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let-f-x-e-x-2-0-x-e-t-2-dt-1-find-a-d-e-verified-by-f-2-developpf-at-integr-serie-

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Question Number 34699 by abdo imad last updated on 10/May/18
let f(x)=e^(−x^2 ) ∫_0 ^x e^t^2 dt 1) find a d.e verified by f 2) developpf at integr serie.
$${let}\:{f}\left({x}\right)={e}^{−{x}^{\mathrm{2}} } \:\int_{\mathrm{0}} ^{{x}} \:{e}^{{t}^{\mathrm{2}} } {dt} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{d}.{e}\:{verified}\:{by}\:{f} \\ $$$$\left.\mathrm{2}\right)\:{developpf}\:{at}\:{integr}\:{serie}. \\ $$
Commented by math khazana by abdo last updated on 11/May/18
we have f^′ (x)= −2x e^(−x^2 ) ∫_0 ^x e^t^2 dt +e^(−x^2 ) e^x^2 =−2xf(x) +1 so f issolution of the d.e. y^′ =−2xy +1 2) let put f(x)=Σ_(n=0) ^∞ a_n x^n ⇒f^′ (x)= Σ_(n=1) ^∞ na_n x^(n−1) f^′ +2xf −1=0 ⇔ Σ_(n=1) ^∞ n a_n x^(n−1) +2x Σ_(n=0) ^∞ a_n x^n −1 =0 ⇔ Σ_(n=0) ^∞ (n+1)a_(n+1) x^n + 2Σ_(n=0) ^∞ a_n x^(n+1) −1 =0⇔ Σ_(n=0) ^∞ (n+1)a_(n+1) x^n +2 Σ_(n=1) ^∞ a_(n−1) x^n −1 =0⇔ a_1 + Σ_(n=1) ^∞ {(n+1)a_(n+1) +2 a_(n−1) }x^n −1=0 ⇔ a_(1 ) =1 and ∀n≥1 (n+1)a_(n+1) +2 a_(n−1) =0 ⇒ (n+1)a_(n+1) = −2 a_(n−1) ⇒ a_(n+1) =((−2)/(n+1)) a_(n−1) ⇒ a_(2n+1) = ((−2)/(2n+1)) a_(2n−1) and a_(2n) =−(2/(2n)) a_(2n−2) =((−1)/n) a_(2n−2) Π_(k=1) ^n a_(2k+1) = (−2)^(n+1) ((Π_(k=1) ^n a_(2k−1) )/(Π_(k=0) ^n (2k+1))) .a_3 .a_5 .....a_(2n+1) = (((−2)^(n+1) )/(1.3.5....(2n+1))) a_1 .a_3 ...a_(2n−1) ⇒ a_(2n+1) = (((−2)^(n+1) )/(1.3.5.....(2n+1))) also we have Π_(k=1) ^n a_(2k) = (((−1)^n )/(n!)) Π_(k=1) ^n a_(2k−2) ⇒ a_2 .a_4 .....a_(2n) = (((−1)^n )/(n!)) a_0 a_2 ....a_(2n−2) ⇒ a_(2n) = (((−1)^n )/(n!)) a_0 f(x) = Σ_(n=0) ^∞ a_(2n) x^(2n) +Σ_(n=0) ^∞ a_(2n+1) x^(2n+1) = Σ_(n=0) ^∞ (((−1)^n )/(n!))a_0 x^(2n) +Σ_(n=0) ^∞ (((−2)^(n+1) )/(1.3.5...(2n+1))) x^(2n+1)
$${we}\:{have}\:{f}^{'} \left({x}\right)=\:−\mathrm{2}{x}\:{e}^{−{x}^{\mathrm{2}} } \:\int_{\mathrm{0}} ^{{x}} \:{e}^{{t}^{\mathrm{2}} } {dt}\:\:+{e}^{−{x}^{\mathrm{2}} } \:{e}^{{x}^{\mathrm{2}} } \\ $$$$=−\mathrm{2}{xf}\left({x}\right)\:+\mathrm{1}\:{so}\:{f}\:{issolution}\:{of}\:{the}\:{d}.