mm-302
===
Subject:
: Re: NDSolve and time-delayed equations?the
equation you posted is solvable in Mathematica 5.0 and i'mnot
sure if it's delay equation. like...Plot[Evaluate[ n[t] /.
NDSolve[{n'[t] == 0.5*n[t]*(1 - n[t]/100),n[0] == 10}, n, {t,
0, 20}] ], {t, 0, 10}]the one with [t, t-lag] is a delay
equation though. what kind of problem results in an equation
as thefirst? would you mind explain it? sean> Can NDSolve be
used to approximate the dynamics of> continuous but >
time-delayed equations? Here is an example of the> standard
continuous > logistic model used in ecology:> NDSolve[{n'[t]
== 0.5*n[t]*(1 - n[t]/100), n[0] ==> 10}, n, {t, 0, 20}]> It
does, of course, have an analytical solution.> A time-delayed
version would make the derivative a> function of two > values:
n'[t,t-lag], but I can't figure out if a> formulation like this
> is possible. The key thing seems to be that while> the
derivative of n > is a simple function of two parameters, n
itself is> not.> Any suggestions, other than iterating as a>
discrete-time model with > very small time-steps?> Gareth
Russell> Columbia
University__________________________________Do you
Yahoo!?Yahoo! Mail - More reliable, more storage, less
spamhttp://mail.yahoo.com
===
Subject:
: Re: mathematica and image...
help me please>What file format for the image?i dont have
preference>What version of Mathematica?5.0
===
Subject:
Creating a symmetric matrixAn equivalent but perhaps even
simpler version is:A + Transpose[A] -
IdentityMatrix[First[Dimensions[A]]*AAndrzej> It seems that I
was the only one who thinks that a triangular matrix> is a
strick triangular matrix, that is with zeros on the diagonal.>
had I not made this assumption I would have suggested:>
A+Transpose[A]*Array[KroneckerDelta[#1 !=
#2,True]&,Dimensions[A]]> Andrzej>Greetings,>How can I
efficiently build a symmetric matrix from an
upper>triangular>one, i.e., extract the upper triangular
elements and insert them into>the lower triangle in such a way
as to make the resulting square>matrix>symmetric?>Mark>> The
most natural way must be>> A+Tranpose[A]>> e.g.>> A =
Array[KroneckerDelta[#1 < #2, True] & , {3, 3}];>> {{0, 1, 1},
{0, 0, 1}, {0, 0, 0}}>> A + Transpose[A]>> {{0, 1, 1}, {1, 0,
1}, {1, 1, 0}}>> Andrzej Kozlowski>> Chiba, Japan>>
http://www.mimuw.edu.pl/~akoz/
===
Subject:
: nonlinear pde systemI
would like to solve a nonlinear parabolic pde systemof two
variables u, v. Each in turn is a function of(x,t). In future,
I would like to extend such a thingto (x, y, z, t). Please let
me know how to do this, ifpossible in
Mathmematica.-Harsha===
'
'
'
'
'
Subject:=The equations look like
this:a*u_t = b*u_xx + c*f(u,v)d*v_t = e*v_xx + h*g(u,v)for-ll
<= x <= ll0 <= t <= 10with boundary conditions: u_x(+/-ll, t),
v_x(+/-ll,t).===
Subject:=_______________________________
___Do you Yahoo!?Yahoo! Mail - More reliable, more storage,
less spamhttp://mail.yahoo.com
===
Subject:
: Re: ReduceTry using a
symbol rather than ?$Versionform=Reduce[{ A Exp[x y]==B Cos[x
y]+C Sin[x y], D Exp[-x y]==B Cos[x y]+C Sin[x y], (D[A Exp[x
y],y]==D[(B Cos[x y]+C Sin[x y]),y]), (D[D Exp[-x y],y]==D[(B
Cos[x y]+C Sin[x y]),y])}/. y->-a/2, {A,B,C,D}]x == 0 && A ==
0 && B == 0 && D == 0 || A != 0 && x == 0 && B == A && D == A
|| ((a*x)/2 - Pi)/(2*Pi) [NotElement] Integers && A == 0 && B
