Taylor's theorem

The exponential function ${\textstyle y=e^{x}}$ (red) and the corresponding Taylor polynomial of degree four (dashed green) around the origin.

In calculus, Taylor's theorem gives an approximation of a Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\textstyle k} -times differentiable function around a given point by a polynomial of degree Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\textstyle k} , called the Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\textstyle k} -th-order Taylor polynomial. For a smooth function, the Taylor polynomial is the truncation at the order Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\textstyle k} of the Taylor series of the function. The first-order Taylor polynomial is the linear approximation of the function, and the second-order Taylor polynomial is often referred to as the quadratic approximation. There are several versions of Taylor's theorem, some giving explicit estimates of the approximation error of the function by its Taylor polynomial.

Taylor's theorem is named after Brook Taylor, who stated a version of it in 1715, although an earlier version of the result was already mentioned in 1671 by James Gregory.

Taylor's theorem is taught in introductory-level calculus courses and is one of the central elementary tools in mathematical analysis. It gives simple arithmetic formulas to accurately compute values of many transcendental functions such as the exponential function and trigonometric functions. It is the starting point of the study of analytic functions, and is fundamental in various areas of mathematics, as well as in numerical analysis and mathematical physics. Taylor's theorem also generalizes to multivariate and vector valued functions. It provided the mathematical basis for some landmark early computing machines: Charles Babbage's difference engine calculated sines, cosines, logarithms, and other transcendental functions by numerically integrating the first 7 terms of their Taylor series.

Motivation

File:E^x with linear approximation.png

Graph of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\textstyle f(x)=e^{x}} (blue) with its linear approximation Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\textstyle P_{1}(x)=1+x} (red) at Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\textstyle a=0} .

If a real-valued function Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\textstyle f(x)} is differentiable at the point ${\textstyle x=a}$, then it has a linear approximation near this point. This means that there exists a function h₁(x) such that

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=f(a)+f'(a)(x-a)+h_{1}(x)(x-a),\quad \lim _{x\to a}h_{1}(x)=0.}

Here

$$P_{1}(x)=f(a)+f'(a)(x-a)$$

is the linear approximation of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\textstyle f(x)} for x near the point a, whose graph ${\textstyle y=P_{1}(x)}$ is the tangent line to the graph Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\textstyle y=f(x)} at x = a. The error in the approximation is: $$R_{1}(x)=f(x)-P_{1}(x)=h_{1}(x)(x-a).$$

As x tends to a, this error goes to zero much faster than Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x-a)} , making Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)\approx P_1(x)} a useful approximation.

Error creating thumbnail:

Graph of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle f(x)=e^x} (blue) with its quadratic approximation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_2(x) = 1 +x + \dfrac{x^2}{2}} (red) at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle a=0} . Note the improvement in the approximation.

For a better approximation to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle f(x)} , we can fit a quadratic polynomial instead of a linear function:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_2(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^2.}

Instead of just matching one derivative of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle f(x)} at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x=a} , this polynomial has the same first and second derivatives, as is evident upon differentiation.

Taylor's theorem ensures that the quadratic approximation is, in a sufficiently small neighborhood of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x=a} , more accurate than the linear approximation. Specifically,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = P_2(x) + h_2(x)(x - a)^2, \quad \lim_{x \to a} h_2(x) = 0.}

Here the error in the approximation is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R_2(x) = f(x) - P_2(x) = h_2(x)(x - a)^2,}

which, given the limiting behavior of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h_2} , goes to zero faster than Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x - a)^2} as x tends to a.

File:Tayloranimation.gif

Approximation of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle f(x)= \dfrac{1}{1+x^2}} (blue) by its Taylor polynomials Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle P_k} of order Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle k=1,\ldots,16} centered at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x=0} (red) and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x=1} (green). The approximations do not improve at all outside Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-1,1)} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle (1-\sqrt{2}, 1+\sqrt{2})} , respectively.

Similarly, we might get still better approximations to f if we use polynomials of higher degree, since then we can match even more derivatives with f at the selected base point.

In general, the error in approximating a function by a polynomial of degree k will go to zero much faster than Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x-a)^k} as x tends to a. However, there are functions, even infinitely differentiable ones, for which increasing the degree of the approximating polynomial does not increase the accuracy of approximation: we say such a function fails to be analytic at x = a: it is not (locally) determined by its derivatives at this point.

Taylor's theorem is of asymptotic nature: it only tells us that the error Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle R_k} in an approximation by a Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle k} -th order Taylor polynomial P_k tends to zero faster than any nonzero Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle k} -th degree polynomial as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x \to a} . It does not tell us how large the error is in any concrete neighborhood of the center of expansion, but for this purpose there are explicit formulas for the remainder term (given below) which are valid under some additional regularity assumptions on f. These enhanced versions of Taylor's theorem typically lead to uniform estimates for the approximation error in a small neighborhood of the center of expansion, but the estimates do not necessarily hold for neighborhoods which are too large, even if the function f is analytic. In that situation one may have to select several Taylor polynomials with different centers of expansion to have reliable Taylor-approximations of the original function (see animation on the right.)

