Rouben Rostamian

MaplePrimes Activity


These are replies submitted by Rouben Rostamian

@Aliocha Ah, I see the reason for the confusion.  The f(n,x) in my answer stands for the antiderivative of your f(n,x),  so I should have called it by some other name, like F:

F := (n,x) -> n*x + 2*sum((n-k)/k*sin(k*x), k=1..n-1);

You wish to integrate your f(n,x) over the interval (0,Pi).  That amounts to evaluating

F(n,Pi) - F(n,0);

But sin(k*x) is zero at 0 and Pi, therefore the summation evaluates to zero and we are left with n*Pi - n*0, that is n*Pi, as you had expected.
 

@Aliocha I don't have Maple 2026 but the formula works just fine in Maple 2024 and 2025.  Here it is:

restart;

kernelopts(version);

`Maple 2024.2, X86 64 LINUX, Oct 29 2024, Build ID 1872373`

f := (n,x) -> n*x + 2*sum((n-k)/k*sin(k*x), k=1..n-1);

proc (n, x) options operator, arrow; x*n+2*(sum((n-k)*sin(k*x)/k, k = 1 .. n-1)) end proc

f(1,x);

x

f(2,x);

2*x+2*sin(x)

f(3,x);

3*x+4*sin(x)+sin(2*x)

f(4,x);

4*x+6*sin(x)+2*sin(2*x)+(2/3)*sin(3*x)

Download mw.mw

 

 

@C_R My comment applies to your worksheet 2d_BVP_Haberman.mw where BC3 specifies zero temperature on the outer boundary.  That's where you need the compatibility condition that I have noted.  That wouldn't be an issue in the case of an insulated boundary.

@janhardo In the solutions that you have provided, you are taking q = q(t), but the original statement says q = q(r,t). 

To make the boundary conditions BC1 and BC3 mutually consistent, you need to have Q(a,t)=0 for all t.  Otherwise BC3 says that the temperature is zero on the outer wall while BC1 prescibes a nonzero rate of change of temperature at r=a and z=0.

Once you impose that condition, you may obtain a series solution in a straightforward way.  I can provide the details if that is of interest.

@dharr Thanks for the detailed explanation.  Voted up!

The additional option known=F fixes the issue:

dchange({tr}, DE, {xi, v(xi)}, known=F);

All is well now.

@sand15 Thanks for pointing out the presence of the "R" factor.  I didn't see it at the first glance.

@vv Thanks for the change of variable suggestion.  I should have seen that :-(

Anyway, here is the complete solution of the problem.

restart;

 

De-nesting radicals

 

The key to the calculations is the elementary algebraic identity
which may be verified by hand:

If A^2-B = C^2,  then  sqrt(A+`&+-`(sqrt(B))) = sqrt((A+C)*(1/2))+`&+-`(sqrt((A-C)*(1/2)))

 

The consequences

 

e__1 := sqrt(3*r^4 + 2*r^2*sqrt(2*r^4 + 1) + 1);

(3*r^4+2*r^2*(2*r^4+1)^(1/2)+1)^(1/2)

Apply the de-nesting identity with A = 3*r^4+1,  B = (2*r^2)^2*(2*r^4+1)","

A := 3*r^4 + 1;
B := (2*r^2)^2*(2*r^4+1);
factor(A^2 - B);
C := sqrt(%) assuming r > 0;

3*r^4+1

4*r^4*(2*r^4+1)

(r^4+1)^2

r^4+1

and therefore e__1 is equivalent to

E__1 := sqrt((A+C)/2) + sqrt((A-C)/2) assuming r > 0;

(2*r^4+1)^(1/2)+r^2

Let's verify:

simplify(E__1^2 - e__1^2) assuming r > 0;

0

Similarly

e__2 := sqrt(3*r^4 - 2*r^2*sqrt(2*r^4 + 1) + 1);

(3*r^4-2*r^2*(2*r^4+1)^(1/2)+1)^(1/2)

is equivalent to

E__2 := sqrt((A+C)/2) - sqrt((A-C)/2) assuming r > 0;

(2*r^4+1)^(1/2)-r^2

 

Integration

 

Following vv's work, we begin by converting the original double integral
to polar coordinates

f := (2*x^2+1)/(x^4+6*x^2*y^2+y^4+1)-(y^2+1)/(x^4+y^4+2);

(2*x^2+1)/(x^4+6*x^2*y^2+y^4+1)-(y^2+1)/(x^4+y^4+2)

g := simplify(eval(f, [x=r*cos(t), y=r*sin(t)]));

(2*r^2*cos(t)^2+1)/(r^4*cos(t)^4+6*r^4*cos(t)^2*sin(t)^2+r^4*sin(t)^4+1)+(-r^2*sin(t)^2-1)/(r^4*cos(t)^4+r^4*sin(t)^4+2)

h := simplify(int(g*r, t=-Pi..Pi)) assuming r>0;

