Taylor's theorem#Explicit formulas for the remainder

File:E^x with linear approximation.png $P\_1(x)=1+x$ (red) at $a=0$.]]

If a real-valued function $f(x)$ is differentiable at the point $x=a$, then it has a linear approximation near this point. This means that there exists a function h₁(x) such that

$$f(x)\; =\; f(a)\; +\; f\text{'}(a)(x\; -\; a)\; +\; h\_1(x)(x\; -\; a),\; \backslash quad\; \backslash lim\_\{x\; \backslash to\; a\}\; h\_1(x)\; =\; 0.$$

Here

$$P\_1(x)\; =\; f(a)\; +\; f\text{'}(a)(x\; -\; a)$$

is the linear approximation of $f(x)$ for x near the point a, whose graph $y=P\_1(x)$ is the tangent line to the graph $y=f(x)$ at {{nowrap|1=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 $(x-a)$, making $f(x)\backslash approx\; P\_1(x)$ a useful approximation.

File:E^x with quadratic approximation corrected.png

For a better approximation to $f(x)$, we can fit a quadratic polynomial instead of a linear function:

$$P\_2(x)\; =\; f(a)\; +\; f\text{'}(a)(x\; -\; a)\; +\; \backslash frac\{f\text{'}\text{'}(a)\}\{2\}(x\; -\; a)^2.$$

Instead of just matching one derivative of $f(x)$ at $x=a$, 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 $x=a$, more accurate than the linear approximation. Specifically,

$$f(x)\; =\; P\_2(x)\; +\; h\_2(x)(x\; -\; a)^2,\; \backslash quad\; \backslash lim\_\{x\; \backslash to\; a\}\; h\_2(x)\; =\; 0.$$

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 $h\_2$, goes to zero faster than $(x\; -\; a)^2$ as x tends to a.

File:Tayloranimation.gif 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 $(x-a)^k$ 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 $R\_k$ in an approximation by a $k$-th order Taylor polynomial P_k tends to zero faster than any nonzero $k$-th degree polynomial as $x\; \backslash to\; a$. 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 $f(x)$ 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 $f(x)$ 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 $f(x)$ to within a given error tolerance. (Given the degree and error tolerance, we find the interval.)

{{clear}}

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

{{math theorem

| name = Taylor's theorem{{ citation|first1=Angelo|last1=Genocchi|first2= Giuseppe|last2=Peano|title=Calcolo differenziale e principii di calcolo integrale|location=(N. 67, pp. XVII–XIX)|publisher=Fratelli Bocca ed.|year=1884}}{{Citation | last1=Spivak | first1=Michael | author1-link=Michael Spivak | title=Calculus | publisher=Publish or Perish | location=Houston, TX | edition=3rd | isbn=978-0-914098-89-8 | year=1994| page=383}}{{springer|title=Taylor formula|id=p/t092300}}

| math_statement = Let k ≥ 1 be an integer and let the function {{nowrap|f : R → R}} be k times differentiable at the point {{nowrap|a ∈ R}}. Then there exists a function {{nowrap|h_k : R → R}} such that

$$f(x)\; =\; \backslash sum\_\{i=0\}^k\; \backslash frac\{f^\{(i)\}(a)\}\{i!\}(x-a)^i\; +\; h\_k(x)(x-a)^k,$$

and

$$\backslash lim\_\{x\backslash to\; a\}\; h\_k(x)\; =\; 0.$$

This is called the Peano form of the remainder.

}}

The polynomial appearing in Taylor's theorem is the $\backslash boldsymbol\{k\}$-th order Taylor polynomial

$$P\_k(x)\; =\; f(a)\; +\; f\text{'}(a)(x-a)\; +\; \backslash frac\{f\text{'}\text{'}(a)\}\{2!\}(x-a)^2\; +\; \backslash cdots\; +\; \backslash frac\{f^\{(k)\}(a)\}\{k!\}(x-a)^k$$

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 {{nowrap|h_k : R → R}} and a $k$-th order polynomial p such that

$$f(x)\; =\; p(x)\; +\; h\_k(x)(x-a)^k,\; \backslash quad\; \backslash lim\_\{x\backslash to\; a\}\; h\_k(x)\; =\; 0\; ,$$

then p = P_k. 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

$$R\_k(x)\; =\; o(|x-a|^\{k\}),\; \backslash quad\; x\backslash to\; a.$$

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.

