mm-246
===
Subject: Linear Algebra: When to convert R(A) to C(A) and
one is this (please forgive me if this seemslike an idiotic
question but I need clarification).My book has a problem that
I'm slightly confused about (I promise thatit's not
homework):Find a basis for the space spanned by: 1 -2 0 -2v1 =
2 v2 = 0 v3 = 4 v4 = -4 -3 4 -2 6and suppose A =
[v1,v2,v3,v4]Without explaining why they're doing this, they
simply say:We write the vectors as rows of a matrix A and then
reduce the matrix torow-echelon form...Ok, so I understand that
they're giving us 4 column (as opposed to row)vectors and that
they're saying we need to Transpose C(A) where A = v1thru v4
in order to get it equivelant to it's row matrix form.Question
1:Why? I mean why do we need to transpose the matrix first? I'm
lostthere and the book does not do a good job in explaining
this. Iunderstand Gaussian Elimination and such. What I don't
understand ishow I know that I need to Transpose the matrix
first prior to reducingit to ref as opposed to simply reducing
it without Transposing it!Question 2:Maybe it's the book or
maybe it's me (probobly me). But in the abovequestion, how did
we know that we need to find the basis of the rowspace rather
than the column space? I *think* I know but maybe you canclear
it up. Is it because the vectors are assigned in columnar
formrather than their row form (ie, v1 = (1,2,3), v2=(-2,0,4),
v3=(0,4,2),v4=(-2,-4,6). Is the book making the assumption that
listing the vectoras a vertical stack implies Row Space and
horizontal stack impliesColumn Space? So even though the
question did not explicity sayFind a basis for the *row* space
spanned by ..., we should have justknown they meant row rather
than column space in this case due tohow they represented it
visually? These are the kinds of things about math textbooks
that drive me crazy which is why I'm hoping someone out there
===
to convert R(A) to C(A) and another
The first one is this (please forgive me if this seems> like
an idiotic question but I need clarification).> My book has a
problem that I'm slightly confused about (I promise that> it's
not homework):> Find a basis for the space spanned by:> 1 -2 0
-2> v1 = 2 v2 = 0 v3 = 4 v4 = -4> -3 4 -2 6[...]If you use the
transposes of the vectors as rows of a matrix then
thetransposes of the non-zero rows of the echelon form are a
basis for thespan.If you use the vectors as columns then the
columns of the originalmatrix which correspond to the
non-pivot columns of the echelon formmake up a basis for the
span.The difference is that if you use the vectors as rows,
you don't haveto back track to the original matrix. On the
other hand, if you use thevectors as columns, you get a subset
of the original vectors whichforms a basis for the span.So,
using the vectors as rows gives (the transposes of) (1, 0 -2)
and(0, 1, -1/2) as a basis. Using the vectors as columns gives
(thetransposes of) (0, 4, -2) and (-2, -4, 6) as a basis.The
row space is the subspace spanned by the transposes of the
non-zerorows of the echelon form. Row refers to matrix rows
not row vectors. Whether vectors are rows or columns is a
non-issue but I miss the good-- Paul SperryColumbia, SC
===
(USA)Subject: Re: Linear Algebra: When to convert R(A) to C(A)
your time and reply. It is very much appreciated thatthere are
helpful folks like you in this newsgroup. It may not havetaken
you long to type in your reply but I can assure you that it
willtake me quite a bit of time to digest what you just said
===
again!!!Subject: Interpreting Standard Deviation as a
percentage by support1.mathforum.org (8.11.6/8.11.6/The Math
Forum, $Revision: 1.9 primary) id i2B1kOa09662;Let us say you
have a mean return of 15.6%. And you have a standarddeviation
of 20.4%. How many percent is one standard deviation, andhow
do I compute such a number? I was talking to someone, and he
said one standard deviation is about5%... how did he get this
===
number, please I need help with this ASAP.Subject: Re:
Interpreting Standard Deviation as a percentage> Let us say
you have a mean return of 15.6%. And you have a standard>
deviation of 20.4%.Your sentence is ambiguous. It's clear that
the mean rate ofreturn is 15.6%, but it is ambiguous as to
whether the std deviationis in absolute or relative terms.In
other words, is the rate of return that is one
standarddeviation above the mean 15.6% + 20.4% = 36% (this
would betaking the 20.4% as an absolute), or is the rate of
returnthat is one std dev above the mean 15.6% + 0.204*15.6% =
18.78%(this would be taking the 20.4% as relative)?So, which
does the person (or the problem) mean when itquotes a std dev
of 20.4%?> How many percent is one standard deviation, and>
how do I compute such a number? If the 20.4% std dev quote
means the std dev is 20.4% of themean, then the std dev is
3.18%. If the 20.4% std dev quoteis absolute, then the std dev
is 20.4%.In this case it gets confusing because the thing
beingmeasured is itself a percentage! Consider it this way --
the mean return is 0.156. Now,is the standard deviation 20.4%
(of 0.156) or is it 0.204?> I was talking to someone, and he
said one standard deviation is about> 5%... how did he get
this number, please I need help with this ASAP.Looks like he's
thinking relative, but he's still off.-- Rich Carreiro
===
rlcarr@animato.arlington.ma.usSubject: Re: Interpreting
Standard Deviation as a percentage> Let us say you have a mean
return of 15.6%. And you have a standard> deviation of 20.4%.>
Your sentence is ambiguous. It's clear that the mean rate of>
return is 15.6%, but it is ambiguous as to whether the std
deviation> is in absolute or relative terms.*However*, it is
standard in financial circles for the standard deviation tobe
in absolute terms.> In other words, is the rate of return that
is one standard> deviation above the mean 15.6% + 20.4% = 36%
(this would be> taking the 20.4% as an absolute), or is the
rate of return> that is one std dev above the mean 15.6% +
0.204*15.6% = 18.78%> (this would be taking the 20.4% as
relative)?> So, which does the person (or the problem) mean
when it> quotes a std dev of 20.4%?> How many percent is one
standard deviation, and> how do I compute such a number?> If
the 20.4% std dev quote means the std dev is 20.4% of the>
mean, then the std dev is 3.18%. If the 20.4% std dev quote>
is absolute, then the std dev is 20.4%.> In this case it gets
confusing because the thing being> measured is itself a
percentage!> Consider it this way -- the mean return is 0.156.
