one hard integral for Maple: int(exp(cos(x))*cos(x-sin(x)),x=0..2*Pi);
Nasser M. Abbasi
Maple 12
Can do this integral numerically, but not analytically:
> restart;
f:=exp(cos(x))*cos(x-sin(x));
int(f,x=0..2*Pi);
------> returns unevaluated
> restart;
f:=exp(cos(x))*cos(x-sin(x));
evalf(Int(f,x=0..2*Pi));
6.283185307
OK.
I do not have Maple13, but was wondering if it can do it in Maple13?
--Nasser
Axel Vogt
Plotting and playing suggests f:= x -> exp(cos(x))*cos(x-sin(x)) is
periodic and symmetric like cos, actually both f(Pi+x) - f(Pi-x) and
f(2*Pi+x) - f(x) are zero, so the task is 2*Int(f(x), x= 0 .. Pi)
Smells like a Fourier thing ... playing with exp(cos(x)+I*sin(x)) and
differentiating gives
Int(f(x),x=0..Pi); expand(%);
Pi
/
|
| exp(cos(x)) cos(x) cos(sin(x))
|
/
0
+ exp(cos(x)) sin(x) sin(sin(x)) dx
And exp(cos(x)+I*sin(-x)); evalc(Im(%)); diff(%, x);
-exp(cos(x)) cos(x) cos(sin(x)) + exp(cos(x)) sin(x) sin(sin(x))
Now -exp(cos(x))*sin(sin(x)) is 0 in 0 and in Pi and an antiderivative,
hence the task is
Int(2*exp(cos(x))*cos(x)*cos(sin(x)),x = 0 .. Pi)
which Maple 12 evaluates to be Pi.
Done.
Nasser M. Abbasi
"Axel Vogt" <&nor...@axelvogt.de> wrote in message
news:7o2i91F...@mid.individual.net...
hi Alex;
But Pi is the WRONG answer. It should be 2 Pi?
--Nasser
Rouben Rostamian
> Maple 12
> Can do this integral numerically, but not analytically:
>
>> restart;
> f:=exp(cos(x))*cos(x-sin(x));
> int(f,x=0..2*Pi);
Maple 11, 12, 13 can do it if we expand f first:
int(expand(f), x=0..2*Pi);
The result is 2*Pi.
--
Rouben Rostamian
Axel Vogt
> "Axel Vogt" <&nor...@axelvogt.de> wrote in message
> news:7o2i91F...@mid.individual.net...
>> Nasser M. Abbasi wrote:
> hi Alex;
>
> But Pi is the WRONG answer. It should be 2 Pi?
>
As said: your original integral is twice of it. Axel
Nasser M. Abbasi
"Nasser M. Abbasi" <n...@12000.org> wrote in message
news:hfh586$84q$1...@aioe.org...
Alex, With the transformation you did, I was not sure if Pi you showed above
is the answer to the same problem I had.
The analytical answer to
f:=exp(cos(x))*cos(x-sin(x));
int(f,x=0..2*Pi);
should be 2 Pi, not Pi.
--Nasser
Axel Vogt
Much better than my way ...
Fırat
I want to ask new integral such as
f:=(t-a)^c/(b-t)^d; #a,b,c,d are contant
How can I found analytical integral solution
Firat
G. A. Edgar
<59c20e76-4591-4522...@v25g2000yqk.googlegroups.com>,
F�rat <agim...@gmail.com> wrote:
Well, int(t^c*(1-t)^d,t=0..1); is the Beta function.
Change variables x=(t-a)/(b-a) to reduce yours to the case a=0, b=1.
