University of Warwick

Coventry

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University of Warwick

Coventry

Certainly not for this.

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.

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

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

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.

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?

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?

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.

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

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.

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)?

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.

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. thanks

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.

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?

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.

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).

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).

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.

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.

hiddengem have a look at e & m again me thinks.

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