Finding the mclauren series forthe following function

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The discussion focuses on finding the Maclaurin series for the function f(x) = xe^(-2x). The derivatives of the function are calculated, yielding values at x=0 for f(0), f'(0), f''(0), f'''(0), fiv(0), and fv(0). The series is expressed as a power series involving these derivatives, but the user struggles to identify a pattern in the coefficients. A potential regularity is observed in the coefficients, suggesting a relationship with powers of 2. The conversation concludes with an acknowledgment of the helpfulness of this insight.
Dell
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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??
 
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0= 0(2^{-1})
1= 1(2^0)
4= 2(2^1)
12= 3(2^2)
32= 4(2^3)
80= 5(2^4)

Does that help?
 
helps a lot, thanks, how did you see that?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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