mm-1983 === Subject: analysis-nested Let Xk -->X be a convergent sequence in a metric space. Let A be a family of closed sets with the property that for each B (is element of ) A. there is an N such that k > N implies Xk (is element of ) B. Prove that X is a element of (intersection A) -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: analysis-nested janet http://mathforum.org/epigone/k12.ed.math/talstreutee > Let Xk -->X be a convergent sequence in a metric space. Let A > be a family of closed sets with the property that for each B > (is element of ) A. there is an N such that k > N implies Xk > (is element of ) B. Prove that X is a element of (intersection A) You need to show that if B belongs to the collection A, then X belongs to B. To this end, let B be an element of A. Can you think of how to use what you're given to prove that X belongs to B? Hint: X is the limit of a certain convergent sequence in B (not necessarily the entire original sequence) and B is closed. Dave L. Renfro -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: help can some help me wiht a math problem its finding the area of a triangle wiht the base od 11 in. and the sides of 5 adn 6 inches -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: help > can some help me wiht a math problem its finding the area of a > triangle wiht the base od 11 in. and the sides of 5 adn 6 inches Assuming you are in trigonometry, I think I'd tackle this by finding an angle. Let's find the smallest angle, which would be between the sides that are 6 and 11 inches. Therefore by the Law of Cosines, 5^2=6^2+11^2-2(6)(11)(cosA)=>25=157-132cosA=>-132=-132cosA=>A=0. Therefore the triangle is a straight line, which means it has no area. David Moran -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: help >can some help me wiht a math problem its finding the area of a >triangle wiht the base od 11 in. and the sides of 5 adn 6 inches The points at which the endpoints of the segments meet are three colinear points; not virtexes (virtices?). That is not a triangle. Notice that the sum of the lengths of two of the sides equal the length of the other side (5 + 6 = 11). A triangle requires three non-colinear points. What is the area? Since the given forms a segment, not a triangle, the area is ZERO. G C -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Word Problem help If a person is changed to it opposite, then the person would have to be hit an odd amount of times. The only people who stay like they were in the first place has been hit an even amount of times. So, the only ones who change are square numbers. For instance, If you take a look at the 5th person, the 1st and 5th person make them change. They were hit twice. If you take a look at a square # such as 9, it will change with the 1st, (changed) 3rd (back to original), and 9th hit.(changed) If the seat # is a perfect square # ( rational square root), then that seat number will be opposite than it was. > Hey..please help me with this word problem: > The auditorium at Centennial High Schoool has one thousand seats. > numbered from 1 to 1000, One day each seat was filled and the 1000 > people followed these directions: > First, each person stood up. > Next, every second person, including the person in seat two sat down. > Then every third person including the person in seat 3, changed to the > opposite. That is, if the person was standing, he or she sat down. If > the person was sitting, he or she stood up. > Following this, every fourth person, including the person in seat 4, > changed to the opposite. Then, every fifth person, including the > person in seat 5, changed to the opposite, and so on. Finally, the one > thousandth person changed to the opposite. After this last change, was > the one thousandth person sitting or was that person standing? > Questions: > 1. Was the person in seat 1 sitting or was that person standing? > 2. For which of the seats 1-20 were people sitting? Standing? > 3. Was the person in seat 1000 sitting or was that person standing? > 4. For which of seats 1-1000 were people standing? -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Word Problem help This was an excellent problem - where did you find it? Also, I think that Glen Prideaux provided exactly the right kind of help toward solving it. -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Word Problem help It all boils down to whether a given number has an odd or even number of distinct factors. Consider, for example, the number 24. Pair off its factors, x and y, where xy = 24, one always is larger than the other: 1, 24 2, 12 3, 8 4, 6 Since 24 has an even number of factors, the person in seat 24 will return to the original seated postion. Most numbers have all their distinct factors pairing off in this way, larger and smaller. The numbers which don't are perfect squares. Consider, for example, 36. 