Higher derivatives and multivariable analyticity

A several-variable Taylor formula is not obtained by writing unrelated one-variable expansions in each coordinate. Mixed derivatives occur because a displacement changes all coordinates at once. Multilinear derivatives organize those interactions, and a remainder estimate tells us whether the infinite Taylor expansion equals the function.

1 Higher derivatives as multilinear maps

Work over the real field with finite $n\ge1$. A scalar function is Ck on an open set when all coordinate partial derivatives through order $k$ exist and are continuous. Identify a j-linear scalar map with its finitely many coefficients on tuples of coordinate basis vectors. The higher derivative is obtained by differentiating this coefficient map; the following norm estimates justify that this agrees with differentiating into the space of multilinear maps.

For a j-linear map $B$, $j\ge1$, multilinearity gives the exact finite expansion

$$B[v_1,\ldots,v_j]=\sum_{i_1,\ldots,i_j=1}^n B[e_{i_1},\ldots,e_{i_j}](v_1)_{i_1}\cdots(v_j)_{i_j}.$$

Put $M_B=\max_{i_1,\ldots,i_j}|B[e_{i_1},\ldots,e_{i_j}]|$. The triangle inequality and Cauchy–Schwarz give

$$|B[v_1,\ldots,v_j]|\le M_B\prod_{r=1}^j\sum_i|(v_r)_i|\le n^{j/2}M_B\prod_{r=1}^j\|v_r\|.$$

Thus the multilinear operator norm

$$\|B\|=\sup\{|B[v_1,\ldots,v_j]|:\|v_r\|\le1\ (1\le r\le j)\}$$

is finite and satisfies $M_B\le\|B\|\le n^{j/2}M_B$. The lower bound tests basis vectors. The finite expansion also makes $B$ continuous on the compact product of closed unit balls, so this supremum is attained. The inequalities show that convergence or differentiability of its finitely many coefficients is equivalent to the corresponding operator-norm statement. In particular continuous coordinate derivatives of a Ck function construct continuous multilinear derivatives through order $k$ by induction, using the finite-dimensional differentiability criterion on the coefficient map.

More explicitly, for $1\le j\le k$ that induction gives

$$D^jf(x)[v_1,\ldots,v_j]=\sum_{i_1,\ldots,i_j=1}^n\partial_{i_1}\cdots\partial_{i_j}f(x)(v_1)_{i_1}\cdots(v_j)_{i_j}.$$

Continuity of mixed partials permits an adjacent pair of differentiation orders to be interchanged by the C2 coordinate-rectangle theorem, applied to the relevant lower derivative. Every permutation is a finite composition of adjacent interchanges. Hence these multilinear maps are symmetric. For $j=2$, the formula is $D^2f(a)[u,v]=u^TH_f(a)v$.

For fixed $h$, let $\phi(t)=f(a+th)$ on a segment inside the open domain. Applying the chain rule successively to the coefficient functions proves

$$\phi^{(j)}(t)=D^jf(a+th)[h,\ldots,h].$$

If $h\ne0$, divide each input by $\|h\|$ in the norm definition; if $h=0$, multilinearity makes the value zero. Therefore $|D^jf(x)[h,\ldots,h]|\le\|D^jf(x)\|\|h\|^j$ in both cases.

For a multi-index $\alpha=(\alpha_1,\ldots,\alpha_n)$ of nonnegative integers, put $|\alpha|=\sum_i\alpha_i$, $\alpha!=\prod_i\alpha_i!$ and $h^\alpha=\prod_i h_i^{\alpha_i}$. In the repeated-input expansion there are exactly $j!/\alpha!$ ordered coordinate choices with multiplicities $\alpha$. Symmetry gives the same mixed partial for every such choice. Consequently

$$\frac1{j!}D^jf(a)[h,\ldots,h]=\sum_{|\alpha|=j}\frac{\partial^\alpha f(a)}{\alpha!}h^\alpha.$$

This counts mixed coefficients rather than guessing them from separate coordinate expansions. In particular the quadratic mixed term is $f_{xy}hk$, not $f_{xy}hk/2$.

