You know how in textbooks these days have a set of exercises? Imagine those, except any question which asked you to prove something would just look like a statement of fact. I think it was to save space and paper which these days isn't as a big a deal(Original post by Zephyr1011)
Really? How did they used to say it? And how did they distinguish between a request to prove something and a statement of fact?

 Follow
 681
 28042013 22:42

The Polymath
 Follow
 78 followers
 16 badges
 Send a private message to The Polymath
 PS Helper
Offline16ReputationRep:PS Helper Follow
 682
 28042013 23:04
(Original post by shamika)
You know how in textbooks these days have a set of exercises? Imagine those, except any question which asked you to prove something would just look like a statement of fact. I think it was to save space and paper which these days isn't as a big a deal 
 Follow
 683
 28042013 23:15
(Original post by metaltron)
I did a sneaky check on wolfram before I posted to check whether 'a' gave three distinct solutions... it did!
One thing that bothers me about the question is that it asks to find the sum of the values, when there seems to be only one value.
Yeah I suppose we shall see! (I'll let you know what the problem turned out to be)
(Original post by Mladenov)
Nope. As the expression, is not correct, the subsequent equalities in your solution are also incorrect. And finally, your conclusion is not true. 
 Follow
 684
 29042013 00:16
(Original post by Mladenov)
Really, I do not know. After the first exercise the authors said: In subsequent propositions, we shall omit the words "Prove that" and "Show that", and thus stating all problems as theorems.
Let me suggest another lovely functional equation. It seems as though you are the only one who enjoys doing functional eqs.
Problem 99**
Find all continuous functions , defined for all , which satisfy , for all .
And I like functional equations too, I'm just nowhere near as good at them. The only widely applicable way of solving them I know is trying special values, like etc.
This problem looks a lot like product rule, although I'm not sure if that has any significance
If you take the case , then .
If , then . As both x and y are positive integers greater than one, the only possible solution is if both constants are 0. Therefore a possible solution is
This is about as far as I've gotten 
 Follow
 685
 29042013 00:16
I got 99 problems but my ***** ain't one
Spoiler:Showhahahahahahahahaha I had to!Spoiler:Show
Problem 100*/**
Rigorously prove that the area of a rectangle of side lengths and is (for real numbers ).Last edited by Jkn; 29042013 at 09:36. 
 Follow
 686
 29042013 00:32
(Original post by Jkn)
I got 99 problems but my ***** ain't oneSpoiler:Showhahahahahahahahaha I had to!Spoiler:Show
Problem 99*/**
Rigorously prove that the area of a rectangle of side lengths and is (for real numbers ).
b) Does using integration count? Or is that circular reasoning? 
 Follow
 687
 29042013 00:40
(Original post by Zephyr1011)
a) Surely it depends upon the units used?
b) Does using integration count? Or is that circular reasoning?
That would form a circular argument I do believe. Do note that in an elementary proof of this type, using calculus is a little bit mad (and one may argue that in order to be rigorous you would have to include a proof, not only of the fundamental theorem of calculus, but of this operation in a 2 dimension plane being equivalent to area when considering positive values.)
Btw, by rigorous I don't mean for it to be of a university level of rigour, but just explanative justification. 
Dirac Spinor
 Follow
 2 followers
 2 badges
 Send a private message to Dirac Spinor
Offline2ReputationRep: Follow
 688
 29042013 00:54
(Original post by Mladenov)
Problem 99**
Find all continuous functions , defined for all , which satisfy , for all .
let and let epsilon tend to zero giving:
. Continuity guarantees existence.
let y=x:
but
so f identically zero is only option.
I suspect there's a flaw as calculus is very often fickle. 
 Follow
 689
 29042013 01:05
(Original post by Jkn)
Yes. You will need to define the concept of area in your proof.
That would form a circular argument I do believe. Do note that in an elementary proof of this type, using calculus is a little bit mad (and one may argue that in order to be rigorous you would have to include a proof, not only of the fundamental theorem of calculus, but of this operation in a 2 dimension plane being equivalent to area when considering positive values.)
Btw, by rigorous I don't mean for it to be of a university level of rigour, but just explanative justification. 
 Follow
 690
 29042013 02:25
(Original post by Jkn)
What makes you think that? :P 
 Follow
 691
 29042013 06:42
(Original post by Jkn)
Oh hmm... But if you were to suppose it tended to a finite limit why is it incorrect to suppose that as ? (and then proceed in construction a contradiction)
I doubt that you can obtain a contradiction for this part.
(Original post by Zephyr1011)
If you don't know why a convention is used, why use it?
(Original post by Zephyr1011)
And I like functional equations too, I'm just nowhere near as good at them. The only widely applicable way of solving them I know is trying special values, like etc.
This problem looks a lot like product rule, although I'm not sure if that has any significance
If you take the case , then .
If , then . As both x and y are positive integers greater than one, the only possible solution is if both constants are 0. Therefore a possible solution is
This is about as far as I've gotten
This is only one possible solution and there is an infinite number of them.
(Original post by bensmith)
The best I can do at this hour of the night:
let and let epsilon tend to zero giving:
. Continuity guarantees existence.
let y=x:
but
so f identically zero is only option.
