McGill University Department of Mathematics and Statistics
MATH 243 Analysis 2, Winter 2023
Assignment 1: Solutions
1. Let f : R → R, f(x) := x1/3. Prove that f is not differentiable at 0. Hint: Considerg◦f whereg:R→R,g(x):=x3.
1. Solution: (Using the hint)
Note that g is differentiable on R with g′(0) = 0. Assume that f is differentiable at 0. We will
compute (g ◦ f)′(0) with two different methods:
(i) By the chain rule we have that g ◦ f is differentiable at 0 and that
(g ◦ f)′(0) = g′(f(0)) · f′(0) = g′(0) · f′(0) = 0 · f′(0) = 0.
(ii) Now we compute (g◦f)′(0) directly. Note that (g◦f)(x) = x for all x ∈ R; thus (g◦f)′(x) =
1 for all x ∈ R; especially, (g ◦ f)′(0) = 1.
So (g◦f)′(0) is both 0 and 1 and we have a contradiction. This implies that f is not differentiable
2. Solution: (Using the limit definition of the derivative)
By the limit definition of the derivative, f is differentiable at 0 iff
x→0 x x→0 :=h(x)
x1/3 −2/3 = lim = lim x
f (x) − f (0) x→0 x − 0
exists. However, we will show that this limit does not exist, which then proves that f is not differentiable at 0. We will use the one-sequence criterion for the non-existence of the limit of a function: we will construct a sequence (xn) of real numbers with lim(xn) = 0 such that lim (h(xn)) does not exist.
Let (x ) := 1 . Then lim(x ) = lim 13 = 0, and (h(x )) = ((n−3)−2/3) = (n2), which nn3 nn n
divergesto+∞,sincen2 ≥nforalln∈Nandlim(n)=+∞.
This proves that lim h does not exist, which in turn proves that f is not differentiable at 0. x→0
2. Let f : R → R,
x2 if x ∈ Q
0 if x ∈ R \ Q
(a) Show that f is differentiable at 0 and find f′(0). (b) Show that f is not differentiable at any c ̸= 0.
Remark: You may use, without proof, that both Q and R \ Q are dense in R. Solution:
(a) We will prove that f is differentiable at 0 with f′(0) = 0. It is most convenient to do this byusingthedefinitionofalimit. Letε>0,δ:=εandlet|x|<δ. Then
f(x) − f(0) f(x) |f(x)| x2
x−0 −0= x = |x| ≤ |x| =|x|<δ=ε
Hence f is differentiable at 0 and f′(0) = lim f(x) − f(0) = 0. x→0 x−0
(b) Let c ̸= 0. We will prove that f isn’t even continuous at c; this will especially prove that f isn’t differentiable at c. We will prove that f isn’t continuous at c by showing that lim f (x)
does not exist:
Since both Q and R \ Q are dense in R, there exist sequences (xn) and (un) with xn ∈ Q,
un ∈ R\Q for all n ∈ N and lim(xn) = c, lim(un) = c. Then lim(f(xn)) = lim(x2n) = c2 and
lim (f(un)) = lim (0) = 0 ̸= c2. Hence lim f(x) does not exist and f is neither continuous x→c
nor differentiable at c.
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3. Let f : R → R,
x2 sin(1/x2) 0
(a) Prove that f is differentiable on R. (b) Prove that f′ is unbounded on [−1,1].
if x ̸= 0 if x = 0
(c) Conclude from (b) or prove otherwise that f′ is discontinuous at 0. Solution:
(a) Let x ̸= 0. Then
f′(x)=2xsin(1/x2)+x2cos(1/x2)·(−2· 1)=2xsin(1/x2)−2cos(1/x2)
Since all functions involved are continuous on R \ {0}, f is even continuously differentiable
onR\{0}. Atx=0wehave
f′(0)= lim f(x)−f(0) = lim x2sin(1/x2) = limxsin(1/x2)=0
x→0 x−0 x→0 x x→0
since lim x = 0 and sin(1/x2) is bounded on R \ {0} (note that | sin(1/x2)| ≤ 1 for all
x ∈ R \ {0}). Thus f is differentiable at 0.
