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February 13

Finding x in logs

log_{2} x + log_{2} (x+5) = log_{2} 9

I use the law of logs and multiply x into x+5, then I raise 2 to both sides and end up with x^2+5x=9 but I can't factor that and get nice numbers, and I know I'm not supposed to use the quadratic formula. What am I doing wrong here? 98.221.85.188 (talk) 03:40, 13 February 2009 (UTC)

Complete the square from first principles like an honest man? Algebraist 03:46, 13 February 2009 (UTC)

5 doesn't divide evenly... 98.221.85.188 (talk) 04:19, 13 February 2009 (UTC)

Your problem does not have a nice round answer. So either you have to accept that or you wrote down the wrong equation above. Dragons flight (talk) 04:24, 13 February 2009 (UTC)

Well it's a webwork problem meaning we have to type in the answer on the web, and then it tells us if we got it right or wrong. So it's supposed to have an exact answer, but I keep getting an approximation of 1.4. The problem looks like it's written correctly. I don't know where the problem is. 98.221.85.188 (talk) 04:27, 13 February 2009 (UTC)

Nvm, I put sqrt15.25-2.5 and it says I was correct 98.221.85.188 (talk) 04:29, 13 February 2009 (UTC)

You missed something: You need to reject the extraneous root. You can't take the logarithm of a negative number. Michael Hardy (talk) 22:52, 13 February 2009 (UTC)

$x^{2}+5x=9\,$

To complete the square, you divide 5 by 2 and then square, and add that amount to both sides:

$x^{2}+5x+{\frac {25}{4}}=9+{\frac {25}{4}}={\frac {61}{4}}.$

Then:

$\left(x+{\frac {5}{2}}\right)^{2}={\frac {61}{4}},$

so that

$x+{\frac {5}{2}}=\pm {\frac {{}\ {\sqrt {61}}\ {}}{2}},$

etc. Now the complication: One of the solutions is negative. You need to reject that one since there is no base-2 logarithm of a negative number. Michael Hardy (talk) 22:51, 13 February 2009 (UTC)

Reals-chauvinist! —Tamfang (talk) 06:42, 18 February 2009 (UTC)

Finding distance given initial speed and friction

I'm making a simple game that involves throwing balls around a complex level. The ground has some friction on the balls that slows them down. I'm not going for a perfect simulation, so as it is I'm just multiplying their velocity to a constant f slightly less than 1. So, given an initial position and speed, and a certain friction, I'm trying to predict where it will stop so I can plug in some AI code. I'm guessing it'll involve some calculus, but it's been a while... What I have looks like:

$P=P_{0}+v\times t$

$v=v_{0}\times f^{t}$

But I'm kinda lost there. If anyone could point out the way on how to solve this thing, I'd be very grateful. Thanks! -- JSF —Preceding unsigned comment added by 189.112.59.185 (talk) 13:34, 13 February 2009 (UTC)

I don't think you want $P=P_{0}+v\times t$, you probably want $P=P_{0}+\int v\,\mathrm {d} t$. That gives you:

$P=P_{0}+\int _{0}^{\infty }v\,\mathrm {d} t$

$=P_{0}+\int _{0}^{\infty }v_{0}f^{t}\,\mathrm {d} t$

$=P_{0}+{\frac {v_{0}}{\log {f}}}{\big [}f^{t}]_{t=0}^{\infty }$

$=P_{0}-{\frac {v_{0}}{\log {f}}}$

(Note, since f<1, log(f) is negative, so that minus sign does make sense.) That's if it's done continuously, if you are actually simulating it using discrete time steps, you'll get a slightly different answer (but not too far off if the time steps are small enough). --Tango (talk) 14:31, 13 February 2009 (UTC)

Watch out for the calculus because it tells you that with constant friction the ball never stops completely. Possibly you want a procedure like this pseudocode:

Enter P0 = start position (distance units e.g. inches)
      V0 = start velocity (distance per time step)
      F  = friction (velocity change per time step)
 p=P0
 v=V0
NEX:
 p = p + v*(1+F)/2
 v = v * F
 if v > 0.01 goto NEX
REM The ball stops at position p.  

From these start values

P0, V0, F = 1, .1, .8

the ball rolled for 11 time steps (loops to NEX) and stopped at position p = 1.41. Cuddlyable3 (talk) 20:52, 13 February 2009 (UTC)

College Math Problem Plea ....

A steel block of weight W rests on a horizontal surface in an inaccessible part of a machine. The coffecient of friction between the block and the surface is "u". To extract the block, a magnetic rod is inserted into the machine and this rod is used to pull the block at a constant speed along the surface with a force F. The magnetic force of attraction between the rod and the block is M. Explain why

(a) M > u X W (b) F = u X W

Math problem was under the chapter : "NEWTON'S THIRD LAW" in Mechanics M1 Book of Mathematics Course PLEASE DO HELP!!!!

—Preceding unsigned comment added by 202.72.235.204 (talk) 21:33, 13 February 2009 (UTC)

We are not going to do your homework for you. J.delanoy^gabs_adds 21:37, 13 February 2009 (UTC)

Actually I was unable to do this one so I thought what better place for help than Wiki and ofcourse you guys!!!! —Preceding unsigned comment added by 202.72.235.204 (talk) 21:49, 13 February 2009 (UTC)

What bit are you stuck on? If you show us your working so far, we'll try and help you with the next bit, but we're not going to do the whole question for you. If you don't even know where to start, you should go and talk to your teacher. --Tango (talk) 21:50, 13 February 2009 (UTC)

Okay, WORKING:

The question's first part (a) suggests that magnetic attraction "M" should be greater than frictional force experienced by steel block "uW". When this happens, infact, the block will start to accelerate as the net force on block exceeds 0 (M > friction). But the block travels with steady speed as stated in the question. But as we see it is for granted that M is always constant thus object accelerates always.

Now, whats the deal with M and F. How does pulling the rod with a force of F change anything?

When the magnetic rod is in contact with the block there is a reactive force between them than partially cancels out M. If you weren't pulling the rod, it would completely cancel it out, by pulling the rod just the right amount you leave just enough resultant force on the block to counteract the friction allowing for constant velocity. Try drawing a diagram showing the block, the surface and the rod with all the forces (I count 7 forces in total). --Tango (talk) 22:43, 13 February 2009 (UTC)

CLARIFICATION...............................................................................................:

Okay, I didn't consider the steel block to be in contact with the magnet. There are few things for clarification though:

Lets take the steel block into consideration: Taking the steel block travels to the left

Forces to the right: The magnetic force M and the pull from the magnetic rod F / uW

Forces to the left (all horizontally): Reaction contact force R = M and friction uW

Resultant force horizontally = 0

My question is that it is logical that the contact reaction force on block from magnet would decrease INCREMENTALLY AS due to the pull increments but mathematically speaking, we can take reaction force constant and workout arithmetic summation to find net force taking that there is granted constant pull from magnet on block.

WHAT REALLY HAPPENS, DOES THE CONTACT FORCE DECREASES AS PULL EXIST OR INCREASE OR CONTACT FORCE REMAINS CONSTANT AS PULL ARISES??? —Preceding unsigned comment added by 202.72.235.208 (talk) 16:54, 14 February 2009 (UTC)

The pulling force from the rod isn't a separate force, it's just the difference between the magnetic attraction and the contact force. I suggest you include the rod in your diagram and consider the forces on it too (F is a force on the rod). --Tango (talk) 17:36, 14 February 2009 (UTC)

You're overthinking this Q. Yes, in reality any force which gets the block moving would also cause it to accelerate, but just ignore that. When they said it moves at a constant speed, what they really meant was "it will move at a slight acceleration, which is minimal enough that you need not consider it in your calculations". Similarly, there isn't a single coefficient of friction, but rather are two, a higher static one and a lower dynamic one. So, you would need to pull with a greater force to "break the block loose", then decrease the pulling force to prevent acceleration. StuRat (talk) 16:35, 15 February 2009 (UTC)

February 14

Mathematical fraction

The term "a third of a mil" in reference to a dollar amount is used in a New Jersey state statute. Please advise what that fraction is and what the decimal number is that should be used when multiplying another larger number. For instance what would I multiply $1,000,000 by to find a "third of a mil" of that amount?

Thank you,

Frank J. Mcmahon Mahwah, NJ [email removed] —Preceding unsigned comment added by 69.127.4.198 (talk) 02:07, 14 February 2009 (UTC)

Try a lawyer? I'd assume offhand "a third of a mil" means ${\frac {1}{3}}\cdot 1000000$ dollars. If there's some special legal meaning of the phrase, then it is something a lawyer would know, not a mathematician. Maybe it would help if we could see the full sentence that "a third of a mil" first appears in. By the way, we never email responses, so I removed yours to lessen visibility to spam-bots.... Eric. 131.215.158.184 (talk) 07:17, 14 February 2009 (UTC)

In some contexts I believe that, just like "1/3 per cent" means X / 300, this could mean "1/3 per thousand" or X / 3000. -- SGBailey (talk) 09:48, 14 February 2009 (UTC)

I would think "a third of a mil" is just short for "a third of a million" or $333,333.33. As SGBailey says, it could mean "a third per mil", or 1/3000 times. Either way, it's not a very common way of saying it, but then lawyers like to make things as confusing as possible - it keeps them in work. --Tango (talk) 12:53, 14 February 2009 (UTC)

I would assume that it refers to mill (currency). "A third of a mil" is 1/3000 of a dollar. -- BenRG (talk) 13:10, 14 February 2009 (UTC)

In light of [1], about library funding in New Jersey, it looks like SGBailey got it right; it's a third per mil. Of course, that also means it's a third of a mill per dollar (in this case, per dollar of assessed property value). So to answer the original question, for a property assessed at $1,000,000, a "third of a mil" would be around $333. —JAO • T • C 14:54, 14 February 2009 (UTC)

Numbers

I have these series of numbers and I know they are relate, but I don't know what they are called.

0,1,3,6,10,15, 21,28...

0,+1,+2,+3,+4,+5...

