# Taylor's theorem: Wikis

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# Encyclopedia

The exponential function y = ex (continuous red line) and the corresponding Taylor polynomial of degree four (dashed green line) around the origin.

In calculus, Taylor's theorem gives a sequence of approximations of a differentiable function around a given point by polynomials (the Taylor polynomials of that function) whose coefficients depend only on the derivatives of the function at that point. The theorem also gives precise estimates on the size of the error in the approximation. The theorem is named after the mathematician Brook Taylor, who stated it in 1712, though the result was first discovered 41 years earlier in 1671 by James Gregory.

## Taylor's theorem in one variable

### Motivation

Taylor's theorem asserts that any sufficiently smooth function can locally be approximated by polynomials. A simple example of application of Taylor's theorem is the approximation of the exponential function ex near x = 0:

$\textrm{e}^x \approx 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!}.$

The approximation is called the n-th order Taylor approximation to ex because it approximates the value of the exponential function by a polynomial of degree n. This approximation only holds for x close to zero, and as x moves further away from zero, the approximation becomes worse. The quality of the approximation is controlled by the remainder term:

$R_n(x) = \textrm{e}^x - \left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!}\right).$

More generally, Taylor's theorem applies to any sufficiently differentiable function ƒ, giving an approximation, for x near a point a, of the form

$f(x)\approx f(a) + f'(a)(x-a) +\frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n.$

The remainder term is just the difference of the function and its approximating polynomial

$R_n(x) = f(x) - \left(f(a) + f'(a)(x-a) +\frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n\right).$

Although an explicit formula for the remainder term is seldom of any use, Taylor's theorem also provides several ways in which to estimate the value of the remainder. In other words, for x near enough to a, the remainder ought to be "small"; Taylor's theorem gives information on precisely how small it actually is.

### Statement

The precise statement of the theorem is as follows: If n ≥ 0 is an integer and ƒ is a function which is n times continuously differentiable on the closed interval [a, x] and n + 1 times differentiable on the open interval (a, x), then

$f(x) = f(a) + \frac{f'(a)}{1!}(x - a) + \frac{f^{(2)}(a)}{2!}(x - a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x - a)^n + R_n(x).$

Here, n! denotes the factorial of n, and Rn(x) is a remainder term, denoting the difference between the Taylor polynomial of degree n and the original function. The remainder term Rn(x) depends on x and is small if x is close enough to a. Several expressions are available for it.

The Lagrange form[1] of the remainder term states that there exists a number ξ between a and x such that

$R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!} (x-a)^{n+1}.$

This exposes Taylor's theorem as a generalization of the mean value theorem. In fact, the mean value theorem is used to prove Taylor's theorem with the Lagrange remainder term.

The Cauchy form[2] of the remainder term states that there exists a number ξ between a and x such that

$R_n(x) = \frac{f^{(n+1)}(\xi)}{n!}(x-\xi)^n(x-a).$

More generally, if G(t) is a continuous function on [a,x] which is differentiable with non-vanishing derivative on (a,x), then there exists a number ξ between a and x such that

$R_n(x) = \frac{f^{(n+1)}(\xi)}{n!}(x-\xi)^n\cdot\frac{G(x)-G(a)}{G'(\xi)}.$

This exposes Taylor's theorem as a generalization of the Cauchy mean value theorem.

The above forms are restricted to the case of functions taking real values. However, the integral form[3] of the remainder term applies as well when the function takes complex values. It is:

$R_n(x) = \int_a^x \frac{f^{(n+1)} (t)}{n!} (x - t)^n \, dt,$

provided, as is often the case, ƒ(n) is absolutely continuous on [a, x]. This shows the theorem to be a generalization of the fundamental theorem of calculus.

In general, a function does not need to be equal to its Taylor series, since it is possible that the Taylor series does not converge, or that it converges to a different function. However, for many functions ƒ(x), one can show that the remainder term Rn approaches zero as n approaches ∞. Those functions can be expressed as a Taylor series in a neighbourhood of the point a and are called analytic.

Taylor's theorem (with the integral formulation of the remainder term) is also valid if the function ƒ has complex values or vector values. Furthermore, there is a version of Taylor's theorem for functions in several variables. For complex functions analytic in a region containing a circle C surrounding a and its interior, there is a contour integral expression for the remainder

$R_n(x) = \frac{1}{2 \pi i}\int_C \frac{f(z)}{(z-a)^{n+1}(z-x)}dz$

valid inside of C.

### Estimates of the remainder

Another common version of Taylor's theorem holds on an interval (ar, a + r) where the variable x is assumed to take its values. This formulation of the theorem has the advantage that it is often possible to explicitly control the size of the remainder terms, and thus arrive at an approximation of a function valid in a whole interval with precise bounds on the quality of the approximation.

