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In calculus, Taylor's theorem gives an approximation of a k times differentiable function around a given point by a k-th order Taylor-polynomial. For analytic functions the Taylor polynomials at a given point are finite order truncations of its Taylor's series, which completely determines the function in some neighborhood of the point. The exact content of "Taylor's theorem" is not universally agreed upon. Indeed, there are several versions of it applicable in different situations, and some of them contain explicit estimates on the approximation error of the function by its Taylor-polynomial.

Taylor's theorem is named after the mathematician Brook Taylor, who stated a version of it in 1712. Yet, an explicit expression of the error was provided much later on by Joseph-Louis Lagrange. An earlier version of the result is already mentioned in 1671 by James Gregory.

Taylor's theorem is taught on introductory level calculus courses and it is one of the central elementary tools in mathematical analysis. Within pure mathematics it is the starting point of more advanced asymptotic analysis, and it is commonly used in more applied fields of numerics as well as in mathematical physics. Taylor's theorem also generalizes to multivariate and vector valued functions f : Rn &rarr; Rm on any dimensions n and m. This generalization of Taylor's theorem is the basis for the definition of so-called jets which appear in differential geometry and partial differential equations.

Motivation
If a real-valued function f is differentiable at the point a then it has a linear approximation at the point a. This means that there exists a function h1 such that


 * $$ f(x) = f(a) + f'(a)(x-a) + h_1(x)(x-a), \qquad \lim_{x\to a}h_1(x)=0.$$

Here


 * $$P_1(x) = f(a) + f'(a)(x-a) \ $$

is the linear approximation of f at the point a. The graph of is the tangent line to the graph of f at. The error in the approximation is
 * $$R_1(x) = f(x)-P_1(x) = h_1(x)(x-a). \ $$

Note that this goes to zero a little bit faster than x &minus; a as x tends to a.

If we wanted a better approximation to f, we might instead try a quadratic polynomial instead of a linear function. Instead of just matching one derivative of f at a, we can match two derivatives, thus producing a polynomial that has the same slope and concavity as f at a. The quadratic polynomial in question is


 * $$P_2(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2. \, $$

Taylor's theorem ensures that the quadratic approximation is, in a sufficiently small neighborhood of the point a, a better approximation than the linear approximation. Specifically,


 * $$f(x) = P_2(x) + h_2(x)(x-a)^2, \qquad \lim_{x\to a}h_2(x)=0.$$

Here the error in the approximation is


 * $$R_2(x) = f(x)-P_2(x) = h_2(x)(x-a)^2 \ $$

which, given the limiting behavior of h2, goes to zero faster than (x &minus; a)2 as x tends to a.

Similarly, we 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 a little bit faster than (x &minus; a)k as x tends to a.

This result is of asymptotic nature: it only tells us that the error Rk in an approximation by a k-th order Taylor polynomial Pk tends to zero faster than any nonzero k-th degree polynomial as x &rarr; 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 formulae 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.)

It is also possible that increasing the degree of the approximating polynomial does not increase the quality of approximation at all even if the function f to be approximated is infinitely many times differentiable. An example of this behavior is given below, and it is related to the fact that unlike analytic functions, more general functions are not (locally) determined by the values of their derivatives at a single point.

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 k-th order Taylor polynomial


 * $$P_k(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(k)}(a)}{k!}(x-a)^k $$

of the function f at the point a. The Taylor polynomial is the unique "asymptotic best fit" polynomial in the sense that if there exists a function hk : R &rarr; R and a k-th order polynomial p such that


 * $$ f(x) = p(x) + h_k(x)(x-a)^k, \qquad \lim_{x\to a}h_k(x)=0,$$

then p = Pk. Taylor's theorem describes the asymptotic behavior of the remainder term


 * $$ \ R_k(x) = f(x) - P_k(x),$$

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


 * $$R_k(x) = o(|x-a|^k), \qquad x\to a.$$

Explicit formulae for the remainder
Under stronger regularity assumptions on f there are several precise formulae for the remainder term Rk 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. 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 a and x, then


 * $$ R_k(x) = \frac{f^{(k+1)}(\xi)}{k!}(x-\xi)^k \frac{G(x)-G(a)}{G'(\xi)} $$

for some number &xi; between a and 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 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)-st derivative of f is continuous on the closed interval [a,x].

