Maple Questions and Posts

Simple Integral of Product Gives Wrong Results...

Asked by: If_you_dont_Mind 10

September 27 2016

2 7

Hello,

I've got the following problem: Sometimes when I let Maple compute an integral it gives a result which is much smaller than it should be. This happens even for rather simple integrals such as the product of a Gauss function and a 1/x-type function. Here's a minimal working example:

restart;
with(plots):
f := x -> exp(-x^2):
g := (x,a) -> 1/(1+abs(x-a)):
plot([f(x),g(x,3)],x=-3..10):
for a from 1 to 100 do
A[a] := int(f(x)*g(x,a),x=-infinity..a-1)
end do:
B:=[seq(A[a],a=1..100)]:
X:=[seq(a,a=1..100)]:
plot(X,B);

The plot it returns looks like this:

which is clearly wrong between around a=35 and a=60...

Does anyone have an idea how this can be repaired?

Use of Explore function on a quiet big trigonometr...

Asked by: Bendesarts 450

September 27 2016

0 9

Hello,

I try to use the Explore function on a trigonometric expressions depending of 8 parameters. My aim is to study the influence of these parameters on the results.

However, i receive an error message and I didn't manage to troubleshoot it.

May you have a look of an extract of my code and see if you see the mistake ?

TestExplore.mw

Thanks a lot for your help.

Error showing as .......

Asked by: shakuntala 25

September 27 2016

0 2

Dear sir

I try the below program and it is not executing because of error and that error is " unable to match delimiters"

Please can you varify it.

sol1 := dsolve([diff(diff(diff(f(eta),eta),eta),eta)+f(eta)*diff(diff(f(eta),eta),eta)+((2*n)/((n+1)))*(1-diff(f(eta),eta)^2) = 0, 1/(Pr) *diff(diff(theta(eta),eta),eta)+f(eta)*diff(theta(eta),eta)-((2*p)/((n+1)))diff(f(eta),eta)*theta(eta) = 0, f(0) = 0, D(f)(0) = 1+lambda*'@@'(D,2)(f)(0), D(f)(10) = 1, theta(0)=0, theta(0)=1+sigma*D(theta(0)),theta(10)=1], numeric, method = bvp); plots[odeplot](sol1, [eta, `@@`(D,2)(f)(eta)], color = red,axes=boxed);

removing rows, or columns from a DataFrame...

Asked by: a_simsim 120

September 27 2016

1 3

Dear Community,

Is there an explicit command to remove rows or columns from a DataFrame? I've only found the Append command. What's the easiest way to do it?

Tx in advance,

best regards

Andras

Extracting a Constant coefficient in trignomentic ...

Asked by: Rajan191 20

September 27 2016

2 3

eq2 := -4*A[2]*cos(2*x)-16*A[4]*cos(4*x)-36*A[6]*cos(6*x)-64*A[8]*cos(8*x)+a*A[0]+cos(8*x)*a*A[8]+cos(6*x)*a*A[6]+cos(4*x)*a*A[4]+2*cos(2*x)*q*A[0]+cos(2*x)*a*A[2]+q*A[8]*cos(6*x)+q*A[8]*cos(10*x)+q*A[6]*cos(4*x)+q*A[6]*cos(8*x)+q*A[4]*cos(2*x)+q*A[4]*cos(6*x)+q*A[2]*cos(4*x)+q*A[2]

How to extract the co-fficent that do not have cos in it like

a*A[0]+q*A[2]

Quantization of the Lorentz Force

by: ecterrab 14605 Product: Maple

September 26 2016

3 0

This presentation is on an undergrad intermediate Quantum Mechanics topic. Tackling the problem within a computer algebra worksheet in the way shown below is actually the novelty, using the Physics package to formulate the problem with quantum operators and related algebra rules in tensor notation.

Quantization of the Lorentz Force

Pascal Szriftgiser¹ and Edgardo S. Cheb-Terrab²

(1) Laboratoire PhLAM, UMR CNRS 8523, Université Lille 1, F-59655, France

(2) Maplesoft

We consider the case of a quantum, non-relativistic, particle with mass m and charge q evolving under the action of an arbitrary time-independent magnetic field where is the vector potential. The Hamiltonian for this system is

H = (`#mover(mi("p",mathcolor = "olive"),mo("→"))`-q*`#mover(mi("A",mathcolor = "olive"),mo("→"))`(X))^2/(2*m)

where is the momentum of the particle, and the force acting in this particle, also called the Lorentz force, is given by

where is the quantized velocity of the particle, and all of , , , , and are Hermitian quantum operators representing observable quantities.

