Advanced Search

Kitonum

21435 Reputation

26 Badges

17 years, 24 days
Contact Kitonum

MaplePrimes Activity


These are answers submitted by Kitonum

expand...

Kitonum 21435
October 05 2020
0 1

Use the  expand  command:


 

> 

restart;
> 

eq_4_34 := x__alpha = C*x__a - C/2 * (x__b + x__c);

x__alpha = C*x__a-(1/2)*C*(x__b+x__c)

 (1)
> 

eq_4_33 := x__0 = 1/3*(x__a + x__b + x__c);

x__0 = (1/3)*x__a+(1/3)*x__b+(1/3)*x__c

 (2)
> 

eq_4_33aux := X__sub = (x__b + x__c);

X__sub = x__b+x__c

 (3)
> 

attempt2 := algsubs((x__b + x__c) = X__sub, eq_4_33);

x__0 = (1/3)*x__a+(1/3)*X__sub

 (4)
> 

attempt3 := X__sub = solve(attempt2, X__sub);

X__sub = -x__a+3*x__0

 (5)
> 

eq_4_34x := x__alpha = algsubs((x__b + x__c) = 3*x__0 - x__a, rhs(eq_4_34));

x__alpha = -(1/2)*C*(-x__a+3*x__0)+C*x__a

 (6)
> 

eq_4_35 := x__a = simplify(solve(eq_4_34x, x__a));

x__a = (1/3)*(3*C*x__0+2*x__alpha)/C

 (7)
> 

expand(eq_4_35);
desired:= x__a = x__0 + (2*x__alpha)/(3*C);

x__a = x__0+(2/3)*x__alpha/C

  

x__a = x__0+(2/3)*x__alpha/C

 (8)
> 

 


 

Download Q20201005_2_new.mw

evalc, plots:-implicitplot...

Kitonum 21435
October 04 2020
5 0 
restart;
z:=x+I*y:
F:=evalc(abs(z+1)*abs(z-1)=1);
 plots:-implicitplot(F, x=-3..3, y=-3..3, gridrefine=3, scaling=constrained, size=[600,300]);

                   

     

Some options...

Kitonum 21435
October 03 2020
1 0

1. To calculate derivatives at specific points, it is better to use the differentiation operator  D  than a combination of  diff  with  subs.
2. For a better perception of what this paraboloid looks like, it is necessary to reduce the ranges on all 3 axes.
3. Calculation confirms that the vertex of this paraboloid is (0, 0, -6).

restart;
f:=(x,y)->2*x^2+3*y^2-x*y-6;
D[1](f)(1,1); # The partial derivative by x in the point (1,1)
D[2](f)(1,1); # The partial derivative by y in the point (1,1)
p1:=plot3d(f(x,y), x=-3..3, y=-3..3, style=patchcontour, view=-7..6):
p2:=plots:-spacecurve([x,1,f(x,1)],x=-3..3, color=red, thickness=3):
p3:=plots:-spacecurve([1,y,f(1,y)],y=-3..3, color=blue, thickness=3):
plots:-display(p1,p2,p3, axes=normal);
minimize(f(x,y), x=-3..3, y=-3..3, location);

             

           

combine, assuming...

Kitonum 21435
October 01 2020
1 1

Download Q20201001_new.mw
 

> 

restart;
> 

Given_eq_3_37 := M = k_ * sqrt(L__1 * L__2);

M = k_*(L__1*L__2)^(1/2)

 (1)
> 

Given_eq_3_30 := k_ = sqrt((L__m)^2 / (L__1 * L__2p));

k_ = (L__m^2/(L__1*L__2p))^(1/2)

 (2)
> 

Given_eq_3_37a := combine(subs([Given_eq_3_30], Given_eq_3_37)) assuming L__1>0, L__2>0;

M = (L__2*L__m^2/L__2p)^(1/2)

 (3)
> 

 


 

Download Q20201001_new.mw

 

free = ......

Kitonum 21435
September 30 2020
2 0

Use the option  free=...  , for example:


 

A := Matrix(4, 4, {(1, 1) = 1, (1, 2) = 3, (1, 3) = 4.5, (1, 4) = 7, (2, 1) = 0, (2, 2) = 2, (2, 3) = 5, (2, 4) = 9, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0, (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 0}); B := Vector(4, {(1) = 0, (2) = 0, (3) = 0, (4) = 0}); C := LinearAlgebra[LinearSolve](A, B, free = 't')

Vector[column](%id = 18446746219697584838)

 (1)

W1 := C[1]; W2 := C[2]; W3 := C[3]; W4 := C[4]; Q := W1*x^4+W2*x^3+W3*x^2+W1

(HFloat(3.0)*t[2]+HFloat(6.5)*t[1])*x^4+(-HFloat(2.5)*t[2]-HFloat(4.5)*t[1])*x^3+t[2]*x^2+HFloat(3.0)*t[2]+HFloat(6.5)*t[1]

 (2)

``

NULL


 

Download MultipleTCoefficients_new.mw

 

Adjustment...

