Nabble - MathGroup

[mg92601] Re: Real and Complex Roots presented in a single plot

2008-10-07T04:07:13Z

On Sep 19, 2:15 pm, Narasimham <mathm...@...> wrote:
> On Sep 12, 2:27 pm, magma <mader...@...> wrote:

> > > We can also recognize and see the real parts of al=
l =

> the
> > > complex roots of z where the curve is nearest to x -axis at {1.4,
> > > 4.2, 6.6, 9.6, 12.3, 15.3, 17.7}. They are near to x-values where the
> > > local maxima/minima occur.But we cannot 'see' their complex parts, as
> > > they need to be computed.
>
> > Please note that your claim that the real part of the complex roots is
> > found at the local minima of the given function as the indipendent
> > variable moves on the real axis, is not valid in general and wrong in
> > this particular instance.
>
> > For example in the interval {1,2} there are 2 local extrema found
> > using the derivative:
>
> > f[z_] := 1.3 Sin[1.7 z] + 0.6 Sin[4 z]
>
> > The derivative:
>
> > In[43]:= fp[z_] := D[f[z], z]
>
> > In[44]:= fp[z]
>
> > Out[44]= 2.21 Cos[1.7 z] + 2.4 Cos[4 z]
>
> > In[73]:= extrema = z /. FindRoot[fp[z], {z, #}] & /@ {1.4, 1.6}
>
> > Out[73]= {1.33751, 1.69029}
>
> > The first is the local minimum (see the plot).
> > But the complex roots in interval {1,2} are
>
> > 1.27946 + 0.374308 I and 1.27946 - 0.374308 I
>
> > The real part is 1.27946, while the minimum was at 1.33751
>
> (I do not have Presentations). My point is that 1.33751 is
> sufficiently near to and corresponding to the root place holder
> 1.27946 so that succesive tangents drawn in complex Newton-Raphson
> procedure the roots would not settle anywhere else.
>
> Narasimham

The point/ claim in a simple case is that e.g. in y = (x - x1)^2 +
a^2 there is a min at x = x1 and a complex root there with real part
x1, the full root being x1 (+/-) a *I .

Narasimham

[mg92600] Re: Axis in two scales

2008-10-07T04:07:02Z

Hi Nacho

You could certainly draw the graphic you want if you spend a lot of
time building it from scratch with graphics primitives -- I suggest
you start by drawing the x-axis you want with Lines.You might be able
to get close to what you want by experimenting with GraphicsGrid. In
either case you risk creating a graphic with a misleading x - axis, so
don' t forget to put some indication of the break of scale at x ==
0.5.

Or, you could use ListLogLinearPlot[t] and get close to what you are
asking for with no fuss or tricky programming, and get on with your
work.

Regards

Mark Westwood

On Oct 6, 9:13 am, Nacho <ncc1701...@...> wrote:

> Hello everyone.
>
> I have a dataset in pairs (x,y) that I would like to plot. The main
> problem is that the data set has interesting parts in the beginning
> that I would like to view clearly.
>
> As an example, create the following table:
>
> t = Table[{x, If[x < .5, Sin[100*x], Sin[x]]}, {x, 0, 5, 0.001}];
> ListPlot[t]
>
> You have a very detailed sine in the beginnng and the rest is less
> interesting.
>
> I would like to "split" the X axis in two, so the left half of the
> plot covers x={0,0.5} and the second half x={0.5,5}] in the same plot
> and with the right ticks.
>
> It is like a piecewise axis... is it possible?
>
> Thanks.

[mg92599] Re: integration

2008-10-07T04:06:51Z

Dear Experts,
I have been trying to simplify(integrate) the following function, but
M6 seems to give a complex answer which i cann't understand.. please
help.

x[s_]=\!$
\*SubsuperscriptBox[\(\[Integral]$, $0$, $s$]$Cos[
\*FractionBox[\(r\ t\ \((\(-\[Kappa]0$ + \[Kappa]1 + r\ \[Kappa]1)\)
+ $(1 + r)$\ S\ $(\[Kappa]0 - \[Kappa]1)$\ $(\(-Log[S]$ + Log[S
+ r\ t])\)\),
SuperscriptBox[$r$, $2$]]] \[DifferentialD]t\)\)

Regards,
RG

Hi, RG,
your problem seems to be in Mathematics, rather than in Mathematica. The integral you need (as it is now) is too cumbersome.
Make the integrand simpler by hiding and simplifying your notations, and the result will become more understandable.
For example, below is your integral along with the result in which I denoted some terms by a, b and c to make it shorter:
In[2]:= \!$
\*SubsuperscriptBox[\(\[Integral]$, $0$, $s$]$Cos[\
t\ a + b + c*Log[S + r\ t]] \[DifferentialD]t$\)

Out[2]= If[(Re[S/(r s)] >= 0 && S/(r s) != 0) || Re[S/(r s)] <= -1 ||
Im[S/(r s)] != 0, -(1/(2 a))
S^(-\[ImaginaryI] c) (-((\[ImaginaryI] a S)/
r))^(-\[ImaginaryI] c) (S^(2 \[ImaginaryI] c)
Gamma[1 + \[ImaginaryI] c, -((\[ImaginaryI] a S)/
r)] (-\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r]) + ((
a^2 S^2)/r^2)^(\[ImaginaryI] c)
Gamma[1 - \[ImaginaryI] c, (\[ImaginaryI] a S)/
r] (\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r])) + (1/(
2 a))(r s + S)^(-\[ImaginaryI] c) (-((\[ImaginaryI] a (r s + S))/
r))^(-\[ImaginaryI] c) ((r s + S)^(2 \[ImaginaryI] c)
Gamma[1 + \[ImaginaryI] c, -((\[ImaginaryI] a (r s + S))/
r)] (-\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r]) + ((
a^2 (r s + S)^2)/r^2)^(\[ImaginaryI] c)
Gamma[1 - \[ImaginaryI] c, (\[ImaginaryI] a (r s + S))/
r] (\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r])),
Integrate[Cos[b + a t + c Log[t r + S]], {t, 0, s},
Assumptions -> ! ((Re[S/(r s)] >= 0 && S/(r s) != 0) ||
Re[S/(r s)] <= -1 || Im[S/(r s)] != 0)]]

This is your result after simplification:

In[3]:= -(1/(2 a))
S^(-\[ImaginaryI] c) (-((\[ImaginaryI] a S)/
r))^(-\[ImaginaryI] c) (S^(2 \[ImaginaryI] c)
Gamma[1 + \[ImaginaryI] c, -((\[ImaginaryI] a S)/
r)] (-\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r]) + ((
a^2 S^2)/r^2)^(\[ImaginaryI] c)
Gamma[1 - \[ImaginaryI] c, (\[ImaginaryI] a S)/
r] (\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r])) +
1/(2 a)(r s + S)^(-\[ImaginaryI] c) (-((\[ImaginaryI] a (r s + S))/
r))^(-\[ImaginaryI] c) ((r s + S)^(2 \[ImaginaryI] c)
Gamma[1 + \[ImaginaryI] c, -((\[ImaginaryI] a (r s + S))/
r)] (-\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r]) + ((
a^2 (r s + S)^2)/r^2)^(\[ImaginaryI] c)
Gamma[1 - \[ImaginaryI] c, (\[ImaginaryI] a (r s + S))/
r] (\[ImaginaryI] Cos[b - (a S)/r] +
Sin[b - (a S)/r])) // FullSimplify

Out[3]= (1/(2 r))\[ExponentialE]^(-((\[ImaginaryI] (b r + a S))/
r)) (\[ExponentialE]^(2 \[ImaginaryI] b) S^(1 + \[ImaginaryI] c)
ExpIntegralE[-\[ImaginaryI] c, -((\[ImaginaryI] a S)/
r)] - \[ExponentialE]^(2 \[ImaginaryI] b) (r s + S)^(
1 + \[ImaginaryI] c)
ExpIntegralE[-\[ImaginaryI] c, -((\[ImaginaryI] a (r s + S))/
r)] + \[ExponentialE]^((2 \[ImaginaryI] a S)/
r) (S^(1 - \[ImaginaryI] c)
ExpIntegralE[\[ImaginaryI] c, (\[ImaginaryI] a S)/
r] - (r s + S)^(1 - \[ImaginaryI] c)
ExpIntegralE[\[ImaginaryI] c, (\[ImaginaryI] a (r s + S))/r]))

It is already comprehensible. Probably you may go still further by checking properties of the integral exponents.
Have a look into the book: Abramowitz, M. & Stegun, I. A. Handbook of Mathematical Functions with formulas,
graphs and mathematical tables. (National Bureau of Standards, 1964). Another way I would go is to try to transform it first,
and to integrate afterwards.

