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

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  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 $a$".

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  • $\begingroup$ Sorry for that wrong formula, I did not notice it, actually messed with the formula of f(x)=f(a)+(x-a)f'(a)...sorry for that and I know that f'(a) means derivative of f at the point a but in that vector formula grad is operating on phi evaluated at the vector r which is variable, that makes me confused. if r is variable then how can we get a value of gradphi at r? the left hand side is symmetric but l is a constant here and r is variable, so should we not write the vector formula as r.gradphi(l) instead of l.gradphi(r) in analogy with the scalar one? $\endgroup$ – Srijan Ghosh Mar 25 at 14:56
  • $\begingroup$ I don't quite understand what you mean. It is an asymptotic formula, i.e. an approximation to the left-hand side with error bounds that can be computed. If you have an analytic function, you can even write down the infinite series and it will give you back your analytic function. In any case, this holds pointwise, so you can plug in whatever you want, no matter if you consider it "constant" or "variable". It is exactly the same as in the one-dimensional case. $\endgroup$ – Klaus Mar 25 at 17:09

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