Question

Taylor Polynomial help, mates. Question will come in next post.

Answer 1

I am having a tough time understanding how I should set the limits for calculating the error after a n-Taylor approximation. The formula says:\[|R_{n}(x)| = (M/(n+1)!)|x-a|^{n+1}\]For some M such that \[|f^{n+1}(x)| \le M\]right? What does these limits mean? I mean, for M and for |x-a| <= some d.

Answer 2

Also, what happens if I take M to be a not too tight boundary for the n+1 derivative of f(x)? The error will raise by a factor of M, and that's all?

Answer 3

you mant M to be an extrema of the given thing there

Answer 4

*want

Answer 5

Indeed, but say it seems that the polynomial is approaching a real number c. If I take M to be c + 1, for instance, does it work? But now I understand, the maxima of a delimited region |x-a| <= d?

Answer 6

is this thing crashing your system as well?

Answer 7

sorry .. posted on wrong thread;

Answer 8

the x-a is just the point on which you want the poly to center at to conform to its representative function

Answer 9

centered at "a" is the usual jargon

Answer 10

as far as "what works"? im not that intimate with it to say for sure; give it a try :)

Answer 11

Also, do I want d to be very small? If so, why? Sorry for all these questions, it's the only thing that is bothering me as I didn't understand it not even to a decent depth.

Answer 12

what does "d" represent? are you trying to use a delta epsilon idea?

Answer 13

Stewart's Calculus states:\[If |f^{n+1}(x)| \le M for |x-a| \le d\]then:\[|R_{n}(x)| \le (M/(n+1)!)|x-a|^{n+1}\]for |x-a| <= d

Answer 14

"The error formula can be considered a generalization of the mean-value theorem."

http://www.math.uiowa.edu/ftp/atkinson/ENA_Materials/Overheads/sec_1-2.pdf

Answer 15

the M there is the extrema on the interval of convergence

Answer 16

http://www.millersville.edu/~bikenaga/calculus/tayerr/tayerr.html

this seems cleaner to read to me

Answer 17

x-a seems to be the interval from which to pick a value for f^n+1(x) from

Answer 18

which i assume is the interval of covergence