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Expert-verified Found in: Page 692 ### Calculus

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

See the step by step solution

## Step 1. Given Information

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 ${x}_{0}$

## Step 2. Explanation

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,

${R}_{n}\left(x\right)=\frac{{f}^{n+1}\left(c\right)}{\left(n+1\right)!}{\left(x-{x}_{0}\right)}^{n+1}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\left(n+1\right)th\mathrm{remainder}\mathrm{is},\phantom{\rule{0ex}{0ex}}{R}_{n+1}\left(x\right)=\frac{{f}^{n+1+1}\left(c\right)}{\left(n+1+1\right)!}{\left(x-{x}_{0}\right)}^{n+1+1}\phantom{\rule{0ex}{0ex}}{R}_{n+1}\left(x\right)=\frac{{f}^{n+2}\left(c\right)}{\left(n+2\right)!}{\left(x-{x}_{0}\right)}^{n+2}$ ### Want to see more solutions like these? 