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Question Number 158469 by zakirullah last updated on 04/Nov/21
find the maclaurin series expension  for the function f(x) = sin^2 x ;      x_o = 0
findthemaclaurinseriesexpensionforthefunctionf(x)=sin2x;xo=0
Answered by TheSupreme last updated on 04/Nov/21
f(x)=sin^2 (x)  f′(x)=2sin(x)cos(x)=sin(2x)  f(x)=Σ_(i=1) ^∞ D^i (sin(2x)) (x^(i+1) /(i+1!))  D^i (sin(2x))=...  i=2n  D^(2n) (sin(2x))_(x=0) =0  i=2n+1  D^(2n+1) (sin(2x))=(−1)^n 2^(n+1) cos(2x)=2(−2)^n   f(x)=Σ(((−2)^n 2x^(2n+2) )/((2n+2)!))
f(x)=sin2(x)f(x)=2sin(x)cos(x)=sin(2x)f(x)=i=1Di(sin(2x))xi+1i+1!Di(sin(2x))=i=2nD2n(sin(2x))x=0=0i=2n+1D2n+1(sin(2x))=(1)n2n+1cos(2x)=2(2)nf(x)=Σ(2)n2x2n+2(2n+2)!
Commented by zakirullah last updated on 04/Nov/21
Great sir G  Sir how 2sin(x)cos(x) = sin(2x)
GreatsirGSirhow2sin(x)cos(x)=sin(2x)
Commented by MJS_new last updated on 05/Nov/21
2sin x cos x =2×((e^(ix) −e^(−ix) )/(2i))×((e^(ix) +e^(−ix) )/2)=       [(a−b)(a+b)=a^2 +b^2 ]  =((e^(2ix) −e^(−2ix) )/(2i))=sin 2x
2sinxcosx=2×eixeix2i×eix+eix2=[(ab)(a+b)=a2+b2]=e2ixe2ix2i=sin2x
Commented by zakirullah last updated on 05/Nov/21
ok sir you′r the great
oksiryourthegreat
Commented by MJS_new last updated on 05/Nov/21
I′m just a tiny snip
Imjustatinysnip
Answered by mr W last updated on 04/Nov/21
we know cos x=Σ_(n=0) ^∞ (((−1)^n x^(2n) )/((2n)!)).  f(x)=sin^2  x=((1−cos (2x))/2)  f(x)=(1/2)−(1/2)Σ_(n=0) ^∞ (((−1)^n (2x)^(2n) )/((2n)!))  f(x)=Σ_(n=1) ^∞ (((−1)^(n+1) 2^(2n−1) x^(2n) )/((2n)!))
weknowcosx=n=0(1)nx2n(2n)!.f(x)=sin2x=1cos(2x)2f(x)=1212n=0(1)n(2x)2n(2n)!f(x)=n=1(1)n+122n1x2n(2n)!

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