Non HSC Maths Marathon (1 Viewer)

[acmilan]

Just like the marathons in the hsc maths forums, except out of scope questions. I'm interested in seeing the types of questions other unis do. This might not work due to lack of people participating, but its worth a shot. (Note: no, im not using this as an attempt to get my assignment questions answered i probably wont post a question if i dont have an answer for it).

Usual rules, answer the question above and post your own. I'll start with a non too difficult one:

Evaluate int (0 to inf) int (0 to inf) e^-x2-y2 dxdy

Hence find int (0 to inf) e^-x2 dx and int (0 to inf) t^-1/2 e^-t dt

(Hint: use polar)

 

[Affinity]

Pi/4, sqrt(Pi)/2 and sqrt(Pi)

Let M be a real square matrix with distinct real eigenvalues. Must there exist a decomposition M=S D S^-1 where D is diagonal and S are matrices with real number entries? Prove your assertion
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What is the volume of an n-dimensional sphere with radius r?
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X is a p*q matrix. prove that (X'X + a*I) is invertible for any positive a.

find vector b such that |y - Xb|^2 + k*|b|^2 is minimized.

here, X' stands for X transposed

 

[acmilan]

Haha no responses for a few days

Well, for
X is a p*q matrix. prove that (X'X + a*I) is invertible for any positive a.

Would you use that if det(X'X + a*I) = 0 then (-a) is an eigenvalue of X'X? So all eigenvalues are negative and thus det(X'X + a*I) =/= 0 for positive a.

 

[Affinity]

Erh.. not quite.. almost there, what you have is that if X'X + a*I is not invertible then X'X has a negative eigenvalue.. it doesn't rule out the possibility of it not having positive numbers or zero as an eigenvalue.

Oh sorry by the way, ' is not exactly transpose.. it's conjugate transpose my bad alternatively you can treat X as real.

 

[darkliight]

Let M be a real square matrix with distinct real eigenvalues. Must there exist a decomposition M=S D S^-1 where D is diagonal and S are matrices with real number entries? Prove your assertion

Hmmm. No? Only if there are n distinct eigenvalues for the nxn matrix M. Take any matrix with only one real distinct eigenvalue (M = {0 1 \ 0 0} say) and we run into problems.

If you meant M was a real nxn matrix, with n distinct eigenvalues, then yes, but the proof of this, that I know anyway, will take more time to type up that I care to. Any LA book should have it though.

What is the volume of an n-dimensional sphere with radius r?

pi^(n/2)r^n/(n/2)!. For odd n, the gamma function is required and is particularly nasty, but on the plus side will cancel out the fractional power of pi in the numerator when you evaluate it out.

 

[Affinity]

Hmmm. No? Only if there are n distinct eigenvalues for the nxn matrix M. Take any matrix with only one real distinct eigenvalue (M = {0 1 \ 0 0} say) and we run into problems.

If you meant M was a real nxn matrix, with n distinct eigenvalues, then yes, but the proof of this, that I know anyway, will take more time to type up that I care to. Any LA book should have it though.


pi^(n/2)r^n/(n/2)!. For odd n, the gamma function is required and is particularly nasty, but on the plus side will cancel out the fractional power of pi in the numerator when you evaluate it out.

It was clearly stated that the eigen values are distinct . The proof is very short.
Care to give a new question?

EDIT: I see what you mean now.. But then the word distinct will serve no purpose.

 

[acmilan]

If i recall correctly the proof of the first question is just a corollary of another theorem

 

[Affinity]

Here's another more or less standard question:

Assuming you all have an idea of what conditional expectation means. The conditional variance Var(X|Y) is defined as E((X - E(X|Y))^2 | Y)

Prove that Var(X) = E(Var(X|Y)) + Var(E(X|Y))

 

[acmilan]

Firstly E(g(X)) = E(E(g(X)|Y))

Proof: Let h(y) = E(g(X)|Y)

E(E(g(X)|Y))
= E(h(y))
= int h(y)f(y) dy
= int (int g(x)f(x|y) dx) f(y) dy
= int int g(x)[f(x,y)/f(y)] f(y) dxdy
= int int g(x)f(x,y) dxdy
= E(g(X))

So

var(X)
= E(X²) - E(X)²
= E(E(X²|Y)) - E(E(X|Y))²
= E(var(X|Y) + E(X|Y)²) - E(E(X|Y))²
= E(var(X|Y)) + E(E(X|Y)²) - E(E(X|Y))²
= E(var(X|Y)) + var(E(X|Y))

hope thats right :/

Edit: i think this proof only works in the continuous case

 

[Slidey]

You can probably solve differential equations. What about their discrete counterparts - difference equations:

2y_k+2+sqrt(2).y_k+1-2y_k=3, k any natural number, with y₀=1, y₁=-1.

 

[darkliight]

You can probably solve differential equations. What about their discrete counterparts - difference equations:

2y_k+2+sqrt(2).y_k+1-2y_k=3, k any natural number, with y₀=1, y₀=-1.

