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Kitonum

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These are answers submitted by Kitonum

allsolutions option...

Kitonum 21550
August 26 2020
2 1


 

> 

restart;
assume(omega>0);assume(zeta>0 and zeta<1);
tf:=omega^2/(s^2+2*zeta*omega*s+omega^2);
y:=tf*1/s;
yt:=inttrans:-invlaplace(y,s,t);
dyt:=diff(yt,t);

solve(dyt=0,t, allsolutions);
about(_Z1);

omega^2/(2*omega*s*zeta+omega^2+s^2)

  

omega^2/((2*omega*s*zeta+omega^2+s^2)*s)

  

1-exp(-omega*zeta*t)*cos((-zeta^2+1)^(1/2)*omega*t)-zeta*exp(-omega*zeta*t)*sin((-zeta^2+1)^(1/2)*omega*t)/(-zeta^2+1)^(1/2)

  

exp(-omega*zeta*t)*(-zeta^2+1)^(1/2)*omega*sin((-zeta^2+1)^(1/2)*omega*t)+zeta^2*omega*exp(-omega*zeta*t)*sin((-zeta^2+1)^(1/2)*omega*t)/(-zeta^2+1)^(1/2)

  

Pi*_Z1/((-zeta^2+1)^(1/2)*omega)

  

Originally _Z1, renamed _Z1~:
  is assumed to be: integer

  
> 

 


 

Download dyt.mw

A way...

Kitonum 21550
August 23 2020
0 0

Let's define new functions  Sin, Cos, Tan, Cot  for which arguments are assumed in degrees:

Sin:=x->sin(x*Pi/180):
Cos:=x->cos(x*Pi/180):
Tan:=x->tan(x*Pi/180):
Cot:=x->cot(x*Pi/180):

# Examples
Sin(30);
Tan(45);
Cos(30);

 

Approximate calculation...

Kitonum 21550
August 19 2020
2 0

Probably this integral cannot be calculated symbolically (even for the specific values of the parameters). So calculate numerically.

Example:


 

> 

restart;
t:=1: A:=2: omega:=3: tau:=4: phi:=5:
int( ((-A*omega*sin(omega*x+phi)*exp(-x/tau) - A*cos(omega*x+phi)*exp(-x/tau)/tau)^2 + 1)^(1/2), x=0..t );
int( ((-A*omega*sin(omega*x+phi)*exp(-x/tau) - A*cos(omega*x+phi)*exp(-x/tau)/tau)^2 + 1)^(1/2), x=0..t, numeric );

int(((-6*sin(3*x+5)*exp(-(1/4)*x)-(1/2)*cos(3*x+5)*exp(-(1/4)*x))^2+1)^(1/2), x = 0 .. 1)

  

3.482744646

 (1)
> 

 


 

Download int.mw

Ways...

Kitonum 21550
August 19 2020
2 1

This can be done in many ways. Here are two options (long and short):


 

> 

restart;
data:=[0,12,0,7,5,3,7,10,0,0,9,3,2,5,0,6]:
N:=0:
for d in data do
if d>=7 then N:=N+1 fi;
od:
N;

5

 (1)
> 

nops(select(`>=`, data, 7));

5

 (2)
> 

 


 

Download 1.mw

match...

Kitonum 21550
August 17 2020
0 1


 

> 

restart;
eq1:= (-k*I + 2*I + m)*sqrt(3) - 3*I*m - 3*k;
eq2:=eq1/2;
eq_given:= a*(k + I*m) + b;
match(eq2 = eq_given, {k,m}, 's');
s;

(-I*k+2*I+m)*3^(1/2)-(3*I)*m-3*k

  

(1/2)*(-I*k+2*I+m)*3^(1/2)-((3/2)*I)*m-(3/2)*k

  

a*(k+I*m)+b

  

true

  

{a = -((1/2)*I)*(3^(1/2)-3*I), a = -((1/2)*I)*3^(1/2)-3/2, b = I*3^(1/2)}

 (1)
> 

 


 

Download Q20200817_new.mw

A way...

Kitonum 21550
August 17 2020
1 0


 

> 

restart;
Expr:=W__1 + W__2 = -sin(-beta + alpha)*((H^2 - h^2)*gamma + h^2*psi)/(2*sin(beta)*sin(alpha));
applyop(p->H^2*p,2,applyop(p->collect(expand(p/H^2),gamma),[2,3],Expr));

W__1+W__2 = -(1/2)*sin(-beta+alpha)*((H^2-h^2)*gamma+h^2*psi)/(sin(beta)*sin(alpha))

  

W__1+W__2 = -(1/2)*H^2*sin(-beta+alpha)*((1-h^2/H^2)*gamma+h^2*psi/H^2)/(sin(beta)*sin(alpha))

 (1)
> 

 


 

Download 1.mw

Another way...

