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What's the better? More RAM or a better processor?...

Asked by: Gabriel Barcellos 65
performance rootfinding fsolve
December 04 2024
3  21

Suppose I want to run several systems of equations as below, however, the processing time exceeds 45 thousand seconds (reaching hours of calculations)

  • What is better, buying a new processor (new notebook) or buying a new stick of RAM?
  • Or does none of this affect Maple? Therefore, it is just a computational disadvantage (the delay time), characterizing the slowness of the program for very large systems as an imminent characteristic.
  • Another question: The newer Maples (new versions) seem to be heavier to run. Is it just me? What seems to me is that the previous versions of Maple, because they contain fewer resources, are lighter and run certain calculations faster. Would this be correct?
  • Last question: Is it possible to do parallel processing in Maple? So that calculations on huge systems can be run faster

Thank you so much! :)

Finding the roots of an equation...

Asked by: MaPal93 170
solve incomplete-question
December 03 2024
2  14

I have a 2-parameters quartic equation in n (where n>=2 and integer) and want to find for which parameter values it is zero, if any exists.

Here I am analyzing the signs of each coefficient, but is there a faster way to solve my problem?
Thanks.

quartic_equation_in_n.mw

Solving equations with matrices and dot products...

Asked by: MaPal93 170
matrix solve linear-algebra
November 26 2024
0  13

I want to solve n nonlinear equations in n unknowns named lambda[1], lambda[2], ..., lambda[n]. For simplicity, I assume n=3. Each of the three equations includes matrices and dot products.

Running solve() takes forever...I assume this is because Maple is trying to find the explicit form. I am not sure, but I have reasons to believe that a solution exists.

Questions:

  1. Did I set up the three equations correctly? Is there a simpler or smarter way to set up my problem?
  2. Is it okay for Maple that lambda[i] sometimes appears isolated and sometimes as a component of a vector?  
  3. Is there a way to keep my quantities (except the lambda vector I guess, since it containts the lambda[i] component I need to solve for) undefined/implicit while making sure Maple correctly deals with matrices and dot products?
    e.g., the solution should display terms like p^%T.R.w as such, rather than the corresponding explicit computation...

Worksheet:

> 

restart;
> 

S:=Matrix([[Sigma[1,1],Sigma[1,2],Sigma[1,3]],[Sigma[2,1],Sigma[2,2],Sigma[2,3]],[Sigma[3,1],Sigma[3,2],Sigma[3,3]]]):

w:=Vector(3,symbol=omega):

r:=Vector([sqrt(S[1,1]/(w^%T.S.w)),sqrt(S[2,2]/(w^%T.S.w)),sqrt(S[3,3]/(w^%T.S.w))]):

R:=Matrix([[r[1],0,0],[0,r[2],0],[0,0,r[3]]]):

p:=Vector(3,symbol=rho):

l:=Vector(3,symbol=lambda):

e[1]:=Vector([1,0,0]):
e[2]:=Vector([0,1,0]):
e[3]:=Vector([0,0,1]):

f:=x->x^2:
w__sq:=f~(w):
l__sq:=f~(l):

 
> 

eq1:='(w[1]*((p^%T.R.w)/(2*(l^%T.w__sq)))*(w^%T.S.e[1]))/(w[1]^2*((p^%T.R.w)^2/(4*(l^%T.w__sq)^2))*(w^%T.S.w)+sigma__delta^2*(w[1]^2*(((l__sq)^%T.w__sq)/(4*(l^%T.w__sq)^2)-lambda[1]/(l^%T.w__sq))+1))';

eq2:='(w[2]*((p^%T.R.w)/(2*(l^%T.w__sq)))*(w^%T.S.e[2]))/(w[2]^2*((p^%T.R.w)^2/(4*(l^%T.w__sq)^2))*(w^%T.S.w)+sigma__delta^2*(w[2]^2*(((l__sq)^%T.w__sq)/(4*(l^%T.w__sq)^2)-lambda[2]/(l^%T.w__sq))+1))';

eq3:='(w[3]*((p^%T.R.w)/(2*(l^%T.w__sq)))*(w^%T.S.e[3]))/(w[3]^2*((p^%T.R.w)^2/(4*(l^%T.w__sq)^2))*(w^%T.S.w)+sigma__delta^2*(w[3]^2*(((l__sq)^%T.w__sq)/(4*(l^%T.w__sq)^2)-lambda[3]/(l^%T.w__sq))+1))';

