Taylor's theorem

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:

Taylor's theorem—Let ${\displaystyle k\ge 1}$ be an integer and let the function ${\displaystyle f:\mathbb{R}\to \mathbb{R}}$ be ${\displaystyle k}$ times differentiable at the point ${\displaystyle a\in \mathbb{R}}$. Then there exists a function ${\displaystyle h_k:\mathbb{R}\to \mathbb{R}}$ such that

$${\displaystyle f(x)=\sum _{i=0}^k\frac{f^{(i)}(a)}{i!}(x-a{)}^i+h_k(x)(x-a{)}^k,}$$

and

$${\displaystyle \underset{x\to a}{lim}{h}_k(x)=0.}$$

This is called the Peano form of the remainder.

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

$${\displaystyle P_k(x)=f(a)+f^{\prime }(a)(x-a)+\frac{f^{″}(a)}{2!}(x-a{)}^2+\cdots +\frac{f^{(k)}(a)}{k!}(x-a{)}^k}$$

of the function ${\displaystyle f}$ at the point ${\displaystyle a}$. The Taylor polynomial is the unique "asymptotic best fit" polynomial in the sense that if there exists a function ${\displaystyle h_k:\mathbb{R}\to \mathbb{R}}$ and a $k$-th order polynomial p such that

$${\displaystyle f(x)=p(x)+h_k(x)(x-a{)}^k,\underset{x\to a}{lim}{h}_k(x)=0,}$$

then ${\displaystyle p=P_k}$. Taylor's theorem describes the asymptotic behavior of the remainder term

$${\displaystyle R_k(x)=f(x)-P_k(x),}$$

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

$${\displaystyle R_k(x)=o(|x-a{|}^k),x\to a.}$$

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.

Mean-value forms of the remainder—Let f : R → R be k + 1 times differentiable on the open interval between $a$ and $x$ with f^(k) continuous on the closed interval between $a$ and $x$. Then

$${\displaystyle R_k(x)=\frac{f^{(k+1)}({\xi }_L)}{(k+1)!}(x-a{)}^{k+1}}$$

for some real number ${\xi }_L$ between $a$ and $x$. This is the Lagrange form of the remainder.

Similarly,

$${\displaystyle R_k(x)=\frac{f^{(k+1)}({\xi }_C)}{k!}(x-{\xi }_C{)}^k(x-a)}$$

for some real number ${\xi }_C$ between $a$ and $x$. This is the Cauchy form of the remainder.

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

$${\displaystyle R_k(x)=\frac{f^{(k+1)}({\xi }_S)}{k!}(x-{\xi }_S{)}^{k+1-p}\frac{(x-a{)}^p}{p}}$$ for some real number ${\xi }_S$ 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

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

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

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].

Integral form of the remainder—Let $f^{(k)}$ be absolutely continuous on the closed interval between $a$ and $x$. Then

$${\displaystyle R_k(x)={\int }_a^x\frac{f^{(k+1)}(t)}{k!}(x-t{)}^{k}dt.}$$

Due to the 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 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

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

throughout I. Then the remainder term satisfies the inequality

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

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

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

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

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

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

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:

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

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

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

$${\displaystyle e^x=1+x+\frac{e^{\xi }}{2}{x}^2<1+x+\frac{e^x}{2}{x}^2,0<x\le 1}$$

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

$${\displaystyle e^x\le \frac{1+x}{1-\frac{x^2}{2}}=2\frac{1+x}{2-x^2}\le 4,0\le x\le 1}$$

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

$${\displaystyle |R_k(x)|\le \frac{4|x{|}^{k+1}}{(k+1)!}\le \frac{4}{(k+1)!},-1\le x\le 1,}$$

so the required precision is certainly reached, when

$${\displaystyle \frac{4}{(k+1)!}<{10}^{-5}⟺4\cdot {10}^5<(k+1)!⟺k\ge 9.}$$

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

$${\displaystyle e^x=1+x+\frac{x^2}{2!}+\cdots +\frac{x^9}{9!}+R_9(x),|R_9(x)|<{10}^{-5},-1\le x\le 1.}$$

For instance, this approximation provides a decimal expression ${\displaystyle e\approx 2.71828}$, 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

$${\displaystyle f(x)=\sum _{k=0}^{\mathrm{\infty }}{c}_k(x-a{)}^k=c_0+c_1(x-a)+c_2(x-a{)}^2+\cdots ,|x-a|<r.}$$

