mm-417 === Subject: Re: Math question I know its sounds a simple question but what are chances of picking the first two greyhounds in a six greyhound race, if all start as equal. They can finish in either top place as long as it is first and second. Is it thirty to one or fifteen to one? Please email me the correct answer You want us to do your homework *and* mail you the answers? I don't think so. === === Subject: Formula for Achievement Meet the tutor here: www.vitalwealthsecrets.com 'Preconceived ideas are the robbers of opportunity' === Subject: Mathcad 2001 Professional: submatrix function bug? I created a worksheet as follows in Mathcad 2001 Professional and got unexpected De-Wei Yin i:=0..3 j:=0..3 f(i,j):=i+j A:=matrix(4,4,f) A=(0 1 2 3 1 2 3 4 2 3 4 5 3 4 5 6) Extract a submatrix numerically using =: submatrix(A,0,1,0,1)=(0 1 1 2) Extract the same submatrix symbolically using ->: submatrix(A,0,1,0,1)->(2 3 3 4) The results are different! If I use A[i,j:=i+j to create the matrix, then the two submatrices are the same. === Subject: More Circle problems correct. 1). Show the common chord to the circle x^2 + y^2 -4x -2y -4 = 0 x^2 +y^2 = 4 (passes through the origin) 2). Prove that y =2x is a tangent to the circle x^2 + y^2 - 8x - y + 5 = 0 & state the co-ordinates of the point of contact. 3). Find the length of the tangents from the centre of the circle x^2 + y^2 -3x +4y - 3 = 0 (Leave as surd) === Subject: Re: More Circle problems correct. 1). Show the common chord to the circle x^2 + y^2 -4x -2y -4 = 0 x^2 +y^2 = 4 (passes through the origin) 2). Prove that y =2x is a tangent to the circle x^2 + y^2 - 8x - y + 5 = 0 & state the co-ordinates of the point of contact. 3). Find the length of the tangents from the centre of the circle x^2 + y^2 -3x +4y - 3 = 0 (Leave as surd) These are cool questions, what course are you taking? Here are some hints. #3 has missing information, but 1 and 2 can be done by solving two equations in two unknowns. They are not linear equations, but you can still solve them. #1 has a even faster solution. http://www.math.fsu.edu/~bellenot bellenot math.fsu.edu +1.850.644.7189 (4053fax) === Subject: Re: More Circle problems sorry I meant to say #1 I haven't yet solved but 2 & 3 I have >correct. >1). Show the common chord to the circle >x^2 + y^2 -4x -2y -4 = 0 >x^2 +y^2 = 4 (passes through the origin) >2). Prove that y =2x is a tangent to the circle x^2 + y^2 - 8x - y + 5 = 0 & >state the co-ordinates of the point of contact. >3). Find the length of the tangents from the centre of the circle x^2 + >y^2 -3x +4y - 3 = 0 (Leave as surd) These are cool questions, what course are you taking? Here are some hints. #3 has missing information, but 1 and 2 can be done by solving two equations in two unknowns. They are not linear equations, but you can still solve them. #1 has a even faster solution. -- http://www.math.fsu.edu/~bellenot === Subject: Re: More Circle problems I am taking A2 P3 Maths, 1 & 3 I am happy with, however number 2 is kinda odd. The teacher did tell us that pythagoras is involved. >correct. >1). Show the common chord to the circle >x^2 + y^2 -4x -2y -4 = 0 >x^2 +y^2 = 4 (passes through the origin) >2). Prove that y =2x is a tangent to the circle x^2 + y^2 - 8x - y + 5 = 0 & >state the co-ordinates of the point of contact. >3). Find the length of the tangents from the centre of the circle x^2 + >y^2 -3x +4y - 3 = 0 (Leave as surd) These are cool questions, what course are you taking? Here are some hints. #3 has missing information, but 1 and 2 can be done by solving two equations in two unknowns. They are not linear equations, but you can still solve them. #1 has a even faster solution. -- http://www.math.fsu.edu/~bellenot === Subject: coordinates problem Sorry for asking this question, but I do need help. I attend this class alone. I have no idea how to solve this. could somebody teach me or help me. thanks a lot. ＝＝＝＝＝＝＝＝＝A 309;＝＝＝＝＝＝＝＝＝ & #65309;＝＝＝＝＝＝＝＝ʍ 0 9;＝＝＝＝＝＝＝＝＝&# 6 5309;＝＝＝＝＝ A point p(1, -1, 2) is rotated 30o about an axis that passes through (1, 2, 1) with a direction vector (3, -2, -1). Please calculate the new coordinates of p after the rotation and describe the 3D transformations process in detail. === Subject: Re: coordinates problem Sorry for asking this question, but I do need help. I attend this class alone. I have no idea how to solve this. could somebody teach me or help me. thanks a lot. ＝＝＝＝＝＝＝＝＝A 309;＝&# 65309;＝＝＝＝＝＝＝＝ᦂ 9;＝A 309;＝＝＝＝＝＝＝＝＝ ＝ᦂ 9;＝＝＝＝＝＝＝＝＝&# 65309; A point p(1, -1, 2) is rotated 30o about an axis that passes through (1, 2, 1) with a direction vector (3, -2, -1). Please calculate the new coordinates of p after the rotation and describe the 3D transformations process in detail. 30 degrees in what direction? At least you could properly transcribe your homework. === Subject: Re: coordinates problem and other do my homework questions No need to apologize. Just stop asking homework questions here. (And for the people answering homework questions here, or giving hints: I suggest you not do so. What you may find is that you will encourage off topic posts by people who think oh, here is a newsgroup where nice people answer random questions.) I suggest all such questions be directed to Doctor Math which actually is set up to provide such help... http://mathforum.org/dr.math/ RJF Sorry for asking this question, but I do need help. I attend this class alone. You mean there is no teacher? I have no idea how to solve this. could somebody teach me or help me. thanks a lot. ＝＝＝＝＝＝＝＝＝A 3 09;＝＝＝＝＝＝＝＝＝& # 65309;＝＝＝＝＝＝＝＝ᦂ 9 ;＝＝＝＝＝＝＝＝＝ 5 309;＝＝＝＝＝ A point p(1, -1, 2) is rotated 30o about an axis that passes through (1, 2, 1) with a direction vector (3, -2, -1). Please calculate the new coordinates of p after the rotation and describe the 3D transformations process in detail. === === Subject: Re: algebraic number theory problem (easy?) Problem: Take a natural number X, that