The Math Thread

[darthbrowser]

[Konradix]

> rry = ln(x)+(sa-1)ln(m)
> e^rry = x+(sa-1)m

Pretty sure this should go to

e^rry = x+m^(sa-1)


*Le Scrooge : D*

[darthbrowser]

> rry = ln(x)+(sa-1)ln(m)
> e^rry = x+(sa-1)m

Pretty sure this should go to

e^rry = x+m^(sa-1)


*Le Scrooge : D*

Why are you raising m to the power of (sa-1)?


Distribute the natural logarithm:

Cancel the natural logithrm:

Factor out the m:

[lordpat]

I made up a math joke.

45 enters the room and sees that 22.5 threw up.
"What happened?" Asked 45
"Pi over eight."

Get it?

[Oranjui]

Can't say that joke is sensible to me [edit: never mind that was actually pretty good], but I can (though ages late) say that I'm pretty sure the intended solution to that thing earlier was

y = [ln(x) + (sa - 1)ln(m)]/r²

Given

rry = ln(x) + (sa - 1)ln(m)

Multiply both sides by r^2

rry = ln(x) + ln(m)*sa - ln(m)

Distribute the ln(m)

e^rry = e^{[ln(x) + ln(m)*sa - ln(m)]}

Raise e to the power of both sides

e^rry = e^{[ln(x) + ln(m^(as)) + ln(1/m)]}

Move the exponents inside of the logs, since a*log(b) = log(b^a)

e^rry = e^ln(x) * e^{ln(m^(as))} * e^ln(1/m)

Break up the exponents, since a^b+c = a^ba^c

e^rry = x * m^as / m

Cancel the logs, since e^ln(x) = x

me^rry = xm^as

Multiply both sides by m

[Supershroom]

The Riddle with the dwarves and King Fratch
This is a fascinating riddle I've learned today in Algebra class. Credits to my Alegbra exercise instructor

An infinite but countable group of dwarves was on the way back home from the mountains when they were surrounded by orcs from all off a sudden, and got all captured and brought to the king of the orcs called Fratch. Being known as extremely brutal against everything he dislikes, he wanted to kill all of the draws, but in a fit of a good mood he gave them a chance to survive:

1) The dwarves got 10 minutes to discuss a strategy.
2) Then Fratch ordered them after their size, starting with the biggest, and everyone got a coloured hat, so every dwarf was able to see all the colours of all following dwarves. There are countably many colours, and the dwarves knew them all.
3) The dwarves had to call out simultaneously which colour their hat has.
4) Every dwarf who successfully guessed his colour would survive, all others would die.

Within the discussion phase, the dwarves managed to find a strategy that granted survival for almost every dwarf, meaning all dwarves except from finitely many.

How did they do this? Let Z be the set of all dwarves and F be the set of all colours, you can imagine both to be the natural numbers. We describe all line-ups of the dwarves with their hats as maps f: Z --> F. Let Map(Z, F) be the set of all possible maps. Then we define an equivalence relation ~ on Map(Z, F) by f ~ g if and only if f and g accord almost everywhere (you can check easily that this is an equivalence relation). By using the axiom of choice we have a map h which maps all equivalence classes [f] relating to ~ to a representative h([f]): Z --> F. During the discussion phase, the dwarves agreed on such a map h. Now, after they've been lined up, each dwarf knew which equivalence class [f] the given order f belongs to by looking at the colours of all following dwarves, so every dwarf z_i shouted out the colour h([f])(z_i). Since the map h([f]) accords to f at all but finitely many places, this granted survival for almost every dwarf.

[lordpat]

Here is a simple probability problem I made up. It is made for like 14 year olds learning probability.

Let's say we have a bag with one blue ball, one red ball and one green ball. You can take a ball at random, but if you do, you must put two balls in the bag, one of each color you didn't chose. If you pick a red ball for example, you must put a green ball and a blue ball in the bag. What are the odds that after repeating this process three times there are no green balls in the bag?

[NanTheDark]

I forgot all the advanced probability stuff I knew, so let's brute force it!

Let's see! I'll draw a ball... ok... I got the blue ball. Now I put in a red ball and a green ball. Now we have two red balls and two green balls. Here, I got a 1/2 chance of getting a red ball or a green ball. If I draw, say, a red ball, I have to add a green and a blue. Then we would have 1 blue ball, 1 red ball and 3 green balls. Now I can draw one blue ball and that way there would be 2 red balls and 4 green balls.

Alright then. So... that tells me nothing really. Let's see how else I can figure it out.

The color is not really important, so I'll try to get rid of blue balls instead, because blue balls are a serious problem. I already got rid of all the blue balls. To do so I follow these steps:

1. Get rid of the blue ball.
2. Get rid of either a red or a green ball, it doesn't matter which.
3. Get rid of the blue ball.

But then I can look at the red ball...

1. Get rid of a different ball.
2. Get rid of the red ball.
3. Get rid of the red ball.

So I can get rid of either one of the colors I get rid of in the first and two steps. I don't think it's possible to do this otherwise since when I draw one ball I don't have that color anymore so... yeah.

I can either:
1. Get rid of the blue ball by discarding blue, then red, then blue
2. Get rid of the blue ball by discarding blue, then green, then blue
3. Get rid of the blue ball by discarding red, then blue, then blue
4. Get rid of the blue ball by discarding green, then blue, then blue
5. Get rid of the red ball by discarding red, then blue, then red
6. Get rid of the red ball by discarding red, then green, then red
7. Get rid of the red ball by discarding blue, then red, then red
8. Get rid of the red ball by discarding green, then red, then red
9. Get rid of the green ball by discarding green, then red, then green
10. Get rid of the green ball by discarding green, then blue, then green
11. Get rid of the green ball by discarding blue, then green, then green
12. Get rid of the green ball by discarding red, then green, then green
13. Get rid of nothing by discarding blue, then red, then green
14. Get rid of nothing by discarding blue, then green, then red
15. Get rid of nothing by discarding red, then blue, then green
16. Get rid of nothing by discarding red, then green, then blue
17. Get rid of nothing by discarding green, then blue, then red
18. Get rid of nothing by discarding green, then red, then blue

So, 18 possibilities if I count them right, and out of them four end with me getting rid of the blue ball. So the probability of losing my blue balls would be of 4/18, or 2/9.

Do I know more than a 14 year old kid? inb4 I got it wrong

[*Emelia K. Fletcher]

each set of three is three hypotheticals

P(G) [1/3, no greens left]
P(B/R) [1, returns a green]
P(G) [1/5, no greens left]
1/15

P(B) [2/3, two greens left]
P(G) [2/4, one green left]
P(G) [1/5, no greens left]
1/15

P(R) [2/3, two greens left]
P(G) [2/4, one green left]
P(G) [1/5, no greens left]
1/15

all other outcomes will lead to at least one green being present by the end

hence, 1/5 that one specific colour has gone, 3/5 that any colour has gone, leaving a 2/5 chance that all colours are present by the end




(i am a 14 year old count me in your sample)

[lordpat]

Agh, that is almost perfect MK, but why arey ou saying the probability of taking a red one is 2/3? The probability of taking either a red one or a blue one is 2/3 but the probability of taking each individual one of the blue and red is 1/3 each.