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12 Nov 2023: b) 2(cos θ θ sin θ)/ cos2 θ. (c) t(1 2 ln t). ... a tan1 y. a). H14. (b) J = 1b(1 eax cos bx) a.

1 Nov 2023: 5. Diagonalise the quadratic form in R2 defined by. F(x, y) = (a cos2 θ b sin2 θ) x2 2(a b)(sin θ cos θ) xy (a sin2 θ b

23 Oct 2023:   cos φ sin φ 0sin φ cos φ 00 0 1. ... ψ̇ φ̇ cos θ. . (b) [optional extra] Calculate the matrix B = R>Ṙ and verify that it is antisymmetric.

3 Oct 2023: 13. Calculate the following sums:. (a)5. n=1 sin(nθ). (b)N. n=1 cos(nθ). ... a) (x,y) = (a cos θ,b sin θ),. (b) (x,y) = (a cosh θ,b sinh θ),.

6 Oct 2023: I0(x) =1. π. π. 0. ex cos θ dθ. Oscillatory Integrals and Stationary Phase. ... a). 1. 0. cos(. x tp). dt , with p > 1 , real ,. (b). π. 2. 0. (. 1 (. 2θ. π. ))γ. cos(.

24 Nov 2023: 26). 4. If a b = 0, and a, b 0 then cos θ = 0, necessarily, i.e. ... We have. a b = ab cos θ,(2, 1, 0) (3, 0, 1) =. 5. 10 cos θ,. cos θ = 6/. 50,. and thus θ = cos1(6/.

1 Dec 2023: X = 0 is given byF′ G = c0, (cos X σ0) G sin XF = 0, (4.4a,b). ... Θ = Ae(1κ2)τ cos κX Z, W = Ae(1κ2)τ cos κX, (5.1a,b)for any amplitude A.

1 Dec 2023: sin α) 2π(2 cos φ, sin φ)/ log Bi1 (see (3.2)). ... Ṽs =U cos(φΦ), ω̃=URH. sin(φΦ). (4.5a,b). Dow. nloa. ded. from. htt.

24 Jul 2023: Ω sed = mg16πa2η. ln(a. b. )(κ̃x sin θ κ̃y cos θ. ... sin θ, (3.40). Usedx = mg. 8πaηln(a. b. )cos θ sin θ, (3.41).

10 Oct 2023: a) σiσj = Iδij iǫijkσk,. (b) σ2σσ2 = σ,. (c) exp(iθn σ/2) = I cos(θ/2) in σ sin(θ/2). ... t(A B) t2. 2[A, B]. t3. 12{[A, [A, B]] [B, [B, A]]}. ).

23 Oct 2023: q1 = Q1 cos λ P2 sin λ, q2 = Q2 cos λ P1 sin λ,. ... p1 = Q2 sin λ P1 cos λ, p2 = Q1 sin λ P2 cos λ.

19 Oct 2023: 0.0380] [0.0177] [0.0083] [0.0052]. Log(maint cos ts /m2) - - - -0.0529- - - [0.0139]. ... Ene rgy clas s FG 0.0003[0.0528]. Log(maint cos ts /m2) 0.1493 [0.0545].

20 Jan 2023: φ(s,v) = α(s) r(n(s) cos v b(s) sin v). where n = n(s) and b = b(s) are the normal and binormal vectors of α. ... φ(u,v) = ((2 v sin(u/2)) sin u, (2 v sin(u/2)) cos u,v cos(u/2)).

17 Jan 2023: V=abc(1-cos2β)1/2=2449.39b=1/b=0.096637a=bcsin(α)/V=bc/V=0.066949c=absin(γ)/V=ab/V=0.064081α=γ=90 cos(β)=-cos( ... β)=0.174249 β=79.965. aa=1=|a||a|cos(ang)=0.06694915.168cos(ang) ang=10.02.

3 Oct 2023: a b = ab cos θ. (23). a. bθ. Figure 10: Angle used in the definition of the scalar product a b. ... 26). 4. If a b = 0, and a, b 0 then cos θ = 0, necessarily, i.e.

