Assignment on mathematics assignment

I have an mathmatic exam, you can choose any time from now to monday,thanks
I have an mathmatic exam,you can choose any time from now to monday,The files are my course note, almost 1 hours,Attached file are example of past year examthanks

Math 321 Exam 2 November 25, 2019
MATH 321 Test 2 Solutions
Problem 1 [15 points]
State whether the following statements are true or false.[Correct answer is enough to
get full credit. If you are not sure about your answer, provide a brief explanation – if you answer
is wrong, your solution may still earn partial credit. ]
(a) E is the intersection of all closed sets that contain E
Let W =
V :V =V , EV V Notice that if E V , V -closed implies E V = V , so E is contained
in any closed set containing E, so E W. On the other hand, E is itself a closed set that
contains E, thus, must contain the intersection of all such sets: E W. Hence, E = W
TRUE
(b) Union of any compact sets is compact.
Counterexample: R = nZ[n, n + 1] is a noncompact union of compact sets
FALSE
(c) f 0(x0) is not defined if x0 is not in the domain of f
By definition. In applications though, some functions can be redefined at one point to become
differentiable. We are not talking about those cases here.
TRUE
(d) f(x) = x is uniformly continuous on R
By definition, since we can pick δ = .
TRUE
(e) for any function f defined everywhere on R and any constant a > 0
lim sup
xx0
f(x) = inf
δ:δ>0
sup
x:0<|xx0|
f(x) = inf
δ:a>δ>0
sup
x:0<|xx0|
f(x)
True, since supx:0<|xx0|f(x) is non-increasing with respect to δ.
TRUE
1
Math 321 Exam 2 November 25, 2019
Problem 2 [20 points]
(a) Show that if a set E R is bounded, then E is compact.
(b) Show that any bounded non-empty open set does not have a minimum(i.e. its
infimum is not contained in the set) (Hint: assume, by contradiction, that the greatest
lower bound is contained in the set. Since the set is open, every point in it is an interior point.
Recall the definition of an interior point and the definition of infimum and lead the argument
to a contradiction.)
Solution(a) Assume E is bounded, i.e. M > 0 s.t. x E |x| ≤ M. Let us show that E0 is
bounded as well: if x E0 E, |x| ≤ M, if x E0 \ E, we have proven(in class + homework)
that ∃{xn} ⊆ E s.t. xn x. [Optional explanation: n N by definition of an accumulation
point xn E s.t. xn (x 1/n, x + 1/n).]
Since for each n |xn| ≤ M, the limit value, by the properties of sequence limits must satisfy
M x M, i.e. |x| ≤ M. As show above, if E is bounded, E0 is also bounded by the same
constant M, and so is E = E E0.
We have also proved that closure of any set is closed.[Optional explanation: consider x E0,
by definition r > 0 (x r, x + r) E contains infinitely many points from E = E E0. Fix
any r > 0. If (x r, x + r) E contains infinitely many points from E, x E0 E, If not, it
contains at least one point y E0, then letting ρ = min{|x y|, r} we observe that the interval
(y ρ, y + ρ) must contain infinitely many points from E, and thus (y ρ, y + ρ) , (x-r, x+r).
Thus, by definition, hence, x E.]
E was proven to be closed and bounded, hence it compact.
(b) Assume that an open set U R is bounded and not empty, then there exists a finite
m = inf U. If m U = R U (as U is open) then r > 0 s.t. (m r, m + r) U. In particular, it
implies that x U s.t. x < m (any x (mr, m), for instance, x = mr/2). This contradicts
the fact that m = min U.
2
Math 321 Exam 2 November 25, 2019
Problem 3 [20 points]
(a) Give an example of a function f(x) that is defined everywhere on R and is not
