<p><strong>Online Nursing Class Course Design Principles</strong></p> <p>In the rapidly evolving landscape of education, online nursing ... 続きを読む

多様体を構成するために、位相空間に完全アトラスを導入するところで質問です。 完全アトラスを導入するメリットとして、この文章の下線部を「異なる座標系を用いたのに同じ計算ができてしまうという問題が解消される」解釈したのですが、そこがよくわかりません。座標系を変えて計算する... 続きを読む

1 Two n-dimensional coordinate systems & and ŋ in S overlap smoothly provided the functions on¯¹ and ŋo §¯¹ are both smooth. Explicitly, if : U → R" and ŋ: R", then ŋ 1 is defined on the open set ε (ur) → ° (UV) V and carries it to n(u)—while its inverse function § 4-1 runs in the opposite direction (see Figure 1). These functions are then required to be smooth in the usual Euclidean sense defined above. This condition is con- sidered to hold trivially if u and do not meet. Č (UV) R" Ĕ(U) n(UV) R" S n(v) Figure 1. 1. Definition. An atlas A of dimension n on a space S is a collection of n-dimensional coordinate systems in S such that (A1) each point of S is contained in the domain of some coordinate system in, and (A2) any two coordinate systems in ✅ overlap smoothly. An atlas on S makes it possible to do calculus consistently on all of S. But different atlases may produce the same calculus, a technical difficulty eliminated as follows. Call an atlas Con S complete if C contains each co- ordinate system in S that overlaps smoothly with every coordinate system in C. 2. Lemma. Each atlas ✅ on S is contained in a unique complete atlas. Proof. If has dimension n, let A' be the set of all n-dimensional coordinate systems in S that overlap smoothly with every one contained in A. (a) A' is an atlas (of the same dimension as ✅).

問2.1の証明が分かりません。 ※1枚目が質問内容、2枚目が仮定

問 2.1 例1 (b), (c) で R" に定義された各種の距離 dp : R" × R” → [0,∞) (p = 1,2,...,∞) において, R” の点列 πm:= (x(m),x(m),...,xmm))∈R(m= R" 2 1,2,･･･) が, 点æ= (π1, 2,...,πn) ∈R" に収束するためには,各k ∈ {1, 2,...,n} に対し (m) →πk (m→8) となることが必要十分であることを示せ.

10,68の答えがどうしてこのようになるか教えてください。 分野は重積分のストークスの定理です

By Green's theorem in space (divergence theorem). Prove that that (V x A) - n ds for any closed surface S. S Prove that 10.66. dS ff n ds = 0. where n is the outward drawn normal to any closed surface S. (Hint: Let A = Oc, SS S where c is an arbitrary vector constant.) Express the divergence theorem in this special case. Use the arbitrary property of c. 10.67. If n is the unit outward drawn normal to any closed surface S bounding the region V, prove that fff div n dv = S V Stokes's theorem 40.68. Verify Stokes's theorem for A = 2yi + 3xj - z²k, where S is the upper half surface of the sphere x² + y² + ² = 9 and C is its boundary. Ans. Common value = 9T 10.65. , y = 0,

こんにちは。線形代数の質問です。 この証明の解き方を教えてもらいたいです。 Prove the following Theorem: Let V ⊂ R^n be a subspace of R^n. Then there exists a subspace W ... 続きを読む

定積分の計算で∫∂/∂x～～～dxというような形の式があるのですが、この場合 ∂/∂xがあるので先に～～～の部分をxで微分してからxで定積分するということでしょうか。

This assumption may be valid because the air column being considered is deeper than the nocturnal turbulent boundary layer. QE is the latent heat flux which is released into the air column with the phase change of water vapor. When there is no phase change of water vapor, QE=0. QR is the heat flux of the long wave radiation and defined by QR=pC * a 0z where (30 /a1), s the warming rate of air due to convergence of long wave radiation, L t(z) and L(2) are the upward and downward long wave radiation at a height of z respectively. Among several kinds of calculation methods for long wave radiation (Stephens, 1984), formulae

