mm-432
===
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
answerYou want us to do your homework *and* mail youthe
answers? I don't think so.
===
Subject: Formula for
AchievementMeet 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 unexpectedDe-Wei Yini:=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 problemscorrect.1). Show the common chord to the
circlex^2 + y^2 -4x -2y -4 = 0x^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 problemscorrect.1). Show the common chord to
the circlex^2 + y^2 -4x -2y -4 = 0x^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
somehints. #3 has missing information, but 1 and 2 can be done
bysolving 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/~bellenotbellenot
math.fsu.edu +1.850.644.7189 (4053fax)
===
Subject: Re: More
Circle problemssorry 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 problemsI am taking A2 P3 Maths, 1 & 3 I am happy
with, however number 2 is kindaodd. 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
problemSorry for asking this question, but I do need help. I
attend thisclass alone.I have no idea how to solve this. could
somebody teach me or help me. thanks a
lot.＝＝＝＝＝＝＝&#
65309;＝＝＝＝＝＝＝
＝＝＝＝＝＝＝&#
65309;＝＝＝＝＝＝＝
＝＝＝＝＝＝＝&#
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 therotation and
describe the 3Dtransformations 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.＝＝＝＝＝＝＝&#
65309;＝＝＝＝＝＝＝
＝＝＝＝＝＝＝&#
65309;＝＝＝＝＝＝＝
＝＝＝＝＝＝＝&#
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 properlytranscribe your
homework.
===
Subject: Re: coordinates problem and other do my
homework questionsNo need to apologize. Just stop asking
homework questionshere.(And for the people answering homework
questions here, orgiving hints: I suggest you not do so. What
you may findis that you will encourage off topic posts by
people whothink oh, here is a newsgroup where nice people
answer randomquestions.) I suggest all such questions be
directed to Doctor Mathwhich 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.
＝＝＝＝＝＝＝&#
65309;＝＝＝＝＝＝＝
＝＝＝＝＝＝＝&#
65309;＝＝＝＝＝＝＝
＝＝＝＝＝＝＝&#
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.
===
===
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 andshow that X squared - 1 is
divisible by 24.Ex. (5sqr - 1)/24=1, (7sqr - 1)/24=2, (11sqr -
1)/24=5The 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 tosatisfy david's criterion (natural number X, that is
not divisible by 3). Isuspect that the exact wording of the
problem is:Given that X is an odd natural number not divisible
by 3, prove that X^2-1is divisible by 24.To the OP, since it's
been a few days since this problem was posted and ifit is
homework, it was probably due before now, a hint: You can
classifyall odd natural numbers not divisible by 3 into two
general categories, onecontaining 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 aparameter, 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 trailof
responses.
===
Subject: formula for roots of quadratic matrix
equation?Content-Length: 217Originator: rusin@vesuviusIs there
a closed form formula forthe roots of the quadratic matrix
equationQAQ - BQ + C = 0where 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: 695Originator:
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} andQ =
sqrt(A)^{-1} sqrt( 4 A - C ) + 2 AI got this from matching
terms on ( sqrt(A) Q + F )^t ( sqrt(A) Q + F ) = F^t F - Cso
it may not give all the roots.- C
===
Subject: Re: formula for
roots of quadratic matrix equation?Content-Length:
1014Originator: rusin@vesuviusIs there a closed form formula
forthe roots of the quadratic matrix equationQAQ - BQ + C =
0where 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 oneconsiders
QAQ' - QB - B'Q' + C = 0,one can generally describe the
solutions, especially ifA 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 setof all solutions of the revised problem.This
address is for information only. I do not claim that these
viewsare those of the Statistics Department or of Purdue
University.Herman Rubin, Department of Statistics, Purdue
Universityhrubin@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 h9BIMWF08909This paper is extension of my
first paper and perhaps it gives a verysimple proof of
GOLDBACHS CONJECTURE. According to my first paperSearch for
Prime Numbers All prime numbers (except 2 & 3) mustsatisfy the
series 6*n-1 or 6*n+1 where n is a positive integerranging one
to infinity. So the necessary condition for the primenumber is
that it can either be expressed as 6*n-1 or as 6*n+1 where nis
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 theprime
number is that it can either be expressed as 6*n-1 or as
6*n+1where n is a positive integer greater than equal to one
(n ranging oneto 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 case1) covers odd numbers
21,27,33,39,[CenterDot].. 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+1In above three situations sum of three prime
numbers x + y + z can beexpressed 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)
coversodd numbers 19,25,31,37,[CenterDot]..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-1In above three situations
sum of three prime numbers x + y + z can beexpressed 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)
coversodd numbers 17,23,29,35,[CenterDot]..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) coversodd numbers
15,21,27,33,[CenterDot]..It is clear from above discussion all
odd numbers greater than equalto fifteen (15)
15,17,19,21,23,25,27,29,31,33,35,37 can beexpressed as sum of
three prime numbers. Odd numbers less than 15cannot be
expressed as sum of three prime numbers covering my seriesbut
there are two prime numbers (2,3) those are not included in
myseries, let us try to express odd numbers less than 15 as
sum of threeprime 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+2Or 9=3+3+3 7=3+2+2So
it is clear from my work all odd numbers greater than 6 can
