L.S.,
How can I produce a Feigenbaum diagram and compute the Feigenbaum-constant
with Maxima.
Thanking you in advance for your answer.
Best regards,
Dr. J. Simons
----- Original Message -----
From: <maxima-request at math.utexas.edu>
To: <maxima at math.utexas.edu>
Sent: Tuesday, June 08, 2010 7:00 PM
Subject: Maxima Digest, Vol 47, Issue 15
> Send Maxima mailing list submissions to
> maxima at math.utexas.edu
>
> To subscribe or unsubscribe via the World Wide Web, visit
> http://www.math.utexas.edu/mailman/listinfo/maxima
> or, via email, send a message with subject or body 'help' to
> maxima-request at math.utexas.edu
>
> You can reach the person managing the list at
> maxima-owner at math.utexas.edu
>
> When replying, please edit your Subject line so it is more specific
> than "Re: Contents of Maxima digest..."
>
>
> Today's Topics:
>
> 1. Re: Double integral of a function (Beginners Question)
> (Doktor Bernd)
> 2. Re: Double integral of a function (Beginners Question)
> (Robert Dodier)
> 3. Re: Double integral of a function (Beginners Question)
> (Doktor Bernd)
> 4. bfloat(sqrt(%i))? (Raymond Toy)
> 5. Re: bfloat(sqrt(%i))? (Dieter Kaiser)
> 6. Can I use maxima for this? (ALEXIOU SPURIDON)
>
>
> ----------------------------------------------------------------------
>
> Message: 1
> Date: Mon, 7 Jun 2010 17:01:15 +0000
> From: Doktor Bernd <doktorbernd at hotmail.com>
> To: <maxima at math.utexas.edu>
> Subject: Re: [Maxima] Double integral of a function (Beginners
> Question)
> Message-ID: <SNT115-W109FBA1DB431EC4E42E7D9B9D50 at phx.gbl>
> Content-Type: text/plain; charset="iso-8859-1"
>
>
>
>
> _________________________________________________________________
> Hotmail: Free, trusted and rich email service.
> https://signup.live.com/signup.aspx?id=60969
> -------------- next part --------------
> URL:
> <http://www.math.utexas.edu/pipermail/maxima/attachments/20100607/a4c8589c/attachment-0001.html>;
>
> ------------------------------
>
> Message: 2
> Date: Mon, 7 Jun 2010 13:06:38 -0600
> From: Robert Dodier <robert.dodier at gmail.com>
> To: Doktor Bernd <doktorbernd at hotmail.com>
> Cc: maxima at math.utexas.edu
> Subject: Re: [Maxima] Double integral of a function (Beginners
> Question)
> Message-ID:
> <AANLkTikp_AVNwP92kCLjbhU4jLbQz-DVNZbq_9_0IuBt at mail.gmail.com>
> Content-Type: text/plain; charset=ISO-8859-1
>
> Since there are a lot of floats in the integrand, I'm guessing
> an approximate result is OK. Instead of
>
>> integrate(integrand(kappa,Cn2,(1-xi)/L,L,Lambda,Theta_inv,xi,k,r),kappa,0,inf);
>> integrate(%,xi,0,1);
>> float(%);
>
> try this:
>
> I : integrand(kappa,Cn2,(1-xi)/L,L,Lambda,Theta_inv,xi,k,r);
> I1 : quad_qags (I, xi, 0, 1)[1];
> I2 : quad_qagi (I1, kappa, 0, inf);
>
> => [3.93347674966401e-20, 9.416575845421898e-22, 3225, 5]
>
> The integrand is very nearly zero almost everywhere,
> which might make it more difficult for the quadrature functions
> to estimate the error. Maybe you can rescale the integrand.
>
> Note that I2 doesn't take into account any error estimate from I1.
> Iterated quadrature is probably not such a great method,
> although Monte Carlo and quasi-Monte Carlo are probably less
> efficient (i.e. need more function evaluations for the same
> accuracy) than iterated quadrature for a 2-d problem.
> Maybe there are better methods especially for low-dimensional
> integration.