解き方教えて欲しいです

問、Determine if each of the following subsets of R* is a subspace of R and explain the reason. [以下の部分集合が R'の部分空間である か否かを判定し,その理由を説明せよ.] E R| 2, + 2,-130 E。 三 1+ (2) W。 E R|ER ニ 2+ 2c E R D, (3) Ws = 20 2 E, ER? E, ミ0 (4) W。 ニ 2 2 ( Ws 2," + 2, +13D0 C,

(ii)と(iii)がどうしてこうなるのか分かりません お願いします

b. [1+] Let t(n) be the number of total partitions of n, as defined in Exam- ple 5.2.5. Let g(n) have the same meaning as in Exercise 5.26. Deduce from (a) that g(n) = 2"t(n) for n >1. c. [2+] Give a simple combinatorial proof of (b). 5,37. a. [2+] Let 1=D po(x), pi(x), be a sequence of polynomials (with coeffi- cients in some field K of characteristic O0), with deg pn=n for all nE N. Show that the following four conditions are equivalent: ) Pn(x + y) =DE>o (") Pe(x)pnーk(y), for all n eN. (i) There exists a power series f(u)=aju+azu'+ E K [[u]] such that と P(x)- un expxf(u). (5.110) n! n>0 仮定 NOTE: The hypothesis that deg pPn=nimplies that aj ¥ 0. () E20 Pa(x) = (E>0 Pn(1)). (iv) There exists a linear operator Q on the vector space K[x] of all poly- nomials in x, with the following properties: ●Ox is a nonzero constant ●Qis a shift-invariant operator, i.e., for all aeK,Qcommutes with the shift operator E4 defined by E® p(x)=D p(x +a). ● We have Qpn(x) =D npn-1(x) for all n e P. NOTE: A sequence po, Pi, .. . of polynomials satisfying the above con- ditions is said to be of binomial tvpe. The operator Q is called a delta

速度の問題

Ifa ball is thrown into the air with a velocity of 40 ft/s, its height in feet after t seconds is given by y= 40t - 162. (a) Find the average velocity for the time period beginning with t =D 2: (1).5 second (2).1 second (3) .05 second (4) .01 second (b) Find the instantaneous velocity when t 3 2. ft/s ft/s ft/s ft/s ft/s

多様体の接空間に関する基底定理の証明です。g(q)=∫〜と定義した関数を微積分学の基本定理を用いながら変形してg(q)=g(0)+∑gᵢuⁱと導出するのですが、これがうまくいきません。 自分は、g(q)の式をまず両辺tで微分して、次に両辺uⁱで積分して、最後に両辺tで積分... 続きを読む

12. Theorem.If{ = (x', , x") is a coordinate system in M at p, then its coordinate vectors d, lp, …… 0,l, forma basis for the tangent space T,(M); and D= E(x) 。 i=1 for all ve T(M). Proof. By the preceding remarks we can work solely on the coordinate neighborhood of G. Since u(c) = Othere is no loss of generality in assuming ど(p) = 0eR". Shrinking W if necessary gives E(W) = {qe R":|q| < } for some 8. Ifg is a smooth function on E(W) then for each 1 <isndefine og (tq) dt du g(9) = for all qe {(W). It follows using the fundamental theorem of calculus that g= g(0) + E&,u' on (W). Thus if fe &(M), setting g = f。' yields f= f(P) + Ex on U. Applying d/ax' gives f(p) = (f /0x)(P). Thus applying the tangent vector e to the formula gives (f) = 0+ E(x'(p) + E Ap)u(x) = E(Px). ず ax Since this holds for all f e &(M), the tangent vectors v and Z Ux') d,l, are equal. It remains to show that the coordinate vectors are linearly independent. But if ) a, o.l, = 0, then application to x' yields dxi 0=24 (P) = 2q d」= 4. In particular the (vector space) dimension of T,(M) is the same as the dimension of M.