(1/2)*w[1]*(`.`(p^%T, R, w))*(`.`(w^%T, S, e[1]))/((l^%T.w__sq)*((1/4)*w[1]^2*(`.`(p^%T, R, w))^2*(`.`(w^%T, S, w))/(l^%T.w__sq)^2+sigma__delta^2*(w[1]^2*((1/4)*(l__sq^%T.w__sq)/(l^%T.w__sq)^2-lambda[1]/(l^%T.w__sq))+1)))

  

(1/2)*w[2]*(`.`(p^%T, R, w))*(`.`(w^%T, S, e[2]))/((l^%T.w__sq)*((1/4)*w[2]^2*(`.`(p^%T, R, w))^2*(`.`(w^%T, S, w))/(l^%T.w__sq)^2+sigma__delta^2*(w[2]^2*((1/4)*(l__sq^%T.w__sq)/(l^%T.w__sq)^2-lambda[2]/(l^%T.w__sq))+1)))

  

(1/2)*w[3]*(`.`(p^%T, R, w))*(`.`(w^%T, S, e[3]))/((l^%T.w__sq)*((1/4)*w[3]^2*(`.`(p^%T, R, w))^2*(`.`(w^%T, S, w))/(l^%T.w__sq)^2+sigma__delta^2*(w[3]^2*((1/4)*(l__sq^%T.w__sq)/(l^%T.w__sq)^2-lambda[3]/(l^%T.w__sq))+1)))

 (1)

 

> 

solve({eq1-lambda[1],eq2-lambda[2],eq3-lambda[3]},{lambda[1],lambda[2],lambda[3]});

Download dot_products.mw

how to get a good curve fitting result?...

Asked by: Aung 65
optimization regression duplicate-question
November 18 2024
1  5
 hello
 how to get curve fitting parameters right?  is that any stragey to get a good estimate.
> 

restart;
> 

with(Statistics):with(plots):with(Optimization):with(LinearAlgebra):
> 


# given data from strain rate curve
E_0[theta] := 7.883352314*10^9;
alpha[theta]:= 0.982

7883352314.

  

.982

 (1)
> 


# experimental creep data under 44 at 100 degree celcius
c_strain := Vector ([<<0>,<0.0284698>,<0.0533808>,<0.0782918>,<0.0996441>,<0.124555>,<0.142349>,<0.156584>,<0.16726>,<0.177936>,<0.181495>,<0.188612>,<0.192171>,<0.19573>,<0.19573>,<0.202847>,<0.206406>,<0.206406>,<0.209964>,<0.209964>,<0.209964>,<0.206406>,<0.209964>>]):

c_time := Vector ([<<0>,<0>,<0.048>,<0.192>,<0.352>,<0.544>,<0.704>,<0.896>,<1.088>,<1.312>,<1.52>,<1.76>,<1.984>,<2.208>,<2.464>,<2.736>,<3.088>,<3.392>,<3.664>,<4.016>,<4.352>,<4.592>,<4.832>>]):
sigma[0] := 44*10^6;
epsilon[0] := sigma[0]/E_0[theta];

44000000

  

0.5581381911e-2

 (2)
> 


# change vector to list
c_strain := convert(c_strain,list):
c_time := convert(c_time,list):

# extract zero from list
c_strain := c_strain [2..-1];
c_time := c_time [2..-1];

[0.284698e-1, 0.533808e-1, 0.782918e-1, 0.996441e-1, .124555, .142349, .156584, .16726, .177936, .181495, .188612, .192171, .19573, .19573, .202847, .206406, .206406, .209964, .209964, .209964, .206406, .209964]

