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\noindent
\fbox{\fbox{\parbox[c][0.25in]{6.85in}{
\parbox{1in}{\bf Math 1110}
\hfill {\Large \bf Taylor Polynomials} \hfill
\parbox{1in}{\bf \hfill Supplement}
}}}

\vspace{0.2in}

Approximation is a major theme of calculus. In particular, we seek to understand a non-linear function in terms of a flattened out {\it linearized} function. If we linearize our function at a point, then the graph of its linearization is called a tangent line (based at that point) and the slope of the tangent line is the derivative of our function. 

Sometimes a linear approximation is not good enough for our purposes. Thus we should develop higher order approximations. A first order approximation is called a linearization and a second order approximation is called a {\it quadratic approximation}. More generally these as well as higher order approximations are called {\it Taylor polynomials}. In second semester calculus these are extended to infinite sums called series. In turns out that in many cases Taylor {\it series} can be used to {\it define}  our most familiar functions. 

We begin by recalling the definition of a linearization and tangent.

\vspace{0.05in}

\noindent \fbox{\bf Definition:} Suppose $f(x)$ is differentiable at $x=a$ (i.e., $f'(a)$ exists). Then $L(x)=f(a)+f'(a)(x-a)$ is the {\bf linearization} of $f(x)$ based (or centered) at $x=a$. The graph of this linearization is the {\bf tangent line} of $f(x)$ based at $x=a$.

\vspace{0.05in}

\noindent {\bf Example:} Let $f(x)=x^2$. Then $f'(x)=2x$. This means $f(3)=9$ and $f'(3)=6$ so that the linearization of $f(x)=x^2$ based at $x=3$ is $L(x)=f(3)+f'(3)(x-3)=9+6(x-3)=6x-9$. Thus the graph of $y=6x-9$ gives us the tangent line of $f(x)=x^2$ at $x=3$. 

\vspace{0.05in}

\noindent {\bf Example:} Let $g(x)=\sin(x)$. Then since $g'(x)=\cos(x)$, we have $L(x)=\sin(0)+\cos(0)(x-0)=0+1(x-0)=x$ is the linearization of the sine function based at $x=0$. This linear approximation is often used by Physicists when an angle is ``small''. In particular, $\sin(0.1) = g(0.1) \approx L(0.1) = 0.1$.

\vspace{0.05in}

The notion of a ``good'' linearization can be turned around to define what we mean by {\it differentiability}. Say $L(x)=A+B(x-a)$. Then we might say $L(x)$ is a good linear approximation of $f(x)$ if $f(x)-L(x)$ is small (in some sense) when $x$ is close to $a$. More precisely we want $\lim\limits_{x \to a} \dfrac{f(x)-L(x)}{x-a}=0$. In other words, we want $f(x)$ and $L(x)$ to match up to first order terms {\it locally}. If this definition is pursued, one can show that a good approximation exists only if $f'(a)$ exists and we have $A=f(a)$ and $B=f'(a)$. Briefly, we only have a ``good'' approximation at $x=a$ when $f(x)$ is differentiable and this good approximation must be the linearization $L(x)=f(a)+f'(a)(x-a)$.

\vspace{0.05in}

Now we turn to a higher order approximation. Notice that if $L(x)=f(a)+f'(a)(x-a)$, then $L(a)=f(a)$ and $L'(a)=f'(a)$. Suppose that we want an approximation, say $Q(x)$, such that $Q(a)=f(a)$, $Q'(a)=f'(a)$, and $Q''(a)=f''(a)$. If we demand that $Q''(x)$ is the number $f''(a)$, then $Q'(x)$ would need to be linear: $Q'(x)=??+f''(a)x$ which could be rewritten as $Q'(x)=???+f''(a)(x-a)$ but if we want $Q'(a)=f'(a)$ we must have $Q'(x)=f'(a)+f''(a)(x-a)$. If we know about antiderivatives, we can conclude that $Q(x)=???+f'(a)(x-a)+\dfrac{f''(a)}{2}(x-a)^2$ and since we want $Q(a)=f(a)$ we must have $Q(x)=f(a)+f'(a)(x-a)+\dfrac{f''(a)}{2}(x-a)^2$. 

\vspace{0.05in}

\noindent \fbox{\bf Definition:} Suppose $f(x)$ is twice differentiable at $x=a$. Then $Q(x)=f(a)+f'(a)(x-a)+\dfrac{f''(a)}{2}(x-a)^2$ is the {\bf quadratic approximation} of $f(x)$ based (or centered) at $x=a$.

