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U...

vv 13992
March 02 2018
0 4
> 

 

> 

restart;
> 

M:=<
r0/(r-b), -r/(r-b),0,0;
-r/(r-b),b*r/r0/(r-b),0,0;
0,0,-b/r0,-1;
0,0,-1,-r0/r>;

C:=<
r/(r-b),0,0,0;
0,-1,0,0;
0,0,(r-b)/r,0;
0,0,0,-1>;

_rtable[18446744074326966870]

  

Matrix(%id = 18446744074331272966)

 (1)

 

> 

U:=< u11,u12,0,  0;     # look for a block-diagonal U, such as M and C
     u12,u13,0,  0;
     0,  0,  u33,u34;
     0,  0,  u34,u44>;

_rtable[18446744074331276462]

 (2)
> 

U^+ . C . U - M:
> 

sol:=solve({entries(%,nolist)}, indets(U), explicit):
nops([sol]);

16

 (3)
> 

U1:=simplify(eval(U,sol[1])) assuming positive; # one of the solutions
> 

 

_rtable[18446744074331290190]

 (4)
> 

simplify(U1^+ . C . U1 - M); # check

_rtable[18446744074793044558]

 (5)

For real symmetric M and C it is easy to obtain a symbolic solution
in terms of sqrt(M1), sqrt(C1)  where M1, C1 are the diagonal matrices containing the eigenvalues.

 

One of the 16 solutions should be your "known" one, but it seems that it's not there. I don't know why; maybe solve did not find all the solutions.

 

> 

 

sort...

vv 13992
March 01 2018
0 3

"non deterministic"  in the sense that the order of the objects in memory cannot (normally) be anticipated by the user.

Example

restart;
b:
sort([a,b], address);
                             [b, a]
restart;
sort([a,b], address);
                             [a, b]

 

symbolic & floats...

vv 13992
February 27 2018
1 0

It is not a good idea to mix symbolics and floats.

Use:

int(int(sin(x^2*y), x = 1 .. Pi), y = 1 .. Pi);  # symbolic

evalf(%);   # approximate the result

Or,

evalf( Int(Int(sin(x^2*y), x = 1 .. Pi), y = 1 .. Pi) );  # numeric integration

 

A procedure...

vv 13992
February 26 2018
1 0 
# The linearization  of F(u)  near u0 in the "direction" h, (h=small)
# (If u=u(x), actually it's about directional derivative in a function space)
Lin:=proc(F, u, u0, h)
local eps;
eval(F, u=u0+eps*h);
series(%, eps,2);
eval(convert(%,polynom),eps=1)
end:

# Your case:
sin(phi2)*sin(psi1(t))*lk*m1*(diff(psi1(t), t, t))+sin(phi2)*(diff(psi1(t), t))^2*cos(psi1(t))*lk*m1:
Lin(%, psi1(t), psi1__0(t), u(t)):   simplify(%);

# This agrees with Rouben's answer.

 

ilog...

vv 13992
February 25 2018
2 0
> 

X:= n ->  ``(3)^n/``(2)^ilog2(3^n):
> 

expand(X(n));

3^n/2^ilog[2](3^n)

 (1)
> 

[seq(X(n),n=1..10)];

[``(3)/``(2), ``(3)^2/``(2)^3, ``(3)^3/``(2)^4, ``(3)^4/``(2)^6, ``(3)^5/``(2)^7, ``(3)^6/``(2)^9, ``(3)^7/``(2)^11, ``(3)^8/``(2)^12, ``(3)^9/``(2)^14, ``(3)^10/``(2)^15]

 (2)
> 

expand(%);

[3/2, 9/8, 27/16, 81/64, 243/128, 729/512, 2187/2048, 6561/4096, 19683/16384, 59049/32768]

 (3)

formal...

vv 13992
February 25 2018
1 1

Unfortunately assuming works well only for simple conditions/properties.

But usually it is easy to obtain sufficient convergence conditions (as you did) and the sum can be computed using
sum(exp(a*k*Ta)*z^(-k), k=0..infinity, formal);

Edit. The convergence condition is actually  abs(exp(a*Ta)/z) < 1
but note that even assuming a<0, Ta<0, z>10  does not work.

 

Letters...

vv 13992
February 24 2018
1 3

Drawing letters is a trivial task in Maple.

It's a matter of defining the polygon(s). It would be nice if someone with artistic abilities would design them.

Note that for all the letters except i and j a single polygon is enough.

For example:

> 

with(plottools):with(plots):
> 

A := [[3,0],[4,0],[.5,10.5],[-.5,10.5],[-4,0],[-3,0],[-2,3],[2, 3],[1.65,4.05],[-1.65,4.05],[0, 9]]:
> 

display(polygon(A), axes=none);

 

After that, in Maple 2017 we can do this:

> 

AA:=extrude(display(polygon([A]), scaling=constrained), 0..1.2,style=surface, axes=none):
> 

animate(display@rotate,[AA,t, [[0,0,0],[0,1,0]] ], t=0..2*Pi,paraminfo=false,orientation=[-90,-15,25]);
> 

 

 

42 indeed...

vv 13992
February 23 2018
0 0
> 

# Next number
> 

f:=proc(n::{posint,name})
local t, k, f1, i;
f1 := (t,k) -> mul(t-i,i=1..k-1)*mul(t-i,i=k+1..nops(f(0)));
convert(add(f(0)[k]*f1(n,k)/f1(k,k),k=1..nops(f(0))),horner)
end:
> 

f(0):=[3,10,2,7,7,42]:
seq(f(k),k=1..7);
f(x);

