Arnold F. answered • 02/12/16

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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

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

If you can't add a comment click on my picture and send me a direct message.

Let me know if this helps.

Let me know if this helps.

Arnold F.

R

_{n}(x) is a function that represents the difference between the approximation and the exact value of f(x). If we know the exact value of the remainder (we don't) we would know the exact value of f(x). But if we can calculate the maximum possible value of R then that can be used as the maximum possible error incurred when using the Taylor approximation for f(x).
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02/13/16

Issac G.

Thank you! That makes a lot more sense. However, how come we use the n+1 and not the n+2 or n+3 derivative?

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02/15/16

Arnold F.

We don't use a higher derivative because then there would be a "discrepancy" between the approximation (which stops at nth derivative) and the maximum error (which you would be calculating using a higher derivative.)

The discrepancy is essentially the n+1 term since it is not part of the approximation and not part of the remainder.

Look at this web page and look at lines 16-18:

http://mathworld.wolfram.com/TaylorSeries.html

Let me know if that helps.

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02/15/16

Issac G.

02/13/16