There are several ways we might use the remainder term:

1. Estimate the error for a polynomial P_k(x) of degree k estimating Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle f(x)} on a given interval (a – r, a + r). (Given the interval and degree, we find the error.)
2. Find the smallest degree k for which the polynomial P_k(x) approximates Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle f(x)} to within a given error tolerance on a given interval (a − r, a + r) . (Given the interval and error tolerance, we find the degree.)
3. Find the largest interval (a − r, a + r) on which P_k(x) approximates Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle f(x)} to within a given error tolerance. (Given the degree and error tolerance, we find the interval.)

Taylor's theorem in one real variable

Statement of the theorem

The precise statement of the most basic version of Taylor's theorem is as follows:

The polynomial appearing in Taylor's theorem is the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle \boldsymbol{k}} -th order Taylor polynomial

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_k(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(k)}(a)}{k!}(x-a)^k }

of the function Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} at the point Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} . The Taylor polynomial is the unique "asymptotic best fit" polynomial in the sense that if there exists a function Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h_k : \R \to \R} and a Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle k} -th order polynomial p such that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = p(x) + h_k(x)(x-a)^k, \quad \lim_{x\to a} h_k(x) = 0 , }

then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p = P_k} . Taylor's theorem describes the asymptotic behavior of the remainder term

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R_k(x) = f(x) - P_k(x), }

which is the approximation error when approximating f with its Taylor polynomial. Using the little-o notation, the statement in Taylor's theorem reads as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R_k(x) = o(|x-a|^{k}), \quad x \to a. }

Explicit formulas for the remainder

Under stronger regularity assumptions on f there are several precise formulas for the remainder term R_k of the Taylor polynomial, the most common ones being the following.

These refinements of Taylor's theorem are usually proved using the mean value theorem, whence the name. Additionally, notice that this is precisely the mean value theorem when Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle k=0} . Also other similar expressions can be found. For example, if G(t) is continuous on the closed interval and differentiable with a non-vanishing derivative on the open interval between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle a} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x} , then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R_k(x) = \frac{f^{(k+1)}(\xi)}{k!}(x-\xi)^k \frac{G(x)-G(a)}{G'(\xi)} }

for some number Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle \xi} between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle a} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x} . This version covers the Lagrange and Cauchy forms of the remainder as special cases, and is proved below using Cauchy's mean value theorem. The Lagrange form is obtained by taking Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G(t)=(x-t)^{k+1}} and the Cauchy form is obtained by taking Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G(t)=t-a} .

The statement for the integral form of the remainder is more advanced than the previous ones, and requires understanding of Lebesgue integration theory for the full generality. However, it holds also in the sense of Riemann integral provided the (k + 1)th derivative of f is continuous on the closed interval [a,x].

Due to the absolute continuity of f^(k) on the closed interval between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle a} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x} , its derivative f^(k+1) exists as an L¹-function, and the result can be proven by a formal calculation using the fundamental theorem of calculus and integration by parts.

Estimates for the remainder

It is often useful in practice to be able to estimate the remainder term appearing in the Taylor approximation, rather than having an exact formula for it. Suppose that f is (k + 1)-times continuously differentiable in an interval I containing a. Suppose that there are real constants q and Q such that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q\le f^{(k+1)}(x)\le Q}

throughout I. Then the remainder term satisfies the inequality

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q\frac{(x-a)^{k+1}}{(k+1)!}\le R_k(x)\le Q\frac{(x-a)^{k+1}}{(k+1)!},}

if x > a, and a similar estimate if x < a. This is a simple consequence of the Lagrange form of the remainder. In particular, if

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |f^{(k+1)}(x)|\le M}

on an interval I = (a − r,a + r) with some Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r > 0} , then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |R_k(x)|\le M\frac{|x-a|^{k+1}}{(k+1)!}\le M\frac{r^{k+1}}{(k+1)!}}

for all x∈(a − r,a + r). The second inequality is called a uniform estimate, because it holds uniformly for all x on the interval (a − r,a + r).

Example

File:Expanimation.gif

Approximation of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle e^x} (blue) by its Taylor polynomials Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_k} of order Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle k=1,\ldots,7} centered at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x=0} (red).

Suppose that we wish to find the approximate value of the function Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle f(x)=e^x} on the interval Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle [-1,1]} while ensuring that the error in the approximation is no more than 10⁻⁵. In this example we pretend that we only know the following properties of the exponential function:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^0=1, \qquad \frac{d}{dx} e^x = e^x, \qquad e^x>0, \qquad x\in\R.}

(★)

From these properties it follows that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle f^{(k)}(x)=e^x} for all Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle k} , and in particular, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle f^{(k)}(0)=1} . Hence the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle k} -th order Taylor polynomial of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle f} at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle 0} and its remainder term in the Lagrange form are given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_k(x) = 1+x+\frac{x^2}{2!}+\cdots+\frac{x^k}{k!}, \qquad R_k(x)=\frac{e^\xi}{(k+1)!}x^{k+1},}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle \xi} is some number between 0 and x. Since e^x is increasing by (★), we can simply use Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle e^x \leq 1} for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x \in [-1,0]} to estimate the remainder on the subinterval Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [-1,0]} . To obtain an upper bound for the remainder on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [0,1]} , we use the property Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle e^\xi <e^x} for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle 0<\xi<x} to estimate

