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ucl past paper

can any1 provide solutions or answers to these papers.
i've been given papers but no answers so i dont knw if im doin them right

http://www.geocities.com/chetlo86/mathm11a_02.pdf

http://www.geocities.com/chetlo86/mathm12a_03.pdf

http://www.geocities.com/chetlo86/mathm13a_02.pdf

REally appreciate answers to any of the Qs. thanx

Reply 1

Jonny W

-------------------------------
mathm11a_02
-------------------------------
Solutions to Questions 1 - 3
-------------------------------

(1)
(a)
Cauchy-Schwarz inequality
If x and y are in R^n then |x.y| <= |x| |y|.

Proof
For any real number k,

0
<= |x + ky|^2
= (x + ky).(x + ky)
= |x|^2 + 2k(x.y) + k^2|y|^2

The quadratic in k on the RHS does not have distinct real roots. So its discriminant, 4|x.y|^2 - 4|x|^2|y|^2, cannot be positive. So |x.y|^2 <= |x|^2 |y|^2. So |x.y| <= |x| |y|.

(b)
(sum over i) (a_i + b_i)^2
= |a|^2 + 2(a.b) + |b|^2
<= |a|^2 + 2|a| |b| + |b|^2 . . . . . by the CS inequality
= (|a| + |b|)^2

sqrt[ (sum over i) (a_i + b_i)^2 ]
<= |a| + |b|
= sqrt[ (sum over i) (a_i)^2 ] + sqrt[ (sum over i) (b_i)^2 ]

(c)
We define two vectors:

x = ( (a_1)^(2/3), (a_2)^(2/3), ..., (a_n)^(2/3) )
y = ( (a_1)^(1/3) b_1, (a_2)^(1/3) b_2, ..., (a_n)^(1/3) b_n )

Then

(sum over i) a_i b_i
= (sum over i) (a_i)^(2/3) [(a_i)^(1/3) b_i]
= x.y
<= |x| |y| . . . . . by the CS inequality
= sqrt[ (sum over i) (a_i)^(4/3) ] * sqrt[ (sum over i) (a_i)^(2/3) (b_i)^2 ]

But, by the CS inequality again,

(sum over i) (a_i)^(2/3) (b_i)^2
<= sqrt[ (sum over i) (a_i)^(4/3) ] * sqrt[ (sum over i) (b_i)^4 ]

Combining the last two results,

(sum over i) a_i b_i
<= sqrt[ (sum over i) (a_i)^(4/3) ] * sqrt[ (sum over i) (a_i)^(2/3) (b_i)^2 ]
<= sqrt[ (sum over i) (a_i)^(4/3) ] * [ (sum over i) (a_i)^(4/3) ]^(1/4) * [ (sum over i) (b_i)^4 ]^(1/4)
= [ (sum over i) (a_i)^(4/3) ]^(3/4) * [ (sum over i) (b_i)^4 ]^(1/4)

--

(2)
(a)
For any epsilon > 0 there is a positive integer N such |x_n - a| < epsilon for all n >= N.

(b)
a is an upper bound for S if x <= a for all x in S.

If
(i) a is an upper bound for S, and
(ii) a <= b whenever b is an upper bound for S
then a is a least upper bound (LUB) for S.

(c)
Let S be a set of real numbers. If S is bounded above then it has an LUB.

(d)
Let S = {x_n : n >= 1}. Then since S is bounded above, S has an LUB -- which we will call a. By definition of an upper bound, x_n <= a for all n. Now we are going to do a proof by contradiction. Suppose that x_n doesn't tend to a. Then there is an epsilon > 0 such that |x_n - a| >= epsilon for infinitely many choices of n. Since x_n <= a for all n, it follows that x_n <= a - epsilon for infinitely many choices of n. Since (x_n) is increasing, it follows that x_n <= a - epsilon for all choices of n. (If the inequality failed for one value of n then it would also fail for all larger values.) So (a - epsilon) is an upper bound for S. That is a contradiction because a is an LUB for S. So x_n -> a.

(e)
Bolzano-Weierstrass Theorem
Let (x_n) be a bounded sequence of real numbers. Then (x_n) has a convergent subsequence.

Proof (by repeated bisection)
We assume that (x_n) is not a constant sequence, because the result is trivial if it is. Since (x_n) is bounded, there are numbers a_1 and b_1 such that all the terms of (x_n) are in the closed interval [a_1, b_1].

Now bisect [a_1, b_1] into two subintervals [a_1, c_1] and [c_1, b_1]. (Thus c_1 = (a_1 + b_1)/2, the midpoint of [a_1, b_1].) Then [a_1, c_1] or [c_1, b_1] (or both) contains infinitely many terms of (x_n). If [a_1, c_1] contains infinitely many terms of (x_n) then let [a_2, b_2] = [a_1, c_1]; otherwise let [a_2, b_2] = [c_1, b_1]. Then [a_2, b_2] contains infinitely many terms of (x_n).