is not divisible by 3 and show that X squared - 1 is divisible by 24. Ex. (5sqr - 1)/24=1, (7sqr - 1)/24=2, (11sqr - 1)/24=5 The solution simply refuses to come to me. It helps to look at this is in base-3 arithmetic. Ask yourself how you can represent an integer N _not_ divisible by 3. There are two possible cases. Then show that for each case, N^2 - 1 = 0 (mod 24) ^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ === Subject: Re: algebraic number theory problem (easy?) Take a natural number X, that is not divisible by 3 and show that X squared - 1 is divisible by 24. Ex. (5sqr - 1)/24=1, (7sqr - 1)/24=2, (11sqr - 1)/24=5 The solution simply refuses to come to me. Perhaps the question also states that X is odd. Department of Mathematics http://www.math.ubc.ca/~israel === Subject: Re: algebraic number theory problem (easy?) >Take a natural number X, that is not divisible by 3 and >show that X squared - 1 is divisible by 24. >Ex. (5sqr - 1)/24=1, (7sqr - 1)/24=2, (11sqr - 1)/24=5 >The solution simply refuses to come to me. Perhaps the question also states that X is odd. .... That's not enough either. It doesn't work for X = 3, 9, 15, 21, .... Would David like to post the exact wording of the problem? === Subject: Re: algebraic number theory problem (easy?) >> Take a natural number X, that is not divisible by 3 and >> show that X squared - 1 is divisible by 24. >> Ex. (5sqr - 1)/24=1, (7sqr - 1)/24=2, (11sqr - 1)/24=5 >> The solution simply refuses to come to me. > Perhaps the question also states that X is odd. > .... That's not enough either. It doesn't work for X = 3, 9, 15, 21, .... Would David like to post the exact wording of the problem? X = 3, 9, 15, and 21 satisfy Robert's criterion (X is odd) but fails to satisfy david's criterion (natural number X, that is not divisible by 3). I suspect that the exact wording of the problem is: Given that X is an odd natural number not divisible by 3, prove that X^2-1 is divisible by 24. To the OP, since it's been a few days since this problem was posted and if it is homework, it was probably due before now, a hint: You can classify all odd natural numbers not divisible by 3 into two general categories, one containing numbers like 7, 13, 19, etc. and one containing 5, 11, 17, etc. Given a representation of numbers X from each category in terms of a parameter, what does that tell you about how X^2-1 factors? === Subject: Re: algebraic number theory problem (easy?) >.... It doesn't work for X = 3, 9, 15, 21, .... X = 3, 9, 15, and 21 satisfy Robert's criterion (X is odd) but fails to satisfy david's criterion (natural number X, that is not divisible by 3).... Of course, thank you. I lost sight of that somewhere along the trail of responses. === Subject: formula for roots of quadratic matrix equation? Content-Length: 217 Originator: rusin@vesuvius Is there a closed form formula for the roots of the quadratic matrix equation QAQ - BQ + C = 0 where A, B, C and Q are all symmetric nxn matrices, n>1? If not, how about for the special case B=I (identity matrix)? === Subject: Re: formula for roots of quadratic matrix equation? Content-Length: 695 Originator: rusin@vesuvius Is there a closed form formula for the roots of the quadratic matrix equation QAQ - BQ + C = 0 where A, B, C and Q are all symmetric nxn matrices, n>1? If not, how about for the special case B=I (identity matrix)? Let sqrt(X) be a symmetric matrix satisfying sqrt(X)^t sqrt(X) = X. Then Q = sqrt(A)^{-1} [ sqrt( F^t F - C ) - F ] where F = -sqrt(A)^{-t} [ I + B ] If B=I, we get F = -2 sqrt(A)^{-t} and Q = sqrt(A)^{-1} sqrt( 4 A - C ) + 2 A I got this from matching terms on ( sqrt(A) Q + F )^t ( sqrt(A) Q + F ) = F^t F - C so it may not give all the roots. -Thomas C === Subject: Re: formula for roots of quadratic matrix equation? Content-Length: 1014 Originator: rusin@vesuvius Is there a closed form formula for the roots of the quadratic matrix equation QAQ - BQ + C = 0 where A, B, C and Q are all symmetric nxn matrices, n>1? If not, how about for the special case B=I (identity matrix)? The equation as written is difficult. However, if one considers QAQ' - QB - B'Q' + C = 0, one can generally describe the solutions, especially if A is positive definite, say A = HH'. The equation then becomes (QH - B'H'^{-1})(QH - B'H'^{-1})' = B'A^{-1}B - C = D, where D would have to be positive semidefinite. The set of all decompositions of D into FF' gives the set of all solutions of the revised problem. This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Proof of GOLDBACH CONJECTURE by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id h9BIMXo13735 by legacy.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.10 $, legacy) id h9BIMWF08909 This paper is extension of my first paper and perhaps it gives a very simple proof of GOLDBACH.89S CONJECTURE. According to my first paper .8bSearch for Prime Numbers.8a All prime numbers (except 2 & 3) must satisfy the series 6*n-1 or 6*n+1 where n is a positive integer ranging one to infinity. So the necessary condition for the prime number is that it can either be expressed as 6*n-1 or as 6*n+1 where n is a positive integer ranging one to infinity. 