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    Just thought I'd make a thread to discuss the January exam/post solutions to past papers etc.

    I'll put links to any full solutions in this original post. There's already a few past paper solutions on this forum at http://www.thestudentroom.co.uk/show....php?t=1126241 as well.
    Full solutions-
    ANALYSIS I:
    2009 Q1 by matt2k8
    True/False Questions from MORSE Society book - by matt2k8

    FOUNDATIONS:
    2010 Q1 by matt2k8
    2006, 2007, 2008, 2009 Q1 by miml
    "Question 1" Questions from MORSE Society book - by matt2k8
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    Have any of you guys seen the e-mail from the MORSE society?
    Pretty damned useful Christmas present they've given us there!
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    (Original post by placenta medicae talpae)
    Have any of you guys seen the e-mail from the MORSE society?
    Pretty damned useful Christmas present they've given us there!
    What's the present? :hat:

    /subbed
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    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>N such that a_n > 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 > 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} >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|<l
    so then a_n > (-l) + l = 0, i.e. a_n > 0 when n > N
    e. false - (\frac{1}{n})\rightarrow 0 but \frac{1}{n} \forall n > 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<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)
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    I'm not 100% sure about h, but the rest seems OK to me?
    Foundations, 2010, Q1:
    Spoiler:
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    a. When n = 1, LHS = 1, RHS = 1^2 = 1 therefore true for n =1.
    Now assume true for n = k that 1 + 3 + 5 + ... + (2k-1) = k^2
    then 1 + 3 + 5 + ... + (2k-1) + (2(k+1) -1) = k^2 + 2k + 1 = (k+1)^2
    Therefore, if true for n =k , true for n = k+1. true for n = 1 so true for all

    natural numbers n by induction.
    b. Every natural number greater than one can be expressed as a product of prime

    numbers, unique up to the order of the terms.
    c. h = 3, a = -4, b = 13
    d. If P \in \mathbb{R}[x], then the remainder on division of P by

    (x-a) is equal to P(a)
    e. (x+1) only
    f. Suppose x is even so x = 2w for some integer w. then y^2 = 6 + 8w^3 = 2



(3+4w^3).
    So then 2|y^2, so 2|y as 2 is prime. So y = 2u for

    some integer u.
    so then 4u^2 = 6 + 8w^3, so 2u^2 = 3 + 4w^3
    But this is a contradiction as the LHS is even but the RHS is odd. So x must be odd.
    g. Only true if a is true, b is false, so equivalent to i. (dunno how to put a truth

    table on here)
    h. True. as g is a surjection A->B, it has a right inverse g^-1 which is an

    injection B->A. So by the Schroeder-Bernstein Theorem, there exists a bijection

    h:A->B.
    i. if p,m \in \mathbb{N}, p is prime and p does not divide m,
    m^{p-1} mod p = 1
    j. 10^{12} mod 11 = 10^{10}.10^2 mod 11
     = 10^2 mod 11 by FLT
     = 100 mod 11 = 1 as 100 = 11*9 + 1
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    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
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    any idea to the solution for 2007 paper foundations 1(i)(i)??
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    (Original post by yeohaikal)
    any idea to the solution for 2007 paper foundations 1(i)(i)??
    I've posted some solutions above, let me know though if you don't agree with the answer though.
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    Really quick analysis question...

    is there a sequence a_n for which |a_n| doesn't converge but a_n does?

    I don't think so...??

    EDIT: Also, is the following a correct definition of a_n is not null
    \displaystyle \exists \epsilon > 0 s.t \forall N \in \mathbb{N}, \exists n > N s.t |a_n|> \epsilon . I don't think so, because it looks like a_n = 1/n might satisfy this definition, but I'm not sure.
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    (Original post by miml)
    Really quick analysis question...

    is there a sequence a_n for which |a_n| doesn't converge but a_n does?

    I don't think so...??

    EDIT: Also, is the following a correct definition of a_n is not null
    \displaystyle \exists \episilon > 0 s.t \forall N \in \mathbb{N}, \exists n > N s.t |a_n|> \episilon . I don't think so, because it looks like a_n = 1/n might satisfy this definition, but I'm not sure.
    I think you're right.

    Time to practice some latex.

    Suppose a_n \rightarrow a

    \forall \epsilon>0, \exists N \in N such that  |a_n-a|<\epsilon ,\forall n>N

    Then by the reverse triangle inequality ||a_n| - |a||< \epsilon, \forall n>N

    Hence  |a_n| \rightarrow |a|

    Is this right?
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    (Original post by PhyMath)
    I think you're right.

    Time to practice some latex.

    Suppose a_n \rightarrow a

    \forall \epsilon>0, \exists N \in N such that  |a_n-a|<\epsilon ,\forall n>N

    Then by the reverse triangle inequality ||a_n| - |a||< \epsilon, \forall n>N

    Hence  |a_n| \rightarrow |a|

    Is this right?
    Yeah, this is the way I did it. Although admittedly I was trying to prove ||a_n|-a|<|a_n-a|, which isn't quite right.
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    I've done all the questions from Foundations/Analysis now from the booklet the MORSE society gave - I think I am gonna upload a few answers to here later but it'll take absolutely ages to latex all of them :\
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    All 89 True/False questions from section 2.1.1 of the booklet from the MORSE Society:
    Spoiler:
    Show

