Taylor's Inequality states:
if |f n+1(x)|<=M
then
|Rn(x)|<=M*|x-a|^(n+1)/(n+1)!
however,
Rn(x)=f n+1(x)*|x-a|^(n+1)/(n+1)!+...
it seems |Rn(x)|>=M*|x-a|^(n+1)/(n+1)! when |f n+1(x)|=M
the inequality is wrong??
Three "Problem Plus" from Stewart Calculus
Homework Statement
Please visit Google Book for online access to the Stewart Calculus 6ed.
http://books.google.com/books?id=EBE3vMmGu50C&pg=PP1&dq=stewart+calculus&ei=gYdPSvXyB4_AlQTg9PCACw
The problem No. 3,4 and 8 that are located on Page 380 and...