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Taylor's theorem shows that any function can be approximated in the vicinity of any convenient point by a series of terms involving the function and its derivatives, Here we look at some applications of the theorem for functions of one and two variables. a Any continuous and differentiable function of a single variable, \(f(x),\) can be approximated near the point \(a\) by the formula \\[ \begin{aligned} f(x)=& f(a)+f(a)(x-a)+0.5 f^{\prime \prime}(a)(x-a)^{2}+\\\ &+\text { terms in } f^{\prime \prime \prime}, f^{\prime \prime \prime}, \ldots . \end{aligned} \\] Using only the first three of these terms results in a quadratic Taylor approximation. Use this approximation together with the definition of concavity to show that any concave function must lie on or below the tangent to the function at point \(a\) b. The quadratic Taylor approximation for any function of two variables, \(f(x, y),\) near the point \((a, b)\) is given by \\[ \begin{aligned} f(x, y)=& f(a, b)+f_{1}(a, b)(x-a)+f_{2}(a, b)(y-b) \\ &+0.5\left[f_{11}(a, b)(x-a)^{2}\right.\\\ &\left.+2 f_{12}(a, b)(x-a)(y-b)+f_{22}(y-b)^{2}\right] \end{aligned} \\] Use this approximation to show that any concave function (as defined by Equation 2.84 ) must lie on or below its tangent plane at \((a, b)\)

Short Answer

Expert verified
Question: Show that any concave function lies on or below its tangent line for a single-variable function and its tangent plane for a two-variable function. Answer: Based on our analysis, for both single-variable and two-variable functions, due to the properties of concavity, all terms containing the second derivative for single-variable or the Hessian matrix for two-variable functions resulted in values less than or equal to zero. Consequently, the actual function will be less than or equal to its quadratic Taylor approximation. Since the quadratic Taylor approximation includes the tangent line or tangent plane equation, it means that the concave function lies on or below its tangent line or tangent plane, respectively.

Step by step solution

01

Write down the quadratic Taylor approximation formula for f(x)

The quadratic Taylor approximation for f(x) near a point a is \\[f(x)=f(a)+f'(a)(x-a)+0.5f''(a)(x-a)^2+[\text{higher-order terms}]\\]
02

Define what it means for a function to be concave

A function is concave if its second derivative \(f''(x)\) is less than or equal to zero for all x, that is, \(f''(x) \le 0\).
03

Compare the approximation with the tangent to f(x) at point a

The tangent to the function f(x) at point a is given by the equation \\[y=f(a)+f'(a)(x-a)\\] Now we need to compare the values of the actual function \(f(x)\) and its quadratic Taylor approximation. For a concave function, we already know that \(f''(x) \le 0\). Hence, all the terms containing the second derivative will either be zero or negative, so the value of actual function \(f(x)\) will be less than or equal to the value of its quadratic Taylor approximation near point a. Since the quadratic Taylor approximation includes the tangent line equation, it means that the actual function \(f(x)\) lies on or below its tangent at point a. b) Two-variable function:
04

Write down the quadratic Taylor approximation formula for f(x, y)

The quadratic Taylor approximation for f(x, y) near point (a, b) is \\[\begin{aligned}f(x, y)=& f(a, b)+f_{1}(a, b)(x-a)+f_{2}(a, b)(y-b) \\ &+0.5\left[f_{11}(a, b)(x-a)^{2}+2 f_{12}(a, b)(x-a)(y-b)+f_{22}(y-b)^{2}\right] \end{aligned}\\]
05

Define the concavity for a function of two variables

A function is concave in two variables if its Hessian matrix is negative semi-definite, that is, the matrix \\[\begin{bmatrix}f_{11}(x, y) & f_{12}(x, y) \\ f_{21}(x, y) & f_{22}(x, y) \end{bmatrix} \\] satisfies the conditions: \\[f_{11}(x, y) \le 0, \quad f_{11}(x, y)f_{22}(x, y)-f_{12}(x, y)f_{21}(x, y) \ge 0\\]
06

Compare the approximation with the tangent plane to f(x, y) at point (a, b)

The tangent plane to the function f(x, y) at point (a, b) is given by the equation \\[z=f(a, b)+f_{1}(a, b)(x-a)+f_{2}(a, b)(y-b)\\] Now we need to compare the values of the actual function \(f(x, y)\) and its quadratic Taylor approximation. For a concave function in two variables, we already know that the Hessian matrix is negative semi-definite, which implies that the terms containing the second-order partial derivatives of the function will result in values less than or equal to zero. Similar to the single-variable case, since the quadratic Taylor approximation includes the tangent plane equation, it means that the actual function \(f(x, y)\) lies on or below its tangent plane at the point \((a, b)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Taylor Approximation
Quadratic Taylor approximation is a powerful tool in microeconomics that helps to understand how a function behaves near a specific point. Imagine you are on a hike, and you want to predict the path ahead based on your immediate surroundings. Just like this, the quadratic Taylor approximation predicts the behavior of functions based on the information around a point called a.