1) Every odd number > 6 is equal to sum of three primes. PROOF: Consider three primes x, y, z. The necessary condition for the prime number is that it can either be expressed as 6*n-1 or as 6*n+1 where n is a positive integer greater than equal to one (n ranging one to infinity). Case 1) say x = 6*a+1, y = 6*b+1, z = 6*c+1 Sum of three prime numbers = x + y + z = 6*a+1 + 6*b+1 + 6*c+1 =6*(a + b + c)+3 Now minimum value of (a + b + c) is three and (a + b + c) assumes all integer value greater than equal to 3 (see note). So case 1) covers odd numbers 21,27,33,39,للللل[Cen terDot]ل.. Case 2) say x = 6*a+1, y = 6*b-1, z = 6*c+1 Or x =6*a+1, y = 6*b+1, z = 6*c-1 Or x = 6*a-1, y = 6*b+1, z = 6*c+1 In above three situations sum of three prime numbers x + y + z can be expressed as Sum of three prime numbers = x + y + z = 6*a+1 + 6*b-1 + 6*c+1 =6*(a + b + c)+1 Now minimum value of (a + b + c) is three and (a + b + c) assumes all integer value greater than equal to 3. So case 2) covers odd numbers 19,25,31,37,للللل[Cen terDot]ل.. Case 3) say x = 6*a+1, y = 6*b-1, z = 6*c-1 Or x =6*a-1, y = 6*b-1, z = 6*c+1 Or x = 6*a-1, y = 6*b+1, z = 6*c-1 In above three situations sum of three prime numbers x + y + z can be expressed as Sum of three prime numbers = x + y + z = 6*a+1 + 6*b-1 + 6*c-1 =6*(a + b + c)-1 Now minimum value of (a + b + c) is three and (a + b + c) assumes all integer value greater than equal to 3. So case 3) covers odd numbers 17,23,29,35,للللل[Cen terDot]ل.. Case 4) say x = 6*a-1, y = 6*b-1, z = 6*c-1 Sum of three prime numbers = x + y + z = 6*a-1 + 6*b-1 + 6*c-1 =6*(a + b + c)-3 Now minimum value of (a + b + c) is three and (a + b + c) assumes all integer value greater than equal to 3. So case 4) covers odd numbers 15,21,27,33,للللل[Cen terDot]ل.. It is clear from above discussion all odd numbers greater than equal to fifteen (15) 15,17,19,21,23,25,27,29,31,33,35,37للل can be expressed as sum of three prime numbers. Odd numbers less than 15 cannot be expressed as sum of three prime numbers covering my series but there are two prime numbers (2,3) those are not included in my series, let us try to express odd numbers less than 15 as sum of three prime numbers (including 2 and 3). 13 = 3+5+5 Or 13 = 7+3+3 11 =5+3+3 Or 11=7+2+2 9=5+2+2 Or 9=3+3+3 7=3+2+2 So it is clear from my work all odd numbers greater than 6 can be expressed as sum of 3 prime numbers. 2) Every even number > 2 is equal to sum of two primes. PROOF: Consider three primes x, y. The necessary condition for the prime number is that it can either be expressed as 6*n-1 or as 6*n+1 where n is a positive integer lies between one to infinity. Case 1) say x = 6*a+1, y = 6*b+1 Sum of two prime numbers = x + y = 6*a+1 + 6*b+1 =6*(a + b)+2 Now minimum value of (a + b) is two and (a + b) assumes all integer value greater than equal to 2. So case 1) covers even numbers 14,20,26,32,للللل[Ce nterDot]ل.. Case 2) say x = 6*a+1, y = 6*b-1 Or x =6*a-1, y = 6*b+1 In above two situations sum of two prime numbers x + y can be expressed as Sum of two prime numbers = x + y = 6*a+1 + 6*b-1 =6*(a + b) Now minimum value of (a + b) is two and (a + b) assumes all integer value greater than equal to 2. So case 2) covers even numbers 12,18,24,30للللل[Cen terDot]ل.. Case 3) say x = 6*a-1, y = 6*b-1 Sum of two prime numbers = x + y = 6*a-1 + 6*b-1 =6*(a + b)-2 Now minimum value of (a + b) is two and (a + b) assumes all integer value greater than equal to 2. So case 3) covers even numbers 10,16,22,28,للللل[Ce nterDot]ل.. It is clear from above discussion all even numbers greater than equal to ten (10) 10,12,14,16,18,20,22,24,26,28,30,32للل can be expressed as sum of two prime numbers. Even numbers less than 10 cannot be expressed as sum of two prime numbers covering my series but there are two prime numbers (2,3) those are not included in my series, let us try to express even numbers less than 10 as sum of two prime numbers (including 2 and 3). 8 = 5+3 6 =3+3 4 =2+2 So it is clear from my work all even numbers greater than 2 can be expressed as sum of 2 prime numbers. NOTE: One important question may comes to mind that the necessary condition for the prime number is that it can either be expressed as 6*n-1 or as 6*n+1 where n is a positive integer ranging one to infinity but it is not sufficient condition. So for all n, 6*n-1 is not prime similar argument can be made for 6*n+1. Then how (a + b + c) assumes all integer value greater than equal to 3. This question can be explained logically. If we concentrate on first few prime numbers expressible in the form 6*n+1 we observe n=1,2,3 (not 4),5,6,7 (not 8,9), 10,11,12,13 (not 14,15),16,للل . According to my prediction (a + b + c) assumes all integer values greater than equal to 3. Now a, b, c assumes specific positive integer values not all positive integer values but still (a +b +c) assumes all integer values greater than equal to 3. Let me explain a assumes specific positive integer values 1,2,3 (not 4),5,6,7 (not 8,9), 10,11,12,13 (not 14,15),16,للل . b assumes specific positive integer values 1,2,3 (not 4),5,6,7 (not 8,9), 10,11,12,13 (not 14,15),16,للل c assumes specific positive integer values 1,2,3 (not 4),5,6,7 (not 8,9), 10,11,12,13 (not 14,15),16,للل. Say we have to obtain (a + b + c) = k (k is any positive integer greater than equal to 3) first off all chose a = 1 so that 6*a+1 is prime number. Our next task is to chose b & c in such way so that 6*b+1 and 6*c+1 are primes and b + c = k-1 (k is any positive integer greater than equal to 3) so b + c has to cover all integer values greater than equal to 2. For b=1,c=1 6*b+1 and 6*c+1 are primes and b + c=2 (here k=3) For b=1,c=2 6*b+1 and 6*c+1 are primes and b + c=3 (here k=4) For b=1,c=3 or b=2,c=2 6*b+1 and 6*c+1 are primes and b + c=4 (here k=5) For b=2,c=3 (b=1,c=4 not possible) 6*b+1 and 6*c+1 are primes and b + c=5 (here k=6) For b=1,c=5 or b=3,c=3 6*b+1 and 6*c+1 are primes and b + c=6 (here k=7) For smaller values of k, a and b are obtained in such a ways that 6*a+1 and 6*b+1 are primes and a + b = k-1. As we go for larger values of k in several ways a, b can be chosen such that a + b = k-1, at least one of these combinations should generate 6*a+1 and 6*b+1 prime. So for large combinations of a, b probability of getting a ,b such that 6*a+1 and 6*b+1are primes is more. Although there is no scope of violation of above requirement still consider a value of k for which it is not possible to find a ,b so that 6*a+1 and 6*b+1 are primes and a + b = k-1. Then we can change the value of a to 2 and try to get a ,b such that 6*a+1 and 6*b+1 are primes and a + b =k-2. If this requirement does not meet we could go for a=3,5,6, 