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allvalues...

Kitonum 21435
February 28 2020
1 11

RootOf  is a way of representing the roots of an equation when explicit formulas are too cumbersome or do not exist at all. The  index=...  indicates the ordinal number of the root if there are several of these roots. To get the roots explicitly, you can use  allvalues  command. But since your equation also contains the parameter  l , in order to get all the roots in explicit form, you need to specify the parameter value:

A:=RootOf(6*_Z^3+(27+3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2))*_Z^2+(3*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^4*RootOf(_Z^2*l^2+3*_Z^4-3)^2-9*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^2+90*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2-18*l^4+6*l^6*RootOf(_Z^2*l^2+3*_Z^4-3)^2-81+45*RootOf(_Z^2*l^2+3*_Z^4-3)^2*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2))*_Z-324-3*l^8+l^10*RootOf(_Z^2*l^2+3*_Z^4-3)^2+108*RootOf(_Z^2*l^2+3*_Z^4-3)^2*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)-3*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^6+sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^8*RootOf(_Z^2*l^2+3*_Z^4-3)^2-63*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^2+30*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^4*RootOf(_Z^2*l^2+3*_Z^4-3)^2+45*l^6*RootOf(_Z^2*l^2+3*_Z^4-3)^2+351*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2-108*l^4, index = 1):

allvalues(RootOf(_Z^2*l^2+3*_Z^4-3));

eval(A, l=1);
allvalues(%);

             

Indexed variables in 1d math...

Kitonum 21435
February 27 2020
2 0 
restart;
f[a,alpha,beta,k];
f[ax,k];
f[c,0,d];

                                            

 

 

Solutions...

Kitonum 21435
February 26 2020
1 1

The numerical solution is better and more accurate than graphical one.
The wage  y  is a function of the sales  x .

# Solution a
# Setting of this function
y:=x->2000+0.1*x;
y(1480);
                   Output:                   2148.0

# Solution b
solve(y(x)=2225, x);
                   Output:                    2250.

 

Visualization...

Kitonum 21435
February 25 2020
1 2

This integral  (=1/24)  is equal to the volume of the following body:

plot3d(x*y, x=0..1, y=0..1-x, style=surface, filled);

                          

 

Syntax...

Kitonum 21435
February 25 2020
2 0

Should be  sin(2*x)  instead of  sin2x . This is just a symbol . The general rule is that the argument of any function must be surrounded by parentheses.

The result in vector form...

Kitonum 21435
February 24 2020
0 1

You can immediately get the result in vector form using  LinearAlgebra:-LinearSolve  command as in the following example:

Sys:={3*x+5*y=21, -2*x+y=-1}:
LinearAlgebra:-LinearSolve(<3,5; -2,1>, <21,-1>);

 

fnormal, simplify(..., zero)...

Kitonum 21435
February 24 2020
1 0

We get the symbolic result containing the imaginary unit  I , but in fact the result is a real number:

Digits:=20:
int(x^5/(exp(x)-1)/(1-exp(-x)), x=0..1);
evalf(%);
fnormal(%,19);
simplify(%, zero);

    

 

fnormal, simplify(...,zero)...

Kitonum 21435
February 24 2020
0 0

We get the symbolic result containing the imaginary unit  I , but in fact this is a real number:

Digits:=20:
int(x^5/(exp(x)-1)/(1-exp(-x)), x=0..1);
evalf(%);
fnormal(%,19);
simplify(%,zero);

                      

 

Solution...

Kitonum 21435
February 24 2020
0 0

First, we get general formulas for the coordinates of the vertices of such a quadrilateral. The vertex A lies at the origin, B lies in the upper half-plane, C to the right of A on the Ox-axis, D lies in the lower half-plane. We assume that  |AB|=a, |BC|=b, |AD|=c, |DC|=b+c-a . We can always inscribe a circle in the quadrilateral  ABCD . It remains for us to find the position of point C so that the sum of the angles  ABC  and  ADC  is 180 degrees.  For all plottings and checking the property of three points lying on one straight line (see the original post), we select specific parameter values  a=5, b=11, c=8 .

 

> 

restart;
local D;
d:=b+c-a:
# e is the first coordinate of the point C, e=|AC|
Eq1:=e^2=a^2+b^2-2*a*b*t:
Eq2:=e^2=c^2+d^2-2*c*d*s:
e:=simplify(solve(solve(Eq1,t)=-solve(Eq2,s), e)[1]);
p1:=(a+b+e)/2: p2:=(c+d+e)/2:

S:=proc(a,b,c)
local p;
p:=(a+b+c)/2;
sqrt(p*(p-a)*(p-b)*(p-c));
end proc:
Dist:=(A,B)->sqrt((A[1]-B[1])^2+(A[2]-B[2])^2):

h1:=simplify(2*S(a,b,e)/e): h2:=simplify(2*S(c,d,e)/e):
# The coordinates of all the points A, B, C, D 
A:=[0,0]; B:=simplify([sqrt(a^2-h1^2),h1]); C:=[e,0]; D:=simplify([sqrt(c^2-h2^2),-h2]);