A precise version of Taylor's theorem in this form is as follows. Suppose ƒ is a function which is n times continuously differentiable on the closed interval [ar, a + r] and n + 1 times differentiable on the open interval (ar, a + r). If there exists a positive real constant Mn such that |ƒ(n+1)(x)| ≤ Mn for all x(ar, a + r), then

$f(x) = f(a) + \frac{f'(a)}{1!}(x - a) + \frac{f^{(2)}(a)}{2!}(x - a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x - a)^n + R_n(x),$

where the remainder function Rn satisfies the inequality (known as Cauchy's estimate):

$|R_n(x)| \le M_n \frac{r^{n+1}}{(n+1)!}$

for all x(ar, a + r). This is called a uniform estimate of the error in the Taylor polynomial centered at a, because it holds uniformly for all x in the interval.

If ƒ is infinitely differentiable on [ar, a + r], then positive constants Mn exist for each n = 1, 2, 3, … such that | ƒ(n+1)(x)| ≤ Mn for all x(ar, a + r). If, in addition, it is possible to select these constants so that

$M_n\frac{r^{n+1}}{(n+1)!} \rightarrow 0$ as $n \rightarrow \infin ,\!$

then ƒ is an analytic function on (ar, a + r). In particular, the remainder term in the Taylor approximation Rn(x) tends to zero uniformly as n→∞. In other words, an analytic function is the uniform limit of its Taylor polynomials on an interval.

## Taylor's theorem for several variables

Taylor's theorem can be generalized to several variables as follows. Let B be a ball in RN centered at a point a, and ƒ be a real-valued function defined on the closure $\bar{B}$ having n+1 continuous partial derivatives at every point. Taylor's theorem asserts that for any $x\in B$,

$f(x)=\sum_{|\alpha|=0}^n\frac{1}{\alpha!}\frac{\partial^\alpha f(a)}{\partial x^\alpha}(x-a)^\alpha+\sum_{|\alpha|=n+1}R_{\alpha}(x)(x-a)^\alpha$

where the summation extends over multi-indices α (this formula uses the multi-index notation).

The remainder terms satisfy the inequality

$|R_{\alpha}(x)|\le\sup_{y\in\bar{B} }\left|\frac{1}{\alpha!}\frac{\partial^\alpha f(y)}{\partial x^\alpha}\right|$

for all α with |α| = n + 1. As was the case with one variable, the remainder terms can be described explicitly. See the proof for details.

## Proof: Taylor's theorem in one variable

### Integral version

We first prove Taylor's theorem with the integral remainder term.[4]

The fundamental theorem of calculus states that

$\int_a^x \, f'(t) \, dt=f(x)-f(a),$

which can be rearranged to:

$f(x)=f(a)+ \int_a^x \, f'(t) \, dt.$

Now we can see that an application of Integration by parts yields:

\begin{align} f(x) &= f(a)+xf'(x)-af'(a)-\int_a^x \, tf''(t) \, dt \ &= f(a)+\int_a^x \, xf''(t) \,dt+xf'(a)-af'(a)-\int_a^x \, tf''(t) \, dt \ &= f(a)+(x-a)f'(a)+\int_a^x \, (x-t)f''(t) \, dt. \end{align}

The first equation is arrived at by letting $u=f'(t)\,$ and dv = dt; the second equation by noting that $\int_a^x \, xf''(t) \,dt = xf'(x)-xf'(a)$; the third just factors out some common terms.

Another application yields:

$f(x)=f(a)+(x-a)f'(a)+ \frac 1 2 (x-a)^2f''(a) + \frac 1 2 \int_a^x \, (x-t)^2f'''(t) \, dt.$

By repeating this process, we may derive Taylor's theorem for higher values of n.

This can be formalized by applying the technique of induction. So, suppose that Taylor's theorem holds for a particular n, that is, suppose that

$f(x) = f(a) + \frac{f'(a)}{1!}(x - a) + \cdots + \frac{f^{(n)}(a)}{n!}(x - a)^n + \int_a^x \frac{f^{(n+1)} (t)}{n!} (x - t)^n \, dt. \qquad(*)$

We can rewrite the integral using integration by parts. An antiderivative of (xt)n as a function of t is given by −(xt)n+1 / (n + 1), so

$\int_a^x \frac{f^{(n+1)} (t)}{n!} (x - t)^n \, dt$
${} = - \left[ \frac{f^{(n+1)} (t)}{(n+1)n!} (x - t)^{n+1} \right]_a^x + \int_a^x \frac{f^{(n+2)} (t)}{(n+1)n!} (x - t)^{n+1} \, dt$
${} = \frac{f^{(n+1)} (a)}{(n+1)!} (x - a)^{n+1} + \int_a^x \frac{f^{(n+2)} (t)}{(n+1)!} (x - t)^{n+1} \, dt.$

Substituting this in (*) proves Taylor's theorem for n + 1, and hence for all nonnegative integers n.

