The Proof is Trivial!

Problem 215**

Prove that

\displaystyle \int_0^{\infty} \int_0^x \frac{\sin(t)}{t} \ dt \ dx

does not converge.

Problem 216*/**

Find a series solution for

\displaystyle \int_0^1 x^{-x} \ dx

(it's fair to say this is going to blow people's minds!)

Original post by MW24595

Lol, thanks.
Ha, I suppose so. xD

Nice, man. Sucks? :tongue:

Why?
So, you guys have Pure Math, Statistics, Mechanics and Decision Math?

Well, it's a strange story. I actually did apply for last year, but due to some passport goof up, I missed the interview last January. Somehow, they sent me an offer anyway, but deferred to 2013. And so, here's where I find myself.

Well the decision maths syllabus is horrendous! The introduce you to beautiful problems like that of the travelling salesman and things like Game Theory and the ruin them by emphasising memorisation of algorithms instead of any of the maths :'( Exams are basically an exercise is how well you can emulate a computer! :/

That's right bud! 7 Pure (****ing only...), 5 Mechanics, 4 Statistics and 2 Decision. This is Edexcel btw but represents the typical setup :tongue:

What's yours like? Didn't you do A-Levels?

What on earth! Firstly, why did they give you an offer anyway without any interview? What things had you done to put in your application that impressed them? Secondly, if they wanted you anyway, why didn't they just give you an offer for the same year? Perhaps they thought you were too young? Thirdly,
was the original offer unconditional or based on STEP I and II? :smile:

(edited 10 years ago)

Reply 1383

I have to be honest guys, I think these problems are wrongly titled in terms of difficulty, I'm about to finish my A-level in maths and further maths and could not even attempt any of these, even those that require 'Only a-level knowledge'.

I also see that the majority answering here are either undergrads or individuals who are preparing for STEP.

Reply 1384

Lord of the Flies

19

Solution 215

Upon seeing the denominator, the obvious step is to differentiate, which gives:

\displaystyle\begin{aligned} \sum_{r=0}^{\infty} \alpha^{2(r-1)}\left(\frac{(-1)^r r}{(2(r+1))!}\right) &=\sum_{r=0}^{\infty} \alpha^{2(r-1)}\left(\frac{(-1)^r}{2(2r+1)!}-\frac{(-1)^r }{(2(r+1))!}\right)\\&=\frac{1}{2\alpha^4}\Big(\alpha \sin \alpha+2\cos \alpha-2\Big)\end{aligned}

Hence the value we seek is the negative half of 212, or in other words

\displaystyle \lim_{\alpha \to \infty} \left( \sum_{r=0}^{\infty} \frac{(-1)^r r\alpha^{2r-1}}{(2r-1)(2(r+1))!} \right)=-\frac{\pi}{24}

Solution 216

\displaystyle \int_0^{\infty} \int_0^x \frac{\sin t}{t}\,dt\,dx\overset{\text{IBP}}=\lim_{x\to \infty}\left(\frac{\pi x}{2}+\cos x-1\right)

Solution 217

\displaystyle \int_0^1 x^{-x}\,dx=\sum_{k\geq 0}\int_0^1 \frac{(-x\ln x)^k}{k!}\,dx\overset{\text{RF}}=\sum_{k\geq 0} \frac{1}{k!}\left( \frac{k!}{(k+1)^{k+1}}\right)= \sum_{k\geq 1} k^{-k}

Looks like the past questions have all been about how well you can do IBP. :biggrin:

Original post by miketree

I have to be honest guys, I think these problems are wrongly titled in terms of difficulty

They're not titled in terms of difficulty.

(edited 10 years ago)

Reply 1385

Original post by miketree

I have to be honest guys, I think these problems are wrongly titled in terms of difficulty, I'm about to finish my A-level in maths and further maths and could not even attempt any of these, even those that require 'Only a-level knowledge'.

I also see that the majority answering here are either undergrads or individuals who are preparing for STEP.