1 36 2 18 3 12 4 9 6 6 Note the repeating 6, the principal square root of 36, which is the essence of being a perfect square. Numbers with an odd number of distinct factors are perfect squares. Those will be left standing. -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: diagonals and vertices Is there a formula or pattern that describes the relationship between the number of vertices of a geometric figure and the number of diagonals for that same figure. A triangle has 3 vertices and 0 diagonals. A square has 4 vertices and 2 diagonals. A pentagon has 5 vertices and 5 diagonals. Any ideas? -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: diagonals and vertices > Is there a formula or pattern that describes the relationship between > the number of vertices of a geometric figure and the number of > diagonals for that same figure. > A triangle has 3 vertices and 0 diagonals. > A square has 4 vertices and 2 diagonals. > A pentagon has 5 vertices and 5 diagonals. > Any ideas? Simple problem in combinatorics: Edges and diagonals are pairs of distinct vertices---count them (this is a simple first exercise in combinatorics). Number of edges=number of vertices---subtract them off. The problem is an easy problem in combinatorics and has little to do with geometry. -- Kevin Karplus karplus@soe.ucsc.edu http://www.soe.ucsc.edu/~karplus life member (LAB, Adventure Cycling, American Youth Hostels) Effective Cycling Instructor #218-ck (lapsed) Professor of Computer Engineering, University of California, Santa Cruz Undergraduate and Graduate Director, Bioinformatics Affiliations for identification only. -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: number combinations I have been trying to figure out all combinations of 9 numbers ive picked.... 2-11-17-20-21-23-37-38-42... These numbers are used in a 6 numberd field from 1-49. I think their is 84 differnt combinations. Can you help? thank you, Carrie -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: number combinations I think I've figured out what you're trying to ask: You're asking how many different 6-element subsets there are in the set of 9 elements that you've chosen, right? If so, the answer in general to the question of how many different k-element subsets there are in set of n elements is n!/(k!(n-k)!) = C(n,k) = C(n,n-k). For n = 9 and k = 6, we end up with (9*8*7)/(3*2) = 84. Paul -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Division Activities Some of you may be interested in an interactive long division module I've been developing. It's not quite finished, but is useable as it is. http://thejuniverse.org/Mathdesign/BNDFT/designnotes.html>design notes : What it is, and why. Dividing Big Numbers is a learning module for the standard long division algorithm. It consists of two interactive tools for exploring and doing long division problems and a website to explain the algorithm, show how to use the tools and provide examples and problems. In essence, the explanation starts with the forgiving method of long division, points out its inefficiencies when using it with big numbers, and then modifies it step by step into the more efficient standard algorithm. The first tool - the LDExplorer - is basically an instantiation of the forgiving method, with a few additional bells and whistles used in the transition to the full method. The second tool - the LDAssistant - is a modification of the first and instantiates the full algorithm. The Dividing Big Numbers website itself is here. This URL is temporary; the final version will have a new URL. Any comments/feedback would be appreciated. -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Division Activities The jimloy website offers some unique ways of solving division problems that may help my struggling students. Two in particular are proficient with their multiplication tables and can do long division with 1 digit divisors. help. > I am a new fifth grade teacher. We are working on long division (ugh!) in > my class and I have several students who are having difficulty with the > concept. What is the best way to teach this concept? Are there any > activities or manipulatives that help the students with this? I have > searched for information on the Internet and in my elementary math ed. > textbooks, but find that the long division algorithm is not explained any > differently from the way I have been attempting to teach it. This is with a > two-digit divisor and a 3-4 digit dividend. Any help you could give would > be greatly appreciated! > Selena > Louisiana > In addition to what Rich recommended, remember that long division is > harder to learn if certain prerequisites are not first mastered. In > this light, here is a link: http://emintsteachers.more.net/lograssc/Math_files/55MathXD/division/divisio nLong.html > Here's another link: > http://www.jimloy.com/arith/longdiv.htm > The concept is simple, in that we're just finding out how many > divisors we can add together without going over the dividend. The > remainder is just that distance between this sum and the dividend. > Long division can be painful with divisors of 2 or more digits because > of all that estimation regarding multi-digit numbers. There is a way > to eliminate most of the painful estimation, but it takes a lot longer > sometimes. The idea to go up powers of 10 of a divisor without going > over the dividend. This might be best reserved for really big > dividends and divisors: > Solve 6,591/29. The