Kitonum

In those cases when isolve does not cope with the problem, brute force method (i.e. the method of enumerating all possible options) can often be successfully used. In your 4-squares example, assuming the variables are positive, it is obvious that each of the variables can vary in the range of 1 .. 12 . Nested for-loops are often used for this method, but I like nested sequences more. We get 76 solutions, but many of them are obtained simply by rearranging one from the other. But a slight modification of this method allows you to immediately get 5 unique solutions.

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remove~(has, A, 0);...

Answered: Kitonum 21525

June 04 2020

1 2

restart;
A := [[1, 0, 9], [5, 0, 6], [4, 0, 0]]; 
remove~(has, A, 0);

[[1, 9], [5, 6], [4]]

Procedure...

Answered: Kitonum 21525

June 04 2020

1 1

Here is a procedure for this:

Addition. It is interesting that if we set the logarithmic scale along the vertical axis, then we get 2 lines close to straight lines:

plots:-logplot([[seq([k,P(k)[1]],k=1..5)],[seq([k,P(k)[2]],k=1..5)]], color=[red,blue]);

Edit.

Download count1.mw

Squaring Equations...

Answered: Kitonum 21525

June 03 2020

2 0

The appearance of extraneous roots when squaring both sides of the equation is in no way related to the presence of square roots in the original equation. Consider the simplest equation x = 2 with an obvious root 2 . But if you square it, you get an equation x^2 = 4 that has 2 roots 2 and -2 . In the general case, let us have the equation f(x) = g(x) and square it. Obviously, the resulting equation f(x)^2 = g(x)^2 is equivalent to the equation (f(x) - g(x))*(f(x) + g(x)) = 0 . The roots of the last equation are the union of the sets of the roots of the original equation f(x) = g(x) and the roots of the equation f(x) = - g(x) , which may have roots that are not the roots of the original equation.

Adjustment...

Answered: Kitonum 21525

June 02 2020

0 1

The function f , by means of which the equation of the blue line is specifed, is given incorrectly. Below is the corrected version:

>

restart;
with(plots):
unprotect(O);
H := [84/37, 14/37]:
f := x-> -6*x+14:
a := 3: A := [a, f(a)]:
O:=[0,0]: zo := [8/3+I*f(8/3)];
ze := [2+I^(eval(diff(f(x), x), x = 2))];
Zo := [8/3, f(8/3)]; Ze := [2, f(2)];
ex := -6*x+14; V := `<,>`(ze-zo);
V := plottools:-arrow(Zo, Ze, .2, .4, .2, color = "Red"):
DR := plot(ex, x = -1 .. 3, color = blue):
Points := pointplot([A[],Zo[],Ze[],H[]], symbol = solidcircle, color = red, symbolsize = 10):
T := textplot([[A[], "A"],[H[],"H"],[O[],"O"]], font = [times, 18], align = {below, right}):
display([DR,V,Points, T], scaling = constrained, size=[800,800]);

Vector[column](%id = 18446746163433603070)

>

Explanation. The direction vector for the blue line is <-2/3, 4> (in fact, you specified it with a complex number v = -2/3+4*I ). Therefore, the slope of this line is k = 4 / (- 2/3) = - 6 and the equation of this line is y=-4 - 6*(x-3) or y=-6*x + 14

Download 1.mw

Adjustment...

Answered: Kitonum 21525

May 29 2020

1 0

I made 3 corrections in your procedure. Now it is working properly. It is only important to note that this construction List1 := [op(List1), p] and so on is extremely inefficient. Below is the same procedure (named FF ), but much faster (of course for large N ):

restart;
F := proc(N)
local p, List1, List3, List5, List7;
List1 := [];
List3 := [];
List5 := [];
List7 := [];
for p from 1 to N do
if isprime(p) and p mod 8 = 1 then List1 := [op(List1), p]; end if;
if isprime(p) and p mod 8 = 3 then List3 := [op(List3), p]; end if;
if isprime(p) and p mod 8 = 5 then List5 := [op(List5), p]; end if;
if isprime(p) and p mod 8 = 7 then List7 := [op(List7), p]; end if;
[nops(List1), nops(List3), nops(List5), nops(List7)];
end do;
end proc:

F(100);
[5, 7, 6, 6]

restart;
FF := proc(N)
local p, k1, k3, k5, k7;
k1:=0; k3:=0; k5:=0; k7:=0;
for p from 1 to N do
if isprime(p) then if p mod 8 = 1 then k1:=k1+1  elif
p mod 8 = 3 then k3:=k3+1  elif
p mod 8 = 5 then k5:=k5+1  elif
p mod 8 = 7 then k7:=k7+1 fi; fi;
[k1,k3,k5,k7];
end do;
end proc:

FF(100);
[5, 7, 6, 6]

Edit.

local...

Answered: Kitonum 21525

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`...

Answered: Kitonum 21525

May 25 2020

0 1

Example:

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

Solution...

Answered: Kitonum 21525

May 24 2020

0 2

Download inflection.mw

Something...

Answered: Kitonum 21525

May 24 2020

0 2

>	restart; f:=t->[3t,4t^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

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

(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

(2)

>

and so on

Download CM.mw

By a procedure...

Answered: Kitonum 21525

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...

Answered: Kitonum 21525

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...

Answered: Kitonum 21525

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

(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

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

(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...

Answered: Kitonum 21525

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

(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);