Thanks --68.231.197.20 (talk) 06:47, 14 February 2009 (UTC)

You should look at the triangular numbers. Eric. 131.215.158.184 (talk) 07:07, 14 February 2009 (UTC)

The OEIS is a good place to answer these questions. Algebraist 14:23, 14 February 2009 (UTC)

they're the perfect squares. 0*0=0, 1*1=1, 2x2=4, 3x3=9, etc. that is +1+3+5 etc. Do you notice a relationship with your series? How would you express that as an equation -- or maybe two? note: this may be BS on my part —Preceding unsigned comment added by 82.120.236.246 (talk) 21:59, 14 February 2009 (UTC)

What are the perfect squares? None of the numbers in the list (other than 0 and 1) are squares... what are you talking about? --Tango (talk) 00:02, 15 February 2009 (UTC)

Perhaps the sum of any two consecutive terms;) hydnjo talk 13:56, 15 February 2009 (UTC)

Dice game problem

I was thinking about a problem this morning and got a bit stuck on figuring out how to calculate the answer. It seemed like the sort of problem that has been calculated before, but I can't seem to find anything on it. Here's how it works:

You are playing a dice game with three six-sided dice. You roll the dice and set aside any that come up 1. You then reroll any dice that aren't one and continue the process of rerolling and setting aside 1's until all three dice have come up 1. The question is how many times on average do you need to roll the dice before they all individually have come up 1?

(Note: The original problem I'm actually trying to solve is very similar, except that instead of the probability of individual success being 1/6 the probability of individual success is 1/11.)

My friend and I worked out a brute force way to approximate it by computer (I think, I haven't typed it up yet), but I'm wondering if there's a more elegant or exact solution. Any suggestions? Thanks for the help. 71.60.89.143 (talk) 20:54, 14 February 2009 (UTC)

Just a quick follow-up - I used Excel to calculate that the expected number of rolls would be approximately 4.878 in order to get at least one success on each of the three dice. For my original problem where the chance of success is 1/11, the expected number of rolls is 8.623. Please feel free to confirm if you like, and if you have a nifty way of solving the problem I'd be interested in reading it. 71.60.89.143 (talk) 22:56, 14 February 2009 (UTC)

Your answers are certainly wrong. The expected number of rolls of one dice until you get a 1 is 6 (or 11), and getting the right number on all three is clearly harder. Algebraist 23:03, 14 February 2009 (UTC)

For three dice, probability p of success each time, I get ${\frac {11-19p+12p^{2}-3p^{3}}{p(2-p)(3-3p+p^{2})}}$, giving 10566/1001 for probability 1/6 and 45727/2317 for probability 1/11. Unfortunately, all I have for the general case (n dice, probability p) is a messy infinite sum, and I don't have time right now to do this properly. Algebraist 23:10, 14 February 2009 (UTC)

Yeah, after I typed the above I realized I made a mistake in my formula in Excel. I'm still not getting the same answer you got above, though. I get 7.38 for the expected value of p=1/6. Hmmm.... 71.60.89.143 (talk) 23:26, 14 February 2009 (UTC)

To clarify what I'm doing in Excel, I let P(k,n) be the probability that at least k of the three dice have succeeded (k=0 to 3) after at least n rolls. So P(2,5) would be the chance that at least two of the three dice hit a success after five rolls. Given that definition for P(k,n), I get the following formula for P(3,n) for n>1.

P(3,n) = P(3,n-1) + (P(2,n-1) - P(3,n-1))*(P(1,1) - P(2,1)) + (P(1,n-1) - P(2,n-1))*(P(2,1) - P(3,1)) + (1 - P(1,n-1))*P(3,1)

Above, the expression P(2,n-1) - P(3,n-1) is the probability that exactly two of the dice succeeded after n-1 rolls. The other expressions are similar. 71.60.89.143 (talk) 23:46, 14 February 2009 (UTC)

Alright, I found the error in my Excel sheet and the formula above.  :) The second term in the last few products is a probability involving rolling all three dice, but it should actually be only partial rolls. I fixed the error and, lo and behold, it matches your answer Algebraist. Good work! 71.60.89.143 (talk) 01:48, 15 February 2009 (UTC)

Just my take. Let X be the number of throws before you get 1 on an n-faced die. Its cdf is

$P(X\leq x)={\frac {1}{n}}\sum _{i=0}^{x-1}\left({\frac {n-1}{n}}\right)^{i}=1-\left({\frac {n-1}{n}}\right)^{x}$

If you try to throw k dice, it is equivalent to look at the maximum of k iid random variables, which has cdf

$P(\max\{X_{1},\ldots ,X_{k}\}\leq x)=\left[1-\left({\frac {n-1}{n}}\right)^{x}\right]^{k}:=F_{k,n}(x)$

Thus the expected value of number of throws should be

$E\,\max\{X_{1},\ldots ,X_{k}\}=\sum _{x=1}^{\infty }x(F_{k,n}(x)-F_{k,n}(x-1))=\sum _{x=1}^{\infty }\left(1-\left[1-\left({\frac {n-1}{n}}\right)^{x}\right]^{k}\right)$

which I am pretty sure is possible to calculate explicitly. (Igny (talk) 02:49, 15 February 2009 (UTC))

Yeah, that's the infinite sum I alluded to above. Algebraist 03:16, 15 February 2009 (UTC)

So if I did not screw up for 3 dice with 6 face the average number of throws is 10.56 and for 11 faces it is 19.74 (Igny (talk) 21:17, 15 February 2009 (UTC))

That's right, Igny. And thanks for breaking down your solution; it's a little simpler and more generalizable than what I used. 63.95.36.13 (talk) 14:56, 16 February 2009 (UTC)

is it possible to fuck up when combining two random number generators?

If you're combining two random number generators that you think are pretty random, but who knows, maybe every so often they aren't random enough, is there any way to do that which looks okay and basically has the correct distribution, but in fact now is way not so random? Thanks.

P.s. this isn't malicious! I wouldn't be asking it this way, and on this forum, if it were —Preceding unsigned comment added by 82.120.236.246 (talk) 22:23, 14 February 2009 (UTC)

This isn't exactly an answer to your question, but: for the most part, any attempt to combine two good PRNGs, or to modify the output of a PRNG to make it "more random", will actually reduce the its quality. That's why there are so few of them that are considered cryptographically strong. By the way, if you do get a straight answer to your question, be sure to remember it in case you ever want to participate in the Underhanded C Contest. « Aaron Rotenberg « Talk « 23:16, 14 February 2009 (UTC)

Yes, it is possible to fuck up. By not defining what distribution you really need. Cuddlyable3 (talk) 23:39, 14 February 2009 (UTC)

If you can reduce the randomness of a PRNG by combining its output in some simple way with another PRNG, then it certainly wasn't cryptographically strong in the first place. However it is usually possible to improve the randomness of even weak PRNGs by combining them. I flatly disagree with Aaron when he claims that it "usually" reduces the quality. It can happen, if there is some unsuspected connection between the two PRNGs (or, of course, if you make some silly mistake in how you "combine" them), but it usually helps rather than hurts.

The simplest way to combine two PRNGs (say, normalized to return a value between 0 and 1) is simply to add their outputs modulo 1. If you do this to two PRNGs of relatively prime period, the period of the new PRNG is the product of the original periods. (Period by itself is not a good measure of randomness, but short period is always a problem.)

An even better way is the McLaren–Marsaglia method, in which you cache values from one of the PRNGs, and use the other one to select a value from the stream. --Trovatore (talk) 23:52, 14 February 2009 (UTC)

"if you can reduce the randomness..by combining its output in some simple way ..then it certainly wasn't ..strong" WTF!! How's this for starters:

perl -we "for(1..10000000){if (int rand 2 + int rand 2){$one++}else{$zero++} }; print qq/got $one ones and $zero zeros\n/"

I suppose the result "got 5831006 ones and 4168994 zeros" means that I just proved Perl's random number generator is way, way insecure!!! —Preceding unsigned comment added by 82.120.236.246 (talk) 00:58, 15 February 2009 (UTC)

I don't actually speak Pathologically Eclectic Rubbish Lister so I'm not quite sure what you're doing here. It looks like you're demanding that two random values chosen from 0 to 2 both be less than 1 in order to increment $zero, in which case I'd expect only 25% zeroes from a good RNG. But I certainly wouldn't be surprised if Perl had a bad RNG — in fact, that seems more likely than not.

In any case you seem to have ignore both of my stipulations — that the two PRNGs be unrelated, and that you not make some silly mistake when combining them (like choosing random bits from a 25-75 proposition). --Trovatore (talk) 03:28, 15 February 2009 (UTC)

int rand 2 + int rand 2 in Perl means int(rand(2 + int(rand(2)))), which should be 0 with probability 5/12 ≈ 0.4167, so Perl's RNG passed this test—though I suspect it wasn't the intended one. -- BenRG (talk) 04:21, 15 February 2009 (UTC)

Yes, cryptologists have studied this problem in great detail and to use your vernacular, you can indeed fuck up badly this way. See for example Ueli Maurer and J. L. Massey, "Cascade ciphers: The importance of being first"[2] and follow some of its references. 207.241.239.70 (talk) 06:41, 18 February 2009 (UTC)

February 15

Formula for partial derivative of A with respect to B given that f(A,B) is constant, applied to vectors

Suppose that we have a vector-valued function $f({\vec {a}},{\vec {b}})$ of the type $(\mathbb {R} ^{n},\mathbb {R} ^{m})\rightarrow \mathbb {R} ^{k}$.

I am trying to find a formula for the value of

$\left({\frac {\partial ({\vec {a}}_{0}...{\vec {a}}_{n})}{\partial ({\vec {b}}_{0}...{\vec {b}}_{m})}}\right)_{f}$.

This is equivalent to the Jacobian matrix of a function $g({\vec {x}})$ such that if $f({\vec {a}},{\vec {b}})={\vec {0}}$, then $g({\vec {b}})={\vec {a}}$.