(((b-c)*a+c*(b+c))*((b-c)*a-b*(b+c))*(a^2+(-b-c)*a-b*c))^(1/2)/(c^2+(-a+b)*c+a*b)

  

[0, 0]

  

[((a^2*b-a^2*c-a*b^2+a*c^2+b^2*c+b*c^2)^2*a^2/((a*b-a*c+b*c+c^2)*(a*b-a*c-b^2-b*c)*(a^2-a*b-a*c-b*c)))^(1/2), (1/2)*(c^2+(-a+b)*c+a*b)*(-((((b-c)*a+c*(b+c))*((b-c)*a-b*(b+c))*(a^2+(-b-c)*a-b*c))^(1/2)+(a+b)*(c^2+(-a+b)*c+a*b))*(-(((b-c)*a+c*(b+c))*((b-c)*a-b*(b+c))*(a^2+(-b-c)*a-b*c))^(1/2)+(c^2+(-a+b)*c+a*b)*(a-b))*(-(((b-c)*a+c*(b+c))*((b-c)*a-b*(b+c))*(a^2+(-b-c)*a-b*c))^(1/2)+(a+b)*(c^2+(-a+b)*c+a*b))*((((b-c)*a+c*(b+c))*((b-c)*a-b*(b+c))*(a^2+(-b-c)*a-b*c))^(1/2)+(c^2+(-a+b)*c+a*b)*(a-b))/(c^2+(-a+b)*c+a*b)^4)^(1/2)/(((b-c)*a+c*(b+c))*((b-c)*a-b*(b+c))*(a^2+(-b-c)*a-b*c))^(1/2)]

  

[(((b-c)*a+c*(b+c))*((b-c)*a-b*(b+c))*(a^2+(-b-c)*a-b*c))^(1/2)/(c^2+(-a+b)*c+a*b), 0]

  

[((a^2*b+a^2*c-a*b^2-4*a*b*c-a*c^2+b^2*c+b*c^2)^2*c^2/((a*b-a*c+b*c+c^2)*(a*b-a*c-b^2-b*c)*(a^2-a*b-a*c-b*c)))^(1/2), -(1/2)*(c^2+(-a+b)*c+a*b)*(-(-(((b-c)*a+c*(b+c))*((b-c)*a-b*(b+c))*(a^2+(-b-c)*a-b*c))^(1/2)+(c^2+(-a+b)*c+a*b)*(a-b))*((((b-c)*a+c*(b+c))*((b-c)*a-b*(b+c))*(a^2+(-b-c)*a-b*c))^(1/2)+(a-b-2*c)*(c^2+(-a+b)*c+a*b))*((((b-c)*a+c*(b+c))*((b-c)*a-b*(b+c))*(a^2+(-b-c)*a-b*c))^(1/2)+(c^2+(-a+b)*c+a*b)*(a-b))*(-(((b-c)*a+c*(b+c))*((b-c)*a-b*(b+c))*(a^2+(-b-c)*a-b*c))^(1/2)+(a-b-2*c)*(c^2+(-a+b)*c+a*b))/(c^2+(-a+b)*c+a*b)^4)^(1/2)/(((b-c)*a+c*(b+c))*((b-c)*a-b*(b+c))*(a^2+(-b-c)*a-b*c))^(1/2)]

 (1)
> 

a:=5: b:=11: c:=8:
e;
simplify([A,B,C,D]);

(1/167)*5118884^(1/2)

  

[[0, 0], [(3655/1279721)*1279721^(1/2), (220/1279721)*1279721^(1/2)*385^(1/2)], [(2/167)*1279721^(1/2), 0], [(2152/1279721)*1279721^(1/2), -(448/1279721)*1279721^(1/2)*385^(1/2)]]

 (2)
> 

E:=[e/(a+b)*a,0]; F:=[e/(c+d)*c,0];

[(5/2672)*5118884^(1/2), 0]

  

[(4/1837)*5118884^(1/2), 0]

 (3)
> 

O1:=simplify(eval([x,y],solve({(y-B[2])/(E[2]-B[2])=(x-B[1])/(E[1]-B[1]), (y-D[2])/(F[2]-D[2])=(x-D[1])/(F[1]-D[1])})));
r:=radnormal(2*S(a,Dist(A,O1),Dist(B,O1))/a);

[(1000/250667)*1279721^(1/2), -(12/250667)*1279721^(1/2)*385^(1/2)]

  

(4/19)*385^(1/2)

