(Original post by gff)
This week seems to be devoted to integrals.
[*] Determine all continuous functions which satisfy
Official TSR Mathematical Society
Announcements  Posted on  

Take our survey to be in with the chance of winning a £50 Amazon voucher or one of 5 x £10 Amazon vouchers  24052016 


(Original post by Farhan.Hanif93)
... 
Spoiler:ShowGuessing this has something to do with tanh(1/2x). If you multiply num and denom by e^1/2x you get root( sinh(x/2)/ cosh(x/2)). Then maybe make the sub t^2=tanh(x/2). Looks like some unfriendly algebra after that though... 
I just used one basic substitution to change the form a bit, and then made some informed guesses about what certain things would integrate to.

(Original post by nohomo)
I've just had a go at this one. I don't think I've found your "amusingly unusual approach" though, and I'm not entirely sure that my answer is right.
If you remove the factor of 2 in the log the answer is correct.
The amusing approach.
(Original post by Blutooth)
[Balkan 1997]
Let m an n be integers greater than 1. Let S be a set with n elements, and let be subsets of S. Assume that for any two elements x and y in S, there is a set such that either x is in and y is not in or x is not in and y is in . Prove that
Spoiler:Show
Denote the set containing all subsets as and define by the following.
Then, let the element be represented over as the set , and note that .
By hypothesis, the representations of two elements , namely the sets and , are not equal and ordered identically.
For if it was true, then for every subset ; e.g. there is no subset that contains , but not , and viceversa.
Now, we would like to know what is the maximum number of representations over a given which satisfy the above conditions.
As already noted, each contains elements that are either or ; all possible permutations are then , and let be contained in the set .
Finally, since , it follows that the map is injective.
Hence, we conclude that .

(Original post by gff)
That's my "good" English  bear with me.
If you remove the factor of 2 in the log the answer is correct.
The amusing approach.
Solution.
Spoiler:Show
Denote the set containing all subsets as and define by the following.
Then, let the element be represented over as the set , and note that .
By hypothesis, the representations of two elements , namely the sets and , are not equal and ordered identically.
For if it was true, then for every subset ; e.g. there is no subset that contains , but not , and viceversa.
Now, we would like to know what is the maximum number of representations over a given which satisfy the above conditions.
As already noted, each contains elements that are either or ; all possible permutations are then , and let be contained in the set .
Finally, since , it follows that the map is injective.
Hence, we conclude that .
Spoiler:Show
Let us associate with each element x in S a number of m binary digitseg for a particular x, .
How we construct a(x)...
If the element x belongs to the set set the mth digit from the left in a(x) to 1. If the element x does not belong to to set set the mth digit from the left in to 0.
Now clearly given m, the total possible number of different numbers is . This is because for each digits column (of which there are m) we can set the value to either 1 or 0==> 2 choices.
Thus if we have at least 2 of the elements of S have the same number (pidginhole principle). So we have an and an where =. What does this mean?
If belongs to a subset so does . If does not belong to a subset neither does . Thus the assumption that we can find a set where there is an but not an does not hold. 
Stupid spammers! Extracted from spammer's thread.
(Original post by oh_1993)
I'm not very experienced with this but if an Abelian set is one where the order of applying operations between elements within a set doesn't matter, then surely you need to know what the operation is that is being applied to the set e.g. multiplication modulo 3, addition...
So g = +/ 1 and prove that 1 * 1 = 1 * 1 ??
Does this even make sense lol
Spoiler:Show
You would probably call it an Abelian group, rather than simply a set, since it has to have an operation associated with it.
However, the good thing about abstract algebra is that you do not need to know all details  you want to be as general as possible given a few assumptions.
Another perhaps confusing thing is that is used not as a number, but as a symbol representing the multiplicative identity.
However, it is the case that the number is the multiplicative identity of familiar groups.
Also, the notation is symbolic. It is a short hand for where the dot is the operation of the group, not necessarily ordinary multiplication.
*****
Here is a good question related to this thread.
It comes from OCR MEI's FP3 Additional Further Maths book.
Spoiler:Show
[*] If and are sets, the symmetric difference is defined by
as shown below in the Venn diagram of the set containing them.
(i) Prove that the set of all subsets of the set forms an Abelian group under the operation .
(ii) Given that and ,
solve the equation
(iii) Prove that if (the identity) for every element of a group, then the group is Abelian. Given an example of such a group.

