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Execute an entire worksheet in a loop?...

Asked by: RK1 160
worksheet incomplete-question
January 23 2023
0  2

Is there a way to execute an entire worksheet for a list of values of a parameter.
Sometimes executing in the standard loop can be cumbersome given you have to add catch statements for errors to prevent the loop from stopping.

How to get polynomial expression...

Asked by: aimLow 50
polynomial
January 23 2023
3  6

Hi,

I'm not able to find a way to bring an equation into it's "polynomial" form:

I'd like this expression in the form of:

c1*v^3+c2*v^2+c1*v+c0 = 0

where c1...c4 are my coefficients

How can I do that? 

How to convert the RootOf function?...

Asked by: Oliveira 190
solve special-function rootof
January 23 2023
0  4

Hello guys
I'm having trouble converting the RootOf function. Attached is a simple notebook.

Sincerely,

Oliveira

Example.mw

how do I use a symbol as a subscript, eg A_*, to l...

Asked by: nscheng 20
subscript label superscript
January 23 2023
2  12

Dear Maple Users:

Could you help in the following question?

How do I use a symbol as a subscript, e.g.  A_*, to label an axis?

Simplifying transfer functions with units and...

by: C_R 3437 Maple
expand collect units
January 22 2023
1  3

Transfer functions are normally not used with units. Involving units when deriving transfer functions can help identify unit inconsistencies and reduce the likelihood of unit conversion errors.

Maple is already a great help in not having to do this manually. However, the final step of simplification still requires manual intervention, as shown in this example.

Given transfer function

H(s) = 60.*Unit('m'*'kg'/('s'^2*'A'))/(.70805*s^2*Unit('kg'^2*'m'^2/('s'^3*'A'^2))+144.*s*Unit('kg'^2*'m'^2/('s'^4*'A'^2))+0.3675e-4*s^3*Unit('kg'^2*'m'^2/('s'^2*'A'^2)))

H(s) = 60.*Units:-Unit(m*kg/(s^2*A))/(.70805*s^2*Units:-Unit(kg^2*m^2/(s^3*A^2))+144.*s*Units:-Unit(kg^2*m^2/(s^4*A^2))+0.3675e-4*s^3*Units:-Unit(kg^2*m^2/(s^2*A^2)))

 (1)

Desired output (derived by hand) where the transfer function is separated in a dimensionless expression and a gain that can be attributed to units with a physical meaning in the context of an application (here displacement per voltage).

H(s) = 60.*Unit('m'/'V')/(.70805*s^2*Unit('s'^2)+144.*s*Unit('s')+0.3675e-4*s^3*Unit('s'^3))

H(s) = 60.*Units:-Unit(m/V)/(.70805*s^2*Units:-Unit(s^2)+144.*s*Units:-Unit(s)+0.3675e-4*s^3*Units:-Unit(s^3))

 (2)

is(simplify((H(s) = 60.*Units[Unit](m*kg/(s^2*A))/(.70805*s^2*Units[Unit](kg^2*m^2/(s^3*A^2))+144.*s*Units[Unit](kg^2*m^2/(s^4*A^2))+0.3675e-4*s^3*Units[Unit](kg^2*m^2/(s^2*A^2))))-(H(s) = 60.*Units[Unit](m/V)/(.70805*s^2*Units[Unit](s^2)+144.*s*Units[Unit](s)+0.3675e-4*s^3*Units[Unit](s^3)))))

true

 (3)

Units to factor out in the denominator are Unit('kg'^2*'m'^2/('s'^5*'A'^2)). Quick check:

Unit('m'*'kg'/('s'^2*'A'))/Unit('kg'^2*'m'^2/('s'^5*'A'^2)) = Unit('m'/'V')

Units:-Unit(m*kg/(s^2*A))/Units:-Unit(kg^2*m^2/(s^5*A^2)) = Units:-Unit(m/V)

 (4)

simplify(Units[Unit](m*kg/(s^2*A))/Units[Unit](kg^2*m^2/(s^5*A^2)) = Units[Unit](m/V))

Units:-Unit(s^3*A/(m*kg)) = Units:-Unit(s^3*A/(m*kg))

 (5)

