# PROBLEM SET 8 (DUE IN LECTURE ON NOV 6 (FRIDAY)) ```PROBLEM SET 8 (DUE IN LECTURE ON NOV 6 (FRIDAY))
(All Theorem and Exercise numbers are references to the textbook by Apostol; for instance
“Exercise 1.15-3” means Exercise 3 in section 1.15.)
Problem 1. Let f : R → R be a differentiable function satisfying
f 0 (x) = f (x) for all x ∈ R.
Prove that f (x) is of the form cex for some c ∈ R. (Hint: consider g(x) =
e−x f (x).)
Problem 2. Prove that
1
lim (1 + x) x = e.
x→0
= 1 by interpreting this limit as a
(Hint: first prove that limx→0 log(1+x)
x
derivative.)
Problem 3. Compute the derivative (for x &gt; 0) of the function
x
f (x) = xx .
Problem 4. The goal of this problem is to construct the sine function. In other words,
we want to find a twice-differentiable function f : R → R such that f (0) = 0,
f 0 (0) = 1, and
f 00 (x) = −f (x) for all x ∈ R.
(a) Let g : [− 12 , 12 ] → R be defined by the integral
Z x
1
√
g(x) =
dt.
1 − t2
0
Prove that g is continuous and strictly increasing.
(b) Let a = g( 12 ) &gt; 0. Let h : [−a, a] → [− 21 , 12 ] be the inverse function of
p
g. Show that h0 (x) = 1 − h(x)2 for all x ∈ [−a, a]. Differentiate this
again to obtain h00 (x) = −h(x) for all x ∈ [−a, a].
(c) Check that h(0) = 0 and h0 (0) = 1. (Now h has all the properties required
of the sine function, but it is only defined on an interval [−a, a].)
(d) Suppose now that b &gt; 0 and j : [−b, b] → R satisfies j(0) = 0, j 0 (0) =
1, and j 00 (x) = −j(x) for all x ∈ [−b, b]. Prove that the function k :
[−2b, 2b] → R defined by
k(x) = 2j(x/2)j 0 (x/2)
satisfies k(0) = 0, k 0 (0) = 1, and k 00 (x) = −k(x) for all x ∈ [−2b, 2b].
(e) Explain how to define the sine function by repeatedly applying part (d)
to the function h.
1
```