In the classic (non-quantum) case, for such a particle in the absence of electrical field is given by

= ,

Problem: Departing from the Hamiltonian, show that in the quantum case the Lorentz force is given by [1]

[1] Photons et atomes, Introduction à l'électrodynamique quantique, p. 179, Claude Cohen-Tannoudji, Jacques Dupont-Roc et Gilbert Grynberg - EDP Sciences janvier 1987.

Solution

We choose to tackle the problem in Heisenberg's picture of quantum mechanices, where the state of a system is static and only the quantum operators evolve in time according to

Also, the algebraic manipulations are simpler using tensor abstract notation instead of the standard 3D vector notation. We then start setting the framework for the problem, a system of coordinates X, indicating the dimension of the tensor space to be 3 and the metric Euclidean, and that we will use lowercaselatin letters to represent tensor indices. In addition, not necessary but for convenience, we set the lowercase latin i to represent the imaginary unit and we request automaticsimplification so that the output of everything comes automatically simplified in size.

>

(1)

Next we indicate the letters we will use to represent the quantum operators with which we will work, and also the standard commutation rules between position and momentum, always the starting point when dealing with quantum mechanics problems

>

Setup(quantumoperators = {F}, hermitianoperators = {A, B, H, p, r, v, x}, realobjects = {`&hbar;`, m, q}, algebrarules = {%Commutator(p[k], p[n]) = 0, %Commutator(x[k], p[l]) = I*`&hbar;`*KroneckerDelta[k, l], %Commutator(x[k], x[l]) = 0})

[algebrarules = {%Commutator(p[k], p[n]) = 0, %Commutator(x[k], p[l]) = I*`&hbar;`*Physics:-KroneckerDelta[k, l], %Commutator(x[k], x[l]) = 0}, hermitianoperators = {A, B, H, p, r, v, x}, quantumoperators = {A, B, F, H, p, r, v, x}, realobjects = {`&hbar;`, m, q, x1, x2, x3, %dAlembertian, Physics:-dAlembertian}]

(2)

Note that we start not indicating as Hermitian, in order to arrive at that result. The quantum operators , , and are explicit functions of X, so to avoid redundant display of this functionality on the screen we use

>

(3)

Define now as tensors the quantum operators that we will use with tensorial notation (recalling: for these, Einstein's sum rule for repeated indices will be automatically applied when simplifying)

>

${A, B, F, p, v, x, Physics:-Dgamma[a], Physics:-Psigma[a], Physics:-d_[a], Physics:-g_[a, b], Physics:-KroneckerDelta[a, b], Physics:-LeviCivita[a, b, c], Physics:-SpaceTimeVector[a](X)}$

(4)

The Hamiltonian,

H = (`#mover(mi("p",mathcolor = "olive"),mo("→"))`-q*`#mover(mi("A",mathcolor = "olive"),mo("→"))`(X))^2/(2*m)

in tensorial notation, is given by

>

H = (1/2)*Physics:-`^`(p[n]-q*A[n](X), 2)/m

(5)

Generally speaking to arrive at what we now need to do is

1) Express this Hamiltonian (5) in terms of the velocity

And, recalling that, in Heisenberg's picture, quantum operators evolve in time according to

2) Take the commutator of with the velocity itself to obtain its time derivative and, from , that commutator is already the force up to some constant factors.

To get in contact with the basic commutation rules between position and momentum behind quantum phenomena, the quantized velocity itself can be computed as the time derivative of the position operator, i.e as the commutator of with

>

I*Physics:-Commutator(H, x[k])/`&hbar;` = (1/2)*(I*q^2*Physics:-AntiCommutator(A[n](X), Physics:-Commutator(A[n](X), x[k]))-I*q*Physics:-AntiCommutator(p[n], Physics:-Commutator(A[n](X), x[k]))-2*(q*A[n](X)-p[n])*Physics:-KroneckerDelta[k, n]*`&hbar;`)/(`&hbar;`*m)

(6)

This expression for the velocity, that involves commutators between the potential , the position and the momentum , can be simplified taking into account the basic quantum algebra rules between position and momentum. We assume that (X) can be decomposed into a formal power series (possibly infinite) of the , hence all the commute between themselves as well as with all the

>

(7)

(Note: in some cases, this is not true, but those cases are beyond the scope of this worksheet.)

