mm-447
Subject: v.difficult (for me) 2D integration by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i0NDiW414044; I've been having
trouble with the following difficult (for me anyway)finite 2D
integration.Integrate f(x,y) =
(x^2+y^2+2)/(sqrt((x^2+y^2)(x^2+y^2+4))) dx dy between x=-1
and 1, y=-1 and 1I need to do this over a square in x-y plane
not a circle (theintegral can be evaluated analytically in
polar coordinates over acircle in x-y but I haven't been able
to do it over a square yet). Iknow the function diverges to
infinity at x=y=0 but since anequivalent integration can be
done in polar coordinates I figure itmust be doable in
cartesians. Diverging to infinity at x=y=0 does notimply it
can't be integrated.I tried mathematica but could not get it
to produce a numerical result(it started giving me back random
eliptical functions and things) so imaking sure it only
evaluated f(x,y) very close (but not at) x=y=0.Iterations
dividing the x-y area into ~15000^2 parts did not get
theanswer to converge with any accuracy. NB - this program
worked forsimpler integrals though took a suprising number of
iterations to getaccurate.I am not looking for an analytical
solution, only a numerical one (ifit exists!) so if anyone
knows a way of doing this I'd be verygrateful for
help!JamiePS. I actually have to do this integration over lots
of diferentlimits of x & y so I would like to know the method
as well as theanswer (if it exists) if possible
===
Subject: Re:
v.difficult (for me) 2D integration .... Integrate f(x,y) =
(x^2+y^2+2)/(sqrt((x^2+y^2)(x^2+y^2+4))) dx dy between x=-1
and 1, y=-1 and 1 .... I know you want the answer numerically
rather than analytically, but FWIW it seems to be an elliptic
integral. Changing to polar coordinates and chopping your
square into eight triangles gives8*((integral from 0 to
pi/4)d(theta))(integral from 0 to sec(theta))drof the
integrand(r^2 + 2)/sqrt(r^2 + 4).The integral with respect to
r needs only elementary functions, so the whole thing
is4*(integral from 0 to pi/4)d(theta)of the integrand(sqrt(1 +
4(cos(theta))^2))/((cos(theta))^2).The substitution t =
tan(theta) then gives4*(integral from 0 to 1)dtof the
integrandsqrt((5 + t^2)/(1 + t^2))which is close to one of the
standard forms of (complete?) elliptic integral. Perhaps
someome more familiar with those can take it from
there.
===
Subject: Re: v.difficult (for me) 2D integrationI've
been having trouble with the following difficult (for me
anyway)finite 2D integration.Integrate f(x,y) =
(x^2+y^2+2)/(sqrt((x^2+y^2)(x^2+y^2+4))) dx dy between x=-1
and 1, y=-1 and 1Derive found the integral in the first
quadrant to be about 2.028775634but it has built in checks
when doing numerical integration and didwarn about dubious
accuracy. Reducing the number of digits of precisionand asking
it to repeat the calculations gave answers that were
slightlysmaller, again with dubious accuracy warnings, but not
wildly varying.So I think your answer is about 4x that
value.
===
Subject: Re: v.difficult (for me) 2D integration>I've
been having trouble with the following difficult (for me
anyway)>finite 2D integration.>Integrate f(x,y) =
(x^2+y^2+2)/(sqrt((x^2+y^2)(x^2+y^2+4))) dx dy> between x=-1
and 1, y=-1 and 1 Derive found the integral in the first
quadrant to be about 2.028775634 but it has built in checks
when doing numerical integration and did warn about dubious
accuracy. Reducing the number of digits of precision and
asking it to repeat the calculations gave answers that were
slightly smaller, again with dubious accuracy warnings, but
not wildly varying. So I think your answer is about 4x that
value.I agree. So the answer seems to be about 8.115
.Mathematica can do one integration symbolically, giving a
result in termsof incomplete elliptic integrals of both first
and second kinds. But itcannot then integrate that result
symbolically. Numerically integratingfrom y = a to y = 1,
using values of a getting progressively closer to 0,and then
doubling that result, I got the answer above.