-r*((2*(r^4+1)^(1/2)*(2*r^4+1)^(1/2)*(r^2+2)*(3*r^4+2*r^2*(2*r^4+1)^(1/2)+1)^(1/2)+(2*r^4+8)^(1/2)*(r^4+2)^(1/2)*(r^4-r^2-(2*r^4+1)^(1/2)+1))*(3*r^4-2*r^2*(2*r^4+1)^(1/2)+1)^(1/2)-(3*r^4+2*r^2*(2*r^4+1)^(1/2)+1)^(1/2)*(2*r^4+8)^(1/2)*(r^4+2)^(1/2)*(r^4-r^2+(2*r^4+1)^(1/2)+1))*Pi/((2*r^4+8)^(1/2)*(r^4+2)^(1/2)*(r^4+1)^(1/2)*(3*r^4+2*r^2*(2*r^4+1)^(1/2)+1)^(1/2)*(2*r^4+1)^(1/2)*(3*r^4-2*r^2*(2*r^4+1)^(1/2)+1)^(1/2))

We convert the nested integrals in h through the de-nesting identities derived above

subs(e__1^2 = E__1^2, e__2^2 = E__2^2, h):
simplify(%) assuming r > 0:
expand(%):
integrand := sort(%, r);

-2*Pi*r^3/((2*r^4+8)^(1/2)*(r^4+2)^(1/2))+2*Pi*r^3/((r^4+1)^(1/2)*(2*r^4+1)^(1/2))-4*Pi*r/((2*r^4+8)^(1/2)*(r^4+2)^(1/2))+2*Pi*r/((r^4+1)^(1/2)*(2*r^4+1)^(1/2))

Ultimately we wish to integrate the above expression over 0 < r and r < infinity.

We examine its individual terms, thus we set

tt := op(integrand);

-2*Pi*r^3/((2*r^4+8)^(1/2)*(r^4+2)^(1/2)), 2*Pi*r^3/((r^4+1)^(1/2)*(2*r^4+1)^(1/2)), -4*Pi*r/((2*r^4+8)^(1/2)*(r^4+2)^(1/2)), 2*Pi*r/((r^4+1)^(1/2)*(2*r^4+1)^(1/2))

It turns out the 3rd and 4th terms integrate to the same value
and therefore they cancel each other.

int(tt[3] + tt[4], r=0..infinity);

0

Therefore we are left with

ans := int(tt[1]+tt[2], r=0..infinity);

int(-2*Pi*r^3/((2*r^4+8)^(1/2)*(r^4+2)^(1/2))+2*Pi*r^3/((r^4+1)^(1/2)*(2*r^4+1)^(1/2)), r = 0 .. infinity)

The integrals of the individual terms diverge, however the integral

of their sum is well-defined;  Following vv's idea, we apply the

change of variable u = r^4 and arrive at the desired answer;

IntegrationTools:-Change(ans, r^4=u, u);

(1/2)*Pi*2^(1/2)*ln(2)

 

 

Download mw2.mw

@vv Your calculation is good but the factor of R is missing in the area element R*dR*dt.  On changing the last line of your code to

evalf(Int(h*R, R=0..infinity));

we arrive at the answer -2.784281108.

@Earl I should have checked for the compatibility of the code with Maple 2020 but I didn't.  Sorry.  To get that Maple 2024 code work in Maple 2020, it suffices to change the line

subs(R=1, m[1]=2, m[2]=3, {de1, de2, ic}):

to

subs(R=1, m[1]=2, m[2]=3, {seq((lhs-rhs)(de1)), seq((lhs-rhs)(de2)), ic}):

@ecterrab That's impressive work on the part of AI.  It explains the step-by-step calculations quite well and every step checks.  What is missing, however, is the motivation behind those calculations.  That's where the human touch is superior in comparison to the current state of AI.

A very nice analysis of the problem is available in https://arxiv.org/pdf/2505.19088 and provides some interesting ideas behind the solution.  It gives 3 different parameterized solutions; see sections 2.2, 2.3, and 2.4.  It is not clear whether those 3 provide all possible solutions.  It will be interesting to see if the AI's solution generates numbers beyond those given by that paper's formulas.  I haven't checked.

Hello Edgardo,

I can't tell whether ExactAI is limited to Physics, or any CAS-related questions.  If the latter, then you may be interested in the following.

I passed this to ChatGPT:

Are there positive integer solutions (a,b,c,u,v,w)
to the following system of equations?
                       a + b + c = u^2 
                        ab + bc + ca = v^2 
                        abc = w^2

It returned one valid solution.  (That's an improvement over last week when it said
there are no solutions.)

Then I asked the same question with an extra sentence at the end:

Are there positive integer solutions (a,b,c,u,v,w)
to the following system of equations?
                       a + b + c = u^2 
                        ab + bc + ca = v^2 
                        abc = w^2
If yes, are there infinitely many solutions?

ChatGpt tried various techniques, and finally concluded that 

"Therefore the system has no positive integer solutions.".  See the attached transcript in response.pdf which contradicts its previous finding.

I don't expect an AI to have answers to all questions, but it is disconcerting when it states a wrong answer as fact.  Perhaps your ExactAI can prevent such incidences.

As hard as it is to believe, the solution x=1633780814400, y=252782198228, z=3474741058973 was actually obtained by Euler himself without help from Maple or even a calculator ;-)

But that's not the smallest solution.  The triplet x=45, y=64, z=180 is also a solution.  There are a lot of other solutions in between.  

By the way, both ChatGPT and Google's Gemni claim that no such x, y, z can exist!

Spoiler alert!

Have a look at https://arxiv.org/pdf/2505.19088 to see how it is done.

I use Maple quite a bit in my daily work, but like you, I find the ribbon interface to be slow and inefficient, and that's why I am staying with Maple 2024.  I would really like to see that former interface returned as an option.  Your suggeston of a user-customizable interface will work for me too.

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