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The Proof is Trivial!

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Problem 238 *

Find all roots of the following polynomial:

∑i=0n((i+1)xi+(i+1)x2n−i)−(n+1)xn\displaystyle\sum_{i=0}^{n} ((i+1)x^i+(i+1)x^{2n-i})-(n+1)x^n

This is a hard one, but can be solved without any advanced techniques.
Reply 1661
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Jkn
11
Original post by bananarama2
That's my permanent state of mind at the minute.

Tell me about it! :/

I mean the whole point of posting solutions is so that others can read, enjoy and occasionally learn from your work and, whilst some solutions cannot be expressed in simple terms, I feel that the idea it to express them is as informative way as possible.

Lately it feels like a lot of the solutions posted here are for some sort of 'point-scoring', as if the more you omit from your solution and the shorter you make it look, the smarter you are. I mean, I see solutions on here that people post in a line that I know for a fact take at least a half page and if questions were being typed up for examiners, then people would undoubtedly put more time and effort into them :/
Original post by james22
Problem 238 *

Find all roots of the following polynomial:

∑i=0n((i+1)xi+(i+1)x2n−i)−(n+1)xn\displaystyle\sum_{i=0}^{n} ((i+1)x^i+(i+1)x^{2n-i})-(n+1)x^n

This is a hard one, but can be solved without any advanced techniques.


This has essentially already been done here.
Original post by Lord of the Flies
This has essentially already been done here.


Ah ok, must have missed that one.
Original post by Jkn

Well if your contradiction meant that the solution was non-trivial, why didn't you... find the solution?


Well. I shall try to give you different idea, as you do not like this or something. But, let me ask you: which part of my proof is not clear? For the nnth time, I suppose that there exists a solution which is not in the set TT. Hence, using (m,n)↦(n−m,m)(m,n) \mapsto (n-m,m), I get that (n−m,m)(n-m,m) is in TT, since the other set (the set of the solutions which are not in TT) is bounded below.

Another thing we can do: continue with (m,n)↦(n−m,m)(m,n) \mapsto (n-m,m) till we get solution of the form (l,1)(l,1). Then l=2l=2.. We can continue, since, if m>1m>1, then m>n−mm>n-m. Otherwise (m,n)(m,n) would not be a solution.
Reply 1665
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Jkn
11
Original post by Mladenov
Well. I shall try to give you different idea, as you do not like this or something. But, let me ask you: which part of my proof is not clear? For the nnth time, I suppose that there exists a solution which is not in the set TT. Hence, using (m,n)↦(n−m,m)(m,n) \mapsto (n-m,m), I get that (n−m,m)(n-m,m) is in TT, since the other set (the set of the solutions which are not in TT) is bounded below.

Another thing we can do: continue with (m,n)↦(n−m,m)(m,n) \mapsto (n-m,m) till we get solution of the form (l,1)(l,1). Then l=2l=2.. We can continue, since, if m>1m>1, then m>n−mm>n-m. Otherwise (m,n)(m,n) would not be a solution.

The question is asking you to find the maximum value of m3+n3m^3+n^3. You haven't done this. And what's more, you don't need to prove anything for lower cases :tongue:

Theres a really nice method without having to know the identities btw :smile:
Original post by Jkn
The question is asking you to find the maximum value of m3+n3m^3+n^3. You haven't done this. And what's more, you don't need to prove anything for lower cases :tongue:

Theres a really nice method without having to know the identities btw :smile:


There probably is. :tongue:

As all the possible pairs consist of consecutive Fibonacci numbers, I found it obvious that the required maximum is 9873+15973987^3+1597^3.
Original post by ukdragon37
I'm now constantly worrying that it wasn't good enough to pass :banghead:


Surely the pass criteria is really low? Do people actually fail on a thesis?
Original post by shamika
Surely the pass criteria is really low? Do people actually fail on a thesis?


I don't know if anyone ever failed but what I wrote was rubbish too :tongue:
Original post by ukdragon37
I don't know if anyone ever failed but what I wrote was rubbish too :tongue:


LOL but it was also 10,000 words of rubbish, you get points for effort :tongue:

Seriously, I don't think it's actually possible to genuinely fail, unless you literally wrote nothing, or it was just a bunch of expletives. Surely you had a supervisor meeting before you submitted?

When you say Pass, does that mean something different in Cambridge?
Original post by ukdragon37
I don't know if anyone ever failed but what I wrote was rubbish too :tongue:


You should have not taken the advice "be the first in everything" literally. (only joking) :tongue:
It may be the case that your 'rubbish' is better than somebody else's masterpiece. :wink:
Original post by shamika
LOL but it was also 10,000 words of rubbish, you get points for effort :tongue:

Seriously, I don't think it's actually possible to genuinely fail, unless you literally wrote nothing, or it was just a bunch of expletives. Surely you had a supervisor meeting before you submitted?

When you say Pass, does that mean something different in Cambridge?


I guess my supervisor said it was "shaping up" to be an OK dissertation. Pass is 60%. You need a pass in the dissertation to get any grade overall.

Original post by jack.hadamard
You should have not taken the advice "be the first in everything" literally. (only joking) :tongue:
It may be the case that your 'rubbish' is better than somebody else's masterpiece. :wink:


Highly doubtful :tongue:
Original post by ukdragon37
I guess my supervisor said it was "shaping up" to be an OK dissertation. Pass is 60%. You need a pass in the dissertation to get any grade overall.


You'll be fine, unless you've managed to piss off your supervisor. When do you find out?
Original post by shamika
You'll be fine, unless you've managed to piss off your supervisor. When do you find out?


25th....
Original post by ukdragon37
25th....


Soon! And that's your entire Part III result? You don't seem worried about the exams, that would be where I would be worried. I was lucky that all years count for the same degree so the only grades I could've got were a first or a fail (and they don't tend to fail as much in Imperial, but the number of thirds and 2:2s is higher than the other good unis I think).

Good luck? Aiming for a Distinction?

BTW: no one tried my question :frown:. Did it get lost in the numbering shuffle? It was around problem 200 I think.
(edited 10 years ago)
Original post by shamika
Problem 200 * / **

Three positive integers are chosen from the first 2k positive integers. What is the probability that the numbers chosen can form the sides of a triangle? What is the limiting probability as k tends to infinity?


Spoiler

Original post by shamika
Soon! And that's your entire Part III result? You don't seem worried about the exams, that would be where I would be worried.


I'm not worried because they've already gone past and I know their results. 25th would be my dissertation and therefore entire Part III result. :tongue:

Original post by shamika

Good luck? Aiming for a Distinction?


I need 62.1 in my dissertation for a distinction in Part III due to exam results, 60 being the passmark (so between that and 62.1 would mean merit overall). I need to get over the passmark in the dissertation to get any grade overall, that's the part I'm worried about :colondollar:
Reply 1677
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Jkn
11
Problem 239*

Define the symbol π\pi (which we can assume to be a new idea) to be equal to the circumference of a circle divided by its diameter. We can assume, without proof, that π\pi is a constant.

Prove that the area of a circle is equal to πr2\pi r^2, where r is the radius.

Hence evaluate sin⁡π\sin \pi.

Note: You'd think that this would go without saying but I will remind people that implicitly suggesting that you have proved something and not actually writing a proof does not constitute a mathematical proof. Note also that theorems cannot be applied unless you offer justification that the object that you are proving is not a necessary condition for that theorem to hold (as this would form a circular argument).
(edited 10 years ago)
Original post by Jkn
Problem 239*

Define the symbol π\pi (which we can assume to be a new idea) to be equal to the circumference of a circle divided by its diameter. We can assume, without proof, that π\pi is a constant.

Prove that the area of a circle is equal to πr2\pi r^2, where r is the radius.

Hence evaluate sin⁡π\sin \pi.

Note: You'd think that this would go without saying but I will remind people that implicitly suggesting that you have proved something and not actually writing a proof does not constitute a mathematical proof. Note also that theorems cannot be applied unless you offer justification that the object that you are proving is not a necessary condition for that theorem to hold (as this would form a circular argument).


Can we use calculus or must it be a geometric argument?
(edited 10 years ago)
Solution 239

Consider a circle of radius r, circumference c.

Enclose the circle by a regular n-gon in such a way that the distance between the center of each face=r (any issues about this not being exactly possible should go when we take the limit).

Each face form a triange with the cente, this triangle has base c/n and height r so it has area cr/(2n).

The total area of the polygon is then cr/2.

Let n->infinity so the polygon->the circle and area=cr/2=pi*2*r*r/2=pi*r^2.

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