Proof Strategy for Multivariable Taylor

A Proof Strategy Built Around the Segment

The Multivariable Taylor Theorem with Lagrange Remainder and the Taylor Formula with Peano Remainder provide useful conclusions, but it is valuable to see another form of the remainder and the proof strategy that produces it. The central move is to restrict the function to the line segment from the point of approximation to the point of evaluation. The resulting one-variable function has derivatives that record the multivariable derivatives in the displacement direction. Repeated use of the Fundamental Theorem of Calculus then gives an exact integral expression for the error.

Throughout, let $U\subseteq\mathbb{R}^m$ be open, let $a\in U$, and choose $h\in\mathbb{R}^m$ so that the segment $\{a+th:0\leq t\leq1\}$ lies in $U$. Suppose $f:U\to\mathbb{R}$ has continuous derivatives through order $r+1$ on a neighborhood of that segment, where $r\geq0$. As in Multivariable Taylor’s Theorem, set $g(t)=f(a+th)$. Repeated application of the chain rule gives $g^{(k)}(t)=D^k f(a+th)[h,\ldots,h]$. Thus the ordinary one-variable Taylor polynomial of $g$ at zero is exactly the multivariable Taylor polynomial evaluated at $a+h$.

Theorem (Multivariable Taylor Formula with Integral Remainder): Under the assumptions above, $$ f(a+h)=\sum_{k=0}^{r}\frac{1}{k!}D^k f(a)[h,\ldots,h] +\frac{1}{r!}\int_0^1(1-t)^rD^{r+1}f(a+th)[h,\ldots,h]\,dt. $$ For $r=0$, the sum is just $f(a)$, $0!=1$, and the integral weight is the constant function $1$.

Deriving the Integral Remainder

Proof. We first prove the corresponding one-variable formula. Let $g$ have continuous derivatives through order $r+1$ on $[0,1]$, and define $$ I_j=\frac{1}{j!}\int_0^1(1-t)^j g^{(j+1)}(t)\,dt $$ for each integer $j\geq0$. For $r=0$, the Fundamental Theorem of Calculus gives $g(1)=g(0)+\int_0^1g'(t)\,dt=g(0)+I_0$. This is the desired formula in the base case.

For an integer $j\geq1$, integration by parts, with $u=(1-t)^j/j!$ and $dv=g^{(j+1)}(t)\,dt$, gives $$ I_j=-\frac{g^{(j)}(0)}{j!}+I_{j-1}. $$ Indeed, the boundary term at $t=1$ is zero, while the boundary term at $t=0$ is $-g^{(j)}(0)/j!$. The remaining integral is $I_{j-1}$, since $du=-(1-t)^{j-1}dt/(j-1)!$. Equivalently, $I_{j-1}=g^{(j)}(0)/j!+I_j$.

Starting from the base case and applying this identity successively shows $g(1)=\sum_{k=0}^{r}g^{(k)}(0)/k!+I_r$. For clarity, if the formula holds through order $j-1$, then $g(1)=\sum_{k=0}^{j-1}g^{(k)}(0)/k!+I_{j-1}$. Substituting $I_{j-1}=g^{(j)}(0)/j!+I_j$ gives the formula through order $j$. This proves the one-variable result for every nonnegative integer $r$. Finally, apply it to $g(t)=f(a+th)$ and replace each $g^{(k)}(t)$ with $D^k f(a+th)[h,\ldots,h]$. At $t=0$, this gives the Taylor terms at $a$; at $t=1$, it gives $f(a+h)$. The stated multivariable formula follows. $\square$

This proof strategy has two distinct parts. First, the segment restriction converts the problem into one variable. Second, integration by parts moves information from the derivative inside the integral into a Taylor coefficient at the starting point. The remainder is what remains after those coefficients have been extracted.

Choose the displacement.
Write the point of evaluation as $a+h$ and verify that the entire segment from $a$ to $a+h$ lies in the domain.

Restrict to one variable.
Define $g(t)=f(a+th)$; the chain rule identifies $g^{(k)}(t)$ with the $k$th derivative of $f$ applied repeatedly to $h$.