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^x = 1 + x + \frac{e^\xi}{2}x^2 < 1 + x + \frac{e^x}{2}x^2, \qquad 0 < x\leq 1 }

using the second order Taylor expansion. Then we solve for e^x to deduce that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^x \leq \frac{1+x}{1-\frac{x^2}{2}} = 2\frac{1+x}{2-x^2} \leq 4, \qquad 0 \leq x\leq 1 }

simply by maximizing the numerator and minimizing the denominator. Combining these estimates for e^x we see that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |R_k(x)| \leq \frac{4|x|^{k+1}}{(k+1)!} \leq \frac{4}{(k+1)!}, \qquad -1\leq x \leq 1, }

so the required precision is certainly reached, when

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{4}{(k+1)!} < 10^{-5} \quad \Longleftrightarrow \quad 4\cdot 10^5 < (k+1)! \quad \Longleftrightarrow \quad k \geq 9. }

(See factorial or compute by hand the values Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle 9! =362880} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle 10! =3628800} .) As a conclusion, Taylor's theorem leads to the approximation

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^x = 1+x+\frac{x^2}{2!} + \cdots + \frac{x^9}{9!} + R_9(x), \qquad |R_9(x)| < 10^{-5}, \qquad -1\leq x \leq 1. }

For instance, this approximation provides a decimal expression Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e \approx 2.71828} , correct up to five decimal places.

Relationship to analyticity

Taylor expansions of real analytic functions

Let I ⊂ R be an open interval. By definition, a function f : I → R is real analytic if it is locally defined by a convergent power series. This means that for every a ∈ I there exists some r > 0 and a sequence of coefficients c_k ∈ R such that (a − r, a + r) ⊂ I and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = \sum_{k=0}^\infty c_k(x-a)^k = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots, \qquad |x-a|<r. }

In general, the radius of convergence of a power series can be computed from the Cauchy–Hadamard formula

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{R} = \limsup_{k\to\infty}|c_k|^\frac{1}{k}. }

This result is based on comparison with a geometric series, and the same method shows that if the power series based on a converges for some b ∈ R, it must converge uniformly on the closed interval Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle [a-r_b,a+r_b]} , where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle r_b=\left\vert b-a \right\vert} . Here only the convergence of the power series is considered, and it might well be that (a − R,a + R) extends beyond the domain I of f.

The Taylor polynomials of the real analytic function f at a are simply the finite truncations

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_k(x) = \sum_{j=0}^k c_j(x-a)^j, \qquad c_j = \frac{f^{(j)}(a)}{j!}}

of its locally defining power series, and the corresponding remainder terms are locally given by the analytic functions

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R_k(x) = \sum_{j=k+1}^\infty c_j(x-a)^j = (x-a)^k h_k(x), \qquad |x-a|<r. }

Here the functions

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} & h_k:(a-r,a+r)\to \R \\[1ex] & h_k(x) = (x-a)\sum_{j=0}^\infty c_{k+1+j} \left(x - a\right)^j \end{align}}

are also analytic, since their defining power series have the same radius of convergence as the original series. Assuming that [a − r, a + r] ⊂ I and r < R, all these series converge uniformly on (a − r, a + r). Naturally, in the case of analytic functions one can estimate the remainder term Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle R_k(x)} by the tail of the sequence of the derivatives f′(a) at the center of the expansion, but using complex analysis also another possibility arises, which is described below.

Taylor's theorem and convergence of Taylor series

The Taylor series of f will converge in some interval in which all its derivatives are bounded and do not grow too fast as k goes to infinity. (However, even if the Taylor series converges, it might not converge to f, as explained below; f is then said to be non-analytic.)

One might think of the Taylor series

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) \approx \sum_{k=0}^\infty c_k(x-a)^k = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots }

of an infinitely many times differentiable function f : R → R as its "infinite order Taylor polynomial" at a. Now the estimates for the remainder imply that if, for any r, the derivatives of f are bounded over (a − r, a + r), then for any order k and for any r > 0 there exists a constant M_k,r > 0 such that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |R_k(x)| \leq M_{k,r} \frac{|x-a|^{k+1}}{(k+1)!} }

(★★)

for every x ∈ (a − r,a + r). Sometimes the constants M_k,r can be chosen in such way that M_k,r is bounded above, for fixed r and all k. Then the Taylor series of f converges uniformly to some analytic function

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} & T_f:(a-r,a+r)\to\R \\ & T_f(x) = \sum_{k=0}^\infty \frac{f^{(k)}(a)}{k!} \left(x-a\right)^k \end{align}}

(One also gets convergence even if M_k,r is not bounded above as long as it grows slowly enough.)

The limit function T_f is by definition always analytic, but it is not necessarily equal to the original function f, even if f is infinitely differentiable. In this case, we say f is a non-analytic smooth function, for example a flat function:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} & f:\R \to \R \\ & f(x) = \begin{cases} e^{-\frac{1}{x^2}} & x>0 \\ 0 & x \leq 0 . \end{cases} \end{align}}

Using the chain rule repeatedly by mathematical induction, one shows that for any order k,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{(k)}(x) = \begin{cases} \frac{p_k(x)}{x^{3k}}\cdot e^{-\frac{1}{x^2}} & x>0 \\ 0 & x \leq 0 \end{cases}}

for some polynomial p_k of degree 2(k − 1). The function Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{-\frac{1}{x^2}}} tends to zero faster than any polynomial as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x \to 0} , so f is infinitely many times differentiable and f^(k)(0) = 0 for every positive integer k. The above results all hold in this case:

• The Taylor series of f converges uniformly to the zero function T_f(x) = 0, which is analytic with all coefficients equal to zero.
• The function f is unequal to this Taylor series, and hence non-analytic.
• For any order k ∈ N and radius r > 0 there exists M_k,r > 0 satisfying the remainder bound (★★) above.

