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## New answers tagged differential-operators

2

Working heuristically, suppose that $f(x,y)=\mathrm{e}^{ax+by}$. Then $$\begin{split}\sum_{n=0}^{4N + 1} (-1)^n f\bigg(x + h\cos\left(\tfrac{2\pi n}{4N + 2}\right), y + h\sin\left(\tfrac{2\pi n}{4N + 2}\right)\bigg)&=\sum_{n=0}^{4N + 1} (-1)^n \mathrm{e}^{ah\cos\left(\tfrac{2\pi n}{4N + 2}\right)+bh\sin\left(\tfrac{2\pi n}{4N + 2}\right)}\\ &=\sum_{... 0 The left-hand side is a vector with ith component \epsilon_{ijk}\partial_j(fG_k), with implicit summation over the repeated indices j,\,k. By the product rule, this splits into \epsilon_{ijk}(\partial_jf)G_k+f\epsilon_{ijk}\partial_jG_k. The first of these terms is the ith component of the vector (\nabla f)\times G; the second term is the ith ... 0$$∇ × (f G) =\sum{\hat{i} \times \frac{\partial}{\partial x}(f G)} =\sum{\hat{i}\times ( \frac{\partial f}{\partial x} G + f \frac{\partial}{\partial x}G)}= (\sum{\hat{i}}\frac{\partial f}{\partial x}) \times G + f (\sum{\hat{i}}\frac{\partial G}{\partial x}) = ∇f × G + f (∇ × G)1 From the left, the product fG is still a vector field, so if you know how to calculate curls, go right ahead, since f just scales the components of G. On the right, \nabla f×G is the cross between the gradient of f (a vector by definition), and G, also a vector, both three-dimensional, so the product is defined; also, f(\nabla × G) is just f,... 2 Why not just write it out? Let G=(G_1, G_2, G_3). Then \begin{align} \nabla \times (fG) &= (\partial_1, \partial_2, \partial_3) \times (fG_1, fG_2, fG_3)\\ &= (\partial_2(fG_3) - \partial_3(fG_2), \partial_3(fG_1)-\partial_1(fG_3), \partial_1(fG_2)-\partial_2(fG_1))\\ &= \ldots \end{align} 0 What you want is the Faà di Bruno's formula. This paper gives some historical background for the formula and this one presents some connections to complex analysis and combinatorics. 0 No. Consider the diagonal vector field\mathbf A=(xy)\frac{\mathbf i+\mathbf j}{\sqrt2}$$for which, clearly, \mathbf B=0. But if we simply rotate the coordinate system by 45^\circ,$$x=\frac{x'-y'}{\sqrt2},\quad y=\frac{x'+y'}{\sqrt2}\mathbf A=\bigg(\frac{x'^2-y'^2}{2}\bigg)\mathbf i' then $\mathbf B=(1)\mathbf i'\neq0$. So this differential ...

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