problem is to find how many 29's we can add > together without going over 6,591. The remainder is the distance > between this sum and 6,591. > Go up powers of 29, getting as close to 6,591 without going over > 6,591, which is 29*100 = 2,900, then subtract: > 6,591-2,900 = 3,691 > We repeat the procedure, going up powers of 29, this time getting as > close to 3,691 without going over 3,691, which is 29*100 = 2,900, then > subtract: > 3,691-2,900 = 791 > We repeat the procedure, this time on 791 with 29*10 = 290: > 791-290 = 501 > Then on 501 with 29*10 = 290: > 501-290 = 211 > Then we do 211/29 the usual way, obtaining > 211 = 29*7 + (remainder) 8. > We add up all those factors of 29 that we used, 100+100+10+10+7 = 227, > and we have our answer, 227 with remainder 8: > 6,591 = 29*227 + 8. > Like I said, it's longer than the usual way, and the usual way is used > at the end anyway, but there's less of painful estimation involving > multi-digit numbers. (Of course, if 211/29 were the problem to begin > with, the above wouldn't be an option. The concept still would be to > find how many 29's we can add together without going over 211, with > the remainder being the distance between this sum and 211.) > Paul -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Division Activities > The jimloy website offers some unique ways of solving division problems that > may help my struggling students. Two in particular are proficient with > their multiplication tables and can do long division with 1 digit divisors. > help. Your students who are proficient with their multiplication tables and with 1-digit divisors might benefit from seeing the shorthand notation version of long division, in case you haven't seen it already. I love sharing it. It shows how simple long division is with 1 divisor, and I think it's great fun - it seems to give many students a sense of empowerment and confidence with regard to defeating this demon of long division. I recommend that it be taught to all students who can handle it (made available to the whole class but not mandatory for the whole class). It turns long division into something that is very short and sweet when the divisor has only 1 digit. It's the same process as long division written out, but because of its shorthand notation, causes some of the work to be done in the head. I recommend it to as many as can handle it because it develops greater mental computational fluency. It looks like this (and please forgive any typo if it exists!): 432 7)4,932,570 704,652 r 6 If you wish, the line normally connected to the ) symbol and drawn above 4,932,570 can here be connected to the ) symbol and drawn below 4,932,570. Below is what we think as we do the division, from left to right on 4,932,570: 7 goes into 4 0 times. 7 goes into 49 7 times with no remainder, so write 7 below 9. 7 goes into 3 0 times, so write 0 below 3. 7 goes into 32 4 times with remainder 4, so write 4 below 2 and remainder 4 above 5. 7 goes into 45 6 times with remainder 3, so write 6 below 5 and remainder 3 above 7. 7 goes into 37 5 times with remainder 2, so write 5 below 7 and remainder 2 above 0. 7 goes into 20 2 times with remainder 6, so write 2 below 0 and r 6 to the right. My students' jaws dropped when I did this on the board without telling what I was doing at first. They were never taught it, and had great fun learning how simple long division is with 1 divisor. Note: Where 7 goes into the target number 0 times, there is a remainder, and we could put it above the next target number, and where there is no remainder, we could put 0 above the next target number. I chose not to do this for the sake of notational efficiency. But this could be done for pedagogical reasons. Paul -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Division Activities I had seen the short division in their textbooks, having not actually seen it myself before now. I will try to bone up on it and perhaps try it with them. > The jimloy website offers some unique ways of solving division problems that > may help my struggling students. Two in particular are proficient with > their multiplication tables and can do long division with 1 digit divisors. > help. > Your students who are proficient with their multiplication tables and > with 1-digit divisors might benefit from seeing the shorthand > notation version of long division, in case you haven't seen it > already. I love sharing it. It shows how simple long division is > with 1 divisor, and I think it's great fun - it seems to give many > students a sense of empowerment and confidence with regard to > defeating this demon of long division. > I recommend that it be taught to all students who can handle it (made > available to the whole class but not mandatory for the whole class). > It turns long division into something that is very short and sweet > when the divisor has only 1 digit. It's the same process as long > division written out, but because of its shorthand notation, causes > some of the work to be done in the head. I recommend it to as many as > can handle it because it develops greater mental computational > fluency. > It looks like this (and please forgive any typo if it exists!): > 432 > 7)4,932,570 > 704,652 r 6 > If you wish, the line normally connected to the ) symbol and drawn > above 4,932,570 can here be connected to the ) symbol and drawn > below 4,932,570. Below is what we think as we do the division, from > left to right on 4,932,570: > 7 goes into 4 0 times. 