Exploring Dual Vector Relationships: f(\vec{x}) and \mbox{d}f_{\vec{x}}(\vec{y})

[LagrangeEuler]

$$f(\vec{x}+\epsilon \vec{y})-f(\vec{x})=\epsilon \mbox{d}f_{\vec{x}}(\vec{y})+O(\epsilon^2)$$.
Is $\mbox{d}f_{\vec{x}}(\vec{y})$ dual vector and why? Is it because $\mbox{d}$ is linear transformation? Also why equality
$$f(\vec{x}+\epsilon \vec{y})-f(\vec{x})=\epsilon \mbox{d}f_{\vec{x}}(\vec{y})+O(\epsilon^2)$$
is correct?

 

[martinbn]

To get a meaningful answer you need to provide more context.

The equality is correct because that is how $df$ is defined. And it is a dual vector because it takes vectors as arguments and gives a number as a result, and it is linear in the argument (the $\vec{y}$ in you expression).

 

[math771]

Yes, $\mbox{d}f_{\vec{x}}(\vec{y})$ is a dual vector because it is a linear transformation from the vector space $\vec{y}$ to the scalar field $\mathbb{R}$. This means that for any vector $\vec{y}$, $\mbox{d}f_{\vec{x}}(\vec{y})$ maps it to a real number, making it a dual vector.

The equality $f(\vec{x}+\epsilon \vec{y})-f(\vec{x})=\epsilon \mbox{d}f_{\vec{x}}(\vec{y})+O(\epsilon^2)$ is correct because it is a representation of the first-order Taylor expansion of the function $f$ around the point $\vec{x}$. The term $\epsilon \mbox{d}f_{\vec{x}}(\vec{y})$ represents the linear approximation of the change in $f$ as we move from $\vec{x}$ to $\vec{x}+\epsilon \vec{y}$, and the term $O(\epsilon^2)$ represents the higher-order terms that account for the non-linear behavior of the function. As $\epsilon$ approaches 0, the higher-order terms become negligible, making the equality accurate.