\vspace{0.05in}

From our discussion above, this is the unique quadratic polynomial which matches $f(x)$ and its first two derivatives at the point $x=a$.

\vspace{0.05in}

\noindent {\bf Example:} Let $f(x)=x^3$. Then $f'(x)=3x^2$ and $f''(x)=6x$. The quadratic approximation of $f(x)=x^3$ centered at $x=1$ is $Q(x)=f(1)+f'(1)(x-1)+\dfrac{f''(1)}{2}(x-1)^2 = 1+3(x-1)+3(x-1)^2$. 

\vspace{0.05in}

\noindent {\bf Example:} Let $g(x)=\cos(x)$. Then $g'(x)=-\sin(x)$ and $g''(x)=-\cos(x)$. So $Q(x)=\cos(0)-\sin(0)x+\dfrac{-\cos(0)}{2}x^2=1-\dfrac{1}{2}x^2$ is the quadratic approximation of cosine based at $x=0$. Since $x=0$ is a critical point for cosine, we have no linear term. Notice that the graph of this quadratic approximation is a parabola opening downward. This reveals that $x=0$ is a local maximum for cosine.

\vspace{0.05in}

Extending the above example gives us the {\it second derivative test}. Suppose that $f(x)$ has a second derivative at $x=a$ and that $x=a$ is a critical point (so $f'(a)=0$). Then $f(x) \approx Q(x)=f(a)+f'(a)(x-a)+\dfrac{f''(a)}{2}(x-a)^2 = f(a)+\dfrac{f''(a)}{2}(x-a)^2$. The graph of $Q(x)$ is a parabola with vertex at $(a,f(a))$. It opens up or down according to the sign of $f''(a)$. Thus since $f(x)\approx Q(x)$ near $x=a$, it is reasonable to conclude that $x=a$ is a local minimum if $f''(a)>0$ and a local maximum if $f''(a)<0$. 

We now generalize to arbitrary order approximations. First, recall that $k!=k(k-1)(k-2)\cdots 3 \cdot 2 \cdot 1$ is the factorial function. For example, $1!=1$, $2!=2$, $3! = 3\cdot 2\cdot 1=6$, $4!=4\cdot 3 \cdot 2 \cdot 1=24$ etc. Since an empty product should be the multiplicative identity, one defines $0!=1$ (admittedly this seems strange at first glance). 

\vspace{0.05in}

\noindent \fbox{\bf Definition:} Let $f(x)$ have derivatives up to order $n$ defined at $x=a$. Then the {\it $n^\text{th}$-order Taylor polynomial centered at $x=a$} of $f(x)$ is 

\vspace{-0.25in}

$$P(x) = f(a)+f'(a)(x-a)+\dfrac{f''(a)}{2}(x-a)^2+\dfrac{f'''(a)}{3!}(x-a)^3+\cdots+\dfrac{f^{(n)}}{n!}(x-a)^n = \sum\limits_{k=0}^n \dfrac{f^{(k)}}{k!}(x-a)^k$$

\vspace{-0.1in}

\noindent
Taylor polynomials centered at $x=0$ are traditionally called {\it Maclaurin polynomials}. Notice that a linearization is just a first order Taylor polynomial and a quadratic approximation is just a second order Taylor polynomial.

\vspace{0.05in}

\noindent {\bf Example:} Let $f(x)=\ln(x)$. Then $f'(x)=1/x=x^{-1}$, $f''(x)=-x^{-2}$, $f'''(x)=2x^{-3}$, and $f^{(4)}(x)=-6x^{-4}$. Thus $f(1)=\ln(1)=0$, $f'(1)=1$, $f''(1)=-1$, $f'''(1)=2$, and $f^{(4)}(1)=-6$. This means that the fourth order Taylor polynomial for the natural logarithm (centered at $x=1$) is:

\vspace{-0.25in}

$$P(x)=0+1(x-1)+\dfrac{-1}{2}(x-1)^2+\dfrac{2}{3!}(x-1)^3+\dfrac{-6}{4!}(x-1)^4 = (x-1)-\dfrac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\dfrac{(x-1)^4}{4}.$$


%\vspace{-0.1in}

\noindent \fbox{\bf Theorem:} If $P(x)$ is the $n^\text{th}$-order Taylor polynomial of $f(x)$ centered at $x=a$, then we have $f(a)=P(a)$, $f'(a)=P'(a)$, \dots, $f^{(n)}(a)=P^{(n)}(a)$. Conversely, suppose $P(x)$ is a polynomial of order $n$ or less and $f(a)=P(a)$, $f'(a)=P'(a)$, \dots, $f^{(n)}(a)=P^{(n)}(a)$. Then $P(x)$ is the $n^\text{th}$-order Taylor polynomial of $f(x)$ centered at $x=a$.

