Goes to Albert Fert and Peter Grünberg, for the discovery of Giant Magnetoresistance.
http://nobelprize.org/nobel_prizes/physics/laureates/2007/
This discovery made the iPod possible.
http://www.nytimes.com/2007/10/09/health/09iht-nobel.4.7820918.html?_r=1&emc=eta1
Tuesday, October 09, 2007
Monday, October 01, 2007
Thursday, August 23, 2007
Queen's Brian May
Just earned his PhD in astrophysics from Imperial College by completing his dissertation on interstellar dust, 35 years after he dropped out to concentrate on his music.
Saturday, April 21, 2007
Diffusion with a source on a semi-infinite line
(Budak/Samarskii/Tikhonov, problem 3.7)
ut = a2uxx + f(x,t)
u(x,0) = φ(x)
u(0,t) = μ(t)
0 ≤ x < ∞ and 0 ≤ t < ∞
The solution of the problem can be represented as a sum
u(x,t) = u1(x,t) + u2(x,t) + u3(x,t)
where u1(x,t) represents the effect of the initial condition u(x,o) = φ(x), u2(x,t) represents the effect of the boundary condition u(0,t) = μ(t), and u3(x,t) represents the effect of the nonhomogeneous term f(x,t).
Solution for u1
ut = a2uxx
u1(x,0) = φ(x)
u1(0,t) = 0
u1(x,t) = ½ (1/√π) ∫0∞ 1/√(a2t) [ e -(x-x')2/4a2t - e -(x+x')2/4a2t ] φ(x') dx'
Solution for u2
ut = a2uxx
u2(x,0) = 0
u2(0,t) = μ(t)
u2(x,t) = ½ (a2/√π) ∫0t x/[a2(t-t')]3/2 e-x2/4a2(t-t') μ(t') dt'
Solution for u3
u3(x,t) = ½ (1/√π) ∫0∞ dx' ∫0t dt' { 1/√[a2(t-t')] [ e -(x-x')2/4a2(t-t') - e -(x+y)2/4a2(t-t') ] f(x',t') }
ut = a2uxx + f(x,t)
u(x,0) = φ(x)
u(0,t) = μ(t)
0 ≤ x < ∞ and 0 ≤ t < ∞
The solution of the problem can be represented as a sum
u(x,t) = u1(x,t) + u2(x,t) + u3(x,t)
where u1(x,t) represents the effect of the initial condition u(x,o) = φ(x), u2(x,t) represents the effect of the boundary condition u(0,t) = μ(t), and u3(x,t) represents the effect of the nonhomogeneous term f(x,t).
Solution for u1
ut = a2uxx
u1(x,0) = φ(x)
u1(0,t) = 0
u1(x,t) = ½ (1/√π) ∫0∞ 1/√(a2t) [ e -(x-x')2/4a2t - e -(x+x')2/4a2t ] φ(x') dx'
Solution for u2
ut = a2uxx
u2(x,0) = 0
u2(0,t) = μ(t)
u2(x,t) = ½ (a2/√π) ∫0t x/[a2(t-t')]3/2 e-x2/4a2(t-t') μ(t') dt'
Solution for u3
u3(x,t) = ½ (1/√π) ∫0∞ dx' ∫0t dt' { 1/√[a2(t-t')] [ e -(x-x')2/4a2(t-t') - e -(x+y)2/4a2(t-t') ] f(x',t') }
Saturday, April 14, 2007
Financial Mathematics
Since finishing my ASA late last year, I've started rediscovering my interest in financial mathematics. I will be posting assorted theorems, problems, observations, etc, from my readings in my finance and investments blog from time to time.
Monday, March 19, 2007
Saturday, February 10, 2007
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