== 0 && C == 0 && D == 0 || 0[1] [Element] Integers && a != 0
&& x == (2*(2*Pi*0[1] + Pi))/a && A == 0 && B == 0 && C == 0
&& D == 0Using D as both a constant and a function appears
very risky. Likewise using C as both a plain constant and a
constant of integration is causing the presence of
0[1]form=Reduce[{ A Exp[x y]==B Cos[x y]+k Sin[x y], d Exp[-x
y]==B Cos[x y]+k Sin[x y], (D[A Exp[x y],y]==D[(B Cos[x y]+k
Sin[x y]),y]), (D[d Exp[-x y],y]==D[(B Cos[x y]+k Sin[x
y]),y])}/. y->-a/2, {A,B,k,d}]x == 0 && A == 0 && B == 0 && d
== 0 || A != 0 && x == 0 && B == A && d == A || ((a*x)/2 -
Pi)/(2*Pi) [NotElement] Integers && A == 0 && B == 0 && k == 0
&& d == 0 || C[1] [Element] Integers && a != 0 && x ==
(2*(2*Pi*C[1] + Pi))/a && A == 0 && B == 0 && k == 0 && d ==
0Bob Hanlon<< I find Reduce in version 5 seems to have lost
some functionality. Inversion 4 the inputform = Reduce[{A
Exp[? y] == B Cos[? y] + C Sin[? y] /. y -> -a/2, D Exp[-? y]
== B Cos[? y] + C Sin[? y] /. y -> a/2, (D[A Exp[? y], y] ==
D[(B Cos[? y] + C Sin[? y]), y]) /. y -> -a/2, (D[D Exp[-? y],
y] == D[(B Cos[ ? y] + C Sin[? y]), y]) /. y -> a/2}, {A, B, C,
D}] produces a perfectly sensible result startingA == 0 && D ==
0 || A == B && D == B && ? == 0 && ? == 0.... In version 5 the
same input causes a long period of cogitation (far longerthan
version 4 took to produce its result), and finally emerges
with asuggestion that I should look for further information on
its failure whichis not yet there. Mathematically, the problem
is relatively straightforward: the equationsreduce to M x =
0,where x={A,B,C,D}, so we just need the conditions underwhich
the determinant of M is zero, and any other special cases. Are
there new controls for Reduce in version 5 which will allow
itsfunctionality to be recovered, or is it irrevocably broken?
>>
===
Subject:
: Graphs of Sequences of Functions:(Hi all, I
am trying to do the graphs of some sequences of
functions,usingmathematica 4.0, in order to see if their
convergence is uniform or not. For example : Fn(x) = x^n , 0=<
x =<1. The problem is that the graphs have to be 2D (not 3D)
but I cannotfind a way to do that. Any suggestions?THANK YOU!
Eleanna
===
Subject:
: Re: Re: Integrate vs NIntegrateThis worked for
me on Mathematica 4.2:f = D[1/(1 + (1 + (a - b)^2)), a];h[a_]
:= NIntegrate[f*b*(1 - b)^2, {b, 0, 1}];FindRoot[h[a] == 0,
{a, 0, 1}]Output:{a ->
0.397861}Reza-------------------------------------------------
------------------------Reza Malek-Madani, Director of
Research and ScholarshipResearch Office, MS 10m, Nimitz Room
17 589 McNair Road U.S. Naval Academy Annapolis MD 21402-5031
Phone: 410-293-2504 (FAX -2507), DSN: 281-2504Email:
research@usna.edu---------------------------------------------
----------------------------->Curt Fischer
03/12/04 11:39 PM >> I am confused why NIntegrate misbehaves
on such a simple function as> mine.> Here is what I have:>
In:> f=D[1/(1+(1+(a-b)^2)),a];>
g=Integrate[f*b*(1-b)^2,{b,0,1},Assumptions->0
FindRoot[g==0,{a,0,1}]> Out:> a->0.397207> However, since
Integrate takes long, I tried to use NIntegrate> instead and
this is what I get> In:> f=D[1/(1+(1+(a-b)^2)),a];>
g:=NIntegrate[f*b*(1-b)^2,{b,0,1}];> FindRoot[g==0,{a,0,1}]>
Out:> a->1> or, FindRoot+NIntegrate give me the upper boundary
of a. If I> abandon the secant method and turn to Newton, i.e.
