Originally Posted by

**johnsomeone** A "term" of the Taylor Series $\displaystyle \sum_{n=0}^{\infty}\frac{f^{(n)}(c)}{n!}(x-c)^n$ are those $\displaystyle \frac{f^{(n)}(c)}{n!}(x-c)^n$ that are being added.

The "general term" is $\displaystyle \frac{f^{(n)}(c)}{n!}(x-c)^n$, so say exactly what that is for this f and c, in terms of n.

The "first three terms" are:

$\displaystyle \frac{ f^{(0)} (c) }{ 0! } (x-c)^0 = f(c)$ (but numerically computed from f and c),

$\displaystyle \frac{ f^{(1)} (c) } { 1! } (x-c)^1 = f'(c)(x-c)$ (but numerically computed from f and c),

$\displaystyle \frac{ f^{(2)} (c) } {2!} (x-c)^2 = \frac{f''(c)}{2}(x-c)^2$ (but numerically computed from f and c).

However, those might not be the first three **non-zero** terms.

The answer wants those f and c references replaced with values as determined by what's given for f and c.

For instance, the first term is: $\displaystyle f(c) = \sin(2c) = \sin\left(2\left(\frac{\pi}{4}\right)\right) = \sin\left(\frac{\pi}{2}\right) = 1.$

The second term is: $\displaystyle f'(c)(x-c) = [2\cos(2c)](x-c) = \left[ 2\cos \left( 2 \left( \frac{\pi}{4} \right) \right) \right] \left( x - \frac{\pi}{4}\right)$

$\displaystyle = \left[2\cos\left(\frac{\pi}{2}\right)\right] \left(x - \frac{\pi}{4}\right) = 0.$

The 3rd term is: $\displaystyle \frac{f''(c)}{2}(x-c)^2 = \left[ \frac{-4\sin(2c)}{2} \right] (x-c)^2 = -2 \sin \left( 2 \left( \frac{\pi}{4} \right) \right) \left( x - \frac{\pi}{4}\right)^2$

$\displaystyle = -2 \sin \left(\frac{\pi}{2}\right) \left(x - \frac{\pi}{4}\right)^2 = -2(1)\left(x - \frac{\pi}{4}\right)^2 = = -2\left(x - \frac{\pi}{4}\right)^2.$