Kitonum 21435
September 30 2020
0 0

In your worksheet the function  F  was not defined correctly. Here's the revised version:


 

> 

restart
> 

with(LinearAlgebra)
> 

F := proc (x) options operator, arrow; x^2 end proc
> 

F(u(t))

u(t)^2

 (1)
> 

for n from 0 while n <= 6 do V[n] := (diff(F(sum(t^i*u[i], i = 0 .. n)), [`$`(t, n)]))/factorial(n) end do
> 

t := 0

0

 (2)
> 

for i from 0 while i <= n-1 do A[i] := V[i] end do

u[0]^2

  

2*u[0]*u[1]

  

2*u[0]*u[2]+u[1]^2

  

2*u[0]*u[3]+2*u[1]*u[2]

  

2*u[0]*u[4]+2*u[1]*u[3]+u[2]^2

  

2*u[0]*u[5]+2*u[1]*u[4]+2*u[2]*u[3]

  

2*u[0]*u[6]+2*u[1]*u[5]+2*u[2]*u[4]+u[3]^2

 (3)
> 

NULL
> 

``


 

Download Adomian_Polynomials_new.mw

Something...

Kitonum 21435
September 27 2020
0 0

Perhaps for your purposes it is sufficient that the index does not take on any values, but from some finite set, for example from  {1, 2, ... ,100} :

for k from 1 to 100 do
alias(F[k]=F[k](x));
od:

whattype(F[1]);
whattype(F[33]);

 

plots:-inequal...

Kitonum 21435
September 27 2020
1 0

The easiest way to do this is with the  plots:-inequal  command:

restart; 
with(plots): 
inequal({x-y-1 <= 0, 0.5*y^2-x-3 <= 0}, x = -4 .. 6, y = -3 .. 5, color = red);

 

plots:-textplot...

Kitonum 21435
September 26 2020
2 0 
restart;
ode:=[diff(x(t),t) = 2*x(t)-y(t), diff(y(t),t) = 3*x(t)-2*y(t)]:
p1:=DEtools:-dfieldplot(ode,[x(t),y(t)],t=-2..2,x=-4..4, y=-4..4,arrows=SLIM,color=coral,labels=[``,``]):
p2:=plot(3*x,x=-4..4,y=-4..4,color=magenta):
p3:=plot(x,x=-4..4,y=-4..4,color=blue):
T:=plots:-textplot([[4.5,-0.2,x],[-0.2,4.5,y]], font=[times,16]):
plots:-display(p1,p2,p3,T, view=[-4.6..4.6,-4.6..4.6], size=[600,600]);

                  

Adjustment...

Kitonum 21435
September 26 2020
1 2


 

> 

restart;

B:=<<39,76,151,301,601>|<7.71E-8,5.43E-9,3.55E-10,2.11E-11,1.32E-12>|
        <26,51,101,201,401>|<6.46E-3,1.17E-4,1.88E-6,2.96E-8,4.46E-10>|
        <26,51,101,201,401>|<2.74E-4,6.34E-6,1.16E-7,1.85E-9,2.92E-11>|
        <26,51,101,201,401>|<6.48E-4,4.39E-5,2.99E-6,1.88E-7,1.18E-8>>;

for i from 1 to 5 do
   B[i, 2] := log[10](B[i, 2]):                                          
   B[i, 4] := log[10](B[i, 4]):                 
   B[i, 6] := log[10](B[i, 6]):                 
   B[i, 8] := log[10](B[i, 8]):        
       

   
end do:  # computing the log of the max-error
B: # This is the table of values we'll plot.
local log:
T:=plot([B[..,[1, 2]],B[1..1,[1, 2]], B[.., [3, 4]],B[1..1,[3, 4]],
         B[..,[5, 6]],B[1..1,[5, 6]],B[.., [7, 8]],B[1..1,[7, 8]]],
        legend = ["","BFFM","", "BHT","", "BHTRKNM", "", "BNM"],
        #title="Efficiency Curve for Example 1",
        style = ["pointline","point","pointline","point","pointline","point","pointline","point"],
        symbolsize = 15,axes = framed,
        symbol = [box,box, circle,circle,diamond,diamond,solidcircle, solidcircle],
        color=[red, red, blue,blue, black, black, green, green],
        axis = [gridlines = [colour = green, majorlines = 1,linestyle = dot]],
        labels = ["NFE", log[10](cat(`Max `,Err))]);

