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### midterm07

Course: M 515, Fall 2009
School: Iowa State
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515 Fall MATH 2007 Midterm Exam Explain your answers carefully! 10 points each. 1. If A, B are measurable sets of finite measure, m(A) = m(B), and AEB show that E is measurable. Solution: Since A, B are measurable, so is A\B and m(B) = m(A) + m(B\A) so that m(B\A) = 0. We have E\A B\A so by monotonicity m (E\A) m (B\A) = 0 so that E\A is measurable. Finally it follows that E = A (E\A) is measurable. 2. Let f...

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515 Fall MATH 2007 Midterm Exam Explain your answers carefully! 10 points each. 1. If A, B are measurable sets of finite measure, m(A) = m(B), and AEB show that E is measurable. Solution: Since A, B are measurable, so is A\B and m(B) = m(A) + m(B\A) so that m(B\A) = 0. We have E\A B\A so by monotonicity m (E\A) m (B\A) = 0 so that E\A is measurable. Finally it follows that E = A (E\A) is measurable. 2. Let f L1 (0, ) and g(y) = 0 f (x) sin (xy) dx Show that g is a bounded and continuous function of y on all of R. Solution: We have |g(y)| 0 |f (x) sin (xy)| dx 0 |f (x)| dx so that g is bounded on R. If yn y, then hn (x) := f (x) sin (xyn ) h(x) := f (x) sin (xy) a.e. The Dominated Convergence Theorem may then be applied, using |hn (x)| |f (x)|, to show g(yn ) = 0 hn (x) dx 0 h(x) dx = g(y) so that g is continuous at y. MATH 515 Fall 2007 Midterm Exam 3. If f BV [a, b], show that f 2 BV [a, b]. Is the converse true? Solution: Solution 1: If f = g - h where g, h are monotone increasing on [a, b] then f 2 = G-H where G = g 2 +h2 , H = 2gh. If g BV then it is bounded below by some constant C, and we can express g = (g - C) + C and therefore g 2 = (g - C)2 + C 2 + 2C(g-C). Since g-C 0 and monotone, it follows that (g-C)2 is monotone, and so g 2 is either monotone or the difference of monotone functions. Similarly for h2 and 2gh, so that f 2 can also be expressed as a difference of monotone functions. Solution 2: Since f BV [a, b] there exists M such |f that (x)| M on [a, b]. If P is any partition of [a, b] we then have n n |f (xk+1 ) - f (xk )| = k=1 k=1 2 2 |f (xk+1 ) - f (xk )||f (xk+1 ) + f (xk )| Therefore V (f 2 , P) 2M V (f, P) 2M V (f ) and so V (f 2 ) 2M V (f ) < The converse is false, for example if f (x) = -1 for x Q, f (x) = 1 otherwise, then f BV but f 2 is. (Thanks to Andy for pointing out the error in the original solution.) 4. If A f dx = 0 for every measurable set A E, show that f = 0 a.e. on E. Solution: Choose A = {x E : f (x) 0}. Then A is measurable, f 0 on A and A f (x) dx = 0 from which it follows that f = 0 a.e. on A. (This was a homework problem). Similarly f = 0 a.e. on the set {x E : f (x) 0}, and so f = 0 a.e. on E. Page 2 MATH 515 Fall 2007 Midterm Exam 5. Prove the Tchebychev inequality: If f L1 (E) and > 0 then m{x E : |f (x)| } 1 |f (x)| dx E Solution: Let E = {x E : |...

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Iowa State - M - 265
MATH 265 Section A Show your work! 1. (30 points) Let f (x, y) = (a) (b) (c) (d) x2Spring 2005 Do not write on this test page!EXAM 22 . +y What is the domain of f ? Sketch the level curve f (x, y) = 1. f f 2f Compute the partial derivatives , a
Iowa State - M - 520
MATH 520 HomeworkSpring 200655. Let A be an m n real matrix, b Rm and dene J(x) = |Ax b|2 for x Rn . (Here |x|2 denotes the 2 norm, the usual Euclidean distance on Rm ). a) What is the Euler-Lagrange equation for the problem of minimizing J?
