mm-1929 === Are we allowed to take his palindrome and expand it (as long as we publish the result here)? A graph of Y = x^2 would be considered exponential because of the exponent on x that is greater than 1, right? What if you graph something like 1/(1-x) as x goes from 0 to 1. The graph looks a little like an exponential curve but it's not really, so what do you call it? -- Chris W Gift Giving Made Easy Get the gifts you want & give the gifts they want http://thewishzone.com That explains something. One often hears/reads in what it seems we have to call the media that something is growing exponentially. This is what they mean. I'd call it asymptotic to x = 1. Actually y = 2^x would be called exponential, but y = x^2 would be called power, or possibly polynomial. y = 1/(1-x) is a rational function of x, the quotient of polynomial functions (often presumed to be reduced to lowest terms, i.e., the numerator and denominator having no non-unit common factor.) X-RFC2646: Original In addition the curve described by y = x^2 is called a parabola While the curve described by y = 1 / (1 - x ) is a hyperbola Both happen to be conic sections. -- -- Geo. Michael Henry No! Bad dog! I said sit! anonymous On Wed, 16 Mar 2005 07:09:42 -0600, In addition, when the dicussion is about growth, then y = x^2 is called quadratic. That is, the function grows quadratically, where exponential growth is much faster. Computer scientists use the big-o function to express growth, very often. They write O(n^2) to mean that there are positive constants c and n0 such that 0 <= O(n^2) <= c * n^2 That is, O(n^2) is an upper bound of n^2. I can't seem to understand this implicit differentiation problem. I'm doing OK until I get to this one with fractions. Please show me the steps to get to the answer. sqrt(xy) = 1 + x^2y Fractions? What fractions? sqrt(xy) = 1 + (x^2)y -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ days. My association with the Department is that of an alumnus. Differentiating the right hand side is easy: we get 2xy + x^2y'. On the left hand side, we have: (1/2*sqrt(xy))*(xy)' = (y + xy')/2*sqrt(xy). So we have: y + xy' --------- = 2xy + x^2y' 2sqrt(xy) Now move all the y' to the left, everything else to the right. Do this by first clearing denominators, then multiplying out: y + xy' = 2*sqrt(xy)*(2xy + x^2y') y + xy' = 4xy*sqrt(xy) + 2x^2*sqrt(xy)*y' xy' - 2x^2*sqrt(xy)y' = 4xy*sqrt(xy) - y Now factor out y' and solve for y' in terms of x and y: y' (x - 2x^2*sqrt(xy)) = y*(4x*sqrt(xy) - 1) y*(4x*sqrt(xy) - 1) y' = -------------------- x - 2x^2*sqrt(xy) What exactly was your problem? -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu fractions How so? x^2y = e^(2y.log x) D(x^2y) = (2y'.log x + 2y/x)e^(2y.log x) D(x^2y) = (2y'.log x + 2y/x)x^2y fractions days. My association with the Department is that of an alumnus. Absent parenthesis, x^2y should be interpreted as x^2*y, not as x^{2y}. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu fractions x^2 y would clarify the notation without having to deal with hair splitting regulations that aren't applicable to printed formulas, where difference in size and half line offset makes it clear if it's x^2y or x^2 y. fractions days. My association with the Department is that of an alumnus. There is certainly no argument that there are ways of writing the above so as to avoid any possible ambiguity. But calling the standard rules of operator precedence hair splitting regulations seems to me to be even sillier. Nonsense. The rules of operator precedent are certainly applicable to printed formulas. Without them, 2+3*5 could be either 25 or 17. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu fractions Yup, the usual, a(b + c) = ab + ac = (ab) + (ac) /= calculator entry ((ab) + a)c However for (mono-spaced) 2 x y 2y x There are no rules of precedence as notation per se makes it clear. -- Riddle of the day Does philosophy cause split hairs? fractions days. My association with the Department is that of an alumnus. Just repeating the same nonsense doesn't stop it from being nonsense. Just because we have notation that does not require us to apply the rules of precedence IN SOME CIRCUMSTANCES does not imply that the rules of precedence do not exist. Riddle of the day: can you talk proper English? -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu No, x^2y = e^(log (x^2y)) = e^(2 log(x) + y) which will work. But I'm curious why would you want to write it that way to differentiate it? 2xy + x^2y' is pretty direct. --Lynn fractions How does x^2y = (x^2)y = x^2 * y = x^2 y? It does not. x^2y = x^(2y) log x^2y = log x^(2y) = 2y x^2y = (e^log x)^2y = e^(2y.log x) That's by definition - exponentiation has higher precedence than multiplication. Of course there's no way to tell what the OP actually meant, but x^2y with no parentheses is (x^2)y. ************************ David C. Ullrich fractions It looks like no such thing. X-RFC2646: Response Complaint from an old-timer. Why don't people want to use parentheses anymore? -- -- Geo. Michael Henry No! Bad dog! I said sit! anonymous fractions Lazy quick speak. Why don't they even want to use spaces for ax^3+5x+y^2+35xy+x(x+y)+7z^xyz+2^x(5+3x) ? fractions days. My association with the Department is that of an alumnus. I find your cryptic telegraphic writing style to be confusing at the best of times, irritating most of the time. This is one of the latter. Nobody says they looked the same. But, since exponentiation has higher precedence than multiplication, you should do exponentiations first, multiplications second. To interpret x^2y as x^{2y} requires you to do the multiplication of 2 established order of precedence. (x^2)y, be contrast, does the exponentiation first, and any remaining products second, the correct order of precedence. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu fractions On Wed, 16 Mar 2005 14:55:28 -0800 I think he considered x^2 * y, instead of x^(2y). But why x^(2y) = e^(2y.log x)? ASsuming base-e logs, log[x^(2y)] = 2y * log(x) log[e^(2y*log(x))] = 2y * log(x) -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ fractions x^(2y) = (e^log x)^2y = e^(2y.log x) Never mind the confusion about precedence of operations, I think Kevin is asking the same question of you that I did, namely, whichever way you interpret what the OP intended, why would you introduce the exponential and logarithm in this problem?? --Lynn days. My association with the Department is that of an alumnus. I must confess that if I were trying to take the derivative of x^{2y}, I would also introduce exponentials and logarithms, since both base and exponent are functions of x. d/dx (x^{2y}) = d/dx (e^{(2y)ln(x)}) =d/dx (2y*ln(x))*x^{2y} = 2y'*ln(x)*x^{2y} + 2y*x^{2y-1}. = 2x^{2y-1}*(y'*x*ln(x) + y). -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu Yeah, I guess that's reasonable, depending on whether derivatives of u(x)^v(x) forms has been studied or not. --Lynn fractions On Wed, 16 Mar 2005 22:58:38 -0800, Is (e^log x)^2y = e^( log x^(2y) ) ? fractions days. My association with the Department is that of an alumnus. [.snip.] exponentiation has higher precedence than multiplication; if you just write x^2y, then you are supposed to do the exponentiation first, so you aren't supposed to do x^{2y}, which requires you to do the product of 2 and y first). In general, (a^b)^c = a^(bc). So (e^log x)^(2y) = e^{(log x)*(2y)} = e^{(2y)*(log x)}. And the properties of log tell you that a*log b = log(b^a), so e^{(2y)*(log x)}= e^{log (x^{2y})}. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu fractions Well, yes, I meant ^2y = ^(2y). I actually didn't write them equally because I copied from William. But I meant that. I don't see how they the equation above holds. You showed below, but I think they are different things. Here: (e^x)^2 != e^(x^2). (1) Right? So, replace x in (1) with log x and replace 2 in (1) with 2y. Then, we have: (e^{log x})^(2y) = e^{log x * 2y}. That's correct. Now, for the other term, we have: e^( [log x]^(2y) ) = e^{2y * log x}. Oops, it is equal. Mmm, it's because we're dealing with logs. I myself. I shouldn't actually post this, since I figured it out (I think so), but perhaps this may help somebody else with a level of knowledge close to mine, so I will leave it. fractions days. My association with the Department is that of an alumnus. No, this is wrong. It is NOT true that [log x]^{2y} = 2y*(log x). What IS true is that log [x^{2y}] = 2y*(log x). I have no real idea of what it is you are trying to do or not do. But the equation above is wrong. So if you somehow were convinced by that equation, then you are still somewhat lost. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu fractions Yeah, I just noticed it... in fact I just posted my clear misunderstanding of these logs rules. Feel free to ignore that other posting, because you just answered my question. fractions days. My association with the Department is that of an alumnus. You think what are different things? (x^2)y and x^{2y} are different things, yes. But (e^{(log x)})^{2y} is the same as e^{log(x^(2y))}. Yes. (e^x)^2 = e^(x*2) = e^(2x). Again, you have to be freaking careful. Do you mean e^{(log x)*(2y)} or do you mean e^{log(x*2y)} ? -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu fractions I was going to explain what ``what'' I was talking about. But I need to re-start. I am not convinced of a lot of things. Allow me to ask questions from scratch, please. e^(t^2) != (e^t)^2 at least for t = 3. Because: e^9 != e^6. Right? Now, my old question: let me ask it again but fixing the notation. Is (e^[log x])^(2y) = e^( log [x^(2y)] ) ? I say no. Because I try x = 3 and y = 1. (e^[log x])^(2y) = e^(1.098612289)^2 = e^1.206948962 e^(log[x^(2y)]) = e^2.197224577 Are we talking about different things or did I make a mistake somewhere? Right. (e^t)^2 = e^(2t) by the laws of exponents. Petty obviously that's different from e^(t^2) because 2t is different from t^2. A counterexample is a valid way to demonstrate that two expressions are unequal. But I submit that it is _far_ better to do the algebra. For one thing, suppose you fail to find a counterexample; you don't know then whether the expressions are equal. I have seen nothing better than the good old-fashioned T: (e^[log x])^(2y) =?