Now,> is the standard deviation 20.4% (of 0.156) or is it
0.204?> I was talking to someone, and he said one standard
deviation is about> 5%... how did he get this number, please I
need help with this ASAP.> Looks like he's thinking relative,
but he's still off.*My* guess is that this someone is
completely ignoring the facts of the caseat hand, and just
blindly plowing standard facts about the stock market(mean
return about 15%, variance about 25%, or std dev about 5%, and
allvarying depending on the time scale you choose to measure --
and themeasurement techniques and even which market you choose
to measure).Given the data, your guess is as good as mine.Jon
===
MillerSubject: URGENT...whats the answer to this riddle please
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i2B1kP009709;EACH RIDDLE HAS AN
ANSWER!!!!! 1. This is the riddle that cannot be answered.
What is the answer tothis riddle? 2. These recreates
themselves if they are placed correctly. What isthe answer to
this riddle? 3. This is a riddle. What is the answer to this
riddle? 4. This riddle is the answer to all five riddles. What
is the answerto this riddle? 5. This is the final riddle. What
is the answer to this riddle? please help me find the answer
===
to this.thank you.Subject: Re: URGENT...whats the answer to
this riddle please> EACH RIDDLE HAS AN ANSWER!!!!!> 1. This is
the riddle that cannot be answered. What is the answer to> this
riddle?This answer is not the answer to any riddle.-- P.A.C.
SmithThe vast majority of Iraqis want to live in a peaceful,
free world.And we will find these people and we will bring
===
them to justice.Subject: Re: URGENT...whats the answer to this
riddle please>EACH RIDDLE HAS AN ANSWER!!!!! >1. This is the
riddle that cannot be answered. What is the answer to>this
riddle? >2. These recreates themselves if they are placed
correctly. What is>the answer to this riddle? >3. This is a
riddle. What is the answer to this riddle? >4. This riddle is
the answer to all five riddles. What is the answer>to this
riddle? >5. This is the final riddle. What is the answer to
this riddle? >please help me find the answer to this.If you
will solve the equation x^4 + y^4 = z^4 in integers, I will
solve your riddle.Oh? Didn't notice this is a math group and
not a f---ing riddles group?-- Stan Brown, Oak Road Systems,
Cortland County, New York, USA http://OakRoadSystems.comAn
expense does not have to be required to be considered
===
necessary. -- IRS Form 1040 line 23 instructionsSubject: Re:
URGENT...whats the answer to this riddle please>EACH RIDDLE
HAS AN ANSWER!!!!!>please help me find the answer to this.> If
you will solve the equation x^4 + y^4 = z^4 in integers, I
will> solve your riddle.> Oh? Didn't notice this is a math
group and not a f---ing riddles> group?Hold your water, direct
her to alt.math.recreational or considering thelogical nature
of the riddle, sci.logic or even this group as it appearsmore
than you can handle gracefully. Indeed, considering the
convolutionof the language, is it a riddle? If not, then such
babbling is best putto alt.math.recreational or something like
===
alt.humor.Riddle of the day: what's so URGENT?Subject: Re:
define recursively the set of bit strings that have more>
zeros than ones.Recursively, I don't know. Possible I suppose
but I suspect not pretty.For n odd there are 2^(n-1) such
strings.For n even there are 2^n - (1/2)*C(n, n/2).Details on
===
request.-- Paul SperryColumbia, SC (USA)Subject: Re: HP49G>
Anybody know a good help site for this thing? This user manual
sucks.> Levihttp://www.hpcalc.org/ is the best website for all
hp calculators. it either has or can point to all useful
information about them, as well as a humongous library of
libraries for the 48/49 series.comp.sys.hp48 is the best
newsgroup for the hp49, hp49+ and other hp calculators,
===
too.Subject: Re: Cubic polynomial> Can you show that there is
some x for which f(x) > 0 and another for> which f(x) < 0?> I
managed to find an x for which f(x) > 0. However, finding an x
such that> f(x) < 0 proved more tricky. I showed that the limit
of the function as x> approaches minus infinity is minus
infinity, and said that as the function> is continuous, then
there must exist some x such that f(x) < 0.> Is there a better
way to find an x for which f(x) < 0?You don't need a better
===
way.Subject: Need HelpI would like to show : lim {x->00}
f(g(x)) = L if lim {x->00} f(x) = L and lim {x->00} g(x) =
L.For every positive number e, there are a,b in Rsuch that
|f(x) - L|<e , |g(y) - L|<e for all x>a, y>b.Let d=max{a,b}
then,For all x>d,|f(x) - L|<e , |g(x) - L|<eBut, with this,I
don't know how can I induce |f(g(x)) - L|<e.If somebody can
===
help me, please post reply.Subject: Re: Need Help> I would
like to show : lim {x->00} f(g(x)) = L> if lim {x->00} f(x) =
L and lim {x->00} g(x) = L.Don't think this is true. For
example take f(x) = e^(-x). Then lim (x->oo)f(x) = 0.g(x) = 0.
