mm-3589 === Subject: Re: bisector construction problem ... > I suspect the 'anti-parallel' triangles may be involved. By > anti-parallel triangle I mean triangles inscribed and tangentially > escribed at three points of a circle. But admittedly this is a bit > vague. The above comment is related to assumptions hyperbolic geometry, may be ignored at undergraduate stage. This is a simple and straight forward trigonometrical calculation. Let in-radius of triangle ABC be r, angle subtended at in-center I by sides AB, BC and CA be GM, AL and BT respectively.Only AL,BT,GM are given in the posed problem. We need to find proportion among IA,IB,IC. Trigonometric relations 2 AL = pi + A & cyclical equations result in the proportion of ray lengthss through I : ( IA/r : IB/r : IC/r ) =( - sec(AL) : - sec(BT) : - sec(GM) ) =( csc(A/2) : csc(B/2) : csc(C/2) ). We can also find sides of triangle ABC. Side 'a' or BC = -r*( Tan[BT]+Tan[GM] ), similarly for b and c by cyclic symmetry , where A + B + C = (AL + BT + GM) /2 = pi ; Homothetic enlargement/reduction with respect to I furnishes other admissible triangles . Please note that if sum of two given angles, say AL + BT > 270 degrees,then the above solution furnishes a situation where the ex-circle contacts two extended sides of a (representatitive of a similar set of the ) solution triangle. Else inner contact. The construction can be done by Ruler and Compass as well. Take an arbitrary length BI and construct a segment of a circle containing C/2 or GM-pi/2 in it in the usual way. Now the ratio of sides CI/CB = - cos(BT)/Cos(AL) is known. The Apollonius circle of this side ratio by harmonic division can be constructed and intersects the first circle at the required point.Other points can be obtained by angle doubling etc. Shall send Mathematica program if someone is interested. Hope that settles it. Narasimham === Subject: Opinions of Myerson's Probability Models for Economic Decisions? Has anyone encountered Probability Models for Economic Decisions by Roger Myerson? If so, do you have any opinion about how good it might be for self-study? I'd appreciate any views. === Subject: Grade 12 math can someone please check my answers? 5 1 / - / x+3 x-1 Answer I got is 4x-8 / (x+3)(x-1) Would 4x-8 need to be simplified further to 2(x-4) Or am I way off on this question? === Subject: Re: Grade 12 math can someone please check my answers? > 5 1 > / - / > x+3 x-1 > Answer I got is > 4x-8 > / > (x+3)(x-1) > Would 4x-8 need to be simplified further to 2(x-4) > Or am I way off on this question? Regardless of the exact meaning of your expression(s), doesn't anyone have a problem with the characterization of this question as Grade 12 math? If indeed the characterization is correct, then secondary math education is in even a worse state than I thought. This looks like a simple problem in first year algebra, i.e. middle school math, not Senior year High School math. === Subject: Re: Grade 12 math can someone please check my answers? > 5 1 > / - / > x+3 x-1 > Answer I got is > 4x-8 > / > (x+3)(x-1) > Would 4x-8 need to be simplified further to 2(x-4) > Or am I way off on this question? Regardless of the exact meaning of your expression(s), doesn't anyone have a problem with the characterization of this question as Grade 12 math? If indeed the characterization is correct, then secondary math education is in even a worse state than I thought. This looks like a simple problem in first year algebra, i.e. middle school math, not Senior year High School math. === Subject: Re: Grade 12 math can someone please check my answers? > Regardless of the exact meaning of your expression(s), > doesn't anyone have a problem with the characterization > of this question as Grade 12 math? > If indeed the characterization is correct, then secondary > math education is in even a worse state than I thought. > This looks like a simple problem in first year algebra, > i.e. middle school math, not Senior year High School math. Although it's not an exact fit for what I want to say, the following, taken from another post of mine (math-teach at The Math Forum), is close enough, given that I don't have any more time to spend here right now. (Basically, I sympathize with what you're saying, but what we and our peers took in the 12'th grade is so very non-representative that it'd be essentially useless to base any judgements such as yours on it.) http://mathforum.org/kb/thread.jspa?messageID=4721354 I think it's very dangerous to base expectations, policy, and curriculum on what worked well for *us*, since the experiences of someone who was the top math student in their school (as I would assume most of us were) is hardly the norm for, and almost certainly not remotely transferable to, the general population (even college bound). Consider, for example, how undergraduate and graduate students who are new at grading papers differ from those who have had several semesters of experience with grading. New graders are almost always tougher than more experienced graders (and usually tougher with things that aren't as important as other things). Despite how obvious this seems to me now, I will admit that in my earliest teaching experiences (early 1980's), I tended to follow the golden rule treat others as you would have them treat you rather than what was probably best for my students. I'd bet that I'm not all that unusual with this evolution of teaching philosophy, also. Dave L. Renfro === Subject: Re: Grade 12 math can someone please check my answers? On 12 Jun 2006 07:15:09 -0700, Pubkeybreaker in alt.math.undergrad: >> 5 1 >> / - / >> x+3 x-1 [...] > Regardless of the exact meaning of your expression(s), > doesn't anyone have a problem with the characterization > of this question as Grade 12 math? Yes and no. As you say below, it's a first-year algebra problem; but if Lisa is encountering it in a 12th grade class, she can't be blamed for describing it as 'Grade 12 math'. > If indeed the characterization is correct, then secondary > math education is in even a worse state than I thought. > This looks like a simple problem in first year algebra, > i.e. middle school math, not Senior year High School > math. I'd have said 8th or 9th grade, which to me means junior high or high school, but in substance I agree. (In the early 60s first year algebra was taught in the 9th grade in the two school systems with which I was involved at the time, one in Indiana and one in Wisconsin, and very little was taught in the 7th and 8th grades.) My impression is that this is still generally the case. Certainly a great many high school