===
Subject: Re:
v.difficult (for me) 2D integration>Integrate f(x,y) =
(x^2+y^2+2)/(sqrt((x^2+y^2)(x^2+y^2+4))) dx dy> between x=-1
and 1, y=-1 and 1> Derive found the integral in the first
quadrant to be about 2.028775634>...> So I think your answer
is about 4x that value.Mathematica can do one integration
symbolically, giving a result in termsof incomplete elliptic
integrals of both first and second kinds. But itcannot then
integrate that result symbolically. Numerically
integratingfrom y = a to y = 1, using values of a getting
progressively closer to 0,and then doubling that result, I got
the answer above.I don't know what eversion of MMA you are
using or what you were givingMMA 5.0 seems happy with the
following, without needing to choose
'a':NIntegrate[(x^2+y^2+2)/Sqrt[(x^2+y^2)(x^2+y^2+4)],{x,0,1},
{y,0,1}]2.02878
===
Subject: Re: v.difficult (for me) 2D
integration>>Integrate f(x,y) =
(x^2+y^2+2)/(sqrt((x^2+y^2)(x^2+y^2+4))) dx dy>> between x=-1
and 1, y=-1 and 1> Derive found the integral in the first
quadrant to be about>>2.028775634 ...>> So I think your answer
is about 4x that value.>Mathematica can do one integration
symbolically, giving a result in>terms of incomplete elliptic
integrals of both first and second kinds.>But it cannot then
integrate that result symbolically. Numerically>integrating
from y = a to y = 1, using values of a getting
progressively>closer to 0, and then doubling that result, I
got the answer above. I don't know what eversion of MMA you
are using or what you were giving MMA 5.0 seems happy with the
following, without needing to choose 'a':
NIntegrate[(x^2+y^2+2)/Sqrt[(x^2+y^2)(x^2+y^2+4)],{x,0,1},{y,
0,1}] 2.02878I had asked MMA 5.0 toNIntegrate[(-(1/Sqrt[4 +
y^2]))*(2*I*((4 + y^2)*EllipticE[I*ArcCsch[y],y^2/(4 + y^2)] -
2*EllipticF[I*ArcCsch[y], y^2/(4 + y^2)])), {y, 0, 1}]and it
did not seem happy with the singularity at y = 0. But I did
not,as MMA suggested, Try using larger values for any of
$MaxExtraPrecisionor the options WorkingPrecision, or
SingularityDepth and MaxRecursion.Perhaps if I had done so,
MMA would have been happy.
===
Subject: Gauss Method
2+4+6+8+...512 by support1.mathforum.org (8.11.6/8.11.6/The
Math Forum, $Revision: 1.9 primary) id i0NG7F525693;My
professor wants me to learn how to do the Gauss
problemsIE:2+4+6+8+...+512 and 1+2+3+4+5+...+1000 using
algebra not any kind
===
Subject: Re: Gauss Method
2+4+6+8+...512 My professor wants me to learn how to do the
Gauss problems IE:2+4+6+8+...+512 and 1+2+3+4+5+...+1000 using
algebra not any kindYou can write 1+2+3+...+1000 as 1+ 2+ 3+
... + 500 +1000+999+998+ ... + 501 Now adding vertically first
gives 500 copies of 1001 to be added.this is, more or less,
what Gauss did.
===
Subject: Re: Gauss Method 2+4+6+8+...512 My
professor wants me to learn how to do the Gauss problems
IE:2+4+6+8+...+512 and 1+2+3+4+5+...+1000 using algebra not
any kindReverse sequence2 + 4 + ... + 510 + 512512+510 +...+ 4
+ 2add514+514 +...+ 514 + 514How may 514 are there? Call that
number n.The total is 514n. Now divide by 2 for answer.Why is
necessary to divide by 2?