Then lim (x->oo)(g(x) = 0.But lim (x->oo)(f(g(x)) = 1.The
problem is that even though f->0, f(0) = 1. Is this what you
===
were trying to do or did you have a typo?Subject: Re: Need
Help> I would like to show : lim {x->00} f(g(x)) = L> if lim
{x->00} f(x) = L and lim {x->00} g(x) = L.> Don't think this
is true. For example take > f(x) = e^(-x). Then lim
(x->oo)f(x) = 0.> g(x) = 0. Then lim (x->oo)(g(x) = 0.> But
lim (x->oo)(f(g(x)) = 1.> The problem is that even though
f->0, f(0) = 1. > Is this what you were trying to do or did
you have a typo?Umm, Maybe my book exercise is wrong.I see
this problem at Korean edition of Introduction to RealAnalysis
written by R.G Bartle& D.R. Sherbert.Your counterexample is
===
right.Subject: quadratic function max-min by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i2BDnda15855;This problem is
related to functions and quadratics - not to calculas,I have a
problem asking for the maximum vertical distance from aparabola
f(x)= -2x^2 + 4x + 3 and a line f(x) = x - 2 The book'sanswer
is 6.125 The x-axis of the vertex of the parabola is 1
thevalue of [-b2a] and the vertex [a maximum since the
parabola opensdownward] becomes (1,5)! However, now I am
stumped. I thought thatthe intersection with the line[whose
slope equals one] would bevertical to the x-axis position and
therefore (1,-1) However, thisgives a distance of 6 and not
===
6.125. Can someone straighten me out???Subject: Re: quadratic
function max-min by support1.mathforum.org (8.11.6/8.11.6/The
Math Forum, $Revision: 1.9 primary) id i2BEKBw20709;>This
problem is related to functions and quadratics - not
to>calculas,>I have a problem asking for the maximum vertical
distance from a>parabola f(x)= -2x^2 + 4x + 3 and a line f(x)
= x - 2 The book's>answer is 6.125 The x-axis of the vertex of
the parabola is 1 the>value of [-b2a] and the vertex [a maximum
since the parabola opens>downward] becomes (1,5)! However, now
I am stumped. I thought that>the intersection with the
line[whose slope equals one] would be>vertical to the x-axis
position and therefore (1,-1) However, this>gives a distance
of 6 and not 6.125. Can someone straighten me>out???the
maximum vertical distance between the parabola and the lineis
not attained at x=1 with vertical distance 6. To see where the
distance reaches its maximum, consider g(x), the difference in
the y-values of the parabola and the line:g(x) = (-2x^2+4x+3)
- (x-2) = -2x^2+3x+5 = -2*(x-0.75)^2 + 6.125.Where does g
attain its maximum, and what is its maximum value ?Best
===
wishesTorsten.Subject: Re: General algorithm for vector
cross-product by support1.mathforum.org (8.11.6/8.11.6/The
Math Forum, $Revision: 1.9 primary) id i2BEKBf20699;You're on
the right track. Make an nxn matrix, n-1 of whose rowsare the
vectors in question, and the last row consists of
markerse1,e2,...,en. (If you like, these are unit vectors.)The
determinant of this matrix is a linear combination
ofe1,e2,...,en; interpret the ei as unit vectors, and you get
a vector orthogonal to the other n-1.(Caveat: if the n-1
vectors are lineraly dependent, then you'regoing to get the 0
vector; in this case, find a maximal subsetof linearly
independent vectors, augment it with some arbitraryvectors
independent of the first guys until you have n-1 of them.)I am
using the cross-product to find a normal that us perpendicular
to two vectors. All of the examples and code snippets that I
have seen deal with the 3D case. Correct me if I am wrong, but
we can find the cross-product of other dimensions (either lower
or higher), can we not?To handle higher dimensions, we could
recursively find the determinent of the vectors, but if we
have n-dimensional vectors, do we need n-1 vectors to fill the
===
n x n matrix?Subject: Looking for a good book on
proofs...Hello.Proofs have always stumped me (which is why I
dropped out of being a mathmajor). I love math and want to
continue studying it. Are there anyproofs-for-dummies type
books out there? All you math purists - pleaseforgive me if
I've offended you by asking this silly-sounding question;
butyou have to appreciate someone who has an interest and
===
recommendations.- RogerSubject: Re: Looking for a good book on
proofs...Hello!I think this book might help:George Polya:
Mathematical discovery, John Wiley & sons, inc., New York
===
-LondonMarinSubject: Re: Looking for a good book on proofs...>
Hello!> I think this book might help:> George Polya:
Mathematical discovery, John Wiley & sons,> inc., New York -
London> MarinCan you explain a little bit why you reccomend
===
this book, before I go outlibrary hunting for it?Subject: Re:
Looking for a good book on proofs...Well, it's not of
proofs-for-dummies type.(in my opinion such a book would be
very harmful)It's a book that will teach you how to use
mathematical principles andintelectual tools by examplesin
elementary mathematics. In short, how to be creative in
mathematics andhow to think with your own head.And, it's
author George Polya is a famous mathematician, known for
teachingother teachers ofmathematics.I like the book.Hope it
===
helps you.MarinSubject: Free math helpWe are an experienced
math tutoring organisation operating out ofCanada. If you
would like a free booklet illustrating how we teach aspecific
===
math concept, email us at mathhelpbooklets@mail.comSubject:
sequence of rationalsHi. My question is how to prove the
following fact:Let {x_n} be a sequence of rational numbers
that converges to anirrational number, x. i.e. x_n -> x. Then
the denominators of eachof the x_n must necessarily grow as n
goes to infinity.This fact is obviously true, but I am having
===
a hard time proving it. JoeSubject: Re: sequence of rationals>
Let {x_n} be a sequence of rational numbers that converges to
an> irrational number, x. i.e. x_n -> x. Then the denominators
of each> of the x_n must necessarily grow as n goes to
infinity.> This fact is obviously true, but I am having a hard
time proving it. Suppose for simplicity x is in [0,1]. How many
rationals in [0,1] are expressible with a denominator <= n?
