Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

I am trying to find a download for the ONEOptimal package for Maple (http://iopscience.iop.org/article/10.1088/0253-6102/61/2/03). I read the article but there is no download link to the file. Does anyone know where I can download the package?

I have a system of equations e.g.

A^2+B*A+C=0

where A,B,C are Matrices and I want to solve for A.

Sure I can write every equations in brakets [..=0], but isn'T it possible to just use the matrix notation?

I've been trying to numerically solve (and plot) this equation. Maple tells me that some matrix is singular - I have no idea, what I can do.

eq := diff(y(x), `$`(x, 3))-(diff(y(x), x))*(diff(y(x), x))+1 = 0;

cond := (D(y))(0) = 0, (D(y))(1) = 1, ((D@@2)(y))(0) = 0

de := dsolve({cond, eq}, y(x), numeric);

Error, (in dsolve/numeric/bvp) matrix is singular

How I can found stabilty of system by Routh, Jury, Liapunov, Nequist on maple?

Hi Maple community

I'm running an algorithm where a non-linear equation system must be solved, in this case is a 26x26 system.

After 16116 succesful previous computations, fsolve stops giving me results.
I checked why and I was first expecting that, for some reason, the 26x26 system had an error and I ended with something like 25x26 or vice versa. But that was not the case.

So I tried the command solve and it not only worked fine but also gave me two results, but I only need one. I guess I could check for the wrong solution and discard it, but I still wondering why fsolve is failing and if there is anything to help fsolve not to fail.

These are the set of equations if somebody wants to check them:

EQ[16117][1] := W[1, 16117]*(-0.3860115660e-1*HRa[1, 16117]-0.1876793978e-1*ga[1, 16117]+0.7836678184e-1) = 2.040147478*10^6*SR[1, 16118], W[1, 16117]*(-0.3915554290e-1*HRa[1, 16117]-0.1903748329e-1*ga[1, 16117]+0.8260795999e-1) = 3.876387504, W[1, 16117]*(-0.1876794098e-1*HRa[1, 16117]-0.9892449327e-2*ga[1, 16117]+0.3810204607e-1) = 2.040147478*10^6*v[1, 16118], HLa[1, 16117] = .9724029753*ga[1, 16117]+HRa[1, 16117], NRa[1, 16117] = 0.7006679273e-1*HRa[1, 16117]-.1803623678*ga[1, 16117]+1.002451672, NLa[1, 16117] = 0.7006679273e-1*HRa[1, 16117]+.2484955248*ga[1, 16117]+1.002451672, SL[2, 16118] = SR[1, 16118], fra[1, 16117] = HRa[1, 16117]-HLa[2, 16117], fra[1, 16117] = .25*NRa[1, 16117]+.25*NLa[2, 16117], ga[1, 16117] = 0.;

EQ[16117][2] := W[2, 16117]*(-0.3860115660e-1*HRa[2, 16117]-0.1876793978e-1*ga[2, 16117]+0.7836678184e-1) = -2.040147478*10^6*SL[2, 16118]+7.152482840, W[2, 16117]*(-0.3915554290e-1*HRa[2, 16117]-0.1903748329e-1*ga[2, 16117]+0.8260795999e-1) = 3.876387504, W[2, 16117]*(-0.1876794098e-1*HRa[2, 16117]-0.9892449327e-2*ga[2, 16117]+0.3810204607e-1) = -1.983845478*10^6*SL[2, 16118]+5.221405977, HLa[2, 16117] = .9724029753*ga[2, 16117]+HRa[2, 16117], NRa[2, 16117] = 0.7006679273e-1*HRa[2, 16117]-.1803623678*ga[2, 16117]+1.002451672, NLa[2, 16117] = 0.7006679273e-1*HRa[2, 16117]+.2484955248*ga[2, 16117]+1.002451672, SL[3, 16118] = 0.3505865589e-5, fra[2, 16117] = HRa[2, 16117]-HLa[3, 16117];