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

$${\displaystyle \frac{1}{R}=\underset{k\to \mathrm{\infty }}{lim sup}|c_k{|}^{\frac{1}{k}}.}$$

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 $[a-r_b,a+r_b]$, where $r_b=\left|b-a\right|$. 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

$${\displaystyle P_k(x)=\sum _{j=0}^k{c}_j(x-a{)}^j,c_j=\frac{f^{(j)}(a)}{j!}}$$

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

$${\displaystyle R_k(x)=\sum _{j=k+1}^{\mathrm{\infty }}{c}_j(x-a{)}^j=(x-a{)}^k{h}_k(x),|x-a|<r.}$$

Here the functions

$${\displaystyle \begin{array}{cc}& h_k:(a-r,a+r)\to \mathbb{R}\\ & h_k(x)=(x-a)\sum _{j=0}^{\mathrm{\infty }}{c}_{k+1+j}{\left(x-a\right)}^j\end{array}}$$

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 $R_k(x)$ 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

$${\displaystyle f(x)\approx \sum _{k=0}^{\mathrm{\infty }}{c}_k(x-a{)}^k=c_0+c_1(x-a)+c_2(x-a{)}^2+\cdots }$$

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

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

$${\displaystyle \begin{array}{cc}& T_f:(a-r,a+r)\to \mathbb{R}\\ & T_f(x)=\sum _{k=0}^{\mathrm{\infty }}\frac{f^{(k)}(a)}{k!}{\left(x-a\right)}^k\end{array}}$$

(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:

$${\displaystyle \begin{array}{cc}& f:\mathbb{R}\to \mathbb{R}\\ & f(x)=\{\begin{array}{ll}{e}^{-\frac{1}{x^2}}& x>0\\ 0& x\le 0.\end{array}\end{array}}$$

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

$${\displaystyle f^{(k)}(x)=\{\begin{array}{ll}\frac{p_k(x)}{x^{3k}}\cdot e^{-\frac{1}{x^2}}& x>0\\ 0& x\le 0\end{array}}$$

for some polynomial p_k of degree 2(k − 1). The function ${\displaystyle e^{-\frac{1}{x^2}}}$ tends to zero faster than any polynomial as $x\to 0$, 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 ${\displaystyle t\in [0,2\pi ]}$ gives

$${\displaystyle f(z)=\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(w)}{w-z}dw,f^{\prime }(z)=\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(w)}{(w-z{)}^2}dw,\dots ,f^{(k)}(z)=\frac{k!}{2\pi i}{\int }_{\gamma }\frac{f(w)}{(w-z{)}^{k+1}}dw.}$$

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

$${\displaystyle |f^{(k)}(z)|\le \frac{k!}{2\pi }{\int }_{\gamma }\frac{M_r}{|w-z{|}^{k+1}}dw=\frac{k!M_r}{r^k},M_r=\underset{|w-c|=r}{max}|f(w)|}$$

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

$${\displaystyle T_f(z)=\sum _{k=0}^{\mathrm{\infty }}\frac{f^{(k)}(c)}{k!}(z-c{)}^k}$$

of f converges uniformly on any open disk $B(c,r)\subset U$ with $S(c,r)\subset U$ into some function T_f. Furthermore, using the contour integral formulas for the derivatives f^(k)(c),

$${\displaystyle \begin{array}{cc}{T}_f(z)& =\sum _{k=0}^{\mathrm{\infty }}\frac{(z-c{)}^k}{2\pi i}{\int }_{\gamma }\frac{f(w)}{(w-c{)}^{k+1}}dw\\ & =\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(w)}{w-c}\sum _{k=0}^{\mathrm{\infty }}{\left(\frac{z-c}{w-c}\right)}^{k}dw\\ & =\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(w)}{w-c}\left(\frac{1}{1-\frac{z-c}{w-c}}\right)dw\\ & =\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(w)}{w-z}dw\\ & =f(z),\end{array}}$$

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

$${\displaystyle f(z)=P_k(z)+R_k(z),P_k(z)=\sum _{j=0}^k\frac{f^{(j)}(c)}{j!}(z-c{)}^j,}$$

