Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Error of a Taylor polynomial

  1. Feb 5, 2013 #1
    the error of a taylor series of order(I think that's the right word) n is given by
    [tex] \frac{f^{n+1} (s)}{n!} (x-a)^n [/tex]

    I think this is right. The error in a linear approximation would simply be

    [tex] \frac{f''(s)}{2} (x-a)^2 [/tex]

    My question is what is s and how do I find it. Use linear approximation of √(47) if you need to use an example because I just did that question so it might be easier to explain.
     
  2. jcsd
  3. Feb 6, 2013 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Actually, it should be
    $$\frac{f^{n+1} (s)}{(n+1)!} (x-a)^{n+1}$$

    I think this is right. The error in a linear approximation would simply be

    [tex] \frac{f''(s)}{2} (x-a)^2 [/tex]

    All that you know for sure is that ##s## is a number between ##a## and ##x##, and it depends on ##x##.

    OK, you didn't specify a value for ##a##, so I'll pick ##a = 49##. Then, if I still remember how to take derivatives, we should have
    $$f(x) = x^{1/2}$$
    $$f'(x) = \frac{1}{2}x^{-1/2}$$
    $$f''(x) = \frac{-1}{4}x^{-3/2}$$
    so
    $$\begin{align}f(a) + f'(a)(x-a) + \frac{1}{2}f''(s)(x-a)^2 &= (49)^{1/2} + \frac{1}{2}(49)^{-1/2}(x-49) + \frac{-1}{8}s^{-3/2}(x-49)^2\\
    &= 7 + \frac{1}{14}(x-49) - \frac{1}{8}s^{-3/2}(x-49)^2\\
    \end{align}$$
    where ##s## is some number between ##a## and ##x##. Setting ##x=47##, we get
    $$\sqrt{47} = 7 - \frac{1}{7} - \frac{1}{2}s^{-3/2}$$
    where ##s## is some number satisfying ##47 \leq s \leq 49##. Putting it another way, if we use only the linear approximation ##7 - 1/7##, then the approximation error is
    ##-\frac{1}{2}s^{-3/2}## for some ##47 \leq s \leq 49##. We can bound this error by taking the worst case ##s## (the value that maximizes the absolute approximation error), which in this case is obtained by choosing ##s=47##. Then the approximation error is guaranteed to satisfy
    $$|\textrm{approximation error}| \leq \frac{1}{2}(47)^{-3/2} = 0.00155176...$$
    Actually, we can also obtain a lower bound for the approximation error by choosing the best-case value of ##s##, namely ##s = 49##. The error will be at least ##\frac{1}{2}(49)^{-3/2} = 0.00145772...##. Thus we in fact have
    $$0.00145772... \leq |\textrm{approximation error}| \leq 0.00155176...$$
    We can check this: the actual value of ##\sqrt{47}## is
    $$\sqrt{47} = 6.8556546...$$
    and our approximation is
    $$7 - 1/7 = 6.85714285...$$
    and the error is therefore ##6.85714285... - 6.8556546... = 0.00148825...## which is indeed within the computed range.
     
    Last edited: Feb 6, 2013
  4. Feb 6, 2013 #3
    Thanks that makes more sense than my book
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Error of a Taylor polynomial
  1. Taylor Polynomials (Replies: 2)

  2. Taylor Polynomials (Replies: 2)

  3. Taylor polynomials (Replies: 1)

Loading...