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December 14, 2025, 12:24:35 pm

Author Topic: TT's Maths Thread  (Read 153814 times)  Share 

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kamil9876

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Re: TT's Maths Thread
« Reply #1275 on: June 20, 2011, 07:43:20 pm »
0
Here is a solution but it may use theorems you are not familiar with:

is an annhilator polynomial for hence the minimal polynomial is either X,(X-1) or X(X-1). Since 0 is not an eigenvalue that means that the minimal polynomial does not have 0 as a root and hence it must be X-1 so E=I.
Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

TrueTears

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Re: TT's Maths Thread
« Reply #1276 on: June 20, 2011, 08:10:32 pm »
0
Here is a solution but it may use theorems you are not familiar with:

is an annhilator polynomial for hence the minimal polynomial is either X,(X-1) or X(X-1). Since 0 is not an eigenvalue that means that the minimal polynomial does not have 0 as a root and hence it must be X-1 so E=I.
lol i dont know what you mean, do you have a more elementary method?
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Re: TT's Maths Thread
« Reply #1277 on: June 21, 2011, 12:45:21 am »
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Actually for my question 4 (the gram schmidt question)

is this way of doing it correct? although it seems a bit too long, i was wondering if theres any shorter ways?

So let be a vector in , then

so then we can basically sub in the respective components for o, compute the respective inner products etc etc and we get:



and so.. the matrix is pretty obvious from that, but yeah that took alot of computation, any slicker ways?

cheers
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kamil9876

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Re: TT's Maths Thread
« Reply #1278 on: June 21, 2011, 10:15:57 am »
0
Here is a solution but it may use theorems you are not familiar with:

is an annhilator polynomial for hence the minimal polynomial is either X,(X-1) or X(X-1). Since 0 is not an eigenvalue that means that the minimal polynomial does not have 0 as a root and hence it must be X-1 so E=I.
lol i dont know what you mean, do you have a more elementary method?

Well a bit hard to say because in some courses the fact that the roots of the minimal polynomial are exactly the eigenvalues is proven pretty early on. What can I use? Jordan Normal Form?


As for the eigenvalue question:




where denotes the usual length.

hence the result. (I am assuming the matrices have real number coefficients?)

I think your third one didn't really take that much computation?
« Last Edit: June 21, 2011, 10:29:03 am by kamil9876 »
Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

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Re: TT's Maths Thread
« Reply #1279 on: June 23, 2011, 09:02:57 pm »
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Another way to obtain the projection matrix is to observe that -every- projection matrix is orthogonally similar to a diagonal matrix with only 1s or 0s on the diagonal (bc projection is just eliminating components of the vector in some orthonormal basis - think geometrically)

So you found your orthonomal basis {u1,u2,u3}. Let U be the matrix with columns u1,u2,u3, then we should have   P= U D U^T  where D=[100,010,000] (because you want to project onto the span of the first 2 vectors).

how'd you go with the exam today?
« Last Edit: June 23, 2011, 09:06:37 pm by zzdfa »

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Re: TT's Maths Thread
« Reply #1280 on: June 23, 2011, 09:04:05 pm »
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edit: double post

TrueTears

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Re: TT's Maths Thread
« Reply #1281 on: June 24, 2011, 08:24:35 am »
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haha yeah, thanks guys i got them in the end anyway, yeah exam was easy as today :)
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Re: TT's Maths Thread
« Reply #1282 on: December 13, 2011, 10:40:33 pm »
+5
Hey guys, I have a few questions that I've been stuck on for a few days, these questions have no solutions provided so I was wondering if anyone can check them for me. The first 3 questions I have completed with full working however I'm not entirely sure if they are correct, the last 4 I have no clue how to solve :3 Any help would be appreciated!



So first we need to work out the amount of each installment, , thus

Now X purchases the annuity on 1 Jan 1986, so we need to work out the price that X bought the annuity for.

Since he purchased it on 1 Jan 1986 and the annuity expires at the end of 2005, the annuity has a maturity left of 19 years.

But since X pays a tax rate of 0.45 on the interest portion of each payment of R, then we need to split up R each period to find the interest payment and capital payment so we can account for the tax.

