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Problem 1.1 (5\%) Consider the function $f:: \mathbb{R}^{3} \rightarrow \mathbb...
Aug 22, 2024
Solution by Steps
step 1
To derive the gradient ∇f(x) \nabla f(x) , we compute the partial derivatives of f(x) f(x) with respect to x1,x2, x_1, x_2, and x3 x_3 . The function is given by f(x)=ex3x12+x22+x32cos⁡(x1) f(x) = e^{x_3} x_1^2 + x_2^2 + x_3^2 \cos(x_1) . Thus, we have:
step 2
The partial derivatives are calculated as follows: ∂f∂x1=2ex3x1−x32sin⁡(x1),∂f∂x2=2x2,∂f∂x3=ex3x12+2x3cos⁡(x1) \frac{\partial f}{\partial x_1} = 2 e^{x_3} x_1 - x_3^2 \sin(x_1), \quad \frac{\partial f}{\partial x_2} = 2 x_2, \quad \frac{\partial f}{\partial x_3} = e^{x_3} x_1^2 + 2 x_3 \cos(x_1) Therefore, the gradient is ∇f(x)=(2ex3x1−x32sin⁡(x1),2x2,ex3x12+2x3cos⁡(x1)) \nabla f(x) = \left(2 e^{x_3} x_1 - x_3^2 \sin(x_1), 2 x_2, e^{x_3} x_1^2 + 2 x_3 \cos(x_1)\right)
step 3
To derive the Hessian ∇2f(x) \nabla^2 f(x) , we compute the second partial derivatives. The Hessian matrix is given by:
step 4
The Hessian matrix is ∇2f(x)=[∂2f∂x12amp;∂2f∂x1∂x2amp;∂2f∂x1∂x3∂2f∂x2∂x1amp;∂2f∂x22amp;∂2f∂x2∂x3∂2f∂x3∂x1amp;∂2f∂x3∂x2amp;∂2f∂x32] \nabla^2 f(x) = \begin{bmatrix} \frac{\partial^2 f}{\partial x_1^2} & \frac{\partial^2 f}{\partial x_1 \partial x_2} & \frac{\partial^2 f}{\partial x_1 \partial x_3} \\ \frac{\partial^2 f}{\partial x_2 \partial x_1} & \frac{\partial^2 f}{\partial x_2^2} & \frac{\partial^2 f}{\partial x_2 \partial x_3} \\ \frac{\partial^2 f}{\partial x_3 \partial x_1} & \frac{\partial^2 f}{\partial x_3 \partial x_2} & \frac{\partial^2 f}{\partial x_3^2} \end{bmatrix} where the entries are computed as follows: ∂2f∂x12=2ex3−x32cos⁡(x1),∂2f∂x22=2,∂2f∂x32=ex3x12−2x3sin⁡(x1) \frac{\partial^2 f}{\partial x_1^2} = 2 e^{x_3} - x_3^2 \cos(x_1), \quad \frac{\partial^2 f}{\partial x_2^2} = 2, \quad \frac{\partial^2 f}{\partial x_3^2} = e^{x_3} x_1^2 - 2 x_3 \sin(x_1) and the mixed derivatives are zero
step 5
Now, we compute f(x),∇f(x) f(x), \nabla f(x) , and ∇2f(x) \nabla^2 f(x) at xˉ=(0,0,0) \bar{x} = (0, 0, 0) and x^=(0,0,1) \widehat{x} = (0, 0, 1) . For xˉ \bar{x} :
step 6
At xˉ=(0,0,0) \bar{x} = (0, 0, 0) : f(0,0,0)=0,∇f(0,0,0)=(0,0,0),∇2f(0,0,0)=[2amp;0amp;00amp;2amp;00amp;0amp;0] f(0, 0, 0) = 0, \quad \nabla f(0, 0, 0) = (0, 0, 0), \quad \nabla^2 f(0, 0, 0) = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix}
step 7
For x^=(0,0,1) \widehat{x} = (0, 0, 1) :
step 8
At x^=(0,0,1) \widehat{x} = (0, 0, 1) : f(0,0,1)=1,∇f(0,0,1)=(0,0,1),∇2f(0,0,1)=[2amp;0amp;00amp;2amp;00amp;0amp;−1] f(0, 0, 1) = 1, \quad \nabla f(0, 0, 1) = (0, 0, 1), \quad \nabla^2 f(0, 0, 1) = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -1 \end{bmatrix}
step 9
The spectral decomposition of the Hessian at both points shows the eigenvalues. For xˉ \bar{x} , the eigenvalues are 2,2,0 2, 2, 0 (indicating a saddle point). For x^ \widehat{x} , the eigenvalues are 2,2,−1 2, 2, -1 (indicating a local maximum). Thus, the conditions for being a local unconstrained minimizer are not satisfied at either point
Answer
The gradient ∇f(x) \nabla f(x) and Hessian ∇2f(x) \nabla^2 f(x) have been derived, and their values at the specified points have been computed. The spectral decompositions indicate that neither point satisfies the conditions for being a local unconstrained minimizer.
Key Concept
The gradient and Hessian are essential for analyzing the behavior of multivariable functions, particularly in optimization problems.
Explanation
The gradient provides the direction of steepest ascent, while the Hessian indicates the curvature of the function, helping to determine local minima, maxima, or saddle points.
Solution by Steps
step 1
To show that if f f is L L -smooth and pk p_k is a descent direction, we start with the Armijo condition:
\( f\left(x_k + \alpha_k p_k\right) < f\left(x_k\right) + \eta \alpha_k \nabla f\left(x_k\right)^{\top} p_k \)
step 2
Since pk p_k is a descent direction, we have \nabla f\left(x_k\right)^{\top} p_k < 0 . Using the L L -smoothness condition, we can express:
\( f\left(x_k + p_k\right) \leq f\left(x_k\right) + \nabla f\left(x_k\right)^{\top} p_k + \frac{L}{2} \|p_k\|_2^2 \)
step 3
Combining these inequalities, we can derive that if αk≤2(η−1)∇f(xk)⊤pkL∥pk∥22 \alpha_k \leq 2(\eta - 1) \frac{\nabla f\left(x_k\right)^{\top} p_k}{L \|p_k\|_2^2} , then the Armijo condition must hold
step 4
To derive an upper bound on the number of backtracking iterations, we note that each iteration reduces αk \alpha_k by a factor of τ \tau . Thus, after m m iterations, we have:
\( \alpha_k = \alpha_{\text{init}} \tau^m \)
step 5
Setting αinit=1 \alpha_{\text{init}} = 1 and solving for m m when the Armijo condition holds gives:
\( m \leq \frac{\log\left(\frac{2(\eta - 1) \nabla f\left(x_k\right)^{\top} p_k}{L \|p_k\|_2^2}\right)}{\log(\tau)} \)
Answer
The upper bound on the number of backtracking iterations is given by m≤log⁡(2(η−1)∇f(xk)⊤pkL∥pk∥22)log⁡(τ) m \leq \frac{\log\left(\frac{2(\eta - 1) \nabla f\left(x_k\right)^{\top} p_k}{L \|p_k\|_2^2}\right)}{\log(\tau)}
Key Concept
The Armijo condition ensures sufficient decrease in function value during optimization.
Explanation
The derived upper bound on backtracking iterations helps in understanding the efficiency of the backtracking line search method in optimization.
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