"Simplification" attempts with the denominator

denom(rhs(H(s) = 60.*Units[Unit](m*kg/(s^2*A))/(.70805*s^2*Units[Unit](kg^2*m^2/(s^3*A^2))+144.*s*Units[Unit](kg^2*m^2/(s^4*A^2))+0.3675e-4*s^3*Units[Unit](kg^2*m^2/(s^2*A^2)))))

s*(.70805*s*Units:-Unit(kg^2*m^2/(s^3*A^2))+144.*Units:-Unit(kg^2*m^2/(s^4*A^2))+0.3675e-4*s^2*Units:-Unit(kg^2*m^2/(s^2*A^2)))

 (6)

collect(s*(.70805*s*Units[Unit](kg^2*m^2/(s^3*A^2))+144.*Units[Unit](kg^2*m^2/(s^4*A^2))+0.3675e-4*s^2*Units[Unit](kg^2*m^2/(s^2*A^2))), Unit('kg'^2*'m'^2/('s'^5*'A'^2)))

s*(.70805*s*Units:-Unit(kg^2*m^2/(s^3*A^2))+144.*Units:-Unit(kg^2*m^2/(s^4*A^2))+0.3675e-4*s^2*Units:-Unit(kg^2*m^2/(s^2*A^2)))

 (7)

is not effective because all units are wrapped in Unit commands. Example:

Unit('kg'^2*'m'^2/('s'^2*'A'^2))

Units:-Unit(kg^2*m^2/(s^2*A^2))

 (8)

Expand does not expand the argument of Unit commands.

> 

expand(Units[Unit](kg^2*m^2/(s^2*A^2))); lprint(%)

Units:-Unit(kg^2*m^2/(s^2*A^2))

  

Units:-Unit(kg^2*m^2/s^2/A^2)

  

NULL

C1: Expanding Unit command

An expand facility could be a solution that expands a Unit command with combined units to a product of separate Unit commands.

When all units are expanded in a separate Unit command, collect or factor can be used to collect units:

.70805*s*Unit('kg')^2*Unit('m')^2/(Unit('A')^2*Unit('s')^3)+144.*Unit('kg')^2*Unit('m')^2/(Unit('A')^2*Unit('s')^4)+0.3675e-4*s^2*Unit('kg')^2*Unit('m')^2/(Unit('A')^2*Unit('s')^2)

.70805*s*Units:-Unit(kg)^2*Units:-Unit(m)^2/(Units:-Unit(A)^2*Units:-Unit(s)^3)+144.*Units:-Unit(kg)^2*Units:-Unit(m)^2/(Units:-Unit(A)^2*Units:-Unit(s)^4)+0.3675e-4*s^2*Units:-Unit(kg)^2*Units:-Unit(m)^2/(Units:-Unit(A)^2*Units:-Unit(s)^2)

 (9)

collect(.70805*s*Units[Unit](kg)^2*Units[Unit](m)^2/(Units[Unit](A)^2*Units[Unit](s)^3)+144.*Units[Unit](kg)^2*Units[Unit](m)^2/(Units[Unit](A)^2*Units[Unit](s)^4)+0.3675e-4*s^2*Units[Unit](kg)^2*Units[Unit](m)^2/(Units[Unit](A)^2*Units[Unit](s)^2), [Unit('A'), Unit('kg'), Unit('m'), Unit('s')])

(.70805*s/Units:-Unit(s)^3+144./Units:-Unit(s)^4+0.3675e-4*s^2/Units:-Unit(s)^2)*Units:-Unit(m)^2*Units:-Unit(kg)^2/Units:-Unit(A)^2

 (10)

factor(.70805*s*Units[Unit](kg)^2*Units[Unit](m)^2/(Units[Unit](A)^2*Units[Unit](s)^3)+144.*Units[Unit](kg)^2*Units[Unit](m)^2/(Units[Unit](A)^2*Units[Unit](s)^4)+0.3675e-4*s^2*Units[Unit](kg)^2*Units[Unit](m)^2/(Units[Unit](A)^2*Units[Unit](s)^2))

0.3675e-4*Units:-Unit(kg)^2*Units:-Unit(m)^2*(19266.66666*s*Units:-Unit(s)+3918367.346+.9999999999*s^2*Units:-Unit(s)^2)/(Units:-Unit(A)^2*Units:-Unit(s)^4)

 (11)

C2: Using the Natural Units Environment

In this environment, no Unit commands are required and the collection of units should work with Maple commands.
However, for the expressions discussed here, this would lead to a naming conflict with the complex variable s of the transfer function and the unit symbol s for seconds.

