Math. 401 (Fulling) 3 October 2003
Test A – Solutions
The limit in question is →0+in all problems, unless otherwise stated.
“Up to [a certain order]” means “stop right before calculating the term of that order.”
“Up through [a certain order]” means “do calculate a term of that order.”
1. (15 pts.) Pronounce each of the following assertions true or false, and write something to
explain your judgment.
(a) 2sin(4)=O(
3
).
TRUE, because sin(4)∼4=O(). (More formally, divide the function by 3, getting
sin(4)
,
which approaches 4 as →0 and hence is less than a constant for sufficiently small. (In fact, it is
less than 4 for all , but that is more than we need to know.))
(b) ω−2ln ω=O(ω−2)asω→+∞.
FA L S E , because the extra factor ln ωgoes to ∞as ω→∞.
(c) t2sin(3t)→0uniformly in t.
FA L S E . Although for each fixed tthe function approaches 0 as does, that limit is not uniform. For
any (no matter how small) we could find many value of tfor which the sine factor equals 1 and the
t2factor is arbitrarily large.
2. (40 pts.) Find approximations to the roots of
x3+x2+4=0.
Stop each series after you have found two nonzero terms.
Regular perturbation theory, x∼x0+x1,leadsto
0=x0
3+O(2)+x
0
2+2x
0
x
2
+O(
2
)+4.
0:x
0
2+4=0 ⇒x
0=±2i.
1:x0
3+2x
0
x
1=0 ⇒x
0
(−4+2x
1
)=0 ⇒x
1=2.
Thus x∼±2i+2.
The other root must be large, so we expect x3≈−x
2
, hence x≈−1/. Therefore, set x≡x/,
getting
x3+x2+4
2=0.
Try x∼x 0+2x 2(since there is no 1term in the equation), getting:
0:x0
3+x0
2=0 ⇒x
0=−1.
1:3x
2
0
x
2
+2x
0
x
2+4=0 ⇒x
2=−4.
Thus x∼−1
−4.
