Practice Final Exam Solutions for Calculus II, Math 5, December 5, 3 Name: Section: Name of TA: This test is to be taken without calculators and notes of any sorts. The allowed time is hours and 5 minutes. Provide exact answers; not decimal approximations! For example, if you mean do not write.44.... Show your work, otherwise credit cannot be given. Write your name, your section number as well as the name of your TA on EVERY PAGE of this test. This is very important. Problem I II III IV V VI VII VIII IX X XI XII Score Total
Name: Section: Name of TA: Problems related to Block : I: (5 points) Compute with an error less than 3 3 e x dx. e y = n k= y k k! + ec y n+ (n + )! where c is some number between and y. Now we set y = x and note that since x ranges between and 3 the variable y ranges between /4 and /9. Hence we know that c can be a number that must be somewhere between and /4 and since e y is monotone increasing we take c = /4 to obtain an upper bound on the remainder of the form e /4 y n+ (n + )!. Now from what we know about the exponential function e < 3 and hence e /4 < 3 /4 which is some number less than. Thus we find that < 3 [e x = (n+)! n k= x k dx 3 k! (n+)! x (n+) dx n+ [ n 3 n < (n+)! n+ n. If we choose n = 3 we find that the remainder of the integral is bounded by 4! 7 7 = Integrating the sum in the integral yields 3 k= II: a) (7 points) Compute the limit 84 8 <. [ k+ 3 k+ k!(k ). lim x e x cos x sin x x 3
3 Using the Taylor expansion for all the functions in the numerator yields e x cos x sin x = + x + x + x3 3! + + x + x x3 3! +... The leading order coefficient is x and hence the limit does not exist. b) (8 points) Does the improper integral x e x dx exist? If yes, compute it. We have to compute and using the substitution lim ε ε x e x dx u = x this integral can be rewritten as ε Clearly the limit as ε does not exist. e u du = e ε e.
4 Name: Section: Name of TA: Problems related to Block : III: a) (7 points) Is the series convergent? Is it absolutely convergent? Using the ratio test we get that which converges to e convergent. k= ( ) k (k!) k k ( ) ((k + )!) k k k k (k + ) k+ (k!) = k + <. Hence the series is absolutely convergent and hence, in particular, b) (8 points) Find the interval of convergence of the power series k=3 We are going to use the ratio test. Set and note that a k+ lim k a k ( ) k ln k k (x )k a k = ln k x k k ln(k + )k = lim x = x k (k + ) ln k hence the interval of convergence contains the interval (, 3). Looking at the end point x = 3 we find the series k=3 ( ) k ln k k which converges since it is alternating and the coefficients decrease monotonically to zero. At x =, however, the series diverges, since it is given by k=3 ln k k and diverges, by the limit comparison test with /k.
5 IV: (5 points) Solve the initial value problem Hence y x y = e x, y() = e. Multiply the equation by the integrating factor e x e x y x e x y = (e x y ) =. y(x) = xe x + ce x where c s a constant. y() = e yields c = and hence our solutions is y(x) = (x + )e x.
6 Name: Section: Name of TA: Problems related to Block 3: V: (5 points) Consider the system of equations x + y + z = b x + y z = x y + az = Determine all values for a and b for which this system has a) non solution, b) exactly one solution, c) infinitely many solutions. In the case b) and c) Compute all the solutions in terms of a and b. The augmented matrix is b a Switching the first and second row leads to b a Row reduction leads to 5 b 4 a 8 5 b If a = 8 and b 5 there is no solution. If a 8 there is always a unique solutions and if a = 8 and b = 5 there are infinitely many solutions. If a 8 we can use back substitution and obtain: z = 5 b a 8, y = 55 b a 8 + 4 b, x = 35 b a 8 + b If a = 8 and b = 5 then the row reduced augmented matrix is 5 3
7 and we find z = t, y = 5t + 3 and x = 3t +. VI: (5 points) Consider the three vectors v =, v =, v 3 = Determine all the values for a and b for which the vector b b is in the span of v, v, v 3. Determine all the values for a and b for which the vector b is a unique linear combination of the vectors v, v, v 3. We have to row reduce the matrix which is row equivalent to b a a a 3 8 a b 5 We have to check for which values of A, b the system is consistent. For all a 8 the vector b is in the span of the vectors v, v, v 3 no matter what the value of b is. If a = 8 then b must be 5/ for the system to be consistent and hence again, b is in the span of the vectors v, v, v 3. In this last case there is a free variable and hence b is not a unique linear combination of the vectors v, v, v 3. If a 8 and only in this case is b a unique linear combination of the vectors v, v, v 3. The first three columns are pivotal. a.
