Solutions of Homework 9, Mathematics 1
Problem 1: Compute the Taylor polynomial of n’th order Pn (x) at x0 = 0 for
the function f (x) = cosh x. Do this
(a) from the definition;
(b) using the known expansions for ex and e−x . Justify why this must coincide
with (a)!
Solution:
(a) Compute the k-th derivative:
f (k) (x) = (cosh x)k =
ex + e−x
2
(k)
=
(ex )(k) + (e−x )(k)
ex + (−1)k e−x
=
2
2
(proof by induction), so
f
(k)
Pn (x) =
(
1 if k is even,
(0) =
0 if k is odd,
n
X
f (k) (0)
k=0
k!
[n/2]
xk =
X
k=0
1 2k
x .
(2k)!
(b) Very similar as with sinh x in the lecture.
Problem 2: Compute the Taylor polynomial P4 (x) of the function tan x at
x0 = 0
(a) from the definition;
(b) using the expansions of sin x and cos x.
Solution:
(a) Differentiate tan x 4 times, plug in 0,... .
(b) As in the hint: let P4 (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 . Then
sin x
cos x
x − 16 x3 + o(x4 )
=
,
1 − 12 x2 + o(x3 )
a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + o(x4 ) = tan x =
as x → 0.
Multiply this equation by the denominator 1 − 21 x2 + o(x3 ), and use that
ak xk o(x3 ) = o(xk+3 ), o(x4 )o(x3 ) = o(x7 ) (easy to check from definition), as
1
well as property (d) from Problem 3:
a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + o(x4 )
1
− x2 (a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + o(x4 ))
2
+ o(x3 )(a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + o(x4 ))
1
= x − x3 + o(x4 ),
6 a0 2 a1 3 a2 4
x + a3 −
x + a4 −
x + a0 o(x3 ) + o(x4 )
a0 + a1 x + a2 −
2
2
2
1
= x − x3 + o(x4 ).
6
Comparing coefficients at x0 , x1 , x2 , and x3 we get a0 = 0, a1 = 1, a2 = 0, and
a3 − 21 = − 16 , i. e. a3 = 31 . Since a0 = 0, the term a0 o(x3 ) on the left disappears
and comparing coeffitients at x4 gives a4 = 0. Hence
x3
.
3
Note: The formal proof that we can simply compare coefficients by expansions
like this is identical to the proof of uniqueness of the Taylor polynomial from
the lecture.
P4 (x) = x +
Problem 3 [Simplifications with the little o]: In the following, the little o
will always refer to behavior as x → 0.
(a) Let a 6= 0 and k ∈ N be constant. Prove that if f (x) = o(axk ), then
f (x) = o(xk ).
(b) Let k, l ∈ N, k ≤ l. Prove that if f (x) = o(xl ), then f (x) = o(xk ) and
f (x)
= o(xk ).
xl−k
(c) Let k, l ∈ N, k ≤ l. Prove that if f (x) = o(xk + xl ), then f (x) = o(xk ).
Hint: We assume limx→0 xfk(x)
= 0 and want to prove limx→0 fx(x)
= 0. Write
k
+xl
f (x)
xk
as a product of xfk(x)
and something.
+xl
(d) [2 points] Let k, l ∈ N, k ≤ l. Prove that if f (x) = o(xk ) + o(xl ), then
f (x) = o(xk ).
(e)* [2* points] Let k, l ∈ N, k ≤ l. Prove that if f (x) = o(xk + o(xl )), then
f (x) = o(xk ).
Solution:
(a) and (b) are easy from definition. (c) follows the hint, see also the detailed
proof of (e).
(d) As in the hint: f (x) = o(xk ) + o(xl ) means that there exist functions
g(x) = o(xk ), h(x) = o(xl ) such that f (x) = g(x) + h(x). Then
lim
x→0
f (x)
g(x) + h(x)
g(x)
h(x) l−k
= lim
= lim k + lim
·x
x→0
x→0 x
x→0 xl
xk
xk
g(x)
h(x)
= lim k + lim
· lim xl−k = 0 + 0 · 0 = 0,
x→0 x
x→0 xl
x→0
2
where we used limx→0 g(x)
= 0, limx→0 h(x)
= 0. The last equality means
xk
xl
f (x)
exactly limx→0 xk = 0, what we wanted to prove.