Due to absolute continuity of f(k) on the closed interval between a and x its derivative f(k+1) exists as an L1-function, and the result can be proven by a formal calculation using 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 a specific form of 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
 * $$q\le f^{(k+1)}(x)\le Q$$

throughout I. Then the remainder term satisfies the inequality


 * $$q\frac{(x-a)^{k+1}}{(k+1)!}\le R_k(x)\le Q\frac{(x-a)^{k+1}}{(k+1)!},$$

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
 * $$|f^{(k+1)}(x)|\leq M$$

on an interval with some r>0, then


 * $$|R_k(x)| \le M\frac{|x-a|^{k+1}}{(k+1)!}\le M\frac{r^{k+1}}{(k+1)!}$$

for all x&isin;(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
Suppose that we wish to approximate the function on the interval [&minus;1,1] while ensuring that the error in the approximation is no more than 10&minus;5. In this example we pretend that we only know the following properties of the exponential function:


 * $$(*) \qquad e^0=1, \qquad \frac{d}{dx} e^x = e^x, \qquad e^x>0, \qquad x\in\mathbb{R}.$$

From these properties it follows that for all k, and in particular,. Hence the k-th order Taylor polynomial of f at 0 and its remainder term in the Lagrange form are given by


 * $$ P_k(x) = 1+x+\frac{x^2}{2!}+\cdots+\frac{x^k}{k!}, \qquad R_k(x)=\frac{e^\xi}{(k+1)!}x^{k+1},$$

where &xi; is some number between 0 and x. Since ex is increasing by (*), we can simply use ex &le; 1 for x &isin; [&minus;1, 0] to estimate the remainder on the subinterval [&minus;1, 0]. To obtain an upper bound for the remainder on [0,1], we use the property e&xi;<ex for 0<&xi;<x to estimate


 * $$ e^x = 1 + x + \frac{e^\xi}{2}x^2 < 1 + x + \frac{e^x}{2}x^2, \qquad 0 < x\leq 1 $$

using the second order Taylor expansion. Then we solve for ex to deduce that


 * $$ e^x \leq \frac{1+x}{1-\frac{x^2}{2}} = 2\frac{1+x}{2-x^2} \leq 4, \qquad 0 \leq x\leq 1 $$

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


 * $$ |R_k(x)| \leq \frac{4|x|^{k+1}}{(k+1)!} \leq \frac{4}{(k+1)!}, \qquad -1\leq x \leq 1, $$

so the required precision is certainly reached, when


 * $$ \frac{4}{(k+1)!} < 10^{-5} \quad \Leftrightarrow \quad 4\cdot 10^5 < (k+1)! \quad \Leftrightarrow \quad k \geq 9. $$

(See factorial or compute by hand the values 9!=362 880 and 10!=3 628 800.) As a conclusion, Taylor's theorem leads to the approximation


 * $$ e^x = 1+x+\frac{x^2}{2!} + \ldots + \frac{x^9}{9!} + R_9(x), \qquad |R_9(x)| < 10^{-5}, \qquad -1\leq x \leq 1. $$

For instance, this approximation provides a decimal expression e≈2.71828, correct up to five decimal places.

Taylor expansions of real analytic functions
Let I⊂R be an open interval. By definition, a function f:I&rarr;R is real analytic if it is locally defined by a convergent power series. This means that for every a &isin; I there exists some r > 0 and a sequence of coefficients ck &isin; R such that (a &minus; r, a + r) ⊂ I and


 * $$ f(x) = \sum_{k=0}^\infty c_k(x-a)^k = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots, \qquad |x-a|<r. $$

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


 * $$ \frac{1}{R} = \limsup_{k\to\infty}|c_k|^\frac{1}{k}. $$

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&isin;R, it must converge uniformly on the closed interval [a &minus; rb, a + rb], where rb = |b &minus; a|. Here only the convergence of the power series is considered, and it might well be that (a &minus; 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


 * $$ P_k(x) = \sum_{j=0}^k c_j(x-a)^j, \qquad c_j = \frac{f^{(j)}(a)}{j!}$$

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


 * $$ R_k(x) = \sum_{j=k+1}^\infty c_j(x-a)^j = (x-a)^k h_k(x), \qquad |x-a|<r. $$

Here the functions


 * $$ h_k:(a-r,a+r)\to \R; \qquad h_k(x) = (x-a)\sum_{j=0}^\infty c_{k+1+j}(x-a)^j $$

are also analytic, since their defining power series have the same radius of convergence as the original series. Assuming that [a &minus; r, a + r] ⊂ I and r < R, all these series converge uniformly on (a &minus; r, a + r). Naturally, in the case of analytic functions one can estimate the remainder term Rk(x) by the tail of the sequence of the derivatives f&prime;(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
There is a source of confusion on the relationship between Taylor polynomials of smooth functions and the Taylor series of analytic functions. One can (rightfully) see the Taylor series