Success, Alexei

--
Alexei Boulbitch, Dr., Habil.
Senior Scientist

IEE S.A.
ZAE Weiergewan
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L-5326 Contern
Luxembourg

Phone: +352 2454 2566
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[mg92598] Re: NDSolve and error

2008-10-07T04:06:41Z

Hello!
Thanks for all your tips. It turned out that was my silly mistake
(should be y[x], not y[t]) but sometimes such mistakes are the most
difficult to find.
Best regards
John

[mg92597] Re: [mg92575] Re: Finding the optimum by repeteadly zooming on the solution space (or something like that)

2008-10-07T04:06:30Z

I appreciate both your help!
I was trying it the other way because I'm more interested in the speed of
the solution than in the precision. Ultimately, what I need to do is to find
the optimum "a" for at least thousands of values of r.
Say, for a couple of values:

Table[NMaximize[{obj, cpoAg1 == 0 && cpoAg2 == 0 && x > 0.1 && y > 0.1}, {a,

x, y}], {r, 0.4, 1, 0.2}]]

It seems that my approach doesn't improve the speed for most values of r, so
I'll try your suggestions.

If someone can suggest me a way of speeding things up, I'd really appreciate
it.

Kind Regards,

cd

2008/10/6 Bill Rowe <readnews@...>

> On 10/5/08 at 6:05 AM, cuak2000@... (Mauricio Esteban Cuak)
> wrote:
>
> >Hello everyone.I know somebody has probably writ mathematica code on
> >this problem. A general answer would probably benefit more people,
> >but I'll do the specific example, 'cause I'm not sure how to explain
> >it in another way.
>
> >I have these restrictions:
>
> >cpoAg1= ( -x + (70*a*(x^0.4 + y^0.4)^0.75)/x^0.6);
>
> >(*and*)
>
> >cpoAg2= ((70*(1 - a)*(x^0.4 + y^0.4)^0.75)/y^0.6 - 2*y);
>
> >(* And I need to find the "a" that will maximise this following
> >function. I don't need and exact solution, but something that is
> >sufficiently near *)
>
> >obj = -x^2/2 + 100*(x^0.4 + y^0.4)^1.75 - y^2;
>
> >(* "A" goes between 0 and 1 so I discretise to obtain the {x,y} that
> >maximise every "a"
>
> >the Table command gives me the list of rules which I can replace on
> >"obj" later to find the maximum. No problem here*)
>
> >rules = Table[
> >FindRoot[{cpoAg1 == 0, cpoAg2 == 0}, {{x, 0.1}, {y, 0.1}}], {a, 0.=
> 1,
> >1, 0.1}
> >];
>
> >(*the maximum is found with this following function *)
>
> >Max[Thread[ReplaceAll[obj, rules]]];
>
> >So far so good...I could discretise "a" as thinly as I want, but I
> >can't afford the luxury of being that inneficient, 'cause I've got
> >to do other things later with this part of the program.
>
> I don't understand why you would take this approach. Why not use
> the built-in function NMaximize, That is:
>
> In[20]:= NMaximize[{obj,
> cpoAg1 == 0 && cpoAg2 == 0 && x > 0.1 && y > 0.1}, {a, x, y}]
>
> Out[20]= {2084.72,{a->0.527688,x->22.6727,y->13.7173}}
>
>

--
Por favor eviten enviarme archivos adjuntos de Word o Powerpoint (
http://www.gnu.org/philosophy/no-word-attachments.es.html )

[mg92596] Re: Axis in two scales

2008-10-07T04:06:19Z

Nacho wrote:

> I have a dataset in pairs (x,y) that I would like to plot. The main
> problem is that the data set has interesting parts in the beginning
> that I would like to view clearly.
>
> As an example, create the following table:
>
> t = Table[{x, If[x < .5, Sin[100*x], Sin[x]]}, {x, 0, 5, 0.001}];
> ListPlot[t]
>
> You have a very detailed sine in the beginnng and the rest is less
> interesting.
>
> I would like to "split" the X axis in two, so the left half of the
> plot covers x={0,0.5} and the second half x={0.5,5}] in the same plot
> and with the right ticks.
>
> It is like a piecewise axis... is it possible?

Not exactly a piecewise axis but the function *ListLogLinearPlot* should
produce something close enough to what you have in mind.

t = Table[{x, If[x < .5, Sin[100*x], Sin[x]]}, {x, 0, 5, 0.001}];
ListPlot[t]
ListLogLinearPlot[t]

Regards,
-- Jean-Marc

[mg92595] Re: an interesting problem of supplying boundary condition for a

2008-10-07T04:06:08Z

Hi,

a) there is no derivative with respect of t, and the variable
is useless
b) Mathematica can't solve hyperbolic equations

Regards
Jens

pratip wrote:

