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Finding quadratic maclaurin polynomial

  • Thread starter kiwifruit
  • Start date
  • #1
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Homework Statement


the question asks to find the quadratic maclaurin polynomial for f(x)
Given f(x) = x sin(x)


The Attempt at a Solution


i know that a maclaurin series is when a=0 in a taylor series.

i did the 1st-5th derivatives of f(x) and then used the formula for taylor polynomial and set a=0.

f'=sin(x) + x cos(x)
f''=2cos(x) -x sin(x)
f'''=-3sin(x) -xcos(x)
f''''=-4cos(x) +xsin(x)
f'''''=5sin(x) +xcos(x)

with the formula for taylor polynomial

Pn(x) = n
Ʃ [f^k (a) (x-a)^k] / k!
k=0

by setting a=0 i got Pn(x)=0+0+x^2 -0- (x^4)/6............


however the answer seems to be P(x)=x^2
anyone able to explain to me how that is so?

i think i got it... is the quadratic maclaurin polynomial just the 4th order maclaurin polynomial??
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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No, the "quadratic" MacLaurin polynomial is NOT fourth degree. It is, as is any quadratic polynomial, second degree. You are correct that the function and first derivative are both 0 at x= 0 so the first non-zero coefficient is the second degree term. The "quadratic MacLaurin polynomial" for this function is just that second degree term, [itex]x^2[/itex].

Graph both y= x sin(x) and [itex]y=x^2[/itex] on the same coordinates. You should find that they remarkably similar close to the origin.
 

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