beexpressed 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 theprime number is that it can
either be expressed as 6*n-1 or as 6*n+1where 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
allinteger value greater than equal to 2. So case 1) covers
even numbers14,20,26,32,[CenterDot].. Case 2) say x = 6*a+1, y
= 6*b-1 Or x =6*a-1, y = 6*b+1In above two situations sum of
two prime numbers x + y can beexpressed 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 allinteger value greater
than equal to 2. So case 2) covers even
numbers12,18,24,30[CenterDot]..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
allinteger value greater than equal to 2. So case 3) covers
even numbers10,16,22,28,[CenterDot]..It is clear from above
discussion all even numbers greater than equalto ten (10)
10,12,14,16,18,20,22,24,26,28,30,32 can be expressed assum of
two prime numbers. Even numbers less than 10 cannot
beexpressed as sum of two prime numbers covering my series but
there aretwo prime numbers (2,3) those are not included in my
series, let ustry 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+2So it is clear from my work all even numbers greater than
2 can beexpressed as sum of 2 prime numbers.NOTE: One
important question may comes to mind that the
necessarycondition for the prime number is that it can either
be expressed as6*n-1 or as 6*n+1 where n is a positive integer
ranging one toinfinity but it is not sufficient condition. So
for all n, 6*n-1 isnot 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 canbe explained logically. If
we concentrate on first few prime numbersexpressible 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, . According tomy 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 integervalues but still (a +b +c) assumes all integer
values greater thanequal to 3. Let me explaina assumes specific
positive integer values 1,2,3 (not 4),5,6,7 (not8,9),
10,11,12,13 (not 14,15),16, .b assumes specific positive
integer values 1,2,3 (not 4),5,6,7 (not8,9), 10,11,12,13 (not
14,15),16,c assumes specific positive integer values 1,2,3
(not 4),5,6,7 (not8,9), 10,11,12,13 (not 14,15),16,.Say we
have to obtain (a + b + c) = k (k is any positive
integergreater than equal to 3)first off all chose a = 1 so
that 6*a+1 is prime number. Our next taskis 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 + chas
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 that6*a+1 and 6*b+1
are primes and a + b = k-1. As we go for larger valuesof k in
several ways a, b can be chosen such that a + b = k-1, atleast
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 suchthat 6*a+1 and 6*b+1are primes is more. Although there
is no scope ofviolation of above requirement still consider a
value of k for whichit 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
thevalue of a to 2 and try to get a ,b such that 6*a+1 and
6*b+1 areprimes and a + b =k-2. If this requirement does not
meet we could gofor a=3,5,6, 7,9,10,11,12,13,16,. Violation of
all thesepossibilities 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 paperSearch 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*5You
can't assume that just because all prime numbers greater than
3 must bein the form 6*n+1 or 6*n-1 with positive integer n
that all numbers in oneof 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 thisquestion 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 h9BIN7n08948This paper is extension of my
first paper and perhaps it gives a verysimple proof of
GOLDBACHS CONJECTURE. According to my first paperSearch for
Prime Numbers All prime numbers (except 2 & 3) mustsatisfy the
series 6*n-1 or 6*n+1 where n is a positive integerranging one
to infinity. So the necessary condition for the primenumber is
that it can either be expressed as 6*n-1 or as 6*n+1 where nis
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 theprime
number is that it can either be expressed as 6*n-1 or as
6*n+1where n is a positive integer greater than equal to one
(n ranging oneto 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 case1) covers odd numbers
21,27,33,39,[CenterDot].. 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+1In above three situations sum of three prime
numbers x + y + z can beexpressed 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)
coversodd numbers 19,25,31,37,[CenterDot]..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-1In above three situations
sum of three prime numbers x + y + z can beexpressed 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)
coversodd numbers 17,23,29,35,[CenterDot]..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) coversodd numbers
15,21,27,33,[CenterDot]..It is clear from above discussion all
odd numbers greater than equalto fifteen (15)
15,17,19,21,23,25,27,29,31,33,35,37 can beexpressed as sum of
three prime numbers. Odd numbers less than 15cannot be
expressed as sum of three prime numbers covering my seriesbut
there are two prime numbers (2,3) those are not included in
myseries, let us try to express odd numbers less than 15 as
sum of threeprime 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+2Or 9=3+3+3 7=3+2+2So
it is clear from my work all odd numbers greater than 6 can
beexpressed 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 theprime number is that it can
either be expressed as 6*n-1 or as 6*n+1where 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
allinteger value greater than equal to 2. So case 1) covers
even numbers14,20,26,32,[CenterDot].. Case 2) say x = 6*a+1, y
= 6*b-1 Or x =6*a-1, y = 6*b+1In above two situations sum of
two prime numbers x + y can beexpressed 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 allinteger value greater
than equal to 2. So case 2) covers even
numbers12,18,24,30[CenterDot]..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
allinteger value greater than equal to 2. So case 3) covers
even numbers10,16,22,28,[CenterDot]..It is clear from above
discussion all even numbers greater than equalto ten (10)