To derive an expression for the interest payment and capital payment of R each period we let L be the lump sum borrowed at the start of an arbitrary period, be the number of periods and be the effective interest rate per period. Then for the first payment period, the loan outstanding at the start of the period is L, the interest payment for the first period is where . Then clearly the capital component paid at the end of the first period is since , we can show in general that for payment period k, the interest component paid at the end of the period is and the capital component paid at the end of the period is

Now back to the question: let and

The capital payments of each period starting from 1986 until the end of 2005 when discounted back to 1986 form the cash flow stream:



The interest payments of each period starting from 1986 until the end of 2005 when discounted back to 1986 form the cash flow stream: (Note we take into consideration the tax rate of 45% which is (1-0.45) = 0.55 of the interest payments)



Now we need to combine both of these cash flow streams and form one single expression to evaluate it.

Let us match together the components:





And so on... so let us now group together the expressions with 0.55 as the coefficient:



Now group together the expressions with 0.45 as the coefficient:







Thus

Now let the lump sum paid on April 1995 be L, L must satisfy the following equation of value: (Let )





Solving the above yields





First we need to work out the amount of each installment paid in advance from 1 Jan 1985 to 2005 (20 years).

Let each installment be R.

First note where d is the effective annual rate of discount.





Now we need to find on 1 Jan 1990 the total outstanding capital left to be repaid.

This is given by

Now from 1 Jan 1990, if B were to increase the yield to 0.12, then we need to find the new installments per period, these new R is given by:

(Note )



Now let A's final payment be L, L must satisfy the following equation of value:











Let be the capital component paid in arrear for the nth period and be the interest component paid in arrear for the nth period. Let be the loan outstanding.

From the question

But

Likewise,

But [This is obvious as represent the loan outstanding at the end of the first period, which is equal to the loan outstanding at the beginning of the period minus the capital paid in the first period]

So

Furthering this pattern,

But

Hence

So we have:







From [1] we have

So [2] becomes

In [3] we have:



This forms a recurrence relation and we can show that

Now we need to find what is. Note that if we assume the person who bought this decreasing annuity does not pay tax, then we should have:

where



Now we can solve for

The price that someone would pay for this decreasing annuity if they do not pay tax is given by:

where

The price is 38252.909

















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TrueTears

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Re: TT's Maths Thread
« Reply #1283 on: June 26, 2012, 08:18:59 pm »
0
I'm trying to show that given and are independent (ie, they are all mutually independent and pairwise independent) then this implies that . And in general I claim that if we have a string of independent 's for i = 1 to n, then where * can be or

So firstly, to prove my initial claim this is what I have so far:









This is where I'm stuck as I want the to cancel but something must have went wrong, but I can't see what?

Any ideas kamil, ahmad?
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Re: TT's Maths Thread
« Reply #1284 on: June 26, 2012, 08:37:42 pm »
+1
If you take A3 and A4 to be something like 'get heads' then your last line is 0

kamil9876

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Re: TT's Maths Thread
« Reply #1285 on: June 26, 2012, 08:38:57 pm »
+1
I'm pretty sure you didn't expand out the numerator correctly. Instead of you should have
Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

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Re: TT's Maths Thread
« Reply #1286 on: June 26, 2012, 08:42:11 pm »
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I'm pretty sure you didn't expand out the numerator correctly. Instead of you should have
Yes I think so too, but I just used the principle of inclusion and exclusion on:



To get

I think the following is wrong, but why?

? Since that's the intersection.
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kamil9876

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Re: TT's Maths Thread
« Reply #1287 on: June 26, 2012, 08:43:59 pm »
+1
Nope, you forgot the and . Note that taking intersections is an associative operation
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Re: TT's Maths Thread
« Reply #1288 on: June 26, 2012, 08:49:25 pm »
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Ah stupid me, what a stupid mistake, thanks it works out now.
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Re: TT's Maths Thread
« Reply #1289 on: June 27, 2012, 02:51:01 am »
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Just wanna verify my solution to this:

Let be a Discrete Uniform random variable over the interval . What is the distribution of ?

The PMF of the Discrete Uniform RV is given by:

Let be the RV such that

Thus

Note that in this case , thus

since as







Thus the PMF of is given by:

if and if

Thanks :)

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