seniors are now taking calculus: the AP calculus exam. This number is far larger than the number of students who took mainstream first-semester calculus in all four-year undergraduate programs in the Fall of 2000. By the time of the next CBMS survey in 2005ÁV06, we can expect that more students will take an AP Calculus exam than will take mainstream Calculus I in the Fall of 2005 in all 2-year and 4-year institutions combined. Mind you, in my experience a fair number even of the kids who have had some calculus in high school are still pretty shaky on basic algebra. Brian === Subject: Re: Grade 12 math can someone please check my answers? On Fri, 09 Jun 2006 21:15:50 EDT, Lisa in alt.math.undergrad: > Hi everyone need help with some math problems these are > the two hardest questions in my review and I just wanted > to see if I was on the right page with these. I donÍt > write equations on the computer much so I hope everyone > who decides to look these over can easily understand what > on this I really appreciate it. : ) > 5 1 > / - / > x+3 x-1 It's understandable, but it would be easier to read in standard single-line format: 5/(x + 3) - 1/(x - 1) (Just be careful to use enough parentheses!) > Answer I got is > 4x-8 > / > (x+3)(x-1) I.e., (4x - 8) / [(x + 3)(x - 1)]; this is correct. > Would 4x-8 need to be simplified further to 2(x-4) You mean to 4(x - 2); 2(x - 4) = 2x - 8, not 4x - 8. That depends entirely on what your teacher wants. All of the following are mathematically correct: (4x - 8) / [(x + 3)(x - 1)] 4(x - 2) / [(x + 3)(x - 1)] (4x - 8) / (x^2 + 2x - 3) 4(x - 2) / (x^2 + 2x - 3) The second is most useful for some purposes; for other purposes the third is a little nicer. And teachers sometimes have their own fairly arbitrary preferences in these things. Brian === Subject: Re: Grade 12 math can someone please check my answers? > Hi everyone need help with some math problems these are the two hardest questions in my review and I just wanted to see if I was on the right page with these. I don't write equations on the computer much so I hope everyone who decides to look these over can easily understand what the appreciate it. : ) > 5 1 > / - / > x+3 x-1 > Answer I got is > 4x-8 > / > (x+3)(x-1) > Would 4x-8 need to be simplified further to 2(x-4) > Or am I way off on this question? as Paul said just treat the two things as symbolic fractions - same rule applies here as to ordinary numerical fractions, that is find a common denominator (the product of (x+3) and (x-1) in this case), and then cross multiply. === Subject: Re: Grade 12 math can someone please check my answers? > 5 1 > / - / > x+3 x-1 Let me educate you on typing math in ascii text. Don't try to break it across lines like that. It does not always format in the reader's window the way you think. Make use of '()' to make explicit the order and grouping of things. So your problem could look like: (5 / (x + 3)) - (x - 1)^{-1} or (5 / (x + 3)) - (1 / (x - 1)) Either is much easier to read. It is also nice to add a little space betweensymbolsotherwiseitmaygethardtoread. I assume you are trying to just get a common denominator and simplify a bit. In which case, you answer is correct. (4x-8)/((x+3)(x-1)) This of course assumes that I have interpreted you garbally goo correctly. If you simplify the numerator futher, you would get 4(x-2) not 2(x-4) as you have written, which I assume is a typo on your part. As to whether you must factor out the 4 is a matter of how your teacher wants it. === Subject: Bernoulli trials (take 2): counterintuitive results ---------------------------------------------------------------------------- ---- I'm teaching myself statistics and am now working through binomial distributions. I can reach the answers in my text. But out of curiousity I did some sensitivity analysis on a couple of parameters, and the results were not at all what I expected . In fact I can't explain them. I've checked my calculations, so the math seems ok. But could anyone help me with the intuition? Here I will explain the problem, show the formula I used, then show the strange results. The Problem: suppose that based on observation of past outcomes, you think there is a 30% chance that one oil well in area will discover oil, and thus a 70% chance that it won't. Assume (perhaps unrealistically) that the outcome of each trial is wholly independent of the others. The formula When p is the observed probability of past success, q is the corresponding probability of failure (1-p) r is the # of successes in question and n is the number of trials the probability of exactly r successes in n trials is Probability (r, n) = (combin n, r) * (p^r) * ( (q^(n-r) ) The problematic solutions I and my book both get an answer of 13%. So far so good. But then I asked, how does the probability of 3 successes change when you vary a) the number of trials? b) the assumed probability of success? Regarding a): the outputs are below. Column (i) is the # of trials and column (ii) is the probability of success in 3 trials. I understand why initially the prob. of 3 successes rises along with the # of trials. But why does it decline after 10 trials -- all the way down to 7% at 20 trials? I have a better than 1-in-4 chance of 3 successes if I try 10 times, but a 7% chance of 3 successes if I try 20 times -- bizarre. (i) (ii) 3 3% 4 8% 5 13% 6 19% 7 23% 8 25% 9 27% 10 27% 11 26% 12 24% 13 22% 14 19% 15 17% 16 15% 17 12% 18 10% 19 9% 20 7% I'm similalry puzzled by b). In my results below, column i) is the probability of success, and column ii) is the probability of 3 succeses over 5 trials. Again, why does it peak? Why, if you have a 100% chance of success, do you have a 0% probability of 3 successes in 3 tries? (i) (ii) 0% 0% 10% 1% 20% 5% 30% 13% 40% 23% 50% 31% 60% 35% 70% 31% 80% 20% 90% 7% 100% 0% === Subject: Re: Bernoulli trials (take 2): counterintuitive results > ---------------------------------------------------------------------------- - --- > I'm teaching myself statistics and am now working through binomial distributions. I can reach the answers in my text. But out of curiousity I did some sensitivity analysis on a couple of parameters, and the results were not at all what I expected . In fact I can't explain them. I've checked my calculations, so the math seems ok. But could anyone help me with the intuition? > Here I will explain the problem, show the formula I used, then show the strange results. > The Problem: suppose that based on observation of past outcomes, you think there is a 30% chance that one oil well in area will discover oil, and thus a 