Multivariate calculus · EC3304
Mean value theorem
& Taylor series
Almost every tractable result in economics rests on replacing a complicated function by a simple polynomial approximation near a point. The mean value theorem makes the idea of a "representative slope" precise, and Taylor's theorem turns it into a systematic expansion, first order for linearisation, second order for curvature and optimisation. This page develops both, in one and several variables, and shows how to linearise a whole system of equations.
The mean value theorem
If \(f\) is \(C^1\) and, for two points \(\mathbf y,\mathbf x\), the domain contains the segment \(\lambda\mathbf y+(1-\lambda)\mathbf x\) for all \(\lambda\in(0,1)\), then there exists a point \(\mathbf w\) on that segment such that
$$f(\mathbf y)-f(\mathbf x)=\nabla f(\mathbf w)\cdot(\mathbf y-\mathbf x).$$
The change in the function between two points equals the gradient evaluated at some intermediate point \(\mathbf w\), dotted with the displacement. In one variable this is the familiar statement that \(f(y)-f(x)=f'(w)(y-x)\): somewhere on the interval the instantaneous slope equals the average slope. The theorem is an existence result, it guarantees such a \(\mathbf w\) exists without telling us where, and it is the bridge from derivatives at a point to differences over an interval.
Taylor's theorem in one variable
If \(f:S\subseteq\mathbb{R}\to\mathbb{R}\) is \(C^{k+1}\) and its domain contains the interval between \(y\) and \(a\), then
$$f(y)=f(a)+\frac{f'(a)}{1!}(y-a)+\frac{f''(a)}{2!}(y-a)^2+\cdots+\frac{f^{(k)}(a)}{k!}(y-a)^k+\frac{f^{(k+1)}(c)}{(k+1)!}(y-a)^{k+1}$$
for some \(c\) between \(y\) and \(a\).
The final term is the Lagrange remainder: it collects everything the degree-\(k\) polynomial leaves out. When \(a\) is close to \(y\), the remainder is small and we approximate \(f(y)\) by the polynomial, effectively replacing the unknown \(c\) with \(a\). Truncating at \(k=1\) gives the tangent-line (linear) approximation; at \(k=2\) it adds the curvature correction. Notice the mean value theorem is exactly the \(k=0\) case of Taylor's theorem.
Second-order Taylor expansion in several variables
If \(f\) is \(C^2\) on \(\mathbb{R}^n\) and its domain contains all convex combinations of \(\mathbf y\) and \(\mathbf a\), then
$$f(\mathbf y)=f(\mathbf a)+\sum_{i=1}^{n}\frac{\partial f(\mathbf a)}{\partial y_i}(y_i-a_i)+\frac12\sum_{i=1}^{n}\sum_{j=1}^{n}\frac{\partial^2 f(\mathbf c)}{\partial y_i\,\partial y_j}(y_i-a_i)(y_j-a_j),$$
which in vector–matrix form is
$$f(\mathbf y)=f(\mathbf a)+\nabla f(\mathbf a)\cdot(\mathbf y-\mathbf a)+\tfrac12(\mathbf y-\mathbf a)^{\!\top}H^f(\mathbf c)(\mathbf y-\mathbf a),$$
for some \(\mathbf c\) on the line segment connecting \(\mathbf y\) and \(\mathbf a\).
The three pieces have clear jobs: \(f(\mathbf a)\) is the value at the base point, the gradient term \(\nabla f(\mathbf a)\cdot(\mathbf y-\mathbf a)\) is the linear (first-order) response, and the quadratic form \(\tfrac12(\mathbf y-\mathbf a)^{\top}H^f(\mathbf c)(\mathbf y-\mathbf a)\) is the curvature correction carried by the Hessian. This is precisely why the definiteness of the Hessian classifies critical points: at a critical point the gradient term vanishes and the sign of the quadratic form decides maximum, minimum, or saddle.
Worked example: a second-order approximation
Approximate \(f(x_1,x_2)=(1+x_1^2+x_2^2)^{-1}\) to second order about \(\mathbf a=(1,1)\). First the value: \(f(1,1)=\tfrac13\). The gradient is \(\nabla f=\big(-2x_1(1+x_1^2+x_2^2)^{-2},\,-2x_2(1+x_1^2+x_2^2)^{-2}\big)\); at \((1,1)\), with \(1+1+1=3\), this is
$$\nabla f(1,1)=\left(-\tfrac{2}{9},-\tfrac{2}{9}\right).$$
The Hessian at \((1,1)\) works out to \(H^f(1,1)=\tfrac{1}{27}\left(\begin{smallmatrix}2&8\\8&2\end{smallmatrix}\right)\), so \(\tfrac12 H^f(1,1)=\tfrac{1}{27}\left(\begin{smallmatrix}1&4\\4&1\end{smallmatrix}\right)\). The second-order Taylor approximation is therefore
$$f(x_1,x_2)\approx \tfrac13-\tfrac29(x_1-1)-\tfrac29(x_2-1)+\tfrac{1}{27}\begin{pmatrix}x_1-1\\x_2-1\end{pmatrix}^{\!\top}\!\begin{pmatrix}1&4\\4&1\end{pmatrix}\!\begin{pmatrix}x_1-1\\x_2-1\end{pmatrix}.$$
Because the matrix \(\left(\begin{smallmatrix}1&4\\4&1\end{smallmatrix}\right)\) has determinant \(1-16=-15<0\), it is indefinite: this bump-shaped function is neither concave nor convex at \((1,1)\), a fact the quadratic term reveals directly.
