mm-1899 === As I was tutoring him in probability recently, a client of mine showed me a cute method of iterated integration by parts which I had never seen before. I assume others in this audience also have not seen this method , which is basically a compact accounting procedure. Suppose we want to integrate x^3 e^(2 x) with respect to x. We will differentiate out the x^3 and repeatedly integrate the exponential factor. We start by writing the x^3 on the left and e^(2 x) a line above it to the right: e^(2x) x^3 We now repeatedly differentiate the left column until we reach 0: e^(2x) x^3 3x^2 6x 6 0 Since the entries in the two columns will be multiplicative factors, we may drop the last row. Now repeatedly integrate the right column to the last row: e^(2x) x^3 [e^(2x)]/2 3x^2 [e^(2x)]/4 6x [e^(2x)]/8 6 [e^(2x)]/16 Starting with +, enter alternating signs before each complete row: e^(2x) + x^3 [e^(2x)]/2 - 3x^2 [e^(2x)]/4 + 6x [e^(2x)]/8 - 6 [e^(2x)]/16 Finally, add or subtract (as indicated) the product d of the two terms in each complete row. We thus get int(x^3 e^(2 x) dx) = x^3 e^(2 x) / 2 - 3/4 x^2 e^(2 x) + 3/4 x e^(2 x) - 3/8 e^(2x) + C = e^(2 x) (x^3 / 2 - 3/4 x^2 + 3 x / 4 - 3/8) + C. We can also modify this algorithm so that it applies to cases where we cannot annihalate one of the factors by iterated differentiation. In such cases, integrate to the penultimate row only. The last term is the integral of the prduct of the last entry in each column. For example, to integrate e^(-x/4) cos 5x, the table is e^(-x/4) + cos 5x -4 e^(-x/4) - -5 sin 5x 16 e^(-x/4) + -25 cos 5x Thus, int(e^(-x/4) cos 5x dx) = -4 e^(-x/4) cos 5x + 80 e^(-x/4) sin 5x - 400 int(e^(-x/4) cos 5x dx) which, by algebra, implies int(e^(-x/4) cos 5x dx) = -4/401 e^(-x/4) cos 5x + 80/401 e^(-x/4) sin 5x + C. -- i have been using this method for the past 10 years or more... This is usually referred to as the tabular method of integration by parts. It is in many texts and Google gets about 94,000 hits. What did you do with these problems? If you show us exactly how far you got and where you got stuck, we can help you most effectively. First, we must assume that that the sales are independent! Then denote by Ac the complement of A; that is, brand A does not sell. Then for part a: P(Ac)P(Bc)P(Cc)=(.7)(.6)(.5). ---------------------------------------------------------------------------- ---- Part b I am lost need help the most here! ---------------------------------------------------------------------------- ---- For c you need to find: P(A)P(Bc)P(Cc)+P(Ac)P(B)P(Cc)+ P(Ac)P(Bc)P(C). ---------------------------------------------------------------------------- ---- For d the answer is (.3)(.4)(.5). [partial solutions, with no indication of the problems] Sigh. I'll answer your questions, but please first read carefull the following advice. Just because these problems are present in YOUR mind, do not assume contain enough context that readers don't have to go searching back any more, may never have reached their servers in the first place. Being the nice guy that I am, I have ferreted out your earlier never, so please quote properly next time. (Quote properly does as bad as quoting too little. See [and are independent] Correct. Since the probabilities are independent, you multiply them together to find the combined probability that all occur: i.e. that A is not a big seller , B is not, AND C is not. Brute-force methods will get the job done. If two brands are big sellers, that means (in your notation) AB(Cc) or A(Bc)C or (Ac) BC If exactly two brands are big sellers then one must be not a big seller. Therefore there are only three ways for exactlyone brand out of three not to be a big seller. No, _you_ need to find it. :-) But yes, that's the correct computation. By the way, this is almost a mirror image of the situation in part (b), where you said you were stumped. Good problem advice is to go back to stumpers after a while, particularly if you've solved a related problem in the meantime. Agreed. One final hint: If there are three brands, then either 0, 1, 2, or 3 of them must be best sellers. Once you have your four numbers for parts a, b, c, d, is there any easy check you can perform to see that the answers ae at least possible? -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ V2 = int[0..2pi] int[0..pi/6] int[ 4 sin(phi) .. 