7 goes into 49 7 times with no remainder, so > write 7 below 9. 7 goes into 3 0 times, so write 0 below 3. 7 goes > into 32 4 times with remainder 4, so write 4 below 2 and remainder 4 > above 5. 7 goes into 45 6 times with remainder 3, so write 6 below 5 > and remainder 3 above 7. 7 goes into 37 5 times with remainder 2, so > write 5 below 7 and remainder 2 above 0. 7 goes into 20 2 times with > remainder 6, so write 2 below 0 and r 6 to the right. > My students' jaws dropped when I did this on the board without telling > what I was doing at first. They were never taught it, and had great > fun learning how simple long division is with 1 divisor. > Note: Where 7 goes into the target number 0 times, there is a > remainder, and we could put it above the next target number, and where > there is no remainder, we could put 0 above the next target number. I > chose not to do this for the sake of notational efficiency. But this > could be done for pedagogical reasons. > Paul -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Division Activities Your students who are proficient with their multiplication tables and with 1-digit divisors might benefit from seeing the shorthand notation version of long division, in case you haven't seen it already. I love sharing it. It shows how simple long division is with 1 divisor, and I think it's great fun - it seems to give many students a sense of empowerment and confidence with regard to defeating this demon of long division. I don't know how this differs from the way I was taught division, with not only one digit but multiple digit divisors, except that I was taught it the French way. So: 4932570 /7 would look like this: 4932570 /7 0 7 03 70 32 704 45 7046 37 70465 20 704652 6 The only difference between your method and mine (except for the positioning of the divisor) is that all the steps are spelled out, but students sitll have to do the mental math (is there any other way to do it?) It may be impressive not to write out the steps, but the procedure itself does not look different. Is there something I'm missing? -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Division Activities > Your students who are proficient with their multiplication tables and > with 1-digit divisors might benefit from seeing the shorthand > notation version of long division, in case you haven't seen it > already. I love sharing it. It shows how simple long division is > with 1 divisor, and I think it's great fun - it seems to give many > students a sense of empowerment and confidence with regard to > defeating this demon of long division. > I don't know how this differs from the way I was taught division, with > not only one digit but multiple digit divisors, except that I was > taught it the French way. > So: > 4932570 /7 > would look like this: > 4932570 /7 > 0 7 > 03 70 > 32 704 > 45 7046 > 37 70465 > 20 704652 > 6 > The only difference between your method and mine (except for the > positioning of the divisor) is that all the steps are spelled out, but > students sitll have to do the mental math (is there any other way to > do it?) It may be impressive not to write out the steps, but the > procedure itself does not look different. > Is there something I'm missing? Your way (the French way) is essentially what I meant with the into the target number 0 times, there is a remainder, and we could put it above the next target number, and where there is no remainder, we could put 0 above the next target number. I chose not to do this for the sake of notational efficiency. The American way has us write out the computations in the vertical rather than doing it mentally. With 1-digit divisors it's easy to do it mentally. But with 2-digit divisors, . . .? I'd like to see the French way on a 2- or 3-digit divisor with much larger dividend. Paul -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Division Activities The American way has us write out the computations in the vertical rather than doing it mentally. With 1-digit divisors it's easy to do it mentally. But with 2-digit divisors, . . .? I'd like to see the French way on a 2- or 3-digit divisor with much larger dividend. In my close to 4 decades living in the US, I've managed not to learn how to do divisions the American way. Here is how I would divide 4932570 by 72: 4932570 72 612 68507,91 365 0570 660 120 48 (decimal comma being equivalent to decimal point in French notation).The French way involves a lot of mental guessing and mulitplication. Basically, one has to guess at how many times the first digit of the divisor will go into the first two digits of the dividend and go on from there. Is that clearer than the American way? mattsmom -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Division Activities > The American way has us write out the computations in the vertical > rather than doing it mentally. With 1-digit divisors it's easy to do > it mentally. But with 2-digit divisors, . . .? I'd like to see the > French way on a 2- or 3-digit divisor with much larger dividend. > In my close to 4 decades living in the US, I've managed not to learn > how to do divisions the American way. Here is how I would divide > 4932570 by 72: > 4932570 72 > 612 68507,91 > 365 > 0570 > 660 > 120 > 48 > (decimal comma being equivalent to decimal point in French > notation).The French way involves a lot of mental guessing and > mulitplication. Basically, one has to guess at how many times the > first digit of the divisor will go into the first two digits of the > dividend and go on from there. Is that clearer than the American way? > mattsmom column. 