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Ronan

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12 years, 214 days
East Grinstead, United Kingdom
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These are answers submitted by Ronan

align and \n...

Ronan 1207
September 22 2024
2 1

Use align and \n (within the text) as an option in textplot or textplot3d. This is just a quick example.

with(plots);with(plottools)
A:=[0,0,0];C[1]:=[1,1,1]
plt1:=point([A,C[1]],symbol=solidcircle,symbolsize=16,color=blue)
plt2:=textplot3d([ [A[],"A",align={above,right}] , [C[1][],typeset("C[1]=\n",C[1]),align={below,left}] ]):

display(plt1,plt2)

Like this...

Ronan 1207
September 17 2024
0 0

One way of doing it.

> 

restart
> 

> 

local D;
> 

BD:=4;
AB:=6;

D

  

4

  

6

 (1)
> 

eq1:=BD*sin(4*theta)=AB*sin(2*theta)

4*sin(4*theta) = 6*sin(2*theta)

 (2)
> 

sol:=solve(eq1,[theta])

[[theta = 0], [theta = (1/2)*Pi], [theta = (1/2)*arctan((1/3)*7^(1/2))], [theta = -(1/2)*arctan((1/3)*7^(1/2))]]

 (3)
> 

assign(sol[3]);
theta:=simplify(theta)

(1/4)*arctan(3*7^(1/2))

 (4)
> 

eq2:=AB*sin(2*theta)=BC*sin(theta)

6*sin((1/2)*arctan(3*7^(1/2))) = BC*sin((1/4)*arctan(3*7^(1/2)))

 (5)
> 

#eq3:=eval(eq2,(sol[3]))
> 

 

> 

simplify(solve(eq2,[BC])[])

[BC = 12*cos((1/4)*arctan(3*7^(1/2)))]

 (6)
> 

assign(%)
> 

BC

12*cos((1/4)*arctan(3*7^(1/2)))

 (7)
> 

DC:=BD*cos(Pi-4*theta)+BC*cos(theta)

-1/2+12*cos((1/4)*arctan(3*7^(1/2)))^2

 (8)
> 

simplify( (8) );

10

 (9)
 

 

Download 2024-09-18_A_Find_DC_geometry.mw

Like this....

Ronan 1207
September 17 2024
1 0

180-2m+60+k = 2n+40+k  implies m+n=100  so m=100-n

m+40+n+x=180

100-n+40+n+x=180

therefore x=40

Doing it visually...

Ronan 1207
September 13 2024
3 2

Looking at it.ABC is half the area and ABE is quater the area. Draw a vertical line from E intersects the diagonal midpoint at say G. Triangle ABF and EGF are similar. with a 4;1 area ratio. So the horizontal coordinate of F is 2/3 AE. also ABF+EGF is quater of the total area.

The triangle EGC area.  is 1/8 of the total area so 3cm^2.

ABF+2* EGF would be 6 (quater of the total area) 

EGF would be 1/6 of that =1. So  total area of 4.

Suggestion:-One Drive or Google docs...

Ronan 1207
August 20 2024
0 3

Could you put the zip file on One Drive Google Docs or Dropbox for example and have a link to it it in your post.

A proof...

Ronan 1207
June 24 2024
0 9

I got curious. I have converted away from the geometry package and used a rational parameterisation for cos and sin

[a cos(t), b sin(t)]  to  [a* (1t^2)/(1+t^2), b*2*t/(1+t^2)]   . In this case t is not the angle.

This proves the angle is constant upto sign changes for signum . But that just changes the reported angle from sfor example 50deg to 130deg.

 

I would like to know the name of this theorem.

