2nd order approx. of barrier function

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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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