7,9,10,11,12,13,16,لل. Violation of all these possibilities is absurd. === Subject: Re: Proof of GOLDBACH CONJECTURE This paper is extension of my first paper and perhaps it gives a very simple proof of GOLDBACH'S CONJECTURE. According to my first paper Search for Prime Numbers All prime numbers (except 2 & 3) must satisfy the series 6*n-1 or 6*n+1 where n is a positive integer ranging one to infinity. So the necessary condition for the prime number is that it can either be expressed as 6*n-1 or as 6*n+1 where n is a positive integer ranging one to infinity. Necessary, yes. Sufficient, no. 6*4+1 = 25 = 5*5 You can't assume that just because all prime numbers greater than 3 must be in the form 6*n+1 or 6*n-1 with positive integer n that all numbers in one of those forms are prime, which is what you're doing. *snip remainder of argument* === Subject: Re: Proof of GOLDBACH CONJECTURE [Steven Lord] | > This paper is extension of my first paper and perhaps it gives a very | > simple proof of GOLDBACH'S CONJECTURE. According to my first paper | Search for Prime Numbers All prime numbers (except 2 & 3) must | > satisfy the series 6*n-1 or 6*n+1 where n is a positive integer | > ranging one to infinity. So the necessary condition for the prime | > number is that it can either be expressed as 6*n-1 or as 6*n+1 where n | > is a positive integer ranging one to infinity. | | Necessary, yes. Sufficient, no. | | 6*4+1 = 25 = 5*5 | | You can't assume that just because all prime numbers greater than 3 must be | in the form 6*n+1 or 6*n-1 with positive integer n that all numbers in one | of those forms are prime, which is what you're doing. | | *snip remainder of argument* You did not read it through, did you? The post is long, and this question is actually adressed. Still wrong, of course. SA === Subject: Proof of GOLDBACH CONJECTURE by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id h9BIN7o13755 by legacy.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.10 $, legacy) id h9BIN7n08948 This paper is extension of my first paper and perhaps it gives a very simple proof of GOLDBACH.89S CONJECTURE. According to my first paper .8bSearch for Prime Numbers.8a All prime numbers (except 2 & 3) must satisfy the series 6*n-1 or 6*n+1 where n is a positive integer ranging one to infinity. So the necessary condition for the prime number is that it can either be expressed as 6*n-1 or as 6*n+1 where n is a positive integer ranging one to infinity. 1) Every odd number > 6 is equal to sum of three primes. PROOF: Consider three primes x, y, z. The necessary condition for the prime number is that it can either be expressed as 6*n-1 or as 6*n+1 where n is a positive integer greater than equal to one (n ranging one to infinity). Case 1) say x = 6*a+1, y = 6*b+1, z = 6*c+1 Sum of three prime numbers = x + y + z = 6*a+1 + 6*b+1 + 6*c+1 =6*(a + b + c)+3 Now minimum value of (a + b + c) is three and (a + b + c) assumes all integer value greater than equal to 3 (see note). So case 1) covers odd numbers 21,27,33,39,للللل[Cen terDot]ل.. Case 2) say x = 6*a+1, y = 6*b-1, z = 6*c+1 Or x =6*a+1, y = 6*b+1, z = 6*c-1 Or x = 6*a-1, y = 6*b+1, z = 6*c+1 In above three situations sum of three prime numbers x + y + z can be expressed as Sum of three prime numbers = x + y + z = 6*a+1 + 6*b-1 + 6*c+1 =6*(a + b + c)+1 Now minimum value of (a + b + c) is three and (a + b + c) assumes all integer value greater than equal to 3. So case 2) covers odd numbers 19,25,31,37,للللل[Cen terDot]ل.. Case 3) say x = 6*a+1, y = 6*b-1, z = 6*c-1 Or x =6*a-1, y = 6*b-1, z = 6*c+1 Or x = 6*a-1, y = 6*b+1, z = 6*c-1 In above three situations sum of three prime numbers x + y + z can be expressed as Sum of three prime numbers = x + y + z = 6*a+1 + 6*b-1 + 6*c-1 =6*(a + b + c)-1 Now minimum value of (a + b + c) is three and (a + b + c) assumes all integer value greater than equal to 3. So case 3) covers odd numbers 17,23,29,35,للللل[Cen terDot]ل.. Case 4) say x = 6*a-1, y = 6*b-1, z = 6*c-1 Sum of three prime numbers = x + y + z = 6*a-1 + 6*b-1 + 6*c-1 =6*(a + b + c)-3 Now minimum value of (a + b + c) is three and (a + b + c) assumes all integer value greater than equal to 3. So case 4) covers odd numbers 15,21,27,33,للللل[Cen terDot]ل.. It is clear from above discussion all odd numbers greater than equal to fifteen (15) 15,17,19,21,23,25,27,29,31,33,35,37للل can be expressed as sum of three prime numbers. Odd numbers less than 15 cannot be expressed as sum of three prime numbers covering my series but there are two prime numbers (2,3) those are not included in my series, let us try to express odd numbers less than 15 as sum of three prime numbers (including 2 and 3). 13 = 3+5+5 Or 13 = 7+3+3 11 =5+3+3 Or 11=7+2+2 9=5+2+2 Or 9=3+3+3 7=3+2+2 So it is clear from my work all odd numbers greater than 6 can be expressed as sum of 3 prime numbers. 2) Every even number > 2 is equal to sum of two primes. PROOF: Consider three primes x, y. The necessary condition for the prime number is that it can either be expressed as 6*n-1 or as 6*n+1 where n is a positive integer lies between one to infinity. Case 1) say x = 6*a+1, y = 6*b+1 Sum of two prime numbers = x + y = 6*a+1 + 6*b+1 =6*(a + b)+2 Now minimum value of (a + b) is two and (a + b) assumes all integer value greater than equal to 2. So case 1) covers even numbers 14,20,26,32,للللل[Ce nterDot]ل.. Case 2) say x = 6*a+1, y = 6*b-1 Or x =6*a-1, y = 6*b+1 In above two situations sum of two prime numbers x + y can be expressed as Sum of two prime numbers = x + y = 6*a+1 + 6*b-1 =6*(a + b) Now minimum value of (a + b) is two and (a + b) assumes all integer value greater than equal to 2. So case 2) covers even numbers 12,18,24,30للللل[Cen terDot]ل.. Case 3) say x = 6*a-1, y = 6*b-1 Sum of two prime numbers = x + y = 6*a-1 + 6*b-1 =6*(a + b)-2 Now minimum value of (a + b) is two and (a + b) assumes all integer value greater than equal to 2. So case 3) covers even numbers 10,16,22,28,للللل[Ce nterDot]ل.. It is clear from above discussion all even numbers greater than equal to ten (10) 10,12,14,16,18,20,22,24,26,28,30,32للل can be expressed as sum of two prime numbers. Even numbers less than 10 cannot be expressed as sum of two prime numbers covering my series but there are two prime numbers (2,3) those are not included in my series, let us try to express even numbers less than 10 as sum of two prime numbers (including 2 and 3). 