> Hi All,
> Down is the equation system
>
> Jc = -\[Rho] Subscript[\[GothicCapitalD],
> k] (D[\[Omega][x, y, t], x] + D[\[Omega][x, y, t], y]) -
> Subscript[\[GothicCapitalD],
> k] (D[Log[T[x, y, t]], x] + D[Log[T[x, y, t]], y]);
>
> ContEq = D[u[x, y, t], x] + D[u[x, y, t], y] == 0;
> MomentumEq = \[Rho] (D[u[x, y, t]^2, x] + D[u[x, y, t]^2, y]) ==
> D[(\[Mu] (D[u[x, y, t], x] + D[u[x, y, t], y]) -
> 2/3 \[Mu] (D[u[x, y, t], x] + D[u[x, y, t], y]) ), x] +
> D[(\[Mu] (D[u[x, y, t], x] + D[u[x, y, t], y]) -
> 2/3 \[Mu] (D[u[x, y, t], x] + D[u[x, y, t], y]) ), y] -
> D[P[x, y, t], x] + D[P[x, y, t], y];
> TranportEq = \[Rho] \[Omega][x, y,
> t] + \[Rho] (D[\[Omega][x, y, t] u[x, y, t], x] +
> D[\[Omega][x, y, t] u[x, y, t], y]) == D[Jc, x] + D[Jc, y];
> EnergyEq =
> Subscript[c,
> p] (D[(\[Rho] u[x, y, t] T[x, y, t]), x] +
> D[(\[Rho] u[x, y, t] T[x, y, t]), y]) ==
> D[(\[Lambda] (D[T[x, y, t], x] + D[T[x, y, t], y])), x] +
> D[(\[Lambda] (D[T[x, y, t], x] + D[T[x, y, t], y])), y];
> {ContEq, MomentumEq, TranportEq, EnergyEq} //
> FullSimplify // TableForm
>
> Now we need a boundary condition that Mathematica can handle. I have freedom to choose any initial values for the unknowns. For this kind of equation I don't know what type of BC I should take. I just took
>
> BC = {u[x, y, 0] == x + y, u[0, y, t] == u[1, y, t],
> u[x, 0, t] == u[x, 1, t], P[x, y, 0] == x, P[0, y, t] == P[1, y, t],
> P[x, 0, t] == P[x, 1, t], T[x, y, 0] == 2 y + 1,
> T[0, y, t] == T[1, y, t],
> T[x, 0, t] == T[x, 1, t], \[Omega][x, y, 0] ==
> x - y, \[Omega][0, y, t] == \[Omega][1, y, t], \[Omega][x, 0,
> t] == \[Omega][x, 1, t]}
>
> Now I Finally insert values to the constants in the PDE so that I can call NDSolve.
>
> Eq = {ContEq, MomentumEq, TranportEq, EnergyEq} /. \[Lambda] -> .1 /.
> Subscript[c, p] -> 1.2 /. \[Rho] -> .01 /. \[Mu] -> 1 /.
> Subscript[\[GothicCapitalD], k] -> 1;
> EQN = Join[Eq, BC]
>
> To solve the PDE I call
>
> Sol = NDSolve[
> EQN, {u, P, T, \[Omega]}, {x, 0, 1}, {y, 0, 1}, {t, 0, 3},
> Method -> {"MethodOfLines",
> "SpatialDiscretization" -> {"TensorProductGrid"}}] // Timing
>
> And I see the following error
>
> *****************
> *****************
> NDSolve::pdord: Some of the functions have zero differential order so \
> the equations will be solved as a system of differential-algebraic \
> equations. >>
>
> NDSolve::mxsst: Using maximum number of grid points 100 allowed by \
> the MaxPoints or MinStepSize options for independent variable x. >>
>
> NDSolve::mxsst: Using maximum number of grid points 100 allowed by \
> the MaxPoints or MinStepSize options for independent variable y. >>
>
> NDSolve::mxsst: Using maximum number of grid points 100 allowed by \
> the MaxPoints or MinStepSize options for independent variable x. >>
>
> General::stop: Further output of NDSolve::mxsst will be suppressed \
> during this calculation. >>
>
> NDSolve::ibcinc: Warning: Boundary and initial conditions are \
> inconsistent. >>
>
> NDSolve::ivcon: The given initial conditions were not consistent with \
> the differential-algebraic equations. NDSolve will attempt to \
> correct the values. >>
> *****************
> *****************
> It runs for long and finally tells LinearSolve out of memory.
> I hope people with your expertise can give me some insight. You can take full freedom to define BC as you wish. We need just a solution. I am mainly interested to find a class of BC that makes the PDE solvable with Mathematica.
> I hope for some prompt reply.
>

[mg92594] Re: Axis in two scales

2008-10-07T04:05:58Z

Hi,

foo[x_?NumericQ] := If[x < .5, Sin[100*x], Sin[x]]

GraphicsRow[
{Plot[foo[x], {x, 0, 0.5}],
Plot[foo[x], {x, 0.5, 5}, Axes -> {True, False}]},
Spacings -> {-10.0, 0}]

??

Regards
Jens

Nacho wrote:

[mg92593] Re: Workbench complains about an invalid LinkObject number

2008-10-07T04:05:47Z

I get that too, but it works ok.

LinkObject::linkn: Argument LinkObject[, 10, 10] in
LinkWrite[LinkObject[, 10, 10], InputNamePacket[In[8]:= ], True] has an
invalid LinkObject number; the link may be closed.

Barry Wardell wrote:

> Hi,
>
> I'm using Workbench version 1.1 on both Linux and Mac with Mathematica
> 6. When I try to run a notebook from within Workbench, the console
> reports the error:
>
> LinkObject::linkn: Argument LinkObject[, 8, 8] in
> LinkWrite[LinkObject[, 8, 8], InputNamePacket[In[5]:= ], True] has an
> invalid LinkObject number; the link may be closed.
>
> Is this a problem I should worry about, or can I safely ignore it. I
> haven't noticed any problems and the notebook seems to open and work
> fine, but just wanted to check in case I missed something.
>
> Barry Wardell
>

[mg92592] Re: Axis in two scales

2008-10-07T04:05:36Z

Nacho wrote:

You can use custom tick marks, take a look at Ticks and FrameTicks.

Here's a way to do it:

First we define a function to transform x coordinates:

trafo = \[Piecewise] {
{#, # <= 0.5},
{Rescale[#, {0.5, 5}, {0.5, 1}], # > 0.5}
} &

In our case this is just a piecewise linear rescaling.

Then apply this function to x coordinates and to the tick positions that
we wish to show:

ListPlot[
{trafo[#1], #2} & @@@ t,
Ticks ->
{{trafo[#], #} & /@ Join[Range[0, 0.5, 0.1], Range[1, 5]],
Automatic}
]

The advantage of this approach is that it is easily generalized: we
could have used trafo = #^(1/3) & for a fancy non-linear scaling, or
just trafo = Log to mimic ListLogLinearPlot (better not try to put a
tick mark at 0 in this case).

[mg92591] Re: [mg92569] Axis in two scales

2008-10-07T04:05:25Z

t = Table[{x, If[x < .5, Sin[100*x], Sin[x]]}, {x, 0, 5, 0.001}];

t1 = Select[t, First[#] < 0.5 &];
t2 = Complement[t, t1];

Grid[{{
ListLinePlot[t1,
ImageSize -> 250,
Axes -> {True, False},
Frame -> {True, True, True, False},
PlotRange -> {{-0.01, 0.5}, {-1.05, 1.05}},
PlotStyle -> Blue],
ListLinePlot[t2,
ImageSize -> 220,
Axes -> {True, False},
Frame -> {True, False, True, True},
PlotRange -> {{0.5, 5.1}, {-1.05, 1.05}},
PlotStyle -> Directive[Thick, Magenta]]}},
Spacings -> -0.25]

Bob Hanlon

---- Nacho <ncc1701zzz@...> wrote:

=============
Hello everyone.

I have a dataset in pairs (x,y) that I would like to plot. The main
problem is that the data set has interesting parts in the beginning
that I would like to view clearly.

As an example, create the following table:

t = Table[{x, If[x < .5, Sin[100*x], Sin[x]]}, {x, 0, 5, 0.001}];
ListPlot[t]

You have a very detailed sine in the beginnng and the rest is less
interesting.

I would like to "split" the X axis in two, so the left half of the
plot covers x={0,0.5} and the second half x={0.5,5}] in the same plot
and with the right ticks.

It is like a piecewise axis... is it possible?

Thanks.

--

Bob Hanlon

[mg92590] RE: [mg92547] RE: [mg92378] Comparison between Mathematica and other

2008-10-07T04:05:14Z

I feel as if I created a lot of confused debate by my recent post. I admit
that my Sort example was not the best. I just wanted to provide my personal
experience on what convinced me to use Mathematica, since there might
somewhere be some persons who think in a similar way to me. My main point
was that the command "Sort" maybe could illustrate the strength of
Mathematica compared to other CAS systems, since it can be used in a very
general way. I am no expert in the other systems.
One could rewrite the same example in a maybe less "scary" way, missing the
chance to tell anything about the pure function notation (#)&, as Murray
suggest. Then we obtain quite typical Mathematica expressions, and if they
still are too frightening, then we maybe cannot claim without cheating that
Mathematica is very accessible. We could also give simpler examples:

Sort strings into dictionary order:

Sort[{"cat", "fish", "catfish", "Cat"}]

Sort by string length

SortBy[{"cat", "fish", "catfish", "Cat"}, StringLength]

but such things maybe are easily done in the other systems also? Sorting
playing-card is another possibility to illustrate the capability of
Mathematica. Some persons, able to generalize, would get the point, while
other would think that they never would desire to sort playing cards by a
CAS system.
(I just have to mention that I googled on the Sort command in another CAS
system, and found a discussion on how to avoid getting variables sorted
according to their machine addresses. I found that quite scaring!)