Clearly only a finite number (namely0, 1, 1/2, 1/3, 2/3, ...,
1/n, 2/n, ..., (n-1)/n). Let d be the minimum distance from x
to this set of rationals. Then d > 0, and no sequence of
rationals with denominators <= n can come within d/2 of
===
x.Subject: Re: sequence of rationals Adjunct Assistant
Professor at the University of Montana.>Hi. My question is how
to prove the following fact:>Let {x_n} be a sequence of
rational numbers that converges to an>irrational number, x.
i.e. x_n -> x. Then the denominators of each>of the x_n must
necessarily grow as n goes to infinity.>This fact is obviously
true, but I am having a hard time proving it. Assume that the
denominators are bounded by some number M>0. Thenwriting each
x_n as a fraction with denominator M, what can you sayabout
the differences|x_n - x_k|for n and k large? They cannot all
be zero (since then the sequencewould converge to a rational
number). So for every K>0, there existsn,k>k such that |x_n -
x_k| is less than zero. Writing both asfractions with
denominator M, is there a minimum for this value?On the other
hand, since the sequence converges, for every e>0 thereexists
K>0 such that if n and k are both larger than K, then
|x_n-x_k|< e. What is your conclusion?--
accept as reality. --- Calvin (Calvin and
===
Non-AP Calculus? by support1.mathforum.org (8.11.6/8.11.6/The
Math Forum, $Revision: 1.9 primary) id i2BLcUv11681;Should
high schools teach Regular Calculus courses for students
whoare not at the level to handle the Advanced Placement
Calculus course?My instinct is to say that those students
would benefit more fromadditional precalculus preparation. Is
that the way collegeprofessors would prefer things, or would
they rather have students whohave seen some calculus in a
watered-down class.I ask this question because I am the
mathematics department chair atmy school, and we are looking
===
you for any insights you can provide!Michael BrownSubject: Re:
Solution of Partial Differntial Equation > I am looking for a
way to solve the following Partial differential>equation
(analytical, numerical-- anything) >
EI[NonBreakingSpace]4Y/[NonBreakingSpace]x4 + M
[NonBreakingSpace]2/[NonBreakingSpace]t2=0> Note that
'[NonBreakingSpace]' represents partial derivative, x and t
are the>independent variables, Y is the dependent variable and
'E' and 'I'and>'M' are constants.So if I correctly interpret
what to me look like weird characters,and what seems to be a
missing Y, EI d^4 Y/dx^4 + M d^2 Y/dt^2 = 0 ?where EI (which
we may as well combine) and M are constants. Positive, I
suppose?> the conditions are : @t=0,Y =0> @x=0, Y=0 (where b
is a constant)But you didn't include b in the conditions!What
domain are you working in, 0 <= x < infinity and 0 <= t <
infinity?You haven't given anywhere near enough conditions to
specify the solution uniquely (of course Y=0 is a solution, so
maybe that's a good thing...)You have, for instance,
separation-of-variables solutions Y = sinh(a x) sin(sqrt(EI/M)
a^2 t)for any complex number a, and of course you can take
linear combinationsor integrals over different values of a.
Robert Israel israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
===
Vancouver, BC, Canada V6T 1Z2Subject: Re: Solution of Partial
Differntial Equation> I am looking for a way to solve the
following Partial differential>equation (analytical,
numerical-- anything)>
EI[NonBreakingSpace]4Y/[NonBreakingSpace]x4 + M
[NonBreakingSpace]2/[NonBreakingSpace]t2=0>> Note that
'[NonBreakingSpace]' represents partial derivative, x and t
are the>independent variables, Y is the dependent variable and
'E' and 'I'and>'M' are constants.> So if I correctly interpret
what to me look like weird characters,> and what seems to be a
missing Y,> EI d^4 Y/dx^4 + M d^2 Y/dt^2 = 0 ?> where EI (which
we may as well combine) and M are constants. Positive,> I
suppose?> the conditions are : @t=0,Y =0> @x=0, Y=0 (where b
is a constant)> But you didn't include b in the conditions!>
What domain are you working in, 0 <= x < infinity and 0 <= t <
infinity?> You haven't given anywhere near enough conditions to
specify the solution> uniquely (of course Y=0 is a solution, so
maybe that's a good thing...)> You have, for instance,
separation-of-variables solutions> Y = sinh(a x)
sin(sqrt(EI/M) a^2 t)> for any complex number a, and of course
you can take linear combinations> or integrals over different
values of a.> Robert Israel israel@math.ubc.ca> Department of
Mathematics http://www.math.ubc.ca/~israel> University of
British Columbia> Vancouver, BC, Canada V6T 1Z2Looks (after
your excellent translation of the wierd symbols)like the
problem of a vibrating elastic rod.-- Julian V. NobleProfessor
^^^^^^^^^^^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ God
is not willing to do everything and thereby take away our free
will and that share of glory that rightfully belongs to us. --
===
N. Machiavelli, The Prince.Subject: Sum evaluation with
Wilf-Zeilberger method by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i2BFJ4l27529;i need some help on the evaluation of sums
containing binomialcoefficients and i want to use the
Wilf-Zeilberger method:For a positive integer q and m=2q or m
=2q+1 define [m/2] = q.For any integer m > 2 define the
