Section 1.3
#6) There are many examples. For instance f : N N given by f (n) = n + 1, andg : N N given by g (n) = n2 . #7) T2 is a denumerable set. There exists a bijection f : T2 N. If T1 is denumerable, then there is a bijection g : T1 N. Thus, f 1 g :
EXAMPLE OF USE OF THE DEFINITION OF LIMIT OF A SEQUENCE 1. Show that lim ( Solution. | 2n ) = 2. n+2 ().
4 4 2n 2| = | | | n+2 n+2 n
Fix > 0. By the Archimedean property, there is N such that 4 <N. We thus have 4 < n K := N . n By plugging this into (), w
MAT 319 SPRING 2008 MIDTERM I
! WRITE YOUR NAME, STUDENT ID BELOW ! NAME : ID :
THERE ARE SIX (6) PROBLEMS. THEY HAVE THE INDICATED VALUE. SHOW YOUR WORK! DO NOT TEAR-OFF ANY PAGE NO CALCULATORS NO PHONES
ON YOUR DESK: ONLY test, pen, pencil eraser.
1 2 3
MAT 319 SPRING 2008 MIDTERM II
! WRITE YOUR NAME, STUDENT ID BELOW ! NAME : ID :
THERE ARE SIX (6) PROBLEMS. THEY HAVE THE INDICATED VALUE. SHOW YOUR WORK! DO NOT TEAR-OFF ANY PAGE NO CALCULATORS NO PHONES
ON YOUR DESK: ONLY test, pen, pencil eraser.
1 2
MAT 320 Fall 2007
Review for Final
Note: Final is cumulative, so use the Midterm 2 Review and the Midterm 1 Practice Exam as well as the material below. Theorem: be able to apply Theorem: and know what goes into the proof Theorem: and be able to prove. 7.
MAT 320 Fall 2007 Theorem: be able to apply
Review for Midterm 2
Theorem: and know what goes into the proof Theorem: and be able to prove. 3.3 Monotone Convergence Theorem. The least upper bound property is crucial. Understand Example 3.3.3(b). 3.4 Theor
MAT 319/320
Solutions of Practice Midterm II
1
Problem 1. Dene a sequence (xn) by : x0 = 1 and xn +1 = 3 xn + 1. Does (xn) have a limit? If yes, what is this limit? Proof. First, notice that if the sequence has a limit l, then it must satisfy the equation
MAT 319/320
Practice Midterm II
1
Problem 1. Dene a sequence (xn) by : x0 = 1 and xn +1 = 3 xn + 1. Does (xn) have a limit? If yes, what is this limit? Problem 2. Is the innite series the limit).
1 + n =1 n + n
convergent? (If yes, you do not need to nd t
MAT 319/320
Correction of Practice Midterm I
Problem 1. Show by induction that for any natural number n Proof. 1. The proposition is true for n = 1 because 3 > 1.
1 one has 3n > n.
2. Assume that 3n > n: then 3n+1 > 3n = n + 2n > n + 1 because n that 3n+1
MAT 319/320
Practice Midterm I
Problem 1. Show by induction that for any natural number n
1 one has 3n > n.
Problem 2. Determine the set A of all x in R such that |5.x + 2| < 8. Problem 3. Is the set B =
1 , n2 + 1
nN
bounded above? bounded below? Does it
y A s 9 B A W 9 8 s A A 6 8 s D A s r D V s r @ 9 89 8 r V f G w V A s r D V @ 9 s D 9 8t s r E A 6 8 f G G s D cfw_ A s r D V w BB 7 t DB @ 9 89 f G P I GF A 7 8 D s A @ W r D t E U 4 T ( ( ' & y A B x C 7 n A W A 8s r D t 7 A X9 u 8 D s E9 89 A X D W x
MAT 319
Practice FINAL
n3 n!
Problem 1. What is the limit of (xn) =
?
n2 1 1
Problem 2. Use the denition of the limit to prove that lim 3n2 + 1 = 3 . Problem 3. Prove that an increasing sequence that is bounded above is necessarily converging. Problem 4.
MAT 319/320 Real Analysis
Correction of Midterm I
Problem 1. (25 points) Let C > 0 be a real number. Show that for any natural number n 1, one has (1 + C)n 1 + n C. Proof. Proof by induction: Lets call P(n) the following proposition: (1 + C)n 1 + n C. 1.