  

[0, 0.48e-1, .192, .352, .544, .704, .896, 1.088, 1.312, 1.52, 1.76, 1.984, 2.208, 2.464, 2.736, 3.088, 3.392, 3.664, 4.016, 4.352, 4.592, 4.832]

 (3)
> 


# for further calculation need to know how many elements are in the list
M := nops(c_strain);
N := nops(c_time);

22

  

22

 (4)
> 


# constitutive equation
creep_strain := proc(t)
local i;
options operator, arrow;
epsilon[0]*(1 + alpha[theta]*add(-(B[i]*(-beta[i]*t + exp(-beta[i] *t) -1))/beta[i],i=1..3))
end proc;

proc (t) local i; options operator, arrow; epsilon[0]*(1+alpha[theta]*add(-B[i]*(-beta[i]*t+exp(-beta[i]*t)-1)/beta[i], i = 1 .. 3)) end proc

 (5)
> 


#define objective function

Digits := 9:
obj := add(
         (
           creep_strain(c_time[j])
           -
           c_strain[j]
         )^2
         , j=1..N
       ):
> 


# curve fitting
opt := NLPSolve(obj,
{
  beta[1] >= 0,
    beta[2] >= 0,
    beta[3] >= 0,
    B[1] >= 0, B[1]<= 0.2,
    B[2] >= 0, B[2] <= 0.3,
    B[3] >= 0, B[3] <= 0.4
}
);

[.467521159408790410, [B[1] = HFloat(0.19999999999999996), B[2] = HFloat(0.30000000000000004), B[3] = HFloat(0.4), beta[1] = HFloat(1.2600715880722035e-9), beta[2] = HFloat(5.152129055134073e-9), beta[3] = HFloat(5.057870561183067e-9)]]

 (6)
> 

# plot the result
display(
   ScatterPlot(c_time, c_strain, symbol=circle, color=blue),
 
  plot(eval(creep_strain(t), opt[2]), t=0.00..max(c_time), color=black)
)

Download at_44_Mpa_at_100C.mw

how to solve this integral? ...

Asked by: Aung 65
integration sum
November 16 2024
0  20
> 

restart:
> 

 

> 

inte_eq := int(-(sum(B[i]*beta[i]* exp(-beta[i] * (t-tau)),i=1..n)),tau=0..t);

int(-(sum(B[i]*beta[i]*exp(-beta[i]*(t-tau)), i = 1 .. n)), tau = 0 .. t)

 (1)

how to solve this ?

Download creep_integral_sol.mw

Infinity limit of expression with summations?...

Asked by: MaPal93 170
sum assumptions limit
November 10 2024
1  15

I want to compute the limit for a parameter (gamma) to infinity of an expression which includes summations over indexes. Below are two approaches. I believe that both should produce the same output, however:

  1. Approach A generates a signum, but I don't understand where this is coming from
  2. Approach B works smoothly once I fix the final value n of the index to some number but runs very slow if I leave n undefined (I had to interrupt the execution)

Maybe approach B is not wrong, and I just need to wait longer...

Any thoughts? 

Approach A: Compute infinity limit while keeping X and lambda undefined (since they are independent of gamma, see below), then define X and lambda
 

problem: what generates signum?

> 

restart;
> 

local gamma;

gamma

 (1)
> 

assume(0 < gamma, 0 < n, 0 < w, 0 < Delta, 0 < vo[i], -1 < rho and rho < 1, 0 < sigma__v, 0 < sigma__d, 0 < sigma__dr, delta[i]::real, delta[j]::real, delta[k]::real, delta__r::real, X[i]::real, X[j]::real, X__r::real);
interface(showassumed=0);