Then since yours is the indefinite integral, it would be called an
"incomplete Beta function" I suppose. Maple reports
Int(x^c*(1-x)^(-d), x) = x^(c+1)*hypergeom([d, c+1], [2+c], x)/(c+1)
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
Fırat
wrote:
> In article
> <59c20e76-4591-4522-b099-c3e09115f...@v25g2000yqk.googlegroups.com>,
>
>
>
>
>
> Fžrat <agimfi...@gmail.com> wrote:
> > On 6 Aralžk, 22:53, Axel Vogt <&nore...@axelvogt.de> wrote:
> > > Rouben Rostamian wrote:
> > > > On 2009-12-06, Nasser M. Abbasi <n...@12000.org> wrote:
> > > >> Maple 12
> > > >> Can do this integral numerically, but not analytically:
>
> > > >>> restart;
> > > >> f:=exp(cos(x))*cos(x-sin(x));
> > > >> int(f,x=0..2*Pi);
>
> > > > Maple 11, 12, 13 can do it if we expand f first:
>
> > > > int(expand(f), x=0..2*Pi);
>
> > > > The result is 2*Pi.
>
> > > Much better than my way ...
>
> > I want to ask new integral such as
>
> > f:=(t-a)^c/(b-t)^d; #a,b,c,d are contant
>
> > How can I found analytical integral solution
>
> > Firat
>
> Well, int(t^c*(1-t)^d,t=0..1); is the Beta function.
> Change variables x=(t-a)/(b-a) to reduce yours to the case a=0, b=1.
> Then since yours is the indefinite integral, it would be called an
> "incomplete Beta function" I suppose. Maple reports
> Int(x^c*(1-x)^(-d), x) = x^(c+1)*hypergeom([d, c+1], [2+c], x)/(c+1)
>
> --
>
> - Alıntıyı göster -
Thanks for your answer. But I have some problem. I wrote my maple file
as you say in the following form:
with(student):
F:=Int((t-a)^c/(b-t)^d,t);
/ c
| (t - a)
F := | -------- dt
| d
/ (b - t)
> changevar(x=(t-a)/(b-a),F,x);
/ c
| (x b - x a) (b - a)
| -------------------- dx
| d
/ (b - a - x b + x a)
> evalf(%);
/ c
| (x b - 1. x a) (b - 1. a)
| -------------------------- dx
| d
/ (b - 1. a - 1. x b + x a)
but it does not give any answer. Main problem is: this integral form
appear in my iteration statement and I have to use this integral value
for new iteration after previous iteration. I want to integrate this
function and get numerical value.
Is this possible?
Firat
Fırat
>
> - Alıntıyı göster -
I forget that the power c and d are non-intager value
Ray Vickson
>
> > - Alıntıyı göster -
>
> Thanks for your answer. But I have some problem. I wrote my maple file
> as you say in the following form:
>
> with(student):
> F:=Int((t-a)^c/(b-t)^d,t);
>
> / c
> | (t - a)
> F := | -------- dt
> | d
> / (b - t)
>
> > changevar(x=(t-a)/(b-a),F,x);
>
> / c
> | (x b - x a) (b - a)
> | -------------------- dx
> | d
> / (b - a - x b + x a)
>
> > evalf(%);
You cannot get a numerical answer unless you specify numerical values
for a, b, c and the integration limits. Maple cannot read your mind.
>
> / c
> | (x b - 1. x a) (b - 1. a)
> | -------------------------- dx
> | d
> / (b - 1. a - 1. x b + x a)
>
> but it does not give any answer. Main problem is: this integral form
> appear in my iteration statement and I have to use this integral value
> for new iteration after previous iteration. I want to integrate this
> function and get numerical value.
You can write answer_above = (b-a)^(c+1-d) * int(x^c/(1-x)^d,x), and
then express this last integral in terms of an incomplete beta
function (or a hypergeometric function, as Maple seems to prefer).
>
> Is this possible?
Of course not. You have not said what are the values of a, b, c and
the integration limits.
R.G. Vickson
>
> Firat
Fırat
>
> - Alıntıyı göster -- Alıntıyı gizle -
>
> - Alıntıyı göster -
In my previous message I said the power c and d are non-integer value.