= e^( log [x^(2y)] ) =?= e^( 2y * log(x) ) =?= e^( log(x) * 2y ) =?= (e^[log x])^(2y) Therefore the expressions _are_ equal. You did not evaluate your expression correctly, because you dropped the parentheses on (e^[log x]). It should have read (e^[log x])^(2y) = (e^(1.098612289))^2 = e^2.197224578 which matches e^(log[x^(2y)]) = e^2.197224577 modulo a rounding error. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ fractions days. My association with the Department is that of an alumnus. Correct. Than you say wrong. Notice that it is NOT the same as saying that (a^b)^c = a^(b^c), because in this case, your exponent is itself a function; that is, we have (a^(f(x))^c = a^(f(x^c)). which is true for SOME functions f. Because I try x = 3 and y = 1. Huh? No, no, no, no, no. You are making the same mistake again. Your left hand side is of the form (a^b)^c. But you are evaluating the middle as if it were a^(b^c). They are NOT the same thing. [e^(1.098612289)]^2 = e^{2.197224578}, You made a mistake; you converted ( e^[log 3] )^(2) into e^[(log 3)^2]. The claim is that (e^[log 3])^2 = e^[log (3^2) ] = e^[log 9]. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu X-RFC2646: Original X-Antivirus: avast! (VPS 0511-0, 15/03/2005), Outbound message X-Antivirus-Status: Clean Hi all! Here's a question that's been bothering me for some time now. I recall reading it years ago but can't remember the source or the answer (if one was given). I'm sure it will be familiar to many people but will lay the problem out in full for those who are unfamiliar with it. Three prisoners are in a cell (for convenience, we'll label them A, B and C). The next morning, two are to be executed. Prisoner A (say) reasons that his chance of survival is 1 in 3. Unable to sleep, he asks the guard if he, prisoner A, is due to be executed. The guard replies that he is not at liberty to answer that question but what he can tell him is that prisoner B is definitely for the chop. Prisoner A then reasons that his chances of survival are 1 in 2 as it is a choice between him and prisoner C. Now, here's the problem: has the guard given him any new information? He knew that one of the other two would be up for the chop and from his point of view it doesn't matter whether it's B or C. So, what are his chances of survival? 1 in 3 or 1 in 2? game theory. I've tried a few other search terms, but can't find what I'm looking for. Any help would be greatly appreciated. This just another version of the Monty Hall problem. There is no solution without additional assumptions. Let's assume that you know in advance the the guard would always name one of B or C. Then he would never have a choice if A was due to be executed. Now there is only a solution if we make assumptions about the guard's potential behavior in case A was *not* due to be executed. If you assume that in this case he would *always* name B, then your chances would be 1 to 2. If you assume that he would *always* name C, then your chances are 0. If you assume that in this case he would toss a coin which of B of C he would name then the chances are 1 to 3. -- Horst recall (if one was the A, B and reasons the guard he is not that prisoner chances of Now, He knew point of chances of led me to what I'm Curse these questions!!!! I think it's still just one in three though. Think about it, every set has at least one of prisoner B or one of prisoner C. So if the guard is going to tell the person only one of the people, and the person he tells is never the person asking, then he just as easily could have said that person C is for sure dead. It doesn't change anything. -- LTP When the llama speaks you listen. Unfortunately the llama hasn't spoken yet. [....] That is for 100% in FAQ. Actually what the guard thinks wont change the fate of B, C, so the question *from the guard point of view is different from B point of view. The guard *knows* who'll be executed.1:1 A prisoner (B or C) knows only about A. 2:3 Why not 1:3 ? Because of guard knowlege - new info the guard *knows* who is for the chop. Without it the answer is different. The standard - and less bloody - version of this game is in a popular TV program 3 gates, one of them keeps a car. I pick a gate, and the presenter (*knowing*) where the car is, opens another empty gate. Keep my first shot, or change ? Antek, Warsaw, Poland. *that's in Europe, for some Americans* Mean you don't Europe in is Americans some for? Where's Europe? Without any information from the guard his chances would be 2 in 3, not 1 in 3. Given the information from the guard, I'd have to say his chances are now 1 in 2. Think of it this way: If prisoner A is NOT chosen first his odds get a little better. If he IS chosen first, then his odds get very poor at once. Interesting question and I'll be waiting others' input. Oops, my bad. 2 in 3 to be executed. Yes, as you say, 1 in 3 for survival. Could someone give me