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metaltron
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 28042013 19:06
(Original post by Jkn)
Which is what makes a good problem. Mmm, hence the possible link to AM
Post some problems! A similar level of difficulty to 86 would be nice (or perhaps BMO2level)

Wonder if anyone recognises where I got this from
Problem 90*
Find the sum of all values of a such that the equation has 3 distinct real solutions.
Last edited by metaltron; 29042013 at 16:30. 
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 28042013 19:06
(Original post by Mladenov)
Problem 98**
Let be a sequence such that and .
Then, the sequence is divergent.
Also, find: .
Assume that the series converges to some value x. Then , and so . This is impossible. There is a contradiction. Therefore the series must diverge 
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 28042013 19:23
(Original post by Zephyr1011)
... 
The Polymath
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(Original post by Mladenov)
Problem 98**
Let be a sequence such that and .
Then, the sequence is divergent.
Also, find: .
Assume the series converges. This implies thats as , . Hence the series approaches the constant relation . This is a contradiction as there exists no solutions such that unless . As , we observe three cases. If is between 0 and 1, the next case >1. If it equals 1 or is greater than 1 the result follows also. Therefore, as these possible solutions cannot occur as n tends to infinity. Hence the series must diverge.
Let m be such that (i.e. m is the required value in this limit). This implies that the sequence becomes as . This yields . Noticing that n approaches n+1, we conclude that . Therefore as (i.e. also divergent).Last edited by Jkn; 28042013 at 20:11. 
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 28042013 19:51
(Original post by The Polymath)
Out of curiosity, why do you keep writing "then, ..." rather than "show that.." or "prove that"? Is it intentional?
(Original post by Jkn)
(This is probably wrong because I've never done divergencetype stuff but I'll give it a go...)
Solution 98
Assume the series converges. This implies thats as , . Hence the series approaches the constant relation . This is a contradiction as there exists no solutions such that unless . As , we observe three cases. If is between 0 and 1, the next case >1. If it equals 1 or is greater than 1 the result follows also. Therefore, as these possible solutions cannot occur as n tends to infinity. Hence the series must diverge.
Let m be such that (i.e. m is the required value in this limit). This implies that the sequence becomes as . This yields . Noticing that n approaches n+1, we conclude that . Therefore as (i.e. also divergent).Last edited by Mladenov; 28042013 at 19:58. 
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 28042013 19:56
(Original post by Mladenov)
Yes, it is. I started writing "then" rather than "prove/show that" after I had read Coxeter's book "Geometry Revisited". 
The Polymath
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 28042013 19:59
(Original post by Mladenov)
Yes, it is. I started writing "then" rather than "prove/show that" after I had read Coxeter's book "Geometry Revisited".
(Original post by Zephyr1011)
Why? Personally, I find it a bit unclear, as it sounds like you're stating it as a fact, rather than as something which must be proven. 
Lord of the Flies
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 28042013 20:12
Wow this thread was on fire today! I just did these very quickly and unfortunately must leave now, so I apologise if I've missed anything  will correct later/tomorrow if so.
Solution 94
Set and since is injective . Plugging into the initial eq. gives
Solution 97
Suppose there exist distinct s.t. then , contradiction hence is injective. Hence has a left inverse and thus . Therefore or
Let then for any fixed we have the wellknown solution of which is . Notice that this implies
Plugging this into we get:
or
Putting all the facts together and considering compositions it shouldn't be too difficult to show that the only possible continuous are or in either case.Last edited by Lord of the Flies; 28042013 at 20:50. 
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 28042013 20:20
But the argument does show that if you assume the expression converges to a finite constant, you would have to deduce that this finite limit was infinite (a contradiction!) Therefore the expression does not tend to a finite limit and either infinitely ascends or infinitely descends (i.e. either 0 or infinity). Is this part correct? 
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metaltron
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 28042013 20:45
(Original post by Jkn)
Dayuuuuuuuuuuum dat some nice algebra
Let's hope the rest doesn't disappoint. 
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 28042013 21:10
(Original post by Zephyr1011)
Why? Personally, I find it a bit unclear, as it sounds like you're stating it as a fact, rather than as something which must be proven. 
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 28042013 21:15
That said, I did have two concerns:
1) I'm wondering if, for logical validity, you need to verify the result (e.g. STEP II, 2004, Question 1). This is always necessarily unless it is evident that no spurious solutions have been produced (though it is almostalways evident to the point of triviality when concerned with more simplistic algebraic manipulations) as I'm sure you know.
2) I sourced this particular question from 'brilliant.org' but it has not accepted the answer than seems logically correct! It ends the question by saying the answer is of the form where c is minimised and asks you to evaluate a+b+c which I got to be 427. It did not accept this answer! Any idea? The solutions will come out in a few days probably. 
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 28042013 21:24
(Original post by shamika)
In fairness, it is a relatively recent innovation to write out exercises with 'show that' or 'prove that', and so if you're reading a book from that era and have got used to it, I can see why its easy to slip into that way of writing. 
Dirac Spinor
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 28042013 21:29
(Original post by Mladenov)
Yes, it is. I started writing "then" rather than "prove/show that" after I had read Coxeter's book "Geometry Revisited". 
metaltron
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 28042013 21:34
(Original post by Jkn)
It certainly didn't!
That said, I did have two concerns:
1) I'm wondering if, for logical validity, you need to verify the result (e.g. STEP II, 2004, Question 1). This is always necessarily unless it is evident that no spurious solutions have been produced (though it is almostalways evident to the point of triviality when concerned with more simplistic algebraic manipulations) as I'm sure you know.
2) I sourced this particular question from 'brilliant.org' but it has not accepted the answer than seems logically correct! It ends the question by saying the answer is of the form where c is minimised and asks you to evaluate a+b+c which I got to be 427. It did not accept this answer! Any idea? The solutions will come out in a few days probably.
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. 
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 28042013 21:38
(Original post by Zephyr1011)
Why? Personally, I find it a bit unclear, as it sounds like you're stating it as a fact, rather than as something which must be proven.(Original post by The Polymath)
What did it say in the book? "Then" doesn't really make sense to me.
(Original post by Lord of the Flies)
...
My approach to problem 97 was quite different (it took me 4 pages A4 to solve it ).
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 .
(Original post by Jkn)
Good point.
But the argument does show that if you assume the expression converges to a finite constant, you would have to deduce that this finite limit was infinite (a contradiction!) Therefore the expression does not tend to a finite limit and either infinitely ascends or infinitely descends (i.e. either 0 or infinity). Is this part correct?
(Original post by bensmith)
I love that book. Projective geometry messes with my mind but it's so awesome.
To summarise: breathtaking and elegant.
I absolutely enjoyed the last two chapters on inversive and projective geometry.Last edited by Mladenov; 28042013 at 21:49.
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