Maple Questions and Posts

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Hello,

I need a little help...

what I want to do, if i have 3 equations  with multiple variables and i want to get derivative of some of these variables wrt other variables while setting others as constants

here in the example

1st i thought the last equation should have partial m / partial f ... how can i get that

2nd if ,another problem, i wanted to set z as a constant so it won't consider getting its derivative, how do i get that

diff_test.mw

f1 := x = y+z

x = y+z

(1)

f2 := y = 2*f^2+3

y = 2*f^2+3

(2)

f3 := z = 22*f^3+33*f+m

z = 22*f^3+33*f+m

(3)

implicitdiff({f1, f2, f3}, {x, y, z}, x, f)

66*f^2+4*f+33

(4)

NULL


Download diff_test.m

 

hope i was clear in my explanation. thank you

 

Hello,

So my questions is fairly simple but I am not sure if Maple can do this. I want to create a Matrix full of partial derivatives (Jacobian), but computing this Jacobian is maxing out my RAM. So I would like to create each row (or column) of the Jacobian seperately and save each row (or column) in a Matrix that has already been saved to a file. Basically I would like to store values in a file (in a loop) without overwriting the file.

Hi every body:

I have to obtain Amax where is the maximum value of the f(t) function at steady state but at first i don't know steady sate what's the mean and the second i don't know how to obtain it with maple. help me plz. tnx. 

f:=(t)->A*sin(x*t)

A := -6000.*x^3*((5.925123867*10^87*I)*x+5.830850605*10^88*x^2+1.752916719*10^90)/((4.241561702*10^89*I)*x^4-2.669934485*10^90*x^5-(3.812926555*10^90*I)*x^2+7.038145625*10^91*x^3+3.174301659*10^89*I-8.376628660*10^90*x)

Sin.mw

I am looking for advice as to avoiding or identifying what may elude to the cause of the error "Error, (in simplify/do) invalid simplification command" in circumstances where i have simply enclosed an expression with simplify(expression), and in most cases an error is not returned.

 

Thankyou in advance

how to find the closed form solution from the given expression in maple?

u = 140-(1/239500800)*x*t^12+(1/479001600)*x^2*t^12-119*t+40*x-(1/4435200)*t^11-(1/19958400)*x*t^11+(1/39916800)*x^2*t^11-(5/72576)*t^9-(1/60480)*x*t^9+(1/120960)*x^2*t^9+(1/2016)*t^8+(1/10080)*x*t^8-(1/20160)*x^2*t^8-(23/2520)*t^7-(1/420)*x*t^7+(1/840)*x^2*t^7+(23/360)*t^6-(1/144)*x^2*t^6+(1/72)*x*t^6-(7/12)*t^5+(1/12)*x^2*t^5-(1/6)*x*t^5+(19/6)*t^4-(3/8)*x^2*t^4+(3/4)*x*t^4-(95/6)*t^3+(5/2)*x^2*t^3-5*x*t^3+54*t^2-7*x^2*t^2+14*x*t^2+21*x^2*t-42*x*t-42*exp(-t)*x+21*exp(-t)*x^2-21*exp(-t)*t-(1/39916800)*t^12-140*exp(-t)-20*x^2

Ah yep so i have been trying to arrange this so that it returns the number of iterations required to reach a single digit value but like everything i do this week completely useless 

 


 

 

restart

unprotect(delta):

delta := proc (x, y) options operator, arrow; piecewise(x = y, 1, x <> y, 0) end proc

then the j th summand of the first digital root of 'a' in base 'b'. can be expressed as:

d := proc (a, b, j) options operator, arrow; sum((delta(j, k)-b*delta(j, k+1))*floor(a*b^(k-floor(ln(a)/ln(b))-1)), k = 1 .. floor(ln(a)/ln(b))+1) end proc:

R[0] := proc (a, b) options operator, arrow; sum(sum((delta(m, k)-b*delta(m, k+1))*floor(a*b^(k-floor(ln(a)/ln(b))-1)), k = 1 .. floor(ln(a)/ln(b))+1), m = 1 .. floor(ln(a)/ln(b))+1) end proc

proc (a, b) options operator, arrow; sum(sum((delta(m, k)-b*delta(m, k+1))*floor(a*b^(k-floor(ln(a)/ln(b))-1)), k = 1 .. floor(ln(a)/ln(b))+1), m = 1 .. floor(ln(a)/ln(b))+1) end proc

