Search results

I suspect that you are talking about an integrating factor?

First solve the characteristic equation r^2 + 8r + 16=0. The general solution is y=Ae^(r_1x) + Be^(r_2x), where r_1 and r_2 are the two solutions to the characteristic eqn, and A and B are constants. Except that in this case, you have only one solution, cause r^2 + 8r + 16 is a perfect...

It is a separable equation. You rearrange the equation to get all the x's on one side and all the y's on the other. Then you integrate both sides, add a constant of integration, and substitute in the particular values you were given to evaluate the constant of integration.

I typed up something about this awhile ago. There are probably other, neater ways to prove it though.

Sometimes one method will give you a much easier integral to evaluate than another. That's why you should know how to use both methods.

That is completely incorrect. I wouldn't have even got an interview for my present job with a transcript full of Ps.

Re: 回复: Re: Do you like maths? Yep - even if you solve 'em using Laplace transforms you still get the characteristic polynomial popping out and you need to know how to solve it. Actually, I am constantly being impressed by just how much you do use what you learn in senior high schools maths...

That proof was interesting, though. I've seen stuff like that using cyclotomic polynomials with cos and sin, but not cot and tan. One of the chapters in Hardy's A Course of Pure Mathematics has a whole heap of questions like that.

Peoples, Read this. Its been floating around on the internet for quite a while, but I only read it for the first time last week. It is rather long (25pp), but I'd like to know what you think of it. http://www.maa.org/devlin/LockhartsLament.pdf

The main advantage of a B something/ BEd. at UNSW vis-a-vis USyd is that it is only four years here instead of 5 over there. Apart from that, they both suck. (As does any education qualification at any other uni, for that matter.) Survivor39 gives good advice - your sister should primarily...

You can also derive it using the substitution x=a tan u, but you will need to know how to integrate sec u.

You can derive it using hyperbolic trig functions, which are not studied at HSC level. If you take maths at uni you should see it in first year. http://en.wikipedia.org/wiki/Hyperbolic_trig_functions

For Q2, any pair of real numbers with x=y (except x=y=0) will satisfy the second of your simultaneous equations. So substitute x=y into the first equation, and you should find that x=y= 2^-6/5 is a solution.

I believe it is a response to the NSW Institute of Teachers trying to *improve* teacher quality - so I guess that makes it a state level thing. Which is why it is happening everywhere, and not just UNSW.

Yes, that is about what happens... Seeing what they are doing to undergraduate teacher training, I am quite glad that I got my qualification when I did. Over the time that I have been at UNSW, they have decreased the number of UOC in maths required for the double degree (from 72 to 60) and...

That is totally screwed up - and its not even a Dawkins University. I just finished BSc/BEd at UNSW - I did about 50% more maths than required for a maths major in the BSc and could have done honours as well, if I'd wanted to. However, they have since screwed up the double degree at UNSW, by...

What uni did you study at?

Unfortunately, the 1 year dip ed at most unis seems to be a dying breed. However, dvse's major argument is correct - you want to waste as little time in (any) school of ed as possible.

I think "by invitation" means that they invite you automatically. TSP in maths always was a bit of a joke at UNSW. It seemed to consist of them writing you this rinky-dink little letter at the end of every year telling you you should do more maths in the future. (I got three of them - they are...

OK, I see what you mean. I didn't actually do the question. As a consequence of the Fundamental Theorem of Algebra and the fact that polynomials with real coefficients have roots that come in complex conjugate pairs, any polynomial with real coefficients can be (theoretically, at least)...