1

 (2)
> 

W:=sum(X[i]*(-lambda*X[i]-lambda*delta[i]-vo[i]),i=1..n)+X__r*(-lambda__r*X__r-lambda__r*delta__r-w*sum(vo[i],i=1..n))+sum((X[i]+w*X__r)*(vo[i]+(1/(w*n))*(Delta)),i=1..n)-gamma/2*sigma__v^2*(1-(1+rho*(n-1))/(n))*sum((X[i]+w*X__r)^2,i=1..n)-gamma/2*sigma__v^2*(rho-(1+rho*(n-1))/(n))*sum(sum((X[i]+w*X__r)*(X[j]+w*X__r),j=1..i-1)+sum((X[i]+w*X__r)*(X[j]+w*X__r),j=i+1..n),i=1..n);
 

sum(X[i]*(-lambda*X[i]-lambda*delta[i]-vo[i]), i = 1 .. n)+X__r*(-lambda__r*X__r-lambda__r*delta__r-w*(sum(vo[i], i = 1 .. n)))+Delta*X__r+sum(Delta*X[i]/(w*n)+vo[i]*X__r*w+X[i]*vo[i], i = 1 .. n)-(1/2)*gamma*sigma__v^2*(1-(1+rho*(n-1))/n)*(n*X__r^2*w^2+sum(2*X__r*w*X[i]+X[i]^2, i = 1 .. n))-(1/2)*gamma*sigma__v^2*(rho-(1+rho*(n-1))/n)*(n^2*X__r^2*w^2-n*X__r^2*w^2+sum(sum(X__r*w*X[j]+X[i]*X[j], j = i+1 .. n)+sum(X__r*w*X[j]+X[i]*X[j], j = 1 .. i-1)+X__r*w*X[i]*n-X__r*w*X[i], i = 1 .. n))

 (3)
> 

limit(W,gamma=infinity);

-signum(-(sum(sum(X[j]*(X__r*w+X[i]), j = i+1 .. n)+sum(X[j]*(X__r*w+X[i]), j = 1 .. i-1)+X__r*w*(n-1)*X[i], i = 1 .. n))+(n-1)*(sum(X[i]*(2*X__r*w+X[i]), i = 1 .. n)))*infinity

 (4)
> 

#convert(%, piecewise);

Approach B: Define X and lambda first, then compute infinity limit

problem: why last execution so slow?

> 

restart;
> 

local gamma;

gamma

 (5)
> 

assume(0 < gamma, 0 < n, 0 < w, 0 < Delta, 0 < vo[i], -1 < rho and rho < 1, 0 < sigma__v, 0 < sigma__d, 0 < sigma__dr, delta[i]::real, delta[j]::real, delta[k]::real, delta__r::real, X[i]::real, X[j]::real, X__r::real);
interface(showassumed=0);

1

 (6)
> 

X[i] := (1/(2*n*w))*sqrt((4*n-3)/(1+rho*(n-1)))*(sigma__d/sigma__v)*Delta-(1/(2*n))*delta[i]-(1/(2*n))*(sum(delta[j],j=1..i-1)+sum(delta[j],j=i+1..n));
X[j] := (1/(2*n*w))*sqrt((4*n-3)/(1+rho*(n-1)))*(sigma__d/sigma__v)*Delta-(1/(2*n))*delta[j]-(1/(2*n))*(sum(delta[k],k=1..j-1)+sum(delta[k],k=j+1..n));
X__r := (1/(2*w*sqrt(n*(1+rho*(n-1)))))*(sigma__dr/sigma__v)*Delta-(1/2)*delta__r;

lambda := sqrt((1+rho*(n-1))/(4*n-3))*(sigma__v/sigma__d);
lambda__r := w*sqrt(n*(1+rho*(n-1)))*(sigma__v/sigma__dr);
 

(1/2)*((4*n-3)/(1+rho*(n-1)))^(1/2)*sigma__d*Delta/(n*w*sigma__v)-(1/2)*delta[i]/n-(1/2)*(sum(delta[j], j = 1 .. i-1)+sum(delta[j], j = i+1 .. n))/n

  

(1/2)*((4*n-3)/(1+rho*(n-1)))^(1/2)*sigma__d*Delta/(n*w*sigma__v)-(1/2)*delta[j]/n-(1/2)*(sum(delta[k], k = 1 .. j-1)+sum(delta[k], k = j+1 .. n))/n

