Bignose

If that is your plan, take as many chemistry classes as your ChemE electives as you can. You should be able to take something like a quantum chemistry, or a inorganic chemistry or additional semester of organic chemistry or a chemistry-based polymers class, or any of the other advanced chemistry classes while getting your B.S. 1) you can see if you like it enough to continue to pursue your goal and 2) you will prepare yourself better for a chemistry graduate degree.

We've had this discussion before: http://www.scienceforums.net/forum/showthread.php?p=341523#post341523

Genetic algorithms seem to have quite a lot of promise in this area. Last time I took in a seminar about them (2,3 years ago now) there weren't any rigorous proofs that they find the absolute maxima or minima, but an awful lot of anecdotal evidence that they did solving some very difficult problems that the traditional solvers usually can't.

"Logic" has almost nothing to do with it. For a long time, it was very "logical" for a lot of people to believe that the earth was resting on the back of a giant tortoise. For a long time, it was very "logical" to believe that heat energy was a fluid called phlogiston. For a long time, it was very "logical" to believe that the moon was made of cheese. All of these have been refuted by evidence. Please cite the evidence that disproves the current model and supports your model. You can't just go by what your gut thinks is "logical". You have to evidence to back it up. End of story.

How about accepting n entries, then hold the drawing every Sunday evening where the odds of each ticket winning a prize is 1/n based on how many entries there were in the previous week? The only other option, is, as you said, to have each code associated with a prize. Most places do it one of those two ways. They are simple to program/understand and usually considered pretty fair. The problem with a changing probability is that not everyone will be given the same chance. As an alternative, you could just set up the system so that every 10,000th entry is a winner, and of course, you wouldn't publish what number the counter is on. The first one should be some random number between 1 and 10,000... say 4,823. But, then make the next winner the 14,823rd person to enter a code, then the 24,823rd, etc. Just set up frequencies for all the prizes like that, and it will be effectively random since none of the people entering the contest will know where in the chain you are.

Again, just because it occurs frequently, doesn't mean it has any specific significance. I'm not going to do the legwork, but I guarantee I can come up with several similar examples for every integer between 1 and 20. Probably every integer between 1 and 100 if you wanted to. Especially, as I said, if you count the "close-to"s. Unless it is actually exactly 12, or you find a compelling reason why 12 is so significant, this is just numerology -- a thoroughly debunked and worthless (in a scientific sense) pursuit.

well, you follow what [math]n^x[/math] means, right? Let's apply some other function to [math]n^x[/math], let's just call it G. [math]G(n^x)[/math] Now, G is a special function that is defined to be the inverse of [math]n^x[/math] so that [math]G(n^x)=x[/math]. The way people normally write out that function [math]G()[/math] is [math]\log_n ()[/math] so that [math]\log_n{n^x} = n^{\log_n{x}} = x[/math] Exponentiation and log are inverses of each other. Log is the operation you do to perform the inverse of exponentiation.

But 99x0.1818181... is equal to 18.0 You can see that from your own equations... You agree that 2/11 is 0.1818181.... yes? and 2/11 is equal to 18/99, right? So 18/99 = 0.181818181818181818..... Multiply both sides by 99 and you get 18 = 99*0.18181818181818181818........ You can do the limit, too. 99*(0.181818181...) = 99*(1/10 + 8/100 + 1/1000 + 8/10000 +...) =99*(1/10) + 99*(8/100) + .... Now let's look at the sum after each term After 1 term, the sum is just 99(1/10) = 9.9 Add the next term, the 99*(8/100) and the sum becomes 17.82 after 3 terms: 17.919 4: 17.9982 5: 17.99919 6: 17.99998 I think you can see what the limit of these sums are. Each additional term brings the sum closer and closer to 18. 99*0.181818181818.... = 18.0 There is no approximation. It is an exact equation. You forgot the infinitely repeating part Sure 99*0.18 is 17.82 but 99*0.181818181818181818.... (where the ... denotes infinitely repeating) is equal to exactly 18.0

Well... do you understand inverse functions? I.e. [math]f^{-1}(f(x)) = x[/math] Like [math]\arcsin{(\sin{x})} = x [/math] or [math]\sqrt{x^2}=x [/math] will logarithms are the inverse functions of exponentiation. [math]\log_n{n^x} = n^{\log_n{x}} = x[/math].