3, 10, 2, 7, 7, 42, 314

  

-197+(6212/15+(-3571/12+(195/2+(-179/12+(13/15)*x)*x)*x)*x)*x

 (1)
> 

f(0):=[3,10,2,7,7,13]:
seq(f(k),k=1..7);
f(x);

3, 10, 2, 7, 7, 13, 140

  

-168+(4175/12+(-5837/24+(1847/24+(-271/24+(5/8)*x)*x)*x)*x)*x

 (2)
> 

f(0):=[1,2,4,8,16];
seq(f(k),k=1..6);
f(x);

[1, 2, 4, 8, 16]

  

1, 2, 4, 8, 16, 31

  

1+(-3/4+(23/24+(-1/4+(1/24)*x)*x)*x)*x

 (3)
> 

f(0):=[1,2,4,8,16,32];
seq(f(k),k=1..9);

[1, 2, 4, 8, 16, 32]

  

1, 2, 4, 8, 16, 32, 63, 120, 219

 (4)
> 

 

Factorization...

vv 13992
February 22 2018
1 0

Your  algebraic fractions al already simplified (but there are of course other possibilities, e.g. partial fractions).
For a factorization, the field K is needed.
For K=Q, the polynomials are already factorized.
If you want K=C (complete factorization) the simplest way is to find the roots with solve. E.g.

p:=(k+2)*(k^2+5)*(k^3-5*k^2+13*k-8)/(6*k*(k^2-3*k+8)):
p1,p2:=(numer,denom)(p):
R1,R2:=[solve(p1,k,explicit)],[solve(p2,k,explicit)]:
mul(k-r,r=R1)/mul(k-r,r=R2);

Solution...

vv 13992
February 22 2018
5 2

Actually, the solution of your ODE is

A(t) = a * exp( I * ( a^2*t + b ) ),  where  a, b are real constants.

This can be obtained with Maple writing A(t) in polar form:

restart;
ode := diff(A(t),t) - I * conjugate(A(t))*A(t)^2:
simplify(eval(ode, A(t)=R(t)*exp(I*phi(t)) )  /exp(I*phi(t)) ) assuming real:
evalc([Re,Im](%)):
dsolve(%);

Workaround...

vv 13992
February 21 2018
1 1

restart;
a := Matrix(3, 3, [[x, y, z], [y, z, x], [z, x, y]]):
ex:= Matrix(3, 3, [[x, y, z], [y, z, x], [z, x, y]]) + B:
#eval(ex, a=A); #does not work
evalindets(ex, 'Matrix', u -> `if`(EqualEntries(u,a),A,u));
                             B + A
evalindets(ex, 'Matrix', u -> `if`(LinearAlgebra[Equal](u,a),A,u));
                             B + A

 

Open Newton–Cotes...

vv 13992
February 19 2018
0 1 
# Open Newton–Cotes (N=n-1)
ONC:=proc(n::posint)
local nodes:=[seq(k/n,k=1..n-1)], B,w,i,k,x;
B:=i -> [seq(`if`(k=i,1,0),k=1..n-1)]:
w:=[seq(int(CurveFitting:-PolynomialInterpolation(nodes,B(i),x),x=0..1),i=1..n-1)] ;
`1/(b-a)`.'c'=w,'p'=2*floor(n/2)+1
end:

ONC(5);
       `1/(b-a)` . c = [11/24, 1/24, 1/24, 11/24], p = 5

Solution...

vv 13992
February 17 2018
1 1 
Optimization:-Minimize(
0,
{
cos(q[1])*cos(q[3])-sin(q[1])*sin(q[3])+sin(q[1])*q[2]+cos(q[1]) - 1 = 0,
sin(q[1])*cos(q[3])+cos(q[1])*sin(q[3])-cos(q[1])*q[2]+sin(q[1]) -1 = 0
},


q[1] = 2 ..  Pi, q[2] = 1 .. 1.5, q[3] = -3 .. -2

#   q[1] = 0 ..  Pi, q[2] = 0.1 .. 1.5, q[3] = -Pi/2 .. 4*Pi/3
# ,  initialpoint = [q[1]=2.0944, q[2]=1.4, q[3]=-2.7925]

);

       [0., [q[1] = 2.30431169727399, q[2] = 1.03640893001776, q[3] = -2.75622305752699]]

Expansion...

vv 13992
February 17 2018
1 2
> 

int_part:=proc(f,h,t,n::integer)
local k,u,v,s;
u:=f;
v:=h;
s:=0;
for k from 1 to n do;
  u:=int(u,t);
  s:=s-(-1)^(k)*u*v;
  v:=diff(v,t);
od;
s;
end:

> 

int_part(exp(-x*t), 1/sqrt(t*(t+1)), t, 6):
A:=unapply(simplify(-eval(%,t=1)),x);

proc (x) options operator, arrow; (1/2048)*exp(-x)*2^(1/2)*(1024*x^5-768*x^4+1216*x^3-3024*x^2+10404*x-46035)/x^6 end proc

 (1)
> 

J:= x -> Int((exp(-x*t))/sqrt(t*(t+1)),t=1..infinity):
> 

evalf[20](eval( [A(x), A(x)-J(x)], x=20));  # check

[0.70333100226385465863e-10, -0.211984348249630e-15]

 (2)
> 

 

workaround...

vv 13992
February 17 2018
0 1

It seems that Maple needs help here.

solve({combine(7*cos(2*t)=7*cos(t)^2-5), t>=0, t<=2*Pi}, t, allsolutions, explicit);

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