10,12,14,16,18,20,22,24,26,28,30,32 can be expressed assum of
two prime numbers. Even numbers less than 10 cannot
beexpressed as sum of two prime numbers covering my series but
there aretwo prime numbers (2,3) those are not included in my
series, let ustry 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+2So it is clear from my work all even numbers greater than
2 can beexpressed as sum of 2 prime numbers.NOTE: One
important question may comes to mind that the
necessarycondition for the prime number is that it can either
be expressed as6*n-1 or as 6*n+1 where n is a positive integer
ranging one toinfinity but it is not sufficient condition. So
for all n, 6*n-1 isnot 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 canbe explained logically. If
we concentrate on first few prime numbersexpressible 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, . According tomy 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 integervalues but still (a +b +c) assumes all integer
values greater thanequal to 3. Let me explaina assumes specific
positive integer values 1,2,3 (not 4),5,6,7 (not8,9),
10,11,12,13 (not 14,15),16, .b assumes specific positive
integer values 1,2,3 (not 4),5,6,7 (not8,9), 10,11,12,13 (not
14,15),16,c assumes specific positive integer values 1,2,3
(not 4),5,6,7 (not8,9), 10,11,12,13 (not 14,15),16,.Say we
have to obtain (a + b + c) = k (k is any positive
integergreater than equal to 3)first off all chose a = 1 so
that 6*a+1 is prime number. Our next taskis 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 + chas
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 that6*a+1 and 6*b+1
are primes and a + b = k-1. As we go for larger valuesof k in
several ways a, b can be chosen such that a + b = k-1, atleast
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 suchthat 6*a+1 and 6*b+1are primes is more. Although there
is no scope ofviolation of above requirement still consider a
value of k for whichit 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
thevalue of a to 2 and try to get a ,b such that 6*a+1 and
6*b+1 areprimes and a + b =k-2. If this requirement does not
meet we could gofor a=3,5,6, 7,9,10,11,12,13,16,. Violation of
all thesepossibilities is absurd.
===
Subject: Maple editorsDoes
anyone know of a decent way to edit Maple code? I've been using
theMaple GUI to edit my programs up until this point, and I
can't stand it anylonger. Has anyone else had problems with
it? Is there an option somewherein the preferences that
orients the environment towards more experiencedcoders? 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 codehitting enter
executes the block of code, I have to hit shift-enter to
pushthings down a lineetc.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 powerfuleditor; I use it for Maple code, for
running Maple, editing andcompiling TeX documents, as well as
for usenet and e-mail client andfile manager. And as diary,
calendar, database, spreadsheet, webbrowser, 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 modulethat 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 onehas to become used to :-) For this
i foundMupad better for handling before i gave it up.More ugly
i imagine MMA: to input functionsi would have to press + 8 or 9 onmy (german) keyboard to get '[' or ']',
brrr.Otherwise stated: the strength (and weakness)of sytems is
beyond user interfaces.
===
Subject: Re: Maple editorsSet
Mathematica's default input format to 'TraditionalForm' you
canthen 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 haveto
use return instead. Despite the documentation which always
refersto enter.
===
Subject: Maple error - illegal use of a
formal parameterI'm trying to run this piece of Maple code,
but it crashes when I try tomodify 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 zeroListdefined 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 toget 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 avalue to 'current' which is
not a formal parameter (it is assigned to alist). 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 byreturn 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 bereplaced 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
questionslike that.
Mihailovshttp://webpages.shepherd.edu/amihailo/
===
Subject:
Brownian Motion ProblemI have done the following numerical
calculation. For a one-dimensionalBrownian motion with steps
x(i) drawn from a normal ditribution of mean 0and unit
variance, the displacements d(i) from zero are computed over
atime interval from t = 0 to 1 with n time steps (n>1). For
some chosenvalue 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 thedistribution
function (PDF) of T.For a = 0, I obtained P(T) = 2(p(T) +
p(1-T)) where p(T) = 1/ (2Pi*SQRT(T)*(1+T) ) and it does agree
with the numerical results. However thegeneral 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 whichd(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. Howeverthe general case for
non-zero a has me stumped. I would greatly appreciate any
suggestions to help solve this problem. Thank youNo need! I've
found it myself.
===
Subject: sign function===Subject: Re: sign
functionPlease 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
readersmay 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 itfails. How may I do
it?? Could someone help me?This is what I
did:x:=matrix([1,2,3]) +- -+ | 1 | | | | 2 | | | | 3 | +-
-+x+1 FAILthis 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 | +- -+ +--+ +--+| |+-|+
Christopher Creutzig (ccr@mupad.de) +--+ Tel.:
05251-60-5525
===
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 | +- -+Hi Christopher,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 Ido it easily with MuPAD?
===
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 Ido 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
perspectivemight have changed during last 5 years. Now we have
siteseer, etc. Anyupdates to the subject, new excellent
textbooks, must read reviews etc?