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Kitonum

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17 years, 160 days
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These are answers submitted by Kitonum

local...

Kitonum 21595
May 27 2020
1 16

In the  seq  command, index is  local  and therefore independent of the previously assigned value:

i:=1: ii:=2:
seq(i, i=1..5);
seq(ii, ii=1..5)

                                     1, 2, 3, 4, 5
                                     1, 2, 3, 4, 5


A workaround:

a:=50:
seq(a,ii=1..5);

 

mul, `if`...

Kitonum 21595
May 25 2020
0 1

Example:

i:=3: n:=8:
mul(`if`(k<>i,lambda[i]-lambda[k],1), k=1 .. n);

           

 

Solution...

Kitonum 21595
May 24 2020
0 2


 

> 

with(plots):
fprime_expr:=x^sin(x)*(cos(x)*ln(x)+sin(x)/x);
RootFinding:-Analytic(diff(fprime_expr,x,x)=0, re=0..10, im=-1..1);
L:=sort(select(type,[%],realcons));
A:=plot(fprime_expr, x=0..10, color=blue):
B:=pointplot([seq([l,eval(fprime_expr,x=l)],l=L)], symbol=solidcircle, symbolsize=12, color=red):
display(A,B);

x^sin(x)*(cos(x)*ln(x)+sin(x)/x)

  

4.78206109907178, 3.48122397295652, 2.16436178885414, .748344075670115-.240667154995571*I, .748344075670115+.240667154995571*I, 9.04887132811185, 6.77657190535075, 7.93089684792655

  

[2.16436178885414, 3.48122397295652, 4.78206109907178, 6.77657190535075, 7.93089684792655, 9.04887132811185]

  

 
> 

 


 

Download inflection.mw

Something...

Kitonum 21595
May 24 2020
0 2


 

> 

restart;
f:=t->[3*t,4*t^2+1]:  # Law of movement
P:=eliminate([x=f(t)[1],y=f(t)[2]],t);  # Elimination of the parameter t
P[2][]=0;  # Implicit equation of trajectory
solve(P[2],y)[]; # Explicit equation of trajectory
M:=f(1/2); # Position of the point at t=1/2

[{t = (1/3)*x}, {4*x^2-9*y+9}]

  

4*x^2-9*y+9 = 0

  

y = (4/9)*x^2+1

  

[3/2, 2]

 (1)
> 

V:=eval(diff(f(t),t),t=1/2); # Velocity of the point at t=1/2 as a vector
sqrt(inner(V,V));  # Magnitude of velocity

[3, 4]

  

5

 (2)
> 

and so on


 

Download CM.mw

By a procedure...

Kitonum 21595
May 24 2020
1 1

It's easier and faster to specify your array  A  through a procedure  (i,j) -> a[i,j] :

A:=Array(1..3,1..4,(i,j)->i^j);

 

ListTools...

Kitonum 21595
May 24 2020
0 5

Use  ListTools  instead of  ArrayTools :

restart;
A:=[[0, 4], [1, 3], [2, 2], [3, 1], [4, 0], [0, 3], [1, 2], [2, 1], [3, 0], [0, 2], [1, 1], [2, 0], [0, 1], [1, 0], [0, 0]];
ListTools:-Reverse(A);

   

A := [[0, 4], [1, 3], [2, 2], [3, 1], [4, 0], [0, 3], [1, 2], [2, 1], [3, 0], [0, 2], [1, 1], [2, 0], [0, 1], [1, 0], [0, 0]]

  [[0, 0], [1, 0], [0, 1], [2, 0], [1, 1], [0, 2], [3, 0], [2, 1], [1, 2], [0, 3], [4, 0], [3, 1], [2, 2], [1, 3], [0, 4]]
 

Visualization...

Kitonum 21595
May 24 2020
1 1


 

> 

restart;
alpha:=Pi/6: V[A]:=1: l:=3: f:=0.2: d:=2.5:
# The movement of the body along the segment AB
de:=m*diff(x1(t),t,t)=G*sin(alpha)-F[fr]:
G:=m*g: F[fr]:=m*g*f*cos(alpha): g:=9.81:
ics:=x1(0)=0, D(x1)(0)=V[A]:
de:=evalf(de/m);
B:=rhs(de);
sol:=dsolve({de,ics}, x1(t)); # The solution of de
x1:=unapply(evalf(rhs(sol)),t);
fsolve(x1(t)=l);
tau:=%[2];
V['B']:=D(x1)(tau);
tau:=evalf[3](tau);
 

diff(diff(x1(t), t), t) = 3.205858157

  

3.205858157

  

x1(t) = (3205858157/2000000000)*t^2+t

  

proc (t) options operator, arrow; 1.602929078*t^2+t end proc

  

-1.715094465, 1.091236545

  

1.091236545

  

4.498349578

  