\vspace{0.05in}

The proof of this theorem is straight forward (but we skip it). It tells us that the Taylor polynomial centered at $x=a$ is {\it only way} to approximate $f(x)$ with a polynomial of order (at most) $n$ so that this polynomial's derivative data at $x=a$ matches $f(x)$ (up to order $n$). 

\vspace{0.05in}

\noindent {\bf Example:} Suppose $P(x)=5+4(x+8)^2-(x+8)^3$ is the third order Taylor polynomial centered at $x=-8$ for some mystery function $f(x)$. Then we must have $f(-8)=P(-8)=5$, $f'(-8)=P'(-8)=0$, $f''(-8)=P''(-8)=4\cdot 2=8$, and $f'''(-8)=P'''(-8)=(-1)\cdot 3!=-6$. And that is all we can say about $f$'s values! We can draw no conclusions about $f^{(4)}(-8)$ or $f(2)$ or $f'(1)$ etc.

\vspace{0.05in}

\noindent {\bf Example:} Find the fourth order Taylor polynomial, centered at $x=2$ for $g(x)=x^4$. This is easy. Answer: $P(x)=x^4$. Why? Well, $g(x)$ {\it is} a fourth order polynomial and it matches all of the derivative data it needs to, so by the above theorem it must be the Taylor polynomial! What is the 100-th order Taylor polynomial of $g(x)=x^4$ centered at $x=\sqrt{\pi}$? Easy, it's just $g(x)=x^4$. 

Likewise, if $h(x)=x^2-3x+1$. Then the zero-th order Maclaurin polynomial for $h(x)$ is $1$, the first order Maclaurin polynomial is $-3x+1$, and the Macluarin polynomials of degree two or higher are just $h(x)=x^2-3x+1$ itself! 

\vspace{0.05in}

Sometimes we need to know exactly how good our estimate is. The following theorem does just that and is called {\it Taylor's Remainder Formula}.

\vspace{0.05in}

\noindent \fbox{\bf Theorem:} Suppose $f(x)$ is $(n+1)$-times differentiable on an interval containing $a$. Then for any $x$ in that interval there is some number between $x$ and $a$, say $\zeta_x$, such that:

\vspace{-0.15in}

$$f(x)-P(x) = f(x)-\left(f(a)+f'(a)(x-a)+\cdots+\dfrac{f^{(n)}(a)}{n!}(x-a)^n\right) = \dfrac{f^{(n+1)}(\zeta_x)}{(n+1)!}(x-a)^{n+1}$$

\vspace{-0.1in}

\noindent where $P(x)$ is the $n$-th order Taylor polynomial of $f(x)$ based at $x=a$. 

\vspace{0.05in}

Possibly surprisingly, this theorem is not terribly hard to prove (yet we still skip the proof here). It follows quickly from Rolle's theorem. What this tells us is that $f(x)$ and $P(x)$ are nearly identical. We just need to add in one more term -- except that term's derivative may need to be evaluated at $\zeta_x$ (possibly near but not quite at $a$).

This formula says if we know how big the $(n+1)$-st derivative can be, we can get an estimate of how far off our Taylor polynomial is relative to how far we have strayed from our base point $x=a$. 

\vspace{0.05in}

\noindent {\bf Example:} The fourth order Maclaurin polynomial for sine is $P(x)=x-\dfrac{x^3}{6}$. Note that the fifth derivative of sine is cosine. So Taylor says $\sin(x)-P(x) = \dfrac{\cos(\zeta_x)}{5!}(x-0)^5 = \dfrac{\cos(\zeta_x)}{120}x^5$ for some $\zeta_x$ between $x$ and $0$. But cosine is at most 1. Thus $|\sin(x)-P(x)| \leq \dfrac{1}{120}|x|^5$. For example, $\sin(0.1) \approx P(0.1) = 0.0998333\cdots$. How good is this approximation? Taylor says its error is less than $\dfrac{(0.1)^5}{120} \approx 8.333\cdots \times 10^{-8}$. So at least 6 decimals must match. In fact, $\sin(0.1)=0.0998334166\cdots$. Thus this cubic polynomial is approximating sine quite well! (At least for small $x$'s.)