NULL

C3: A type declaration or unit assumptions on names

A type declaration as an option of commands like in

Units[TestDimensions](s*(.70805*s*Units[Unit](kg^2*m^2/(s^3*A^2))+144.*Units[Unit](kg^2*m^2/(s^4*A^2))+0.3675e-4*s^2*Units[Unit](kg^2*m^2/(s^2*A^2))), {s::(Unit(1/s))})

true

 (12)

could help Maple in simplification tasks (in its general meaning of making expressions shorter or smaller).
Alternatively, assumptions could provide information of which "unit type" a name is

`assuming`([simplify(H(s) = 60.*Units[Unit](m*kg/(s^2*A))/(.70805*s^2*Units[Unit](kg^2*m^2/(s^3*A^2))+144.*s*Units[Unit](kg^2*m^2/(s^4*A^2))+0.3675e-4*s^3*Units[Unit](kg^2*m^2/(s^2*A^2))))], [s::(Unit(1/s))]); `assuming`([combine(H(s) = 60.*Units[Unit](m*kg/(s^2*A))/(.70805*s^2*Units[Unit](kg^2*m^2/(s^3*A^2))+144.*s*Units[Unit](kg^2*m^2/(s^4*A^2))+0.3675e-4*s^3*Units[Unit](kg^2*m^2/(s^2*A^2))), 'units')], [s::(Unit(1/s))])

Error, (in assuming) when calling 'property/ConvertProperty'. Received: 'Units:-Unit(1/s) is an invalid property'

  

On various occasions (beyond transfer functions) I have looked for such a functionality.

 

C4: DynamicSystems Package with units

C4.1: The complex variable s could be attributed to the unit 1/s (i.e. Hertz) either by default or as an option. This could enable using units within the dynamic system package which is not possible in Maple 2022. An example what the package provides currently can be found here: help(applications, amplifiergain)
The phase plot shows that the package is already implicitly assuming that the unit of s is Hertz. A logical extension would be to have magnitude plots with units (e.g. m/V, as in this example).

 

C4.2: A dedicated "gain" command that takes units into account and that could potentially simplify the transfer function to an expression like (2) in SI units. In such a way the transfer function is separated into a dimensionless (but frequency depended) term and a gain term with units.
This would make the transfer of transfer functions to MapleSim easy and avoid unit conversion errors.

 

Download Collecting_and_expanding_units.mw

Covariant Derivative of Riemann, Ricci and Weyl te...

Asked by: Dark Energy 100
January 22 2023
1  2

hi guys,

suppose we have general metric form in 4-D. I want to calculate Covariant derivative of Riemann, Ricci and Weyl tensors.

please help me.

with best,

Error (in f) when datatype float[8]...

Asked by: ajahyaya 10
ode syntax homework
January 21 2023
2  9

I dont know why I could not solve this problem.

I have attached my worksheet.

Please anyone help me to get solution to this problem.

Thank you so much

fypppp.mw

Is there a way to import functions form NIST Digit...

Asked by: C_R 3437
January 21 2023
3  2

DLMF offers different encodings for mathematical expressions. Example:

 

I was wondering if TeX or pMML (never seen before)  can be imported into Maple and subsequently be used as  Maple Input.

Is there a way to insert a matrix determinant (ie,...

Asked by: zenterix 405
matrix palette determinant
January 21 2023
2  1

When taking notes I sometimes use the palette to insert a matrix into a worksheet. When I do this, the main aspect that is useful to me is being able to visualize matrix expressions as I would write them.

I would like to do the same but for determinants of matrices. Is there a way to get almost the same thing as with the matrix palette, but with vertical bars denoting a determinant rather than the brackets used for matrices?

How do we compute integrals of functions or expres...

Asked by: zenterix 405
integration units int
January 21 2023
0  1

How do we compute integrals of functions or expressions that have units attached?
For example

v__1 := proc (t) options operator, arrow; 3*Unit('m'/'s') end proc

proc (t) options operator, arrow; 3*Unit('m'/'s') end proc

 (1)

v__2 := 4*Unit('m'/'s')

4*Units:-Unit(m/s)

 (2)

s__1 := int(v__1(t), t = 0 .. 5)

15*Units:-Unit(m/s)

 (3)

s__2 := int(v__2, t = 0 .. 5)

20*Units:-Unit(m/s)

 (4)

NULL


The results of the integrals have the wrong units.

