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I'm learning optimization techniques and came accross the gradient (nabla) operator : $\nabla$.

If I'm right, the $\nabla$ operator of a function means the vector of all its partial derivatives. Then, if for example I'm talking about this specific partial derivative $\left(\frac{\partial}{\partial y}\right) F(x,y,z)$ is it a right terminology to write something like : $\nabla^{y}$ (since we can't use the standard i-th-element-in-vector notation like $\nabla^{2}$ as it would mean the Laplacian).

If it is not a correct terminology, is there a correct terminlogy to specify a specific component of the partial derivatives vector?

3 Answers 3

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Nomenclatures are uncountable and is a matter of definition, but I would say that if $y$ is a scalar then it is more common to see $\partial_y F$. Another one I have seen is $F_y$ for simplicity. In this case a double derivative would be $F_{yz}$ for example.

If $y$ was a vector and $F(\textbf{x},\textbf{y},\textbf{z}) = F(x_1,...,x_m,y_1,...,y_l,z_1...,z_k)$ then $\nabla_\textbf{x} F$ would mean $(\dfrac{\partial F}{\partial{x_1}}, ...,\dfrac{\partial F}{\partial{x_m}} )$.

Note that it seems you are thinking about gradients, that is the meaning of nabla in this case, but it can have other meanings.

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No, it's not the correct terminology. If you want to specify a specific component, just use the I,J,K form for specifiying the x, y, or z direction and put the partial derivative you are interested in next to I,J, or K.

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    Can you give me an example of what you are saying?2017-02-12
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    Sure! So, to specify the y-component of the gradient of f, just write (δf/dy)J2017-02-12
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    Got it, thanks!2017-02-12
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I don't know that anyone calls this the nabla operator. The symbol is nabla, but the operator is called the gradient. There is no need for a special notation for an individual partial derivative, you can just write $\partial f/\partial y$ or $\partial/\partial y$ for the operator.

You do sometimes see $\nabla_x f$ for a function of many variables $f(x_1, x_2,..., x_n, y_1, y_2, ..., y_n)$ to mean the gradient w.r.t. the $x$ variables. (But this isn't what the question was asking about)

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    Yes, but even though the question was not about it, it's something very useful for me to know.2017-02-12