70% chance that it won't. Assume (perhaps unrealistically) that the outcome of each trial is wholly independent of the others. > The formula > When > p is the observed probability of past success, > q is the corresponding probability of failure (1-p) > r is the # of successes in question and > n is the number of trials > the probability of exactly r successes in n trials is > Probability (r, n) = (combin n, r) * (p^r) * ( (q^(n-r) ) > The problematic solutions > I and my book both get an answer of 13%. So far so good. > But then I asked, how does the probability of 3 successes change when you vary > a) the number of trials? > b) the assumed probability of success? > Regarding a): the outputs are below. Column (i) is the # of trials and column (ii) is the probability of success in 3 trials. Huh? You mean the probability of 3 successes? > I understand why initially the prob. of 3 successes rises along with the # of trials. But why does it decline after 10 trials -- all the way down to 7% at 20 trials? I have a better than 1-in-4 chance of 3 successes if I try 10 times, but a 7% chance of 3 successes if I try 20 times -- bizarre. I think you are getting confused between the probability of *exactly* three successes and the probability of *at least* three successes. Your formula above gives the probability of *exactly* the given number of successes, as you noted. If you have a 30% probability of success at each trial then, as the number of trials increases, it becomes more and more likely that you will get *more* than three successes. For example, imagine you ran a million trials. What is the chance that you will get exactly three successes? It should be obvious that the chance is very small because almost always you will have many more than three successes. On the other hand, the probability of getting *at least* three successes will be very close to one. > (i) (ii) > 3 3% > 4 8% > 5 13% > 6 19% > 7 23% > 8 25% > 9 27% > 10 27% > 11 26% > 12 24% > 13 22% > 14 19% > 15 17% > 16 15% > 17 12% > 18 10% > 19 9% > 20 7% > I'm similalry puzzled by b). In my results below, column i) is the probability of success, and column ii) is the probability of 3 succeses over 5 trials. Again, why does it peak? Why, if you have a 100% chance of success, do you have a 0% probability of 3 successes in 3 tries? I assume you mean a 0% probability of 3 successes in 5 tries. If you have a 100% chance of success in each trial, then in five trials you are certain to have five successes. Therefore the probability of having exactly three successes is zero. > (i) (ii) > 0% 0% > 10% 1% > 20% 5% > 30% 13% > 40% 23% > 50% 31% > 60% 35% > 70% 31% > 80% 20% > 90% 7% > 100% 0% === Subject: Bernoulli trials s === Subject: Sorry, please ignore my earlier Bernoulli trials post I figured it out already -- doh! === Subject: Distance between two parallel lines Hi folks, can anyone tell me how I go about finding the distance between the two parallel lines with parametric equations shown; line 1: (x=2-t, y=2t, z=1+t) line 2: (x=1+2t, y=3-4t, z=5-2t) Sandy === Subject: Re: Distance between two parallel lines Sandy === Subject: Re: Distance between two parallel lines >...finding the distance between the two parallel lines : > line 1: (x=2-t, y=2t, z=1+t) > line 2: (x=1+2t, y=3-4t, z=5-2t) Hi. Here's G.E's solution using a Math program. v1 = {2, 0, 1}; v2 = {17/6, -2/3, 19/6}; Norm[v2 - v1] Sqrt[35/6] You may find this link interesting... http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html Pick 1 point from 1 line: x0 = {2*t + 1, 3 - 4*t, 5 - 2*t} /. t -> 0 {1, 3, 5} Pick 2 points from the other line: x1 = {2 - t, 2*t, t + 1} /. t -> 0 {2, 0, 1} x2 = {2 - t, 2*t, t + 1} /. t -> 1 {1, 2, 2} Then... Norm[Cross[x2 - x1, x1 - x0]]/ Norm[x2 - x1] Sqrt[35/6] Same answer. :>) Here are the definitions of the functions used: Cross[{a, b, c}, {x, y, z}] {b*z - c*y, c*x - a*z, a*y - b*x} Norm[{a, b, c}] Sqrt[a^2 + b^2 + c^2] -- HTH. :>) Dana > Hi folks, > can anyone tell me how I go about finding the distance between the two > parallel lines with parametric equations shown; > line 1: (x=2-t, y=2t, z=1+t) > line 2: (x=1+2t, y=3-4t, z=5-2t) > Sandy === Subject: Re: Distance between two parallel lines >...finding the distance between the two parallel lines : > line 1: (x=2-t, y=2t, z=1+t) > line 2: (x=1+2t, y=3-4t, z=5-2t) Let U be either unit vector parallel to the lines. Let V be any vector from any point on one line to any point on the other. (For example take the point for which t = 0 on each line) Let 'A*B' denote the scalar product of vectors 'A' and 'B' Let 'a.B' denote the scalar times vector multiplication of scalar 'a' times vector 'B'. Then '(V*U).U' is the vector resolution or component of 'V' parallel to the lines and 'V - (V*U).U' is the vector resolution or component of 'V' perpendicular to the lines and '|| V - (V*U).U ||', or the length of 'V - (V*U).U', is the desired distance. === Subject: Re: Distance between two parallel lines > Hi folks, > can anyone tell me how I go about finding the distance between the two parallel lines with parametric equations shown; > line 1: (x=2-t, y=2t, z=1+t) > line 2: (x=1+2t, y=3-4t, z=5-2t) > Sandy Vector wise: Pick any t value on line 1 to get any point on it...say t=0 and we get v=(2,0,1). Pick any t value on line 2 to get a point on it...say t=0 and we get w=(1,3,5). Consider the vector vw = w-v = (1,3,5) - (2,0,1) = (-1,3,4). Notice that line 1 has a vector along the line: (-1,2,1) (the coefficients of the t's). Project vw (-1,3,4) onto this vector x (-1,2,1). Proj_v(u) = ((u.v) / ||u|| ||v||) v (for general v and u) = ((1+6+4)/ sqrt(26) sqrt(6)) (-1,2,1) = (11 / sqrt(156)) (-1, 2, 1) = (11 / 2 sqrt(39)) (-1, 2, 1) So now you have two sides of a triangle, the third side of which you are trying to find: The triangle is from v to proj_x(vw) to w. It's a right angle triangle, so the length of the side we want, L, satisfies, L^2 + ||(proj_x(vw) - v)||^2 = ||vw||^2 Work out the details to get your answer. man I never realized how hard it was to write vector stuff on the computer like this...maybe next time I'll just write in latex code, and assume the reader will understand it. :) Rob === Subject: Re: Distance between two parallel lines > Hi folks, > can anyone tell me how I go about finding the > distance between the two parallel lines with > parametric equations shown; > line 1: (x=2-t, y=2t, z=1+t) > line 2: (x=1+2t, y=3-4t, z=5-2t) > Sandy Choose a value for t in order to find a point on line 1: Taking t= 0 gives (2, 0, 1). Let (1+2t, 3-4t, 5-2t) be any point on line 2. The distance (squared) between (2, 0, 1) and (1+2t, 3-4t, 5-2t) is, of course,(2t-1)^2+ (3-4t)^2+ (4- 2t)^2. The derivative of that is 4(2t-1)- 8(3-4t)- 4(4-2t)= 8t- 4- 24+ 32t- 16+ 8t= 48t- 44 which is equal to 0 at the minimum value: 48t- 44= 0 so t= 11/12. Remember that that is the t value for the second line: when t= 11/12, the point is (1+ 11/6, 3- 11/3, 5- 11/6)= (17/6, -2/3,19/6). You might note that the vector from (2, 0, 1) to (17/6, -2/3, 19/6) is given by (5/6)i-(2/3)j+ (13/6) and the dot product of that with 2i- 4j- 2k (the vector parallel to the two original lines) is 10/6+ 8/3- 26/6= (10+ 16- 26)/6= 0. Yes, that vector is normal to both lines. The distance between the two parallel lines is the distance between (2, 0, 1) and (17/6, -2/3, 19/6). === Subject: Variance A friend of mine asked me this the other day, and I was stuck not knowing the answer. I was slightly embarassed as I am in the senior year of a BS in math. In the formula for sample variance, s^2 = sum(x - mean(x))^2/(n-1) why do you subtract one from the sample size? Dustin === Subject: Re: Variance > A friend of mine asked me this the other day, and I was stuck not knowing > the answer. I was slightly embarassed as I am in the senior year of a BS in > math. > In the formula for sample variance, > s^2 = sum(x - mean(x))^2/(n-1) > why do you subtract one from the sample size? > Dustin If memory serves, that S^2 thing is termed the sample unbiased estimate of the population variance: if you use the bog-standard sample variance, V, where V = {sum(x - mean(x))^2}/n, you find that the expectation of V, E[V], comes out as E[V] = sigma^2 * (n - 1) / n, where sigma^2 is the *actual* population variance. (The proof isn't too bad.) Hence V is *not* an unbiased estimator for the true population variance (remember that in order for a sample statistic, T, to be an unbiased estimator for some population parameter, theta, its expectation must be equal to theta, i.e. E[T] = theta). But, by using S^2 = V * n / (n - 1) instead of plain old V, we get E[S^2] = E[V] * n / (n - 1) = sigma^2, which is just what the doctor ordered. And as for being embarrassed, don't worry - better to look a bit daft now rather than at exam time. I remember my own chagrin at being the only student in my quantum mechanics tutorial who didn't know what the idea of valency was all about. My tutor, a kindly man, turned to me with a beneficent look and began slowly, well, you can think of it as if atoms have little sticks poking out of them.... Oh dear. Best wishes, Paul. === Subject: Re: Variance >A friend of mine asked me this the other day, and I was stuck not knowing >the answer. I was slightly embarassed as I am in the senior year of a BS in >math. >In the formula for sample variance, > s^2 = sum(x - mean(x))^2/(n-1) >why do you subtract one from the sample size? The reason there's a correction is that the variance in the sample is going to be less than the real variance. (Suppose you just take one sample. Then there's no variation at all in your sampled data. Does that mean you want to estimate the actual variance to be 0? No, taking just one sample gives no information at all about the real variance.) To see that the correction should be exactly what it is you do the math: Let's suppose to simplify things that the real mean is 0 and the real variance in the underlying distribution is 1. Say we take N independent samples X_1, .. X_N. Now, we don't know that the real mean is 0. Our best guess for the real mean is m = (X_1 + ... + X_N)/N. So our estimate for the variance is going to have _something_ to do with the expected value of the sum of (X_j - m)^2. Let's see what that is. First, E(X_1 - m)^2 = E((N-1)X_1 - X_2 - X_3 ... - X_N)^2/N^2 = ((N-1)^2 + 1 + 1 + ... + 1)/N^2 = ((N-1)^2 + (N-1))/N^2 = N(N-1)/N^2 = (N-1)/N. The other terms are all the same, so we get E(sum(X_j-m)^2) = N-1. So if we define the sample variance to be sv = sum(X_j-m)^2/(N-1) then we get E[sv] = 1. So. Defining the sample variance in that funny way makes the expected value of the sample variance exactly equal to the real variance. >Dustin ************************ === Subject: Re: Variance > A friend of mine asked me this the other day, and I was stuck not knowing > the answer. I was slightly embarassed as I am in the senior year of a BS in > math. > In the formula for sample variance, > s^2 = sum(x - mean(x))^2/(n-1) > why do you subtract one from the sample size? That's the formula for predicting the variance of a population given only a sample of the population, and since every sample is somewhat biased the -1 compensates for that bias. With a BS in math you may be able to proove that, I did only one year of stat and recall that expllanation from the first semester. -- Bye. Jasen === Subject: Re: Variance <2c30.448c8134.238d3@clunker.homenet A friend of mine asked me this the other day, and I was stuck not knowing > the answer. I was slightly embarassed as I am in the senior year of a BS in > math. > In the formula for sample variance, > s^2 = sum(x - mean(x))^2/(n-1) > why do you subtract one from the sample size? > That's the formula for predicting the variance of a population given only a > sample of the population, and since every sample is somewhat biased the -1 > compensates for that bias. It's not that the *sample* is (systematically) biased - we assume here that it isn't. The point is that, even with an unbiased sampling technique, the sample *variance* is systematically biased if you use the sample mean to calculate it. > With a BS in math you may be able to proove that, I did only one year of stat > and recall that expllanation from the first semester. > -- > Bye. > Jasen === Subject: Re: Variance : > A friend of mine asked me this the other day, and I was stuck not knowing > the answer. I was slightly embarassed as I am in the senior year of a BS in > math. > In the formula for sample variance, > s^2 = sum(x - mean(x))^2/(n-1) > why do you subtract one from the sample size? Short answer: to adjust for the fact that this is a sample and not the whole population. Longer answer: http://www.childrens-mercy.org/stats/ask/df.asp -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ === Subject: Re: Variance > A friend of mine asked me this the other day, and I was stuck not knowing > the answer. I was slightly embarassed as I am in the senior year of a BS in > math. > In the formula for sample variance, > s^2 = sum(x - mean(x))^2/(n-1) > why do you subtract one from the sample size? > Dustin If you calculate the mean from the sample, and then use this sample mean to calculate the sample variance, then the