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Question Number 139708 by qaz last updated on 30/Apr/21
Prove: 1+2Σ_(n=1) ^∞ a^n cos (nx)=((1−a^2 )/(1−2acos x+a^2 )), (∣a∣<1) And calculate:: ∫_0 ^π (x^2 /(1+8sin^2 x))dx=((5π^3 )/(36))−(π/6)ln^2 2
$${Prove}:\:\:\:\mathrm{1}+\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}^{{n}} \mathrm{cos}\:\left({nx}\right)=\frac{\mathrm{1}−{a}^{\mathrm{2}} }{\mathrm{1}−\mathrm{2}{a}\mathrm{cos}\:{x}+{a}^{\mathrm{2}} },\:\:\:\:\:\:\left(\mid{a}\mid<\mathrm{1}\right) \\ $$$${And}\:{calculate}::\:\:\int_{\mathrm{0}} ^{\pi} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{8sin}\:^{\mathrm{2}} {x}}{dx}=\frac{\mathrm{5}\pi^{\mathrm{3}} }{\mathrm{36}}−\frac{\pi}{\mathrm{6}}{ln}^{\mathrm{2}} \mathrm{2} \\ $$
Answered by Dwaipayan Shikari last updated on 30/Apr/21
1+Σ_(n=1) ^∞ a^n e^(inx) +a^n e^(−inx) =1+Σ_(n=1) ^∞ (ae^(ix) )^n +(ae^(−ix) )^n =1+((ae^(ix) )/(1−ae^(ix) ))+((ae^(−ix) )/(1−ae^(−i) )) =1+((ae^(ix) −a^2 +ae^(−ix) −a^2 )/(1−2acos(x)+a^2 )) =1+((2acos(x)−2a^2 )/(1−2acosx+a^2 ))=((1−a^2 )/(1−2acos(x)+a^2 ))
$$\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}^{{n}} {e}^{{inx}} +{a}^{{n}} {e}^{−{inx}} =\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({ae}^{{ix}} \right)^{{n}} +\left({ae}^{−{ix}} \right)^{{n}} \\ $$$$=\mathrm{1}+\frac{{ae}^{{ix}} }{\mathrm{1}−{ae}^{{ix}} }+\frac{{ae}^{−{ix}} }{\mathrm{1}−{ae}^{−{i}} }\:=\mathrm{1}+\frac{{ae}^{{ix}} −{a}^{\mathrm{2}} +{ae}^{−{ix}} −{a}^{\mathrm{2}} }{\mathrm{1}−\mathrm{2}{acos}\left({x}\right)+{a}^{\mathrm{2}} } \\ $$$$=\mathrm{1}+\frac{\mathrm{2}{acos}\left({x}\right)−\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{1}−\mathrm{2}{acosx}+{a}^{\mathrm{2}} }=\frac{\mathrm{1}−{a}^{\mathrm{2}} }{\mathrm{1}−\mathrm{2}{acos}\left({x}\right)+{a}^{\mathrm{2}} } \\ $$
Commented by qaz last updated on 30/Apr/21
thank you.sir
$${thank}\:{you}.{sir} \\ $$
Answered by mindispower last updated on 30/Apr/21
:: =∫_0 ^π (x^2 /(1+8(1−cos^2 (x))))dx =∫_0 ^π (x^2 /((3−2(√2)cos(x))(3+2(√2)cos(x)))) =(1/6)(∫_0 ^π (x^2 /(3−2(√2)cos(x)))dx+∫_0 ^π (x^2 /(3+2(√2)cos(x)))dx) =(1/6)(A+B) 6A=∫_0 ^π ((1−((1/( (√2))))^2 )/(1+(1/(((√2))^2 ))−2.(1/( (√2)))cos(x)))x^2 dx 6B=∫_0 ^π ((1−(−(1/( (√2))))^2 )/(1+(−(1/( (√2))))^2 +2.−(1/( (√2)))cos(x)))x^2 dx A=(1/6)∫_0 ^π (x^2 +2Σ_(n≥1) ((1/( (√2))))^n x^2 cos(nx)) B=(1/6)∫_0 ^π (x^2 +2Σ_(n≥1) (−(√2))^n x^2 cos(nx)) ∫_0 ^π x^2 cos(nx)dx=−(2/n)∫_0 ^π xsin(nx)dx =(2/n^2 )[π(−1)^n ] A=(π^3 /(18))+(1/3)Σ_(n≥1) ((1/( (√2))))^n .((2π)/n^2 )((−1)^n ) B=(π^3 /(18))+(1/3)Σ_(n≥1) (−(1/( (√2))))^n .(2/n^2 )((−1)^n ) A+B=(π^3 /9)+(1/3)Σ_(n≥1) ((1/2))^n .(π/n^2 ) li_2 (z)=Σ_(n≥1) (z^n /n^2 );∀∣z∣≤1 =(π^3 /9)+(π/3)Li_2 ((1/2)),li_2 ((1/2))=(π^2 /(12))−((ln^2 (2))/2) =(π^3 /9)+(π/3)((π^2 /(12))−((ln^2 (2))/2))=((5π^3 )/(36))−((πln^2 (2))/6)
$$:: \\ $$$$=\int_{\mathrm{0}} ^{\pi} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{8}\left(\mathrm{1}−{cos}^{\mathrm{2}} \left({x}\right)\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \frac{{x}^{\mathrm{2}} }{\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}{cos}\left({x}\right)\right)\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}{cos}\left({x}\right)\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left(\int_{\mathrm{0}} ^{\pi} \frac{{x}^{\mathrm{2}} }{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}{cos}\left({x}\right)}{dx}+\int_{\mathrm{0}} ^{\pi} \frac{{x}^{\mathrm{2}} }{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}{cos}\left({x}\right)}{dx}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left({A}+{B}\right) \\ $$$$\mathrm{6}{A}=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}−\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }−\mathrm{2}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{cos}\left({x}\right)}{x}^{\mathrm{2}} {dx} \\ $$$$\mathrm{6}{B}=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}−\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }{\mathrm{1}+\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\mathrm{2}.