However, as k increases for fixed r, the value of M_k,r grows more quickly than r^k, and the error does not go to zero.

Taylor's theorem in complex analysis

Taylor's theorem generalizes to functions f : C → C which are complex differentiable in an open subset U ⊂ C of the complex plane. However, its usefulness is dwarfed by other general theorems in complex analysis. Namely, stronger versions of related results can be deduced for complex differentiable functions f : U → C using Cauchy's integral formula as follows.

Let r > 0 such that the closed disk B(z, r) ∪ S(z, r) is contained in U. Then Cauchy's integral formula with a positive parametrization γ(t) = z + re^it of the circle S(z, r) with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t \in [0,2 \pi]} gives

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(w)}{w-z}\,dw, \quad f'(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(w)}{(w-z)^2} \, dw, \quad \ldots, \quad f^{(k)}(z) = \frac{k!}{2\pi i}\int_\gamma \frac{f(w)}{(w-z)^{k+1}} \, dw.}

Here all the integrands are continuous on the circle S(z, r), which justifies differentiation under the integral sign. In particular, if f is once complex differentiable on the open set U, then it is actually infinitely many times complex differentiable on U. One also obtains Cauchy's estimate

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |f^{(k)}(z)| \leq \frac{k!}{2\pi}\int_\gamma \frac{M_r}{|w-z|^{k+1}} \, dw = \frac{k!M_r}{r^k}, \quad M_r = \max_{|w-c|=r}|f(w)| }

for any z ∈ U and r > 0 such that B(z, r) ∪ S(c, r) ⊂ U. The estimate implies that the complex Taylor series

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_f(z) = \sum_{k=0}^\infty \frac{f^{(k)}(c)}{k!}(z-c)^k }

of f converges uniformly on any open disk Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle B(c,r) \subset U} with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle S(c,r) \subset U} into some function T_f. Furthermore, using the contour integral formulas for the derivatives f^(k)(c),

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} T_f(z) &= \sum_{k=0}^\infty \frac{(z-c)^k}{2\pi i}\int_\gamma \frac{f(w)}{(w-c)^{k+1}} \, dw \\ &= \frac{1}{2\pi i} \int_\gamma \frac{f(w)}{w-c} \sum_{k=0}^\infty \left(\frac{z-c}{w-c}\right)^k \, dw \\ &= \frac{1}{2\pi i} \int_\gamma \frac{f(w)}{w-c}\left( \frac{1}{1-\frac{z-c}{w-c}} \right) \, dw \\ &= \frac{1}{2\pi i} \int_\gamma \frac{f(w)}{w-z} \, dw \\ &= f(z), \end{align}}

so any complex differentiable function f in an open set U ⊂ C is in fact complex analytic. All that is said for real analytic functions here holds also for complex analytic functions with the open interval I replaced by an open subset U ∈ C and a-centered intervals (a − r, a + r) replaced by c-centered disks B(c, r). In particular, the Taylor expansion holds in the form

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(z) = P_k(z) + R_k(z), \quad P_k(z) = \sum_{j=0}^k \frac{f^{(j)}(c)}{j!}(z-c)^j, }

where the remainder term R_k is complex analytic. Methods of complex analysis provide some powerful results regarding Taylor expansions. For example, using Cauchy's integral formula for any positively oriented Jordan curve Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle \gamma} which parametrizes the boundary Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle \partial W \subset U} of a region Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle W \subset U} , one obtains expressions for the derivatives f^(j)(c) as above, and modifying slightly the computation for T_f(z) = f(z), one arrives at the exact formula

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R_k(z) = \sum_{j=k+1}^\infty \frac{(z-c)^j}{2\pi i} \int_\gamma \frac{f(w)}{(w-c)^{j+1}} \, dw = \frac{(z-c)^{k+1}}{2\pi i} \int_\gamma \frac{f(w) \, dw}{(w-c)^{k+1}(w-z)} , \qquad z\in W. }

The important feature here is that the quality of the approximation by a Taylor polynomial on the region Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle W \subset U} is dominated by the values of the function f itself on the boundary Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle \partial W \subset U} . Similarly, applying Cauchy's estimates to the series expression for the remainder, one obtains the uniform estimates

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |R_k(z)| \leq \sum_{j=k+1}^\infty \frac{M_r |z-c|^j}{r^j} = \frac{M_r}{r^{k+1}} \frac{|z-c|^{k+1}}{1-\frac{|z-c|}{r}} \leq \frac{M_r \beta^{k+1}}{1-\beta}, \qquad \frac{|z-c|}{r} \leq \beta < 1. }

Example

File:Function with two poles.png

Complex plot of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle f(z)=\frac{1}{1+z^2}} . Modulus is shown by elevation and argument by coloring: cyan = Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle 0} , blue = Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle \frac{\pi}{3}} , violet = Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle \frac{2\pi}{3}} , red = Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi} , yellow = Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle \frac{4\pi}{3}} , green = Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle \frac{5\pi}{3}} .