26 Jan 2023: The general solution is. x(t) = X R sin(!Bt ) and y(t) = Y R cos(!Bt ) (1.2). ... m!B. Y = y(t) R cos(!Bt ) = y ẋ! B= y.

1 Dec 2023: the boundary condition to be imposed is. u=cos θ and τrθ =%Bi sgn(y) at r =1 (2.7a,b). ... balance. But FEAE′ =2FAE, where FAE =rEI Bi[2(πβ2) cos β2 sin β2] is the force on the section AE (see appendix B.1).

1 Dec 2023: for transverse waves and. ψ =0, ψy =aπ cos πx0, on x=x0 a sin πx0 and y=0, (2.17a,b). ... 12 π. 2y= 14 π sin(2πy) cos(2πx)12 π. 2y, (3.11). Dow.

16 Oct 2023: λk(A) =λk(C). λk(B)=. 12µsin2 12πkh1 2µsin2 12πkh. |λk(A)| 1, k = 1.M. ... iγ where γ = 2 cos(πh) for = 1,.

4 Feb 2023: a) U = {(u,v) R2 |u > v}, σ(u,v) = (u v, 2uv,u2 v2);(b) U = {(r,z) R2 |r > 0}, σ(r,z) = (r cos(z),r ... σ(u,v) = (sech u cos v, sech u sin v, tanh u).

5 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.

1 Dec 2023: u= eσ cos(2πy). I0(σ)1, v=0. (3.2a,b). Given this fixed, known velocity the concentration simply evolves via the advection–diffusion equation (2.7). ... u(x, y, t)= eσ cos(2πy)Rc′(x,y,t) 1. 0eσ cos(2πs)Rc. ′(x,s,t) ds1. (4.4).

17 Oct 2023: A B) = B ( A) A ( B). (A B) = (B )A (A )B A( B) B( A). ( ... Bound orbit in point mass potential:. r =a(1 e2. )1 e cos ϕ.

29 Sep 2023: b) Starting from M = E e sin E, r = a(1 e cos E) and h = na2. ... cos g =(1 ra) e. 2. e(1 ra) e, (6). and so that cos g = cos M O(e2).

1 Dec 2023: pxpy. )=S(Q;B, 1). (cos θsin θ. )and. (ψxψy. )=Q. (sin θcos θ. ), ... x, y)= 13 B2a2(cos 3θ 3 cos θ, sin 3θ 3 sin θ).

26 Mar 2023: β s. [h3(. hs. cos α sin α)]. (3.11). We note that (3.11) reduces to (2.1a,b) only under the assumption |h/s| | tan α|. ... sk1N = skN vk1N t, vk1N =(. hk1n1)3. cos α. 3(skN sn1), (A3a,b).

1 Dec 2023: tan δ = (Ws 1) sin Φ(Ws 1) cos Φ Q, (2.18). ... Fx(δ, n, Bi) Bifx(|δ|) sgn(δ) Fz(δ, n, Bi) Bifz(|δ|) sgn(cos δ), (3.6a,b)for some functions fx and fz.

31 Aug 2023: cos(. kx φ1 φ22. )︸ ︷︷ ︸. g(x). = f (t) g(x). ... 0. q(t) = Aeω02Q t cos(ωt ϕ) q(t) = (At B)eω0t q(t) = Ae.

2 Aug 2023: 10. sin 2𝜔𝑓 2 cos 𝜔𝑓 = sin 2𝜔0 2 cos 𝜔0 , (2.5). ... volume 𝑉 = 𝑉0, where. 𝑉 = 22(sin 𝜔1 sin 𝜔3) cos 𝜔1 = 𝑉0.

6 Feb 2023: a) U = {(u,v) R2 |u > v}, σ(u,v) = (u v, 2uv,u2 v2);(b) U = {(r,z) R2 |r > 0}, σ(r,z) = (r cos(z),r ... σ(u,v) = (sech u cos v, sech u sin v, tanh u).

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.