2] rho^2 sin(phi) dV where dV = drho dphi dtheta. this is what I thought I was supposed to do and yes my teacher was, by golly, going to solve this come hell or high water with spherical coordinates. What he did, and what I noticed after going over this myself after class, was make the assumption that fixing phi second would give him some weird integral. He was using his drawing, which wasn't very good, to determin what to do instead of looking at the equation. Actually, you can integrate phi first, it just isn't the most natural way to do the problem. Just like when doing rho first you get one of the limits is a function of phi, if you integrate phi first then one of its limits is a function of rho: V2 = int[0..2pi] int[0..2] int[0..arcsin(rho/4)] dV where dV = rho^2 sin(phi) dphi drho dtheta. Try it. You might find it educational. You can work it by hand and get the same answer as the other methods. You just need to understand a little trig and do a little more work on the middle drho integral. --Lynn what r the patterns in pascals triangle? Pascal's triangle can be explained using the binomial coefficient xCy = x!/[y!(x-y)!]. Here x! is the factorial of x and is given by x! = 1*2*3*...*(x-1)*x, and by definition 0! = 1. For example, the factorial of four is 4! = 1*2*3*4 = 24. So how does this relate to Pascal's triangle? Well we know that Pascal's triangle looks like this ووووو1 وووو1 1 ووو1 2 1 وو1 3 3 1 و1 4 6 4 1 ..and that each element is obtained from adding the previous two above it. It just so happens that we can express this relationship via the binomial coefficient mentioned above. Using xCy we can write Pascal's triangle as ووووو[Capi talEHat]وووو0C0 ووووو[Capi talEHat]وو1C0 1C1 ووووو[Capi talEHat]2C0 2C1 2C2 وووو3C0 3C1 3C2 3C3 وو4C0 4C1 4C2 4C3 4C4 So by using the binomial coefficient xCy we see that x denotes what row we are in and y denotes what element we are reffering to. This is extreamly helpful because it cuts down on all the addition previously used to obtain the new elements! For instance, note that in the 5th row the 3rd element has the value of 6. We know that this should correspond to the binomial coefficient of 4C2 (since our index starts at 0 and not 1). We can check this by writing 4C2 = 4!/[2!(4-2)!] = 4!/(2!*2!) = 24/(2*2) = 24/4 = 6. I encourage you to check out http://mathworld.wolfram.com/PascalsTriangle.html and http://en.wikipedia.org/wiki/Pascal%27s_Triangle for more information. ^_^ Kyle P.S. It's worthwhile to note that the sum of the shallow diagonals of Pascal's triangle correspond to the Fibonacci numbers. Moreover, the factorial function is actually just the gamma function at integer arguments, which explains why 0! = 1 and the beautiful identity (x + 1/2)! = Sqrt[Pi]*(2x + 1)!!/2^(x + 1) There are too many patterns to give all of them. Here are a couple of them: If empty spaces are counted as zeros, and the numbers are arranged in a pyramid shape, then each number except the top 1 is the sum of the numbers immediately above left and above right of it. Any row reads the same from either end. It's not clear from the statement of the problem, but it's awesomely clear from the statement of the problem plus the assumption that the OP is not a total moron - if we were allowed to roll three times and then take the largest of the three rolls it's hard to imagine how anyone could possibly be unclear about the optimal strategy. I mean really. ************************ David C. Ullrich in in alt.math.undergrad: [...] Obviously not, since I didn't offer a general strategy. Whether the optimal strategy for the first throw is the same as that for the second throw would have to be determined by analysis, and I made no such analysis. I've now done so, however. The expected value of a single throw is 3.5. Your strategy applied to a two-throw game has an expected value of (1/2) * 3.5 + (1/2) * 5 = 4.25. Your strategy applied to a three-throw game has an expected value of (1/2) * 4.25 + (1/2) * 5 = 4.625, which is not optimal: there's a strategy with an expected value of 14/3. [...] Brian posting-account=YU8ZnAwAAABFplvVaRgMUYVYk_DRcg4s I wasn't claiming that it was the optimal strategy. All I was saying was that the optimal strategy would be maximising the average score per gane. If someome was parolling the gane, i.e paying out the final score in dollars, then the best strategy for the player is the one that maximises the average loss per game of the payroller. If you played a 1-throww game 6k times (k a large number) the aveage payout per game would be (1 +2+3+4+5+6)/6 = 7/2 If you played a 2-throw game 36k times and stooped on a 6 after the 1st throw the average pauout per game would be (6*6 + 5(1+2+3+4+5+6))/36 = 141/36 If you stopped on a 6 or a 5 (6*(6+5) + 4(1+2+3+4+5+6))/36 = 150/36 6,5,4 (6*(6+5+4) + 3(1+2+3+4+5+6))/36 =153/36 6,5,4,3 (6*(6+5+4+3) + 2(1+2+3+4+5+6))/36 =126/36 So you should stop on 6 or 5 for a 3-throw game. posting-account=Tt3RvA0AAABur-q9NAlJK5fGGXUbcJWq Brian, On 2 Apr 2005 17:11:34 -0800, Joseph A. in alt.math.undergrad: The basic idea is straightforward. Suppose that you've just you should throw again if and only if the expected value of the remaining k-throw game is greater than the number that you just threw. The expected value of a 1-throw game is the expected value of a roll of the die, i.e., 3.5. Thus, if you have one throw remaining, you should keep what you have if it's 4 or more and roll again if it's 3 or less. The probability that you just rolled 4 or more is 1/2, as is the probability that you just rolled 3 or less. In the former case you rolled 4, 5, or 6 with equal probability, so your expected payoff is (4 + 5 + 6)/3 = 5, and in the latter case your expected payoff is 3.5, the value of the 1-throw game. Overall, therefore, the expected value of the 2-throw game is (1/2) * 5 + (1/2) * 3.5 = 8.5/2 = 4.25. For the 3-throw game the reasoning is similar. After the first roll, you're playing a 2-throw game if you choose to continue, so you should continue if and only if the expected value of the 2-throw game -- 4.25 -- is greater than your first roll. You should be able to finish it off yourself from there. Brian No, it's called quadratic convergence because of epsilon_{i+1} = -epsilon_i^2 f''(x)/(2 f'(x)) So the error epsilon_{i+1} is proportional to the square of the previous error epsilon_i. It looks as if the first and second derivatives must exist at the root x. Sorry my first post had two errors in it. I'm sure you don't actually mean that. :-) It's not an asumption, it's based on observed data. But don't worry -- since you don't like reading what I write, I will spare you any further responses in future. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ Derivative of function of must be continuous near root, (as you noted), and first trial value must be nearer to root than any maximum or minimum. Each successive result doubles number of significant figures. Known as quadratic convergence. error (i+1)=- {[error(i)]^2} x f(x)/2f'(x) I hope this is what you are after - Ian Hutcheson It's pretty obvious, isn't it, that, if x= 1, sqrt(log(1))= sqrt(0)= 0 = log(1)= log(sqrt(1)). Consider x = e^16, for example: log(sqrt(x)) = 8, while sqrt(log(x)) = 4. And so the equation has another positive real solution: x = e^4 or roughly x = 54.59815 . Ah, I see _now_ that the OP was using base-ten logarithms, and I was instead using natural logarithms above. The other positve real solution to the equation, with the base of the logarithms being 10, is then x = 10^4 = 10000. David Consider x = e^16, for example: log(sqrt(x)) = 8, while sqrt(log(x)) = 4. And so the equation has another positive real solution, roughly x = 54.59815 . David If you are not allowed to use calculus, then use the fact that, for a any real number, there exist an increasing sequence of rational numbers {r_n} such that r_n converges to a. It follows from the fact that x^a is increasing for a rational (which you have already proved) that x^a is increasing for any real a: Suppose a_1 and a_2 are real numbers with a_1< a_2. Let {r_n} be an increasing sequence of rational numbers converging to a_1, {t_1} be an increasing sequence of rational numbers converging to a_2. Then for all i, r_i<= a_1. For some j, a_1< t_j<= a_2 (t_j< a_2 and r_i< a_1 are true because {t_n} and {r_n} are INCREASING functions. To show a_1< t_j for SOME j, let epsilong= a_2- a_1.) posting-account=hXoQ7w0AAADSdZxlCHUb3O5k5oe4oxj9 please: law and they are i.i.d ( independent and identical in distribution) then have Normal[0,1] law and are also i.i.d. This is known as the Box-Muller transformation. Googling this gives quite a few hits, including http://mathworld.wolfram.com/Box-MullerTransformation.html which reproduces what I understand to be the standard proof. posting-account=8jB0Vw0AAACJmu18OPPftv6igzbfKPoN Sequences and series are published in mathematics dictionary. It can be seen in subject algebra or in keywords of the following web page : Home page - www.b192907.com/math123 Then goto By subjects or By keywords. Example : The proof of Sum[n*(n+1)/2] = n*(n+1)*(n+2)/6 Answer : Use keyword is easy to locate the question. It is my hope that more sequences and series can be added to my web site. It would be appreciated if you could provide your questions and comments.