8 Name: Section: Name of TA: VII: (5 points) Consider the matrix 3 5 6 A = 3 4 5 4 7 Find a basis for Col(A) and for Nul(A) as well as for Col(A T ) and for Nul(A T ). Try do this with a little computation as possible. Row reducing A yields (.) as a basis vector for the nullspace of A. Thus we can say that the dimension of Col(A) equals 3. The row reduced matrix is 3 and since the first three columns are pivotal, the first three column vectors of the matrix A form a basis for Col(A). Since the orthogonal complement of Col(A T ) is the nullspace of A we have we have that the orthogonal complement of Nul(A) is Col(A T ). Thus, we have to find three linearly independent vectors that are perpendicular to the vector in (.). Thus we have to solve w + x y + z = which leads to the one-one parametrization z = t, y = s, x = r and w = [r s + t. Hence we get,,
9 as a basis for Col(A T ). To find a basis for Nul(A T ) we remember that this space is the orthogonal complement of Col(A). Hence, we have to find a vector perpendicular to the first three column vectors which can be found by row reducing the system 4 3 5 5 4 which yields 4 3 4 and from which we get that the basis for the nullspace of A T is 3 VIII: (5 points) Diagonalize the matrices 4 a) 4. 4 The eigenvalue 6 has the eigenvector Perpendicular to this is the vector 3 which is an eigenvector with eigenvalue 3. Thus the vector 6
which is perpendicular to both must be an eigenvector, since the matrix is symmetric. The associated eigenvalue is 3 also. [ 6 9 b) 4 The characteristic polynomial is µ 7µ + 3 = (µ 5)(µ ). The iegenvector associated with µ = 5 is the vector [ and the one associated to µ = is [ 9 4
Name: Section: Name of TA: IX: ( points) Solve the recursive relation, i.e., find a n for arbitrary values of n, a n+ = 8a n + 9a n with a = a =. Writing we can write the recursion as x n = [ an a n x n+ = A x n where The solution can be gotten via [ 8 9 A =. x n = A n x. A is diagonalized by A = [ 9 [ 9 [ 9. Hence [ [ [ 9 9 n A n = ( ) n 9 [ Applying this to the initial condition x = we find that a n = 5 (9n + 4( ) n ).
X: ( points) Find the least square solution for the system A x = b where 4 A = 3 3 6 and b =. Solve the problem in two ways, once using the normal equations and then using the QR factorization. The normal equation A T A x = A T b: and and the solution is A T A = A T b = x = 4 4 4 4 4 56 The QR factorization: The Gram-Schmidt procedure applied to the vectors in the column space leads to the matrix The R matrix is then given by Q = R = 4 4 6
3 It remains to solve the equation R x = Q T b. Since Q T b = the solution is as before. x =
4 Name: Section: Name of TA: XI: (5 points) Graph the curve given by the equation x 6xy + 9y =. The associated matrix is given by [ 3 3 9 whose characteristic polynomial is µ 3µ + = (µ )(µ ). Hence the eigenvalues are µ =, µ =. The associated eigenvectors are [ u = 3 and In the u v plane the cureve is given by u = [ 3. u + v = which is an ellipse whose semiaxis in the u direction has length and whose semiaxis in the direction v has length /. To get the picture in the x y plane we have to rotate the u v picture by the rotation matrix U = [ u, u = [ 3 3.
5 XII: (4 points each) Prove or find a counterexample: a) Row operations do not change the column space of a matrix. This is false. Take the matrix [ [ The column space is obviously one dimensional and spanned by the vector reduction the matrix is [ After row [ whose column space is spanned by the vector b) Row operations do not change the null space of a matrix This is true. Since the row operations do not change the solutions set. c) A 3 3 matrix has the eigenvalues,,. Is it necessarily diagonalizable? No, it is not. The matrix The eigenvalue has the corresponding eigenvector and for the eigenvalue we get a single eigenvector (up to a non-zero multiple) d) A real 3 3 matrix must always have at least one real eigenvalue. This is true. The characteristic polynomial of a 3 3 matrix A is given by p(λ) := det(a λi) = λ 3 + a λ + a λ + a 3
6 where a, a, a 3 are numbers. As λ tends to +, p(λ) tends to and when λ tends to, p(λ) tends to +. By the intermediate value theorem p(λ must have a real zero and hence A must have at least one real eigenvalue. e) If a matrix A has the QR factorization A = QR then the equation R x = Q T b has a solution for any b. This is true. If A = QR then we know that QQ T is the projection onto Col(A). Hence the equation QR x = QQ T b has always a solution. Multiplying this expression from the left by Q T and using the Q T Q = I we find that R x = Q T b must have a solution.