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 $\gamma$ which parametrizes the boundary $\mathrm{\partial }W\subset U$ of a region $W\subset U$, 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

$${\displaystyle R_k(z)=\sum _{j=k+1}^{\mathrm{\infty }}\frac{(z-c{)}^j}{2\pi i}{\int }_{\gamma }\frac{f(w)}{(w-c{)}^{j+1}}dw=\frac{(z-c{)}^{k+1}}{2\pi i}{\int }_{\gamma }\frac{f(w)dw}{(w-c{)}^{k+1}(w-z)},z\in W.}$$

The important feature here is that the quality of the approximation by a Taylor polynomial on the region $W\subset U$ is dominated by the values of the function f itself on the boundary $\mathrm{\partial }W\subset U$. Similarly, applying Cauchy's estimates to the series expression for the remainder, one obtains the uniform estimates

$${\displaystyle |R_k(z)|\le \sum _{j=k+1}^{\mathrm{\infty }}\frac{M_r|z-c{|}^j}{r^j}=\frac{M_r}{r^{k+1}}\frac{|z-c{|}^{k+1}}{1-\frac{|z-c|}{r}}\le \frac{M_r{\beta }^{k+1}}{1-\beta },\frac{|z-c|}{r}\le \beta <1.}$$

Example

The function

$${\displaystyle \begin{array}{cc}& f:\mathbb{R}\to \mathbb{R}\\ & f(x)=\frac{1}{1+x^2}\end{array}}$$

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

$${\displaystyle \begin{array}{cc}& f:\mathbb{C}\cup \{\mathrm{\infty }\}\to \mathbb{C}\cup \{\mathrm{\infty }\}\\ & f(z)=\frac{1}{1+z^2}\end{array}}$$

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 $B(1,\sqrt{2})$ and does not converge for any z ∈ C with $\left|z-1\right|>\sqrt{2}$.

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

$${\displaystyle f(\mathit{x})=f(\mathit{a})+L(\mathit{x}-\mathit{a})+h(\mathit{x})\Vert \mathit{x}-\mathit{a}\Vert ,\underset{\mathit{x}\to \mathit{a}}{lim}h(\mathit{x})=0.}$$

If this is the case, then $L=df(\mathit{a})$ 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

$${\displaystyle df(\mathit{a})(\mathit{v})=\frac{\mathrm{\partial }f}{\mathrm{\partial }{x}_1}(\mathit{a})v_1+\cdots +\frac{\mathrm{\partial }f}{\mathrm{\partial }{x}_n}(\mathit{a})v_n.}$$

Introduce the multi-index notation

$${\displaystyle |\alpha |={\alpha }_1+\cdots +{\alpha }_n,\alpha !={\alpha }_1!\cdots {\alpha }_n!,{\mathit{x}}^{\alpha }=x_1^{{\alpha }_1}\cdots x_n^{{\alpha }_n}}$$

for α ∈ Nⁿ and x ∈ Rⁿ. If all the $k$-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

$${\displaystyle D^{\alpha }f=\frac{{\mathrm{\partial }}^{|\alpha |}f}{\mathrm{\partial }{\mathit{x}}^{\alpha }}=\frac{{\mathrm{\partial }}^{{\alpha }_1+\dots +{\alpha }_n}f}{\mathrm{\partial }{x}_1^{{\alpha }_1}\cdots \mathrm{\partial }{x}_n^{{\alpha }_n}}}$$

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.

Multivariate version of Taylor's theorem—Let f : Rⁿ → R be a k-times continuously differentiable function at the point a ∈ Rⁿ. Then there exist functions h_α : Rⁿ → R, where ${\displaystyle |\alpha |=k,}$ such that

$${\displaystyle \begin{array}{cc}& f(\mathit{x})=\sum _{|\alpha |\le k}\frac{D^{\alpha }f(\mathit{a})}{\alpha !}(\mathit{x}-\mathit{a}{)}^{\alpha }+\sum _{|\alpha |=k}{h}_{\alpha }(\mathit{x})(\mathit{x}-\mathit{a}{)}^{\alpha },\\ & \text{and}\underset{\mathit{x}\to \mathit{a}}{lim}{h}_{\alpha }(\mathit{x})=0.\end{array}}$$

