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Taylor polynomial of f(x) = 1/(1-x) and the estimate of its remainder

  • Thread starter Ryker
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  • #1
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Homework Statement


Find the Taylor polynomial for f(x) = 1/(1-x), n = 5, centered around 0. Give an estimate of its remainder.

The Attempt at a Solution


I found the polynomial to be 1 + x + x2 + x3 + x4 + x5, and then tried to take the Lagrange form of the remainder, say, for x in [-1/2, 1/2].

Then I get
[tex]R_{5}(x) = \frac{f^{6}(a)}{6!}x^{6}[/tex]

But
[tex]f^{6}(a) = \frac{720}{(1-a)^{7}},[/tex]
so for the largest values of a and x, that is 1/2, you get that
[tex]|R_{5}(x)| \leq \frac{(\frac{1}{2})^{6}}{(\frac{1}{2})^{7}} = 2,[/tex]
which means the best estimate we can give is that the remainder is less than 2. But that doesn't make much sense to me, because the 5-th degree polynomial is fairly accurate on [-1/2, 1/2] and really close to f(x), so how I am getting such results? Have I made a mistake somewhere?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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That looks fine to me. The remainder 'estimate' isn't really an estimate of the remainder. It's a maximum upper bound for the remainder. In this case it's a pretty pessimistic estimate, probably because you're nearing the singularity at x=1.
 
  • #3
1,086
2
Hey, thanks for the reply. Yeah, I figured it's only the maximum upper bound, so should I have taken a smaller interval? I mean, we aren't really given any specific range, so I guess we can choose our own, it's just that I didn't want to choose one, smaller than [-1/2, 1/2]. Or am I right in thinking none of this really matters, as long as you show how you got to that estimate?
 
  • #4
Dick
Science Advisor
Homework Helper
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Hey, thanks for the reply. Yeah, I figured it's only the maximum upper bound, so should I have taken a smaller interval? I mean, we aren't really given any specific range, so I guess we can choose our own, it's just that I didn't want to choose one, smaller than [-1/2, 1/2]. Or am I right in thinking none of this really matters, as long as you show how you got to that estimate?
I don't see anything wrong with what you've got. Maybe they just want the expression for R_5, not a specific number.
 
  • #5
1,086
2
I put down both just in case anyway, so I guess I should be covered then :smile: Thanks again.
 

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