Warwick First Year Maths-Past paper questions

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*

Reply 43

OP

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...

Reply 44

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...

Reply 45

OP

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...

Reply 46

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?

Reply 47

JohnnySPal

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 :rolleyes:

). 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.

Reply 48

OP

Ah yeah that could be a problem. I wish I never asked this damn question...worst 2 marks ever :shifty:

Reply 49

Kolya

17

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.

See this: http://www.thestudentroom.co.uk/showthread.php?p=10493413&highlight=bounded+sequence#post10493413

Reply 50

OP

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.

Reply 51

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

Reply 52

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 :smile:

v, vi, x
i think for xii you could give your wrong counterexample you gave abother (it's right anyways)

Reply 53

amo17

thanks for the quick reply.

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

Reply 54

amo17

thanks for the quick reply.

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

noooes. think about it.

Reply 55

amo17