 (4)
> 

P:=simplify((A+B)/2); Q:=simplify((B+C)/2);
O2:=simplify(eval([x,y],solve({(B-A)[1]*(x-P[1])+(B-A)[2]*(y-P[2]),(B-C)[1]*(x-Q[1])+(B-C)[2]*(y-Q[2])})));
R:=Dist(A,O2);

[(3655/2559442)*1279721^(1/2), (110/1279721)*1279721^(1/2)*385^(1/2)]

  

[(18981/2559442)*1279721^(1/2), (110/1279721)*1279721^(1/2)*385^(1/2)]

  

[(1/167)*1279721^(1/2), -(57/514360)*1279721^(1/2)*385^(1/2)]

  

(1/3080)*492692585^(1/2)

 (5)
> 

S:=simplify(eval([x,y],solve({y=0,(y-B[2])/(D[2]-B[2])=(x-B[1])/(D[1]-B[1])})));

[(40/16199)*1279721^(1/2), 0]

 (6)
> 

# All the plottings
with(plots): with(plottools):
L1:=line(B,D,color=brown):
L2:=line(S,O2,color=green):
T:=textplot([[A[],"A"],[B[],"B"],[C[],"C"],[D[],"D"],[O1[],"O1"],[O2[],"O2"],[S[],"S"]],align=left,font=[times,18]):
PP:=pointplot([O1,O2,S],color=[red,blue,brown],symbol=solidcircle):
display(plot([A,B,C,D,A]),circle(O1,r,color=red),circle(O2,R,color=blue),PP,L1,L2,T, scaling=constrained, size=[500,500]);

 
> 

 


 

Download problem.mw

  

Ways...

Kitonum 21435
February 23 2020
2 2 
# Without parenthesis
`%+`(seq(k,k=1..5))%/5;

# Or with parenthesis
`%+`(seq(k,k=1..5))/5;

# Or
1%/5*`%+`(seq(k,k=1..5));

                                

 

 

Adjustment...

Kitonum 21435
February 23 2020
0 1

In fact, after calculating the derivatives, we get a linear system:

NULL

> 

restart
> 

Jestar := .222222222222222222222222222223*x[1]+3.98408963290776835548901841189+.222222222222222222222222222223*u[1]^2+1.20728063844142489971746563410*u[1]+.555555555555555555555555555557*x[2]+.555555555555555555555555555557*u[2]^2+2.19736562704190744843616437593*u[2]+.222222222222222222222222222218*x[3]+.222222222222222222222222222218*u[3]^2+.894156749261387141869195542552*u[3]+lambda[1]*(1.2*10^(-29)*x[1]-2.2*10^(-29)*x[2]+7.8*10^(-30)*x[3])+lambda[2]*(-1.61439242336161127378313801260*x[1]+.45420549105153639573117683365*x[2]+.16018693231007487805196117895*x[3]-1.*u[2])+lambda[3]*(-.95012749286786619995420043248*x[1]-.51534877135614818507764030330*x[2]+.46547626422401438503184073577*x[3]-1.*u[3])

.222222222222222222222222222223*x[1]+3.98408963290776835548901841189+.222222222222222222222222222223*u[1]^2+1.20728063844142489971746563410*u[1]+.555555555555555555555555555557*x[2]+.555555555555555555555555555557*u[2]^2+2.19736562704190744843616437593*u[2]+.222222222222222222222222222218*x[3]+.222222222222222222222222222218*u[3]^2+.894156749261387141869195542552*u[3]+lambda[1]*(0.1200000000e-28*x[1]-0.2200000000e-28*x[2]+0.7800000000e-29*x[3])+lambda[2]*(-1.61439242336161127378313801260*x[1]+.45420549105153639573117683365*x[2]+.16018693231007487805196117895*x[3]-1.*u[2])+lambda[3]*(-.95012749286786619995420043248*x[1]-.51534877135614818507764030330*x[2]+.46547626422401438503184073577*x[3]-1.*u[3])

 (1)
> 

Sys := {seq(diff(Jestar, lambda[i]) = 0, i = 1 .. 3), seq(diff(Jestar, u[i]) = 0, i = 1 .. 3), seq(diff(Jestar, x[i]) = 0, i = 1 .. 3)}

{.4444444444*u[1]+1.20728063844142489971746563410 = 0, 1.111111111*u[2]+2.19736562704190744843616437593-1.*lambda[2] = 0, .4444444444*u[3]+.894156749261387141869195542552-1.*lambda[3] = 0, 0.1200000000e-28*x[1]-0.2200000000e-28*x[2]+0.7800000000e-29*x[3] = 0, .222222222222222222222222222218+0.7800000000e-29*lambda[1]+.16018693231007487805196117895*lambda[2]+.46547626422401438503184073577*lambda[3] = 0, .222222222222222222222222222223+0.1200000000e-28*lambda[1]-1.61439242336161127378313801260*lambda[2]-.95012749286786619995420043248*lambda[3] = 0, .555555555555555555555555555557-0.2200000000e-28*lambda[1]+.45420549105153639573117683365*lambda[2]-.51534877135614818507764030330*lambda[3] = 0, -1.61439242336161127378313801260*x[1]+.45420549105153639573117683365*x[2]+.16018693231007487805196117895*x[3]-1.*u[2] = 0, -.95012749286786619995420043248*x[1]-.51534877135614818507764030330*x[2]+.46547626422401438503184073577*x[3]-1.*u[3] = 0}

