Competition - post a photo you've taken of nature >>

A Summer of Maths (ASoM) 2016

Scroll to see replies

What I meant is that thats what I would like to do but the cycles aren't disjoint in the first place so all I could see is checking each element separatly (even though there were only 4 elements so it wasnt too bad) :tongue:

Ah, yes. A case by case check is really the best way forward.

Posted from TSR Mobile

Reply 261

wlog54

Original post by Jellymath

Does anyone have the link for the example sheet that cambridge send as summer work? I remember that it had a question on proving the irrationality of pi and the sum of inverse squares.

I think what you are looking for is the Introductory sheet A0 in this http://www.damtp.cam.ac.uk/user/examples/

Reply 262

Zacken

Original post by Jellymath

Does anyone have the link for the example sheet that cambridge send as summer work? I remember that it had a question on proving the irrationality of pi and the sum of inverse squares.

It is not summer work, rather it is used in Jesus as an example sheet before they cover enough content in lectures for them to be able to attempt a proper example sheet and have a proper supervision. It can be found here.

Reply 263

tridianprime

Original post by Jellymath

Does anyone have the link for the example sheet that cambridge send as summer work? I remember that it had a question on proving the irrationality of pi and the sum of inverse squares.

I think (not certain) you are right that Cambridge do send out some work to do after offers are confirmed, though I am not certain if you can find it anywhere online easily. It is probably better to wait to receive it anyway.

At least, I have heard of it before.

Similarly, this website has some "Introductory Example Sheets" designed for the very beginning of the course - kind of like a zeroth example sheet. I just did the N&S one and isn't really difficult at all, but the Groups one may be better (I haven't looked).

Reply 264

username1162972

There is a booklet but tis really easy.

Posted from TSR Mobile

Reply 265

EricPiphany

The worksheet referred to above is here: https://www.maths.cam.ac.uk/sites/www.maths.cam.ac.uk/files/pre2014/undergrad/admissions/workbook.pdf

Spoiler

Original post by EricPiphany

Spoiler

That moment when uve been looking at the signs of permutations for the past 2 hours and go on tsr to have a break...

Reply 268

EricPiphany

Original post by EnglishMuon

That moment when uve been looking at the signs of permutations for the past 2 hours and go on tsr to have a break...

My whole life is a break.

Reply 269

EnglishMuon

Original post by EricPiphany

My whole life is a break.

pls no moar XD

Reply 270

drandy76

Original post by EnglishMuon

pls no moar XD

ImageUploadedByStudent Room1467400465.687095.jpg

Posted from TSR Mobile

Reply 271

Imperion

Original post by EnglishMuon

lol u know wat i mean :tongue:

Muons Theorem: Every number = 0

Spoiler

Reply 272

EnglishMuon

Original post by Imperion

Spoiler

hahaha who knows could be a typo :wink:

(edited 7 years ago)

Reply 273

jkhan9625

👍

Posted from TSR Mobile

Reply 274

Oxmath

Guys after reading this thread, I went and got Algebra and Geometry by Beardon. Its been interesting so far (although I've just begun chapter 2).I was reading the page on "The Principle of Induction II", and I didn't understand something.When he says: "If not, then we can write

m+1=ab

and so (obviously)

P(m+1)=P(a)P(b)

".Shouldn't that be

P(m+1)=P(a)+P(b)

?Edit: How the **** do I put paragraphs in?

(edited 7 years ago)

Reply 275

EricPiphany

Original post by Oxmath

m+1=ab

and so (obviously)

P(m+1)=P(a)P(b)

".Shouldn't that be

P(m+1)=P(a)+P(b)

?Edit: How the **** do I put paragraphs in?

I don't think the alternative you gave is true either, consider m = 3. Also it's used again later in the proof.
Removing the restriction the the factors counted by P(n) have to be prime seems to solve the problem.

Reply 276

Zacken

Original post by Oxmath

m+1=ab

and so (obviously)

P(m+1)=P(a)P(b)

".Shouldn't that be

P(m+1)=P(a)+P(b)

?Edit: How the **** do I put paragraphs in?

I agree with you,

\mathcal{P}

(assuming its counts multiplicity) should be a completely additive arithmetic function.

Anywho, surely the example given is completely asinine. It folds quickly to contradiction. i.e: assume that the number of prime factors is

\geq n

, then since every prime factor is

\geq 2

we get a number

\geq 2^n

which is easily shown to be bigger than

n

for all

n \geq 2

(via induction, if you wish) - so contradiction, hence

\mathcal{P}(n) < n

(edited 7 years ago)

Reply 277

EnglishMuon

For Q 4 (ii) page 15 of the Beardon book is this supposed to be an obvious solution? From i I just showed any cycle

( a_{1} a_{2}...a_{m}) = (1\ a_{1})(1\ a_{m})(1\ a_{m-1}) ... (1\ a_{1})

so should I be reapplying the identity

(1\ a)(1\ b)(1\ a) = (ab)

to this expression to get terms of the form

(1\ a-1)(1\ a)(1\ a-1)

but this seems fiddly.

Reply 278

Zacken

Original post by EnglishMuon

For Q 4 (ii) page 15 of the Beardon book is this supposed to be an obvious solution? From i I just showed any cycle

( a_{1} a_{2}...a_{m}) = (1\ a_{1})(1\ a_{m})(1\ a_{m-1}) ... (1\ a_{1})

so should I be reapplying the identity

(1\ a)(1\ b)(1\ a) = (ab)

to this expression to get terms of the form

(1\ a-1)(1\ a)(1\ a-1)

but this seems fiddly.

It is meant to be obvious (by which I mean short, it wasn't obvious to me the first time around :tongue:

):

Every permutation in

S_n

can be written as the product of disjoint cycles. Every cycle of length of at least 2 can be written as a product of transpositions. Every transposition (a b) can be written as the product of 3 transpositions of the form (1 a) (1 b) (1 a), so any element in