Mathematics

[Chokladkakan]

Good evening fellow Internet,

Surely I cannot be the only one on these forums who find beauty in a delicately formulated derivative equation or the intricate identities of the trigonometric properties?

Mathematics is, has always been and shall remain art. Why else did the thinkers of past times spend equal amounts of time painting as they formulated theorems and dictated their wonderful ideas?

Anyone is welcome, whatever level of mathematics you happen to be pursuing for the moment. For those interested I am sure I won't be the only one happy to help with explaining things or helping out when met with a problem.

If you happen to be stuck with homework, I don't mind leading you in the right direction either!

Since I have no substantial opening topic, I recommend exploring the Cambridge University's Nrich math site (http://nrich.maths.org/) for some entertaining brain teasers.

PS. Let us all mourn a bit over the passing away of Benoît Mandelbrot (http://www.guardian.co.uk/science/2010/oct/17/benoit-mandelbrot-obituary), the pioneer in fractal geometry and a man who most certainly saw his math as art.

[Cthulhu]

http://z0r.de/L/z0r-de_1976.swf

[Chokladkakan]

Background music! Excellent! The only problem is that I will now have to have this thread open all the time... I also wish it would go on longer.

I thought I'd introduce some problems to solve in here, problems to which I do have the answer if confirmation would be asked for. I'll have one problem for higher sort of levels and one for lower ones. If you feel that either one is too hard or too easy, please say.

Higher level:
Given that the area of any triangle can be computed with http://upload.wikimedia.org/math/2/2/2/222ecf4bb160025a3cf7cebf0dbd7b64.png where C angles is the angle included between the sides a and b, prove the much older method of calculating a triangles area: Heron's Formula, http://upload.wikimedia.org/math/6/8/a/68a3ad9c82dc198e451b8bad4155ab7a.png, where s is the semi-perimeter of the triangle, http://upload.wikimedia.org/math/9/4/e/94e13c4f066fd8802476f8646c1a5134.png.

If you wish, you may also explain the logic behind Heron's Formula, thus explaining how Heron of Alexandria might have come up with it. Keep in mind the one we commonly use, the above mentioned, wasn't yet invented.

Lower level:
This problem will require a bit of knowledge about probability, however not more than I reckon you cover in middle school or the like.

We have a lottery set up in front of us. The lottery consists of four balls, numbered 1 through 4, all of which are placed in a bag.
To enter the lottery you have to pick a number. Consequentially to win, the number you picked has to be the same as the one picked.

What are the chances of winning?

If the lottery turns out to be a tad bit too easy to win in, imagine instead that you have to pick two numbers. Two numbers from the bag are then picked and you will win only if both of the numbers are your numbers in no particular order.

How hard is it to win with this new setup? Would you play?

[Xaozaddiction]

Man, hard problem is hard.

Lottery 1: you have a 25% chance to win. Good odds yes?
Lottery 2: 1/6 (around 17%) chance to win. Not as good of odds. I think i will stick with the first.

[Cthulhu]

Nothing to see here folks, move along <_<

[Chokladkakan]

Man, hard problem is hard.

Lottery 1: you have a 25% chance to win. Good odds yes?
Lottery 2: 1/6 (around 17%) chance to win. Not as good of odds. I think i will stick with the first.

The answers are correct. Would you like a follow up question, or perhaps something harder (yet easier than the hard one (or rather less dependent on trigonometric skills))?

About the hard one: it isn't actually very hard, provided one has the required background knowledge. It requires you to understand some of the trigonometric identities of the common trigonometric functions, as well as how to apply some of the usual rules of trigonometry with sort of non-ordinary inputs. I don't know what sort of math you are doing, but it might very well be that you haven't encountered that sort of thing yet.


I'll also throw out a follow up for the lottery be it you want it or not:

Let us increase the number of balls with unique numbers in the bag to ten. Decide yourself whether the player should pick two or three numbers now, or perhaps both, it is up to you. Furthermore it will not cost 4 coins to participate.

