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Q. 10

Expert-verifiedFound in: Page 692

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

If *f(x)* is an *nth*-degree polynomial and ${P}_{n}\left(x\right)$ is the *nth* Taylor polynomial for *f *at ${x}_{0}$, what is the *nth* remainder ${R}_{n}\left(x\right)$? What is ${R}_{n+1}\left(x\right)$?

Thus, the required remainders are ${R}_{n}\left(x\right)=\frac{{f}^{n+1}\left(c\right)}{(n+1)!}{(x-{x}_{0})}^{n+1}and{R}_{n+1}\left(x\right)=\frac{{f}^{n+2}\left(c\right)}{(n+2)!}{(x-{x}_{0})}^{n+2}$

The given data is If *f(x)* is an *nth*-degree polynomial and ${P}_{n}\left(x\right)$ is the *nth* Taylor polynomial for *f *at ${x}_{0}$

Consider a function *f *that can be differentiated *(n+1)* times in some open interval *I* that contains the point ${x}_{0}$ and ${R}_{n}\left(x\right)$ be the *n* remainder for *f* at $x={x}_{0}$

Hence, for each point $x\in I$, there is at least one c between ${x}_{0}andx$ such that,

${R}_{n}\left(x\right)=\frac{{f}^{n+1}\left(c\right)}{(n+1)!}{(x-{x}_{0})}^{n+1}\phantom{\rule{0ex}{0ex}}\mathrm{Now},(n+1)th\mathrm{remainder}\mathrm{is},\phantom{\rule{0ex}{0ex}}{R}_{n+1}\left(x\right)=\frac{{f}^{n+1+1}\left(c\right)}{(n+1+1)!}{(x-{x}_{0})}^{n+1+1}\phantom{\rule{0ex}{0ex}}{R}_{n+1}\left(x\right)=\frac{{f}^{n+2}\left(c\right)}{(n+2)!}{(x-{x}_{0})}^{n+2}$

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