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Math Help - Bound of error(Lagrange Interpolation)

  1. #1
    Member javax's Avatar
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    Bound of error(Lagrange Interpolation)

    Hello everybody

    Given the Lagrange polynomial with the table:

    \begin{tabular} {|c|c|c|c|}<br />
\hline<br />
x & 8.3 & 8.6 & 8.7 \\<br />
\hline<br />
f(x) & 17.56492 & 18.50515 & 18.32091 \\<br />
\hline<br />
\end{tabular}

    Find P_2(x) for x = 8.4, the actual error and the bound of error for the function f(x) = x \ln x

    Ok now I've found P_2(x) and I'm approximating P_2(8.4) which from my polyinomial is 17.877155 (close enough I think).
    But I don't know how to find the bound of error using the remainder term! Anyone please show me the steps at least. Thanks in advance.
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  2. #2
    MHF Contributor arbolis's Avatar
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    Maybe this thread can help you : http://www.mathhelpforum.com/math-he...-integral.html.
    Post #10
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  3. #3
    Member javax's Avatar
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    Ok thanks...I want to know if this works.
    Applying the formula:

    |f(x)-P_2(x)|\leq \frac{sup_{[a;b]}|f^{(n+1)}(x)|}{(n+1)!}\prod_{i=0}^{n}{(x-x_i)}

    now setting a=8.3, b=8.7, I've got f^{(3)}(x)=-\frac{1}{x^2}

    Is it okay to choose sup_{[a;b]}|f^{(3)}(x)| = \frac{1}{8.3^2} since |f^{(3)}(x)| is decreasing on [a, b] ?
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  4. #4
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by javax View Post

    Is it okay to choose sup_{[a;b]}|f^{(3)}(x)| = \frac{1}{8.3^2} since |f^{(3)}(x)| is decreasing on [a, b] ?
    I think so.
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