Question Number 30764 by abdo imad last updated on 25/Feb/18
![let I_n = ∫_0 ^(1/2) (1−2t)^n e^(−t) dt with n integr not 0 1) prove that ∀t∈[0,(1/2)] (1/(√e))(1−2t)^n ≤ (1−2t)^n e^(−t) ≤(1−2t)^n then find lim_(n→∞ ) I_n 2) prove that I_(n+1 ) =1−2(n+1)I_n 3) calculate I_1 ,I_2 , and I_3 .](Q30764.png)
$${let}\:\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\left(\mathrm{1}−\mathrm{2}{t}\right)^{{n}} \:{e}^{−{t}} {dt}\:\:{with}\:{n}\:{integr}\:{not}\:\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:\forall{t}\in\left[\mathrm{0},\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$$\frac{\mathrm{1}}{\sqrt{{e}}}\left(\mathrm{1}−\mathrm{2}{t}\right)^{{n}} \leqslant\:\left(\mathrm{1}−\mathrm{2}{t}\right)^{{n}} \:{e}^{−{t}} \:\leqslant\left(\mathrm{1}−\mathrm{2}{t}\right)^{{n}} \:{then}\:{find}\:{lim}_{{n}\rightarrow\infty\:} {I}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:{I}_{{n}+\mathrm{1}\:} =\mathrm{1}−\mathrm{2}\left({n}+\mathrm{1}\right){I}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:{I}_{\mathrm{1}} ,{I}_{\mathrm{2}} ,\:{and}\:{I}_{\mathrm{3}} . \\ $$
Commented by abdo imad last updated on 01/Mar/18
![1) for 0≤t≤(1/2) we have( 1−2t)^n ≥0 1≤e^t ≤ (√e) ⇒(1/(√e))≤e^(−t) ≤1 ⇒ (1/(√e))(1−2t)^n ≤(1−2t)^n e^(−t) ≤ (1−2t)^n ⇒ (1/(√e)) ∫_0 ^(1/2) (1−2t)^n dt ≤ ∫_0 ^(1/2) (1−2t)^n e^(−t) dt ≤ ∫_0 ^(1/2) (1−2t)^n dt but (1/(√e))∫_0 ^(1/2) (1−2t)^n dt=(1/(√e)) [ ((−1)/(2(n+1))) (1−2t)^(n+1) ]_0 ^(1/2) = (1/(2(n+1)(√e))) ⇒ (1/(2(n+1)(√e)))≤ I_n ≤ (1/(2(n+1))) and its clear that lim_(n→∞) I_n =0 2) let integrate by parts u^′ =(1−2t)^n and v=e^(−t) I_n =[((−1)/(2(n+1)))(1−2t)^(n+1) e^(−t) ]_0 ^(1/2) +∫_0 ^(1/2) ((−1)/(2(n+1)))(1−2t)^(n+1) e^(−t) dt =(1/(2(n+1))) −(1/(2(n+1))) I_(n+1) ⇒2(n+1)I_n =1−I_(n+1) ⇒ I_(n+1) =1−2(n+1)I_n 3)I_1 =1−2I_0 but I_0 = ∫_0 ^(1/2) e^(−t) dt =[ −e^(−t) ]_0 ^(1/2) =1−e^((−1)/2) =1−(1/(√e)) I_2 =1−4I_1 =1−4(1−(1/(√e)))=−3 +(4/(√e)). I_3 = 1−6 I_(2 ) =1−6(−3 +(4/(√e)))=19 −((24)/(√e)) .](Q30938.png)
$$\left.\mathrm{1}\right)\:{for}\:\mathrm{0}\leqslant{t}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\:{we}\:{have}\left(\:\mathrm{1}−\mathrm{2}{t}\right)^{{n}} \geqslant\mathrm{0}\:\:\mathrm{1}\leqslant{e}^{{t}} \leqslant\:\sqrt{{e}}\:\Rightarrow\frac{\mathrm{1}}{\sqrt{{e}}}\leqslant{e}^{−{t}} \leqslant\mathrm{1} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\sqrt{{e}}}\left(\mathrm{1}−\mathrm{2}{t}\right)^{{n}} \:\leqslant\left(\mathrm{1}−\mathrm{2}{t}\right)^{{n}} \:{e}^{−{t}} \leqslant\:\left(\mathrm{1}−\mathrm{2}{t}\right)^{{n}} \:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\sqrt{{e}}}\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−\mathrm{2}{t}\right)^{{n}} {dt}\:\leqslant\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−\mathrm{2}{t}\right)^{{n}} \:{e}^{−{t}} {dt}\:\leqslant\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\left(\mathrm{1}−\mathrm{2}{t}\right)^{{n}} {dt}\:{but} \\ $$$$\frac{\mathrm{1}}{\sqrt{{e}}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−\mathrm{2}{t}\right)^{{n}} {dt}=\frac{\mathrm{1}}{\sqrt{{e}}}\:\left[\:\frac{−\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\:\left(\mathrm{1}−\mathrm{2}{t}\right)^{{n}+\mathrm{1}} \right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} =\:\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)\sqrt{{e}}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)\sqrt{{e}}}\leqslant\:{I}_{{n}} \leqslant\:\:\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\:{and}\:{its}\:{clear}\:{that}\: \\ $$$${lim}_{{n}\rightarrow\infty} \:{I}_{{n}} =\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{let}\:{integrate}\:{by}\:{parts}\:{u}^{'} \:=\left(\mathrm{1}−\mathrm{2}{t}\right)^{{n}} \:{and}\:{v}={e}^{−{t}} \\ $$$${I}_{{n}} =\left[\frac{−\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\left(\mathrm{1}−\mathrm{2}{t}\right)^{{n}+\mathrm{1}} \:{e}^{−{t}} \right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:+\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{−\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\left(\mathrm{1}−\mathrm{2}{t}\right)^{{n}+\mathrm{1}} \:{e}^{−{t}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\:−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\:{I}_{{n}+\mathrm{1}} \:\:\Rightarrow\mathrm{2}\left({n}+\mathrm{1}\right){I}_{{n}} =\mathrm{1}−{I}_{{n}+\mathrm{1}} \:\Rightarrow \\ $$$${I}_{{n}+\mathrm{1}} =\mathrm{1}−\mathrm{2}\left({n}+\mathrm{1}\right){I}_{{n}} \\ $$$$\left.\mathrm{3}\right){I}_{\mathrm{1}} =\mathrm{1}−\mathrm{2}{I}_{\mathrm{0}} \:{but}\:{I}_{\mathrm{0}} =\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:{e}^{−{t}} {dt}\:=\left[\:−{e}^{−{t}} \right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{1}−{e}^{\frac{−\mathrm{1}}{\mathrm{2}}} =\mathrm{1}−\frac{\mathrm{1}}{\sqrt{{e}}} \\ $$$${I}_{\mathrm{2}} =\mathrm{1}−\mathrm{4}{I}_{\mathrm{1}} =\mathrm{1}−\mathrm{4}\left(\mathrm{1}−\frac{\mathrm{1}}{\sqrt{{e}}}\right)=−\mathrm{3}\:+\frac{\mathrm{4}}{\sqrt{{e}}}. \\ $$$${I}_{\mathrm{3}} =\:\mathrm{1}−\mathrm{6}\:{I}_{\mathrm{2}\:} =\mathrm{1}−\mathrm{6}\left(−\mathrm{3}\:+\frac{\mathrm{4}}{\sqrt{{e}}}\right)=\mathrm{19}\:−\frac{\mathrm{24}}{\sqrt{{e}}}\:\:. \\ $$$$ \\ $$