\vspace{0.05in}  

\noindent {\bf Example:} Approximate $e$ (Euler's number) out to 3 decimals. We have $P(x)=1+x+\dfrac{x^2}{2}+\cdots+\dfrac{x^n}{n!}$ is the $n$-th order Maclaurin polynomial for $f(x)=e^x$. Suppose we at least know $e<3$. Then $f^{(n+1)}(\zeta_1)=e^{\zeta_1} \leq e^1 < 3$ for $0 \leq \zeta_1 \leq 1$. Thus if we wish to approximate $f(1)=e^1=e$ with $P(1)$, Taylor says our error is no more than $\dfrac{f^{(n+1)}(\zeta_1)}{(n+1)!}(1-0)^{n+1} < \dfrac{3}{(n+1)!}$. We want the error to be smaller than $10^{-3}$ so our answer is correct out to 3 digits. We need $\dfrac{3}{(n+1)!} < 10^{-3}$ so that $3 \times 10^3 = 3000 < (n+1)!$. Since $7!=5040$, $n=6$ should work for us. So $e \approx 1+1+\dfrac{1}{2}+\cdots+\dfrac{1}{6!}=1+1+\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{24}+\dfrac{1}{120}+\dfrac{1}{720} = 2.7180555\cdots$. {\it Note:} $e=2.7182818\cdots$ so our approximation is correct to 3 digits exactly!

\vspace{0.05in}

Taylor polynomials also give us a way to extend our second derivative test. Recall that the second derivative test does not apply when the second derivative is zero at a critical point. We now offer a test that overcomes this problem.

\vspace{0.05in}

\noindent \fbox{\bf Theorem:} ($n$-th Derivative Test) Let $f(x)$ be a function such that $f^{(n)}(a)$ exists and $f'(a)=f''(a)=\cdots=f^{(n-1)}(a)=0$ (i.e., the first $n-1$ derivatives of $f$ vanish at $x=a$) but $f^{(n)}(a) \not=0$. We have the following: 
\begin{itemize}
\item If $n$ is odd, then $x=a$ is not a local extreme (it is neither a local minimum nor a local maximum).
\item If $n$ is even and $f^{(n)}(a)>0$, then $x=a$ is a local minimum.
\item If $n$ is even and $f^{(n)}(a)<0$, then $x=a$ is a local maximum.
\end{itemize}

\vspace{-0.1in}

\noindent {\it Proof idea:} The $n$-th order Taylor polynomial of $f(x)$ based at $x=a$ is $P(x)=f(a)+\dfrac{f^{(n)}(a)}{n!}(x-a)^n$ (since all lower order derivatives are zero).
Near $x=a$, higher powers of $x-a$ are much closer to zero than lower powers, so $(x-a)^n$ dominates $(x-a)^{n+1}$. Thus the Taylor remainder (a multiple of $(x-a)^{n+1}$) is small enough that it does not matter. Therefore, the graphs of $f(x)$ and $P(x) = f(a)+\dfrac{f^{(n)}(a)}{n!}(x-a)^n$ are almost the same near $x=a$. Notice that $x-a$, $(x-a)^3$, $(x-a)^5$, etc. do not have an extrema at $x=a$ while $(x-a)^2$, $(x-a)^4$, etc. do. Thus if $n$ is odd, $x=a$ is not an extremum. If $n$ is even, $x=a$ is a local extreme. Moreover, this extrema is a minimum or maximum according to whether the coefficient of $(x-a)^n$ is positive or negative.

\vspace{0.05in}

\noindent {\bf Example:} $f(x)=(x-3)^4$ has a critical point at $x=3$. Notice that $f'(x)=4(x-3)^3$, $f''(x)=12(x-2)^2$, and $f'''(x)=24(x-2)$ are zero at $x=3$. However, $f^{(4)}(x)=24$ so $f^{(4)}(3)=24>0$ (our first non-zero derivative has even order and is a positive number). Thus $x=3$ is a local minimum. On the other hand, $f'(10)=4(10-3)^3 \not=0$. Thus since our first non-zero derivative at $x=10$ is of odd order (i.e., order 1), our test says $x=10$ is neither a minimum nor a maximum. This really is not a surprise since $x=10$ is not even a critical point!

