Bignose

Well, there is a generalized matrix inverse also known as pseudoinverse for non-square matrices. The pseduoinversed matrix R would be a 3x4 matrix, and you can right mulitply a 3x3 matrix by a 3x4 matrix.

Assuming that that answer for a is right (I didn't check, and I assume you have checked), you've almost done the entire answer. Let me call the first matrix Q and second matrix (your answer) R. These are related by Q = H*R where here H is the transformation matrix. Well, to isolate H, you just right multiply both sides of the this equation by the inverse of R, R^-1. Since a matrix times its inverse is the identity matrix on the right hand side of the equation you get H times the identity matrix. Then, any matrix times the identity matrix is itself, so the right hand side is just H. The equation becomes Q*R^-1 = H. So, you just need to get R^-1 and multiply Q by R^-1 to get H, the transformation matrix you seek.

One of the ways you can write a solution to the (1-D scalar linear) wave equation is u(x,t) (the solution) = F(x-c*t) + G(x+ct) where c is the speed of the wave. The exact form of the functions F&G are based on the boundary and initial conditions. But, you can see how they behave from their form. At time t=0, you have F(x). At time t=(1/c), you have F(x-1), or to look at it in another way, the information from F(x) has now been transferred over to x-1. Exactly how a wave moves from one side to the other. In the same way, at time t=(1/c) the information that was at G(x) is now at G(x+1). Information moves both ways, depending on whether it is information on F or G. This is because the wave equation is linear, so the information that is formally seperated into F & G can be summed to get the final answer. Finally, what I wrote is just a discussion about the mathematics behind the wave equation, but there are many, many situations where the mathematical abstraction is an exceptionally good model for real physics. Any medium where dissipation is very low the wave equation will be applicable. Also, for fast time scales, the wave equation is good -- basically so long as the time you are modeling is short dissipation won't have time to take effect.

This piece of information is correct, but you are not answering the question at hand. Which is, on average, how many coin flips will it take to get one heads. Or, in math-speak, what is the expected value of the number of flips until you get a head. That 75% number is correct, in that it is the cumulative probability for two flips. That is, 75% is the probability of taking 2 or less flips. You have to use the frequency to get the average number. That is 75% (for two flips) - 50% (for one flips) = 25% to take exactly two flips. Which can also be calculated since you have to go tails then heads on the flips, each of which is a 1/2 chance, so the chances of going tails then heads is 1/2*1/2 = 25%. Next, you have to weight each of these possibilities correctly to get the expected value: The expected value of a discrete distribution is SUM x*p(x) where the sum is over the entire domain of all values of x. p(x) is the probability. One of the issues here is that you have been calculating the cumulative distributions. I.e. 0.75 is the probability that is takes at most 2 flips, i.e. the chances of it taking one flip PLUS the chances of it taking two flips. p(x) in the above expected value is the probability it takes exactly x flips to get one heads. So the coin flip problem is the sum of 1*(probability it only takes 1 coin flip to get a heads=0.5) + 2*(probability it only takes 2 flips to get a heads = 0.75-0.50=0.25) + 3*(probability it only takes 3 flips to get a heads = 0.875-0.75=0.125 ) + 4*(0.063) + etc. This is SUM (from n=1 to inifnity) of n*(1/2)^n Instead of infiinty, if we sum to a set number of terms, a pattern emerges: Fisrt column=number of terms in the sum, second colunm = result of the sum 5 1.781 6 1.875 7 1.93 8 1.961 9 1.979 10 1.988 11 1.994 12 1.997 13 1.998 14 1.999 15 1.999 Now, if you keep adding more and more terms to that sum, it is going to converge to 2. Meaning that the average number of flips it is going to take until you get one heads is two. That is from the very definition of the expected value. Now, let's do the exact same thing with the dice rolling again, let's use the negative binomial distribution nb(x; r,p) = ( (x+r-1) choose (r-1))*p^r*(1-p)^x x in this case would be number of rolls before rolling that first four. That is, x=0 corresponds to rolling a 4 on the very first try. p=1/12, and r=1. nb(0,1,1/12)=1/12=0.083 This is the change of rolling a 4 on the first try nb (1,1,1/12)=0.076 (I'm going to skip typing the ,1,1/12 in the rest) nb (2) = 0.07 nb (3) = 0.064 nb (4) = 0.059 This is the chance it takes exactly 5 rolls to roll the sum to 4 on the last roll. nb(5) = 0.054 and so on... Now, let's compute the expected value of this E(x)= SUM (i=0 to infinity) of i*nb(i,1,1/2) Again, we can't sum to infinity, so we take a discrete number of terms and see if a pattern emerges. This series is going to take a lot longer to converge than the coin flipping one, so I'm not going to write every term. 10 terms' SUM = 2.552 20 = 5.853 30 = 8.17 40 = 9.532 50 = 10.267 60 = 10.643 70 = 10.83 80 = 10.92 90 = 10.963 100 = 10.983 110 = 10.992 120 = 10.996 130 = 10.998 140 = 10.999 150 = 10.9999 I think we can see that this series is converging to 11. Meaning that the expected value is 11 failed rolls before 1 successful rolls, or 12 rolls in total to average 1 roll of 4 on the dice. Again, taken from the exact definition of an expected value. On average, 12 rolls to get one roll that sums to 4 on two dice, and on average two flips to get one heads on a coin. I don't know anything more I can do to convince you, Sisyphus. If you need more, please don't hesitate to ask, but this last post I hope is pretty conclusive.

