Question

Show that the function $$ \phi(\mathbf{x})=x_{1}^{2}-x_{2}^{4}+1 $$ has a saddle point at the origin, i.e., the origin is a critical point that is neither a minimum nor ? maximum. What happens if Newton's method is applied to find this saddle point?

Short answer

If so, can Newton's method be used to find this saddle point? Answer: Yes, the origin (0,0) is a saddle point for the given function. However, due to the determinant of the Hessian matrix at the origin being 0, Newton's method cannot be directly applied to find this saddle point.

Step-by-step solution

Find the gradient of the function

To find the gradient, we need to calculate the partial derivatives of \(\phi(\mathbf{x})\) with respect to \(x_1\) and \(x_2\). $$ \nabla\phi(\mathbf{x}) = \begin{bmatrix} \frac{\partial \phi}{\partial x_1} \\ \frac{\partial \phi}{\partial x_2} \end{bmatrix} = \begin{bmatrix} 2x_1 \\ -4x_2^3 \end{bmatrix} $$

Find the critical points by setting the gradient to zero

Now that we have the gradient, we can find the critical points by setting each component of the gradient to zero: $$ 2x_1 = 0 \\ -4x_2^3 = 0 $$ Solving these equations, we get that the critical point is \((0,0)\), which is the origin.

Find the Hessian matrix of the function

In order to classify the critical point, we need to calculate the Hessian matrix of our function. The Hessian matrix contains the second partial derivatives: $$ H(\phi(\mathbf{x})) = \begin{bmatrix} \frac{\partial^2 \phi}{\partial x_1^2} & \frac{\partial^2 \phi}{\partial x_1 \partial x_2} \\ \frac{\partial^2 \phi}{\partial x_2 \partial x_1} & \frac{\partial^2 \phi}{\partial x_2^2} \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & -12x_2^2 \end{bmatrix} $$

Classify the critical point

Now, let's evaluate the Hessian matrix at the origin \((0,0)\): $$ H(\phi(0,0)) = \begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix} $$ Since the determinant of the Hessian matrix at the origin is 0 (\(\text{det}(H(\phi(0,0))) = 2\times0 - 0\times0 = 0\)), we can't directly use the second partial derivative test to classify the critical point. However, we can still conclude that the origin is a saddle point because it is not a local minimum or maximum for both variables due to the following reasons: - For \(x_1\), the second partial derivative with respect to \(x_1\) at the origin is \(\frac{\partial^2 \phi}{\partial x_1^2}(0,0) = 2 > 0\), meaning \(x_1 = 0\) is a local minimum. - For \(x_2\), there is no change in the second partial derivative with respect to \(x_2\) near the origin, meaning \(x_2 = 0\) is not a local minimum nor maximum as \(\frac{\partial^2 \phi}{\partial x_2^2}(0,0) = 0\). Thus, the origin is a saddle point since it has different character for different variables.

Apply Newton's method

In order to use Newton's method to find the saddle point, at each iteration, we need to update the current estimate of the position, \(\mathbf{x}^{(k)}\), using the following expression: $$ \mathbf{x}^{(k+1)} = \mathbf{x}^{(k)} - H(\phi(\mathbf{x}^{(k)}))^{-1}\nabla \phi(\mathbf{x}^{(k)}) $$ However, since the determinant of the Hessian matrix at the origin is 0, meaning it's not invertible, Newton's method cannot be directly applied to find this saddle point.

Key concepts

Gradient of a Function

The gradient of a function represents the direction and rate at which the function increases most rapidly. It is a vector that points in the direction of the greatest rate of increase of the function, and its magnitude corresponds to the rate of the increase. To find the gradient, one must take the partial derivatives with respect to each variable. For the function \( \phi(\mathbf{x})=x_{1}^{2}-x_{2}^{4}+1 \), the gradient is calculated as \( abla\phi(\mathbf{x}) = \begin{bmatrix}2x_1 \-4x_2^3\end{bmatrix} \).

Critical points occur when the gradient is zero. For our function, setting \( 2x_1 = 0 \) and \( -4x_2^3 = 0 \) reveals that the origin \( (0,0) \) is a critical point. At this point, the function neither increases nor decreases, indicating a potential extremum or saddle point.

Hessian Matrix

The Hessian matrix is a square matrix comprised of the second-order partial derivatives of a function. This matrix is instrumental in assessing the curvature and, therefore, the local behavior—whether it's a minimum, maximum, or saddle point—around a critical point of the function. The Hessian for \( \phi(\mathbf{x}) \) is \( H(\phi(\mathbf{x})) = \begin{bmatrix}2 & 0 \0 & -12x_2^2\end{bmatrix} \). Evaluating the Hessian at the origin gives \( H(\phi(0,0)) = \begin{bmatrix}2 & 0 \0 & 0\end{bmatrix} \).

Since the determinant is zero, we cannot conclusively determine the nature of the critical point with the second derivative test alone. However, examining the sign of the second-order partial derivatives, we can infer that the origin exhibits properties of both a minimum and a non-extremum, heralding it as a saddle point.

Newton's Method

Newton's method, also known as the Newton-Raphson method, is an iterative approach to find successively better approximations to the roots (or zeroes) of a real-valued function. Using this method, an initial guess is improved upon by applying the formula\( \mathbf{x}^{(k+1)} = \mathbf{x}^{(k)} - H(\phi(\mathbf{x}^{(k)}))^{-1}abla \phi(\mathbf{x}^{(k)}) \).

In the context of our function \( \phi(\mathbf{x}) \), Newton's method encounters a problem because the Hessian matrix at the critical point is not invertible—its determinant is zero. Therefore, we cannot apply the iteration formula to find the saddle point at the origin, highlighting a limitation of Newton's method when dealing with non-invertible Hessians.

Saddle Point

A saddle point of a function is a critical point at which the function does not exhibit a clear maximum or minimum. Instead, it behaves differently in various directions; it might increase in one direction while decreasing in another. In our case, the origin is such a point for \( \phi(\mathbf{x}) \). While \( x_1 = 0 \) corresponds to a local minimum (\( \frac{\partial^2 \phi}{\partial x_1^2}(0,0) = 2 \), which is positive), the second partial derivative with respect to \( x_2 \) at the origin is zero, indicating a flat curvature. Hence, the point \( (0,0) \) is neither a maximum nor a minimum for \( x_2 \), qualifying it as a saddle point.