I'd say that a good chunk of these problems are roughly between BMO1 and BMO2 level of difficulty (the non-insane ones at least) which correspond to questions being for the 0.1 -0.001 percentile typically :colone:

(perhaps something like 0.5-0.005 of those taking A-Level maths?)

The asterisks are purely based on assumed knowledge that would be typically required (without independently inventing a new area of maths that already exists of course :lol:

) Many * problems are far far harder than *** problems!

To be fair, a lot of the problems are related to previous problems and so would seem much harder to somebody who hadn't been participating in this thread regularly! So, in that sense, it's almost as though the problems are evolving as we collectively learn new pieces of mathematics as well as develop our problem solving abilities (well... for me at least :colondollar:

) Personally, if I was only just started to post on here , I would've known where to start on half of the problems I tend to be able to do or have a go at now!

If you want to get involved, don't be put off! Whilst people certainly won't appreciate discussion of A-Level 'exercises', that doesn't mean problems of a similar difficulty cannot be discussed :smile:

Here's a fun one if you are interested:

Problem 217*

Simplify

\displaystyle \sum_{r=1}^{n} \sin(rx)

Hint:

Spoiler

(edited 10 years ago)

Reply 1386

Original post by miketree

I have to be honest guys, I think these problems are wrongly titled in terms of difficulty, I'm about to finish my A-level in maths and further maths and could not even attempt any of these, even those that require 'Only a-level knowledge'.

I also see that the majority answering here are either undergrads or individuals who are preparing for STEP.

I'm not preparing for STEP....or an undergrad.

Reply 1387

Original post by Lord of the Flies

Comment on 214

In my opinion this is not a very interesting question. The whole thing is algebra. Jkn can you replace it with something else? Otherwise, I'll post a solution later.

Sorry bro! I wrote 2r+1 instead of 2r on the denominator :lol:

I've changed it to an r now! :lol:

Solution 216

\displaystyle \int_0^1 x^{-x}\,dx=\sum_{k\geq 0}\int_0^1 \frac{(-x\ln x)^k}{k!}\,dx\overset{\text{RF}}=\sum_{k\geq 0} \frac{1}{k!}\left( \frac{k!}{(k+1)^{k+1}}\right)= \sum_{k\geq 1} k^{-k}

Impressively quick, as usual! :biggrin:

Are you as delighted by the result as I am? :colone:

It was a result found by Johann Bernoulli and is nicknames the "Sophmore's Dream" (given the fact that it represents one of those integrals that make a student think "what if integration was that straightforward" :lol:

)

Looks like the past questions have all been about how well you can do IBP. :biggrin:

Well the great skill that you are underestimating is, I believe, the ability to have an immediate insight into a question and follow what you are "supposed" to do. This is incidentally the exact same skill that STEP aims to develop! Whilst you may have spotted the use of IBP straight away, people that have not been through months (or years!) worth of having ideas that fail (and, hence, developing insight into what techniques are appropriate) will likely struggle with tho type of question :tongue:

Also, I fear I made 216 a little too straightforward because the most obvious thing to try is Maclaurins :lol:

I just thought that, if I didn't put "series solution", everyone would get pissed off with me :lol:

Original post by bananarama2

I'm not preparing for STEP....or an undergrad.

I think you going to Cambridge to study Physics doesn't exactly make you a great counterexample :lol:

(edited 10 years ago)

Reply 1388

Original post by Jkn

Problem 217*

Simplify

\displaystyle \sum_{r=1}^{n} \sin(rx)

Hint:

Spoiler

Whilst I thank you for your efforts in providing a problem, I have never come across any methods in trying to work this out.

Only summation I have learnt is those of the arithmetic and geometric series, using general sums (eg.

\displaystyle \sum_{r = 1}^{n} r^{2} = \frac{n}{6}(n + 1)(2n + 1)

) and others, and then method of differences.

The only way i'd be able to attempt this is by using a taylor approximation and then using general sums.

Once again thanks for your efforts though!

(edited 10 years ago)

Reply 1389

Original post by bananarama2

I'm not preparing for STEP....or an undergrad.