EQ[16117][3] := W[3, 16117]*(-0.3860115660e-1*HRa[3, 16117]-0.1876793978e-1*ga[3, 16117]+0.7836678184e-1) = -2.040147478*10^6*SL[3, 16118]+10.82168541, W[3, 16117]*(-0.3915554290e-1*HRa[3, 16117]-0.1903748329e-1*ga[3, 16117]+0.8260795999e-1) = 3.876387504, W[3, 16117]*(-0.1876794098e-1*HRa[3, 16117]-0.9892449327e-2*ga[3, 16117]+0.3810204607e-1) = -1.983845478*10^6*SL[3, 16118]+8.751240594, HLa[3, 16117] = .9724029753*ga[3, 16117]+HRa[3, 16117], NRa[3, 16117] = 0.7006679273e-1*HRa[3, 16117]-.1803623678*ga[3, 16117]+1.002451672, NLa[3, 16117] = 0.7006679273e-1*HRa[3, 16117]+.2484955248*ga[3, 16117]+1.002451672, SL[4, 16118] = 0.5304364281e-5, fra[3, 16117] = HRa[3, 16117];

And after these the solving command that I used was:

SOL[j]:=fsolve({seq(EQ[j][n],n=1..N)},indets({entries(EQ[j],nolist)},assignable(name)));

Which returns

SOL[j]:=

As I said, then I tried the solve command:

SOL[j]:=solve({seq(EQ[j][n],n=1..N)},indets({entries(EQ[j],nolist)},assignable(name)));

which returns:

SOL[16117] :=

{HLa[1, 16117] = 1.011251860, HLa[2, 16117] = .5007913055, HLa[3, 16117] = -0.4240068535e-1, HRa[1, 16117] = 1.011251860, HRa[2, 16117] = .8728245835, HRa[3, 16117] = .2686716410, NLa[1, 16117] = 1.073306847, NLa[2, 16117] = .9685353734, NLa[3, 16117] = .9417827567, NRa[1, 16117] = 1.073306847, NRa[2, 16117] = 1.132612831, NRa[3, 16117] = 1.078974668, SL[2, 16118] = 0.1737463747e-5, SL[3, 16118] = 0.3505865589e-5, SL[4, 16118] = 0.5304364281e-5, SR[1, 16118] = 0.1737463747e-5, W[1, 16117] = 90.12372195, W[2, 16117] = 69.57451714, W[3, 16117] = 49.58407210, fra[1, 16117] = .5104605550, fra[2, 16117] = .9152252689, fra[3, 16117] = .2686716410, ga[1, 16117] = 0., ga[2, 16117] = -.3825916698, ga[3, 16117] = -.3199006320, v[1, 16118] = 8.447574110*10^(-7)},

{HLa[1, 16117] = 3.043461992, HLa[2, 16117] = 2.386862361, HLa[3, 16117] = -0.4240068535e-1, HRa[1, 16117] = 3.043461992, HRa[2, 16117] = 1.087485894, HRa[3, 16117] = .2686716410, NLa[1, 16117] = 1.215697293, NLa[2, 16117] = 1.410701230, NLa[3, 16117] = .9417827567, NRa[1, 16117] = 1.215697293, NRa[2, 16117] = .8376385519, NRa[3, 16117] = 1.078974668, SL[2, 16118] = 0.2032780481e-5, SL[3, 16118] = 0.3505865589e-5, SL[4, 16118] = 0.5304364281e-5, SR[1, 16118] = 0.2032780481e-5, W[1, 16117] = -106.0268094, W[2, 16117] = 265.7250566, W[3, 16117] = 49.58407210, fra[1, 16117] = .6565996307, fra[2, 16117] = 1.129886580, fra[3, 16117] = .2686716410, ga[1, 16117] = 0., ga[2, 16117] = 1.336253076, ga[3, 16117] = -.3199006320, v[1, 16118] = 9.883410782*10^(-7)}

Thanks in advance for any recommendations and suggestions.
 

