## how to fix this error in Equilibrium point for non...

rhs; with(DEtools);# to find the equilibrium points
fx1 = sigma1*y-beta1*x1*y-v1*x1-mu1*x1+(2*alphad+alphaa)*k*A*s; fx2 = alpha*v2*x1*z-beta2*x2*y-mu2*x2+v1*x1; fy = sigma2*(beta1*x1*y+beta2*x2*y)-mu4*y-mu5*y^2; fz = sigma3+y*(kz+y)-mu3*z-v2*z*x2; fs = (k*(alphas-alphad)-mus)*s;

eqs := solve({fs, fx1, fx2, fy, fz}, {s, x1, x2, y, z})

L := map(subs, [eqs], [x1, x2, y, z, s])

J := unapply(VectorCalculus:-Jacobian([fx1, fx2, fy, fz, fs], [x1, x2, y, z, s]), x1, x2, y, z, s); J(x1, x2, y, z, s)

## Simplify fails to operate...

Verification_known_T_R_30_07_2024.mw That`s the worksheet.

For some strange reason the command simplify doesn`t work. Beforehand it managed to simplify quite bulky expressions.

Thank you in advance!

## How to solve coupled nonlinear differential equati...

I would like to solve this system of differential equations y_1 , y_2. However, there should be no exact solutions for this problem. Is there a way to get a numerical solution of such coupled equations in maple and if so, how? I know the dsolve() command, but it did not work here. The boundary conditions are f(0)-1 = K(0) = 0 and f'(inf) = K'(inf) = 0. It would be nice if someone could help me or tell me where to look to solve something like this.

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## Why does the units standard environement not overl...

Recently there have been some questions about the unit packages. I would like to add another one that has been on my desk for a while.

The Simple environement overloads the command frem and piecewise. The "most powerful" (can we say so?) environement Standard does not

```{with(Units:-Simple)[]} minus {with(Units:-Standard)[]}
{frem, piecewise}

```

For example calculating the remainder of a length in mm does not work in the Standard environment

```restart;
with(Units:-Simple):
convert(frem(1.234*Unit(m),Unit(cm)),units,mm);
4.000000000 Unit(mm)

restart;
with(Units:-Standard):
convert(frem(1.234*Unit(m),Unit(cm)),units,mm)
Error, invalid input: frem received 1.234*Units:-Unit(m), which is not valid for its 1st argument, x

```

Why is that?

## Why is sin^(-1) the reciprocal of sin but sin^(-1...

In the Programming Guide, Ch. 3 "Maple Expressions", subsection 3.13 "Other Expressions" there is a section called "Composition".

There is the following snippet

In particular, although the inverses of the circular functions are commonly denoted by a power-like notation in written mathematics, in Maple, for example, sin^(-1) denotes the reciprocal of the sin function, while sin@@(-1) denotes the arcsine (arcsin).

I opened a new worksheet to check this. I found the results confusing.

 (1)

 (2)

sin^(-1)

 (3)

sin^(-1)(x)

 (4)

(sin^(-1))(x)

 (5)

 (6)

 (7)

sin@@(-1)

 (8)

sin@@(-1)(x)

 (9)

(sin@@(-1))(x)

 (10)

Consider (1) and (2).

(1) is in accordance with the quoted snippet: sin^(-1) is the reciprocal of the sine function. But when we use this same expression as a function call, the function that is called is arcsine. Does this make sense to be this way?

Next, consider (3), (4), and (5), which I expected to be the same expressions as (1), (2) and (2), respectively. The only difference is that (1) and (2) use 2D math and (3), (4), and (5) use Maple input.

Both results using Maple input give as output the reciprocal of sine.

Why is there this difference between the 2D version and the Maple input version?

Consider (4) and (5). Why is it that we need to add parentheses for the argument x to be applied to the function?

Finally, what is the reasoning behind the syntax sin@@(-1) denoting arcsine?