The topic of the thread was to compare Mathematica with other CAS systems,
not to persuade newbies to use a CAS system. Then Paolo B. should
concentrate on the strong sides of Mathematica.
The strong side of Mathematica is not that it is easily accessible, the
strong side is that it is useful and from start designed in a clever, logic
and systematic way to provide a unified approach to both symbolic and
numerical math, wiith a design which provides freedom and power to the user.
New and occasional users could find their way into the system via the
numerous live examples in the help system. Do not forget the arbitrary
numerical precision, the huge number range, the freedom to choose
programming style and the pattern matching!

And do not forget to mention MathGroup, with 92576 letters and almost always
pleasant answers: "We should not do YOUR homework, but anyway the answer
is..."

I do not think neither my wife, my daughter nor I would have been impressed
by a beautiful woman recommending Mathematica. And manipulate is a nice
command, the cream on the top, but it was not our reason to choose
Mathematica from the beginning. Or?

Ingolf Dahl
Sweden

> -----Original Message-----
> From: peter lindsay [mailto:pl.0@...]
> Sent: den 4 oktober 2008 12:17
> To: mathgroup@...
> Subject: [mg92547] Re: [mg92537] Re: [mg92527] RE: [mg92378]
> Comparison between Mathematica and other
>
> I agree. The object is to persuade people how accessible
> Mathematica is, not how incredibly learned and expert the
> users need to be.
> Peter
>
> 2008/10/3 Murray Eisenberg <murray@...>
>
> > I'd exercise caution before presenting such an example to
> people whom
> > you're trying to convince to turn to Mathematica. all the vertical
> > strokes, including those in the brackets, make this look
> like a jumble
> > of incomprehensible symbols. The effect could be precisely the
> > opposite of what you intend.
> >
> > Ingolf Dahl wrote:
> >
> > > I do not know how this is example is interesting or how
> it is done
> > > in
> > other
> > > systems, but once, 14 years back, it convinced me that I
> should use
> > > Mathematica for a specific problem. I wanted to sort the
> > > eigenvectors of
> > a
> > > real matrix according to decreasing size of the real part of the
> > > eigenvalues. For complex eigenvalue pairs I wanted the eigenvalue
> > > with positive imaginary part sorted first. I just wanted
> to specify
> > > the
> > sorting
> > > rule, and did not feel for writing the sorting algorithm
> from scratch.
> > This
> > > is how I solved it in Mathematica
> > >
> > > Transpose[
> > > Sort[Transpose[Eigensystem[RandomReal[{-1, 1}, {6, 6}]]],
> > >
> > > Re[#2[[1]]] < Re[#1[[1]]] ||
> > > Re[#2[[1]]] == Re[#1[[1]]] && Im[#2[[1]]] < Im[#1[[1]]] &]]
> > >
> > > I think this also is a good example of the use of functional
> > > programming, and it helped me to get in to it.
> > > In Mathematica we are thus able to sort any kind of
> "objects" with
> > > any
> > kind
> > > of sorting criteria, thanks to the generality of the language.
> >
> > --
> > Murray Eisenberg murray@...
> > Mathematics & Statistics Dept.
> > Lederle Graduate Research Tower phone 413 549-1020 (H)
> > University of Massachusetts 413 545-2859 (W)
> > 710 North Pleasant Street fax 413 545-1801
> > Amherst, MA 01003-9305
> >
> >
>
>
>

[mg92589] Re: Axis in two scales

2008-10-07T04:05:03Z

On 10/6/08 at 4:13 AM, ncc1701zzz@... (Nacho) wrote:

>I have a dataset in pairs (x,y) that I would like to plot. The main
>problem is that the data set has interesting parts in the beginning
>that I would like to view clearly.

>As an example, create the following table:

>t = Table[{x, If[x < .5, Sin[100*x], Sin[x]]}, {x, 0, 5, 0.001}];
>ListPlot[t]

>You have a very detailed sine in the beginnng and the rest is less
>interesting.

>I would like to "split" the X axis in two, so the left half of the
>plot covers x={0,0.5} and the second half x={0.5,5}] in the same
>plot and with the right ticks.

>It is like a piecewise axis... is it possible?

There are a number of ways that should achieve something
suitable. Based on my interpretation of what you are asking for,
I suggest the following:

func = Piecewise[{{Sin[100 x], x < .5}, {Sin[x], x >= .5}}];

GraphicsRow[{Plot[func, {x, 0, .5},
Frame -> {True, True, False, False}, Axes -> None],
Plot[func, {x, .5, 5}, Frame -> {True, False, False, False},
Axes -> None]}, Spacings -> 0]

I've used Piecewise rather than creating a list of data points
to apply since that allows me to take advantage of Mathematica's
adaptive sampling algorithm. I see no point to explicitly
computing values and using ListPlot if the only purpose is to
produce a graphic.

Literally, have created two separate plots and use GraphicsRow
to combine them in an appropriate fashion to make a single
graphic. The net effect is a single graphic with a break in the x-axis.

This particular approach does introduce one aspect in the final
graphic that could be seen as misleading. As the function is
defined, there start of the graph on the right hand side should
appear directly above the end of the graph on the left hand side
rather than being displaced along the horizontal axis.

But given the scaling changes, it isn't clear to me making the
start of the right graphic align vertically with the end of the
left graphic is any less misleading.

[mg92588] Why is Mathematica assuming k==l and how do I make it not to?

2008-10-07T04:04:51Z

As in

Assuming[Element[{k, l}, Integers] ,
Integrate[Cos[k alpha] Cos[l alpha], {alpha, -Pi, Pi}]]

I get 0 whereas the answer is Pi if k=l;

Thanks!

Aaron

[mg92587] Re: integration

2008-10-07T04:04:38Z

Dear RG,

Which part of the result do you have difficulties with? The result is
an expression which is true if certain conditions (listed in the first
part of the output) are true.

The expression is indeed complex, but there isn't a law against the
existence of complex expressions. Nor is it the case that they can
always be simplified. Have you tried some values for the various
constants in the expression to see if the result make any sense?

Cheers -- Sjoerd

[mg92586] Re: Print A Math Formula In Formatted Notebook

2008-10-07T04:04:27Z

John wrote:

> Add a cell to a notebook that prints a math formula in the evaluated
> notebook.
>
> Use appropriate pallet to create the formula.
>
> It is easy to create the formula, but I have not yet found a way to
> print the formula in the evaluated notebook. On some tries, the word
> null printed In the location where the formula should be. On others,
> null is replaced with a rectangular box that is tinted pink.
>
> Why is this so hard? Anyone, who uses Mathematica to write a math
> paper, has to do it over and over again.
>
> The evaluated notebook is sent to the publisher, and the formulas have
> to be in the evaluated notebook.
>
> John
>
>
>
>

At the risk of sounding a bit school master-ish, how about structuring
your query with some consideration of English grammar! Your first two
sentences seem to be imperatives (instructions to the reader do
something) rather than questions!

The question you are asking seems utterly trivial - type in any formula
and press Shift-Return (or press the Enter key on a large keyboard) and
your formula will be evaluated as far as possible and printed out!

Nobody here seems to know what it is you want to do!