There is a technique for finding this kind of partial derivative of scalar functions, and I tried to generalize it to vector-valued functions:

$df({\vec {a}},{\vec {b}})=\left({\frac {\partial ({\vec {f}}_{0}...{\vec {f}}_{k})}{\partial ({\vec {a}}_{0}...{\vec {a}}_{n})}}\right)_{\vec {b}}d{\vec {a}}+\left({\frac {\partial ({\vec {f}}_{0}...{\vec {f}}_{k})}{\partial ({\vec {b}}_{0}...{\vec {b}}_{m})}}\right)_{\vec {a}}d{\vec {b}}$

Holding $f({\vec {a}},{\vec {b}})$ constant, we get:

$\left({\frac {\partial ({\vec {f}}_{0}...{\vec {f}}_{k})}{\partial ({\vec {a}}_{0}...{\vec {a}}_{n})}}\right)_{\vec {b}}d{\vec {a}}=-\left({\frac {\partial ({\vec {f}}_{0}...{\vec {f}}_{k})}{\partial ({\vec {b}}_{0}...{\vec {b}}_{m})}}\right)_{\vec {a}}d{\vec {b}}$

Multiplying both sides by $(d{\vec {b}})^{-1}$ and $\left({\frac {\partial ({\vec {f}}_{0}...{\vec {f}}_{k})}{\partial ({\vec {a}}_{0}...{\vec {a}}_{n})}}\right)_{\vec {b}}^{-1}$:

$(d{\vec {a}})(d{\vec {b}})^{-1}=-\left({\frac {\partial ({\vec {f}}_{0}...{\vec {f}}_{k})}{\partial ({\vec {a}}_{0}...{\vec {a}}_{n})}}_{\vec {b}}\right)^{-1}\left({\frac {\partial ({\vec {f}}_{0}...{\vec {f}}_{k})}{\partial ({\vec {b}}_{0}...{\vec {b}}_{m})}}_{\vec {a}}\right)$

I am not sure, but I believe that $(d{\vec {a}})(d{\vec {b}})^{-1}$ is equivalent to $\left({\frac {\partial ({\vec {a}}_{0}...{\vec {a}}_{n})}{\partial ({\vec {b}}_{0}...{\vec {b}}_{n})}}\right)_{f}$, which, if it is correct, gives me an answer to my original question. Is this valid, or did I make a mistake somewhere?

On a side note, I'm new to both Wikipedia's formatting and LaTeX; if you have any comments on my formatting, I'd like to hear them.

24.130.128.99 (talk) 02:14, 15 February 2009 (UTC)

 —Preceding unsigned comment added by 24.130.128.99 (talk) 02:12, 15 February 2009 (UTC) 

I believe you are right, see implicit function theorem. Upd: k must be equal to n for the inverse matrix to make sense. (Igny (talk) 04:58, 15 February 2009 (UTC))

The Problem I am stuck with

Question

---

A car of mass 1200 kg, towing a caravan of mass 800 kg, is travelling along a motorway at a constant speed of 20 m/s. There are air resistance forces on the car and the caravan, of magnitude 100N and 400 N respectively. Calculate the magnitude of the force on the caravan from the towbar, and the driving force on the car.

The car brakes suddenly , and begins to decelerate at a rate of 1.5 m/s2. Calculate the force on the car fro the towbar. What effect will the driver notice?

---

I did get the first part which is quite easy. Given the objects travel at constant speed, pull on them must equal to the resistive forces to achieve equilibrium(constant speed).

Thus: Magnitude of the force on the caravan from towbar = 400N

     Driving force = 500N

MY PROBLEM

I am totally lost at the second part of the question:

---

The car brakes suddenly , and begins to decelerate at a rate of 1.5 m/s². Calculate the force on the car from the towbar. What effect will the driver notice?

---

The book says the answer to this part is: 800N forwards; it will appear that the car is being pushed from behind.

But it doesn't say anything about why this is so. My question or help required is that why this is so? —Preceding unsigned comment added by 202.72.235.208 (talk) 08:54, 15 February 2009 (UTC)

If the 800 kg caravan is decelerating at 1.5 m/s², what net force must be acting on it ? 400N of this force comes from air resistance - where does the rest come from ? The car exerts a force on the caravan through the towbar - what does Newton's third law then tell you about the force exerted by the caravan on the car ? Gandalf61 (talk) 09:24, 15 February 2009 (UTC)

First, a translation for US readers: "caravan" = "trailer". Next, there is an apparent assumption that only the car is braking. Next, we need a diagram:

         400N -> +------+
           __    |      |
         _/  \_  | 800kg|
100N -> |1200kg|-|      |
        +-O--O-+ +-O--O-+ 

Now calculate the total deceleration force needed on the trailer:

F = ma = (800kg)1.5m/s2 = 1200kg•m/s2 = 1200N
                

Now, if there's a 1200N deceleration force on the trailer, and 400N of that is initially provided by wind resistance, the additional 800N must be provided by the tow bar. Note that either the rate of deceleration will decrease, or the braking force and force transmitted by the tow bar must increase, as the speed (and therefore wind resistance) decreases. StuRat (talk) 16:17, 15 February 2009 (UTC)

...and the driver may notice that he must apply increasing force on the brake pedal to keep the deceleration constant, and that the wind, engine and tire noises decrease. Cuddlyable3 (talk) 21:01, 15 February 2009 (UTC)

StuRat, that's a simply phenomenal piece of AsciiArt. Kudos! --DaHorsesMouth (talk) 21:28, 15 February 2009 (UTC)

Thanks ! StuRat (talk) 17:41, 16 February 2009 (UTC)

Thanks people for your help and specially thankful to StuRat for his visual description, really appreciate that kind of helping hand. By the way, air resistance was told to be kept constant in this problem unless otherwise stated in this one, it was written at the top of the chapter, SORRY FORGOT TO MENTION. The answer is that the car had to provide extra 800N braking force despite its own 1700N needed braking for such deceleration.

                                  400N ->+------+
           __                            |      |
         _/  \_                          | 800kg|
100N -> |1200kg|<-800N-------------800N->|      |
        +-O--O-+                         +-O--O-+ 

Braking(1700 + 800)N ->

Towbar acts as a couple, meaning 800N forces bothways parallely. Thus the reaction force is 800N. —Preceding unsigned comment added by 202.72.235.202 (talk) 18:34, 16 February 2009 (UTC)

Collatz-like sequence

Does anyone know of any results concerning the Collatz-like sequences generated by

$f(n)={\begin{cases}n/2&{\mbox{if }}n\equiv 0{\pmod {2}}\\3n-1&{\mbox{if }}n\equiv 1{\pmod {2}}\end{cases}}$

I have found papers on various generalisations of the Collatz conjecture, but I haven't found any results on this specific case.

As far as I can tell, there are three loops:

$1\rightarrow 2\rightarrow 1$

$5\rightarrow 14\rightarrow 7\rightarrow 20\rightarrow 10\rightarrow 5$

$17\rightarrow 50\rightarrow 25\rightarrow \cdots \rightarrow 68\rightarrow 34\rightarrow 17$

and every sequence I have tested eventually enters one of these loops. Having found three loops, I was surprised not to find more - why just three ? Gandalf61 (talk) 09:14, 15 February 2009 (UTC)

Never mind - I have just realised that these are essentially the same as Collatz sequences if we replace n with −n. Gandalf61 (talk) 11:37, 15 February 2009 (UTC)

Resolved

StuRat (talk) 15:31, 15 February 2009 (UTC)

Name the curve

What is the name of the curve(of four cusps)described by the enclosure of a moving straight line of length A, wherein the end points of line A move along their respective X and Y axis?

The equation given for this curve is: x^2/3 + y^2/3 = A^2/3. The curve is similar to the hypocycloid of four cusps,(the astroid), however the line generation appears to be uniquely different.

Vaughnadams (talk) 19:26, 15 February 2009 (UTC)

According to our article, that is an astroid. Algebraist 19:44, 15 February 2009 (UTC)

It looks a bit like this. Cuddlyable3 (talk) 20:51, 15 February 2009 (UTC)

Maths: discovery or invention?

Are mathematical developments discoveries or inventions, and does someone's answer to this question effect the conclusions they can draw? Thanks in advance. 86.8.176.85 (talk) 19:46, 15 February 2009 (UTC)

This depends on your Philosophy of mathematics. That article has some positions that various people have held. Algebraist 19:53, 15 February 2009 (UTC)

This is a perennial source of discussion for philosophers; there is no clearly correct answer. It's analogous to solving a crossword puzzle – would you say that you created the solution, or that you discovered it? Even usage among mathematicians is varied. I typically say that I discover a new mathematical object but I invent a new technique. — Carl (CBM · talk) 19:58, 15 February 2009 (UTC)

One often speaks of constructing a new object also. Algebraist 20:45, 15 February 2009 (UTC)

Mathematics is discovery in an abstract universe that does not exist but is useful to invent. I think the questioner means affect not effect the conclusions one can draw. The answer to the first question does not affect the conclusions one can draw, only the mathematics can prove or disprove a mathematical conclusion. Cuddlyable3 (talk) 20:46, 15 February 2009 (UTC)

It can, actually. A realist (about mathematical objects) is forced to conclude that the continuum hypothesis must be either true or false, and may be able to convince himself one way or the other. Some types of antirealist, on the other hand, are able to conclude that CH is without truth value. Algebraist 21:36, 15 February 2009 (UTC)

A related question would be whether mathematical concepts or techniques are copyrightable or patentable? It's a relevant question when you consider cryptology and compression technologies? I wonder what would would be the effect of someone having a patent on the Pythagorean theorem? -- Tcncv (talk) 07:58, 16 February 2009 (UTC)

I believe there is prior art. 76.126.116.54 (talk) 08:11, 16 February 2009 (UTC)

Theorems are just statements of fact, I don't think they are copyrightable or patentable. Mathematical algorithms can be patented, but are not copyrightable, as I recall (a given software implementation of the algorithm is copyrightable, though). --Tango (talk) 11:59, 16 February 2009 (UTC)

Mathematical algorithms are really no different from other mathematical facts and therefore should not be patentable. Indeed, generally they are not patentable outside the United States, see software patent. — Emil J. 13:48, 16 February 2009 (UTC)

Any copyrighted image or text that can be digitised becomes just a big binary number that may be judged to be an Illegal number. Psssst the secret number is 42 but you didn't hear that from me. Cuddlyable3 (talk) 15:11, 16 February 2009 (UTC)

Not serious series

1. Counters of beansFine mathematicians with too much free time, can you supply the last term of this series:

3 , 3 , 5 , 4 , 4 , 3 , 5 , ?