obvious calculation s^2 = sum(x - mean(x))^2/n will be biased, in the sense that if you took a sample of size n very many times, calculated the variance in this way for each sample, and then averaged the variances, the result would not converge to the true variance of the population. Intuitively, the reason is that the sample mean will tend to be biased towards the centre of the sample values, so the deviations from the sample mean will tend to be lower than from the true mean. The n - 1 correction adjusts for this underestimation, and gives an unbiased estimate of the true population variance. However, if in the variance calculation you use the true mean (as, say, known from theory) rather than the sample mean, then the n - 1 correction should *not* be used. === Subject: Re: Variance > A friend of mine asked me this the other day, and I was stuck not knowing > the answer. I was slightly embarassed as I am in the senior year of a BS in > math. > In the formula for sample variance, > s^2 = sum(x - mean(x))^2/(n-1) > why do you subtract one from the sample size? > Dustin > If you calculate the mean from the sample, and then use this sample > mean to calculate the sample variance, then the obvious calculation s^2 > = sum(x - mean(x))^2/n will be biased, in the sense that if you took a > sample of size n very many times, calculated the variance in this way > for each sample, and then averaged the variances, the result would not > converge to the true variance of the population. > Intuitively, the reason is that the sample mean will tend to be biased > towards the centre of the sample values, so the deviations from the > sample mean will tend to be lower than from the true mean. > The n - 1 correction adjusts for this underestimation, and gives an > unbiased estimate of the true population variance. > However, if in the variance calculation you use the true mean (as, say, > known from theory) rather than the sample mean, then the n - 1 > correction should *not* be used. Note also, that when using this correction for the standard deviation instead of the variance, it introduces a slight bias of its own, which is quite small and is usually ignored. === Subject: Non-Euclid Geometry I am a Japanese and I am trying to find English books about non-Euclid geometry. Also, I want to hear many opinions from those who are interested in non-Euclid geometry. I am looking forward to reading messages from all over the world. === Subject: Re: Non-Euclid Geometry On 11 Jun 2006 06:16:00 -0700, balljuggling >I am a Japanese and I am trying to find English books about non-Euclid >geometry. that when you do those searches you will need to use the standard English spelling: Non-euclidean geometry . >Also, I want to hear many opinions from those who are >interested in non-Euclid geometry. I am looking forward to reading >messages from all over the world. ************************ === Subject: Distance from a Sphere Hi everyone, can anyone tell me how to locate the point on the sphere with the equation x^2 + y^2 + z^2 + 2x - 2y - 4z - 3 = 0 which is closest to the origin ?? Any help much appreciated !!! Brian === Subject: Re: Distance from a Sphere They have helped ! Brian === Subject: Re: Distance from a Sphere <2545127.1150051755803.JavaMail.jakarta@nitrogen.mathforum.org>, Brian > Hi everyone, > can anyone tell me how to locate the point on the sphere with the equation > x^2 + y^2 + z^2 + 2x - 2y - 4z - 3 = 0 which is closest to the origin ?? > Any help much appreciated !!! > Brian If you know the technique, this is a more or less classic Lagrange multiplier problem: Minimize x^2 + y^2 + z^2 subject to the constraint x^2 + y^2 + z^2 + 2x - 2y - 4z - 3 = 0. G. E. Ivey's suggestion also works fine. -- Paul Sperry Columbia, SC (USA) === Subject: Re: Distance from a Sphere > Hi everyone, > can anyone tell me how to locate the point on the > sphere with the equation > x^2 + y^2 + z^2 + 2x - 2y - 4z - 3 = 0 which is > closest to the origin ?? > Any help much appreciated !!! > Brian Do you know how to find the center of the sphere? Write this as (x^2+ 2x+ )+ (y^2- 2y+ )+ (z^2-4z+ )= 3 and complete the square in each part so you get (x- x_0)^2+ (y- y_0)^2+ (z- z_0)^2= R^2. The line from the center, (x_0,y_0,z_0), to (0,0,0) passes through the sphere at the point closest to (0,0,0). === Subject: Re: But, what if you are wrong? Oh! Never mind. I just figured out the real question/subject. I'm joining the party a little late. :~( -- Dana >> Can A^4+B^4+C^4= D^3 ?, where A+B+C=0 === Subject: Proof that sum 1/k from 1...n isn't an integer for n > 1 ways to improve it and show the best way, of course! First, a lemma. (*) If a prime p evenly divides all but one term of the sum S = x_1 + x_2 + ... + x_n then p doesn't evenly divide S. Proof of (*) Let S = x_1 + x_2 + ... + x_n and wlog let x_n be the term not divisible by p. Since the first (n - 1) terms are divisible by p, for some integer k they factor into: p*k = x_1 + x_2 + ... + x_(n-1) Now if S = (p*k + x_n) is evenly divisible by p it can be factored as S = p*(k + x_n') where p*x_n' = x_n, contradicting the hypothesis that p doesn't evenly divide x_n. So p must not evenly divide S. Proposition: sum[k=1 to n] 1/k isn't an integer for n > 1. Proof of Proposition: sum[k=1 to n] 1/k = 1/1 + 1/2 + 1/3 + ... + 1/(n - 1) + 1/n Since n > 1 showing that this sum isn't integer is equivalent to showing that the sum minus 1 isn't an integer so remove the first term leaving: 1/2 + 1/3 + ... + 1/(n - 1) + 1/n Now multiply each kth term in the sum above by the fraction (n!/(k + 1)) / (n!/(k + 1)) to obtain [ (n!/2) + (n!/2) + ... + (n!/(n - 1)) + (n!/n) ] / n! Let p be the largest prime less than n, this prime exists because n is at least 2. Since p is a factor of n! and does not evenly divide the pth term in the numerator, it follows from (*) that p does not evenly divide the numerator. Therefore the sum can't be an integer. === Subject: Re: Proof that sum 1/k from 1...n isn't an integer for n > 1 > First, a lemma. > (*) If a prime p evenly divides all but one term of the sum > S = x_1 + x_2 + ... + x_n > then p doesn't evenly divide S. > Proof of (*) > Let S = x_1 + x_2 + ... + x_n and wlog let x_n be > the term not divisible by p. > Since the first (n - 1) terms are divisible by p, > for some integer k they factor into: > p*k = x_1 + x_2 + ... + x_(n-1) > Now if S = (p*k + x_n) is evenly divisible by p it > can be factored as S = p*(k + x_n') where p*x_n' = x_n, > contradicting the hypothesis that p doesn't evenly > divide x_n. So p must not evenly divide S. > Proposition: sum[k=1 to n] 1/k isn't an integer for n > 1. > Proof of Proposition: > sum[k=1 to n] 1/k = 1/1 + 1/2 + 1/3 + ... + 1/(n - 1) + 1/n > Since n > 1 showing that this sum isn't integer is > equivalent to showing that the sum minus 1 isn't > an integer so remove the first term leaving: > 1/2 + 1/3 + ... + 1/(n - 1) + 1/n > Now multiply each kth term in the sum above by > the fraction (n!