ID #:
Math 162 (41) - Midterm Exam
Winter 2020
Monday February 10, 2019 - 11:30 to 12:20
• Read everything carefully. • Write readable.
• Write your answers in the space provided below each of the problems. If you need additional space, use the opposite side.
• Give full proofs of all your statements, indicate if you use a result from class or homework.
• If a problem has multiple parts, you may assume any part for all succeeding parts.
• In the first problem no proofs are required. • There are four problems
Problem 1. (2+2+2+2+2 points) Decide for each of the following statements if they are true or false:
(i) Let f : [0, 1] → R be a Riemann integrable function and n ∈ N. Then also fn: [0, 1] → R given by fn(x) = (f (x))n is Riemann integrable.
True False
(ii) Let f : [0, 1] → R be Riemann integrable. Then there is a continuous function F : [0, 1] → R and a dense set C ⊂ (0, 1) such that F is differentiable at any x ∈ C with F0(x) = f (x).
True False
(iii) Let f : R → R be a twice differentiable function with f + f00= 0 and f (0) = 0, f0(π/2) = 1, then f (x) = sin(x) for any x ∈ R.
True False
(iv) Let p : R → R be a polynomial of degree at most n. Then p = Pn,0, where Pn,0
denotes the Taylor polynomial of p at 0 of degree n. True
False
(v) Let a : N → R be a Cauchy sequence such that a(n) ∈ Q for any n ∈ N. Then a converges to a limit l ∈ Q.
True False Solution:
(i) You proved on the homework that the composition of a continuous and a Riemann integrable function is again Riemann integrable. Since x 7→ xn is
continuous on all of R this in particular applies to this problem
(ii) Given such f , we showed that F : [0, 1] → R defined by F (x) =´0xf (t) dt
defines a continuous (in fact Lipschitz continuous) function. The fundamental theorem of calculus states that F is differentiable at any point at which f is continuous. Since you proved that any Riemann integrable function is continuous at a dense set, it follows that F has all the desired properties. (iii) Note that sin0(π/2) = cos(π/2) = 0.
(iv) That was one of the motivations for the exact definition Taylorpolynomials: Recall that if we write p(x) =Pn
j=0ajx
j, then p(j)(0) = j! a j.
(v) We proved in class that any Cauchy sequence in R converges but (exactly as for suprema) you really need all the real numbers for that. An explicit example is to use that Q is dense to choose an ∈ (
√
Problem 2. (10 points) Let f : R → R be a continuous and bounded function. Define g : [0, 1] → R by g(x) = ( f (1/x) if x 6= 0 0 else.
Show that g is Riemann integrable.
Solution Let’s start from the very beginning. We defined for a bounded function g the two quantities
ˆ 1∗ 0 g(x) dx = inf ˆ 1 0 ψ : ψ ∈ S(0, 1), f ≤ ψ and ˆ 1 0∗ g(x) dx = sup ˆ 1 0 φ : φ ∈ S(0, 1), f ≥ φ ,
where S(0, 1) is the space of step functions and integral of a step functions is defined by a finite sum. We then called g Riemann integrable if
ˆ 1∗ 0 g(x) dx = ˆ 1 0∗ g(x) dx,
in which case we can use either of them as definition of the Riemann integral. It follows from the definitions that one can equivalently characterize Riemann integrability as follows:
(1) ∀ε > 0 ∃ φ, ψ ∈ S(0, 1) : φ ≤ f ≤ ψ, ˆ 1
0
(ψ − φ) ≤ ε.
Typically, that is unless your function g is a step function anyways, you have to choose different step functions for different ε.
One of the most important results on Riemann integrable functions was that continuous functions on closed intervals are Riemann integrable. (In fact, for functions on open intervals, the notion of being Riemann integrable is not even defined and one has to refer to improper Riemann integrals.)
