All-order multivariable Taylor theorem

1. Notation and the segment hypothesis

Let $d$ be a positive integer and $m$ a nonnegative integer. Let $U\subset\mathbb R^d$ be open, $a\in U$, and $h\in\mathbb R^d$, and assume

$$[a,a+h]=\{a+th:0\le t\le1\}\subset U.$$

For the integral and Lagrange remainder formulas below, suppose $f:U\to\mathbb R$ is $C^{m+1}$ on an open neighborhood $V\subset U$ of this segment. Thus all iterated coordinate partial derivatives through order $m+1$ exist and are continuous on $V$. No convexity of all of $U$ is required. The displacement $h$ is not required to be a unit vector.

A multi-index is $\alpha=(\alpha_1,\ldots,\alpha_d)\in\mathbb N_0^d$, where $\mathbb N_0=\{0,1,2,\ldots\}$. Put

$$|\alpha|=\sum_{j=1}^d\alpha_j,\qquad \alpha!=\prod_{j=1}^d\alpha_j!,\qquad h^\alpha=\prod_{j=1}^dh_j^{\alpha_j},\qquad \partial^\alpha f=\partial_1^{\alpha_1}\cdots\partial_d^{\alpha_d}f.$$

A zero exponent contributes the factor 1 even when the coordinate is zero; $0!=1$, $h^0=1$ and $\partial^0f=f$. Write $e_j$ for the $j$th coordinate unit vector. The degree-$m$ Taylor polynomial, written in displacement coordinates, is

$$T_{m,a}(h)=\sum_{|\alpha|\le m}\frac{\partial^\alpha f(a)}{\alpha!}h^\alpha.$$

Its degree can be less than $m$. The continuous-partial hypothesis makes the order of differentiation immaterial; a justification is given below.

2. One-variable tools and closed endpoints

In the one-variable integral theorem, continuous derivatives on the closed interval mean that the derivatives on its interior extend continuously to its endpoints, with the corresponding one-sided derivatives there. The derivative values in the Taylor polynomial are those endpoint values. Extension to an open interval is a sufficient, but stronger, way to satisfy this convention. In the present multivariable setting it is automatic: the open set $\{t\in\mathbb R:a+th\in V\}$ contains $[0,1]$.

Fermat's necessary condition. At a differentiable interior local maximum or minimum, $f'(a)=0$.

Proof. At a local maximum, sufficiently small positive increments give nonpositive difference quotients and negative increments give nonnegative ones. A two-sided derivative can satisfy both sign limits only by being zero. Negating the function treats a minimum. $\square$

The interior hypothesis matters: $f(x)=x$ on $[0,1]$ has extrema at endpoints despite derivative 1. The converse fails: $x^3$ has zero derivative at zero but no extremum there.

Rolle's theorem. If $a<b$, $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f(a)=f(b)$, then $f'(c)=0$ for some $c\in(a,b)$.

Proof. The extreme-value theorem supplies maximum and minimum. If the function is constant, every interior point works. Otherwise a value differs from the equal endpoint values, so either a maximum or a minimum occurs in the interior. Fermat's condition applies there. $\square$

Mean value theorem. Under the same continuity and differentiability hypotheses but without equal endpoint values, some $c\in(a,b)$ satisfies $f(b)-f(a)=f'(c)(b-a)$.

Subtract the secant line $f(a)+[f(b)-f(a)](x-a)/(b-a)$ from $f$ and apply Rolle. This proves the theorem without assuming $f'$ is continuous. Consequences include: derivative zero on an interval makes the function constant; nonnegative derivative makes it nondecreasing; and $|f'|\le M$ on an interval gives $|f(y)-f(x)|\le M|y-x|$ for points in it. An interval-wide derivative bound cannot be inferred from one derivative value.

The order-$n$ Taylor polynomial at $a$ is $T_n(x)=\sum_{k=0}^nf^{(k)}(a)(x-a)^k/k!$, whose degree may be smaller than $n$ if leading derivatives vanish.

