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f-n-1-x-f-n-x-x-n-x-n-f-0-x-1-f-5-x-f-4-2x-0-x-

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Question Number 1873 by 123456 last updated on 19/Oct/15
f_(n+1) (x)=f_n (x)(x−n)(x+n) f_0 (x)=1 f_5 (x)=? f_4 (2x)=0,x=?
fn+1(x)=fn(x)(x−n)(x+n)f0(x)=1f5(x)=?f4(2x)=0,x=?
Answered by Rasheed Soomro last updated on 19/Oct/15
f_(n+1) (x)=f_n (x)(x−n)(x+n)...............I f_0 (x)=1 f_5 (x)=? f_4 (2x)=0,x=? −−−−−−−−−−−−−−− •Let n=0, substuting in I f_1 (x)=f_0 (x)(x−0)(x+0)=x^2 •Let n=1,substituting in I f_2 (x)=f_1 (x)(x−1)(x+1) =x^2 (x−1)(x+1) •Let n=2 ,substituting in I f_3 (x)=f_2 (x)(x−2)(x+2) =x^2 (x−1)(x+1)(x−2)(x+2) =x^2 (x^2 −1)(x^2 −4) •Let n=3 , substituting in I f_4 (x)=f_3 (x)(x−3)(x+3) =x^2 (x^2 −1)(x^2 −4)(x^2 −9) f_4 (2x)=4x^2 (4x^2 −1)(4x^2 −4)(4x^2 −9)=0 4x^2 =0 ∣ 4x^2 −1=0 ∣ 4x^2 −4=0 ∣ 4x^2 −9=0 x=0 ∣ x=±(1/2) ∣ x=±1 ∣ x=±(3/2)....................A •Let n=4 , substituting in I f_5 (x)=f_4 (x)(x−4)(x+4) =x^2 (x^2 −1)(x^2 −4)(x^2 −9)(x^2 −16)..........B A and B are answers.
fn+1(x)=fn(x)(x−n)(x+n)……………If0(x)=1f5(x)=?f4(2x)=0,x=?−−−−−−−−−−−−−−−∙Letn=0,substutinginIf1(x)=f0(x)(x−0)(x+0)=x2∙Letn=1,substitutinginIf2(x)=f1(x)(x−1)(x+1)=x2(x−1)(x+1)∙Letn=2,substitutinginIf3(x)=f2(x)(x−2)(x+2)=x2(x−1)(x+1)(x−2)(x+2)=x2(x2−1)(x2−4)∙Letn=3,substitutinginIf4(x)=f3(x)(x−3)(x+3)=x2(x2−1)(x2−4)(x2−9)f4(2x)=4x2(4x2−1)(4x2−4)(4x2−9)=04x2=0∣4x2−1=0∣4x2−4=0∣4x2−9=0x=0∣x=±12∣x=±1∣x=±32………………..A∙Letn=4,substitutinginIf5(x)=f4(x)(x−4)(x+4)=x2(x2−1)(x2−4)(x2−9)(x2−16)……….BAandBareanswers.

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