The function

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} & f : \R \to \R \\ & f(x) = \frac{1}{1+x^2} \end{align}}

is real analytic, that is, locally determined by its Taylor series. This function was plotted above to illustrate the fact that some elementary functions cannot be approximated by Taylor polynomials in neighborhoods of the center of expansion which are too large. This kind of behavior is easily understood in the framework of complex analysis. Namely, the function f extends into a meromorphic function

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} & f:\Complex \cup \{\infty\} \to \Complex \cup \{\infty\} \\ & f(z) = \frac{1}{1+z^2} \end{align}}

on the compactified complex plane. It has simple poles at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle z=i} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle z=-i} , and it is analytic elsewhere. Now its Taylor series centered at z₀ converges on any disc B(z₀, r) with r < |z − z₀|, where the same Taylor series converges at z ∈ C. Therefore, Taylor series of f centered at 0 converges on B(0, 1) and it does not converge for any z ∈ C with |z| > 1 due to the poles at i and −i. For the same reason the Taylor series of f centered at 1 converges on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle B(1, \sqrt{2})} and does not converge for any z ∈ C with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle \left\vert z-1 \right\vert>\sqrt{2}} .

Generalizations of Taylor's theorem

Higher-order differentiability

A function f: Rⁿ → R is differentiable at a ∈ Rⁿ if and only if there exists a linear functional L : Rⁿ → R and a function h : Rⁿ → R such that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(\boldsymbol{x}) = f(\boldsymbol{a}) + L(\boldsymbol{x}-\boldsymbol{a}) + h(\boldsymbol{x})\lVert\boldsymbol{x}-\boldsymbol{a}\rVert, \qquad \lim_{\boldsymbol{x}\to\boldsymbol{a}} h(\boldsymbol{x})=0. }

If this is the case, then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle L = df(\boldsymbol{a})} is the (uniquely defined) differential of f at the point a. Furthermore, then the partial derivatives of f exist at a and the differential of f at a is given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle df( \boldsymbol{a} )( \boldsymbol{v} ) = \frac{\partial f}{\partial x_1}(\boldsymbol{a}) v_1 + \cdots + \frac{\partial f}{\partial x_n}(\boldsymbol{a}) v_n. }

Introduce the multi-index notation

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\alpha| = \alpha_1+\cdots+\alpha_n, \quad \alpha!=\alpha_1!\cdots\alpha_n!, \quad \boldsymbol{x}^\alpha=x_1^{\alpha_1}\cdots x_n^{\alpha_n} }

for α ∈ Nⁿ and x ∈ Rⁿ. If all the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle k} -th order partial derivatives of f : Rⁿ → R are continuous at a ∈ Rⁿ, then by Clairaut's theorem, one can change the order of mixed derivatives at a, so the short-hand notation

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D^\alpha f = \frac{\partial^{|\alpha|}f}{\partial\boldsymbol x^\alpha} = \frac{\partial^{\alpha_1 + \ldots + \alpha_n}f}{\partial x_1^{\alpha_1}\cdots \partial x_n^{\alpha_n}}}

for the higher order partial derivatives is justified in this situation. The same is true if all the (k − 1)-th order partial derivatives of f exist in some neighborhood of a and are differentiable at a. Then we say that f is k times differentiable at the point a.

Taylor's theorem for multivariate functions

Using notations of the preceding section, one has the following theorem. Template:Math theorem

If the function f : Rⁿ → R is k + 1 times continuously differentiable in a closed ball Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B = \{ \mathbf{y} \in \R^n : \left\|\mathbf{a}-\mathbf{y}\right\| \leq r\}} for some Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r > 0} , then one can derive an exact formula for the remainder in terms of (k+1)-th order partial derivatives of f in this neighborhood. Namely,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} & f( \boldsymbol{x} ) = \sum_{|\alpha|\leq k} \frac{D^\alpha f(\boldsymbol{a})}{\alpha!} (\boldsymbol{x}-\boldsymbol{a})^\alpha + \sum_{|\beta|=k+1} R_\beta(\boldsymbol{x})(\boldsymbol{x}-\boldsymbol{a})^\beta, \\ & R_\beta( \boldsymbol{x} ) = \frac{|\beta|}{\beta!} \int_0^1 (1-t)^{|\beta|-1}D^\beta f \big(\boldsymbol{a}+t( \boldsymbol{x}-\boldsymbol{a} )\big) \, dt. \end{align} }

In this case, due to the continuity of (k+1)-th order partial derivatives in the compact set B, one immediately obtains the uniform estimates

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|R_\beta(\boldsymbol{x})\right| \leq \frac{1}{\beta!} \max_{|\alpha|=|\beta|} \max_{\boldsymbol{y}\in B} |D^\alpha f(\boldsymbol{y})|, \qquad \boldsymbol{x}\in B. }