This time, however, you must not get all of your numbers right to win, instead you will get more money the more right ones you have got.

For example: suppose you picked 3, 5 and 7. The numbers drawn are 2, 3 and 7. That would mean you got two out of three. The number of corrects will reward you with this many coins respectively:


4 coins
12 coins
24 coins


Given this, what will your average reward be for playing one time? Is it sensible to play?

[Ivernus]

@Chok, first question is interesting...still mucking around with it but not really sure where to start, trig. was never my strong point


Well considering I just finished an exam on Discrete Mathmatics, I figure adding something to this thread would be fitting. So i'll do one of the things I remember better from the exam (so I don't have to go get a book), and that is Coding Theory!

I'll gloss over most of the details, but the idea behind coding theory is to be able to send a message between Parties 'A' and 'B' so even if Party 'C' intercepts them, they cannot interpret the message. One of the most common ways to do this is to encode messages into numerical data (numbers) which are really easy to manipulate and apply a 'encoding' pattern. This encoding pattern scrambles the message into a bunch of letters that make no sense unless you know the 'decoding' pattern.

The most commonly used way of encoding the letters is to make A = 0, B = 1, C = 2, D = 3 ... X = 23, Y = 24, Z = 25. The reason for this is to allow the use of Modulo 26. Not going into too much detail the Modulo function returns the remainder of a number when its divided by another number. So for example 17 Modulo 5 = 2, as 5x3 + 2 = 17 (2 is the remainder part). The reason this is important for codes is, if a number goes over 25, you need a way to loop it back round to the start, and Modulo 26 is an incredible easy way to do this.

Anyway I don't want to go too deep into the concepts here (I know thanks to emma doing that is a great way to Alienate myself =D), so now I'll go into some actually examples

The Shift Cipher
Possibly one of the most basic codes, the idea is you 'shift' all of the letters along in the alphabet by a set number. Using an example to demonstrate, Using a shift Cipher of 1:
Hello World (which numerical is: 7,4,11,11,14,22,14,17,11,3)
Becomes
IfmmpXpsme (8,5,12,12,15,23,15,18,12,4)

Decrypting a shift Cipher is rather easy, and can be done by trail and error for small messages because there are only 26 combinations, which is why this code is no longer really used as its so easy to break

The Affine Cipher
The next step up from the Shift Cipher, the affine code is alot more complicated. The idea behind the Affine Cipher is to use a permutation to re-arrange the letters in a more complicated way. Now I won't go into the technical details of this but to explain it in simple terms, when encoding a message, you take the numerical value, multiple it by a number and then add a second number.
So, encoding the letter 'B' with an an Affine Cipher (3,5):
B = 1
Encoding B = ((1)*3)+5 (Modulo 26) = 8
Encoded B = I

Now due to the fact we use multiplication, decoding is not as simple as the Shift Cipher. In order to decode the message we need to inverse the process of the encoding, without using division. To do this, we rely on the fact that under multiplication in Modulo 26, each number has an inverse. What this means is that
'B' * a * a-1 = 'B' (where a-1 is the inverse of a)
To find an inverse of a, we need a number such that:
a * a-1 = 1 (Modulo 26)
So for the example of 3
3 * 9 = 27, and 27 = 1 Modulo 26 (26 + 1)
So 9 would be our inverse

So now that we have our inverse, we can decode the letter
H = 8
Decoding H = (8 - 5) * 9 = 1
Or
Decoding H = (8 + 21) * 9 = 261 (Mod 26) = 1 (The reason for this is 21 and 5 are inverses, as 21 + 5 = 26. You really only need to use this if the number is to small and is going to go into negatives)

1 = B
Anywaaay, I think I'll stop there now, and leave some examples for people to try, although I doubt people would XD.