Now repeat. Bisect [a_2, b_2] into the two subintervals [a_2, c_2] and [c_2, b_2]. Then [a_2, c_2] or [c_2, b_2] (or both) contains infinitely many terms of (x_n). If [a_2, c_2] contains infinitely many terms of (x_n) then let [a_3, b_3] = [a_2, c_2]; otherwise let [a_3, b_3] = [c_2, b_2]. Then [a_3, b_3] contains infinitely many terms of (x_n).

Continuing in this way, we get a nested sequence [a_1, b_1], [a_2, b_2], [a_3, b_3], ... of intervals such that:
(i) for all i, [a_i, b_i] contains infinitely many terms of (x_n);
(ii) (a_i) is increasing and (b_i) is decreasing;
(iii) b_i - a_i = (b_1 - a_1) / 2^(i - 1) -> 0 as i -> infinity.

(a_i) converges because it is increasing and bounded above. Let L be the limit. Since b_i - a_i -> 0 as i -> infinity, we also have b_i -> L as i -> infinity.

Finally we choose a subsequence of (x_n(i)) of (x_n) such that x_n(i) is in [a_i, b_i] for all i. Then x_n(i) -> L as i -> infinity.

--

(3)
(a)
A real sequence (x_n) is Cauchy if for any epsilon > 0 there is a positive integer N such |x_n - x_m| < epsilon for all n, m >= N.

(b)
A real sequence converges if and only if it is Cauchy.

(c)
Comparison test
Suppose that (x_n) and (y_n) are real sequences with 0 <= x_n <= y_n for all n. Suppose also that (sum) y_n converges. Then (sum) x_n converges.

Proof
We define the partial sums of the two series:

s_N = (sum from n = 1 to N) x_n
S_N = (sum from n = 1 to N) y_n

Then for all positive integers N and M with N > M we have

|s_N - s_M|
= (sum from n = M + 1 to N) x_n
<= (sum from n = M + 1 to N) y_n . . . . . since x_n <= y_n for all n
= |S_N - S_M| . . . . . (*)

We know that (S_n) converges. By the GPoC, (S_n) is Cauchy. Hence by (*), (s_n) is Cauchy. Hence by the GPoC again, (s_n) converges. So (sum) x_n converges.

(d)
Let (x_n) be a real sequence, and L be a number. Suppose that x_(n + 1)/x_n -> L as n -> infinity. Then:
(i) if |L| < 1 then (sum) x_n converges;
(ii) if |L| > 1 then (sum) x_n diverges.

(e)
(i)
3^n/(4^n n^2) <= 1/n^2 for all n. The sum converges by the comparison test.

(ii)
Let x_n = 5^n (n!)^2/(2n)!. Then

x_(n + 1)/x_n
= 5 (n + 1)^2/[(2n + 1)(2n + 2)]
-> 5/4 as n -> infinity

The sum diverges by the ratio test (since 5/4 > 1).

(iii)
Let x_n = n!/n^(n + 2). Then

x_(n + 1)/x_n
= [(n + 1)!/(n + 1)^(n + 3)] * [n^(n + 2)/n!]
= (n + 1) n^(n + 2) / (n + 1)^(n + 3)
= n^(n + 2) / (n + 1)^(n + 2)
= [(n + 1)/n]^[-(n + 2)]
= [1 + 1/n]^[-(n + 2)]
= [1 + 1/n]^(-2) * [1 + 1/n]^(-n)
-> 1 * 1/e
= 1/e as n -> infinity

The sum converges by the ratio test (since 1/e < 1).

Reply 2

john !!

for the third link, the first q:

Probability of red ball being drawn first time = 5/8
so the probability that 2 red balls are added is 5/8
probability that 2 greens are added is 3/8

so the probability that the probability of chosing a red on the second time is 7/10 is 5/8
and the probability that the probability of chosing a red on the second time is 5/10 is 3/8

erm... *scream*
sometimes I wish I did S1...

Reply 3

Jonny W

-------------------------------
mathm13a_02
-------------------------------
Solutions to Questions 1 - 3
-------------------------------

(1)
(a)
(i)
Let X and Y be the respective results of the first and second draws.

If X = R then there are 7 red and 3 green balls just before the second draw.
If X = G then there are 5 red and 5 green balls just before the second draw.

So

P(Y = R)
= P(Y = R | X = R).P(X = R) + P(Y = R | X = G).P(X = G)
= (7/10)(5/8) + (5/10)(3/8)
= 5/8

(ii)
P(X = R, Y = G)
= P(Y = G | X = R).P(X = R)
= (3/10)(5/8)
= 3/16

P(X = G, Y = R)
= P(Y = R | X = G).P(X = G)
= (5/10)(3/8)
= 3/16

P(one ball of each colour in the two draws)
= P(X = R, Y = G) + P(X = G, Y = R)
= 3/8

(iii)
P(X = R | Y = R)
= P(X = R, Y = R)/P(Y = R)
= (5/8)(7/10) / (5/8)
= 7/10

(b)
P(wins all three games) = (1/2)(1/4)(1/8) = 1/64
P(wins all but the first game) = (1/2)(1/2)(1/4) = 1/16
P(wins all but the second game) = (1/2)(3/4)(1/4) = 3/32
P(wins all but the third game) = (1/2)(1/4)(7/8) = 7/64

P(wins at least two games) = 1/64 + 1/16 + 3/32 + 7/64 = 9/32

--

(2)
We can represent the results of the n trials by a string of length n, with each letter in the string being either "F" (failure) or "S" (success). There are nCr result strings that contain exactly r "S"s. Each such string has probability p^r (1 - p)^(n - r). So

P(exactly r successes) = nCr * p^r (1 - p)^(n - r)

Let X_1, ..., X_n be the Bernoulli random variables. Let Y = X_1 + ... + X_n. Then mu = E(Y) = E(X_1) + ... + E(X_n) = n*E(X_1) = np.