Hi,

I did some hypothesis testing exercises and I cross checked the result with Maple. I just used following vectors for an unpaired test

a := [88, 89, 92, 90, 90];
b := [92, 90, 91, 89, 91];

I ended up with the following solution:

HFloat(1.5225682336585966)
HFloat(-3.122568233658591)
for a 0.95 confidence interval.

 

Using

TwoSampleTTest(a, b, 0, confidence = .95, summarize = embed)

and

TwoSampleTTest(a, b, 0, confidence = .975, summarize = embed)

I get following results:

-2.75177 .. 1.15177

-3.13633 .. 1.53633

respectively. I can not explain the discrepancy.

 

Best regards,

Oliver

 

PS:

Maple Code in case files won´t be attached.

 

 

Unpaired t Test
restart;
Unpaired test-test dataset
a := [88, 89, 92, 90, 90];
b := [92, 90, 91, 89, 91];
The se² estimate is given by:
se²=var(a)+var(b)+2*cov(a*b)=var(a)+var(b)
se²=
sigma[a]^2/Na+sigma[b]^2/Nb;
with Na, Nb being the length of vector a and b respectively.
                             2                              2
  sigma[[88, 89, 92, 90, 90]]    sigma[[92, 90, 91, 89, 91]]
  ---------------------------- + ----------------------------
               Na                             Nb             
sigma[a]^2;
 and
sigma[b]^2;
 are approximated by
S[a]^2;
 and
S[b]^2;
                                             2
                  sigma[[88, 89, 92, 90, 90]]
                                             2
                  sigma[[92, 90, 91, 89, 91]]
                                           2
                    S[[88, 89, 92, 90, 90]]
                                           2
                    S[[92, 90, 91, 89, 91]]
with
S[X]^2;
 defined as
S[X]*`²` = (sum(X[i]-(sum(X[j], j = 1 .. N))/N, i = 1 .. N))^2/N;
                                 2
                             S[X]
                                                 2
                      /      /         N       \\
                      |      |       -----     ||
                      |  N   |        \        ||
                      |----- |         )       ||
                      | \    |        /    X[j]||
                      |  )   |       -----     ||
                      | /    |       j = 1     ||
                      |----- |X[i] - ----------||
                      \i = 1 \           N     //
             S[X] ᅡᄇ = ----------------------------
                                   N              
with(Statistics);
Sa := Variance(a);
                   HFloat(2.1999999999999993)
Sb := Variance(b);
                   HFloat(1.3000000000000003)
Now we are ready to do hypothesis testing (0.95).
We have (with k=min(Na,Nb)=5):
C = mean(a)-mean(b); Deviation := t_(alpha/a, k-1)*se(Sa/k-Sb/k);
c := Mean(a)-Mean(b); deviation := 2.776*sqrt((1/5)*Variance(a)+(1/5)*Variance(b));
                  HFloat(-0.7999999999999972)
                   HFloat(2.3225682336585938)
upperlimit := c+deviation; lowerlimit := c-deviation;
                   HFloat(1.5225682336585966)
                   HFloat(-3.122568233658591)

Execution of built in student test
TwoSampleTTest(a, b, 0, confidence = .95, summarize = embed);

 

 

 

 Ibragimova Evelina, the 6th form

 school # 57, Kazan, Russia

The Units Converter

restart:
`Conversion formula`;
d:=l=n*m:
d;

                    Conversion formula
                    l = n m

m - shows how many minor units in one major one (coefficient)
`Coefficient`;
m:=1000;
                   Coefficient
                   m:=1000

n - the number of major units
n:=7/3;
                   n := 7/3

l - the number of minor units
l:=;

The result (the desired value)
solve(d);
                   7000/3
evalf(solve(d));
                   2333.333333

That is : there are 2333.3 (or 7000/3 ) minor units in 7/3 major units .

 

Other example

m - shows how many minor units in one major one (coefficient) 
`Coefficient`;
m:=4200;
                   Coefficient
                   m:=4200

n - the number of major units 
n:=;
                 
l - the number of minor units
l:=100;

                  l:=100

The result (the desired value)
solve(d);
                   1/42
evalf(solve(d));
                   0.02380952381

That is : there are 0.02 (or 1/42) major units in 100 minor units .