  

(1/2)*sigma__dr*Delta/(w*(n*(1+rho*(n-1)))^(1/2)*sigma__v)-(1/2)*delta__r

  

((1+rho*(n-1))/(4*n-3))^(1/2)*sigma__v/sigma__d

  

w*(n*(1+rho*(n-1)))^(1/2)*sigma__v/sigma__dr

 (7)
> 

W:=sum(X[i]*(-lambda*X[i]-lambda*delta[i]-vo[i]),i=1..n)+X__r*(-lambda__r*X__r-lambda__r*delta__r-w*sum(vo[i],i=1..n))+sum((X[i]+w*X__r)*(vo[i]+(1/(w*n))*(Delta)),i=1..n)-gamma/2*sigma__v^2*(1-(1+rho*(n-1))/(n))*sum((X[i]+w*X__r)^2,i=1..n)-gamma/2*sigma__v^2*(rho-(1+rho*(n-1))/(n))*sum(sum((X[i]+w*X__r)*(X[j]+w*X__r),j=1..i-1)+sum((X[i]+w*X__r)*(X[j]+w*X__r),j=i+1..n),i=1..n):
 

W limit for n=3

> 

Winf := collect(expand(simplify(limit(eval(W,n=3),gamma=infinity))),[Delta,Delta^2]):

quad_term := collect(expand(simplify(coeff(Winf,Delta^2))),[sigma__d,sigma__dr]);
linear_term := expand(simplify(coeff(Winf,Delta)));
coeff_term := collect(expand(simplify(eval(Winf,Delta=0))),delta__r^2);

(-(1/4)/(sigma__v*(1+2*rho)^(1/2)*w^2)+(1/2)*(1/(1+2*rho))^(1/2)/(w^2*sigma__v))*sigma__d+(1/12)*3^(1/2)*sigma__dr/((1+2*rho)^(1/2)*sigma__v*w)

  

-(1/2)*delta__r-(1/6)*delta[1]/w-(1/6)*delta[2]/w-(1/6)*delta[3]/w

  

((1/4)*3^(1/2)*sigma__v*w/((1+2*rho)^(1/2)*sigma__dr)+(1/2)*3^(1/2)*sigma__v*w*rho/((1+2*rho)^(1/2)*sigma__dr))*delta__r^2+(1/36)*sigma__v*delta[1]^2/((1+2*rho)^(1/2)*sigma__d)+(1/36)*sigma__v*delta[2]^2/((1+2*rho)^(1/2)*sigma__d)+(1/36)*sigma__v*delta[3]^2/((1+2*rho)^(1/2)*sigma__d)+(1/18)*sigma__v*delta[1]^2*rho/((1+2*rho)^(1/2)*sigma__d)+(1/18)*sigma__v*delta[2]^2*rho/((1+2*rho)^(1/2)*sigma__d)+(1/18)*sigma__v*delta[3]^2*rho/((1+2*rho)^(1/2)*sigma__d)+(1/18)*sigma__v*delta[2]*delta[1]/((1+2*rho)^(1/2)*sigma__d)+(1/18)*sigma__v*delta[1]*delta[3]/((1+2*rho)^(1/2)*sigma__d)+(1/18)*sigma__v*delta[2]*delta[3]/((1+2*rho)^(1/2)*sigma__d)+(1/9)*sigma__v*delta[2]*delta[1]*rho/((1+2*rho)^(1/2)*sigma__d)+(1/9)*sigma__v*delta[1]*delta[3]*rho/((1+2*rho)^(1/2)*sigma__d)+(1/9)*sigma__v*delta[2]*delta[3]*rho/((1+2*rho)^(1/2)*sigma__d)

 (8)

W limit for generic n

> 

Winf := collect(expand(simplify(limit(W,gamma=infinity))),[Delta,Delta^2]):

quad_term := collect(expand(simplify(coeff(Winf,Delta^2))),[sigma__d,sigma__dr]);
linear_term := expand(simplify(coeff(Winf,Delta)));
coeff_term := collect(expand(simplify(eval(Winf,Delta=0))),delta__r^2);
 

NULL

Download Infinity_limit_and_summations.mw

How to diff() on summations and double summations?...

