# Solving nonlinear least-squares with first order Taylor expansion

I'm referencing this Wikipedia article.

I understand that a Taylor expansion of a function $$f(x)$$ around $$x = a$$ can be given by

$$f(x) = \sum_{n = 0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n$$

According to the linked section of the article, we can approximate the model function as $$f(x_i, \beta)$$ for some data point $$x_i$$ and a vector of weights $$\beta$$.

So, looking a the Taylor expansion above, and evaluating up to $$n = 1$$, I get this...

$$f(x_i, \beta) = f^k(x_i, \beta) + \frac{\partial{f(x_i, \beta)}}{\partial \beta_j}(\beta _j - \beta _j^k)$$

as the first order Taylor expansion of $$f$$ evaluated around $$\beta _j$$.

I'm confused about why we can just sum over all the $$j$$ when expressing the Taylor expansion wrt to the vector $$\beta ^k$$.

Thanks for any assistance.

• The Jacobian of a function is a matrix, and the derivative approximation term can be considered as a dot product of two vectors for fixed $x_i$. Think about the case of a single variable function and extend it to multi-variables. – Tab1e May 24 at 0:53
• @Tab1e sorry I don't quite see the dot product that you mentioned – Carpetfizz May 24 at 22:05