If the function f : Rⁿ → R is k + 1 times continuously differentiable in a closed ball ${\displaystyle B=\{\mathbf{y}\in {\mathbb{R}}^n:\Vert \mathbf{a}-\mathbf{y}\Vert \le r\}}$ for some ${\displaystyle r>0}$, 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,

$${\displaystyle \begin{array}{cc}& f(\mathit{x})=\sum _{|\alpha |\le k}\frac{D^{\alpha }f(\mathit{a})}{\alpha !}(\mathit{x}-\mathit{a}{)}^{\alpha }+\sum _{|\beta |=k+1}{R}_{\beta }(\mathit{x})(\mathit{x}-\mathit{a}{)}^{\beta },\\ & R_{\beta }(\mathit{x})=\frac{|\beta |}{\beta !}{\int }_0^1(1-t{)}^{|\beta |-1}{D}^{\beta }f(\mathit{a}+t(\mathit{x}-\mathit{a}))dt.\end{array}}$$

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

$${\displaystyle \left|R_{\beta }(\mathit{x})\right|\le \frac{1}{\beta !}\underset{|\alpha |=|\beta |}{max}\underset{\mathit{y}\in B}{max}|D^{\alpha }f(\mathit{y})|,\mathit{x}\in B.}$$

Example in two dimensions

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

$${\displaystyle \begin{array}{cc}{P}_3(\mathit{x})=f(\mathit{a})+& \frac{\mathrm{\partial }f}{\mathrm{\partial }{x}_1}(\mathit{a})v_1+\frac{\mathrm{\partial }f}{\mathrm{\partial }{x}_2}(\mathit{a})v_2+\frac{{\mathrm{\partial }}^{2}f}{\mathrm{\partial }{x}_1^2}(\mathit{a})\frac{v_1^2}{2!}+\frac{{\mathrm{\partial }}^{2}f}{\mathrm{\partial }{x}_1\mathrm{\partial }{x}_2}(\mathit{a})v_1{v}_2+\frac{{\mathrm{\partial }}^{2}f}{\mathrm{\partial }{x}_2^2}(\mathit{a})\frac{v_2^2}{2!}\\ & +\frac{{\mathrm{\partial }}^{3}f}{\mathrm{\partial }{x}_1^3}(\mathit{a})\frac{v_1^3}{3!}+\frac{{\mathrm{\partial }}^{3}f}{\mathrm{\partial }{x}_1^2\mathrm{\partial }{x}_2}(\mathit{a})\frac{v_1^2{v}_2}{2!}+\frac{{\mathrm{\partial }}^{3}f}{\mathrm{\partial }{x}_1\mathrm{\partial }{x}_2^2}(\mathit{a})\frac{v_1{v}_2^2}{2!}+\frac{{\mathrm{\partial }}^{3}f}{\mathrm{\partial }{x}_2^3}(\mathit{a})\frac{v_2^3}{3!}\end{array}}$$

Proofs

Proof for Taylor's theorem in one real variable

Let

$${\displaystyle h_k(x)=\{\begin{array}{ll}\frac{f(x)-P(x)}{(x-a{)}^k}& x\ne a\\ 0& x=a\end{array}}$$

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

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

It is sufficient to show that

$${\displaystyle \underset{x\to a}{lim}{h}_k(x)=0.}$$

The proof here is based on repeated application of L'Hôpital's rule. Note that, for each $j=0,1,...,k-1$, ${\displaystyle f^{(j)}(a)=P^{(j)}(a)}$. Hence each of the first $k-1$ derivatives of the numerator in ${\displaystyle h_k(x)}$ vanishes at ${\displaystyle x=a}$, 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

$${\displaystyle \begin{array}{cc}\underset{x\to a}{lim}\frac{f(x)-P(x)}{(x-a{)}^k}& =\underset{x\to a}{lim}\frac{\frac{d}{dx}(f(x)-P(x))}{\frac{d}{dx}(x-a{)}^k}\\ & =\cdots \\ & =\underset{x\to a}{lim}\frac{\frac{d^{k-1}}{d{x}^{k-1}}(f(x)-P(x))}{\frac{d^{k-1}}{d{x}^{k-1}}(x-a{)}^k}\\ & =\frac{1}{k!}\underset{x\to a}{lim}\frac{f^{(k-1)}(x)-P^{(k-1)}(x)}{x-a}\\ & =\frac{1}{k!}(f^{(k)}(a)-P^{(k)}(a))=0\end{array}}$$