 * $$ f(x) \approx \sum_{k=0}^\infty c_k(x-a)^k = c_0 + c_1(x-a) + c_2(x-a)^2 + \ldots $$

of an infinitely many times differentiable function f:R&rarr;R as its "infinite order Taylor polynomial" at a. Now the estimates for the remainder of a Taylor polynomial implies that for any order k and for any r>0 there exists a constant Mk,r>0 such that


 * $$(*) \quad |R_k(x)|\leq M_{k,r}\frac{|x-a|^{k+1}}{(k+1)!} $$

for every x&isin;(a-r,a+r). Sometimes these constants can be chosen in such way that Mk,r &rarr; 0 when k &rarr; ∞ and r stays fixed. Then the Taylor series of f converges uniformly to some analytic function


 * $$ T_f:(a-r,a+r)\to\mathbb R; \qquad T_f(x) = \sum_{k=0}^\infty \frac{f^{(k)}(a)}{k!}(x-a)^k. $$

Here comes the subtle point. It may well be that an infinitely many times differentiable function f has a Taylor series at a which converges on some open neighborhood of a, but the limit function Tf is different from f. An important example of this phenomenon is provided by


 * $$ f:\mathbb R \to \mathbb R; \qquad f(x) = \begin{cases} e^{-\frac{1}{x^2}} &, x>0, \\ 0 &, x\leq 0.\end{cases} $$

Using the chain rule one can show inductively that for any order k,


 * $$ f^{(k)}(x) = \begin{cases} \frac{p_k(x)}{x^{2k}}e^{-\frac{1}{x^2}} &, x>0 \\ 0 &, x\leq 0\end{cases}$$

for some polynomial pk. The function $$e^{-\frac{1}{x^2}}$$ tends to zero faster than any polynomial as x &rarr; 0, so f is infinitely many times differentiable and for every positive integer k. Now the estimates for the remainder for the Taylor polynomials show that the Taylor series of f converges uniformly to the zero function on the whole real axis. Nothing is wrong in here:


 * The Taylor series of f converges uniformly to the zero function Tf(x)=0.
 * The zero function is analytic and every coefficient in its Taylor series is zero.
 * The function f is infinitely many times differentiable, but not analytic.
 * For any k&isin;N and r>0 there exists Mk,r>0 such that the remainder term for the k-th order Taylor polynomial of f satisfies (*).

Taylor's theorem in complex analysis
Taylor's theorem generalizes to functions $$f:\mathbb C\to\mathbb C$$ which are complex differentiable in an open subset U ⊂ C of the complex plane. However, its usefulness is diminished by other general theorems in complex analysis. Namely, stronger versions of related results can be deduced for complex differentiable functions f : U &rarr; 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 &gamma;(t)=reit of the circle S(z,r) with t &isin; [0,2&pi;] gives


 * $$ f(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(w)}{w-z}dw, \quad f'(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(w)}{(w-z)^2}dw, \quad \ldots, \quad f^{(k)}(z) = \frac{k!}{2\pi i}\int_\gamma \frac{f(w)}{(w-z)^{k+1}}dw. $$

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 the Cauchy's estimates


 * $$ |f^{(k)}(z)| \leq \frac{k!}{2\pi}\int_\gamma \frac{M_r}{|w-z|^{k+1}}dw = \frac{k!M_r}{r^k},

\qquad M_r = \max_{|w-c|=r}|f(w)| $$

for any z &isin; U and r > 0 such that B(z, r) ∪ S(c, r) ⊂ U. These estimates imply that the complex Taylor series


 * $$ f(z) \approx \sum_{k=0}^\infty \frac{f^{(k)}(c)}{k!}(z-c)^k $$

of f converges uniformly on any open disk B(c, r) ⊂ U with S(c, r) ⊂ U into some function Tf. Furthermore, using the contour integral formulae for the derivatives f(k)(c),


 * $$\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  \Big(\frac{z-c}{w-c}\Big)^k dw \\ = \ & \frac{1}{2\pi i} \int_\gamma \frac{f(w)}{w-c}\Big( \frac{1}{1-\frac{z-c}{w-c}} \Big) 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 &isin; C and a-centered intervals (a &minus; r, a + r) replaced by c-centered disks B(c, r). In particular, the Taylor expansion holds in the form