Extract Taylor coefficients.
Start with the Fundamental Theorem of Calculus and use integration by parts to move successive derivatives into coefficients at $t=0$.

Estimate what remains.
Bound the derivative inside the integral and integrate the nonnegative weight to obtain an error bound.

Worked Example: The Degree-Zero Formula

Consider $f(x,y)=x^2+3y$, $a=(1,-1)$, and $h=(u,v)$. Take $r=0$. The segment function is $g(t)=f(1+tu,-1+tv)$, so direct substitution gives $$ g(t)=(1+tu)^2+3(-1+tv)=-2+(2u+3v)t+u^2t^2. $$ Consequently $g'(t)=2u+3v+2tu^2$. The integral remainder formula says that $f(a+h)-f(a)=\int_0^1g'(t)\,dt$, and evaluating this integral gives $$ \int_0^1(2u+3v+2tu^2)\,dt=2u+3v+u^2. $$ Directly, $f(1+u,-1+v)=-2+2u+u^2+3v$ and $f(1,-1)=-2$, so their difference is also $2u+3v+u^2$. This verifies the formula in the order-zero case, where the weight is $1$ throughout the interval.

Worked Example: A Cubic Polynomial and an Exact Remainder

Let $f(x,y)=x^3+xy^2$, take $a=(0,0)$, and write $h=(u,v)$. Along the segment, $g(t)=f(tu,tv)=t^3(u^3+uv^2)$. The Taylor polynomial of degree two at the origin is zero: $g(0)=g'(0)=g''(0)=0$. Also, $g'''(t)=6(u^3+uv^2)$. The integral remainder for $r=2$ is therefore $$ \frac{1}{2}\int_0^1(1-t)^2\,6(u^3+uv^2)\,dt =3(u^3+uv^2)\int_0^1(1-t)^2\,dt =u^3+uv^2. $$ Here $\int_0^1(1-t)^2dt=1/3$, as follows by integrating $1-2t+t^2$. Direct substitution gives $f(u,v)=u^3+uv^2$, so the integral remainder recovers the function exactly.

From an Integral to an Error Bound

The integral formula also makes quantitative estimates direct. The derivative in the integrand is evaluated at points on the segment, but the repeated argument $h$ is fixed. If the $(r+1)$st derivative is bounded there, multilinearity supplies a factor of $\|h\|_2^{r+1}$. The remaining integral is a scalar weight whose value can be computed exactly.

Theorem (Uniform Derivative Bound for the Taylor Remainder): Under the hypotheses of the Multivariable Taylor Formula with Integral Remainder, suppose that $$ \bigl|D^{r+1}f(a+th)[v_1,\ldots,v_{r+1}]\bigr| \leq M\prod_{i=1}^{r+1}\|v_i\|_2 $$ for every $t\in[0,1]$ and all $v_1,\ldots,v_{r+1}\in\mathbb{R}^m$. Then $$ \left|f(a+h)-\sum_{k=0}^{r}\frac{1}{k!}D^k f(a)[h,\ldots,h]\right| \leq \frac{M}{(r+1)!}\|h\|_2^{r+1}. $$

Proof. Denote the difference inside the absolute value by $R_r(h)$. The integral remainder formula and the assumed bound, applied with every $v_i=h$, give $$ |R_r(h)|\leq \frac{1}{r!}\int_0^1(1-t)^r M\|h\|_2^{r+1}\,dt. $$ The weight is nonnegative on $[0,1]$, and $\int_0^1(1-t)^r\,dt=1/(r+1)$. Thus $$ |R_r(h)|\leq \frac{M}{r!(r+1)}\|h\|_2^{r+1} =\frac{M}{(r+1)!}\|h\|_2^{r+1}. $$ This includes $r=0$: the integral of the constant weight $1$ is $1$, and the bound becomes $M\|h\|_2$. $\square$