{{math theorem

| name = Mean-value forms of the remainder

| math_statement = Let {{nowrap|f : R → R}} be k + 1 times differentiable on the open interval between $a$ and $x$ with f{{i sup|(k)}} continuous on the closed interval between $a$ and $x$.The hypothesis of f{{i sup|(k)}} being continuous on the closed interval between $a$ and $x$ is not redundant. Although f being k + 1 times differentiable on the open interval between $a$ and $x$ does imply that f{{i sup|(k)}} is continuous on the open interval between $a$ and $x$, it does not imply that f{{i sup|(k)}} is continuous on the closed interval between $a$ and $x$, i.e. it does not imply that f{{i sup|(k)}} is continuous at the endpoints of that interval. Consider, for example, the function {{nowrap|f : [0,1] → R}} defined to equal $\backslash sin(1/x)$ on $(0,1]$ and with $f(0)=0$. This is not continuous at 0, but is continuous on $(0,1)$. Moreover, one can show that this function has an antiderivative. Therefore that antiderivative is differentiable on $(0,1)$, its derivative (the function f) is continuous on the open interval $(0,1)$, but its derivative f is not continuous on the closed interval $[0,1]$. So the theorem would not apply in this case. Then

$$R\_k(x)\; =\; \backslash frac\{f^\{(k+1)\}(\backslash xi\_L)\}\{(k+1)!\}\; (x-a)^\{k+1\}$$

for some real number $\backslash xi\_L$ between $a$ and $x$. This is the Lagrange form{{harvnb|Kline|1998|loc=§20.3}}; {{harvnb|Apostol|1967|loc=§7.7}}. of the remainder.

Similarly,

$$R\_k(x)\; =\; \backslash frac\{f^\{(k+1)\}(\backslash xi\_C)\}\{k!\}(x-\backslash xi\_C)^k(x-a)$$

for some real number $\backslash xi\_C$ between $a$ and $x$. This is the Cauchy form{{harvnb|Apostol|1967|loc=§7.7}}. of the remainder.

Both can be thought of as specific cases of the following result: Consider $p>0$

$$R\_k(x)\; =\; \backslash frac\{f^\{(k+1)\}(\backslash xi\_S)\}\{k!\}(x-\backslash xi\_S)^\{k+1-p\}\backslash frac\{(x-a)^p\}\{p\}$$

for some real number $\backslash xi\_S$ between $a$ and $x$. This is the Schlömilch form of the remainder (sometimes called the Schlömilch-Roche). The choice $p=k+1$ is the Lagrange form, whilst the choice $p=1$ is the Cauchy form.

}}

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 $k=0$. 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 $a$ and $x$, then

$$R\_k(x)\; =\; \backslash frac\{f^\{(k+1)\}(\backslash xi)\}\{k!\}(x-\backslash xi)^k\; \backslash frac\{G(x)-G(a)\}\{G\text{'}(\backslash xi)\}$$

for some number $\backslash xi$ 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 Lagrange form is obtained by taking $G(t)=(x-t)^\{k+1\}$ and the Cauchy form is obtained by taking $G(t)=t-a$.

{{anchor|Integral form of the remainder}}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].

{{math theorem|name=Integral form of the remainder{{harvnb|Apostol|1967|loc=§7.5}}. |math_statement=Let $f^\{(k)\}$ be absolutely continuous on the closed interval between $a$ and $x$. Then

$$R\_k(x)\; =\; \backslash int\_a^x\; \backslash frac\{f^\{(k+1)\}\; (t)\}\{k!\}\; (x\; -\; t)^k\; \backslash ,\; dt.$$

}}

Due to the absolute continuity of f{{i sup|(k)}} on the closed interval between $a$ and $x$, its derivative f{{i sup|(k+1)}} exists as an L{{i sup|1}}-function, and the result can be proven by a formal calculation using the fundamental theorem of calculus and integration by parts.

