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Consider the function ƒ(11, 12) = (2+1 − 1)² + (x1 + x2 − 1)². (a) Find the global minimum of f, and justify your answer. (b) Starting at z(0) = (0,0), perform two steps of Newton's method to find 7(²). Show your work.

Consider the function ƒ(11, 12) = (2+1 − 1)² + (x1 + x2 − 1)². (a) Find the global minimum of f, and justify your answer. (b) Starting at z(0) = (0,0), perform two steps of Newton's method to find 7(²). Show your work.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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1. Consider the function ƒ(#₁, #2) = (2x₁ − 1)² + (x1 + x2 − 1)². (a) Find the global minimum of f, and justify your answer. (b) Starting at z(0) = (0,0), perform two steps of Newton's method to find 7(2), Show your work. (c) We are going to repeat this problem using gradient descent with backtracking line-search. i. Starting at z) = (0,0) with learning rate µ0), write down the gradient descent equation for z(¹), ii. Suppose we want to set (0) using backtracking line search with y = 0.2 and Armijo's condition f(x(¹) ≤ f(x) − µ©y||▼ƒ(x®)||2. Find a value of (0) that satisfies this. iii. Suppose instead you started with µ(0) = 1 and an update of 0 ← ½μ(0) (i.c. =). In the worst case, how many steps of back-tracking would you have to take before accepting (¹)?
Transcribed Image Text:1. Consider the function ƒ(#₁, #2) = (2x₁ − 1)² + (x1 + x2 − 1)². (a) Find the global minimum of f, and justify your answer. (b) Starting at z(0) = (0,0), perform two steps of Newton's method to find 7(2), Show your work. (c) We are going to repeat this problem using gradient descent with backtracking line-search. i. Starting at z) = (0,0) with learning rate µ0), write down the gradient descent equation for z(¹), ii. Suppose we want to set (0) using backtracking line search with y = 0.2 and Armijo's condition f(x(¹) ≤ f(x) − µ©y||▼ƒ(x®)||2. Find a value of (0) that satisfies this. iii. Suppose instead you started with µ(0) = 1 and an update of 0 ← ½μ(0) (i.c. =). In the worst case, how many steps of back-tracking would you have to take before accepting (¹)?
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