# What is the vector form of Taylor series expansion?

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 $$\vec \nabla\phi(\vec l)$$]

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 $$a$$".