Advanced Search

Kitonum

21435 Reputation

26 Badges

17 years, 24 days
Contact Kitonum

MaplePrimes Activity


These are answers submitted by Kitonum

Procedure...

Kitonum 21435
August 28 2020
1 4

To avoid repeated calls to Typesetting:-Settings , you can enter your equations through the procedure  P  below:

> 

restart;
P:=proc(ode)
local t;
t:=indets(ode, name)[];
Typesetting:-Settings(usedot=false,prime=t,typesetprime=true):
ode;
end proc:

ode1:=P(diff(y(t),t$2)+diff(y(t),t)+y(t)= 0);
ode2:=P(diff(y(x),x$2)+diff(y(x),x)+y(x)= 0);

diff(diff(y(t), t), t)+diff(y(t), t)+y(t) = 0

  

diff(diff(y(x), x), x)+diff(y(x), x)+y(x) = 0

 (1)
> 

 


 

Download prime.mw

assign...

Kitonum 21435
August 28 2020
2 0

Use  assign  for this instead of  :=  :

restart;
for a from 1 to 2 do
assign(convert(cat("k",a),symbol)=a);
od;
k1, k2;

                                             1, 2

Here is another way of assignment through the seq command (multiple assignment), where  :=  doesn't work, but  assign  does:

restart;
seq(k||i, i=1..2) := seq(a, a=1..2);  # an error
assign(seq(k||i=i, i=1..2));  # OK
k1, k2;  # check

   Error, invalid left hand side in assignment
                              1, 2

allsolutions option...

Kitonum 21435
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 21435
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 21435
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 21435
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 21435
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 21435
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 21435
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 21435
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 21435
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 21435
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 21435
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 21435
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 21435
August 07 2020
0 0

Example:

convert(1.0*a+b, rational);

                           a + b

First 52 53 54 55 56 57 58 Last Page 54 of 289