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

matt2k8
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 16122010 13:38
Last edited by matt2k8; 04012011 at 21:53. 
placenta medicae talpae
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 17122010 01:14
Have any of you guys seen the email from the MORSE society?
Pretty damned useful Christmas present they've given us there! 
electriic_ink
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 17122010 01:32
(Original post by placenta medicae talpae)
Have any of you guys seen the email from the MORSE society?
Pretty damned useful Christmas present they've given us there!
/subbed 
matt2k8
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 17122010 12:31
Analysis I, 2009, Q1  missed out f though as I'm not sure what to do? I can prove but not what the question says :\
Spoiler:Show
a. is not increasing as but tends to infinity as
b. none of these
c. Let be a sequence that is not bounded above.
Note that this means given any real number C and any , we can find n with such that . (*)
WTS it is possible to construct an increasing subsequence .
First choose . Due to property (*), we can choose with so that .
Then due to property (*) we can continue this process inductively, choosing at each stage such that and to create the increasing subsequence.
d. Because l>0 and , we know that st when n>N,
so then , i.e. when n > N
e. false  but
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 )
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 converge in
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)Last edited by matt2k8; 17122010 at 12:54. 
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 17122010 18:08
I'm not 100% sure about h, but the rest seems OK to me?
Foundations, 2010, Q1:Spoiler:Show
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 + ... + (2k1) = k^2
then 1 + 3 + 5 + ... + (2k1) + (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 , then the remainder on division of P by
(xa) is equal to P(a)
e. (x+1) only
f. Suppose x is even so x = 2w for some integer w. then .
So then , so as 2 is prime. So y = 2u for
some integer u.
so then , so
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 SchroederBernstein Theorem, there exists a bijection
h:A>B.
i. if , p is prime and p does not divide m,
j.
by FLT
as 100 = 11*9 + 1 
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 23122010 14:05
FOUNDATIONS
January 2006January 2007
1) ab means 'a divides b' ie . If ab and bc then and
2) Every can be written as a product of prime numbers. This factorisation is unique up to order.
3)
is not a subgroup.
4) Venn diagrams
5)
6) If P(x) = (xa)Q(x) + R(x) then R is the constant polynomial P(a)
7) (x+4) and (x1)
8) Coefficient of is
Let a=1, bx = 1 in
9) where
where
where
10) There is a bijection but no bijectionJanuary 2008
Last edited by miml; 28122010 at 19:24. 
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 28122010 19:05
any idea to the solution for 2007 paper foundations 1(i)(i)??

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 28122010 19:26
(Original post by yeohaikal)
any idea to the solution for 2007 paper foundations 1(i)(i)?? 
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 28122010 19:46
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
. I don't think so, because it looks like a_n = 1/n might satisfy this definition, but I'm not sure.Last edited by miml; 28122010 at 20:06. 
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 28122010 20:16
(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
. I don't think so, because it looks like a_n = 1/n might satisfy this definition, but I'm not sure.
Time to practice some latex.
Suppose
such that
Then by the reverse triangle inequality
Hence
Is this right? 
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 28122010 20:25
(Original post by PhyMath)
I think you're right.
Time to practice some latex.
Suppose
such that
Then by the reverse triangle inequality
Hence
Is this right? 
matt2k8
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 04012011 16:58
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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 04012011 17:20
All 89 True/False questions from section 2.1.1 of the booklet from the MORSE Society:
Spoiler:Show
Q1  False: consider , then so has a convergent subsequence, but so is unbounded
Q2: False, if then is strictly increasing, but tends to 0 so cannot tend to infinity.
Q3: False, as if , (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
Q7: True
Q8: True
Q9: False, consider
Q10: False, the harmonic series diverges to infinity therefore
Q11: False, consider ,
Q12: False, consider ,
Q13: True
Q14: False, consider
Q15: True
Q16: False, a sequence is Cauchy iff it is convergent
Q17: True
Q18: True
Q19: False, consider
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
Q23: False, consider the case when x = y
Q24: True
Q25: True
Q26: False, consider
Q27: True
Q28: True
Q29: True
Q30: False, as the harmonic series does not converge
Q31: True
Q32: True
Q33: False if is chosen so the series is absolutely convergent
Q34: False, consider
Q35: False, consider
Q36: True
Q37: True
Q38: False, consider
Q39: False, consider
Q40: False as it tends to 1
Q41: False, consider
Q42: True
Q43: True
Q44: True
Q45: False, consider
Q46: True
Q47: False, consider
Q48: True
Q49: True
Q50: False, consider
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
Q58: True
Q59: False, consider
Q60: False, consider
Q61: False, consider
Q62: False, consider
Q63: False, consider
Q64: True
Q65: True
Q66: True
Q67: False, e.g.
Q68: True
Q69: False, consider
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 using ratio lemma)
Q78: True
Q79: True
Q80: True
Q81: True
Q82: False, consider
Q83: False, consider
Q84: True
Q85: False, consider
Q86: False (Integral test, or comparison with 1/n)
Q87: False, consider
Q88: False, consider the set
then
but sup A = 3
Q89: TrueLast edited by matt2k8; 05012011 at 17:54. 
placenta medicae talpae
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 04012011 17:22
(Original post by matt2k8)
it'll take absolutely ages to latex all of them :\ 
matt2k8
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 04012011 18:30
(Original post by placenta medicae talpae)
Lol, be grateful you don't have to do the ST113 programming mathematica/LaTeX assignment this Christmas! :P 
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 04012011 18:45
Be grateful you're not latexing the booklet? :P

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 04012011 18:45
(Original post by Narev)
Be grateful you're not latexing the booklet? :P 
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 04012011 19:56
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.

matt2k8
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 04012011 20:20
(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.Last edited by matt2k8; 04012011 at 20:21. 
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 04012011 20:26
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?