Section4.3Taylor's theorem

Note: less than a lecture (optional section)

Subsection4.3.1Derivatives of higher orders

When $$f \colon I \to \R$$ is differentiable, we obtain a function $$f' \colon I \to \R\text{.}$$ The function $$f'$$ is called the first derivative of $$f\text{.}$$ If $$f'$$ is differentiable, we denote by $$f'' \colon I \to \R$$ the derivative of $$f'\text{.}$$ The function $$f''$$ is called the second derivative of $$f\text{.}$$ We similarly obtain $$f'''\text{,}$$ $$f''''\text{,}$$ and so on. With a larger number of derivatives the notation would get out of hand; we denote by $$f^{(n)}$$ the $$n$$th derivative of $$f\text{.}$$

When $$f$$ possesses $$n$$ derivatives, we say $$f$$ is $$n$$ times differentiable.

Subsection4.3.2Taylor's theorem

Taylor's theorem 1  is a generalization of the mean value theorem. Mean value theorem says that up to a small error $$f(x)$$ for $$x$$ near $$x_0$$ can be approximated by $$f(x_0)\text{,}$$ that is

\begin{equation*} f(x) = f(x_0) + f'(c)(x-x_0), \end{equation*}

where the “error” is measured in terms of the first derivative at some point $$c$$ between $$x$$ and $$x_0\text{.}$$ Taylor's theorem generalizes this result to higher derivatives. It tells us that up to a small error, any $$n$$ times differentiable function can be approximated at a point $$x_0$$ by a polynomial. The error of this approximation behaves like $${(x-x_0)}^{n}$$ near the point $$x_0\text{.}$$ To see why this is a good approximation notice that for a big $$n\text{,}$$ $${(x-x_0)}^n$$ is very small in a small interval around $$x_0\text{.}$$

Definition4.3.1.

For an $$n$$ times differentiable function $$f$$ defined near a point $$x_0 \in \R\text{,}$$ define the $$n$$th order Taylor polynomial for $$f$$ at $$x_0$$ as

\begin{equation*} \begin{split} P_n^{x_0}(x) & := \sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}{(x-x_0)}^k \\ & = f(x_0) + f'(x_0)(x-x_0) + \frac{f''(x_0)}{2}{(x-x_0)}^2 + \frac{f^{(3)}(x_0)}{6}{(x-x_0)}^3 + \cdots + \frac{f^{(n)}(x_0)}{n!}{(x-x_0)}^n . \end{split} \end{equation*}

See Figure 4.8 for the odd degree Taylor polynomials for the sine function at $$x_0=0\text{.}$$ The even degree terms are all zero, as even derivatives of sine are again a sine, which are zero at the origin.

Taylor's theorem says a function behaves like its $$n$$th Taylor polynomial. The mean value theorem is really Taylor's theorem for the first derivative.

The term $$R_n^{x_0}(x):=\frac{f^{(n+1)}(c)}{(n+1)!}{(x-x_0)}^{n+1}$$ is called the remainder term. This form of the remainder term is called the Lagrange form of the remainder. There are other ways to write the remainder term, but we skip those. Note that $$c$$ depends on both $$x$$ and $$x_0\text{.}$$

Proof.

Find a number $$M_{x,x_0}$$ (depending on $$x$$ and $$x_0$$) solving the equation

\begin{equation*} f(x)=P_{n}^{x_0}(x)+M_{x,x_0}{(x-x_0)}^{n+1} . \end{equation*}

Define a function $$g(s)$$ by

\begin{equation*} g(s) := f(s)-P_n^{x_0}(s)-M_{x,x_0}{(s-x_0)}^{n+1} . \end{equation*}

We compute the $$k$$th derivative at $$x_0$$ of the Taylor polynomial $${(P_n^{x_0})}^{(k)}(x_0) = f^{(k)}(x_0)$$ for $$k=0,1,2,\ldots,n$$ (the zeroth derivative of a function is the function itself). Therefore,

\begin{equation*} g(x_0) = g'(x_0) = g''(x_0) = \cdots = g^{(n)}(x_0) = 0 . \end{equation*}

In particular, $$g(x_0) = 0\text{.}$$ On the other hand $$g(x) = 0\text{.}$$ By the mean value theorem there exists an $$x_1$$ between $$x_0$$ and $$x$$ such that $$g'(x_1) = 0\text{.}$$ Applying the mean value theorem to $$g'$$ we obtain that there exists $$x_2$$ between $$x_0$$ and $$x_1$$ (and therefore between $$x_0$$ and $$x$$) such that $$g''(x_2) = 0\text{.}$$ We repeat the argument $$n+1$$ times to obtain a number $$x_{n+1}$$ between $$x_0$$ and $$x_n$$ (and therefore between $$x_0$$ and $$x$$) such that $$g^{(n+1)}(x_{n+1}) = 0\text{.}$$

Let $$c:=x_{n+1}\text{.}$$ We compute the $$(n+1)$$th derivative of $$g$$ to find

\begin{equation*} g^{(n+1)}(s) = f^{(n+1)}(s)-(n+1)!\,M_{x,x_0} . \end{equation*}

Plugging in $$c$$ for $$s$$ we obtain $$M_{x,x_0} = \frac{f^{(n+1)}(c)}{(n+1)!}\text{,}$$ and we are done.