How can these two statements be compatible? They say the same thing, but with two different numbers in the key spot. How can it be both 8 rolls and 12 rolls, one has to be correct. My vote is still for the 12, not only from my first posts, but also from the expected value of the negative binomial distribution, which describes this situation exactly. And how does the simulation where it counts the number of rolls until a 4 is rolled, which resulted in 12 rolls being needed on average, prove you right, when I keep saying 12 and you keep saying 8?

Edit: I deleted this post since I had some pretty significant errors in it. I came back and fixed it later, but in terms of full disclosure, I wanted to let people know that this post is not the same as I posted before. OK, Dug out my old statistics book, and really, what this situation is, is a negative binomial distribution. From my statistics book (Devore's Probability and Statisitcs for Engineering and the Sciences, 4th ed., 1995) The negative binomial random variable and distribution are based on an experiment satisfying the following conditions: 1) The experiment consists of a sequence of independent trials. 2) Each trial can result in either a success (S) or failure (F). 3) The probability of success is constant from trial to trial, so P(S on trial i) = p for i=1,2,3,... 4) The experiment continues (trials are performed) until a total of r successes have been observed, where r is a specificed positive integer. This is exactly what we are trying to describe here, where Sucess S is defined as a roll of the dice where the sum is 4, p=1/12, and we are looking for r=1 success until the trials are stopped. The negative binomial distribution is: nb(x; r,p)= ( (x+r-1) choose (r-1) )*(p^r)*(1-p)^x (Sorry I didn't type this into Latex, I don't have the time to lok up how to do the x choose y properly, I'll try to come back and fix it later tonight) The expected value of the negtive binomial is E(X) - r*(1-p)/p and the variance V(X)= r*(1-p)/p^2 Using the values from above, r=1 & p=1/12, E(x)=11, meaning there the expected number, or average number, of failures before a success occurs is 11. Meaning, 12 rolls, on average, are needed until rolling a 4. This is what I calculated above, but not from the distribution. The difference between this calculation and what Sysiphus is calculating, is that this is the average number of rolls, and Sys is caclulating the probability of needing 12 rolls or less. The probabilty of needing 12 rolls or less is not 1, I agree with that, because there are very good chances it will take more than 12 rolls to get a 4. But, on average, it is 12 rolls.

