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 = π/2

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 the center: here, $$ f(x) = \sin x $$ and $$ a = \frac{\pi}{2} . We $$need to find the $$ (n+1) -th $$derivative of $$ \sin x . $$Determine the pattern of derivatives of $$ \sin x $$:

$$ f(x) = \sin x $$
$$ f'(x) = \cos x $$
$$ f''(x) = -\sin x $$
$$ f^{(3)}(x) = -\cos x $$
$$ f^{(4)}(x) = \sin x $$

Notice the derivatives repeat every 4 steps. Use this cyclic pattern to express $$ f^{(n+1)}(x)  in $$terms of $$ \sin x  or$$ $$ \cos x $$ with appropriate signs. Evaluate $$ f^{(n+1)}(c)  at $$some $$ c $$ between $$ a = \frac{\pi}{2} $$ and $$ x . $$Since $$ c  is $$unknown but lies in this interval, keep $$ f^{(n+1)}(c)  in $$the expression to represent the remainder. Write the remainder term explicitly as:

$$ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x - \frac{\pi}{2})^{n+1} $$

where $$ f^{(n+1)}(c) $$ follows the derivative pattern of $$ \sin x $$ and $$ c  is $$between $$ \frac{\pi}{2} $$ and $$ x . $$This expression represents the remainder for the nth-order Taylor polynomial centered at $$ \frac{\pi}{2} .$$

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 = π/2

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Key Concepts

Taylor Polynomial and Taylor Series

Remainder Term (Lagrange Form)

Derivatives of sin x and their Patterns