Warwick First Year Maths-Past paper questions

Certainly not for this.

Reply 102

silent ninja

19

4(a) 2007 answered by irrational above, do you state the Absolute version of the ratio test or the other one which states for a_n>0 ? Also in your definition you forgot to mention a_n can't be zero.

Reply 103

JohnnySPal

2

amo17

thanks ok, i will take that advice. but seriously out of interest i want to know how to do the question. is it just a rearrangement?

thanks

Just do some simple jiggling with the fraction. It's equivalent to finding an N such that:

|10/n| < 1/10

and because you're considering natural numbers N you can even remove the abslute value signs.

Okay, I'm sure the question I faced was a bit harder than this. Consider my last post to be a stupid reply :p:

Reply 104

amo17

2

so the answers 100, not 1000 lol, do not ask me how i got that!

thanks!

Reply 105

IrrationalNumber

silent ninja

4(a) 2007 answered by irrational above, do you state the Absolute version of the ratio test or the other one which states for a_n>0 ? Also in your definition you forgot to mention a_n can't be zero.

I was using the absolute version and yes you're quite right, a_n zero means the ratio doesn't exist which sort of kills the test.

Reply 106

OP

Just so you all know (because I didn't till just now) the Foundations 2005 answers are on MathStuff.
One of the parts to question 1 had something about partitions of a set, is this still on the syllabus because I've not seen it in the notes?

Reply 107

Swayum

Can someone please link me to/upload the first year analysis notes and/or question sheets? Someone posted them around July but the links don't work anymore.

Reply 108

OP

Swayum

Can someone please link me to/upload the first year analysis notes and/or question sheets? Someone posted them around July but the links don't work anymore.

Most the work was hosted on the lecturers sites which are down at the moment for unknown reasons. And there is some stuff on mathstuff but you need to be a student to view it.

Reply 109

Lili!

1

on the foundations jan 04 paper question 4c..uhh i don't know how to do it. if a disjoint cycle is s=(137)(2456), how do you write s^-1 and s^2 etc? i've seen the answers but am confused as to how they got it! doh. merci beaucoup :smile:

Reply 110

OP

Lili!

on the foundations jan 04 paper question 4c..uhh i don't know how to do it. if a disjoint cycle is s=(137)(2456), how do you write s^-1 and s^2 etc? i've seen the answers but am confused as to how they got it! doh. merci beaucoup :smile:

s^-1 is the inverse. In that s do you know that it defines a function that sends 1 to 3, 3 to 7 and 7 to 1? So the inverse would send 3 to 1, 1 to 7, and 7 to 3. You can do a similar thing to do (2456) aswell.
There's also an easy way to do it, just write the cycles backwards, it's the same thing.
For s^2, this is the same as the composition ss, do you know how to do composition of permutations? If not it's worth checking the lecture notes but the idea is not too difficult if you think of each two numbers having a function between them.

Reply 111

Totally Tom

what does the identity of S_n actually look like? does it just map 1-1 2-2 3-3 4-4 5-5 so it's basically (empty)?

Reply 112

Lili!

1

benwellsday

s^-1 is the inverse. In that s do you know that it defines a function that sends 1 to 3, 3 to 7 and 7 to 1? So the inverse would send 3 to 1, 1 to 7, and 7 to 3. You can do a similar thing to do (2456) aswell.
There's also an easy way to do it, just write the cycles backwards, it's the same thing.
For s^2, this is the same as the composition ss, do you know how to do composition of permutations? If not it's worth checking the lecture notes but the idea is not too difficult if you think of each two numbers having a function between them.

ahhh awesome. :smile:

thanks

Reply 113

Totally Tom

benwellsday

s^-1 is the inverse. In that s do you know that it defines a function that sends 1 to 3, 3 to 7 and 7 to 1? So the inverse would send 3 to 1, 1 to 7, and 7 to 3. You can do a similar thing to do (2456) aswell.
There's also an easy way to do it, just write the cycles backwards, it's the same thing.
For s^2, this is the same as the composition ss, do you know how to do composition of permutations? If not it's worth checking the lecture notes but the idea is not too difficult if you think of each two numbers having a function between them.

is s^n simply not s? i mean (132) no matter how many times you apply it, it'll always be the same pattern won't it?

Reply 114

OP

Totally Tom

is s^n simply not s? i mean (132) no matter how many times you apply it, it'll always be the same pattern won't it?

I don't think so. For example s^2 in the case of s = (132) is the composition of s and s, so first 1 is sent to 3 by s, then 3 is sent to 2 by s. Similarly 2 goes to 1 then to 3, and 3 goes to 2 then to 1, so s^2 = (123). s^3 is the same as s though, so it repeats from there.

Reply 115

estel

Oh no. The avatar >.<

Reply 116

OP

Question time, analysis 2005, 4c)
Estimate the number of terms required to approximate pi within 10^-10 for each formula below

\pi = 4 \sum ^{\infty}_{k=0} \frac{(-1)^k}{2k+1}

and

\pi = \sum ^{\infty}_{k=0} \frac{1}{16^k} (\frac{4}{8k+1} - \frac{2}{8k+4} - \frac{1}{8k+5} -\frac{1}{8k+6})

with the hint that the bracketed term is always between 0 and 4 (which need not be proved).

Reply 117

hiddengem

alright guys, this is my answers to 2008 analysis question 1. i don't think anyone posted this before.
could someone compare answers
cheers

a. true

b. false a(n)=(-1/n)

c. false a(n)=(-1-1/n) a=-1

d. false a(n)=((-1)^n)

e. false a(n)=(((-1)^n)n)

f. true

g. true

h. true

i. limsup question (i think won't appear this year)

j. true

k. false a(n) = (((-1)^(n+1))/n)

l. true

m. false a(n) = 1, 0.5, 3, 4, 5, 6......

n. another limsup question

o. true

thanks

could someone confirm whether im right in thinking limsup is off the test. i'm sure the class teacher said so but lol just to make sure.

Reply 118

estel

Yes, limsup is back off of A1 this year.

Reply 119

Totally Tom