I give some values as you say but it does not evulate integral for
example
>c:=0.3, d:=0.5;
>int(x^c/(1-x)^d,x);
Ray Vickson
restart;params:={c=0.3,d=0.5};
params := {c = 0.3, d = 0.5}
> J:=int(x^c/(1-x)^d,x);
(c + 1)
x hypergeom([d, c + 1], [2 + c], x)
J := ------------------------------------------
c + 1
J1:=subs(params,J);
1.3
J1 := 0.7692307692 x hypergeom([0.5, 1.3], [2.3], x)
Of course, if we give numerical values for the integration limits we
will get a numerical evaluation:
J2:=subs(params,int(x^c/(1-x)^d,x=0 .. .97)): value(J2);
1.362551855
R.G. Vickson
Fırat
>
> - Alıntıyı göster -
Thanks for your kindly help and understanding. I am beginner in Maple.
I want to ask a question. I want to use the first example in Maplesoft
webpage
I wrote as follow
>with(Student(NumericalAnalysis));
`Error, NumericalAnalysis is not a command in the Student package\n`
and give above error message.
I want write this in Maple 10 classical worksheet. Please let me know
where I did wrong.
Firat
Robert Israel
> Thanks for your kindly help and understanding. I am beginner in Maple.
> I want to ask a question. I want to use the first example in Maplesoft
> webpage
>
>
>http://www.maplesoft.com/support/help/AddOns/view.aspx?path=3DStudent%2fNum=
> ericalAnalysis%2fInitialValueProblem
>
> I wrote as follow
>
> >with(Student(NumericalAnalysis));
> `Error, NumericalAnalysis is not a command in the Student package\n`
I hope what you wrote was actually
with(Student[NumericalAnalysis]);
> and give above error message.
>
> I want write this in Maple 10 classical worksheet. Please let me know
> where I did wrong.
The NumericalAnalysis subpackage was first introduced in Maple 13. It
does not exist in Maple 10.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Bill
> > Maple 11, 12, 13 can do it if we expand f first:
>
> > int(expand(f), x=0..2*Pi);
>
> > The result is 2*Pi.
Actually, a little simplify will produce the desired result directly:
y := exp(cos(x))*cos(x-sin(x)):
ti := int(y, x=0..2*Pi):
simplify(ti, [trig]);
2 Pi
Bill
Fırat
<isr...@math.MyUniversitysInitials.ca> wrote:
> =?UTF-8?B?RsSxcmF0?= <agimfi...@gmail.com> writes:
> > Thanks for your kindly help and understanding. I am beginner in Maple.
> > I want to ask a question. I want to use the first example in Maplesoft
> > webpage
>
> > ericalAnalysis%2fInitialValueProblem
>
> > I wrote as follow
>
> > >with(Student(NumericalAnalysis));
> > `Error, NumericalAnalysis is not a command in the Student package\n`
>
> I hope what you wrote was actually
>
> with(Student[NumericalAnalysis]);
>
> > and give above error message.
>
> > I want write this in Maple 10 classical worksheet. Please let me know
> > where I did wrong.
>
> The NumericalAnalysis subpackage was first introduced in Maple 13. It
> does not exist in Maple 10.
> --
> Robert Israel isr...@math.MyUniversitysInitials.ca
> Department of Mathematics http://www.math.ubc.ca/~israel
> University of British Columbia Vancouver, BC, Canada
Thanks for your answer
Archimedes
Using simplify with the trig option is square brackets is equivalent
to issuing the command "expand" first, and then simplifying the
output. So simplify(ti, trig); is not the same as simplify(ti,
[trig]); Is this documented somewhere? I did not see this in the
help pages for simplify, or simplify/siderels
Regards,
Georgios
Bill
>
> Using simplify with the trig option is square brackets is equivalent
> to issuing the command "expand" first, and then simplifying the
> output. So simplify(ti, trig); is not the same as simplify(ti,
> [trig]); Is this documented somewhere? I did not see this in the
> help pages for simplify, or simplify/siderels
>
> Regards,
> Georgios
I agree that the help pages don't explain this well. If you want to
see what the difference is, try executing
> trace(simplify)
and then run the other commands. It's quite a bit different. Also try
'expand. to see how it calls 'simplify.'
Bill
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