Calculus On Manifolds: A Modern Approach To Classical Theorems Of Advanced CalculusAvalon Publishing, 1965 - 160 ページ This little book is especially concerned with those portions of ”advanced calculus” in which the subtlety of the concepts and methods makes rigor difficult to attain at an elementary level. The approach taken here uses elementary versions of modern methods found in sophisticated mathematics. The formal prerequisites include only a term of linear algebra, a nodding acquaintance with the notation of set theory, and a respectable first-year calculus course (one which at least mentions the least upper bound (sup) and greatest lower bound (inf) of a set of real numbers). Beyond this a certain (perhaps latent) rapport with abstract mathematics will be found almost essential. |
この書籍内から
検索結果1-3 / 63
47 ページ
... Proof . Each subrectangle S of P is divided into several sub- rectangles S1 , . . . , Sa of P ' , so v ( S ) = v ... proof for upper sums is similar . 3-2 Corollary . If P and P ' are any two partitions , then L ( f , P ' ) ≤ U ( ƒ , P ) ...
... Proof . Each subrectangle S of P is divided into several sub- rectangles S1 , . . . , Sa of P ' , so v ( S ) = v ... proof for upper sums is similar . 3-2 Corollary . If P and P ' are any two partitions , then L ( f , P ' ) ≤ U ( ƒ , P ) ...
59 ページ
... Proof . Let PA be a partition of A and PB a partition of B. Together they give a partition P of AXB for which any ... proof of the last inequality is entirely analogous to the proof of the first . Since f is integrable , sup { L ( ƒ , P ) ...
... Proof . Let PA be a partition of A and PB a partition of B. Together they give a partition P of AXB for which any ... proof of the last inequality is entirely analogous to the proof of the first . Since f is integrable , sup { L ( ƒ , P ) ...
67 ページ
... proof is very simple : if F ' = f , then ( Fo - = ( fog ) · g ' ; thus the left side is F ( g ( b ) ) F ( g ( a ) ) , while the right side is - Fog ( b ) Fog ( a ) = F ( g ( b ) ) − F ( g ( a ) ) . - We leave it to the reader to show ...
... proof is very simple : if F ' = f , then ( Fo - = ( fog ) · g ' ; thus the left side is F ( g ( b ) ) F ( g ( a ) ) , while the right side is - Fog ( b ) Fog ( a ) = F ( g ( b ) ) − F ( g ( a ) ) . - We leave it to the reader to show ...
他の版 - すべて表示
多く使われている語句
boundary bounded function calculus called closed curves closed rectangle continuously differentiable coordinate system Define f defined by f(x,y definition denoted Df(a Dif(a differentiable function div F dx¹ equation f(a¹ f(x¹ Figure finite number Fubini's theorem function f ƒ and g ƒ is differentiable ƒ is integrable Hint inner product interior Jordan-measurable k-dimensional manifold k-form k-tensor least upper bound Let A CR Let f lim h→0 linear transformation matrix Michael Spivak Möbius strip ms(f n-chain notation open cover open rectangle open set open set containing orientation-preserving partial derivatives partition of unity Problem Proof prove Show that ƒ singular n-cube Stokes subrectangle subset Suppose Theorem 2-2 unique usual orientation V₁ vector field vector space volume element Ʌ dx Ʌ dxi Σ Σ