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TOEIC・英語 大学生・専門学校生・社会人

「教科書のこの部分の要点を日本語で書け」 なんて書けばいいでしょうか😵‍💫😵‍💫

10-2 Expressing the Past: Necessity, Advisability, Expectation PRESENT:(a) Julia has to get a visa. (b) Julia has got to get a visa. (c) Julia must get a visa. Past necessity: had to In (d): had to needed to: Julia needed to get a visa. There is no other past form for must (when it rmeans neceasity) or have got to. PAST: (d) Julia had to get a visa. PRESENT:(e) I should study for the test. I want to Past advisability: pass it. (f) lought to study for the test. (g) T had better study for the test. should have ought to have + past participle In the past, should is more common than ought to. The past form of had better (had better have) is almost never used. PAST: Ifailed the test. (h) Ishould have studied for it. (i) lought to have studied for it. (i) I shouldn't have gone to the movies the night before. The meaning in (h) and (i): Studying was a good idea, but 1 didn't do it. I made a mistake. The meaning in (j): It was a bad idea to go to the movies. I made a mistake. Usual pronunciation of should have: “should-ev" or “should-e." lao was/were supposed to: unfulfilled expectation or obligation in the past PRESENT:(k) We are supposed to leave now. PAST: (1) We were supposed toleave last week. PRESENT:(m)The mail should be here. Should have + past participle: past expectation The speaker expected something to happen%; it may or may not have occurred, as in (n).. PAST: (n) The mail should have been here by now.

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数学 大学生・専門学校生・社会人

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.

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