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Mathematics Thread

What is the hardest math you learned?

  • Calculus (specify in comments)

    Votes: 67 35.6%
  • Trigonometry

    Votes: 24 12.8%
  • (Enriched) Geometry

    Votes: 17 9.0%
  • Equations (specify)

    Votes: 7 3.7%
  • Graphing Equations (specify)

    Votes: 4 2.1%
  • Basic Algebra (specify)

    Votes: 9 4.8%
  • I can do better!

    Votes: 47 25.0%
  • What kind of math is that?!

    Votes: 13 6.9%

  • Total voters
    188

cmhardw

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My wife and I just did an interesting math problem. I post it here because we had fun thinking about this, and perhaps others will too.

My wife is working on a project making maps in one of her classes. Her class is split up into teams of 5 people, and everyone in a team is working on their own project making a map. When working on a project, each member in a team of 5 people must ask another member in the team for feedback on their project.

Question: If everyone in a team of 5 randomly asks another member in their team for help, then what is the probability that everyone in the team is asked to review someone else's project?

For example, one outcome is that the other four team members might ask my wife to review their projects, and she would ask someone to review her project. This would be two people reviewing others' projects. Stating the original question a different way, what is the probability that there are five project reviewers in my wife's team?
 

lerenard

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My wife and I just did an interesting math problem. I post it here because we had fun thinking about this, and perhaps others will too.

My wife is working on a project making maps in one of her classes. Her class is split up into teams of 5 people, and everyone in a team is working on their own project making a map. When working on a project, each member in a team of 5 people must ask another member in the team for feedback on their project.

Question: If everyone in a team of 5 randomly asks another member in their team for help, then what is the probability that everyone in the team is asked to review someone else's project?

For example, one outcome is that the other four team members might ask my wife to review their projects, and she would ask someone to review her project. This would be two people reviewing others' projects. Stating the original question a different way, what is the probability that there are five project reviewers in my wife's team?

11/256?
I labeled the people A through E and calculated 256 possible combinations based on the assumption that B reviews A (assuming that is one "orbit" and that other possibilites where someone else reviews A would be parallel and so aren't worth calculating individually). If two pairs of people review the person who reviewed them, then not everyone has been a reviewer, so I first paired each group of people up and found there were 6 combinations where one pair reviewed each other, and then another 6 where no one reviewed the person who reviewed them. Is this right?

Oh, and in response to your response about the mean versus the mode, I vaguely remember that from AP Statistics... The thing about that class is that I never actually learned anything, I just zoned the teacher out and used the example problems in the textbook to figure out how to do my homework and then it was fresh enough in my memory that I could figure out how to make stuff work on test day, but as far as applying stuff from way back then to a random problem now, there's no hope.
 
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Cale S

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Question: If everyone in a team of 5 randomly asks another member in their team for help, then what is the probability that everyone in the team is asked to review someone else's project?
I think this is right...

Let's say the 5 people in the team are called A, B, C, D, and E.
There are 5 people and each can choose one of the 4 others. This means a chart showing all possible selections would have 4^5 = 1024 different possibilities.
Let's also say that we represent the choices made by a sequence of 5 letters, first with who A chose, then who B chose, etc. If we had BCAAD, that means C and D both chose A, B chose C, D chose E, etc.
For everyone to be chosen, the 5 letter sequence has to be an anagram of ABCDE. Also, A cannot be in the first position, B cannot be in the second, C cannot be in the third, etc.
The number of anagrams of ABCDE (including itself) is 5! = 120. Only 42 of these have A not in the first position, B not in second, etc.
42/1024 ≈ 4.10%



Edit: ok apparently I was wrong (I counted 42 by hand) and it should actually be 44 so 44/1024 ≈ 4.29%
 
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Stefan

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Your "my wife" sounds so weird to me. Not sure whether it's because you're married or because I'm used to you using her name :)

Of the 4^5 possibilities, we're interested in every 5-cycle (there are 4!) and every 2-cycle+3-cycle (there are 5c2*2):

11/256

Or with simple counting: http://ideone.com/VuEkvR
 
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cmhardw

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Your "my wife" sounds so weird to me. Not sure whether it's because you're married or because I'm used to you using her name :)

I do usually use her name on the forum. I thought to post more generally for those on the forum who don't know me and who I haven't met. I'll probably just use her first name in the future :)

My solution:
I also got 11/256

I counted 4^5 total ways that 5 people can choose someone else to review their project.

I counted the derangements of 5 elements ways that all five people can be assigned as reviewer to someone else (since no one can review themselves).

5!*(1/2!-1/3!+1/4!-1/5!)=44

44/1024

11/256
 

Ickathu

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Wow, it isn't everyday that you hear someone say that! :tu
(Though I suppose being on the math thread in a cubing forum helps, lol)

Most number theory beyond the level of your class would probably no longer actually involve numbers... Though I think if you enjoy other parts of higher math (linear algebra? calculus? real analysis? group theory?) you'll enjoy more advanced number theory as well.

