Carrotsticks
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Re: 2012 HSC MX2 Marathon
HSC 2012 MX2 Marathon (archive) (3 Viewers)
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Re: 2012 HSC MX2 Marathon
looking at the solution one of my students put forth, not too long
mirakon
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Carrotsticks
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Re: 2012 HSC MX2 Marathon
The question is, how DID they get the result?
Suppose I ask you "Give me a bound for the series
", you can't use 'proof by induction' because induction doesn't actually give you the formula or the expression. Rather, it is a means of verifying an identity or a conjecture.
If I wrote a question and received to proofs, one by induction and the other by deriving it from scratch, of course I will consider the latter to have a 'better proof' although both are perfectly valid. Do you know what I'm saying?
I don't like 'proof by induction' because it really should be called 'Verification by induction'.
Similar to Epsilon Delta arguments for the limit. It usually doesn't give you the actual limit, but it lets you VERIFY whether your conjectured limit is actually the limit. You NEED a limit to test otherwise the epsilon/delta proof won't work.
Similarly, you need a formula or expression to prove in order to use induction, otherwise it won't work.
Haha I know. But the inductive proof as seanieg demonstrated, is very straightforward.
The question is, how DID they get the result?
Suppose I ask you "Give me a bound for the series
If I wrote a question and received to proofs, one by induction and the other by deriving it from scratch, of course I will consider the latter to have a 'better proof' although both are perfectly valid. Do you know what I'm saying?
I don't like 'proof by induction' because it really should be called 'Verification by induction'.
Similar to Epsilon Delta arguments for the limit. It usually doesn't give you the actual limit, but it lets you VERIFY whether your conjectured limit is actually the limit. You NEED a limit to test otherwise the epsilon/delta proof won't work.
Similarly, you need a formula or expression to prove in order to use induction, otherwise it won't work.
jet
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Re: 2012 HSC MX2 Marathon
Every lecturer I've ever had has mentioned that mathematicians work backwards. They start with a result they think is true and try to find a proof for it.
P.s. You need to have a read over what you have written. Some of it doesn't make sense.
I think you're greatly underestimating the skill of professional mathematiciansHaha I know. But the inductive proof as seanieg demonstrated, is very straightforward.
The question is, how DID they get the result?
Suppose I ask you "Give me a bound for the series
If I wrote a question and received to proofs, one by induction and the other by deriving it from scratch, of course I will consider the latter to have a 'better proof' although both are perfectly valid. Do you know what I'm saying?
I don't like 'proof by induction' because it really should be called 'Verification by induction'.
Similar to Epsilon Delta arguments for the limit. It usually doesn't give you the actual limit, but it lets you VERIFY whether your conjectured limit is actually the limit. You NEED a limit to test otherwise the epsilon/delta proof won't work.
Similarly, you need a formula or expression to prove in order to use induction, otherwise it won't work.
Every lecturer I've ever had has mentioned that mathematicians work backwards. They start with a result they think is true and try to find a proof for it.P.s. You need to have a read over what you have written. Some of it doesn't make sense.
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Carrotsticks
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Re: 2012 HSC MX2 Marathon
Regarding induction, I understand what you are saying.
Perhaps the HSC has killed my opinion of Induction as it did for many other aspects of education. I haven't done anything extra-curricular, so my only exposure to it was the HSC.
Which part for example?I think you're greatly underestimating the skill of professional mathematicians
P.s. You need to have a read over what you have written. Some of it doesn't make sense.
Regarding induction, I understand what you are saying.
Perhaps the HSC has killed my opinion of Induction as it did for many other aspects of education. I haven't done anything extra-curricular, so my only exposure to it was the HSC.
jet
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Re: 2012 HSC MX2 Marathon
Well the entire bit between "This means that for all..." and "Suppose...". There isn't any other place that you mention a_c so why do you introduce it?
Carrotsticks
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Re: 2012 HSC MX2 Marathon
Haha that was to give proper justification for the strict inequalities.
jet
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Re: 2012 HSC MX2 Marathon
a) you say for every c in N, i.e. for EVERY natural number, a_c is between the two products. This makes no sense because there aren't an infinite number of a_k's and there's no guarantee that any of the a's will be between the two products.
b) You're relying on the density of the real numbers which you haven't even mentioned.
c) It doesn't justify the strict inequalities because you have already explained why they are strict. If instead you went like this:
Here are two products, A and B.
Fiddle with them a little.
Oh look, I've got an a_c s.t. A > a_c > B
So A > B strict.