(1)

DigitalRoot := proc (a, b, N) options operator, arrow; foldl(R[0], R[0](a, b), seq(b, k = 1 .. N)) end proc

proc (a, b, N) options operator, arrow; foldl(R[0], R[0](a, b), seq(b, k = 1 .. N)) end proc

(2)

MAIN BASIC EXAMPLE

SELECT THE NUMBER:

a := 345346462

345346462

(3)

 

 

SELECT THE NUMBER BASE FOR WHICH YOU WISH TO REPRESENT a:

b := 10

10

(4)

The output will be the digits of the  base b representation of a, (or the coefficients of it's p-adic expansion?):

seq(d(a, b, j), j = 1 .. floor(ln(a)/ln(b))+1)

3, 4, 5, 3, 4, 6, 4, 6, 2

(5)

DigitalRoot(a, b, 0), DigitalRoot(a, b, 1), DigitalRoot(a, b, 2)

37, 10, 1

(6)

n := 0:

37

 

0

(7)

``

``


Sorry EDIT its ok i got it old_garbage_2015002.mw

Download old_garbage_2015.mw

Dear Users!

Hope you would be fine. In the following maple code, I want to write the derivative of psi in term of psi like it did manually in red portion. For higher M and k it very hard to do it manully. It there any command to fix my problem for any value of k and M.

restart; k := 2; M := 4;

with(linalg); with(LinearAlgebra);

printlevel := 2;

for i while i <= 2^(k-1) do

for j from 0 while j <= M-1 do

psi[M*i+j-M+1] := simplify(2^((1/2)*k)*sqrt(GAMMA(j+1)*(j+alpha)*GAMMA(alpha)^2/(Pi*2^(1-2*alpha)*GAMMA(j+2*alpha)))*(sum((-1)^i1*GAMMA(j-i1+alpha)*(2*(2^k*x-2*i1+1))^(j-2*i1)/(GAMMA(alpha)*factorial(i1)*factorial(j-2*i1)), i1 = 0 .. floor((1/2)*j))));

`&psi;&psi;`[M*i+j-M+1] := simplify(diff(psi[M*i+j-M+1], x))

end do

end do; r := 2^(k-1)*M;

VV := Vector[column](r, proc (i) options operator, arrow; psi[i] end proc);

DV := Vector[column](r, proc (i) options operator, arrow; `&psi;&psi;`[i] end proc);

``&psi;&psi;`[2] := 8*sqrt((alpha+1)*(1/2))*sqrt(2)*sqrt(alpha*GAMMA(alpha)^2*4^alpha/GAMMA(2*alpha))/sqrt(Pi) = 8*sqrt((alpha+1)*(1/2))*psi[1];

`&psi;&psi;`[3] := 16*sqrt((2+alpha)*(alpha+1)/(1+2*alpha))/sqrt(2)*(2*sqrt(2)*sqrt((alpha+1)*GAMMA(alpha)^2*4^alpha/GAMMA(1+2*alpha))*alpha*(4*x+1)/sqrt(Pi)) = 16*sqrt((2+alpha)*(alpha+1)/(1+2*alpha))*psi[2]/sqrt(2)

I am waiting your response. Thanks

I'm working with tensors using the DifferentialGeometry package (not yet checked out the Physics package since I plan on working with Lie algebras), and in particular I'm computing the Riemann tensor of a metric.

Raising the first index of CurvatureTensor(g), the Riemann tensor has the symmetries of a symmetric product of two 2-forms, and this is how I would like to present it, identifying terms like dw1dw2dw1dw2 with terms like dw1dw2dw2dw1 with the appropriate sign. Ideally I would arrive at an expression that is written in terms of (dw1^dw2) (dw1^dw2) (etc.) basis elements, provided that my tensor has these symmetries.

Is there some readily available functionality to do this, or a simple, standard method in Maple?