Here's my favourite theorem. It's easy to state but its proofs can be very involved:

Let p,q be distinct odd prime numbers. Consider the two congruences

x^2=p (mod q)
y^2=q (mod p),

where x,y are integers. Then either both congruences have a solution, or neither of them do,
unless
p=q=3 (mod 4), in which case exactly one of the congruences has a solution in integers.

Shoutout to all math students out there: what's your favourite theorem?

Calc is pretty cool. I'm doing statistics now, but I might also do linear algebra this semester. Group theory fascinates me, but I've never formally done it.

Favorite theorem? That's a tough question. I have no idea.
 

cmhardw

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Man, your solution is a harder challenge that the question :). Please explain or give a hint.

Let (x,y) be an ordered pair denoting
x=Project Owner
y=Project reviewer

x may not equal y (a person cannot review their own project)

Let there be five people named 1,2,3,4,5

I am interested in all ordered pairs:
(1,_)
(2,_)
(3,_)
(4,_)
(5,_)

where y is an element of {1,2,3,4,5} and x/=y

Each person has a choice of 4 reviewers, so 4^5 counts the number of ways that the whole group may ask people to review their project.

To count the number of ways that all five members of the group may be a reviewer I want to know the number of functions I can create out of these five ordered pairs where the range is the entire set {1,2,3,4,5} and x/=y

If x may not equal y, then there are the derangements of 5 elements ways to create ordered pairs such that x/=y and the range is the set {1,2,3,4,5}

5!*(1/0!-1/1!+1/2!-1/3!+1/4!-1/5!) uses inclusion/exclusion to count the number of ways to derange five elements and is equal to 44.

44/1024=11/256 is the probability that every member of the group is asked to be a reviewer.
 
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Stefan

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5!*(1/0!-1/1!+1/2!-1/3!+1/4!-1/5!) uses inclusion/exclusion
Ah, I didn't realize I could add those first two summands and they'd cancel out. I still had to google to find the meaning of the summands, and then cancel and factor out. I got this longer formula where i is the minimum number of "self-reviewers". From this I get to your formula as well. And it's certainly the neatest solution, we others all seemed to have non-scalable solutions (I thought mine was neat, but for larger than 5 I'd need to reanalyze and it'd just get messy).
 

cmhardw

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Ah, I didn't realize I could add those first two summands and they'd cancel out. I still had to google to find the meaning of the summands, and then cancel and factor out. I got this longer formula where i is the minimum number of "self-reviewers". From this I get to your formula as well. And it's certainly the neatest solution, we others all seemed to have non-scalable solutions (I thought mine was neat, but for larger than 5 I'd need to reanalyze and it'd just get messy).

I took a combinatorics class in college and my professor for that class really liked the inclusion exclusion technique. I think that's partly why I like it so much, because we used it a lot in class.

He started with the general version of your formula and showed us how after factoring out n! that it reduces to this and if n>1 it reduces even further to this. I use this last formula when calculating derangements when n>1.
 
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Stefan

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Holy ****, now that both you and Ravi called it derangements, I found out that not only is that a well-known term, but there's even a notation for it so I can say !5. Wow. That's gonna come in handy for those riddles where you have to reach certain numbers by combining certain others.
 

cmhardw

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Holy ****, now that both you and Ravi called it derangements, I found out that not only is that a well-known term, but there's even a notation for it so I can say !5. Wow. That's gonna come in handy for those riddles where you have to reach certain numbers by combining certain others.

I learned the term derangement in my courses, but I did not know about the notation !n for the derangements of n items, cool! Yes, derangements are fun and they do seem to come up from time to time in combinatorial problems and riddles.
 

Stefan

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Lol. Should've said speedsolved by a person

When I scramble and solve, I go through roughly 80 positions (even more when I'm "hand-scrambling" for a while). Do all of those count, or only those "during the solve", or or only the one "between scrambling and solving"? Also, does FMC count (asking because it's not really "speed"solving)?
 

obelisk477

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When I scramble and solve, I go through roughly 80 positions (even more when I'm "hand-scrambling" for a while). Do all of those count, or only those "during the solve", or or only the one "between scrambling and solving"? Also, does FMC count (asking because it's not really "speed"solving)?
I'll only answer what I'm looking for if you commit to actually giving it a go once you're satisfied you have all of the stipulations
 

Artic

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hey guys,

Hopefully the math gurus can help me on a problem. I need to fit a 5th degree polynomial through two points, P0 (x0,y0) and P1 (x1,y1). Obviously, there are an infinite number of 5th degree polynomials that pass through the same points P0 and P1. What I need is a way of controlling which polynomial is created. For example, if I have the function defined as:

2zy8z1c.png


I want to be able to control which curve is created by changing the time t. In that way, I can create a family of 5th degree curves all of which pass through the same two points.

I should get something that looks like the following:

2w57n1d.png


Any ideas how to do this?
 

Christopher Mowla

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Some might recall that earlier in this thread, I posted about a new method I invented regarding how to derive formulas like \( \sum\limits_{i=1}^{n}{i}=\frac{n\left( n+1 \right)}{2} \) from scratch using only a simple image.

Yesterday, I created and uploaded a three part video series on YouTube. I present the method in "Part 1", abstract it in "Part 2", and prove it in "Part 3". For those interested, enjoy!:)
 
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