Questions
1) The following message is encoded using a shift cipher 9, try to decode it:

cqjcrblxaanlc


2) The following message was encoded using an affine cipher (7,7) try to decode it:

hgdbvbwwjvk

I sure hope I did those two right.

Edit: Typo found by chok, cause he is awesome like that

[screwbaII]

(I know thanks to emma doing that is a great way to Alienate myself =D)

:| That's only because when you start talking about math your speech turns into something like blah blah blah numbers numbers blah blah and my eyes kind of start to feel like this after a while:
http://1.bp.blogspot.com/_COQNR9EHR28/So1tnZrc8_I/AAAAAAAAAjs/Wh6LM_EgvGc/s320/googly-eyes.jpg

This post isn't math related at all, but I'm just pointing out maths is the only thing that can make me feel really stupid, thus I am not going to post in this thread :| (anymore)

[leaves the smart people to their smart subjects and goes back to writing a lab report on retinotopic and spatiotopic relevence in gender specific face aftereffects ;~;]

[Chokladkakan]

Thanks for bringing a very interesting topic up! I presume you have read Simon Singh's The Code Book as well? :)

Question 1:


I will solve this visually for the sake of presenting another method, if you don't mind:


Input: CQJCRBLXAANLC

Plaintext:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Cypher text:
J K L M N O P Q R S T U V W X Y Z A B C D E F G H I
^
^
^
^
^
^
^
^
^
^
^
^
^

Deciphered text:
THATISCORRECT


Question 2:


You did in fact make a mistake on this one:


A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

With 7*(number of letter)+7 mod 26 goes to

7 14 21 2 9 16 23 4 11 18 25 6 13 20 1 8 15 22 3 10 17 24 5 12 19 0
H O V C J Q X E L S Z G N U B I P W D K R Y F M T A

Which with the ciphered input HQDBVBWWJVK gives AFSOCORRECT. I believe your input should be HGDBVBWWJVK instead.

[Ivernus]

Haha, not surprised I made a typo on that code, my brain is so fried at the moment, that said I've editted it so other people can try it if they want. Unfortunately I haven't read Simon Singh's Codebook, but I have been interested in it, whats your opinion on it?

Also just for fun, I'll pose another coding question, but I'll see how many people can solve it before I explain it.

RSA Code Question
The following message was intercepted:
7, 4, 11, 11, 14
However it has been encoding using the RSA Code Format, meaning you only know that a = 7 and n = 77.

I'll edit in how this code works tomorrow when I'm not so tired

[Chokladkakan]

I find it to be a rather wonderful read, as with all of Singh's books! I strongly recommend it.

I have to take care of some choirs, but I shall have it solved for you by afternoon, I hope. :)

[Ivernus]

Mmm, in that case when I have resecured a job and have some extra cash, I will seek to acquire it.

Also to keep you entertained throughout the day, here is an old classic puzzle I remember from years ago.

The Missing Equation

The following equation has had all the numbers removed from it, however you remember that every digit between 0 to 9 appeared exactly once in the equation. Can you place the numbers to fit the equation?

_ _ + _ _ = _ _, x _ = _ _ _ (Forums make this hard to lay out :(


If thats hard to understand. A 2 digit number + another 2 digit number, gives a 2 digit number as a result. This resultant, when times by a 1 digit number, will give a 3 digit number answer

[Chokladkakan]

I find the RSA cryptography very interesting. Your problem I cannot seem to solve just yet, despite having been able to solve other problems I made up. It could be that I am misinterpreting your notation, for it doesn't seem to be the standard. By a, do you mean the the decoder or encoder constant, and by n do you mean pq or Euler's totient of pq?

[Ivernus]

Ah sorry, should have specified that. The way we were taught is
n = pq
a = encoder
b = decoder
I'll put up some more information on it abit later today.

[Chokladkakan]

I am afraid I have not been able to decode your message, Ivernus.

