An exact 1-D limit challenge - 24
Vladimir Bondarenko
Is there a lady or a gentleman who can show *the steps* to
calculate - as always, using a computer algebra system -
the exact value of this limit?
limit(int(1/(2+sin(a*z)), z= 0..Pi), a= infinity);
(Please do not send the answer only, that cat won't jump ;)
Best wishes,
Vladimir Bondarenko
VM and GEMM architect
Co-founder, CEO, Mathematical Director
http://www.cybertester.com/ Cyber Tester, LLC
http://maple.bug-list.org/ Maple Bugs Encyclopaedia
http://www.CAS-testing.org/ CAS Testing
Axel Vogt
>
> Hello the computer algebra admirers,
>
> Is there a lady or a gentleman who can show *the steps* to
> calculate - as always, using a computer algebra system -
> the exact value of this limit?
>
> limit(int(1/(2+sin(a*z)), z= 0..Pi), a= infinity);
>
> (Please do not send the answer only, that cat won't jump ;)
>
It is Pi/sqrt(3).
Maple 10.03 tells about its problem on integrating, so
I use assumptions adapted to its problem
int(1/(2+sin(a*z)), z= 0..Pi)
assuming (0<a, -1/2 < z, z < -1/2+1/2*a);
% assuming 1 < a;
For that Maple gives me
sol:=2/3*floor(1/2+1/2*a)*Pi/a*3^(1/2)
-1/9*Pi/a*3^(1/2)+2/3*arctan(2/3*3^(1/2)*tan(1/2*Pi*a)
+1/3*3^(1/2))/a*3^(1/2)
If it ever converges than against its fixpoint which are
the even integers:
s:=[op(sol)];
eval(s, a = 2*n) assuming n::posint;
Summing gives the result. Ok, convergence proof is missing.
C W
assume(a>0);Limit(Int(1/(2+sin(a*z)), z= 0..Pi), a= infinity);
I1:=assume(u>0,u<2*Pi),student[changevar](a*z=u,Int(1/(2+sin(a*z)), z=
0..Pi),u);
val1:=floor(a/2)*int(op(1,I1),u=0..2*Pi)+int(op(1,I1),u=0..2*Pi*frac(Pi*a/(2*Pi)));
que:=assume(r>0,r<1),subs(floor(1/2*a)=1/2*a-r,frac(1/2*a)=r,val1);
map(que,a=infinity);
Chris
C W
C W wrote:
>
> Yawn..
>
> assume(a>0);Limit(Int(1/(2+sin(a*z)), z= 0..Pi), a= infinity);
> I1:=assume(u>0,u<2*Pi),student[changevar](a*z=u,Int(1/(2+sin(a*z)), z=
> 0..Pi),u);
> val1:=floor(a/2)*int(op(1,I1),u=0..2*Pi)+int(op(1,I1),u=0..2*Pi*frac(Pi*a/(2*Pi)));
> que:=assume(r>0,r<1),subs(floor(1/2*a)=1/2*a-r,frac(1/2*a)=r,val1);
> map(que,a=infinity);
^^^^^^ Correct it to :
assume(a>0,r>0,r<1):
Limit(Int(1/(2+sin(a*z)), z= 0..Pi), a= infinity):
I1:=assume(u>0,u<2*Pi),student[changevar](a*z=u,Int(1/(2+sin(a*z)), z=
0..Pi),u):
val1:=floor(a/2)*int(op(1,I1),u=0..2*Pi)+int(op(1,I1),u=0..2*Pi*frac(Pi*a/(2*Pi))):
que:=assume(r>0,r<1),subs(floor(1/2*a)=1/2*a-r,frac(1/2*a)=r,val1):
limit(que,a=infinity);
1/3*3^(1/2)*Pi
Chris
Brad Cooper
"Vladimir Bondarenko" <v...@cybertester.com> wrote in message
news:1144520425.6...@v46g2000cwv.googlegroups.com...
> Hello the computer algebra admirers,
>
> Is there a lady or a gentleman who can show *the steps* to
> calculate - as always, using a computer algebra system -
> the exact value of this limit?
