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Hi
Two new things recently added to the latest version of Physics available on Maplesoft's R&D Physics webpage are worth mentioning outside the framework of Physics.

  • automaticsimplification. This means that after "Physics:-Setup(automaticsimplification=true)", the output corresponding to every single input (literally) gets automatically simplified in size before being returned to the screen. This is fantastically convenient for interactive work in most situations.

  • Add Physics:-Library:-Assume, to perform the same operations one typically performs with the  assume command, but without the side effect that the variables get redefined. So the variables do not get redefined, they only receive assumptions.

This new Assume implements the concept of an "extended assuming". It permits re-using expressions involving the variables being assumed, expressions that were entered before the assumptions were placed, as well as reusing all the expressions computed while the variables had assumptions, even after removing the variable's assumptions. None of this is possible when placing assumptions using the standard assume. The new routine also permits placing assumptions on global variables that have special meaning, that cannot be redefined, e.g. the cartesian, cylindrical or spherical coordinates sets, or the coordinates of a coordinate spacetime system within the Physics package, etc.

Examples:

 

with(Physics):

This is Physics from today:

Physics:-Version()[2]

`2014, December 9, 16:51 hours`

(1.1)
• 

Automatic simplification is here. At this point automaticsimplification is OFF by default.

Setup(automaticsimplification)

[automaticsimplification = false]

(1.2)

Hence, for instance, if you input the following expression, the computer just echoes your input:

Physics:-`*`(a, c)+Physics:-`*`(a, d)+Physics:-`*`(b, c)+Physics:-`*`(b, d)

a*c+a*d+b*c+b*d

(1.3)

There is however some structure behind (1.3) and, in most situations, it is convenient to have these structures
apparent, in part because they frequently provide hints on how to proceed ahead, but also because a more
compact expression is, roughly speaking, simpler to understand. To see this
automaticsimplification in action,
turn it ON:

Setup(automaticsimplification = true)

[automaticsimplification = true]

(1.4)

Recall this same expression (you could input it with the equation label (1.3) as well) 

Physics:-`*`(a, c)+Physics:-`*`(a, d)+Physics:-`*`(b, c)+Physics:-`*`(b, d)

(c+d)*(a+b)

(1.5)

What happened: this output, as everything else after you set automaticsimplification = true and with no
exceptions, is now further processed with simplify/size before being returned. And enjoy computing with frankly
shorter expressions all around! And no need anymore for "simplify(%, size)" every three or four input lines.

Another  example, typical in computer algebra where expressions become uncomfortably large and difficult to
read: convert the following input to 2D math input mode first, in order to compare what is being entered with the
automatically simplified output on the screen

-Physics:-`*`(Physics:-`*`(Physics:-`*`(3, sin(x)^(1/2)), cos(x)^2), sin(x)^m)+Physics:-`*`(Physics:-`*`(Physics:-`*`(3, sin(x)^(1/2)), cos(x)^2), cos(x)^n)+Physics:-`*`(Physics:-`*`(Physics:-`*`(4, sin(x)^(1/2)), cos(x)^4), sin(x)^m)-Physics:-`*`(Physics:-`*`(Physics:-`*`(4, sin(x)^(1/2)), cos(x)^4), cos(x)^n)

-4*(cos(x)^n-sin(x)^m)*sin(x)^(1/2)*cos(x)^2*(cos(x)^2-3/4)

(1.6)

You can turn automaticsimplification OFF the same way

Setup(automaticsimplification = false)

[automaticsimplification = false]

(1.7)
• 

New Library:-Assume facility; welcome to the world of "extended assuming" :)

 

Consider a generic variable, x. Nothing is known about it

about(x)

x:

  nothing known about this object

 

Each variable has associated a number that depends on the session, and the computer (internally) uses this
number to refer to the variable.

addressof(x)

18446744078082181054

(1.8)

When using the assume  command to place assumptions on a variable, this number, associated to it, changes,
for example:

assume(0 < x and x < Physics:-`*`(Pi, 1/2))

addressof(x)

18446744078179060574

(1.9)

Indeed, the variable x got redefined and renamed, it is not anymore the variable x referenced in (1.8).

about(x)

Originally x, renamed x~:

  is assumed to be: RealRange(Open(0),Open(1/2*Pi))

 


The semantics may seem confusing but that is what happened, you enter x and the computer thinks x~, not x 
anymore.This means two things:

1) all the equations/expressions, entered before placing the assumptions on x using assume, involve a variable x 
that is different than the one that exists after placing the assumptions, and so these previous expressions
cannot
be reused
. They involve a different variable.