On a different note, I have a problem for you guys:

Imagine there is a conference coming up somewhere in the world. The conference will have thirty participants sitting around a round table. The round table is prepared in beforehand with name tags on each seat, one for each booked participant.

Come the day of the conference, and for some reason or the other everyone's terribly confused. This results in no one sitting at their appropriate seat.

The chief of the conference then says that if this happens again tomorrow, they will rotate the table and at least two people have their tags in front of them.

Is that claim correct? Are you able to prove it? If so, please do show your working!

Oh, an answer for the missing equation by the way:

39 + 46 = 85, * 2 = 170

About the RSA code:

This is what I have got so far:

p = 11
q = 7
n = 77
m = 10 * 6 = 60

a = 7

b * a mod 60 = 1
Hence
b = 43

Problem is, with b being 43, I get the following:

C(1 - 5) = 7, 4, 11, 11, 14

M(1) = C(1) ^ b mod n
= 7 ^ 43 mod 77 = 35
M(2) = C(2) ^ b mod n
= 4 ^ 43 mod 77 = 53
M(3) = C(3) ^ b mod n
= 11 ^ 43 mod 77 = 11
M(4) = C(4) ^ b mod n
= 11 ^ 43 mod 77 = 11
M(5) = C(5) ^ b mod n
= 14 ^ 43 mod 77 = 49

Hence
M(1 - 5) = 35, 53, 11, 11, 49

Which makes little to no sense...

[thebunni15]

The claim probably will not be correct, since it's happening the next day, people will probably not be confused and sit at the appropriate designated seats.
But, if by chance, it does happen, the probability that the claim is correct is about 6%.
Then again, what is the chance that the conference will happen again tomorrow?


That's my answer...I am entitled to it myself.

[Chokladkakan]

The claim has nothing to do with whether or not they are confused tomorrow, since the chief of the conference explicitly says that IF it happens again before stating his hypothesis.

The claim is correct, though:

The train of thought is as follows:
If thirty people are to get seated around a table with explicit name tags, yet no one sits in front of their own, every single person will be off from their seat by some amount of chairs.

Now if you are sitting right next to your name tag, you would be one (1) step away. If you're sitting slightly further off you might be 5, 6 or 7 seats away, with the maximum possible being 29, if you are right next to your seat but on the "wrong side."

If you assign such numbers to everyone around the table, there will be 30 people with numbers between 1 and 29, and here's the trick: for that to be true, at least two people must share numbers.

This is known as the Pigeonhole Principle and it applies when any number option, n, is less than the number of participants, m.

[Chokladkakan]

On the topic of mathematics being beautiful, have the butterfly curve:
http://i53.tinypic.com/33lcgvs.gif
http://i51.tinypic.com/s4ynvc.png

[thebunni15]

Math can look very nice, especially yours.

[Cthulhu]

http://www.superliminal.com/cube/applet.html

[Sbacity_Boy]

Sorry, I'm not here to answer any of these problems. I myself am just in the basics and have trouble doing Logic Puzzles. But I found this cool 4x4 box that the rows, columns, diagonal crosses, and every 2x2 squares in this 4x4 add up to the same number. For example:

______________
34l___l___l___l___l
34l___l___l___l___l
34l___l___l___l___l
34l___l___l___l___l
34 34 34 34 34 34

As you can see, it is a 4x4 box. And in those boxes you must put the number 1-16 using each number only once. And then all the columns, rows, diagonal crosses(bottom left to top right; Top left to bottom right), and 2x2 boxes will equal 34. I have the answer and if you can give me the number from top left to bottom right then I can tell you if you got it right. So start filling in, each box has the possibility of 16 numbers!!

[Chokladkakan]

These patterns are known as magic squares and have been a recreational mathematics branch for ages. People often play sudoku nowadays, which has the same origin. :)

The one you present there is this:


16 | 3 | 2 | 13
-----------------
5 | 10 | 11 | 8
-----------------
9 | 6 | 7 | 12
-----------------
4 | 15 | 14 | 1


It was created by Albrecht Dürer in 1514, which is why the two center pieces at the bottom will say 15, 14. The numbers 4 and 1 on the remaining bottom cells represent the letters D and A, as in Dürer, Albrecht.