/(k + 1)) / (n!/(k + 1)) to obtain That's a complicated way of expressing it. You could just say that the previous sum is clearly equal to the following one. > [ (n!/2) + (n!/2) + ... + (n!/(n - 1)) + (n!/n) ] / n! You mean n!/3 instead of the second n!/2 > Let p be the largest prime less than n, this prime exists > because n is at least 2. Since p is a factor of n! and > does not evenly divide the pth term in the numerator, > it follows from (*) that p does not evenly divide the > numerator. Therefore the sum can't be an integer. There's a problem here. How do you know that _p_ does not divide evenly the p-th term in the numerator? First of all, my guess is that you mean (p - 1)-th term here. Anyway, what you're saying is that _p_ does not divide evenly n!/p. Why? Suppose that n >= 2p; then it is not true that _p_ does divide evenly n!/p. Jose Carlos Santos === Subject: Re: Proof that sum 1/k from 1...n isn't an integer for n > 1 >> First, a lemma. >> (*) If a prime p evenly divides all but one term of the sum >> S = x_1 + x_2 + ... + x_n >> then p doesn't evenly divide S. >> Proof of (*) >> Let S = x_1 + x_2 + ... + x_n and wlog let x_n be >> the term not divisible by p. >> Since the first (n - 1) terms are divisible by p, >> for some integer k they factor into: >> p*k = x_1 + x_2 + ... + x_(n-1) >> Now if S = (p*k + x_n) is evenly divisible by p it >> can be factored as S = p*(k + x_n') where p*x_n' = x_n, >> contradicting the hypothesis that p doesn't evenly >> divide x_n. So p must not evenly divide S. >> Proposition: sum[k=1 to n] 1/k isn't an integer for n > 1. >> Proof of Proposition: >> sum[k=1 to n] 1/k = 1/1 + 1/2 + 1/3 + ... + 1/(n - 1) + 1/n >> Since n > 1 showing that this sum isn't integer is >> equivalent to showing that the sum minus 1 isn't >> an integer so remove the first term leaving: >> 1/2 + 1/3 + ... + 1/(n - 1) + 1/n >> Now multiply each kth term in the sum above by >> the fraction (n!/(k + 1)) / (n!/(k + 1)) to obtain > That's a complicated way of expressing it. You could just say > that the previous sum is clearly equal to the following one. >> [ (n!/2) + (n!/2) + ... + (n!/(n - 1)) + (n!/n) ] / n! > You mean n!/3 instead of the second n!/2 Yes. >> Let p be the largest prime less than n, this prime exists >> because n is at least 2. Since p is a factor of n! and >> does not evenly divide the pth term in the numerator, >> it follows from (*) that p does not evenly divide the >> numerator. Therefore the sum can't be an integer. > There's a problem here. How do you know that _p_ does not divide evenly > the p-th term in the numerator? First of all, my guess is that you mean > (p - 1)-th term here. Yes, I meant (p - 1). > Anyway, what you're saying is that _p_ does not > divide evenly n!/p. Why? Suppose that n >= 2p; then it is not true that > _p_ does divide evenly n!/p. You're right. Can I get a hint for a good way to do this proof. === Subject: Re: Proof that sum 1/k from 1...n isn't an integer for n > 1 >> Anyway, what you're saying is that _p_ does not >> divide evenly n!/p. Why? Suppose that n >= 2p; then it is not true that >> _p_ does divide evenly n!/p. > You're right. Can I get a hint for a good way to do this proof. Sure. Instead of working with the greatest prime which divides _n_, work with the greatest power of 2 which divides _n_. Jose Carlos Santos === Subject: New Problem How do you factor 2m^2 - m - 7 = 0 The question is asking me to find the roots, and I assume you have to factor it out first. === Subject: Re: New Problem >How do you factor 2m^2 - m - 7 = 0 >The question is asking me to find the roots, and I assume you have to factor it out first. If you have a quadratic equation and aren't sure how to approach it, the best general way is to use the quadratic formula. If you can easily find factors, fine, but spending much time looking for them is not so good. Just go directly to the general method, the quadratic equation. bob === Subject: Re: New Problem > The question is asking me to find the roots I think you are looking for the following... http://mathworld.wolfram.com/QuadraticFormula.html > ...I assume you have to factor it out first? Some can be factored, but this one in particular doesn't seem to factor into anything useful...as far as I can tell. Good luck... -- HTH. :>) Dana > How do you factor 2m^2 - m - 7 = 0 > The question is asking me to find the roots, and I assume you have to > factor it out first. === Subject: Re: New Problem nothing I do works..(2x-7)(x+1) (2x+7) (x-1) nothing works what am I doing wrong? === Subject: Re: New Problem Try completing the square or the quadratic formula. You can use both to double check your self. If you try either and are still unsure... post what you have done and someone (hopefully less sarcastic then previous posts) will show you your mistakes. Paul === Subject: Re: New Problem <5417622.1150132873797.JavaMail.jakarta@nitrogen.mathforum.org nothing I do works..(2x-7)(x+1) (2x+7) (x-1) nothing works what am I doing wrong? Perhaps the fault lies in the assumption that it can be factored??????? === Subject: Re: New Problem Lisa a .8ecrit : > How do you factor 2m^2 - m - 7 = 0 > The question is asking me to find the roots, and I assume you have to factor it out first. 1) open a textbook 2) read it 3) apply the appropriate result to your problem Cody === Subject: Re: New Problem > Lisa a .8ecrit : > How do you factor 2m^2 - m - 7 = 0 > The question is asking me to find the roots, and I > assume you have to factor it out first. > 1) open a textbook > 2) read it > 3) apply the appropriate result to your problem > Cody Sarcasm is really unnecessary. === Subject: Re: New Problem > How do you factor 2m^2 - m - 7 = 0 > The question is asking me to find the roots, and I assume you have to factor it out first. May I suggest that you show us what you have already done? We are not here to do your homework for you. Also, I thought this newsgroup was devoted to undergraduate level math and not freshman high school level math??? I will give hints in socratic style: (1) If one root is rational, what can you say about the other one? (2) If m = a/b, for integers a and b, what can you say about a and b with respect to the coefficients of the polynomial? (3) If the