrm{2}\right){x}−{sin}\left(\mathrm{2}{nx}\right)=\mathrm{2}{sin}\left({x}\right){coos}\left({n}+\mathrm{1}\right){x} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{2}{cos}\left({n}+\mathrm{1}\right){xcos}\left({x}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} {cos}\left({nx}\right)−{cos}\left({n}+\mathrm{2}\right){x}\:{dx}=\mathrm{0} \\ $$$$\forall{n}\in\mathbb{N}^{\ast} \:{I}_{\mathrm{2}{n}+\mathrm{2}} ={I}_{\mathrm{2}{n}} \Leftrightarrow{I}_{\mathrm{2}{n}} ={I}_{\mathrm{2}} \\ $$$$\Leftrightarrow{I}_{\mathrm{2}{n}} =\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{sin}\left(\mathrm{2}{nx}\right){cos}\left({x}\right)}{{sin}\left({x}\right)}{dx}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{sin}\left(\mathrm{2}{x}\right){cos}\left({x}\right)}{{sin}\left({x}\right)}{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{2}{cos}^{\mathrm{2}} \left({x}\right){dx}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left(\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)\right){dx}=\mathrm{2}\pi \\ $$$${we}\:{get}\:\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{a}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}.\pi \\ $$$$=\pi\left(\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{{a}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}−\mathrm{1}\right)=\pi\left({ch}\left({a}\right)−\mathrm{1}\right) \\ $$$$ \\ $$

Commented by HongKing last updated on 20/Jan/22

perfect my dear Sir thank you

$$\mathrm{perfect}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com