Iowa State - M - 265
MATH 265Spring 2005Practice Final from Fall 20041. (a) Find parametric equations for the line through (2, -3, 1) parallel to the vector from (2, 0, 0) to (4, 3, 1). (b) Find an equation for the plane through the origin that is perpendicular to
Iowa State - M - 267
Assignment 11 answers 1 -2 1 0 2 1 0 1 #1. y(t) = C1 e-6t + C2 e-3t + C3 e10t + C4 e15t 2 1 0 -1 1 -2 -1 0 2 cos t - sin t 2 sin t + cos t + C2 e-t 5 cos t 5 sin t#2. y(t) = C1 e-t#3. If the geometric multiplicity is less than the
Iowa State - M - 516
MATH 516SPRING 2008Supplementary Homework 11. If , are signed measures, show that if and only if | . 2. If is a signed measure and = 1 - 2 where 1 , 2 are positive measures, show that 1 + 2 - (Here 1 + means that 1 - + is a positi
Iowa State - M - 267
MATH 267Section D2FALL 2004EXAM 11. (20 points) Find the solution of y4 y = 2x e y(0) = 1What is the existence interval for this solution? 6 2. (15 points) Find the general solution of x + x = t2 1. t 3. (15 points) Use the substitution v
Iowa State - M - 267
Exam 2 answers1 1. y(t) = 9 et + 24 e-3t - 1 cos 3t - 1 sin 3t 8 6 32. The period is T =2 3and the amplitude is A =19 33. det A = 6 which implies that the null space of A contains only the zero vector; Ax = [7 8 0]T 4. F (s) = 1 + e s-s
Iowa State - M - 267
Assignment 5 answers Page 674, #10. y(x) = C1 x5 + C2 x-1 Page 674, #24. y(x) = |x|1 -2C1 cos (3 2ln |x|) + C2 sin (5 + 4 - 1t 43 2ln |x|)Page 207, #22. y(t) = - 9 e-2 t - 4t Page 207, #32. yp (t) = 2 e-t - t17 -2 t 4 tePage 21
Iowa State - M - 267
Assignment 6 answers Page 223, #18. Transient solution is 17 -t 1 e sin t + e-t cos t 10 10 and the steady state solution is 1 1 sin 2t - cos 2t 5 10 Page 239, #8. F (s) =1 (s+3)2 1-e-3s s2Page 239, #28. F (s) = Page 246, #4. Y (s) =3 s-5 s2
Iowa State - M - 265
Outline of sections covered on Exam 1 1. Chapter 13 13.1 Parametric curves in the plane: nonparametric forms, slope and arc length 13.2 Vectors in the plane: geometric point of view, magnitude and direction 13.3 Algebraic calculations with vectors
Iowa State - M - 307
MATH 307FALL 2002QUIZ 2Name:1. (5 points) Compute Ax if 1 3 A= 0 2 1 4 0 2 1 -1 1 2 1 1 x= -2 5 6 1 2. (5 points) What is the rank of the matrix A = 2 0 0 2 2 ? 4 43. (10 points) Let T (x) = a) Find the matrix of T .3x1 - 5x2
Iowa State - M - 267
Exam 1 answers 1. y(x) = 2. x(t) =t3 9 3 -2x 2e-1 1 -3 2for x &lt;ln 3 2t C - 7 + t6t3. y(t) = e(C-t)e4. Check that y1, y2 both satisfy the equation and that they are linearly independent; y(x) = x - ex-2. 5. y(t) = et -4C1 cos (3
Iowa State - M - 267
Assignment 1 answers Page 28, #10: Interval of existence is11 6&lt; t &lt; .Page 28, #14: Interval of existence is - &lt; t &lt; . Page 39, #4: y = tan (ex + C) Page 39, #8: y = C(x - 1)ex Page 39, #16: y = - ln (2 - ex ) for x &lt; ln 2 Page 39, #36: T 56.18
Iowa State - M - 267
Assignment 2 answers Page 62, #8: y(x) = (1 + x3 ) Page 62, #24: y(t) =1 1+Cet C-x cos x ln |1+x3 | 3+CPage 85, #10: exact, y = Page 85, #20: exact, y = Cx-2 - x2 Page 85, #36: tan-1y x- 1 ln (1 + 2y2 x2 )- ln |x| = C1
Iowa State - M - 267
Assignment 8 answers Page 286, #6. e(t) = e-2t sin t Page 286, #11. L{x} =e-ps s= L{H(t - p)}, so that x(t) = H(t - p).Page 296, #28. y(t) = ys (t) + yi (t), where 1 ys (t) = 3t(e-(t-u) - e-4(t-u) )g(u) du04 1 yi (t) = e-t - e-4t 3 31
Iowa State - M - 414
MATH 414SPRING 2004 Homework for January 12-14 (due January 21)1. If x, y R, show that |xy| = |x|y|. (Suggestion: check all of the cases.) x2 + y 2 2. If x, y R show that xy . (Suggestion: start with the 2 fact that (x y)2 0.) 3. Let S be th
Iowa State - M - 501