Matrix(5, 8, {(1, 1) = 39, (1, 2) = 0.7710000000e-7, (1, 3) = 26, (1, 4) = 0.646e-2, (1, 5) = 26, (1, 6) = 0.274e-3, (1, 7) = 26, (1, 8) = 0.648e-3, (2, 1) = 76, (2, 2) = 0.5430000000e-8, (2, 3) = 51, (2, 4) = 0.117e-3, (2, 5) = 51, (2, 6) = 0.634e-5, (2, 7) = 51, (2, 8) = 0.439e-4, (3, 1) = 151, (3, 2) = 0.3550000000e-9, (3, 3) = 101, (3, 4) = 0.188e-5, (3, 5) = 101, (3, 6) = 0.1160000000e-6, (3, 7) = 101, (3, 8) = 0.299e-5, (4, 1) = 301, (4, 2) = 0.2110000000e-10, (4, 3) = 201, (4, 4) = 0.2960000000e-7, (4, 5) = 201, (4, 6) = 0.1850000000e-8, (4, 7) = 201, (4, 8) = 0.1880000000e-6, (5, 1) = 601, (5, 2) = 0.1320000000e-11, (5, 3) = 401, (5, 4) = 0.4460000000e-9, (5, 5) = 401, (5, 6) = 0.2920000000e-10, (5, 7) = 401, (5, 8) = 0.1180000000e-7})

  

Warning, A new binding for the name `log` has been created. The global instance of this name is still accessible using the :- prefix, :-`log`.  See ?protect for details.

  

 
> 

 


 

Download Help_data_points_new1.mw

 

Plotting of a vertical line...

Kitonum 21435
September 25 2020
0 0

In the example below, we construct 2 vertical lines (directrixes for an ellipse) in two ways:


 

> 

restart;
a:=5: b:=3:
P:=plots:-implicitplot(x^2/a^2+y^2/b^2=1, x=-a..a, y=-b..b, color=blue):
c:=sqrt(a^2-b^2);
e:=c/a;
L1:=plot([[-a/e,t,t=-4..4],[a/e,t,t=-4..4]], color=red): # The first way
L2:=plots:-implicitplot([x=-a/e, x=a/e], x=-7..7, y=-4..4, color=red): # The second way
plots:-display(P, L1, scaling=constrained, view=[-7..7,-4..4]);
plots:-display(P, L2, scaling=constrained, view=[-7..7,-4..4]);

4

  

4/5

  

 

 
> 

 


Edit. We can also build any straight lines using the plottools:-line  command (the third way).

Download vl-2.mw

D...

Kitonum 21435
September 22 2020
1 0

Example:

restart;
f:=x->x^3;
(D@@2)(f)(x);
(D@@2)(f)(3);

 

The proof of 2) and 3)...

Kitonum 21435
September 22 2020
0 0

Let (x0,y0) is a point on the ellipse E.


 

> 

restart;
eq:=x0*x/a^2+y0*y/b^2=1: # The equation of the tangent in the point(x0,y0) (from wiki)
OT:=solve(eval(eq,y=0),x);
Om:=x0:
Om*OT;

k:=-coeff(lhs(eq),x)/coeff(lhs(eq),y); # The slope of the line eq
eq1:=y-y0=-1/k*(x-x0); # The equation of the normal in the point(x0,y0)
ON:=solve(eval(eq1,y=0),x);
simplify(ON,{a^2-b^2=c^2});
 

a^2/x0

  

a^2

  

-x0*b^2/(a^2*y0)

  

y-y0 = a^2*y0*(x-x0)/(x0*b^2)

  

(a^2-b^2)*x0/a^2

  

c^2*x0/a^2

 (1)
> 

 


Edit.

Download prop_23.mw

 

plottools:-reflect...

Kitonum 21435
September 22 2020
0 0

Or (an example):

restart;
p:=plot(exp(x), x=-2..2);
plottools:-reflect(p, [[0,0],[1,1]]);

 

Procedure for this...

Kitonum 21435
September 21 2020
1 0

Probably the procedures  IntegerPoints (or  IntegerPoints1) from the post  https://www.mapleprimes.com/posts/202437-Integer-Points-In-Polyhedral-Regions  will be useful for solving your problems.

First 50 51 52 53 54 55 56 Last Page 52 of 289