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 {{nowrap|(k + 1)}}-times continuously differentiable in an interval I containing a. Suppose that there are real constants q and Q such that

$$q\backslash le\; f^\{(k+1)\}(x)\backslash le\; Q$$

throughout I. Then the remainder term satisfies the inequality{{harvnb|Apostol|1967|loc=§7.6}}

$$q\backslash frac\{(x-a)^\{k+1\}\}\{(k+1)!\}\backslash le\; R\_k(x)\backslash le\; Q\backslash frac\{(x-a)^\{k+1\}\}\{(k+1)!\},$$

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

$$|f^\{(k+1)\}(x)|\backslash le\; M$$

on an interval {{nowrap|1=I = (a − r,a + r)}} with some $r\; >\; 0$ , then

$$|R\_k(x)|\backslash le\; M\backslash frac\{|x-a|^\{k+1\}\}\{(k+1)!\}\backslash le\; M\backslash frac\{r^\{k+1\}\}\{(k+1)!\}$$

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

File:Expanimation.gif Suppose that we wish to find the approximate value of the function $f(x)=e^x$ on the interval $[-1,1]$ 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:

{{NumBlk|:|$e^0=1,\; \backslash qquad\; \backslash frac\{d\}\{dx\}\; e^x\; =\; e^x,\; \backslash qquad\; e^x>0,\; \backslash qquad\; x\backslash in\backslash R.$|{{EquationRef|★}}}}

From these properties it follows that $f^\{(k)\}(x)=e^x$ for all $k$, and in particular, $f^\{(k)\}(0)=1$. Hence the $k$-th order Taylor polynomial of $f$ at $0$ and its remainder term in the Lagrange form are given by

$$P\_k(x)\; =\; 1+x+\backslash frac\{x^2\}\{2!\}+\backslash cdots+\backslash frac\{x^k\}\{k!\},\; \backslash qquad\; R\_k(x)=\backslash frac\{e^\backslash xi\}\{(k+1)!\}x^\{k+1\},$$

where $\backslash xi$ is some number between 0 and x. Since e^x is increasing by ({{EquationNote|★}}), we can simply use $e^x\; \backslash leq\; 1$ for $x\; \backslash in\; [-1,0]$ to estimate the remainder on the subinterval $[-1,0]$. To obtain an upper bound for the remainder on $[0,1]$, we use the property

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

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

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

$$|R\_k(x)|\; \backslash leq\; \backslash frac\{4|x|^\{k+1\}\}\{(k+1)!\}\; \backslash leq\; \backslash frac\{4\}\{(k+1)!\},\; \backslash qquad\; -1\backslash leq\; x\; \backslash leq\; 1,$$

so the required precision is certainly reached, when

(See factorial or compute by hand the values $9!\; =362880$ and $10!\; =3628800$.) As a conclusion, Taylor's theorem leads to the approximation

For instance, this approximation provides a decimal expression $e\; \backslash approx\; 2.71828$, correct up to five decimal places.

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 {{nowrap|(a − r, a + r) ⊂ I}} and

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

$$\backslash frac\{1\}\{R\}\; =\; \backslash limsup\_\{k\backslash to\backslash infty\}|c\_k|^\backslash frac\{1\}\{k\}.$$

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 $[a-r\_b,a+r\_b]$, where $r\_b=\backslash left\backslash vert\; b-a\; \backslash right\backslash vert$. Here only the convergence of the power series is considered, and it might well be that {{nowrap|(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

$$P\_k(x)\; =\; \backslash sum\_\{j=0\}^k\; c\_j(x-a)^j,\; \backslash qquad\; c\_j\; =\; \backslash frac\{f^\{(j)\}(a)\}\{j!\}$$

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

Here the functions

$$\backslash begin\{align\}$$

& 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 {{nowrap|[a − r, a + r]}} ⊂ I and r < R, all these series converge uniformly on {{nowrap|(a − r, a + r)}}. Naturally, in the case of analytic functions one can estimate the remainder term $R\_k(x)$ 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.