Hello everyone,

I have a project that uses several machines to make a computation. I wrote a package (called here "package") with procedures, "proc".  I wrote several .mpl files structured as:

read "package.mpl":

W:= Read("file1.mpl"):    #I wrote a procedure to read and assign to a variable. This is not the problem.

Export("output.mpl", proc(W));

The situation is as follows: in all but one of my machines, this works fine, and proc(W) is exported correctly. However, in this ugly duckling (which is running the GUI version of Maple18), maple prints as an output, in blue:

Export("output.mpl",...)

and in place of ... it gives the answer expected. But it does not export automatically. I can, of course, manually correct this, but this is not optimal. Moreover, if the output is too big, there will be display issues.

Help? Any light is appreciated, thanks.

Marcelo

PS: I do not expect an answer for the PS; but I also do not know how to make my package "official", in the sense that I always read the .mpl file with all the procedures. Is there a way to call it like the LinearAlgebra and so on and use the package:-proc syntax? 

The pH levels of a lake controls the conc. of harmless ammonium ions(NH4+) and toxic ammonia (NH3) in the lake.

For pH levels <8, conc. of ammonium ions are little affected by pH changes, but decline over many orders of magnitude as pH levels increase beyond pH 8.

Toxic ammonia are negligible at low pH , increase over many orders of magnitude as the pH level increases and reach a high plateau at about pH =10 (after which, NH3 are little affected by pH changes).

 

How do I illustrate this graphically?

 

Percentage survivorship is a function of time; there ia negligible decline in survivorship at pH=6, marked decline in survivorship at pH=3.5, resulting in no survivors after just 8 hours.

What is the function for this problem and how do I plot it in Maple?

 

Hi all,

I have the following expression,

Typesetting[delayDotProduct](b, K[P], true)+k*K[D]-(Typesetting[delayDotProduct](b, m[1], true)+Typesetting[delayDotProduct](b, m[2], true)+K[D]*m[2])*k*K[P]/(Typesetting[delayDotProduct](b, K[D], true)+k*m[1]+k*m[2]+K[P]*m[2]-m[1]*m[2]*(Typesetting[delayDotProduct](b, K[P], true)+k*K[D])/(Typesetting[delayDotProduct](b, m[1], true)+Typesetting[delayDotProduct](b, m[2], true)+K[D]*m[2]))

where K and K_P are the controller gains, k is the stiffness, b is the damper, m1 and m2 are masses.

How can I find the condition on variables such that the numerator of this expression is greater that zero?

The conditions should appear as inequalities.


 

b*K[P]+k*K[D]-(b*m[1]+b*m[2]+K[D]*m[2])*k*K[P]/(b*K[D]+k*m[1]+k*m[2]+K[P]*m[2]-m[1]*m[2]*(b*K[P]+k*K[D])/(b*m[1]+b*m[2]+K[D]*m[2]))

b*K[P]+k*K[D]-(b*m[1]+b*m[2]+K[D]*m[2])*k*K[P]/(b*K[D]+k*m[1]+k*m[2]+K[P]*m[2]-m[1]*m[2]*(b*K[P]+k*K[D])/(b*m[1]+b*m[2]+K[D]*m[2]))

(1)

``


 

Download Routh-Hurwitz_stability.mw

 

with(PDEtools):
V:=-exp(I*(k*x+omega*t+theta))*sqrt((k^2+omega)/(k*sigma))*sech(sqrt(k^2+omega)*(-2*k*t+x));
pde[1] := simplify(I*(diff(V, t, t))+diff(V, x, x)-I*sigma*V*(conjugate(V)*(diff(V, x))-V*(diff(conjugate(V), x)))) = 0;

 

What wouldn't work in Maple 2018 if I removed the Microsoft Visual C++ 2015 redistributable?  I have older versions of the C++ redistributable packages (ie 2013).  I hadn't noticed anything unusual when I initially removed it but maybe there's something that's affected in Maple?  Code generation package routines maybe?  What commands in Maple would be affected?

Reason is, I'm getting errors with another software and re-installing the 2015 C++ redistributable isn't installing properly.  So I'm just hoping there's no issues using Maple without the C++ 2015 redistributable being installed properly. 

 

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