2 Taylor's theorem with an integral remainder

Suppose $f$ is $C^k$ near the segment from $a$ to $a+h$, with $k\ge1$. Repeated one-variable integration gives

$$f(a+h)=\sum_{j=0}^{k-1}\frac{D^jf(a)[h,\ldots,h]}{j!} +\frac1{(k-1)!}\int_0^1(1-t)^{k-1}D^kf(a+th)[h,\ldots,h]dt.$$

Here the $j=0$ term is $f(a)$. A proof is induction using integration by parts on the remainder, beginning with the fundamental theorem for $k=1$. Each integration introduces one further power of $(1-t)$ and one derivative value at the starting point.

Subtracting the constant value $D^kf(a)$ inside the integral and using its continuity yields

$$f(a+h)=\sum_{|\alpha|\le k}\frac{\partial^\alpha f(a)}{\alpha!}h^\alpha+o(\|h\|^k).$$

For a quantitative degree-$k$ error, assume $C^{k+1}$ regularity and $\|D^{k+1}f\|\le M$ on the segment. The integral formula with $k+1$ gives

$$|R_k(h)|\le\frac{M\|h\|^{k+1}}{(k+1)!}.$$

The bound concerns derivatives along the whole segment, not merely derivatives at its starting point.

Here are the actual induction steps. Write $I_k=\int_0^1(1-t)^{k-1}\phi^{(k)}(t)dt/(k-1)!$. The fundamental theorem gives $\phi(1)=\phi(0)+I_1$. If $\phi$ is C(k+1), integration by parts gives

$$I_{k+1}=\left[\frac{(1-t)^k\phi^{(k)}(t)}{k!}\right]_0^1+I_k=-\frac{\phi^{(k)}(0)}{k!}+I_k.$$

Thus replacing $I_k$ adds exactly $\phi^{(k)}(0)/k!$ and leaves $I_{k+1}$; this proves the finite formula by induction. For the degree-k Peano conclusion, assume Ck on a fixed open neighborhood of $a$ and subtract $D^kf(a)$ under the displayed integral. The added constant integral is $D^kf(a)[h,\ldots,h]/k!$, and the remaining error is bounded by

$$\frac{\|h\|^k}{k!}\sup_{0\le t\le1}\|D^kf(a+th)-D^kf(a)\|=o(\|h\|^k).$$

This is an operator-norm continuity argument on a fixed local ball. It does not require C(k+1) for the Peano statement. With C(k+1), integration of $(1-t)^k/k!$ gives the separate quantitative factor $1/(k+1)!$. The multi-index finite-order theorem and its complete integral/Lagrange proof below preserve these distinct regularity assumptions; the scalar Lagrange intermediate point is not a vector-valued mean value assertion.

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.

3 Worked mixed expansion

Let $f(x,y)=e^x\cos y$ and expand at zero. Through total degree 3,

$$T_3=1+x+\frac{x^2-y^2}{2}+\frac{x^3}{6}-\frac{xy^2}{2}.$$

One way to obtain it is to multiply $1+x+x^2/2+x^3/6+\cdots$ by $1-y^2/2+\cdots$ and retain all terms of total degree at most 3. A derivative-based check gives $f_{xyy}(0,0)=-1$; its coefficient is $-1/(1!2!)=-1/2$.

For a concrete remainder bound, restrict along $(tx,ty)$. The fourth derivative is a sum of binomial-weighted terms bounded in absolute value by $e^{|x|}(|x|+|y|)^4$ for $0\le t\le1$. Thus

$$|f(x,y)-T_3(x,y)|\le\frac{e^{|x|}(|x|+|y|)^4}{24}.$$

This bound is deliberately simple rather than sharp. It is valid for simultaneous coordinate changes and reveals the fourth-order error scale.

4 A sufficient analytic criterion

Derivative-growth criterion. Let $n\ge1$, $r_0,C,R>0$, and $f\in C^\infty(B(a,r_0))$ be real scalar-valued. For $j=0$, set $\|D^0f(x)\|=|f(x)|$. Assume for every integer $j\ge0$ and every $x$ in this same ball that $\|D^jf(x)\|\le Cj!R^{-j}$. Then for $\|h\|<\min(r_0,R)$,

$$f(a+h)=\sum_{j=0}^\infty\frac{D^jf(a)[h,\ldots,h]}{j!}.$$

Proof and precise convergence regions. The closed segment is inside the ball. The degree-k integral remainder has bound $C(\|h\|/R)^{k+1}$, tending to zero. A homogeneous term has absolute value at most $C(\|h\|/R)^j$. For every $0<\rho<\min(r_0,R)$ these estimates are uniform for $\|h\|\le\rho$, and the geometric majorant proves absolute and uniform convergence of the series of homogeneous terms there.