Then that justifies strict inequalities. You went in the reverse and said that the inequality is strict so I can find an a_c which means that the inequality is strict.
You don't need to assume that each successive product is smaller than the last because all of the a's are less than 1. So you can naturally deduce that
a_1 > a_1a_2 > a_1a_2a_3 > ...
It's probably not a good idea because
a) you say for every c in N, i.e. for EVERY natural number, a_c is between the two products. This makes no sense because there aren't an infinite number of a_k's and there's no guarantee that any of the a's will be between the two products.
b) You're relying on the density of the real numbers which you haven't even mentioned.
c) It doesn't justify the strict inequalities because you have already explained why they are strict. If instead you went like this:
Here are two products, A and B.
Fiddle with them a little.
Oh look, I've got an a_c s.t. A > a_c > B
So A > B strict.
Then that justifies strict inequalities. You went in the reverse and said that the inequality is strict so I can find an a_c which means that the inequality is strict.
You don't need to assume that each successive product is smaller than the last because all of the a's are less than 1. So you can naturally deduce that
a_1 > a_1a_2 > a_1a_2a_3 > ...
Carrotsticks
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Re: 2012 HSC MX2 Marathon
In Analysis, we have had to prove many 'obvious' things, so I figured this would have to be the case.
a) This was my mistake. I assumed that there were an infinite number of terms a_k such that there would exist SOME a_c (dunno why I said 'for all' lol) existing between two products.
That is great! If that is the case, then that little section is to be ignored.
In Analysis, we have had to prove many 'obvious' things, so I figured this would have to be the case.
a) This was my mistake. I assumed that there were an infinite number of terms a_k such that there would exist SOME a_c (dunno why I said 'for all' lol) existing between two products.
jet
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Re: 2012 HSC MX2 Marathon
If it was infinite, you would have some problems with your proof, which you'll realise in a couple of weeks when you do infinite series with Florica!
Carrotsticks
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Re: 2012 HSC MX2 Marathon
Could not believe my eyes when I saw the Basel Problem for my Q8.
Was even more proud of the fact that I had in fact done the exact same question whilst doing my own research into infinite series (procrastinating and whatnot during HSC).
Looking forward to it. They have always fascinated me.
Could not believe my eyes when I saw the Basel Problem for my Q8.
Was even more proud of the fact that I had in fact done the exact same question whilst doing my own research into infinite series (procrastinating and whatnot during HSC).
Carrotsticks
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Re: 2012 HSC MX2 Marathon
I just realised that in my proof i have many typos with the limits! Especially in the w.l.o.g part. But the idea is still the same. Too much copy pasting and changing variable names.
I just realised that in my proof i have many typos with the limits! Especially in the w.l.o.g part. But the idea is still the same. Too much copy pasting and changing variable names.
seanieg89
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Re: 2012 HSC MX2 Marathon
Don't let the HSC course make induction seem like a triviality, it is often a powerful way to prove results concisely. And indeed most non-obvious results are "guessed" well before they are proven.
By the way, I don't entirely follow what you are saying in the early half of that proof but I don't think what you are trying to do works...
> \sum_{1\leq i_1 \leq \ldots \leq i_{k-1} \leq n} \left(\prod_{j=1}^{k-1} a_{i_j}\right)$\\ if and only if $k>(n-2)/3$. So in general when you pair your symmetric sums up not every bracketed term will be positive...the early ones will be negative. $)
Don't let the HSC course make induction seem like a triviality, it is often a powerful way to prove results concisely. And indeed most non-obvious results are "guessed" well before they are proven.
By the way, I don't entirely follow what you are saying in the early half of that proof but I don't think what you are trying to do works...
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Re: 2012 HSC MX2 Marathon
It makes sense in my mind, but I'll try my best to explain what I was doing in words:
1. When we expand the product (1-a_k), we get the the expression 1 - sum a_i_1 + sum a_i_2 - sum a_i_3 + ... and this continues on until we have sum a_i_n, which is essentially the same thing as prod a_k.
2. Recognise that the first two terms are the RHS of the question, so it follows that if we prove sum a_i_2 - sum a_i_3 + ... +/- prod a_k is positive and nonzero, then the strict inequality is proved.
3. Now, we have to prove that the cyclic sums satisfy a_i_2 > sum a_i_3 > sum a_i_4 > ... > sum a_i_n (which is the same as product in this case).
4. We know trivially that
will be less than 
etc because a_k is between 0 and 1, so the more a_k we have, the smaller the value. Now we have to prove that the CYCLIC sum of such terms preserves the inequality.
