UoM Maths Thread

[hobbitle]

^
not really

Ha yea I phrased it badly, my bad.

[clueless123]

Hey guys can someone clarify this for me? I'll go to a consultation today but just thought I would get a different perspective/explanation if possible. Cheers

In the lec notes:
Vectors ( 1,-1,2,1),(-2,2,-4,-2),(1,0,3,0)


M~B. So I learnt that because the rank of B = 2, the set of vectors can only span a R^2..?
If thats true, how come a basis for the subspace of R^4 spanned by these vectors exists?
I thought the basis only existed if the vectors spanned the space

Cheers

[Oilerian]

Hey guys can someone clarify this for me? I'll go to a consultation today but just thought I would get a different perspective/explanation if possible. Cheers

In the lec notes:
Vectors ( 1,-1,2,1),(-2,2,-4,-2),(1,0,3,0)


M~B. So I learnt that because the rank of B = 2, the set of vectors can only span a R^2..?
If thats true, how come a basis for the subspace of R^4 spanned by these vectors exists?
I thought the basis only existed if the vectors spanned the space

Cheers

A basis for a subspace can of course have different elements (and therefore span different dimension) than the entire R^4 vector space. Say vector (0,0,0,1) alone is clearly a basis for a subspace (0,0,0,n) in R^4 but not a basis for vector space R^4.

[Bardia Saeedi]

hi guys.

I did specialist maths last year.

My mark for is was below that of required for Calc2.

I was wondering how much different Calc1 is from specialist maths (VCE), especially since I have already done it.

EDIT1:

I consider myself fairly strong in maths. My mark in Methods was higher than the calc1.

I am considering doing calc1 in sem 2, along with data analysis.

[LeviLamp]

Hey

Having both tried to complete VCE Specialist Maths exams and completed Calculus 1, I'd say the content is similar in many, many aspects of the subject, but that the level of mathematical operation required in Calculus 1 is much higher; the problems set in the practice book and especially on the exam are largely much more difficult than those I found in the Specialist Maths VCE exams. So it's up to you, but if you want a solid foundation going into Calculus 2 (IF you plan on eventually taking it), I don't think Calculus 1 is a waste of a subject. It's pretty difficult to do well in (as are, purportedly, all the first year maths subjects) and the department doesn't scale marks if cohorts do badly or too well, but if you put in some hard work for either subject you'll probably be at least OK. That's just my two cents worth, anyway; other people probably think differently!

[hobbitle]

I didn't do specialist maths but I'd say from talking to others and having taken Calculus 1 (and Calc 2 and Linear Algebra) that you really really need a solid foundation in Calc1/Specialist before doing Calc2 which is (arguably I suppose) a very difficult Level 1 subject. So even if you do repeat a lot of content, that's ok, because a good mark in Calc1 will put you in good stead for Calc2 etc.

[notveryasian]

If you are fairly confident in your mathematical capabilities and feel that you have a good knowledge of the Calc 1/ specialist maths material then see (probably from a student centre) if you are able to enrol in Calc 2. You may be asked to do an entrance test or something along those lines.

[Bardia Saeedi]

Thanks for the advice.

I will NOT do Calc2, as it will not benefit my in anyway, and it seems to be way too hard

[Deleted User]

Can someone explain to me why the sum of n*0.5^(n-1) from n=1 to infinity is equal to 4? Thanks

[hobbitle]

Can someone explain to me why the sum of n*0.5^(n-1) from n=1 to infinity is equal to 4? Thanks

This explanation will be terrible so hopefully someone else will step in.
But if you substitute the first few integers in the series for n you will see how the series converges to 4.
When n = 1, a_n = 1
When n = 2, a_n = 1
When n = 3, a_n = 0.75
When n = 4, a_n = 0.5
When n = 5, a_n = 0.3125

Adding them all together you can see you are approaching 4.
You know that you will never actually get to 4 because as n approaches infinity, n/2^(n-1) (which is the same as what you wrote) gets closer and closer to zero.  Because the demoninator is becoming exponentially bigger than the numerator as n increases.

There will be a technical more mathy explanation but maybe that helps for now.

[kinslayer]

Can someone explain to me why the sum of n*0.5^(n-1) from n=1 to infinity is equal to 4? Thanks

You can see a geometric series hiding underneath the "n" term there. You can make a guess that the sum converges since the sequence clearly goes to zero: the geometric term will go to zero much more quickly than the linear "n" term goes to infinity. You can prove this analytically by using the ratio test.

I don't know how technical this needs to be but I have attached one possible way of computing the sum (getting rid of that "n") using composition. Another way would be to start off with a geometric series and then differentiate both sides.

[Deleted User]

Thanks. Do you mind elaborating steps 3 and 4? I'm not too familiar with summations so I'm not sure if they're rules that I'm assumed to know?

[kinslayer]

Steps 3 onward are using the definition of the geometric series and algebraic manipulation. See:

http://en.wikipedia.org/wiki/Geometric_series

Under "geometric power series" it actually has the general form of the series we computed. To get that general form, start with and differentiate both sides. That's actually simpler than what I've done so you should try it and see what you get

Note that you can only take the derivative inside an infinite sum when the sum is absolutely convergent, which that is because it's a geometric sequence with .

[notveryasian]

today's AM1 lecture recording is twice as slow as it should be...submitted a report to the IT services, hopefully it gets fixed soon...also, Paul Norbury's voice is pretty funny in slow motion 

[hobbitle]

Just download it and watch it in VLC at double speed.