2. Riddle me this: why does six fear seven ? Cuddlyable3 (talk) 21:19, 15 February 2009 (UTC)

The answer to 1. is 5. Algebraist 21:23, 15 February 2009 (UTC)

2: Because 7 8 9. As for 1, I say the answer is pi. -mattbuck (Talk) 21:24, 15 February 2009 (UTC)

1. is A005589 at the OEIS. Algebraist 21:26, 15 February 2009 (UTC)

Interesting how they stopped at "one hundred", nicely sidestepping the issue of one hundred one versus one hundred and one (though still vulnerable to the challenge from a hundred). --Trovatore (talk) 23:02, 15 February 2009 (UTC)

The value of 4 for the noughth entry is also arguable. Algebraist 00:01, 16 February 2009 (UTC)

Mattbuck, the mission I give you is to take a LARGE piece of paper and write down the exact answer in very small print. Cuddlyable3 (talk) 14:29, 16 February 2009 (UTC)

3. I gave a junior class an introduction to using x as a variable (which some of them found baffling) by getting them to evaluate these expressions:

a) $2x+7-8/4$ when $x=3$

b) $0.5+5x+2$ when $x=5$

One student evaluated a) as 28 which is wrong. He got b) wrong too but at least he was consistent in his method. What was his answer to b) ? Cuddlyable3 (talk) 14:29, 16 February 2009 (UTC)

57.5 --Tango (talk) 14:46, 16 February 2009 (UTC)

That's not really wrong, is it? Concatenation can be denoted by juxtaposition just like multiplication. — Emil J. 15:02, 16 February 2009 (UTC)

You can denote anything you like however you like, but it's not standard. If you want to concatenate digits you would usually use some kind of symbol to denote it (perhaps $2\circ x$). --Tango (talk) 15:11, 16 February 2009 (UTC)

But the standard notation for concatenation of strings over a finite alphabet in combinatorics, theoretical computer science, and related areas is simple juxtaposition, that's the point. People also use stuff like $x\frown y$ when they want to be extra explicit, but then again, there is explicit notation $x\cdot y$ for multiplication as well. I've never seen $\circ$ used for concatenation. — Emil J. 15:35, 16 February 2009 (UTC)

In those areas the symbols usually just represent an arbitrary alphabet, not actual digits (they may use the same symbols, but they aren't actually representing numbers). If you are doing addition and subtraction with them then it is clear that they are numbers. Juxtaposition of numbers (with at least one represented by a variable - obviously the juxtaposition of '2' and '3' means twenty-three) almost universally denotes multiplication. --Tango (talk) 16:15, 16 February 2009 (UTC)

I have seen || used as well when there are other operations floating around using juxtaposition, as in $x||y$ GromXXVII (talk) 19:56, 16 February 2009 (UTC)

Mattbuck, almost finished are you? Cuddlyable3 (talk) 20:02, 16 February 2009 (UTC)

February 16

CASINOS

hello:D

i have a high school assignment on probability.. basically, i and my group have to design a simple, creative and original casino game. and we have to explain the winning/losing probabilities of the game and the concept of the game.. how do u suggest we go about creating the game? any possible suggestions? and what are key things we have to consider when creating a new game?

thanks. please respond asap. thanks:D --218.186.12.201 (talk) 09:20, 16 February 2009 (UTC)Pearl

I suggest something involving dice - they are a very easy way of getting randomness. (You could use coins, but they have fewer options which will probably make the game more boring.) The key thing to remember when making a casino game is that there needs to be a "house edge" - the odds have to favour the casino, otherwise you won't make any money. A very simple game would be for the player to bet some money, then the player and someone from the casino each roll a die, if the player gets a higher roll the casino gives him money equal to his bet, if he gets a lower or equal roll, he losses the bet. The fact that the casino wins when there is a tie is what gives the house edge. I suggest you come up with something a little more interesting, though! --Tango (talk) 11:45, 16 February 2009 (UTC)

Roulette where the wheel is a Realeaux polygon and the ball is a Meissner tetrahedron. You can bet that's original.Cuddlyable3 (talk) 14:53, 16 February 2009 (UTC)

Simply giving the house an advantage is one thing, but the most successful games will give the house an advantage without appearing to do so. Or, better yet, they will appear to give the players the advantage. This is why so many carnival games are "fixed". So, let's use this quote about coin tossing from an early post on this Desk: ("It is six times as likely that HTT will be the first of HTT, TTH, and TTT to occur than either of the others" ). So, we could give the players a 2X payout if they manage to flip the TTH or TTT sequence before the HTT sequence, and keep their money otherwise, and they will be certain the odds are in their favor, when they really aren't.

You could also give the players the option to try for HHT or HHH before the THH sequence, with the same 2X payout when they win. This will help if they think the game is fixed in some way. Note that this game requires that they flip the coin continuously and count every sequence of 3. So, don't just count the 1st-3rd sequences and 4th-6th sequences as possible matches, but also look at the 2nd-4th and 3rd-5th coin tosses.

Also note that this doesn't need to use a coin, but any binary event will work. You could use a roulette wheel with black being one outcome and red the other (you'd also need to assign the green zero and double zero to act as either red or black). You could use card draws being red or black, as well. You could use dice throws with odds or evens. StuRat (talk) 18:00, 16 February 2009 (UTC)

I believe casinos are required by law to publish the odds for their games (in many jurisdictions, probably not all), so that kind of trick doesn't really work. Carnivals are far less regulated. --Tango (talk) 18:12, 16 February 2009 (UTC)

Yes, but gamblers either don't understand odds or don't care about them, or they wouldn't play at all, would they ? Gamblers tend to go on "instinct", which is frequently wrong, allowing them to be fleeced of their money. StuRat (talk) 18:22, 16 February 2009 (UTC)

Well, yes, many gamblers are just idiots. Some do know they are likely to lose but consider the enjoyment of playing to be worth it. --Tango (talk) 18:36, 16 February 2009 (UTC)

Someone said earlier to use dice, which is a good idea. You can design some really cool games with big jackpots and things like that. Like a 6 could pay you $5 AND give you an extra free roll (you could make a lot if you go on a run of 6's). You can do almost anything you want as far as payouts, the only trick is that once you design the game you have to properly calculate the expected payouts and then make the fee to play slightly higher (maybe round up to the nearest dollar for your profit). That is the core of the assignment is being able to calculate what the fair price of the game is and then charge a little more. Having fixed payouts like this in my opinion is easier, but it is more like a carnival game (where you charge $x dollars a try) than a casino game, because most casino games allow you to bet whatever amount you like. Anythingapplied (talk) 22:01, 16 February 2009 (UTC)

To make the game interesting the player needs to have choices to make along the way. Perhaps the extra roll could cost something (less than the usual bet), so you have to decide whether to take it or not. (That's a pretty rubbish choice - if you're playing the game, then obviously you think the usual bet is a fair price so, of course, you'll take it if it is cheaper, but you get the idea.) --Tango (talk) 22:14, 16 February 2009 (UTC)

Ideally you want a format where one event influences the likelihood of a second. Card games have this, and the varying nature of the game and the mental activity required to make decisions makes them more involving than games like roulette or craps where you're just betting on chance. At least a semblance of control can make a game much more addictive, as players think skill can recover the losses that luck is responsible for, rather than crediting everything to fortune. 86.8.176.85 (talk) 18:38, 17 February 2009 (UTC)

If you can pull it off, then great, but it's very hard to make a game like that and ensure that the house will always win (in the long run). For example, Blackjack works like that but casinos have to use multiple decks to reduce the effectiveness of card counting (by making it closer to completely random) because allowing people playing cleverly to make a difference can easily result in the house losing money. --Tango (talk) 18:56, 17 February 2009 (UTC)

oh my goshh:DD thanks soo much. i found some of the information a lil' hard to understand but im sure a lil reading up will totally help me... thank you so much for ur time and useful help. i knw i can depend on wiki helpdesk anytime..lol :DPearl--218.186.12.201 (talk) 11:36, 18 February 2009 (UTC)

Didn't understand

I couldn't understant the mechanical dialect in this mechanics question.

A drop-forge hammer, of mass 1500kg, falls under gravity on to a piece of hot metal which rests in a fixed die. From the instant the hammer strikes the piece of metal until it comes to rest, the hammer decelerates at 1.5 m/s².

Find the magnitude of the force exerted by the hammer on the piece of metal (a) while the hammer is decelerating (b) after the hammer has come to rest

Answers written in the book : (a) 17250 N !!!!!!!!!!!!!!WOWOWOWOWOW!!!!!!!!!! (b) 15000 N

PLEASE NOTE THE BOLD PART OF THIS QUESTION, IT MADE ME THINK HARD AND I AM AT ODDS

My View: I understand that there would be a normal reaction force when the hammer hits the metal. Naturally this reaction force would be much greater than the hammer's weight to provide upward acceleration against motion (hammer travels down to hit metal piece) as the flexibility of hot metal allows it.

My question is, why would the downward contact force on piece of metal be 17250 N (force exerted by hammer on metal).

It should be, the force exerted by metal on hammer is 17250N (2250N > weight of hammer) thus exactly what is needed to provide that deceleration ( or upward acceleration against motion). If contact force on metal by hammer is 17250N, reaction would have to be more ( but I don't see how and why this is so).

