Recent content by infinitylord

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    Proving that three closed orbits must contain a fixed point

    A smooth vector field on the phase plane is known to have exactly three closed orbit. Two of the cycles, C1 and C2 lie inside the third cycle C3. However C1 does not lie inside C2, nor vice-versa. What is the configuration of the orbits? Show that there must be at least one fixed point bounded...
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    Showing f is a solution to quantum oscillator SWE

    Homework Statement For a 1-dimensional simple harmonic oscillator, the Hamiltonian operator is of the form: H = -ħ2/2m ∂xx + 1/2 mω2x2 and Hψn = Enψn = (n+1/2)ħωψn where ψn is the wave function of the nth state. defining a new function fn to be: fn = xψn + ħ/mω ∂xψn show that fn is a...
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    Relativistic mechanics Taylor expansion

    Oh so the c's cancel out leaving the desired first term! Thanks for the help
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    Relativistic mechanics Taylor expansion

    Thanks for the help and sorry it has taken me so long to respond. Okay, here's where I got with this. v = (p/m)/(1+(p/mc)2)1/2 multiply the top and bottom by mc/p to get c/(1+(mc/p)2)1/2 Then I tried using the known expansion (1+x)n = 1+nx+1/2(n-1)nx^2... where n=-1/2 and x=(mc/p)2 (and c...
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    Relativistic mechanics Taylor expansion

    I understand that it means that momentum is very large compared to mass. However, I'm still not really sure where to begin with the actual expansion. Am I expanding the (c-v)/c?
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    Relativistic mechanics Taylor expansion

    Homework Statement For a particle traveling near the speed of light, find the first non-vanishing term in the expansion of the relative difference between the speed of the particle and the speed of light, (c-v)/c, in the limit of very large momentum p>>mc. Hint: Use (mc/p) as a small parameter...
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    Proving ||∇h||^2 for h=fog using Differentiable Functions and the Dot Product

    Honestly because I believe I got the answer and I had to turn in the homework the following morning. I've been extremely busy with homework and preparing for my first college midterms.
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    Proving ||∇h||^2 for h=fog using Differentiable Functions and the Dot Product

    Thank you! I actually immediately figured it out when you mentioned the matrices.
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    Proving ||∇h||^2 for h=fog using Differentiable Functions and the Dot Product

    Homework Statement Let f:R2−>R be a differentiable function at any point, and g be the function g:R3−>R2defined by: g(u,v,w)=(g1,g2)=(u2+v2+w2,u+v+w) consider the function h=fog and prove that ||∇h||^2 = 4(∂f/∂x)^2*g1 + 4(∂f/∂x)(∂f/∂y)*g2 + 3(∂f/∂y)^2.The Attempt at a Solution...
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    Boundary points and limit of f(x,y)

    I tried looking up the domain and limit of the function via wolfram alpha. It gave me that my domain was correct, so I'm not sure what to do about that. I still don't know what to do about the boundary point either. And I suspected that the limit DOES exist. And Wolfram alpha again put that the...
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    Boundary points and limit of f(x,y)

    Let f(x,y) be defined by f(x,y) = [x2y2]/[x2y2 + (x-y)2] a) Find the domain of the function f. b) show that (0,0) is a boundary point of the domain of f c) Compute the following limit if it exists: lim (x,y) ---> (0,0) f(x,y) The Attempt at a Solution a) I first change the value (x-y)2 to...
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    Min(|v|) and max(|v|) in relation to norms of a vector

    Thanks guys! I got it now I believe. I used the triangle inequality with v=(1,0) and w=(0,1). That way (after some simplification) it turns into: v1+w2\leqv1-v1+w2-w2. Therefore, 2<0. Which is completely untrue.
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    Min(|v|) and max(|v|) in relation to norms of a vector

    Homework Statement I have a homework problem in honors calculus III that I'm having a little trouble with. Given these three qualities of norms in Rn: 1) f(v)\geq0, with equality iff v=0 2) f(av)=|a|f(v) for any scalar a 3) f(v+w) \leq f(v)+f(w) we were given a set of 3 functions and told...
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    Linear Algebra orthogonal basis and orthogonal projection

    Oh! you're right, I hadn't even realized it didn't belong to the subspace. I just re-did it and got \vec{V_{2}}=(1,1,-1). it satisfies a+b+2c=0, and <\vec{V_{1}},\vec{V_{2}}>=-1+1+0=0. Therefore, they are orthogonal. This would then make the projection for b onto \vec{V_{2}}=(1,1,-1). And the...
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    Linear Algebra orthogonal basis and orthogonal projection

    Thank you very much! I believe I understand it all now. a) For \vec{V_{1}}, \vec{V_{2}}\inP. \vec{V_{1}} can be arbitrary so long as it satisfies the condition a+b+2c=0. So I chose \vec{V_{1}}=(-1,1,0) \vec{V_{2}} must be orthogonal to \vec{V_{1}}, so I chose it to be (1,1,0). And that is the...