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March 4

Summation of K^2

This is a homework assignment, that by the time this is answered will have already been due, but i want to know. What would be the first line of an algebraic proof (i.e. not mathematical induction) of $\sum _{i=1}^{n}i^{2}={\frac {n(n+1)(2n+1)}{6}}={\frac {n^{3}}{3}}+{\frac {n^{2}}{2}}+{\frac {n}{6}}$ Thanks, --Omnipotence407 (talk) 02:50, 4 March 2008 (UTC)[reply]

Here's one way. Suppose that the 1² + 2² + ··· + n² is some polynomial function P(n) of n. Then P satisfies: P(X)-P(X-1)=X², and P(0)=0. But there's exactly one such polynomial, and those two conditions clearly ensure that, conversely, P(n) is the sum of the first n squares. Bikasuishin (talk) 03:01, 4 March 2008 (UTC)[reply]

My favorite proof starts with

\sum _{i=1}^{n}i^{2}=\sum _{i=1}^{n}\sum _{k=1}^{i}i=\sum _{k=1}^{n}\sum _{i=k}^{n}i=\cdots

. -- Meni Rosenfeld (talk) 08:43, 4 March 2008 (UTC)[reply]

Nice one! Bikasuishin (talk) 13:16, 4 March 2008 (UTC)[reply]

See Graham, Knuth, Patashnik, Concrete Mathematics, which uses this as an example and as such shows several ways to solve it. – b_jonas 13:56, 5 March 2008 (UTC)[reply]

statistics question

a friend gave me this problem and i am struggling to solve it! assistance please:D

When John and Larry play a game of tennis against each other the probabilities of them winning are John (2/3) and Larry (1/3). John's is exactly twice that of larry's. The tennis rules are followed, win with atleast four points by 2 points. (due to time constraints) if a deuce is called, both with 3 points, the next point determines the winner.

-Show that there are 70 possible ways such a game may be played.

- to assist let X be the number of pts played, what values can x take?

- for each value of X find number of ways such game could be played and show a probability model for such a game.

I need this almost immediately! Thanks if you need anything else feel free to email me: [e-mail address removed to reduce spamming opporunities]

Tim

I ve tried everything, but i just cant get the 70 possible ways... you can have it like this

John w w w l l l  | w
Larry l l l w w w | l

and loads of diff ways, but i can only find a few!!!

~ anon ~

Work through one point at a time working out the options (you'll have to split it into multiple cases after the 1st 4 points). To start you off, there are 2 options for the first point (John can win, or Larry can win), 2 for the 2nd and 2 for the 3rd and 2 for the 4th, they are all independent, so that's 2*2*2*2=16 cases. After that, you have to take into account the possibility that the game will have already finished, so it's gets a little more complicated - you may wish to use some kind of probability tree to help. Good luck! --Tango (talk) 10:25, 4 March 2008 (UTC)[reply]

I assume (it is not clear from the formulation) that the probabilities 2/3 and 1/3 are the probabilities of winning the next point. A game might go like JLJJJ or LJLJJLJ, in which the letters indicate the successive winners. Then, I do find exactly 70 possible games. It does help to make a matrix in which the index for the rows gives the number of points John has won in the course of a possibly unfinished game (0 through 4, more is not necessary), and the other index for the columns likewise for Larry. So this is a 5×5 matrix, with one corner indexed by (0,0) and the opposite corner by (4,4). The position (4,4) in the matrix cannot be reached; the others can. The cell (0,0) is the starting position. In each cell of the matrix you can determine whether the game is still undecided, or who has won. You can also determine the number of ways of getting to that cell if you have already determined in how many ways you can get to the cells with fewer total points. (See also Pascal's triangle). Finally, and this is what makes it all work, for each given cell, all ways of reaching it have the same probability. (See also the binomial theorem; the probabilities in a diagonal with a constant total number of points correspond to the terms of such a sum.) Hope this helps. --Lambiam 13:19, 4 March 2008 (UTC)[reply]

Why are the rules so complicated? Couldn't you just say, "first person to 4 points wins"? Black Carrot (talk) 03:09, 5 March 2008 (UTC)[reply]

In this case, that's exactly what the rules are, since only having one point after a deuce completely eliminates the "must win by 2 points" rule. The idea for that rule is to reduce the effect of luck - if two people of equal ability play 7 points, one of them is sure to get 4 points, even though they aren't actually any better. By requiring the win to be by 2 points, they have to actually get a lead and keep it. It's not very important for a game, but the same rule applies to sets (must win by 2 games) and that is important because it removes the advantage of going first. --Tango (talk) 10:50, 6 March 2008 (UTC)[reply]

A needing question...