--

We represent P_1's first 2N choices of direction by a string of length 2N, with each letter in the string being either "N" (North) or "E" (East).

Q
= P(P_1 reaches the city centre)
= P(string contains exactly N "N"s and N "E"s)
= (2N)C(N) (1/2)^(2N)
= (2N)! / [(N!)^2 2^(2N)]

--

Let Z be the number of people who meet at the city centre. Then Z ~ Binomial(4, Q). So

P(Z >= 2)
= 1 - P(Z = 0) - P(Z = 1)
= 1 - (1 - Q)^4 - 4Q(1 - Q)^3
= 1 - (1 - 4Q + 6Q^2 - 4Q^3 + Q^4) - 4Q(1 - 3Q + 3Q^2 - Q^3)
= 6Q^2 - 8Q^3 + 3Q^4
= Q^2(3Q^2 - 8Q + 6)

--

(3)
Let y be any position vector. Then the resultant moment about the point with position vector y is

G_y
= (sum over i) [(r_i - y) x F_i]
= (sum over i) (r_i x F_i) - (sum over i) (y x F_i)
= (sum over i) (r_i x F_i) - y x [(sum over i) F_i]
= (sum over i) (r_i x F_i) - y x F

Since (y x F) is perpendicular to F,

F.(G_y) = F.[(sum over i) (r_i x F_i)]

The RHS does not depend on y.

--

Suppose that F.G = 0 and that F is not 0. We want a position vector r such that a force F applied at r is equivalent to the set of forces. That is, we want a position vector r such that

r x F = G . . . . . (*)

A solution r to (*) must be perpendicular to G. Because F and (F x G) are perpendicular to G, it follows that r must be a linear combination of F and (F x G). Since

F x F = 0

(F x G) x F
= -F(G.F) + G(F.F) . . . . . http://mathworld.wolfram.com/VectorTripleProduct.html
= G|F|^2

the general solution to (*) is

r = k F + (F x G)/|F|^2 . . . . . (**)

where k is any scalar.

The set of forces is equivalent to a force F applied at r, for any r that is given by (**).

--

F = F0 (1, 1, 1)

G
= (a, 0, c) x (0, F0, 0) + (0, b, 0) x (0, 0, F0)
= F0 [(a, 0, c) x (0, 1, 0) + (0, b, 0) x (0, 0, 1)]
= F0 [(-c, 0, a) + (b, 0, 0)]
= F0 (b - c, 0, a)

F.G = F0^2 (a + b - c)

Suppose that c = a + b. Then F.G = 0. So the forces are equivalent to a single force.

--

G = F0 a (-1, 0, 1)

Using (**), the point of application of the single force can be

r
= k F + (F x G)/|F|^2
= k F0 (1, 1, 1) + F0^2 a (1, 1, 1) x (-1, 0, 1)/(3 F0^2)
= k F0 (1, 1, 1) + (1/3) a (1, 1, 1) x (-1, 0, 1)
= k F0 (1, 1, 1) + (1/3) a (1, -2, 1) . . . . . (1)

for any scalar k. The line described by (1) as k varies intersects the plane z = 0 at the point (1/3) a (0, -3, 0) = (0, -a, 0).

[Check. F0 (0, -a, 0) x (1, 1, 1) = F0 (-a, 0, -a) = G.]

Reply 4

madhapper

thanx guys ne more solutions to these wud b great. but thanx for all ya help so far

Reply 5

madhapper

The other Qs to mathm11_02 which couldnt be read can be found here:

http://www.geocities.com/chetlo86/mathm11a_02_Q4_.pdf

Reply 6

Jonny W

-------------------------------
mathm11a_02
-------------------------------
Solutions to Questions 4 - 6
-------------------------------

(4)
(a)
exp(x) = (sum from n = 0 to infinity) x^n / n!

(b)
A real series (sum from n = 1 to infinity) x_n converges if there is a number L such that [(sum] -> L as N -> infinity.

A real series (sum from n = 1 to infinity) x_n converges absolutely if (sum from n = 1 to infinity) |x_n| converges.

(c)
For any real nonzero x,

|x^(n + 1) / (n + 1)!| / |x^n / n!|
= |x| / (n + 1)
-> 0 as n -> infinity.

The series (sum from n = 0 to infinity) |x^n / n!| therefore converges by the ratio test. In other words, the series (sum from n = 0 to infinity) x^n / n! converges absolutely.