I assumed you meant y_1 = 1, then solved for y_2, giving 1/2-1/sqrt(2). I setup a recurrance relation and after pages of mess (maybe there is a nicer way of solving these ), I got this mess:

y_k = -1/3(5+2*sqrt(2))*(1/sqrt(2))^k + 1/6*(4-5*sqrt(2))*(-sqrt(2))^k + 3/sqrt(2)

Works for y_0 = -1, y_1 = 1, y_2 = 1/2-1/sqrt(2) and then I tested it for y_3 = 3-1/(2*sqrt(2)) and seemed to work for all of them. Seems like a harsh answer though so I'm still not convinced I'm right.

P.S. WTB latex support ...... still.

EDIT: Fixed now that I used the right y_2 value... I think.

 

[Slidey]

Sorry, y_1=-1. That's k=1, y=-1.

Latex support... I'll bring it up right now with the admins.

Anyway, you look like you're on the right track (not quite there - no need to work out y_2, just find a single solution to the equation* and then solve it as if it were homogenous, and then combine solutions). Sorry, I suppose it is a bit of a nasty question to put online. I was just interested in the relationship between difference equations (recurrence, whatever) and differential equations. You needn't solve for the initial conditions - a general solution is fine.

*Hint: the right-hand side function is constant, so...

 

[darkliight]

Ok, solving it as if it's homogenous I get y_k = C(1/sqrt(2))^k + D(-sqrt(2))^k.

Now, with the constant 3 in the original question, do we just need to add another constant E to our solution?

 

[acmilan]

Now, with the constant 3 in the original question, do we just need to add another constant E to our solution?

Yes, but that constant has a particular value, which should be 3/sqrt(2) i think (correct me if wrong, havent done these before).

 

[darkliight]

Agreed, I get:

C = -1/3(1+4sqrt(2))
D = 1/6(8-sqrt(2))
E = 3/sqrt(2).

So, the same as my original solution except for the change in initial conditions.

y_k = -1/3(1+4sqrt(2))*(1/sqrt(2))^k + 1/6(8-sqrt(2))*(-sqrt(2))^k + 3/sqrt(2)

 

[acmilan]

Some random discrete ones:

1. Prove there are infinitely many primes
2. Prove is a = b (mod m) then gcd(a,m) = gcd(b,m) (note = means congruent here in the mod case)
3. Prove that if n is prime, then for all integers a and b, if n|ab then n|a or n|b
4. If a and n are integers, prove aⁿ-1 is prime only if a = 2 and n is prime
5. Show 2ⁿ+1 is prime only if n is a power of 2

 

[darkliight]

1. Suppose not, then there is some finite number of, say p, prime numbers. Mupltiplying all of these prime numbers together yields a number divisible by every prime number. Now add 1 to this number, but this new number can not be divisible by any of the p prime numbers, since division will always result in a remainder of 1 and 1 is not divisible by any prime number. Thus, this new number is itself prime or divisible by some other prime number not included in our list. Contradiction. <3 Euclid.

I need to get some study done. Have fun with the rest.

 

[KeypadSDM]

4: Note for n = 1, a = p+1 is a solution. Thus consider n > 1

a^n - 1 = (a-1)*(a^(n-1) + a^(n-2) + ... + 1) for n>1
Thus a = 2 as a-1 can't be non-unitary, and the RH bracket is clearly non unitary [at least 1 term + 1]
Assume n is divisible by k
n = km
a^km - 1 = (a^m - 1)*(a^m(k-1) + a^m(k-2) ... + 1)
Contradiction.

Fill the proof out yourself, it's fairly straighforward, the main contradiction steps are there. The last step I forgot how to do, but you just prove you can decompose into more than 1 non-unitary factors.

3: n is prime, and n|ab
Assume n !|a
Thus kn = ab for some k.
Hence a = k/q for some integral q, as n can't be factorised and a is not divisible by n.
Hence b = qn
=> n | q

2) Using: gcd(a + bm, m) = gcd(a, m)
Trivial.
[gcd(a,m) = gcd(b + mk,m) = gcd(b,m)]

 

[Affinity]

Prove that there is an explicit way to disect a cube into n cubes (perhaps of different sizes) for say... n>100

Prove that one can do the same with cyclic quadrilaterals, that is given a cyclic quadrilateral one can divide it into n cyclic quadrilaterals when n>100

Here's one which I don't have the solution to (maybe there isn't a nice solution, but seems like a sensible question)

n children sit in a circle, the teachers starts numbering them clockwise until she reach number k, and that child is removed from the circle. the process then repeats itself, starting from the child next to the one who left. Who will be the last to remain in the circle? express your answer in terms of n and k

 

[Li0n]

Four distinct points including 0;, 0, z1, z2, z3 lie on a circle. Show that the points 1/z1, 1/z2, 1/z3 are collinear, i.e lie on a straight line.

Taken from 2520 maths book