Example in two dimensions

For example, the third-order Taylor polynomial of a smooth function Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f:\mathbb R^2\to\mathbb R} is, denoting Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{x}-\boldsymbol{a}=\boldsymbol{v}} ,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} P_3(\boldsymbol{x}) = f ( \boldsymbol{a} ) + {} &\frac{\partial f}{\partial x_1}( \boldsymbol{a} ) v_1 + \frac{\partial f}{\partial x_2}( \boldsymbol{a} ) v_2 + \frac{\partial^2 f}{\partial x_1^2}( \boldsymbol{a} ) \frac {v_1^2}{2!} + \frac{\partial^2 f}{\partial x_1 \partial x_2}( \boldsymbol{a} ) v_1 v_2 + \frac{\partial^2 f}{\partial x_2^2}( \boldsymbol{a} ) \frac{v_2^2}{2!} \\ & + \frac{\partial^3 f}{\partial x_1^3}( \boldsymbol{a} ) \frac{v_1^3}{3!} + \frac{\partial^3 f}{\partial x_1^2 \partial x_2}( \boldsymbol{a} ) \frac{v_1^2 v_2}{2!} + \frac{\partial^3 f}{\partial x_1 \partial x_2^2}( \boldsymbol{a} ) \frac{v_1 v_2^2}{2!} + \frac{\partial^3 f}{\partial x_2^3}( \boldsymbol{a} ) \frac{v_2^3}{3!} \end{align}}

Proofs

Proof for Taylor's theorem in one real variable

Let

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h_k(x) = \begin{cases} \frac{f(x) - P(x)}{(x-a)^k} & x\not=a\\ 0&x=a \end{cases} }

where, as in the statement of Taylor's theorem,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(k)}(a)}{k!}(x-a)^k.}

It is sufficient to show that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to a} h_k(x) =0. }

The proof here is based on repeated application of L'Hôpital's rule. Note that, for each Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle j=0,1,...,k-1} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{(j)}(a)=P^{(j)}(a)} . Hence each of the first Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle k-1} derivatives of the numerator in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h_k(x)} vanishes at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=a} , and the same is true of the denominator. Also, since the condition that the function Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle f} be Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle k} times differentiable at a point requires differentiability up to order Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle k-1} in a neighborhood of said point (this is true, because differentiability requires a function to be defined in a whole neighborhood of a point), the numerator and its Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle k-2} derivatives are differentiable in a neighborhood of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle a} . Clearly, the denominator also satisfies said condition, and additionally, doesn't vanish unless Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x=a} , therefore all conditions necessary for L'Hôpital's rule are fulfilled, and its use is justified. So

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \lim_{x\to a} \frac{f(x) - P(x)}{(x-a)^k} &= \lim_{x\to a} \frac{\frac{d}{dx}(f(x) - P(x))}{\frac{d}{dx}(x-a)^k} \\[1ex] &= \cdots \\[1ex] &= \lim_{x\to a} \frac{\frac{d^{k-1}}{dx^{k-1}}(f(x) - P(x))}{\frac{d^{k-1}}{dx^{k-1}}(x-a)^k}\\[1ex] &= \frac{1}{k!}\lim_{x\to a} \frac{f^{(k-1)}(x) - P^{(k-1)}(x)}{x-a}\\[1ex] &=\frac{1}{k!}(f^{(k)}(a) - P^{(k)}(a)) = 0 \end{align}}

where the second-to-last equality follows by the definition of the derivative at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x=a} .

Alternate proof for Taylor's theorem in one real variable

Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} be any real-valued continuous function to be approximated by the Taylor polynomial.

Step 1: Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle F} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle G} be functions. Set Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle F} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle G} to be

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} F(x) = f(x) - \sum^{n-1}_{k=0} \frac{f^{(k)}(a)}{k!}(x-a)^{k} \end{align}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} G(x) = (x-a)^{n} \end{align}}

Step 2: Properties of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle F} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle G} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} F(a) & = f(a) - f(a) - f'(a)(a - a) - ... - \frac{f^{(n-1)}(a)}{(n-1)!}(a-a)^{n-1} = 0 \\ G(a) & = (a-a)^n = 0 \end{align}}

Similarly,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} F'(a) = f'(a) - f'(a) - \frac{f''(a)}{(2-1)!}(a-a)^{(2-1)} - ... - \frac{f^{(n-1)}(a)}{(n-2)!}(a-a)^{n-2} = 0 \end{align}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} G'(a) &= n(a-a)^{n-1} = 0\\ &\qquad \vdots\\ G^{(n-1)}(a) &= F^{(n-1)}(a) = 0 \end{align}}

Step 3: Use Cauchy Mean Value Theorem

Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_{1}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g_{1}} be continuous functions on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a, b]} . Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a < x < b} so we can work with the interval Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a, x]} . Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_{1}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g_{1}} be differentiable on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a, x)} . Assume Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g_{1}'(x) \neq 0} for all Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in (a, b)} . Then there exists Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{1} \in (a, x)} such that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{f_{1}(x) - f_{1}(a)}{g_{1}(x) - g_{1}(a)} = \frac{f_{1}'(c_{1})}{g_{1}'(c_{1})} \end{align}}

Note: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G'(x) \neq 0} in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a, b)} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F(a), G(a) = 0} so

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{F(x)}{G(x)} = \frac{F(x) - F(a)}{G(x) - G(a)} = \frac{F'(c_{1})}{G'(c_{1})} \end{align}}

for some Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{1} \in (a, x)} .

