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Math Help - Taylor polynomial optimal value

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    Taylor polynomial optimal value

    Given an nth order Taylor polynomial of f(x) about a number b, the error estimate is given as
    | \text{error} | \leq \frac{M}{(n + 1)!}(x - b)^{n + 1}

    provided that |f^{(n + 1)}(x) | \leq M.

    If given a greatest bound for the error I can figure out an interval J containing b such that the error is less than or equal to the bound.

    Is there any way to find an "optimal" value for M, in the sense that we find an interval J such that the error takes its maximum possible (given) value somewhere in the interval. (I guess this would also mean, "How can we find the largest possible interval J?")

    I think I'm being clear here. If not I'll give an example to work with to make things clearer.

    -Dan
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    Quote Originally Posted by topsquark View Post
    Given an nth order Taylor polynomial of f(x) about a number b, the error estimate is given as
    | \text{error} | \leq \frac{M}{(n + 1)!}(x - b)^{n + 1}

    provided that |f^{(n + 1)}(x) | \leq M.

    If given a greatest bound for the error I can figure out an interval J containing b such that the error is less than or equal to the bound.

    Is there any way to find an "optimal" value for M, in the sense that we find an interval J such that the error takes its maximum possible (given) value somewhere in the interval. (I guess this would also mean, "How can we find the largest possible interval J?")

    I think I'm being clear here. If not I'll give an example to work with to make things clearer.

    -Dan
    I do not think that \frac{M}{(n + 1)!}(x - b)^{n + 1} is the optimal upper bound. It is just an upper bound. Proofs of Taylor's theorem never say it is the best bound, just a bound. The point of it as if you make n large then the denominator will get large. And you can claim that the error is good enough (where 'good enough' depends on the context of the problem).
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