4932570 72 612 68507,91 365 0570 660 120 48 Next, to do it the American way, I just write out in the column all the multiplication and subtraction that I'm referring to, which is just inserted between the numbers in your column. I'll keep the divisor and quotient off to the right where you are used to seeing them, and I'll stop when I hit the decimal point: 4932570 72 432 68507,91 612 576 365 360 570 504 66 Obviously, where the inserted numbers come from: 432 = 6*72, 576 = 8*72, 360 = 5*72, 504 = 7*72. The numbers in your column come from the subtraction that's suggested with the insertions. That's the American way. Is the French way such that all this multiplication and subtraction is done entirely in the head, and not written off to the side? Paul -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Division Activities >>In my close to 4 decades living in the US, I've managed not to learn >>how to do divisions the American way. Here is how I would divide >>4932570 by 72: >>4932570 72 >> 612 68507,91 >> 365 >> 0570 >> 660 >> 120 >> 48 >>(decimal comma being equivalent to decimal point in French >>notation).The French way involves a lot of mental guessing and >>mulitplication. Basically, one has to guess at how many times the >>first digit of the divisor will go into the first two digits of the >>dividend and go on from there. Is that clearer than the American way? .... > Next, to do it the American way, I just write out in the column all > the multiplication and subtraction that I'm referring to, which is > just inserted between the numbers in your column. I'll keep the > divisor and quotient off to the right where you are used to seeing > them, and I'll stop when I hit the decimal point: > 4932570 72 > 432 68507,91 > 612 > 576 > 365 > 360 > 570 > 504 > 66 > Obviously, where the inserted numbers come from: 432 = 6*72, 576 = > 8*72, 360 = 5*72, 504 = 7*72. The numbers in your column come from the > subtraction that's suggested with the insertions. That's the American > way. Is the French way such that all this multiplication and > subtraction is done entirely in the head, and not written off to the > side? numbers are arranged a bit differently. I.e. I was taught to write it as: 68507 _________ 72 | 4932570 612 365 570 66 The subtractions that Paul shows explicitly were combined with the multiplication mentally with the aid of writing small 'borrow' numbers where needed. E.g., in the first step 72 is estimated to go into 493 6 times. Then 6 multiplied by 2 is 12 which is subtracted from the '3' in 493. To do this we need to 'borrow' 10 from the ten's place in 493, so we put a little '1' by the 9 as a reminder and then put the result of the subtraction (13-12=1) under the 3 in 493. Next multiply 6 times 7 and add the 'borrowed' 1 to get 43 and subtract that from 49 to get 6 which is put under the 9. Finally drop the 2 down to get a new intermediate dividend of 612 and repeat the process. -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Division Activities Obviously, where the inserted numbers come from: 432 = 6*72, 576 = 8*72, 360 = 5*72, 504 = 7*72. The numbers in your column come from the subtraction that's suggested with the insertions. That's the American way. Is the French way such that all this multiplication and subtraction is done entirely in the head, and not written off to the side? show their work. They can, of course, scribble on scrap paper the intermediate steps. More likely, they do as I do, mutter as they go along. But the kind of work you show would not appear in a French math notebook. My husband does the same as I, but in Chinese. It works beautifully. Between French and Chinese, we can check each other's math without getting tangled in each other's computations. Of course, there's the calculator (wielded by Matt). mattsmom -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Good support in maths I have come across a source providing good support to those experiencing difficulties with doing homework in maths including arithmetic(http://www.bymath.com/studyguide/ari/form1.htm), geometry (http://www.bymath.com/studyguide/geo/pro/pro1/pro1.htm), algebra (http://www.bymath.com/studyguide/alg/pro/pro1/pro1.htm), functions and graphics (http://www.bymath.com/studyguide/fun/pro/pro.htm), principles of analysis(http://www.bymath.com/studyguide/ana/pro/pro1/pro1.htm), all this is also supported by a good deal of examples and illustrations. === Subject: I need a 2-D Triangle mesh