roots are not rational, then what method(s) are available to solve it? (4) What is the required condition on the discriminant for the roots to be rational? What is the value of the discriminant? === Subject: Re: New Problem > How do you factor 2m^2 - m - 7 = 0 > The question is asking me to find the roots, and I > assume you have to factor it out first. > May I suggest that you show us what you have already > done? > We are not here to do your homework for you. > Also, I thought this newsgroup was devoted to > undergraduate level > math and not freshman high school level math??? Perhaps this is college pre-algebra or algebra at a University; hence, by definition undergraduate. > I will give hints in socratic style: > (1) If one root is rational, what can you say about > the other one? > (2) If m = a/b, for integers a and b, what can > you say about a and > with respect to the coefficients of the polynomial? > (3) If the roots are not rational, then what > method(s) are available to > solve it? > (4) What is the required condition on the > discriminant for the roots to > be rational? > What is the value of the discriminant? === Subject: Quick Probability/Counting Technique Questions Hey there, I've got a few questions that are giving me trouble. 1. How many even numbers greater than 4000 can be formed using some or all of the digits 1,2,3,4,5,6 if each digit must feature no more than once in a number? 2. Security codes are to be allocated with each code being either 2 digits followed by 3 letters OR 3 digits followed by 2 letters. eg. 12AQT or 129PT Any of the digits 0 to 9 can be used and any of the letters A-Z can be used but no digit or letter may be repeated. Determine how many codes are possible in each of the following cases e)A code must contain a 9 f)A code must contain a 5 g)A code must contain at least one of the digits 5 and 9 -These problems relate to permutations and combinations.. Counting Techniques. 3. A child is told she can bring 5 toys with her on a holiday. The child decides to choose the 5 from 6 jigsaws 8 dolls 4 balls 2 trucks How many of these sets have at least one from each of the four categories listed above? Any at all help is much appreciated, === Subject: Re: Quick Probability/Counting Technique Questions On Mon, 12 Jun 2006 14:45:11 EDT, Cherokee in alt.math.undergrad: > Hey there, I've got a few questions that are giving me trouble. In general it's best to give some indication of what you've already tried; quite apart from the fact that people like to know that you *have* made a serious effort, it can save you from being told a lot of things that you already know. None of these requires any kind of tricky technique or analysis. In each you have a set of things to be counted, and the problem can be solved by some combination of three standard techniques: * breaking up that set into non-overlapping smaller sets that can be counted more easily and then adding the subtotals; * counting some bigger set, counting the number of things in that bigger set that *aren't* in the one that you're trying to count, and taking the difference; or * counting the things in each of two overlapping subsets, adding these partial results, and then subtracting from this preliminary total the number of things that are in the overlap and so have been counted twice. > 1. How many even numbers greater than 4000 can be formed > using some or all of the digits 1,2,3,4,5,6 if each digit > must feature no more than once in a number? Every such number will obviously have four, five, or six digits. Imagine choosing one of them one digit at a time, starting at the righthand end. Since the number is to be even, it must end in 2, 4, or 6, so there are 3 ways to choose the last digit. Once that's chosen, there are 5 digits still available, and you can use any of them to fill the 2nd position from the right; altogether that's 5*3 = 15 ways to fill the rightmost two slots. You've now used two of the six available digits, so you have only 4 possible choices for the 3rd slot; this means that there are 4*5*3 = 60 ways to fill the rightmost three slots. At this point it's easy to go astray by trying to do too much at once. Don't try to take care of the 'greater than 4000' requirement at this point; for now just count all of the even numbers having four or more digits. We've used up three of the six available digits, so there are 3 ways to fill the 4th slot, and hence 3*4*5*3 = 180 four-digit even numbers that use only the digits 1 through 6 and use each digit at most once. If you continue in this fashion, you can easily calculate the number of five-digit and six-digit numbers of this type; all of them are clearly greater than 4000 and should therefore be included in your final total. However, the 180 four-digit numbers that we counted include some that are less than 4000; these have to be excluded from the final total, so we need to count them. Clearly they all begin with 1, 2, or 3, and they all end with 2, 4, or 6. If one of them begins with 1 or 3, it can end in any of 2, 4, and 6; if it begins with 2, however, it can end only in 4 or 6, so it's simplest to split the counting into two cases. If the number begins with 1 or 3, there are 2 ways to fill the leftmost slot and 3 ways (2, 4, or 6) to fill the rightmost slot. That leaves four digits, so there are 4 possible choices for the 2nd slot from the left and 3 possible choices for the remaining slot, for a total of 2*3*4*3 = 72 four-digit even numbers beginning with 1 or 3, using only the digits 1 through 6, and using each of them at most once. In similar fashion you can calculate the number that begin with a 2, but I'll let you do that. > 2. Security codes are to be allocated with each code being > either 2 digits followed by 3 letters OR 3 digits > followed by 2 letters. eg. 12AQT or 129PT > Any of the digits 0 to 9 can be used and any of the > letters A-Z can be used but no digit or letter may be > repeated. > Determine how many codes are possible in each of the > following cases > e)A code must contain a 9 Break the count into two parts, just as you would if there were no such restriction. How many codes are there with three digits and two letters? If we were allowed to use a 9, there would be 10*9*8 = 720 possible combinations of digits. How many of them include a 9? There are 9*8 = 72 three-digit strings of the form 9xy (with x not equal to y). There are 72 more of the form x9y and another 72 of the form xy9. No three-digit string is in more than one of these categories, so there are 3*72 = 216 'bad' strings, and hence 720 - 216 = 504 usable three-digit