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visits member for 3 years, 4 months
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Apr
4
comment How can a piece of A4 paper be folded in exactly three equal parts?
@StevenTaschuk: Ah, that makes sense and is pretty obvious now you say it and makes a lot of sense.
Apr
4
comment How can a piece of A4 paper be folded in exactly three equal parts?
@StevenTaschuk: What does it mean then?
Apr
3
comment Fibonacci trick and proving it.
+1 for the geometric interpretation (and of course the correct proof).
Mar
28
comment Coin problem with 6 and 10
@MarcvanLeeuwen: Perhaps I should have said the correct response is that there is no valid answer.
Mar
28
comment Coin problem with 6 and 10
I think technically the answer is that there is no answer (ie no largest value). Giving you an infinite amount of 10s would give you infinity so you should likely avoid answers like that (and indeed any answer that pretends that infinity is an answer).
Mar
26
comment Probability that random moves in the game 2048 will win
I think your number of moves is wrong. ie its 1024 moves to get to 2048. Each move spawns a two or a four, if you assume its a two every time then the total on the board increases by two each move which means that 1024 moves gives a total of 2048 on the board.
Dec
20
comment Show that a continuous function with a certain integral property must be f(x)=x.
Ah yes, of course. I was thinking that it relied on the fact f(x)=x but it just relies on the fact that if f(x)=x then the integral is the same which is different.
Dec
20
comment Show that a continuous function with a certain integral property must be f(x)=x.
Where does the first line come from?
Nov
10
comment Is my proof correct for: $\sqrt[7]{7!} < \sqrt[8]{8!}$
Is his proof correct? Isn't there a mistake when he takes the 8! to the power of thing outside of the thing? Certainly the fact that you have 7!<8^7 and he gets 7!<8! makes me think there is a problem somewhere... Though of course this does demonstrate why getting rid of the roots is such a good idea because they confuse things. :)
Nov
1
comment How to weigh up to 100kg with 5 weights
@BrianM.Scott: Cool. I should have known it was a well studied thing. A name means I can read more about it. :)
Oct
31
comment How to weigh up to 100kg with 5 weights
@Cruncher: yeah, its more the mathematical analogy that is easy to imagine. I imagine I would get completely muddled if I tried to actually do maths like that... :)
Oct
31
comment How to weigh up to 100kg with 5 weights
@self.: No. This can be thought of as an expression of base 3 in a strange way. When counting in base 3 you have for each power of 3 a value of either zero, one or two. With this solution we are doing the same thing effectively but we give them values of one (its on the oposite side to the item being weighed), zero (not on the scales) or minus one (its on the same side as the weight being measured. If you wanted higher powers you'd need multiples of some weights. eg if you always had two of each size then powers of five would work. Powers of four would work as well but be inefficient.
Oct
7
comment Fastest way to check if $x^y > y^x$?
so can you do similar optimisations for if x and y are both less than e?
Oct
6
comment Proof that $n^3 + 3n^2 + 2n$ is a multiple of $3$.
Its not a bad way. I just got to the end and thought "oooh, that looks like a nice result. Oooh, look it factorises nicely to a really nice result! oh wait, that's what we started with..."
Oct
6
comment Proof that $n^3 + 3n^2 + 2n$ is a multiple of $3$.
You could have made that algebra much easier in the middle by just splitting the factor of k+3 and leaving the other two factors alone.
Oct
6
comment I don't understand how this works? induction proof
$n^3+3n^2+2n+1$ should be $n^3+3n^2+3n+1$
Oct
6
comment Derivation of the quadratic equation
@String: This is definitely an agree to disagree moment. :) I think the other answer shows reasoning much better. Especially since this one requires you to remember the form of the answer as well. :) But yes, as you say clearly a subjective opinion and I didn't mean to suggest otherwise in my first comment. :)
Oct
6
comment Derivation of the quadratic equation
@Jonathan: I'd also add that the shifting by -b/2 is effectively the same as is done in completing the square...
Oct
6
comment Derivation of the quadratic equation
@String: That looks really clumsy to me compared to completing the square method described by "눉송이".
Oct
6
comment Can the limit of a product exist if neither of its factors exist?
@PeterLeFanuLumsdaine: Ah yes, of course. Cheers. :)