Add these rules to the algebra rules already set so that they are all taken into account when simplifying things

>

(8)

>

I*Physics:-Commutator(H, x[k])/`&hbar;` = (-A[k](X)*q+p[k])/m

(9)

The right-hand side of (9) is then the k^th component of the velocity tensor quantum operator, the relationship is the same as in the classical case

>

(10)

and with this the Hamiltonian (5) can now be rewritten in term of the velocity completing step 1)

>	$simplify(H = (1/2)Physics[`^`](p[n]-qA[n](X), 2)/m, {SubstituteTensorIndices(k = n, (rhs = lhs)(v[k] = (-A[k](X)*q+p[k])/m))})$

H = (1/2)*m*Physics:-`^`(v[n], 2)

(11)

For step 2), to compute

`#mover(mi("F",mathcolor = "olive"),mo("→"))` = m*(diff(v(t), t)) and m*(diff(v(t), t)) = I*m*Physics:-Commutator(H, v(t)[k])/`&hbar;`

we need the commutator between the different components of the quantized velocity which, contrary to what happens in the classical case, do not commute. For this purpose, take the commutator between (10) with itself after replacing the free index

>

Physics:-Commutator(v[k], v[n]) = -q*(Physics:-Commutator(A[k](X), p[n])+Physics:-Commutator(p[k], A[n](X)))/m^2

(12)

To simplify (12), we use the fact that if f is a commutative mapping that can be decomposed into a formal power series in all the complex plan (which is assumed to be the case for all (X)), then

where is the momentum operator along the axis. This relation reads in tensor notation:

>

(13)

Add this rule to the rules previously set in order to automatically take it into account in (12)

>

$[algebrarules = {%Commutator(p[k], p[n]) = 0, %Commutator(p[k], A[n](X)) = -I*`&hbar;`*Physics:-d_[k](A[n](X), [X]), %Commutator(x[k], p[l]) = I*`&hbar;`*Physics:-KroneckerDelta[k, l], %Commutator(x[k], x[l]) = 0, %Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0}]$

(14)

>

Physics:-Commutator(v[k], v[n]) = -I*q*`&hbar;`*(Physics:-d_[n](A[k](X), [X])-Physics:-d_[k](A[n](X), [X]))/m^2

(15)

Also add this other rule so that it is taken into account automatically

>

$[algebrarules = {%Commutator(p[k], p[n]) = 0, %Commutator(p[k], A[n](X)) = -I*`&hbar;`*Physics:-d_[k](A[n](X), [X]), %Commutator(v[k], v[n]) = -I*q*`&hbar;`*(Physics:-d_[n](A[k](X), [X])-Physics:-d_[k](A[n](X), [X]))/m^2, %Commutator(x[k], p[l]) = I*`&hbar;`*Physics:-KroneckerDelta[k, l], %Commutator(x[k], x[l]) = 0, %Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0}]$

(16)

Recalling now the expression of the Hamiltonian (11) as a function of the velocity, one can compute the components of the force operator "()Component(v*B,k)=m (v[k])=(i m [H,v[k]][-])/`&hbar;`"

>

F[k](X) = I*m*%Commutator((1/2)*m*Physics:-`^`(v[n], 2), v[k])/`&hbar;`

(17)

Simplify this expression for the quantized force taking the quantum algebra rules (16) into account

>

F[k](X) = (1/2)*q*(-Physics:-`*`(Physics:-d_[n](A[k](X), [X]), v[n])+Physics:-`*`(Physics:-d_[k](A[n](X), [X]), v[n])-Physics:-`*`(v[n], Physics:-d_[n](A[k](X), [X]))+Physics:-`*`(v[n], Physics:-d_[k](A[n](X), [X])))

(18)

It is not difficult to verify that this is the antisymmetrized vector product . Departing from expressed using tensor notation,

>

(19)

and taking into acount that

multiply both sides of (19) by , getting

>

(20)

>

(21)

Finally, replacing the repeated index m by n

>

(22)

Likewise, for

multiplying (19), this time from the right instead of from the left, we get

>

(23)

>

(24)