    Q1 - False: consider a_n = (1+(-1)^n)n, then a_{2n-1} = 0 \rightarrow 0 so (a_n) has a convergent subsequence, but a_{2n} = 4n \rightarrow \infty so (a_n) is unbounded
    Q2: False, if a_n = \frac{-1}{n} then (a_n) is strictly increasing, but tends to 0 so cannot tend to infinity.
    Q3: False, as if (a_n) \rightarrow a, (|a_n|) \rightarrow |a| (shown in post above this one, cba to type it out :P)
    Q4: True
    Q5: Not sure on this one, can't think of a counter example but the last inequality shouldn't be strict I think?
    Q6: False: consider a_n = \frac{1+(-1)^n}{n}
    Q7: True
    Q8: True
    Q9: False, consider a_n = (-1)^n
    Q10: False, the harmonic series diverges to infinity therefore (s_n) \rightarrow \infty
    Q11: False, consider a_n = 1/n^2, b_n = 1/n^3
    Q12: False, consider a_n = -1/n, b_n = 1/n^2
    Q13: True
    Q14: False, consider a_n = (-1)^n
    Q15: True
    Q16: False, a sequence is Cauchy iff it is convergent
    Q17: True
    Q18: True
    Q19: False, consider a_n = (-1)^n
    Q20: False, can show that series tends to 5 by the sandwhich rule, so it cannot tend to 3 due to uniqueness of limits.
    Q21: False, can show it tends to 5 by the sandwhich rule, so cannot tend to 1 due to uniqueness of limits.
    Q22: False: consider a_n = \frac{1+(-1)^n}{n}
    Q23: False, consider the case when x = y
    Q24: True
    Q25: True
    Q26: False, consider a_n = b_n = 1/n
    Q27: True
    Q28: True
    Q29: True
    Q30: False, as the harmonic series does not converge
    Q31: True
    Q32: True
    Q33: False if (a_n) is chosen so the series is absolutely convergent
    Q34: False, consider a_n = n -1
    Q35: False, consider a_n = -1
    Q36: True
    Q37: True
    Q38: False, consider a_n = -n
    Q39: False, consider a_n = (-1)^nn
    Q40: False as it tends to 1
    Q41: False, consider a_n = 1 - 1/n
    Q42: True
    Q43: True
    Q44: True
    Q45: False, consider a_n = (-1)^n with a = 1
    Q46: True
    Q47: False, consider a_n = \sqrt{n}
    Q48: True
    Q49: True
    Q50: False, consider A = \{x \in \mathbb{R} : 2 \leq x \leq 3\}
    Q51: True
    Q52: False, consider 0
    Q53: False, consider any conditionally convergent series
    Q54: False, consider the harmonic series
    Q55: True
    Q56: True
    Q57: False, consider a_n = 1/n
    Q58: True
    Q59: False, consider a_n = -1/n
    Q60: False, consider a_n = -1 -1/n
    Q61: False, consider a_n = 2 + (-1)^n
    Q62: False, consider a_n = (-1)^n
    Q63: False, consider a_n = (1 + (-1)^n)n
    Q64: True
    Q65: True
    Q66: True
    Q67: False, e.g. a_n = 10 + 1/n
    Q68: True
    Q69: False, consider a_n = \frac{(-1)^n}{\sqrt{n}}
    Q70: True
    Q71: True
    Q72: True (but not on syllabus anymore)
    Q73: True
    Q74: False, all Cauchy sequences of real numbers converge to a real number.
    Q75: True
    Q76: True
    Q77: False (ratio test for series, or just show (a_n) \not\rightarrow 0 using ratio lemma)
    Q78: True
    Q79: True
    Q80: True
    Q81: True
    Q82: False, consider a_n = 1/n
    Q83: False, consider a_n = (0.5)^n
    Q84: True
    Q85: False, consider a_n = 1/n
    Q86: False (Integral test, or comparison with 1/n)
    Q87: False, consider a_n = (-1)^{n+1}
    Q88: False, consider the set A = \{ 3 - \frac{1}{n} : n \in \mathbb{N} \}
    then  a &lt; 3 \forall a \in A
    but sup A = 3
    Q89: True
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    (Original post by matt2k8)
    it'll take absolutely ages to latex all of them :\
    Lol, be grateful you don't have to do the ST113 programming mathematica/LaTeX assignment this Christmas! :P
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    (Original post by placenta medicae talpae)
    Lol, be grateful you don't have to do the ST113 programming mathematica/LaTeX assignment this Christmas! :P
    Be grateful I'm not a stats student or that I don't have to do loads of LaTeX'ing?
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    Be grateful you're not latexing the booklet? :P
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    (Original post by Narev)
    Be grateful you're not latexing the booklet? :P
    Good point
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    2.72 is false. let a(n)=1 for n even and a(n)=-1 for n odd, then lim inf(a) is not eq to lim sup(a) but it's sum is conditional convergent.
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    (Original post by yeohaikal)
    2.72 is false. let a(n)=1 for n even and a(n)=-1 for n odd, then lim inf(a) is not eq to lim sup(a) but it's sum is conditional convergent.
    In your example the sequence of partial sums just oscillated between -1 and 0 so the series doesn't converge - I said it was true as it's required that (a_n) \rightarrow 0 for the series \Sigma a_n to converge, but if  lim \sup \ a_n \not= \lim \inf \ a_n then (a_n) does not converge to any limit
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    Q5 is true.. by giving a strict inequality, it just gives a stronger (sufficient) definition of not null, though definitely not necessary. is that right?
 
 
 
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