David Bailey
http://www.dbaileyconsultancy.co.uk

[mg92585] Re: Axis in two scales

2008-10-07T04:04:16Z

On Oct 6, 1:13 am, Nacho <ncc1701...@...> wrote:

Transform the x-scale. For instance, define

W[x_] := If[x < .5, x, .5 + (x-.5)/9]

then plot using

ListPlot[{W@#[[1]],#[[2]]}&/@t,
Ticks->{{W@#,#}&/@{.25,.5,2,3,4,5},Automatic}]

For more general situations, consider "nicer" functions such as

With[{c = 9}, ListPlot[{ArcSinh[c*#[[1]]],#[[2]]}&/@t,
Ticks->{{ArcSinh[c*#],#}&/@{.1,.2,.5,1,2,3,4,5},Automatic}]]

with the parameter (in this case, c) and the list of tick values
adjusted to taste.

[mg92584] Re: integration

2008-10-07T04:04:05Z

Jean-Marc Gulliet wrote:

> RG wrote:
>
>> I have been trying to simplify(integrate) the following function, but
>> M6 seems to give a complex answer which i cann't understand.. please
>> help.
>>
>> x[s_]=\!$
>> \*SubsuperscriptBox[\(\[Integral]$, $0$, $s$]$Cos[
>> \*FractionBox[\(r\ t\ \((\(-\[Kappa]0$ + \[Kappa]1 + r\ \[Kappa]1)\)
>> + $(1 + r)$\ S\ $(\[Kappa]0 - \[Kappa]1)$\ $(\(-Log[S]$ + Log[S
>> + r\ t])\)\),
>> SuperscriptBox[$r$, $2$]]] \[DifferentialD]t\)\)
>
> First, notice that if we use the *InputForm* of the above expression, we
> can easily add assumptions on the parameters of the integral (or we
> could use *Assuming*), for instance that S, r, and s are real and r != 0
> or s > 0.
>
> However, it seems that the above integral has no solution if the
> parameter S is positive. On the other hand, ff we allow S to be negative
> (or complex) then the integral has a symbolic complex solution.
>
> In[49]:= Integrate[
> Cos[(r*t*(-\[Kappa]0 + \[Kappa]1 + r*\[Kappa]1) + (1 + r)*
> S*(\[Kappa]0 - \[Kappa]1)*
> (-Log[S] + Log[S + r*t]))/r^2], {t, 0, s},
> Assumptions -> S > 0]
>
> Out[49]= Integrate[
> Cos[(r t (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1) + (1 +
> r) S (\[Kappa]0 - \[Kappa]1) (-Log[S] + Log[S + r t]))/r^2], {t,
> 0, s}, Assumptions -> S > 0]
>
> In[46]:= Integrate[
> Cos[(r*t*(-\[Kappa]0 + \[Kappa]1 + r*\[Kappa]1) + (1 + r)*
> S*(\[Kappa]0 - \[Kappa]1)*
> (-Log[S] + Log[S + r*t]))/r^2], {t, 0, s},
> Assumptions -> {Element[{S, r, s}, Reals], r != 0, s > 0}]
>
> Out[46]= If[r S > 0 || s + S/r <= 0, (1/(
> 2 (\[Kappa]0 - \[Kappa]1 - r \[Kappa]1)))
> r S^(-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)) (r s + S)^(-((
>
> [... output partially deleted ...]
>
> r^2)] Sin[(S \[Kappa]0)/r^2 - (S \[Kappa]1)/r^2 - (
> S \[Kappa]1)/r]),
> Integrate[
> Cos[(r t (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1) + (1 +
> r) S (\[Kappa]0 - \[Kappa]1) (-Log[S] + Log[S + r t]))/
> r^2], {t, 0, s},
> Assumptions ->
> r != 0 && s > 0 && r S <= 0 && r (r s + S) > 0 &&
> S \[Element] Reals]]
>
>
> You can manipulate further the integral thanks to *FullSimplify* and
> some assumptions on the parameters.
>
> Assuming[r S > 0 || s + S/r <= 0,
> FullSimplify[
> 1/(2 (\[Kappa]0 - \[Kappa]1 - r \[Kappa]1))
> r S^(-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)) (r s + S)^(-((
>
> [... input partially deleted ...]
>
> S \[Kappa]1)/r])]]

It took a long time, but the last expression returned the following
result (which is valid only for r S > 0 || s + S/r <= 0):

(1/(2 r))E^(-((I S (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1))/
r^2)) (S ExpIntegralE[-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/
r^2), -((I S (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1))/r^2)] -
S^(-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)) (r s + S)^(
1 + (I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)
ExpIntegralE[-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2), -((
I (r s + S) (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1))/r^2)] +
E^((2 I S (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1))/
r^2) (S ExpIntegralE[(I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2, (
I S (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1))/r^2] - (S^3)^((
I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2) (r s + S)^(
1 - (I (1 + r) S (\[Kappa]0 - \[Kappa]1))/
r^2) ((\[Kappa]0 - (1 + r) \[Kappa]1)^2)^((
I (1 + r) S (\[Kappa]0 - \[Kappa]1))/
r^2) (S^2 (r s + S)^2 (\[Kappa]0 - (1 + r) \[Kappa]1)^2)^(-((
I (1 + r) S (\[Kappa]0 - \[Kappa]1))/
r^2)) ((r s + S) (S + r Conjugate[s]))^((
I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)
ExpIntegralE[(I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2, (
I (r s + S) (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1))/r^2]))

Regards,
-- Jean-Marc

[mg92583] Re: Axis in two scales

2008-10-07T04:03:54Z

David Park's Presentation Package can do this. David has a response a
little while ago:

Re: How can I create a two-axis graph in Mathematica v6

You might check this out. I highly recommend his package.

Kevin

Nacho wrote:

[mg92582] Re: integration

2008-10-07T04:03:43Z

RG wrote:

> I have been trying to simplify(integrate) the following function, but
> M6 seems to give a complex answer which i cann't understand.. please
> help.
>
> x[s_]=\!$
> \*SubsuperscriptBox[\(\[Integral]$, $0$, $s$]$Cos[
> \*FractionBox[\(r\ t\ \((\(-\[Kappa]0$ + \[Kappa]1 + r\ \[Kappa]1)\)
> + $(1 + r)$\ S\ $(\[Kappa]0 - \[Kappa]1)$\ $(\(-Log[S]$ + Log[S
> + r\ t])\)\),
> SuperscriptBox[$r$, $2$]]] \[DifferentialD]t\)\)

First, notice that if we use the *InputForm* of the above expression, we
can easily add assumptions on the parameters of the integral (or we
could use *Assuming*), for instance that S, r, and s are real and r != 0
or s > 0.

However, it seems that the above integral has no solution if the
parameter S is positive. On the other hand, ff we allow S to be negative
(or complex) then the integral has a symbolic complex solution.

In[49]:= Integrate[
Cos[(r*t*(-\[Kappa]0 + \[Kappa]1 + r*\[Kappa]1) + (1 + r)*
S*(\[Kappa]0 - \[Kappa]1)*
(-Log[S] + Log[S + r*t]))/r^2], {t, 0, s},
Assumptions -> S > 0]

Out[49]= Integrate[
Cos[(r t (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1) + (1 +
r) S (\[Kappa]0 - \[Kappa]1) (-Log[S] + Log[S + r t]))/r^2], {t,
0, s}, Assumptions -> S > 0]

In[46]:= Integrate[
Cos[(r*t*(-\[Kappa]0 + \[Kappa]1 + r*\[Kappa]1) + (1 + r)*
S*(\[Kappa]0 - \[Kappa]1)*
(-Log[S] + Log[S + r*t]))/r^2], {t, 0, s},
Assumptions -> {Element[{S, r, s}, Reals], r != 0, s > 0}]

Out[46]= If[r S > 0 || s + S/r <= 0, (1/(
2 (\[Kappa]0 - \[Kappa]1 - r \[Kappa]1)))
r S^(-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)) (r s + S)^(-((

[... output partially deleted ...]

r^2)] Sin[(S \[Kappa]0)/r^2 - (S \[Kappa]1)/r^2 - (
S \[Kappa]1)/r]),
Integrate[
Cos[(r t (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1) + (1 +
r) S (\[Kappa]0 - \[Kappa]1) (-Log[S] + Log[S + r t]))/
r^2], {t, 0, s},
Assumptions ->
r != 0 && s > 0 && r S <= 0 && r (r s + S) > 0 &&
S \[Element] Reals]]

You can manipulate further the integral thanks to *FullSimplify* and
some assumptions on the parameters.