This is not yet an assertion that the sum of the absolute values of all individual monomials has that same Euclidean-ball range. Testing $D^jf(a)$ on coordinate basis vectors instead gives $|\partial^\alpha f(a)|\le Cj!R^{-j}$ for $|\alpha|=j$. The multinomial theorem gives the explicit bound

$$\sum_{|\alpha|=j}\left|\frac{\partial^\alpha f(a)}{\alpha!}h^\alpha\right|\le C R^{-j}\sum_{|\alpha|=j}\frac{j!}{\alpha!}\prod_i|h_i|^{\alpha_i}=C\left(\frac{\sum_i|h_i|}{R}\right)^j.$$

Hence the ungrouped multi-index series is absolutely convergent at least when $\|h\|<r_0$ and $\sum_i|h_i|<R$. Grouping its finitely many monomials of each degree recovers the already proved homogeneous equality. To exhibit an actual neighborhood, choose $0<r<\min(r_0/\sqrt n,R/n)$. On the closed cube $|h_i|\le r$, all required bounds hold with a summable geometric majorant. This proves a convergent local power-series representation, hence real analyticity at $a$, without conflating the two convergence regions.

Derivative bounds only at $a$ cannot replace these neighborhood bounds. The flat function from the power-series chapter has all derivatives zero at zero while remaining positive elsewhere. Composing it with the first coordinate gives the same obstruction in every finite dimension; the one-variable zero-extension proof below supplies its actual smoothness justification.

For the following converse, take a positive integer $n$, a center $a\in\mathbb R^n$, $s>0$ and real coefficients $c_\alpha$. Define $f(a+h)=\sum_\alpha c_\alpha h^\alpha$ on $|h_i|<s$ under the stated finite absolute coefficient bound. This defines the function whose regularity is to be proved; $C^\infty$ regularity is not assumed.

Converse: absolute coefficient control permits differentiation. Suppose for some $s>0$,

$$A=\sum_{\alpha\in\mathbb N_0^n}|c_\alpha|s^{|\alpha|}<\infty.$$

Interpret this absolute sum as the supremum of finite subsums, or enumerate multi-indices in increasing total degree. On every closed cube $|h_i|\le r<s$, the original series is uniformly convergent by the M-test. For one partial derivative and $m=|\alpha|\ge1$, the term is zero when $\alpha_j=0$; otherwise, for $0<r<s$,

$$|\alpha_jc_\alpha h^{\alpha-e_j}|\le\frac{|c_\alpha|s^m}{s}\,m(r/s)^{m-1}.$$

The factor $m q^{m-1}$ is uniformly bounded for $q=r/s<1$, for instance by the convergent scalar sum $\sum_{m\ge1}m q^{m-1}=1/(1-q)^2$. Thus the differentiated terms have a summable coefficient majorant and converge uniformly on the cube.

Here is why convergence really gives a derivative. Let $P_N$ be the finite sum of monomials of total degree at most $N$. Fix all coordinates except the jth, and use that coordinate as a variable on $[-r,r]$. Both $P_N$ and $\partial_jP_N$ converge uniformly; the values at coordinate zero converge as well. The one-variable uniform-derivative theorem proves that the limit has that partial derivative. The uniform limit of the derivative polynomials is continuous on the whole cube. Doing this for every coordinate proves C1 regularity on its interior by the continuous-partials differentiability criterion.