---

                 |     |
                 |  ?  |  HAMMER
                 |   ^ |
            |---------------|
            |               |
            |     METAL     |

---

WHAT ARE YOUR VIEWS PEOPLE? —Preceding unsigned comment added by 202.72.235.202 (talk) 19:04, 16 February 2009 (UTC)

The key to this kind of question is not to worry about the scenario but just to extract the relevant information. The relevant information here is that the hammer has a mass of 1500kg and is decelerating at 1.5 m/s². Those are the only two facts you need to know. --Tango (talk) 19:15, 16 February 2009 (UTC)

Item b is the weight of the hammer only. Item a is the weight (W=mg) plus the force required to decelerate that mass at that rate (F=ma). It's also apparent from the answers that they use the rather imprecise value of g=10m/s², rather than 9.81. Also, while not part of the problem, you should realize that not all of the force exerted on the metal is passed on to the die. Some of the force is converted into heat and noise or used to deform the metal. StuRat (talk) 03:05, 17 February 2009 (UTC)

---

Also, while not part of the problem, you should realize that not all of the force exerted on the metal is passed on to the die. Some of the force is converted into heat and noise or used to deform the metal. NO. --76.167.241.45 (talk) 06:42, 18 February 2009 (UTC)

Are you going to explain ? Do you not believe that metal is heated by the stamping process ? Do you not believe that sound results ? Where do you think the energy comes from to create this heat and noise ? StuRat (talk) 16:32, 19 February 2009 (UTC)

First of all, "force" and "heat and noise" are not even the same kind of thing. The latter is energy; force does not "convert" into heat. That's like saying my sister converted into two hours. Second, force is the time derivative of momentum, and momemtum is conserved; so as long as the same momemtum is transmitted over the same period of time, the average force is completely determined. So the notion that somehow "force" can be "lost" lacks any merit. --71.106.173.110 (talk) 05:02, 20 February 2009 (UTC)

It sounds like you think that forces and energy are completely unrelated. They are, in fact, closely related. From our energy article it says that energy "describes the amount of work that can be performed by a force". When talking about a time derivative, a better analogy would be velocity and acceleration. In this case, changing the rate of acceleration will affect the final velocity. And, again, where do you say that the energy to produce the heat and noise magically comes from ? StuRat (talk) 05:34, 20 February 2009 (UTC)

February 17

Finding unknown constant in function

Disclaimer: I'm not at all good at math, so this may be a stupid question.

I have no idea if the title I chose is at all descriptive, but it was the best I could think of. This isn't homework; in fact I asked my school math teacher and he couldn't tell me. My problem in practical terms would take forever to explain, but I've formalized it (hopefully correctly) as this:

$f(x)=-x^{4}+k$ and $c=\int _{a}^{b}f(x)$. Given $c$, $a$ and $b$, how can I find $k$?.

Aforementioned math teacher assures me it's possible; he just couldn't tell me how off the top of his head.

I could easily write a program to brute-force it, but that would be too slow for what I want to use it for (even given OCaml or C) (and besides it's really ugly and I want to know how to do it properly).

Thanks a lot, and apologies if I've screwed up the problem description - let me know if I've missed anything. --Aseld ^talk 12:47, 17 February 2009 (UTC)

$c=-{\frac {b^{5}}{5}}+bk+{\frac {a^{5}}{5}}-ak$. Rearrange and solve. Algebraist 13:16, 17 February 2009 (UTC)

Ah, using the indefinite integral. Thanks a lot. --Aseld ^talk 13:25, 17 February 2009 (UTC)

Well, no, it's a definite integral, but you do them by finding the antiderivative (which is the indefinite integral) and substituting in. --Tango (talk) 13:48, 17 February 2009 (UTC)

What is the difference between an antiderivative and an indefinite integral? Or are they just two names for the same thing? 92.1.184.208 (talk) 15:01, 17 February 2009 (UTC)

They're the same thing (well, almost. Some authors define the indefinite integral to be the set of all antiderivatives). Antiderivative is a much better term, to my mind. Algebraist 15:04, 17 February 2009 (UTC)

What I meant was, you're using the indefinite integral, which I had all but forgotten existed. --Aseld ^talk 19:39, 17 February 2009 (UTC)

It's not indefinite, though, it's definite. There are limits of integration. Indefinite and definite integrals are essentially the same thing, just with definite ones you substitute in the limits and with indefinite you leave it in terms of x (and add a constant). The one in your problem is definite. --Tango (talk) 20:57, 17 February 2009 (UTC)

What concerns me is that a "math teacher" couldn't do this. It was obvious on inspection that a linear function of k would arise. Is he qualified for the job he has?86.160.105.65 (talk) 19:47, 17 February 2009 (UTC)

It sounds like the OP hasn't learnt calculus properly and is just learning it themselves, so the teacher probably doesn't teach at a level that does calculus so it may have been years since they've used it. --Tango (talk) 20:57, 17 February 2009 (UTC)

Pretty much. I'm sure he could have worked it out, he just couldn't tell me off the top of his head. --Aseld ^talk 23:36, 17 February 2009 (UTC)

There's nothing to work out. Either you know how to integrate polynomials or you don't (they are the first integrals anyone learns). --Tango (talk) 00:16, 18 February 2009 (UTC)

Sorry, you can't solve it without a dx in the integral so it is not solvable(unless f(x) can be simplified into g(x)dx where g(x) is an arbitrary function with no lone differential in it).The Successor of Physics 11:55, 18 February 2009 (UTC)

Good point - I think Algebraist and I were just assuming the dx being missing with just a typo, but we should have stated that. --Tango (talk) 14:52, 18 February 2009 (UTC)

He didn't state what measure he was integrating with respect to, either. As always, you have to infer things from context. Algebraist 16:02, 18 February 2009 (UTC)

Yes. Apologies; the missing dx was an oversight. Algebraist, do you mean that I should have stated that I was integrating with respect to x? Again, apologies; I'll remember that next time. --Aseld ^talk 08:24, 19 February 2009 (UTC)

I meant that notational omissions that don't cause confusion are not worth worrying about. Algebraist 12:15, 19 February 2009 (UTC)

Simplifying an Equation

This is for part of my homework on potential dividers, I've got so far, but I can't seem to simplify this equation. The equation I'm trying to simplify is this: Potential_divider#Voltage_source. I've got these two equations:

$V_{\text{out}}={\frac {R_{2}\|R_{\text{L}}}{R_{1}+R_{2}\|R_{\text{L}}}}V_{\text{in}}$ and $\quad R_{2}\|R_{\text{L}}\triangleq \left({\frac {1}{R_{2}}}+{\frac {1}{R_{\text{L}}}}\right)^{-1}={\frac {R_{2}R_{\text{L}}}{R_{2}+R_{\text{L}}}}\,,$

and they are supposed to simplify to this equation $V_{\text{out}}={\frac {R_{2}}{R_{1}\left(1+{\frac {R_{2}}{R_{\text{L}}}}\right)+R_{2}}}V_{\text{in}}\quad$

Any help with the steps of simplifying this would help, thanks Jammie (talk) 17:31, 17 February 2009 (UTC)

Start by dividing top and bottom by $R_{2}\|R_{\text{L}}$ and it should be obvious from there. --Tango (talk) 17:44, 17 February 2009 (UTC)

Principal Ideal Ring, not an Integral Domain

Hi. I'm studying some algebra, and noting that a P.I.D. is defined as a Principal Ideal Ring that is also an Integral Domain. This suggests to me that there exist Principal Ideal Rings that are not Integral Domains, but I can't seem to find one or dream one up.

Can anyone give me a push in the right direction? Thanks in advance. -GTBacchus^(talk) 17:35, 17 February 2009 (UTC)

How about $\mathbb {Z} _{4}$? --Tango (talk) 17:41, 17 February 2009 (UTC)

I guess that works. Thanks. When I need a ring with zero divisors, I should probably think of $\mathbb {Z} _{n}$ for composite n.

Now let me see if I can take it another step. If we want a ring that is a P.I.R., but lacks unity, then we can use the ideal generated by 2 in $\mathbb {Z} _{4}$! I think I'm getting the hang of it; thanks again. -GTBacchus^(talk) 18:15, 17 February 2009 (UTC)

Indeed, $\mathbb {Z} _{4}$ is always my first choice when looking for a counterexample involving zero divisors ($\mathbb {Z} _{6}$ is useful sometimes too, as having distinct prime factors can make a difference). The ideal generated by 2 is rather a trivial example of a P.I.R - it only has one non-zero element, so there is no way you could even try and make a non-principle ideal - but I guess it is an example. --Tango (talk) 18:38, 17 February 2009 (UTC)

Maximal LFSRs

If I want a maximum length sequence from an LFSR I have to connect the right feedback taps, usually 2 or 4 of them. The article gives a table of equations that correspond to the tap numbers for up to a 19-stage register and I can confirm these are right by simulating them in software. When I need a longer sequence [3] tells me "There is no quick way to determine if a tap sequence is maximal length." while this reference tabulates tap numbers for up to 168 stages. That gives a huge maximal sequence that is impractical to "run" (simulate) on any computer known. How can anyone find (test) tap numbers for such big registers? Cuddlyable3 (talk) 20:18, 17 February 2009 (UTC)

If one can factor 2^n-1, then one can calculate the order of an (irreducible) n-stage LFSR fairly easily. One can check irreducibility fairly quickly, and if the lfsr has maximal length, then it is irreducible.

Basically, the irreducible polynomial of degree n over a field of size q defines a finite field of size q^n, and "x" is a nonzero element of these field, so generates a cyclic subgroup of a cyclic group of order q^n-1. To calculate the order of x just means checking the prime divisors p of q^n-1 to see if x^((q^n-1)/p) = 1, which can be done in time polynomial in n (assuming q fixed and p known) using exponentiation by squaring. Finding primitive polynomials is basically easy, but for computer science applications people tend to want sparse primitive polynomials which are a bit harder to find and not known to exist. You said you only needed to connect 2 to 4 taps, but if you let me choose your maximal length LFSR you probably need to connect about half of them. JackSchmidt (talk) 21:32, 17 February 2009 (UTC)

It will help me if you would be kind enough to take me step by step through finding taps for a maximal 21 (say) -stage LFSR. I wrote "usually 2 or 4" taps because I gather from the mysterious table that 2 or 4 taps always suffices. I don't know how one proves that. Cuddlyable3 (talk) 12:34, 18 February 2009 (UTC)

"2 or 4 taps always suffices" is roughly speaking an open problem in the theory of finite fields as far as I know.