Assuming[r S > 0 || s + S/r <= 0,
FullSimplify[
1/(2 (\[Kappa]0 - \[Kappa]1 - r \[Kappa]1))
r S^(-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)) (r s + S)^(-((

[... input partially deleted ...]

S \[Kappa]1)/r])]]

HTH,
-- Jean-Marc

[mg92581] Re: Using an locator and Rotating 3D graphics

2008-10-07T04:03:32Z

Manipulate is like a set-piece dynamic display. It tries to be very versitle
but for most custom dynamic presentations it becomes a pain-in-the-neck. It
is better to learn how to construct your own dynamic presentations. Here is
more direct code for your display:

Module[
{pt1 = {1, 0}, pt2 = {0, 1}},
Row[{
Graphics[
{Dynamic@Arrow[{pt1, pt2}],
Locator[Dynamic[pt1]],
Locator[Dynamic[pt2]]},
PlotRange -> 2,
ImageSize -> 200],
Dynamic@
ParametricPlot3D[{Part[pt2 - pt1, 1] Cos[u],
Part[pt2 - pt1, 2] Sin[u], u}, {u, 0, 4 Pi},
PlotRange -> {{-2, 2}, {-2, 2}},
ColorFunction -> Function[{x, y, z, u}, Hue[u]],
PlotStyle -> Thick,
ImageSize -> 350,
BoxRatios -> {1, 1, 1}]}]
]

You could use DynamicModule, but it really isn't necessary here. The
Locators are confined to the lhs plot and not imposed on the entire display.
In the code above I bypassed calculating the v variable you used. Let's
reintroduce v. Quite often we might have a set of primary dynamic variables,
such as pt1 and pt2 here, which are manipulated by the mouse or sliders or
other dynamic elements, and a set of dependent dynamic variables that depend
on the primary variables, such as v here. We can handle this situation as in
the following code. Here we use the two argument form of Dynamic and when a
primary dynamic variable is adjusted the routine calcAll is called to
calculate all the dependent quantities (only v in this case). I've also made
some stylistic changes to the display.

Module[
{(* Primary dynamic variables *)
pt1 = {1, 0}, pt2 = {0, 1},
(* Dependent dynamic variable *)
v,
(* Other variables *)
calcAll},

calcAll[p1_, p2_] := (v = p2 - p1);

(* Initialize dependent variables *)
calcAll[pt1, pt2];

(* Display *)
Panel[
Row[{
Graphics[
{Arrowheads[.1],
Dynamic@Arrow[{pt1, pt2}],
Locator[Dynamic[pt1, (pt1 = #; calcAll[pt1, pt2]) &],
Graphics[{AbsolutePointSize[8], Point[{0, 0}]},
ImageSize -> 10]],
Locator[Dynamic[pt2, (pt2 = #; calcAll[pt1, pt2]) &], None]},
PlotRange -> 2,
ImageSize -> 200],
Dynamic@
ParametricPlot3D[{v[[1]] Cos[u], v[[2]] Sin[u], u}, {u, 0,
4 Pi}, PlotRange -> {{-2, 2}, {-2, 2}},
ColorFunction -> Function[{x, y, z, u}, Hue[u]],
PlotStyle -> Thick,
Axes -> False,
SphericalRegion -> True, RotationAction -> Clip,
ImageSize -> 350,
Boxed -> False,
BoxRatios -> {1, 1, 1}]}],
Style["Custom Dynamics", 16]]
]

--
David Park
djmpark@...
http://home.comcast.net/~djmpark/

"KB" <KennethLeeBaker@...> wrote in message
news:gccheo$s5t$1@......

> When I try to use locators as inputs to 3D graphics, I end up not
> being able to rotate 3d graphics. Clicking on the graphic moves the
> nearest locator rather than the graphic. Switching to a collection of
> 2D sliders works in a pinch, but does not achieve the desired effect.
> Here is an example:
>
> Manipulate[v = pt2 - pt1;
> {Graphics[Arrow[{pt1, pt2}], PlotRange -> 2],
> ParametricPlot3D[{v[[1]] Cos[u], v[[2]] Sin[u], u}, {u, 0, 4 Pi},
> PlotRange -> {{-2, 2}, {-2, 2}},
> ColorFunction -> Function[{x, y, z, u}, Hue[u]],
> PlotStyle -> Thick]}, {{pt1, {0, 1}}, Locator}, {{pt2, {1, 0}},
> Locator}]
>
>
> Any ideas?
>

[mg92580] Format Type of new Output Cells to OutputForm

2008-10-07T04:03:21Z

I need to set the Format Type of new Output Cells to OutputForm.

This can be done with menu commands: Edit>Preferences...>Evaluation
and then select
Format type of new Output Cells to OutputForm

This is ok, but I would like to do it programmatically.
I guess I have to set some option in the notebook, but how?

[mg92579] Using an locator and Rotating 3D graphics

2008-10-06T01:14:57Z

When I try to use locators as inputs to 3D graphics, I end up not
being able to rotate 3d graphics. Clicking on the graphic moves the
nearest locator rather than the graphic. Switching to a collection of
2D sliders works in a pinch, but does not achieve the desired effect.
Here is an example:

Manipulate[v = pt2 - pt1;
{Graphics[Arrow[{pt1, pt2}], PlotRange -> 2],
ParametricPlot3D[{v[[1]] Cos[u], v[[2]] Sin[u], u}, {u, 0, 4 Pi},
PlotRange -> {{-2, 2}, {-2, 2}},
ColorFunction -> Function[{x, y, z, u}, Hue[u]],
PlotStyle -> Thick]}, {{pt1, {0, 1}}, Locator}, {{pt2, {1, 0}},
Locator}]

Any ideas?

[mg92578] NET/Link Thread in C#

2008-10-06T01:14:46Z

Dear All-

Can anyone give me a hint on how to start C# threads from
Mathematica ?

I tried:
====
Needs["NETLink`"]
InstallNET[];

LoadNETType["System.Threading.Thread"]

myThread = NETNew[ThreadStart]

and I got:
=======
NETType["System.Threading.Thread", 8]
NETNew::args: Improper count or type of arguments. >>
$Failed

I also tried:
========
LoadNETType["System.Threading.Thread"]
t = NETNew[`Thread, Console`Out@WriteLine["Hello from .NET"] Start]

and I got:
======
NETType["System.Threading.Thread", 8]
NETNew::args: Improper count or type of arguments. >>
$Failed

I believe class ThreadStart needs to be instantiated before class
Thread, as the former is an argument of the latter's constructor.

Please help Thank you.