Theorem: Taylor's integral remainder

Let $n\ge0$ and let $f$ have continuous derivatives through order $n+1$ on the closed interval between $a$ and $x$. Then

$$f(x)=T_n(x)+\frac1{n!}\int_a^x f^{(n+1)}(t)(x-t)^n\,dt.$$

Consequently, if $|f^{(n+1)}(t)|\le M$ there, then $|f(x)-T_n(x)|\le M|x-a|^{n+1}/(n+1)!$.

Proof. At $n=0$, the identity is the fundamental theorem $f(x)-f(a)=\int_a^xf'(t)dt$. For $n\ge1$, integrate the remainder by parts, taking $u=(x-t)^n/n!$ and $dv=f^{(n+1)}(t)dt$. The boundary at $x$ is zero and at $a$ contributes $-f^{(n)}(a)(x-a)^n/n!$. The remaining integral is the order-$(n-1)$ remainder. Induction therefore yields the identity. This uses oriented integrals, so it also holds when $x<a$. Taking absolute values, reversing integration limits if needed, and integrating $|x-t|^n$ gives the stated bound. Continuity of the highest derivative makes every use of parts and the bound valid.

This theorem provides a sufficient regularity setting, not the weakest possible hypotheses. For a Taylor series to represent $f$ at a fixed $x$, the remainders must tend to zero as $n\to\infty$. A finite-order error bound alone does not prove that without control of its dependence on $n$.

3. Repeated directional differentiation, with the coefficient count

For a fixed $h$, define $D_hq(x)=\left.\frac{d}{ds}q(x+sh)\right|_{s=0}$ wherever this derivative exists. Put $D_h^0f=f$ and $D_h^{k+1}f=D_h(D_h^kf)$. These are repeated derivatives in the same fixed direction, not derivatives with a varying direction.

The first directional derivative. If $q$ is $C^1$ near $x$, split the increment from $x$ to $x+sh$ into the $d$ successive coordinate increments. For sufficiently small $s$, all of these small segments lie in a ball contained in the domain. The one-variable mean value theorem gives, for each $j$ with $h_j\ne0$, a point $z_{j,s}$ on the $j$th small segment such that

$$q(x+sh)-q(x)=\sum_{j:h_j\ne0}s h_j\partial_jq(z_{j,s}).$$

Here $z_{j,s}\to x$ as $s\to0$; coordinates with $h_j=0$ make no increment. Divide by $s\ne0$ and use continuity of the partial derivatives to obtain

$$D_hq(x)=\sum_{j=1}^dh_j\partial_jq(x). \tag{1}$$

This argument also handles negative $s$, since the mean value theorem uses the interval between the two coordinate endpoints.

Why the mixed partials commute. For a $C^2$ function $q$, fix different coordinates $i,j$ and take small positive $u,v$ so a coordinate rectangle based at $x$ lies in the domain. Its double difference is

$$\Delta=q(x+ue_i+ve_j)-q(x+ue_i)-q(x+ve_j)+q(x).$$

Apply the mean value theorem first to $s\mapsto q(x+se_i+ve_j)-q(x+se_i)$ on $[0,u]$, then to the difference of $\partial_iq$ in the $j$th coordinate. It gives

$$\frac{\Delta}{uv}=\partial_j\partial_iq(x+\xi e_i+\eta e_j),\qquad 0<\xi<u,\quad0<\eta<v.$$

Reversing the two coordinate roles gives $\Delta/(uv)=\partial_i\partial_jq(x+\widetilde\xi e_i+\widetilde\eta e_j)$ for another point in the same rectangle. Let $u,v\downarrow0$. Continuity gives $\partial_j\partial_iq(x)=\partial_i\partial_jq(x)$. For derivatives of higher order, apply this result to the inner derivative of $f$ preceding an adjacent pair of differentiations, and then differentiate the resulting identity by any remaining outer derivatives. All orders involved are at most $m+1$. Every permutation is a finite succession of adjacent swaps, so all orderings with the same coordinate multiplicities agree.

