FOR DUTCH STUDENTS! ENGLISH VERSION NEXT PAGE

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1 FOR DUTCH STUDENTS! ENGLISH VERSION NEXT PAGE Tentamen Analyse 8 december 203, duur 3 uur. Voeg aan het antwoord van een opgave altijd het bewijs, de berekening of de argumentatie toe. Als jeeen onderdeel van een opgave niet kunt maken, ga dan toch door met de volgende onderdelen. Je mag daarbij de in eerdere onderdelen verschafte informatie gebruiken. De (grafische) rekenmachine mag gebruikt worden bij het tentamen. De rekenmachine kan gebruikt worden om numerieke berekeningen uit te voeren. De rekenmachine kan echter NIET gebruikt worden om argumenten te geven voor verkregen oplossingen. Opgave Toon met behulp van de (ε,δ)-definitie aan dat de functie f : IR\{ 2} IR, gedefinieerd door continu is in. f(x) = 3 2x+4 voor iedere x IR\{ 2}, Opgave 2 Beschouw een continue functie f : IR IR waarvoor geldt dat f(0) = 0 en f(x) +x 2 voor iedere x IR. a) Bewijs dat de vergelijking f(x) = 203 minstens twee oplossingen heeft. b) Bewijs dat de functie f een minimum heeft. zie ommezijde voor de overige opgaven Cijferbepaling: Puntentelling tentamen: opgave : 5 ptn; opgave 2: 7 ptn; opgave 3: 6 ptn; opgave 4: 6 ptn; opgave 5: 5 ptn; opgave 6: 6 ptn; opgave 7: 5 ptn. Het cijfer voor het tentamen is de som van de punten gedeeld door 4. De toets telt voor 20% mee.

2 Opgave 3 Beschouw een differentieerbare functie f : [0,4] IR waarvoor geldt dat f(2) = 2 en f (x) 3 4 voor iedere x [0,4]. a) Toon aan dat f een contractie van [0,4] is. b) Toon aan dat de rij (t n ) n=, gedefinieerd door t = 3 en t n+ = f(t n ) voor iedere n IN, een iet heeft en bepaal deze iet. Opgave 4 Beschouw de functie f : [0,+ ) IR gedefinieerd door f(x) = x voor iedere x [0,+ ). a) Gebruik het eerste orde Taylor polynoom van f in 36 om een schatting te geven van 37. b) Gebruik de stelling van Taylor om een bovengrens (d.w.z. een getal) te geven van de fout die gemaakt wordt bij deze schatting. Opgave 5 Bepaal, indien mogelijk, x x 7 x 5 +x 3 x x 8 +x 6 +x 4 +x 2 4. Opgave 6 Beschouw de rij (t n ) n= gedefinieerd door t n = ln( ) voor iedere (n+) 2 n IN. Laat (S n ) n= de bijbehorende rij van partiële sommen zijn. a) Toon aan dat S n = ln2+ln(n+2) ln(n+) voor iedere n IN. b) Is de rij (t n ) n= sommeerbaar? Zo ja, bepaal de som van deze rij. Hint: Voor iederex > 0,y > 0geldt ln(xy) = lnx+lny enln( x y ) = lnx lny. Bovendien mag je gebruiken dat ln een continue functie is. Opgave 7 Bereken 0 x 2 + x+ dx.

3 FOR NON-DUTCH STUDENTS! Exam Mathematical Analysis 2 December 8, 203, duration 3 hours. For every exercise you have to provide the full proof, calculation and/or arguments. If you can t solve an item of some exercise, continue with the next items. It is allowed to use the information that you have obtained before. The (graphical) calculator can be used during the exam. However, the graphical calculator can only be used for numerical calculations. It CANNOT be used to provide arguments for the solutions obtained. Exercise Show, by using the (ε,δ)-definition, that the function f : IR\{ 2} IR, defined by f(x) = 3 for every x IR\{ 2}, 2x+4 is continuous in. Exercise 2 Consider a continuous function f : IR IR with the property that f(0) = 0 and f(x) +x 2 for every x IR. a) Show that the equation f(x) = 203 has at least two solutions. b) Show that the function f has a minimum. Exercise 3 Consider a differentiable function f : [0, 4] IR with the property that f(2) = 2 and f (x) 3 4 for every x [0,4]. a) Show that f is a contraction of the interval [0,4]. b) Show that the sequence (t n ) n=, defined by t = 3 and t n+ = f(t n ) for every n IN, has a it and determine this it. Turn this page for the other exercises. 2 Grading: Grading exam: exercise : 5 pts; exercise 2: 7 pts; exercise 3: 6 pts; exercise 4: 6 pts; exercise 5: 5 pts; exercise 6: 6 pts; exercise 7: 5 pts. The grade for the exam is the sum of the points scored, divided by 4. The midterm counts for 20%.

4 Exercise 4 Consider the function f : [0,+ ) IR defined by f(x) = x for every x [0,+ ). a) Use the first order Taylor polynomial of f in 36 in order to give an estimate of 37. b) Use Taylor s Theorem in order to give an upper bound (i.e. a number) of the error made for this estimate. Exercise 5 Determine, if possible, x x 7 x 5 +x 3 x x 8 +x 6 +x 4 +x 2 4. Exercise 6 Considerthesequence(t n ) n= definedbyt n = ln( )forevery n IN. (n+) 2 Let (S n ) n= be the corresponding sequence of partial sums. a) Show that S n = ln2+ln(n+2) ln(n+) for every n IN. b) Is (t n ) n= summable? If yes, determine the sum of this sequence. Hint: For every x > 0,y > 0 we have ln(xy) = lnx + lny and ln( x y ) = lnx lny. Moreover you can use the fact that ln is a continuous function. Exercise 7 Determine 0 x 2 + x+ dx.