sumf(n) = Sum(k=0,...,n) F(n,k) = F(0,k)+F(1,k)+...+F(n,n)
withF(n,k)=(-1)^(n-k) * m/(m-k) *
binomial(m-k,k)*binomial(m-1-2k, n-k)and the usual binomial
coefficient binomial(n,k) = n(n-1)(n-2)...(n-k+1)/k!with k in
N and n in Z.Lemma: For any integer m with m > 2 we have
f(n)=1 for n = 0,...,[m/2].I want to use the Wilf-Zeilberger
method to prove the lemma.The first step is to find a
recurrence for F. This can by doneby using the Zeilberger
algorithm that gives a representationSum(j=0,...,J)
a(j,n)*F(n+j,k) = G(n,k+1)-G(n,k) (*)with a rational function
G(n,k), some finite integer Jand coefficients a(j,n)
independent of k.Unfortunately, i do not have an
implementation of the Zeilbergeralgorithm (but there exist
implementations for Maple and Mathematica,both computer
algebra systems that i do not own)Can anybody help me - by
computing the recurrence (*) and thefunction G(n,k), or by
giving me hints how to get animplementation of the Zeilberger
algorithm that runs ona windows platform (without Maple or
===
Mathematica)?Frank Haefnerfrankdothaefner@web.deSubject: Re:
Sum evaluation with Wilf-Zeilberger method> i need some help
on the evaluation of sums containing binomial> coefficients
and i want to use the Wilf-Zeilberger method:> For a positive
integer q and m=2q or m =2q+1 define [m/2] = q.> For any
integer m > 2 define the sum> f(n) = Sum(k=0,...,n) F(n,k) =
F(0,k)+F(1,k)+...+F(n,n) with> F(n,k)=(-1)^(n-k) * m/(m-k) *
binomial(m-k,k)*binomial(m-1-2k, n-k)> and the usual binomial
coefficient > binomial(n,k) = n(n-1)(n-2)...(n-k+1)/k!> with k
in N and n in Z.> Lemma: For any integer m with m > 2 we > have
f(n)=1 for n = 0,...,[m/2].> I want to use the Wilf-Zeilberger
method to prove the lemma.> The first step is to find a
recurrence for F. This can by done> by using the Zeilberger
algorithm that gives a representation> Sum(j=0,...,J)
a(j,n)*F(n+j,k) = G(n,k+1)-G(n,k) (*)> with a rational
function G(n,k), some finite integer J> and coefficients
a(j,n) independent of k.> Unfortunately, i do not have an
implementation of the Zeilberger> algorithm (but there exist
implementations for Maple and Mathematica,> both computer
algebra systems that i do not own)> Can anybody help me - by
computing the recurrence (*) and the> function G(n,k), or by
giving me hints how to get an> implementation of the
Zeilberger algorithm that runs on> a windows platform (without
Maple or Mathematica)?> Frank Haefner>
frankdothaefner@web.deMaple now includes the Wilf-Zeilberger
algorithms.Here it is in Maple 9:>
with(SumTools[Hypergeometric]):> FF := (-1)^(n-k) * m/(m-k) *
binomial(m-k,k)*binomial(m-1-2*k, n-k);FF :=
(-1)^(n-k)*m*binomial(m-k, k)*binomial(m-1-2*k, n-k)/(m-k)> F
:= unapply(FF,n,k);F := proc (n, k) options operator, arrow;
(-1)^(n-k)*m*binomial(m-k,k)*binomial(m-1-2*k, n-k)/(m-k) end
proc> Zpair := Zeilberger(FF,n,k,En):> L := Zpair[1];L :=
1-En> G := unapply(Zpair[2],n,k);G := proc (n, k) options
operator, arrow;
-k*(-1/2*m+k)*binomial(m-k,k)*m*((-1)^(n-k)*binomial(m-1-2*k,
n-k)-(-1)^(n-k+1)*binomial(m-1-2*k,n-k+1))/((m*n-2*n-n^2-1+m)*
k+(1/2*n^2+n-1/2*m*n-1/2*m+1/2)*n-(-1/2*n^2-n+1/2*m*n+1/2*m-1/
2)*m+(-1/2*n^2-n+1/2*m*n+1/2*m-1/2)*n) end proc> checkit :=
F(n,k)-F(n+1,k) = G(n,k+1)-G(n,k);checkit :=
(-1)^(n-k)*m*binomial(m-k,
k)*binomial(m-1-2*k,n-k)/(m-k)-(-1)^(n-k+1)*m*binomial(m-k,
k)*binomial(m-1-2*k, n-k+1)/(m-k)
=-(k+1)*(k+1-1/2*m)*binomial(m-k-1,
k+1)*m*((-1)^(n-k-1)*binomial(m-2*k-3,n-k-1)-(-1)^(n-k)*
binomial(m-2*k-3,n-k))/((m*n-2*n-n^2-1+m)*(k+1)+(1/2*n^2+n-1/2
*m*n-1/2*m+1/2)*n-(-1/2*n^2-n+1/2*m*n+1/2*m-1/2)*m+(-1/2*n^2-n
+1/2*m*n+1/2*m-1/2)*n)+k*(-1/2*m+k)*binomial(m-k,k)*m*((-1)^(
n-k)*binomial(m-1-2*k,
n-k)-(-1)^(n-k+1)*binomial(m-1-2*k,n-k+1))/((m*n-2*n-n^2-1+m)*
k+(1/2*n^2+n-1/2*m*n-1/2*m+1/2)*n-(-1/2*n^2-n+1/2*m*n+1/2*m-1/
2)*m+(-1/2*n^2-n+1/2*m*n+1/2*m-1/2)*n)>
simplify(expand(rhs(checkit)-lhs(checkit)));0If you need help
reading the results, let me know.Here it is in a different
format, turn off wrapping to see it properly:> with(SumTools[>
with(SumTools[Hypergeometric]):> FF := (-1)^(n-k) * m/(m-k) *
binomial(m-k,k)*binomial(m-1-2*k, n-k); FF := (n - k) (-1) m
binomial(m - k, k) binomial(m - 1 - 2 k, n - k)
-------------------------------------------------------------
m - k > F := unapply(FF,n,k); F := (n, k) -> (n - k) (-1) m
binomial(m - k, k) binomial(m - 1 - 2 k, n - k)
-------------------------------------------------------------
m - k > Zpair := Zeilberger(FF,n,k,En):> L := Zpair[1]; L := 1
- En> G := unapply(Zpair[2],n,k);G := (n, k) -> - 1
--------------------------------------------------------------
--------------- / 2 /1 2 1 1 1 m n - 2 n - n - 1 + m/ k + |- n
+ n - - m n - - m + -| n - (%1) m + (%1) n 2 2 2 2/ / / 1 / (n
- k) |k |- - m + k| binomial(m - k, k) m (-1) 2 / binomial (m
- 1 - 2 k, n - k) (n - k + 1) - (-1) binomial(m - 1 - 2 k, n -
k + 1)/| / 1 2 1 1 1 %1 := - - n - n + - m n + - m - - 2 2 2 2
> checkit := F(n,k)-F(n+1,k) = G(n,k+1)-G(n,k); (n - k) (-1) m
binomial(m - k, k) binomial(m - 1 - 2 k, n - k) checkit :=
-------------------------------------------------------------
m - k (n - k + 1) (-1) m binomial(m - k, k) binomial(m - 1 - 2
k, n - k + 1) -
--------------------------------------------------------------
