MATH2039 Analysis 2020/21
Hello, dear friend, you can consult us at any time if you have any questions, add WeChat: daixieit
MATH2039W1
SEMESTER 1 EXAMINATION 2020/21
MATH2039 Analysis
1. (a) [6 marks] For the set S defined as S = [2, 3] u {5} u {8}, determine whether it has an infimum and/or a supremum and justify your answers.
(b) [10 marks] Let S = { ln, m e N}. Determine inf(S) and sup(S), if they
exist. Using the definitions of infimum and supremum, justify your answers.
(c) [9 marks] Prove the following statement: Let S be a set with an infimum inf(S). For every ε > 0 there is a point a e S such that
inf(S) < a < inf(S) + ε.
2. (a) [9 marks] Show that if a sequence of real numbers (an )neN is non-increasing and bounded below, then it converges and
lim an = inf{an , n e N}.
(b) [8 marks] Prove that the sequence (an )neN defined by
an = n2 + 1, n e N
diverges to infinity.
(c) [8 marks] Prove that the sequence (an )neN defined by
an = (-1)n + , n e N
does not converge.
3. (a) Prove, using the appropriate definitions, that the function f (x) = x3 is
(i) [8 marks] continuous at all points of 皿.
(ii) [5 marks] uniformly continuous on the interval [1, 2].
(iii [5 marks] not uniformly continuous on 皿.
(b) [7 marks] Using the mean value theorem, or otherwise, show that the function f (x) = sin x is decreasing on the interval [ , π].
4. (a) [8 marks] Evaluate the following limit if it exists and justify all steps in your
calculation:
(b) [5 marks] Show that the improper integral
diverges, but that lim M-o |
M
-M |
dx = 0. 1 + x4 |
(c) [12 marks] Determine the radius and interval of convergence for the power series
(3x - 4)n .
n=0
1. (a) (Similar to exercises) [扌 m之sdo] Set S is bounded below by 2 (since [2, 3] = {x e 皿l2 < x < 3}, 2 < 5 and 2 < 8. Therefore, by the
completeness axiom it has an infimum. Since 2 e S and 2 is a lower bound, it follows that 2 = inf(S).
Similarly, S is bounded above by 8 and so it has a supremum. Since 8 is an upper bound and 8 e S, we have 8 = sup(S).
Since S is bounded above and below, it is bounded.
(b) (Similar to exercises) [←尸 m之sdo] Set S = { ln, m e N} is not bounded above. Taking m = 1 we see that, for example S contains N as a subset, which is not bounded. Therefore S has no supremum.
Since > 0 for all m, n e N, S is bounded below by 0, and hence by
completeness axiom, it has an infimum. Considering the subset of S in which n = 1, we have that S contains elements for all m e N. It is enough to show that 0 is the infimum of this subset. We have already seen that 0 is a lower bound. Suppose that 1 > 6 > 0 is another lower bound. By the Archimedean property there exists m0 e N such that
m0 > - log3 6
Now,
3m0 > 3- log3 e = 1
Rearranging the inequality, we get < 6 contradicting the assumption that 6 is a lower bound. Therefor inf(S) = 0 and since is never 0, inf(S) S.
(c) (Similar to exercises) [9 m之sdo] Suppose, for a contradiction, that there is no such a e S. Then, there is an ε > 0 such that for all a e S either a 2 inf(S) + ε or a < inf(S). The case a < inf(S) contradicts the first property of the infimum. Hence for all a e S we have a 2 s0 = inf(S) + ε. Thus s0 is a lower bound for S. Hence, by the second property of infimum, inf(S) 2 s0 = inf(S) + ε implying ε = 0, which gives a contradiction.
2. (a) (Similar to exercises) [9 m之sdo] The set {an , n e N} is bounded below and hence has an infimum. Assume L = inf{an , n e N}.
By the approximation property for infimum, for any given ε > 0 3n0 e N such that
L < an0 < L + ε.
Now, for n > n0 we have
lan - Ll = an - L < an0 - L < ε.
Since ε > 0 was arbitrary, we have proved that
Vε > 0 3n0 e N such that lan - Ll < ε, Vn > n0 .
That is, we have shown that L = limn-o an .
(b) (Similar to exercises) [扌 m之sdo] Let K > 0 be given. By the Archimedean property
there exists n0 e N such that n0 > ,K. Now, for n > n0 we have n2 + 1 > n0(2) + 1 > n0(2) > K.
Since K > 0 was arbitrary, we have proved that the sequence (n2 + 1)neN diverges to infinity and we write limn-o (n2 + 1) = o.
(c) (Similar to exercises) [扌 m之sdo]
We prove that (an ) is a not Cauchy sequence. That is,
3ε > 0 such that Vn0 e N 3p, q e N with p, q > n0 and la夕 - ap l 2 ε.
Taking ε = 1, n0 e N, p = n0 , q = n0 + 1, we get
lan0 +1 - an0 l = l(-1)n0 +1 + l - (-1)n0 - l =
l2(-1)n0 +1 + - l = l2(-1)n0 +1 - l.
For n0 even, this becomes
l - 2 - l = 2 + > 1
and for n0 odd, it becomes l2 - l > 1.
Since the sequence (an ) is not a Cauchy sequence, it doesn’t converge. Alternatively, one can identify the subsequence (a2k)k eN which converges to 1 and the subsequence (a2k -1 )k eN which converges to -1. Since (an )neN has two convergent subsequences which converge to different limits, the sequence (an )neN diverges.