[mg92577] Re: Comparison between Mathematica and other

2008-10-06T01:14:35Z

On 10/5/08 at 6:06 AM, siegman@... (AES) wrote:

>2) Now it's certainly true that if you want to modify the default
>behavior of Sort[ ], you do indeed have to be able to do (or more
>likely copy and modify) what I'd agree is some very modest level of
>procedural programming, as well illustrated in the example above:

>Re[#2[[1]]] < Re[#1[[1]]] || Re[#2[[1]]] == Re[#1[[1]]] &&
>Im[#2[[1]]] < Im[#1[[1]]] &]]

But this need not have been done this way. It could have been done:

Ordering@(Re/@{##}) || Equal@@(Re/@{##}) && Ordering@(Im/@{##})&

From the perspective of someone not familiar with Mathematica
and its notation, this version is undoubtedly more opaque. But
for me, this is much clearer and easier to read. The issue of
determining what is being grouped by the numerous square
brackets has been eliminated.

[mg92576] Re: Comparison between Mathematica and other symbolic systems

2008-10-06T01:14:25Z

On 10/5/08 at 6:06 AM, siegman@... (AES) wrote:

>In article <gc7fsf$eo7$1@...>,
>Bill Rowe <readnews@...> wrote:

>>On 10/3/08 at 6:41 AM, awnl@... (Albert Retey) wrote:

>>>When choosing a system, I think one needs to answer these
>>>questions:

>>>1) can the system solve the problem at hand 2) how much effort is
>>>it to feed the problem to the system 3) how efficient is the
>>>system in calculating the solution

>>Which really says 2) is the most important consideration. But I
>>would expand 2) to be the amount of time to input the problem *and*
>>verify the input has been done correctly. In my experience, the
>>time to verify/debug input is by far where most of the effort is
>>spent.

>No mention at all of "how easy it is to learn to use the system"
>(and remember how to use it between infrequent uses) ?!?!?!?

The current version of Mathematica has more than 2900 built-in
symbols, with many of these having long lists of options. I
would expect any other system with capabilities equivalent to
Mathematica to have a similarly long list of built-in commands
and options. This number of commands and options is far too long
for infrequent users to remember. It would require a significant
reduction in this number to enable easy recall by infrequent
users. But that also implies a significant reduction in capability.

It is very hard to see how another system with equivalent
capability to Mathematica could be significantly easier for
infrequent users.

[mg92575] Re: Finding the optimum by repeteadly zooming on the solution space (or something like that)

2008-10-06T01:14:14Z

On 10/5/08 at 6:05 AM, cuak2000@... (Mauricio Esteban Cuak)
wrote:

>Hello everyone.I know somebody has probably writ mathematica code on
>this problem. A general answer would probably benefit more people,
>but I'll do the specific example, 'cause I'm not sure how to explain
>it in another way.

>I have these restrictions:

>cpoAg1= ( -x + (70*a*(x^0.4 + y^0.4)^0.75)/x^0.6);

>(*and*)

>cpoAg2= ((70*(1 - a)*(x^0.4 + y^0.4)^0.75)/y^0.6 - 2*y);

>(* And I need to find the "a" that will maximise this following
>function. I don't need and exact solution, but something that is
>sufficiently near *)

>obj = -x^2/2 + 100*(x^0.4 + y^0.4)^1.75 - y^2;

>(* "A" goes between 0 and 1 so I discretise to obtain the {x,y} that
>maximise every "a"

>the Table command gives me the list of rules which I can replace on
>"obj" later to find the maximum. No problem here*)

>rules = Table[
>FindRoot[{cpoAg1 == 0, cpoAg2 == 0}, {{x, 0.1}, {y, 0.1}}], {a, 0.=
1,
>1, 0.1}
>];

>(*the maximum is found with this following function *)

>Max[Thread[ReplaceAll[obj, rules]]];

>So far so good...I could discretise "a" as thinly as I want, but I
>can't afford the luxury of being that inneficient, 'cause I've got
>to do other things later with this part of the program.

I don't understand why you would take this approach. Why not use
the built-in function NMaximize, That is:

In[20]:= NMaximize[{obj,
cpoAg1 == 0 && cpoAg2 == 0 && x > 0.1 && y > 0.1}, {a, x, y}]

Out[20]= {2084.72,{a->0.527688,x->22.6727,y->13.7173}}

[mg92574] Re: Comparison between Mathematica and other

2008-10-06T01:14:03Z

In article <gc9mh8$2br$1@...>, "peter lindsay" <pl.0@...>
wrote:

> I agree. The object is to persuade people how accessible Mathematica is, not
> how incredibly learned and expert the users need to be.
> Peter

Not certain what earlier words Peter is agreeing with, but I agree with
what Peter himself said -- perhaps edited to say:

"The [primary objectives should be] to persuade people how accessible
Mathematica [can be], not how incredibly learned and expert [its]
users need to be -- **and then to make it that accessible**."

[mg92573] Re: Expressing values in the legend in terms of original data

2008-10-06T01:13:53Z

In addition to my previous reply I should have taken the automatic
ranging of ContourPlot into account as it scales values to give a 0..1
range.

So, if you want to make sure the legend maps GrayLevels 0..1 to the
-2..2 range (in case the ends of this range are not within the plot
region) you have to adjust the scaling of ContourPlot accordingly:

contourPlotFun =
ContourPlot[
2 Sin[x y], {x, -\[Pi]/2, \[Pi]/2}, {y, -\[Pi]/2, \[Pi]/2},
ColorFunctionScaling -> False,
ColorFunction -> (GrayLevel[1 - ((# + 2)/4)] &)];

ShowLegend[contourPlotFun, {GrayLevel[1 - #] &, 10, " 2", "-2",
LegendPosition -> {1.1, -.4}, LegendShadow -> None}]

Cheers -- Sjoerd

On Oct 1, 1:12 pm, er <erwann.rog...@...> wrote:

> Hi,
>
> Considering
>
> contourPlotFun=
> ContourPlot[2 Sin[x y],{x,-\[Pi]/2,\[Pi]/2},{y,-\[Pi]/2,\[Pi]/2},
> DisplayFunction\[Rule]Identity];
>
> ShowLegend[
> contourPlotFun,{GrayLevel[1-#]&,10," 1","-1",
> LegendPosition\[Rule]{1.1,-.4}}];
>
> The legend here is misleading, because the min and max values are -2
> and 2, respectively, not -1 and 1. So how can I modify the input to
> ShowLegend to reflect that?
>
> Thanks!

[mg92572] Re: Expressing values in the legend in terms of original data

2008-10-06T01:13:42Z

Er,

In my version of Mathematica (6.03 WinXP) the ContourPlot is in color
by default. It doesn't make sense to add a grayscale legend. So either
you color the legend or you add a greyscale ColorFunction to the
ContourPlot:

contourPlotFun =
ContourPlot[
2 Sin[x y], {x, -\[Pi]/2, \[Pi]/2}, {y, -\[Pi]/2, \[Pi]/2},
ColorFunction -> (GrayLevel[1 - #] &)];

ShowLegend[contourPlotFun, {GrayLevel[1 - #] &, 10, " 2", "-2",
LegendPosition -> {1.1, -.4}}]

The legend doesn't adapt to the range of your plot. You have to
provide the range yourself. In the expression above I put in 2 and -2
and that does the trick.

Cheers,

Sjoerd

On Oct 1, 1:12 pm, er <erwann.rog...@...> wrote:
> Hi,
>
> Considering
>
> contourPlotFun=
> ContourPlot[2 Sin[x y],{x,-\[Pi]/2,\[Pi]/2},{y,-\[Pi]/2,\[Pi]/2},
> DisplayFunction\[Rule]Identity];
>

> ShowLegend[
> contourPlotFun,{GrayLevel[1-#]&,10," 1","-1",
> LegendPosition\[Rule]{1.1,-.4}}];
>
> The legend here is misleading, because the min and max values are -2
> and 2, respectively, not -1 and 1. So how can I modify the input to
> ShowLegend to reflect that?
>
> Thanks!