{e}. \\ $$$${y}^{'} =−\mathrm{2}{xy}\:+\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{let}\:{put}\:{f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{{n}} {x}^{{n}} \Rightarrow{f}^{'} \left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\infty} {na}_{{n}} {x}^{{n}−\mathrm{1}} \\ $$$${f}^{'} \:+\mathrm{2}{xf}\:−\mathrm{1}=\mathrm{0}\:\Leftrightarrow\:\sum_{{n}=\mathrm{1}} ^{\infty} {n}\:{a}_{{n}} \:{x}^{{n}−\mathrm{1}} \\ $$$$+\mathrm{2}{x}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{{n}} {x}^{{n}} \:−\mathrm{1}\:=\mathrm{0}\:\Leftrightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left({n}+\mathrm{1}\right){a}_{{n}+\mathrm{1}} {x}^{{n}} \:\:+\:\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{{n}} {x}^{{n}+\mathrm{1}} \:−\mathrm{1}\:=\mathrm{0}\Leftrightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left({n}+\mathrm{1}\right){a}_{{n}+\mathrm{1}} \:{x}^{{n}} \:\:\:+\mathrm{2}\:\sum_{{n}=\mathrm{1}} ^{\infty} {a}_{{n}−\mathrm{1}} {x}^{{n}} \:−\mathrm{1}\:=\mathrm{0}\Leftrightarrow \\ $$$${a}_{\mathrm{1}} \:\:+\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left\{\left({n}+\mathrm{1}\right){a}_{{n}+\mathrm{1}} \:+\mathrm{2}\:{a}_{{n}−\mathrm{1}} \right\}{x}^{{n}} \:−\mathrm{1}=\mathrm{0}\:\Leftrightarrow \\ $$$${a}_{\mathrm{1}\:} =\mathrm{1}\:{and}\:\:\forall{n}\geqslant\mathrm{1}\:\:\:\left({n}+\mathrm{1}\right){a}_{{n}+\mathrm{1}} \:+\mathrm{2}\:{a}_{{n}−\mathrm{1}} =\mathrm{0}\:\Rightarrow \\ $$$$\left({n}+\mathrm{1}\right){a}_{{n}+\mathrm{1}} \:=\:−\mathrm{2}\:{a}_{{n}−\mathrm{1}} \:\Rightarrow\:{a}_{{n}+\mathrm{1}} =\frac{−\mathrm{2}}{{n}+\mathrm{1}}\:{a}_{{n}−\mathrm{1}} \:\Rightarrow \\ $$$${a}_{\mathrm{2}{n}+\mathrm{1}} =\:\frac{−\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:{a}_{\mathrm{2}{n}−\mathrm{1}} \:\:\:{and}\:{a}_{\mathrm{2}{n}} \:\:=−\frac{\mathrm{2}}{\mathrm{2}{n}}\:{a}_{\mathrm{2}{n}−\mathrm{2}} =\frac{−\mathrm{1}}{{n}}\:{a}_{\mathrm{2}{n}−\mathrm{2}} \\ $$$$\prod_{{k}=\mathrm{1}} ^{{n}} {a}_{\mathrm{2}{k}+\mathrm{1}} =\:\left(−\mathrm{2}\right)^{{n}+\mathrm{1}} \:\:\frac{\prod_{{k}=\mathrm{1}} ^{{n}} \:{a}_{\mathrm{2}{k}−\mathrm{1}} }{\prod_{{k}=\mathrm{0}} ^{{n}} \left(\mathrm{2}{k}+\mathrm{1}\right)}\: \\ $$$$.{a}_{\mathrm{3}} .{a}_{\mathrm{5}} …..{a}_{\mathrm{2}{n}+\mathrm{1}} \:=\:\frac{\left(−\mathrm{2}\right)^{{n}+\mathrm{1}} }{\mathrm{1}.\mathrm{3}.\mathrm{5}….\left(\mathrm{2}{n}+\mathrm{1}\right)}\:\:{a}_{\mathrm{1}} .{a}_{\mathrm{3}} \:…{a}_{\mathrm{2}{n}−\mathrm{1}} \\ $$$$\Rightarrow\:{a}_{\mathrm{2}{n}+\mathrm{1}} =\:\:\frac{\left(−\mathrm{2}\right)^{{n}+\mathrm{1}} }{\mathrm{1}.