If you were to analyze a magic square you would realize that there are internal symmetries everywhere (start from the bottom line of the diagram as you read it):

| | | 13 +2 16 | | | +6
----------------- -----------------
| | 11 | +5 | 10 | | +3
----------------- -----------------
| 6 | | +2 | | 7 | +6
----------------- -----------------
4 | | | | | | 1

(Note how the change between 2, 5, 2 is +3, -3 and the change between 6, 3, 6 is -3, +3)


| 3 | 2 | | | |
----------------- -----------------
| | | | 10 | 11 |
----------------- -----------------
| | | | 6 | 7 |
----------------- -----------------
| 15 | 14 | | | |

-1 +1


| | | | | |
----------------- -----------------
5 | | | 8 -4 | 10 | 11 | +4
----------------- -----------------
9 | | | 12 | 6 | 7 |
----------------- -----------------
| | | | | |


16 | | | 13 +12
-----------------
| | |
-----------------
| | |
-----------------
4 | | | 1
-3


You will also find an intriguing pattern if you draw lines between 1 - 4, 5 - 8, 9 - 12 and 13 - 16.

These symmetries and patterns are obvious because they have to exist for the square to be magic, which is what makes them rather interesting.

Another interesting square is this:


96 | 11 | 89 | 69
-----------------
88 | 69 | 91 | 16
-----------------
61 | 86 | 18 | 99
-----------------
19 | 98 | 66 | 81


Which is you physically rotate it 180 degrees turns into this:


18 | 99 | 86 | 61
-----------------
66 | 81 | 98 | 19
-----------------
91 | 16 | 69 | 88
-----------------
89 | 68 | 11 | 96


Look at it closely and you'll realize that they are both magic with the same sum: 264.

PS. There are problems in here suitable for people who have studied less mathematics too: this post by Ivernus (http://bunnibunni.com/forums/showpost.php?p=87517&postcount=12) is a good example!

[Sbacity_Boy]

That's weird, cause on mine it did? Add up I mean.

The answer I have is this. Unless I made missed something I am pretty sure they all do.

_1_l_12_l_7_l_14_
_8_l_13_l_2_l_11_
_10_l_3_l_16_l_5_
_15_l_6_l_9_l_4_

[Chokladkakan]

Huh, fancy that!

[Sbacity_Boy]

xD Yeah, when I first saw it our teacher said that it added up to a "special number" when you took the four numbers in the columns and rows, and later when we knew the number(34) we found that every 2x2 box did to. We thought that was awesome! ^.^

[Chokladkakan]

The only downside of magic squares is that they are, like I said, only recreational. Much like Fermat's Last Theorem they give us no insight into deeper things.

[Chokladkakan]

I have a new problem, this time for people who have studied or are studying calculus:

As we start studying calculus I reckon most start with differentiation. You probably get to explore a few functions and come up with ideas as of how the gradient of the function changes, but then you are simply given a table to refer too, without too much of explanation or, more importantly (for me, anyway), proof.

So, what I propose you do is prove http://latex.codecogs.com/gif.latex?\frac{\mathrm{d}%20}{\mathrm{d}%20x}(e^{ x})=e^{x}

Hint of you get stuck:
You will most likely have to prove http://latex.codecogs.com/gif.latex?\frac{\mathrm{d}%20}{\mathrm{d}%20x}(\ma thrm{ln}%20x)=\frac{1}{x} in order to have a solid proof of http://latex.codecogs.com/gif.latex?\frac{\mathrm{d}%20}{\mathrm{d}%20x}(e^{ x})=e^{x}.

I will gladly present my way of solving it either upon request or after someone proposes their approach.

