Then at x 0 we have f x 1 1 3 x X n 0 1 n 3 x n X n 0 1 n 3 n x n x 1 3 At x 2

# Then at x 0 we have f x 1 1 3 x x n 0 1 n 3 x n x n 0

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Then atx= 0, we havef(x) =11 + 3x=Xn=0(-1)n(3x)n=Xn=0(-1)n3nxn,∀|x|<13,Atx= 2, we havef(x) =11 + 3x=17 + 3(x-2)=17·11 +3(x-2)7=17Xn=0(-1)n3(x-2)n7n149
=Xn=0(-1)n3n7n+1(x-2)n,∀|x-2|<73.[Example]Find Taylor series forf(x) =sin(3x+ 1) atx= 1.Solution:By the formulasin(A+B) =sin A cos B+cos A sin B, we havef(x) =sin(3x+ 1) =sin[3(x-1) + 4] =sin[3(x-1)]cos4 +cos[3(x-1)]sin4=cos4Xn=0(-1)n[3(x-1)]2n+1(2n+ 1)!+sin4Xn=0(-1)n[3(x-1)]2n(2n)!=cos4Xn=0(-1)n32n+1(2n+ 1)!(x-1)2n+1+sin4Xn=0(-1)n32n(2n)!(x-1)2n,xR.[Example]Find the coefficients, up to degree 4, of the Taylor series forf(x) =exsin(3x)atx= 0.Proof:f(x) =1 +x+x22!+x33!+....3x-(3x)33!+(3x)55!-...= 3x+ 3x2+32!-333!x3+-333!+33!x4+...= 3x+ 3x2-62x3-4x4+......Some problems discussed in the classA.Find the Taylor series ofsin(x2)atx= 0.Solution:sin(x2) = (x2)-(x2)33!+(x2)55!-(x2)77!+......=Xn=0(-1)n(x2)2n+1(2n+ 1)!=Xn=0(-1)nx2(2n+1)(2n+ 1)!.B.Find the Taylor series ofsin xxatx= 0.Solution:sin xx=x-(x2)33!+(x2)55!-(x2)77!+......x=n=0(-1)n(x)2n+1(2n+1)!x=Xn=0(-1)nx2n(2n+ 1)!.150
C.Find the Taylor series ofcos2xatx= 0.Solution:We noticecos2x=1 +cos(2x)2.Sincecos(2x) = 1-(2x)22!+(2x)44!-(2x)66!+......=Xn=0(-1)n(2x)2n(2n)!=Xn=0(-1)n22nx2n(2n)!.Thereforecos2x=1 +cos(2x)2=12+12Xn=0(-1)n22nx2n(2n)!= 1 +Xn=1(-1)n22n-1x2n(2n)!.D.Find the Taylor series ofx2exatx= 0.Solution:x2ex=x2Xn=0xnn!=Xn=0xn+2n!.E.Find the coefficients, up to degree 2, of Taylor series forex1-xatx= 0.Solution:ex1-x=1+x+x22!+...1+x+x2+x3+...= 1+2x+1+1+12!x2+[higher degree terms].F. Fund Taylor series off(x) = (1 +x)e-xatx= 0.Solution:f(x) = (1+x)Xn=0(-x)nn!=Xn=0(-1)nxnn!+Xn=0(-1)nxn+1n!= 1+Xn=1(-1)nxnn!+Xn=1(-1)n+1xn(n-1)!= 1 +Xn=1(-1)nn!+(-1)n+1(n-1)!xn.G. Find coefficients up to degree 6 forf(x) =ln cos xatx= 0.Solution:f(x) =ln1-x22!+x44!-x66!+...=ln1-x22!-x44!+x66!-...151
=-x22!-x44!+x66!-...-12x22!-x44!

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