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I am trying to understand what this notation means... $\nabla_y$. For example:

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Specifically I guess I am confused how about why $\nabla_y v(y)=r^{n-1}\nabla_x u(x+r^{n-1}y)$

MathIsHard
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    It looks like $\nabla_y = \partial/\partial y$. – Gregory Apr 15 '18 at 01:11
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    Specifically, $\nabla_y = (\partial/\partial y_1, \dots,\partial/\partial y_n)$ where $y \in \Bbb{R}^n$, since $y$ is a vector parameter. – B. Mehta Apr 15 '18 at 01:12
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    Generally $\nabla$ refers to the gradient operator. This turns a multi-variable scalar function into a vector function. The vector denotes the direction of greatest ascent. – CogitoErgoCogitoSum Apr 15 '18 at 01:12
  • Thank you. Where does the $\nabla_x$ come in though? because a partial y wouldn't cause that would it? I don't see why $\nabla_y v(y)=r^{n-1}\nabla_x u(x+r^{n-1}y)$ – MathIsHard Apr 15 '18 at 01:13
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    Yes, why wouldn't it be $\nabla_y$ on the right-hand side? – saulspatz Apr 15 '18 at 01:18

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The subscript is to stress which variable we are taking the gradient with respect to. So $\nabla_y v(y) = \nabla_y u(x + r^{n - 1}y)$ is the gradient with respect to $y$ of the composition of functions $u$ with $y \mapsto x + r^{n - 1}y$. Thus, the chain rule applies, and we get $r^{n - 1}\nabla u(x + r^{n - 1}y)$.

The $\nabla_x$ appearing in the text is to stress that this last gradient is with respect to $x$, so that in particular, we do not have to do any more chain-ruling with this. So $\nabla_x u(x + r^{n - 1}y)$ is just the gradient of the function $u$, evaluated at the point $x + r^{n - 1}y$ (rather than the gradient of the function of $y$, $u(x + r^{n - 1}y)$, which involves (as is calculated here) the chain rule.

Chris
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