8 = 5+3 6 =3+3 4 =2+2 So it is clear from my work all even numbers greater than 2 can be expressed as sum of 2 prime numbers. NOTE: One important question may comes to mind that the necessary condition for the prime number is that it can either be expressed as 6*n-1 or as 6*n+1 where n is a positive integer ranging one to infinity but it is not sufficient condition. So for all n, 6*n-1 is not prime similar argument can be made for 6*n+1. Then how (a + b + c) assumes all integer value greater than equal to 3. This question can be explained logically. If we concentrate on first few prime numbers expressible in the form 6*n+1 we observe n=1,2,3 (not 4),5,6,7 (not 8,9), 10,11,12,13 (not 14,15),16,للل . According to my prediction (a + b + c) assumes all integer values greater than equal to 3. Now a, b, c assumes specific positive integer values not all positive integer values but still (a +b +c) assumes all integer values greater than equal to 3. Let me explain a assumes specific positive integer values 1,2,3 (not 4),5,6,7 (not 8,9), 10,11,12,13 (not 14,15),16,للل . b assumes specific positive integer values 1,2,3 (not 4),5,6,7 (not 8,9), 10,11,12,13 (not 14,15),16,للل c assumes specific positive integer values 1,2,3 (not 4),5,6,7 (not 8,9), 10,11,12,13 (not 14,15),16,للل. Say we have to obtain (a + b + c) = k (k is any positive integer greater than equal to 3) first off all chose a = 1 so that 6*a+1 is prime number. Our next task is to chose b & c in such way so that 6*b+1 and 6*c+1 are primes and b + c = k-1 (k is any positive integer greater than equal to 3) so b + c has to cover all integer values greater than equal to 2. For b=1,c=1 6*b+1 and 6*c+1 are primes and b + c=2 (here k=3) For b=1,c=2 6*b+1 and 6*c+1 are primes and b + c=3 (here k=4) For b=1,c=3 or b=2,c=2 6*b+1 and 6*c+1 are primes and b + c=4 (here k=5) For b=2,c=3 (b=1,c=4 not possible) 6*b+1 and 6*c+1 are primes and b + c=5 (here k=6) For b=1,c=5 or b=3,c=3 6*b+1 and 6*c+1 are primes and b + c=6 (here k=7) For smaller values of k, a and b are obtained in such a ways that 6*a+1 and 6*b+1 are primes and a + b = k-1. As we go for larger values of k in several ways a, b can be chosen such that a + b = k-1, at least one of these combinations should generate 6*a+1 and 6*b+1 prime. So for large combinations of a, b probability of getting a ,b such that 6*a+1 and 6*b+1are primes is more. Although there is no scope of violation of above requirement still consider a value of k for which it is not possible to find a ,b so that 6*a+1 and 6*b+1 are primes and a + b = k-1. Then we can change the value of a to 2 and try to get a ,b such that 6*a+1 and 6*b+1 are primes and a + b =k-2. If this requirement does not meet we could go for a=3,5,6, 7,9,10,11,12,13,16,لل. Violation of all these possibilities is absurd. === Subject: Maple editors Does anyone know of a decent way to edit Maple code? I've been using the Maple GUI to edit my programs up until this point, and I can't stand it any longer. Has anyone else had problems with it? Is there an option somewhere in the preferences that orients the environment towards more experienced coders? Does anyone know of an emacs module for Maple code? Some of the annoyances I've encountered are: tab doesn't actually indent things, it skips you to the next block of code hitting enter executes the block of code, I have to hit shift-enter to push things down a line etc. Please let me know if you have any suggestions. - Chris === Subject: Re: Maple editors [Chris Bebbington] | Does anyone know of a decent way to edit Maple code? I've been using the | ... | | Please let me know if you have any suggestions. there are several maple-modes around. Emacs is the most powerful editor; I use it for Maple code, for running Maple, editing and compiling TeX documents, as well as for usenet and e-mail client and file manager. And as diary, calendar, database, spreadsheet, web browser, html editor, ... hth, SA === Subject: Re: Maple editors there are several maple-modes around. Emacs is the most powerful editor; I use it for Maple code, for running Maple, editing and compiling TeX documents, as well as for usenet and e-mail client and file manager. And as diary, calendar, database, spreadsheet, web browser, html editor, ... -- Yes, I am familiar with emacs, but I've had no luck finding a Maple module that i could get to work. Do you knwo where I could find one? - Chris Bebbington === Subject: Re: Maple editors Yes, I am familiar with emacs, but I've had no luck finding a Maple module that i could get to work. Do you knwo where I could find one? http://www.k-online.com/~joer/ === Subject: Re: Maple editors ... Some of the annoyances I've encountered are: tab doesn't actually indent things, it skips you to the next block of code hitting enter executes the block of code, I have to hit shift-enter to push things down a line etc. Sounds like the usual behaviour to which one has to become used to :-) For this i found Mupad better for handling before i gave it up. More ugly i imagine MMA: to input functions i would have to press + 8 or 9 on my (german) keyboard to get '[' or ']', brrr. Otherwise stated: the strength (and weakness) of sytems is beyond user interfaces. === Subject: Re: Maple editors Set Mathematica's default input format to 'TraditionalForm' you can then use round brackets () instead of square []. Sounds like the usual behaviour to which one has to become used to :-) For this i found