------- = m - k 1 / / 1 -
--------------------------------------- |(k + 1) |k + 1 - - m|
binomial(m - k - 1, k + 1) m (%1) (k + 1) + (%2) n - (%3) m +
(%3) n 2 / / (n - k - 1) (-1) binomial(m - 2 k - 3, n - k - 1)
(n - k) 1 / / - (-1) binomial(m - 2 k - 3, n - k)/| +
--------------------------------- |k | / (%1) k + (%2) n -
(%3) m + (%3) n 1 / (n - k) - - m + k| binomial(m - k, k) m
(-1) binomial(m - 1 - 2 k, n - k) 2 / (n - k + 1) - (-1)
binomial(m - 1 - 2 k, n - k + 1)/| / 2 %1 := m n - 2 n - n - 1
+ m 1 2 1 1 1 %2 := - n + n - - m n - - m + - 2 2 2 2 1 2 1 1 1
%3 := - - n - n + - m n + - m - - 2 2 2 2 >
simplify(expand(rhs(checkit)-lhs(checkit))); 0-- G. A. Edgar
===
http://www.math.ohio-state.edu/~edgar/Subject: Re:
am looking at either Mathematica or Maple 9. Any help or>
advice you offer would be greatly appreciated.> Consider also
MathCad. It's a bit pricey, but is a> workhorse for data
analysis and presentation. It> interfaces reasonably well to
Excel, too.Note also that there is an academic discount for
MathCad.-- Julian V. NobleProfessor Emeritus of
^^^^^^^^^^^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ God
is not willing to do everything and thereby take away our free
will and that share of glory that rightfully belongs to us. --
===
N. Machiavelli, The Prince.Subject: Re: Recommendation between
two programs...> I am looking at getting a math software
program for school. I am> currently a college student, and I
was just introduced to Maple 9 by> my Calculus instructor.
WOW! I never knew such a product existed.:-)> [...] > Right
now, I am looking at either Mathematica or Maple 9. Any help
or> advice you offer would be greatly appreciated.If you're
running Windows, there's a trial version of Maple 9
available;please see http://www.maplesoft.com/trial.asp--
Thomas RichardMaple SupportScientific Computers
===
GmbHhttp://www.scientific.deSubject: Re: Recommendation
between two programs...
~^>Pn0&%&Ux8>1=w8P?^q%:g?%]2+oVLC;x!s,~MYjl!j>x`k>b9B5_NaM'4_X
:z Zw76--> I am looking at getting a math software program for
school. I am> currently a college student, and I was just
introduced to Maple 9 by> my Calculus instructor. WOW! I never
knew such a product existed. > Thinking I would do a little
more research on Maple 9, I actually> discovered Mathematica,
IDL, and Matlab (actually I was introduced to> Matlab as well,
but we didnt tool around with it). > Here is a little about me:
I am working toward a degree in> Biochemistry; I want to take
as many forms of higher maths as I can> (Calc II, III, etc.);
I have a little computer knowledge; price> doesnt matter too
much (Im not rich but I could justify spending> about two
hundred dollars if I could find a neat and useful software>
application);A couple of comments.First, the full version of
Mathematica sells for considerably more than $200. The student
version is less. I don't know how much less. Maybe, the student
version sells for as little as $200.Maple, Matlab and
Mathematica are more or less equivalent. By that I mean most
problems you would want to use these programs can be handled
by any of them. For the most part, whichever you are more
comfortable with is the one to go with. Since your comments
imply you have little experience with any of these programs, I
suggest one of the main criteria should be what is being used
predominantly by your school, colleagues etc. All of these
programs are quite capable and will take significant
study/usage to fully master.Another approach to selecting an
appropriate package is to find out what others in your field
of study primarily use. One way to do this might be to search
Amazon for books showing how to use one of these software
packages in your field of study. For example, Mathematica for
Physics by Zimmerman and Olness describes how to use
Mathematica to solve a variety of common physics problems. You
might even find similar books at your school library. Reading a
couple of books even if you buy them from Amazon will be
considerably cheaper than the software and are probably
something you will want eventually.FWIW, I use Mathematica.
Other software with similar functionality I've used include
Mathcad and Matlab. IMHO, either Matlab or Mathematica are far
more caple than Mathcad.-- To reply via email subtract one
===
hundred nineSubject: Re: Recommendation between two
programs...Matlab (except for the symbolic toolkit) and
Mathcadare not generally considered computer algebra
systems.They are, I think, simply interactive mathematical
programs. If theydo what you want to do, fine. Can they (say)
differentiatex^2 and get 2*x? The Matlab symbolic toolkit
provides a smallsubset of Maple to Matlab users.RJF> Maple,
Matlab and Mathematica are more or less equivalent> FWIW, I
use Mathematica. Other software with similar functionality
I've > used include Mathcad and Matlab. IMHO, either Matlab or
===
Mathematica are > far more caple than Mathcad.Subject: Re:
Recommendation between two programs...> Matlab (except for the
symbolic toolkit) and Mathcad> are not generally considered
computer algebra systems.> They are, I think, simply
interactive mathematical programs. If they> do what you want
to do, fine. Can they (say) differentiate> x^2 and get 2*x?