 

Another example

m - shows how many minor units in one major one (coefficient) 
`Coefficient`;
m:=;
                   Coefficient

n - the number of major units 
n:=2;

                    n := 2
                 
l - the number of minor units
l:=200;

                  l:=200

The result (the desired value)
solve(d);
                   100
evalf(solve(d));
                  100

That is : Coefficient is 100 .

How can I modify the appearance of the arrowheads on the vectors displayed in phaseportrai? In particular, how can I "fill in" the arrowheads so that the arrowhead is not just an outline?

My code is:

phaseportrait([D(x)(t)=-0.4*x(t)+(0.5+4*x(t))*y(t),D(y)(t)=0.4*x(t)-(4.5+4*x(t))*y(t)],[x(t),y(t)],t=0..100,[[x(0)=1,y(0)=0.0]],x=0..1,y=0..0.1,stepsize=0.01,scaling=UNCONSTRAINED,linecolour=BLACK,dirgrid=[17,17],linestyle=1,arrows=SLIM,axes=BOXED);

Thank you

  The geometry of the triangle
  Romanova Elena,  8 class,  school 57, Kazan, Russia

       Construction of triangle and calculation its angles

       Construction of  bisectors
      
       Construction of medians
      
       Construction of altitudes


> restart:with(geometry):      

The setting of the height of the triandle and let's call it "Т"
> triangle(T,[point(A,4,6),point(B,-3,-5),point(C,-4,8)]);

                                  T

        Construction of the triangle
> draw(T,axes=normal,view=[-8..8,-8..8]);

Construction of the triangle АВС

> draw({T(color=gold,thickness=3)},printtext=true,axes=NONE);     
Calculation of the distance between heights А and В - the length of a side АВ

> d1:=distance(A,B);

                           d1 := sqrt(170)

        
        Calculation of the distance between heights В and С - the length of a side ВС
> d2:=distance(B,C);

                           d2 := sqrt(170)

       The setting of line which passes through two points А and В
> line(l1,[A,B]);

                                  l1

       Display the equation of line l1
> Equation(l1);
> x;
> y;

                         -2 + 11 x - 7 y = 0

        The setting of line which passes through two points А and С
> line(l2,[A,C]);

                                  l2

       Display the equation of line l2
> Equation(l2);
> x;
> y;

                          56 - 2 x - 8 y = 0

         The setting of line which passes through two points В and С
> line(l3,[B,C]);

                                  l3

        Display the equation of line l3
> Equation(l3);
> x;
> y;

                          -44 - 13 x - y = 0

        Check the point А lies on line l1
> IsOnLine(A,l1);

                                 true

        Check the point А lies on line l1
> IsOnLine(B,l1);

                                 true

        Calculation of the andle between lines l1 and l2
> FindAngle(l1,l2);

                              arctan(3)

        The conversion of result to degrees
> b1:=convert(arctan(97/14),degrees);

                                      97
                               arctan(--) degrees
                                      14
                     b1 := 180 ------------------
                                       Pi

        Calculation of decimal value of this angle
> b2:=evalf(b1);

                      b2 := 81.78721981 degrees

        Calculation of the andle between lines l1 and l3
> FindAngle(l1,l3);

                             arctan(3/4)

       The conversion of result to degrees
> b3:=convert(arctan(97/99),degrees);

                                      97
                               arctan(--) degrees
                                      99
                     b3 := 180 ------------------
                                       Pi

        Calculation of decimal value of this angle
> b4:=evalf(b3);

                      b4 := 44.41536947 degrees

       Calculation of the angle between lines l2 and l3
> FindAngle(l2,l3);

                              arctan(3)

       The conversion of  result to degrees
> b5:=convert(arctan(97/71),degrees);

                                      97
                               arctan(--) degrees
                                      71
                     b5 := 180 ------------------
                                       Pi

