Math Is Fun Forum

Thank you bobbym, that's great. I've been playing with the numbers myself and I'm completely with you on 1 & 2 . 3 & 4 are utterly beyond me, though. Someone suggested the formula in the wikipedia article under collision counting as a possible approach to me, but i've no idea, really, what to do with it

i think you're right, thank you very much bobbym

Hmmm...that's what i was trying to do, well, I really need to get off to bed, but for what it's worth, our old friend Wolfram gives, as its answer to

This:

Which I can get from my answer, so I guess it's just a question of laws of logs, I don't suppose anyone has any ideas?

Sorry bobbym, yes, i agree, i get the same for both.

So, surely, I should keep them? The answer book simply has

I had thought that maybe the answer book was giving a simplified answer, it's an old A-level book, but the syllabus was very different back then, so I'm never sure what I'm expected to know But looking at it, if I draw a graph, I don't think T can ever - on this graph - be above N on the y-axis, so - presumably - the distance can always be given by N - T, with no need to worry about what would happen if T were to occur above N, giving a negative distance?

It's great to be back, bob bundy Thanks so much. I have a feeling you're right, I just didn't have enough confidence in my answer, but the book agrees with me that the answer to part (ii) is:

{7/3} and anonimnystefy, you're absolutely right. We know that h varies with t, specifically, it varies at the uniform rate of:

(As the question tells us.)

And we know that V varies with h, specifically, it varies at a rate of:

(According to my calculations)

We know, then, that V also varies with t, since it varies with h, which varies with t. Specifically, by the chain rule, it varies at a rate of:

Ahhh thank you anonimnystefy, now I see, because of course

And then all the terms in between cancel That's a great help

I'm not sure what exactly your approach would be. Originally, I just posted my approach:

If 2 is a zero of the polynomial, then (x - 2) is a factor.

(By polynomial long-division)

Therefore, the other zero is -1

But having re-read your first post, I think that might be what you would have done anyway. I'm not sure that I know any more efficient method. But I suggest you just divide once and then factorise, that - at least - might make things a little faster?

Oh yes, that's absolutely true and a very nice fact. In the universe of mathematics this is fine, the only problem would be if you tried to do it yourself!

Well, if we do away with aleph numbers and the like and just concentrate on this abstract concept ∞, then we have to think about what infinity is. This is perhaps not the most clear-cut question in the world, but - really - we can think of infinity by playing a counting game. As we count the numbers:

1, 2, 3, 4, 5, 6, 7, 8, 9...

We add one to each number to get the next number. We can go on counting:

...1,000, 1,001, 1,002, 1,003, 1,004, 1,005...

And on:

...1,000,000, 1,000,001, 1,000,002, 1,000,003...

If we kept on and on counting (although, if we counted one number per second, it would take us over 31.6 years of non-stop counting to reach one billion, so it would not be practically possible to do this, just theoretically) we can reach numbers of immense size, such as Graham's number, which far far outstrips the number of atoms in the universe. In fact, I don't think it would even be possible to imagine how much bigger Graham's number is than that. And we can carry on and what it's pretty easy to realise here is that it can go on forever, we can always keep adding one to make the next bigger number, with no end to the counting. This is where the concept of ∞ comes in.

In the theories of calculus and many other areas of mathematics, we use the symbol to denote unboundedness, really, doing it over and over without end. It is not really treated as a number. If we want to try to treat it as a number and do operations on it, we really need something like Cantor's cardinal arithmetic. But, without going into all of that, we can make it a little more intuitive and just say that ∞ is the end of counting. This is not all that precise, really, but if we just want to get our heads around what this symbol: ∞ means, we have to look back to counting. Essentially ∞ says that I can go on adding 1 for ever and ever and ever and it will not end (in-finite), so I will use this symbol ∞ to mean the point beyond which I cannot keep counting. For this reason 1 + ∞= ∞  and 1 - ∞ = -∞

Hehe! Useful trick, though, I got rather good at it when I was doing complex numbers, but that was a while ago now and - after staring at 2x^2 - x - 16 = 0 for a little while before realising it wouldn't factorise, I set about trying trawl the method up from the bottom of my brain