MATH34011/44011 Solutions to Examples 1
1)(i)
q
1
(1 − ) = O( 2 ) since
q
(1 − )
1
2
Also
q
→ 1 as → 0 + .
(1 − ) = o(1) since
q
(1 − )
1
→ 0 as → 0 + .
(ii) 4π2 = O(2 ). 4π2 = o()
1
1
1
(iii) 1000 2 = O( 2 ), 1000 2 = o(α ), α < 1/2.
(iv) log(1 + ) = O(), log(1 + ) = o(1).
1−cos = O(2 ), 1−cos
= o().
(v) 1+cos
1+cos (vi)
1
1 1
1
cosh = (e + e− ).
2
1
1
1
Hence e− cosh = O(e− 2 e ).
(vii) etan → 1 as → 0 hence etan = O(1).
1
3
(viii) 2 /(1 − cos ) = O(− 2 ).
2)
1
1
3
3
3
−1
h = (1 − A02 ) − e A1 A0 2 − e2 A02
2
4
2
A2
A2
− 12
2A0 8A0
!
+ O(e3 ),
where
A0 = 1 − 3µ0 (1 − µ0 ),
A1 = 3µ1 (2µ0 − 1),
A2 = 3(µ2 (2µ0 − 1) + µ21 ).
3) We need to check that the definitions for an asymptotic sequence are satisfied,
ie
φn+1 = o(φn ) as
→ x0 .
a)
φn = x−rn eix
φn+1
eix x−rn+1
= ix −rn = x−(rn+1 −rn ) → 0 as x → ∞.
φn
e x
b) φn = xrn sin(αx)
φn+1
xrn+1 sin(αx)
= rn
= xrn+1 −rn ,
φn
x sin(αx)
1
provided x 6= 0 and we are close to x = 0 so that sin(αx) 6= 0. Then
φn+1
→ 0 as x → 0.
φn
c)
φn = e−rn x
φn+1
e−xrn+1
= −xrn = e−(rn+1 −rn )x → 0 as x → ∞.
φn
e
d)
φn = ern x
φn+1
exrn+1
= xrn = ex(rn+1 −rn ) → 0 as x → −∞.
φn
e
4)a)
φn = xrn eix
φn+1
eix xrn+1
= ix rn = x(rn+1 −rn ) → ∞ as x → ∞,
φn
e x
and so φn is not an asymptotic sequence.
b)
φn = x−rn sin(nx)
φn+1
sin((n + 1)x)x−rn+1
n + 1 −(rn+1 −rn )
=
∼
x
→ ∞ as x → 0,
−r
φn
sin(nx)x n
n
and so φn is not an asymptotic sequence.
(
c)
φn =
x−n cos x,
x−n sin x,
neven
nodd
x−n+1 cos(x)
φn+1
= −n
= cot(x),
φn
x sin(x)
if n is even.
φn+1
x−n+1 cos(x)
= −n
= tan(x),
φn
x sin(x)
if n is odd.
Similarly
There are values of x = (2k + 1)π/2, k an integer, for which the limit as x → ∞
does not exist. So φn is not an asymptotic sequence.
5)
J0 (2x) =
∞
X
0
(−1)n x2n
x4 x6
2
=
1
−
x
+
−
+ ....
(n!)2
4
36
2
Suppose that
J0 (2x) ∼
∞
X
an (sin x)n ,
as x → 0
0
then
J0 (2x)
a0 = lim
x→0 (sin x)0
!
= lim (J0 (2x)) = 1.
x→0
J0 (2x) − a0
a1 = lim
x→0
(sin x)1
!
= 0,
J0 (2x) − a0 − a1 sin x
a2 = lim
x→0
(sin x)2
!
= −1.
Hence
J0 (2x) ∼ 1 − (sin x)2 + . . . .
b) Calculation gives a0 = 1, a1 = −1.
c) We find a0 = 1, and
J0 (2x) − a0
a1 = lim
x→0
(sin x)3
!
→ ∞.
Thus it is not possible to expand J0 (2x) in terms of (sin x)3n .
3