Search

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).

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)).

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.

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).

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.

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.

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.

1 Feb 2023: Let B be a ball in R2, and letf Cc(R2) be supported in B. ... Let x0 B and let L0 be the tangentline to B through x0.

31 Aug 2023: 11. CHAPITRE 3. SÉRIES NUMÉRIQUES. IV Quelques applications. Application 3.17. (cos 1) ̸ Q. ... n 0, d’où :. cos 1 =n=0. (1)n. (2n)! Comme (1)n.