Linearisation of a system of functions
Keeping only the first-order term gives the linear approximation
$$f(\mathbf y)\approx f(\mathbf a)+\nabla f(\mathbf a)\cdot(\mathbf y-\mathbf a).$$
Applied to every equation of a system \(f_1(\mathbf y)=c_1,\dots,f_n(\mathbf y)=c_n\), i.e. \(\mathbf f(\mathbf y)=\mathbf c\), this linearises the whole system around a point \(\mathbf y^*\):
$$f_i(\mathbf y^*)+\nabla f_i(\mathbf y^*)\cdot(\mathbf y-\mathbf y^*)\approx c_i\ \text{ for each }i,\qquad\text{i.e.}\qquad J^f(\mathbf y^*)\,(\mathbf y-\mathbf y^*)\approx \mathbf c-\mathbf f(\mathbf y^*),$$
where \(J^f(\mathbf y^*)\) is the Jacobian of the system at \(\mathbf y^*\). A hard non-linear system has been replaced, locally, by a linear one that can be solved with matrix algebra, the standard first step in analysing equilibria and their stability.
Why approximations matter in economics
First-order Taylor expansions are the basis of log-linearisation, the workhorse technique for solving dynamic macroeconomic models: equilibrium conditions are linearised around a steady state so that the model can be studied with linear methods. Second-order expansions matter wherever curvature carries economic meaning: the Hessian term underpins second-order conditions for utility- and profit-maximisation, and a second-order Taylor expansion of a utility function is exactly how the Arrow–Pratt measure of risk aversion and the certainty-equivalent approximation are derived. Whenever an economist writes "to a first-order approximation" or "up to second order", this is the theorem being invoked.
Check your understanding
Find the second-order Taylor expansion of \(f(x)=\ln(1+x)\) about \(a=0\), and use it to approximate \(\ln(1.1)\).
Here \(f(0)=0\), \(f'(x)=(1+x)^{-1}\) so \(f'(0)=1\), and \(f''(x)=-(1+x)^{-2}\) so \(f''(0)=-1\). The second-order expansion is
$$\ln(1+x)\approx x-\tfrac12 x^2.$$
At \(x=0.1\): \(0.1-\tfrac12(0.01)=0.1-0.005=0.095\). The true value is \(\ln(1.1)\approx0.09531\), so the quadratic approximation is accurate to about three decimal places. (This is also the basis of the common approximation \(\ln(1+x)\approx x\) for small growth rates.)
How does the mean value theorem relate to Taylor's theorem?
The mean value theorem is the zeroth-order case of Taylor's theorem with the Lagrange remainder. Taking \(k=0\) in the univariate expansion gives \(f(y)=f(a)+f'(c)(y-a)\) for some \(c\) between \(a\) and \(y\), which rearranges to \(f(y)-f(a)=f'(c)(y-a)\), exactly the mean value theorem. Each higher-order Taylor expansion generalises the same idea: a difference in function values is captured exactly by a derivative evaluated at an intermediate point.
At a critical point \(\mathbf a\) of a \(C^2\) function, what does the second-order Taylor expansion tell you about whether \(\mathbf a\) is a maximum or minimum?
At a critical point the gradient vanishes, \(\nabla f(\mathbf a)=\mathbf 0\), so the expansion reduces to
$$f(\mathbf y)-f(\mathbf a)\approx\tfrac12(\mathbf y-\mathbf a)^{\top}H^f(\mathbf a)(\mathbf y-\mathbf a).$$
The sign of the right-hand side is governed by the definiteness of the Hessian. If \(H^f(\mathbf a)\) is negative definite the difference is negative for all nearby \(\mathbf y\), so \(\mathbf a\) is a local maximum; if positive definite, a local minimum; if indefinite, a saddle point. This is the second-order condition of optimisation, read straight off the Taylor expansion.
Download the full lecture slides
These notes cover the mean value theorem and Taylor series strand of the multivariate calculus material. The full deck also covers the implicit function theorem, the Hessian and integration.
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