I always thought that The Incredible Machine was decent. The results weren't grossly non-physical, for sure. But, I wouldn't bank on it for any kind of accuracy if that's what you are looking for. I.e. if you put a baseball on the end of a teeter totter and dropped a bowling ball on the other end, I wouldn't turn to the game for a prediction of how high or far the baseball would fly.

A function of two variables is just like a function of one variable, it just has two inputs instead of one. It may be easier to see this if you fix several values of one of that variables. "Lock" in a value of y at Y1, and look at f(x,Y1). This looks just like an f(x) value once you plug in Y1. Now, plug in a different value Y2. And another Y3. The function f(x) (which will really be f(x,Y2) or f(x,Y3)) will change, but still look just like a single variable function. It is just the natural extension of the single variable function. Because nature and life and physics have things that depend on more than one variable. The velocity of air flowing along the ground is going to depend on many variables -- at the very least it is going to depend on the longitude and latitude of where you measure it. And, it is going to depend on how high above the ground you measure it (altitude) as well. So the wind velocity, v, is a function of at least three variables. v(long., lat., alt.) Now, if you wanted to model it, there could even be more variables. Like the temperature and pressure. And the gradients thereof. I.e. if a high pressure front is moving into a low pressure front, there is a steep gradient of pressure, so the wind is going to be higher than if there is a gentle gradient of pressure. So, wind velocity so far could be a funciton of at least 7 variables, now: v(long., lat., alt., temp., temp. gradient, press., press. gradient) And there are potentially many others (like humidity) but I think you see the point. Let me give you a very different example. What variables would go into writing a function of the average income of a person? There is pretty much no way to capture this with only one variable because of all the competing influences. You'd have to include level of education (a college grad on average makes more than a high school dropout), you'd have to include geographic location (the cost of living in a coastal state is higher than a Midwestern state), you'd have to include gender, race, maybe religion. You may consider including a variable that includes one's parents (i.e. a child whose parents both went to college usually ends up with a better job that a child whose parents both dropped out of high school). And there are probably many more. But, again, the point is that the result -- the average income of a person -- is going to be a function of many, many different things. These cannot be captured by only one variable f(x). Usually, you are going to need f(x,y,z,a,b,c,...). The big thing is that it actually is a pretty special case where only one variable drives the resulting output. Of course, you study the one variable cases extensively to start with to learn the basics and keep confusion to a minimum. But, the f(x,y,z,...) is just the natural extension of f(x) to acknowledge that very often in real life more than one variable influences the final outcomes.

No, it doesn't. How can there be an infinite number of zeros... and then a digit after it? It's an infinite number! There is no end! There is no end to put a digit after -- it has no meaning.

absolute1, PLEASE find and read this reference I posted above. It talks about exactly the situation you are talking about -- inelastic collisions where you often have to use BOTH conservation of momentum and energy to solve problems. And, again, at the end of the section there is a completely worked out example involving a bullet impacting a block. You can find this physics text at any university library (it is a very popular text for physics courses) and any library can get you a copy through Interlibrary Loan. For that matter, almost any university level physics text will have a section on inelastic collisions. Please study that section and learn how to solve problems of these kind and then come back and see if you still think that there is something wrong with conservation of energy & momentum (which may be two of the most verified theories ever).

Please read Physics For Scientists and Engineers 4th edition by Serway, p. 244, the section called Elastic Collisions. This talks about how you have to use both the conservation of momentum and energy to solve problems when the collision isn't perfectly inelastic. In fact, on p. 246, there is a worked example of a bullet impacting a block.