 

> 

restart;
with(plots):with(LinearAlgebra):

 
> 

x0 := 100;
y0 := 40;
a := 7;
b := 5;
c := sqrt(a^2 - b^2);
e1:= x^2/a^2 + y^2/b^2 - 1;
F1:=[ -c, 0];
F2:=[ c, 0];
eq := simplify((a^2 - x0^2)*(y - y0)^2 + (b^2 - y0^2)*(x - x0)^2 + 2*x0*y0*(x - x0)*(y - y0)) = 0;
sol := solve({eq}, {y});
tang1:=(lhs-rhs)~(op(sol[1]));

tang2:= (lhs-rhs)~(op(sol[2]));
sol2 := op(solve({op(sol[1]), x^2/a^2 + y^2/b^2 - 1 }, {x, y}));
xM2 := rhs(sol2[1]);
yM2 := rhs(sol2[2]);
A:=[ xM2, yM2];
sol3 := op(solve({op(sol[2]), x^2/a^2 + y^2/b^2 - 1 }, {x, y}));
xM3 := rhs(sol3[1]);
yM3 := rhs(sol3[2]);
B:=[ xM3, yM3];

slpvecAB:=<(B[2]-A[2]),B[1]-A[1]>;  #slpvec
Pol:=-slpvecAB[1]*(x-A[1])+slpvecAB[2]*(y-A[2]);
simplify(Pol);
isolate(%, y);

TANG := proc(t) local xM, yM; xM :=a*(1-t^2)/(1+t^2);# a*cos(t);
               yM :=b*2*t/(1+t^2);# b*sin(t);
               return(expand(1/49*x*xM + 1/25*y*yM - 1 ));
        end proc;

#t := -0.25;

TT:= TANG(t);
#Equation(TT);
M:=[a*(1-t^2)/(1+t^2),b*2*t/(1+t^2)];#[ a*cos(t), b*sin(t)];

100

  

40

  

7

  

5

  

2*6^(1/2)

  

(1/49)*x^2+(1/25)*y^2-1

  

[-2*6^(1/2), 0]

  

[2*6^(1/2), 0]

  

-1575*x^2+(8000*y-5000)*x-9951*y^2-3920*y+328400 = 0

  

{y = (4000/9951)*x-1960/9951+(5/9951)*13087^(1/2)*x-(500/9951)*13087^(1/2)}, {y = (4000/9951)*x-1960/9951-(5/9951)*13087^(1/2)*x+(500/9951)*13087^(1/2)}

  

y-(4000/9951)*x+1960/9951-(5/9951)*13087^(1/2)*x+(500/9951)*13087^(1/2)

  

y-(4000/9951)*x+1960/9951+(5/9951)*13087^(1/2)*x-(500/9951)*13087^(1/2)

  

x = (49/1642)*13087^(1/2)+1225/3284, y = -(125/3284)*13087^(1/2)+245/1642

  

(49/1642)*13087^(1/2)+1225/3284

  

-(125/3284)*13087^(1/2)+245/1642

  

[(49/1642)*13087^(1/2)+1225/3284, -(125/3284)*13087^(1/2)+245/1642]

  

x = -(49/1642)*13087^(1/2)+1225/3284, y = (125/3284)*13087^(1/2)+245/1642

  

-(49/1642)*13087^(1/2)+1225/3284

  

(125/3284)*13087^(1/2)+245/1642

  

[-(49/1642)*13087^(1/2)+1225/3284, (125/3284)*13087^(1/2)+245/1642]

  

slpvecAB := Vector(2, {(1) = (125/1642)*13087^(1/2), (2) = -(49/821)*13087^(1/2)})

  

-(125/1642)*13087^(1/2)*(-(49/1642)*13087^(1/2)+x-1225/3284)-(49/821)*13087^(1/2)*(y+(125/3284)*13087^(1/2)-245/1642)

  

-(1/6568)*13087^(1/2)*(500*x-245+392*y)

  

y = -(125/98)*x+5/8

  

proc (t) local xM, yM; xM := a*(1-t^2)/(1+t^2); yM := 2*b*t/(1+t^2); return expand((1/49)*x*xM+(1/25)*y*yM-1) end proc

  

-(1/7)*x*t^2/(t^2+1)+(1/7)*x/(t^2+1)+(2/5)*y*t/(t^2+1)-1

  

[7*(-t^2+1)/(t^2+1), 10*t/(t^2+1)]

 (1)
> 

 

> 

P:=rhs~(solve( [tang1, TT],[x,y])[]);

[-7*(93*t+56+13087^(1/2))/(93*t-56-13087^(1/2)), -(5/107)*(56+13087^(1/2))*(107*t-13087^(1/2)+56)/(93*t-56-13087^(1/2))]

