Exercise Sheet 3 Solutions

1.

(a)

Let

\[ f : \mathbb{R}^2 \to \mathbb{R}, \quad f(x, y) = 9x^2 - y^3 + 9xy \]

We are asked to compute the Hessian matrix at the point \( (x, y) = (3, -3) \).

Step 1: Compute second-order partial derivatives

To compute the Hessian matrix, we first compute the first-order partial derivatives:

\[ \frac{\partial f}{\partial x} = 18x + 9y, \quad \frac{\partial f}{\partial y} = -3y^2 + 9x \]

Then we compute the second-order partial derivatives:

\[ \frac{\partial^2 f}{\partial x^2} = 18, \quad \frac{\partial^2 f}{\partial y^2} = -6y, \quad \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = 9 \]

At the point \( (3, -3) \), we evaluate:

\[ \frac{\partial^2 f}{\partial x^2}(3, -3) = 18, \quad \frac{\partial^2 f}{\partial y^2}(3, -3) = -6(-3) = 18, \quad \frac{\partial^2 f}{\partial x \partial y}(3, -3) = 9 \]

Step 2: Form the Hessian matrix

\[\begin{split} H_{(3, -3)} = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix} = \begin{bmatrix} 18 & 9 \\ 9 & 18 \end{bmatrix} \end{split}\]

(b)

We recall the following definitions and propositions:

• Definition: A symmetric matrix \( A \) is positive definite if for all non-zero vectors \( a \), we have \( a^T A a > 0 \).

• Proposition: A symmetric matrix is positive definite if and only if all its eigenvalues are positive.

To compute the eigenvalues, solve the characteristic equation:

\[\begin{split} \det(H - \lambda I) = 0 \Rightarrow \begin{vmatrix} 18 - \lambda & 9 \\ 9 & 18 - \lambda \end{vmatrix} = (18 - \lambda)^2 - 81 = 0 \end{split}\]

Simplifying:

\[ (18 - \lambda)^2 = 81 \Rightarrow 18 - \lambda = \pm 9 \Rightarrow \lambda = 9, \ 27 \]

Since both eigenvalues are positive, the Hessian matrix at the point \( (3, -3) \) is positive definite. Therefore, \( f(x, y) \) has a local minimum at this point.

2.

Let \( f(x) = x \cdot \ln(x) \) defined on the interval \( [1, e^2] \).

(a)

To apply the Mean Value Theorem (MVT), we must verify that:

• \( f \) is continuous on \( [1, e^2] \)

• \( f \) is differentiable on \( (1, e^2) \)

Since \( f(x) = x \ln(x) \) is a product of continuous and differentiable functions for \( x > 0 \), both conditions are satisfied.

(b)

We compute:

\[ f(e^2) = e^2 \cdot \ln(e^2) = e^2 \cdot 2 = 2e^2 \]

\[ f(1) = 1 \cdot \ln(1) = 0 \]

Hence, the average rate of change is:

\[ \frac{f(e^2) - f(1)}{e^2 - 1} = \frac{2e^2}{e^2 - 1} \]

Next, compute the derivative:

\[ f'(x) = \frac{d}{dx}[x \ln(x)] = \ln(x) + 1 \]

We solve:

\[ f'(c) = \ln(c) + 1 = \frac{2e^2}{e^2 - 1} \Rightarrow \ln(c) = \frac{2e^2}{e^2 - 1} - 1 = \frac{e^2 + 1}{e^2 - 1} \Rightarrow c = \exp\left( \frac{e^2 + 1}{e^2 - 1} \right) \]

(c)

The Mean Value Theorem states that there exists a point \( c \in (1, e^2) \) where the instantaneous rate of change \( f'(c) \) equals the average rate of change over the interval:

\[ f'(c) = \frac{f(e^2) - f(1)}{e^2 - 1} \]

Geometrically, this means the tangent line to the curve at \( x = c \) is parallel to the secant line connecting the endpoints \( (1, f(1)) \) and \( (e^2, f(e^2)) \).

3.

(a)

We compute the derivatives at \( x = 0 \):

• \( f(x) = \arctan(x) \)

• \( f'(x) = \frac{1}{1+x^2} \Rightarrow f'(0) = 1 \)

• \( f''(x) = \frac{-2x}{(1+x^2)^2} \Rightarrow f''(0) = 0 \)

• \( f^{(3)}(x) = \frac{6x^2 - 2}{(1 + x^2)^3} \Rightarrow f^{(3)}(0) = -2 \)

Hence, the third-degree Taylor polynomial is:

\[ P_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 = 0 + x + 0 - \frac{2}{6}x^3 = x - \frac{1}{3}x^3 \]

(b)

The Lagrange remainder is:

\[ R_3(x) = \frac{f^{(4)}(c)}{4!} x^4 = \frac{f^{(4)}(c)}{24} x^4 \quad \text{for some } c \in (0, x) \]

We previously found:

\[ f^{(4)}(c) = \frac{24c(1 - c^2)}{(1 + c^2)^4} \]

So,

\[ R_3(x) = \frac{c(1 - c^2)}{(1 + c^2)^4} x^4 \quad \text{for some } c \in (0, x) \]

(c)

Our goal is to show:

\[ |R_3(x)| = \left| \frac{c(1 - c^2)}{(1 + c^2)^4} x^4 \right| \leq \frac{x^4}{4(1 + c^2)^2} \quad \text{for } c \in (0, 1) \]

We simplify the absolute value of the remainder:

\[ |R_3(x)| = \frac{c(1 - c^2)}{(1 + c^2)^4} x^4 \]

So we now want to prove that:

\[ \frac{c(1 - c^2)}{(1 + c^2)^4} \leq \frac{1}{4(1 + c^2)^2} \]

Now we multiply both sides by \( (1 + c^2)^4 \) (which is strictly positive):

\[ 4c(1 - c^2) \leq (1 + c^2)^2 \]

Now expand both sides:

Left-hand side:

\[ 4c(1 - c^2) = 4c - 4c^3 \]

Right-hand side:

\[ (1 + c^2)^2 = 1 + 2c^2 + c^4 \]

So we want to show:

\[ 4c - 4c^3 \leq 1 + 2c^2 + c^4 \quad \Leftrightarrow \quad c^4 + 4c^3 + 2c^2 - 4c + 1 \geq 0 \]

Now factor the left-hand side:

\[ c^4 + 4c^3 + 2c^2 - 4c + 1 = (c^2 + 2c - 1)^2 \geq 0 \]

This inequality clearly holds for all c including \( c \in (0, 1) \), so the bound is valid.