>
> plot3d (I, [kappa, 0.1, 1], [xi, 0, 1], [grid, 50, 50]);
> seems to be informative.
>
> By the way, what is the origin of this problem?
>
> best
>
> Robert Dodier
>
>
> ------------------------------
>
> Message: 3
> Date: Mon, 7 Jun 2010 20:59:17 +0000
> From: Doktor Bernd <doktorbernd at hotmail.com>
> To: <maxima at math.utexas.edu>
> Subject: Re: [Maxima] Double integral of a function (Beginners
> Question)
> Message-ID: <SNT115-W5813DE12E594EB49D9DA7DB9D50 at phx.gbl>
> Content-Type: text/plain; charset="iso-8859-1"
>
>
>>Since there are a lot of floats in the integrand, I'm guessing
>>an approximate result is OK. Instead of
>
> Yes, an approximation is fine.
>
>>> integrate(integrand(kappa,Cn2,(1-xi)/L,L,Lambda,Theta_inv,xi,k,r),kappa,0,inf);
>>> integrate(%,xi,0,1);
>>> float(%);
>>
>>try this:
>>
>>I : integrand(kappa,Cn2,(1-xi)/L,L,Lambda,Theta_inv,xi,k,r);
>>I1 : quad_qags (I, xi, 0, 1)[1];
>>I2 : quad_qagi (I1, kappa, 0, inf);
>>
>> => [3.93347674966401e-20, 9.416575845421898e-22, 3225, 5]
>
> Great, I was still waiting for my accurate result (>6 hours), so I guess
> an approximation is the only practicable way for me.
> I used the QUADPACK functions before, but did not succeed. I was missing
> [1] in I1. Argh, now I know.
>
>>The integrand is very nearly zero almost everywhere,
>>which might make it more difficult for the quadrature functions
>>to estimate the error. Maybe you can rescale the integrand.
>
> Defenitely a good idea.
>
>>Note that I2 doesn't take into account any error estimate from I1.
>>Iterated quadrature is probably not such a great method,
>>although Monte Carlo and quasi-Monte Carlo are probably less
>>efficient (i.e. need more function evaluations for the same
>>accuracy) than iterated quadrature for a 2-d problem.
>>Maybe there are better methods especially for low-dimensional
>>integration.
>
> I see, there is still a lot of documentation left for me to read.
>
>>plot3d (I, [kappa, 0.1, 1], [xi, 0, 1], [grid, 50, 50]);
>>seems to be informative.
>
> Thanks for this hint, plotting would have been my next question.
>
>>By the way, what is the origin of this problem?
>
> I am trying to calculate the scintillation index of a laser beam
> propagating through the atmosphere.
> spec is the atmospheric spectrum and because its influence on the beam
> changes with propagation distance
> I have to do the double integration along frequency kappa and normalized
> propagation distance xi.
>
> Thank you very very much,
> Bernd
>
>
> _________________________________________________________________
> Hotmail: Trusted email with powerful SPAM protection.
> https://signup.live.com/signup.aspx?id=60969
> -------------- next part --------------
> URL:
> <http://www.math.utexas.edu/pipermail/maxima/attachments/20100607/b9d59762/attachment-0001.html>;
>
> ------------------------------
>
> Message: 4
> Date: Mon, 07 Jun 2010 17:57:42 -0400
> From: Raymond Toy <toy.raymond at gmail.com>
> To: Maxima List <maxima at math.utexas.edu>
> Subject: bfloat(sqrt(%i))?
> Message-ID: <4C0D6B56.6050100 at gmail.com>
> Content-Type: text/plain; charset=ISO-8859-1
>
> Currently maxima says bfloat(sqrt(%i)) is 1b0*(-1)^(1/4). That doesn't
> seem very nice. I was expecting something along the lines of
> bfloat(rectform(sqrt(%i))) -> .707b0 + %i*.707b0.