For an example of finding a maximal LFSR, let's start with 4 stage. We first need an irreducible polynomial of degree 4 over the binary field GF(2). Let's use f = x^4+x^3+x^2+x+1 = (x^5-1)/(x-1). It is then easy to see that x has order 5 in the finite field F = GF(2)[x]/(x^4+x^3+x^2+x+1) of size 16, since x^5=1 mod x^5-1. We need a polynomial where x has order 2^4-1 = 15, not 5. We then compute the minimal polynomial of random elements of F until we find an element with order 15. Let's check A=x+1. Now (x+1)^3 = x^3+x^2+x+1 mod f is not 1, so x+1 does not have order 3, and (x+1)^5 = x^3+x^2+1 mod f is not 1, so x+1 does not have order 5. By Lagrange's theorem, x+1 must have order 15 (the only divisor of 15 left). Hence we wan to compute its minimal polynomial, that is we want to find a relation of linear dependence amongst (x+1)^i for i=0,1,...,4. The powers mod f are just: 1, x+1, x^2+1, x^3+x^2+x+1, x^3+x^2+x. We check that A^4 + A^3 + 1 = 0 mod f, and so our primitive polynomial is x^4 + x^3 + 1, agreeing with the table in LFSR.

Ok, now for n=21, we first need an irreducible polynomial of degree 21 over GF(2). Let's choose f = x^21 + x^7 + 1. Now 2^21 - 1 = 7*7*127*337, and we check that x^49 = 1 mod f yet x^7 is not 1 mod f, so x has order exactly 49 in F = GF(2)[x]/(f). We want it to have order 2^21-1, so this is not good enough. We choose a random element of the field F like A=x+1, and check its powers. A^((2^21-1)/7) mod f is not 1, A^((2^21-1)/127)) mod f is not 1, and A^((2^21-1)/337) mod f is not 1, so the order of A in F is exactly 2^21-1. Now we just need to find its minimal polynomial. Its powers A^0,...,A^21 are: 1, x+1, x^2+1, x^3+x^2+x+1, x^4+1, x^5+x^4+x+1, x^6+x^4+x^2+1, x^7+x^6+x^5+x^4+x^3+x^2+x+1, x^8+1, x^9+x^8+x+1, x^10+x^8+x^2+1, x^11+x^10+x^9+x^8+x^3+x^2+x+1, x^12+x^8+x^4+1, x^13+x^12+x^9+x^8+x^5+x^4+x+1, x^14+x^12+x^10+x^8+x^6+x^4+x^2+1, x^15+x^14+x^13+x^12+x^11+x^10+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1, x^16+1, x^17+x^16+x+1, x^18+x^16+x^2+1, x^19+x^18+x^17+x^16+x^3+x^2+x+1, x^20+x^16+x^4+1, x^20+x^17+x^16+x^7+x^5+x^4+x. We then see that A^21 + A^20 + A^17 + A^16 + A^7 + A^6 + A^3 + A^2 + 1 = 0 mod f, so our primitive polynomial is x^21 + x^20 + x^17 + x^16 + x^7 + x^6 + x^3 + x^2 + 1. Notice that it has 7ish taps, and that my method expects to get about 10 taps for the 21 stage LFSR.

However, many people are not interested in any old 21-stage LFSR of length 2^21-1. Many people want one with very few taps. I suspect that most people would want to use x^21 + x^2 + 1 as it is also primitive and has fewer terms.

The lexicographically first primitive polynomial for n=21,22,23,24 are:

• x^21 + x^2 + 1
• x^22 + x + 1
• x^23 + x^5 + 1
• x^24 + x^4 + x^3 + x + 1.

These take less than a second to compute by a brute force search, but using reasonably smart irreducibility and primitivity tests. JackSchmidt (talk) 17:36, 18 February 2009 (UTC)

Finding the lexicographically first primitive polynomial with 2 or 4 taps (3 or 5 terms) of degrees n=2..169 took 4 minutes, for what it is worth. It made no attempt to get 2 taps instead of 4, but often did find a 2 tap first. JackSchmidt (talk) 17:50, 18 February 2009 (UTC)

It is worth my thanks.Cuddlyable3 (talk) 14:30, 19 February 2009 (UTC)

CHain rule question

A viscous liquid is poured onto a flat surface. It forms a circular patch whose area grows at a constant rate of 5 cm/s. Find in terms of π, the radius after 20 seconds

Using the chain rule I get dA/dt = dA/dr x dr/dt
5 = 2πr x dr/dt

That's as far as I can get. Any hints? --RMFan1 (talk) 23:03, 17 February 2009 (UTC)

I don't think you need the chain rule - the questions asks for the radius after 20 seconds, not the rate of growth of the radius. After 20 seconds we have A = 100 cm² (I think that rate of growth should be 5 cm²/s), and A = πr², so r is ... Gandalf61 (talk) 23:11, 17 February 2009 (UTC)

I was so focused on using the chain rule (as that's the chapter I'm on) that I completely missed the obvious. Thanks. --RMFan1 (talk) 23:34, 17 February 2009 (UTC)

February 18

Making n the subject

If A= P(1+r)n^{Superscript text} Then make n the subject of formula —Preceding unsigned comment added by 66.36.215.172 (talk) 10:21, 18 February 2009 (UTC)

Assuming you mean A = P(1+r)ⁿ look into power rules etc. 89.240.5.5 (talk) 10:38, 18 February 2009 (UTC)

(added header) Consider taking logarithms of both sides. Confusing Manifestation(Say hi!) 22:34, 18 February 2009 (UTC)

A rant about inequalities

Don't get them at all, no sir.

Find all x that satisfy:

${\frac {-2x+5}{x+3}}<1$

Now I was told I have to consider two cases, x > 3 and x < 3. x = 3 is obviously undefined.

x > 3:

$-2x+5<x+3\,$

$-3x<-2\,$

$x>{\frac {2}{3}}$

So x > 3 and $x>{\frac {2}{3}}$. I guess that means $x>{\frac {2}{3}}$?

x < 3:

$-2x+5>x+3\,$

(Since x+3 must be negative)

$-3x>-2\,$

$x<{\frac {2}{3}}$

We get the same kind of logic as before. This is how it was explained to me. I won't even start on the fact that there's a numerator that has an x in it as well, not to mention that you can probably rearrange simple inequalities as rational ones and no doubt create a world of pain for yourself anyway. Someone is playing a joke on me.

Anyhow I was told that the method I'm using sucks and that I should solve the inequality by getting the right hand side to zero first.

${\frac {-2x+5}{x+3}}-1<0$

${\frac {-3x+2}{x+3}}<0$

Okay, so above is negative for x > 0, but also for x < -3. That's what I've ascertained from examining the equation with my head, which is not a nice algorithmic process for solving problems, and there's probably something I've overlooked. I'm skeptical now because all the other answers in this useless book are of the form a < x < b. But I wouldn't be surprised if it's part of the elaborate joke to make all the even-numbered questions of a completely different type to the odds. Ha. Ha. HA. Help me before I kill myself. 82.32.49.186 (talk) 10:56, 18 February 2009 (UTC)

You are approaching the problem correctly. Both of the methods you described work and can be used to solve this problem, but unfortunately you made numerical mistakes in both approaches.

In your first approach: under what conditions is x+3 positive? You use that x+3 is positive if and only if x > 3, but that is incorrect. Similarly for negative.

Also in your first approach, you have inadvertantly switched "and" and "or". When is it true that both x > 3 and x > 2/3? The answer is x > 3, not x > 2/3. (However, it was supposed to be x > -3 anyhow.)

Although your first approach does indeed work, I agree that the second approach is easier. You correctly reach the inequality ${\frac {-3x+2}{x+3}}<0$, but then incorrectly state that that holds for x > 0 and x < -3. As you haven't explained your reasoning in detail, I can't help you find your mistake, but maybe you should try x = 1/2 as an example.

Finally, the answer will not be of the form a < x < b. Please let us know if you would further help with this problem; it looks to me like you are understanding this material well, and just making easy-to-make numerical errors. Eric. 131.215.158.184 (talk) 11:41, 18 February 2009 (UTC)

Through 5 or 6 mintues of tedious trial and error, I examined the points where the sign of the function (the rational expression, treat it as a function) switched. The function, according to these calculations is negative for all x < -3 and all x > (2/3), if I got it right. I suspect that the trick is to examine the cases less than or equal to where the denominator is undefined, and where the numerator is = 0. Does this trick hold for every rational, polynomial expression? Brute force is not my idea of a good time 82.32.49.186 (talk) 12:10, 18 February 2009 (UTC)

You are (roughly) correct: for any rational expression a / b (whether polynomial or otherwise), the sign of a / b can be determined by knowing the signs of a and b. Thus the "critical points" to examine are when the numerator a or denominator b are zero, for those are the only points where the expression a / b can change sign. I am sure others can pick it up from here if you have further questions, I am going to bed. (And by the way, your answer is correct.) Eric. 131.215.158.184 (talk) 12:34, 18 February 2009 (UTC)

My book gives -3 < x < 2/3 as the answer. What the hell? 82.32.49.186 (talk) 12:48, 18 February 2009 (UTC)

Did you give us the problem correctly? Then the book is wrong. Try x = 0 if you have doubts. Eric. 131.215.158.184 (talk) 18:54, 18 February 2009 (UTC)

An obvious thing to do to get a quick feel whether your results are the right ones is to use a spreadsheet to calculate and graph f(x) against x. In your case maybe plot values from x = -5 to +5 in steps of 0.1? and just see which values of x have f(x) < 1. That will help you spot gross blunders (eg +3 instead of -3 and so on easily. -- SGBailey (talk) 12:46, 18 February 2009 (UTC)

If you say x > 2/3 AND x > 3, then that means x > 3. You cannot ignore the fact that the word AND means AND, nor the fact that 3 > 2/3.

Next, where you say x < 2/3 AND x < 3, then that means x < 2/3.

So the solution is x > 3 OR x < 2/3. Any number greater than 3 is a solution; any number less than 2/3 is a solution; no other numbers are solutions.