Please show your work on how you got these different values. If you show all your work, we can probably show you where you made a mistake. My experience with these problems is that usually you have to use BOTH conservations of energy and momentum. Usually just one or the other is not sufficient. In order to "pick" which method to use is usually determined by what information is available. There aren't really any better "rules" to determine which one to use. Sometimes you just have to dig in a try one method and see what happens. And then compare with the other. Or, as in the above example, usually you will need both. It all is determined by what information is available. Have you tried working through a university level calculus based physics text? There are many examples in the homework problems where if you work through them you will develop an intuition for which method (or methods) is better in a given situation.

absolute, you basically asked for one post to be closed (by writing "I'm sorry but I just lost all interest bringing up topics for discussion in here.") so the mods obliged you. In the other, all you did was post a "conversation" with yourself -- if there was something to actually have other people discuss, why just post and finish a conversation with only yourself? Regarding the "problem" you had with the first thread (the one you basically asked to be closed), your calculations were correct, but the interpretation was way off. Your problem was basically that when two objects of different masses are given the same amount of momentum, they should also have the same amount of energy. But, that isn't true at all -- as your calculations showed. Momentum and energy are not the same thing, that's why there are two different conservation principles. This isn't a "problem", it is why there are two different quantities. Conservation of momentum does not imply conservation of energy or vice versa. In addition, equivalence of momentums does not imply equivalence of energies or vice versa. That is, just because two objects have equivalent momentums in no way whatsoever does it imply that they have to have equal energies. If have something to discuss, I'd hope you take mooeypoo to heart here, and actually start presenting some evidence, logic, and substantiation.

Please note that this is just a shot in a dark and may be very off base, but the phrase I quoted here caught my attention. It made me want to bring up at least a question or too -- in that is the temperature variation being taken into account? If the mixture has to be so vigorously mixed, there is going to be a temperature rise from the viscous dissipation in the fluid. Then, if you had another fluid, which may not be at the same temperature, the temperature change can change the pH, maybe even more so than the basic nature of the new fluid? Or, is the solution unstable? That is, the mixing overcomes the stability of the droplets of the indicator by breaking the droplets, but the addition of a new fluid again causes the indicator to go back to droplets and thus not working like a true indicator should because it is re-coalescing? Like I said, these are just guesses, and may be way off.

You might want to get your hands on a copy of J.P. Boyd's book Chebyshev and Fourier Spectral Methods -- I found the second edition from Dover an exceptionally good book with plenty of theory (and my memory may not be 100% correct, but I am pretty sure that a pretty detailed derivation of the quadrature points is presented) and plenty of applications. The best thing is that the author's style made the book very readable to me. Very highly recommended -- and since it is a Dover book, the price really cannot be beat should you choose to own your own copy of it.

Wonderful, we're going to start this again. Please answer this question which cuts right to the heart of the matter. If 0.99999... (infinitely repeating 9's) is NOT equal to 1, what is the number that comes between them? If there is no number that comes between them, then they are equal. If they are unequal, then there must be a number between them. So... since you say they aren't equal, what is the number between them? Same thing with your 0.0000 (infinitely repeating 0's)1 number. First of all, there really isn't any such number. If there are infinitely repeating 0's, how can there be a 1 on the "end"? It can't, there are infinitely many 0's. But, even if it were possible? What number comes between 0.00000 ... etc. and 0? Again, there isn't any such number, so they are equal.

Seems like http://en.wikipedia.org/wiki/Logarithm and http://en.wikipedia.org/wiki/Matrix_%28mathematics%29 is a pretty good place to start.

Just to back up what swansont said here, let's look at the math a little closer. So, you have [math]F_g[/math] the gravity force. [math]F_g = mg[/math]. [math]F_g[/math] becomes the [math]F[/math] in [math]F = ma[/math], so [math]ma = mg[/math] or you can cancel the masses out and you get [math]a=g[/math]. Now, acceleration is the second derivative of position with respect to time so [math]a = \frac{d^2 x}{dt^2} = g[/math] and that's what you'll integrate to get how long it takes to get to the bottom. Note that the last equation there is completely independent of mass.