===
Subject: Gauss Method by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i0O1ZU606812;The story goes that
Karl Frederich Gauss, as a young boy, add up 1 + 2+ 3 + ... +
100by noting that the end-numbers have a sum of: 1 + 100 =
101then the next end-numbers have: 2 + 99 = 101and the next
end-numbers have: 3 + 98 = 101 ... and so on.Since there are
50 such sums: 50 x 101 = 5,050 is the total.Your problems can
be handled in a similar fashion.2 + 4 + 6 + 8 + ... + 508 +
510 + 512The end-sum is: 2 + 512 = 514And the next is: 4 + 510
= 514And the next is: 6 + 508 = 514 ... and so on.There are 128
such pairs to add up.The sum is: 128 x 514 = 65,792You should
be able to the last problem yourself.
===
Subject: Gauss Method
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i0O4fRv19907;Thank you Thank you
Thank you!
===
Subject: Tell me ...... by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i0O5IpT22077;Spare me qim. Why the hell is it wrong?? Just
because some poxy bookhasn't got those angles in *their*
answers??Don't you realise that when solving an equation, you
can always checkyour answer(s) by substituting into the
equation and checking that youget equality!? (Damn I should
have shouted that). Try doing somethingintelligent like this
next time qim. Had you done it here, you wouldhave immediately
seen that your answersor tan A = 2 arctan = 63.4.bc1st =
63.4.bc and 3rd 180.bc + 63.4.bc = 243.4.bcare in fact
correct.Good god man, why assume *you* are the wrong one if
you get an answerdifferent to some poxy text book. Use a bit
of common sense next time- check your answers fer cryin' out
loud (as I suggested above).And have a bit of faith in *your*
calculations. Looks to me like yourbook is developing quite a
history for screwing up its answers (as innot including *all*
solutions).
===
Subject: Challenging the algebra by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i0O61El25397;Funny you should
mention that, for I've got a result here that I thinkwill
allow me to prove that there is an inherent flaw in...ALGEBRA!
I'm thinking about it... thinking... thinking....Yes! I'm
right!! Proof to follow, but first, let's see if any of youcan
prove it first...
===
Subject: Time better spent fishing by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i0OEswH28094;Give it up and move
on!
===
Subject: exercise at algorithm analysis by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i0OKRuN18952;hi to allhow/were i
can get exercise at algorithm
analysis?thankssomebody
===
Subject: How much does Fred weigh by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i0Q05C304899;Fred is a meat cutter
who sells chicken breast for $3.88 per pound,ground beef for
$2.75 per pound, and pork chops for $3.50 per pound. Fred
wears size 36 jeans, is aobut five foot five, wears baggy
shirtsand has size 12 shoes. What does Fred weigh?
===
Subject:
Re: How much does Fred weigh Fred is a meat cutter who sells
chicken breast for $3.88 per pound, ground beef for $2.75 per
pound, and pork chops for $3.50 per pound. Fred wears size 36
jeans, is aobut five foot five, wears baggy shirts and has
size 12 shoes. What does Fred weigh?And more importantly,
what's a Henway?
===
Subject: Re: How much does Fred weigh> Fred
is a meat cutter who sells chicken breast for $3.88 per pound,>
ground beef for $2.75 per pound, and pork chops for $3.50 per
pound. > Fred wears size 36 jeans, is aobut five foot five,
wears baggy shirts> and has size 12 shoes. What does Fred
weigh? And more importantly, what's a Henway?About 3
pounds.
===
Subject: Re: How much does Fred weigh Fred is a meat
cutter who sells chicken breast for $3.88 per pound, ground
beef for $2.75 per pound, and pork chops for $3.50 per pound.
Fred wears size 36 jeans, is aobut five foot five, wears baggy
shirts and has size 12 shoes. What does Fred weigh?Chicken,
beef and pork?
===
Subject: regexI want to match
mailto:null@void.com......gif in a string that contains
manysimilar patterns.I used mailto:(.[^]*) to match each
mailto: ... string and parsed outthe email address.But now I
can't seem to extend the match to the first 'gif'.I have tried
a number of different combinations but without the match
Ineed.I know what I need. I need to do as I did in the example
above except Ineed to not include a group of characters instead
of just one. 