Kitonum 21550
August 15 2020
1 1

Example:

restart;
N:=8: # The number of the points
plot([seq([n/(N-1),ln(1+sin(Pi*n/(N-1)))], n=0..N-1)], legend = numerical, style = point, symbol = box, color = blue, symbolsize = 15);

 

A way...

Kitonum 21550
August 12 2020
2 1

1. Don't use square brackets to group terms, only parentheses.
2. In this example, use  simplify  with  siderels :


 

> 

restart;
> 

with(Student[LinearAlgebra]):
> 

eq4:= v__a(t) = (v__an(t)-v__ap(t))/2 - L__arm/2*diff((i__ap(t)-i__an(t)), t) - R__arm/2*(i__ap(t)-i__an(t));

v__a(t) = (1/2)*v__an(t)-(1/2)*v__ap(t)-(1/2)*L__arm*(diff(i__ap(t), t)-(diff(i__an(t), t)))-(1/2)*R__arm*(i__ap(t)-i__an(t))

 (1)
> 

eq4_2:= simplify(eq4, {i__ap(t)-i__an(t) = i__a(t)});

v__a(t) = (1/2)*v__an(t)-(1/2)*v__ap(t)-(1/2)*L__arm*(diff(i__a(t), t))-(1/2)*R__arm*i__a(t)

 (2)
> 

 


 

Download Maximum_Modulation_Index_for_MMC_with_CCC_new.mw
 

Better simplification...

Kitonum 21550
August 12 2020
1 0

In fact, your simplification requires no assumptions. You can get an even better simplification through  tan(x/2) :


 

> 

restart;
u1:=-ln(csc(x)-cot(x));
simplify(u1);
convert(u1, tan);
eval(%, {tan(x)=2*tan(x/2)/(1-tan(x/2)^2)});
applyop(normal,[2,1],%);

-ln(csc(x)-cot(x))

  

-ln((1-cos(x))/sin(x))

  

-ln((1/2)*(1+tan((1/2)*x)^2)/tan((1/2)*x)-1/tan(x))

  

-ln((1/2)*(1+tan((1/2)*x)^2)/tan((1/2)*x)-(1/2)*(1-tan((1/2)*x)^2)/tan((1/2)*x))

  

-ln(tan((1/2)*x))

 (1)
> 

 


 

Download simpl.mw

LinearSolve...

Kitonum 21550
August 10 2020
0 1

Your system only has a zero solution:


 

> 

restart; with(LinearAlgebra)
> 

M := Matrix(Matrix(7, 7, {(1, 1) = -mu, (1, 2) = a[1]*beta[x]*k[1]*PI[h]/mu, (1, 3) = omega[o]*beta[x]*a[1]*PI[h]/mu, (1, 4) = sigma, (1, 5) = -beta[2]*PI[h]/mu, (1, 6) = 0, (1, 7) = a[1]*beta[x]*k[2]*PI[h]/mu, (2, 1) = 0, (2, 2) = -(mu+r[o]+alpha[o])*a[1]*beta[x]*k[1]*PI[h]/mu, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (2, 6) = 0, (2, 7) = a[1]*beta[x]*k[2]*PI[h]/mu, (3, 1) = 0, (3, 2) = 0, (3, 3) = -(mu+r[1]+alpha[1]+k[o])*a[1]*beta[x]*k[1]*PI[h]/mu, (3, 4) = 0, (3, 5) = 0, (3, 6) = 0, (3, 7) = omega[o]*beta[x]*a[1]*k[2]*PI[h]/mu, (4, 1) = 0, (4, 2) = r[o], (4, 3) = r[1], (4, 4) = -mu-sigma, (4, 5) = 0, (4, 6) = 0, (4, 7) = 0, (5, 1) = 0, (5, 2) = alpha[o], (5, 3) = alpha[1], (5, 4) = 0, (5, 5) = -e[o], (5, 6) = 0, (5, 7) = 0, (6, 1) = 0, (6, 2) = -a[2]*beta[y]*k[1]*PI[m]/mu[b], (6, 3) = -a[2]*beta[y]*k[2]*PI[m]/mu[b], (6, 4) = 0, (6, 5) = -a[2]*beta[y]*PI[m]/mu[b], (6, 6) = -mu[b], (6, 7) = 0, (7, 1) = 0, (7, 2) = a[2]*beta[y]*k[2]*PI[m]/mu[b], (7, 3) = a[2]*beta[y]*k[2]*PI[m]/mu[b], (7, 4) = m[7, 4], (7, 5) = a[2]*beta[y]*PI[m]/mu[b], (7, 6) = 0, (7, 7) = -mu[b]})); V := Vector([`$`(0, 7)])