(b) Let xn := √ 1 for all n ∈ N. Then xn ∈ [−1, 1] and xn ̸= 0 for all n ∈ N. Furthermore, 2πn
f′(xn) = √ 2 sin(2πn) − 2√2πn cos(2πn) = −2√2πn = −2√2π · √n 2πn
Consequently (f′(xn)) is unbounded (in fact, we even have that lim(f′(xn)) = −∞) which means that f′ is unbounded on [−1,1].
(c) 1. Solution: Since every continuous function on a closed and bounded interval is bounded but f′ is unbounded on [−1,1] by part (b) it follows immediately that f′ cannot be continuous at all x ∈ [−1,1]. However, it was shown in part (a) that f′ is continuous at all x ̸= 0. Consequently, f′ is discontinuous at 0.
2. Solution: Consider the sequence (xn) defined in part (b). We have that lim(xn) = 0
and it was shown in (b) that lim(f′(xn)) = −∞ which by the sequential criterion especially
means that lim f′(x) does not exist. That, in turn, means especially that f′ is discontinuous
Github
Here is a graph of f:
And here is a graph of f′ (note that this function is difficult to sketch—the y-range was cut off (rather arbitrarily) at y = ±50 in the sketch below):
程序代写 CS代考 加QQ: 749389476
4. Let I ⊆ R be an interval, let c ∈ I, and let f : I → R be a function which is differentiable at c. Let n ∈ N and define fn : I → R, fn(x) := [f(x)]n for all x ∈ I. Use Caratheodory’s theorem
and the binomial formula
n n n n − j j (a+b)= jab
to prove that fn is differentiable at c and to compute (fn)′(c). Don’t use the chain rule!
Since f is differentiable at c, it follows from Caratheodory’s theorem that there exits a function φ : I → R, which is continuous at c, such that
∀x∈I: f(x)=f(c)+φ(x)(x−c) Thus we have for all x ∈ I that
fn(x) = [f(x)]n = [f(c) + φ(x)(x − c)]n
j f (c)φ(x)(x−c) nnnn−jj j
nnn−jj j
=f(c)+ jf (c)φ(x)(x−c) j=1
= f n (c) +
n n j
f n−j (c) φj (x)(x − c)j −1 ·(x − c) j=1
where the function ψ : I → R is continuous at c as a sum of products of functions continuous at c. We can now conclude from Caratheodory’s theorem that fn is differentiable at c. Furthermore,
n n j
(fn)′(c)=ψ(c)= fn−j(c)φj(x)(x−c)j−1
n n
= fn−1(c)φ(x)+ fn−j(c)φj(x)(x−c)j−1
1 j=2 j
= nfn−1(c)φ(c) = nfn−1(c)f′(c)
Note that we obtain the exact same result via the chain rule.
5. (Optional) Consider the function f : R → R,
xn if x ≥ 0
where n ∈ N. Prove that f ∈ Cn−1(R) \ Cn(R) i.e. prove that f is (n − 1)-times continuously
differentiable but not n-times continuously differentiable. Solution:
dkn n−k n!n−k
dxk x =n(n−1)…(n−k+1)x = (n−k)!x forall0≤k≤n(notethatthezeroth
derivative of a function is just the function itself). Thus
n! xn−k ifx>0
f(k)(x) = (n−k)! ∀0 ≤ k ≤ n
0 if x < 0
We still need to analyze differentiability of f(k) at 0. We start by proving that f(k)(0) = 0 for
all0≤k≤n−1. Thisisdonebyinductionoverk:
Base case k = 0: f(0)(0) = f(0) = 0.
Inductivestepk→k+1: Letk
We have now proven that f ∈ Cn−1(R). It remains to show that f ∈/ Cn(R). We will do this by showing that f(n)(0) does not exist:
lim f(n−1)(x)−f(n−1)(0)= lim 0−0=0but x→0− x−0 x→0− x
lim f(n−1)(x) − f(n−1)(0) = lim n!x = n! ̸= 0 and thus x→0+ x−0 x→0+ x
lim f(n−1)(x) − f(n−1)(0) = f(n)(0) DNE x→0 x−0
This proves that f ∈ Cn−1(R) \ Cn(R).