In the proof, we have computed $${(P_n^{x_0})}^{(k)}(x_0) = f^{(k)}(x_0)$$ for $$k=0,1,2,\ldots,n\text{.}$$ Therefore, the Taylor polynomial has the same derivatives as $$f$$ at $$x_0$$ up to the $$n$$th derivative. That is why the Taylor polynomial is a good approximation to $$f\text{.}$$ Notice how in Figure 4.8 the Taylor polynomials are reasonably good approximations to the sine near $$x=0\text{.}$$

We do not necessarily get good approximations by the Taylor polynomial everywhere. Consider expanding the function $$f(x) := \frac{x}{1-x}$$ around 0, for $$x < 1\text{,}$$ we get the graphs in Figure 4.9. The dotted lines are the first, second, and third degree approximations. The dashed line is the 20th degree polynomial, and yet the approximation only seems to get better with the degree for $$x > -1\text{,}$$ and for smaller $$x\text{,}$$ it in fact gets worse. The polynomials are the partial sums of the geometric series $$\sum_{n=1}^\infty x^n\text{,}$$ and the series only converges on $$(-1,1)\text{.}$$ See the discussion of power series Section 2.6.

If $$f$$ is infinitely differentiable, that is, if $$f$$ can be differentiated any number of times, then we define the Taylor series:

\begin{equation*} \sum_{k=0}^\infty \frac{f^{(k)}(x_0)}{k!}{(x-x_0)}^k . \end{equation*}

There is no guarantee that this series converges for any $$x \not= x_0\text{.}$$ And even where it does converge, there is no guarantee that it converges to the function $$f\text{.}$$ Functions $$f$$ whose Taylor series at every point $$x_0$$ converges to $$f$$ in some open interval containing $$x_0$$ are called analytic functions. Most functions one tends to see in practice are analytic. See Exercise 5.4.11, for an example of a non-analytic function.

The definition of derivative says that a function is differentiable if it is locally approximated by a line. We mention in passing that there exists a converse to Taylor's theorem, which we will neither state nor prove, saying that if a function is locally approximated in a certain way by a polynomial of degree $$d\text{,}$$ then it has $$d$$ derivatives.

Taylor's theorem gives us a quick proof of a version of the second derivative test. By a strict relative minimum of $$f$$ at $$c\text{,}$$ we mean that there exists a $$\delta > 0$$ such that $$f(x) > f(c)$$ for all $$x \in (c-\delta,c+\delta)$$ where $$x\not=c\text{.}$$ A strict relative maximum is defined similarly. Continuity of the second derivative is not needed, but the proof is more difficult and is left as an exercise. The proof also generalizes immediately into the $$n$$th derivative test, which is also left as an exercise.

Proof.

As $$f''$$ is continuous, there exists a $$\delta > 0$$ such that $$f''(c) > 0$$ for all $$c \in (x_0-\delta,x_0+\delta)\text{,}$$ see Exercise 3.2.11. Take $$x \in (x_0-\delta,x_0+\delta)\text{,}$$ $$x \not= x_0\text{.}$$ Taylor's theorem says that for some $$c$$ between $$x_0$$ and $$x\text{,}$$

\begin{equation*} f(x) = f(x_0) + f'(x_0) (x-x_0) + \frac{f''(c)}{2}{(x-x_0)}^{2} = f(x_0) + \frac{f''(c)}{2}{(x-x_0)}^{2} . \end{equation*}

As $$f''(c) > 0\text{,}$$ and $${(x-x_0)}^2 > 0\text{,}$$ then $$f(x) > f(x_0)\text{.}$$

Subsection4.3.3Exercises

Exercise4.3.1.

Compute the $$n$$th Taylor polynomial at $$0$$ for the exponential function.

Exercise4.3.2.

Suppose $$p$$ is a polynomial of degree $$d\text{.}$$ Given $$x_0 \in \R\text{,}$$ show that the $$d$$th Taylor polynomial for $$p$$ at $$x_0$$ is equal to $$p\text{.}$$

Exercise4.3.3.