(e) As in the hint: f (x) = o(xk + o(xl )) means that there exists g(x) = o(xl )
(x)
such that limx→0 xkf+g(x)
= 0. Then
f (x)
f (x)
xk + g(x)
= lim k
·
k
x→0 x + g(x)
x→0 x
xk
f (x)
g(x)
= lim k
= 0 · (1 + 0) = 0,
· lim 1 + k
x→0 x + g(x) x→0
x
lim
which we wanted to prove.
Problem 4: Use Taylor’s formula to compute the limits:
(a) [2 points]
lim
x→0
x − sin x
ex − 1 − x −
x2
2
2
,
ln(1 + x + x2 ) + ln(1 − x − x )
,
x→0
x sin x
1
sin x x
lim
,
x→0
x
(b)
lim
(c) [3 points]
cos x − 1 + 12 sin2 x
,
x→0
x4
x
−x
cos(xe ) − cos(xe )
lim
.
x→0
x3
(d)
lim
(e) [3 points]
Solutions/answers:
(a)
lim
x→0
ex
x − sin x
−1−x−
x− x−
x2
2
= lim
x→0
=
1+x+
x3
lim 63
x→0 x
6
x2
2
+ o(x4 )
+
o(x3 )
(b) −1
3
+
x3
6
+ o(x4 )
+ o(x3 ) − 1 − x −
x3
6
1
6
x→0 1
6
= lim
+ o(x)
= 1.
+ o(1)
x2
2
(c)
1
sin x
1
sin x x
= lim eln( x )· x ,
lim
x→0
x→0
x
sin x
1
x2
1
lim ln
· = lim ln 1 −
+ o(x3 ) ·
x→0
x
x x→0
6
x
2
2
x
x
1
= lim
− + o(x3 ) + o − + o(x3 )
x→0
6
6
x
x
= lim − + o(x) = 0,
x→0
6
1
sin x x
=⇒ lim
= e0 = 1.
x→0
x
Note that from the same computation, we get limx→0
(d) − 81
(e)
sin x
x
1
x2
1
= e− 6 .
cos(xex ) − cos(xe−x )
x→0
x3
3
3
cos(x + x2 + x2 + o(x3 )) − cos(x − x2 + x2 + o(x3 ))
= lim
x→0
x3
3
1
x3
2
3 2
1 − 2 (x + x + 2 + o(x )) + o((x + x2 + x2 + o(x3 ))3 )
= lim
x→0
x3
3
1
x3
2
3 2
1 − 2 (x − x + 2 + o(x )) + o((x − x2 + x2 + o(x3 ))3 )
−
x3
1 2
3
3
1 − 2 x − x + o(x ) + o(x3 ) − 1 + 21 x2 − x3 + o(x3 ) + o(x3 )
= lim
x→0
x3
= − 2.
lim
Problem 5: Consider the function f (x) = arctan
x+1
x−1
.
(a) [2 points] Compute its derivative, wherever it exists.
(b) [1 point] Why does it NOT follow from (a) and Corollary 2.2.13 that
f (x) = − arctan x + C for some constant C?
(c) [2 points] Find the connection between f (x) and arctan x, expressing any
constants explicitely by numbers.
Solution:
For x 6= 1, after differentiation (chain rule + quotient rule) and simplifi1
0
cations, we get f 0 (x) = − 1+x
Hence, by Corollary 2.2.13,
2 = (− arctan x) .
the functions f (x) and − arctan x must differ only by a constant, on any interval where both are defined. Since R \ {1} is not an interval, we do not get
f (x) = − arctan x+C on the full definition domain of f . However, we can apply
4
Corollary 2.2.13 separately on intervals (−∞, 1) and (1, +∞), and get possibly
different constants on each of the intervals:
(
− arctan x + C1 for x < 1,
x+1
(1)
=
f (x) = arctan
x−1
− arctan x + C2 for x > 1.