Claim. For every $0\le k\le m+1$ and every $x\in V$,

$$D_h^kf(x)=\sum_{|\alpha|=k}\frac{k!}{\alpha!}h^\alpha\partial^\alpha f(x). \tag{2}$$

Complete induction. At $k=0$, the only multi-index is $0$, and the right side is $f(x)$. Suppose (2) holds for an integer $k$ with $0\le k\le m$. Each $\partial^\alpha f$ in that finite sum is $C^1$. By (1), linearity of differentiation and commutation of partial derivatives,

$$\begin{aligned} D_h^{k+1}f(x) &=\sum_{|\alpha|=k}\sum_{j=1}^d\frac{k!}{\alpha!}h^\alpha h_j\partial^{\alpha+e_j}f(x)\\ &=\sum_{|\beta|=k+1}\left(\sum_{j:\beta_j\ge1}\frac{k!}{(\beta-e_j)!}\right)h^\beta\partial^\beta f(x). \end{aligned}$$

For a fixed $\beta$, the pairs producing it are exactly $(\alpha,j)=(\beta-e_j,j)$ with $\beta_j\ge1$; no others occur. Moreover,

$$(\beta-e_j)!=\frac{\beta!}{\beta_j},\qquad \sum_{j:\beta_j\ge1}\frac{k!}{(\beta-e_j)!} =\frac{k!}{\beta!}\sum_{j:\beta_j\ge1}\beta_j =\frac{k!(k+1)}{\beta!} =\frac{(k+1)!}{\beta!}.$$

The omitted coordinates have $\beta_j=0$ and contribute zero to the sum of coordinates. This is precisely (2) at $k+1$, completing the induction. In counting terms, $k!/\alpha!$ is the number of ordered lists of $k$ coordinate choices in which coordinate $j$ occurs $\alpha_j$ times; the displayed recurrence proves that count rather than assuming it.

Now set $g(t)=f(a+th)$. Differentiating at any $t$ is differentiation of $f$ at $a+th$ in direction $h$. Starting with $g^{(0)}(t)=f(a+th)$ and using this observation successively gives

$$g^{(k)}(t)=D_h^kf(a+th)=\sum_{|\alpha|=k}\frac{k!}{\alpha!}h^\alpha\partial^\alpha f(a+th),\qquad0\le k\le m+1. \tag{3}$$

All these derivatives are continuous on an open interval containing $[0,1]$.

4. Taylor's theorem and both exact remainders

Theorem. Under the hypotheses in Section 1,

$$f(a+h)=T_{m,a}(h)+R_m(a,h),$$

where

$$\begin{aligned} R_m(a,h) &=\frac1{m!}\int_0^1(1-t)^mD_h^{m+1}f(a+th)\,dt\\ &=\sum_{|\alpha|=m+1}\frac{m+1}{\alpha!}h^\alpha\int_0^1(1-t)^m\partial^\alpha f(a+th)\,dt. \tag{4} \end{aligned}$$

If $h\ne0$, there also exists one $\theta\in(0,1)$ such that

$$\begin{aligned} R_m(a,h) &=\frac{D_h^{m+1}f(a+\theta h)}{(m+1)!}\\ &=\sum_{|\alpha|=m+1}\frac{\partial^\alpha f(a+\theta h)}{\alpha!}h^\alpha. \tag{5} \end{aligned}$$

Integral proof. The one-variable theorem above applies to $g$ from 0 to 1. To make its induction explicit in these coordinates, put

$$I_r=\frac1{r!}\int_0^1(1-t)^rg^{(r+1)}(t)\,dt\qquad(0\le r\le m).$$

The fundamental theorem gives $I_0=g(1)-g(0)$. For $1\le r\le m$, integration by parts gives exactly

$$\begin{aligned} I_r&=\left[\frac{(1-t)^r}{r!}g^{(r)}(t)\right]_0^1+\frac1{(r-1)!}\int_0^1(1-t)^{r-1}g^{(r)}(t)\,dt\\ &=-\frac{g^{(r)}(0)}{r!}+I_{r-1}. \end{aligned}$$

Thus the identity $I_0=g(1)-\sum_{k=0}^0g^{(k)}(0)/k!$ is the base case. If $I_{r-1}=g(1)-\sum_{k=0}^{r-1}g^{(k)}(0)/k!$, the displayed recurrence gives $I_r=g(1)-\sum_{k=0}^rg^{(k)}(0)/k!$. Induction proves it up to $r=m$, including $m=0$. Formula (3) at 0 now gives

$$\sum_{k=0}^m\frac{g^{(k)}(0)}{k!}=\sum_{k=0}^m\sum_{|\alpha|=k}\frac{\partial^\alpha f(a)}{\alpha!}h^\alpha=T_{m,a}(h).$$

Formula (3) at order $m+1$ gives (4), since $(m+1)!/m!=m+1$ and the sum is finite. In particular the degree is $m$ and the derivative in the remainder has order $m+1$.