===
Subject: Re: JSH: Algebra is supreme Adjunct
Assistant Professor at the University of Montana....> Dedekind
has shown that the algebraic integers form a gcd domain
(actually> a bit more: it is a Bezout domain), so the gcd
exists. It is quite> similar to the standard gcd in the
integers, except that this definition> is not concerned with
the sign, so in the integers, the gcd of 6 and 14> is +2 or
-2.> Now a little theorem:> gcd(a * b, c) | gcd(a, c) * gcd(b,
c).> You might try to prove it. The prove is easy in a unique
factorisation> domain, less easy in other domains.This is true
in any gcd domain. Don't know if it is true in apre-Schreier
domain by itself, but here goes my proof for
gcd-domain.(Seefor a table of implications; gcd implies
pre-Schreier).In a pre-Schreier ring, any two factorizations
of the same elementhave common refinements; and both gcd and
pre-Schreier domain implyProp. If gcd(a,b)=1, then gcd(a,bc) =
gcd(a,c).Proof. Let x = gcd(a,c), y = gcd(a,bc). Clealry x|y.
Write a = yu, bc= yv. Then the factorizations bc=yv have a
common refinement, so bc = yv = p*q*r*s, where b = pq, c = rs,
y=pr, v=qs. Thusp|gcd(b,y). But if p|y then p|a, so p|gcd(b,y)
= 1. Therefore, p is aunit. But that means that y=pr divides
c=rs, so y|a, y|c, hencey|gcd(a,c)=x. Thus, x|y and y|x,
giving that, up to units, they areequal. QED. Now back to the
condition we have: we want to prove that gcd(ab,c)divides
gcd(a,c)*gcd(b,c).Let r = gcd(a,c), s = gcd(b,c). We want to
show that gcd(ab,c) dividesrs.Write a = ru, c = rv with
gcd(u,v) = 1; b = sw c = sy with gcd(w,y) = 1.Using the
pre-Schreier property, we can take the two factorizations ofc,
rv = sy and find a common refinement, that is, writec =
h*k*m*n, wherer = hk, v = mns = hm, y = kn.We haveab =
(ru)(sw) = (hku)(hmw) = h^2*k*m*u*w. c = h*k*m*n.rs = h^2*k*m;
and we see that the gcd of ab and c is h*k*m*gcd(huw,n).Since
n|y and gcd(n,y) = 1, we have gcd(huw,n) = gcd(hu,n). Since
n|vand gcd(u,v)=1, we have gcd(n,u)=1, so gcd(huw,n)
=gcd(hu,n)=gcd(h,n). So gcd(ab,c) = h*k*m*gcd(h,n), which
divides h*k*m*h = h^2*k*m = rs =
gcd(a,c)*gcd(b,c).QED
===
Subject: Re: JSH: Algebra is supreme
Adjunct Assistant Professor at the University of Montana....>
Dedekind has shown that the algebraic integers form a gcd
domain (actually> a bit more: it is a Bezout domain), so the
gcd exists. It is quite> similar to the standard gcd in the
integers, except that this definition> is not concerned with
the sign, so in the integers, the gcd of 6 and 14> is +2 or
-2.> Now a little theorem:> gcd(a * b, c) | gcd(a, c) * gcd(b,
c).> You might try to prove it. The prove is easy in a unique
factorisation> domain, less easy in other domains.This got me
thinking and actually it is not always true in arbitrarygcd
domains (I was too much working with the ideas I had rather
thanreal proofs). It is, however, true in Bezout domains
(which thealgebraic integers are).Proof: gcd(a, c) = p * a + q
* c and gcd(b, c) = r * b + s * c for some p, q, r, s, by being
Bezout. gcd(a, b) * gcd(b, c) = (p * r) * a * b + (p * a * s +
