University of Warwick
Coventry

# Warwick First Year Maths-Past paper questions

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Yeah I was just making up what I'd do at that point because the dividng by sin(f(n)) weren't there. It made me think of trying some sort of alternating sum but seeing as the easiest one converges ((-1)^n (1/n)) I gave up. If something like that comes up I'll skip it, leave it for the end, then just write in some words that might get me 1 mark.
University of Warwick
Coventry
JohnnySPal
I was trying to think for a good 15 minutes of such a sequence. Then I realised that was the paper I sat, and that was the question only one person in our whole year got right (I think - it was either one or nobody anyway).

I'm fairly sure Dave McC told me this ridiculous sequence that satisfied the conditions, but it one you couldn't possibly come up with in an exam. Ask him if he can remember the example if you really care that much, but I wouldn't bother.

EDIT: I highly doubt sin(n) will do the trick. I'll eat my non-existent hat if the ratio of terms converges to anything at all. Bold claim there

Wow, I'm glad I'm not alone! I spent a good 15 minutes trying to find such a sequence thinking "Aaah, those first years and their piss easy questions". When I realised I couldn't find one, I left this thread feeling slightly thick and wondering why I'm thicker now than I was back when I was a fresher!

I reckon guys like Swaran and Dave found it pretty quickly...
Yeh, I also fell into the trap of sitting here for 15 minutes looking for a solution. *searches for PHD student to ask*.
Edit: a cleverer person than I suggests that it might be... errr
Er, split up[0,1] into 2 partitions going up, then 3 going down, then 4 going up, the 5 coming down etc

*shrug*
Partitions of [0,1]?
I can't explain what I think that means but it gave me an idea.
Let the sequence be something that does this
1, 0.9, 0.8, ...., 0.2, 0.1, 0, 0.05, 0.1, 0.15, ..., 0.9, 0.95, 1, 0.975, 0.95, ..., 0.05, 0.025, 0, 0.025, ..., etc
Is there an easy way to define a sequence like that? Something like, for the first 11 terms a_n = 1 - (n-1)(0.1), then for the next 20 a_n = (n-11)(0.05), and then try to generalise whats going on. Not the easiest thing to explain in an exam...
benwellsday
Partitions of [0,1]?
I can't explain what I think that means but it gave me an idea.
Let the sequence be something that does this
1, 0.9, 0.8, ...., 0.2, 0.1, 0, 0.05, 0.1, 0.15, ..., 0.9, 0.95, 1, 0.975, 0.95, ..., 0.05, 0.025, 0, 0.025, ..., etc
Is there an easy way to define a sequence like that? Something like, for the first 11 terms a_n = 1 - (n-1)(0.1), then for the next 20 a_n = (n-11)(0.05), and then try to generalise whats going on. Not the easiest thing to explain in an exam...

I was thinking something along these lines yesterday but then how would you go about showing that the limit of this sequence tends to one as you're always going to have that cross over point...
It kind of looks like the ratio tends to one, and you don't have to prove it for the question, so that's all that really matters. If you could define some formula for any term in the sequence then it'd make life a lot easier.
Also ignore the 0 in my sequence, that wouldn't go down to well when you do the ratio...
benwellsday
It kind of looks like the ratio tends to one, and you don't have to prove it for the question, so that's all that really matters. If you could define some formula for any term in the sequence then it'd make life a lot easier.
Also ignore the 0 in my sequence, that wouldn't go down to well when you do the ratio...

i just had a 'better' idea for a sequence.
starting at 2, take off 1, now add a half now add a third, now take off a quarter... i.e. when ever you get within the boundry of [1.75,2] start taking off the next term and when you hit the boundry [1,1.25] start adding the next term.

since these terms are going to be 1/n we are guaranteed that it will always hit both these targets....

what yall think?
Totally Tom
I was thinking something along these lines yesterday but then how would you go about showing that the limit of this sequence tends to one as you're always going to have that cross over point...

