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 №  Condition free/or 0.5$
m111. Let f, g: A ->R be integrable. a. For any partition of and any subrectangle of , show that and and therefore and . doc
m1314. If is a closed rectangle, show that is Jordan measurable if and only if for every there is a partition of such that , where consists of all subrectangles intersecting and consists of allsubrectangles contained in .
m54(a) Let A C Rn be an open set such that boundary A is an (n - 1) -dimensional manifold. Show that N = AU boundary A is an -dimensional manifold with boundary. (It is well to bear in mind the following example: if A = {x &#1028;Rn}: |x| < 1 or 1 < |x| < 2}, then N = AU boundary A is a manifold with boundary, but &#8706; N &#8800; boundary A. (b) Prove a similar assertion for an open subset of an n-dimensional manifold. buy
m159a. If c: [0, 2&#960;] x [-1, 1] -> c: [0 , 2&#960;] x [-1, 1] -> R3 is defined by c (u,v) = (2 eos (u) + vsin (u/2) eos (u), 2sin (u) + vsin (u/2) sin (u), veos (u/2)). buy
m160a. If C is a set of content 0, show that the boundary of C also has content 0. b. Give an example of a bounded set C of measure 0 such that the boundary of C does not have measure 0. buy
m161a. If f: |a, b|-> R is non-negative and the graph of f in the x,y -plane is revolved around the -axis in R3 to yield a surface M, show that the area of M is buy
m162a. If f is a differentiable vector field on M C Rn, show that there is an open set A&#1069;M and a differentiable vector field F on A with F(x) = F (x) for x&#1028;M. b. If M is closed, show that we can choose A = Rn. buy
m163a. If f : R -> R satisfies f1 (a) &#8800; 0 for all a €R, show that f is 1-1 on all of R. buy
m164a. If M is a k-dimensional manifold in Rn and k < n, show that M has measure 0. b. If M is a closed -dimensional manifold with boundary in Rn, show that the boundary of M is &#8706;M. Give a counter-example if M is not closed. c. If M is a compact -dimensional manifold with boundary in Rn, show that M is Jordan-measurable. buy
m165a. Let &#906;: Rn -> Rn be self-adjoint with matrix A = (aij), so that aij = aji. If f (x) = <Tx, x> =&#931; aij xixj, show that Dkf (x) = 2 &#931;j = 1 akjxj. By considering the maximum of <Tx, x>on Sn-1 show that there is x&#1028;Sn-1 and ^ &#1028;R with Tx = ^x. b. If V = {y&#1028;Rn: <x, y> = 0}, show that &#906;(v) CV and &#906;: V and &#906;: V -> V is self-adjoint. c. Show that &#906; has a basis of eigenvectors. buy
m166a. Let f : R -> R be defined by F (x) = {x 2 sin 1/x) x &#8800; 0, 0 x = 0. buy
m167a. Let f: Rn -> R be a continuously differentiable function. Show that f is not 1-1. b. Generalize this result to t the case of a continuously differentiable function f: Rn -> Rm with m < n. buy
m168a. Let g: Rn -> Rn be a linear transformation of one of the following types: buy
m169a. Show that an unbounded set cannot have content 0. buy
m170a. Show that the set of all rectangles [a1, b1] x . x [an, bn] where each ai and each bi are rational can be arranged into a sequence (i.e. form a countable set). b. If A C Rn is any set and O is an open cover of A, show that there is a sequence U1, U2, U3,. of members of O which also cover A. buy
m171a. Show that Theorem 5-5 is false if M is not required to be compact. b. Show that Theorem 5-5 holds for noncom-pact M provided that w vanishes outside of a compact subset of M. buy
m172a. Show that this length is the least upper bound of lengths of inscribed broken lines. buy
m173a. Suppose that f: (0, 1) -> R is a non-negative continuous function. Show that &#8747; (0, 1) exists if and only if lim &#1028;-> &#8747; c 1-c f exists. b. Let An = [1 - 1/2n, 1 - 1/2n +1] Suppose that f: (0, 1) ->R satisfies &#8747;Arf = (-1)n/n and f(x) = 0 for all x &#1028; Un An. Show that &#8747;(0,1)f does not exist, but lim&#1028;->&#8747;(&#1028;, 1 - &#1028;)f = log 2. buy
m175An absolute k-tensor on v is a function Vk ->R of the form |w| for w &#1028; Ak (V). An absolute k-form on M is a function such that n (x) is an absolute k-tensor on Mx. Show that &#8747;Mn can be defined, even if M is not orientable. buy
m219Applying the generalized divergence theorem to the set M = {X&#1028;rN: |x| < a} and , find the volume of Sn – 1 = {x&#1028;Rn: |x| = 1} in terms of the -dimensional volume of Bn = {x &#1028; Rn: |x| <1}. (This volume is &#960; n/2 / (n/2)! if is even and 2(n+1)/2&#960; (n-1)/2 1&#8729;3&#8729;5&#8729;. &#8729; N if is odd.) buy
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