Mupad better for handling before i gave it up. More ugly i imagine MMA: to input functions i would have to press + 8 or 9 on my (german) keyboard to get '[' or ']', brrr. Otherwise stated: the strength (and weakness) of sytems is beyond user interfaces. === Subject: Re: Maple editors hitting enter executes the block of code, I have the opposite problem on Mac, enter doesn't do anything, I have to use return instead. Despite the documentation which always refers to enter. === Subject: Maple error - illegal use of a formal parameter I'm trying to run this piece of Maple code, but it crashes when I try to modify a list that was passed as a parameter. Here is the code: Utils := module() export getMonomials: local numMonos, getMonomials_recur, zeroList: getMonomials := proc(S::integer, d::integer) local i, current, monos: i := d: numMonos := 0: if d = 0 then monos[1] := copy(zeroList(S)): numMonos := 1: return monos: end if: current := zeroList(S): while i > 0 do current[1]:= i: getMonomials_recur(S, d-i, current, 2, monos): i:=i-1: end do: return monos: end: getMonomials_recur := proc(S, d, current, pos, ARR) local max, i: if pos > S then if d > 0 then return: end if: ARR[numMonos+1]:=copy(current): numMonos:=numMonos + 1: return: end if: max:=min(d, current[pos-1]): if max=0 then ARR[numMonos+1]:=copy(current): numMonos:=numMonos + 1: return: end if: for i from 1 to max do current[pos]:=i: getMonomials_recur(S, d-i, current, pos + 1, ARR): end do: current[pos]:=0: end: zeroList:=proc(S) return [seq(0,i=1..S)]: end: end: Is there any way to get around this? I used to have it set up with zeroList defined as: zeroArray:=proc(S) local i, ARR: ARR:=array(1..S): for i from 1 to S do ARR[i]:= 0; end do: return ARR: end: But, then I it's return value us an array, not a list, and I am unable to get the size of the array using: T := zeroArray(): size := nops(T): which is a problem for me. Any suggestions? === Subject: Re: Maple error - illegal use of a formal parameter ... Any suggestions? The error occurs because the procedure getMonomials_recur is assigning a value to 'current' which is not a formal parameter (it is assigned to a list). It can be corrected by removing 'current' from the list of parameters (in 3 places) and moving this procedure inside the getMonomials procedure, so that the module started as follows: Utils := module() export getMonomials: local zeroList: getMonomials := proc(S::integer, d::integer) local i, current, monos, numMonos, getMonomials_recur: getMonomials_recur := proc(S, d, pos, ARR) Also, since 'monos' is a table, 'return monos' should be replaced by return eval(monos) (in 2 places). After that, the module works as follows: Utils:-getMonomials(3,5); table([1 = [5, 0, 0], 2 = [4, 1, 0], 3 = [3, 1, 1], 4 = [3, 2, 0], 5 = [2, 2, 1] ]) It looks as if it contains the partitions of d of length <=S, so it can be replaced with the following Maple procedure: gm:=(S,d)->select(p->(nops(p)<=S),combinat[partition](d)): gm(3,5); [[1, 2, 2], [1, 1, 3], [2, 3], [1, 4], [5]] Zeroes can be added and lists can be reversed if necessary. I think that comp.soft-sys.math.maple would be a better place for questions like that. Mihailovs http://webpages.shepherd.edu/amihailo/ === Subject: Brownian Motion Problem I have done the following numerical calculation. For a one-dimensional Brownian motion with steps x(i) drawn from a normal ditribution of mean 0 and unit variance, the displacements d(i) from zero are computed over a time interval from t = 0 to 1 with n time steps (n>>1). For some chosen value of a, I then calculate the total amount of time T (<=1) for which d(i) a. My question is, if this procedure is repeated many times, what is the distribution function (PDF) of T. For a = 0, I obtained P(T) = 2(p(T) + p(1-T)) where p(T) = 1/ (2 Pi*SQRT(T)*(1+T) ) and it does agree with the numerical results. However the general case for non-zero a has me stumped. I would greatly appreciate any suggestions to help solve this problem. === Subject: Re: Brownian Motion Problem I have done the following numerical calculation. For a one-dimensional Brownian motion with steps x(i) drawn from a normal ditribution of mean 0 and unit variance, the displacements d(i) from zero are computed over a time interval from t = 0 to 1 with n time steps (n>>1). For some chosen value of a, I then calculate the total amount of time T (<=1) for which d(i) > a. My question is, if this procedure is repeated many times, what is the distribution function (PDF) of T. For a = 0, I obtained P(T) = 2(p(T) + p(1-T)) where p(T) = 1/ (2 Pi*SQRT(T)*(1+T) ) and it does agree with the numerical results. However the general case for non-zero a has me stumped. I would greatly appreciate any suggestions to help solve this problem. Thank you No need! I've found it myself. === Subject: sign function === Subject: Re: sign function Please do not post the same question _separately_ to different newsgroups. If you feel that you _must_ post the same question to more than one group, crossposting is much better. I already responded to your question in k12.ed.math . Interested readers may see the response there, once its moderator posts my message. David Cantrell === Subject: MuPAD: Vector (or matrix) + scalar addition don't work?! Hi! I just want to add a scalar to a vector (or matrix) in MuPAD... but it fails. How may I do it?? Could someone help me? This is what I did: x:=matrix([1,2,3]) +- -+ | 1 | | | | 2 | | | | 3 | +- -+ x+1 FAIL this don't work too: matrix([1,2,3])+1 FAIL === Subject: Re: MuPAD: Vector (or matrix) + scalar addition don't work?! I just want to add a scalar to a vector (or matrix) in MuPAD... but it fails. How may I do it?? Could someone help me? Works for me: > matrix([[1,2,3], [4,5,6], [7,8,9]]) + 10 +- -+ | 11, 2, 3 | | | | 4, 15, 6 | | | | 7, 8, 19 | +- -+ matrix([1,2,3])+1 FAIL What should the result be, mathematically speaking? Maybe you mean > map(matrix([1,2,3]), `+`, 