Your other method also works. You have

${\frac {-3x+2}{x+3}}<0$

and then you can divide both sides by −3, which requires changing "<" to ">" since what you're dividing by is negative:

${\frac {x-(2/3)}{x+3}}>0.$

Now either x > 3 or x < 2/3 or x is either between those two or equal to one of them. If x > 3 then both the numerator and denominator are positive, so you've got a solution. If x < 2/3 then both are negative, so you've got a solution. If x is between the two, then the denominator is negative and the numerator is positive, so you haven't got a solution. Michael Hardy (talk) 21:38, 18 February 2009 (UTC)

An alternative approach involves the rule that "multiplying both sides of an inequality by a strictly positive value does not affect the inequality". Instead of testing individual cases, multiply everything by $(x+3)^{2}$. For $x\neq 3$, this is guaranteed to be positive, and you can then rearrange the inequality to give you (quadratic) > 0. Then a factorisation of the quadratic and a quick check of where it's positive should spit out the answer (just remember to check for any funny business around the x=3 point as well). Confusing Manifestation(Say hi!) 22:32, 18 February 2009 (UTC)

Problem from Linear Algebra

The following question is from Kluwer's Foundations of Linear Algebra and has been driving me crazy for the last day. I imagine the solution is simple and that I am just missing it, whatever the case, any help would be deeply appreciated:

    Let A be a nonsingular 4 x 4 matrix so and B, C, D, and E be 2 x 2 matrices with A = (B C)
                                                                                         (D E)
    can BE - DC, BE - CD, EB - DC, and EB - CD all be singular?

The best result I can get is that if B and D commute or C and E, then no; and if the commutator of B and E or of D and C is singular, then no (not 100% sure that I didn't make a mistake on the last one though.) Again, thanks in advance for any help:) Phoenix1177 (talk) 11:18, 18 February 2009 (UTC)

With the help of the partial results you provided, I found a counterexample with a little trial and error (i.e., a nonsingular four by four matrix A such that BE - DC etc. are all singular). However, in my example the commutator of B and E and of D and C are singular, so I'd suggest checking your proof of that fact. I'll leave the joy of hunting for counterexamples to you. Eric. 131.215.158.184 (talk) 11:51, 18 February 2009 (UTC)

I assumed that I made a mistake there, it's late here. At any rate, I was hoping that it wouldn't come down to hunting out a counterexample...66.202.66.78 (talk) 12:16, 18 February 2009 (UTC)

I (or someone else) can give you a hint in the morning if you want one. As usual, when there exist counterexamples, there is often a very simple counterexample. Eric. 131.215.158.184 (talk) 12:38, 18 February 2009 (UTC)

Thank you, I actually found a counterexample as soon as you mentioned there was one. This is more a case of bad textbook logic, on my part, than anything else; since this was in a chapter on determinants, and finding a counterexample didn't seem to have much to do with determinants, I assumed that a proof was being looked for...obviously, the answer is in the negative, so no proof. Thank you :) P.S. The stuff about the commutators was assuming one of them was invertible. Phoenix1177 (talk) 13:15, 18 February 2009 (UTC)

February 19

Orthogonal

Hi, is an orthogonal a T shape or an L shape? ~ R.T.G 06:43, 19 February 2009 (UTC)

In mathematics, orthogonal is an adjective meaning "perpendicular to" (in other fields, it is still an adjective, but may have a slightly different meaning). So you can have a pair of orthogonal lines or orthogonal planes, or you can say that one line is orthogonal to another, but you can't have "an orthogonal". In both a T shape and in an L shape, the two lines are orthogonal to one another in each case. Gandalf61 (talk) 09:09, 19 February 2009 (UTC)

You can have an orthogonal relationship but clears it up now thanks ~ R.T.G 14:40, 19 February 2009 (UTC)

Surd Questions

Maybe its just the lack of sleep or something but i havent done these in a while and i have an assessment on these soon..im probably missing something pretty simple but any help would be appreciated =) !

[Ill show the answers i received]

Q.1

${\frac {2{\sqrt {8}}a^{3}}{a}}+4{\sqrt {2a}}-{\frac {5{\sqrt {3}}2a^{5}}{a^{2}}}$

$={\frac {4{\sqrt {2}}a^{3}}{a}}+4{\sqrt {2a}}-{\frac {20{\sqrt {2}}a^{5}}{a^{2}}}(Factoring)$

$=4{\sqrt {2a^{2}}}+4{\sqrt {2a}}-20{\sqrt {2}}a^{3}(Division/Subtracttopfrombottom)$

(i also thought about factoring the pronumerals[and move them infront of the radical then subtract/divide]but that didnt work out either..

$=4a{\sqrt {2a}}+4{\sqrt {2a}}-20{\sqrt {2}}a$

Answer is: $-12{\sqrt {2a}}$

Q.2

$1{\frac {4}{5}}{\sqrt {7}}-{\frac {2{\sqrt {7}}}{3}}$

i dont have much of an idea what to do here..anything to help me learn would be appreciated

Answer is ${\frac {19}{12}}{\sqrt {7}}$

Q.3

Just a few small things here...

I expand this by multiplying the equation in the first set of brackets by whats out side then.. multiplying it by the other one? Or by expanding the brackets first and..doing ..something i dont know

$2{\sqrt {3}}(1+4{\sqrt {3}})(1-{\sqrt {3}})$

$(2{\sqrt {3}}+7{\sqrt {3}})(1-{\sqrt {3}})$

$2{\sqrt {3}}-3{\sqrt {3}}+7{\sqrt {3}}-7{\sqrt {3}}$

Correct answer is $18-22{\sqrt {3}}$

(also how do i expand if theres 3 brackets :S)

Any help would be appreciated.

You're doing some rather strange things. You seem to be equating

$2{\sqrt {3}}(1+4{\sqrt {3}})\,$

with

$2{\sqrt {3}}+7{\sqrt {3}}.\,$

That is wrong. You need the distributive law:

$2{\sqrt {3}}(1+4{\sqrt {3}})=(2{\sqrt {3}}\cdot 1)+(2{\sqrt {3}}\cdot 4{\sqrt {3}})\,$

The first term in the last expression simplifies to

$2{\sqrt {3}}.\,$

The second term is

$2{\sqrt {3}}\cdot 4{\sqrt {3}}=2\cdot 4\cdot {\sqrt {3}}\cdot {\sqrt {3}}=2\cdot 4\cdot 3=24.\,$

And you've done something similar repeatedly throughout what you've done above. I haven't figured out how you got 7, but it shouldn't be there.

If you don't know that

${\sqrt {3}}\cdot {\sqrt {3}}=3,\,$

then you have no idea what square roots are. That is what they are. Michael Hardy (talk) 11:37, 19 February 2009 (UTC)

Also, you wrote:

${\frac {2{\sqrt {8}}a^{3}}{a}}+4{\sqrt {2a}}-{\frac {5{\sqrt {3}}2a^{5}}{a^{2}}}$

Are you ABSOLUTELY sure that it didn't say

${\sqrt {8a^{3}}}\,$

rather than

${\sqrt {8}}a^{3}\,$

and

${\sqrt {32a^{5}}}\,$

rather than

${\sqrt {3}}2a^{5}\,$

or

${\sqrt {32}}a^{5},\,$

the former being what you wrote?

If you're not paying close attention to that sort of thing, then it's useless to get into the other parts of the problem.

The following are two DIFFERENT things:

${\sqrt {32}}a^{5}\,$

${\sqrt {32a^{5}}}\,$

Michael Hardy (talk) 11:43, 19 February 2009 (UTC)

Akin to absolute value

Is there a name, or standard notation, for the function $f:(0,\infty )\rightarrow (0,\infty )$ defined by

$f(x)={\begin{cases}x,&{\mbox{if }}x\geq 1,\\1/x,&{\mbox{otherwise}}\end{cases}}$?

I'm thinking of it as being to multiplication what absolute value is to addition, also noting that the two are connected through exponentials and logarithms:

$f(x)=\mathrm {e} ^{|\ln x|}$

—Bromskloss (talk) 16:18, 19 February 2009 (UTC)

Help with LaTeX

Hey everyone. I'm having trouble getting something to compile correctly and my google-fu is failing me. Perhaps you can help.

What I want to have is a sentence appear on the next line, centered, and I want to be able to "label" the line all the way at the right with an asterisk (where an equation number might normally show up, say). Can anyone tell me how to do this?

Thanks in advance. –King Bee ^{(τ • γ)} 19:27, 19 February 2009 (UTC)

\hfill is basically what you want.

Wrapped up nicely in a command:

 \newcommand{\kingbee}[1]{~\newline\phantom{($\ast$)}\hfill$\displaystyle{#1}$\hfill($\ast$)\newline}

And a demo tex document:

\documentclass{article}
\newcommand{\kingbee}[1]{~\newline\phantom{($\ast$)}\hfill$\displaystyle{#1}$\hfill($\ast$)\newline}
\begin{document}
Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum.
\kingbee{x^2+1=0}
Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum.

\kingbee{x^2+1=0}

Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum.
\kingbee{x^2+1=0}

Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum.

\kingbee{x^2+1=0}
Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum.
\end{document}

Probably other ways to do it too, but the latex code eqnarray and equation was complicated. JackSchmidt (talk) 19:56, 19 February 2009 (UTC)

I recall using something from CTAN some time ago which allowed for arbitrary equation labels, say an asterisk or daggar, but it escapes me now. You could check there. Baccyak4H (Yak!) 19:59, 19 February 2009 (UTC)

Finite multiplicative subgroup of direct product of fields

A finite subgroup of the group of units of a commutative field is cyclic. Is a finite subgroup of the group of units of a direct product of n commutative fields generated by at most n elements? It is fine to assume all the fields are equal, so just a direct power.

I have an abelian groups proof, but I was interested in something with polynomials or varieties or tori or something. JackSchmidt (talk) 19:36, 19 February 2009 (UTC)

Your question is equivalent to: for all integers $m_{1},\ldots ,m_{n}$, any subgroup of $\mathbb {Z} /m_{1}\mathbb {Z} \times \cdots \mathbb {Z} /m_{n}\mathbb {Z}$ can be generated by at most n elements. But I do not know if that is true or not. Eric. 131.215.158.184 (talk) 07:35, 20 February 2009 (UTC)

Showing my working

Find the coordinates of the stationary point on the curve $y=xe^{-3x}$

I began by finding dy/dx: $dy/dx=(1-3x)e^{-3x}$

Then to find the x coordinate of the stationary point$dy/dx=(1-3x)e^{-3x}=0$

So for the expression on the left to equal 0, either (1-3x) or e^{-3x} must be equal to 0. One obvious solution is (1-3x)=0 when x=1/3 but what about the e^{-3x}? I know it doesn't equal 0 because no value of x could make it equal to 0.

