Question

Remainders Find the remainder Rₙ for the nth−order Taylor polynomial centered at a for the given functions. Express the result for a general value of n.f(x) = sin x, a = 0

Accepted Answer

Recall that the remainder term $$ R_n(x) $$ for the nth-order Taylor polynomial of a function $$ f(x) $$ centered at $$ a  is $$given by the Lagrange form of the remainder:

$$ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x - a)^{n+1} $$

where $$ c  is $$some value between $$ a $$ and $$ x . $$Identify the function and center: here, $$ f(x) = \sin x $$ and $$ a = 0 . So $$the remainder term becomes:

$$ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1} $$

with $$ c $$ between 0 and $$ x . $$Determine the $$ (n+1) th $$derivative of $$ \sin x . $$Since derivatives of $$ \sin x $$ cycle every 4 steps:

$$ f^{(1)}(x) = \cos x, \quad f^{(2)}(x) = -\sin x, \quad f^{(3)}(x) = -\cos x, \quad f^{(4)}(x) = \sin x, \quad \ldots $$

Use this cyclic pattern to express $$ f^{(n+1)}(c)  in $$terms of $$ \sin c  or$$ $$ \cos c $$ with appropriate sign. Substitute the expression for $$ f^{(n+1)}(c) $$ back into the remainder formula:

$$ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1} $$

where $$ f^{(n+1)}(c)  is $$one of $$ \sin c, \cos c, -\sin c, -\cos c $$ depending on $$ n+1 $$ modulo 4. Summarize the remainder term for general $$ n  as$$:

$$ R_n(x) = \frac{\pm \sin c 	ext{ or } \pm \cos c}{(n+1)!} x^{n+1} \quad 	ext{for some } c \in (0, x) $$

This expresses the remainder for the nth-order Taylor polynomial of $$ \sin x $$ centered at 0.

Remainders Find the remainder Rₙ for the nth−order Taylor polynomial centered at a for the given functions. Express the result for a general value of n.

f(x) = sin x, a = 0

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Key Concepts

Taylor Polynomial and Taylor Series

Remainder Term (Lagrange Form)

Derivatives of sin x