I suspect there's a flaw as calculus is very often fickle.
Your first equation may not be true for , since it is possible that is injective.
Secondly, continuity does not imply differentiability. 
 Follow
 692
 29042013 09:47
(Original post by Mladenov)
Well. If we take , from the recurrent relation, we obtain , which is .
I doubt that you can obtain a contradiction for this part.
There are hundreds of techniques when it comes to functional equations. It is annoying that there are no books on functional eqs 
 Follow
 693
 29042013 11:25
Just realised I never posted that alternate solution I thought of for problem 88. It seems way too elaborate and, of course, I did it the other way initially, but it seems creative so I'll post it for people's interest...
Solution 88 (2)
AMGM yields .
Let . Substituting yields . Noting that N is a constant we note that it is easy shown that this function has a stationary point at x=e and that this point is a maximum. Hence we deduce that is at a maximum when n is chosen such that .
This implies two possibilities. Defining [.] as the greatest integer smaller than or equal to "." (arithmetic function) we have or (as the maximum when specifying integer solutions must occur either with the integers above or below).
Assume there exists a subset of the values of a (denoted by ) such that . This implies that this particular subset (of size p) is such that which implies however this is a contradiction as it can be easily shown that the function f(p) defined such that is greater than or equal to zero for all real p (and hence the positive integers). Therefore, as no such subset can exist (for all p), the optimal value for a is and clearly the second most optimal value would be 2 (notice that we need not consider any values greater than that of the optimal, as has had been shown. Note also that we could have guessed that 3 was optimal as it is closer to e than 2).
As every positive integer can be expressed as the sum of 2s and 3s except for 1 we get the following result for P(N)...
For some positive integer m,
(I was going to leave my answer in terms of the arithmetic function (i.e. all one case) but it seemed tedious).Last edited by Jkn; 29042013 at 13:00. 
 Follow
 694
 29042013 13:55
(Original post by Jkn)
Why is it not valid to omit the limit notation if accompanied with the assumption that n is very large? (sorry if these are stupid questions)
You write either: , or: for each positive number there exists such that for all , .
This is the concept, you have as , not for all sufficiently large ; in other words, for all sufficiently large , is arbitrarily close to but it is not equal to .
(Original post by Jkn)
I've never really tried functional equations questions. For me these methods seem nonrigourous (not justifying that all solutions are found). Similar to curve sketching
Solution 100
Define be the unit square with area . Let be the area of the rectangle . Obviously, the function is completely additive, and if is isomorphic to , we have .
It is clear that every square which has side of length is isomorphic to .
Suppose is a rectangle with sides of length and , where . We partition into unit squares. Hence .
Next, let be a square with side of length , where is a positive integer. We partition the unit square into squares with side of length , which are isomorphic to . So, .
We are ready to prove the formula for rational numbers. Suppose we have with sides of length , and . We partition this rectangle into squares with side of length . Hence .
Finally, let be a rectangle with sides of length and . I suppose that both and are irrational. Let be a sequence of rational numbers such that . We define similarly. Now, let be the sequence of rectangles with side of length and . Also is the sequence of rectangles with sides of length , .
Notice . Therefore .Last edited by Mladenov; 29042013 at 14:02. 
 Follow
 695
 29042013 15:08
(Original post by Mladenov)
There are two possibilities:
You write either: , or: for each positive number there exists such that for all , .
This is the concept, you have as , not for all sufficiently large ; in other words, for all sufficiently large , is arbitrarily close to but it is not equal to .
I shall post my solution to problem 97 in order to demonstrate rigorous solution to functional equation. Although, I would say that Lord of the Flies' solution is perfectly justified.
Solution 100
Your solution is excellent to my knowledge. I posted it in the hope that people would notice that consideration must be given to irrationality and I was glad to see you had considered this in ample detail.
More problems please!
Btw, have you done your team selection test yet?

I'm gonna be a complete **** and post a problem I can't solve (and have no intention of trying to )
Problem 101**
Prove that for positive real numbers x, y and z.Spoiler:Show
Last edited by Jkn; 29042013 at 15:12. 
 Follow
 696
 29042013 16:31
Solution 90 (part 2)
We require that there are 3 distinct real roots. We must conclude, based on the fact that the polynomial is of degree 4 (i.e. cannot be reduced to a polynomial of a lower degree) and that all imaginary roots come in pairs (i.e. theory of conjugate pairing), that there are 4 real roots. From this the deduction has been made that there exists exactly one repeated root. This is correct.
What isn't correct is that we have assumed that, in the quadratic factorisation found, the repeated roots must be that of those in the same quadratic. This is incorrect! (I believe siklos addresses issues like this in two different questions in A.P.I.M.)
One way to address this is to find all other ways in which to factories the polynomial. Allow me to propose another strategy. Starting from the factorisation found in your solution, we deduce two possibilities either the roots are shared between one of the polynomials (leading to only) or one root from one is equal to one root from another. There is only one such root for which this could occur and the other roots must be equal and opposite (there are a number of strategies that could be used but we only need to find one so observation is sufficient). A keen eye will spot that both quadratics have the root x=0 when 1a=0 (i.e. a=1).