@@ represents repeated composition. sin@@3 represents the function sin(sin(sin))).

Is sin@@(-1) equal to arcsine simply by convention or is there some logical reason?

## What is the difference between m[f] and m:-f where...

I am reading the Maple Programming Guide chapter 3 "Maple Expressions". There are subsections on "Indexed Expressions" and "Member Selection".

The following is based on reading these sections.

Consider the following module definition

m := module() export e, f:=2; end module:

What is the difference between m[e] and m:-e or m[f] and m:-f?

The documentation says that both evaluate module m's export e but the difference is that the index selection form will evaluate e before resolving the export.

What does this mean exactly?

When I try out in a worksheet both m[f] and m:-f return 2 and m[e] and m:-e return the name e.

## How choose the case after finding parameter ...

when i finding parameter i want just choose a case for example a_1=a_1  and any other case a_2=0,and remove other case how i can do in maple

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## System of PDE's error in solve and plot the grap...

dear maple user help me to correct this error .

I want to solve the coupled pdes  plot the graphs but unable to get it. please help me. Here are the maple codes

restart:
L:=0.4:
d1:=0.3: d2:=0.3:

h:=z->piecewise( z<=d1,   1,
z<=d1+L,   1-(delta/(2))*(1 + cos((2*(Pi)/L)*(z - d1 - L/2))),
z<=d1+L+d2,  1 ):

a:=x4*S*Gr/(4*x1*x5)*sin(alpha):
b:=1/Da+(x3*M/(x1*(1+m^2))):
c:=(1/x1)*DP:
x1:=1/((1-phi1)^2.5*(1-phi2)^2.5):
x5:=khnf/kf:
x3:=shnf/sf:
x4:=(1-phi2)*((1-phi1)+phi1*(RBs1)/(RBf))+phi2*RBs2/RBf:
shnf:=sbf*((ss2+2*sbf-2*phi2*(sbf-ss2))/(ss2+2*sbf+phi2*(sbf-ss2))):
sbf:=sf*((ss1+2*sf-2*phi1*(sf-ss1))/(ss1+2*sf+phi1*(sf-ss1))):
ss2:=2.7*10^(-8):
ss1:=59.6*10^(6):
sf:=6.67*10^(-1):
khnf:=kbf*((ks2+2*kbf-2*phi2*(kbf-ks2))/(ks2+2*kbf+phi2*(kbf-ks2))):
kbf:=kf*((ks1+2*kf-2*phi1*(kf-ks1))/(ks1+2*kf+phi1*(kf-ks1))):
ks1:=401:
ks2:=76.5:
kf:=0.4972:
RBs1:=8933*16.7*10^(6):
RBs2:=6320*18*10^(6):
RBf:=1063*1.8*10^(6):
DP:=(1/192)*(18432*F-77*b^2*h(z)^8+308*b*h(z)^8*a+768*h(z)^6*a+12*b*h(z)^8)/(h(z)^4*(b*h(z)^2-6)):
conds:= w(h(z),0)=0, theta(h(z),0)=0, D[1](w)(r,0)=0, D[1](theta)(r,0)=0:
pdes:= diff(w(r, z), r, r)+(1/r)*(diff(w(r, z), r))-(1/Da+(x3/x1)*M)*w(r, z)-(1/x1)*(DP+theta*x4*sin(alpha))=0,
diff(theta(r, z), r,r)+(1/r)*(diff(theta(r, z), r))+(S/x5)=0:
pars:= { F=1.5,Da=0.1,phi1=0.01,phi2=0.02,alpha=Pi/4,S=0.1,Gr=2,m=0.5}:

PrVals:=[2, 5, 7,9]:
colors:=[red, green, blue, black]:
for j from 1 to numelems(PrVals) do
pars1:=`union`( pars, {M=PrVals[j]}):
pdSol:= pdsolve( eval([pdes], pars1),
eval([conds], pars1),
numeric
);
plt[j]:=pdSol:-plot( w(r,z), z=1, r=0..1, numpoints=200, color=colors[j]);
od:
plots:-display( [seq(plt[j], j=1..numelems(PrVals))]);
Error, final value in for loop must be numeric or character
Error, unable to execute seq
PrVals:=[2, 5, 7,9]:
colors:=[red, green, blue, black]:
for j from 1 to numelems(PrVals) do
pars1:=`union`( pars, {M=PrVals[j]}):
pdSol:= pdsolve( eval([pdes], pars1),
eval([conds], pars1),
numeric
);
plt[j]:=pdSol:-plot( theta(r,z), z=1, r=0..1, numpoints=200, color=colors[j]);
od:
plots:-display( [seq(plt[j], j=1..numelems(PrVals))]);
Error, final value in for loop must be numeric or character
Error, unable to execute seq

PrVals:=[2, 5, 7,9]:
colors:=[red, green, blue, black]:
for j from 1 to numelems(PrVals) do
pars1:=`union`( pars, {M=PrVals[j]}):
pdSol:= pdsolve( eval([pdes], pars1),
eval([conds], pars1),
numeric
);
plt[j]:=pdSol:-plot( DP,  z=0..1, numpoints=200, color=colors[j]);
od:
plots:-display( [seq(plt[j], j=1..numelems(PrVals))]);

## simplify the multiply exponential term ...

i am looking for simplify this type of simplifying assume beta is Real and there is any stuf package for working with complex and conjugate automaticaly

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## The shortest distance between the green and red po...

A little continuation of topics 1 and 2. This is a very similar cube from 2

but with a different equation:

f1 := (x1-sin(x1))^2+(x2-sin(x2))^2+(x3-sin(x3))^2-0.02513144866;
And other point coordinates (-.8283302152, -.8283302152, .8283302152) and (.8283302152, .8283302152, -.8283302152).

## Limitrule not easy to understand ...

I'm a little busy with the Rule feature in Maple
Just started with the limits and can't get 1 limit rule working in my FSimp procedure yet
As a limit example I saw this in Maple help

```with(Student:-Calculus1);
infolevel[Student[Calculus1]] := 1;

Rule[lhopital, ln(x)](Limit(x*ln(x), x = 0, right));
Creating problem #2

lim   (x ln(x)) =   lim   (-x)
x -> 0+             x -> 0+

```

A meaningless example of how one limit of a function is equal to another limit of a function, when I think there is no connection between the two functions. ( I don't see it )
You might as well calculate the limit value right away.

It gets more interesting when you analyze the function/expression : x*ln(x) and start rewriting it, to a [0/0] or [infinity/infinity] form in this case and from this it shows the need to start using lhopital rule.

The Rule feature has a Hint capability and gives as a hint, lhopital that this rule is applied
Of course, you also have a series of standard limits, beyond the lhopital limits
The Rule feature for the limit is not yet step by step, because the need to start using a lhopital limit rule has not yet been demonstrated by the user

## How could I install the update package of Differen...

I'm interested in installing the package update from I M Anderson's site, the Differential Geometry Software Project. This corresponds to the 2022 update. There are many new features:
- Submanifold theory
- Differential operators and  Pseudo-Differential Operators
- Principal connections
- Integrable System Library
- Cartan normal connections
and more.
It's unfortunate that this update isn't included in this version of Maple 2024

## How to extract the coefficient of a complicated po...

Hi all guys, how can I extract the coefficient of order 6 of expand(SUMY-T) as below? I use the coeff but no use and I search last posts in mapleprime but no fit for my case, and substitute the parematers of a,b,c....

Thank you!

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Error, unable to compute coeff

## Understanding the ScientificConstants package....

I am going through the documentation on ScientificConstants and trying things out in a worksheet.

 (1)

Copy and paste from the list above.

 (2)

Manually try to type in .

Why doesn't the latter work?