Imperial College London
M2PM1 Analysis 2
Progress Test 2
10 December 2010
1. [8 marks]
i. Dene what it means for a collection of subsets of R to be an algebra and what it
means for to be a -algebra.
ii. What is the Borel -algebra on R ?
iii. Prove that th
M2PM1 Comments on Sheet 9
1. We have |x| x2 + y 2 hence |f (x, y )| 1 for any (x, y ) R2 . The partial derivatives
for the function f for any (x, y ) R2 , (x, y ) = (0, 0) are
y2
f
(x, y ) =
3;
x
(x2 + y 2 ) 2
We have
f
xy
(x, y ) =
3.
y
( x2 + y 2 ) 2
M2PM1 Comments on Sheet 7
1. It is enough to show that f is bounded for (x, y ) = (0, 0), since f (0, 0) is just 0. But for
(x, y ) = (0, 0), we have that |f (x, y )| 1 . Indeed, multiplying by x2 + y 2 , we see that this
2
inequality is equivalent to 2|x
M2PM1 Comments on Sheet 5
1. The form of the remainder in Taylors theorem tells us that the expression we have to
show tends to 0 is equal to hf (n+1) (a + h)/(n +1)! for some (0, 1) (depending on h). But
f (n+1) is differentiable, hence continuous, at a,
M2PM1 Comments on Sheet 1
1. Given any
> 0, let =
/3. Then > 0, and if 0 < |x| < (where x R3 ), then
|f (x) 1| = 3|x2 + x2 | 3(x2 + x2 + x2 ) = 3|x|2 < 3 2 = .
1
3
1
2
3
By definition, this means f (x) 1 as x 0.
2. Suppose there is some l such that f (x)
M2PM1 Comments on Sheet 2
1. For this function the expression [f (h) f (0)]/h is equal to h sin(1/h), and we know from
the lectures that this 0 as h 0. By the definition of the derivative, this means that
f (0) = 0.
2. The first part can be done using the
M2PM1 Comments on Sheet 3
1. By contradiction. Suppose that l < C , that is, C l > 0. Then we can take
the definition of the limit, which gives us: there is some > 0 such that
= C l in
0 < |x a| < = |f (x) l| < C l,
and hence in particular 0 < |x a| < = f
M2PM1 Comments on Sheet 6
1. Choose partitions 1 and 2 such that s(f, 1 ) > j (f ) /2 and S (f, 2 ) < J (f ) + /2.
Let = 1 2 ; then by Proposition 3.2 from lectures, s(f, ) s(f, 1 ) and S (f, )
S (f, 2 ). It follows that
S (f, ) s(f, ) S (f, 2 ) s(f, 1 )
M2PM1 Comments on Sheet 4
1. If f and g satisfy the conditions stated in the question, then f (n1) and g (n1) satisfy
the conditions required to appy the version of LHpitals rule given in the lectures, hence
o
(n1)
(n1)
(x)/g
(x) l as x a. Now repeat this