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I am trying to derive the taylor expansion for the potential $U(\vec r + \delta \vec r)$.

  The general expression for the taylor expansion is: $f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$.

I am trying to derive the same expression as in the wikipedia article : https://en.wikipedia.org/wiki/Fine_structure#:~:text=This%20yields%20a,can%20be%20estimated%3A

But my derivation is not proper and it's messy, with multiple punctuality mistakes etc. From my understanding the variable is $\delta\vec r$. So we need to derivate with respect to it.

$U(\vec r + \delta \vec r)=U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec r'-\delta \vec r)^0 + \sum_i \frac{\partial}{\partial x_i}U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec x'_i-\delta x_i)^1|_{\delta x_i=0}+\frac 1 2\sum_{i,j}\frac{\partial}{\partial x_i}\frac{\partial}{\partial x_j}U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec x'_i-\delta x_i)|_{\delta x_i=0}(\delta \vec x'_j-\delta x_j)|_{\delta x_j=0} +...$

I have the following 3 questions for what I wrote above:

1- If we want to be precise, shouldn't I write $\frac{\partial}{\partial \delta x_i}$ instead of $\frac{\partial}{\partial x_i}$ ? Because if am writing $\frac{\partial}{\partial x_i}$ than it looks as if I am derivating with respect to $x_i$.

2- In the wikipedia article the 2nd term is written as the product of the Nabla operator and the potential, so a gradient. How do I cleanly produce that expression?

I am trying to derive the Taylor expansion for the potential $U(\vec r + \delta \vec r)$. The general expression for the Taylor expansion is: $$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n.$$

I am trying to derive the same expression as in the Wikipedia article on Fine structure, but my derivation is not proper and it's messy, with multiple punctuality mistakes etc. From my understanding the variable is $\delta\vec r$. So we need to take the derivative with respect to it.

\begin{align} U(\vec r + \delta \vec r)=& U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec r'-\delta \vec r)^0 \\ &+ \sum_i \frac{\partial}{\partial x_i}U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec x'_i-\delta x_i)^1|_{\delta x_i=0}\\ &+\frac 1 2\sum_{i,j}\frac{\partial}{\partial x_i}\frac{\partial}{\partial x_j}U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec x'_i-\delta x_i)|_{\delta x_i=0}(\delta \vec x'_j-\delta x_j)|_{\delta x_j=0} +\cdots \end{align}

I have the following questions for what I wrote above:

1. If we want to be precise, shouldn't I write $\frac{\partial}{\partial \delta x_i}$ instead of $\frac{\partial}{\partial x_i}$ ? Because if am writing $\frac{\partial}{\partial x_i}$ than it looks as if I am taking the derivative with respect to $x_i$.

2. In the Wikipedia article, the 2nd term is written as the product of the Nabla operator and the potential, so a gradient. How do I cleanly produce that expression?

I am trying to derive the taylor expansion for the potential $U(\vec r + \delta \vec r)$.

The general expression for the taylor expansion is: $f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$.

I am trying to derive the same expression as in the wikipedia article : https://en.wikipedia.org/wiki/Fine_structure#:~:text=This%20yields%20a,can%20be%20estimated%3A

But my derivation is not proper and it's messy, with multiple punctuality mistakes etc. From my understanding the variable is $\delta\vec r$. So we need to derivate with respect to it.

$U(\vec r + \delta \vec r)=U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec r'-\delta \vec r)^0 + \sum_i \frac{\partial}{\partial x_i}U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec x'_i-\delta x_i)^1|_{\delta x_i=0}+\frac 1 2\sum_{i,j}\frac{\partial}{\partial x_i}\frac{\partial}{\partial x_j}U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec x'_i-\delta x_i)^2|_{\delta x_i=0}(\delta \vec x'_j-\delta x_j)^2|_{\delta x_j=0} +...$

I have the following 3 questions for what I wrote above:

1- If we want to be precise, shouldn't I write $\frac{\partial}{\partial \delta x_i}$ instead of $\frac{\partial}{\partial x_i}$ ? Because if am writing $\frac{\partial}{\partial x_i}$ than it looks as if I am derivating with respect to $x_i$.

2- In the wikipedia article the 2nd term is written as the product of the Nabla operator and the potential, so a gradient. How do I cleanly produce that expression?

3- As you can see, in my 3rd term (order 2) the $\delta$'s are in power 2, but in the formula in wikipedia they are not. But if I am to abide by what the Taylor expansion formula writes than I do need to use power 2. So what's the deal?

I am trying to derive the taylor expansion for the potential $U(\vec r + \delta \vec r)$.

The general expression for the taylor expansion is: $f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$.

I am trying to derive the same expression as in the wikipedia article : https://en.wikipedia.org/wiki/Fine_structure#:~:text=This%20yields%20a,can%20be%20estimated%3A

But my derivation is not proper and it's messy, with multiple punctuality mistakes etc. From my understanding the variable is $\delta\vec r$. So we need to derivate with respect to it.

$U(\vec r + \delta \vec r)=U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec r'-\delta \vec r)^0 + \sum_i \frac{\partial}{\partial x_i}U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec x'_i-\delta x_i)^1|_{\delta x_i=0}+\frac 1 2\sum_{i,j}\frac{\partial}{\partial x_i}\frac{\partial}{\partial x_j}U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec x'_i-\delta x_i)|_{\delta x_i=0}(\delta \vec x'_j-\delta x_j)|_{\delta x_j=0} +...$

I have the following 3 questions for what I wrote above:

1- If we want to be precise, shouldn't I write $\frac{\partial}{\partial \delta x_i}$ instead of $\frac{\partial}{\partial x_i}$ ? Because if am writing $\frac{\partial}{\partial x_i}$ than it looks as if I am derivating with respect to $x_i$.

2- In the wikipedia article the 2nd term is written as the product of the Nabla operator and the potential, so a gradient. How do I cleanly produce that expression?