(Original post by gff)
I like questions which can teach you new ideas  I hope you do as well.
[*] Let be a nonempty set and let be an increasing function on the set of all subsets of , meaning that
Prove that there exists , a subset of , such that .
I use to mean x is a subset of y, and can even be equal to y.
Spoiler:Show
Define Where
Statement to be proved
Assuming
Proof
Taking f to both sides this
This .
Since statement is true for n=0 because and maps onto itself, the statement is true for for all
Note if we have succeeded in tackling the question. If not we can find a set where . Proof
and also .
Thus for each n either we succeed in finding a set that satisfies or we generate a set which has fewer elements than .We keep on repeating this procedure(eg is A_0 the set, If yes stop, if no is A_1 the set with the quality f(T)=T... ) until we have found a set satisfying f(T)=T or until we have a set of one element. This set must be equal to its image, else we have a sets whose image is the emptyset. But this contradicts the original assumptions.

Here is an interesting question to solve. An even number of people are sitting down at a table for breakfast. When they come back in the evening for dinner, they are not necessarily seated in the same order. Whatever the new seating arrangement prove that there are at least 2 people who are sitting with the same number of people in between them at both dinner and breakfast.

You're assuming the set is finite.

(Original post by gff)
That's my "good" English  bear with me.
If you remove the factor of 2 in the log the answer is correct.
The amusing approach.
Solution.
Spoiler:Show
Denote the set containing all subsets as and define by the following.
Then, let the element be represented over as the set , and note that .
By hypothesis, the representations of two elements , namely the sets and , are not equal and ordered identically.
For if it was true, then for every subset ; e.g. there is no subset that contains , but not , and viceversa.
Now, we would like to know what is the maximum number of representations over a given which satisfy the above conditions.
As already noted, each contains elements that are either or ; all possible permutations are then , and let be contained in the set .
Finally, since , it follows that the map is injective.
Hence, we conclude that .
Sorry I haven't been posting in a while  been a bit ill . 
(Original post by SimonM)
You're assuming the set is finite. 
(Original post by Blutooth)
Good point. I suppose the result doesn't hold for infinite sets.
Can't resist sharing the nice solution.
Spoiler:Show
Consider the family of sets
We observe that and conclude that the family is nonempty.
Hence, let be the intersection of all sets in , and our aim is to show that .
If , then , and by considering the intersection over all , we deduce that and that .
Next, because is increasing by hypothesis, it follows that , and hence .
Finally, since is included in every element in , we also conclude that .
The double inclusion proves that , as desired.

^^ awesome soln. Exposed me to a newish idea in maths.

Check out this documentary about IMO candidates. To sum it up in 3 words: interesting, saddening and inspiring.

(Original post by Blutooth)
Check out this documentary about IMO candidates. To sum it up in 3 words: interesting, saddening and inspiring.
Good documentary though  it introduced me to "upper" level mathematics, perhaps even had a role in me studying it currently. 
(Original post by Oh I Really Don't Care)
In my first term (perhaps second) I met Jos. Seemed socially acceptable (well, to the mathmos at least ...).
Good documentary though  it introduced me to "upper" level mathematics, perhaps even had a role in me studying it currently.
I was just wondering quite a few of those talented mathmos had asperges. Have you noticed any people with asperges while studying for your maths degree?
I have a friend who is very good at maths, and sometimes I wonder if he has an autistic spectrum disorderthough it might be that he's just a bit socially awkward (as some mathmos often are :P).
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following: