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The Proof is Trivial!

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Original post by Zacken
And I'll have to make less mistakes. :tongue:


Me too!
Problem 573*:

Compute the 100100th derivative of x2+1x3x.\displaystyle \frac{x^2 + 1}{x^3 - x}.
Original post by Zacken
Problem 573*:

Compute the 100100th derivative of x2+1x3x.\displaystyle \frac{x^2 + 1}{x^3 - x}.

Spoiler

Original post by joostan

Spoiler



Yep! :biggrin:
Original post by Lord of the Flies
Well the concept of winning/losing moves presupposes knowledge of the game's outcome, which some sense answers your question. It requires that the winning player be perfect (i.e. will not make mistakes beyond said move), and that one player plays only winning moves, the other losing moves. That should be quite clear.So it is more whether the idea of a winning move makes sense in any game where you can only win or lose, or equivalently, whether there is a "correct" outcome to any such game. The answer is essentially yes under appropriate, almost obvious conditions, and is given in full detail by Zermelo's theorem in game theory:For every (finite) game of perfect information (that is, one where all useable information is available to both players at all times; chess, go, or reversi are examples), not involving chance (so not backgammon for instance), between two players who take moves in alternation, there is either a winning strategy for one of the players, or both players can force a draw. If we disregard the possibility of draws (as in our painting game), then we have existence of winning & losing moves/strategies.


I love stuff like this. I wonder, do you get to delve into some game theory as a Cambridge mathmo? I might consider economics if not!
Problem 574* :

Define fn(a)f^{n}(a) to be the nn-th derivative of ff evaluated at aa. Find f2005(0)f^{2005}(0) where

Unparseable latex formula:

\displaystyle[br]\begin{equation*} f(x) = \frac{1}{1 + 2x + 3x^2 + \cdots + 2005x^{2004}}\end{equation*}

Original post by Mathstatician
I love stuff like this. I wonder, do you get to delve into some game theory as a Cambridge mathmo? I might consider economics if not!


Not at undergrad level no, but I am sure there is in part III! :biggrin:

To answer the ω\omega-extension of that problem, for those interested: we know that player 1 wins n×mn\times m for all n,mn,m (excluding the trivial case). Player 2 actually wins 2×ω2\times \omega (and this implies player 1 wins n×ωn\times \omega for all n2n\neq 2). Explanation in spoiler.

Spoiler

(edited 7 years ago)
Original post by Zacken
Problem 573*:

Compute the 100100th derivative of x2+1x3x.\displaystyle \frac{x^2 + 1}{x^3 - x}.


These are great questions for quickly learning a new rearranging technique/idea, how do you come up with them? :smile:
Original post by EnglishMuon
These are great questions for quickly learning a new rearranging technique/idea, how do you come up with them? :smile:


I've come across these, can't take the credit for coming up with them myself, unfortunately!
Original post by Lord of the Flies
Not at undergrad level no, but I am sure there is in part III! :biggrin:


Oh. I find that surprising! I'd expect the mathematicians approach to be more rigorous and so feature later than in economics but my understanding is Warwick maths teach game theory in years two and three. Shame!
Original post by Zacken
Problem 574* :

Define fn(a)f^{n}(a) to be the nn-th derivative of ff evaluated at aa. Find f2005(0)f^{2005}(0) where

Unparseable latex formula:

\displaystyle[br]\begin{equation*} f(x) = \frac{1}{1 + 2x + 3x^2 + \cdots + 2005x^{2004}}\end{equation*}

Spoiler

Pilfered from Putnam, to replace the one I did:
Problem 575:*/**
Prove that the limit:
limn1n2k=1n(n2+k2)1n\displaystyle \lim_{n \to \infty} \dfrac{1}{n^2}\prod_{k=1}^n(n^2+k^2)^{\frac{1}{n}} exists and find its value.
Original post by joostan

Spoiler


Daaamn, nice!
Original post by joostan

Spoiler

Pilfered from Putnam, to replace the one I did:
Problem 575:*/**
Prove that the limit:
limn1n2k=1n(n2+k2)1n\displaystyle \lim_{n \to \infty} \dfrac{1}{n^2}\prod_{k=1}^n(n^2+k^2)^{\frac{1}{n}} exists and find its value.

Forgive my ignorance but are we allowed to assume things like the arithmetic geometric mean inequality? (not saying I have found a solution using it yet :P)
(edited 7 years ago)
Original post by EnglishMuon
Forgive my ignorance but are we allowed to assume things like the arithmetic geometric mean inequality? (not saying I have found a solution using it yet :P)


I don't see why not, but I'm not convinced it's going to help. . .
Original post by EnglishMuon
Forgive my ignorance but are we allowed to assume things like the arithmetic geometric mean inequality? (not saying I have found a solution using it yet :P)


It is Putnam so AM-GM is trivial. Results that you can assume in the IMO can be assumed in Putnam for sure (Have not put pen to paper so I don't know if it works)


Posted from TSR Mobile
Original post by physicsmaths
It is Putnam so AM-GM is trivial. Results that you can assume in the IMO can be assumed in Putnam for sure (Have not put pen to paper so I don't know if it works)


Posted from TSR Mobile


Thanks :smile: I knew Putnam is supposed to be tough but I wasn't sure if it was one of those prove everything you use type exams.
Original post by joostan
Pilfered from Putnam, to replace the one I did:
Problem 575:*/**
Prove that the limit:
limn1n2k=1n(n2+k2)1n\displaystyle \lim_{n \to \infty} \dfrac{1}{n^2}\prod_{k=1}^n(n^2+k^2)^{\frac{1}{n}} exists and find its value.

Spoiler

(edited 7 years ago)
I like this because it looks wrong.

Problem 576

Show that there is a constant a>1a>1 such that for any sequence (αn)(\alpha_n) of positive reals:

lim supn(1+αn+1αn)na\displaystyle\limsup_{n\to\infty}\left(\frac{1+\alpha_{n+1}}{ \alpha_{n}}\right)^n\geqslant a

What is the best possible value for a?a?
Original post by Smaug123

Spoiler



This is nice but annoying - I was trying to do it this way last night but was too tired to get it to work. However, have you actually proved that the limit exists? I'm not sure what they expect in Putnam problems. I assumed that they wanted to see some convergence test applied.
Original post by atsruser
This is nice but annoying - I was trying to do it this way last night but was too tired to get it to work. However, have you actually proved that the limit exists? I'm not sure what they expect in Putnam problems. I assumed that they wanted to see some convergence test applied.


The point is that the Riemann sum converges to the integral in the limit as nn \to \infty, so the limit exists iff the integral exists.
Granted a little exposition might be necessary to justify, but for the purposes of this thread I imagine this is sufficient.

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