Abstract

A new three-term conjugate gradient method is proposed in this article. The new method was able to solve unconstrained optimization problems, image restoration problems, and compressed sensing problems. The method is the convex combination of the steepest descent method and the classical LS method. Without any linear search, the new method has sufficient descent property and trust region property. Unlike previous methods, the information for the function f x is assigned to d k . Next, we make some reasonable assumptions and establish the global convergence of this method under the condition of using the modified Armijo line search. The results of subsequent numerical experiments prove that the new algorithm is more competitive than other algorithms and has a good application prospect.

Highlights

  • IntroductionConsider the following unconstrained optimization: min􏼈f(x) | x ∈ Rn􏼉, f: Rn ⟶ R

  • Consider the following unconstrained optimization: min􏼈f(x) | x ∈ Rn􏼉, f: Rn ⟶ R. (1)According to the research of other scholars, there are abounding effective forms to solve unconstrained optimization problems

  • Where αk is the step size of the k th iteration obtained by a certain line search rule and dk is search direction of step k, which are two important factors for solving unconstrained optimization problems

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Summary

Introduction

Consider the following unconstrained optimization: min􏼈f(x) | x ∈ Rn􏼉, f: Rn ⟶ R. E three-term conjugate gradient algorithm is easy to converge because it automatically has sufficient descent. E conjugate gradient method with three terms was first proposed by Zhang et al [27]; they defined dk as. A three-term conjugate gradient algorithm was proposed in Yuan’s paper [28]. 2. Conjugate Gradient Algorithm and Convergence Analysis is section will give a new modified Liu–Storey method combining (7) and (12), as shown below: Lemma 1. Assumption 1 is kept true, and there is a positive constant α, which satisfies (12) in Algorithm 1. From Algorithm 1, for the obtained step size αk, it is clear that αk/ρ does not satisfy (12), and by (13) and (14), f􏼠xk αk ρ dk􏼡. We get the contradiction. us, result (24) is true, and this completes the proof

Numerical Experiments
Conclusions

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