Recent content by chipotleaway

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    Deriving spherical unit vectors in terms of cartesian unit vectors

    Thanks, that makes sense.I was following this .pdf https://www.csupomona.edu/~ajm/materials/delsph.pdf [Broken]
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    Deriving spherical unit vectors in terms of cartesian unit vectors

    I'm trying to find the azimuthal angle unit vector \vec{\phi} in the cartesian basis by taking the cross product of the radial and \vec{z} unit vectors. \vec{z} \times \vec{r} = <0, 0, 1> \times <sin(\theta)cos(\phi), sin(\theta)sin(\phi), cos(\theta)> = <-sin(\theta)sin(\phi)...
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    Rotation of coordinates (context of solving simple PDE)

    My lecturer did the change of coordinates for a more general constant coefficient PDE \sum_{j=1}^n a_j\frac{\partial f}{\partial x_j}=b(x,u) in n-variables by defining the new variables as: y_1=\frac{x_1}{a_1} and y_j=x_j-\frac{a_j}{a_1}x_1 How do you get this?
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    Rotation of coordinates (context of solving simple PDE)

    So I need an arbitrary constant?
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    Rotation of coordinates (context of solving simple PDE)

    If you rotate your rectangular coordinate system (x,y) so that the rotated x'-axis is parallel to a vector (a,b), in terms of the (x,y) why is it given by x'=ax+by y'=bx-ay I got x'=ay-bx, y'=by+ax from y=(b/a)x. By the way this is from solving the PDE aux+buy=0 by making one of the...
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    Prove this vector identity

    Sorry, my mistake. IT should be ##(\vec{r}\cdot\nabla)\nabla \times \vec{F}##
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    Prove this vector identity

    Now I've got: (G_1\frac{\partial r}{\partial x}+G_2\frac{\partial r}{\partial y}+G_3\frac{\partial r}{\partial z})-(G\frac{\partial r_1}{\partial x}+G\frac{\partial r_2}{\partial y}+G\frac{\partial r_3}{\partial z})+2G_1+2G_2+2G_3. When I expand out the vectors(\frac{\partial r}{\partial x}...
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    Prove this vector identity

    0! Which gives ((\nabla \times F).\nabla)r-(\nabla.r)(\nabla \times F)+2\nabla \times F or (G.\nabla)r-(\nabla.r)G+2G One term less = a bunch of less components to deal with - I'll try expanding it out now and see where I get.
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    Prove this vector identity

    Homework Statement Show that: curl(r \times curlF)+(r.\nabla)curlF+2curlF=0, where r is a vector and F is a vector field. (Or letting G=curlF=\nabla \times F i.e. \nabla \times (r \times G) + (r.\nabla)G+2G=0) The Attempt at a Solution I used an identity to change it to reduce (?) it to...
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    Showing a metric space is complete

    Homework Statement Show the space of all space of all continuous real-valued functions on the interval [0, a] with the metric d(x,y)=sup_{0\leq t\leq a}e^{-Lt}|x(t)-y(t)| is a complete metric space. The Attempt at a Solution Spent a few hours just thinking about this question, trying to prove...
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    Circle is a set of a discontinuities?

    So then the rectangle is an infinite set of discontinuous points as well, but we can integrate over it because we just use g(x,y) instead? Why can't we do the same thing with a circle by changing to polar coordinates so that the domain can be defined like the rectangle in Cartesian? (I think...
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    Circle is a set of a discontinuities?

    I'm not sure I understand - what about a rectangle? Why is the characteristic function over that not discontinuous?
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    Circle is a set of a discontinuities?

    Why is the characteristic function* of a ball in Rn continuous everywhere except on its surface?My lecturer said that a circle is a 'set of discontinuities' - what exactly does that mean? (some context: we're looking at how we can integrate over a ball. Previously we've only looked at Riemann...
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    2nd order ODE - undetermined coefficients?

    Wow, thanks a lot! Would never have thought of that - trig was one of my extra weaker points back in high school. I've solved the non-homogenous case, gotten the resonance frequencies and now doing the homogenous case to get the general solution. The solution I've got for y''=\omega_0^2y=0...
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    Variable coefficient 2nd order DE

    Unfortunately the solution I got doesn't satisfy the original DE Here's how they did it in the text: 2tv''-v'=- Letting w=v' 2tw'-w=0 Separating variables and solving for w(t) \therefore w(t)=v'(t)=ct^{\frac{1}{2}} \therefore v(t)=\frac{2}{3}ct^{\frac{3}{2}}+k So the annihilator...
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