統計の問題です。 この問題の解き方がどうしてもわからないです💦 答えだけでもいいので教えていただけませんか？

Assignment 3 (From Lecture 4, 5.)< Suppose we get samples from a population with a distribution of 300 mean and 20 standard deviations. In this case, answer the following questions.< ← (1) What will be the sample distribution of the sample mean when the sample size is 400?< (2) What will be the sample distribution of the sample mean when the sample size is 1600?< ←

こちらの穴埋め分かる方いらっしゃいますか？

[1] Monopoly Suppose a pharmaceutical firm as a patented monopoly that has a plan to supply a newly developed good q. This good requires R&D cost of k> 0 as a fixed cost but is supplied at a constant marginal cost of c> 0. Once launched in a market, the market demand is estimated as shown in (1) with a > 0. Then, answer the quizzes [1.1] and [1.2]. 4a ・(1) 3√9 [1.1] If the firm commercializes good q, how much would q and p be set? Select numbers appropriate to the blanks (i) - (iv) in (2). (iii) (i) • a C and P = (iv) . C (2) (ii) 9 = P = [1.2] Show an interval of k satisfying that the firm creates this market. Select numbers appropriate to the blanks (v) - (vii) in (3). (vii) (v) a = (0, (vi) . C ] (3)

大問3の解答はこれでいいでしょうか？ また、大問2の(2)はおそらくm=5なのですが、ひとつひとつ合成して答えを出すのはいけないでしょうか？

[2] 複素数を成分とする3次正方行列全体のなす複素ベクトル空間をVとする. NEV をN2 ≠ 0 および №3 = 0 をみたすものとする. ただし, 0は3次零行列を表す. (1) 3項列ベクトルu∈C3 で, ベクトル N2u, Nu, u がC上線形独立となるものが 存在することを示せ. (2) 線形写像 f: V→Vを f(x) = NX - XN (X EV) で定めるとき, fm = 0 となる最小の正の整数m を求めよ.ただし, fm は f の m m 回合成写像 fm = fo... of を表す.

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,

問題としてはこのURLのやつでexercise2.2.9の問題です。 2.2.9. Define T : ℓ^2(Zn ) → ℓ^2(Zn ) by (T(z))(n) =z(n + 1) − z(n). Find all eigenvalues of T.... 続きを読む

16:22マ l 全 の Exerc: 164/520 matrices, convolution operators, and Fourier r operators. 2.2.9. Define T:l'(Zn) - → e°(ZN) by ニ Find all eigenvalues of T. 2.2.10. Let T(m):e'(Z4) → '(Z) be the Fourier multipliei (mz)' where m = (1,0, i, -2) defined by T (m)(2) = i. Find be l(Z4) such that T(m) is the convolutior Tb (defined by Th(Z) = b*z). ii. Find the matrix that represents T(m) with resp standard basis. 2.2.11. i. Suppose Ti, T2:l(ZN) → e(ZN) are tra invariant linear transformations. Prove that th sition T, o T, is translation invariant. ii. Suppose A and B are circulant NxN matric directly (i.e., just using the definition of a matrix, not using Theorem 2.19) that AB is Show that this result and Theorem 2.19 imp Hint: Write out the (m + 1,n+1) entry of the definition of matrix multiplication; compare hint to Exercise 2.2.12 (i). iii. Suppose b,, bz e l'(Zn). Prove that the cor Tb, o Tb, of the convolution operators Tb, and convolution operator T, with b = 2 bz * b.. E Exercise 2.2.6. iv. Suppose m,, mz € l"(Z). Prove that the cor T(m2) ° T(m) and T(m) is the Fourier multiplier operator T) m(n) = m2(n)m」(n) for all n. v. Suppose Ti, T2:l"(Zw) → e'(Zn) are linear tra tions. Prove that if Ti is represented bya matri respect to the Fourier basis F (i.e., [T; (z)]F =A Tz is represented by a matrix Az with respect t the composition T20T, is represented by the ma with respect to F. Deduce part i again. Remark:ByTheerem 2.19, we have just proved of the Fourier multiplier operat Aresearchgate.net - 非公開