1) +- -+ | 2 | | | | 3 | | | | 4 | +- -+ === Subject: Re: MuPAD: Vector (or matrix) + scalar addition don't work?! > I just want to add a scalar to a vector (or matrix) in MuPAD... but it > fails. How may I do it?? Could someone help me? Works for me: >> matrix([[1,2,3], [4,5,6], [7,8,9]]) + 10 +- -+ | 11, 2, 3 | | | | 4, 15, 6 | | | | 7, 8, 19 | +- -+ sorry here you are wrong, in MuPAD 2.5.x this will not work. For me it works too, but we are using the developer version of MuPAD :-) === Subject: Re: MuPAD: Vector (or matrix) + scalar addition don't work?! I just want to add a scalar to a vector (or matrix) in MuPAD... but it fails. How may I do it? Excactly the same way you'd do it with pencil & paper: Not at all. What result would you expect? Andre' === Subject: Re: MuPAD: Vector (or matrix) + scalar addition don't work?! I just want to add a scalar to a vector (or matrix) in MuPAD... but it > fails. How may I do it? Excactly the same way you'd do it with pencil & paper: Not at all. What result would you expect? With: x:=matrix([1,2,3]) I expect to have a matrix with [2,3,4] (1 added to each element). This works great with GNU Octave, and Maxima (for example). How may I do it easily with MuPAD? === Subject: Re: MuPAD: Vector (or matrix) + scalar addition don't work?! I expect to have a matrix with [2,3,4] (1 added to each element). Let me prepare you for the next caveat: exp(matrix(...)) is *not* matrix(exp(...),..., exp(...)). MuPAD sticks to mathematical notation here, which seems to differ from that of many engineers. This works great with GNU Octave, and Maxima (for example). How may I do it easily with MuPAD? As I said: map(matrix([1,2,3]), `+`, 1). === Subject: Re: MuPAD: Vector (or matrix) + scalar addition don't work?! With: x:=matrix([1,2,3]) I expect to have a matrix with [2,3,4] (1 added to each element). This works great with GNU Octave, and Maxima (for example). How may I do it easily with MuPAD? Christopher gave you the answer (using map).=== Subject: Implementation of a symbolic computation system In 1998 there was a thread with the same subject. I assume that perspective might have changed during last 5 years. Now we have siteseer, etc. Any updates to the subject, new excellent textbooks, must read reviews etc? === Subject: Maple question - simplify with orthogonal polynomials When I type this in Maple (V9.01) int(sum((2*l+1)/2*f[l]*LegendreP(l,mu),l=0..10)*LegendreP(2,mu),mu=-1..1); I get as a result f[2] which I would expect. However, when I type this int(sum((2*l+1)/2*f[l]*LegendreP(l,mu),l=0..N)*LegendreP(2,mu),mu=-1..1); (changed the upper limit of the sum from 10 to a symbolic N) the simplifcation doesn't work. How can I force Maple to carry out the simplification? gert === Subject: Re: Maple question - simplify with orthogonal polynomials If N=0 or N=1, the answer is not f[2], is it? With a symbolic limit, you would want the answer if N>=2 then f[2] else 0 or similar. This is more of a user-recognized simplification, not one that is generally pre-programmed into a CAS. When I type this in Maple (V9.01) int(sum((2*l+1)/2*f[l]*LegendreP(l,mu),l=0..10)*LegendreP(2,mu),mu=-1..1); I get as a result f[2] which I would expect. However, when I type this int(sum((2*l+1)/2*f[l]*LegendreP(l,mu),l=0..N)*LegendreP(2,mu),mu=-1..1); (changed the upper limit of the sum from 10 to a symbolic N) the simplifcation doesn't work. How can I force Maple to carry out the simplification? gert === Subject: Algorithms to solve equations involving radicals Are there methods to solve equations such as: sqrt(2*x + y) + sqrt(x + y) = 3 or sqrt(x+1)+sqrt(x+2)+sqrt(x+3)=3 That is to eliminate the radicals, to convert this to a polynomial expression? === Subject: Re: Algorithms to solve equations involving radicals |>Are there methods to solve equations such as: |>sqrt(2*x + y) + sqrt(x + y) = 3 |>or |>sqrt(x+1)+sqrt(x+2)+sqrt(x+3)=3 |>That is to eliminate the radicals, to convert this to a polynomial |>expression? Solve for x? In Maple: solve(sqrt(2*x+y) + sqrt(x+y)=3, x); 1/2 1/2 27 + 6 (18 + y) , 27 - 6 (18 + y) Of course the first answer can't be right, at least if y > 0. It's actually wrong for all real y. The second answer is right if -9 <= y <= 18, I think. It shows you have to be careful about branches. The resultant of the polynomials z^2 - (2*x+y) and (3-z)^2 - (x+y) with respect to z is x^2 - 54 x + 81 - 36 y. This being 0 is necessary for a solution, but not sufficient. solve(sqrt(x+1)+sqrt(x+2)+sqrt(x+3)=3,x); 2 4 3 2 -1 + RootOf(-12 _Z + 72 _Z - 28 + 3 _Z - 12 _Z , 0.4319724644) With an equation in a single variable, Maple seems to be smart enough to check the roots of the polynomial and select the ones that work. Department of Mathematics http://www.math.ubc.ca/~israel === Subject: Re: Algorithms to solve equations involving radicals Are there methods to solve equations such as: sqrt(2*x + y) + sqrt(x + y) = 3 or sqrt(x+1)+sqrt(x+2)+sqrt(x+3)=3 That is to eliminate the radicals, to convert this to a polynomial expression? The second example is ugly: you'll get a sixth or eighth order polynomial even if you can eliminate all the radicals. These you'll have to do numerically anyway, so use an equation solver. Mine gives one root in the range 0 < x < 2 , whose value is .186600209943 ^^^^^^^^^^^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ === Subject: Re: Algorithms to solve equations involving radicals One must validate and verify the solutions at each stage to ensure that spurious roots are not introduced. The Grobner basis method is suited to that task for multivariate polynomials. In the presence of radicals, you will have to oversee this by checking the results. You must also interpret the branch of the radicals if required. In your second example, there is a root near x=-.813. Spurious roots occur when you interpret the sqrt with a negative value in one of the terms, which disappears when you square