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

$$f(x)\; \backslash approx\; \backslash sum\_\{k=0\}^\backslash infty\; c\_k(x-a)^k\; =\; c\_0\; +\; c\_1(x-a)\; +\; c\_2(x-a)^2\; +\; \backslash cdots$$

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 known to be bounded over (a − r, a + r), then for any order k and for any r > 0 there exists a constant {{nowrap|M_k,r > 0}} such that

{{NumBlk|:|$|R\_k(x)|\; \backslash leq\; M\_\{k,r\}\; \backslash frac\{|x-a|^\{k+1\}\}\{(k+1)!\}$|{{EquationRef|★★}}}}

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

$$\backslash begin\{align\}$$

& 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 {{nowrap|M_k,r}} is not bounded above as long as it grows slowly enough.)

The limit function {{nowrap|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:

$$\backslash begin\{align\}$$

& 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,

$$f^\{(k)\}(x)\; =\; \backslash begin\{cases\}$$

\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 $e^\{-\backslash frac\{1\}\{x^2\}\}$ tends to zero faster than any polynomial as $x\; \backslash to\; 0$, so f is infinitely many times differentiable and {{nowrap|1=f{{i sup|(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 ({{EquationNote|★★}}) 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 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 {{nowrap|1=γ(t) = z + re^it}} of the circle S(z, r) with $t\; \backslash in\; [0,2\; \backslash pi]$ gives

$$f(z)\; =\; \backslash frac\{1\}\{2\backslash pi\; i\}\backslash int\_\backslash gamma\; \backslash frac\{f(w)\}\{w-z\}\backslash ,dw,\; \backslash quad\; f\text{'}(z)\; =\; \backslash frac\{1\}\{2\backslash pi\; i\}\backslash int\_\backslash gamma\; \backslash frac\{f(w)\}\{(w-z)^2\}\; \backslash ,\; dw,\; \backslash quad\; \backslash ldots,\; \backslash quad\; f^\{(k)\}(z)\; =\; \backslash frac\{k!\}\{2\backslash pi\; i\}\backslash int\_\backslash gamma\; \backslash frac\{f(w)\}\{(w-z)^\{k+1\}\}\; \backslash ,\; dw.$$

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{{harvnb|Rudin|1987|loc=§10.26}}

$$|f^\{(k)\}(z)|\; \backslash leq\; \backslash frac\{k!\}\{2\backslash pi\}\backslash int\_\backslash gamma\; \backslash frac\{M\_r\}$$

\leq \frac{M_r \beta^{k+1}}{1-\beta}, \qquad \frac

File:Function with two poles.png

The function

$$\backslash begin\{align\}$$

& 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

$$\backslash begin\{align\}$$

& 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 $z=i$ and $z=-i$, 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 $B(1,\; \backslash sqrt\{2\})$ and does not converge for any z ∈ C with $\backslash left\backslash vert\; z-1\; \backslash right\backslash vert>\backslash sqrt\{2\}$.

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

$$f(\backslash boldsymbol\{x\})\; =\; f(\backslash boldsymbol\{a\})\; +\; L(\backslash boldsymbol\{x\}-\backslash boldsymbol\{a\})\; +\; h(\backslash boldsymbol\{x\})\backslash lVert\backslash boldsymbol\{x\}-\backslash boldsymbol\{a\}\backslash rVert,$$

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

If this is the case, then $L\; =\; df(\backslash boldsymbol\{a\})$ is the (uniquely defined) differential of {{math|f}} at the point {{math|a}}. Furthermore, then the partial derivatives of {{math|f}} exist at {{math|a}} and the differential of {{math|f}} at {{math|a}} is given by

$$df(\; \backslash boldsymbol\{a\}\; )(\; \backslash boldsymbol\{v\}\; )\; =\; \backslash frac\{\backslash partial\; f\}\{\backslash partial\; x\_1\}(\backslash boldsymbol\{a\})\; v\_1\; +\; \backslash cdots\; +\; \backslash frac\{\backslash partial\; f\}\{\backslash partial\; x\_n\}(\backslash boldsymbol\{a\})\; v\_n.$$

