Deriving the Quadratic Taylor Formula

From a Line to a Two-Variable Formula

A quadratic Taylor polynomial for a function of two variables records its value, slopes, and curvature at a chosen point. Its terms can look like a collection of coefficients to memorize. A useful way to derive them is to examine the function along a single line through that point.

On a line, a function of two variables becomes a function of one variable. Its first derivative records the change in the direction of the line, and its second derivative records the curvature in that direction. Because the line can point in any direction, these one-variable derivatives determine all the quadratic terms.

We will derive the formula at a point \((a,b)\). Write a nearby point as \((a+h,b+k)\), where \(h\) and \(k\) are its horizontal and vertical displacements. Assume that \(f\) has continuous second partial derivatives on an open neighborhood of \((a,b)\). This condition ensures that the line restrictions used below are twice differentiable and that the mixed partial derivatives agree.

Restricting the Function to a Line

Fix a displacement \((h,k)\), and consider the line segment from \((a,b)\) to \((a+h,b+k)\). A parameter \(t\) runs from \(0\) to \(1\) along this segment. Define the one-variable function

At \(t=0\), the line is at the base point, so \(g(0)=f(a,b)\). At \(t=1\), it reaches the point being approximated, so \(g(1)=f(a+h,b+k)\). The chain rule gives the derivatives of \(g\).

Differentiating again gives

Since the second partial derivatives are continuous, \(f_{xy}=f_{yx}\) near the base point. Combining the two mixed terms therefore yields

In particular, the first two derivatives at \(t=0\) are

and

These quantities give the linear and quadratic contributions to the approximation. The next step explains why the second derivative enters with a factor of one-half.

Deriving the Quadratic Taylor Formula

For a twice continuously differentiable one-variable function, the fundamental theorem of calculus, applied twice, gives

To see the identity, first write \(g(1)=g(0)+\int_0^1g'(s)\,ds\). For each \(s\), write \(g'(s)=g'(0)+\int_0^s g''(t)\,dt\). Substitution and reversing the order of integration over \(0\leq t\leq s\leq1\) give \(g(1)=g(0)+g'(0)+\int_0^1(1-t)g''(t)\,dt\).

Separate the constant value \(g''(0)\) from the integrand. Since \(\int_0^1(1-t)\,dt=1/2\), this gives

As the displacement \((h,k)\) approaches \((0,0)\), continuity of the second partial derivatives makes the final integral small compared with \(h^2+k^2\). Indeed, the difference \(g''(t)-g''(0)\) is a sum of differences in second partial derivatives, multiplied by \(h^2\), \(hk\), or \(k^2\). Along the whole segment, those differences tend uniformly to zero as the displacement tends to zero. Also, \(|hk|\leq(h^2+k^2)/2\). Thus the integral, divided by \(h^2+k^2\), tends to zero.

Here \(o(h^2+k^2)\) means a remainder whose ratio to \(h^2+k^2\) tends to zero as \((h,k)\to(0,0)\), with \((h,k)\ne(0,0)\). Substituting the expressions for \(g(0)\), \(g'(0)\), and \(g''(0)\) into the one-variable identity produces the displayed formula. In particular, the mixed term has coefficient \(f_{xy}(a,b)\): the two equal mixed contributions in \(g''(0)\) combine, and the factor \(1/2\) then cancels their factor of \(2\).

Worked Examples

A Common Mixed-Term Error

A frequent mistake is to include the mixed partial derivative with a factor of one-half. The error often comes from seeing the \(\frac12g''(0)\) in the one-variable formula but forgetting that \(g''(0)\) contains two mixed contributions.

When the Formula Applies

The continuity assumption on second partial derivatives is important: it ensures that the line restriction has the required second-order behavior as the displacement shrinks. The polynomial can still be computed from derivative values in other situations, but those values alone do not guarantee that the function has the stated remainder behavior. Do not infer the approximation just from the existence of first and second partial derivatives at one point.

The line method also clarifies what the polynomial captures. The constant term gives the function value at the base point; the linear terms record the initial change along every direction; and the quadratic terms record the initial curvature along each direction. In applications, this gives a local approximation that can retain curvature information the tangent-plane approximation omits. The polynomial is an approximation in general, not an identity: equality throughout a neighborhood occurs for functions that have no higher-order variation there, such as the quadratic function in the first example.