Strict-intermediate-point proof. Let $R=g(1)-\sum_{k=0}^mg^{(k)}(0)/k!$ and define

$$H(t)=g(t)-\sum_{k=0}^m\frac{g^{(k)}(0)}{k!}t^k-Rt^{m+1}.$$

Then $H(1)=0$ and $H^{(r)}(0)=0$ for $0\le r\le m$. Rolle on $[0,1]$ gives a point $c_1\in(0,1)$ with $H'(c_1)=0$. If $m\ge1$, Rolle on $[0,c_1]$, using $H'(0)=0$, gives $c_2\in(0,c_1)$ with $H''(c_2)=0$. More precisely, whenever $1\le r\le m$ and $H^{(r)}(c_r)=0$ with $c_r>0$, the additional zero $H^{(r)}(0)=0$ allows Rolle on $[0,c_r]$ to produce $c_{r+1}\in(0,c_r)$ with $H^{(r+1)}(c_{r+1})=0$. After $m+1$ applications, $\theta=c_{m+1}\in(0,1)$ satisfies

$$0=H^{(m+1)}(\theta)=g^{(m+1)}(\theta)-(m+1)!R.$$

For $m=0$, the first application already is the last. Substitute (3) to obtain (5).

The same $\theta$ in every term of (5) comes from applying a scalar theorem once to the single line function $g$. Applying separate mean value arguments to individual partial derivatives or to separate coordinate paths can give different points; those points cannot be silently replaced by a common one. The common-point claim here is for a real scalar function, not a componentwise common-point claim for a vector-valued function.

If $h=0$, then $g$ is constant, $T_{m,a}(0)=f(a)$ and $R_m(a,0)=0$; (4) remains valid. No strict geometric intermediate point on the singleton segment is required. Even for $h\ne0$, the line function can be constant; the Rolle proof still applies. At $m=0$, the formulas reduce to $f(a+h)-f(a)=\int_0^1\nabla f(a+th)\cdot h\,dt$ and the scalar segment mean value formula.

5. Bounds and a weaker local Taylor statement

If $|\partial^\alpha f(x)|\le M$ for every $|\alpha|=m+1$ and every $x$ on the segment, (4) gives

$$|R_m(a,h)|\le M\sum_{|\alpha|=m+1}\frac{|h^\alpha|}{\alpha!} =\frac{M}{(m+1)!}\left(\sum_{j=1}^d|h_j|\right)^{m+1}. \tag{6}$$

Indeed $\int_0^1(1-t)^m\,dt=1/(m+1)$, and expanding the product of $m+1$ copies of $\sum_j|h_j|$ groups ordered coordinate choices with exactly the coefficient count proved in (2). A bound at $a$ alone does not imply (6). Alternatively, a direct bound $|D_h^{m+1}f(a+th)|\le K\|h\|^{m+1}$ gives $|R_m|\le K\|h\|^{m+1}/(m+1)!$.

The $C^{m+1}$ assumption is sufficient for (4) and (5); it is not a requirement for every local Taylor approximation. If $m\ge1$ and $f$ is only $C^m$ on a neighborhood of $a$, then

$$f(a+h)=T_{m,a}(h)+o(\|h\|^m)\qquad(h\to0). \tag{7}$$

To prove this weaker assertion without adding another derivative, apply (4) with degree $m-1$ to sufficiently short segments. The integral containing the constant $D_h^mf(a)$ equals $D_h^mf(a)/m!$, exactly the degree-$m$ term by (2). After subtracting that constant inside the integral, the error is bounded by

$$\frac{\varepsilon(h)}{m!}\left(\sum_{j=1}^d|h_j|\right)^m,\qquad \varepsilon(h)=\max_{|\alpha|=m}\sup_{0\le t\le1}|\partial^\alpha f(a+th)-\partial^\alpha f(a)|\longrightarrow0.$$

There are finitely many such $\alpha$, and continuity at $a$ controls all segment points because $\|th\|\le\|h\|$. Cauchy–Schwarz gives $\sum_j|h_j|\le\sqrt d\|h\|$, proving (7). At $m=0$, continuity alone gives $f(a+h)=f(a)+o(1)$. These weaker conclusions do not assert an order-$(m+1)$ remainder or an infinite series representation.

