Suppose we’re working with a function $f(x)$ that is continuous and has $n+1$ continuous derivatives on an interval about $x=0$. We can approximate $f$ near $0$ by a polynomial $P_n(x)$ of degree $n$:

- For $n=0$, the best constant approximation near $0$ is
\[P_0(x)=f(0)\]
which matches $f$ at $0$.

- For $n=1$, the best linear approximation near $0$ is
\[P_1(x)=f(0)+f'(0)x.\]
Note that $P_1$ matches $f$ at $0$ and $P_1’$ matches $f’$ at $0$.

- For $n=2$, the best quadratic approximation near $0$ is \[P_2(x)=f(0)+f'(0)x+\frac{f”(0)}{2!}x^2.\] Note that $P_2$, $P_2’$, and $P_2”$ match $f$, $f’$, and $f”$, respectively, at $0$.

Continuing this process, \[P_n(x)=f(0)+f'(0)x+\frac{f”(0)}{2!}x^2+\ldots +\frac{f^{(n)}(0)}{n!}x^n.\] This is the **Taylor polynomial of degree $n$ about $0$** (also called the **Maclaurin series of degree $n$**). More generally, if $f$ has $n+1$ continuous derivatives at $x=a$, the **Taylor series of degree $n$ about $a$** is \[\sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k=f(a)+f'(a)(x-a)+\frac{f”(a)}{2!}(x-a)^2+\ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n.\] This formula approximates $f(x)$ near $a$. Taylor’s Theorem gives bounds for the error in this approximation:

#### Taylor’s Theorem

Suppose $f$ has $n+1$ continuous derivatives on an open interval containing $a$. Then for each $x$ in the interval, \[f(x)=\left[\sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k\right]+R_{n+1}(x)\] where the error term $R_{n+1}(x)$ satisfies $\displaystyle R_{n+1}(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ for some $c$ between $a$ and $x$.

This form for the error $R_{n+1}(x)$, derived in 1797 by Joseph
Lagrange, is called the Lagrange formula for the remainder. The
*infinite* Taylor series converges to $f$,
\[f(x)=\sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{k!}(x-a)^k,\]
if and only if $\displaystyle \lim_{n\to\infty} R_n(x)=0$.

#### Examples of Taylor Series about $0$

- For $f(x)=e^x$,
\[f^{(k)}(x)=e^x \quad\Longrightarrow\quad f^{(k)}(0)=1.\]
So
\begin{eqnarray*}
e^x&=&1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots\\
&=&\sum_{k=0}^{\infty}\frac{x^k}{k!}
\end{eqnarray*}
which converges for all $x$ since $\displaystyle\lim_{n\to\infty}
R_n(x)=\lim_{n\to\infty} \frac{e^cx^{(n+1)}}{(n+1)!}=0$ for all $c$
between $0$ and $x$.

- For $f(x)=\ln (1+x)$,
\begin{eqnarray*}
\left.\begin{array}{l}
f(x)=\ln(1+x)\\
f'(x)=\frac{1}{1+x}\\
f”(x)=\frac{-1}{(1+x)^2}\\
f”'(x)=\frac{2}{(1+x)^3}\\
f^{(4)}(x)=\frac{-3\cdot 2}{(1+x)^4}\\
\vdots
\end{array}\right\}\qquad\Longrightarrow\qquad
\left\{ \begin{array}{l}
f(0)=0\\
f'(0)=1\\
f”(0)=-1\\
f”'(0)=2\\
f^{(4)}(x)=-6\\
\vdots
\end{array} \right.
\end{eqnarray*}
So
\begin{eqnarray*}
\ln (1+x)&=&x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots\\
&=&\sum_{k=0}^\infty (-1)^k\frac{x^{k+1}}{k+1}
\end{eqnarray*}
which converges only for $-1
The Taylor Series in $(x-a)$ is the

*unique*power series in $(x-a)$ converging to $f(x)$ on an interval containing $a$. For this reason,- By Example 1,
\[e^{-2x}=1-2x+2x^2-\frac{4}{3}x^3+\ldots\]
where we have substituted $-2x$ for $x$.

- By Example 2, since $\displaystyle \frac{d}{dx}[\ln (1+x)]=\frac{1}{1+x}$, we can differentiate the Taylor series for $\ln (1+x)$ to obtain \[\frac{1}{1+x}=1-x+x^2-x^3+\ldots.\] Substituting $-x$ for $x$, \[\frac{1}{1-x}=1+x+x^2+x^3+\ldots.\]

In the Exploration, compare the graphs of various functions with their first through fourth degree Taylor polynomials.

Exploration

### Key Concepts

**Taylor’s Theorem**Suppose $f$ has $n+1$ continuous derivatives on an open interval containing $a$. Then for each $x$ in the interval, \[f(x) = \left[\sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x-a)^k\right]+ R_{n+1}(x)\]

where the error term $R_{n+1}(x)$ satisfies $R_{n+1}(x) = \left[\frac{f^{(n+1)}(c)}{(n+1)!}\right] (x-a)^{n+1}$ for some $c$ between $a$ and $x$.

[I’m ready to take the quiz.] [I need to review more.]

- By Example 1,
\[e^{-2x}=1-2x+2x^2-\frac{4}{3}x^3+\ldots\]
where we have substituted $-2x$ for $x$.