r * b * q + q * s * c) * c, which is linear in a * b and c,
and so is divisible by gcd(a * b, c).There are gcd-domains
where the theorem is false: Q(sqrt(-5)) is one.Set a = 2, b =
3, c = 1 + sqrt(-5), we now have: gcd(a, c) * gcd(b, c) = 1 *
1 = 1, gcd(a * b, c) = c.Ehr, Q(sqrt(-5)) is a field, so
everything divides everything. Iassume you meant Z[sqrt(-5)],
but in that case, you are incorrect: itis not a gcd
domain.Consider the numbers (1+sqrt(-5))*(4-sqrt(-5)) =
9+3*sqrt(-5) and (1+sqrt(-5))*(-1-2sqrt(-5)) = 9-3*sqrt(-5)I
claim they do not have a gcd. A number that divides both must
havenorm dividing both their norms. Since they are conjugates,
both normsare equal, and equal to:N(9+3*sqrt(-5)) =
N(3)*N(3+sqrt(-5)) = 3^2*(9+5)= 2*3^2*7Both are divisible by
3, so any gcd is a multiple of 3, hence has norma multiple of
3^2. They are also both divisible by 1+sqrt(-5), so agcd is a
multiple of 1+sqrt(-5), hence of norm a multiple of
2*3(therefore, of norm a multiple of 2*3^2). So the only
possibilitiesare norm 18 or norm 126.If a gcd had norm 126,
then it would be an associate of both9+3*sqrt(-5) and of
9-3*sqrt(-5); so 9+3*sqrt(-5) and 9-3*sqrt(-5)would be
associates. But this is false, since the only units here are1
and -1. So any gcd would necessarily have norm 18. So if a gcd
is ofthe form a+b*sqrt(-5), a and b integers, then we have a^2
+ 5b^2 =18. We may assume b is nonnegative. So 0<=b<=1; but
b=0 implies a^2=18,impossible, and b=1 implies a^2=13, also
impossible. Therefore, thereare no elements of norm 18, and in
particular, 9+3*sqrt(-5) and9-3*sqrt(-5) do not have a gcd in
Z[sqrt(-5)]. Therefore, Z[sqrt(-5)]is not a UFD.(In case you
are wondering how I got the example: since the classnumber is
5, find five nonprincipal ideals P,Q,R,S,T and consider
PQRSand PQRT. A gcd would be a multiple of the actual numbers
whichgenerate PQ, PR, and QR, so its prime factorization would
have to beof the form PQR*I; and I would have to be trivial by
uniquefactorization into primes. But PQR is not principal.
Having that as agoal, I just took the easiest examples I knew:
I took the elements1+sqrt(-5), 1-sqrt(-5), and 7. Factoring as
prime ideals, we have(1+sqrt(-5)) = PQ(1-sqrt(-5)) = PR (7) =
STwith none of P,Q,R,S,T principal. QR=(3), and it was just a
matter offinding out what RS and RT were.)
===
Subject: Re: JSH:
Algebra is supreme Adjunct Assistant Professor at the
University of Montana. [.snip.]Oops; sorry, a typo:(In case
you are wondering how I got the example: since the classnumber
is 5, find five nonprincipal ideals P,Q,R,S,T and consider
PQRSand PQRT. That should be class number is 2, of course.
Sorry...
===
Subject: recurrence relationmay you help me to
solve the recurrence relation below? / | ( nabla^{k+1} f )
(Y_1, ...,Y_{k+1}) = | = Y_{k+1} (nabla^k f (Y_1, ...,Y_{k}) )
| - sum_{i=1}^k nabla^k f (Y_1, .,nabla_{Y_{k+1}}Y_i ,..,Y_k)/
| nabla^1 f (X)= df X | | Y_i denotes a tangent vector field
over a differentiable manifold M,f is a real valued function
over M, f: M--> R, nabla denotes a linear (affine) connession;
nabla^{k+1} f := nabla (nabla^k f).Thank you very much ,