generator with source code in FORTRAN by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3CFkaH11719; Hi there, I am dealing with a remeshing problem which needs generate a new mesh each time step. I 've been searching on internet for a while but haven't found a good one. I need the triangle mesh can be very nununiform and the nodes on boundary are inputs. Peter === Subject: Re: I need a 2-D Triangle mesh generator with source code in FORTRAN > Hi there, > I am dealing with a remeshing problem which needs generate a new mesh > each time step. I 've been searching on internet for a while but > haven't found a good one. > I need the triangle mesh can be very nununiform and the nodes on > boundary are inputs. > Peter Triangle http://www-2.cs.cmu.edu/~quake/triangle.html the best general 2D triangular mesh generator out there, and the page says it won the It's in C, but should be able to be coupled with Fortran code. Matt === Subject: Re: I need a 2-D Triangle mesh generator with source code in FORTRAN > I am dealing with a remeshing problem which needs generate a new mesh > each time step. I 've been searching on internet for a while but > haven't found a good one. > I need the triangle mesh can be very nununiform and the nodes on > boundary are inputs. Jonathan Shewchuk's program (called simply Triangle) at http://www-2.cs.cmu.edu/~quake/triangle.html will probably do what you want. The source code is in C, but C and Fortran can be made binary compatible in both directions, with perhaps a little work writing wrapper functions. --- Roy === Subject: Re: I need a 2-D Triangle mesh generator with source code in FORTRAN > Hi there, > I am dealing with a remeshing problem which needs generate a new mesh > each time step. I 've been searching on internet for a while but > haven't found a good one. > I need the triangle mesh can be very nununiform and the nodes on > boundary are inputs. > Peter Although I have never done exactly this, I have done something similar. Probably you can generate the mesh recursively, and that will be fairly fast. If I were doing it, I would subdivide an existing triangle with a line from the center of the longest side to the vertex opposite that -- Julian V. Noble Professor Emeritus of Physics ^^^^^^^^^^^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ For there was never yet philosopher that could endure the toothache patiently. -- Wm. Shakespeare, Much Ado about Nothing. Act v. Sc. 1. === Subject: Re: Vector Crossproduct in polar coordinates >I know you can't just do this by stuffing the vector's elements into the >Cartesian cross-product formula, and I'm also curious if there's a way >to do it that doesn't involve going all the way back to Cartesian >coordinates< >>Exactly!! That is my point. How does one perform the crossproduct of a 3d >>polar coordinate pair of vectors in the polar coordinate system WITHOUT > using >>a conversion to Cartesian coordinates. In reviewing a dozen or so calculus > or >>vector analysis texts and Google, I have found no information >>GC > As GORDO said you do it in exactly the same way as for rectangular > coordinates. Use the determinant method with the top row (er, etheta, ez) as > unit vectors instead of (i, j, k). The second row is (ar, atheta, az) for > vector a and the third row is (br, btheta, bz) for vector b. This gives a X > b where a and b are the components of the vectors in polar coordinates and > the cross product is also in polar coordinates. Ditto for spherical > coordinates. So how is the determinant method of representing a vector not cartesian? If I'm doing the in-head calculations correctly you are getting unit vectors that are exactly parallel to the (cartesian) i, j and k vectors, and could be labeled as such with no loss of generality. Certainly _I_ would say that by the looks like a duck, quacks like a duck rule you're just using cartesian coordinates. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com === Subject: Re: Vector Crossproduct in polar coordinates >I know you can't just do this by stuffing the vector's elements into the >Cartesian cross-product formula, and I'm also curious if there's a way >to do it that doesn't involve going all the way back to Cartesian >coordinates< >Exactly!! That is my point. How does one perform the crossproduct of a 3d >>polar coordinate pair of vectors in the polar coordinate system WITHOUT > using >>a conversion to Cartesian coordinates. In reviewing a dozen or so calculus > or >>vector analysis texts and Google, I have found no information >>GC > As GORDO said you do it in exactly the same way as for rectangular > coordinates. Use the determinant method with the top row (er, etheta, ez) as > unit vectors instead of (i, j, k). The second row is (ar, atheta, az) for > vector a and the third row is (br, btheta, bz) for vector b. This gives a X > b where a and b are the components of the vectors in polar coordinates and > the cross product is also in polar coordinates. Ditto for spherical > coordinates. > So how is the determinant method of representing a vector not cartesian? > If I'm doing the in-head calculations correctly you are getting unit > vectors that are exactly