[Chokladkakan]

I have been pondering about the 196-Algorithm (http://mathworld.wolfram.com/196-Algorithm.html) for quite a while. For that matter, I wanted to formulate an efficient, generic and scalable way of getting the inverse of any whole number.
Long story short: I succeeded last night, and I dare say it is rather elegant:
http://latex.codecogs.com/gif.latex?%5Clarge%20%5Ctextup%7BNumber%20%7Dn%20% 5Cin%20%5Cmathbb%7BN%7D,%20%5Ctextup%7B%20%7D%20%5 Ctextup%7Blength%20%7D%20m=1+%5Ctextup%7Bint%7D(%5 Clog_%7B10%7D(n))
http://latex.codecogs.com/gif.latex?\large%20\textup{Reverse%20}r(n)=9.9\cdo t%2010^{m-2}\cdot%20\sum_{l=1}^{m}(\frac{1}{10^{2\cdot%20(l-1)}}\cdot%20(n%20\textup{%20mod%20}%2010^{l}))+\fr ac{n}{10^{m+1}}
http://latex.codecogs.com/gif.latex?%5Clarge%20%5Ctextup%7BExample:%20%7D%20 r(429183920482)=284029381924

By the way: would someone be interested in the proof of the last problem?

[Chokladkakan]

Proof of http://latex.codecogs.com/gif.latex?\frac{\mathrm{d}%20}{\mathrm{d}%20x}(e^{ x})=e^{x}:

There are multiple ways of doing this, depending on what definition of http://latex.codecogs.com/gif.latex?e you use. I chose to pursue it as being whatever value http://latex.codecogs.com/gif.latex?x necessary for http://latex.codecogs.com/gif.latex?\ln%20x=1:

http://latex.codecogs.com/gif.latex?%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cma thrm%7Bd%7D%20x%7D(%5Cmathrm%7Bln%7De%5E%7Bx%7D)

Method one (using exponential properties of logarithms):
http://latex.codecogs.com/gif.latex?%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cma thrm%7Bd%7D%20x%7D(%5Cmathrm%7Bln%7De%5E%7Bx%7D)=% 5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cmathrm%7Bd%7D %20x%7D(x%5Ccdot%20%5Cmathrm%7Bln%7De)
http://latex.codecogs.com/gif.latex?%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cma thrm%7Bd%7D%20x%7D(%5Cmathrm%7Bln%7De%5E%7Bx%7D)=1

Method two (using the derivative of the natural logarithm):
http://latex.codecogs.com/gif.latex?%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cma thrm%7Bd%7D%20x%7D(%5Cmathrm%7Bln%7De%5E%7Bx%7D)=% 5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cmathrm%7Bd%7D %20x%7D(e%5E%7Bx%7D)%5Ccdot%20%5Cfrac%7B1%7D%7Be%5 E%7Bx%7D%7D=1,%5Ctextup%7B%20%7D%5Cbecause%20%5Cte xtup%7Bmethod%20one%7D
Hence, solving for http://latex.codecogs.com/gif.latex?\frac{\mathrm{d}%20}{\mathrm{d}%20x}(e^x )
http://latex.codecogs.com/gif.latex?%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cma thrm%7Bd%7D%20x%7D(e%5E%7Bx%7D)=1%5Ccdot%20e%5E%7B x%7D=e%5E%7Bx%7D