Asked by: MaPal93 170
sum differentiation diff
November 06 2024
2  14
I am trying to compute partial derivatives of some complicated expression which include summations. First, I noticed that sum behaves differently if I use 1D vs. 2D math. Why?

Questions:
  1. Partial derivative of a summation: why is it not just 2*X[i]?
  2. Partial derivative of a double summation: how to define the nested structure of a double summation with j<>i?
  3. System of n+1 equations: how to define and solve for it?

For 3., each i equation is the partial derivative of my complicated expression with summations with respect to X[i], where i ranges from 1 to n. The last equation is the partial derivative with respect to X_r (a fixed variable).

Thanks.

> 

restart

> 

A := sum(X[i]^2, i = 1 .. n); eq[1] := diff(A, X[i]) = 0

sum(X[i]^2, i = 1 .. n)

  

sum(2*X[i], i = 1 .. n) = 0

 (1)

> 

B__wrong := sum(sum((X__r*w+X[i])*(X__r*w+X[j]), j = 1 .. n), i = 1 .. n); B__correct := 'sum(sum((X__r*w+X[i])*(X__r*w+X[j]), j = 1 .. n), i = 1 .. n)'

n^2*X__r^2*w^2+sum(sum(X__r*w*X[j]+X[i]*X[j], j = 1 .. n)+n*X__r*w*X[i], i = 1 .. n)

  

sum(sum((X__r*w+X[i])*(X__r*w+X[j]), j = 1 .. n), i = 1 .. n)

 (2)

> 

eqs := seq(eq[i], i = 1 .. n); vars := seq(X[i], i = 1 .. n)

Error, range bounds in seq must be numeric or character

  

Error, range bounds in seq must be numeric or character

  
 

NULL

Download equations_with_summations.mw

How do i define a tensor with an equation of three...

Asked by: Steve255 15
tensor differentialgeometry physics
October 30 2024
3  4

I am studying linear operators that have vanishing Nijenhuis Torsion, that is a (1-1) tensor L whose corresponding (1-2) tensor N given by a tensor equation of L (see attached) is identically zero.

I am new to maple, i have used it to plot vector fields and solve systems of equations in the past but i am unfamiliar with the DifferentialGeometry and Physics packages.

Attached is my best effort at solving this problem directly for a simple 2d case without the use of any packages and i am wondering if it is possible to do it all in one line without having to define tensor components one by one.

nijenhuis_torsion.mw

Singular vectors of a 2x2 matrix ...

Asked by: rzarouf 45
linearalgebra eigenvectors singularvalues
October 24 2024
3  10

Hi, I have a very concrete task, which is to find the singular vector associated with the largest singular matrix of the matrix M below. Here, a is an arbitrary complex number, l is of magnitude stritcly less than 1, and k>0. The command 'SingularValues' returns the 2 singular values of M. But when I search for the corresponding singular vectors, there is a bug. Those vectors are the eigenvectors of M*M. When I use the command Eigenvectors, the second argument returned is the 2 by 2 null matrix. And when I use the command eigenvects I have an error message (see below). Anyone could help me to fix this issue please?

Many thanks in advance. 

> 

with(LinearAlgebra); assume(k > 0); assume(abs(l) < 1); M := Matrix(2, 2, {(1, 1) = l-(1-l*conjugate(l))*k, (1, 2) = -a*(1-l*conjugate(l))*k, (2, 1) = 0, (2, 2) = l+(1-l*conjugate(l))*k})

Matrix(%id = 36893488151898158132)

 (1)
> 

simplify(SingularValues(M)[1])