(d)
Theorem about multiplication of series
Suppose that (sum from n = 0 to infinity) a_n and (sum from n = 0 to infinity) b_n are absolutely convergent real series. Then

[(sum] * [(sum]
= (sum from n = 0 to infinity) c_n

where

c_n = (sum from i = 0 to n) a_i b_(n - i)

--

By the theorem,

exp(x)exp(y) = (sum from n = 0 to infinity) c_n

where

c_n
= (sum from i = 0 to n) [x^i / i!][y^(n - i) / (n - i)!]
= (1/n!) (sum from i = 0 to n) nCi x^i y^(n - i)
= (1/n!) (x + y)^n . . . . . using the binomial theorem

So

exp(x)exp(y)
= (sum from n = 0 to infinity) (1/n!)(x + y)^n
= exp(x + y)

(e)
For any x >= 0,

exp(x)
= 1 + x + x^2/2 + (sum from n = 3 to infinity) x^n / n!
>= 1 + x + x^2/2 . . . . . because x^n / n! >= 0 for all n >= 3

--

Method 1 (not using the hint)
For any x <= 0,

(d/dx) [(1 + x) - exp(x)] = 1 - exp(x) >= 0

But (1 + x) - exp(x) = 0 when x = 0. So for any x <= 0,

(1 + x) - exp(x) <= 0

For any x <= 0,

(d/dx) [(1 + x + x^2/2) - exp(x)] = 1 + x - exp(x) <= 0

But (1 + x + x^2/2) - exp(x) = 0 when x = 0. So for any x <= 0,

(1 + x + x^2/2) - exp(x) >= 0
exp(x) <= 1 + x + x^2/2

Method 2 (using the hint)
For any x with -1 <= x <= 0,

exp(x) = 1 + x + x^2/2 + x^3 * [ 1/3! + x/4! + x^2/5! + x^3/6! + ... ]

and

1/3! + x/4! + x^2/5! + x^3/6! + ...
>= 1/6 - (1/4! + 1/6! + 1/8! + 1/10! + ...)
= 1/6 - [ (1/2)(exp(1) + exp(-1)) - 1 - 1/2 ]
= 5/3 - (1/2)(exp(1) + exp(-1))
>= 0

and so, since also x^3 <= 0,

exp(x) <= 1 + x + x^2/2 . . . . . (1)

For any x <= -1,

1 + x + x^2/2
= 1/2 + (1/2)(x + 1)^2
>= 1/2 + 0
= 1/2
>= exp(-1)
>= exp(x) . . . . . (2)

(1) and (2) together show that, for any x <= 0,

exp(x) <= 1 + x + x^2/2

--

(5)
(a)
f: R -> R is continuous at a c in R if: for any epsilon > 0 there is a delta > 0 such that |f(x) - f(c)| < epsilon for all x with |x - c| < delta.

(b)
Intermediate Value Theorem
Let f: R -> R be continuous. Suppose that p and q are in R, with p < q. Let k be any number between f(p) and f(q). Then there is an L in [p, q] such that f(L) = k.

Proof (by repeated bisection)
Without loss of generality, we assume that f(p) <= f(q). Then f(p) <= k <= f(q). Let a_1 = p and b_1 = q. Then f(a_1) <= k <= f(b_1).

Let c_1 be the midpoint (a_1 + b_1)/2 of [a_1, b_1].
(i) If f(c_1) <= k then let [a_2, b_2] be the right-hand half [c_1, b_1] of [a_1, b_1].
(ii) If f(c_1) > k then let [a_2, b_2] be the left-hand half [a_1, c_1] of [a_1, b_1].
Then f(a_2) <= k <= f(b_2).

Now repeat. Let c_2 be the midpoint (a_2 + b_2)/2 of [a_2, b_2].
(i) If f(c_2) <= k then let [a_3, b_3] be the right-hand half [c_2, b_2] of [a_2, b_2].
(ii) If f(c_2) > k then let [a_3, b_3] be the left-hand half [a_2, c_2] of [a_2, b_2].
Then f(a_3) <= k <= f(b_3).

Continuing in this way we get a nested sequence [a_1, b_1], [a_2, b_2], [a_3, b_3], ... of intervals such that:
(i) for all i, f(a_i) <= k <= f(b_i);
(ii) (a_i) is increasing and (b_i) is decreasing;
(iii) b_i - a_i = (b_1 - a_1) / 2^(i - 1) -> 0 as i -> infinity.

(a_i) converges because it is increasing and bounded above. Let L be the limit. Since b_i - a_i -> 0 as i -> infinity, we also have b_i -> L as i -> infinity.

Since f is continuous at L, and a_i -> L as i -> infinity,

f(a_i) -> f(L) as i -> infinity . . . . . (1)

[Proof of (1). Suppose that we are given an epsilon > 0. By the definition of continuity, there is a delta > 0 such that |f(x) - f(L)| < epsilon for all x with |x - L| < delta. By the definition of a limit, there is a positive integer N such that |a_i - L| < delta for all i >= N. Putting those two statements together, |f(a_i) - f(L)| < epsilon for all i >= N.]

Similarly,

f(b_i) -> f(L) as i -> infinity . . . . . (2)

Since f(a_i) <= k <= f(b_i) for all i, it follows from (1) and (2) that f(L) = k.