Right, but you just said it yourself, a set of two will contain one heads. In other words, on average you will have to flip a coin 2 times to get at least one heads. The frequency of getting heads is 1 heads per two flips. Multiply that by 2 flips and you get one heads. (1 heads/2 flips)*(2 flips) = 1 heads Just like the frequency of rolling two dice that sum to 4 is 1 sum to four per 12 rolls. Mulitply that by 12 rolls and you get, on average, one sum to four. Write yourself a program in your favorite language, or you can even do this in Excel. Here is the psuedo code of the simulation I wrote: begin dicerollingprogram totalrolls=0 do i=1,number of trials rolls=0 count=1 sumtofour=.false. do while (sumtofour=.false.) roll two dice, d1 & d2 if (d1+d2=4) then sumtofour=.true. rolls=count else count=count+1 endif enddowhile totalrolls=totalrolls+rolls enddo avgrollstofour=totalrolls/(number of trials) I wrote this up in FORTRAN and ran it for 100000 trials several times avgrollstofour? I got 11.994, 12.002, 12.001, 11.998, 11.989, 12.001, 12.002 when I kept re-running it. It takes an average of 12 rolls to to get one roll where the dice sum to 4.

No, Sisyphus, you are not right here, because you've broken an independence assumption here. Namely, that each roll is brand new and not dependent upon the values before it. I think that the key assumption/phrase here is "on average" You are assuming that the player is re-rolling until he gets his 4. Using F as a Failure (not rolling a 4) and S as a Success (rolling a 4), you are calculating the distribution of FFFS, FFFFFFS, FS, FFFFS, FFFFFFFFS, FFFFS, FFFFFS You are calculating the probability of each of those results, since you stop as soon as you get that S. And, you are right in that you cannot ever guarantee an S, i.e. there is a probability for the result FFF...100 billion more F's ...FFFS that is not zero. Whereas my calculations are calculating FFFFSFFFFSFFFFFFFFFFSFFFFSFFFFFFFFFFFFFFFFFFSFFFFFSFFFFFSFFFFSFFFFFFFFFFFFSFFFFFFFFFFS and calculating the average number of F's between each S. Which will be 11 (=12-1) meaning, on average, you will have to roll the pair of dice 12 times to have one sum to 4. Or, put another way, if you look at any 12 letters in a row there, on average, it will contain at least one S. There will be 12 letter strings were there are no S's, and there will be strings where there are many S's, but on average, each 12 letter string will have 1 S. I'll make you a friendly wager. I get 8 rolls (approx. your 50% point), and if I get 1 or more 4's in those 8 rolls, You give me 1 penny. If I don't get at least 1 4, I give you a penny. If we play this game long enough (thereby invoking the "on average" phrase), I will have much more money than you. The independence of each roll allows me to add together the probabilities. Though, by invoking the phrase "on average" the word probabilities really shouldn't be used -- I should use the word frequencies. Maybe this is the source of the confusion. I have been casual with the use of the language, and for that I apologize, I should have been more careful. I shouldn't be so quick to mix the words since the difference between the words is subtle, but pertinent. So, by invoking the phrase "on average" the frequencies can be summed together, and again I can state that on average it will only be 12 rolls until at least 1 sum of 4 has occured.

Well, there are two different points of view, both valid so long as you take care of the details. The first point of view is the one I advocated above, that 1&3 and 3&1 are seperate events. I find this point of view easier. Bascially, it treats each single die as it's own seperate event. Then 1&3 would be the probability that the first die reads 1 and the second die reads 3. The probability for each die to read any one number is 1/6, so the probability for the first die to read 1 and the second to read 3 is (1/6)*(1/6) = 1/36. And, the entry 1&3 is one of 36 entries on the table. So, it all comes out the same. So, the chances of 1&3 or 3&1 are just 2/36 I find this way the easiest and most straightforward way to think about the problems. But, the second way would to treat each die as indeterminable. That is, you don't call one die the 'first' and the other the 'second'. So, one die needs to roll either a 1 or a 3, the chances of that is 2/6, and the other has to roll a specific number (a 3 or 1 resepctively), the chances of the other die rolling any specific number is 1/6. The product of these two is 2/36, the same as the above. So, you can see how either way of thinking gets the same answer. It is good to know how to do either method, but each way, the 1&3 and 3&1 comes up with 2/36. The first method I think is by far the easiest to see, since, in effect, you just count the results off of the table like the one in my previous post. But, you can imagine how much more difficult it gets with larger numbers of random variables. The table above gets much more complicated if you roll 3 dice, just imagine how complicated it would be if you rolled 20 dice. That's where the second method can be more useful. Here is a good test, do you see how 7 is the most common sum to roll? Can you caclucate the odds using both methods I listed above?

