2nd order approx. of barrier function

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SUMMARY

The discussion focuses on deriving the second-order Taylor series for the log barrier function defined as $$f(x)=-\sum_{i=1}^{m}\log(b_{i}-a_{i}^{T}x)$$. The user attempts to calculate the gradient and Hessian but expresses uncertainty regarding the correctness of their solution. The gradient is proposed as $$\nabla f(x)=f(x_{0})-\sum_{i=1}^{m}\bigg(\dfrac{a_{i}}{b_{i}-a_{i}^{T}x}\bigg)(x-x_{0})$$, while the Hessian is identified as incorrect. The discussion highlights the challenges of calculating gradients and Hessians for vector-valued functions.

PREREQUISITES
  • Understanding of Taylor series expansion
  • Familiarity with log barrier functions in optimization
  • Knowledge of gradient and Hessian calculations
  • Basic linear algebra concepts, particularly vector operations
NEXT STEPS
  • Study the derivation of Taylor series for multivariable functions
  • Learn about the properties and applications of log barrier functions in optimization
  • Explore gradient and Hessian computation techniques for vector-valued functions
  • Review linear algebra, focusing on vector calculus and matrix operations
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Students and professionals in mathematics, optimization, and machine learning who are working with barrier functions and require a deeper understanding of Taylor series expansions and their applications.

FOIWATER
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Homework Statement


write the 2nd order taylor series for the log barrier function $$-\sum_{i=1}^{m}(b_{i}-a_{i}^{T}x)$$

Homework Equations


See Above

The Attempt at a Solution


Here is my attempt at a solution
$$\nabla f(x)=f(x_{0})-\sum_{i=1}^{m}\bigg(\dfrac{a_{i}}{b_{i}-a_{i}^{T}x}\bigg)(x-x_{0})-\dfrac{1}{2}(x-x_{0})^{T}\sum_{i=1}^{m}\bigg(\dfrac{a_{i}^{T}a_{i}}{(b_{i}-a_{i}^{T}x)^{2}}\bigg)(x-x_{0})$$

So, my problem is I've never really encountered calculating gradients and hessians for vector valued functions of only one variable (x, in this case). Am I way out to lunch?

Thanks.
 
Perhaps it would help if you were to say exactly what the "log barrier function" is!
 
oh I typed it wrong

$$f(x)=-\sum_{i=1}^{m}log(b_{i}-a_{i}^{T}x)$$

But I know now that what I have above is not correct. I think the gradient might be right actually, but the hessian is certainly not.
 

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