We can determine the constants by plugging in any point from the intervals:
for example, for x = 0, f (0) = arctan(−1) = − π4 , while − arctan(0) = 0, so
C1 = − π4 . We can also obtain C2 by taking limit in (1):
π
x+1
π
= arctan 1 = lim arctan
= lim − arctan x + C2 = − + C2 ,
x→+∞
x→+∞
4
x−1
2
so C2 =
3π
4
and
f (x) = arctan
x+1
x−1
(
− arctan x −
=
− arctan x +
π
4 for x < 1,
3π
4 for x > 1.
Note that the jump of +π at the point x = 1 on the right hand side is consistent
with the jump on the left hand side of the equation, since
x+1
π
lim arctan
= lim arctan y = − ,
y→−∞
x→1−
x−1
2
x+1
π
lim arctan
= lim arctan y = .
y→+∞
x→1+
x−1
2
Problem 6: Compute
the following indefinite integrals on (−∞, +∞):
R
(a) [2
points]
sinh
x
dx;
R
(b) R cosh x dx;
(c) |x| dx;
Hint: Find a possible primitive F on R by considering f on (0, ∞) and (−∞, 0)
separately. Then make sure that the constants match so that F is differentiable
also at 0 (and FR0 (0) = f (0)).
(d) [3 points] f (x) dx, where
(
x for x ≥ 1,
f (x) =
1 for x < 1.
Hint: Find a possible primitive F on R by considering f on (1, ∞) and (−∞, 1)
separately. Then make sure that the constants match so that F is differentiable
also at 1 (and F 0 (1) = f (1)).
Solution:
(a)
Z
Z
Z
1
ex − e−x
dx =
ex dx + −e−x dx
2
2
1 x
=
e + e−x + C = cosh x + C
2
Z
sinh x dx =
5
R
(b) Analogously, cosh x = sinh x + C.
(c) This is very similar to the detailed solution of (d), with the result of
( 2
Z
x
+ C for x ≥ 0,
f (x) dx = 2 x2
− 2 + C for x < 0.
(d) Let us look for an antiderivative F of f , i. e. a differentiable (and hence
continuous) function F such that F 0 = f on R. Since F 0 (x) = x on (1, +∞)
and F is continuous on [1, +∞), we must have
F (x) =
x2
+ C1 on [1, +∞).
2
Analogously, since F 0 (x) = 1 on (−∞, 1) and F is continuous on (−∞, 1], we
must have
F (x) = x + C2 on (−∞, 1].
Hence 21 + C1 = F (1) = 1 + C2 , C2 = C1 − 12 . On the other hand, for any
constant C1 = C and matching C2 = C − 12 , the function
( 2
x
+ C for x ≥ 1,
F (x) = 2
x + C − 21 for x < 1
will be an antiderivative of f : this is clear for x 6= 1 and at 1, both
F+0 (1) = xbx=1 = 1
and
x + C − 12 − F (1)
x→1−
x−1
x + C − 21 − ( 12 + C)
= lim
x→1−
x−1
x−1
= lim
= 1,
x→1− x − 1
F−0 (1) = lim
so F 0 (1) exists and equals 1 = f (1). Hence,
( 2
Z
x
+ C for x ≥ 1,
f (x) dx = 2
x + C − 21 for x < 1.
Problem 7: Explain the second equality in
Z
dx
√
= arcsin x + C = − arccos x + C.
1 − x2
Deduce that arcsin x = − arccos x + π2 .
6
Solution: The equality when an indefinite integral is present is taken as equality
of sets, i. e.
Z
dx
√
= {arcsin x + C : C ∈ R} = {− arccos x + C : C ∈ R}.
1 − x2
The function arcsin x + C1 corresponding to C = C1 in the left set must equal
some function − arccos x + C2 from the right set, but the the constants C1 and
C2 might be different. In particular, if we take C1 = 0, there must exist some
C2 such that
arcsin x = − arccos x + C2 for all x ∈ R.
Taking x = 0 in this identity, we find out the constant:
0 = arcsin 0 = − arccos 0 + C2 = −
so C2 =
π
2
π
+ C2 ,
2
and
arcsin x = − arccos x +
7
π
2
for all x ∈ R.