use> FindRoot[g==0,{a,0.5}] instead then I get the message
that Jacobian is> singular at a=0.5 and get no solution.
Perturbing the starting value> of a does not help.> What is
going on here?My guess is that NIntegrate does not do problems
with symbolicparameters,and yet you are trying to make it
calculate the integral of anexpressioninvolving a, with no
numerical value for a defined.-- Curt Fischer> PS. Is there a
way to get M5 to tackle the problem where:> (1) I define some
function f[x]:=NIntegrate[g[x,y],{y,0,1}], then> (2) take the
derivative of f[x] with respect to x, say h[x]:=f'[x]> and
then (3) Use FindRoot to find x such that h[x]==0
===
Subject:
: Re:
Undiscovered Bug?MathGroup: When expr is a list, expr[[{i1,
i2, ... in ]] gives a list ofparts. In general, the head of
expr is applied to the list of parts.ReplacePart's behavior,
however, produces a different result ... Recall, we had for
Part ... In[5]:= (x/y)[[{2, 1, 1, 0, 2, 2, 0}]] 2 2 Times x
Out[5]= --------- 3 y But, we have for ReplacePart ... In[1]:=
ReplacePart[(x/y), a, {2, 1, 1, 0, 2, 2, 0}] x
ReplacePart::partw: Part {2, 1, 1, 0, 2, 2, 0} of - does
notexist. yMaybe I'm still missing something, but it seems to
me Part should berevised to detect the same errors as
ReplacePart.Harold
===
Subject:
: Accounting of downloadsSomeone sent
an email to me that said people who have never heard of
Mathematica are afraid of MathReader because they think it
might have viruses or something, so, if they aren't afraid of
looking at web sites they can look
at:http://users.adelphia.net/~cnelson9/It turns out that if
you save a Mathematica notebook as html with the sections
closed, it chops it up into bite sized pieces with links to
the sections. Cliff NelsonThe second volume of Buckminster
Fuller's books Synergetics was published in 1979, twenty five
years ago. All of humanity had to learn things like the
synergetics coordinate system to do more with less soon or it
would be curtains for humanity according to Bucky, but look at
these low numbers of downloads. And life still goes on.Even
Bucky Fuller fans are just not interested. Cliff NelsonBegin
forwarded message:Dear Clifford Nelson,At the moment, I only
have access to the top 150 downloads. You have two notebooks
in the top 150. (I only have one, myself).--Ed Pegg Jr.Could
you give me an accounting of the number of downloads of my
notebooks from MathSource
please?http://library.wolfram.com/infocenter/search/?search_
results=1;search_person_id=607? Buckminster Fuller Notebooks
[in /MathSource: Packages and
Programs/]http://library.wolfram.com/infocenter/MathSource/600
/#129 in all time downloads. In the past year, notebook
downloaded 1118 times? Bucky Number Mandelbrot [in
/MathSource: Packages and
Programs/]http://library.wolfram.com/infocenter/MathSource/428
/#86 in all time downloads. In the past year, notebook
downloaded 1335? Four Triangle Fractals using Bucky Numbers
and Synergetics Coordinates [in /MathSource: Packages and
Programs/]http://library.wolfram.com/infocenter/MathSource/754
/Not in top 150? Solving Matrix Problems Using Bucky Numbers
[in /MathSource: Packages and
Programs/]http://library.wolfram.com/infocenter/MathSource/
4277/Not in top 150Cliff Nelson