This can also be performed for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a, c_{1})} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{F'(c_{1})}{G'(c_{1})} = \frac{F'(c_{1}) - F'(a)}{G'(c_{1}) - G'(a)} = \frac{F''(c_{2})}{G''(c_{2})} \end{align}}

for some Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{2} \in (a, c_{1})} . This can be continued to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{n}} .

This gives a partition in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a, b)} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a < c_{n} < c_{n-1} < \dots < c_{1} < x}

with

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{F(x)}{G(x)} = \frac{F'(c_{1})}{G'(c_{1})} = \dots = \frac{F^{(n)}(c_{n})}{G^{(n)}(c_{n})} .}

Set Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c = c_{n}} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{F(x)}{G(x)} = \frac{F^{(n)}(c)}{G^{(n)}(c)}}

Step 4: Substitute back

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{F(x)}{G(x)} = \frac{f(x) - \sum^{n-1}_{k=0} \frac{f^{(k)}(a)}{k!}(x-a)^{k}}{(x-a)^{n}} = \frac{F^{(n)}(c)}{G^{(n)}(c)} \end{align}}

By the Power Rule, repeated derivatives of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x - a)^{n}} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G^{(n)}(c) = n(n-1)...1} , so:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{F^{(n)}(c)}{G^{(n)}(c)} = \frac{f^{(n)}(c)}{n(n-1)\cdots1} = \frac{f^{(n)}(c)}{n!}.}

This leads to:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} f(x) - \sum^{n-1}_{k=0} \frac{f^{(k)}(a)}{k!}(x-a)^{k} = \frac{f^{(n)}(c)}{n!}(x-a)^{n} \end{align}.}

By rearranging, we get:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} f(x) = \sum^{n-1}_{k=0} \frac{f^{(k)}(a)}{k!}(x-a)^{k} + \frac{f^{(n)}(c)}{n!}(x-a)^{n} \end{align},}

or because Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{n} = a} eventually:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = \sum^{n}_{k=0} \frac{f^{(k)}(a)}{k!}(x-a)^{k}.}

Derivation for the mean value forms of the remainder

Let G be any real-valued function, continuous on the closed interval between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle a} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x} and differentiable with a non-vanishing derivative on the open interval between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle a} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x} , and define

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F(t) = f(t) + f'(t)(x-t) + \frac{f''(t)}{2!}(x-t)^2 + \cdots + \frac{f^{(k)}(t)}{k!}(x-t)^k. }

For Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t \in [a,x] } . Then, by Cauchy's mean value theorem,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{F'(\xi)}{G'(\xi)} = \frac{F(x) - F(a)}{G(x) - G(a)}}

(★★★)

for some Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle \xi} on the open interval between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle a} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x} . Note that here the numerator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle F(x)-F(a)=R_k(x)} is exactly the remainder of the Taylor polynomial for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle y=f(x)} . Compute

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} F'(t) = {} & f'(t) + \big(f''(t)(x-t) - f'(t)\big) + \left(\frac{f^{(3)}(t)}{2!}(x-t)^2 - \frac{f^{(2)}(t)}{1!}(x-t)\right) + \cdots \\ & \cdots + \left( \frac{f^{(k+1)}(t)}{k!}(x-t)^k - \frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1}\right) = \frac{f^{(k+1)}(t)}{k!}(x-t)^k, \end{align}}

plug it into (★★★) and rearrange terms to find that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R_k(x) = \frac{f^{(k+1)}(\xi)}{k!}(x-\xi)^k \frac{G(x)-G(a)}{G'(\xi)}.}

This is the form of the remainder term mentioned after the actual statement of Taylor's theorem with remainder in the mean value form. The Lagrange form of the remainder is found by choosing Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G(t) = (x-t)^{k+1} } and the Cauchy form by choosing Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G(t) = t-a} .

Remark. Using this method one can also recover the integral form of the remainder by choosing

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G(t) = \int_a^t \frac{f^{(k+1)}(s)}{k!} (x-s)^k \, ds,}

but the requirements for f needed for the use of mean value theorem are too strong, if one aims to prove the claim in the case that f^(k) is only absolutely continuous. However, if one uses Riemann integral instead of Lebesgue integral, the assumptions cannot be weakened.

Derivation for the integral form of the remainder

Due to the absolute continuity of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{(k)}} on the closed interval between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle a} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x} , its derivative Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{(k+1)}} exists as an Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L^1} -function, and we can use the fundamental theorem of calculus and integration by parts. This same proof applies for the Riemann integral assuming that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{(k)}} is continuous on the closed interval and differentiable on the open interval between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle a} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x} , and this leads to the same result as using the mean value theorem.