Download integration_with_units.mw

power of 1 PlusMinus 1...

Asked by: gharouce 5
January 21 2023
1  11

I have X=+-1,

and I want to calculate X^n.

 Thank's

How to calculate a multiple series in Maple?...

Asked by: Alex0099 60
sum summation
January 20 2023
2  12

Dear members of the forum, please tell me if it is possible to calculate the series presented below by Maple 2022. As far as I understand, first you need to calculate the inner and then the outer sum, but I don’t know how to do this with the help of the program, this series does not converge, as it seems to me, but I can be wrong, if the series diverges, then I need to show it.

Sorry for my ignorance, but maybe I wrong apply such commands for calculation this sums:

Sum(F, a = 1 .. infinity, b = 1 .. infinity) = DefiniteSummation(F, a = 1 .. infinity, b = 1 .. infinity)

I understand, that Sum () not be able to recive more than one args, but I don't understand how to make this calculation...

Any idea? Thanks for advices and help!

How to run an executable program from Maple...

Asked by: fnavarro 45
programming linux define_external
January 20 2023
0  2

Dear All,

I have an executable program which I have generated with Fortran. Is it possible to run such a program from Maple? It sounds a bit weird but that would simplify my computation, since I would not need to use scripts in Linux. Thank you very much.

Cheers.

What is the best way to make the results of callin...

Asked by: zenterix 405
solve unapply
January 20 2023
1  4

Let's say we have four equations and four unknowns and we use solve to find a solution.

The return value of solve is a set.

Here is an example

solve({T__1 = m__1*a, a = R*alpha, -R*T__1+R*T__2 = I__s*a/R, g*m__2-T__2 = m__2*a}, {T__1, T__2, a, alpha})

{T__1 = R^2*g*m__2*m__1/(R^2*m__1+R^2*m__2+I__s), T__2 = m__2*g*(R^2*m__1+I__s)/(R^2*m__1+R^2*m__2+I__s), a = R^2*g*m__2/(R^2*m__1+R^2*m__2+I__s), alpha = R*g*m__2/(R^2*m__1+R^2*m__2+I__s)}

 (1)

NULL


If we want to make the four expressions above procedures, the manual way is to basically copy the right-hand side of each expression and then write

T__1 := (R,m__1, m__2, g, I__s) -> ...

T__2 := (R,m__1, m__2, g, I__s) -> ...

a := (R,m__1, m__2, g, I__s) -> ...

alpha := (R,m__1, m__2, g, I__s) -> ...

Is there a way to do this automatically from the return value of solve?

If the return value were a list, I would use something like

T__1 := unapply(result[1], R, m__1, m__2, g, I__s)

T__2 := unapply(result[2], R, m__1, m__2, g, I__s)

a := unapply(result[3], R, m__1, m__2, g, I__s)

alpha := unapply(result[4], R, m__1, m__2, g, I__s)

But the return value is not a list.

So, in summary my quesions are

1) what, in general, is the best way to obtain the desired procedures?

2) is there a way to use the strategy I suggested if the result were a list, but for sets?

Download solveEqs.mw

problem with differentiation...

Asked by: dtarp 5
syntax differentiation
January 20 2023
1  2

i need to differentiate this equation 

Potential energy eq:
p := (m1*a1 + m2*(a123 + z(t)))*sin(theta(t))*g + m3*(sin(theta(t))*(a1234 + z(t)) + cos(theta(t))*a5)*g;


  p := (m1 a1 + m2 (a123 + z(t))) sin(theta(t)) g    + m3 (sin(theta(t)) (a1234 + z(t)) + cos(theta(t)) a5) g

i have tried these 2 methods

diff(p, theta)

diff(p, theta(t))

but both of them comes out 0 

where they should give 

diff((m1*a1 + m2*(a123 + z))*sin(theta)*g + m3*(sin(theta)*(a1234 + z) + cos(theta)*a5)*g, theta);


       (m1 a1 + m2 (a123 + z)) cos(theta) g + m3 (cos(theta) (a1234 + z) - sin(theta) a5) g

how is the syntax for partial differentiating a pre defined equation dependent on time.

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