parallel to the (cartesian) i, j and k vectors, > and could be labeled as such with no loss of generality. Certainly _I_ > would say that by the looks like a duck, quacks like a duck rule > you're just using cartesian coordinates. > -- > Tim Wescott > Wescott Design Services > http://www.wescottdesign.com I think you mean the determinant method of representing the cross product. It is still cartesian in polar coordinates but the (er, ethta, ez) unit vectors are not the same as the (i, j, k) unit vectors, in general, except for ez being in the same direction as k of course. Otherwise er = Cos(Theta)*i + Sin(theta)*j eTheta = -Sin(Theta)*i + Cos(Theta)*j === Subject: Re: Vector Crossproduct in polar coordinates >To add to what Gordo says, the easiest way to see why what you propose does >not work is to look at the units. The (r,Theta, z) coordinates of a position >vector are not all in the same units. r and z are in length units while >Theta is in radians (or degrees). The cross product of two position vectors >in these units would give the Theta component of the product as length >squared and the other two components as Length X radians, which is obviously >invalid. >For a position vector the change from x,y,z coordinates to r,Theta,z >coordinates can be looked at as a coordinate rotation about the z-axis so >that r is the vector sum of x and y. The theta component in the rotated >system is in this sense zero. However if you take a different vector and >look at its cylindrical coordinates they define a different local coordinate >system incompatible with the first vector, so you can't just find the cross >product of (ra,0,za) and (rb,0,zb) because the r and theta directions are >different for the two cases. Yes ,I now understand. I mis-understood the previous suggestions. As you can tell, my vector analysis knowledge is somewhat shaky. GC . === Subject: Re: Dealing with variable diffusion coefficients and an advection term > If your actual problem can be expressed as a variable-coefficient > diffusion problem: > df/dt = -dQ/dx , Q = -D*df/dx > Then you should use the conservative finite-difference form, i.e. > dQ/dx = ( Q(i+0.5)-Q(i-0.5) )/dx , Q(i) = -D(i)*(f(i+0.5)-f(i-0.5))/dx > --> df/dt = -dQ/dx = ( D(i+0.5)*(f(i+1)-f(i)) - D(i-0.5)*(f(i)-f(i-1) > )/dx^2 > where the diffusivities at mid-points are evaluated using some > interpolation, i.e. linear: D(i+0.5) = 0.5*(D(i)+D(i+1)) As I noted in my reply to Bill Frensley, I have tried this method but large gradients in D, and a cross term have rendered it useless, f goes negative! I have printed up the reference you gave below. Maybe it will give me some insight or possibly even solve my problems... If you know of a flux conservative method that can handle large cross terms and steep gradients in D I would appreciate any help you can give me. I am looking into Flux corrected transport methods, it appears they were created to handle the problems I am experiencing, but I am a little concerned about some of the seemingly adhoc fixes. The papers I am looking at are by Boris, Book and Hain. They appear to be some of the original papers, if not *the* original. Do you have any experience with FCT methods? If so, have they changed much since the early 70's? Are there any good recent review papers or texts? Shawn > If your problem cannot be expressed as df/dt = d(D*df/dx)/dx, but is > actually an > advection diffusion problem: > df/dt = D*df/dx^2 + V*df/dx > then a good method is given here > http://www.ldeo.columbia.edu/~mspieg/ > check out the pre-print (available in PDF) > A semi-lagrangian crank-nicholson algorithm for the numerical solution > of advection-diffusion problems. > This is a robust and fairly simple method, applicable to a wide range > of > Peclet Numbers (i.e. L*V/D, with L a characteristic length, i.e. > domain size), which is the ratio of diffusion velocity to advective > velocity. It should maintain 2nd order as well. > Hope this helps, > Matt === Subject: Re: Dealing with variable diffusion coefficients and an advection term > It's not quite clear to me whether you also have a real advection > (or drift or streaming) term in the original problem or not. If not, > the simple answer is to use the flux conservative method and pay no > attention whatever to other formulations. In the future I may add a real advection term, but we currently believe that this will only be a small correction and so do not have one for now. A useful heurestic principle is that you should strictly respect any known > integral relations in the problem, if you want an integral quantity to converge > quickly. This is the reason for only considering flux-conservative > discretizations. In a diffusion problem, the integal you are likely to want is > the the total flux through some surface. If your discretization does not obey > the continuity relation (flux conservation), you will obviously have built in > a significant error term, for which the only cure is an absurdly large number of > mesh points. > Another useful heurestic is: No good can ever come from carrying out the > chain-rule differentiation of a Sturm-Liouville differential operator (or the > higher-dimension