The fundamental theorem of calculus states that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=f(a)+ \int_a^x \, f'(t) \, dt.}

Now we can integrate by parts and use the fundamental theorem of calculus again to see that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} f(x) &= f(a)+\Big(xf'(x)-af'(a)\Big)-\int_a^x tf''(t) \, dt \\ &= f(a) + x\left(f'(a) + \int_a^x f''(t) \,dt \right) -af'(a)-\int_a^x tf''(t) \, dt \\ &= f(a)+(x-a)f'(a)+\int_a^x \, (x-t)f''(t) \, dt, \end{align} }

which is exactly Taylor's theorem with remainder in the integral form in the case Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k=1} . The general statement is proved using induction. Suppose that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = f(a) + \frac{f'(a)}{1!}(x - a) + \cdots + \frac{f^{(k)}(a)}{k!}(x - a)^k + \int_a^x \frac{f^{(k+1)} (t)}{k!} (x - t)^k \, dt. }

(eq1)

Integrating the remainder term by parts we arrive at

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int_a^x \frac{f^{(k+1)} (t)}{k!} (x - t)^k \, dt = & - \left[ \frac{f^{(k+1)} (t)}{(k+1)k!} (x - t)^{k+1} \right]_a^x + \int_a^x \frac{f^{(k+2)} (t)}{(k+1)k!} (x - t)^{k+1} \, dt \\ = & \ \frac{f^{(k+1)} (a)}{(k+1)!} (x - a)^{k+1} + \int_a^x \frac{f^{(k+2)} (t)}{(k+1)!} (x - t)^{k+1} \, dt. \end{align}}

Substituting this into the formula in (eq1) shows that if it holds for the value Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} , it must also hold for the value Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k+1} . Therefore, since it holds for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k=1} , it must hold for every positive integer Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} .

Derivation for the remainder of multivariate Taylor polynomials

We prove the special case, where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f:\mathbb R^n\to\mathbb R} has continuous partial derivatives up to the order Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k+1} in some closed ball Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B} with center Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{a}} . The strategy of the proof is to apply the one-variable case of Taylor's theorem to the restriction of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} to the line segment adjoining Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{x}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{a}} . Parametrize the line segment between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{a}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{x}} by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{u}(t)=\boldsymbol{a}+t(\boldsymbol{x}-\boldsymbol{a})} We apply the one-variable version of Taylor's theorem to the function Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(t) = f(\boldsymbol{u}(t))} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(\boldsymbol{x})=g(1)=g(0)+\sum_{j=1}^k\frac{1}{j!}g^{(j)}(0)\ +\ \int_0^1 \frac{(1-t)^k }{k!} g^{(k+1)}(t)\, dt.}

Applying the chain rule for several variables gives

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} g^{(j)}(t)&=\frac{d^j}{dt^j}f(\boldsymbol{u}(t))\\ &= \frac{d^j}{dt^j} f(\boldsymbol{a}+t(\boldsymbol{x}-\boldsymbol{a}))\\ &= \sum_{|\alpha| =j} \left(\begin{matrix} j\\ \alpha\end{matrix} \right) (D^\alpha f) (\boldsymbol{a}+t(\boldsymbol{x}-\boldsymbol{a})) (\boldsymbol{x}-\boldsymbol{a})^\alpha \end{align}}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tbinom j \alpha} is the multinomial coefficient. Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tfrac{1}{j!}\tbinom j \alpha=\tfrac{1}{\alpha!}} , we get:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(\boldsymbol{x})= f(\boldsymbol{a}) + \sum_{1 \leq |\alpha| \leq k}\frac{1}{\alpha!} (D^\alpha f) (\boldsymbol{a})(\boldsymbol{x}-\boldsymbol{a})^\alpha+\sum_{|\alpha|=k+1}\frac{k+1}{\alpha!} (\boldsymbol{x}-\boldsymbol{a})^\alpha \int_0^1 (1-t)^k (D^\alpha f)(\boldsymbol{a}+t(\boldsymbol{x}-\boldsymbol{a}))\,dt.}

See also

• Hadamard's lemma
• Laurent series – Power series with negative powers
• Padé approximant
• Newton series
• Approximation theory
• Function approximation

References

• Apostol, Tom (1967), Calculus, Wiley, ISBN 0-471-00005-1.
• Apostol, Tom (1974), Mathematical analysis, Addison–Wesley.
• Bartle, Robert G.; Sherbert, Donald R. (2011), Introduction to Real Analysis (4th ed.), Wiley, ISBN 978-0-471-43331-6.
• Hörmander, L. (1976), Linear Partial Differential Operators, Volume 1, Springer, ISBN 978-3-540-00662-6.
• Kline, Morris (1972), Mathematical thought from ancient to modern times, Volume 2, Oxford University Press.
• Kline, Morris (1998), Calculus: An Intuitive and Physical Approach, Dover, ISBN 0-486-40453-6.
• Pedrick, George (1994), A First Course in Analysis, Springer, ISBN 0-387-94108-8.
• Stromberg, Karl (1981), Introduction to classical real analysis, Wadsworth, ISBN 978-0-534-98012-2.
• Rudin, Walter (1987), Real and complex analysis (3rd ed.), McGraw-Hill, ISBN 0-07-054234-1.
• Tao, Terence (2014), Analysis, Volume I (3rd ed.), Hindustan Book Agency, ISBN 978-93-80250-64-9.

External links

• Taylor Series Approximation to Cosine at cut-the-knot
• Trigonometric Taylor Expansion interactive demonstrative applet
• Taylor Series Revisited at Holistic Numerical Methods Institute