The example doesn't work (says the always correct JohnnySPal ). Considering the sequence $(b_n) = \left( \frac{a_{n+1}}{a_n} \right)$ I can choose an arbitrarily large N such that $b_N = 0$ - namely choose N such that $a_{N+1} = 0$. All you have to do then is choose epsilon = 1/2 in the definition of convergence.

EDIT: Alright so you've already thought about the zero issue, albeit for a different reason. It could very well work then.

EDIT to EDIT: No, the modified sequence doesn't work either. Just leave the zeros in for a second to make this explanation simpler, but remember they shouldn't really be there. Consider the two terms that immediately precede each zero. When you take the ratio of these terms you'll always get 0.5. So take epsilon = 1/4 in the definition of convergence of b_n, and bear in mind you'll be able to find an arbitrarily large N s.t. $|b_N - 1| > 1/4$.

DISCLAIMER: Because JohnnySPal wrote this, the sequence in question is almost certain to in fact be a perfect valid solution.
Ah yeah that could be a problem. I wish I never asked this damn question...worst 2 marks ever
benwellsday
I had thought of that but thought it'd be too hard to show that the ratio tends to 1. $\frac{\sin(n+1)}{\sin(n)} \to 1$. It doesn't seem to be that obvious.
Makes sense, but not sure how you'd go about showing it works.
$\frac {5+ \sin \sqrt{n+1}}{5+ \sin \sqrt{n}} \to 1$ is the problem, (it's obviously bounded and not convergent).

At a guess, $\sqrt{n+1} - \sqrt{n} \to 0 \Rightarrow -\epsilon + \sqrt{n} \leq \sqrt{n+1} \leq \epsilon + \sqrt{n}$ for all n > N, and as you choose epsilon to be small $sin {\sqrt{n} + \epsilon} \approx sin{\sqrt{n}}$. Eurgh, too many questions, I should really work on something, preferably past paper related.
hey guys, this is my solution for 2005 paper anlaysis question 1(found http://mathstuff.warwick.ac.uk/files/MA1/MA129/2005/MA129_exq_0_0_05.pdf . Could someone please compare it to their solutions.

i. false (as cauchy iff conv.) is this exp enough?

ii. true

iii. true

iv. false (tends to 5)

v. true

vi. true

vii. true

viii. true

ix. false (counterexample is an=bn=(1/n))

x. false (an= - (1/x))

xi. false (as divergent)

xii. false need c>1

any help greatly appreciated.
thanks
amo17
hey guys, this is my solution for 2005 paper anlaysis question 1(found http://mathstuff.warwick.ac.uk/files/MA1/MA129/2005/MA129_exq_0_0_05.pdf . Could someone please compare it to their solutions.

i. false (as cauchy iff conv.) is this exp enough?

ii. true

iii. true

iv. false (tends to 5)

v. true

vi. true

vii. true

viii. true

ix. false (counterexample is an=bn=(1/n))

x. false (an= - (1/x))

xi. false (as divergent)

xii. false need c>1

any help greatly appreciated.
thanks

these 3 are wrong, go check them v, vi, x
i think for xii you could give your wrong counterexample you gave abother (it's right anyways)

i can sorta see why x is wrong, but could you explain your reasoning behind v? and is vi. wrong as there isn't any of the whole natural N stuff?

thanks
amo17

i can sorta see why x is wrong, but could you explain your reasoning behind v? and is vi. wrong as there isn't any of the whole natural N stuff?

thanks

lol, is it just because x>y-100 doesn't imply x>y.

but i don't get v? is it sommat to do with the whole not for all natural numbers?
amo17
lol, is it just because x>y-100 doesn't imply x>y.

but i don't get v? is it sommat to do with the whole not for all natural numbers?

consider x=y

and consider the sequence 1,10,1,1,1,1...
amo17

xii. false need c>1

any help greatly appreciated.
thanks