MATH 2B Lecture Notes - Lecture 16: Wart

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

ey table is seen below xeecions tor Chapter 1 12, Fori, b € Ζ.el * b-aa (0 (D 14 ForVEQ r the operation defined in Example 1 on the wr John Sue Henry. P iDFor 15ine if is ass ociative by determining if the final six engu defined ang property (ii)of Theorem 1 . So no element i if the operntion debined in Example 1 of Emle if each element has an inverse Section 1.3 Groups iw comutatityn ild Chapter 3 where the Funda the exercises that follow etermining why they are not Write out the Cayley table for the gru(o)nud lensify thee element the 10 elements of the group Z 18 its operation uses ts in the first coordinnte and +2 in the secotu Ldessity the Zn and write out the Cayley table. Reeall th of each element does Theorem 1.14 tell us aont the entries in the Cayley table ol s xrewp that is set with exactly one element. A {n), will always form group since there 19. What nly one way to define an operation on the set. Be sure to shhow that the properi combination of the usual notice what set the rule is be of a group all hold. 21. We can create even create groupe with games! Consider four cups placed in a ssare patter table. If we have a penny in one of the cups there are four ways we can move it to t cup: Horizontally, Vertically, Diagonally, or Stay nehere it is. We will p, S. To define nn operation, consider two movements in a row, v.e.,m ove the penny as r tells us to, then after that move the penny as y instruets. the set of integers the set of real numbers n the set of positive integers. e set of positive integers. n the set of positive integers. on the set of integers e set of negative integers. we ple, HV D since if we first move it horizontally and then vertically nltogether e moved it diagonally. Create the Cayley table for this group, identify the identity and the inverse of cach element 22. Prove that the set A 0 under matrix multiplication. Is it abelian? 23. Verify that 3(R) is a group using the addition of funetions as defined in Example 13 e set of nonzero rational n he set of nonzero real num n on the set R (-1 Section 1.4 Basic Algebra in Groups 24. Prove the Cancellation Law (Theorem 1.20). Do not use any theorems that occur after it in the text ssociative, commutative, has roof or counterexample fo 25. Assume G is a group and a e G. Consider a fixed integer k and use PMI (Theorem 0.3) to prove that for all nEN, a aYou will need to consider cases here since k can be positive, negative, or o 26. Prove Theorem 1.23 (ii) for the case of n > 0 and m

must understand functions to be able linearize. First let's review what graphs of certain functions looks like. Sketch the shape of each type of y vs. x function below. k is listed as a generic constant of proportionally. Linear y = kx Inverse y = k/x Inverse Square y = k/x^2 Power y = kx^2 You will notice that only the linear function is a straight line. We can easily find the slope of our line by measuring the rise and dividing it by the run of the graph or calculating it using two points. The value of the slope should equal the constant, k, from the equation. Finding k is a bit more challenging in the last three graphs because the slope isn't constant. This should make sense since your graphs aren't linear. So how do you calculate the constant, k? You need to transform the non-linear graph into a linear graph in order to calculate a constant slope. You can accomplish this by transforming one or both of the axes for the graph. The hardest part is figuring out which axes to change and how to change them. The easiest way to accomplish this task is to solve your equation for the constant. Note in the examples from the last page there is only one constant, but this process could be done for other equations with multiple constants. Instead of solving for a single constant, put all of the constants on one side of the equation. When you solve for the constant, the other side of the equation should be in fraction form. This fraction gives the rise and run of the linear graph. Whatever is in the numerator is the vertical axis and the denominator is the horizontal axis. If the equation is not in fraction form, you will need to inverse one or more of the variables to make a fraction. First let's solve each equation to figure out what we should graph. Then look below at the example and complete the last one, a sample AP question, on your own. State what should be graphed in order to produce a linear graph to solve for k. Inverse Graph Vertical Axis: _____ Horizontal Axis: _____ Inverse Square Graph Vertical Axis: _____ Horizontal Axis: _____ Power (Square) Graph Vertical Axis: _____ Horizontal Axis: _____