\vspace{0.1in}

\noindent \fbox{\bf Homework:}

\begin{enumerate}
\item Compute the quadratic approximation of $f(x)=e^x$ based at $x=0$ (i.e., the second order Maclaurin polynomial).

\item Find the third order Taylor polynomial of $f(x)=\ln(x-2)$ based at $x=3$.

\item Find the third order Maclaurin polynomial of $f(x)=\mathrm{arctan}(x)$.

\item Find Taylor polynomials based at $x=-1$ of $f(x)=x^3$ of all possible orders.

\item Compute the fifth order Maclaurin polynomials for sine, cosine, and $e^x$.

\item Suppose $f(x)=x^4+3x^2-x+1$. What is the 10-th order Taylor polynomial of $f(x)$ based at $x=-7$?

\item Suppose $P(x)=5-4(x+1)+2(x+1)^3-(x+1)^4$ is the fifth order Taylor polynomial based at $x=-1$ for $f(x)$. Find $f(-1)$, $f'(-1)$, $f''(-1)$, $f'''(-1)$, and $f^{(4)}(-1)$.

\item Suppose $P(x)=x^3$ is the fourth order Taylor polynomial based at $x=2$ for $f(x)$. Find  $f(2)$, $f'(2)$, $f''(2)$, $f'''(2)$, and $f^{(4)}(2)$. What about $f^{(5)}(2)$? How about $f'(3)$?

\item Use Taylor's remainder formula to explain why if $P(x)$ is a Taylor polynomial of order $k$ where $k \geq n$ based at $x=a$ for a polynomial $f(x)$ of order $n$, then $P(x)=f(x)$.

\item Approximate $\cos(0.1)$ using a fourth order Maclaurin polynomial. Use Taylor's remainder formula to estimate the possible error.

\end{enumerate}

\newpage

\noindent \fbox{\bf Answers:}

\begin{enumerate}
\item All derivatives of $f(x)=e^x$ are just itself. Thus $f^{(k)}(0)=e^0=1$. So $Q(x)=1+1\cdot x+\dfrac{1}{2}\cdot x^2 = \dfrac{1}{2}x^2+x+1$.

\item We have $f'(x)=\dfrac{1}{x-2}=(x-2)^{-1}$, $f''(x)=-(x-2)^{-2}$, and $f'''(x)=2(x-2)^{-3}$. Thus $f(3)=\ln(3-2)=\ln(1)=0$, $f'(3)=(3-2)^{-1}=1$, $f''(3)=-(3-2)^{-2}=-1$, and $f'''(3)=2(3-2)^{-3}=2$. Therefore, $P(x) = 0+1(x-3)+\dfrac{-1}{2}(x-3)^2+\dfrac{2}{3!}(x-3)^3 = (x-3)-\dfrac{1}{2}(x-3)^2+\dfrac{1}{3}(x-3)^3$. 

One might (correctly) guess that the $n$-th order Taylor polynomial based at $x=3$ would be:\\ $P(x)=(x-3)-\dfrac{1}{2}(x-3)^2+\dfrac{1}{3}(x-3)^3-\dfrac{1}{4}(x-3)^4+\cdots \pm \dfrac{1}{n}(x-3)^n$.

\item Recall that a Maclaurin polynomial is just a Taylor polynomial based at $x=0$. Taking derivatives we get: $f'(x)=\dfrac{1}{x^2+1}$, $f''(x)=\dfrac{-2x}{(x^2+1)^2}$, and $f'''(x)=\dfrac{-2(x^2+1)^2+2x(2)(x^2+1)2x}{(x^2+1)^4}$ so that $f(0)=\mathrm{arctan}(0)=0$, $f'(0)=\dfrac{1}{0^2+1}=1$, $f''(x)=\dfrac{-2(0)}{(0^2+1)^2}$, and $f'''(0)=\dfrac{-2(0^2+1)^2+0}{(0^2+1)^4}=-2$. 

Therefore, $P(x)=0+1\cdot x+\dfrac{0}{2}\cdot x^2+\dfrac{-2}{3!}\cdot x^3 = x-\dfrac{1}{3}x^3$.

