Let Delta =|(sin theta cos phi, sin theta sin phi, cos theta), (cos theta cos phi, cos theta sin phi, -sin theta), (-sin theta sin phi, sin theta cos phi, 0)|
formulae of quantum mechanics is Heisenberg's uncertainty principle, which shows that the uncertainty in the measurement of a particle's position (Δx) and momentum (Δp) cannot both be arbitrarily small at the same time (where h is the Planck constant): displaystyleDeltax,Deltapgeqfrach4pi....
1. a) Verify that x=A cos (\omega t+\phi )+Bsin(\omega t+\phi) (for A and B arbitrary constants) is a solution to the differential equation: \ddot{x}=-\omega ^{2}x b) Show that this solution can be rewritten as x=C cos(\omega t+\phi ) and express C and Find the v...
y));if(ix <0x41100000)/* small x: normal case */return(cpackf(sinhf(x) * cosf(y), coshf(x) * sinf(y)));/* |x| >= 9, so cosh(x) ~= exp(|x|) */if(ix <0x42b17218) {/* x < 88.7: expf(|x|) won't overflow */h = expf(fabsf(x)) *0.5...
sin(self.phi) self.z = torch.cos(self.theta) self.grid3d = torch.from_numpy(np.zeros( [self.height, self.width, 4], dtype=np.float32)) self.grid3d[:,:,0] = self.x self.grid3d[:,:,1] = self.y self.grid3d[:,:,2] = self.z self.grid3d[:,:,3] = self.grid[:,:,...
(20*360) z=t*3 02 蝴蝶曲线球坐标 PRO/E 方程:rho = 8 * t theta = 360 * t * 4 phi = -360 * t * 8 03 Rhodonea 曲线 采用笛卡尔坐标系 theta=t*360*4 x=25+(10-6)*cos(theta)+10*cos((10/6-1)*theta) y=25+(10-6)*sin(theta)-6*sin((10/6-1)*theta) *** 04 圆...
dLon = deltaLatlong[1] lat1 = latlong1[0] lat2 = latlong2[0] a = (sin(dLat/2) *sin(dLat/2) +sin(dLon/2) *sin(dLon/2) * cos (lat1) * cos (lat2)) c =2* arctan2 (sqrt (a), sqrt (1-a)) d = r * c# initial bearingy =sin(dLon) * cos (lat2) ...
[Phi]]^4 + 0.0035264*Cos[\[Theta]]^4 + 0.0101456*Cos[\[Theta]]^2 *Cos[\[Phi]]^2*Sin[\[Theta]]^2 + 0.0290355*Cos[\[Theta]]^2*Sin[\[Theta]]^2*Sin[\[Phi]]^2 + 0.0065072*Cos[\[Phi]]^2*Sin[\[Theta]]^4* Sin[\[Phi]]^2), {\[Theta], 0, Pi}, {\[Phi], 0, ...
四维球坐标(三维球面)中的拉普拉斯方程一般式delta u=d^2u/dx^2+d^u/dy^2+d^2u/dz^2+d^2u/dw^2.这个是拉普拉斯在四维空间中的一般式,通过x=rcos(phi1)y=rsin(phi1)cos(phi2)z=rsin(phi1)sin(phi2)cos(theta)w=r sin(phi1)si
(delta,n) Ce2_Cg2.*((m/(2*pi*kB*T))^1.5)*(4*pi*c^3/omega_eg^3*delta^2)/(exp(0.5*m*c^2/(kB*T)*delta^2/omega_eg^2)-(g/ntot*(2*pi*m*kB*T/(h^2))^1.5)^-1); for n = 1:1:50; i=i+1; A(i)=quadl(@(delta)F(delta,n),0,delta_up) 分享回复赞 漆黑...