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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).

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.

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).

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.

1 Nov 2023: 10. The linear map R3 R3 defined by. x 7 x′ = cos θ x (x n) (1 cos θ) n sin θ (x n) (). is a rotation by angle θ in ... Hence show that. Rii = 2 cos θ 1 , εijkRjk = 2ni sin θ.

7 Jul 2023: a) Prove that Tn(cos(y). )= cos(ny) for all integers n > 0. ... b) Prove that cos(π/n) is algebraic for all integers n > 1.(c) For each integer n > 1, determine whether cos(π/n) is rational or not.

12 Jul 2023: p(m)i |i〉|θi〉(cos(θi)|0〉 sin(θi)|1〉) as an O(poly log(N)) sized circuit. of 1- and 2-qubit gates. ... Suppose wehave the n-qubit starting state |ψst〉 = sin(θ)|g〉 cos(θ)|b〉, where |g〉 G, |b〉 G andthe value of θ (0,π/2) is known.

26 Mar 2023: 7); cos θ h3/r F 2h4/r2, whichleads to the requirement that. ... obtain the function of integration (for details, see Appendix B),. G(θ ) = (4 cos θ )1/4, (54)for π /2 < θ < 3π /2.

7 Jul 2023: φ(r,θ) = a0 c0lnr. n=1. (anr. n cnrn) cos nθ. n=1. ... f(θ) =. n=1. An cos nθ,. for some coefficients An. (i) Solve Laplace’s equation 2φ = 0 in the annulus 1 < r < e2 with boundaryconditions:.

11 Jan 2023: on the two-sphere. Show that. V θ(φ) = A cos αφ B sin αφ,. ... V φ(φ) = C cos αφ D sin αφ,. where A, B, C and D are constants and α = cos θ0.

17 Jan 2023: P : primitive, A,B,C : face centered. I : body centered F : (all-)face centered. ... Miller indices of lattice planes. a. b. 110. a. b. b = 1/k = 1  k = 1.

23 Aug 2023: c)f = t2 t sin t t2 , f′ = (2 cos t)t sin t 6 2t as t. ... cos(β. 2). β. I(x) =. (bc)λβ(ac)λβ. f(c. y. λβ. )exp. (iλψ.

17 Jan 2023: bb = 1 bc = 0 cc = 1. This means a is perpendicular to the bc face b is perpendicular to the ac face and c is perpendicular to the ab face. ... Theta vs 2Theta. Homework 5. a=15.168 b=10.348 c=15.847 alpha=90.0 beta=100.035 and gamma=90.0.

29 Nov 2023: 10if (ii) holds then (i)(a) is equivalent to (i)(b).11(ii) gives that (x,x) R. ... a2 3b2 2c2). Thus we see quickly that b = 0, a 2c = 0 and a2 2c2 = 6.