Let’s solve the actual problem. For ε > 0 given we have to construct step functions φ, ψ as in (1). Since f is assumed to be continuous, and x 7→ 1/x is continuous in R \ {0}, we find that g is continuous in (0, 1]. In a first step we want to exploit this, but as remarked above we can only do so on any subinterval of the form [a, 1] for 0 < a < 1. (You could specify to use a = ε at this point, but if it is not that obvious to you, simply keep going with a and specify it at the end, just make sure that you are aware that you choose everything in the proper order eventually.) Since f is bounded, so is g and we can choose some C > 0 such that |g(x)| ≤ C for any x ∈ [0, 1]. If we take the step functions on [0, a] defined constantly by +C and −C respectively, we clearly have that
ˆ a 0
if we take a = ε/(2C). Now that a is fixed, we can use that g : [a, 1] → R is continuous, hence integrable. Therefore, we find φ, ψ ∈ S(a, 1) such that φ ≤ g ≤ ψ and
ˆ 1 a
(ψ − φ) ≤ ε. Define new step functions ˜φ, ˜ψ ∈ S(0, 1) by
˜ φ(x) = ( φ(x) if x ∈ [a, 1] −C if x ∈ [0, a) and ˜ ψ(x) = ( ψ(x) if x ∈ [a, 1] C if x ∈ [0, a). Then ˜φ ≤ g ≤ ˜ψ by construction and
ˆ 1 0 ( ˜ψ − ˜φ) = ˆ a 0 2C + ˆ 1 a (ψ − φ) ≤ 2ε.
Problem 3. (10 points) Let p : R → R be a polynomial. Show that ˆ ∞ 0 p(x)e−2xdx exists. Solution
Let us first recall what the definition of improperly Riemann integrable means in this case. The function f : R → R defined by f (x) = p(x)e−2x is the product of two
continuous functions, hence itself continuous. In particular, f is Riemann integrable on any interval [a, b] for a, b ∈ R. This implies that the only issue about the improper integral in question is that we want to integrate along the entire positive axis, i.e. we only have to check if the limit
lim
b→∞
ˆ b 0
p(x)e−2xdx
exists. First of all note that the integrand does not necessarily have a sign, hence it does not suffice to show that´0bp(x)e−2xdx is bounded independently of b. (Although there is a trick to reduce to this here: Since p is a polynomial, you can find R > 0 such that p has a sign in [R, ∞), then consider ´Rbp(x)e−2x instead.) Recall that you proved on homework that it suffices to check, that for ε > 0 given, you can find a > 0
such that ˆ
b
a
p(x)e−2x dx ≤ ε for any b ≥ a. Let us check that condition. Write
p(x)e−2x = (p(x)e−x)e−x
and note that the first term is bounded on [0, ∞), since p(x)e−x is a continuous function with limx→∞p(x)e−x= 0 as proved in class. Hence we find some C > 0 such
that
|p(x)e−2x| ≤ Ce−x for any x > 0. Therefore,
Problem 4. (10 points) Let f : (−1, 1) → R be a smooth function (i.e. f is l-times differentiable for any l ∈ N) such that f(k)(0) = 0 and |f(k)|(x) ≤ 1 for any
x ∈ (−1, 1), k ∈ N. Show that f = 0.
Solution There are two ways to solve this problem. The first one is to inductively use the mean value theorem if f 6= 0 to produce points eventually contradicting the
assumption |f(k)| ≤ 1 for k large. The second one is to use Taylorpolynomials. Note
that both assumptions are important, since the function e−1/x2 extended by 0 to 0 has all its derivatives vanishing at 0 but does not vanish identically.
Given the assumptions on all the derivatives at 0 on should try to use the Taylor expansion centered at 0. By assumption, we have that
Pn,0 = 0
for any n ∈ N. As the example mentioned above shows this might not give any information on the behaviour of the function far away from 0, but we know how to estimate the difference. Recall that we proved in class that
f (x) − Pn,0(x) = Rn(x) with Rn(x) = 1 n! ˆ x 0 (x − t)nf(n+1)(t) dt. We can estimate the remainder term (very brutally) as follows
|Rn(x)| ≤
1 n! by assumption also using that x ∈ (−1, 1). Since
lim
n→∞