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

Alternate proof for Taylor's theorem in one real variable

Let ${\displaystyle f(x)}$ 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

$${\displaystyle \begin{array}{c}F(x)=f(x)-\sum _{k=0}^{n-1}\frac{f^{(k)}(a)}{k!}(x-a{)}^k\end{array}}$$

$${\displaystyle \begin{array}{c}G(x)=(x-a{)}^n\end{array}}$$

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

$${\displaystyle \begin{array}{cc}F(a)& =f(a)-f(a)-f^{\prime }(a)(a-a)-...-\frac{f^{(n-1)}(a)}{(n-1)!}(a-a{)}^{n-1}=0\\ G(a)& =(a-a{)}^n=0\end{array}}$$

Similarly,

$${\displaystyle \begin{array}{c}{F}^{\prime }(a)=f^{\prime }(a)-f^{\prime }(a)-\frac{f^{″}(a)}{(2-1)!}(a-a{)}^{(2-1)}-...-\frac{f^{(n-1)}(a)}{(n-2)!}(a-a{)}^{n-2}=0\end{array}}$$

$${\displaystyle \begin{array}{cc}{G}^{\prime }(a)& =n(a-a{)}^{n-1}=0\\ & ⋮\\ G^{(n-1)}(a)& =F^{(n-1)}(a)=0\end{array}}$$

Step 3: Use Cauchy Mean Value Theorem

Let ${\displaystyle f_1}$ and ${\displaystyle g_1}$ be continuous functions on ${\displaystyle [a,b]}$. Since ${\displaystyle a<x<b}$ so we can work with the interval ${\displaystyle [a,x]}$. Let ${\displaystyle f_1}$ and ${\displaystyle g_1}$ be differentiable on ${\displaystyle (a,x)}$. Assume ${\displaystyle g_1^{\prime }(x)\ne 0}$ for all ${\displaystyle x\in (a,b)}$. Then there exists ${\displaystyle c_1\in (a,x)}$ such that

$${\displaystyle \begin{array}{c}\frac{f_1(x)-f_1(a)}{g_1(x)-g_1(a)}=\frac{f_1^{\prime }(c_1)}{g_1^{\prime }(c_1)}\end{array}}$$

Note: ${\displaystyle G^{\prime }(x)\ne 0}$ in ${\displaystyle (a,b)}$ and ${\displaystyle F(a),G(a)=0}$ so

$${\displaystyle \begin{array}{c}\frac{F(x)}{G(x)}=\frac{F(x)-F(a)}{G(x)-G(a)}=\frac{F^{\prime }(c_1)}{G^{\prime }(c_1)}\end{array}}$$

for some ${\displaystyle c_1\in (a,x)}$.

This can also be performed for ${\displaystyle (a,c_1)}$:

$${\displaystyle \begin{array}{c}\frac{F^{\prime }(c_1)}{G^{\prime }(c_1)}=\frac{F^{\prime }(c_1)-F^{\prime }(a)}{G^{\prime }(c_1)-G^{\prime }(a)}=\frac{F^{″}(c_2)}{G^{″}(c_2)}\end{array}}$$

for some ${\displaystyle c_2\in (a,c_1)}$. This can be continued to ${\displaystyle c_n}$.

This gives a partition in ${\displaystyle (a,b)}$:

$${\displaystyle a<c_n<c_{n-1}<\cdots <c_1<x}$$

with

$${\displaystyle \frac{F(x)}{G(x)}=\frac{F^{\prime }(c_1)}{G^{\prime }(c_1)}=\cdots =\frac{F^{(n)}(c_n)}{G^{(n)}(c_n)}.}$$

Set ${\displaystyle c=c_n}$:

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

Step 4: Substitute back

$${\displaystyle \begin{array}{c}\frac{F(x)}{G(x)}=\frac{f(x)-\sum _{k=0}^{n-1}\frac{f^{(k)}(a)}{k!}(x-a{)}^k}{(x-a{)}^n}=\frac{F^{(n)}(c)}{G^{(n)}(c)}\end{array}}$$