\mathrm{3}.\mathrm{5}…..\left(\mathrm{2}{n}+\mathrm{1}\right)}\:\:{also}\:{we}\:{have} \\ $$$$\prod_{{k}=\mathrm{1}} ^{{n}} \:{a}_{\mathrm{2}{k}} \:=\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\prod_{{k}=\mathrm{1}} ^{{n}} \:{a}_{\mathrm{2}{k}−\mathrm{2}} \:\Rightarrow \\ $$$${a}_{\mathrm{2}} .{a}_{\mathrm{4}} …..{a}_{\mathrm{2}{n}} \:=\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\:{a}_{\mathrm{0}} \:{a}_{\mathrm{2}} ….{a}_{\mathrm{2}{n}−\mathrm{2}} \Rightarrow \\ $$$${a}_{\mathrm{2}{n}} \:\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:{a}_{\mathrm{0}} \\ $$$${f}\left({x}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{\mathrm{2}{n}} \:{x}^{\mathrm{2}{n}} \:\:\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{\mathrm{2}{n}+\mathrm{1}} {x}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}{a}_{\mathrm{0}} \:{x}^{\mathrm{2}{n}} \:\:\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{2}\right)^{{n}+\mathrm{1}} }{\mathrm{1}.\mathrm{3}.\mathrm{5}…\left(\mathrm{2}{n}+\mathrm{1}\right)}\:{x}^{\mathrm{2}{n}+\mathrm{1}} \\ $$
Commented by math khazana by abdo last updated on 11/May/18
we have f(x) = e^(−x^2 ) ∫_0 ^x e^t^2 dt ⇒ f(−x) = e^(−x^2 ) ∫_0 ^(−x) e^t^2 dt =_(t=−u) e^(−x^2 ) ∫_o ^x e^u^2 (−du) = − e^x^2 ∫_0 ^x e^u^2 du =−f(x) so f is odd ⇒ a_(2n) =0 ⇒ f(x) = Σ_(n=0) ^∞ (((−2)^(n+1) )/(1.3.5....(2n+1))) x^(2n+1) .
$${we}\:{have}\:{f}\left({x}\right)\:=\:{e}^{−{x}^{\mathrm{2}} } \:\:\int_{\mathrm{0}} ^{{x}} \:\:{e}^{{t}^{\mathrm{2}} } {dt}\:\Rightarrow \\ $$$${f}\left(−{x}\right)\:=\:{e}^{−{x}^{\mathrm{2}} } \:\:\:\int_{\mathrm{0}} ^{−{x}} \:{e}^{{t}^{\mathrm{2}} } {dt} \\ $$$$=_{{t}=−{u}} \:{e}^{−{x}^{\mathrm{2}} } \:\:\int_{{o}} ^{{x}} \:\:{e}^{{u}^{\mathrm{2}} } \left(−{du}\right)\:=\:−\:{e}^{{x}^{\mathrm{2}} } \:\int_{\mathrm{0}} ^{{x}} \:{e}^{{u}^{\mathrm{2}} } {du}\:=−{f}\left({x}\right) \\ $$$${so}\:{f}\:{is}\:{odd}\:\:\Rightarrow\:{a}_{\mathrm{2}{n}} =\mathrm{0}\:\:\Rightarrow \\ $$$${f}\left({x}\right)\:\:=\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\:\frac{\left(−\mathrm{2}\right)^{{n}+\mathrm{1}} }{\mathrm{1}.\mathrm{3}.\mathrm{5}….\left(\mathrm{2}{n}+\mathrm{1}\right)}\:{x}^{\mathrm{2}{n}+\mathrm{1}} \:\:. \\ $$

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