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First year Analysis/Foundations January Exam 2011

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Reply 40
Avatar for yeohaikal
yeohaikal
Original post by placenta medicae talpae
Lawl, is this a desperate attempt to hide who you are?! :biggrin:


LOL. Obviously it is. As everybody can tell, I'm the LaTeX-iliterate guy here!
Students on campus at the University of Warwick
University of Warwick
Coventry
Reply 41
Avatar for yeohaikal
yeohaikal
Original post by matt2k8
Have let one of my friends borrow my answers until tommorow so no more will be typed tonight


Q33 - an=1/na_n = 1/n
Q45 - thanks, don't know what I was thinking with that one haha :smile: fixed
Q67 - thanks, fixed it now, wasn't paying much attention as I don't know much about lim sup


Wait. I'm still confused over Q33. an=1/na_n = 1/n makes the series summation of an=1/na_n = 1/n convergent right? Am I missing something here?
Original post by Narev
It'll probably come out in Analysis II...I wish I knew how to use LaTeX...


Christ, then what is this MSA booklet I have in front of me :tongue:

Incidentally, I think this year's module was lectured from the 2009-10 notes, along with the mistakes that were in there.
Reply 43
Avatar for Narev
Narev
1
Original post by yeohaikal
LOL. Obviously it is. As everybody can tell, I'm the LaTeX-iliterate guy here!


Yes, and wait till the revision lectures. Hmpf.

*I think I was responding halfway, and then going "I wish I knew how to ---" and then my mind put in LaTeX. Or you can see it as "I wish I knew how to use latex."

PS. WMS and MathPhys have given the revision lecture timings to the members right?
(edited 13 years ago)
Reply 44
Avatar for Narev
Narev
1
Original post by TheTallOne
Christ, then what is this MSA booklet I have in front of me :tongue:

Incidentally, I think this year's module was lectured from the 2009-10 notes, along with the mistakes that were in there.


Can't be surely..I'm sure the lecturer would have spotted the mistakes.

On the other hand, the MSA lecture notes on the course webpage are extremely outdated, and the revision guide has typos fixed on a "report" basis - updated copies of both can be found on the relevant pages on the MORSE website.

I've kind of fixed the Downloads page to be a 'just for fun' page too - non members and members can download random stuff from there.
Reply 45
Avatar for miml
miml
17
Original post by yeohaikal
Wait. I'm still confused over Q33. an=1/na_n = 1/n makes the series summation of an=1/na_n = 1/n convergent right? Am I missing something here?

∑an→a  ⟹  an→0\sum{a_n} \rightarrow a \implies a_n \rightarrow 0 . But not the other way round, hence the counterexample.

http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)
Original post by Narev
Can't be surely..I'm sure the lecturer would have spotted the mistakes.

On the other hand, the MSA lecture notes on the course webpage are extremely outdated, and the revision guide has typos fixed on a "report" basis - updated copies of both can be found on the relevant pages on the MORSE website.

I've kind of fixed the Downloads page to be a 'just for fun' page too - non members and members can download random stuff from there.


Sometimes. If there was a mistake, sometimes it passed through, sometimes she wrote it on the board and then realised, taking a while to then correct them again. In any case, the lectured notes seem very Jon Warren/MORSE Soc.
Reply 47
Avatar for Apex9
Apex9
8
doesn't it say consider (bn-an)? and then the previous lemma which you should have proved (or if not assumed true?)
Reply 48
Avatar for Narev
Narev
1
Original post by TheTallOne
Sometimes. If there was a mistake, sometimes it passed through, sometimes she wrote it on the board and then realised, taking a while to then correct them again. In any case, the lectured notes seem very Jon Warren/MORSE Soc.


Jon Warren has that effect. I still like his Analysis II notes.
Reply 49
Avatar for yeohaikal
yeohaikal
Original post by miml
FOUNDATIONS

January 2006



January 2007



January 2008



January 2009



6) Yes, a surjection A->B has a right inverse (which is also an injection) B->A.

7) αβ \displaystyle\alpha\beta = (1234)
βα \displaystyle \beta \alpha = (1243)

8) Closure: ∀a,b∈G,ab=ba∈G\displaystyle \forall a,b \in G , ab=ba\in G
Associativity: ∀a,b,c∈G,(ab)c=a(bc)\displaystyle \forall a,b,c \in G, (ab)c=a(bc)
Identity Element: ∃e∈G:∀a∈Gae=ea=a\displaystyle \exists e \in G : \forall a \in G ae = ea = a
Inverse Element: ∀a∈G∃a−1∈G:aa−1=a−1a=e\displaystyle \forall a \in G \exists a^{-1} \in G: aa^{-1} = a^{-1}a = e

9) 1517=15(mod17)\displaystyle 15^{17} = 15 (mod 17) by FLT


Ok. Just going through Jan 2006 at the moment.