Stony Brook University - MAT 320 Midterm II
Solutions 1. (25 points) The functions f and g are continuous on the interval [a, b] and dierentiable on (a, b). If f (a) = g(a) and f (b) = g(b), prove that there is a point c, with a < c < b, where f (c) = g (
Stony Brook University MAT 320 Introduction to Analysis Final Examination with Solutions
December 21, 2007 Work any five problems. Tell us clearly which ones you have chosen. 1. Show how the Bolzano-Weierstrass Theorem (Every bounded sequence contains a c
MAT 319
Solutions for HW12
Exercise 1. Section 6.3, #7c. Proof. Just write x3ln x = 1/x3 = g(x) and apply LHospitals rule on (0, + ) (notice that g (x) = 3x 4 is well-dened and non-zero on that interval and that both functions are dierentiable). f (x) 1/x
MAT 319
Solutions for HW11
Exercise 1. Section 6.1, #16. Proof. Since the function tan is continuous and strictly increasing from ( 2 , 2 ) to R, we know that the inverse function arctan exists, and is continuous and strictly increasing from R to (cos x).
MAT 319/320
Solutions for HW10
Exercise 1. Section 6.1, #1a. Proof. As usual we have to go back to the denition, so we write the quotient
( x + h )3 x 3 h x3 + 3x2h + 3x h2 + h3 x3 h
=
= 3x2 + h.(3x + h2) 3x2 when h 0
Exercise 2. Section 6.1, #4. Proof. T
MAT 319/320
Solutions for HW9
Exercise 1. Section 5.3, #3. Proof. Start with any point, x0 = 1/2 for example. Then we know that there exists x1 such that f (x1)
1 2
f (x0) . If f (x1) = 0, then we are done; if not then necessarily x1
1 2
x0, and
there is
MAT 319/320
Solutions for HW8
Exercise 1. Section 4.3, #5a. Proof. We have that limx 1+ x 1 = + . Indeed, for any given > 0 if we take x (1, 1 + ), 1 then we have that x 1 > . Now since limx 1+ x = 1, we know that by the comparison thex 1 orem, limx 1+ x
MAT 319/320
Solutions for HW7
Exercise 1. Section 4.1, #2. |x 4| 1 x 2 = x + 2 x 4 , therefore if you take x 4 < 1 you will get Proof. Notice that 2 x 2 < 1/2 and for the second inequality it is sucient to take x 4 < 2.10 2. that
Exercise 2. Section 4.1,
MAT 319/320
Correction of HW6
Exercise 1. Section 3.6, #1. Proof. Pick 1 = 1. Since the sequence is unbounded, one can nd xn1 > 1. Pick 2 = max (2, xn1 + 1). For the same reason, one can nd xn2 > 2. Continue like this: by construction, the subsequence is
MAT 319/320
Correction of HW5
Exercise 1. Page 80, #1. Proof. Take for example the following sequence: x2k = k, x2k+1 = 1.
Exercise 2. Page 80, #3. Proof. Recall that fn satises the relation fn+2 = fn+1 + fn and that all the fn are > 0. f f f therefore we
MAT 319/320
Correction of HW4
Exercise 1. Page 67, #6a. Proof. By the sum rule, the limit of (2 + 1/n) is equal to 2. By the product rule, the limit of (2 + 1/ n)2 is 2.2 = 4.
Exercise 2. Page 67, #9. 1 Proof. One has yn = n + 1 n = (multiply the numerato
MAT 319/320
Correction of HW3
Exercise 1. Page 50, #2. Proof. 1. If S is bounded then there exists a lower bound m and an upper bound M . By denition, they are such that any x in S satises m x M . But this means x [m, M ].Therefore S [m, M ]. 2. Conversel
MAT 319/320
Correction of HW2
Exercise 1. Page 38, #2. Proof. The set S2 is not empty and is bounded below (for example by 0), so it has an inmum. Lets prove that 0 = inf (S2): 1. For any x in S2, one has 0 x; x < 0 + (take /2 for example). 2. For any > 0
MAT 319/320
Correction of HW1
Exercise 1. Page 15, #2. Proof. By induction: 1. The property is true for n = 1, because 13 =
2 1 .1.2 2
2. Assume that the property is true for n, and prove that its true for n + 1: 13 + 13 + 13 + 13 + 13 + + n3 = 1 .n.(n +
1) Cauchy Convergence Criterion: A sequence (xn ) is Cauchy if and only if it is convergent. Proof. Suppose (xn ) is a convergent sequence, and lim(xn ) = x. Let > 0. We can nd N N such that for all n N , |xn x| < /2. Therefore, by the triangle inequality
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