Let $$f(x) := \abs{x}^3\text{.}$$ Compute $$f'(x)$$ and $$f''(x)$$ for all $$x\text{,}$$ but show that $$f^{(3)}(0)$$ does not exist.

Exercise4.3.4.

Suppose $$f \colon \R \to \R$$ has $$n$$ continuous derivatives. Show that for every $$x_0 \in \R\text{,}$$ there exist polynomials $$P$$ and $$Q$$ of degree $$n$$ and an $$\epsilon > 0$$ such that $$P(x) \leq f(x) \leq Q(x)$$ for all $$x \in [x_0,x_0+\epsilon]$$ and $$Q(x)-P(x) = \lambda {(x-x_0)}^n$$ for some $$\lambda \geq 0\text{.}$$

Exercise4.3.5.

If $$f \colon [a,b] \to \R$$ has $$n+1$$ continuous derivatives and $$x_0 \in [a,b]\text{,}$$ prove $$\lim\limits_{x\to x_0} \frac{R_n^{x_0}(x)}{{(x-x_0)}^n} = 0\text{.}$$

Exercise4.3.6.

Suppose $$f \colon [a,b] \to \R$$ has $$n+1$$ continuous derivatives and $$x_0 \in (a,b)\text{.}$$ Prove: $$f^{(k)}(x_0) = 0$$ for all $$k = 0, 1, 2, \ldots, n$$ if and only if $$\lim\limits_{x\to x_0} \frac{f(x)}{{(x-x_0)}^{n+1}}$$ exists.

Exercise4.3.7.

Suppose $$a,b,c \in \R$$ and $$f \colon \R \to \R$$ is differentiable, $$f''(x) = a$$ for all $$x\text{,}$$ $$f'(0) = b\text{,}$$ and $$f(0) = c\text{.}$$ Find $$f$$ and prove that it is the unique differentiable function with this property.

Exercise4.3.8.

(Challenging)   Show that a simple converse to Taylor's theorem does not hold. Find a function $$f \colon \R \to \R$$ with no second derivative at $$x=0$$ such that $$\abs{f(x)} \leq \abs{x^3}\text{,}$$ that is, $$f$$ goes to zero at 0 faster than $$x^3\text{,}$$ and while $$f'(0)$$ exists, $$f''(0)$$ does not.

Exercise4.3.9.

Suppose $$f \colon (0,1) \to \R$$ is differentiable and $$f''$$ is bounded.

1. Show that there exists a once differentiable function $$g \colon [0,1) \to \R$$ such that $$f(x) = g(x)$$ for all $$x \not= 0\text{.}$$ Hint: See Exercise 4.2.14.

2. Find an example where the $$g$$ is not twice differentiable at $$x=0\text{.}$$

Exercise4.3.10.

Prove the $$n$$th derivative test. Suppose $$n \in \N\text{,}$$ $$x_0 \in (a,b)\text{,}$$ and $$f \colon (a,b) \to \R$$ is $$n$$ times continuously differentiable, with $$f^{(k)}(x_0) = 0$$ for $$k=1,2,\ldots,n-1\text{,}$$ and $$f^{(n)}(x_0) \not= 0\text{.}$$ Prove:

1. If $$n$$ is odd, then $$f$$ has neither a relative minimum, nor a maximum at $$x_0\text{.}$$

2. If $$n$$ is even, then $$f$$ has a strict relative minimum at $$x_0$$ if $$f^{(n)}(x_0) > 0$$ and a strict relative maximum at $$x_0$$ if $$f^{(n)}(x_0) < 0\text{.}$$

Exercise4.3.11.

Prove the more general version of the second derivative test. Suppose $$f \colon (a,b) \to \R$$ is differentiable and $$x_0 \in (a,b)$$ is such that, $$f'(x_0) = 0\text{,}$$ $$f''(x_0)$$ exists, and $$f''(x_0) > 0\text{.}$$ Prove that $$f$$ has a strict relative minimum at $$x_0\text{.}$$ Hint: Consider the limit definition of $$f''(x_0)\text{.}$$

Named for the English mathematician Brook Taylor 2  (1685–1731). It was first found by the Scottish mathematician James Gregory 3  (1638–1675). The statement we give was proved by Joseph-Louis Lagrange 4  (1736–1813)
https://en.wikipedia.org/wiki/Brook_Taylor
https://en.wikipedia.org/wiki/James_Gregory_(mathematician)
https://en.wikipedia.org/wiki/Lagrange
For a higher quality printout use the PDF versions: https://www.jirka.org/ra/realanal.pdf or https://www.jirka.org/ra/realanal2.pdf