5 Solutions Exam Mathematical Analysis, December 8, 203 Exercise Note that f() = 2. We have to show that for every ε > 0 there is a δ > 0 such that for every x IR\{ 2} with x < δ we have f(x) f() < ε. So let ε > 0. Choose δ = min{, 4ε}. Let x IR\{ 2} be such that x < δ. Then x < δ, hence x > 0. Moreover 3 f(x) f() = 2x+4 2 = 3 (x+2) 2(x+2) x = 2(x+2) 4 x < 4 δ 4 4ε = ε. Exercise 2 a) Note that f(0) = 0 < 203 < 000 f(00) and f(0) = 0 < 203 < 000 f( 00). Applying the Intermediate Value Theorem twice to the continuous function f and the compact intervals [ 00, 0] and [0, 00] yields the existence of at least one solution in ( 00,0) and at least one solution in (0,00). b) Note that for every x / [ 4,4] we have f(x) +x 2 > 7. The Theorem of Weierstrass, applied to the continuous function f and the compact interval [ 4,4], tells us that there is a minimum on [ 4,4], i.e. there is a c [ 4,4] such that f(c) f(x) for every x [ 4,4]. In particular f(c) f(0) = 0 hence f(c) f(x) for every x / [ 4,4] as well. Hence f(c) is the minimum of f on IR. Exercise 3 a) (i) Let x [0,4]. If x = 2 then obviously f(2) = 2 [0,4]. If x 2 then the Mean Value Theorem, applied to f and the interval [2,x] or [x,2] yields the existence of a ξ between 2 and x such that f(x) f(2) = f (ξ)(x 2). So f(x) 2 = f(x) f(2) = f (ξ) x x = 3 2. Hence 2 f(x) 7 2 so f(x) [0,4]. (ii) Let x,y [0,4],x y. Again according to the Mean Value Theorem, applied to f and the interval [x,y] or [y,x], there is a ξ between x and y such thatf(x) f(y) = f (ξ)(x y). Hence f(x) f(y) = f (ξ) x y 3 4 x y. From (i) and (ii) it follows that f is a contraction of [0,4] with contraction constant 3 4. b) According to the Contraction Theorem the sequence is convergent and its it is the unique fixed point of f on [0,4]. Since f(2) = 2 we infer that

6 the it is 2. Exercise 4 Note that f (x) = 2 x and f (x) = 4x for every x > 0. x a) The first order Taylor polynomial of f in 36 is given by p(x) = f(36)+ f (36)(x 36) = 6+ 2 (x 36) for every x [0,+ ). An estimate for 37 is therefore p(37) = 6 2. b) According to Taylor s Theorem (with c = 36 and x = 37) there is a ξ (36,37) such that f(37) = p(37)+ f (ξ) 2 (37 36) 2 or 37 = 6 2 8ξ ξ. So the absolute value of the error is 8ξ ξ which is smaller than = 728. Exercise 5 Define the functions f and g on (0, ) by f(x) = x 7 x 5 + x 3 x and g(x) = x 8 +x 6 +x 4 +x 2 4 for every x (0, ). Note that f() = g() = 0, that f and g are differentiable on (0, ), that f (x) = 7x 6 5x 4 +3x 2 and g (x) = 8x 7 +6x 5 + 4x 3 +2x > 0 for every x (0, ). According to l Hôpital s Theorem we derive that x x 7 x 5 +x 3 x x 8 +x 6 +x 4 +x 2 4 = f () g () = 4 20 = 5. Exercise 6 a) Denote the statement S n = ln2+ln(n+2) ln(n +) by P(n). We will use induction in order to show that P(n) is true for every n IN. Step : P() reads S = ln2+ln3 ln2 = ln3 2ln2. Now S = t = ln( ) = ln = ln3 ln4 = ln3 2ln2, so P() is true. Let n IN and assume that P(n) is true. Now S n+ = S n +t n+ = ln2+ln(n+2) ln(n+)+ln( (n+2) 2) = ln2+ln(n+2) ln(n+)+ln( (n+2)2 (n+2) 2 ) = ln2+ln(n+2) ln(n+)+ln( (n+)(n+3) (n+2) 2 ) = ln2+ln(n+2) ln(n+)+ln(n+)+ln(n+3) 2ln(n+2) = ln2+ln(n+3) ln(n+2), so P(n+) is true. b) n S n = n ( ln2+ln(n+2) ln(n+)) = n ( ln2+ln(n+2 n+ )) 2 = ( ln2+ln(+ n)) n + n = ln2+ln = ln2.

7 So (t n ) is summable and its sum equals ln2. Exercise 7 Using the substitution z = x+ (hence x = z and dz = dx) we get 0 x 2 + x+ dx = = = (z ) 2 + dz z z 2 2z +2 dz z (z 2+ 2 z )dz = [ 2 z2 2z +2lnz] z=2 z= = (2 4+2ln2) ( 2 2+2ln) = 2ln2 2.