 (2)
> 

solve(Sys)

{lambda[1] = -0.3405136588e31, lambda[2] = -73.32501333, lambda[3] = 81.81631382, u[1] = -2.716381437, u[2] = -67.97014107, u[3] = 182.0748534, x[1] = -8572.737031, x[2] = -17682.28255, x[3] = -36684.27842}

 (3)
> 

NULL

NULL


 

Download example-new.mw

multiplication sign...

Kitonum 21435
February 22 2020
0 1

y = x*(x^4 - 10*x^2 + 39)/30

Statistics:-NonlinearFit...

Kitonum 21435
February 22 2020
1 0

For best results, I replaced the constant  2  with a third parameter  a3 . We specify the segment  [a,b]  and the step  h  to obtain data  X  and  Y  and then use the  Statistics:-NonlinearFit  command. When reducing the segment  [a,b] , we get better results:
 

> 

restart;
a, b:=0, 5:
f1:=x->(2*x)/(3+5*x):  
f2:=x->a1*(a3-exp(-a2*x)):
h:=0.1: n:=(b-a)/h;
X:=[seq(i*h,i=0..n)];
Y:=[seq(f1(i*h),i=0..n)];
F:=Statistics:-NonlinearFit(f2(x), X, Y, x);
plot([f1(x),F], x=0..5, color=[blue,red],legend=["f1","f2"]);

50.00000000

  

[0., .1, .2, .3, .4, .5, .6, .7, .8, .9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 3.0, 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 3.7, 3.8, 3.9, 4.0, 4.1, 4.2, 4.3, 4.4, 4.5, 4.6, 4.7, 4.8, 4.9, 5.0]

  

[0., 0.5714285714e-1, .1000000000, .1333333333, .1600000000, .1818181818, .2000000000, .2153846154, .2285714286, .2400000000, .2500000000, .2588235294, .2666666666, .2736842106, .2800000000, .2857142858, .2909090910, .2956521740, .3000000000, .3040000000, .3076923076, .3111111112, .3142857142, .3172413794, .3200000000, .3225806452, .3250000000, .3272727272, .3294117648, .3314285714, .3333333334, .3351351352, .3368421052, .3384615384, .3400000000, .3414634146, .3428571428, .3441860466, .3454545454, .3466666666, .3478260870, .3489361702, .3500000000, .3510204082, .3520000000, .3529411764, .3538461538, .3547169812, .3555555556, .3563636364, .3571428572]

  

HFloat(0.3473297773329595)-HFloat(0.31980563612041574)*exp(-HFloat(1.1646240098951652)*x)

  

 
> 

a, b:=0, 2:
h:=0.1: n:=(b-a)/h;
X:=[seq(i*h,i=0..n)];
Y:=[seq(f1(i*h),i=0..n)];
F:=Statistics:-NonlinearFit(f2(x), X, Y, x);
plot([f1(x),F], x=0..2, color=[blue,red],legend=["f1","f2"]);

20.00000000

  

[0., .1, .2, .3, .4, .5, .6, .7, .8, .9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0]

  

[0., 0.5714285714e-1, .1000000000, .1333333333, .1600000000, .1818181818, .2000000000, .2153846154, .2285714286, .2400000000, .2500000000, .2588235294, .2666666666, .2736842106, .2800000000, .2857142858, .2909090910, .2956521740, .3000000000, .3040000000, .3076923076]

  

HFloat(0.31271326321044796)-HFloat(0.30377027346591473)*exp(-HFloat(1.6468916879789395)*x)

  

 
> 

 


Edit.

Download nonlinfit2.mw

plot...

Kitonum 21435
February 21 2020
1 0

Here is another way. The first and second points almost coincide, so they block one another:

restart;
plot([[[28,.6481496576]],[[28, .648149657615473]],[[28, .6512873548]]],style='point',color=["Blue","Orange","Red"], symbol=solidcircle, symbolsize=12, labels = ["k", "y(k)"], legend = ["10-digit precision", "15-digit precision", "Floating-point iteration"] ,legendstyle = [font = ["HELVETICA", 9], location = right]);

                             

 

Single quotes...

Kitonum 21435
February 21 2020
0 2

Use single quotes for single use (in 2d math only):

'a >= b'

                                  

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