(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

多様体の接ベクトルに関する性質の証明なんですけど、(1)が分かりません。 素直に訳すと、(1)は「多様体上の滑らかな関数fとgが点pの座標近傍で同じなら、その接ベクトルv(f)とv(g)は同じになる」だと思うんですけど、証明でf(p)=0,g(p)=1と明らかに異なる値をと... 続きを読む

11. Lemma. Let ve T,(M). (1) If f, ge F(M) are equal on a neighbor- hood of p, then Uf) = v(g). (2) If he F(M) is constant on a neighborhood of p, then u(h) 0. 三 Proof.(1) By linearity it suffices to show that if f = 0 on a neighbor- hood W of p, then u(f) = 0. Let g be a bump function at p with support in W; then fg = 0onall of M. But v(0)= u(0 + 0) = u(0) + u(0) implies v(0) = 0. Thus 0= (fg) = v(S)g(p) + f(p)v{g) = u(f), since f(p) (2) By(1) we can assume that h has constant value c on all of M. If 1 is the constant function of value 1, then 0 and g(p) 1. ニ (1) = (1.1) = (1)1 + lu(1) = 2v(1). Hence v(1) = 0, and v(h) = u(c· 1) = cu{1) = 0.

(2)教えてください。(1.1)、(1.2)どちらでも大丈夫なので教えていただけるの嬉しいです。幾何です。

3| 曲面のパラメータ表示 p:U→ R3 (p € C®(U))を与え,座標曲面 S= p(U) を考える。また,曲線c = c(s) : I→ U (ce C®(I)) を考え, 7(s):= (po c)(s) :I→Sを測地線とする.このとき次の問に答えよ。 (1) (s) の速度ベクトルの大きさ ||(s)|| は, dy (S)|| = const for Vt EI ds を満たすことを示せ、ここで, const とは定数(constant) の略記号のことで ある。 注:したがって,パラメータsは,?の孤長パラメータの定数倍となる。 (2) パラメータ変換s= £(t) (t € I)を行うと,曲線う(t):= 7(E(t)) は,あ る関数 p(t) e C(1) が存在して、 霊 - d的 ()= A))or t eī = p(t (1t) for tei dt2 を満たすことを示せ、ここで(…)"は,(…)のS接成分を表す。これを座 標曲面Sのパラメータ表示を用いた方程式で表すと, d'ck de dek -(t) = p(t). dt -(t)(k= 1,2) for tef dt dt? dt を満たすことと同値である。(式(1.1), (1.2) のどちらを示してもよい。)

有識者の方解説お願いしたいです。

曲面のパラメータ表示 p:U→ R° (p e C®(U)を与え,座標曲面 S= 9(U) を考える.また,曲線c= c(s) :I→ U (ce C®(I)) を考え, 7(5):= (poc)(s) : I→Sを測地線とする.このとき次の問に答えよ。 (1) (s) の速度ベクトルの大きさ |会(s)|| は, dy = Const for Vt E I ds を満たすことを示せ、ここで,const とは定数 (constant) の略記号のことで ある。 注:したがって,パラメータ sは, yの弧長パラメータの定数倍となる。 (2) パラメータ変換s= {(t) (t e Ii) を行うと,曲線(t) := (E(t)) は,あ る関数 p(t) e Co (ī) が存在して, ds (()) = p()() for tei T dy dt を満たすことを示せ、ここで(…)" は,(…)のS-接成分を表す。これを座 標曲面Sのパラメータ表示を用いた方程式で表すと, dck ( (%3D 1,2) for teI dPck dc dei -(t) =D p(t). dt? dt dt dt を満たすことと同値である.(式(1.1), (1.2) のどちらを示してもよい.) 注:測地線y=(s) は, 弧長パラメータの定数倍を用いて求められるが,上 記の(1)より,式(1.1) または式(1.2) を測地線の定義としてもよいことが分 かる。ただしこの場合,(t) のパラメータtは,もはや一般に弧長パラメー タの定数倍としては与えられない.また式 (1.1) は,「測地線とは,座標曲面 S上の加速度が速度に各点で比例している曲線」とも解釈出来ることを表し ている。