Originally Posted by

**BoombaWoomba** ...Step-by-step explanations if you will. I'd really appreciate it. Anyways, here are some of the problems:

"Find the Taylor polynomial of order four for the function at x=0, and use it to approximate the value of the function at x=0.2."

1. ) 5 sin ( - x )

...

You know the general form of the Taylor polynomial:

$\displaystyle f(x) = f(a)+\frac{x-a}{1!}f'(a) + \ldots + \frac{(x-a)^{n-1}}{(n-1)!}f^{(n-1)} + R_n$ ......... $\displaystyle f^{(n-1)}$ is the (n-1)th derivative of f.

With your question a = 0.

Calculate the derivatives: Code:

f(x) = 5sin(-x) --> f(0) = 0
f'(x) = -5cos(x) --> f'(0) = -5
f''(x) = 5sin(x) --> f''(0) = 0
f'''(x) = 5cos(x) --> f'''(0) = 5
f''''(x) = -5sin(x) --> f''''(0) = 0

Now plug in all values:

$\displaystyle f(x) = 0 + (-5)x + 0 + \frac{x^3}{3!} \cdot 5 = -5x + \frac56 x^3$

Calculate $\displaystyle f(0.2) = -5 \cdot 0.2 + \frac56 (0.2)^3 = -1 + \frac1{150} = -\frac{149}{150} \approx -0.9933333..$

Check this result by calculating $\displaystyle 5 \sin(-0.2) \approx -0.993346...$