The Proof is Trivial!

OP

Given that in the second limit you're going to take the limit to infinity anyway, how about in the first limit you...

(The limits have to be this way around, because otherwise it causes issues when you have irrational x)

I mentioned that I took m->∞ when I was finding the first limit; is that not enough? Or why can't we just write m=n?

Reply 221

shamika

16

Original post by und

I mentioned that I took m->∞ when I was finding the first limit; is that not enough? Or why can't we just write m=n?

Ordinarily, you would care what m is doing in the first (inner) limit, because as you've pointed out, the result you've stated only holds for

m \geq q

. However, this isn't a problem overall, since when you look at the second limit...

Reply 222

Mladenov

1

Problem 44***

Let

P(n)

be a polynomial with coefficients in

\mathbb{Z}

. Suppose that

\deg(P)=p

, where

p

is a prime number. Suppose also that

P(n)

is irreducible over

\mathbb{Z}

. Then there exists a prime number

q

such that

q

does not divide

P(n)

for any integer

n

.

Problem 45*

Let

p

be a prime number,

p> 3

. Given that the equation

p^{k}+p^{l}+p^{m}=n^{2}

has an integer solution, then

p \equiv -1 \pmod 8

.

(edited 11 years ago)

Reply 223

und

OP

16

Original post by shamika

Ordinarily, you would care what m is doing in the first (inner) limit, because as you've pointed out, the result you've stated only holds for

m \geq q

. However, this isn't a problem overall, since when you look at the second limit...

Can we make m arbitrarily large in the first place?

Reply 224

shamika

16

Original post by und

Can we make m arbitrarily large in the first place?

Yes. I'm trying to hint strongly as to why but clearly I'm being useless at hinting :tongue:

Reply 225

Felix Felicis

13

Original post by DJMayes

...

Original post by bananarama2

...

Original post by joostan

...

Yeah, as has been said, the problem with that method is the partial fraction decomposition, it's not an identity

Reply 226

und

OP

16

Original post by shamika

Yes. I'm trying to hint strongly as to why but clearly I'm being useless at hinting :tongue:

The only thing that I see in that limit is m->∞, but I know nothing about repeated limits so I don't know what the conditions are for using this fact in the first limit.

Reply 227

shamika

16

Original post by und

The only thing that I see in that limit is m->∞, but I know nothing about repeated limits so I don't know what the conditions are for using this fact in the first limit.

In this case, assume you can use that fact :smile:

I think the thing that's confusing you is that you might be thinking that one limit is applied first and then the other. They're both being applied at the same time, that is, the following doesn't happen:

1) fix an m
2) calculate the inner limit
3) then calculate the outer limit.

Think of it as both occurring simultaneously. To be honest, I was envisaging the difficulty of trying to swap the limits around, but in the case x is rational, it's ok to do so (at least, the result is the same either way)

However, I think you've sufficiently understood to be able to credit yourself with the answer. Let me know if you want me to write something up, but it'll essentially be what I've already tried to explain :tongue:

EDIT: this happened to be on a tripos paper - and Gowers also has a crack at explaining it (see the comments in: http://gowers.wordpress.com/2012/04/28/a-look-at-a-few-tripos-questions-ii/#comments

(edited 11 years ago)

Reply 228

Noble.

17

Original post by Felix Felicis

Problem 42*/**

Show that

\displaystyle\sum_{r=0}^{88} \dfrac{1}{\cos r \cdot \cos (r+1)} = \dfrac{\cos 1}{\sin^{2} 1}

(angles measured in degrees)

Solution 42

\displaystyle\sum_{r=0}^{88} \dfrac{1}{\cos r \cdot \cos (r+1)} = \dfrac{1}{\sin(1)} \displaystyle\sum_{r=0}^{88} \dfrac{\sin(1)}{\cos r \cdot \cos (r+1)}

= \dfrac{1}{\sin(1)} \displaystyle\sum_{r=0}^{88} \dfrac{\sin(1)\cdot \cos(0) - \sin(0) \cdot \cos(1)}{\cos r \cdot \cos (r+1)}

= \dfrac{1}{\sin(1)} \displaystyle\sum_{r=1}^{89} \tan(r) - \dfrac{1}{\sin(1)} \displaystyle\sum_{r=0}^{88} \tan(r)

= \dfrac{\tan(89)}{\sin(1)} = \dfrac{\cot(1)}{\sin(1)} = \dfrac{\cos(1)}{\sin^2(1)}

(edited 11 years ago)

Reply 229

Felix Felicis

13

Original post by Noble.

...

Awesome

Reply 230

ukdragon37

14

Original post by bananarama2

I've got it. It you plot lateral binary graph associated with the symbolic mathematical solutions, you can reduced the problem down to simple SNF's. And by using Dalek's contraction algorithm the answer is: the physical quantity of dimensions m to the ml to the lq to the qt
to the ttheta to the x has planck quantity root g minus m plus l minus 2q
plus t hbar to the m plus l minus q plus t minus xe to the minus m minus l minus 7q minus 3t plus 6xk to the minus 2x epsilon 0 to the q.

You know the answer to a set theory question is wrong when it contains the word "planck" :pierre:

Reply 231

bananarama2

Original post by ukdragon37

You know the answer to a set theory question is wrong when it contains the word "planck" :pierre:

tehee

(edited 11 years ago)

Reply 232

username937760

16

Solution 43:

Firstly, we have to apply the limit as n becomes arbitrarily large. As the power increases, there are one of three possibilities:

-1<cosx<1

, in which case it will approach zero.

cosx = 1

, in which case it will remain at one.

cosx = -1

, in which case it will oscillate between -1 and 1.

Next, look at the different possibilities of m with regards to x. If x is rational, then sufficiently large m will contain the denominator of x and a multiple of 2, making x a multiple of

2\pi

so that the overall thing will reach a limit of 1. However, if x is irrational then it is impossible for it to become an integer multiple of

\pi

. Because of this, cosx will approach zero as n tends towards infinity as the magnitude of cosx will be less than one.

Reply 233

Llewellyn

14

My solution to #42 is absolutely horrendous :lol:

even the latex code hates it.

Original post by bananarama2

..And by using Dalek's contraction algorithm the answer is:

Reply 234

shamika

16

Original post by DJMayes

Solution 43:

Firstly, we have to apply the limit as n becomes arbitrarily large. As the power increases, there are one of three possibilities:

-1<cosx<1

, in which case it will approach zero.

cosx = 1

, in which case it will remain at one.

cosx = -1

, in which case it will oscillate between -1 and 1.

Next, look at the different possibilities of m with regards to x. If x is rational, then sufficiently large m will contain the denominator of x and a multiple of 2, making x a multiple of