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17 Jan 2023: For isotropic atoms:. A= fj cos[2π(hxky+z)] exp(–8π2U sin2θ/λ2). B= fj sin[2π(hxky+z)] exp(–8π2U sin2θ/λ2). ... Fc2 = A2 B2; φc = tan–1(B/A). Usually R = |Fo–Fc| / |Fo| serves as a quality control.

17 Jan 2023: Ahk = Σj q { (fjfj’) cos[2π(hxjkyj+zj)] fj”sin[2π(hxjkyj+zj)] }. Bhk = Σj q { (fjfj’) sin[2π(hxjkyj+zj)] fj”cos[2π(hxjkyj+zj)] }. Fhk = A iB; tan(φhk) = ... esd(xj) = [ (B–1)jj w(Fo2–Fc2)2 / (NR–NP) ]. under the assumption that

7 Dec 2023: r3; (12b). _c % 32r3ð1þB cos 2cÞ%a. rsin c; (12c). where we recall that B = (a2 % 1)/(a2 1) A (0,1) is the shapefactor of the spheroid. ... Fig. 3A). Otherwise, it will escape due tothe fact that :r p cos c.

1 Dec 2023: GF′′=α2F, σG+ iαA[G cos X F′ sin X]F′′=G′′α2G, (4.1a,b). where F and G are functions of X alone. ... which have the leading-order solution. F =a(Y)[sin X cos ξY X π], G=a(Y) sin X cos ξY (4.5a,b).

13 Sep 2023: and b =pI1. (3.66). Then we can write. ̇ =b a cos. ... sin2 (3.67). and. ̇ =I1a. I3. (b a cos) cos. sin2 (3.68).

6 Mar 2023: Prove that itscircumference is 2π sin(ρ) and that it bounds a disc on S2 of area 2π(1 cos(ρ)). ... sin(A) sinh(c) = sinh(a). (b) Deduce that for a general hyperbolic triangle, one has.

12 Jul 2023: i) Λ = R(φ) :=. . 1 0 0 00 cos φ sin φ 00 sin φ cos φ 00 0 0 1. ...  , ii) Λ = B(β) :=. . cosh β sinh β 0 0sinh β cosh β 0 0.

9 May 2023: taking the opportunity to remind ourselves of this. In Electromagnetism, the magnetic field obeys r B = 0 which means that it can be. ... magnetic field B in terms of the current density. B(x) =µ04.

12 Jul 2023: g(θ) =. . cos θ sin θ 0sin θ cos θ 0. ... Aµ 7 A′µ = gAµg1 (µg)g1. (b) Given that Aµ(x) g, show that A′µ(x) g.

14 Feb 2023: r3; (12b). _c 32r3ð1þB cos 2cÞa. rsin c; (12c). where we recall that B = (a2 1)/(a2 1) A (0,1) is the shapefactor of the spheroid. ... 3A). Otherwise, it will escape due tothe fact that :r p cos c.