MATH 501 1. If f (x) =FALL 2006Supplementary Homework 21 nd f (x) directly from the denition of derivative. x2 x2 x Q 0 x Qc . Show that f is dierentiable at x = 02. Let f (x) = and nd f (0).f (x + h) f (x h) h0 2h (sometimes called the
Iowa State - M - 516
MATH 516SPRING 2008Supplementary Homework 513. If is a Radon measure on a locally compact Hausdorff space X, L1 (), 0 and (E) = E d, show that is a also a Radon measure on X. 14. If is a finite positive Radon measure and is a finite si
Iowa State - M - 307
MATH 307FALL 2002End of semester calendarNovember 18: Sections 4.2-4.3 November 20: Sections 4.3-5.1 November 22: Section 5.1; go over Chapter 4 homework; last hand-in homework assigned, due December 6December 2: Sections 5.1-5.2; quiz on Chap
Iowa State - M - 267
MATH 267 FALL 2004 HOMEWORK ASSIGNMENTSWEEK 1August 23: Page 6, #1-4, page 16, #1-5,15-17August 24: page 28, #3,6,11,13,14August 26: page 39, #2-5,9-11,13-16,35b,36August 27: page 39, #26,27WEEK 2August 30: Page 50, #2,3,6,10,12; Page 134,
Iowa State - M - 515
MATH 515Fall 2007 Explain your answers carefully!Final Exam1. If E Rn has nite measure, and f is a bounded, measurable function on E, dene the Lp integral average Ap = 1 m(E) |f (x)|p dxE1 pfor 1 p &lt; . Show that if p &lt; q then Ap Aq . Sol
Iowa State - M - 166
MATH 166 REFERENCE SHEETINTEGRATION TRICKS 1. Find a direct formula. Consider ln, inv. trig. Formulas, f ( x) d 2nd Fnd. Thrm. Of Calc./Corollary f (t ) dt dx a 2. Try Substitution 3. Expand. (1 + ex 2) dx = (1 + 2e x + e 2 x )dx4. Separ
Iowa State - M - 166
ANCTIL, MICHAEL MATH 166 FINAL REVIEW W ORKSHEET Topics to Know: I. Applications of the Integral a) Area of a plane region b) Volume of a solid of revolution c) Length of a plane curve d) Area of a surface of revolution e) Work done by a variable for
Iowa State - M - 166
Name:Math 166 Calculus II Dec 13, 2004 FinalAnswers1. Let u = 1 + x2 , then du = 2xdx. Now du -1 x=a = lim |x=1 2 a 2u x=1 u -1 1 = lim = a 2(1 + x2 ) 4 1 = lim a 2x=a 1x 1 dx = lim 2 )2 a 2 (1 + xa 12x dx (1 + x2 )22. 0.125 =125 10
Iowa State - M - 166
Iowa State University Math 166 Instructor:Final Exam May 6, 2004 TA:Name: Section:Instructions: Answer each question completely. Show all work. No credit for mere answers with no work shown. Show the steps of calculations. State the reasons tha
Iowa State - M - 165
MIKE ANCTIL MATH 165: LIMITS PRACTICE1.limx2x2 - 4 x-2Direct substitution gives0 , which is an indeterminate. 0But the numerator can be factored:limx2( x - 2 )( x + 2 )x-2Simplifying the limit gives a form that is solvable by dire
Iowa State - M - 166
Improper IntegralsIndeterminate Forms00 - f (x )dx = llim l f (x )dx - - Taylor polynomial for f about x = ab b+af (x )dx = liml + a f (x )dxl0 0 0p n (x ) = f (a ) + f ' (a )(x - a ) +Taylor series for k =0f(k
Iowa State - M - 165
MIKE ANCTIL MATH 165, SOLVING MOTION PROBLEMSOne of the important applications of calculus is solving motion problems, that is, problems involving position, velocity, and acceleration of an object. There is a general procedure that can normally be
Iowa State - M - 165
MIKE ANCTIL MATH 165, MAKING THE PRODUCT RULE WITH IMPLICIT DIFFERENTIATION EASIERM AKING THE PRODUCT RULE WITH IMPLICIT DIFFERENTIATION EASIER When the Product Rule is combined with Implicit Differentiation, it often causes headaches. Derivative p
Iowa State - M - 165
MIKE ANCTIL MATH 165, CURVE SKETCHINGCurve sketching is an important skill to know for quick analysis of an equation. By looking at a few different aspects of the equation, even obscure functions can be sketched with some accuracy. There are six ge
Iowa State - M - 165
