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# The Proof is Trivial! watch

1. (Original post by Mladenov)
Is it not just the number of perfect squares in the interval ?
2. Problem 98**

Let be a sequence such that and .
Then, the sequence is divergent.
Also, find: .
3. (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 BMO2-level)
------------
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.
Solution 90

4. (Original post by Mladenov)
Problem 98**

Let be a sequence such that and .
Then, the sequence is divergent.
Also, find: .
Well, assuming that part of your question was to prove that the series is divergent:

Assume that the series converges to some value x. Then , and so . This is impossible. There is a contradiction. Therefore the series must diverge
5. (Original post by Zephyr1011)
...
Exactly. You have to find the limit to complete the solution.
6. (Original post by Mladenov)
Then, the sequence is divergent.
Out of curiosity, why do you keep writing "then, ..." rather than "show that.." or "prove that"? Is it intentional?
7. (Original post by Mladenov)
Problem 98**

Let be a sequence such that and .
Then, the sequence is divergent.
Also, find: .
(This is probably wrong because I've never done divergence-type stuff but I'll give it a go...)
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).
8. (Original post by The Polymath)
Out of curiosity, why do you keep writing "then, ..." rather than "show that.." or "prove that"? Is it intentional?
Yes, it is. I started writing "then" rather than "prove/show that" after I had read Coxeter's book "Geometry Revisited".

(Original post by Jkn)
(This is probably wrong because I've never done divergence-type 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).
The second part is wrong. You can't assume . It is senseless.
9. (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".
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.
10. (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".
What did it say in the book? "Then" doesn't really make sense to me.

(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.
This
11. 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 well-known 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.
12. (Original post by Mladenov)
The second part is wrong. You can't assume . It is senseless.
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?
13. (Original post by metaltron)
Solution 90

Dayuuuuuuuuuuum dat some nice algebra
14. (Original post by Jkn)
Dayuuuuuuuuuuum dat some nice algebra
Thank you

Let's hope the rest doesn't disappoint.
15. (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.
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.
16. (Original post by metaltron)
Thank you

Let's hope the rest doesn't disappoint.
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 almost-always 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.
17. (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.
Really? How did they used to say it? And how did they distinguish between a request to prove something and a statement of fact?
18. (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".
I love that book. Projective geometry messes with my mind but it's so awesome.
19. (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 almost-always 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.
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.
20. (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.
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.

(Original post by Lord of the Flies)
...
Your solutions are correct. Solution 97 is pretty nice.
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?
Nope. As the expression, is not correct, the subsequent equalities in your solution are also incorrect. And finally, your conclusion is not true.

(Original post by ben-smith)
I love that book. Projective geometry messes with my mind but it's so awesome.
I second!
To summarise: breathtaking and elegant.
I absolutely enjoyed the last two chapters on inversive and projective geometry.

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