and expand. You will also have to deal with products of radicals, and eliminate them, in the second example, from an equation and its square: sqrt(a)*sqrt(b)=C*sqrt(a)+D*sqrt(b)+E and the square of this now involves sqrt(a)*sqrt(b), which must be replaced with the rhs from the above form. You must then verify each solution of the resulting 8th order equation in x (it factors into 2 quartics) and reject the invalid ones. The quartic's general solution in radicals leads to numerical results that may not cancel the coefficient of sqrt(-1) exactly, due to round-off; I obtained answers of the form x=-.813+%I*10^-11. Interpolation to polish the real root of the quartic responsible for that solution gives the best final answer. There is a lot of effort and thinking in this, I am not certain that you can clean it up to make an algorithm that always gets the right answer. Numerically unstable cases will always present a significant challenge. Are there methods to solve equations such as: sqrt(2*x + y) + sqrt(x + y) = 3 or sqrt(x+1)+sqrt(x+2)+sqrt(x+3)=3 That is to eliminate the radicals, to convert this to a polynomial expression? === Subject: Re: Algorithms to solve equations involving radicals I have come up with the following also: for the first, for example assume t, u > 0 and substitute 2*x + y=t^2 ,x + y=u^2 into the equation. This gives a third equation, u + t = 3. as a solution. This is making the assumption that these equations have real roots, however. One must validate and verify the solutions at each stage to ensure that spurious roots are not introduced. The Grobner basis method is suited to that task for multivariate polynomials. In the presence of radicals, you will have to oversee this by checking the results. You must also interpret the branch of the radicals if required. In your second example, there is a root near x=-.813. Spurious roots occur when you interpret the sqrt with a negative value in one of the terms, which disappears when you square and expand. You will also have to deal with products of radicals, and eliminate them, in the second example, from an equation and its square: sqrt(a)*sqrt(b)=C*sqrt(a)+D*sqrt(b)+E and the square of this now involves sqrt(a)*sqrt(b), which must be replaced with the rhs from the above form. You must then verify each solution of the resulting 8th order equation in x (it factors into 2 quartics) and reject the invalid ones. The quartic's general solution in radicals leads to numerical results that may not cancel the coefficient of sqrt(-1) exactly, due to round-off; I obtained answers of the form x=-.813+%I*10^-11. Interpolation to polish the real root of the quartic responsible for that solution gives the best final answer. There is a lot of effort and thinking in this, I am not certain that you can clean it up to make an algorithm that always gets the right answer. Numerically unstable cases will always present a significant challenge. > Are there methods to solve equations such as: > sqrt(2*x + y) + sqrt(x + y) = 3 > or > sqrt(x+1)+sqrt(x+2)+sqrt(x+3)=3 > That is to eliminate the radicals, to convert this to a polynomial > expression? > === Subject: Riemann's Hypothesis by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id h9CDQJo22136 by legacy.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.10 $, legacy) id h9CDQJd28873 People claimed that they have prooved or disprooved Riemann's Hypothesis but usually their is always a mathematical error that drops the whole, or people like Berliochi come up with something, but we don't know why it should be a proof or a disproof. New ways to prove the problem are being created frequently (numerical verification and spectral interpretation seem to be the best) but personally I think that we are still far of from a proof. However everyday zeros are being calculated to verify RH at a certain range, like zetagrid, an open project where anybody who has a computer can participate. They have already verified this problem for more than 500 ion zeros and their's also a 1 million dollar prize if zetagrid manages to falsify Riemann's Hypothesis (Indeed, RH can only be falsified using numerival methods) and a lot of other prizes, if interested you can go to www.zetgrid.net === Subject: a new and natural lifestyle by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id h9E49to17519 by legacy.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.10 $, legacy) id h9E49s924694 Nudism,what is this?Have you ever know it's real meaning?It has no relation to porn and eroticism.It is a natural and comfortable lifestyle ,it makes me so free.There are around 1m people enjoy this! want to look what they say or make friends with them? http://www.nudistfriendfinder.com/i/1 have a look! === Subject: Re: a new and natural lifestyle There are around 1m people enjoy this! It seems they also enjoy newsgroup spamming! === Subject: spss v12.0 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id h9EAUPo10252 by legacy.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.10 $, legacy) id h9EAUPI32658 I want information... === === Subject: Running around in circles Find the equation of the circles that have the x axis as a tangent, a radius of 5 and they pass through the point (0,8) === Subject: Re: Running around in circles Find the equation of the circles that have the x axis as a tangent, a radius of 5 and they pass through the point (0,8) Hint: what's the y coordinate of the centre? Department of Mathematics http://www.math.ubc.ca/~israel === Subject: Re: Running around in circles Thats all thats in the question, my teacher says that it is quite simple after u have drawn it but I can't seem to grasp the connections. Also it's two circles. >Find the equation of the circles that have the x axis as a tangent, a radius >of 5 and they pass through the point (0,8) Hint: what's the y coordinate of the centre? Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2