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    (Original post by placenta medicae talpae)
    Lawl, is this a desperate attempt to hide who you are?!
    LOL. Obviously it is. As everybody can tell, I'm the LaTeX-iliterate guy here!
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    (Original post by matt2k8)
    Have let one of my friends borrow my answers until tommorow so no more will be typed tonight


    Q33 - a_n = 1/n
    Q45 - thanks, don't know what I was thinking with that one haha 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. a_n = 1/n makes the series summation of a_n = 1/n convergent right? Am I missing something here?
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    (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

    Incidentally, I think this year's module was lectured from the 2009-10 notes, along with the mistakes that were in there.
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    (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?
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    (Original post by TheTallOne)
    Christ, then what is this MSA booklet I have in front of me

    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    On the other hand, the MSA lecture notes on the course webpage are extremely outdated, and the revision guide has typos fixed on a "report" basis - updated copies of both can be found on the relevant pages on the MORSE website.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.
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    (Original post by yeohaikal)
    Wait. I'm still confused over Q33. a_n = 1/n makes the series summation of a_n = 1/n convergent right? Am I missing something here?
    \sum{a_n} \rightarrow a \implies a_n \rightarrow 0 . But not the other way round, hence the counterexample.

    http://en.wikipedia.org/wiki/Harmoni...s_(mathematics)
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    (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.
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    doesn't it say consider (bn-an)? and then the previous lemma which you should have proved (or if not assumed true?)
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    (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.
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    (Original post by miml)
    FOUNDATIONS

    January 2006

    1) Every \displaystyle n>1 \in \mathbb{N} can be written as a product of prime numbers. This factorisation is unique up to order.

    2) \displaystyle 7\mathbb{Z} \cap 3\mathbb{Z}=21\mathbb{Z}

    3) \displaystyle (A \cap B)^c = A^c \cup B^c and \displaystyle (A \cup B)^c = A^c \cap B^c

    4)  \displaystyle (\not P) \land Q

    5) (x-3) and (x-1)

    6) \displaystyle \binom{n}{k-1} + \binom{n}{k} = \frac{n!}{(k-1)!(n-k+1)!}+\frac{n!}{(k)!(n-k)!}=\dfrac{n!k + n!(n-k+1)}{k!(n-k+1)!} =\displaystyle \frac{n!(k+n-k+1)}{k!(n+1-k)!} = \frac{(n+1)!}{k!(n+1-k)!} = \binom{n+1}{k}

    7) \displaystyle f:\mathbb{Z}\rightarrow\mathbb{Z  } where \displaystyle f(x)=(x+1)x(x-1)
    \displaystyle f:\mathbb{Z}\rightarrow\mathbb{Z  } where \displaystyle f(x)=?
    \displaystyle f:\mathbb{Z}\rightarrow\mathbb{Z  } where  \displaystyle f(x)=x^2
    \displaystyle f:\mathbb{Z}\rightarrow\mathbb{Z  } where \displaystyle f(x)=2x

    8) There is a bijection \displaystyle f:A \rightarrow B

    9) (134)(25)

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

    11) \displaystyle 17^{18} = 17^{(19-1)} = 1 (mod 19) by FLT.


    January 2007

    1) a|b means 'a divides b' ie \displaystyle \exists k \in \mathbb{N} : b=ak . If a|b and b|c then \displaystyle k_1, k_2 \in \mathbb{N} : b=ak_1 and \displaystyle c=bk_2 \implies c = ak_1k_2 \implies a|c

    2) Every \displaystyle n>1 \in \mathbb{N} can be written as a product of prime numbers. This factorisation is unique up to order.

    3) \displaystyle  5\mathbb{Z} \cap 3\mathbb{Z} = 15\mathbb{Z}
    \displaystyle  5\mathbb{Z} + 3\mathbb{Z} = {Z}
    \displaystyle  5\mathbb{Z} \cup 3\mathbb{Z} is not a subgroup.

    4) Venn diagrams

    5)  \diplaystyle\begin{matrix}

(P & \wedge  & Q) & \vee   & (\setminus   & P  & \wedge  & Q)  & \Leftrightarrow   & Q \\ 

 1& 1 &1  &1  &0  &1  &0  &1  &1  &1 \\ 

 1& 0& 0 & 0 & 0 & 1 & 0 &  0&1  &0 \\ 

 0& 0& 1 & 1&  1&  0&  1&  1&  1& 0\\ 

 0& 0& 0 & 0&  1&  0&  1& 0 &  1&0 

\end{matrix}

    6) If P(x) = (x-a)Q(x) + R(x) then R is the constant polynomial P(a)

    7) (x+4) and (x-1)

    8) Coefficient of x^k is \displaystyle \binom{n}{k}a^{n-k}b^k
    Let a=1, bx = -1 in \displaystyle (a+bx)^n = \sum_{k=0}^{\infty} \binom{n}{k}a^{n-k}(bx)^k \implies (-1+1)^n = 0 = \sum_{k=0}^{\infty} (-1)^k \binom{n}{k}