Well, the first issue here is that you do not have the correct probability for getting a sum of 4. There are not 19 combinations from rolling 2 dice, there are in fact 36. That comes from 6*6. Your error comes from not recognizing that 3&1, 2&2, and 1&3 are 3 separate events. 3&1 and 1&3 are different. For small set like the two dice, it can be very helpful to make up a table of all the possible events: 1&1 2&1 3&1 4&1 5&1 6&1 1&2 2&2 3&2 4&2 5&2 6&2 1&3 2&3 3&3 4&3 5&3 6&3 1&4 2&4 3&4 4&4 5&4 6&4 1&5 2&5 3&5 4&5 5&5 6&5 1&6 2&6 3&6 4&6 5&6 6&6 From the table, you can see all 36 possibilities, and how 3 possibilities yield a sum of 4 (I bolded those for you). So, the actual probability of rolling a sum of 4 is 3/36 or 1/12. Now, in terms of how many rolls it would take, on average (which is a very key phrase here!), we know that on any one roll the probability of rolling a sum of 4 is 1/12. Now, imagine if we rolled the two dice twice in a row. What are the chances that there is a sum of 4 on either roll? Well, you get 1/12 on the first roll, and 1/12 on the second roll. So 1/12 + 1/12 = 2/12 or 1/6. I think you can see how 3 rolls of the pair of dice increases of rolling one sum to 4 is 1/4. And if you keep increasing the number of rolls, how you keep increasing the probability. Finally, you have to recognize that when the probability is 1, that means on average you can expect that even to occur. So, I bet you can calculate how many rolls it would take, on average, until a sum to 4 occurs.

Assuming that those bold letters are vectors, division as you have used it there really isn't defined. i.e. if u = [1 0 13] and if v = [-4 7 2] what is u/v or v/u ? See: http://mathworld.wolfram.com/VectorDivision.html

Any of the classic handbooks will list a full description of the coordinate systems -- the transformation equations and the metric, etc. I have Schaum's Outlines Mathematicql Handbook by Murray R. Spiegel that is literally falling apart I use it so often. I count 12 different coordinate systems listed. MathWorld (http://www.mathworld.com) will have much if not all the same information as the handbook. I don't know if that is as "comprehensive learning" as you want, but it is a good place to start.

