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Question Number 50375 by prof Abdo imad last updated on 16/Dec/18
let f(x)=(1/(cosx)) prove that f^()n)) (x)=((p_n (sinx))/(cos^(n+1) x)) with p_n is apolynom 2) calculate p_1 ,p_2 and p_3 3) detdrmine p_n (1).
letf(x)=1cosxprovethatf)n)(x)=pn(sinx)cosn+1xwithpnisapolynom2)calculatep1,p2andp33)detdrminepn(1).
Answered by kaivan.ahmadi last updated on 10/Jan/19
f′(x)=((sinx)/(cos^2 x))⇒p_1 (x)=x f^(′′) (x)=((cos^3 x+2sin^2 xcosx)/(cos^4 x))=((cos^2 x+2sin^2 x)/(cos^3 x))= ((1+sin^2 x)/(cos^3 x))⇒p_2 (x)=x^2 +1 f^((3)) (x)=((2sinxcos^4 x+3sinxcos^2 x(1+sin^2 x))/(cos^6 x))= ((2sinxcos^4 x+3sinxcos^2 x+3sin^3 xcos^2 x)/(cos^6 x))= ((2sinxcos^2 x+3sinx+3sin^3 x)/(cos^4 x))= ((5sinx+sin^3 x)/(cos^4 x))⇒p_3 (x)=x^3 +5x ⋮ f^((4)) (x)=((5cos^5 x+3sin^2 xcos^5 x+20sin^2 xcos^3 x+4sin^4 xcos^3 x)/(cos^8 x))= ((5cos^2 x+3sin^2 xcos^2 x+20sin^2 x+4sin^4 x)/(cos^5 x))= ((5−5sin^2 x+3sin^2 x−3sin^4 x+20sin^2 x+4sin^4 x)/(cos^5 x))= ((5+18sin^2 x+sin^4 x)/(cos^5 x))⇒p_4 (x)=x^4 +18x^2 +5 ⋮ p_1 (1)=1=1! p_2 (1)=2=2! p_3 (1)=6=3! p_4 (1)=24=4! ⋮ p_n (1)=n!
f′(x)=sinxcos2x⇒p1(x)=xf″(x)=cos3x+2sin2xcosxcos4x=cos2x+2sin2xcos3x=1+sin2xcos3x⇒p2(x)=x2+1f(3)(x)=2sinxcos4x+3sinxcos2x(1+sin2x)cos6x=2sinxcos4x+3sinxcos2x+3sin3xcos2xcos6x=2sinxcos2x+3sinx+3sin3xcos4x=5sinx+sin3xcos4x⇒p3(x)=x3+5x⋮f(4)(x)=5cos5x+3sin2xcos5x+20sin2xcos3x+4sin4xcos3xcos8x=5cos2x+3sin2xcos2x+20sin2x+4sin4xcos5x=5−5sin2x+3sin2x−3sin4x+20sin2x+4sin4xcos5x=5+18sin2x+sin4xcos5x⇒p4(x)=x4+18x2+5⋮p1(1)=1=1!p2(1)=2=2!p3(1)=6=3!p4(1)=24=4!⋮pn(1)=n!

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