2) Also, because, after placing the assumptions using assume, x refers to a different object, programs that depend
on the
x that existed before placing the assumptions will not recognize the new x redefined by assume .

 

For example, if x was part of a coordinate system and the spacetime metric g[mu, nu]depends on it, the new variable x
redefined within assume, being a different symbol, will not be recognized as part of the dependency of "g[mu,nu]." This
posed constant obstacles to working with curved spacetimes that depend on parameters or on coordinates that
have a restricted range. These problems are resolved entirely with this new
Library:-Assume, because it does not
redefine the variables. It only places assumptions on them, and in this sense it works like
assuming , not assume .
As another example, all the
Physics:-Vectors commands look for the cartesian, cylindrical or spherical coordinates
sets
[x, y, z], [rho, phi, z], [r, theta, phi] in order to determine how to proceed, but these variables disappear if you use
assume to place assumptions on them. For that reason, only assuming  was fully compatible with Physics, not assume.

 

To undo assumptions placed using the assume command one reassigns the variable x to itself:

x := 'x'

x

(1.10)

Check the numerical address: it is again equal to (1.8) 

addressof(x)

18446744078082181054

(1.11)

·All these issues get resolved with the new Library:-Assume, that uses all the implementation of the existing 
assume command but with a different approach: the variables being assumed do not get redefined, and hence:
a) you can reuse expressions/equations entered before placing the assumptions, you can also undo the
assumptions and reuse results obtained with assumptions. This is the concept of an
extended assuming. Also,
commands that depend on these assumed variables will all continue to work normally, before, during or after
placing the assumption, because
the variables do not get redefined.

Example:

about(x)

x:

  nothing known about this object

 

So this simplification attempt accomplishes nothing

simplify(arccos(cos(x)))

arccos(cos(x))

(1.12)

Let's assume now that 0 < x and x < (1/2)*Pi

Library:-Assume(0 < x and x < Physics:-`*`(Pi, 1/2))

{x::(RealRange(Open(0), Open((1/2)*Pi)))}

(1.13)

The new command echoes the internal format representing the assumption placed.

a) The address is still the same as (1.8)

addressof(x)

18446744078082181054

(1.14)

So the variable did not get redefined. The system however knows about the assumption - all the machinery of the
assume command is being used

about(x)

Originally x, renamed x:

  is assumed to be: RealRange(Open(0),Open(1/2*Pi))

 


Note that the renaming is to the variable itself - i.e. no renaming.

Hence, expressions entered before placing assumptions can be reused. For example, for (1.12), we now have

simplify(arccos(cos(x)))

x

(1.15)

To clear the assumptions on x, you can use either of Library:-Assume(x=x) or Library:-Assume(clear = {x, ...}) in
the case of many variables being cleared in one go, or in the case of a single variable being cleared:

Library:-Assume(clear = x)

about(x)

x:

  nothing known about this object

 


The implementation includes the additionally functionality, for that purpose add the keyword
additionally 
anywhere in the calling sequence. For example:

Library:-Assume(x::positive)

{x::(RealRange(Open(0), infinity))}

(1.16)

about(x)

Originally x, renamed x:

  is assumed to be: RealRange(Open(0),infinity)

 

Library:-Assume(additionally, x < 1)

{x::(RealRange(Open(0), Open(1)))}

(1.17)

Library:-Assume(x = x)

In summary, the new Library:-Assume command implements the concept of an extended assuming, that can be
turned ON and OFF at will at any moment without changing the variables involved.