Simplifying now the expression (18) for the quantized force taking into account (22) and (24) we get

>

$simplify(F[k](X) = (1/2)*q*(-Physics[`*`](Physics[d_][n](A[k](X), [X]), v[n])+Physics[`*`](Physics[d_][k](A[n](X), [X]), v[n])-Physics[`*`](v[n], Physics[d_][n](A[k](X), [X]))+Physics[`*`](v[n], Physics[d_][k](A[n](X), [X]))), {(rhs = lhs)(Physics[LeviCivita][b, c, k]*Physics[`*`](v[b], B[c](X)) = Physics[`*`](v[n], Physics[d_][k](A[n](X), [X]))-Physics[`*`](v[n], Physics[d_][n](A[k](X), [X]))), (rhs = lhs)(Physics[LeviCivita][b, c, k]*Physics[`*`](B[c](X), v[b]) = Physics[`*`](Physics[d_][k](A[n](X), [X]), v[n])-Physics[`*`](Physics[d_][n](A[k](X), [X]), v[n]))})$

F[k](X) = (1/2)*q*Physics:-LeviCivita[b, c, k]*(Physics:-`*`(v[b], B[c](X))+Physics:-`*`(B[c](X), v[b]))

(25)

i.e.

in tensor notation. Finally, we note that this operator is Hermitian as expected

>

(26)

Download: Quantization_of_the_Lorentz_force.mw, Quantization_of_the_Lorentz_force.pdf

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft
Editor, Computer Physics Communications

Finite Difference Method for PDE...

Asked by: amrramadaneg 76

September 26 2016

0 2

Dear all

I have a PDE and its analytical solution. I want to find the numerical solution by Finite Difference Method.

I duscratize the PDE and boundary condition and Could not able to solve them togethe.

Here is the file FEM-Nu.mw

How to find the Taylor expansion...

Asked by: Markiyan Hirnyk 7819

September 26 2016

0 10

of the implicit function sin(x+y)+sin(x) = y at x = Pi , y=0 of order 15? Here is one of the difficulties

restart; eval(implicitdiff(sin(x+y)+sin(x) = y, y, x$15), [x = Pi, y = 0]);
Error, (in expand/bigprod) Maple was unable to allocate enough memory to complete this computation. Please see ?alloc

How does odeplot sample the solution of an ODE ?...

Asked by: sand15 1102

September 26 2016

1 9

Hi everybody,

I solve an ODE system numerically with the command

Tend := ….. :
sol := dsolve(MySys, numeric, method=rosenbrock, range=0..Tend): # or rkf45

One of my colleagues wants the solution in array form for using it in a spreadsheet (namely Excel).
Here is an example of what I do :

data := plottools:-getdata( plots:-odeplot(sol, […..], 0..Tend, refine = Q) ): # Q rational
result := data[3]:
ExcelTools:-Export (result, …)

To understand the following, you need to know that the time step varies by several decades (at least 3 or 4). and that numfun is typically in the order of 10^5

When I ask Maple to construct data , Maple seems to sample the solution solaccording to the local time step dsolve has selected (and probably some other parameters ?)
Because this time step varies in very large proportions, the result is sometimes a concentration of points in regions where the solution is rapidly changing (which is desirable) … with some lacunarity in slow varying regions.

To obviate this, sometimes poor appearance, I used to force Maple to compute the solution by regularly sampling the range ]0, Tend[ and combining this array with the original one (result).
But this not satisfactory for I do not avoid lacunarity region (unless I proceed in some ad hoc way)

Here is my question :
Could you give me some elements about the way odeplot adapts its sampling strategy according to the local time step, the budget N (the Q value I guess) and maybe numfun ?

Thank you in advance

Why don't i assign the result of a optimization to...

Asked by: miguelbravo 70

September 26 2016

0 6

I have excuted a loop for i from.....by...while.....do l2:=NLPSolve(.......) with variables, t1,th,tl, etc. I need get the numerical result of theses variables and assign them to t1,th, etc. After execute a function with them. I have tried with op, map and assign, but the program don't understand the numerical value of variables. Why?. How can i assign theses numerical vaules?. Thanks!

how i can pdsolve this equation?...

Asked by: farahnaz 15

September 26 2016

0 6

hi

please help to me for solve this equation via pdsolve?

thanks

dsove2.mw

Download dsove2.mw

not answer found for dsolve??...