let me try to separate the two issues here. One the one hand, a tensor is a mathematical construct whose real motivation is for ease of notation. It is far nicer to write a bold T rather than writing out all the components. For example, in the mometum equation I wrote above, these is a velocity component in each direction. That means there are 3 separate velocity equations, one for each component (the velocity in the x direction, the velocity in the y direction, the velocity in the z direction). The stress tensor, T, has 9 components (xx,xy,xz,yx,yy,yz,zx,zy,zz). I could write each of them out for each equation, but, it is far easier to write out that one line above. Secondly, in tensor formation, the equations are indifferent about what coordinate system used to describe the situation. This is desirable, since nature is indifferent about what coordinate system we as humans choose to impart. I.e. instead of the x,y,z components of velocity, maybe it would be easier to study the system in the cylindrical coordinates (r, theta, z), or spherical (r,theta,phi), or any other ones. The rules of tensor calculus set up exactly how to translate from one coordinate system to another, and the really nice thing is that the basic equation remains the same. The momentum equation I wrote above is correct for al systems of coordinates. This is the real benefit of tensor notation. Now, on the issue of time. Let me write a very basic equation: [math]\frac{df(t)}{dt} = g(t)[/math] The time derivative, [math]\frac{df(t)}{dt}[/math], is the limit as the time increment goes to zero (delta t). This represents the instantaneous rate of change of the function f at a given time t. It has no other information at all. The same statement can be said about the right hand side, which the rate of change is equal to. It is a function of t, but the evaluation of that function happens only at one specific point in time, t. Not a range. Assuming the equation can be solved (which it can for a lot of interesting phenomena), it is solved for all t. It is not solved for a period of time that you specify, you get a solution f(t)= something. That function again only gives the value at one specified t and does not necessarily say anythng about the times around it. Now, that is the mathematical world's ideal. Obviously, in the real world, we never have the entire continuous data. We never know g(t) for all t. We take readings in time, whether by hand (looking at an instument) or by computer (which samples at a fixed rate, cannot be infinite). In fact, in solving differential equations numerically, we assume g(t) is not constantly changing, but is fixed for a short time. But, and here is the really important part, this is only an approximation. It does not describe the curves perfectly. One of the major issues with computational solutions is to choose small enough periods of time that g(t) would not vary too much over that period, yet large enough periods of time so the computer program can end. I.e. if g(t) changes slowly, say 0.1% every second, there is not much advantage to simulating on the microsecond scale. But, the physics of the equations determine this. Again, I cannot emphasize this enough, they are only approximations. So, I hope I have explained well enough that time is not trapped in a tensor formulation. On the one hand you have a mathematical construct that has some very nice properties to make the math easier to handle. On the other, you have phenomena that vary in time. The two are actually very different subjects. All the way back to the fluid mechanic equation, the velocity field, v, changes in time and those changes in time are written using a tensor. But, I could have also just written out each component term by term. Time is not wrapped up in the tensor forumation.

Rumplefloople?!? I am exceptionally confused about what you mean by "...describe them currently into tensor..." and oceans and etc. But, I'm going to at least take a stab. The fluid mechanics equation describe the motion of the fluid in the ocean. They can be written like: [math]\frac{D \mathbf{v}}{D t} = \nabla \cdot \mathbf{T} [/math] where [math]\mathbf{T}[/math] is the stress tensor of the fluid. The terms that make up the components of the stress tensor can be functions of time, as well as position. But, any time later, the components can change. There are times when you can approximate the stress tensor as unchanging in time -- that depends on the situation, but in general is can be a function of time and position. A tensor is a mathematical construct so that the equations can be written like the one above rather than write out each term explictly. The above equations in spherical coordinates take an entire page to write out, but in vector/tensor notations it is one short line. Time may or may not be a coordinate to be included, but either way, the tensor notation is just a convenience not whatever you were trying to say above.

Yeah, the tree said it right, pi has an exact number, it just isn't rational. Proofs of its irrationality have been around for quite some time, I am sure you can find several by using your favorite web search engine.

Way back when, they would calculate the area of regular polygon inscribed and circumscribed to the circle. There are formulas you should be able to find or look up to calcuate the areas of regular polygons. I know I read about some monk who spend his whole life doing the calculations for something like a 20 million sided regular polygon. However, on a more practical note, I have an interesting idea. Find a piece of paper that you have several of the same sheets of the exact same kind. Use a ruler and a corner to cut out exactly one unit square of that paper, say exactly one inch squared or one cm squared. Actually, you probably want to do this several times, like 10 so that you get a good average (since it is difficult to get exactly one square unit). Now, weigh those squares so that you know the weight of one unit square of that paper. You're going to need a pretty accurate scale, some kitchen scales are acurate enough. Next, trace out lots of circles using a compass, all the same radius, and weight those. The average weight of those cut out circle divided by the average weight of the unit square will give you the average area of the circle. Then you can get pi.