(c)
Define a continuous function g: R -> R by g(x) = f(x) - f(x + 1). Then g(0) = f(0) - f(1) = f(2) - f(1) = -[f(1) - f(2)] = -g(1). So either g(0) and g(1) are both zero, or one is positive and the other negative. The Intermediate Value Theorem implies that there is a y in [0, 1] such that g(y) = 0. Then f(y) = f(y + 1).

--

(6)
(a)
Suppose that f is not bounded. Then there is a sequence (x_n) in [a, b] with |f(x_n)| >= n for all n. By the Bolzano-Weierstrass Theorem, (x_n) has a convergent subsequence (x_n(i)). Let L be the limit of the subsequence. Then L is in [a, b]. Since f is continuous at L, f(x_n(i)) -> f(L) as i -> infinity. That is a contradiction -- because |f(x_n(i))| >= n(i) >= i for all i, f(x_n(i)) can't converge to anything as i -> infinity. So f is bounded.

(b)
By definition of the supremum, there is a sequence (x_n) in [a, b] with f(x_n) -> M as n -> infinity. By the Bolzano-Weierstrass Theorem, (x_n) has a convergent subsequence (x_n(i)). Let L be the limit of the subsequence. Then L is in [a, b]. Since f is continuous at L, f(x_n(i)) -> f(L) as i -> infinity. Since any subsequence of a convergent subsequence converges to the same limit, f(x_n(i)) -> M as i -> infinity. So f(L) = M.

(c)
f/g is a continuous function on [a, b]. By (a), f/g is bounded. So there is a number c such that |f(x)/g(x)| <= c for all x in [a, b]. So f(x) <= c g(x) for all x in [a, b].

Reply 7

evariste_3086

madhapper

5) f cont a c if lim f(x)=f(c) as x->c (or delta epsilon def)
IMV
if f [a,b]->R is cont and y is between f(a) and f(b) then there is a point c in [a,b] with f(c)=y
proof
let A={x in [a,b]:f(x)<y}
A is bounded above by b
A is non-empty since f(a)<y so a in A
let c=sup A
suppose f(c)>y
let e=f(c)-y>0
f is cont at c so there exits d st |x-c|<d |f(x)-f(c)|<e for x in [a,b]
hence if |x-c|<d x in [a,b]
f(x)>f(c)-e=y
in particular
c-d<x<c+d for all x in[a,b]
f(x)>y
this implies c-d/2 is an upper bound for y. This follows since x in A x<c and f(x)<y so x cannot satisfy c-d<x as f(x)>y. which contradicts c=sup A.(ie lowest upperbound)
similar argument shows in the case f(c)<y c is not an upperbound.Just show there is an element of A greater than c.

c) f(x)=f(x+2) for all x
let g(x)=f(x)-f(x+1)
since f is cont then so is g
g(0)=f(0)-f(1)
g(1)=f(1)-f(2)
but f(0)=f(2)
so g(1)=f(1)-f(0)=-(f(0)-f(1))=-g(0)
therefore g changes sign on [0,1] so g(0)<0<g(1)
hence by IVT there exists y in [0,1] such that g(y)=0 ie f(y)=f(y+1)

6)
a) suppose f were not bounded on [a,b].
for all n there exits xn in [a,b] st |f(xn)|>n
this gives a bounded sequence xn in [a,b]
by Bolzano it has a convergent sub-sequence xnk->x say and since
a<xnk<b
a<x<b
f is cont so f(xnk)->f(x)
but |f(xnk)|>nk
contradiction since the sequence f(xnk)->f(x) but the terms get larger by assumption that |f(xn)|>n
b) let M=sup{f(x):x in[a,b]}
suppose no c exists st f(c)=M
so f(x)<M for all x in [a,b]
so the function x->M-f(x) is continuous and +ve and non-zero in [a,b]
then so is
g:x->1/(M-f(x))
so it is bounded on [a,b]
ie there is k st
1/(M-f(x))<k for all x in [a,b]
so 1/k<M-f(x) for all x in [a,b]
ie f(x)<M-1/k for all x in [a,b]
contradiction since M is least upper bound.
c) let h(x)=f(x)/g(x)
since f and g>0 and cont then h is cont for all x in [a,b].
let M=sup{h(x):x in[a,b]}
by (ii) there exists c st h(c)=M for all x in [a,b]
so h(x)<M for all x in [a,b]
ie f(x)/g(x)<M for all x in [a,b]
so f(x)<Mg(x) for all x in [a,b]

Reply 8

evariste_3086

madhapper

can any1 provide solutions or answers to these papers.
i've been given papers but no answers so i dont knw if im doin them right

http://www.geocities.com/chetlo86/mathm12a_03.pdf

4) if a,b are co-prime then there exists h,k st ah+bk=1
use euclids algo
53=17.3+2
17=8.2+1
work back
1=17-8.2
1=17-8.(53-17.3)
1=25.17-8.53
so mod 53 we have 25.17=1
so 17^(-1)=25 mod 53

b) flt if p is prime a an integer a^(p-1)=1 mod p
17^(518) mod 53
=17^(-1).17^(-1).17.17.17^(518) mod 53
=25.25.17^520
=(530+95).{17^(52)}^10 mod 53
=42 mod 53 (by flt)

6) lagrange
for this need several facts.
1) the cosets determined by H partition G ie G=UaH (for a in G)
proof
let H be a subgroup
clearly for a in G {C=contained in}
aH C G for all a by closure so UaH C G
conversely for any b in G
b=be CbH C UaH as a goes through all a in G
which shows G C UaH
so result.