You are one of the minority, and how much extra reading have you done, along with your A-Level(s)? I am not saying you guys should stop, I'm just saying that the one star question is wrongly titled.

Reply 1390

Original post by Lord of the Flies

They're not titled in terms of difficulty.

I apologise, I used the wrong word, they are wrongly titled in terms of knowledge needed in order to attempt to comprehend the question(s).

Reply 1391

Original post by miketree

Whilst I thank you for your efforts in providing a problem, I have never come across any methods in trying to work this out.

Only summation I have learnt is those of the arithmetic and geometric series, using general sums eg.

\displaystyle \sum_{r = 1}^{n} r^{2} = \frac{n}{6}(n + 1)(2n + 1)

and others, and then method of differences.

The only way i'd be able to attempt this is by using a taylor approximation and then using general sums.

Once again thanks for your efforts though!

Well, the aim is to challenge! :biggrin:

The level of difficulty of this problem is similar to that of a STEP question (in fact, evaluating this and the equivalent result for cosines comprised the first half of a STEP question in the past!) so you should be very pleased with yourself if you can make an headway! Do not be afraid to spend over an hour on a question if necessary! :smile:

The key here is to think about what formula you can use to transform the summation into something that looks more familiar. Is there some clever trick that can be made to form a type of series you are familiar with?

The types of series you you'll have met are arithmetic series, geometric series, sums of squares, cubes, etc.. and those that can be evaluated using the method of differences :smile:

(edited 10 years ago)

Reply 1392

Original post by Jkn

...

With all due respect:

a) If you must have STEP knowledge for this question, it must be titled with ** because that does not "require only A-level knowledge" quoted from your original post.

b) I don't believe transforming a summation into an integral is in any A-level specification, apologies if I'm wrong (maybe MEI).

c) Yes, that is correct.

Reply 1393

Original post by miketree

You are one of the minority, and how much extra reading have you done, along with your A-Level(s)? I am not saying you guys should stop, I'm just saying that the one star question is wrongly titled.

I can see your view point, but the * problems can be done with a-level knowledge no matter how stretched it is :smile:

Reply 1394

Original post by bananarama2

I can see your view point, but the * problems can be done with a-level knowledge no matter how stretched it is :smile:

Quoted from OP:

* = requires only A-level knowledge.

Original post by Jkn

Problem 217*

Simplify

\displaystyle \sum_{r=1}^{n} \sin(rx)

This is not A-level knowledge.

(edited 10 years ago)

Reply 1395

Original post by miketree

With all due respect:

a) If you must have STEP knowledge for this question, it must be titled with ** because that does not "require only A-level knowledge" quoted from your original post.

b) I don't believe transforming a summation into an integral is in any A-level specification, apologies if I'm wrong (maybe MEI).

c) Yes, that is correct.

STEP knowledge is a-level knowledge.

Lots of these things can be extracted from a-level knowledge.

Reply 1396

Original post by miketree

Quoted from OP:

* = requires only A-level knowledge.

Then we don't disagree. Stretched a-level knowledge is a-level knowledge nonetheless.

Reply 1397

hollieeilloh

2

Original post by miketree

I have to be honest guys, I think these problems are wrongly titled in terms of difficulty, I'm about to finish my A-level in maths and further maths and could not even attempt any of these, even those that require 'Only a-level knowledge'.

I also see that the majority answering here are either undergrads or individuals who are preparing for STEP.

I have to disagree with you, I mean I think this thread is just a bit of fun and obviously, yeah, you have to be pretty good (understatement) at maths to do most of the problems - but hey what's the point of doing something if it's not challenging? From the looks of things, some **/*** problems are less complicated than * problems - it all just depends on what you know :smile:

I only watch this thread to have my mind blown on a regular basis, but felt the need to post. :colone:

Reply 1398

Felix Felicis

13

Original post by miketree

This is not A-level knowledge.

It can be done with A-level knowledge, surely you know de Moivre's theorem?

Reply 1399