1.09

 (1)
> 

# The movement of the body along the curve BC
sys:=m*diff(x(t),t,t)=0, m*diff(y(t),t,t)=m*g;
ics_sys:=x(0)=0, y(0)=0, D(x)(0)=V['B']*cos(alpha), D(y)(0)=V['B']*sin(alpha):
Sol:=evalf(dsolve({sys, ics_sys},{x(t),y(t)})); # The solution of sys
x:=unapply(eval(x(t),Sol),t); y:=unapply(eval(y(t),Sol),t);
T:=evalf(solve(x(t)=d));
h:=evalf(y(T));
T:=evalf[2](T);
h:=evalf[3](h);   
 

m*(diff(diff(x(t), t), t)) = 0, m*(diff(diff(y(t), t), t)) = 9.81*m

  

{x(t) = 3.895685011*t, y(t) = 4.905000000*t^2+2.249174789*t}

  

proc (t) options operator, arrow; 3.895685011*t end proc

  

proc (t) options operator, arrow; 4.905000000*t^2+2.249174789*t end proc

  

.6417356621

  

3.463375629

  

.64

  

3.46

 (2)


Visualization

> 

with(plots): with(plottools):
bg:=display(curve([[-2.6,-1],[-2.6,-4],[0,-4],[0,-0.1],[3,3*tan(Pi/6)-0.1]], color=black, thickness=2)):
P:=s->piecewise(s>=0 and s<=tau,[(3-x1(s))*cos(Pi/6),(3-x1(s))*sin(Pi/6)],[-x(s-tau),-y(s-tau)]):
Trajectory:=t->display(plot([P(s)[1],P(s)[2],s=0..t],linestyle=3,color=red), pointplot(`if`(t>=0 and t<=tau,[(3-x1(t))*cos(Pi/6),(3-x1(t))*sin(Pi/6)], [-x(t-tau),-y(t-tau)]),symbol=solidcircle, symbolsize=20, color=red)):
animate(Trajectory,[t], t=0..tau+T, background=bg, scaling=constrained, size=[500,500], axes=none);

 
> 

 


Edit.

Download sol_Maple_new1.mw

Calculations in Maple...

Kitonum 21595
May 23 2020
1 1

All calculations are made with 10 significant digits (by default). In the final answers, the number of digits is left as in your document:


 

> 

restart;
alpha:=Pi/6: V[A]:=1: l:=3: f:=0.2: d:=2.5:
# The movement of the body along the segment AB
de:=m*diff(x1(t),t,t)=G*sin(alpha)-F[fr]:
G:=m*g: F[fr]:=m*g*f*cos(alpha): g:=9.81:
ics:=x1(0)=0, D(x1)(0)=V[A]:
de:=evalf(de/m);
B:=rhs(de);
sol:=dsolve({de,ics}, x1(t)); # The solution of de
fsolve(rhs(sol)=l);
tau:=evalf[3](%[2]);
V['B']:=B*tau+V[A];
 

diff(diff(x1(t), t), t) = 3.205858157

  

3.205858157

  

x1(t) = (3205858157/2000000000)*t^2+t

  

-1.715094465, 1.091236545

  

1.09

  

4.494385391

 (1)
> 

# The movement of the body along the curve BC
sys:=m*diff(x(t),t,t)=0, m*diff(y(t),t,t)=m*g;
ics_sys:=x(0)=0, y(0)=0, D(x)(0)=V['B']*cos(alpha), D(y)(0)=V['B']*sin(alpha):
Sol:=evalf(dsolve({sys, ics_sys})); # The solution of sys
T:=evalf(solve(eval(x(t),Sol)=d));
h:=evalf(eval(eval(y(t),Sol),t=T));
T:=evalf[2](T);
h:=evalf[3](h);   
 

m*(diff(diff(x(t), t), t)) = 0, m*(diff(diff(y(t), t), t)) = 9.81*m

  

{x(t) = 3.892251925*t, y(t) = 4.905000000*t^2+2.247192696*t}

  

.6423016927

  

3.466940605

  

.64

  

3.47

 (2)
> 

 


 

Download sol_Maple.mw

Or (shorter)...

Kitonum 21595
May 22 2020
0 0

use 2-argument  arctan :

arctan(cos(beta),sin(beta)) assuming beta>0, beta<Pi/2;

                                    


See help on the  arctan  function. The two-argument  actan(y,x)  is a very useful function because it for a point  (x,y)  returns its polar angle in the range  -Pi<phi<=Pi  , for example  arctan(-1, -sqrt(3)) = -5*Pi/6 . 

Use Matrix instead of linalg[matrix]...

Kitonum 21595
May 20 2020
0 1

Your assignments do not work, probably due to the fact that you are using the deprecated command  linalg[matrix] . See

D__CoR := linalg[matrix](4, 4, [1, 0, 0, x(t), 0, 1, 0, 0, 0, 0, 1, z(t), 0, 0, 0, 1]);
whattype(D__CoR);
D__CoR := Matrix(4, 4, [1, 0, 0, x(t), 0, 1, 0, 0, 0, 0, 1, z(t), 0, 0, 0, 1]);
whattype(D__CoR);

                        

 

 

CurveFitting:-PolynomialInterpolation...