 * $$ f(z) = P_k(z) + R_k(z), \qquad P_k(z) = \sum_{j=0}^k \frac{f^{(k)}(c)}{k!}(z-c)^k, $$

where the remainder term Rk 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 &gamma; which parametrizes the boundary ∂W ⊂ U of a region W ⊂ U, one obtains expressions for the derivatives f(j)(c) as above, and modifying slightly the computation for, one arrives at the exact formula


 * $$ 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 W ⊂ U is dominated by the values of the function f itself on the boundary ∂W ⊂ U. Similarly, applying Cauchy's estimates to the series expression for the remainder, one obtains the uniform estimates


 * $$ |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
The function f:R&rarr;R defined by


 * $$ f(x) = \frac{1}{1+x^2} $$

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


 * $$ f:\mathbb C\cup\{\infty\} \to \mathbb C\cup\{\infty\}; \quad f(z) = \frac{1}{1+z^2}$$

on the compactified complex plane. It has simple poles at z=i and z=&minus;i, and it is analytic elsewhere. Now its Taylor series centered at z0 coverges on any disc B(z0,r) with r<|z-z0|, where the same Taylor series converges at z&isin;C. Therefore Taylor series of f centered at 0 converges on B(0,1) and it does not converge for any z&isin;C with |z|>1 due to the poles at i and &minus;i. For the same reason the Taylor series of f centered at 1 converges on B(1,√2) and does not converge for any z&isin;C with |z-1|>√2.

Higher order differentiability
A function f:Rn &rarr; R is differentiable at a &isin; Rn if and only if there exists a linear functional L : Rn &rarr; R and a function h : Rn &rarr; R such that


 * $$f(\boldsymbol{x}) = f(\boldsymbol{a}) + L(\boldsymbol{x}-\boldsymbol{a}) + h(\boldsymbol{x})|\mathbf{x}-\mathbf{a}|,

\qquad \lim_{\boldsymbol{x}\to\boldsymbol{a}}h(\boldsymbol{x})=0. $$

If this is the case, then L = df(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


 * $$ df( \boldsymbol{a} )( \boldsymbol{v} ) = \frac{\partial f}{\partial x_1}(\boldsymbol{a})v_1 + \cdots + \frac{\partial f}{\partial x_n}(\boldsymbol{a})v_n. $$

Introduce the multi-index notation


 * $$ |\alpha| = \alpha_1+\cdots+\alpha_n, \quad \alpha!=\alpha_1!\cdots\alpha_n!, \quad \boldsymbol{x}^\alpha=x_1^{\alpha_1}\cdots x_n^{\alpha_n} $$

for &alpha; &isin; Nn and x &isin; Rn. If all the k-th order partial derivatives of f : Rn &rarr; R are continuous at a &isin; Rn, then by Clairaut's theorem, one can change the order of mixed derivatives at a, so the notation


 * $$ D^\alpha f = \frac{\partial^{|\alpha|}f}{\partial x_1^{\alpha_1}\cdots \partial x_n^{\alpha_n}}, \qquad |\alpha|\leq k $$

for the higher order partial derivatives is justified in this situation. The same is true if all the (k &minus; 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
If the function f : Rn &rarr; R is k+1 times continuously differentiable in the closed ball B, 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,


 * $$ f( \boldsymbol{x} ) = \sum_{|\alpha|=0}^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, \qquad

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. $$

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


 * $$\big|R_\beta(\boldsymbol{x})| \leq \frac{|\beta|}{\beta!} \max_{|\alpha|=|\beta|} \max_{\boldsymbol{y}\in B} |D^\alpha f(\boldsymbol{y})|, \qquad \boldsymbol{x}\in B. $$

Proof for Taylor's theorem in one real variable
Let


 * $$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,


 * $$P(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(k)}(a)}{k!}(x-a)^k.$$

It is sufficient to show that


 * $$\lim_{x\to a} h_k(x) =0. \, $$

The proof here is based on repeated application of L'Hôpital's rule. Note that, for each, $$f^{(j)}(a)=P^{(j)}(a)$$. Hence each of the first k&minus;1 derivatives of the numerator in $$h_k(x)$$ vanishes at $$x=a$$, and the same is true of the denominator. So


 * $$\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} = \cdots = \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}\\ &=\frac{1}{k!}\lim_{x\to a} \frac{f^{(k-1)}(x) - P^{(k-1)}(x)}{x-a}\\ &=\frac{1}{k!}(f^{(k)}(a) - f^{(k)}(a)) = 0 \end{align}$$

where the second to last equality follows by the definition of the derivative at x = a.