Substituting this in yields and hence, for a=1, the quadratic has roots x=0, x=3 and x=3. Therefore . QED.
Spoiler:ShowSolutions on 'brilliant' come out midweek. I'll let you know if they have anything different 
metaltron
 Follow
 4 followers
 10 badges
 Send a private message to metaltron
Offline10ReputationRep: Follow
 697
 29042013 17:20
(Original post by Jkn)
Right! I have finally found the golden nugget!
Solution 90 (part 2)
We require that there are 3 distinct real roots. We must conclude, based on the fact that the polynomial is of degree 4 (i.e. cannot be reduced to a polynomial of a lower degree) and that all imaginary roots come in pairs (i.e. theory of conjugate pairing), that there are 4 real roots. From this the deduction has been made that there exists exactly one repeated root. This is correct.
What isn't correct is that we have assumed that, in the quadratic factorisation found, the repeated roots must be that of those in the same quadratic. This is incorrect! (I believe siklos addresses issues like this in two different questions in A.P.I.M.)
One way to address this is to find all other ways in which to factories the polynomial. Allow me to propose another strategy. Starting from the factorisation found in your solution, we deduce two possibilities either the roots are shared between one of the polynomials (leading to only) or one root from one is equal to one root from another. There is only one such root for which this could occur and the other roots must be equal and opposite (there are a number of strategies that could be used but we only need to find one so observation is sufficient). A keen eye will spot that both quadratics have the root x=0 when 1a=0 (i.e. a=1).
Substituting this in yields and hence, for a=1, the quadratic has roots x=0, x=3 and x=3. Therefore . QED.
Spoiler:ShowSolutions on 'brilliant' come out midweek. I'll let you know if they have anything different 
 Follow
 698
 29042013 17:35
(Original post by metaltron)
Ok, I will add the last part to my post too. That site brilliant.org is fantastic! When you complete a whole week's solution can you have a go at the next level or do I have to wait until next week now?
Unfortunately not! I managed to complete all the number theory and geo/combinatorics last week on level 4 and so found out that way
My advice to you if you have just started on the site is to try and work out what level you want to be at and make sure you don't rush their "initial test" (take time and get some paper!) What I did was to fire through the first 3 questions on each test and then pretty much guessed the 4th on each one. This resulted in me being placed in level 3 for both olympiad sets (though working your way up is good training). I am such that you would be suited to starting on level 4 or 5 so make sure you do well on the test!
Also note that the format/style of the questions are not similar to british olympiad questions but are very similar to the US's. 
metaltron
 Follow
 4 followers
 10 badges
 Send a private message to metaltron
Offline10ReputationRep: Follow
 699
 29042013 19:44
(Original post by Jkn)
Okay awesome. I hope people besides those who post are doing/trying some of these problems. It's such an awesome problemsolution collection that has been created We've completely gone into olympiad territory now though. Would be really nice to have some more like problem 66 don't you think? I was so chuffed with that one! It's so nice you the irrationality of pi is a consequence of the convergence of the maclaurin expansion for e
Unfortunately not! I managed to complete all the number theory and geo/combinatorics last week on level 4 and so found out that way
My advice to you if you have just started on the site is to try and work out what level you want to be at and make sure you don't rush their "initial test" (take time and get some paper!) What I did was to fire through the first 3 questions on each test and then pretty much guessed the 4th on each one. This resulted in me being placed in level 3 for both olympiad sets (though working your way up is good training). I am such that you would be suited to starting on level 4 or 5 so make sure you do well on the test!
Also note that the format/style of the questions are not similar to british olympiad questions but are very similar to the US's.
May I ask what cyc on the bottom of a sigma means? 
bananarama2
 Follow
 12 followers
 0 badges
 Send a private message to bananarama2
Offline0ReputationRep: Follow
 700
 29042013 20:25
Reply
Submit reply
Related discussions:
 The Proof is 'notso' Trivial  Physics Edition
 Matrices: detA=0 > there exists a nontrivial solution to Ax=0 ...
 Stuck on a proof!
 Slight ambiguity in STEP question
 Extremely difficult lower triangular matrices question proof ...
 Proof by Induction ( Algebra Year 1 Uni)
 Is there a bottom line to what should be proven?
 Recursive unprovability
 Algebraic proof help! Maths GCSE
 Preparing for proofbased mathematics at university
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
 SherlockHolmes
 Notnek
 charco
 Mr M
 TSR Moderator
 Nirgilis
 usycool1
 Changing Skies
 James A
 rayquaza17
 RDKGames
 randdom
 davros
 Gingerbread101
 Kvothe the Arcane
 The Financier
 The Empire Odyssey
 Protostar
 TheConfusedMedic
 nisha.sri
 Reality Check
 claireestelle
 Doonesbury
 furryface12
 Amefish
 harryleavey
 Lemur14
 brainzistheword
 Rexar
 Sonechka
 LeCroissant
 EstelOfTheEyrie
 CoffeeAndPolitics
 an_atheist
 Moltenmo
 Labrador99
Updated: February 22, 2018
Share this discussion:
Tweet