How does the Lagrange Form of the Error Bound give you the maximum expected error?

Question

The Lagrange form of the Error Bound states that the n+1 term of a Taylor Approximation gives you the remainder such that F(x)=P_n(x)+R_n(x), but what does that mean? I understand that the goal is to find the maximum error that can be expected, but the process does not seem logical. Why is it the next term that gives you the error? Why cant it be the one after? Also, when attempting to evaluate M such that it is greater than or equal to the n+1 derivative of F(z) where z is some value between c (where the function is centered) and x, how does finding the largest value for z in the interval give you the maximum value? I see that the biggest value is obviously going to maximize the expected error, but I don't see how evaluating the n+1 derivative of the function at z correlates with the Taylor series since all the derivatives beforehand were evaluated at c.

Answer 1

If you evaluate the LaGrange form at c you are not getting the Maximum possible error but only the next term in a converging series.

 

The remainder can be written as  an integral (which is a little cumbersome to write here); but the mean value theorem states that there is some value (z) of x within the limits of integration that can be used to get the exact value of the integral and hence the exact remainder.

 

Of course we don't necessarily know the exact value of z that will work so we choose the value of z that maximizes the expression of R_n(x) and that becomes the Maximum Error.

 

It is certainly possible that z=c but not necessarily so. Remember you are not finding the largest value of z but the value of z that maximizes the (n+1)st derivative.

 

 

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Isaac,

 

It's not letting me add another comment. So I am adding this here.

 

(I added this comment already but it didn't appear so I am adding it again.)

If you use the n+2 derivative there will be a "discrepancy" between the approximation and the calculation of the maximum error. The n+1 derivative term will not be part of the approximation and will not be part of the error calculation.

See lines 16-18 of this web page:
http://mathworld.wolfram.com/TaylorSeries.html

 

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Let me know if this helps.