Introduce the multi-index notation

$$|\backslash alpha|\; =\; \backslash alpha\_1+\backslash cdots+\backslash alpha\_n,\; \backslash quad\; \backslash alpha!=\backslash alpha\_1!\backslash cdots\backslash alpha\_n!,\; \backslash quad\; \backslash boldsymbol\{x\}^\backslash alpha=x\_1^\{\backslash alpha\_1\}\backslash cdots\; x\_n^\{\backslash alpha\_n\}$$

for {{math|α ∈ Nⁿ}} and {{math|x ∈ Rⁿ}}. If all the $k$-th order partial derivatives of {{math|f : Rⁿ → R}} are continuous at {{math|a ∈ Rⁿ}}, then by Clairaut's theorem, one can change the order of mixed derivatives at {{math|a}}, so the short-hand notation

$$D^\backslash alpha\; f\; =\; \backslash frac\{\backslash partial^$$

for the higher order partial derivatives is justified in this situation. The same is true if all the ({{math|k − 1}})-th order partial derivatives of {{math|f}} exist in some neighborhood of {{math|a}} and are differentiable at {{math|a}}.This follows from iterated application of the theorem that if the partial derivatives of a function {{math|f}} exist in a neighborhood of {{math|a}} and are continuous at {{math|a}}, then the function is differentiable at {{math|a}}. See, for instance, {{harvnb|Apostol|1974|loc=Theorem 12.11}}. Then we say that {{math|f}} is {{math|k}} times differentiable at the point {{math|a}}.

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

{{math theorem|name=Multivariate version of Taylor's theoremKönigsberger Analysis 2, p. 64 ff.|math_statement= Let {{math|f : Rⁿ → R}} be a {{math|k}}-times continuously differentiable function at the point {{math|a ∈ Rⁿ}}. Then there exist functions {{math|h_α : Rⁿ → R}}, where $|\backslash alpha|=k,$ such that

$$\backslash begin\{align\}$$

& f(\boldsymbol{x}) = \sum_

For example, the third-order Taylor polynomial of a smooth function $f:\backslash mathbb\; R^2\backslash to\backslash mathbb\; R$ is, denoting $\backslash boldsymbol\{x\}-\backslash boldsymbol\{a\}=\backslash boldsymbol\{v\}$,

$$\backslash begin\{align\}$$

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}

Let{{harvnb|Stromberg|1981}}

$$h\_k(x)\; =\; \backslash begin\{cases\}$$

\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,

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

It is sufficient to show that

$$\backslash lim\_\{x\backslash to\; a\}\; h\_k(x)\; =0.$$

The proof here is based on repeated application of L'Hôpital's rule. Note that, for each $j=0,1,...,k-1$, $f^\{(j)\}(a)=P^\{(j)\}(a)$. Hence each of the first $k-1$ derivatives of the numerator in $h\_k(x)$ vanishes at $x=a$, and the same is true of the denominator. Also, since the condition that the function $f$ be $k$ times differentiable at a point requires differentiability up to order $k-1$ 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 $k-2$ derivatives are differentiable in a neighborhood of $a$. Clearly, the denominator also satisfies said condition, and additionally, doesn't vanish unless $x=a$, therefore all conditions necessary for L'Hôpital's rule are fulfilled, and its use is justified. So

$$\backslash begin\{align\}$$

\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 $x=a$.

Let $f(x)$ be any real-valued continuous function to be approximated by the Taylor polynomial.

Step 1: Let $F$ and $G$ be functions. Set $F$ and $G$ to be

$$\backslash begin\{align\}$$

F(x) = f(x) - \sum^{n-1}_{k=0} \frac{f^{(k)}(a)}{k!}(x-a)^{k}

\end{align}

$$\backslash begin\{align\}$$

G(x) = (x-a)^{n}

\end{align}

Step 2: Properties of $F$ and $G$:

$$\backslash begin\{align\}$$

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,

$$\backslash begin\{align\}$$

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}

$$\backslash begin\{align\}$$

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 $f\_\{1\}$ and $g\_\{1\}$ be continuous functions on $[a,\; b]$. Since so we can work with the interval $[a,\; x]$. Let $f\_\{1\}$ and $g\_\{1\}$ be differentiable on $(a,\; x)$. Assume $g\_\{1\}\text{'}(x)\; \backslash neq\; 0$ for all $x\; \backslash in\; (a,\; b)$.

