Distances to planes and lines
1. Using vector methods, nd the distance from the point (1,0,0) to the plane 2x + y
Include a cartoon sketch illustrating your solution.
2z = 0.
Answer: The sketch shows the plane and the point P = (1, 0, 0). Q = (0, 0, 0) is
18.100B Problem Set 1
Due Friday September 15 2006 by 3 PM
1) (10 pts) Prove that there is no rational number whose square is 12.
2) (10 pts) Let S be a non-empty subset of the real numbers, bounded above. Show that if
u = sup S, then for every
The Tangent approximation
4. Critique of the approximation formula.
First of all, the approximation formula for functions of two or three variables
x 0, y 0 .
x, yz 0 .
is not a precise m
Functions of two variables
Examples: Functions of several variables
f (x, y) = x2 + y 2 ) f (1, 2) = 5 etc.
f (x, y) = xy 2 ex+y
f (x, y, z) = xy log z
Ideal gas law: P = kT /V .
Dependent and independent variables
In z = f (x, y) we say x, y are independ
18.100B Problem Set 2
Due Friday September 22 2006 by 3 PM
1) (10 pts) Prove that the empty set is a subset of every set.
2) (10 pts) If x, y are complex, prove that
(Hint This is equivalent to proving the following two inequalities
18.100B Problem Set 4
Due Friday October 6 2006 by 3 PM
1) Give an example of an open cover of the set E = cfw_(x , x2 ) 2 R2 : x2 + x2 < 1 R2 which
has no nite subcover. (As usual, R2 is equipped with standard Euclidean metric.)
SOLUTIONS TO PS 10
Solution/Proof of Problem 1. From the denition, we can nd that
fn (t) = ( )n f0 (t) +
Notice that limn
|( 2 )n | + |
( 2 )k =
( )k .
= 3 and since |f0 (t)| = | sin t| 1 we have
SOLUTIONS TO PS2
Proof. It is true that for any two sets A, B, the intersection A \ B is a subset of
A. Now consider = A \ Ac . So is a subset of A for any set A.
Proof. Notice that
y| , |x|
So we only need to p
18.100B Problem Set 5
Due Friday October 20 2006 by 3 PM
1) Let M be a complete metric space, and let X M. Show that X is complete if and only if X
2) a) Show that a sequence in an arbitrary metric space (xn ) converges if and only if
18.100B Problem Set 3
Due Friday September 29 2006 by 3 PM
1) (10 pts) In vector spaces, metrics are usually de ned in terms of nor s which measure the length
of a vector. If V is a vector space de ned over R, then a norm is a function from vecto
18 100B Problem Set 9 Solutions
1) First we need to show that
nx + 1
converges pointwise but not uniformly on (0, 1). If we x some x 2 (0, 1), we have that
lim fn (x) = lim
n!1 nx + 1
Thus the fn converge pointwise. However, cons
The Tangent Approximation
1. The tangent plane.
For a function of one variable, w = f (x), the tangent line to its graph
at a point (x0 , w0 ) is the line passing through (x0 , w0 ) and having slope
For a function of two va
18 100B Problem Set 5 Solutions
1) We have X M, with M complete. X is complete if and only if every Cauchy sequence of X
converges to some x 2 X. Let (xi ) be Cauchy, with xi 2 X. M being complete implies that
xi ! y 2 M. Therefore y is a lim
18.100B Problem Set 9
Due Friday December 1 2006 by 3 PM
1) Let fn (x) = 1/(nx+1) and gn (x) = x/(nx+1) for x 2 (0, 1) and n 2 . Prove that fn converges
pointwise but not uniformly on (0, 1), and that gn converges uniformly on (0, 1).
2) Let fn
SOLUTIONS TO PS 8
Solution/Proof of Problem 1 From f (x) = f (x2 ), we have
f (x) = f (x 2 ) = f (x 4 ) = = f (x 2 )
Now let yn = x 2 , and assume x = 0, so lim yn = 1. Since f is continuous, we
have if x 6= 0
f (x) = lim f (x)
18 100B Problem Set 3 Solutions
1) We begin by dening d : V V ! R such that d(x, y) = kx yk. Now to show that this
function satises the denition of a metric. d(x, y) = kx yk 0 and
d(x, y) = 0 () kx
So the function is positive denite.
18.100B Problem Set 6
Due Friday October 27 2006 by 3 PM
1) Prove that if
|an | is converges, then
|an |2 also converges.
2) Prove that
n(n + 1)(n + 2)
Hint: Use a telescope trick, i. e. an argument of the form
18 100B Problem Set 1 Solutions
1) The proof is by contradiction. Assume 9r 2 Q such that r2 = 12. Then we may write r as a
with a, b 2 Z and we can assume that a and b have no common factors. Then
a 2 a2
12 = r2 =
2 = a2 .
18 100B Problem Set 7 Solutions
1) We have ai > 0 and ai
ai for all i = 0, 1, 2, ., and lim ai = 0, and we want to show the
( 1)i ai = a0
a1 + a2
So we de ne sn to be the partial sums of the rst n + 1 terms of
18.100B Problem Set 8
Due Thursday November 9 2006 by 3 PM
1) Let f : [0, 1) !
be continuous, and suppose
f (x2 ) = f (x)
holds for every x 0. Prove that f has to be a constant function.
Hint: Show that f (0) = f (x) if 0 x < 1, and f (1) = f (x)
SOLUTIONS TO PS6
Solution/Proof of Problem 1. Since
|an | converges, lim |an | = 0. So 9N 2
such that for n > N , |an | < 1. Thus for n P N we have |a2 | |an | and
by the comparison theorem and the convergence of
|an | we conclude
Our text states and proves Stokes' Theorem in 12.11, but ituses
the scalar form for writing both the line integral and the surface integral
involved. In the applications, it is the vector form of the theorem that is
most likely to be quote
18.100B Problem Set 7
Due Friday November 3 2006 by 3 PM
1) Consider an in nite series with alternating signs
a1 + a2
a3 + a4
( 1)n an .
Prove the Leibnitz criteria for convergence:
If ai > 0 for all i, ai 1 ai and P i 1 ai = 0, then the
SOLUTIONS TO PS4
Solution/Proof of Problem 1 Consider the open set
(x1 , x2 ) 2 R2 : x2 + x2 < 1
Then we can see that E [Bn because for any point (x, y) 2 E, x2 + y 2 < 1, we
can nd an n big enough such that x2 + y 2 < 1 n , i.
18.100B Problem Set 10
Due Friday December 8 2006 by 3 PM
1) Let (fn ) be the sequence of functions on R dened as follows.
f0 (t) = sin t and fn 1 (t) = fn (t) + 1 for n 2 .
Show that fn ! 3 uniformly on R. What can you say if we choose f0 (