>
> limit(int(1/(2+sin(a*z)), z= 0..Pi), a= infinity);
MuPAD 3.0 gives
int(1/(2+sin(a*z)), z= 0..PI,Continuous);
2/(a*sqrt(3))*arctan((2*tan(PI*a/2)+1)/sqrt(3))-PI*sqrt(3)/(9*a)
As a -> infinity, the second term -> 0
In the first term there is 2*tan(PI*a/2)+1)/sqrt(3)
As a -> infinity, this oscillates between +infinity and -infinity
consequently arctan((2*tan(PI*a/2)+1)/sqrt(3))
oscillates between +PI/2 and -PI/2
This means that arctan((2*tan(PI*a/2)+1)/sqrt(3)) is
bounded so that the first term -> 0 as a -> infinity
Therefore limit(int(1/(2+sin(a*z)), z= 0..Pi), a= infinity);
is zero.
This is easy to verify using MuPAD to plot the integral.
Cheers,
Brad
Vladimir Bondarenko
writes on Mon, Apr 10 2006 4:11 am
http://groups.google.com/group/sci.math.symbolic/msg/7e621416a65d84dd
BC> MuPAD 3.0 gives
BC> int(1/(2+sin(a*z)), z= 0..PI,Continuous);
Hello,
It's great to hear from you again!
Yesterday, having seen the terrific MuPAD bug you discovered
"MuPAD 3.0 code gives strange result"
http://groups.google.com/group/sci.math.symbolic/msg/7d1883bc6af628ac
I was in a total delight - and I have comments about it to be
uploaded soon.
The computer algebra systems of the 4th generation will have
no trouble whatsoever given such simple tasks like
limit(int(1/(2+sin(a*z)), z= 0..Pi), a= infinity);
and return the correct answer to the user instantly.
As for today's MuPAD....
Alas, our effort to get the integral directly fails
int(1/(2+sin(a*z)), z= 0..PI);
Warning: Found potential discontinuities of the antiderivative.
Try option 'Continuous' or use properties (?assume). [intlib::
antiderivative]
int(1/(sin(a*z) + 2), z = 0..PI)
Thus, at this point me or you the users are tremendously tempted...
I confess that it is almost beyond my power not to run and apply
the Continuous option there! Well... I have a vague recollection...
(According to MuPAD's help
http://www.mupad.com/STATIC/DOC25/eng/stdlib/int.shtml
The option Continuous is a technical option to tell the system that
it may assume that g is continuous. With the option Continuous, int
suppresses the search for discontinuities of g in the interval of
integration and uses the fundamental theorem of calculus without
checking whether it applies mathematically.
There is a bright wording about such stuff I adore. As Carl Love
(aka Carl DeVore) has it, it's
http://groups.google.com/group/comp.soft-sys.math.maple/msg/5f1c64c4c5d43bd7?hl=en&
"a dangerous last-ditch effort at simplification".
So to say, praying to Euler, "and lead us not into temptation" ;)
But this cannot stop me as I am really brave (I fear nobody, save
my Mom).
So I try
> int1 := int(1/(2+sin(a*z)), z= 0..PI,Continuous);
2/3/a*3^(1/2)*arctan(1/3*3^(1/2)*(2*tan(1/2*a*PI) + 1)) -
1/9/a*PI*3^(1/2)
and it is only my 15+ years QA experience that drives me, right
before the very moment of uploading, to try this...
> float(subs(int1, a=2));
-2.710505431e-20
> numeric::int(1/(2+sin(2*z)), z= 0..PI);
1.813799364 // humm...
> numeric::int(1/(2+sin(101*z)), z= 0..PI);
numeric::int(1/(2+sin(102*z)), z= 0..PI);
numeric::int(1/(2+sin(103*z)), z= 0..PI);
numeric::int(1/(2+sin(104*z)), z= 0..PI);
numeric::int(1/(2+sin(105*z)), z= 0..PI);
numeric::int(1/(2+sin(106*z)), z= 0..PI);
numeric::int(1/(2+sin(107*z)), z= 0..PI);
numeric::int(1/(2+sin(108*z)), z= 0..PI);
numeric::int(1/(2+sin(109*z)), z= 0..PI);
numeric::int(1/(2+sin(110*z)), z= 0..PI);
1.807813228
1.813799364
1.807929463
1.813799364
1.808041271
1.813799364
1.808148899
1.813799364
1.808252577
1.813799364
> float(int(1/(2+sin(1000*z)), z= 0..PI));
float(int(1/(2+sin(1001*z)), z= 0..PI));
float(int(1/(2+sin(1002*z)), z= 0..PI));
float(int(1/(2+sin(1003*z)), z= 0..PI));
1.813799364
1.813195368
1.813799364
1.813196573
So we together encountered the fact that MuPAD is not able to
produce a correct output...