For a prescribed higher derivative order $\ell$, the falling-factorial coefficient is at most $m^\ell$, and its terms are bounded on $|h_i|\le r$ by $|c_\alpha|s^m s^{-\ell}m^\ell(r/s)^{m-\ell}$, with zero terms when the relevant exponents are too small. This polynomial-times-geometric factor is uniformly bounded; apply the same argument successively on nested cubes strictly between $r$ and $s$. Every derivative order exists and is continuous. Evaluating at $h=0$ leaves just the matching monomial, proving $\partial^\alpha f(a)=\alpha!c_\alpha$. This argument supplies the termwise rule rather than assuming it from formal notation.

Define

$$f(x)=\begin{cases}e^{-1/x^2},&x\ne0,\\0,&x=0.\end{cases}$$

Here are all steps proving smoothness and failure of analyticity. For $x\ne0$, put $t=1/x$ and define polynomials recursively by

$$P_0(t)=1,\qquad P_{j+1}(t)=2t^3P_j(t)-t^2P_j'(t).$$

Since $dt/dx=-t^2$, the product and chain rules show that the derivative of $P_j(1/x)e^{-1/x^2}$ is $P_{j+1}(1/x)e^{-1/x^2}$. To control the origin, for every fixed power $p\ge0$ and integer $N>p$ the nonnegative exponential series gives $e^u\ge u^N/N!$ for $u>0$, whence $u^pe^{-u}\le N!u^{p-N}\to0$ as $u\to\infty$. Thus every polynomial in $1/x$ times $e^{-1/x^2}$ tends to zero; multiplying by one further $1/x$ still has that property.

Let $g_j(x)=P_j(1/x)e^{-1/x^2}$ off zero and $g_j(0)=0$. Each $g_j$ is continuous at zero. Its difference quotient there is

$$\frac{g_j(x)-g_j(0)}x=\frac1xP_j(1/x)e^{-1/x^2}\longrightarrow0.$$

Away from zero its derivative is $g_{j+1}$; at zero the quotient just proved gives the same derivative value $g_{j+1}(0)=0$. Starting with $g_0=f$, induction proves $f\in C^\infty(\mathbb R)$ and $f^{(j)}(0)=0$ for every $j$. Every Taylor coefficient vanishes, but $f(x)>0$ at every nonzero $x$. No neighborhood representation by its zero Taylor series is possible. The exponential identities and positive-series bound used here are constructed in the elementary-functions chapter independently of this counterexample.

Exercises with complete solutions

E1 Foundation. Find the total-degree-2 Taylor polynomial of $\log(1+x+y)$ at zero and specify a neighborhood where the logarithm is defined.

S1. Put $u=x+y$. The polynomial is $u-u^2/2=x+y-x^2/2-xy-y^2/2$. The function is defined wherever $1+x+y>0$; for example $|x|+|y|<1$ is a suitable neighborhood. The series itself converges there because $|u|<1$.

E2 Mixed terms. Expand $(x+2y)^3$ and express its coefficients using the multi-index rule for the degree-3 Taylor formula.

S2. The expansion is $x^3+6x^2y+12xy^2+8y^3$. At zero the relevant third derivatives are 6,12,24,48, which are divided respectively by $3!$, $2!1!$, $1!2!$, $3!$, giving coefficients 1,6,12,8.

E3 Remainder. If $\|D^3f\|\le12$ on the segment from $a$ to $a+h$, bound the error of its degree-2 Taylor polynomial when $\|h\|=0.1$.

S3. The bound is $12(0.1)^3/3!=0.002$. A pointwise bound at $a$ alone would not authorize the same conclusion.

E4 Analyticity. Explain why $f(x,y)=e^{-1/x^2}$ for $x\ne0$ and zero for $x=0$ is not analytic at $(0,0)$, though every mixed derivative exists and is continuous.

S4. All derivatives involving $y$ are zero; derivatives in $x$ extend continuously by zero from the exponential-decay argument. Thus every Taylor coefficient at $(0,0)$ is zero. Yet values are positive at arbitrarily nearby points with $x\ne0$, so the Taylor series cannot equal the function on any neighborhood. Restricting an alleged analytic representation to $y=0$ would give the same contradiction in one variable.

Summary

Multilinear derivatives organize simultaneous directional changes; multi-indices count the mixed terms correctly. Integral Taylor remainders provide both small-o and quantitative bounds. Analyticity requires an infinite-order neighborhood control or a convergent-series argument, not merely existence of derivatives at one point.