Download AutomaticSimplificationAndAssume.mw

 

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

Hi all

Assume that we have a function, say f(t) and we want to substitute t in it where t is:

t=[0,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1]

by subs or other better command, how can we do it?

best wishes

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

Hi all

I want to produce following c_nm's(which are differentiation based formula) . assume that N and M are known and f(t) is arbitrary. also n=1,2,...,N and m=0,1,..,M-1

how can we do this?

regards


Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

Just got Maple 18 on academic license. I've used Maple 15, 16 and 17 before this.

It wont write simple math symbols as + - * and := 

I have tried both document- and worksheet, reinstalling and so on...

 

What is the problem?

Thanks

I am using Maple 12 in Win 64. I have some difficulties in evaluating expressions in Maple. For example simply writing cos(x) then differentiating it w.r.t x. After writing cos(x) in 2D Math mode, I do right click for selecting diff but it does not work. The pointer turns in to busy mode and nothing comes to select. Also, I tried this operation in the tutorial file teaching differentiation, however problem persists.

I have some lengthy formulas in the maple. I don't want to waste time on rewritting them in a word document.
Is there a way to import those equations in a clean and tidy form to a word document or the mathtype program or something else! :)

Hi Again

Assume that we have known matrix namely, Q, of order (m+1)*(m+1) and we want to construct following matrix

where 0(bar) is zero matrix of orde (m+1)*(m+1) and New matrix should be of order {N*(m+1)}*{N*(m+1)} where N is known constant.

thanks for any guide


Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

Maplesoft regularly hosts live webinars on a variety of topics. Below you will find details on an upcoming webinar we think may be of interest to the MaplePrimes community.  For the complete list of upcoming webinars, visit our website.

Hollywood Math 2

In this second installment of the Hollywood Math webinar series, we will present some more examples of mathematics being used in Hollywood films and popular hit TV series. For instance, have you wondered how Ben Campbell solved his professor’s challenge so easily in the movie “21”? Or about the details of the Nash equilibrium that John Nash first developed in a “A Beautiful Mind”? We’ve got the answers! These relevant, and exciting examples can be used as material to engage your students with examples familiar to them, or you can just attend the webinar for its entertainment value.

Anyone with an interest in mathematics, especially high school and early college math educators, will be both entertained and informed by attending this webinar. At the end of the webinar you’ll be given an opportunity to download an application containing all of the examples that we demonstrate.

To join us for the live presentation, please click here to register.

If you missed the first webinar in this two part series, you can view the 'Hollywood Math' recording on our website.

Hi All. Hope all is well.

Assume that we have partitioned [0,a], into N equidistant subintervals and in each subinterval we have M sets of polynomials of arbitrary form[say bij(t)](a.e Taylor series, or Bernstein series,…)

for Example with N=4, M=3 and by Taylor series we have:

 

now we want to approximate a function, asy f(t), in this interval with following form:

 

If we have:

(Tau is a constant number)
then: How can  we find L and Z matrices using maple? Is it any way? (or other softwares?)

Regards

 

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

 

This is the first presentation of updates for the DE and Mathematical Functions programs of Maple 18. It includes several improvements, all in the Mathematical Functions sector, as well as some fixes. The update and instructions for its installation are available on the Maplesoft R&D webpage for DEs and mathematical functions. Some of the items below were mentioned here in Mapleprimes - you are welcome to present suggestions or issues; if possible they will be addressed right away in the next update.

  • Filling gaps in the FunctionAdvisor regarding all the 6 complex components: abs, argument, conjugate, Im, Re, signum, as well as regarding Heaviside (step function), Dirac, min and max.
  • Fix the simplification and differentation rule for doublefactorial
  • Make convert(..., hypergeometric) work the same way as convert(blabla, hypergeom)
  • Implement integral forms for Heaviside(z) and JacobiAM(z, k) via convert(..., Int)
  • Implement appropriate display for the inert %intat function as well as its conversion to the inert Int
  • Make the FunctionAdvisor/DE return not just the PDE system satisfied by f(z, k) = JacobiAM(z, k)and also (new) the ODE satisfied by f(z) = JacobiAM(z, k)
  • Fix conversion rule from Heaviside(z) to Sum
  • Fix unexpected error interruption when differentiating min(...) and max(...) containing more than three arguments
  • Fix issue in simplify/conjugate
  • Improvement in expand/int: factors in disguise are put outside the integration sign
  • Various improvements in the case of multiple integrals involving the Dirac function
  • Make Intat fully inert (before it was evaluating its arguments)
  • Make value of inert indexed objects work

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

Hi all.