By the Power Rule, repeated derivatives of ${\displaystyle (x-a{)}^n}$, ${\displaystyle G^{(n)}(c)=n(n-1)...1}$, so:

$${\displaystyle \frac{F^{(n)}(c)}{G^{(n)}(c)}=\frac{f^{(n)}(c)}{n(n-1)\cdots 1}=\frac{f^{(n)}(c)}{n!}.}$$

This leads to:

$${\displaystyle \begin{array}{c}f(x)-\sum _{k=0}^{n-1}\frac{f^{(k)}(a)}{k!}(x-a{)}^k=\frac{f^{(n)}(c)}{n!}(x-a{)}^n\end{array}.}$$

By rearranging, we get:

$${\displaystyle \begin{array}{c}f(x)=\sum _{k=0}^{n-1}\frac{f^{(k)}(a)}{k!}(x-a{)}^k+\frac{f^{(n)}(c)}{n!}(x-a{)}^n\end{array},}$$

or because ${\displaystyle c_n=a}$ eventually:

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

Derivation for the mean value forms of the remainder

Let G be any real-valued function, continuous on the closed interval between $a$ and $x$ and differentiable with a non-vanishing derivative on the open interval between $a$ and $x$, and define

$${\displaystyle F(t)=f(t)+f^{\prime }(t)(x-t)+\frac{f^{″}(t)}{2!}(x-t{)}^2+\cdots +\frac{f^{(k)}(t)}{k!}(x-t{)}^k.}$$

For ${\displaystyle t\in [a,x]}$. Then, by Cauchy's mean value theorem,

for some $\xi$ 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

$${\displaystyle \begin{array}{cc}{F}^{\prime }(t)=& f^{\prime }(t)+(f^{″}(t)(x-t)-f^{\prime }(t))+\left(\frac{f^{(3)}(t)}{2!}(x-t{)}^2-\frac{f^{(2)}(t)}{1!}(x-t)\right)+\cdots \\ & \cdots +\left(\frac{f^{(k+1)}(t)}{k!}(x-t{)}^k-\frac{f^{(k)}(t)}{(k-1)!}(x-t{)}^{k-1}\right)=\frac{f^{(k+1)}(t)}{k!}(x-t{)}^k,\end{array}}$$

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

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

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

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

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

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 ${\displaystyle f^{(k)}}$ on the closed interval between $a$ and $x$, its derivative ${\displaystyle f^{(k+1)}}$ exists as an ${\displaystyle L^1}$-function, and we can use the fundamental theorem of calculus and integration by parts. This same proof applies for the Riemann integral assuming that ${\displaystyle f^{(k)}}$ 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

$${\displaystyle f(x)=f(a)+{\int }_a^x{f}^{\prime }(t)dt.}$$

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

$${\displaystyle \begin{array}{cc}f(x)& =f(a)+(x{f}^{\prime }(x)-a{f}^{\prime }(a))-{\int }_a^{x}t{f}^{″}(t)dt\\ & =f(a)+x\left(f^{\prime }(a)+{\int }_a^x{f}^{″}(t)dt\right)-a{f}^{\prime }(a)-{\int }_a^{x}t{f}^{″}(t)dt\\ & =f(a)+(x-a)f^{\prime }(a)+{\int }_a^x(x-t)f^{″}(t)dt,\end{array}}$$

which is exactly Taylor's theorem with remainder in the integral form in the case ${\displaystyle k=1}$. The general statement is proved using induction. Suppose that

Integrating the remainder term by parts we arrive at

$${\displaystyle \begin{array}{cc}{\int }_a^x\frac{f^{(k+1)}(t)}{k!}(x-t{)}^{k}dt=& -{\left[\frac{f^{(k+1)}(t)}{(k+1)k!}(x-t{)}^{k+1}\right]}_a^x+{\int }_a^x\frac{f^{(k+2)}(t)}{(k+1)k!}(x-t{)}^{k+1}dt\\ =& \frac{f^{(k+1)}(a)}{(k+1)!}(x-a{)}^{k+1}+{\int }_a^x\frac{f^{(k+2)}(t)}{(k+1)!}(x-t{)}^{k+1}dt.\end{array}}$$

Substituting this into the formula in (eq1) shows that if it holds for the value ${\displaystyle k}$, it must also hold for the value ${\displaystyle k+1}$. Therefore, since it holds for ${\displaystyle k=1}$, it must hold for every positive integer ${\displaystyle k}$.