 (2)
> 

Q:=rhs~(solve( [tang2, TT],[x,y])[]);

[-7*(93*t+56-13087^(1/2))/(93*t-56+13087^(1/2)), (5/107)*(-56+13087^(1/2))*(107*t+13087^(1/2)+56)/(93*t-56+13087^(1/2))]

 (3)
> 

slpvecPF2:=simplify(<(P[2]-F2[2]),P[1]-F2[1]>);
PF2:=simplify(-slpvecPF2[1]*(x-P[1])+slpvecPF2[2]*(y-P[2]));
> 

 

slpvecPF2 := Vector(2, {(1) = -5*(56+13087^(1/2))*(-107*t+13087^(1/2)-56)/(-9951*t+5992+107*13087^(1/2)), (2) = ((-186*t+2*13087^(1/2)+112)*6^(1/2)-651*t-7*13087^(1/2)-392)/(93*t-56-13087^(1/2))})

  

(2*(465+(-5*t+y)*13087^(1/2)+(-93*y-280)*t+56*y)*6^(1/2)+(5*t*x-7*y)*13087^(1/2)+7*(40*x-93*y)*t-465*x-392*y)/(93*t-56-13087^(1/2))

 (4)
> 

slpvecQF2:=simplify(<(Q[2]-F2[2]),Q[1]-F2[1]>);

QF2:=simplify(-slpvecQF2[1]*(x-Q[1])+slpvecQF2[2]*(y-Q[2]));

slpvecQF2 := Vector(2, {(1) = 5*(-56+13087^(1/2))*(107*t+13087^(1/2)+56)/(9951*t-5992+107*13087^(1/2)), (2) = ((-2*6^(1/2)+7)*13087^(1/2)+(-186*t+112)*6^(1/2)-651*t-392)/(93*t-56+13087^(1/2))})

  

(((10*t-2*y)*13087^(1/2)+(-186*y-560)*t+112*y+930)*6^(1/2)+(-5*t*x+7*y)*13087^(1/2)+(280*x-651*y)*t-465*x-392*y)/(93*t-56+13087^(1/2))

 (5)
> 

 

> 

 

> 

alpha :=(simplify(VectorAngle(slpvecQF2,slpvecPF2),assume=real))

(1/2)*Pi-signum(93*t-56+13087^(1/2))*signum(93*t-56-13087^(1/2))*arcsin((93/14)*(1302*6^(1/2)*t^2+2057*t^2+1498*6^(1/2)-2743)/(((-100*t^2+112*6^(1/2)-292)*13087^(1/2)+8649*6^(1/2)*t^2+37034*t^2-16223*6^(1/2)+50018)^(1/2)*((100*t^2-112*6^(1/2)+292)*13087^(1/2)+8649*6^(1/2)*t^2+37034*t^2-16223*6^(1/2)+50018)^(1/2)))

 (6)
> 

Setof:=frontend(indets,[alpha])

{Pi, arcsin((93/14)*(1302*6^(1/2)*t^2+2057*t^2+1498*6^(1/2)-2743)/(((-100*t^2+112*6^(1/2)-292)*13087^(1/2)+8649*6^(1/2)*t^2+37034*t^2-16223*6^(1/2)+50018)^(1/2)*((100*t^2-112*6^(1/2)+292)*13087^(1/2)+8649*6^(1/2)*t^2+37034*t^2-16223*6^(1/2)+50018)^(1/2))), signum(93*t-56-13087^(1/2)), signum(93*t-56+13087^(1/2))}

 (7)
> 

(sin(Setof[2]))^2

(8649/196)*(1302*6^(1/2)*t^2+2057*t^2+1498*6^(1/2)-2743)^2/(((-100*t^2+112*6^(1/2)-292)*13087^(1/2)+8649*6^(1/2)*t^2+37034*t^2-16223*6^(1/2)+50018)*((100*t^2-112*6^(1/2)+292)*13087^(1/2)+8649*6^(1/2)*t^2+37034*t^2-16223*6^(1/2)+50018))