I have a YouTube account, and I was wondering about the average rate of viewers on my videos. So, how can I find out the average rate of viewers that watched my video with a calculator? —Preceding unsigned comment added by Sirdrink13309622 (talk • contribs) 11:11, 4 March 2008 (UTC)[reply]

You mean something like average number of viewers per hour? Just take the total number of viewers and divide by the number of hours the video has been available. --Tango (talk) 12:26, 4 March 2008 (UTC)[reply]

Uniqueness of reduced matrix

Hi. I'm working through Michael Artin's Algebra, because I want to learn about rings and fields, but I'm embarrassingly stuck by a problem about matrix operations in chapter 1 (problem 2.19, for anyone with a copy). The problem is to show that the reduced row echelon form of a matrix is uniquely determined, or equivalently, that if two matrices in RREF are row equivalent, then they are equal.

If you follow the link under "reduced row echelon form" above, it leads to a section in Row echelon form, where we read, "every matrix reduces to a unique matrix in reduced row echelon form by elementary row operations (see Elementary matrix transformations)."

I'd like to prove this assertion, using the definitions of matrix operations and the elementary row operations, which is basically all we've got to work with so far in the book. Vector spaces haven't been introduced, so I don't think I can talk about the row space of the matrix. I have an idea that I should use induction, and show that two row equivalent RREF matrices have their last rows identical, but I'm not getting how the details work out. Can anyone help me, please? -GTBacchus^(talk) 16:33, 4 March 2008 (UTC)[reply]

It's worth noting that both row echelon forms are related to the solutions of linear equations.. ie a_xy (a value in the matrix) could be the coefficient of z_x in the y^th linear equation

eg for y sets of equations of the type a_1,3z₁ + a_2,3z₂ + a_3,3z₃ +etc =0 (z 's are the unknown quantities here)

(More explicity 4a+2b+3c+ etc =0 , 5a+6b+0c+ etc = 0 as the unreduced form giving a matrix

(4 2 3 etc

5 6 0 etc

etc

)

The reduced form gives a solution to this set of linear equations .. the last row gives the most reduced (less variables) solution.. ((Swapping columns can give the values of other variables))

Now assuming you believed me when I said that - all you need to do (excluding a few tidies) is to show that a set of linear equations can only have one set of solutionshope I've used the word 'set' correctly here - hopefully my meaning was obvious (this seems obvious - but maybe you will want to prove that too - in which case - back to square one!)

Apologies If I seem to have simply restated your question.. Depends on how much you already knew (more than me is my guess)

In the meantime I too look forward to a PROPER solution to this....87.102.44.156 (talk) 18:54, 4 March 2008 (UTC)[reply]

It seems to me that row equivalent may have the answer you seek. Any row operation applied to an RREF matrix either destroys its RREF property or has no effect at all. The only trick is ruling out the possibility that two distinct RREF matrices are row-equivalent by a sequence of operations that remove and then restore the RREF property. I think you could do that with some elementary matrix analysis, perhaps. --Tardis (talk) 19:24, 4 March 2008 (UTC)[reply]

In particular, look at what multiplying by an elementary matrix does to the rank of that matrix. 67.103.157.66 (talk) 03:44, 5 March 2008 (UTC)[reply]

Log^2 n

What does $log^{2}n$ mean? Does it mean log(log(n))? --Dacium (talk) 23:37, 4 March 2008 (UTC)[reply]

It almost always means

(\log n)^{2}

, in the same way as

\sin ^{2}x=(\sin x)^{2}

. Not a very good notation, but rather common nonetheless. Bikasuishin (talk) 23:48, 4 March 2008 (UTC)[reply]

(edit conflict)It either means log(log(n)) or (log(n))². I don't think the latter is particularly commonplace (it is for trig functions, but not for log), but taking the log of a log seems an odd thing to do - what's the context? --Tango (talk) 23:50, 4 March 2008 (UTC)[reply]

Re: log of log: see iterated logarithm, although that's not exactly what's being discussed. --Tardis (talk) 00:41, 5 March 2008 (UTC)[reply]

A double log isn't that uncommon - for example, it is known that

\sum _{n=2}^{\infty }{\frac {1}{n\log n(\log \log n)^{\alpha }}}

converges iff

\alpha >1

, and the complexity of the fastest widely known multiplication algorithm is

O(n\log n\log \log n)\;\!

(though apparently this will change that). -- Meni Rosenfeld (talk) 10:48, 5 March 2008 (UTC)[reply]

The usual shorthand notation for log(log(n)) is log⁽²⁾n, with parentheses around the "exponent". Without the parentheses, it means take the log first and then square. In cases where one is repeating the log more than two times, this notation might be more useful, but for log(log(n)) I'd stick with spelling it out to avoid confusion. —David Eppstein (talk) 01:04, 5 March 2008 (UTC)[reply]

Writing 2 in parentheses can also be interpreted as a second derivative. -- Meni Rosenfeld (talk) 10:48, 5 March 2008 (UTC)[reply]