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Question Number 36926 by maxmathsup by imad last updated on 07/Jun/18

let f(x) = e^(−x^2 ) 1) prove that f^((n)) (x)=p_n (x)e^(−x^2 ) with p_n is a polynom 2) find a relation of recurrence between the p_n 3) calculate p_1 ,p_2 ,p_3 ,p_4

$${let}\:{f}\left({x}\right)\:=\:{e}^{−{x}^{\mathrm{2}} } \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{f}^{\left({n}\right)} \left({x}\right)={p}_{{n}} \left({x}\right){e}^{−{x}^{\mathrm{2}} } \:\:{with}\:{p}_{{n}} \:{is}\:{a}\:{polynom} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{a}\:{relation}\:{of}\:{recurrence}\:{between}\:{the}\:{p}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:{p}_{\mathrm{1}} ,{p}_{\mathrm{2}} ,{p}_{\mathrm{3}} ,{p}_{\mathrm{4}} \\ $$

Answered by math khazana by abdo last updated on 10/Jun/18

we have f^′ (x) =−2x e^(−x^2 ) f^((2)) (x) =−2 e^(−x^2 ) +4x^2 e^(−x^2 ) =(4x^2 −2)e^(−x^2 ) let suppose that f^((n)) (x) =p_n (x) e^(−x^2 ) with p_n a polynome ⇒ f^((n+1)) (x)= (p_n (x)e^(−x^2 ) )^′ =p_n ^′ (x) e^(−x^2 ) −2x p_n (x) e^(−x^2 ) ={ p_n ^′ (x) −2x p_n (x)}e^(−x^2 ) =p_(n+1) (x)e^(−x^2 ) with p_(n+1) (x)= p_n ^′ (x) −2x p_n (x) and its clear that deg(p_n ) =n 2) p_(n+1) (x)= −2xp_n (x) +p_n ^′ (x) 3) p_1 (x) =−2x and p_2 (x) =−2xp_1 (x) +p_1 ^′ (x) =4x^2 −2 p_3 (x) =−2xp_2 (x) +p_2 ^′ (x) =−2x(4x^2 −2) +8x=−8x^3 +4x +8x =−8x^3 +12x p_4 (x) =−2xp_3 (x) +p_3 ^′ (x) =−2x(−8x^3 +12x) −24x^2 +12 =16 x^4 −24x^2 −24x^2 +12 =16 x^4 −48x^2 +12 .

$${we}\:{have}\:{f}^{'} \left({x}\right)\:=−\mathrm{2}{x}\:{e}^{−{x}^{\mathrm{2}} } \\ $$$${f}^{\left(\mathrm{2}\right)} \left({x}\right)\:=−\mathrm{2}\:{e}^{−{x}^{\mathrm{2}} } \:+\mathrm{4}{x}^{\mathrm{2}} \:{e}^{−{x}^{\mathrm{2}} } \:=\left(\mathrm{4}{x}^{\mathrm{2}} \:−\mathrm{2}\right){e}^{−{x}^{\mathrm{2}} } \\ $$$${let}\:{suppose}\:{that}\:{f}^{\left({n}\right)} \left({x}\right)\:={p}_{{n}} \left({x}\right)\:{e}^{−{x}^{\mathrm{2}} } \:{with}\:{p}_{{n}} \\ $$$${a}\:{polynome}\:\Rightarrow\:{f}^{\left({n}+\mathrm{1}\right)} \left({x}\right)=\:\left({p}_{{n}} \left({x}\right){e}^{−{x}^{\mathrm{2}} } \right)^{'} \\ $$$$={p}_{{n}} ^{'} \left({x}\right)\:{e}^{−{x}^{\mathrm{2}} } \:−\mathrm{2}{x}\:{p}_{{n}} \left({x}\right)\:{e}^{−{x}^{\mathrm{2}} } \\ $$$$=\left\{\:{p}_{{n}} ^{'} \left({x}\right)\:−\mathrm{2}{x}\:{p}_{{n}} \left({x}\right)\right\}{e}^{−{x}^{\mathrm{2}} \:} \:\:\:={p}_{{n}+\mathrm{1}} \:\left({x}\right){e}^{−{x}^{\mathrm{2}} } \\ $$$${with}\:\:{p}_{{n}+\mathrm{1}} \left({x}\right)=\:{p}_{{n}} ^{'} \left({x}\right)\:−\mathrm{2}{x}\:{p}_{{n}} \left({x}\right)\:{and}\:{its}\:{clear} \\ $$$${that}\:{deg}\left({p}_{{n}} \right)\:={n} \\ $$$$\left.\mathrm{2}\right)\:{p}_{{n}+\mathrm{1}} \left({x}\right)=\:−\mathrm{2}{xp}_{{n}} \left({x}\right)\:+{p}_{{n}} ^{'} \left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{p}_{\mathrm{1}} \left({x}\right)\:=−\mathrm{2}{x}\:\:{and}\:{p}_{\mathrm{2}} \left({x}\right)\:=−\mathrm{2}{xp}_{\mathrm{1}} \left({x}\right)\:+{p}_{\mathrm{1}} ^{'} \left({x}\right) \\ $$$$=\mathrm{4}{x}^{\mathrm{2}} \:−\mathrm{2} \\ $$$${p}_{\mathrm{3}} \left({x}\right)\:=−\mathrm{2}{xp}_{\mathrm{2}} \left({x}\right)\:+{p}_{\mathrm{2}} ^{'} \left({x}\right)\: \\ $$$$=−\mathrm{2}{x}\left(\mathrm{4}{x}^{\mathrm{2}} \:−\mathrm{2}\right)\:+\mathrm{8}{x}=−\mathrm{8}{x}^{\mathrm{3}} \:+\mathrm{4}{x}\:+\mathrm{8}{x} \\ $$$$=−\mathrm{8}{x}^{\mathrm{3}} \:+\mathrm{12}{x} \\ $$$${p}_{\mathrm{4}} \left({x}\right)\:=−\mathrm{2}{xp}_{\mathrm{3}} \left({x}\right)\:+{p}_{\mathrm{3}} ^{'} \left({x}\right) \\ $$$$=−\mathrm{2}{x}\left(−\mathrm{8}{x}^{\mathrm{3}} \:+\mathrm{12}{x}\right)\:−\mathrm{24}{x}^{\mathrm{2}} \:+\mathrm{12} \\ $$$$=\mathrm{16}\:{x}^{\mathrm{4}} \:\:−\mathrm{24}{x}^{\mathrm{2}} \:−\mathrm{24}{x}^{\mathrm{2}} \:+\mathrm{12} \\ $$$$=\mathrm{16}\:{x}^{\mathrm{4}} \:−\mathrm{48}{x}^{\mathrm{2}} \:+\mathrm{12}\:. \\ $$

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