Advanced Search

Kitonum

20084 Reputation

26 Badges

17 years, 29 days
Contact Kitonum

MaplePrimes Activity


These are answers submitted by Kitonum

Procedure...

Kitonum 20084
June 04 2020
1 1


Here is a procedure for this:

> 

restart;
P:=proc(k)
local n, k1, k2;
n:=2; k1:=0; k2:=0;
while n<10^k do
if isprime(10*n+1) then k2:=k2+1 else k1:=k1+1 fi;
n:=nextprime(n);
od;
[k1,k2];
end proc:

Examples of use

> 

seq(P(k), k=1..6);

[2, 2], [18, 7], [131, 37], [1021, 208], [8240, 1352], [69231, 9267]

 (1)
> 

plot([[seq([k,P(k)[1]],k=1..5)],[seq([k,P(k)[2]],k=1..5)]], color=[red,blue]);

 
> 

 

 

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...

Kitonum 20084
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...

Kitonum 20084
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]);

zo := [8/3-2*I]

  

ze := [1]

  

Zo := [8/3, -2]

  

Ze := [2, 2]

  

ex := -6*x+14

  

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...

Kitonum 20084
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...

Kitonum 20084
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 20084
May 25 2020
0 1

Example:

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

           

 

Solution...

Kitonum 20084
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 20084
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 20084
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 20084
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 20084
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 20084
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 20084
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 20084
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 20084
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

First 47 48 49 50 51 52 53 Last Page 49 of 280