What is the vector form of Taylor series expansion?

Question

What is the expression for expansion of $\phi(\vec r+ \vec l)$ where $\vec r$ is variable and $\vec l$ is a constant vector. I think it can be expanded as a vector form of taylor series as $\phi(\vec r+\vec l)=\phi(\vec r)+\vec l.\vec \nabla\phi(\vec r)+....$ in analogy with general taylor series expansion of $f(x-a)$=$f(a)+xf'(a)+\frac {x^2}{2!}f''(a)+....$. But I can't be sure about it. I have seen this formula in a book. If that vector formula is true, (for further reference I denote this formula $f(x+a)$=$f(a)+xf'(a)+\frac {x^2}{2!}f''(a)+....$ as formula 1) then why the constant vector $\vec l$ is multiplied with the gradient of $\phi$ and why the gradient is evaluated at $\vec r$ where $\vec r$ is variable in oppose to formula 1 where variable $x$ is multiplied with derivative of $f(a)$ and derivative of $f(x)$ is evaluated at $a$ where $a$ is constant? And isn't it obvious that $f'(x)$ is evaluated at $a$? if a was a variable, then how could we evaluate $f'(x)$ at $a$? If r is variable then how can we get a value of gradphi at r? the left hand side is symmetric but $\vec l$ is a constant here and $\vec r$ is variable, so should we not write the vector formula as $\vec r.\vec \nabla\phi(\vec l)$ instead of $\vec l.\vec \nabla\phi(\vec r)$ in analogy with the scalar one?[I want to mean the value of $\vec \nabla\phi(\vec r)$ at point $\vec l$ by writing $\vec \nabla\phi(\vec l)$]

Answer 1

1. Your vector formula is correct whereas the scalar one is wrong (should be $+$ instead of $-$).

2. The left-hand side is symmetric in the two variables, so it doesn't matter which one you consider "constant". Also, $f'(a)$ does not mean "derivative of $f$ w.r.t. $a$" but "derivative of $f$ evaluated at the point $a$".