Kitonum 21595
May 19 2020
0 8


For work, it is convenient to set  lambda  as a function of  x :

> 

restart
> 

lambda := proc (x) options operator, arrow; (-1.565845910+2.393779704*x^2+1.564996800*x^4+1.800900000*x^6)/(x^2+2)^4 end proc

proc (x) options operator, arrow; (-1.565845910+2.393779704*x^2+1.564996800*x^4+1.800900000*x^6)/(x^2+2)^4 end proc

 (1)
> 

data := [seq([x, lambda(x)], x = 0 .. 1, .1)]

[[0, -0.9786536938e-1], [.1, -0.9445602702e-1], [.2, -0.8473253124e-1], [.3, -0.7004169612e-1], [.4, -0.5215959052e-1], [.5, -0.3283205351e-1], [.6, -0.1345044703e-1], [.7, 0.5065808511e-2], [.8, 0.2221891338e-1], [.9, 0.3780341321e-1], [1.0, 0.5177568635e-1]]

 (2)
> 

P:=CurveFitting:-PolynomialInterpolation(data, x);

-0.6488921957e-3*x^10-.1643827022*x^9+.8708102427*x^8-1.816961080*x^7+1.747871420*x^6-.4949702068*x^5-.3193961523*x^4-0.1962083349e-1*x^3+.3469960911*x^2-0.5683119e-4*x-0.9786536938e-1

 (3)
> 

plot([lambda(x),P], x=-0.5..1.5, color=[blue,red]);

 
> 

 


We see that on the segment  [0,1] the graphs practically coincide, but outside the segment the approximation is much worse.

Download help_new.mw

allsolutions, convert(... , piecewise)...

Kitonum 21595
May 19 2020
2 4

We can easily get the result as  piecewise  function:


 

> 

restart;
int(cos(j*t)*cos(k*t), t = -Pi..Pi, allsolutions) assuming posint:
convert(%, piecewise, j);

piecewise(j = k, Pi, 0)

 (1)
> 

 


 

Download int.mw

Sum...

Kitonum 21595
May 18 2020
0 1

If you want an undisclosed sum, then use  Sum  instead of  sum :

E:=(2*h^2*Sum(x[i]^2,i=1..4))*alpha+(2*h^2*Sum(x[i]^2*y[i],i=1..4))*beta+2*h*Sum(x[i]^2,i=1..4)-2*h*Sum(x[i]*x[i+1],i=1..4);
map(t->2*t,collect(E/2, h));

    

 

Visualization and simplified answer...

Kitonum 21595
May 17 2020
0 0

The visualization shows that Maple returns the correct result, which can be simplified (see the last line of the code):


 

NULL

> 

restart;
f := 1-4*(x-1)*(1+y)/(x^2*(1-y)):
solve(f>0, useassumptions) assuming  1<=x, x<=4, 0<y,y<1;
y:=x->solve(f, y);
A:=plot3d(f, x=1..4, y=0..1):
B:=plots:-spacecurve([x,y(x),0], x=1..4, color=red, thickness=3):
C:=plot3d(0, x=1..4, y=0..1, color=yellow, transparency=0.5):
plots:-display(A, B, C, view=[0..4,0..1,-5..5], axes=normal);
plots:-inequal([{0 < y, 1 <= x, x < 2, y < (x^2-4*x+4)/(x^2+4*x-4)}, {0 < y, 2 < x, x <= 4, y < (x^2-4*x+4)/(x^2+4*x-4)}], x=1..4, y=0..1, color=green, nolines); # Visualization of the answer
{y>0, 1 <= x, x < 2, y < (x^2-4*x+4)/(x^2+4*x-4)},{0 < y, 2 < x, x <= 4, y < (x^2-4*x+4)/(x^2+4*x-4)};  # This is the answer (the domain is the union of 2 flat regions)

{x = 1, 0 < y, y < 1}, {0 < y, 1 < x, x < 2, y < (x^2-4*x+4)/(x^2+4*x-4)}, {0 < y, 2 < x, x < 4, y < (x^2-4*x+4)/(x^2+4*x-4)}, {x = 4, 0 < y, y < 1/7}

  

proc (x) options operator, arrow; solve(f, y) end proc

  

 

 

{1 <= x, 0 < y, x < 2, y < (x^2-4*x+4)/(x^2+4*x-4)}, {x <= 4, 0 < y, 2 < x, y < (x^2-4*x+4)/(x^2+4*x-4)}

 (1)

``


 

Download positive_function_new2.mw

2 ways...

Kitonum 21595
May 17 2020
1 0

This can be done in many ways. Here are 2 ways:


 

Example: Using a Formula

 

Height of a Cylinder:

> 

ex6v := 600

> 

ex6r := 4

> 

ex6h := ex6v/(Pi*ex6r^2); plot3d([0, ex6h], x = -4 .. 4, y = -sqrt(4^2-x^2) .. sqrt(4^2-x^2), style = surface, filled); plots:-display(plottools:-cylinder([0, 0, 0], ex6r, ex6h, style = surface, strips = 100))

(75/2)/Pi

 

 

 
> 

NULL


 

Download heightofcylinder-new.mw

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