Given f(x) = 1/2x^T*Ax + c^T*x, where A E R^n*n is symmetric positive definite. How many steps are needed for Newton's method to obtain a global minimizer for from an arbitrary point 
?

In this problem, we are dealing with a quadratic function of the form:

\[
f(x) = \frac{1}{2}x^T A x + c^T x
\]

where \( A \in \mathbb{R}^{n \times n} \) is a symmetric positive definite matrix, and \( c \in \mathbb{R}^n \). We want to apply **Newton's method** to minimize this quadratic function and determine how many steps are needed to obtain the global minimizer starting from an arbitrary point.

### Step 1: First and Second Derivatives

1. **Gradient**: The gradient of \( f(x) \) with respect to \( x \) is given by:
   \[
   \nabla f(x) = A x + c
   \]

2. **Hessian**: Since \( A \) is symmetric and positive definite, the Hessian matrix \( \nabla^2 f(x) \) is constant and equal to:
   \[
   \nabla^2 f(x) = A
   \]
   Therefore, the second derivative of the function is the matrix \( A \), which does not depend on \( x \).

### Step 2: Newton's Method

Newton's method iteratively updates \( x \) using the following rule:
\[
x_{k+1} = x_k - (\nabla^2 f(x_k))^{-1} \nabla f(x_k)
\]
Substituting the expressions for the gradient and Hessian:

\[
x_{k+1} = x_k - A^{-1} (A x_k + c)
\]
Simplifying this expression:
\[
x_{k+1} = x_k - (x_k + A^{-1}c) = - A^{-1}c
\]

This shows that Newton's method converges to the exact solution \( x^* = -A^{-1} c \) in **one step**, because the Hessian is constant, and the function is quadratic.

### Conclusion

For this quadratic problem with a symmetric positive definite matrix \( A \), Newton's method reaches the global minimizer in exactly **one iteration** from any arbitrary starting point. This is a well-known result for quadratic optimization problems with Newton's method.

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Would you like further clarification on any part of the process?

Here are 5 related questions you may find interesting:

1. What is the difference between Newton's method and gradient descent for quadratic functions?
2. Why does Newton's method converge in one step for quadratic functions?
3. What are the conditions for Newton's method to converge globally in other cases?
4. How does the positive definiteness of matrix \( A \) affect the convergence?
5. Can Newton's method fail for non-quadratic functions?

**Tip**: For quadratic functions with symmetric positive definite Hessians, Newton's method is very efficient because it uses second-order information (curvature) to jump directly to the minimum in one step.

Given f(x) = 1/2x^T*Ax + c^T*x, where A E R^n*n is symmetric positive definite. How many steps are needed for Newton's method to obtain a global minimizer from an arbitrary point?

Discover how Newton's method solves a quadratic optimization problem with a symmetric positive definite matrix in one iteration. Learn the key formulas, such as the gradient and Hessian, and see how the method directly reaches the global minimizer. Suitable for graduate-level students.

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