The Matlab symbolic toolkit provides a small> subset of Maple
to Matlab users.I don't know about Mathcad, but MATLAB's
Symbolic Math Toolbox 3.0.1 (thelatest version) provides quite
a bit of the functionality of the Maple 8kernel. To
differentiate x^2 and obtain 2*x, the commands you need are:%
Declare x as a symbolic variablesyms x% Define the symbolic
expression -- you don't need to do this, but I usuallydo sof =
x^2;% Differentiate -- can omit x if that's the only symbolic
variable in fdf = diff(f, x)% Integrate df -- can omit x if
that's the only symbolic variable in dff2 = int(df, x)You can
plot this using MATLAB's graphics capabilities:ezplot(f)There
are similar functions for performing surface, mesh, or contour
plotson symbolic objects.The Symbolic Math Toolbox does not
include the Maple GUI or any of Maple'sgraphics capabilities,
and (if I remember correctly) restricts four or fiveof the
Maple programming commands (PROC and WITH are the two big ones
I canremember; you can include some libraries with
maple('with', libraryname) butnot all.) The Extended Symbolic
Math Toolbox removes the limitations on theprogramming
commands, but does not remove the restriction on
Maple'sgraphics capabilities. [If the Extended toolbox had
Maple's graphicscapabilities, it would essentially _be_
Maple.]If anyone's still reading this and would like more
information on theSymbolic Math Toolbox or the Extended
Symbolic Math Toolbox, take a look atthese
pages:http://www.mathworks.com/products/symbolic/http://
www.mathworks.com/products/extsymbolic/-- Steve
Lordslord@mathworks.com> RJF> Maple, Matlab and Mathematica
are more or less equivalent> FWIW, I use Mathematica. Other
software with similar functionality> I've used include Mathcad
and Matlab. IMHO, either Matlab or> Mathematica are far more
===
caple than Mathcad.Subject: Re: Recommendation between two
programs...I have looked at Mathlab version 12 or 13.
Thestudent version for $99 includes the symbolic
toolkit.Without it, the program is not symbolic at all.I think
the commitment to symbolic interchangeof data with maple was
(on the product I tested) eitherpoorly done or dubiously
designed. E.g.I don't mind declaring symbols assyms xbut you
would then expect that syms would interactwith numbers from
matlab.My recollection is that x+1 fails to work, because1 is
not a symbol. You must create a symbolic objectthat is also 1.
Otherwise you get a type error.Since I reported this to
MathWorks, and my trial versionexpired and I haven't gotten a
new one, I don't knowwhat they did with this. My impression
was that Mathworkstreated my report as a user error and
discarded it.Why do you think that plotting makes a numeric
systeminto a symbolic system? If you believe that, then
thisnewsgroup should be discussing any program capable
ofoutput :)>Matlab (except for the symbolic toolkit) and
Mathcad>are not generally considered computer algebra
systems.>They are, I think, simply interactive mathematical
programs. If they>do what you want to do, fine. Can they (say)
differentiate>x^2 and get 2*x? The Matlab symbolic toolkit
provides a small>subset of Maple to Matlab users.> I don't
know about Mathcad, but MATLAB's Symbolic Math Toolbox 3.0.1
(the> latest version) provides quite a bit of the
functionality of the Maple 8> kernel. To differentiate x^2 and
obtain 2*x, the commands you need are:> % Declare x as a
symbolic variable> syms x> % Define the symbolic expression --
you don't need to do this, but I usually> do so> f = x^2;> %
Differentiate -- can omit x if that's the only symbolic
variable in f> df = diff(f, x)> % Integrate df -- can omit x
if that's the only symbolic variable in df> f2 = int(df, x)>
You can plot this using MATLAB's graphics capabilities:>
ezplot(f)> There are similar functions for performing surface,
mesh, or contour plots> on symbolic objects.> The Symbolic Math
Toolbox does not include the Maple GUI or any of Maple's>
graphics capabilities, and (if I remember correctly) restricts
four or five> of the Maple programming commands (PROC and WITH
are the two big ones I can> remember; you can include some
libraries with maple('with', libraryname) but> not all.) The
Extended Symbolic Math Toolbox removes the limitations on the>
programming commands, but does not remove the restriction on
Maple's> graphics capabilities. [If the Extended toolbox had
Maple's graphics> capabilities, it would essentially _be_
Maple.]> If anyone's still reading this and would like more
information on the> Symbolic Math Toolbox or the Extended
Symbolic Math Toolbox, take a look at> these pages:>
http://www.mathworks.com/products/symbolic/>
===
http://www.mathworks.com/products/extsymbolic/Subject: Re:
Recommendation between two programs...> I have looked at
Mathlab version 12 or 13. The> student version for $99
includes the symbolic toolkit.> Without it, the program is not
symbolic at all.> I think the commitment to symbolic
interchange> of data with maple was (on the product I tested)
either> poorly done or dubiously designed. E.g.> I don't mind
declaring symbols as> syms x> but you would then expect that
syms would interact> with numbers from matlab.> My
recollection is that x+1 fails to work, because> 1 is not a
symbol. You must create a symbolic object> that is also 1.