MIKE ANCTIL MATH 165 SECTION 2.22.4 PRACTICE: EARLY DERIVATIVE RULESFind the derivative: 1)y = 2 x sin x x 2 cos xThis problem uses the product rule twice (once for each product), the power rule, and definitions of trigonometric derivatives. Pr
Iowa State - M - 166
Wangsness Math 166Practice for Test 4 SolutionsName:In order to obtain full credit on this exam, you must show all necessary work when solving each problem. In each problem that requires graphing an equation, you must indicate the vertices, di
Iowa State - M - 166
Standard Form: _ Parabola: Vertex: _ Focus: _ Directrix: _ Circle: Center: _ Radius: _ Ellipse: Center: _ a: _ b: _ Foci: _ and _ Ends of Major Axis: _ and _ Ends of Minor Axis: _ and _ Hyperbola: Center: _ a: _ b: _ Foci: _ and _ Vertices: _ and _ A
Iowa State - M - 165
MIKE ANCTIL MATH 165 SECTION 2.8 RELATED RATES PRACTICE PROBLEMA train is moving to the left (left on the x-axis) along a curved track. The direction its headlight points is described by the functiony = ( x 2 ) + 1 . Joe is standing at the point
Iowa State - M - 166
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Iowa State - M - 165
MIKE ANCTIL MATH 165 SECTION 2.8 RELATED RATES PRACTICE PROBLEMA train is moving to the left (left on the x-axis) along a curved track. The direction its headlight points is described by the functiony = ( x - 2 ) + 1 . Joe is standing at the poin
Iowa State - M - 166
Iowa State University Math 166 Instructor:Midterm Exam February 26, 2004 TA:AnswersSection:Instructions: Answer each question completely. Show all work. No credit for mere answers with no work shown. Show the steps of calculations. State the r
Iowa State - M - 165
Iowa State - M - 166
MIKE ANCTIL THE UNIT CIRCLEThe Unit Circle
Iowa State - M - 166
MIKE ANCTIL x MATH 166: GRAPHS OF y = eGraphs of y = exy = e x and y = ln x can be confusing functions, especially when evaluating limits of them. Alwaysremember that while it frequently works out to be this simple, a limit is not just pluggin
Iowa State - M - 166
MATH 166 IN FOXTROT5/22/05Paige says: &quot;RELAX. IT WAS How much of a discount did she receive? We can use Calculus II to find out!1 3k k =1OFF.&quot;continued on next page.MATH 166 IN FOXTROT1 3k k =1is a geometric series. But it is no
Iowa State - M - 166
ANCTIL, MICHAEL MATH 166 FINAL REVIEW WORKSHEET Topics to Know: I. Applications of the Integral a) Area of a plane region b) Volume of a solid of revolution c) Length of a plane curve d) Area of a surface of revolution e) Work done by a variable forc
Iowa State - M - 166
Objectives for Math 1661Applications of the IntegralSet up and evaluate integrals to calculate 1. Area of a plane region 2. Volume of a solid of revolution 3. Length of a plane curve 4. Area of a surface of revolution 5. Work done by a variable
Iowa State - M - 166
MIKE ANCTIL TH MATH 166 REVIEW, VARBERG 9 ED., CHAPTERS 5, 75.1: The Area of a Plane Region Find the area bounded by the function y = x and y = x - 2 .2Step 1: GraphStep 2: Find the intercepts This is done by setting the equations equal to eac
Iowa State - M - 166
Correction to the solution to #10: The last two rows of the posted solution should be as follows: 27 27 36 3 9 3 27 3 = + = 9 2 6 2 2 21
Iowa State - M - 166
MIKE ANCTIL TH MATH 166 REVIEW, VARBERG 9 ED., CHAPTERS 5, 75.1: The Area of a Plane Region Find the area bounded by the function y = x and y = x - 2 .25.2: Volumes of Solids: Slabs, Disks, and Washers Find the volume of the solid whose base is
Iowa State - M - 166
MIKE ANCTIL 12.4 PRACTICE PROBLEMSChoose the Best Graph:(x + 3)2 + ( y + 2)24 16A) B)=1A)9 x 2 - 16 y 2 + 54 x + 64 y - 127 = 0B)C)D)C)D)MIKE ANCTIL 12.4 PRACTICE PROBLEMSDetermine the Correct Equation: Horizontal Ellipse Cent
