🏷️ LIMITED TIME OFFER: GET 20% OFF GRADE+ YEARLY SUBSCRIPTION →

HomeStudy Guides420,000US230,000

MATH 13 Study Guide - Midterm Guide: Intermediate Value Theorem, Disjoint Sets

83 views4 pages

turquoisegnu128

15 Oct 2018

School

Department

Course

Professor

For unlimited access to Study Guides, a Grade+ subscription is required.

Document Summary

Problem 1. (a) for n = 1, 2, 3, . de ne the interval an = ( 1/n, n). Rewrite the intersection of n n an as an interval. For this problem you do not need to prove that your intervals(cid:84) answer is correct. [0, 1). (b) give an example of sets a and b with the property that p(a b) (cid:54)= p(a) p(b), and show that your sets have this property. (p means power set. ) Take a = {1, 2} and b = {1, 3}. Then the set {2, 3} is in p(a b). P q p = q (p = q) = p s. 1more generally, take any a and b such that neither is a subset of the other. Say x a b and y b a. Then {x, y} is in p(a b) but not in p(a) or p(b).

Related textbook solutions

Calculus

ISBN: 9781319050733

Single Variable Calculus: Early Transcendentals

4th Edition, 2018

ISBN: 9781337687805

CALCULUS:EARLY TRANSCENDENTALS

ISBN: 9781319050740

Related Documents

MATH 13 Study Guide - Midterm Guide: Joule, Contraposition, Subset

turquoisegnu128

MATH 13 Study Guide - Midterm Guide: Contraposition

turquoisegnu128

MATH 13 Study Guide - Final Guide: Disjoint Sets, Mathematical Induction, Distributive Property

turquoisegnu128

Related Questions

I have the answer for the following question. But some of theanswers are not correct.

Please do the following question by yourself and check which arethe wrong answers,please post your answers and compare to it.

Let r in and u =(1 1 0), v = (1 4 1), w = (r2 1 r2), b = (3+2r 5+12r 2r). For which values r in is the set {u. v, w} linearly independent? For which values r in is the vector b a linear combination of u. v and w? For which of these values of r can b be written as a linear combination of u. v and w in more than one way? The set 3 of all column vectors of length three, with real entries, is a vector space. Is the subset B = {(x y z) in 3 | xy + yz = 0} a subspace of 3? Justify your answer. Show that the set of all twice differentiable functions f : satisfying the differential equation sin(x) f"(x) + x2f(x) = 0 is a vector space with respect to the usual operations of addition of functions and multiplication by scalars. Here, f" denotes the second derivative of f. Let S be the following subset of the vector space 3 of all real polynomials p of degree at most 3: S = {p in 3 | p(l) = 0, p'(l) = 0}, where p' is the derivative of p. Determine whether S is a subspace of 3. Determine whether the polynomial q(x) = x - 2x2 + x3 is an element of S. u = (1 1 0), v = (1 4 1), w = (r2 1 r2) if {u, v, w} is linearly independent set 1(4r2 - 1)-1(r2 - r2) ne 0 4r2 ne 1 r ne plusmn1/2 this set is linearly independent set for values of e ne plusmn1/2 - - - - - - - (3+2r 5+12r 2r) = a1 (1 1 0) + a2 (1 4 1) + a3 (r2 1 r2) this forms system of equations in a1, a2, a3 if 4r2 - 1 ne 0 then system has unique solution if 4r2 - 1 = 0 and -2 - 4r = 0 rArr r = -1/2, the system has more than one solution if 4r2 - 1 ne 0 b can be written as linear combination of u,v,w as uniquely if r = -1/2 then b can be written as linear combination of u,v,w more than one way - - - - - - - - - (1 1 -1) in B and (1 0 1) in B (1 1 -1) + (1 0 1) = (2 1 0) not in B since 2times1+1times0 ne 0 so addition of vector is failed, this set is not forms subspace consider the following differential equation (sinx)f"(x) + x2f (x) = 0 each solution is continuous function the set of all solutions is subset of space of set of all continuous functions n from R to R let f,g are two solutions (sinx)f"(x) + x2f (x) = 0 (sinx)g"(x)+x2g(x) = 0 ccheck whether af+bg is solution or not (sinx)f"(x) + x2f(x) = 0 rArr (sinx) af"(x) + x2?f (x) = 0 (sinx)g"(x) + x2g (x) = 0 rArr (sinx)bg" (x) + x2bg (x) = 0 (sinx)(af"(x)+bg"(x)) +x2 (af (x) + bg(x)) = 0 af+bg is also solution to differential equation the set of all solutions to differential equation forms subspace - - - - - - - - - - consider the following subset of P3 {p in P3/.p(l)=0, p'(l) = 0} let p, q in S rArr p(l)= 0, p'(l) = 0 and q(l) = 0, q'(l) = 0 (ap + bq) (l) = ap (l) +b q(l) = a (0) + b (0) = 0 (ap + bq)'(1) = ap'(l) +bq'(1) = a (0) + b(0) = 0 p, q in S rArr ap +bq in S then S is subspace q(x) = x - 2x2 + x3 q(1) = 1 - 2 + 1 = 0 q'(x)=1 - 4x + 3x2 q'(l) = 1 - 4 + 3 = 0 q is element of S

Full step-by-step solutions will be chosen.