−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{cos}\left({x}\right)}{x}^{\mathrm{2}} {dx} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\pi} \left({x}^{\mathrm{2}} +\mathrm{2}\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{{n}} {x}^{\mathrm{2}} {cos}\left({nx}\right)\right) \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\pi} \left({x}^{\mathrm{2}} +\mathrm{2}\underset{{n}\geqslant\mathrm{1}} {\sum}\left(−\sqrt{\mathrm{2}}\right)^{{n}} {x}^{\mathrm{2}} {cos}\left({nx}\right)\right) \\ $$$$\int_{\mathrm{0}} ^{\pi} {x}^{\mathrm{2}} {cos}\left({nx}\right){dx}=−\frac{\mathrm{2}}{{n}}\int_{\mathrm{0}} ^{\pi} {xsin}\left({nx}\right){dx} \\ $$$$=\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\left[\pi\left(−\mathrm{1}\right)^{{n}} \right] \\ $$$${A}=\frac{\pi^{\mathrm{3}} }{\mathrm{18}}+\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{{n}} .\frac{\mathrm{2}\pi}{{n}^{\mathrm{2}} }\left(\left(−\mathrm{1}\right)^{{n}} \right) \\ $$$${B}=\frac{\pi^{\mathrm{3}} }{\mathrm{18}}+\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}\geqslant\mathrm{1}} {\sum}\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{{n}} .\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\left(\left(−\mathrm{1}\right)^{{n}} \right) \\ $$$${A}+{B}=\frac{\pi^{\mathrm{3}} }{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} .\frac{\pi}{{n}^{\mathrm{2}} } \\ $$$${li}_{\mathrm{2}} \left({z}\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{z}^{{n}} }{{n}^{\mathrm{2}} };\forall\mid{z}\mid\leqslant\mathrm{1} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{9}}+\frac{\pi}{\mathrm{3}}{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right),{li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{9}}+\frac{\pi}{\mathrm{3}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}\right)=\frac{\mathrm{5}\pi^{\mathrm{3}} }{\mathrm{36}}−\frac{\pi{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{6}} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 30/Apr/21
bravo bravo sir power...
$${bravo}\:{bravo}\:{sir}\:{power}… \\ $$
Commented by mindispower last updated on 01/May/21
withe pleasur Sir
$${withe}\:{pleasur}\:{Sir} \\ $$
Answered by mathmax by abdo last updated on 01/May/21
Σ_(n=1) ^∞ a^n cos(nx) =Re(Σ_(n=0) ^∞ a^n e^(inx) −1) Σ_(n=0) ^∞ (ae^(ix) )^n =(1/(1−ae^(ix) )) =(1/(1−acosx−iasinx)) =((1−acosx+iasinx)/((1−acosx)^2 +a^2 sin^2 x)) ⇒Σ_(n=1) ^∞ a^n cos(nx) =((1−acosx)/(1−2acosx+a^2 ))−1 ⇒ 1+2Σ_(n=1) ^∞ a^n cos(nx) =1+((2−2acosx)/(1−2acosx+a^2 ))−2 =((2−2acosx)/(1−2acosx+a^2 ))−1 =((2−2acosx−1+2acosx−a^2 )/(a^2 −2acosx +1)) =((1−a^2 )/(a^2 −2acosx +1))
$$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{a}^{\mathrm{n}} \:\mathrm{cos}\left(\mathrm{nx}\right)\:=\mathrm{Re}\left(\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\:\mathrm{a}^{\mathrm{n}} \:\mathrm{e}^{\mathrm{inx}} \:−\mathrm{1}\right) \\ $$$$\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(\mathrm{ae}^{\mathrm{ix}} \right)^{\mathrm{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{ae}^{\mathrm{ix}} }\:=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{acosx}−\mathrm{iasinx}}\:=\frac{\mathrm{1}−\mathrm{acosx}+\mathrm{iasinx}}{\left(\mathrm{1}−\mathrm{acosx}\right)^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \mathrm{x}} \\ $$$$\:\Rightarrow\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{a}^{\mathrm{n}} \:\mathrm{cos}\left(\mathrm{nx}\right)\:=\frac{\mathrm{1}−\mathrm{acosx}}{\mathrm{1}−\mathrm{2acosx}+\mathrm{a}^{\mathrm{2}} }−\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{1}+\mathrm{2}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{a}^{\mathrm{n}} \mathrm{cos}\left(\mathrm{nx}\right)\:=\mathrm{1}+\frac{\mathrm{2}−\mathrm{2acosx}}{\mathrm{1}−\mathrm{2acosx}+\mathrm{a}^{\mathrm{2}} }−\mathrm{2} \\ $$$$=\frac{\mathrm{2}−\mathrm{2acosx}}{\mathrm{1}−\mathrm{2acosx}+\mathrm{a}^{\mathrm{2}} }−\mathrm{1}\:=\frac{\mathrm{2}−\mathrm{2acosx}−\mathrm{1}+\mathrm{2acosx}−\mathrm{a}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} −\mathrm{2acosx}\:+\mathrm{1}}\:=\frac{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} −\mathrm{2acosx}\:+\mathrm{1}} \\ $$

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