So my question really is whether it's ok to show my working as follows: ...

$(1-3x)e^{-3x}=0$

$(1-3x)=0$

$3x=1$

$x=1/3$

...

Or is there a way to show why the e^{-3x} cannot equal 0? --RMFan1 (talk) 20:45, 19 February 2009 (UTC)

It is, of course, possible to prove that e^x never takes the value zero, but whether you're expected to do so will depend on the situation. In my education, for example, at school level (age 16-18 say) I was expected to take the important properties of e^x for granted, while as a first year undergraduate (age 19 or so) I was expected to be able to prove them all from the definitions, and ever since then I've been expected to take them for granted again. At a guess, if you're allowed to assume the chain rule and the fact that the derivative of e^x is e^x, you're probably allowed to assume it's never zero. Algebraist 21:03, 19 February 2009 (UTC)

In "showing your work", I would at least state that the step where e^x disappears is justified by the fact that e^x ≠ 0. I wouldn't do that in a research paper since "everybody knows" it, but in a homework assignment you're supposed to demonstrate that you know what you're doing. In an exposition within a Wikipedia article I might explicitly state it—that depends on the context. Michael Hardy (talk) 22:57, 19 February 2009 (UTC)

While we're on the subject of showing work, I'll repeat something that one of my supervisors drummed into me: in a fully written-out argument, in all but the most obvious contexts, every line should begin with actual English text, even if it's just 'thus' or 'so'. If you habitually just write down a sequence of equations or similar, it can be very unclear whether the equations are being asserted, assumed, inferred, claimed to be equivalent, taken as a definition, or whatever. Algebraist 00:06, 20 February 2009 (UTC)

If you don't like words, though, you can always use $\implies$ and $\therefore$ and similar. --Tango (talk) 00:38, 20 February 2009 (UTC)

Theory about graphs

Is there any theory about, or perhaps is it known, whether any graph can be described mathematically? For example, if I was to randomly scribble something onto some graph paper, could any such a graph definitely be described mathematically? --RMFan1 (talk) 23:56, 19 February 2009 (UTC)

When you say "graph", which of these two kinds are you referring to?

—Bromskloss (talk) 00:15, 20 February 2009 (UTC)

And what do you mean by "described mathematically"? --Tango (talk) 00:17, 20 February 2009 (UTC)

Im referring to graphs of functions. And I mean is it known whether anything someone could possibly scribble on paper could be defined by an equation. Or if there's a theory, general belief among mathematicians etc. on this --RMFan1 (talk) 00:26, 20 February 2009 (UTC)

I'm thinking kind of. Assuming you draw a continuous function, there is a finite polynomial that approximates the function to the point that the polynomial will never leave your graphite line. However, in the more mathematical sense, no. The number of possible mathematical equations is denumerable, and the number of possible functions that can be visualized is nondenumerable. In other words, there are more graphs you can make than there are equations to describe them. Indeed123 (talk) 03:54, 20 February 2009 (UTC)

It is too bad we don't have an article on mechanical curves. These were studied in the 16th centurye or so, and were found not to be described well by the standard mathematics of the time which was based either on classical geometry or polynomial/rational functions. I think an example curve is a catenary, but I don't have a reference handy. At any rate, these curves helped usher in the use of calculus into geometry and made analytic geometry more analytic and less algebraic. JackSchmidt (talk) 04:20, 20 February 2009 (UTC)

Possibly also the cycloid. JackSchmidt (talk) 04:25, 20 February 2009 (UTC)

I'm thinking it's more of a physics question: Are all possible movements of the pen such that they lie within what can be described by a mathematical expression? I have no answer to that and it seems to me that you would need an infinitely accurate description of physics to answer it. Do we even necessarily have well-defined paths of the pen in quantum mechanics? (I know to little about it.) —Bromskloss (talk) 07:52, 20 February 2009 (UTC)

February 20

Analysis on an interesting function

Hi there - just finished a last bit of work and was wondering if anyone extremely kind with a little time to spare could spot anything stupid I've done in the process (highly likely) - I may well have botched it up so pull no punches;

If $f_{0}:\mathbb {R} \to \mathbb {R}$ is defined by $f_{0}(x)=|x|$ for $|x|\leq {\frac {1}{2}}$ and $f_{0}(x+z)=f_{0}(x)\forall z\in \mathbb {Z}$. We define $f_{n}(x)=4^{-n}f_{0}(4^{n}x)$.

Sketching f₀, f₁ and f₀+f₁ as follows: Media:graphfunction101.png

Showing that, for each $x\in \mathbb {R}$, $\sum _{n\geq 0}f_{n}(x)$ converges: if I've drawn my graphs right, the maximum value attained is must be less than or equal to (less than in this case) the sum of the maximum values of each f_n, i.e. $\sum _{n\geq 0}f_{n}(x)\leq {\frac {1}{2}}+{\frac {1}{8}}+{\frac {1}{32}}+...={\frac {1}{2}}({\frac {1}{1-{\frac {1}{4}}}})={\frac {2}{3}}$ and so since the sum is monotonically increasing & bounded must converge.

Defining $f:\mathbb {R} \to \mathbb {R}$ as $f(x)=\sum _{n\geq 0}f_{n}(x)$, we want to show that f is continuous everywhere. By the definition of continuity I'm using we want $\lim _{x\to a}f(x)=f(a)\forall a$; then $\lim _{x\to a}\sum _{n\geq o}f_{n}(x)=\sum _{n\geq o}f_{n}(a)$. Since we know (or I'm fine showing) the f_n are all convergent, it seems to be simply a matter of swapping the sum and the limit - I'm not completely sure how to prove it's alright to do this? Once that's done we have continuity of f.

Lastly, we want to know where f is differentiable - it seems the non-differentiable points are ${\frac {1}{2}},{\frac {1}{4}},{\frac {3}{4}},{\frac {1}{8}},...$ and so I'd say the function is non-differentiable over the dyadic rationals $\mathbb {D}$, so differentiable over $\mathbb {R} \backslash \mathbb {D}$ - is this correct? If so, is there any way to make the argument more rigorous than 'by inspection' or do I have to go back to the fundamental definition of 'differentiable'?

Obviously I fully understand you're not here to just 'check my working' and I've asked/rambled an awful lot here, but any small help would be greatly appreciated if you have any time spare at all - thanks very much! Also, I tried to find the rules on uploading/linking to images but I had no luck - apologies if I broke any rules! Thanks very much again - Mathmos6 (talk) 04:36, 20 February 2009 (UTC)Mathmos6

Setting up Maekawa's Algorithm

For an assignment, I had to come up with an algorithm to set up Maekawa's Algorithm. You don't need to read the algorithm. For N elements, I had to come up with N sets of K elements such that for any two sets you pick, at least one element will be shared between them. In other words, every set intersects with every other set. Now that I'm done and I've submitted it, I thought I'd Google for what the "best" algorithm is and, surprisingly, I can't find one anywhere. So, I have two questions:

1) What is the "best" algorithm for doing this? Does it have a name that I can Google and find? Keep in mind that by "best", I am implying that it is an algorithm that is best when implemented on a computer. So, big-O complexity is what is normally used to rate algorithms.

2) I know I didn't invent a new algorithm. So, does it have a name? To do this, I have to describe the algorithm. I do that below. I hope I can get a collapse box to work properly so I don't flood the RD:

algorithm

Begin with N sets of K elements, all empty (meaning all elements are null). Assume N=7 and K=3 for this example. Set SNi*k to [0, ik+1, ik+2] for i=0 to 2. So, S0=[0,1,2], S3=[0,3,4], S5=[0,5,6]. I just noticed that there is an error in my description and result. I'll have to go back and fix that on my submission. All I did was fill the first element with 0 and then fill the rest, in order with 1, 2, 3, 4, 5, 6.

Now, take any empty set (S1 for example). Si must contain i, so put a 1 in S1. Then, create an array of N boolean values, all false. For every Sj that intersects with S1, take each element e of Sj and set Ue to true. So, S0 intersects because it has a 1. Set U0=true, U1=true, U2=true. No other sets intersect. This process will, from now on, be called "set intersects in U." After setting intersects in U, find any Sj that doesn't intersect with S1. S3 doesn't. Pick an element e in S3 such that Ue is false. U0 is true, so that won't work. U3 is false. Add 3 to S1. Now, S1 is [1, 3, null]. Set intersects in U again. Because S1 is not full, find another set that doesn't intersect (S5). Pick an e in S5 such that Ue is false. 5 will work. Add 5 to S1.

Repeat for S2. Add 2 to S2. Set all U to false. Set intersects in U. Find a set that doesn't intersect. Pick an e such that Ue is false. Set intersects in U. Pick another e. Then, repeat for the next empty set. In the end, you will have N sets such that each one will intersect, assuming that it is possible for the given N and K values to start with.

Thanks. -- kainaw™ 05:01, 20 February 2009 (UTC)

Could you describe the problem you were trying to solve more carefully? It seems trivially solvable to me - simply give all of the N sets the same element. Then all pairs of sets share that element, and the other K-1 elements are irrelevent. —Preceding unsigned comment added by Maelin (talk • contribs) 06:18, 20 February 2009 (UTC)

That would work if all you wanted was N sets. However, this is for Maekawa's Algorithm. Each of the N elements must be located in the sets. To do that fairly, consider N=7 and K=3 (7 sets of 3 elements). There are 7 unique elements to place in 21 positions, so each of the 7 unique elements should appear 3 (21/7) times and every set should intersect every other set. -- kainaw™ 06:22, 20 February 2009 (UTC)

Unreachable chess positions

What chess positions are unreachable by any legal series of moves from the starting position, besides those that would require an illegal amount of material, those that have both kings in check or the same king in check by two knights, and those that have pawns on the first or eighth rank? NeonMerlin 05:57, 20 February 2009 (UTC)

Any where the pawns have changed columns without an appropriate number of opponent pieces having been captured. -- SGBailey (talk) 07:02, 20 February 2009 (UTC)