    9) \displaystyle f:\mathbb{N}\rightarrow\mathbb{N  } where  \displaystyle \left\{\begin{matrix}

f(1)=1

\\ f(x)=x-1 \forall x>1



\end{matrix}\right.
    \displaystyle f:\mathbb{N}\rightarrow\mathbb{N  } where  \displaystyle f(x)=x^2
    \displaystyle f:\mathbb{N}\rightarrow\mathbb{N  } where  \displaystyle f(x)=5

    10) There is a bijection \displaystyle f:\mathbb{N} \rightarrow \mathbb{Q} but no bijection \displaystyle f:\mathbb{N} \rightarrow \mathbb{R}


    January 2008

    1) \displaystyle 8\mathbb{Z} + 12\mathbb{Z}=4\mathbb{Z}

    2) \displaystyle \varnothing , {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c}

    3) A\(B u C)

    4) ~P V ~Q

    5) (x+3) and (x-2)

    6) Let a=b=1 in \displaystyle (a+b)^n = \sum_{k=0}^{n} a^kb^{n-k} \binom{n}{k} \implies (1+1)^n = 2^n = \sum_{k=0}^{n} \binom{n}{k}

    7) ?

    8) There is an injection (but no bijection)  \displaystyle f: A \rightarrow B

    9) (1324)

    10)  \displaystyle 15^{22} = (15^2)^{11} = 15^2 (mod 11) by FLT  \displaystyle 15^2 = 4^2 (mod 11) = 16 (mod 11) = 5 (mod 11)


    January 2009

    1)  \displaystyle 9\mathbb{Z} + 15\mathbb{Z} = 3\mathbb{Z}

    2) A\(B u C)

    3) (~P) ^ (~Q)

    4) (2x+1) and (x+4)

    5)  f:\mathbb{N}^3 \rightarrow \mathbb{N} where f(a,b,c) = a(b+1)(c+2) [/latex]

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

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

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

    9) \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.
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    (Original post by miml)
    \sum{a_n} \rightarrow a \implies a_n \rightarrow 0 . But not the other way round, hence the counterexample.

    http://en.wikipedia.org/wiki/Harmoni...s_(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.
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    (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.
    The reason it follows from the previous lemma is that b_n \geq a_n eventually so b_n - a_n \geq 0 eventually therefore b-a \geq 0 (as (b_n - a_n) \rightarrow b - a)
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    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) become (135)(2467)? :confused:
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    (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) become (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)
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    (Original post by matt2k8)
    Analysis I, 2009, Q1 - missed out f though as I'm not sure what to do? I can prove \frac{1}{|a_n|}\rightarrow \infty but not what the question says :\
    Spoiler:
    Show

    a. a_n = (1 + (-1)^n)n is not increasing as (a_n) = (1,4,3,...) but tends to infinity as a_n \geq n \rightarrow \infty
    b. none of these
    c. Let (a_n) be a sequence that is not bounded above.
    Note that this means given any real number C and any N\in\mathbb{N}, we can find n with n&gt;N such that a_n &gt; C. (*)
    WTS it is possible to construct an increasing subsequence a_{n_k}.
    First choose n_1 = 1. Due to property (*), we can choose n_2 with n_2 &gt; 1 so that a_{n_2} \geq a_1.
    Then due to property (*) we can continue this process inductively, choosing n_{k+1} at each stage such that n_{k+1} &gt;n_k and a_{n_{k+1}} \geq a_{n_k} to create the increasing subsequence.
    d. Because l>0 and (a_n) \rightarrow l, we know that \exists N \in \mathbb{N} st when n>N,
    |a_n - l|&lt;l
    so then a_n &gt; (-l) + l = 0, i.e. a_n &gt; 0 when n > N
    e. false - (\frac{1}{n})\rightarrow 0 but \frac{1}{n} \forall n &gt; 0
    g. irrational. suppose we can write a/b = p/q where p,q are integers with q nonzero. then a = bp/q which is a contradiction as a is irrational.
    h. 5. (sandwhich rule + product rule and using that 0&lt;2^n,3^n\leq 5^n)
    i. -1. (this is the inteval (-1,1))
    j. -n is bounded above by -1, but all subsequences tend to minus infinity
    k. there cannot, as all cauchy sequences in \mathbb{R} converge in \mathbb{R}
    l. it is always convergent
    m. is convergent. (can prove it's absolutely convergent by comparing modulus of it with 1/10^n)
    n. not convergent. (general term does not tend to zero)
    o. it does. (any absolutely conergent series is convergent)
    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.
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    (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

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

    \frac{sinx}{x} \rightarrow 0 as x \rightarrow \infty, you are thinking of that \frac{sinx}{x} \rightarrow 1 as x \rightarrow 0.
    Of course \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?
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    is permutation tested???
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    (Original post by yeohaikal)
    is permutation tested???
    *looks this up*

    Appaz 9.9 and 9.10 are not being tested

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

    Btw, you doing discrete this term?
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    Added the rest of the question 1 foundations
    (Original post by electriic_ink)
    Of course \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
 
 
 
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