The second method, however, utilized the derivative of the natural logarithm and thus I will have to prove that too. There are multiple ways of doing that as well, again depending on what formal definition one uses, but I did as follows using first principle:

http://latex.codecogs.com/gif.latex?%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cma thrm%7Bd%7D%20x%7D(%5Cmathrm%7Bln%7Dx)=%5Clim_%7B% 5Ctriangle%20x%5Crightarrow%200%7D(%5Cfrac%7B%5Cma thrm%7Bln%7D(x+%5Ctriangle%20x)-%5Cmathrm%7Bln%7Dx%7D%7B%5Ctriangle%20x%7D)
What follows is all just simplifying using the properties of logarithms in reverse and basic rules of division:
http://latex.codecogs.com/gif.latex?=%5Clim_%7B%5Ctriangle%20x%5Crightarrow% 200%7D(%5Cfrac%7B1%7D%7B%5Ctriangle%20x%7D%5Ccdot% 20%5Cmathrm%7Bln%7D(%5Cfrac%7Bx+%5Ctriangle%20x%7D %7Bx%7D))
http://latex.codecogs.com/gif.latex?=%5Clim_%7B%5Ctriangle%20x%5Crightarrow% 200%7D(%5Cmathrm%7Bln%7D((%5Cfrac%7Bx+%5Ctriangle% 20x%7D%7Bx%7D)%5E%7B%5Cfrac%7B1%7D%7B%5Ctriangle%2 0x%7D%7D))
http://latex.codecogs.com/gif.latex?=%5Clim_%7B%5Ctriangle%20x%5Crightarrow% 200%7D(%5Cmathrm%7Bln%7D((%5Cfrac%7Bx%7D%7Bx%7D+%5 Cfrac%7B%5Ctriangle%20x%7D%7Bx%7D)%5E%7B%5Cfrac%7B 1%7D%7B%5Ctriangle%20x%7D%7D))
http://latex.codecogs.com/gif.latex?=%5Clim_%7B%5Ctriangle%20x%5Crightarrow% 200%7D(%5Cmathrm%7Bln%7D((1+%5Cfrac%7B%5Ctriangle% 20x%7D%7Bx%7D)%5E%7B%5Cfrac%7B1%7D%7B%5Ctriangle%2 0x%7D%7D)) http://latex.codecogs.com/gif.latex?%5Ctextup%7Blet%20%7D%5Cfrac%7B%5Ctriang le%20x%7D%7Bx%7D=u,%20%5Ctherefore%20%5Ctriangle%2 0x=u%5Ccdot%20x

http://latex.codecogs.com/gif.latex?=%5Clim_%7Bu%5Crightarrow%200%7D(%5Cmath rm%7Bln%7D((1+u)%5E%7B%5Cfrac%7B1%7D%7Bu%5Ccdot%20 x%7D%7D))
http://latex.codecogs.com/gif.latex?=%5Clim_%7Bu%5Crightarrow%200%7D(%5Cmath rm%7Bln%7D((1+u)%5E%7B%5Cfrac%7B1%7D%7Bu%7D%7D)%5E %7B%5Cfrac%7B1%7D%7Bx%7D%7D)
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[Chokladkakan]

You folks have forgotten about this thread. That is ok: I'm sure it was an honest mistake. :)

Either way, I have a problem most of you should be able to fathom:

There are numbers called triangle numbers. They represent the amount of stones or beads of whatever needed to form equilateral triangles. As such, the first triangle number is 1, the second is 3 and the third is 6. The fourth number is 10 and the fifth is 15.

I now pose a small challenge: what is the 167th triangle number?

[Josephzu]

I really like football, Beck explained. I really like taking part in the game. I really like standing in there on 3rd down and finding whacked within the chin and completing a pass and quieting the crowd. When youre enjoying on the road and its 3rd down and theyre screaming, and you may rip a pass in there and get it and shut em all up? I really like it.
I really like scoring a touchdown and figuring out precisely what that man or woman that I threw the touchdown pass to put into it. I like knowing that my O-line by no means gets to touch the football, but theyre busting their butts due to the fact they need to win, and theyre okay with not touching the soccer....
I really like seeing every person fired up. I like the feeling after the sport if you set in all that work and also you won. And occasionally, though I hate losing, at times I like the sensation after a loss, when you say you realize what, that sucked, Theres A great number of emotions that I have missed because I havent been in a position to get the guy. Ive been the backup.

[MurfillerbMugmovie]

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