Then there exists $c\_\{1\}\; \backslash in\; (a,\; x)$ such that

$$\backslash begin\{align\}$$

\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: $G\text{'}(x)\; \backslash neq\; 0$ in $(a,\; b)$ and $F(a),\; G(a)\; =\; 0$ so

$$\backslash begin\{align\}$$

\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 $c\_\{1\}\; \backslash in\; (a,\; x)$.

This can also be performed for $(a,\; c\_\{1\})$:

$$\backslash begin\{align\}$$

\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 $c\_\{2\}\; \backslash in\; (a,\; c\_\{1\})$.

This can be continued to $c\_\{n\}$.

This gives a partition in $(a,\; b)$:

with

$$\backslash frac\{F(x)\}\{G(x)\}\; =\; \backslash frac\{F\text{'}(c\_\{1\})\}\{G\text{'}(c\_\{1\})\}\; =\; \backslash dots\; =\; \backslash frac\{F^\{(n)\}(c\_\{n\})\}\{G^\{(n)\}(c\_\{n\})\}\; .$$

Set $c\; =\; c\_\{n\}$:

$$\backslash frac\{F(x)\}\{G(x)\}\; =\; \backslash frac\{F^\{(n)\}(c)\}\{G^\{(n)\}(c)\}$$

Step 4: Substitute back

$$\backslash begin\{align\}$$

\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 $(x\; -\; a)^\{n\}$, $G^\{(n)\}(c)\; =\; n(n-1)...1$, so:

$$\backslash frac\{F^\{(n)\}(c)\}\{G^\{(n)\}(c)\}\; =\; \backslash frac\{f^\{(n)\}(c)\}\{n(n-1)\backslash cdots1\}\; =\; \backslash frac\{f^\{(n)\}(c)\}\{n!\}.$$

This leads to:

$$\backslash begin\{align\}$$

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:

$$\backslash begin\{align\}$$

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 $c\_\{n\}\; =\; a$ eventually:

$$f(x)\; =\; \backslash sum^\{n\}\_\{k=0\}\; \backslash frac\{f^\{(k)\}(a)\}\{k!\}(x-a)^\{k\}.$$

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

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

For $t\; \backslash in\; [a,x]$. Then, by Cauchy's mean value theorem,

{{NumBlk|:|$\backslash frac\{F\text{'}(\backslash xi)\}\{G\text{'}(\backslash xi)\}\; =\; \backslash frac\{F(x)\; -\; F(a)\}\{G(x)\; -\; G(a)\}$|{{EquationRef|★★★}}}}

for some $\backslash xi$ on the open interval between $a$ and $x$. Note that here the numerator $F(x)-F(a)=R\_k(x)$ is exactly the remainder of the Taylor polynomial for $y=f(x)$. Compute

$$\backslash begin\{align\}$$

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 ({{EquationNote|★★★}}) and rearrange terms to find that

$$R\_k(x)\; =\; \backslash frac\{f^\{(k+1)\}(\backslash xi)\}\{k!\}(x-\backslash xi)^k\; \backslash frac\{G(x)-G(a)\}\{G\text{'}(\backslash xi)\}.$$

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

$$G(t)\; =\; \backslash int\_a^t\; \backslash frac\{f^\{(k+1)\}(s)\}\{k!\}\; (x-s)^k\; \backslash ,\; ds,$$

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{{i sup|(k)}} is only absolutely continuous. However, if one uses Riemann integral instead of Lebesgue integral, the assumptions cannot be weakened.

