Let's discuss Mathematics

[Mise]

Yeah, I got that far - maybe I've just lost the ability to differentiate, but I get something that looks nothing like what I wanted...

EDIT: Okay, yeah, I did lose the ability to differentiate... I thought d(1/x)/dx was ln(x)... (interestingly, whilst simultaneously thinking that d(ln(x))/dx was 1/x!) It all works now!

 

[Atticus]

Before using that as proof, you should check what are your definitions for e, exponent function and logarithm, and also how their derivatives are proved to be what they are: the proofs may use the very fact that lim (1+1/n)^n =e (it is equivalent to the thing you're proving).

Of course that depends on what purpose you are trying to prove the thing. If it's just for yourself, you may not want to be such a purist (but then again excell would have been enough of proof).

I suppose usually e is defined to be 1/0! + 1/1! + 1/2! + 1/3!+...., that is
\sum_{k=0}^{\infty} 1/k!.
You should get the result e= lim (1+ 1/n)^n from that: use binomial theorem to express (1+1/n)^n as sum, and then compare it to the n:th partial sum of the series that defines e. It does require some extra effort, I suppose.

 

[ParadigmShifter]

My book defines log x as the integral from 1 to x of dt/t.

It does say you can use the power series definition for e instead.

 

[Atticus]

Well, that's the difficulty: there's more than one definitions you can use. So if Mise wants to do this as easily as possible, he can define e=lim (1+1/n)^n, and then conclude that
(1+1/n)^n * (1-1/n)^n -->1, so the latter part goes to 1/e.

That would be little dishonest though, and would not increase much of the understanding either.

 

[Mise]

I really did it to see how on Earth you get from "if the probability of an event occuring is 1/n, what is the probability of that event occuring after n trials?" to "e"... So yeah, using a def'n of e or ln that presented that answer on the plate would have left me asking "why is e defined like that?"

 

[ParadigmShifter]

exp(x) can be defined from the differential equation

dy/dx = y

i.e. which function is it's own derivative.

 

[Atticus]

Yes, the sad thing about maths is that often definitions are chosen so that the poofs would become easy, not because of their historical origin or the easiness to grasp them.

I suppose the lay man definition of e is "the number for which De^x =e^x".

So you can use the derivative argument, and if you want to go into more detail, find a book that uses series or limes as the definition, and invert it's proof for e^x:s derivative.

 

[ParadigmShifter]

Limes?

 

[Atticus]

Finnish word for "limit" (raja-arvo) doesn't start with lim, so it's generally called limes -or just lim. It becomes reflex, you start to think "limes" whenever you write lim.

 

[dutchfire]

Interesting little maths problem I encountered in class recently:

ParadigmShifter and dutchfire enter in a big scrabble tournament. There are 2^n players, and it's a knock-out tournament. However, since everyone got so drunk before the tournament started, they all lost their ability to play scrabble, and in every match, each player just has a 1/2 chance of winning and going through to the next round. Assuming the draw for each round is fair, what's the chance p_n ParadigmShifter and dutchfire will have to play each other eventually?

Hint 1: You may want to express p_n as a function of p_(n-1) first.
Hint 2: You may then want to use Excel to formulate an induction hypothesis.

 

[ParadigmShifter]

Good question, I'm drinking now though so will try and answer it later on or tomorrow.

And I won't be using Excel

 

[ParadigmShifter]

P(1) = 1

EDIT: I'm not that drunk yet I am playing poker though.

 

[Atticus]

Maybe you should use empirical methods to find it out?

 

[ParadigmShifter]

Nah, mathematicians use pencils rather than computers

 

[ParadigmShifter]

P(2)

Well, we are 50% to meet in the first round. If we don't meet in the first round we both need to win (25% chance) to meet in the final, so that's a probability of 0.75 I believe, since those events are independent (3 beers though, may change my mind in a while).

3rd place in the 18 man tourney for poker, was chip leader for most of the game

 

[LulThyme]

I really did it to see how on Earth you get from "if the probability of an event occuring is 1/n, what is the probability of that event occuring after n trials?" to "e"... So yeah, using a def'n of e or ln that presented that answer on the plate would have left me asking "why is e defined like that?"

This responses pre-supposes a definition for "e" which you still havn't told us and is what Atticus is getting at.
Some people DO define "e" as the limit above and some students even learn it this way. For them, the result you are discussing is NOT surprising but other results which are not surprising for you (because they easily follow from your definition) might be more surprising for them.

That's why it would have been enlightening for you to state which definition of e you are using or, at the very least, which properties of e you consider "fair game" for proving this limit.

 

[Atticus]

Nah, mathematicians use pencils rather than computers

No, I meant Scrabble tournaments, not computers...

p_n = 1/(n-1) + (1/2)² p_n-1
to begin with.

 

[LulThyme]

Interesting little maths problem I encountered in class recently:

ParadigmShifter and dutchfire enter in a big scrabble tournament. There are 2^n players, and it's a knock-out tournament. However, since everyone got so drunk before the tournament started, they all lost their ability to play scrabble, and in every match, each player just has a 1/2 chance of winning and going through to the next round. Assuming the draw for each round is fair, what's the chance p_n ParadigmShifter and dutchfire will have to play each other eventually?

Hint 1: You may want to express p_n as a function of p_(n-1) first.
Hint 2: You may then want to use Excel to formulate an induction hypothesis.

Let x be the round they will meet in if they always win.
We have that P(x=i)=2^(i-1)/(2^n-1)
(Fix one of the two players. There are 2^n-1 other players and he has 2^(i-1) "new potential opponents" in each round.)
If they are to meet in round x, they have to win (x-1) rounds each, so that the final answer is the sum over i from 1 to n of 2^(i-1)/(2^n-1) (1/2)^(2i-2). This can be simplified to the sum of 2^(1-i)/(2^n-1).

Either by hand (its a geometric sum), or using Wolfram alpha, you find that the answer is 2^(1-n).

This seems easier than to do it with a recurrence.


EDIT: sorry for the edits, forgot a (-1) in the number of opponents.

 

[Atticus]

Well done, sir!

 

[LulThyme]

P(2)

Well, we are 50% to meet in the first round.

1/3 of chance to meet in first round of a 2 round tournament.