some code snippets or just explain how i could reduce these two functions to lookup tables for use in software? Claz Lighten up! Sum is often used, colloquially, and frequently in NZ, to mean arithemtic problem. Were you never told off at school of playing around instead of Doing your sums? sum. (AA x AHA x AHHHA )^2 = ADMIT ONE NOT IMDA., Jim Spriggs What about sum as in total of the multiplication performed? I find probability quite difficult........... Is there any easy (and free) ebook for probability??? Any suggestions are welcomed!!!!! Thx~~~ If I know that: and if x increases, z decreases if y increases, z increases. Can I simulate what z would be for any x and y? I want to know the formula between the variables. Janos from three data points, you can just define an interpolating plane: z=a+b*x+c*y compute a,b,c from a linear systems of equations and thats all you can do here. this gives [a,b,c] = 1.8571 -0.3286 2.0571 hence your mononicity requirement is satisfied. hth peter It could be lots of things. Do you have any reason to think it could have a particular form such as z being a polynomial in x and y? Do you seek a precise formula or merely a best fit? X-RFC2646: Original I appreciate it. Hej baba, I suppose that you know how to derive f(x)=exp(x) or g(x)=ln(x). Let's suppose : f(x) is a function on an interval [a,b], with a graph in the (x,y) coordinate system. Now a point on the graph has coordinates (x0,y0) = (x0,f(x0)). The tangent in this point has equation (Straight line!): (y-y0) = (x-x0) * f'(x0). It is the VALUE of the derivative for x=x0 that defines the direction of the tangent in the point (x0,y0). Does this help you? Anyone know of any good software and ebooks for adults who aren't very skilled in basic math? I'm looking for an ebook similar to Merriam Websters Pocket Guide to Business & everyday math. (But maybe with a few more illustrations and a slightly more description, but not too much more) A good convenient ebook for a palm tungsten is priority #1 to me. If it's good, I'd pay for it, but every commercial ebook is PDF only. (I assume protected) I'm also looking for a calculator that can do fractions, EasyCalc is really good: http://sourceforge.net/projects/easycalc/ But.. it doesn't display fractions (well actually it does, but they aren't proper fractions. 1.25 shows 5/4) The thing is, going through the steps to convert to proper would be a distraction at this point. And a scribble pad type of program to scratch out practice stuff on, so far, the best is the one that comes with palm, but it only pans up/down. Plus, when with calculator and quickly switch to note pad before I forget what the correct answer is. :-/ ) I was looking for one that doesn't have a lot of artsy features, just an electronic version of paper & pencil. The ones I've seen have a very limited paper size. Something designed for scribbling math stuff on would be nice, memo pad works but kind of difficult to draw some of the symbols. (Kind of weird, I get a lot of stuff wrong due to sloppy handwriting) I've found a LOT of really nice web sites and some good paper books, but nothing for a palm. Protected PDF's (so far) aren't really an option for me, as I'm using Linux and a very ancient OS X laptop. (Older than Adobe supports) So, the concern is that any protected document may not be convertable. (Actually, I've not yet managed to find anything that can fetch several PDF's and convert to 1 ebook, the web sites I've seen seem mix PDF and HTML, iSilo obviously can't work with these. The reason for all this, is so I can read and study on a palm when in public places, (while waiting, etc..) All that stuff just doesn't seem to stick in my head very long and it's embarassing to be seen with basic math books. A Palm is nice because no-one knows what you're doing :-) Any suggestions on any of these? Particularly ebooks? Consider log(b_k) and try to give not only the definition, but also some useful theorems... Vitaliy posting-account=YqlKJg0AAAByPc8EKAi7NZGYqJuLUI30 O O O O O O if 3 circles in a row are placed directly opposite 3 other circles also in a row as shown above, is it possible to join each each circle in one row to each of the circles in the other row with curves which never cross each other (in 2 dimensions of course)? any help with this would be very much appreciated. i've been trying to solve it for almost a week now and i always end up having one line that has to cross. Nawahl This cannot be done. If we label the top circle T and the bottom circles B, and join two top circles to each of the three bottom circles, we always get (after rearranging a bit): /-------- B -------- / T---------- B ----------T / -------- B --------/ Now if you try and add a third T, then wherever you place it on the plane, there will obviously always be one of the Bs you cannot reach... Mike. posting-account=e363Sg0AAAC1QkLRICcuI6fjffHORe0q The subject is pretty self-explanatory, I'd like to know if anyone could give me a step-by-step method of solving the above, as I've tried every trick I can muster, to no avail.