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Question-48091

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Question Number 48091 by peter frank last updated on 19/Nov/18
Commented by maxmathsup by imad last updated on 19/Nov/18
we have e^(−x) sinx =Im(e^(−x+ix) )=Im(e^((−1+i)x) ) but e^((−1+i)x) =Σ_(n=0) ^∞ (((−1+i)^n x^n )/(n!)) but (−1+i)=(√2)(−(1/( (√2))) +(i/( (√2)))) =(√2)(cos(((3π)/4))+isin(((3π)/4))) =(√2)e^(i((3π)/4)) ⇒(−1+i)^n =((√2))^n e^(i((3nπ)/4)) =((√2))^n { cos(((3nπ)/4))+isin(((3nπ)/4))} ⇒e^(−x) sin(x)=Σ_(n=0) ^∞ ((((√2))^n )/(n!)) sin(((3nπ)/4))x^n x=(π/3) ⇒ e^(−(π/3)) sin((π/3))=Σ_(n=0) ^∞ ((((√2))^n )/(n!)) sin(((3nπ)/4))((π/3))^n .
$${we}\:{have}\:{e}^{−{x}} {sinx}\:={Im}\left({e}^{−{x}+{ix}} \right)={Im}\left({e}^{\left(−\mathrm{1}+{i}\right){x}} \right)\:{but} \\ $$$${e}^{\left(−\mathrm{1}+\boldsymbol{{i}}\right)\boldsymbol{{x}}} =\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}+{i}\right)^{{n}} {x}^{{n}} }{{n}!}\:\:{but}\:\left(−\mathrm{1}+{i}\right)=\sqrt{\mathrm{2}}\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$=\sqrt{\mathrm{2}}\left({cos}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)+{isin}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)\right)\:=\sqrt{\mathrm{2}}{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\Rightarrow\left(−\mathrm{1}+{i}\right)^{{n}} =\left(\sqrt{\mathrm{2}}\right)^{{n}} \:{e}^{{i}\frac{\mathrm{3}{n}\pi}{\mathrm{4}}} \\ $$$$=\left(\sqrt{\mathrm{2}}\right)^{{n}} \left\{\:{cos}\left(\frac{\mathrm{3}{n}\pi}{\mathrm{4}}\right)+{isin}\left(\frac{\mathrm{3}{n}\pi}{\mathrm{4}}\right)\right\}\:\Rightarrow{e}^{−{x}} {sin}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(\sqrt{\mathrm{2}}\right)^{{n}} }{{n}!}\:{sin}\left(\frac{\mathrm{3}{n}\pi}{\mathrm{4}}\right){x}^{{n}} \: \\ $$$${x}=\frac{\pi}{\mathrm{3}}\:\Rightarrow\:{e}^{−\frac{\pi}{\mathrm{3}}} \:{sin}\left(\frac{\pi}{\mathrm{3}}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(\sqrt{\mathrm{2}}\right)^{{n}} }{{n}!}\:{sin}\left(\frac{\mathrm{3}{n}\pi}{\mathrm{4}}\right)\left(\frac{\pi}{\mathrm{3}}\right)^{{n}} \:. \\ $$
Commented by peter frank last updated on 19/Nov/18
thanks
$$\mathrm{thanks}\: \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 19/Nov/18
p=e^(−x) cosx q=e^(−x) sinx p+iq=e^(−x) (cosx+isinx)=e^(−x) .e^(ix) =e^(x(−1+i)) =e^(tx) e^(tx) =1+tx+((t^2 x^2 )/(2!))+((t^3 x^3 )/(3!))+... =1+(−1+i)x+(−1+i)^2 ×(x^2 /(2!))+(−1+i)^3 ×(x^3 /(3!)) other terms not considered as to find coefficient of upto x^3 =1+(−x+ix)+(−ix^2 )+(−2i)(−1+i)×(x^3 /(3!)) =1+(−x+ix)+(−ix^2 )+(2i+2)(x^3 /6) =(1−x+(x^3 /3))+i(x−x^2 +(x^3 /3)) so e^(tx) =e^((−1+i)x) =(1−x+(x^3 /3))+i×(x/3)(x^2 −3x+3) e^(−x) ×e^(ix) =(1−x+(x^3 /3))+i×(x/3)(x^2 −3x+3) e^(−x) (cosx+isinx)=(1−x+(x^3 /3))+i×(x/3)(x^2 −3x+3) hence e^(−x) sinx=(x/3)(x^2 −3x+3) proved e^(−(π/3)) sin((π/3))=(π/9)((π^2 /9)−π+3)
$${p}={e}^{−{x}} {cosx} \\ $$$${q}={e}^{−{x}} {sinx} \\ $$$${p}+{iq}={e}^{−{x}} \left({cosx}+{isinx}\right)={e}^{−{x}} .{e}^{{ix}} ={e}^{{x}\left(−\mathrm{1}+{i}\right)} ={e}^{{tx}} \\ $$$${e}^{{tx}} =\mathrm{1}+{tx}+\frac{{t}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{t}^{\mathrm{3}} {x}^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$$=\mathrm{1}+\left(−\mathrm{1}+{i}\right){x}+\left(−\mathrm{1}+\mathrm{i}\right)^{\mathrm{2}} ×\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\left(−\mathrm{1}+{i}\right)^{\mathrm{3}} ×\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:{other} \\ $$$${terms}\:{not}\:{considered}\:{as}\:{to}\:{find}\:{coefficient}\:{of} \\ $$$${upto}\:{x}^{\mathrm{3}} \\ $$$$=\mathrm{1}+\left(−{x}+{ix}\right)+\left(−{ix}^{\mathrm{2}} \right)+\left(−\mathrm{2}{i}\right)\left(−\mathrm{1}+{i}\right)×\frac{{x}^{\mathrm{3}} }{\mathrm{3}!} \\ $$$$=\mathrm{1}+\left(−{x}+{ix}\right)+\left(−{ix}^{\mathrm{2}} \right)+\left(\mathrm{2}{i}+\mathrm{2}\right)\frac{{x}^{\mathrm{3}} }{\mathrm{6}} \\ $$$$=\left(\mathrm{1}−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right)+{i}\left({x}−{x}^{\mathrm{2}} +\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right) \\ $$$${so}\:{e}^{{tx}} ={e}^{\left(−\mathrm{1}+{i}\right){x}} =\left(\mathrm{1}−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right)+{i}×\frac{{x}}{\mathrm{3}}\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{3}\right) \\ $$$${e}^{−{x}} ×{e}^{{ix}} =\left(\mathrm{1}−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right)+{i}×\frac{{x}}{\mathrm{3}}\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{3}\right) \\ $$$${e}^{−{x}} \left({cosx}+{isinx}\right)=\left(\mathrm{1}−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right)+{i}×\frac{{x}}{\mathrm{3}}\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{3}\right) \\ $$$${hence} \\ $$$${e}^{−{x}} {sinx}=\frac{{x}}{\mathrm{3}}\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{3}\right)\:\:{proved} \\ $$$${e}^{−\frac{\pi}{\mathrm{3}}} {sin}\left(\frac{\pi}{\mathrm{3}}\right)=\frac{\pi}{\mathrm{9}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{9}}−\pi+\mathrm{3}\right) \\ $$

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