MATH34011 Solution to Coursework 1
1a)
tan x ∼
2
X
an φn (x),
0
with φn = x2n+1 as x → 0. Hence
"
#
tan x
tan x
a0 = lim
= lim
=1
x→0
x→0
φ0
x
[2 marks] Next
tan x − x
sec2 x − 1
tan x − a1 φ0
= lim
=
lim
a1 = lim
,
x→0
x→0
x→0
φ1
x3
3x2
#
"
"
#
using l’Hospital’s rule. Rearranging gives
1
tan2 x
= .
a1 = lim
2
x→0
3x
3
#
"
[4 marks]
Finally
#

x4
24

3

tan x − x − 31 x3
sin x − (x + x3 ) cos x 
tan x − a0 φ0 − a1 φ1

a2 = lim
= lim
= lim
,
x→0
x→0
x→0
φ2
x5
x5 cos x
"
#

lim 
(x −
x3
3
+
x5
5!
x→0
"
x3
)(1
3
+ . . .) − (x +
−
x2
5
x (1 − 2 + . . .)
x2
2
+
+ . . .) 
=
2
.
15
[4 marks]
b) φn = (sin x)2n+1 as x → 0. Hence
"
#
tan x
tan x
a0 = lim
= lim
= 1.
x→0
x→0
φ0
sin x
[2 marks]
Next
tan x − sin x
1 − cos x
1
a1 = lim
= lim
= .
3
2
x→0
x→0 cos x sin x
2
sin x
[4 marks]
Finally
tan x − sin x − 12 sin3 x
1 − cos x − 12 sin2 x cos x
a2 = lim
=
lim
,
x→0
x→0
sin5 x
sin4 x cos x
"
#
1
"
#

= lim
x→0
1 − (1 −


= lim
x→0
x2
 2
−
x4
24
x2
2
x4
24
+
− 12 (x2 −
3
+ . . .) − 12 (x − x6 + . . .)2 (1 −
2
(1 − x2 + . . .)(x4 + . . .)
x4
+
3
4
x (1
. . .)(1 −
+ ..)
x2
2
+
x4
24
x2
2

+ . . .) 
=−
+
x4
24

+ . . .) 
,
1
1 1
3
+ + = .
24 4 6
8
[4 marks]
2)
Z ∞
t
e−t (1 + )−λ dt.
λ
0
The form of the integrand suggests to put t = λT. Then
I(λ) =
I(λ) = λ
Z ∞
e−λT (1 + T )−λ dT = λ
0
Z ∞
e−λP (T ) dT
0
where P (T ) = T + log(1 + T ).
Now
P 0 (T ) = 1 +
[4 marks].
1
,
1+T
P 00 (T ) = −
1
.
(1 + T )2
Note that P 0 (T ) > 0 for all T > 0 and so P (T ) is an increasing function of T .
Thus the minimum of P (T ) is at T = 0 for T ≥ 0 and P (0) = 0.
[4marks]
So put
u = P (T ) − P (0) = T + log(1 + T ).
Hence
I(λ) = λ
Z ∞
−λP (T )
e
dT = λ
e−λu
0
0
For small T
Z ∞
dT
du.
du
T2 T3 T4
+
−
+ ...,
2
3
4
T2 T3
u ∼ 2T −
+
+ ....
2
3
u∼T +T −
[5 marks]
We can now use Watson’s lemma and we need dT /du for small u. Inverting, the
leading term is T ∼ u/2 + . . . and so either put
T ∼
u
+ c1 uα + c2 uβ + . . . ,
2
T =
u
+ c1 u 2 + c2 u 3 + . . . .
2
or try
Then
u ∼ 2T −
T2 T3
+
+ ...
2
3
2
u
1 u
1 u
u ∼ 2( +c1 u2 +c2 u3 +. . .)− ( +c1 u2 +c2 u3 +. . .)2 + ( +c1 u2 +c2 u3 +. . .)3 +. . . ,
2
2 2
3 2
c1
1
1
u ∼ u + u2 (2c1 − ) + u3 (2c2 − + ) + O(u4 ).
8
2
24
[7 marks]
Equating the coefficients if like powers of uto zero gives
1
c1 = ,
8
Hence
T ∼
c2 =
c1
1
1
−
=−
.
4
48
192
u
1
1 3
+ u2 −
u + O(u4 ),
2 16
192
as u → 0.
[5 marks]
Thus
1 u u2
dT
∼ + −
+ ....
du
2 8 64
Using Watson’s Lemma gives
I(λ)λ
Z ∞
0
1 u u2
1
Γ(2) Γ(3)
( + −
+ . . .) du ∼ λ
+
−
+ ... .
2 8 64
2λ
8λ2
64λ3
"
−λu
e
Hence
I(λ) ∼
1
1
1
+
−
+ ...
2 8λ 32λ2
#
as λ → ∞.
[5 marks]
3