In contrast to a sweet CAS of the 4th generation.
Cheers,
Vladimir
Peter Luschny
> Yesterday, having seen the terrific MuPAD bug you discovered [...]
> I was in a total delight
I think you cannot better describe the difference
between you and a MuPad, Maple, or Mathematica user:
The latter mourn about what thrills you with delight.
Regards Peter
Vladimir Bondarenko
writes on Mon, Apr 10 2006 3:00 pm
http://groups.google.com/group/sci.math.symbolic/msg/0c1ca3b56164343f
PL> I think you cannot better describe the difference...
Hello Peter,
Hurrah!! I am very, very glad to hear your voice again,
even if in a form of characters!!!
http://groups.google.com/group/comp.soft-sys.math.maple/msg/276b5ef697a9e015
PL> I allow a 90 per cent discount :)
No, I am not a politician still, - so 50:50! ;)
http://groups.google.com/group/sci.math.symbolic/msg/0c1ca3b56164343f
PL> I think you cannot better describe the difference
PL> between you and a MuPad, Maple, or Mathematica user:
PL> The latter mourn about what thrills you with delight.
I was wondering could I better describe the happy chime
between me (AXIOM, Derive, MuPAD, Maple, and Mathematica
user) and a MuPAD, Maple, or Mathematica user than this
way?
Bugs which thrill me with delight, have higher chances
to be published quickly, so the chances that they will
be fixed in the next release are presumably higher, so
in fact both me and an AXIOM, Derive, MuPAD, Maple, or
Mathematica user should rather rejoice that there is
a approach to discover bugs quickly ;)
(Now the CAS manufacturers should also rejoice and to
stand in a long line to use this opportunity to the
total delight of their customers - or reinvent our
algorithms - do you personally like this picture? ;))
"Maple - unsafe at any evaluation!" -- Peter Luschny
;-)
How, how, how much you are right!
We calculated SUCH Turkish delight ;)
I was wondering what could be a right email address
to send you some this ambrosia? I promise you, a person
who knows what Beauty is, a theoretician, you will feel
yum-yum-yum! :-)
Just in case, my most usable address is
v b @ c y b e r t e s t e r . c o m
(I wrote it with blank spaces to fool the Google Groups
truncation...)
http://groups.google.com/group/comp.soft-sys.math.maple/msg/276b5ef697a9e015
PL> And if you succeed and eventually a better Maple is
PL> build, you will become famous like Nader and might
PL> some day be candidate for the presidential election
PL> of the Ukraine :)
Preparing for the elections '08 ! ;-)
Best, Vladimir
Peter Luschny
>> Peter Luschny writes
> PL> I think you cannot better describe the difference
> PL> between you and a MuPad, Maple, or Mathematica user:
> PL> The latter mourn about what thrills you with delight.
> I was wondering could I better describe the happy chime
> between me (AXIOM, Derive, MuPAD, Maple, and Mathematica
> user) and a MuPAD, Maple, or Mathematica user than this
> way?
I said, no, you could not.
> (Now the CAS manufacturers should also rejoice and to
> stand in a long line to use this opportunity to the
> total delight of their customers - or reinvent our
> algorithms - do you personally like this picture? ;))
They really ought to know all about it. And if they
do not, they should invest in your software, sure.
> "Maple - unsafe at any evaluation!" -- Peter Luschny
Well, this is a little bit unfair, especially if taken
out of context, so let me be more precise:
Every CAS is unsafe at any evaluation.
However, apart from freshmen, every user should know that
and exercise accordingly great caution in using this
kind of software.
> PL> And if you succeed and eventually a better Maple is
> PL> build, you will become famous like Nader
Yes, but I observed that your campaign lost rather
than gained support. You forgot another quotation
from me: 'Vladimir's marketing is not perfect yet.'
Sadly, this is still true.
> Preparing for the elections '08 ! ;-)
You are not the only one:
http://pandagon.net/2006/04/04/let-me-be-your-hillary-clinton/
Regards Peter