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Trouble with taylor's Inequality

  1. Aug 8, 2007 #1
    1. The problem statement, all variables and given/known data
    Ok so I have been having trouble computing [tex] |R_{n}|[/tex]
    in problems.

    2. Relevant equations
    [tex]|R_{n}|\leq \frac{M}{(n+1)!}|x-a|[/tex]^(n+1)
    [tex] M\geq f[/tex]^(n+1) (h)

    3. The attempt at a solution
    Can some one walk me through calculatng some generic solution.
    Like finding M and all that? I need more insight on what it actually does too.
    Last edited: Aug 9, 2007
  2. jcsd
  3. Aug 9, 2007 #2


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    if f is differentiable, you can use the methods of calculus to find the extremas of f and hence its absolute max. such a number is as M in your post.

    in other cases, you can find such an M w/o calculus. If foor instance, f(x)=4sin(x). Well, |sin(x)|[itex]\leq[/itex]1, so |f(x)|[itex]\leq[/itex]4. So M=4 will do
  4. Aug 9, 2007 #3


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    If you use the nth Taylor polynomial for ex, about x= 0, the error is less than
    where M is an upperbound on the n+1 derivative between 0 and x. Of course, every derivative of ex is ex itself so we just need an upperbound on that. For example if x= 1, We know the derivative is less than e so we can use that as an upper bound. The error is less than

    Calculating errors for the Taylor Polynomials for sine and cosine is even easier: all derivatives are either sine or cosine and 1 is always an upper bound.
  5. Aug 9, 2007 #4
    Ok so suppose I have my function [tex]f(x)=\sqrt(x)[/tex] and the interval [.5,1.5] using n=2 centered at 1. So:
    [tex] |R_{2}|\leq \frac{M}{3!} |x-1|^3[/tex]
    [tex] f^3(x)= \frac{3}{8x^(5/2)}\leq M[/tex]. My upperbound on [tex]f^3(x)[/tex] is 1.5 so [tex] f^3(1.5)=\frac{\sqrt(3)}{72}[/tex]
    So: [tex] |R_{2}|\leq \frac{\sqrt(3)}{72*3!} |x-1|^3[/tex] X=1.5
    So: So: [tex] |R_{2}|\leq \frac{\sqrt(3)}{72*3!} |.5|^3[/tex]
    How does that sound?
    Last edited: Aug 9, 2007
  6. Aug 9, 2007 #5
    Say [tex]f(x) = \sqrt{x+1}[/tex] centered at 0, the reason why I am doing it this way is because you can get same result upon a basic substitution. On [tex](-1,1)[/tex].

    [tex]f(x) = (x+1)^{1/2} \implies f'(0)=1[/tex]
    [tex]f'(x) = \frac{1}{2}(x+1)^{-1/2} \implies f'(0)=\frac{1}{2}[/tex]
    [tex]f''(x) = - \frac{1}{4}(x+1)^{-3/2} \implies f''(0) = -\frac{1}{4}[/tex]

    That means for some [tex]y[/tex] between [tex]x\not =0[/tex] and [tex]0[/tex] we have:
    [tex]R_2(x) = f(x) - T_2(x) = \frac{f'''(y)}{3!}x^3[/tex]

    [tex]\left| \frac{f'''(y)}{3!}x^3 \right| = \left| \frac{3}{3!\cdot 8}\cdot \frac{(y+1)^{-5/2}}{1} \cdot x^3 \right| \leq \frac{3}{3!\cdot 8}[/tex]
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