Calculator/Math Thread

**cmowla** · 07-22-2012

Originally Posted by IanTheCuber

Pascal's Triangle FTW

[...]

I think a similar method can be applied to Fibonacci Sequence, eh?

Not exactly. I mean it's even more simple to apply Pascal's Triangle to calculate the Fibonacci numbers: you don't have to adjust the pascal's triangle at all. Just look at Wikipedia. They show that

$F_{n}=\sum\limits_{k=0}^{\left\lfloor \frac{n-1}{2} \right\rfloor }{\left( \begin{matrix}n-k-1 \\k \\\end{matrix} \right)}$

and they show this image which represents that:

**IanTheCuber** · 07-29-2012

I just wanted to make it more "in depth"

**Ranzha V. Emodrach** · 08-06-2012

I got some errors that look like this:
Warning: touch() [function.touch]: Utime failed: Permission denied in [path]/includes/class.latex-vb.php on line 167

Warning: touch() [function.touch]: Utime failed: Permission denied in [path]/includes/class.latex-vb.php on line 167

Warning: touch() [function.touch]: Utime failed: Permission denied in [path]/includes/class.latex-vb.php on line 167

Anyways, that's not the problem I have for you guise.

Backstory: So there's this game my friend Shiaohan showed me in middle school involving a shuffled deck of cards. It was a pretty stupid game, looking back, but the gameplay was simple--draw the top card off the deck, turn it over on the table, and say "Ace."
If the card is an Ace, you get zero points, your game is over, and you must shuffle the cards and start over from zero points.
If the card isn't an Ace, you get one point, and you repeat the process of drawing cards, each successive time proclaiming the next value ("Two", "Three",... "King", and wrap around to "Ace" again), racking up as many points as you can before you have to start over.
The goal is to get 52 points, aka going through the entire deck without fail.
It quickly occurred to me that you can troll this game by having a sorted deck of cards and then placing one card from the top to the bottom. In this way, the result obviously comes out to 52 points, even if the deck wasn't completely shuffled.
Problem: What is the probability of playing this game and receiving a result of 52 points?

**ben1996123** · 08-06-2012

Originally Posted by Ranzha V. Emodrach

I got some errors that look like this:
Warning: touch() [function.touch]: Utime failed: Permission denied in [path]/includes/class.latex-vb.php on line 167

Warning: touch() [function.touch]: Utime failed: Permission denied in [path]/includes/class.latex-vb.php on line 167

Warning: touch() [function.touch]: Utime failed: Permission denied in [path]/includes/class.latex-vb.php on line 167

Anyways, that's not the problem I have for you guise.

Backstory: So there's this game my friend Shiaohan showed me in middle school involving a shuffled deck of cards. It was a pretty stupid game, looking back, but the gameplay was simple--draw the top card off the deck, turn it over on the table, and say "Ace."
If the card is an Ace, you get zero points, your game is over, and you must shuffle the cards and start over from zero points.
If the card isn't an Ace, you get one point, and you repeat the process of drawing cards, each successive time proclaiming the next value ("Two", "Three",... "King", and wrap around to "Ace" again), racking up as many points as you can before you have to start over.
The goal is to get 52 points, aka going through the entire deck without fail.
It quickly occurred to me that you can troll this game by having a sorted deck of cards and then placing one card from the top to the bottom. In this way, the result obviously comes out to 52 points, even if the deck wasn't completely shuffled.
Problem: What is the probability of playing this game and receiving a result of 52 points?

I think it would just be $\left(\frac{12}{13}\right)^{52}\approx1.56\%$

**Ranzha V. Emodrach** · 08-06-2012

Originally Posted by ben1996123

I think it would just be $\left(\frac{12}{13}\right)^{52}\approx1.56\%$

Um, no.

**ben1996123** · 08-06-2012

Originally Posted by Ranzha V. Emodrach

Um, no.

oh yeah. i just realised it's a deck of cards.

**Ranzha V. Emodrach** · 08-06-2012

Originally Posted by ben1996123

oh yeah. i just realised it's a deck of cards.

facehoof

It's okay xD

I think it's more of a binary pass/fail intuition system that I have to figure out.

**5BLD** · 08-28-2012

So over lunch I thought of this problem, and instead of solving it with just, the volormula of a sphere I decided to have a crack at integration. Of course, being me, I accidentally did it for a cylinder, not a sphere. So I chatted to ben about rotating it about the x axis, because I forgot how to do it and didn't think of doing it, and yeah here's the solutions.

He brought up another interesting point about chopping the sphere into n segments. Anyway enough build up:

question:

solution:

then we talked about chopping into more/fewer pieces. Weirdly it turns out than for -1<=x<=1 one of the cuts should be

Any thoughts?

**Endgame** · 08-28-2012

I'm hungry as ****, but s'graven haeg dude.

**StachuK1992** · 09-11-2012

Anyone care to help me figure these two problems out from Abstract Algebra I?

If a|(b + c) and gcd(b, c) = 1, prove that gcd(a, b) = 1 = gcd(a, c).

and

Show that if n lines are drawn on the plane so that none of them are parallel, and so that
no three lines intersect at a point, then the plane is divided by those lines into (n^2+n+2)/2 regions.

The second one, I simply have no idea how to approach it. We've done nothing similar to this that I can recall.
The first one, we've done stuff similar, but I'm not sure how to start it. Help me just start this one?
Thanks

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