The remainder term in the Lagrange form can be derived by the mean value theorem for integrals in the following way:

$R_n = \int_a^x \frac{f^{(n+1)} (t)}{n!} (x - t)^n \, dt =f^{(n+1)}(\xi) \int_a^x \frac{(x - t)^n }{n!} \, dt,$

where ξ is some number from the interval [a, x]. The last integral can be evaluated immediately, which leads to

$R_n = \frac{f^{(n+1)}(\xi)}{(n+1)!} (x-a)^{n+1}.$

More generally, for any function G(t), the mean value theorem asserts the existence of ξ in the interval [a, x] satisfying

$R_n = \int_a^x \frac{f^{(n+1)} (t)}{n!} (x - t)^n \frac{G'(t)}{G'(t)}\, dt =\frac{f^{(n+1)}(\xi)}{n!} (x-\xi)^n \frac{1}{G'(\xi)} \int_a^x G'(t) \, dt$
$= \frac{f^{(n+1)}(\xi)}{n!} (x-\xi)^n \cdot \frac{G(x)-G(a)}{G'(\xi)}.$

### Mean value theorem

An alternative proof, which holds under milder technical assumptions on the function ƒ, can be supplied using the Cauchy mean value theorem. Let G be a real-valued function continuous on [a, x] and differentiable with non-vanishing derivative on (a, x). Let

$F(t) = f(t) + \frac{f'(t)}{1!}(x-t) + \cdots + \frac{f^{(n)}(t)}{n!}(x-t)^n.$

By Cauchy's mean value theorem,

$\frac{F'(\xi)}{G'(\xi)} = \frac{F(x) - F(a)}{G(x) - G(a)}$ (1)

for some ξ ∈ (a, x). Note that the numerator F(x) − F(a) = Rn is the remainder of the Taylor polynomial for ƒ(x). On the other hand, computing F′(t),

$F'(t) = f'(t) - f'(t) + \frac{f''(t)}{1!}(x-t) - \frac{f''(t)}{1!}(x-t) + \cdots + \frac{f^{(n+1)}(t)}{n!}(x-t)^n = \frac{f^{(n+1)}(t)}{n!}(x-t)^n.$

Putting these two facts together and rearranging the terms of (1) yields

$R_n = \frac{f^{(n+1)}(\xi)}{n!}(x-\xi)^n\cdot\frac{G(x)-G(a)}{G'(\xi)},$

which was to be shown.

Note that the Lagrange form of the remainder comes from taking G(t) = (xt)n+1, the given Cauchy form of the remainder comes from taking G(t) = (ta), and the integral form of the remainder comes from taking

$G(t) = \int_a^t \frac{f^{(n+1)}(s)}{n!} (x-s)^n ds.$

## Proof: several variables

Let x = (x1,...,xN) lie in the ball B with center a. Parametrize the line segment between a and x by u(t) = a + t(xa). We apply the one-variable version of Taylor's theorem to the function ƒ(u(t)):

$f(x)=f(u(1))=f(a)+\sum_{k=1}^n\left.\frac{1}{k!}\frac{d^k}{dt^k}\right|_{t=0}f(u(t))\ +\ \int_0^1 \frac{(1-t)^n }{n!} \frac{d^{n+1}}{dt^{n+1}} f(u(t))\, dt.$

By the chain rule for several variables,

$\frac{d^k}{dt^k}f(u(t)) = \frac{d^k}{dt^k} f(a+t(x-a)) = \sum_{|\alpha|=k} \left(\begin{matrix} k \\ \alpha\end{matrix} \right) (D^\alpha f) (a+t(x-a))\cdot (x-a)^\alpha$

where $\left(\begin{matrix}k \\ \alpha\end{matrix}\right)$ is the multinomial coefficient for the multi-index α. Since $\frac{1}{k!}\left(\begin{matrix}k\\ \alpha\end{matrix}\right)=\frac{1}{\alpha!}$, we get

$f(x)= f(a)+\sum_{|\alpha|=1}^n\frac{1}{\alpha!} (D^\alpha f) (a)(x-a)^\alpha+\sum_{|\alpha|=n+1}\frac{n+1}{\alpha!} (x-a)^\alpha \int_0^1 (1-t)^n (D^\alpha f)(a+t(x-a))\,dt.$

The remainder term is given by

$\sum_{|\alpha|=n+1}\frac{n+1}{\alpha!} (x-a)^\alpha \int_0^1 (1-t)^n (D^\alpha f)(a+t(x-a))\,dt.$

The terms of this summation are explicit forms for the Rα in the statement of the theorem. These are easily seen to satisfy the required estimate.

## Notes

1. ^ Klein (1998) 20.3; Apostol (1967) 7.7.
2. ^ Apostol (1967) 7.7.
3. ^ Apostol (1967) 7.5.
4. ^ Note that this proof requires ƒ(n) to be absolutely continuous on [a, x] so that the fundamental theorem of calculus holds. Except at the end when the mean value theorem is invoked, differentiability of ƒ(n) need not be assumed since absolute continuity implies differentiability almost everywhere as well as the validity of the fundamental theorem of calculus, provided the integrals involved are understood as Lebesgue integrals. Consequently, the integral form of the remainder holds with this particular weakening of the assumptions on ƒ.

## References

• Apostol, Tom (1967). Calculus. Jon Wiley & Sons, Inc.. ISBN 0-471-00005-1.
• Klein, Morris (1998). Calculus: An Intuitive and Physical Approach. Dover. ISBN 0-486-40453-6.