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Taylor's Theorem

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, \[P_n(x)=f(0)+f'(0)x+\frac{f''(x)}{2!}x^2+\ldots +\frac{f^{(n)}(0)}{n!}x^n.\] 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 $0$

  1. 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 $x$.

  2. 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 $x$.