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- Thread starter jordanl122
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- #2

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[tex]\frac{f^{(n+1)}(c)}{(n+1)!}(x-x_{0})^{n+1}[/tex]

This is almost the same as the expression for the [itex](n+1)[/itex]th term, but (as you seem to have noticed) the derivative of [itex]f[/itex] is evaluated at some [itex]c[/itex] between [itex]x[/itex] and [itex]x_0[/itex].

Basically the [itex]c[/itex] can come about as a result of using the mean value theorem to justify the remainder term. Actually finding [itex]c[/itex] may be tricky, but because you know it's between x and [itex]x_{0}[/itex], you can find the maximum value of the error based on that. For example, if you have an expansion which gives an error term [itex]R_{5}[/itex] of

[tex] \frac{\cos c}{3!}x^{5}[/tex]

you know that the maximum value of the error is

[tex] \frac{1}{5!}x^{5}[/tex]

since [itex]|\cos c|[/itex] is at most equal to unity. Furthermore, you could consider an expansion where [itex]x[/itex] is only between 0 and 1. Then the biggest the error term could be is

[tex]\frac{1}{120}[/tex]

Finding the maximum error over a certain range doesn't give you the true remainder for a particular value of x, but it gives a "worst case scenario".

I hope this makes some sense.

- #3

Hurkyl

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Differential approximation says:

f(y) ~ f(x) + f'(x) (y - x)

And the mean value theorem says (after a little rearrangement) that there exists a c in [x, y] such that:

f(y) = f(x) + f'(c) (y - x)

A Taylor series is just a vast generalization of differential approximation -- it includes higher order derivatives. The error term is simply the corresponding generalization of the mean value theorem.

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