Otherwise you get a type error.No, that works. The toolbox
should convert the 1 to a symbolic objectinternally and return
x+1. I've just verified that it does so. There weresome bugs
relating to having the result of a computation be the
singlesymbol n or t; however if I remember correctly those
have been fixed in thelatest version.> Since I reported this
to MathWorks, and my trial version> expired and I haven't
gotten a new one, I don't know> what they did with this. My
impression was that Mathworks> treated my report as a user
error and discarded it.I checked our records. The problem you
reported was basically thatsomething like simplify(34) did not
call the Symbolic SIMPLIFY functionthat is only defined for
symbolic objects. This is because 34 is not asymbolic object
-- it is a double array and MATLAB treats it as such untilyou
explicitly convert it to something else. Either
simplify(sym(34)) orsimplify(34+sym(0)) would work by
converting the expression inside theSIMPLIFY call into a
symbolic object, then calling the symbolic SIMPLIFYfunction on
it.There is a problem with doing an automatic cast type of
operation on 34 ina situation like this, which is what I
gather you were expecting.Currently, SIMPLIFY is only defined
for symbolic objects. What if someoneelse creates a class (say
a polynomial class) and implements the SIMPLIFYfunction for
that class? MATLAB would no longer be able to do theautomatic
cast to symbolic -- it would have to guess whether you wanted
34to be a symbolic object or 34 to be an object of that other
class. Now it'slikely that in some cases MATLAB would guess
correctly ... and in some casesit wouldn't.If you want this to
work, there is a way to do it ... but be aware of theproblem I
listed above. Create a directory, @sym, inside a directory
onyour MATLAB path. The @sym directory should not be on your
path, just itsparent. Put the following function inside that
directory as simplify.m:function y=simplify(x)y =
simplify(sym(x));> Why do you think that plotting makes a
numeric system> into a symbolic system? If you believe that,
then this> newsgroup should be discussing any program capable
of> output :)The example with EZPLOT was simply intended to
show that MATLAB could notonly manipulate symbolic objects but
could also display them graphically.-- Steve
===
Lordslord@mathworks.comSubject: Re: Recommendation between two
programs... by support1.mathforum.org (8.11.6/8.11.6/The Math
Forum, $Revision: 1.9 primary) id i2B4G3U23958;My suggestion
is that before buying you would try Mathematicawww.wolfram.com
and Derive www.derive.com , free fully-functional30 days trial
versions are available. There is also a
free-trialfully-functional MuPAD 3 at www.mupad.com but now
I'd say it'ssomewhat raw as for my taste; however in the years
to come whereyour career hopefully will go upward, MuPAD most
probably willbe essentially much more influential. I use all
of them as wellas Maple over a decade for scientific
computing.As for Maple 9.03, this applications is a savage
perfidious attackon the Maple users. Some spits and shrieks at
the most activeMaple group comp.soft-sys.math.maple include>
Maplesoft is ing around with Macintosh users.> I can't believe
they released software that has such poor> performance.
Shouldn't that have been spotted during development> What
about calling 9.00 Beta-1, 9.01 Beta-2,> 9.02
ReleaseCandidate-1, 9.03 ReleaseCandidate-2?> Do you know how
much time for installation, re-installation,>
re-re-installation etc. you are thieving from your customers?>
And: we are still waiting for the 'true' 9. release.> Are you
kidding? Of course they should not sell software> which is at
that level of maturity. These are quite common> terms to
describe /internal/ stages of development.> The soap opera
they are showing right now with the> releases of 9.00, 9.01,
9.02, 9.03> I just don't want to buy big, substantial bugs>
which hinder the use of basic features.> Did you watch
Maplesoft over the years? The price went> up, their hype went
up, their arrogance went up -> and the quality went down.and
finally, a yet another Maple expert's opinion:> I think
Maplesoft's days are numberedMy personal impression is that a
fairly buggy Maple 8 was rewrittenhurry-scurry by loony grad
students, or else how to explain thediscovered by GEMM fact
that Maple 9 is even more buggy that notholy Maple 8 was? Over
my career I've tested dozens applications,and no doubt, Maple 9
and the Guinness Record Book'04 are a well-matched couple.Okay,
let's come back to the Maple bugs facts. Maple 9.03 has atleast
several thousands distinct bug manifestations; please
realizethat Maplesoft's now selling a BETA VERSION under the
price stickerof a fully tested commercial version.I claim this
because my company Cyber Tester, LLC is in the processof
deployment of the first world's automated testing
AI-drivenenvironment, it is a kind of intelligent bug
microscope.But even a worse point is that Maplesoft not only
lost the lastdroplet of shame buying a beta version but that,
as our diagramsshow, Maplesoft has already lost _control_ over
the whole Mapledevelopment; to see this clearly you may wish to
visit our MapleBugs Encyclopaedia http://maple.bug-list.org/
during the severalnext months. By the way, within the next 10
days we will uploada next major upgrade of the Maple bugs
database.For you the above-said lost of control means that if
you wouldbuy Maple 9, there is none guarantee that you would
ever buyMaple 10 while most probably in the years to come the
other3 companies will survive.Had you have any other questions
please feel absolutely free to drop me a message at
vb@cybertester.com . On a personal note,biochemistry and
molecular biology is my first love, and theisoelectric point
of proteins'.Best wishes,Vladimir BondarenkoGEMM
architectCo-founder, CEO, Mathematical DirectorCyber Tester,
LLChttp://www.cybertester.com/http://maple.bug-list.org/http:/
/www.CAS-testing.org/.........................................
===
..........................Subject: Free Math HelpWe are an
experienced math tutoring organisation operating out
ofCanada.To receive a free booklet from us, giving you an
example of how weteach specific math concepts, email
===
mathhelpbooklets@mail.comSubject: Re: some questions about
Hensel lifting> U = (Bk C) mod Gk mod pk (about mod Gk and mod
Hk see below)> V = (Ak C) mod Hk mod pk> G2k = Gk + pk U> H2k
= Hk + pk V> C' = D + Ak U + Bk V> A2k = Ak - pk C' Ak> B2k
= Bk - pk C' Bk> If my recollection is correct, it is
possible to divide C' Ak> by Gk mod pk, and that will keep the
degree of A2k < deg(G2k).> This is compensated by the division
of C' Bk and Hk.> Question 2: How are linear and quadratic
combined in practice?> In giac/xcas, I use quadratic lift
for polynomial factorization,> except for the last step if a
linear lift is sufficient> (in combination with true factor
tries). Note that the> real situation is a little bit more
complicated since you have> to lift a product of n terms
instead of 2.> You can have also a look in mupad source code>
in lib/POLYLIB/FACTOR/plift.mu> the bound2list seems to find a
best strategy in mixing> linear/quadratic lifts. But I'm not
sure it is really> better than quadratic except last step>
since you need to compute the Ak and Bk.I just performed a
little experimentation following up your idea and it seems to
work, assuming you meant to pair As with Hs and Bs with
Gs:take (C' Ak) mod Hk mod pktake (C' Bk) mod Gk mod pkBut
I'll have to scribble a little to prove to myself that this
works.........................................................
>> at http://www.TitanNews.com <<<<-=Every Newsgroup -
Anonymous, UNCENSORED, BROADBAND Downloads=-