Assume that we have partitioned [0,a], into N equidistant subintervals and in each subinterval we have M sets of poly nomials of the following form:

where Tm(t)=tm( namely Taylor Series) and tf is a(final point)
for Example with N=4, M=3 we have:

now we want to approximate a function, asy f(t), in this interval with following form:

How can we do this with maple????

how can we find the ci's?????

Thanks a lot

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

I have just begun thinking of trying to make some mathematically defined objects using a 3d printer. I would be happy to hear from anyone who has done this using Maple to prepare input. Pointers for a novice in 3d printing would be appreciated.  I have access to a MakerBot Replicator 2. But the people who have it have only used it to scan objects and make 3d copies of them. 

---Edwin

Hi all;

Hope all of you  be in good health

I want to construct a special function b_{nm}(t) like:

with piecewise command i did it but the result is incorrect.

any one can help me to do it?

Best wishes

 

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

Hi all 

I have the following segment of maple program which belongs to time delay systems dynamic. here C=X-X0-G.Z-X.Dtau.P+X.Dtau.Z-U.P, is a matrix(vector) which comes from reordering the system terms and my goal is to minimizing J:=X.E.Transpose(X)+U.E.Transpose(U), subject to constraint C=0, but i don't know how to do so.

I will be so grateful if anyone can guide me

best wishes

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department


restart:
with(Optimization):
with(LinearAlgebra):
macro(LA= LinearAlgebra):
L:=1:  r:=2:  tau:= 1:
interface(rtablesize= 2*r+1):

Z:= Matrix(
     2*r+1, 2*r+1,
     [tau,
      seq(evalf((L/(2*(iz-1)*Pi))*sin(2*(iz-1)*Pi*tau/L)), iz= 2..r+1),
      seq(evalf((L/(2*(iz-1-r)*Pi))*(1-cos(2*(iz-1-r)*Pi*tau/L))), iz= r+2..2*r+1)
      ],
     scan= columns,
     datatype= float[8]
);
                        
Dtau00:= < 1 >:
Dtau01:= Vector[row](r):
Dtau02:= Vector[row](r):
Dtau10:= Vector(r):
Dtau20:= Vector(r):

Dtau1:= LA:-DiagonalMatrix([seq(evalf(cos(2*i*Pi*tau/L)), i= 1..r)]):
Dtau2:= LA:-DiagonalMatrix([seq(evalf(sin(2*i*Pi*tau/L)), i= 1..r)]):
Dtau3:= -Dtau2:
Dtau4:= copy(Dtau1):

Dtau:= < < Dtau00 | Dtau01 | Dtau02 >,
         < Dtau10 | Dtau1  | Dtau2  >,
         < Dtau20 | Dtau3  | Dtau4  > >;
 
P00:= < L/2 >:
P01:= Vector[row](r):
P02:= Vector[row](r, j-> evalf(-L/j/Pi), datatype= float[8]):
P10:= Vector(r):
P20:= Vector(r, i-> evalf(L/2/i/Pi)):
P1:= Matrix(r,r):
P2:= LA:-DiagonalMatrix(P20):
P3:= LA:-DiagonalMatrix(-P20):
P4:= Matrix(r,r):

P:= < < P00 | P01 | P02 >,
      < P10 | P1  | P2  >,
      < P20 | P3  | P4  > >;

interface(rtablesize=2*r+1):    # optionally
J:=Vector([L, L/2 $ 2*r]):      # Matrix([[...]]) would also work here

E:=DiagonalMatrix(J);

X:=  Vector[row](2*r+1,symbol=a);
U:=Vector[row](2*r+1,symbol=b);