continuous at infinitely many points. Prove your statement.
(b) Assume that f is a monotone increasing continuous function defined on R
with the range not bounded from either above or below. Carefully show that
lim
x+f(x) = +, limx→−∞ f(x) = −∞ and the range of f is equal to R, i.e. it
attains every real value. Cite or state all definitions and theorems you use in your
argument.
Solution (a) for example, the characteristic function of rational numbers f(x) = χQ(x) is discontinuous at every point a R. Indeed, take a sequence {xn} ⊂ Q s.t. xn a, then f(xn) = 1
for all n, so f(xn) 1. On the other hand, if we take a sequence {x0n} ⊂ R \ Q s.t. x0n a,
then f(x0n) = 0 for all n, so f(x0n) 0, 0 6= 1, thus, limxa f(x) fails to exist.
Many other examples exist.
(b) Fix any y R. Since f is not bounded from below, there exists a R s.t. f(a) < y. Since f
is not bounded from above, there exists b R s.t. f(b) > y. Thus, f(a) < y < f(b), and, since f
is monotone increasing, a < b. Applying the intermediate value theorem (Darboux property of
continuous functions) to f on [a, b] we conclude that x (a, b) s.t. f(x) = y. Thus, f attains
all real values.
Consider an arbitrary M > 0 since f is not bounded from above there exists x0 R s.t.
f(x) > M, then x > x0 f(x) > f(x0) > M. By definition it means that limx→∞ f(x) = +.
The other limit lim
x→−∞ f(x) = −∞ is proven analogously.
3
Math 321 Exam 2 November 25, 2019
Problem 4 [20 points]
(a) Show that the following limit does not exist
lim
x0
sin x 1
(b) Consider f(x) = x4 sin x 1 on R, assume that f(0) = 0.
Show that f is differentiable on R. Find the explicit formula for f0, explain your
answer.(You may assume that (sin t)0 = cos t and (1/t)0 = 1/t2.)
Solution (a) Let us use the sequential definition. Consider xn = πn 1 0, then sin x1 n = 0,
so f(xn) 0. On the other hand, x0n = 2πn+ 1 π/2 0, then sin x1 0 n = 1, so f(x0n) 1. Since
lim
n→∞ f(xn) 6= limn→∞ f(x0n) while both xn and x0n converge to 0, so limx0 f(x) does not
exist.
(b) For any x 6= 0 we can apply the product and chain rules to obtain
f0(x) = 4x3 sin x 1 + x4 cos x 1 x 12 = 4x3 sin x 1 x2 cos x 1
Let us evaluate the derivative at x = 0:
lim
x0
f(x) f(0)
x 0
= lim
x0
x3 sin x 1 = 0
the last equality is implied by the squeeze theorem, since
x3 sin x 1 ≤ |x3| → 0
4
Math 321 Exam 2 November 25, 2019
Problem 5 [20 points]
(a) Show that any periodic function that is differentiable everywhere on R must have
infinitely many points where its derivative is equal to 0. (Hint: use Rolle’s theorem).
(b) Assume that a continuous function f: [a, b] R additionally satisfies the positivity constraint: x [a, b] f(x) > 0. Show that f is actually ‘bounded away from zero’:
there exists a constant α > 0 s.t. x [a, b] f(x) α > 0. (Hint: use compactness of [a, b]).
Solution (a) Assume that T > 0 is the value of the period of f. For any n N let an = nT,
b
n = (n + 1)T = an + T. Since f is periodic, f(an) = f(bn). By Rolle’s theorem there exists
x
n (an, bn) s.t. f 0(xn) = 0. Since the open intervals (an, bn) are disjoint, xn 6= xm when
n 6= m, therefore, we have a sequence of distinct points in which f 0 is zero.
(b) The shortest way to show it is to notice that a continuous function f attains its minimum
on the compact set [a, b]: xmin [a, b] s.t. f(xmin) = infx[a,b] f(x). By the assumption of the
problem, f(xmin) > 0. Thus, we may pick α = f(xmin).
Alternatively, we can use a compactness argument to prove it.
5

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