>
> Ray
>
>
>
> ------------------------------
>
> Message: 5
> Date: Tue, 08 Jun 2010 00:48:06 +0200
> From: Dieter Kaiser <drdieterkaiser at web.de>
> To: Raymond Toy <toy.raymond at gmail.com>
> Cc: Maxima List <maxima at math.utexas.edu>
> Subject: Re: [Maxima] bfloat(sqrt(%i))?
> Message-ID: <1275950887.1720.31.camel at dieter>
> Content-Type: text/plain; charset="UTF-8"
>
> Am Montag, den 07.06.2010, 17:57 -0400 schrieb Raymond Toy:
>> Currently maxima says bfloat(sqrt(%i)) is 1b0*(-1)^(1/4). That doesn't
>> seem very nice. I was expecting something along the lines of
>> bfloat(rectform(sqrt(%i))) -> .707b0 + %i*.707b0.
>
> That is the case for float too:
>
> (%i6) float(sqrt(%i));
> (%o6) 1.0*(-1)^(1/4)
>
> This is because Maxima simplifies:
>
> (%i7) sqrt(%i);
> (%o7) (-1)^(1/4)
>
> We get numerical values the following way:
>
> (%i8) sqrt(1.0*%i);
> (%o8) .7071067811865475*%i+.7071067811865476
> (%i9) sqrt(1.0b0*%i);
> (%o9) 7.071067811865475b-1*%i+7.071067811865475b-1
>
> But not when we use numer, float or bfloat:
>
> (%i10) sqrt(%i),float;
> (%o10) (-1)^0.25
>
> (%i11) sqrt(%i),numer;
> (%o11) 1.0*(-1)^0.25
>
> (%i12) sqrt(%i),bfloat;
> (%o12) 1.0b0*(-1)^(1/4)
>
> We have similar problems with all even powers:
>
> (%i1) float((%i)^(1/6));
> (%o1) 1.0*(-1)^(1/12)
> (%i2) bfloat((%i)^(1/6));
> (%o2) 1.0b0*(-1)^(1/12)
>
> More problems with (-1):
>
> (%i4) (sqrt(-1)),numer;
> (%o4) 1.0*(-1)^0.5
>
> (%i5) float(sqrt(-1));
> (%o5) %i
>
> (%i6) sqrt(-1),bfloat;
> (%o6) %i
>
> When we go into the details we will see two problems:
>
> Maxima does not follow the convention to evaluate numerically the power
> function if one of the arguments is a float or bigfloat value.
>
> Maxima does not preserve the type consequently.
>
> Dieter Kaiser
>
>
>
>
> ------------------------------
>
> Message: 6
> Date: Tue, 8 Jun 2010 10:17:33 +0300 (EEST)
> From: ALEXIOU SPURIDON <moka1 at otenet.gr>
> To: maxima at math.utexas.edu
> Subject: Can I use maxima for this?
> Message-ID:
> <4377562.129.1275981453210.JavaMail.tomcat at vermeer.otenet.gr>
> Content-Type: text/plain; charset="UTF-8"; format=flowed
>
> I want to find the maximum(in fact just the value of the function at the
> max will do)
> of a function that is of the form
>
> F(x)=(int_-infty^x g(y) dy)^2-(int_-infty^x q(y) dy)
>
> Taking the derivative one gets
> g(x) int_-infty^x g(y) dy-q(x) int_-infty^x q(y) dy=0
>
> The problem with this is that the LHS is not a monotonous function
> and it may be complicated to locate all zeros and then check for the
> max.
> One could take the second derivatives to check, but the function is not
> analytically integrable
>
> I'm just curious if maxima could be of any help here.
> I have done the (fortran) routines to compute the function, it's now a
> question of locating the maxima
>
>
> ------------------------------
>
> _______________________________________________
> Maxima mailing list
> Maxima at math.utexas.edu
> http://www.math.utexas.edu/mailman/listinfo/maxima
>
>
> End of Maxima Digest, Vol 47, Issue 15
> **************************************
>
> __________ Informatie van ESET Smart Security, versie van database
> viruskenmerken 5182 (20100608) __________
>
> Het bericht is gecontroleerd door ESET Smart Security.
>
> http://www.eset.com
>
>
>