Asked by: torabi 90

September 26 2016

0 7

hi

in attached file below not answer found for dsolve?

please help me

thanks..

dsove.mw

Download dsove.mw

KroneckerProduct() function not performing as it d...

Asked by: jojokaila 5

September 25 2016

1 5

I'm trying to compute the tensor product of two column vectors as

with(LinearAlgebra):

A:=Matrix([[1/sqrt(2)],[0],[0],[1/sqrt(2)]]);

KroneckerProduct(A,A);

And the output is a column vector with entries: "16 x 1 Matrix", "Data Type: Anything", "Storage: rectangular", "Order: Fortran_order"

The Maple documentation indicates that this function should output the result of the kronecker tensor product of the input matrices, and I've followed the same form as the examples in the documentation... Does anyone know why this isn't working as it should?

time dependent variable...

Asked by: shakuntala 25

September 25 2016

1 1

> restart;
> libname = [shootlib, libname];
> with(shoot);
Error, invalid input: with expects its 1st argument, pname, to be of type {`module`, package}, but received shoot
> with(plots);
Pr := 10; s = -.1; lambda := 0; Gr := 1.0; Gm := 1.0; beta := -1.20;
10
s = -0.1
0
1.0
1.0
-1.20
> M := 0.; z := .1; Xi := .5; Nt := .5; Nb := .2; l := 5; Nr := .5; epsilon1 := .2; epsilon2 := .2;
0.
0.1
0.5
0.5
0.2
5
0.5
0.2
0.2
> Prff := Pr/(1+4.*N*(1/3));
10
-----------------
1 + 1.333333333 N
> FNS := {f(eta), h(eta), r(eta), u(eta), v(eta), theta(eta), `ϕ`(eta)};
{f(eta), h(eta), r(eta), u(eta), v(eta), theta(eta), ϕ(eta)}
> ODE := {diff(h(eta), eta)+.75*l*f(eta)*h(eta)-(1/4)*l*u(eta)*epsilon2-Nt*(.75*f(eta)*r(eta)-(1/4)*u(eta)*epsilon1+Nb*r(eta)*h(eta)+Nt*r(eta)*r(eta))/Nb = 0, .75*f(eta)*r(eta)+diff(r(eta), eta)-(1/4)*u(eta)*epsilon1+Nb*r(eta)*h(eta)+Nt*r(eta)*r(eta) = 0, diff(v(eta), eta)+3*(f(eta)*v(eta)-u(eta)*u(eta))/(4*Pr)-(M+lambda)*u(eta)+theta(eta)-Nr*`ϕ`(eta) = 0, diff(f(eta), eta) = u(eta), diff(u(eta), eta) = v(eta), diff(theta(eta), eta) = r(eta), diff(`ϕ`(eta), eta) = h(eta)};
/ / d \
{ 0.75 f(eta) r(eta) + |----- r(eta)| - 0.05000000000 u(eta)
\ \ deta /

2 / d \ 3
+ 0.2 r(eta) h(eta) + 0.5 r(eta) = 0, |----- v(eta)| + -- f(eta) v(eta)
\ deta / 40

3 2 / d \
- -- u(eta) + theta(eta) - 0.5 ϕ(eta) = 0, |----- h(eta)|
40 \ deta /

+ 3.75 f(eta) h(eta) - 0.1250000000 u(eta) - 1.875000000 f(eta) r(eta)

2
- 0.5000000000 r(eta) h(eta) - 1.250000000 r(eta) = 0,

d d d
----- f(eta) = u(eta), ----- u(eta) = v(eta), ----- theta(eta) = r(eta),
deta deta deta

d \
----- ϕ(eta) = h(eta) }
deta /
> IC := {f(0) = s, h(0) = xi, r(0) = tau, u(0) = 0, v(0) = alpha(0), theta(0) = 1-(1/4)*epsilon1, `ϕ`(0) = (1/4)*epsilon2};
{f(0) = s, h(0) = xi, r(0) = tau, u(0) = 0, v(0) = alpha(0),

theta(0) = 0.9500000000, ϕ(0) = 0.05000000000}
> L := 2;
2
> BC = {u(L) = 0, theta(L) = 0, `ϕ`(L) = 0};
BC = {u(2) = 0, theta(2) = 0, ϕ(2) = 0}
> S := Shoot(ODE, IC, BC, FNS, [alpha = .42453091564332, tau = -.21166705749821127, xi = -.4944583739651814]);
/ / / d \
Shoot|{ 0.75 f(eta) r(eta) + |----- r(eta)| - 0.05000000000 u(eta)
\ \ \ deta /

