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Finding the mclauren series forthe following function

  1. May 17, 2009 #1
    f(x)=xe-2x

    fx=xe-2x
    f'x=e-2x-2xe-2x
    f''x=-2e-2x-2(e-2x-2xe-2x)=-4e-2x+4xe-2x
    f'''x=8e-2x+4(e-2x-2xe-2x)=12e-2x-8xe-2x
    fivx=-24e-2x-8(e-2x-2xe-2x)=-32e-2x+16xe-2x
    fvx=64e-2x+16(e-2x-2xe-2x)

    f(0)=0
    f'(0)=1
    f''(0)=-4
    f'''(0)=12
    fiv(0)=-32
    fv(0)=80

    fx=0+x/(1!)-4x2/(2!)+12x3/(3!)-12x4/(4!)+80x5/(5!)......
    =0+x-2x2+2x3-4x4/3+2x5/3....


    i need to express this series as a progression but cannot find the regularity of it, can anyone see anything or help me to see it??
     
  2. jcsd
  3. May 17, 2009 #2

    HallsofIvy

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    Science Advisor

    [itex]0= 0(2^{-1})[/itex]
    [itex]1= 1(2^0)[/itex]
    [itex]4= 2(2^1)[/itex]
    [itex]12= 3(2^2)[/itex]
    [itex]32= 4(2^3)[/itex]
    [itex]80= 5(2^4)[/itex]

    Does that help?
     
  4. May 17, 2009 #3
    helps a lot, thanks, how did you see that?
     
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