Ph.D. Katarína Bellová Page Mathematics (0-PHY-BIPMA) RETAKE EXAM, 4 April 08, 0:00 :00 Problem [4 points]: Prove that for any positive integer n, the following equality holds: + 4 + 7 + + (3n ) = n(3n ). Solution: We will prove the equality by mathematical induction. For n =, the left-hand side equals, while the right hand side equals (3 ) = and the equality is true. Assume now that the equality is true for some n = k N, we will prove it for n = k +. Using the induction hypothesis in the first line, we have + 4 + 7 + + (3k ) + (3(k + ) ) = k(3k ) + (3(k + ) ) = 3k k + (3k + ) = 3k + 5k + (k + )(3k + ) = (k + )(3(k + ) ) =. Since this is eactly the equality for n = k +, the second induction step is proved and so is the statement. Problem [4 points]: Is the sequence (a n ) n, where a n = cos(n π), convergent? If yes, compute its it. If no, compute sup n a n and inf n a n. Solution: For even n, n is even, and cos(n π) =. For odd n, n is also odd, and hence cos(n π) =. Thus, for any n N we have cos(n π) = ( ) n, which is not convergent, sup n a n = and inf n a n =. Problem 3 [4 points]: Prove that there eists at least one solution R of the equation tan + =. Solution: We want to use intermediate value theorem. For that, we need to make sure to work on an interval [a, b] where the function f() = tan + is continuous, for eample on some closed subinterval of ( π, π ). Checking some values of the function, we find out that for eample interval [0, π 3 ] works: f(0) = tan 0 + 0 = 0; f( π 3 ) = tan π 3 + ( π 3 ) = 3 + ( π 3 ) > + = ; f() is continuous on [0, π 3 ]. Hence, as the value lies between the values f(0) and f( π 3 ), by intermediate value theorem there must eist some (0, π 3 ) such that f() =. Problem 4 [4 points]: Compute the it ( e ).
Ph.D. Katarína Bellová Page Mathematics (0-PHY-BIPMA) e Solution: First, = > 0, so the epression is well defined and the it is of type (which can be anything). Due to continuity of the eponential function, we will have ( e ) = e ln ( e ) = e ln ( e ), if the it in the eponent eists. Let us compute it: by using l Hospital rule twice (indicating the type each time; second use applies only to the second it in the epression) and algebraic properties of its, ln ( ) e 0 0 = = (e ) (e )() ( e ) e 0 0 = = e + e e = e =. e e + (e ) + e () (e ) + () The calculation is justified backwards: since the its and epressions (we do not divide by 0) on the right eist, also the its on the left eist and equal the right-hand sides. Getiing back to our it, we get ( e ( ) ) = e / = e. Problem 5 [5 points]: For which values of a R is the following function f continuous at = 0? For which a R is it differentiable at = 0? { sin(a) f() = for 0, for = 0. Solution: For continuity at 0, we must have f() = f(0). Here sin(a) f() = = { sin(a) a a = a = a if a 0, sin(0 ) = 0 = 0 = a if a = 0. In the first case we used substitution y = a: y 0 as 0 and y 0 if 0; then sin y y 0 y =. In any case, we got f() = a. Hence, the function will be continuous at 0 if and only if a = f(0), i.e. a =. If f is differentiable at 0, it must be continuous at 0. Hence, f can not be differentiable for
Ph.D. Katarína Bellová Page 3 Mathematics (0-PHY-BIPMA) any a. Let us check from definition whether f is differentiable at 0 for a = : f f() f(0) (0) = 0 = sin() = sin() 0 0 = cos() cos() = 0 0 = sin() = 0. Here we used l Hospital rule twice for its of type 0 0, the calculations are justified backwards as the right hand sides eist. We see that for a =, f (0) eists, and f is differentiable at 0 if and only if a =. Problem 6 [4 points]: Give the definition domain of the function f() = sin(ln ) and compute its first and second derivative. Solution: Logarithm is defined for > 0, sin y is defined for any y R. Hence, the definition domain of f is (0, + ). For (0, + ), by chain rule and then quotient and chain rule, f () = cos(ln ) cos(ln ) =, f () = (cos(ln )) cos(ln ) () = ( sin(ln )) cos(ln ) sin(ln ) cos(ln ) =. Problem 7 [4 points]: What is the maimal value of a > 0 such that the function f() = ln is invertible on (0, a]? Solution: For > 0, f () = = is positive if and only if < =, and negative if and only if > =. We see that f is continuous on (0, + ), while it is ( ] [ ( ] increasing on 0, and decreasing on )., + Thus, f is invertible on 0,, but not on any interval (0, a] for any a >, and the maimal a we are looking for is a =. Problem 8 [5 points]: Find all local maima of the function f() = cos + on R. Is any of them a global maimum? Solution: As the function is differentiable, in any local maimum we must have f (0) = 0, i.e. sin + = 0, sin = /, = { π 6 5π 6 + kπ, k Z, + kπ, k Z. Furthermore, as f is twice differentiable, in any local maimum we must have f () 0. We