2 / d \ 3
+ 0.2 r(eta) h(eta) + 0.5 r(eta) = 0, |----- v(eta)| + -- f(eta) v(eta)
\ deta / 40

3 2 / d \
- -- u(eta) + theta(eta) - 0.5 ϕ(eta) = 0, |----- h(eta)|
40 \ deta /

+ 3.75 f(eta) h(eta) - 0.1250000000 u(eta) - 1.875000000 f(eta) r(eta)

2
- 0.5000000000 r(eta) h(eta) - 1.250000000 r(eta) = 0,

d d d
----- f(eta) = u(eta), ----- u(eta) = v(eta), ----- theta(eta) = r(eta),
deta deta deta

d \
----- ϕ(eta) = h(eta) }, {f(0) = s, h(0) = xi, r(0) = tau, u(0) = 0,
deta /

v(0) = alpha(0), theta(0) = 0.9500000000, ϕ(0) = 0.05000000000}, BC,

{f(eta), h(eta), r(eta), u(eta), v(eta), theta(eta), ϕ(eta)}, [

alpha = 0.42453091564332, tau = -0.21166705749821127,

\
xi = -0.4944583739651814]|
/
RungeKutta(ODE, BC, alpha = .42453091564332, tau = -.21166705749821127, xi = -.4944583739651814, output=plot);
/ / / d \
RungeKutta|{ 0.75 f(eta) r(eta) + |----- r(eta)| - 0.05000000000 u(eta)
\ \ \ deta /

2 / d \ 3
+ 0.2 r(eta) h(eta) + 0.5 r(eta) = 0, |----- v(eta)| + -- f(eta) v(eta)
\ deta / 40

3 2 / d \
- -- u(eta) + theta(eta) - 0.5 ϕ(eta) = 0, |----- h(eta)|
40 \ deta /

+ 3.75 f(eta) h(eta) - 0.1250000000 u(eta) - 1.875000000 f(eta) r(eta)

2
- 0.5000000000 r(eta) h(eta) - 1.250000000 r(eta) = 0,

d d d
----- f(eta) = u(eta), ----- u(eta) = v(eta), ----- theta(eta) = r(eta),
deta deta deta

d \
----- ϕ(eta) = h(eta) }, BC, alpha = 0.42453091564332,
deta /

\
tau = -0.21166705749821127, xi = -0.4944583739651814, output = plot|
/
>

Dear sir

in the above problem im geiitng the problem with , with(shoot) command and even it is not executing at

S := Shoot(ODE, IC, BC, FNS, [alpha = .42453091564332, tau = -.21166705749821127, xi = -.4944583739651814]) this command, here alpha,tau and zi variable should change.

shooting method...

Asked by: shakuntala 25

September 25 2016

1 28

> restart;
> with(plots);
> Eql := diff(f(eta), eta, eta, eta)+.5*f(eta)*(diff(f(eta), eta, eta)) = 0;
/ d / d / d \\\ / d / d \\
|----- |----- |----- f(eta)||| + 0.5 f(eta) |----- |----- f(eta)|| = 0
\ deta \ deta \ deta /// \ deta \ deta //
> blt := 10;
10
> bcs1 := f(0) = f0, (D(f))(0) = 0, (D(f))(blt) = 1;
f(0) = f0, D(f)(0) = 0, D(f)(10) = 1
> L := [0];
[0]
> for k to 1 do R := dsolve(eval({Eql, bcs1}, f0 = L[k]), f(eta), numeric, output = listprocedure); X1 || k := rhs(R[3]); X2 || k := rhs(R[4]) end do;
[
[eta = proc(eta) ... end;, f(eta) = proc(eta) ... end;,
[

d
----- f(eta) = proc(eta) ... end;,
deta

d / d \ ]
----- |----- f(eta)| = proc(eta) ... end;]
deta \ deta / ]
proc(eta) ... end;
proc(eta) ... end;
> print([X2], [1 .. 1, 0]);

dear sir/madam

in the above problem i should get the asnser (at print line) but its not getting so please can you tell me why it is not getting.

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