2) for any a,b in G aH=bH or aH n bH is empty.(if cosets share an element they are equal)
proof.
assume z is in aH n bH
so z=ahi=bhj for some hi,hj in H
which gives a=bhjhi^(-1) (inverse exists since G a group)
let x=ahk be in aH for some hk in H
so x=bhjh^(-1)hk
this gives aH C bH since hjhi^(-1)hk C H
sim it can be shown bH C aH so aH=bH as required.

3) |aH|=|H| for all a in G
proof
define map f :h:

->aH by f(x)=ax
so any ahi in aH can be formed by f(hi) so f surjective
if f(hi)=f(hj) then ahi=ahj =>hi=hj so f injective and hence a bijection.

4) lagrange.
let H1,H2,...Hk denote the distinct cosets of H
then
G=H1+H2+...HK+H by (1) (2) above
so |G|=k|H| by (3) above
ie |H| divides |G|.
b) if G has prime order then any subgroup has order 1 or p by lagrange.
since the subgroup generated by powers of any element must have order 1 or p.if G has more than 1 element there exists an element of order p then this element generates the group.so G is cyclic.
c) since s3 has order 6 only possible orders for subgroups are 1,2,3,6
so subgroups are {i} ,{i,a,b},{i,c},{i,d} {i,e} S3,
with
i =123->123
a=123->231
b=123->312
c=123->132
d=123->321
e=123->213.

Reply 9

evariste_3086

madhapper

.

http://www.geocities.com/chetlo86/mathm12a_03.pdf

1)
1100|1000
1110|0100
0111|0010
0011|0001

1100|1000
00-10|1-100
0111|0010
0011|0001

1100|1000
00-10|1-100
0100|001-1
0011|0001

1100|1000
0100|001-1
00-10|1-100
0011|0001

1000|10-11
0100|001-1
0010|-1100
0011|0001

1000|1 0 -1 0
0100|0 0 1 -1
0010|-1 1 0 0
0001|1 -1 0 1

the last matrix after | is A^-1
consider
1 1 =A
1 1
and
1 1 =B
0 1
(AB)^(t)=
1 1
2 2
and
A^(t)B^(t)=
2 1
2 1
ij th entry of [A*B]=aijbij
so jith entry of [A*B]^(t)=aijbij
so ijth entry of [A*B]^(t)=ajibji
ij th entry of A^(t=aji
ijth entry of B^(t)=bji
so ijth entry of [A^(t)*B^(t)]=ajibji
so same as ijth entry of [A*B]^(t).

Reply 10

evariste_3086

madhapper

can any1 provide solutions or answers to these papers.
i've been given papers but no answers so i dont knw if im doin them right

http://www.geocities.com/chetlo86/mathm12a_02.pdf

2 denote A triangle B by ATB and A^c by A'
A'|A| B| B'| AnB'| A'nB| ATB
n | y| y| n| n | n | n
n | y| n| y| y | n | y
y | n| y| n| n | y | y
y | n| n| y| n | n | n

b)
A| B| C| ATB| (ATB)nC| AnC| BnC| (AnC)T(BnC)
y| y| y| n| n| y| y| n|
y |y |n| n| n| n| n| n|
y| n| y| y| y| y| n| y|
y |n| n| y| n| n| n| n|
n| y| y| y| y| n| y| y|
n| y| n| y| n| n| n| n|
n| n| y| n| n| n| n| n|
n| n| n| n| n| n| n| n|
as columns (ATB)nC and (AnC)T(BnC) are equal the expressions are the same.
A| B| C| AT(BnC)| BnC| ATB| ATC| (ATB)n(AnC)|
y| y| y| n| y| n| n| n|
y| y| n| y| n| n| y| n|
y| n| y| y| n| y| n| n|
y| n| n| y| n| y| y| y|
n| y| y| y| y| y| y| y|
n| y| n| n| n| y| n| n|
n| n| y| n| n| n| y| n|
n| n| n| n| n| n| n| n|
so columns AT(BnC) and (ATB)n(AnC) different so the expressions not equal.
i) let (x,y) be in (AxB)n(CxD)
then (x,y) in AxB and (x,y) in CxD
so x in A and x in C
y in B and y in D
=> x in AnC
y in BnD
=> (x,y) in (AnC)x(BnD).

ii)
let A={1} B={1,2} C= {1,2} D={1}
AxB={(1,1) (1,2)}
CxD={(1,1) (2,1)}
so (AXB)T(CxD)={(1,2) (2,1)}
ATC={2}
BTD={2}
so (ATC)x(BTD)={(2,2)}
hence two are not the same.

