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Question Number 207109 by MATHEMATICSAM last updated on 06/May/24

If y = (1 + x)(1 + x^2 )(1 + x^4 ) .... (1 + x^(2n) )  then find (dy/dx) at x = 0.

Ify=(1+x)(1+x2)(1+x4)....(1+x2n)thenfinddydxatx=0.

Answered by Berbere last updated on 06/May/24

y(x)=(1+x)Π_(k=1) ^n (1+x^(2k) )  ((y′)/y)=(1/(1+x))+Σ_(k=1) ^n ((2kx^(2k−1) )/(1+x^(2k) ))∣_(x=0)   ((y′(0))/(y(0)))=1⇒y′(0)=y(0)=1

y(x)=(1+x)nk=1(1+x2k)yy=11+x+nk=12kx2k11+x2kx=0y(0)y(0)=1y(0)=y(0)=1

Answered by mr W last updated on 06/May/24

y=(((1−x)(1+x)(1+x^2 )(1+x^4 )...(1+x^(2n) ))/((1−x)))  y=(((1−x^2 )(1+x^2 )(1+x^4 )...(1+x^(2n) ))/((1−x)))  ...  y=((1−x^(4n) )/(1−x))  y=1+x+x^2 +x^3 +x^4 +...+x^(4n−1)   y′=1+2x+3x^2 +4x^3 +...+(4n−1)x^(4n−2)   y′(0)=1  y′(1)=1+2+3+4+...+(4n−1)=2n(4n−1)

y=(1x)(1+x)(1+x2)(1+x4)...(1+x2n)(1x)y=(1x2)(1+x2)(1+x4)...(1+x2n)(1x)...y=1x4n1xy=1+x+x2+x3+x4+...+x4n1y=1+2x+3x2+4x3+...+(4n1)x4n2y(0)=1y(1)=1+2+3+4+...+(4n1)=2n(4n1)

Commented by manxsol last updated on 19/May/24

   ⇊

Commented by mr W last updated on 19/May/24

thanks!

thanks!

Answered by mathzup last updated on 08/May/24

ln∣y∣=ln∣1+x∣+Σ_(k=1) ^n ln(1+x^(2k) )   by derivation we get  (y^′ /y)=(1/(1+x)) +Σ_(k=1) ^n ((2k x^(2k−1) )/(1+x^(2k) )) ⇒  y^′ (x)=y(x)((1/(1+x)) +Σ_(k=1) ^n 2k(x^(2k−1) /(1+x^(2k) )))  ⇒y^′ (0)=y(o)(1+0)=y(0)=1

lny∣=ln1+x+k=1nln(1+x2k)byderivationwegetyy=11+x+k=1n2kx2k11+x2ky(x)=y(x)(11+x+k=1n2kx2k11+x2k)y(0)=y(o)(1+0)=y(0)=1

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