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Original post by gff
Somebody who didn't manage to do it has negged you. :tongue:
This solution will look nicer in a spoiler, in my opinion. Give strangers the chance to have a go with it. :biggrin:


Haha, haters gonna hate. But yeah, I can certainly spoiler the soln.
Reply 3581
Avatar for gff
gff
5
This week seems to be devoted to integrals. :tongue:

[*] Determine all continuous functions f:[0,1]→Rf : [0, 1] \to \mathbb{R} which satisfy

∫01f(x)(x−f(x)) dx=112\displaystyle \int_{0}^{1} f(x)(x - f(x))\ dx = \frac{1}{12}
Original post by gff
This week seems to be devoted to integrals. :tongue:

[*] Determine all continuous functions f:[0,1]→Rf : [0, 1] \to \mathbb{R} which satisfy

∫01f(x)(x−f(x)) dx=112\displaystyle \int_{0}^{1} f(x)(x - f(x))\ dx = \frac{1}{12}

Spoiler

Reply 3583
Avatar for gff
gff
5
Original post by Farhan.Hanif93
...


Yup, it is a nice little question. :tongue: I better go back to sleep. :biggrin:
Reply 3584
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gff
5
A rather amusingly unusual approach works for this one. :tongue:


[*] Compute

∫ex−1ex+1 dx,     x>0\displaystyle \int \sqrt{\frac{e^x - 1}{e^x + 1}}\ dx, \ \ \ \ \ x > 0
[Balkan 1997]
Let m an n be integers greater than 1. Let S be a set with n elements, and let A1,A2,...Am A_1, A_2,...A_m be subsets of S. Assume that for any two elements x and y in S, there is a set Ai A_i such that either x is in Ai A_i and y is not in Ai A_i or x is not in Ai A_i and y is in Ai A_i . Prove that n≤2m n \leq2^m
(edited 12 years ago)
Original post by gff
A rather amusingly unusual approach works for this one. :tongue:


[*] Compute

∫ex−1ex+1 dx,     x>0\displaystyle \int \sqrt{\frac{e^x - 1}{e^x + 1}}\ dx, \ \ \ \ \ x > 0


Hypothesis, since i'm about to sleep.

Spoiler

(edited 12 years ago)
Reply 3587
Avatar for nohomo
nohomo
5
Original post by gff
A rather amusingly unusual approach works for this one. :tongue:


[*] Compute

∫ex−1ex+1 dx,     x>0\displaystyle \int \sqrt{\frac{e^x - 1}{e^x + 1}}\ dx, \ \ \ \ \ x > 0


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.

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.

Spoiler

(edited 12 years ago)
Reply 3588
Avatar for gff
gff
5
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.


That's my "good" English - bear with me. :tongue:
If you remove the factor of 2 in the log the answer is correct.

The amusing approach.

Spoiler



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 A1,A2,...Am A_1, A_2,...A_m be subsets of S. Assume that for any two elements x and y in S, there is a set Ai A_i such that either x is in Ai A_i and y is not in Ai A_i or x is not in Ai A_i and y is in Ai A_i . Prove that n≤2m n \leq2^m


Solution.

Spoiler

(edited 12 years ago)
Original post by gff
That's my "good" English - bear with me. :tongue:
If you remove the factor of 2 in the log the answer is correct.

The amusing approach.

Spoiler





Solution.

Spoiler



Very good, the above is a nice, rigorous soln. However, I think I would phrase my answer to the above question slightly differently- partly because I'm not too conversant with standard set theory notation, and also because there are others in this forum as ignorant as myself.

Spoiler

(edited 12 years ago)
Reply 3590
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gff
5
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


I don't see a reason why not to answer your question here, since the OP's request was answered.

Spoiler



*****
Here is a good question related to this thread.
It comes from OCR MEI's FP3 Additional Further Maths book.

Spoiler

(edited 12 years ago)
Reply 3591
Avatar for gff
gff
5
I like questions which can teach you new ideas - I hope you do as well. :tongue:


[*] Let A\mathbb{A} be a non-empty set and let f:P(A)→P(A)f : \mathcal{P}(\mathbb{A}) \to \mathcal{P}(\mathbb{A}) be an increasing function on the set of all subsets of A\mathbb{A}, meaning that

   f(X)⊂f(Y),    if X⊂Y.\ \ \ f(X) \subset f(Y), \ \ \ \ \text{if} \ X \subset Y.


Prove that there exists TT, a subset of A\mathbb{A}, such that f(T)=Tf(T) = T.
Original post by gff
I like questions which can teach you new ideas - I hope you do as well. :tongue:


[*] Let A\mathbb{A} be a non-empty set and let f:P(A)→P(A)f : \mathcal{P}(\mathbb{A}) \to \mathcal{P}(\mathbb{A}) be an increasing function on the set of all subsets of A\mathbb{A}, meaning that

   f(X)⊂f(Y),    if X⊂Y.\ \ \ f(X) \subset f(Y), \ \ \ \ \text{if} \ X \subset Y.


Prove that there exists TT, a subset of A\mathbb{A}, such that f(T)=Tf(T) = T.



I use x⊂y x \subset y to mean x is a subset of y, and can even be equal to y.

Spoiler

(edited 12 years ago)
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.
(edited 12 years ago)
Reply 3594
Avatar for SimonM
SimonM
18
You're assuming the set is finite.
Original post by gff
That's my "good" English - bear with me. :tongue:
If you remove the factor of 2 in the log the answer is correct.

The amusing approach.

Spoiler





Solution.

Spoiler


I think I just got lucky with that integration question you posted. Just thought you know what the heck let's try that substitution - didn't realise it would actually work.

Sorry I haven't been posting in a while - been a bit ill :frown:.
Original post by SimonM
You're assuming the set is finite.


Good point. I suppose the result doesn't hold for infinite sets.
Reply 3597
Avatar for gff
gff
5
Original post by Blutooth
Good point. I suppose the result doesn't hold for infinite sets.


PRSOM for the post.

Can't resist sharing the nice solution. :tongue:

Spoiler

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