Iowa State - M - 166
12.3: #13 Find points of tangency. Ellipse9x 2 + 4 y 2 = 1y-intercept: (0, 6)x y2 + =1 4 92(A)Equation of a tangent line for an ellipse (given in the book) at (x0, y0):xx0 yy 0 + = 1 (B) 4 9Plugging the y-intercept (0, 6) into equation
Iowa State - M - 166
MIKE ANCTIL MATH 166 PRACTICE PROBLEMThis problem is similar to problem #22 in the Integration by Substitution Section of the Techniques of Integration chapter of the book.x4x dx + 16Initially this problem appears to not match any direct fo
Iowa State - M - 166
Iowa State University Math 166 Instructor:Midterm Exam October 7, 2004 TA:Name: Section:Instructions: Answer each question completely. Show all work. No credit for mere answers with no work shown. Show the steps of calculations. State the reaso
Iowa State - M - 166
MIKE ANCTIL APRIL 28 FINAL REVIEW PROBLEMS1) Find the Sum of the Geometric Series:1 1 1 1 1 + 2 + 3 + 4 + . + n 7 7 7 7 7You should also be able to identify this series if given in this form:1 1 1 1 + + + + . 7 49 343 2401From this we can ea
Iowa State - M - 166
MATH 166 REFERENCE SHEETSUMMARY OF TESTS FOR SERIES Test Series nth - TermConvergesDivergesComment This test cannot be used to show convergence. Sum: S =an =1 n =0limnnan 0Geometric Series ar (bn =1 nnr &lt;1lim n bn =
Iowa State - MATH - 166
1 Exercises 6.2 1. R is the region bounded by the graphs of y = x2+1, y = 0, x = 0, x = 2. Let S be the solid of revolution generated by revolving R about the x-axis. Find the volume of S. y R 2 y = x2 + 1 or x = y 1
Iowa State - MATH - 166
Math 166 Final Exam Spring 2006Name &amp; Section: Instructor:Answer each question completely. Show all work and all steps, and state reasons for conclusions. No credit is allowed for mere answers with no work shown. Give numerical answers exactly, n
Iowa State - MATH - 166
1 Here let f(x) be a real-valued function with continuous derivatives of all orders on an open interval I of the x-axis. Let x0 be a real number in I. The Taylor series for f(x) about x0 is the power series (1)k =0f ( k ) ( x0 ) k ( x - x0 )
Iowa State - MATH - 166
1 Return now to the above power series (1)an =0n ( x x0 ) n .Suppose now that one of the following two cases occurs. (i) The power series (1) converges for all real numbers x. In this case, let R = . (ii) There is a positive real number R
Iowa State - MATH - 166
1Some ObservationsNow let be the curve which is the graph of the quadratic function y = f(x) = ax2 + bx + c where a0. Note as before that 2 b ax2 + bx + c = a x + x + c a 2 2 b b b a x2 + x + + c a = a 2a 2a
Iowa State - MATH - 166
Exercises 6.1 26. ( (1,-1) y (8,4) R (3,-1)xIntersection of x = y 2 - 2 y and x = y + 4 : y 2 - 2 y = y + 4. y 2 - 3 y - 4 = 0. ( y + 1) ( y - 4) = 0. y = -1 y = 4 or x=3 x = 8. Area of R = 4 -1( ( y
Iowa State - MATH - 166
Exercises 8.3 3.tdt 3t + 4Let u = 3t + 4. u 2 = 3t + 4. 2udu = 3dt dt = t= 1 2 ( u - 4). 3 tdt 2 udu. 31 2 ( u - 4) 2 udu 3 3 3t + 4 = u 3 2 ( u - 4u ) = du 9 u 2 = ( u 2 - 4 )du 9 2 1 = u 3 - 4u + C 9 3 3 1 2 8 = ( 3t + 4 ) 2 -
Iowa State - MATH - 166
1 Departmental Final Solution Spring 2006 1. ( x + 1)213dx= lim bb 21 dx ( x + 1) 3b 1 -2 = lim - ( x + 1) b 2 2 1 -2 = ( 2 + 1) 2 1 = . 18 2. k + 5 - k + 6 k =1n kk +1 n n n +1 1 2 2 3 3 4 n - 2 n -1
Iowa State - MATH - 166
1 Exercises 8.3 27. y = f ( x) = f ( x) = = -2 (21 . x + 2x + 5 x 2 + 2 x + 5 0 - 1 ( 2 x + 2)2. 2 + 2x + 5 f ( x) = 0 for x = -1. f ( x) &gt; 0 for x &lt; -1, f ( x) &lt; 0 for x &gt; -1. 1 Maximum value of f(x) is f(-1) = . 4(xx +1(x)2+ 2