How to get the derivative of this convex quadratic

In summary, the derivative of the function f(x) is equal to [(Q+Q^T)/2]x- b, where Q is a matrix and b is a vector. The multiplication of matrices is not commutative, so the derivative must be symmetrized. This can be seen in a simple two-dimensional example.
  • #1
justin_huang
13
0
[itex]\frac{d}{dx}f(x)=\frac{d}{dx}[ \frac{1}{2}x_{}^{T}Qx-b_{}^{T}x][/itex]

how to get this derivative, what is the answer? is there textbook describe it?
 
Physics news on Phys.org
  • #3
This equation is analogous to a second degree polynomial, so I think the derivative is simply Qx - b.
 
  • #4
Not quite. Because multiplication of matrices is not commutative, taking the derivative you have to "symmetrize" Q- the derivative is [itex][(Q+Q^T)/2]x- b[/itex].

Look at the simple two dimensional situation:
[tex]\frac{1}{2}\begin{bmatrix}x & y \end{bmatrix}\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}- \begin{bmatrix}e & f\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}[/tex]
[tex]= \frac{1}{2}ax^2+ \frac{b+ c}{2}xy+ \frac{1}{2}dy^2- ex- ey[/tex]

The "derivative with respect to x", since x is a vector, is the gradient:
[tex]\begin{bmatrix}ax+ \frac{b+c}{2}y- e & dy+ \frac{b+c}{2}x- f\end{bmatrix}[/tex]
[tex]= \begin{bmatrix}a & \frac{b+c}{2} \\ \frac{b+c}{2} & d\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}- \begin{bmatrix}e & f\end{bmatrix}[/tex]
 
  • #5

The derivative of a convex quadratic function can be obtained by using the chain rule and the properties of matrix derivatives. The first step is to rewrite the function in matrix form as f(x) = 1/2 x^T Q x - b^T x, where Q is a symmetric matrix and b is a vector.

Next, we can use the properties of matrix derivatives to find the derivative of the first term, 1/2 x^T Q x. This can be written as x^T Q x = x^T (Qx) = (Qx)^T x. Using the product rule, we can find the derivative as follows:

d/dx (1/2 x^T Q x) = (Qx)^T x + x^T (Qx)

= x^T (Q^T + Q) x

= x^T (2Q) x

= 2 x^T Q x

Similarly, the derivative of the second term, b^T x, can be found as b.

Therefore, the derivative of the convex quadratic function f(x) is given by:

d/dx f(x) = 2 x^T Q x - b

This is the general formula for the derivative of a convex quadratic function. It is important to note that this formula can also be derived using the definition of the derivative and the properties of matrix multiplication.

There are many textbooks and online resources that describe the process of finding the derivative of a convex quadratic function. Some examples include "Introduction to Linear Algebra" by Gilbert Strang and "Matrix Differential Calculus with Applications in Statistics and Econometrics" by Jan R. Magnus and Heinz Neudecker. It is also helpful to consult with a mathematics or statistics expert for further clarification and understanding.
 

1. What is a convex quadratic function?

A convex quadratic function is a mathematical function that has the form f(x) = ax^2 + bx + c, where a, b, and c are constants and x is the independent variable. The graph of a convex quadratic function is a parabola that opens upwards, meaning that it is always increasing or staying constant.

2. What is the purpose of finding the derivative of a convex quadratic function?

The derivative of a convex quadratic function helps us determine the slope of the function at any given point. This is useful in understanding the behavior and properties of the function, such as identifying its maximum or minimum values. It is also used in optimization problems to find the optimal value of the function.

3. How do you find the derivative of a convex quadratic function?

To find the derivative of a convex quadratic function, we can use the power rule, which states that the derivative of a function of the form f(x) = ax^n is f'(x) = nax^(n-1). In the case of a convex quadratic function, the derivative is f'(x) = 2ax + b. We can also use the product rule and chain rule to find the derivative of more complex convex quadratic functions.

4. Are there any special cases when finding the derivative of a convex quadratic function?

Yes, there are two special cases when finding the derivative of a convex quadratic function. The first case is when the quadratic function has a constant term of c = 0. In this case, the derivative simplifies to f'(x) = 2ax. The second case is when the quadratic function has a coefficient of a = 0, meaning that it is a linear function. In this case, the derivative is simply the slope of the line, given by f'(x) = b.

5. Can the derivative of a convex quadratic function ever be negative?

No, the derivative of a convex quadratic function can never be negative. This is because the slope of a convex quadratic function is always positive or zero. If the slope is positive, the function is increasing, and if the slope is zero, the function is constant. This is because a convex quadratic function always opens upwards, and the derivative represents the slope of the function at any given point.

Similar threads

Replies
7
Views
827
  • Linear and Abstract Algebra
Replies
4
Views
2K
  • Linear and Abstract Algebra
Replies
21
Views
1K
Replies
5
Views
584
Replies
5
Views
1K
  • Linear and Abstract Algebra
Replies
8
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
591
  • Linear and Abstract Algebra
Replies
3
Views
1K
Replies
4
Views
2K
  • Linear and Abstract Algebra
Replies
6
Views
2K
Back
Top