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HSC 2013 MX2 Marathon (archive) (4 Viewers)

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seanieg89

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Re: HSC 2013 4U Marathon

Can you please criticise this proof? I am starting to have doubts about it.

===

Start with the polynomial,



We need to make the lowest reducible rational polynomial that has that as a root, so we simply:



The above polynomial has the root
It is the simplest most reducible rational polynomial that has p + root(q) as a root.

Now consider:



Where a_1, a_2 ... are all rational polynomial functions of lowest reducibility.

If is a root then among the series a_1, a_2... there must exist a_k for some k such that is a root.
Since a_k is irreducible and rational, then that polynomial must be:



And that polynomial also has the root

Therefore that is also a root of P(x).
Is there a gap or something in this? Or is it all good?
Its pretty good, certainly everything you have said is true and this method extends nicely to more complicated situations.

The main question is why a given algebraic number must be the root of a unique minimal (with respect to degree) monic irreducible polynomial with rational coeffs? (note we need uniqueness as well as existence here, think about why.)

If you resolve this question, it is clear that the minimal polynomial of this particular root must have degree at least 2 (else it would be rational). And therefore the polynomial you have constructed is the "minimal polynomial" of p+sqrt(q). The rest of the argument is fine.


My argument is less general but works well here. (And is what I hoped a student would find by thinking of the complex conjugate root theorem and trying to mimic it.)



At the beginning I should probably have proven that two numbers of that form are equal iff their rational and surdic parts are equal, which is easy to do but am tired and lazy now. Other than that it is a pretty rigorous argument.
 

Sy123

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Re: HSC 2013 4U Marathon

In this proof I will assume that converges to (Taylor Series).






























==================================================



Nice work, it can also be done by considering the geometric series:



Substituting cis theta, equating real parts for a series in cosine. Then integrating with bounds alpha and pi yields the LHS and the RHS plus a limit to infinity of an integral which one must prove converges to zero.


Its pretty good, certainly everything you have said is true and this method extends nicely to more complicated situations.

The main question is why a given algebraic number must be the root of a unique minimal (with respect to degree) monic irreducible polynomial with rational coeffs? (note we need uniqueness as well as existence here, think about why.)

If you resolve this question, it is clear that the minimal polynomial of this particular root must have degree at least 2 (else it would be rational). And therefore the polynomial you have constructed is the "minimal polynomial" of p+sqrt(q). The rest of the argument is fine.


My argument is less general but works well here. (And is what I hoped a student would find by thinking of the complex conjugate root theorem and trying to mimic it.)



At the beginning I should probably have proven that two numbers of that form are equal iff their rational and surdic parts are equal, which is easy to do but am tired and lazy now. Other than that it is a pretty rigorous argument.
Ah alright, is that not a definition of an algebraic number? To be a root of a rational polynomial?

Also your method is very clever.
 

Fus Ro Dah

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Re: HSC 2013 4U Marathon

Also your method is very clever.
It is a good method, but why do you say this as if it is alien to you? You can imagine that the proof for this would be very similar to the proof of the existence of complex conjugate pairs in real polynomials, since the complex conjugate is analogous to the irrational conjugate p-sqrt(q).
 

Sy123

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Re: HSC 2013 4U Marathon

It is a good method, but why do you say this as if it is alien to you? You can imagine that the proof for this would be very similar to the proof of the existence of complex conjugate pairs in real polynomials, since the complex conjugate is analogous to the irrational conjugate p-sqrt(q).
Yes that is true.

===

Also Registered User, are you sure your series is correct?

Doesn't look very nice because of the sines of integers, is it a typo?
 
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Re: HSC 2013 4U Marathon

It can also be done by considering the geometric series:



Substituting cis theta, equating real parts for a series in cosine. Then integrating with bounds alpha and pi yields the LHS and the RHS plus a limit to infinity of an integral which one must prove converges to zero.
Didn't think of that. Seems like an easier method.
Did you make up this question?
 

Sy123

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Re: HSC 2013 4U Marathon

Didn't think of that. Seems like an easier method.
Did you make up this question?
I made up the solution (with help from sean in its rigour), the problem is a discarded one from STEP.
 

hayabusaboston

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Re: HSC 2013 4U Marathon

I found fun question. I wonder how u guys will solve it.
"Show that there is a multiple of 2013 whose last four digits are 2012"

I dont know how to find a nice elegant solution. Mines messy as hell. U guys have something elegant?
 
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Sy123

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Re: HSC 2013 4U Marathon

Using the result from the previous question:



Integrating both sides from pi to 2, and using the double angle rule, the result from the Basel problem and so on, we arrive at:



Then noticing that

We can notice that the LHS of what I have right now, is the LHS of the question - 1 (since at k=0 the function approaches 1). This then cancels out the term on the RHS, therefore leaving it with:





I know that it should evaluate to zero, however the same problem from when I encountered trying the Basel problem comes up where I cannot justify interchanging the theta integral and the limit to infinity as the alpha integral converges to zero due to by parts. However I cannot justify putting it in there unfortunately :/
 
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Makematics

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Re: HSC 2013 4U Marathon

I found fun question. I wonder how u guys will solve it.
"Show that there is a multiple of 2013 whose last four digits are 2012"

I dont know how to find a nice elegant solution. Mines messy as hell. U guys have something elegant?
I still can't get it :( Got the first question though :) The junior one. took about 90 minutes though -.- very satisfying though.
 

Sy123

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Re: HSC 2013 4U Marathon

I found fun question. I wonder how u guys will solve it.
"Show that there is a multiple of 2013 whose last four digits are 2012"

I dont know how to find a nice elegant solution. Mines messy as hell. U guys have something elegant?
IF there does exist a multiple of 2013 like that, then for some pair of integers (n,m) The following equality must hold:




And for some value k, if m has last four digits 2012 then k is integer.



On the k-m plane it is a linear function and hence there must exist some pair of integers (k,m) which satisfy the above equality as the domain of the function is all real k and m.

Hence there exists an integer value of m for which k is an integer, hence there must exist a value m with last four digits 2012.

EDIT: Done
 
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Makematics

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Re: HSC 2013 4U Marathon

EDIT: Find an example of a function f(x) such that its inverse is the same as its derivative.
 
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hayabusaboston

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Re: HSC 2013 4U Marathon

IF there does exist a multiple of 2013 like that, then for some pair of integers (n,m) The following equality must hold:




And for some value k, if m has last four digits 2012 then k is integer.



On the k-m plane it is a linear function and hence there must exist some pair of integers (k,m) which satisfy the above equality as the domain of the function is all real k and m.

Hence there exists an integer value of m for which k is an integer, hence there must exist a value m with last four digits 2012.

EDIT: Done
I am inexpressibly mesmerised by your brilliance...
 

Sy123

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Re: HSC 2013 4U Marathon

After completing Sy123's question on conjugate roots, it is a nice challenge exercise to consider the analogous problem for cube roots:

So is this how complex numbers are involved:



Is the minimal rational polynomial with roots alpha + cuberoot(beta).

Following the same template as the square root conjugate proof we can prove that the roots of this polynomial above can be the root of P(x).

Solving the cubic as roots of unity.







And of course this can be extended to nth roots (right?)
 
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