X0:= Vector[row](2*r+1,[1]);
G:=Vector[row](2*r+1,[1]);
C:=simplify(X-X0-G.Z-X.Dtau.P+X.Dtau.Z-U.P);

Z := Matrix(5, 5, {(1, 1) = 1., (1, 2) = 0., (1, 3) = 0., (1, 4) = 0., (1, 5) = 0., (2, 1) = 0., (2, 2) = 0., (2, 3) = 0., (2, 4) = 0., (2, 5) = 0., (3, 1) = 0., (3, 2) = 0., (3, 3) = 0., (3, 4) = 0., (3, 5) = 0., (4, 1) = 0., (4, 2) = 0., (4, 3) = 0., (4, 4) = 0., (4, 5) = 0., (5, 1) = 0., (5, 2) = 0., (5, 3) = 0., (5, 4) = 0., (5, 5) = 0.})

Dtau := Matrix(5, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (2, 1) = 0, (2, 2) = 1., (2, 3) = 0, (2, 4) = 0., (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 1., (3, 4) = 0, (3, 5) = 0., (4, 1) = 0, (4, 2) = -0., (4, 3) = -0., (4, 4) = 1., (4, 5) = 0, (5, 1) = 0, (5, 2) = -0., (5, 3) = -0., (5, 4) = 0, (5, 5) = 1.})

P := Matrix(5, 5, {(1, 1) = 1/2, (1, 2) = 0, (1, 3) = 0, (1, 4) = -.318309886100000, (1, 5) = -.159154943000000, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = .1591549430, (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0, (3, 4) = 0, (3, 5) = 0.7957747152e-1, (4, 1) = .1591549430, (4, 2) = -.159154943000000, (4, 3) = 0, (4, 4) = 0, (4, 5) = 0, (5, 1) = 0.7957747152e-1, (5, 2) = 0, (5, 3) = -0.795774715200000e-1, (5, 4) = 0, (5, 5) = 0})

E := Matrix(5, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (2, 1) = 0, (2, 2) = 1/2, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 1/2, (3, 4) = 0, (3, 5) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 1/2, (4, 5) = 0, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = 0, (5, 5) = 1/2})

X := Vector[row](5, {(1) = a[1], (2) = a[2], (3) = a[3], (4) = a[4], (5) = a[5]})

U := Vector[row](5, {(1) = b[1], (2) = b[2], (3) = b[3], (4) = b[4], (5) = b[5]})

X0 := Vector[row](5, {(1) = 1, (2) = 0, (3) = 0, (4) = 0, (5) = 0})

G := Vector[row](5, {(1) = 1, (2) = 0, (3) = 0, (4) = 0, (5) = 0})

C := Vector[row](5, {(1) = 1.500000000*a[1]-2.-.1591549430*a[4]-0.7957747152e-1*a[5]-.5000000000*b[1]-.1591549430*b[4]-0.7957747152e-1*b[5], (2) = a[2]+.1591549430*a[4]+.1591549430*b[4], (3) = a[3]+0.7957747152e-1*a[5]+0.7957747152e-1*b[5], (4) = a[4]+.3183098861*a[1]-.1591549430*a[2]+.3183098861*b[1]-.1591549430*b[2], (5) = a[5]+.1591549430*a[1]-0.7957747152e-1*a[3]+.1591549430*b[1]-0.7957747152e-1*b[3]})

(1)

J:=X.E.Transpose(X)+U.E.Transpose(U);

J := a[1]^2+(1/2)*(a[2]^2)+(1/2)*(a[3]^2)+(1/2)*(a[4]^2)+(1/2)*(a[5]^2)+b[1]^2+(1/2)*(b[2]^2)+(1/2)*(b[3]^2)+(1/2)*(b[4]^2)+(1/2)*(b[5]^2)

(2)

Minimize(J,{C=0});






Error, (in Optimization:-NLPSolve) invalid arguments

 

#XP:=-.015+X[1]+add(X[l+1]*f1(l)+X[r+l+1]*f2(l), l= 1..r):
#plot([XP,T1], t= 0..1);#,legend= "Solution Of x(t) with r=50"):

 

 

 

 

 

 

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