Harder HSC Papers for MX2 (1 Viewer)

[glittergal96]

Re: Harder HSC Papers

I don't think a single value like that is enough to uniquely specify a function using the recurrence.

 

[glittergal96]

Re: Harder HSC Papers

Yep.

 

[InteGrand]

Re: Harder HSC Papers

I don't think a single value like that is enough to uniquely specify a function using the recurrence.

But it is true that P(0,0) would be 1, isn't it? And once P(r, s) is clearly defined, wouldn't a general formula for it come out simply from its definition? The two values they gave are not arbitrarily chosen by them, but come from the definition they meant. But yeah, it may be an inconvenience for their general formula.

 

[dan964]

Re: Harder HSC Papers

here is all the papers back to 1967
https://9eeba4054ee764a743f35bb25b5...Year 12 - HSC/Maths/hscpapers_extension2.html

 

[glittergal96]

Re: Harder HSC Papers

But it is true that P(0,0) would be 1, isn't it? And once P(r, s) is clearly defined, wouldn't a general formula for it come out simply from its definition? The two values they gave are not arbitrarily chosen by them, but come from the definition they meant. But yeah, it may be an inconvenience for their general formula.

I would say so yes, but this is of course a matter of convention.

And no, I think that that recurrence formula requires two specified values to give you a unique solution. A general formula does not follow just from knowledge of P(0,0).

 

[InteGrand]

Re: Harder HSC Papers

I would say so yes, but this is of course a matter of convention.

And no, I think that that recurrence formula requires two specified values to give you a unique solution. A general formula does not follow just from knowledge of P(0,0).

 

[glittergal96]

Re: Harder HSC Papers

My point is that the give you a physical problem from which you can deduce both a recurrence relation and initial values.

In passing to the problem of solving a recurrence, we need to obtain two pieces of initial data, in this case the values of P(1,0) and P(0,1).

As you need two, and the question mentions these two, I don't see why discussion of P(0,0) is important.

 

[glittergal96]

Re: Harder HSC Papers

And "by convention" means it is a matter of convention what domain to define P on (whether or not to include (0,0)), given that main interest is for positive integers clearly.

 

[braintic]

Re: Harder HSC Papers

I honestly don't think P(0,0) is meaningful. I can see why it could be equal to zero as much as why it could be equal to one (though I don't think I can put it into words).
I see it in kind of a similar way to 0^0 being undefined.

 

[InteGrand]

Re: Harder HSC Papers

I honestly don't think P(0,0) is meaningful. I can see why it could be equal to zero as much as why it could be equal to one (though I don't think I can put it into words).
I see it in kind of a similar way to 0^0 being undefined.

I was thinking that if P(1,0) is going to be 0 (and so P(2,0), P(3,0), etc. all equal 0), then P(0,0) would have to be 1? As glittergal96 said:

''But you have thrown zero tails to start with, without throwing any heads.''

which makes it sound like given the 'zeroth' tails has 'just' been tossed, we mean that we are just starting, which means we must have 0 heads, i.e. P(0,0) = 1. If P(1,0) is meaningful, then P(0,0) should be too. Maybe we should restrict the tails count to be positive.

(Of course this is all irrelevant for the problem at hand.)

Also, for any fixed positive integer N, the sum over all non-negative integers k of P(k,N) is going to be 1, so it is probably natural to have P(0,0) to be 1, if we were interested in assigning a value to it.