[mg92571] integration

2008-10-06T01:13:31Z

[mg92570] Re: Comparison between Mathematica and other symbolic systems

2008-10-06T01:13:19Z

On Oct 3, 12:41 pm, Albert Retey <a...@...> wrote:
> Hi,
>
> > As part of a presentation to students, I will have to support the
> > claim that "Mathematica is better than other systems when it comes to
> > symbolic computations".
>
> isn't this quite a stupid claim?

Albert,

yes it is, but that's an edited version, necessary for the post to
appear on this list. Still, I have to disagree with you, as only a
minority of the users switch from system to another depending on their
needs. Most of us probably master one or at most two systems. That's
it.

Paolo

[mg92569] Axis in two scales

2008-10-06T01:13:08Z

Hello everyone.

I have a dataset in pairs (x,y) that I would like to plot. The main
problem is that the data set has interesting parts in the beginning
that I would like to view clearly.

As an example, create the following table:

t = Table[{x, If[x < .5, Sin[100*x], Sin[x]]}, {x, 0, 5, 0.001}];
ListPlot[t]

You have a very detailed sine in the beginnng and the rest is less
interesting.

I would like to "split" the X axis in two, so the left half of the
plot covers x={0,0.5} and the second half x={0.5,5}] in the same plot
and with the right ticks.

It is like a piecewise axis... is it possible?

Thanks.

[mg92568] Re: [mg92562] Re: Comparison between Mathematica and other

2008-10-06T01:12:57Z

Still, I would avoid such a highly condensed bit of code and would
probably first try something more like the following:

complexGreater[z_, w_] :=
(Re[z] > Re[w]) || (Re[z] == Re[w] && Im[z] > Im[w])
evalGreater[edata1_, edata2_] :=
complexGreater[First@edata1, First@edata2]

esys = Eigensystem[RandomReal[{-1, 1}, {6, 6}]]
Transpose[Sort[Transpose[esys], evalGreater]]

That separates the data and operation upon the data from the definition
of the underlying comparison functions. And yes, in complexGreater I'd
put in the redundant parentheses, or at least the second pair.

Once I did that, I _might_ condense the whole thing if I were going to
do it just once, but certainly not if it were going to be re-used.
Then, of course, I'd define a function...

sortByEigenvalues[esys_]:=
Transpose[Sort[Transpose[esys], evalGreater]]

and then probably redo it so as to eliminate the need for evalGreater:

sortByEigenvalues[esys_]:=
Transpose[Sort[Transpose[esys],complexGreater[First@#1,First@#2]&]]

I'd still surely want to keep the complexGreater function, as that seems
a separate "chunk".

One thing I would still _not_ want to do would be to use the #1[[1]] and
#2[[1]] forms, which my eye finds too difficult to understand at a
glance -- too hard to separate the argument number from the Part number.

AES wrote:

>>>> Transpose[
>>>> Sort[Transpose[Eigensystem[RandomReal[{-1, 1}, {6, 6}]]],
>>>>
>>>> Re[#2[[1]]] < Re[#1[[1]]] ||
>>>> Re[#2[[1]]] == Re[#1[[1]]] && Im[#2[[1]]] < Im[#1[[1]]] &]]
>>>>
>>>> I think this also is a good example of the use of functional programming,
>>>> and it helped me to get in to it.
>>>> In Mathematica we are thus able to sort any kind of "objects" with any
>>> kind
>>>> of sorting criteria, thanks to the generality of the language.
>
> I'm really not attempting to re-open the functional vs procedural
> programming battle here, but I'd offer a couple or three observations on
> the above:
>
> 1) Commands like Sort[ ], Transpose[ ], Re[ ] are evidently "functions"
> and I suppose these functions contain massive levels of sophisticated
> functional programming (and pattern matching) in their internals -- --
> but just using them as functions doesn't, it seems to me, really
> constitute hard-core "functional programming" by those users who do
> this.
>
> Their names are ordinary English or mathematical terms; what they do (or
> are likely to do) will seem obvious to even the most unsophisticated
> user; and they can be readily used even by unsophisticated (or entirely
> procedurally oriented) users, at least in their default modes, without
> even knowing what functional programming is -- that is, without such
> users learning (or having to remember) any kind of arcane symbols or
> functional programming notations or structures, and especially any
> symbols and notations that are not used in and familiar from, say,
> elementary algebra or calculus.
>
> Just using -- or even defining for yourself -- a named function (macro,
> subroutine, whatever) doesn't seem to me "functional programming" at any
> very meaningful level; and doing this certainly doesn't depend on the
> "generality" of Mathematica as a language. These same capabilities are
> certainly present, and commonplace, in what I believe are labeled as
> crudely "procedural languages" (FORTRAN, BASIC, etc).
>
> 2) Now it's certainly true that if you want to modify the default
> behavior of Sort[ ], you do indeed have to be able to do (or more likely
> copy and modify) what I'd agree is some very modest level of procedural
> programming, as well illustrated in the example above:
>
> Re[#2[[1]]] < Re[#1[[1]]] ||
> Re[#2[[1]]] == Re[#1[[1]]] && Im[#2[[1]]] < Im[#1[[1]]] &]]
>
> But even I (a resolutely procedural type) would have little trouble with
> this particular example, since the level of 'arcanity' is really pretty
> limited: The "#n" notation for arguments is familiar from TeX (and what
> other languages?); the "||" and "&&" notations are familiar from
> low-level logic classes; and the "==" and "[[ ]]" notations are pretty
> basic Mathematica. The only 'arcane' thing (for me) is the single "&"
> at the end, which I have no idea why is there, but I'll dutifully copy
> it.
>
> 3) As a fallout from this, I'll continue to argue that if Mathematica
> really wants to expand and serve the widest possible customer base --
> e.g., high school math and physics students, college students (and grad
> students) in nontechnical fields, working engineers in industry, etc. --
> for those audiences (aka customers) it should:
>
> 1) Substantially sharpen its focus on making and keeping Mathematica
> simpler and easier to use and much easier to learn for those audiences,
> including a focus on better documentation and fewer complexities and
> "gotchas" (which is not imply taking away any of the pattern matching
> and functional programming underpinnings that are so crucial to other,
> more sophisticated users of Mathematica); and
>
> 1) Drastically reduce its prices.
>
> --AES, Oct 2008
>

--
Murray Eisenberg murray@...
Mathematics & Statistics Dept.
Lederle Graduate Research Tower phone 413 549-1020 (H)
University of Massachusetts 413 545-2859 (W)
710 North Pleasant Street fax 413 545-1801
Amherst, MA 01003-9305

[mg92567] Re: [mg92564] Re: Comparison between Mathematica and other symbolic

2008-10-06T01:12:47Z

And "remember how to use it between infrequent uses" is a good criterion
to note when comparing the long-to-type Mathematica names of functions,
on the one hand, and the terse but often cryptic names of functions in
some other systems.

AES wrote:

> In article <gc7fsf$eo7$1@...>,
> Bill Rowe <readnews@...> wrote:
>
>> On 10/3/08 at 6:41 AM, awnl@... (Albert Retey) wrote:
>>
>>> When choosing a system, I think one needs to answer these questions:
>>> 1) can the system solve the problem at hand
>>> 2) how much effort is it to feed the problem to the system
>>> 3) how efficient is the system in calculating the solution
>> Which really says 2) is the most important consideration. But I
>> would expand 2) to be the amount of time to input the problem
>> *and* verify the input has been done correctly. In my
>> experience, the time to verify/debug input is by far where most
>> of the effort is spent.
>
> No mention at all of "how easy it is to learn to use the system" (and
> remember how to use it between infrequent uses) ?!?!?!?
>