3)
a) book-work ill let you find these. good revision!
b)
i) not reflexive since the empty set is a subset and the intersection of the empty set with the empty set is clearly empty.
symmetric since if A,B are 2 subsets and ARB then there is an element in the intersect so BRA
not transitive. consider
A={1} B={1,2} C={2,3}
then AnB={1} so ARB
BnC={2} so BRC
but AnC=empty set so A not Related to C

ii) reflexive: aRa since |a-a|=0<2
symmetric since if aRB then |b-a|=|a-b|<2
not transitive eg consider
a=2 b=3 c=4.5
then
|a-b|=1<2 so aRb
|3-4.5|=1.5<2 so bRc
but |2-4.5|=2.5>2.5 so a not related to c

iii)
reflexive: (a,b)=(a,b)+(0,0) so (a,b)R(a,b)
symmetric
if (a,b)R(c,d) then (c,d)=(a,b)+(x,y) with x,y integers
so (a,b)=(c,d)+(-x,-y) and -x,-y integers
so (c,d)R(a,b)
transitive
if (a,b)R(c,d) then (c,d)=(a,b)+(x,y) with x,y integers and if (c,d)R(e,f) then (e,f)=(c,d)+(x',y') with x',y' integers
then
(e,f)=(a,b)+(x+x',y+y') and x+x', y+y' are integers which gives
(a,b)R(e,f)
so this is an equivalence relation.
the classes are
[a,b]={(c,d) st (c,d)=(a,b)+(x,y)}
with a,b,c,d reals and x,y integers.
THIS IM NOT SURE but i think a class of reps is given by
{[0,N],[N,0],[N,N]}
where 0<N=<1 or N is irrational.

Reply 11

evariste_3086

[QUOTE="W" Jonny="Jonny"]-------------------------------
mathm11a_02
-------------------------------
Solutions to Questions 4 -

(c)
f/g is a continuous function on [a, b]. By (a), f/g is bounded. So there is a number c such that |f(x)/g(x)| <= c for all x in [a, b]. So f(x) <= c g(x) for all x in [a, b].
just a small point about this. The question asked for less than or equal to. Just by quoting part(a) does not show you can get equality for any x in [a,b].by picking any bound c we can only say f/g<c for all x. part(b) ensures we have equality if we pick the supremum since f/g attains the sup for some x in [a,b] .

Reply 12

Jonny W

evariste

just a small point about this. The question asked for less than or equal to. Just by quoting part(a) does not show you can get equality for any x in [a,b].by picking any bound c we can only say f/g<c for all x. part(b) ensures we have equality if we pick the supremum since f/g attains the sup for some x in [a,b] .

The question doesn't ask us to prove that f/g attains its supremum.

Reply 13

evariste_3086

Jonny W

The question doesn't ask us to prove that f/g attains its supremum.

it asks for a constant so f(x)< or = cg(x)
by picking just any bound c we can only say f(x)/g(x)<c for all x in [a,b]
by picking the sup, M say, we can say f(x)/g(x)< or = M since there is an x in [a,b] such that f/g attains its supremum by part b. ie there is x in [a,b] st
f(x)/g(x)=M for that x and
f(y)/g(y)< or=M for all other y in [a,b] but we do have equality for at least 1 x.

Reply 14

Jonny W

evariste

Surely you would agree that the following statement is true:

"sin(x) <= 2 for all real x"

Reply 15

evariste_3086

Jonny W

Surely you would agree that the following statement is true:

"sin(x) <= 2 for all real x"

i never meant your answer was wrong. of course the above statement is correct
using that example then,if the the question was along the lines:
1) prove sinx is bounded by 2 for all x in [0,pi]
2) prove sup{sin x}=1 for x in [0,pi]
3) find a constant,c, such that sinx<=c for all x in[0,pi]
would a neater answer be c=2 by part(1) or 1 by part (2)?
i didnt write the question but the fact they ask you to show the sup is attained and ask for equality in f(x)<=cg(x) suggests that would prefer the sup result to be quoted.This is not meant to say your answer is wrong, apologies if this is how you have taken my comment.
Anyway have you done the mechanics question? i would like to see the answers, mechanics is beyond me.

Reply 16

madhapper

mechanics questions is the 1 im most stuck on. hopefully someone can help me out ere

Reply 17

Jonny W

-------------------------------
mathm13a_02
-------------------------------
Solutions to Questions 4 - 6
-------------------------------

(4)
See the attached diagram.

--

Let C be the point where the intrinsic angle is psi. Let N be the normal force exerted by the curve at C on the rope element AB.

Resolving along the tangent at C,

(T - deltT/2) cos(deltpsi/2) + mu N = (T + deltT/2) cos(deltpsi/2)
mu N = deltT cos(deltpsi/2)

(We say "deltT" rather than "deltaT" or "delta T" to emphasize that we are referring to a single variable, not to the product of delta and T.) Resolving perpendicular to the tangent at C,

N
= (T - deltT/2) sin(deltpsi/2) + (T + deltT/2) sin(deltpsi/2)
= 2T sin(deltpsi/2)

Eliminating N,

deltT cos(deltpsi/2) = 2T mu sin(deltpsi/2)
deltT / [2tan(deltpsi/2)] = T mu

Finally we let A and B get close together, so that deltT and deltpsi become small:

dT/d(psi) = T mu

--

T = K e^(mu psi) for some constant K

We now allow A and B to be any points on the curve (not necessarily close together). Let T_A and T_B be the tensions at A and B. Let psi_A and psi_B be the intrinsic angles at A and B.