combinations. Each can be combined with any pair of distinct letters, for a total of 504*26*25 codes with three digits and two letters. The calculation of the number of codes of the other type is similar. The two types don't overlap, so you can simply add the two partial results to get the desired answer. > f)A code must contain a 5 This is of course the same kind of problem as (e). In fact, if you think about it, you should be able to see why it's essentially the *same* problem, not just the same kind, and why it has the same answer. > g)A code must contain at least one of the digits 5 and 9 Start by adding your answers to (e) and (f). Of course this doesn't quite work, because every code that contains both a 5 and a 9 gets counted twice. To correct this double counting, you need to subtract from that total the number of codes containing both a 5 and a 9; this number can be calculated using exactly the same techniques that are used to answer (e) and (f). > -These problems relate to permutations and combinations.. > Counting Techniques. > 3. A child is told she can bring 5 toys with her on a > holiday. The child decides to choose the 5 from > 6 jigsaws > 8 dolls > 4 balls > 2 trucks > How many of these sets have at least one from each of the > four categories listed above? The problem statement is ambiguous: the answer depends on whether the items of each type are identical (so that the four balls, for instance, are completely interchangeable) or distinguishable. I'm going to assume that they're distinguishable. Clearly she must take two of one kind of toy and one of each of the others. The most straightforward solution, if not the quickest, is to divide the problem into four cases according to the type of toy of which she takes two. Two jigsaws: There are C(6, 2) = 6!/(2!*4!) = 15 ways to choose 2 of the 6 jigsaws. There are then 8 ways to choose a doll, 4 ways to choose a ball, and 2 ways to choose a truck, for a grand total of 15*8*4*2 possible selections. Do the other cases similarly and then add them up. Brian === Subject: Re: Problems with proof (propositional calculus) >> Ensuing discussion on Pierce's axiom and it's independence. >> Prove: |- ((A -> B) -> A) -> A >> The proof should be done with assumptions (there is no axioms, >> only inference rules). > That is Pierce's axiom which is independent of natural deduction. Depends on *which* natural deduction system. Using Fitch natural deduction for classical logic: 1 | (p->q)->p hyp 2 || ~ p hyp 3 ||| p hyp 4 |||| ~ q hyp 5 |||| p 3, reit 6 |||| ~ p 2, reit 7 ||| q 4-6, ~ elim 8 || p -> q 3-7, -> int 9 || p 3, 8, -> elim 10 | p 2-9, ~ elim 11 ((p -> q) -> p) -> p) 1-10, -> int The ~ elim steps are the key (not intuitionistically admissible; double negation elimination is an equivalent rule). I think this is the nearest you can get intuitionistically: 1 | (p -> ~~ q) -> p hyp 2 || ~ p hyp 3 ||| p hyp 4 |||| ~ q hyp 5 |||| p 3, reit 6 |||| ~ p 2, reit 7 ||| ~~ q 4-6, ~ int 8 || p -> ~~ q 3-7, -> int 9 || p 3, 8, -> elim 10 | ~~ p 2-9, ~ int 11 ((p -> ~~ q) -> p) -> ~~p) 1-9, -> int (Stronger hypothesis and weaker conclusion). Is there a proof of the classical result that only applies ~ elim or ~~ elim once? Jack Campin: 11 Third St, Newtongrange EH22 4PU, Scotland | tel 0131 660 4760 for CD-ROMs and free | fax 0870 0554 975 stuff: Scottish music, food intolerance, & Mac logic fonts | mob 07800 739 557 === Subject: CRITICAL TORUS !!! In observing math , ive noticed , and im not alone, that many (proof)problems in math (especially the harder ones ) can be transformed into some kind of critical line problem. The most famous example is of course Riemann's Hypothese. But also the tanc conjecture can be rewritten as a critical line problem. Representions theorems can be rewritten as a critical line problem. For all clarity , i consider critical line problems as follows 1) a subgroup of zeros all lie on the critical line 2) none of the zeros of a subgroep lie on the critical line 3) no 3 zeros lie on a line (probably reducable to 1) or 2) so less important i guess ) Every upperbound or lowerbound conjecture is equivalent to some critical line problem You can just literally bend functions to get a critical line , if ya already have a simple critical function/figure/path SO if ya have a critical monotonic increasing polynoom like x^a , you can easily transform it to a line , bye transforming ( bending ) the function And don't forget that e.g. circels are the product of 2 complex lines. So critical problems in 2D are often easily transformable into critical line problems in 2D So the 2D equivalent problems are mainly and often simply reducable to critical lines A bit like belonging to an NP class to make a comparison.( but for proofs instead of computations ) However !!! some problems might not be expressible in 2D criticals , but in 3D criticals with the concept 3D being most general ! i mentioned before transforming polynoom to a line but in 3D we have topology joining the party; we could have a critical torus or polysigned critical torus or even more complicated stuff , not easily , or not at all , transformable to a line , or even 2D disk we all know topology is strongly related to number theory and metamathematics is getting popular i hope ive made clear why i believe so strongly in this critical theory importance. consider it a bit like NP=P , extended Riemann Hypothese and metamathematics. so i am deeply intrested in the critical torus since it is relating so much math concepts. and it relates them in only 2 words, nice :-) i do not claim to be a genius or anything , but as far as i know , this has never been studied. and as an intuitionist and constructivist , this is very appealing to me. i could continue with more than 3 dimensions , but i think that we should start with the simplest extension ( 2D --> 3D ; holes 0 --> +1 ) so does anybody know functions that have or might have a critical torus ? also this appears to me as a nice puzzle would be nice to see this online , e.g. at websites and forums this mail is not complete in the sence that the critical torus can be considered in different ways; do the zeros lie on the surface of the torus , or in the torus ? the most logical and intresting is according to me the critical surface of a torus , since other wise the condition is very loose and its more of a generalization to a strip , not a line. there is also probably a connection with eigenvalues and cellular automata , but thats confusing me at the moment. hope you think about it. i find it a logical question anyway.. greetz tommy1729