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Taylor's theorem

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In calculus, Taylor's theorem gives a sequence of increasingly better approximations of a differentiable function near 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 is named after the mathematician Brook Taylor, who stated it in 1712, even though the result was first discovered 41 years earlier in 1671 by James Gregory.

Image:Taylorspolynomialexbig.svg
The exponential function Failed to parse (Missing texvc executable; please see math/README to configure.): y=e^x (continuous red line) and the corresponding Taylor polynomial of degree four around the origin (dashed green line).

Contents

Taylor's theorem in one variable

A simple example of Taylor's theorem is the approximation of the exponential function Failed to parse (Missing texvc executable; please see math/README to configure.): \textrm{e}^x

near x = 0:
Failed to parse (Missing texvc executable; please see math/README to configure.): \textrm{e}^x \approx 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!}.


The precise statement of the theorem is as follows: If n ≥ 0 is an integer and f 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 we have

Failed to parse (Missing texvc executable; please see math/README to configure.): 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


Here, n! denotes the factorial of n, and Rn is a remainder term, denoting the difference between the Taylor polynomial of degree n and the original function. The remainder term Rn 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

Failed to parse (Missing texvc executable; please see math/README to configure.): R_n = \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

Failed to parse (Missing texvc executable; please see math/README to configure.): R_n = \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

Failed to parse (Missing texvc executable; please see math/README to configure.): R_n = \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 integral form[3] of the remainder term is

Failed to parse (Missing texvc executable; please see math/README to configure.): R_n(x) = \int_a^x \frac{f^{(n+1)} (t)}{n!} (x - t)^n \, dt,


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

For some functions f(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 f 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, we have a contour integral expression for the remainder

Failed to parse (Missing texvc executable; please see math/README to configure.): 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 (a-r,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 f is a function which is n times continuously differentiable on the closed interval [a-r, a+r] and n + 1 times differentiable on the open interval (a-r, a+r). If there exists a positive real constant Mn such that |f(n+1)(x)| ≤ Mn for all x ∈ (a-r,a+r), then

Failed to parse (Missing texvc executable; please see math/README to configure.): 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 Rn satisfies the inequality (known as Cauchy's estimate):

Failed to parse (Missing texvc executable; please see math/README to configure.): |R_n(x)| \le M_n \frac{r^{n+1}}{(n+1)!}


for all x ∈ (a-r,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, in addition, f is infinitely differentiable on [a-r,a+r] and

Failed to parse (Missing texvc executable; please see math/README to configure.): M_n\frac{r^{n+1}}{(n+1)!} \rightarrow 0
as Failed to parse (Missing texvc executable; please see math/README to configure.): n \rightarrow \infin ,\!


then f is analytic on (a-r,a+r). In other words, an analytic function is the uniform limit of its Taylor polynomials on an interval. This makes precise the idea that analytic functions are those which are equal to their Taylor series.

Taylor's theorem for several variables

Topics in calculus

Fundamental theorem
Limits of functions
Continuity
Vector calculus
Tensor calculus
Mean value theorem

Differentiation

Product rule
Quotient rule
Chain rule
Implicit differentiation
Taylor's theorem
Related rates
Table of derivatives

Integration

Lists of integrals
Improper integrals
Integration by:
parts, disks, cylindrical
shells
, substitution,
trigonometric substitution,
partial fractions

Taylor's theorem can be generalized to several variables as follows. Let B be a ball in RN centered at a point a, and f be a real-valued function defined on the closure Failed to parse (Missing texvc executable; please see math/README to configure.): \bar{B}

having n+1 continuous partial derivatives at every point. Taylor's theorem asserts that for any Failed to parse (Missing texvc executable; please see math/README to configure.): x\in B

,

Failed to parse (Missing texvc executable; please see math/README to configure.): f(x)=\sum_{|\alpha|=0}^n\frac{D^\alpha f(a)}{\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

Failed to parse (Missing texvc executable; please see math/README to configure.): |R_{\alpha}(x)|\le\sup_{y\in\bar{B} }\left|\frac{D^\alpha f(y)}{\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

Failed to parse (Missing texvc executable; please see math/README to configure.): \int_a^x \, f'(t) \, dt=f(x)-f(a),


which can be rearranged to:

Failed to parse (Missing texvc executable; please see math/README to configure.): f(x)=f(a)+ \int_a^x \, f'(t) \, dt.


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

Failed to parse (Missing texvc executable; please see math/README to configure.): \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 Failed to parse (Missing texvc executable; please see math/README to configure.): u=f'(t)

and Failed to parse (Missing texvc executable; please see math/README to configure.): dv=dt
the second equation by noting that Failed to parse (Missing texvc executable; please see math/README to configure.): \int_a^x \, xf''(t) \,dt = xf'(x)-xf'(a)
the third just factors out some common terms.)

Another application yields:

Failed to parse (Missing texvc executable; please see math/README to configure.): 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

Failed to parse (Missing texvc executable; please see math/README to configure.): 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 (x − t)n as a function of t is given by −(xt)n+1 / (n + 1), so

Failed to parse (Missing texvc executable; please see math/README to configure.): \int_a^x \frac{f^{(n+1)} (t)}{n!} (x - t)^n \, dt


Failed to parse (Missing texvc executable; please see math/README to configure.): {} = - \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


Failed to parse (Missing texvc executable; please see math/README to configure.): {} = \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 in the following way:

Failed to parse (Missing texvc executable; please see math/README to configure.): 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.


The last integral can be solved immediately, which leads to

Failed to parse (Missing texvc executable; please see math/README to configure.): R_n = \frac{f^{(n+1)}(\xi)}{(n+1)!} (x-a)^{n+1}.


Mean value theorem

An alternative proof, which holds under milder technical assumptions on the function f, 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

Failed to parse (Missing texvc executable; please see math/README to configure.): F(t) = f(t) + \frac{f'(t)}{1!}(x-t) + \dots + \frac{f^{(n)}(t)}{n!}(x-t)^n.


By Cauchy's mean value theorem,

Failed to parse (Missing texvc executable; please see math/README to configure.): \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 f(x). On the other hand, computing F′(t),

Failed to parse (Missing texvc executable; please see math/README to configure.): F'(t) = f'(t) - f'(t) + \frac{f''(t)}{1!}(x-t) - \frac{f''(t)}{1!}(x-t) + \dots + \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

Failed to parse (Missing texvc executable; please see math/README to configure.): 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) = (t-a)n+1, and the given Cauchy form of the remainder comes from taking G(t) = (t-a).

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(x-a). We apply the one-variable version of Taylor's theorem to the function f(u(t)):

Failed to parse (Missing texvc executable; please see math/README to configure.): 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,

Failed to parse (Missing texvc executable; please see math/README to configure.): \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 Failed to parse (Missing texvc executable; please see math/README to configure.): \left(\begin{matrix}k\\ \alpha\end{matrix}\right)

is the multinomial coefficient for the multi-index α.  Since Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{1}{k!}\left(\begin{matrix}k\\ \alpha\end{matrix}\right)=\frac{1}{\alpha!}

, we get

Failed to parse (Missing texvc executable; please see math/README to configure.): 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

Failed to parse (Missing texvc executable; please see math/README to configure.): \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.

See also

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 f(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 f(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 f.


References

External links

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