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HSC 2013 MX2 Marathon (archive) (5 Viewers)

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RealiseNothing

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Re: HSC 2013 4U Marathon

The second part is relatively simple to prove. When we get the Harmonic Series which . You can construct the following bound to show this:



The RHS and so the result falls out.

Now when we get and so for we get
 
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RealiseNothing

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Re: HSC 2013 4U Marathon

Let's light a fire here. No-one knows what school I teach at, right? But if you THINK I might be a teacher at your school, then there is a good chance some questions from this forum might be in your trial. Only 51 pages to sift through.
I feel sorry for your students.
 

Sy123

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Re: HSC 2013 4U Marathon



so for we get for some value L.

Now all I need is the case where , but I want to try do a more holistic proof rather than one by cases.
You're first part is correct, for the second though, I'd advise against the case by case basis, it might work though.

There is an easier method however.
 

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Re: HSC 2013 4U Marathon



so for we get for some value L.

Now all I need is the case where , but I want to try do a more holistic proof rather than one by cases.
Consider the curve y=1/x^p in the interval [1,infinity) and partition it into rectangles of unit sub-interval, and construct upper bound rectangles.
 

RealiseNothing

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Re: HSC 2013 4U Marathon

Constructing rectangle bounds for the curve gives the inequality:



Consider the RHS, let where





As









There may be an algebraic mistake somewhere as I haven't double checked my working, but it's the same idea nonetheless.
 

Sy123

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Re: HSC 2013 4U Marathon

Constructing rectangle bounds for the curve gives the inequality:



Consider the RHS, let where





As









There may be an algebraic mistake somewhere as I haven't double checked my working, but it's the same idea nonetheless.
Yep that is pretty much it.
 
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Re: HSC 2013 4U Marathon

Can students be expected to use the fundamental theorem of calculus freely?
 

Sy123

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Re: HSC 2013 4U Marathon

Can students be expected to use the fundamental theorem of calculus freely?
I don't see a need to but feel free to do so if it will help.
 

seanieg89

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Re: HSC 2013 4U Marathon

For something like part i) of your most recent question, you would need to provide the students with a rigorous definition of integration from which to work ...as this is not assumed knowledge in 4U. You also need to state what kind of functions f you want this to be proven for.

(Unless by "prove" you mean "provide a heuristic argument for why this is true using the informal mx2 notion of an integral").
 

RealiseNothing

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Re: HSC 2013 4U Marathon

The first part is taking infinite rectangles between the limits of 1 and 2. Their height will be with width

Just sum this from 1 to infinity to get the area under the curve.
 

Sy123

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Re: HSC 2013 4U Marathon

For something like part i) of your most recent question, you would need to provide the students with a rigorous definition of integration from which to work ...as this is not assumed knowledge in 4U. You also need to state what kind of functions f you want this to be proven for.

(Unless by "prove" you mean "provide a heuristic argument for why this is true using the informal mx2 notion of an integral").
Is there a problem with the argument; that considering finite n, the sum of the lower rectangles of width 1/n is approximately the area (which is the integral), and then saying as n increases without bound the rectangles all sum up to become the area?
Similar logic is there when deriving integrals for Volumes question, is this not rigorous/not in syllabus ?

I took the first part only from a STEP paper, it said 'Justify by means of a sketch', is using the word prove changing the meaning the original question had?
 

seanieg89

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Re: HSC 2013 4U Marathon

Is there a problem with the argument; that considering finite n, the sum of the lower rectangles of width 1/n is approximately the area (which is the integral), and then saying as n increases without bound the rectangles all sum up to become the area?
Similar logic is there when deriving integrals for Volumes question, is this not rigorous/not in syllabus ?
That's about as good as you can do with mx2 definitions and notions, but it is certainly not rigorous as the object you are proving things about (the Riemann integral) is not rigorously defined. Other things to think about:


1. What is the definition of area itself? (If you choose to define the integral via area).

2. How do you know that as n->inf the rectangles "area" approaches the "area" under the curve? Beyond the fact that the pictures look more and more alike.

3. You should state an assumption like continuity on f to guarantee those expressions make sense. If you don't assume something like this and your function is highly irregular, then the heuristic of measuring the "area" of things by rectangular approximations breaks down (and indeed the Riemann integral is not defined for such functions).



The main reason I am stressing these things is that typically, the integral is DEFINED via sums of the sort on the RHS, and "area under curves" is defined via this rigorous notion of integration. This puts things on a solid logical basis.
 
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Sy123

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Re: HSC 2013 4U Marathon

That's about as good as you can do with mx2 definitions and notions, but it is certainly not rigorous as the object you are proving things about (the Riemann integral) is not rigorously defined. Other things to think about:


1. What is the definition of area itself? (If you choose to define the integral via area).

2. How do you know that as n->inf the rectangles "area" approaches the "area" under the curve? Beyond the fact that the pictures look more and more alike.

3. You should state an assumption like continuity on f to guarantee those expressions make sense.



The main reason I am stressing these things is that typically, the integral is DEFINED via sums of the sort on the RHS, and "area under curves" is defined via this rigorous notion of integration. This puts things on a solid logical basis.
Ah alright then :s

Second part is still open though.
 
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Re: HSC 2013 4U Marathon

Yay Laplace transforms!





 
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