Matrix(7, 7, {(1, 1) = -mu, (1, 2) = a[1]*beta[x]*k[1]*PI[h]/mu, (1, 3) = omega[o]*beta[x]*a[1]*PI[h]/mu, (1, 4) = sigma, (1, 5) = -beta[2]*PI[h]/mu, (1, 6) = 0, (1, 7) = a[1]*beta[x]*k[2]*PI[h]/mu, (2, 1) = 0, (2, 2) = -(mu+r[o]+alpha[o])*a[1]*beta[x]*k[1]*PI[h]/mu, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (2, 6) = 0, (2, 7) = a[1]*beta[x]*k[2]*PI[h]/mu, (3, 1) = 0, (3, 2) = 0, (3, 3) = -(mu+r[1]+alpha[1]+k[o])*a[1]*beta[x]*k[1]*PI[h]/mu, (3, 4) = 0, (3, 5) = 0, (3, 6) = 0, (3, 7) = omega[o]*beta[x]*a[1]*k[2]*PI[h]/mu, (4, 1) = 0, (4, 2) = r[o], (4, 3) = r[1], (4, 4) = -mu-sigma, (4, 5) = 0, (4, 6) = 0, (4, 7) = 0, (5, 1) = 0, (5, 2) = alpha[o], (5, 3) = alpha[1], (5, 4) = 0, (5, 5) = -e[o], (5, 6) = 0, (5, 7) = 0, (6, 1) = 0, (6, 2) = -a[2]*beta[y]*k[1]*PI[m]/mu[b], (6, 3) = -a[2]*beta[y]*k[2]*PI[m]/mu[b], (6, 4) = 0, (6, 5) = -a[2]*beta[y]*PI[m]/mu[b], (6, 6) = -mu[b], (6, 7) = 0, (7, 1) = 0, (7, 2) = a[2]*beta[y]*k[2]*PI[m]/mu[b], (7, 3) = a[2]*beta[y]*k[2]*PI[m]/mu[b], (7, 4) = m[7, 4], (7, 5) = a[2]*beta[y]*PI[m]/mu[b], (7, 6) = 0, (7, 7) = -mu[b]})

  

Vector[column](%id = 18446745476990704686)

 (1)
> 

LinearSolve(M, V)

Vector[column](%id = 18446745477001216958)

 (2)
> 

 


 

Download trc_new.mw

 

Procedure...

Kitonum 21550
August 10 2020
0 8

Here is the procedure to write explicitly this double sum:

> 

restart;
S:=n->sum(sum(a[i,j]*x^i*y^j, j=0..n-i), i=0..n);

# Example
S(3);

proc (n) options operator, arrow; sum(sum(a[i, j]*x^i*y^j, j = 0 .. n-i), i = 0 .. n) end proc

  

x^3*a[3, 0]+x^2*y*a[2, 1]+x*y^2*a[1, 2]+y^3*a[0, 3]+x^2*a[2, 0]+x*y*a[1, 1]+y^2*a[0, 2]+x*a[1, 0]+y*a[0, 1]+a[0, 0]

 (1)
> 

 


 

Download double_sum.mw

args, nargs...

Kitonum 21550
August 10 2020
0 1

OP wrote "Never understood the proc() without parameters"

In fact, everything is clear here. Parameterless procedures are often used when the number of actual parameters (called arguments) is not known in advance. In this case they are referred with  args (argument sequence)  and  nargs (number of arguments) :

Your generalized example (finds the Euclidean norm for any dimension):

 norm:=proc()
sqrt(sum(args[i]^2,i=1..nargs)) end:

norm(3,4,5);

                                      

 

In the following example, we find the arithmetic mean of a sequence of numbers for any number of these numbers:

restart;
M:=proc()
`+`(args)/nargs;
end proc:

# Example:

M(2,3,5,6);

                                        4

convert( ... , rational);...

Kitonum 21550
August 07 2020
0 0

Example:

convert(1.0*a+b, rational);

                           a + b

About evalb...

Kitonum 21550
August 07 2020
0 2

From the help:
"Important: The evalb command does not simplify expressions. It may return false for a relation that is true. In such a case, apply a simplification to the relation before using evalb."

Your example:

restart;
f := x + 1 = (x^2 - 1)/(x - 1):
evalb(simplify(f));

                                     true


See help for details.
 

This is a curve...

Kitonum 21550
August 06 2020
0 0

This is not a surface but a curve as the intersection of two surfaces (the cylinder  (y-1)^2+z^2=1  and the plane  x=1 ):


 

> 

restart;
plots:-intersectplot((y-1)^2+z^2=1, x=1, x=0..2, y=0..2, z=0..1, thickness=2, axes=normal, scaling=constrained);

 
> 

 


 

Download plot.mw

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