\item Computing derivatives: $f'(x)=3x^2$, $f''(x)=6x$, $f'''(x)=6$, and $f^{(k)}(x)=0$ for $k>3$ so that $f(-1)=-1$, $f'(-1)=3$, $f''(-1)=-6$, $f'''(-1)=6$, and $f^{(k)}(-1)=0$ for $k>3$. Therefore, the zero-th order polynomial is $P(x)=-1$, first order is $P(x)=-1+3(x-1)$, second order is $P(x)=-1+3(x-1)+\dfrac{-6}{2}(x-1)^2=-1+3(x-1)-3(x-1)^2$, and third order is $P(x)=-1+3(x-1)+\dfrac{-6}{2}(x-1)^2+\dfrac{6}{3!}(x-1)^3=-1+3(x-1)-3(x-1)^2+(x-1)^3$ which is actually equal to $x^3$ itself! Higher order terms contribute nothing since $f^{(k)}(-1)=0$ for $k>3$. In fact, all Taylor polynomials of order 3 or more based at any point are just $f(x)=x^3$ itself!

\item Derivatives of $\sin(x)$ are $\cos(x)$, $-\sin(x)$, $-\cos(x)$, $\sin(x)$ and so on. Evaluating $\sin(x)$ itself and these derivatives at $x=0$ yields $0,1,0,-1,0,$ etc. The same data is needed for cosine. We have that the fifth order Maclaurin polynomial for $\sin(x)$ is $P(x)= x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}$. Likewise, the fifth order Maclaurin polynomial for $\cos(x)$ is $P(x)=1-\dfrac{x^2}{2}+\dfrac{x^4}{4!}$. 

Recall from problem \#1, the derivatives of $e^x$ are just itself and so at $x=0$ we just get $e^0=1$. Thus the fifth order Maclaurin polynomial of $e^x$ is $P(x)=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}$. 

It should be easy to guess what higher order Maclaurin polynomials for sine, cosine, and $e^x$ are.

\item Any Taylor polynomial of order 4 or more based at any point is just going to be $f(x)=x^4+3x^2-x+1$ itself. 

\item In a Taylor polynomial of $f(x)$ based at $x=-1$, the coefficient of $(x+1)^k$ is $\dfrac{f^{(k)}(-1)}{k!}$. So if the coefficient of $(x+1)^k$ is $c$, we have $f^{(k)}(-1)=c \cdot k!$. Thus $f(-1)=5$, $f'(-1)=-4$, $f''(-1)=0\cdot 2=0$ (there is no $(x+1)^2$ term), $f'''(-1)=2\cdot 3!=12$, and $f^{(4)}(-1)=-1\cdot 4!=-24$.

\item First, note that $P'(x)=3x^2$, $P''(x)=6x$, $P'''(x)=6$, and $P^{(4)}(x)=0$. Since a function and its fourth order Taylor polynomial based at $x=2$ must have matching data at $x=2$ (up to the fourth derivative), we get that $f(2)=P(2)=2^3=8$, $f'(2)=P'(2)=3\cdot 2^2=12$, $f''(2)=P''(2)=6\cdot 2=12$, $f'''(2)=P'''(2)=6$, and $f^{(4)}(2)=P^{(4)}(2)=0$.

This is as much as we can say. While $P^{(5)}(2)=0$, there is no guarantee that $f^{(5)}(2)$ matches this value. In the end, $f^{(5)}(2)$ is completely unknowable from the data given. Likewise, this Taylor polynomial does not tell us about data at other inputs, so $f'(3)$ is also unknowable from our given information.

\item Notice that when $f(x)$ is a polynomial of order $n$, $f^{(k)}(x)=0$ for $k>n$. Thus the Taylor remainder formula is zero for polynomials of order $n$ or more.

\item Using work from problem \#5 to get the fourth order Maclaurin polynomial for cosine, we have $\cos(0.1) \approx 1-\dfrac{0.1^2}{2}+\dfrac{0.1^4}{4!} = 0.9550041\overline{666}$. Since the fifth derivative of cosine is $-\sin(x)$ and $|-\sin(x)| \leq 1$, the Taylor remainder formula caps the error at $\dfrac{1}{5!}(0.1)^5 = 0.00000008\overline{333}$. {\it Note:} $\cos(0.1) = 0.995004165278\cdots$.

\end{enumerate}



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