 (8)
> 

simpascn:=arcsin(sqrt(factor( (8) ))); # the t's eliminate

arcsin((1/58693852)*(1451954478752954-68582717957500*6^(1/2))^(1/2))

 (9)
> 

 alpha1:=(subs(Setof[2]=simpascn,alpha))*180/Pi

180*((1/2)*Pi-signum(93*t-56+13087^(1/2))*signum(93*t-56-13087^(1/2))*arcsin((1/58693852)*(1451954478752954-68582717957500*6^(1/2))^(1/2)))/Pi

 (10)
> 

plot(alpha1,t=-10..10,view=[-10..10,0..180])

 
> 

 
> 

 

> 

 

> 

 


 

Download 2024-06-24_constant_angle.mw

One possible way...

Ronan 1207
June 15 2024
2 0

 Hope this helps. There are probably more efficient ways. 

restart

N := 5

5``

 (1)

for n to N do M || n := Matrix(n); if n <> 1 then for i to n do for j to n+1-i do M || n[i, j] := f[i+j-1]-add(M || n[i-a, j+a-1], a = i-1 .. 1, -1) end do end do else M1[1] := f__1 end if; print(M || n) end do

Matrix(%id = 36893491047679352452)

 (2)

``NULL

Download 2024-06-15_A_Binonm_coeffs.mw

How about using the Backup...

Ronan 1207
May 15 2024
0 5

Click in Restore Backup and list the files by date saved,

Use fsolve...

Ronan 1207
May 13 2024
1 0

The set of equations; as far as I can see, cannot be solved explicitly do to the trig. functions.
I rearranged your equations so you only need to solve eq5 and eq6 for y and z.

I randomly made up values for  t, T, a,c,e,f just to show the use of fsolve. 

Hope this helps you.

Edit I re-attached the worksheet because I had a mistake in it.


 

> 

restart
> 

t:=Pi/2;
T:=t;

(1/2)*Pi

  

(1/2)*Pi

 (1)
> 

a:=3;c:=2;e:=4;f:=2

3

  

2

  

4

  

2

 (2)
> 

#z:=1
> 

 w := sqrt(a^2 - (a - y)^2);
v := cot(t + arcsin(w/a)) ;
 u := (sqrt(c^2 - (c - z)^2) - v*x - w)/x^2;
 x := y + z + e;
eq5 := v + cot(T + arcsin((u*x^2 + v*x + w)/c)) + 2*u*x = 0;
eq6 :=f = Pi/30*(6*u^2*x^5 + 15*u*v*x^4 + (20*u*w + 10*v^2)*x^3 + 30*v*w*x^2 + 30*x*w^2) - Pi/3*(-z^3 + 3*z^2*c - y^3 + 3*y^2*a);

(9-(3-y)^2)^(1/2)

  

-(9-(3-y)^2)^(1/2)/((3-y)^2)^(1/2)

  

((4-(2-z)^2)^(1/2)+(9-(3-y)^2)^(1/2)*x/((3-y)^2)^(1/2)-(9-(3-y)^2)^(1/2))/x^2

  

y+z+4

  

-(9-(3-y)^2)^(1/2)/((3-y)^2)^(1/2)-(4-(2-z)^2)^(1/2)/((2-z)^2)^(1/2)+2*((4-(2-z)^2)^(1/2)+(9-(3-y)^2)^(1/2)*(y+z+4)/((3-y)^2)^(1/2)-(9-(3-y)^2)^(1/2))/(y+z+4) = 0

  

2 = (1/30)*Pi*(6*((4-(2-z)^2)^(1/2)+(9-(3-y)^2)^(1/2)*(y+z+4)/((3-y)^2)^(1/2)-(9-(3-y)^2)^(1/2))^2*(y+z+4)-15*((4-(2-z)^2)^(1/2)+(9-(3-y)^2)^(1/2)*(y+z+4)/((3-y)^2)^(1/2)-(9-(3-y)^2)^(1/2))*(y+z+4)^2*(9-(3-y)^2)^(1/2)/((3-y)^2)^(1/2)+(20*((4-(2-z)^2)^(1/2)+(9-(3-y)^2)^(1/2)*(y+z+4)/((3-y)^2)^(1/2)-(9-(3-y)^2)^(1/2))*(9-(3-y)^2)^(1/2)/(y+z+4)^2+10*(9-(3-y)^2)/(3-y)^2)*(y+z+4)^3-30*(9-(3-y)^2)*(y+z+4)^2/((3-y)^2)^(1/2)+30*(y+z+4)*(9-(3-y)^2))-(1/3)*Pi*(-y^3-z^3+9*y^2+6*z^2)

