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Finding the mclauren series forthe following function

  • Thread starter Dell
  • Start date
  • #1
590
0
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??
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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[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?
 
  • #3
590
0
helps a lot, thanks, how did you see that?
 

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