Derivation for the mean value forms of the remainder
Let G be any real-valued function, continuous on the closed interval between a and x and differentiable with a non-vanishing derivative on the open interval between a and x, and define



F(t) = f(t) + f'(t)(x-t) + \frac{f''(t)}{2!}(x-t)^2 + \cdots + \frac{f^{(k)}(t)}{k!}(x-t)^k. $$

Then, by Cauchy's mean value theorem,



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

for some &xi; on the open interval between a and x. Note that here the numerator F(x) − F(a) = Rk(x) is exactly the remainder of the Taylor polynomial for f(x). Compute


 * $$\begin{align}

F'(t) = & f'(t) + \big(f''(t)(x-t) - f'(t)\big) + \Big(\frac{f^{(3)}(t)}{2!}(x-t)^2 - \frac{f^{(2)}(t)}{1!}(x-t)\Big) +  \cdots \\ & \cdots + \Big( \frac{f^{(k+1)}(t)}{k!}(x-t)^k - \frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1}\Big) = \frac{f^{(k+1)}(t)}{k!}(x-t)^k, \end{align}$$

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



R_k = \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 $$ \ G(t)=(x-t)^{k+1} \ $$ and the Cauchy form by choosing $$ \ G(t) = t-a$$.

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



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 absolute continuity of f(k) on the closed interval between a and x its derivative f(k+1) exists as an L1-function, and we can use fundamental theorem of calculus and integration by parts. This same proof applies for the Riemann integral assuming that f(k) is continuous on the closed interval and differentiable on the open interval between a and x, and this leads to the same result than using the mean value theorem.

The fundamental theorem of calculus states that


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

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


 * $$ \begin{align}

f(x) &= f(a)+\Big(xf'(x)-af'(a)\Big)-\int_a^x tf''(t) \, dt \\ &= f(a) + x\Big(f'(a) + \int_a^x f(t) \,dt \Big) -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 k=1. The general statement is proved using induction. Suppose that
 * $$ (*) \quad f(x) = f(a) + \frac{f'(a)}{1!}(x - a) + \cdots + \frac{f^{(k)}(a)}{k!}(x - a)^k + \int_a^x \frac{f^{(k+1)} (t)}{k!} (x - t)^k \, dt. $$

Integrating the remainder term by parts we arrive at


 * $$\begin{align}

\int_a^x \frac{f^{(k+1)} (t)}{k!} (x - t)^k \, dt = & - \Big[ \frac{f^{(k+1)} (t)}{(k+1)k!} (x - t)^{k+1} \Big]_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 (*) shows that if it holds for the value k, it must also hold for the value k + 1. Therefore, since it holds for k = 1, it must hold for every positive integer k.

Derivation for the remainder of multivariate Taylor polynomials
We prove the special case, where f : Rn &rarr; R has continuous partial derivatives up to the order k+1 in some closed ball B with center a. The strategy of the proof is to apply the one-variable case of Taylor's theorem to the restriction of f to the line segment adjoining x and 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 :


 * $$f(x)=g(1)=g(0)+\sum_{j=1}^k\frac{1}{j!}g^{(j)}(0)\ +\ \int_0^1 \frac{(1-t)^k }{k!} g^{(k+1)}(t)\, dt.$$

Applying the chain rule for several variables gives


 * $$\begin{align}

g^{(j)}(t)&=\frac{d^j}{dt^j}f(u(t)) = \frac{d^j}{dt^j} f(\mathbf{a}+t(\mathbf{x}-\mathbf{a})) \\ &= \sum_{|\alpha|=j} \left(\begin{matrix} j \\ \alpha\end{matrix} \right) (D^\alpha f) (\mathbf{a}+t(\mathbf{x}-\mathbf{a})) (\mathbf{x}-\mathbf{a})^\alpha \end{align}$$

where $$\left(\begin{matrix}j \\ \alpha\end{matrix}\right)$$ is the multinomial coefficient. Since $$\frac{1}{j!}\left(\begin{matrix}j\\ \alpha\end{matrix}\right)=\frac{1}{\alpha!}$$, we get


 * $$f(\mathbf x)= f(\mathbf a)+\sum_{|\alpha|=1}^k\frac{1}{\alpha!} (D^\alpha f) (\mathbf a)(\mathbf x-\mathbf a)^\alpha+\sum_{|\alpha|=k+1}\frac{k+1}{\alpha!}

(\mathbf x-\mathbf a)^\alpha \int_0^1 (1-t)^k (D^\alpha f)(\mathbf a+t(\mathbf x-\mathbf a))\,dt.$$