The formula looks like this:
\[ f(x) = f(a) + f'(a)(x-a) + 0.5f''(a)(x-a)^2+\text{[higher-order terms]}\]

In essence, this is creating a parabolic shape that best fits the curve of the original function at point a, using the function's value, slope, and curvature (second derivative). The beauty of this approximation is its simplicity and usefulness in analysis, allowing us to create a 'best guess' scenario of a function's behavior in a local area without complex calculations.
Concave Functions
A concave function is shaped like an upside-down bowl, think of a satellite dish or a skateboard half-pipe. This shape has a critical property that if you pick any two points on the curve and connect them with a straight line, the line will always lie above the curve. In mathematical terms, a function is concave if its second derivative is less than or equal to zero across its domain:\[ f''(x) \le 0 \].

This characteristic is particularly meaningful in microeconomics, reflecting decreasing marginal returns or preferences with diminishing marginal utility. For example, the more you consume a good, the less additional utility you get from each new unit consumed. When we map this to a graph, we get our concave function, showcasing at each point along the graph how utility changes with consumption.
Hessian Matrix Negative Semi-definite
In the realm of functions with two variables, concavity is detected through the Hessian matrix. The Hessian is a square matrix of second-order partial derivatives of a function. It's like our multidimensional satellite dish curvature test. If a function is concave, the Hessian matrix is negative semi-definite. This sounds complex, but it simply means that for any vector v, when multiplied by the Hessian, the resulting value is non-positive. Here’s the Hessian for a two-variable function:\[\begin{bmatrix}f_{11}(x, y) & f_{12}(x, y) \ f_{21}(x, y) & f_{22}(x, y)\end{bmatrix}\].

The conditions \(f_{11}(x, y) \le 0\), and \(f_{11}(x, y)f_{22}(x, y) - f_{12}(x, y)f_{21}(x, y) \ge 0\) must hold for the function to be concave. In other words, this measure tells us about the 'bowl shape' of the graph. Applying this in an economic context helps to determine the nature of returns to scale in production or the level of risk aversion in utility functions.

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Most popular questions from this chapter

Here are a few useful relationships related to the covariance of two random variables, \(x_{1}\) and \(x_{2}\) a. \(S h o w \quad t h\) at \(\quad \operatorname{Cov}\left(x_{1}, x_{2}\right)=E\left(x_{1} x_{2}\right)-E\left(x_{1}\right) E\left(x_{2}\right)\) An important implication of this is that if \(\operatorname{Cov}\left(x_{1}, x_{2}\right)=0, E\left(x_{1} x_{2}\right)=E\left(x_{1}\right) E\left(x_{2}\right) .\) That is, the expected value of a product of two random variables is the product of these variables' expected values. b. Show that \(\operatorname{Var}\left(a x_{1}+b x_{2}\right)=a^{2} \operatorname{Var}\left(x_{1}\right)+b^{2} \operatorname{Var}\left(x_{2}\right)+\) \\[ 2 a b \operatorname{Cov}\left(x_{1}, x_{2}\right) \\] c. In Problem \(2.15 \mathrm{d}\) we looked at the variance of \(X=k x_{1}+(1-k) x_{2} 0 \leq k \leq 1 .\) Is the conclusion that this variance is minimized for \(k=0.5\) changed by considering cases where \(\operatorname{Cov}\left(x_{1}, x_{2}\right) \neq 0 ?\) d. The corrdation cocfficicnt between two random variables is defined as \\[ \operatorname{Corr}\left(x_{1}, x_{2}\right)=\frac{\operatorname{Cov}\left(x_{1}, x_{2}\right)}{\sqrt{\operatorname{Var}\left(x_{1}\right) \operatorname{Var}\left(x_{2}\right)}} \\] Explain why \(-1 \leq \operatorname{Corr}\left(x_{1}, x_{2}\right) \leq 1\) and provide some intuition for this result. e. Suppose that the random variable \(y\) is related to the random variable \(x\) by the linear equation \(y=\alpha+\beta x\). Show that \\[ \beta=\frac{\operatorname{Cov}(y, x)}{\operatorname{Var}(x)} \\] Here \(\beta\) is sometimes called the (theoretical) regression coefficient of \(y\) on \(x\). With actual data, the sample analog of this expression is the ordinary least squares (OLS) regression coefficient.