Ph.D. Katarína Bellová Page 4 Mathematics (0-PHY-BIPMA) have f ( () = cos, f π ) 6 + kπ = 3 < 0, f ( 5π 6 + kπ ) = 3 > 0. We see that the only possible local maima occur at = π 6 +kπ, k Z; as f () = 0, f () < 0 is also a sufficient condition for a local maimum, these are indeed local maima. Since f() = + (as cos is bounded and = + ), function f does not have any global maimum. Problem 9 [4 points]: Compute the indefinite integral arctan d. Solution: Integrating by parts, arctan d = arctan + d = arctan + d = arctan ( ) d + = arctan ( arctan ) + C = (( + ) arctan ) + C. Problem 0 [4 points]: Compute the definite integral 3 π 0 sin( 3 ) d. Solution: Substituting y = 3, dy = 3 d, we get 3 π 0 sin( 3 ) d = π 0 3 sin y dy = 3 cos y π Problem [4 points]: Does the following series converge? ( ) tan k 3 + 5 k=0 0 = 3 ( ) = 3. Solution: We compare the series with the series + k= (the convergence does not depend k 3/ on any finitely many first terms of the series, so it does not matter that our series starts with
Ph.D. Katarína Bellová Page 5 Mathematics (0-PHY-BIPMA) k = 0 and the second series with k = ): ( ) tan k + = k 3/ k + = k + ( ) tan tan ( tan y = y 0 + y k + = =. ) k 3/ k 3 + 5 k 3/ k + k 3 + 5 + 5 k 3 Since the it is finite and series + k= ( ) k=0 tan k. 3 +5 k 3/ converges (as 3/ > ), so does the series Problem [3 points]: Write the comple number i 3 i in the algebraic form. Solution: i 3 i = i 3 i 3 + i 3 + i = 3 + i 3i + = 4 i = 9 + 0 5 5 i. Problem 3 [5 points]: Prove that 3 vectors in space are linearly dependent if and only if they are coplanar. Solution: First assume that vectors v, v, v 3 R 3 are linearly dependent. Then there eist coefficients α, α, α 3, not all of them zero, such that α v + α v + α 3 v 3 = 0. Without loss of generality, assume that α 0 (otherwise, just rename the vectors). Then v = α α v α 3 α v 3. This means that if vectors v and v 3 span a plane, then v lies in this plane. If v and v 3 lie in one line, then v also lies in this line. If v = v 3 = 0, then also v = 0. In any case, vectors v, v, v 3 are coplanar (if v and v 3 are colinear, we just have many options to choose the plane). For the reverse implication, assume that v, v, v 3 R 3 are coplanar. Assume first that the three vectors span a plane P. As P is a -dimensional subspace of R 3, there must eist two of the vectors v, v, v 3 which span P - assume these are v, v 3 (otherwise rename the vectors). Then (v, v 3 ) form a basis of P, and we can epress v P in this basis: v α v α 3 v 3 = 0. v = α v + α 3 v 3, This is a non-trivial linear combination of v, v, v 3 which gives 0 (the first coefficient 0), so v, v, v 3 are linearly dependent. If v, v, v 3 do not span a plane, they all lie in a single line. If at least one of the vectors v, v, v 3 is non-zero, e.g. v 3 0, then vectors v, v are scalar multiples of v 3, and v = αv 3, v αv 3 = 0.
Ph.D. Katarína Bellová Page 6 Mathematics (0-PHY-BIPMA) This is again a non-trivial linear combination of v, v, v 3 which gives 0. The remaining case when v = v = v 3 = 0 is trivial: then e.g. v = 0 is non-trivial linear combination of v, v, v 3 which gives 0. Problem 4 [3 points]: Consider the set V of all polynomials with real coefficients of degree at most 3. With the usual addition and multiplication by real numbers, V forms a vector space over R. What is its dimension? Justify your answer! Solution: We have V = {α 3 3 + α + α + α 0 : α 0, α, α, α 3 R}. We see that polynomials,,, 3 form a basis of V (they span V by the above equality, and they are linearly independent, as α 3 3 + α + α + α 0 = 0 as a polynomial implies α 0 = α = α = α 3 = 0). Hence, the dimension of V is the number of its basis vectors, i.e. 4.