(1/2)*(2*(-abs(l)^2+1)*k*(k^2*(abs(l)-1)^2*(abs(l)+1)^2*abs(a)^4+(4*k^2*abs(l)^4+(-8*k^2+4)*abs(l)^2+4*k^2)*abs(a)^2+4*l^2+8*abs(l)^2+4*conjugate(l)^2)^(1/2)+2*k^2*(abs(a)^2+2)*abs(l)^4+4*(-abs(a)^2*k^2-2*k^2+1)*abs(l)^2+2*k^2*(abs(a)^2+2))^(1/2)

 (2)
> 

MatrixMatrixMultiply(HermitianTranspose(M), M)

Matrix(%id = 36893488151996380692)

 (3)
> 

Eigenvectors(MatrixMatrixMultiply(HermitianTranspose(M), M))

Vector[column](%id = 36893488151975984532), Matrix(%id = 36893488151975984652)

 (4)
> 

with(linalg)
> 

eigenvectors(Matrix(2, 2, {(1, 1) = (l-(1-l*conjugate(l))*k)*conjugate(l-(1-l*conjugate(l))*k), (1, 2) = -conjugate(l-(1-l*conjugate(l))*k)*a*(1-l*conjugate(l))*k, (2, 1) = -k*conjugate(a*(1-l*conjugate(l)))*(l-(1-l*conjugate(l))*k), (2, 2) = a*(1-l*conjugate(l))*k^2*conjugate(a*(1-l*conjugate(l)))+(l+(1-l*conjugate(l))*k)*conjugate(l+(1-l*conjugate(l))*k)}))

Error, (in linalg:-eigenvectors) eigenvects only works for a matrix of rationals, rational functions, algebraic numbers, or algebraic functions at present

  
 

``

Download singular_vect_2_by_2_matrix.mwsingular_vect_2_by_2_matrix.mw

How can I make maple compute faster?...

Asked by: GFY 40
performance incomplete-question
October 21 2024
0  2

Recently a new configuration of a computer, with a better cpu - AMD Ryzen 7 8700G w/ Radeon 780M Graphics, but maple computation is particularly slow, cpu call special less computation is particularly slow, is there any way to improve?

what does this result from Maple mean?...

Asked by: jrive 195
rootof summation
October 08 2024
0  3

I do not understandwhat this result from Maple means.

I don't know what it means when a result has "Root of", and in this case, there is also a big summation sign for which I can't tell if the variables underneath it are part of the summartion, or what.  I'd appreciate any guidance to make sense of what this means..

restart

with(inttrans)

assump := R1::real, R2::real, L1::real, C1::real, C2::real, L2::real, omega::real, M::real, R1 > 0, R2 > 0, L1 > 0, C1 > 0, L2 > 0, C2 > 0, omega > 0, M > 0

R1::real, R2::real, L1::real, C1::real, C2::real, L2::real, omega::real, M::real, 0 < R1, 0 < R2, 0 < L1, 0 < C1, 0 < L2, 0 < C2, 0 < omega, 0 < M

 (1)

expr := Vin/(s*(R1+s*L1-1/(s*C1)+(omega*M)^2/(R2+s*L2-1/(s*C2))))

Vin/(s*(R1+s*L1-1/(s*C1)+omega^2*M^2/(R2+s*L2-1/(s*C2))))

 (2)

invlaplace(expr, s, t)

Vin*C1*(sum(exp(_alpha*t)*(-1+C2*_alpha*(L2*_alpha+R2))/(4*C1*C2*L1*L2*_alpha^3+2*C1*C2*M^2*_alpha*omega^2+3*C1*C2*L1*R2*_alpha^2+3*C1*C2*L2*R1*_alpha^2+2*C1*C2*R1*R2*_alpha-2*C1*L1*_alpha-2*C2*L2*_alpha-C1*R1-C2*R2), _alpha = RootOf(1+C1*C2*L1*L2*_Z^4+(C1*C2*L1*R2+C1*C2*L2*R1)*_Z^3+(C1*C2*M^2*omega^2+C1*C2*R1*R2-C1*L1-C2*L2)*_Z^2+(-C1*R1-C2*R2)*_Z)))

 (3)

NULL

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