4. I think this is wrong. It should be (not P) union Q
7. Don't think your solutions are right.
For part (i), the integer 5 has no pre-image and hence is not surjective. Try f(x)=x if x<1, f(x)=x+1 if x> and equal to 1
(ii) f(x)=2x is a solution because the odd numbers have no pre-image and hence not surjective. Injective because every integer corresponds to a different even number.
(iii) f(x)=|x| is a solution. Negative integers have no pre-image. Every positive integer has 2 pre-images.
(iv) For bijective functions, the most reliable is f(x)=x.
Reply 50
Avatar for yeohaikal
yeohaikal
Original post by miml
∑an→a  ⟹  an→0\sum{a_n} \rightarrow a \implies a_n \rightarrow 0 . But not the other way round, hence the counterexample.

http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)


Lol. I'm talking about Q33 here. I know the Harmonic Series diverges. summation of ((-1)^n)(a_n) is conditionally convergent and hence can be made to converge to 1. So it's not the required counterexample.
Original post by jakash
*rips hair out*

I swear this should be so simple.

Corollary 2.23 (page 6) in the revision guide, Assignment 11 in Workbook 3:



The inequalities are less than or equal to btw, and the 'eventually' I can handle.

I just can't prove it.

I fail at analysis. :frown:

The reason it follows from the previous lemma is that bn≥anb_n \geq a_n eventually so bn−an≥0b_n - a_n \geq 0 eventually therefore b−a≥0b-a \geq 0 (as (bn−an)→b−a(b_n - a_n) \rightarrow b - a)
(edited 13 years ago)
Right, sorry if this is reaaaally dim of me, but I don't get how to re-write rubbish cycle notation into disjoint cycle notation.

E.g. how does (123456)(1234567)(123456)(1234567) become (135)(2467)(135)(2467)? :confused:
Reply 53
Avatar for miml
miml
17
Original post by placenta medicae talpae
Right, sorry if this is reaaaally dim of me, but I don't get how to re-write rubbish cycle notation into disjoint cycle notation.

E.g. how does (123456)(1234567)(123456)(1234567) become (135)(2467)(135)(2467)? :confused:


Starting from the second set of brackets, we start with 1
1 -> 2 in the second bracket and 2->3 in the first one, so 1->3 ie (13
now we carry on from 3 in the second bracket
3 -> 4 in the second bracket and 4->5 in the first one, so 3 ->5 ie (135
now 5-> 6 ->1 thus (135)

then carry on from the smallest number not in (135) ie 2.
2->3->4 gives (135)(24
4->5->6 gives (135)(246
6->7 (and it ends here, because 7 isn't in the first bracket) gives (135)(2467 we can finish here, because we've accounted for all the elements, but to check 7->1->2 which closes the final bracket...

(135)(2467)
Original post by matt2k8
Analysis I, 2009, Q1 - missed out f though as I'm not sure what to do? I can prove 1∣an∣→∞\frac{1}{|a_n|}\rightarrow \infty but not what the question says :\

Spoiler



Thanks very much for these, although for part b) as n is natural, musn't tan n be bounded ?

For f, I don't know how you'd do this either but is wrong. As tends to zero but its inverse, does not.
(edited 13 years ago)
Original post by electriic_ink
Thanks very much for these, although for part b) as n is natural, musn't tan n be bounded ?

For f, I don't know how you'd do this either but is wrong. As tends to zero but its inverse, does not.

Nah, you'll always be able to find an natural number n such that tan n > 1000 for example, it just might be hard to find

sinxx→0\frac{sinx}{x} \rightarrow 0 as x→∞x \rightarrow \infty, you are thinking of that sinxx→1\frac{sinx}{x} \rightarrow 1 as x→0x \rightarrow 0.
(edited 13 years ago)
Original post by matt2k8
Nah, you'll always be able to find an natural number n such that tan n > 1000 for example, it just might be hard to find


I see. Ty.

sinxx→0\frac{sinx}{x} \rightarrow 0 as x→∞x \rightarrow \infty, you are thinking of that sinxx→1\frac{sinx}{x} \rightarrow 1 as x→0x \rightarrow 0.


Of course 1∣an∣→∞\frac{1}{|a_n|} \rightarrow \infty . I really shouldn't have been doing this at 4am XP

Can't you then just take a subsequence of positive terms and say they tend to infinity and that the subsequence of negative terms tends to minus infinity?
Reply 57
Avatar for yeohaikal
yeohaikal
is permutation tested???
Original post by yeohaikal
is permutation tested???


*looks this up*

Appaz 9.9 and 9.10 are not being tested :frown:

Which is a shame, because I really like 9.9.
And how 9.10 could be tested anyway beats me! :biggrin:

Btw, you doing discrete this term? :smile:
(edited 13 years ago)
Added the rest of the question 1 foundations
Original post by electriic_ink
Of course 1∣an∣→∞\frac{1}{|a_n|} \rightarrow \infty . I really shouldn't have been doing this at 4am XP

Can't you then just take a subsequence of positive terms and say they tend to infinity and that the subsequence of negative terms tends to minus infinity?


Yeah I think if it has infinitely many positive terms the subsequence made up of them would tend to positive infinity and if it had infinitely many negative ones the subsequence made up of them would tend to negative infinity
(edited 13 years ago)

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