 (3)
> 

 fsolve({ eq5,eq6},complex);

{y = -0.5302494488e-1+0.8720213598e-1*I, z = .1023238140-.3150972913*I}

 (4)
> 

 


 

Download 2024-05-13_A_An_approach.mw

One way...

Ronan 1207
April 01 2024
0 0 
f:=x-> (1-k*x)/(1+x^2)
f := proc (x) options operator, arrow; (1-k*x)/(1+x^2) end proc

slpf:=diff(f(x),x)
                           k      2 (-k x + 1) x
                slp := - ------ - --------------
                          2                 2   
                         x  + 1     / 2    \    
                                    \x  + 1/    

m:=eval(slpf,x=3);# slope of tangent line at x=3

                              2      3 
                         m := -- k - --
                              25     50

f(3)
                            3      1 
                          - -- k + --
                            10     10

line:=unapply(m*x+c,x)
 line := proc (x) options operator, arrow; ((2/25)*k-3/50)*x+c 

    end proc




line(3)=f(3);# x=3 is tangent point
                  6      9          3      1 
                  -- k - -- + c = - -- k + --
                  25     50         10     10

c:=solve(line(3)=f(3),c)
                               27     7 
                        c := - -- k + --
                               50     25

line(x)
                   /2      3 \     27     7 
                   |-- k - --| x - -- k + --
                   \25     50/     50     25

 

Check under Options Display...

Ronan 1207
March 24 2024
3 1

Check under Tools Options Display Typesetting level to see if it is set to extended.  

It looks like I have it working correctl...

Ronan 1207
March 23 2024
0 0

@dharr  I appear to have the setup working in principal. I created seperate worksheets. One for the Main Package and one for the Sub Package. Both attached.  I have hilighted the differences in both in red. Then I copied both help files across to the correct location. They now list correctly in Help. 

 

New_RT_Edit_Dadabase_2024.mw

New_UHG_Edit_Dadabase_2024.mw

One way...

Ronan 1207
March 20 2024
1 2

Interesting problem.  I found this method in the RealDomain help page.

restart;

n:=3;m:=2;
eqx:=x^(n/m)=a;
use RealDomain in (maple_sol:=solve(eqx,[x]))[] end use;  #also tried solve()
F:=map(X->eval(eqx,X),maple_sol);
map(X->evalb(X),F);
                             n := 3

                             m := 2

                               (3/2)    
                       eqx := x      = a

                          [     (2/3)]
                          [x = a     ]

                          F := [a = a]

                             [true]

A way....

Ronan 1207
March 14 2024
2 2

This is rather similar to the 1st question I asked on MaplePrimes. I just happened to look at it today

Dual sum loop with a condition - MaplePrimes  

> 

restart
 
> 

for n from 1 to 10 do
add(add(add(`if`(i+j+k<=n, 1/(i*j*k),0),i=1..n),j=1..n),k=1..n);
end do;

0

  

0

  

1

  

5/2

  

17/4

  

49/8

  

967/120

  

801/80

  

4523/378

  

84095/6048

 (1)
> 

 

 

Download A_2024-03-15_Sum_with_Condition.mw

Would this do?...

Ronan 1207
March 11 2024
1 1

This is a way using seq

M := Matrix(4, 4, [[m[1, 1], m[1, 2], m[1, 3], m[1, 4]], 
                   [m[2, 1], m[2, 2], m[2, 3], m[2, 4]],
                   [m[3, 1], m[3, 2], m[3, 3], m[3, 4]],
                   [m[4, 1], m[4, 2], m[4, 3], m[4, 4]]]);

L:=[seq(M[1 .. 4, i], i = 1 .. 4)];

whattype(L[3]);

 

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