Then

T_A = K e^(mu psi_A)
K = T_A/e^(mu psi_A)

so

T = T_A exp[mu (psi - psi_A)]

and

T_B = T_A exp[mu (psi_B - psi_A)]

--

Let A and B be the points where the parts of the string going to P and Q leave contact with the cylinder. Then |psi_B - psi_A| = pi (the sign depends on which side A is on).

Before the extra weight is added to Q,

P = Q e^(mu pi) . . . . . P is heavier than Q, the friction just stops P sliding down

After the extra weight, which we will call W, is added to Q,

Q + W = P e^(mu pi) . . . . . Q + W is heavier than P, the friction just stops Q + W sliding down

So

Q + W = Q e^(2mu pi)
W = Q[e^(2mu pi) - 1]

--

(5)
The particle's acceleration is

-kv^2 - f(x) when x > 0 and v > 0
kv^2 - f(x) when x > 0 and v < 0

--

Before the particle first comes to rest, x > 0 and v > 0. So

d/dx (v^2)
= 2v dv/dx
= 2 dv/dt
= 2[-kv^2 - f(x)]

d/dx (v^2) + 2k v^2 = -2f(x)
e^(2kx) [d/dx (v^2)] + 2k e^(2kx) v^2 = -2f(x) e^(2kx) . . . . . using an integrating factor
d/dx [e^(2kx) v^2] = -2f(x) e^(2kx)

Integrating wrt x,

[e^(2kx) v^2] (from 0 to a) = -2 (int from 0 to a) f(x) e^(2kx) dx
-u^2 = -2 (int from 0 to a) f(x) e^(2kx) dx . . . . . since v = 0 when x = a
u^2 = 2 (int from 0 to a) f(x) e^(2kx) dx . . . . . (*)

--

While the particle returns to O, x > 0 and v < 0. So

d/dx (v^2)
= 2v dv/dx
= 2 dv/dt
= 2[kv^2 - f(x)]

d/dx (v^2) - 2k v^2 = -2f(x)
e^(-2kx) [d/dx (v^2)] - 2k e^(-2kx) v^2 = -2f(x) e^(-2kx)
d/dx [e^(-2kx) v^2] = -2f(x) e^(-2kx)

Integrating wrt x,

[e^(-2kx) v^2] (from 0 to a) = -2(int from 0 to a) f(x) e^(-2kx) dx
-(u1)^2 = -2(int from 0 to a) f(x) e^(-2kx) dx
(u1)^2 = 2(int from 0 to a) f(x) e^(-2kx) dx

--

(u1)^2
= 2(int from 0 to a) Ak e^(-2kx) dx
= A[-e^(-2kx)] (from 0 to a)
= A(1 - e^(-2ka))

--

Adapting (*), the particle next comes to rest at a displacement y, where

A(1 - e^(-2ka))
= (u1)^2
= 2(int from 0 to -y) f(x) e^(2kx) dx
= 2(int from 0 to -y) Ak e^(2kx) dx
= A[e^(2kx)](from 0 to -y)
= A(e^(-2ky) - 1)

e^(-2ky) = 2 - e^(-2ka)

--

(6)
See the attached graph.

--

V(-sqrt(7/2))
= V(sqrt(7/2))
= V0 (7/2)(1/2) / (1/2)
= (7/2)V0 . . . . . (*)

By the graph and (*), V(x) > (7/2)V0 whenever |x| > sqrt(7/2). So the particle can never be outside the interval [-sqrt(7/2), sqrt(7/2)].

Also by the graph and (*), V(x) < (7/2)V0 whenever |x| < sqrt(7/2). So the particle cannot be at rest in the interval (-sqrt(7/2), sqrt(7/2)), and hence cannot change direction in that interval.

Finally, the gradient of V is not zero at x = -sqrt(7/2) and x = sqrt(7/2). So when the particle approaches one of those points, it gets there after a finite time and then immediately starts moving in the opposite direction.

So the paricle moves between the positions x = -sqrt(7/2) and x = sqrt(7/2).

--

The energy equation is

(1/2)m(x')^2
= KE
= -V(x)
= -V0 x^2(x^2 - 3)/(4 - x^2)
= (3/4)V0 x^2 (1 - x^2/3) / (1 - x^2/4)

(x')^2 = [3V0 x^2/(2m)] * [(1 - x^2/3) / (1 - x^2/4)]

Since [(1 - x^2/3) / (1 - x^2/4)] is nearly equal to 1, the energy equation can be approximated by

(x')^2 = 3V0 x^2/(2m)

--

x' = -sqrt[3V0/(2m)] x . . . . . negative root because the particle is moving towards the origin

x(t) = x(0) exp{ -sqrt[3V0/(2m)]t }

x(t) -> 0 as t -> infinity, but x(t) is nonzero for all t.