Worked Example: A Trigonometric Error Bound

Let $f(x,y)=\sin(x-2y)$ at the origin, and put $h=(u,v)$ and $s=u-2v$. The segment restriction is $g(t)=\sin(ts)$. Its first three derivatives are $g'(t)=s\cos(ts)$, $g''(t)=-s^2\sin(ts)$, and $g'''(t)=-s^3\cos(ts)$. Since $g(0)=0$, $g'(0)=s$, and $g''(0)=0$, the degree-two Taylor polynomial is $s$. The integral remainder gives $$ \sin(s)-s=-\frac{s^3}{2}\int_0^1(1-t)^2\cos(ts)\,dt. $$ Because $|\cos(ts)|\leq1$, its absolute value is at most $\frac{|s|^3}{2}\int_0^1(1-t)^2dt=|s|^3/6$. Finally, Cauchy–Schwarz gives $|s|=|(u,v)\cdot(1,-2)|\leq\sqrt{5}\,\|(u,v)\|_2$. Hence $$ |\sin(u-2v)-(u-2v)|\leq\frac{5^{3/2}}{6}\|(u,v)\|_2^3. $$ The estimate holds for every $u,v$, because the bound on cosine holds along the entire segment for every real $s$.

Worked Example: A Rational Function on Its Domain

Let $f(x,y)=1/(1+x+y)$ at $a=(0,0)$, and set $s=u+v$. If $|s|<1$, then $1+ts>0$ for every $t\in[0,1]$, so the segment lies in the domain. The restriction is $g(t)=1/(1+ts)$, with $g(0)=1$, $g'(0)=-s$, and $g''(t)=2s^2/(1+ts)^3$. For $r=1$, the formula gives $$ \frac{1}{1+s}=1-s+\int_0^1(1-t)\frac{2s^2}{(1+ts)^3}\,dt. $$ For $s\neq0$, substitute $q=1+ts$. Then $dt=dq/s$ and $1-t=(1+s-q)/s$, so the integral becomes $$ 2\int_1^{1+s}\frac{1+s-q}{q^3}\,dq =\left[-\frac{1+s}{q^2}+\frac{2}{q}\right]_{1}^{1+s} =\frac{1}{1+s}-(1-s) =\frac{s^2}{1+s}. $$ The same value holds when $s=0$, since both the integral and $s^2/(1+s)$ are zero. Thus the remainder after the linear Taylor polynomial $1-s$ is exactly $s^2/(1+s)$.

Reading the Weight and Avoiding a Common Pitfall

The weight $(1-t)^r$ is nonnegative, and its integral is $1/(r+1)$. For $r\geq1$, it decreases from its maximum at the starting point $t=0$ to zero at the endpoint $t=1$. For $r=0$, it is constant and equal to $1$ throughout $[0,1]$; it does not decrease to zero. Keeping this edge case in view prevents an incorrect geometric interpretation of the weight.

The integral remainder is an average, with a specific weight, of directional derivatives along the segment. In general, it should not be replaced by the derivative at a single point unless a separate result justifies that step. The Multivariable Taylor Theorem with Lagrange Remainder supplies a different exact representation under its hypotheses; the integral formula does not itself say that its integrand is constant or that the remainder is evaluated at one chosen point. What the integral representation gives especially clearly is a route to estimates: control the derivative everywhere on the segment, then integrate the weight.

When applying the strategy, check the domain before differentiating the segment function, identify the Taylor coefficients at $t=0$, and only then estimate the integral remainder. The segment condition is essential: knowing that $a$ and $a+h$ lie in the domain does not by itself guarantee that every intermediate point does. Once the segment and regularity hypotheses are verified, the proof is systematic and works for every order, including degree zero.

Check Your Understanding

Use the formula and proof strategy in this tutorial to answer the following questions.

Why does the restriction $g(t)=f(a+th)$ encode the multivariable derivatives in the direction $h$?
In the integration-by-parts step, how is $I_{j-1}$ related to $I_j$ and $g^{(j)}(0)$?
What is the value of the remainder weight when $r=0$, and how does that case enter the error-bound theorem?
If the $(r+1)$st derivative is bounded by $M$ on the segment, why does the Taylor error contain the factor $1/(r+1)!$?
Why can the integral remainder not automatically be replaced by a derivative evaluated at one intermediate point?