1 thing cumulative graphs are good for are more easier comparisons between curves. That is, because the cumulative graphs have already been normalized between 0 and 1, you can directly compare two sets of data, no matter if the actual ranges of the data are very different or not. Some of the numbers of interest are more easily found on a cumulative graph. Like the mean is the point when the graph crosses 0.5. Doesn't matter if the actual data is bimodal or trimodal or non-modal at all. Some of those multimodal graphs can be hard to guess what the mean is, with the different sized peaks. Finally, the integral under the cumulative distribution has to be one. That has some nice properties, especially when converting variables or means. For example, you use the integral properties when you want to calculate the geometric mean or the harmonic mean from a set of data that is set up to calculate the number mean. Conversion between data sets comes into play, too. For example, say you have a particle diameter distribution, and you want to convert that to a particle mass distribution. Integral properties come into play there, also, and it is usually much easier to work with the cumulative functions rather than the frequency distributions. All that said, I think that people in general have a much greater intuition about the frequency distributions. For example, they can look at your length of word frequency distribution and see that the mean length of the word is 3 1/2 letters for example because that is where the peak will be in the middle of that curve. It would probably take a person a long time to look at the cumulative distribution and gather than info. But, both forms of the curve have their uses.

I read this thread and I think that a quote from Indiana Jones is pretty pertinent: "Archaeology is the search for facts... not truth. If it is truth you are looking for the philosophy class is right down the hall" Put in Science for Archaeology, and you got yourself a winner.

My advice would be to consult as many different sources as possible. Unless you are trying to learn something cutting edge, there are always many different ways to approach and hence lots of different things have been written about the subject. For example, you mentioned learning calculus. There are probably over 100 different calculus textbooks out there -- it is a very well studied and written about subject. The point is that if one text, or how one author writes about a topic doesn't jive with you, go and find another. Chances are someone out there has written a book that will approach a subject just like you would. You just have to find them. Again, in the realm of calculus texts, they range from your typical univeristy level text, to proof-driven and more abstract advanced calculus back down to Calculus for Cats by Amdahl and Loats (which actually covers the material for a calc 1 class pretty well... I was skeptical from first glance at the title, too). So, use your local library and inter-library loan well. Try out as many different points-of-view as you can. Like I said, chances are at least one of the books will really click for you.

Please don't take this as in a snarky or mean or putting you down tone. It is not what I mean. But, lboogy, did you read the wili link John posted? Specifically, "Molarity (M) denotes the number of moles of a given substance per liter of solution." Nothing about mole fraction, which is a very different quantity. Molarity has units of moles/liter. Mole fraction would be moles/moles or unitless.

??? Are we looking at the same equation here, guys? if y=ct, then y'=c and y''=0. Plugging all that back in, you get (ct)*0 + ©^2 = c c has to equal 0 or 1 (not determined by the initial conditions). Not only that, but shouldn't there be two constants of integration, because of the second derivative?

I would be highly impressed if someone finds the analytic solution to this. I am not saying that there isn't one, but personally, if I had to have a solution to that non-linear second order beast, I would use a computer to give me the discrete approximate solution.

Part of the problem here spunnery is that you went from Classical Mechanics to Relativity pretty quickly there. Physics on the Classical/Newtonian scale do not work when you start getting into discussions about velocities near the speed of light. 0.99c definately falls into that area. So, the same logic that does work on the Newtonian scale (trains) does not apply on the relativistic scale.

There is no reason whatsoever you can't call or email the local university and ask to speak to an advisor to discuss your options. You can't expect them to drop everything and help you immediately, they have their own full time jobs and are probably advising 100's of other students as well, but they should take the time to answer your questions and if necessary try to find a time the two of your can meet. At the very least, if you call and talk to the advisor and they can't help you, they should know how to direct you to someone who can. There are many people at the university who will be able to answer your questions, advisors, recruiters, admissions personnel, professors in the department you are interested in. There are probably more people than you ever wanted to meet. Go find them. Call up the admissions office and schedule a campus visit -- tell them what department(s) you want to visit specifically. There are so many options here, there is no excuse not to have all the questions you want answered as soon as possible.

I would seriously suggest talking with your guidance counselor/advisor. Or, if you don't have one, talk with an advisor at your local college. Or, write an email to an advisor. The reason I keep pushing taking to an advisor is that it is basically the advisor's job to answer these questions for you, and can be very specific about opportunities at their university, and should be able to talk about opportunities in their industry. Sounds like a comp sci advisor would be a good place to start, and maybe also talking with a physics advisor.