The height of a ball that is thrown straight up with a certain force is a function of the time ( \(t\) ) from which it is released given by \(f(t)=-0.5 g t^{2}+40 t\) (where \(g\) is a constant determined by gravity). a. How does the value of \(t\) at which the height of the ball is at a maximum depend on the parameter \(g\) ? b. Use your answer to part (a) to describe how maximum height changes as the parameter \(g\) changes. c. Use the envelope theorem to answer part (b) directly. d. On the Earth \(g=32,\) but this value varies somewhat around the globe. If two locations had gravitational constants that differed by \(0.1,\) what would be the difference in the maximum height of a ball tossed in the two places?

Suppose a firm's total revenues depend on the amount produced \((q)\) according to the function \\[ R=70 q-q^{2} \\] Total costs also depend on \(q\) \\[ C=q^{2}+30 q+100 \\] a. What level of output should the firm produce to maximize profits \((R-C) ?\) What will profits be? b. Show that the second-order conditions for a maximum are satisfied at the output level found in part (a). c. Does the solution calculated here obey the "marginal revenue equals marginal cost" rule? Explain.

Suppose that a firm has a marginal cost function given by \(M C(q)=q+1\) a. What is this firm's total cost function? Explain why total costs are known only up to a constant of integration, which represents fixed costs. b. As you may know from an earlier economics course, if a firm takes price \((p)\) as given in its decisions, then it will produce that output for which \(p=M C(q)\). If the firm follows this profit-maximizing rule, how much will it produce when \(p=15 ?\) Assuming that the firm is just breaking even at this price, what are fixed costs? c. How much will profits for this firm increase if price increases to \(20 ?\) d. Show that, if we continue to assume profit maximization, then this firm's profits can be expressed solely as a function of the price it receives for its output. e. Show that the increase in profits from \(p=15\) to \(p=20\) can be calculated in two ways: (i) directly from the equation derived in part (d) and (ii) by integrating the inverse marginal cost function \(\left[M C^{-1}(p)=p-1\right]\) from \(p=15\) to \(p=20 .\) Explain this result intuitively using the envelope theorem.

Because we use the envelope theorem in constrained optimization problems often in the text, proving this theorem in a simple case may help develop some intuition. Thus, suppose we wish to maximize a function of two variables and that the value of this function also depends on a parameter, \(a: f\left(x_{1}, x_{2}, a\right) .\) This maximization problem is subject to a constraint that can be written as: \(g\left(x_{1}, x_{2}, a\right)=0\) a. Write out the Lagrangian expression and the first-order conditions for this problem. b. Sum the two first-order conditions involving the \(x\) 's. c. Now differentiate the above sum with respect to \(a-\) this shows how the \(x\) 's must change as a changes while requiring that the first-order conditions continue to hold. A. As we showed in the chapter, both the objective function and the constraint in this problem can be stated as functions of \(a: f\left(x_{1}(a), x_{2}(a), a\right), g\left(x_{1}(a), x_{2}(a), a\right)=0 .\) Dif ferentiate the first of these with respect to \(a\). This shows how the value of the objective changes as \(a\) changes while kecping the \(x^{\prime}\) s at their optimal values. You should have terms that involve the \(x\) 's and a single term in \partialflaa. e. Now differentiate the constraint as formulated in part (d) with respect to \(a\). You should have terms in the \(x^{\prime}\) 's and a single term in \(\partial g / \partial a\) f. Multiply the results from part (e) by \(\lambda\) (the Lagrange multiplier), and use this together with the first-order conditions from part (c) to substitute into the derivative from part (d). You should be able to show that \\[ \frac{d f\left(x_{1}(a), x_{2}(a), a\right)}{d a}=\frac{\partial f}{\partial a}+\lambda \frac{\partial g}{\partial a^{\prime}} \\] which is just the partial derivative of the Lagrangian expression when all the \(x^{\prime}\) 's are at their optimal values. This proves the enyclope theorem. Explain intuitively how the various parts of this proof impose the condition that the \(x\) 's are constantly being adjusted to be at their optimal values. 8\. Return to Example 2.8 and explain how the cnvelope theorem can be applied to changes in the fence perimeter \(P-\) that is, how do changes in \(P\) affect the size of the area that can be fenced? Show that, in this case, the envelope theorem illustrates how the Lagrange multiplier puts a value on the constraint.

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