5. So I set up an array
. But now, I want to set up a parallel array but with different terms in each inequality such that when added, the different combinations from these arrays will construct the cyclic sum.
6. Firstly, note that the a's range from a_1 to a_n. My first parallel array (the inequalities will still hold) will be
(Note that it 'cycles' back to the original a_1).
7. My second parallel array will be
8. This process repeats n times so our last array will be
9. So if we add up all the arrays, we get something like what we need. But we have only considered consecutive groupings! What about product groupings like
?? We will need those in order to properly construct the cyclic sum.
10. So as you can imagine we must now count the number of possible combinations in the cyclic sum and consider the MAX of them (let this be M) and this means we have to make M arrays in order to satisfy all term of the expansion. By doing so, it means we have all bases covered ranging from sum a_i_1 (which has the fewest combinations, namely C(n,1) all the way up to sum a_i_n (which will consequently have the same number of combinations as sum a_i_1 due to symmetry of Pascal's Triangle). So obviously, this will mean that we will have 'repeats' of the sums with fewer terms (like sum a_i_1), but this is okay because that means once we add them sum, we will just have some constant multiple of the bare minimum that we need for the sum of the arrays to construct a cyclic sum.
*Now I just realised where things can go wrong. This means we must prove that EVEN with this constant multiple, the inequality is preserved because the multiple could very easily imbalance the inequality and actually FLIP it around!*. I don't think the proof for this will be easy. It surely MUST be true (otherwise the identity doesn't work, and by induction we know it to be true) but the question now is HOW.
Sigh.
It makes sense in my mind, but I'll try my best to explain what I was doing in words:
1. When we expand the product (1-a_k), we get the the expression 1 - sum a_i_1 + sum a_i_2 - sum a_i_3 + ... and this continues on until we have sum a_i_n, which is essentially the same thing as prod a_k.
2. Recognise that the first two terms are the RHS of the question, so it follows that if we prove sum a_i_2 - sum a_i_3 + ... +/- prod a_k is positive and nonzero, then the strict inequality is proved.
3. Now, we have to prove that the cyclic sums satisfy a_i_2 > sum a_i_3 > sum a_i_4 > ... > sum a_i_n (which is the same as product in this case).
4. We know trivially that
5. So I set up an array
6. Firstly, note that the a's range from a_1 to a_n. My first parallel array (the inequalities will still hold) will be
7. My second parallel array will be
8. This process repeats n times so our last array will be
9. So if we add up all the arrays, we get something like what we need. But we have only considered consecutive groupings! What about product groupings like
10. So as you can imagine we must now count the number of possible combinations in the cyclic sum and consider the MAX of them (let this be M) and this means we have to make M arrays in order to satisfy all term of the expansion. By doing so, it means we have all bases covered ranging from sum a_i_1 (which has the fewest combinations, namely C(n,1) all the way up to sum a_i_n (which will consequently have the same number of combinations as sum a_i_1 due to symmetry of Pascal's Triangle). So obviously, this will mean that we will have 'repeats' of the sums with fewer terms (like sum a_i_1), but this is okay because that means once we add them sum, we will just have some constant multiple of the bare minimum that we need for the sum of the arrays to construct a cyclic sum.
*Now I just realised where things can go wrong. This means we must prove that EVEN with this constant multiple, the inequality is preserved because the multiple could very easily imbalance the inequality and actually FLIP it around!*. I don't think the proof for this will be easy. It surely MUST be true (otherwise the identity doesn't work, and by induction we know it to be true) but the question now is HOW.
Sigh.
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Re: 2012 HSC MX2 Marathon
Hey guys,
In our 3U test today, we got this question:
<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{-2}^{2}\frac{x}{1-x^4}dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{-2}^{2}\frac{x}{1-x^4}dx" title="\int_{-2}^{2}\frac{x}{1-x^4}dx" /></a>
Everyone else wrote that it equals zero (odd function etc.) but I thought that you can only integrate something if it is defined in the domain so I wrote that it diverges as x can't equal 1 or -1. Is that right?
EDIT: Should probably have posted in 3U but you guys will probably know better...
Hey guys,
In our 3U test today, we got this question:
<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{-2}^{2}\frac{x}{1-x^4}dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{-2}^{2}\frac{x}{1-x^4}dx" title="\int_{-2}^{2}\frac{x}{1-x^4}dx" /></a>
Everyone else wrote that it equals zero (odd function etc.) but I thought that you can only integrate something if it is defined in the domain so I wrote that it diverges as x can't equal 1 or -1. Is that right?
EDIT: Should probably have posted in 3U but you guys will probably know better...
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