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Analysis: Taylor Polynomial Approximation

  1. Nov 15, 2011 #1
    For #4, I'm mostly confident I did it correctly. In determining the error, we're supposed to find the maximum absolute value on an interval I. I set I = (0,2pi). Is that right?

    http://i111.photobucket.com/albums/n149/camarolt4z28/4-1.png [Broken]

    For #5,

    http://i111.photobucket.com/albums/n149/camarolt4z28/5.png [Broken]

    For #5, again, I'm mostly confident I did it correctly. But, again, in determining the error, we're supposed to find the maximum absolute value on an interval I. I set I = (0, .2). Is that right? I think I'm okay on b, too. I just want to make sure. I was a bit delinquent in taking notes in class when we covered this. Thanks.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 15, 2011 #2
    For number 4, why did you pick [itex]x_0 = 0[/itex] in your equation for p3?
     
  4. Nov 15, 2011 #3
    For 4(b)? I didn't know what to assign x0, so I picked zero. What should I use? How do I know what to use?
     
  5. Nov 15, 2011 #4
    What did you use for the first half of this question? Why change?
     
  6. Nov 15, 2011 #5
    Good question. I didn't make the connection in incorporating the two problems. I have to redo the problem.

    Also, am I correct in identifying the proper intervals for determining maximum absolute value of the n+1 derivative?
     
    Last edited: Nov 15, 2011
  7. Nov 15, 2011 #6
    Looks like your using the Lagrange form for error estimates. Your interval for problem 5b is correct, but review problem 4b again based on the other most recent corrections.
     
  8. Nov 15, 2011 #7
    For 5(a), I set x0 = 0. Is that a correct assumption? I set the interval I = [0, .2]. In general, what do I make the interval? Is it between x and x0?

    For 4(a), I redid the polynomial with x0 = pi/4 and x = .266pi. What do I make the interval I for the error estimation?
     
  9. Nov 15, 2011 #8
    You should review your notes, text, or explore the web. The interval is such that [itex]x_0 < c < x[/itex] (sometimes expressed as [itex]a < \xi_L < x[/itex]).
     
    Last edited: Nov 15, 2011
  10. Nov 15, 2011 #9
    What do you think I've been doing? The answer isn't clear to me. That's why I'm asking it.

    According to what I've read, you want to find the maximum value of the derivative in between x and x0. In 4(b), it would be between pi/4 and .266pi.
     
    Last edited: Nov 15, 2011
  11. Nov 15, 2011 #10
    Shackleford - Calm down, take a deep breath - your answer for 5a is correct. But I guess you are having some difficulty understanding why, so I'l try again. Please be patient.

    I think you know the general form of the Taylor Series so I won't waste time going through that. But I want to make sure you understand what we are really saying when we look for "a Taylor expansion about [itex]x_0[/itex]."

    The resulting expression is a local approximation of the function such that it approximates the function within an interval about the point [itex]x_0[/itex]. By definition, the approximation is exact at [itex]x_0[/itex].

    Also by definition, the Taylor series is an infinite sum and the function equates to this sum if it converges. When we take a truncation of this infinite series, the function is approximated because we are taking a finite number of terms.

    The error term (ie, remainder term) resulting from the difference between the function and the approximation by the truncated series is typically a generalization based on certain restrictions on the function ([itex]C^n[/itex] continuity, [itex]f^{(n+1)}[/itex] exists, etc) and the Mean Value Theorem that allow us to construct the remainder term in such an abbreviated expression as the one you have been using.

    Let's restate the Theorem:
    Suppose [itex]f\in C^n[a,b][/itex] and [itex]f^{(n+1)}[/itex] exists on [itex][a,b][/itex]. Let [itex]x_0\in[a,b][/itex]. For every [itex]x \in [a,b][/itex], there exist [itex]\xi(x)[/itex] between [itex]x_0[/itex] and [itex]x[/itex] with
    [tex]
    f(x) = P_n(x)+R_n(x)
    [/tex]
    where
    [tex]
    P_n(x) = \sum_{k=0}^n{ \frac{f^k(x_0)}{k!}\,(x-x_0)^k}
    [/tex]
    and
    [tex]
    R_n(x)=\frac{f^{(n+1)}(\xi(x))}{(n+1)!} \, (x-x_0)^{n+1}
    [/tex]

    Now, the [itex]f^{(n+1)}(\xi(x))[/itex] term in the expression for [itex]R_n[/itex] is written that way since we really don't know where that point [itex]\xi[/itex] is since the Mean Value Theorem was used to generalize all the remaining terms (we didn't show these steps). We do know it is somewhere between [itex]x_0[/itex] and [itex]x[/itex].

    Now lets look at your example problems.

    Problem 4a: you recognized correctly that, since we wanted a polynomial in terms of [itex](x-\pi/4)[/itex], the function needs to be expanded about [itex]x_0=\pi/4[/itex]. Your [itex]p_4[/itex] expansion is correct.

    Problem 4b: Continuing with the development of 4a, approximate [itex]f(x)[/itex] with [itex]p_3[/itex] and [itex]x=48^\circ[/itex]. Since the value of [itex]x[/itex] is very close to [itex]x_0=45^\circ[/itex] (recall [itex]45^\circ[/itex] corresponds to [itex]\pi/4[/itex]radians) from problem 4a, use the terms of the polynomial you developed before up to and including the third order term to construct [itex]p_3[/itex]. The terms evaluate to powers of [itex](x-\pi/4)[/itex] as before or [itex](48-45) \pi / 180= \pi/60[/itex] radians.

    Now, since [itex]x=(48/180)\pi = 0.2667 \pi[/itex] and [itex]x_0=0.2500 \pi[/itex], we know somewhere in that interval there is a value [itex]\xi[/itex] such that [itex]f^{(n+1)}(\xi)[/itex]will yield the correct term to summarize the error in the expression for [itex]R_n[/itex]. But we don't have to know the exact value if we just want to bound the error, that is, to find the maximum magnitude of the error will only require us to find the maximum value (or approximation) of [itex]f^{(n+1)}(\xi)[/itex] over the interval [itex]x_0<\xi<x[/itex].

    For your problem, we need to bound [itex]f^{(4)}(\xi) = \cos(\xi)[/itex]. Thus, we should use [itex]f^{(4)}(\xi) = \cos(\xi)< \cos(\pi/4)[/itex] since the cosine function is monotonically decreasing for increasing [itex]x[/itex] in this interval (ie, [itex]\cos(x)[/itex] gets smaller with increasing values of [itex]x[/itex]). Hence, the error term is approximated as
    [tex]
    R_3 \le \frac{\cos(\pi/4)}{4!} \, (\pi/60)^4 \approx 2.21 \times 10^{-7}
    [/tex]

    I know this is a bit much to swallow, but I hope it helps. Let me know if it doesn't and I try some more.
     
  12. Nov 15, 2011 #11
    Yeah, I was able to write down the correct expression for 4(b).

    For 5(a), it says denote the Taylor polynomial of degree n in powers of x for f. Of course, the general form is x - x0, so that's how I know we're looking at the function around x0 = 0.

    You're right. I needed just a little bit of clarification on the theory and how it ties into the computation. Thanks. I'm stumped on the last problem, so I may post it here sometime tonight.
     
  13. Nov 15, 2011 #12
    Do I actually have to prove anything? Just by looking at the notes and the definition, I can clearly see that f is uniformly continuous since the derivative is bounded.

    http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled.png

    According to the notes:

     
  14. Nov 16, 2011 #13
    Please, disregard the previous post. The professor went over this problem today in class.
     
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