3. (a) (i) (Similar to exercises) [扌 m之sdo] To show that f (x) = x3 is a continuous function on 皿, we show that f is continuous at a e 皿.
That is, Vε > 0 3δ > 0 such that for x e S,
lx - al < δ =÷ lf (x) - f (a)l < ε. Let ε > 0 be given. Consider lx3 - a3 l = lx - allx2 + ax + a2 l and restrict lx - al < , getting x < .
Now, lx3 - a3 l = lx - allx2 + ax + a2 l < lx - al( + + lal2 ) = lx - ala2 = Clx - al, with C = a2 . Choosing δ < min{ , } we get lx3 - a3 l < Clx - al < ε.
Since ε > 0 was arbitrary, we have shown that the function f (x) = x3 is continuous at a e 皿. Since a e 皿 was arbitrary, we have proved that the function is continuous on 皿.
(ii) (Similar to exercises) [θ m之sdo] Let ε > 0, choose δ = . For
x, y e [1, 2], lx - yl < δ follows
lx3 -y3 l = lx-yllx2 +yx+y2 l < lx-yl3 max{lxl2 , lyl2 } < 12lx-yl < ε. Since ε > 0 was arbitrary, we have shown that the function is uniformly continuous on [1, 2].
(iii) (Similar to exercises) [θ m之sdo] To see that f is not uniformly continuous on 皿, we argue by contradiction. We start with a bit of algebra, namely
lf (x) - f (a)l = lx3 - a3 l = lx - al lx2 + ax + a2 l. To say that f is NOT uniformly continuous is to say that there exists ε > 0 such that for all δ > 0 there exist x, a in 皿, with lx - al < δ and lf (x) - f (a)l 2 ε .
We will show this for ε = 1 (in fact it works for any ε > 0). Fix δ > 0. We need to find x, a with lx - al < δ but with lx3 - a3 l 2 ε = 1. To ensure that lx - al < δ let us choose x to be a + . We consider what happens as a - o. We have
lx3 - a3 l = lx - al lx2 + xa + a2 l = ((a + )2 + a(a + ) + a2 ) 2 a2 δ . For a 2 1/,δ we have a2 δ 2 ε = 1.
Thus, we see that for any fixed δ, if a is large enough, that is, for a 2 1/,δ we can find a value of x with lx - al < δ, but with lx3 - a3 l 2 1.
(b) (Similar to exercises) [玄 m之sdo] Since f/ (x) = cos(x) < 0 for x e ( , π) we conclude that f (x) = sin x is decreasing on the interval [ , π].
Alternatively, using the MVT : Let x1 , x2 e [ , π] and x1 < x2 . We want to show that f (x1 ) > f (x2 ). Consider the interval [x1 , x2] and apply the MVT with the function f and this interval. Since f is differentiable on 皿 it is continues on
[x1 , x2] and differentiable on (x1 , x2 ). By the MVT there exists c e (x1 , x2 ) such
that
f (x2 ) - f (x1 )
x2 - x1 .
Since f/ (c) = cos(c) < 0 and x2 > x1 , it must be f (x2 ) < f (x1 ) as required.
4. (a) (Similar to exercises) [扌 m之sdo]
This is an example of the indeterminate form 1o which, using some algebraic manipulations, is transformed into the indeterminate form , where l’Hopital’s rule is used to calculate the limit.
We write
g(l)im-o ╱ 、g =g(l)im-o ╱exp ln( )、g =g(l)im-o exp x ln().
Because of continuity of the exponential function, we can write
x - 2 x - 2
g-o x g-o x
Now, the exponent has indeterminate form o . 0. We convert the exponent into the indeterminate form by writing
exp g(l)im-o x ln() = exp g(l)im-o .
Finally, applying the l’Hopital’s rule, we get
exp lim ln( ) = exp lim () = e-2 .
g g2
(b) (Similar to exercises) [θ m之sdo] We first need to write -o(o) dx as the sum of two improper integrals, for instance
o x3 0 x3 o x3
-o 1 + x4 -o 1 + x4 0 1 + x4
and then evaluate the two resulting improper integrals separately. So,
o x3 M x3
0 1 + x4 M-o 0 1 + x4
M x3
M-o 0 1 + x4
= lMim-o ╱ ln(1 + M4 ) - ln 1、= o,
and so the original improper integral -o(o) dx diverges.
However, when we evaluate limM-o -M(M) dx, we get
lMim-o -M(M) dx = lMim-o ╱ln(1 + M4 ) - ln(1 + (-M)4、= 0.
(c) (Similar to exercises) [←… m之sdo] The radius of convergence is , and interval of convergence is (-7, ) :
Apply the ratio test and calculate:
│(3x - 4)n+1/52(n+1) │ 1 3 │ 4 │
Hence, this series converges absolutely for │x - │ < 1, that is for
│x - │ < , and so the radius of convergence is . We now need to check the endpoints of the interval (-7, ).
At x = -7, the series becomes (3(-7) - 4)n /52n = which diverges by the n-th term test for divergence.
At x = , the series becomes (3(29/3) - 4)n /52n = diverges by the n-th term test for divergence.
(-1)n ,
1, which
So, the series converges absolutely for all x in the open interval (-7, ), and diverges elsewhere.
2022-01-18