This is done primarily to make the math easier. Nature itself does not really care about what coordinate system/set of basis vector we use to describe something with. In fact, to be considered a "good" physical law, it has to be completely valid in all coordinate systems. Let me give you an example of why the resolving is done. Consider two identical cannonballs. You're atop a tower defending your castle from invaders, and you fire your cannon straight out horizontally. (so that the cannonball from the gun initially only has horizontal velocity.) Further say that since you forgot to call out "fire in the hole" when you shot your cannon off, you startled the guy manning the cannon next to yours, causing him to drop his cannonball he was trying to put into the end of his cannon. So, both cannonballs are let go at the exact same time, the only difference between them is that one has no horizontal velocity (the dropped one) and the other has some horizontal velocity (the fired one). Now, further assuming that the ground up to the tower is perfectly flat, which cannonball hits the ground first? Obviously, you can use physics equations to determine the position, the velocity, the acceleration, etc. of the cannonballs are any time. But, the really neat thing is that by choosing a coordinate system that has one coordinate in the direction of gravity and another perpendicular to gravity along the line of the fired cannon, you can write separate equations for how far away from the tower the cannon ball is, and how high off the ground the cannon ball is. That is to say, with a good choice of coordinate systems, you can make those two quantities completely independent of each other. The distance away from the tower has no effect on the ball's height above the ground and vice versa. In this way, you should find that the equations for the height above the ground for the two balls is exactly the same. Only differences in the two ball's vertical components change the ball's equation for height above the ground. In this case, both vertical velocities were zero, even though they had very different horizontal velocities, so both vertical equations are the same. This is the advantage of breaking vectors into components. You can quickly identify what affects what parts of the velocity. Choice of coordinate system is usually done to make the problem as simple to answer as possible. For example, consider the flow of fluid in a tube. Almost always, the cylindrical coordinate system is used. This is a coordinate system that instead of the common x,y,z (also known as the Cartesian coordinate system) used r, [math]\theta[/math] ("theta") and z. r is the distance from the center of the pipe and [math]\theta[/math] is the angle around the pipe (like the hands on a clock face. The minute hand at 2:15 makes a different angle with the 12 than it does at 2:18 that angle is [math]\theta[/math]). z is the same. The reason this coordinate system is chosen is because it makes the math much, much easier. The governing equations of fluid mechanics are still perfectly valid if you wanted to use the x,y,z coordinates, but they are much harder to solve. So, in the two examples here, the common theme is picking coordinates that make the problem easier to solve. In the first, we picked a coordinate to be in the direction of gravity. In the second, we picked a coordinate to go in the direction of the flow. In the first, we could have picked a coordinate direction that was 45 degrees to gravity. It wouldn't have been as easy to solve, but it can be done. I hope that the ideas I've tried to present are clear. If not, please don't hesitate to ask more questions. But the main points are that using vectors to separate the components makes things easier. It may not seem so at first, but in the long run, it is the much preferred method because it is significantly easier.

That's all this is... You can do something similar probably with almost any number if you account for all of the known physical constants and combinations thereof. Especially if you start counting "close to"s, like you do here. Unless you can provide a really compelling reason with supporting evidence as to why 12 is special, this isn't anything more than numerology.

Any space with a null metric line running through it. It may be faster to go to the null line -- travel up the null line (which by definition takes no distance) -- and then travel to the second point than to take the straight line between points. For example, imagine a space with a null metric line running along the x=y, 45 degree line. To travel from (3,3.5) to (0,0) you can either go directly to (0,0) for a total of 4.609 units or you can travel to any point on that line, say (3,3) and then travel the null line to (0,0). I.e. the distance between (3,3) and (3,3.5) is 0.5 In this case, the shortest distance between (3,3.5) and (0,0) is along the line that intersects (3,3.5) and is perpendicular to x=y. Depending on the metric of the space, straight lines aren't always fastest. On the surface of a sphere, the shortest distances are great circles. On the surface of a cylinder, the shortest distances are helices. It all depends on the metric of the space.

1) we (the forum members) don't do your homework for you. We'll check your answers and help correct you where you make mistakes, but we won't just give you an answer 2) this question doesn't make a lot of sense to me. You can choose u to be anything. You can let u=x, let u=x^2, let u = sqrt{x}.... so you can "convert" it to anything. Is there something more specific you were supposed to do? I am unfamiliar with the term "converted substitution" as the question seems to imply. So, explain why you chose b, and why (i.e. show your work) and I'll comment on if I see any mistakes or not.