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

The fundamental theorem of calculus states that

$$f(x)=f(a)+\; \backslash int\_a^x\; \backslash ,\; f\text{'}(t)\; \backslash ,\; dt.$$

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

$$\backslash begin\{align\}$$

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 $k=1$. The general statement is proved using induction. Suppose that

{{NumBlk|:|$f(x)\; =\; f(a)\; +\; \backslash frac\{f\text{'}(a)\}\{1!\}(x\; -\; a)\; +\; \backslash cdots\; +\; \backslash frac\{f^\{(k)\}(a)\}\{k!\}(x\; -\; a)^k\; +\; \backslash int\_a^x\; \backslash frac\{f^\{(k+1)\}\; (t)\}\{k!\}\; (x\; -\; t)^k\; \backslash ,\; dt.$|{{EquationRef|eq1}}}}

Integrating the remainder term by parts we arrive at

$$\backslash begin\{align\}$$

\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 {{nowrap|in ({{EquationNote|eq1}})}} 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 $k$.

We prove the special case, where $f:\backslash mathbb\; R^n\backslash to\backslash mathbb\; R$ has continuous partial derivatives up to the order $k+1$ in some closed ball $B$ with center $\backslash boldsymbol\{a\}$. 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 $\backslash boldsymbol\{x\}$ and $\backslash boldsymbol\{a\}$.{{harvnb|Hörmander|1976|pp=12–13}} Parametrize the line segment between $\backslash boldsymbol\{a\}$ and $\backslash boldsymbol\{x\}$ by $\backslash boldsymbol\{u\}(t)=\backslash boldsymbol\{a\}+t(\backslash boldsymbol\{x\}-\backslash boldsymbol\{a\})$ We apply the one-variable version of Taylor's theorem to the function $g(t)\; =\; f(\backslash boldsymbol\{u\}(t))$:

$$f(\backslash boldsymbol\{x\})=g(1)=g(0)+\backslash sum\_\{j=1\}^k\backslash frac\{1\}\{j!\}g^\{(j)\}(0)\backslash \; +\backslash \; \backslash int\_0^1\; \backslash frac\{(1-t)^k\; \}\{k!\}\; g^\{(k+1)\}(t)\backslash ,\; dt.$$

Applying the chain rule for several variables gives

$$\backslash begin\{align\}$$

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 $\backslash tbinom\; j\; \backslash alpha$ is the multinomial coefficient. Since $\backslash tfrac\{1\}\{j!\}\backslash tbinom\; j\; \backslash alpha=\backslash tfrac\{1\}\{\backslash alpha!\}$, we get:

$$f(\backslash boldsymbol\{x\})=\; f(\backslash boldsymbol\{a\})\; +\; \backslash sum\_\{1\; \backslash leq\; |\backslash alpha|\; \backslash leq\; k\}\backslash frac\{1\}\{\backslash alpha!\}\; (D^\backslash alpha\; f)\; (\backslash boldsymbol\{a\})(\backslash boldsymbol\{x\}-\backslash boldsymbol\{a\})^\backslash alpha+\backslash sum\_\{|\backslash alpha|=k+1\}\backslash frac\{k+1\}\{\backslash alpha!\}$$

(\boldsymbol{x}-\boldsymbol{a})^\alpha \int_0^1 (1-t)^k (D^\alpha f)(\boldsymbol{a}+t(\boldsymbol{x}-\boldsymbol{a}))\,dt.

{{Portal|Mathematics}}

• {{annotated link|Hadamard's lemma}}
• {{annotated link|Laurent series}}
• {{annotated link|Padé approximant}}
• {{annotated link|Newton series}}
• {{annotated link|Approximation theory}}
• {{annotated link|Function approximation}}

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• [http://www.cut-the-knot.org/Curriculum/Calculus/TaylorSeries.shtml Taylor Series Approximation to Cosine] at cut-the-knot
• [http://cinderella.de/files/HTMLDemos/2C02_Taylor.html Trigonometric Taylor Expansion] interactive demonstrative applet
• [http://numericalmethods.eng.usf.edu/topics/taylor_series.html Taylor Series Revisited] at [http://numericalmethods.eng.usf.edu Holistic Numerical Methods Institute]

{{Calculus topics}}

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Category:Articles containing proofs

Category:Theorems in calculus

Category:Theorems in real analysis

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