6. Worked example: third order and a fourth-order remainder

Let $f(x,y)=e^{x+2y}$, $a=(0,0)$ and $h=(u,v)$. For $\alpha=(p,q)$, $\partial^\alpha f(0,0)=2^q$. Therefore

$$\begin{aligned} T_{3,0}(u,v) ={}&1+u+2v+\frac{u^2}{2}+2uv+2v^2\\ &+\frac{u^3}{6}+u^2v+2uv^2+\frac43v^3. \end{aligned}$$

For example, the $u^2v$ coefficient is $\partial_x^2\partial_yf(0,0)/(2!1!)=2/2=1$; in $D_h^3f(0,0)$ the same term has coefficient $(3!/(2!1!))\,2=6$, which becomes 1 only after division by $3!$.

Put $s=u+2v$. The line function is $g(t)=e^{ts}$, so $g^{(4)}(t)=s^4e^{ts}$. Thus

$$e^{u+2v}-T_{3,0}(u,v)=\frac{s^4}{6}\int_0^1(1-t)^3e^{ts}\,dt,$$

and for a nonzero displacement one common $\theta\in(0,1)$ gives $R_3=s^4e^{\theta s}/24$. In particular $|R_3|\le e^{|s|}|s|^4/24$. If $s=0$, the line is constant and the error is exactly zero, even when $h\ne0$. This cancellation is compatible with the general bound, and illustrates why total degree, rather than a separate degree limit in each coordinate, is used.

7. Practice and full solution

P01. For $f(x,y,z)=5+x^2y+2xyz-yz^2$, expand about $a=(1,0,1)$ with displacement $h=(u,v,w)$.

1. Find $T_{2,a}(h)$ and the exact remaining terms.
2. Verify every nonzero third-order mixed coefficient using $\partial^\alpha f/\alpha!$, and verify the directional integral and Lagrange remainders.
3. Decide whether $T_{3,a}$ is exact, including at $h=0$.

Full solution. Substitution and collection by total degree give

$$\begin{aligned} f(1+u,v,1+w) &=5+(1+u)^2v+2(1+u)v(1+w)-v(1+w)^2\\ &=5+2v+4uv+u^2v+2uvw-vw^2. \end{aligned}$$

Hence $T_{2,a}=5+2v+4uv$ and $R_2=u^2v+2uvw-vw^2$. The third partials are constant. Their nonzero multi-indices and values are

$$\partial^{(2,1,0)}f=2,\qquad\partial^{(1,1,1)}f=2,\qquad\partial^{(0,1,2)}f=-2.$$

All other third-order multi-index derivatives vanish. Division by $2!1!0!=2$, $1!1!1!=1$ and $0!1!2!=2$ gives respectively the coefficients $1,2,-1$. Formula (2) gives

$$D_h^3f=\frac{3!}{2!1!0!}\,2u^2v+\frac{3!}{1!1!1!}\,2uvw+\frac{3!}{0!1!2!}\,(-2)vw^2 =6u^2v+12uvw-6vw^2.$$

This is independent of the point on the segment. The integral remainder is $\frac12\int_0^1(1-t)^2D_h^3f\,dt=D_h^3f/6=R_2$, because the integral of $(1-t)^2$ is $1/3$. For $h\ne0$, the Lagrange form has the same value for every $\theta\in(0,1)$, so in particular for the point supplied by the theorem. Finally all fourth-order derivatives vanish; the degree-3 polynomial is the full displayed expansion, with zero remainder for every displacement. At $h=0$ both remainders are zero directly.