Shackleford - Calm down, take a deep breath - your answer for 5a is correct. But I guess you are having some difficulty understanding why, so I'l try again. Please be patient.
I think you know the general form of the Taylor Series so I won't waste time going through that. But I want to make sure you understand what we are really saying when we look for "a Taylor expansion about x_0."
The resulting expression is a local approximation of the function such that it approximates the function within an interval about the point x_0. By definition, the approximation is exact at x_0.
Also by definition, the Taylor series is an infinite sum and the function equates to this sum if it converges. When we take a truncation of this infinite series, the function is approximated because we are taking a finite number of terms.
The error term (ie, remainder term) resulting from the difference between the function and the approximation by the truncated series is typically a generalization based on certain restrictions on the function (C^n continuity, f^{(n+1)} exists, etc) and the Mean Value Theorem that allow us to construct the remainder term in such an abbreviated expression as the one you have been using.
Let's restate the Theorem:
Suppose f\in C^n[a,b] and f^{(n+1)} exists on [a,b]. Let x_0\in[a,b]. For every x \in [a,b], there exist \xi(x) between x_0 and x with
<br />
f(x) = P_n(x)+R_n(x)<br />
where
<br />
P_n(x) = \sum_{k=0}^n{ \frac{f^k(x_0)}{k!}\,(x-x_0)^k}<br />
and
<br />
R_n(x)=\frac{f^{(n+1)}(\xi(x))}{(n+1)!} \, (x-x_0)^{n+1}<br />
Now, the f^{(n+1)}(\xi(x)) term in the expression for R_n is written that way since we really don't know where that point \xi is since the Mean Value Theorem was used to generalize all the remaining terms (we didn't show these steps). We do know it is somewhere between x_0 and x.
Now let's look at your example problems.
Problem 4a: you recognized correctly that, since we wanted a polynomial in terms of (x-\pi/4), the function needs to be expanded about x_0=\pi/4. Your p_4 expansion is correct.
Problem 4b: Continuing with the development of 4a, approximate f(x) with p_3 and x=48^\circ. Since the value of x is very close to x_0=45^\circ (recall 45^\circ corresponds to \pi/4radians) from problem 4a, use the terms of the polynomial you developed before up to and including the third order term to construct p_3. The terms evaluate to powers of (x-\pi/4) as before or (48-45) \pi / 180= \pi/60 radians.
Now, since x=(48/180)\pi = 0.2667 \pi and x_0=0.2500 \pi, we know somewhere in that interval there is a value \xi such that f^{(n+1)}(\xi)will yield the correct term to summarize the error in the expression for R_n. But we don't have to know the exact value if we just want to bound the error, that is, to find the maximum magnitude of the error will only require us to find the maximum value (or approximation) of f^{(n+1)}(\xi) over the interval x_0<\xi<x.
For your problem, we need to bound f^{(4)}(\xi) = \cos(\xi). Thus, we should use f^{(4)}(\xi) = \cos(\xi)< \cos(\pi/4) since the cosine function is monotonically decreasing for increasing x in this interval (ie, \cos(x) gets smaller with increasing values of x). Hence, the error term is approximated as
<br />
R_3 \le \frac{\cos(\pi/4)}{4!} \, (\pi/60)^4 \approx 2.21 \times 10^{-7}<br />
I know this is a bit much to swallow, but I hope it helps. Let me know if it doesn't and I try some more.