analogs). This point I would like to agree with, but have been very unsuccessful in trying to implement up until now. There seem to be two problems that are forcing us to expand the derivative, but possibly your method noted below can handle them. Before we expanded the derivatives we had regions where f went negative with time. The solutions suggested to us were, 1) The cross terms, d/dx(Dxy*df/dx) and d/dy(Dxy*df/dy), (which maybe I should have written out in my original posting) required us to expand the derivatives, and then use a special discretization for the mixed derivative d2f/dxdy which depended on the sign of Dxy, and 2) The large gradients in all the diffusion coefficients required us to upwind the advection term. Once we used both of these suggestions we got results that at least remained positive and looked reasonable. However, when we used this method in a similar problem where the cross-diffusion was small and compared the results to those we got when solveing the problem using a simple flux conservative method, the solutions were qualitatively the same, but quantitatively very different. Because of that we started trying to find higher order upwinding schemes and have been looking into possibly using flux corrected transport methods for the advection part. The notion that the eigenvalues of the discretized operator should be confined > to the same region of the complex plane as those of the continuum operator > appears nowhere that I know of in introductory works on numerical analysis. > (I presume that it is hidden in the notion of a consistent discretization.) > In particular, order-of-error analysis does not address this problem at all, > and in fact it frequently directs our attention to schemes which fail this > simple test. > To return to your problem: If, in addition to the position-dependent diffusivity > you also have a real advection term, the optimum way to proceed is to use a > as a continuity equation at each node, so as to get a flux-conserving scheme. > The flux between each pair of neighboring nodes, however, is calculated by > analytically solving the full (steady-state) equation along the line between > the two nodes, making simple approximations for the behavior of the diffusivity > and velocity between the nodes. If you are interested in this, I can give you > more details. scheme is not really intended for my problem, which I think is supposed to be simpler. However, since my problem is already in conservative form I would guess that it would be simple to implement. Possibly if the Scharfetter-Gummel scheme is equipt to handle large advection, then the advection like term in my problem will not be a concern. Will it also easily handle cross diffusion? I should look into this scheme anyway, if we do add a real advection term this method may be useful. Another thought, is there such thing as a flux-corrected method which will allow us to solve the problem in flux conservative form? I am just introducing myself to the subject by reading some papers by Jay P. Boris, David L. Book and K. Hain from the early to mid 70's. Have the methods improved since then and is there any that you might recommend for my problem? Shawn > - Bill Frensley === Subject: Re: Solving extremely large linear systems >>I need some suggestions. I have to solve a linear system which has >>almost two million variables. Obviously I cannot solve by taking the >>inverse of the matrix, because I cannot construct the matrix. As usual >>in all large-scale systems, the matrix to be solved is highly sparse. >>Can you recommend me some (fast and easily implementable:) ) methods >>for solving this problem. Pointers to papers, books, URLs are highly >>appreciated. > A list of Fortran, C, and C++ codes for linear algebra, including > problems with sparse matrices, is at > http://www.netlib.org/utk/people/JackDongarra/la-sw.html . Links to For some reason this webpage does not mention about CLAPACK package. It is the lapack package f2ced into C. You can find it in netlib. Just a random thought... === Subject: solution of polynomials with real coeffs I would like to know if there is a condition using which I can determine if a given polynomial with real coeffs has real roots? For my case, I know that the polynomial order never exceeds 15. Ex:- 2nd order polynomial has real roots if b^2 - 4ac > 0. I am looking something in this line of thought for a general polynomial. Any help is greatly appreciated. raju === Subject: Re: solution of polynomials with real coeffs For a polynomial of degree 5 and above, there is no method to find root as proved by Abel. I don't know if there are any algorithmic method for a general degree polynomial. But if you have a polynomial of odd degree, then at least one real root exists since complex roots comes in pairs. > I would like to know if there is a condition using which I can determine > if a given polynomial with real coeffs has real roots? For my case, I > know that the polynomial order never exceeds 15. > Ex:- 2nd order polynomial has real roots if b^2 - 4ac > 0. I am looking > something in this line of thought for a general polynomial. > Any help is greatly appreciated. > raju