        Calculation of decimal value of  this angle
> b6:=evalf(b5);

                      b6 := 53.79741070 degrees

        Check the sum of all the angles of the triangle
> b2+b4+b6;

                         180.0000000 degrees

        Analytical information about the point А
> detail(A);
   name of the object: A
   form of the object: point2d
   coordinates of the point: [4, 6]
          Analytical information about the point В
> detail(B);
   name of the object: B
   form of the object: point2d
   coordinates of the point: [-3, -5]
          Analytical information about the point С
> detail(C);
   name of the object: C
   form of the object: point2d
   coordinates of the point: [-4, 8]

   The setting of heights of the triangle points A,B,C and let's call it "Т"

   with(geometry):
> triangle(ABC, [point(A,7,8), point(B,6,-7), point(C,-6,7)]):
        The setting of the bisector of angle А in triandle АВС
> bisector(bA, A, ABC);

                                  bA

        Analytical information about the bisector of angle А in the triandle
> detail(bA);
   name of the object: bA
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: (15*170^(1/2)+226^(1/2))*_x+(-13*226^(1/2)-170^(1/2))*_y+97*226^(1/2)-97*170^(1/2) = 0

        Construction of the triangle
> draw(ABC,axes=normal,view=[-8..8,-8..8]);

 Construction of the triangle ABC

> draw({ABC(color=gold,thickness=3)},printtext=true,axes=NONE);     

 Construction of the bisector of angle А

> draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3)},printtext=true,axes=NONE);    

The setting of the bisector of angle В in the triangle АВС

> bisector(bB, B, ABC);

                                  bB

       Analytical information about the bisector of angle B in the triandle
> detail(bB);
   name of the object: bB
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: (-15*340^(1/2)-14*226^(1/2))*_x+(-12*226^(1/2)+340^(1/2))*_y+97*340^(1/2) = 0

         Construction of the bisector of angle В
>draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3),bB(color=red,thickness=3)},printtext=true,axes=NONE);    



    The setting of the bisector of angle С in the triangle АВС

> bisector(bC, C, ABC);

                                  bC

        Analytical information about the bisector of angle С in the triangle
> detail(bC);
   name of the object: bC
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: (14*170^(1/2)-340^(1/2))*_x+(13*340^(1/2)+12*170^(1/2))*_y-97*340^(1/2) = 0

        Construction of the bisector of angle С
>draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3),bB(color=red,thickness=3),bC(color=blue,thickness=3)},printtext=true,axes=NONE);  

 Calculation of the point of intersection of the bisectors and let's call it "О"

> intersection(O,bA,bB,bC);coordinates(O);

                                  O


     7 sqrt(85) - 3 sqrt(2) sqrt(113) + 3 sqrt(85) sqrt(2)
  [2 -----------------------------------------------------,
       sqrt(85) sqrt(2) + sqrt(2) sqrt(113) + 2 sqrt(85)

          -16 sqrt(85) - 7 sqrt(2) sqrt(113) + 7 sqrt(85) sqrt(2)
        - -------------------------------------------------------]
             sqrt(85) sqrt(2) + sqrt(2) sqrt(113) + 2 sqrt(85)

       Construction of the bisectors and  marking of the point of intersection  "О" in the triandle
>draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3),bB(color=red,thickness=3),bC(color=blue,thickness=3),O},printtext=true,axes=NONE);
> restart:
> with(geometry):
       The setting of the heights of the triangle points A,B,C and let's call it "Т"
> point(A,7,8),point(B,6,-7),point(C,-6,7);

                               A, B, C

        Let's call "Т1"
> triangle(T1,[A,B,C]);

                                  T1

        Construction of "Т1"
> draw(T1(color=gold,thickness=3),axes=NONE,printtext=true);
  The setting of the median from the point В in the trianglemedian(mB,B,T1,B1);
> median(mb,B,T1);

                                  mB


                                  mb

        Construction of the median from the point В
> draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mb},printtext=true,axes=NONE);

The setting of the median from the point А in the trianglemedian(mA,A,T1,A1);
> median(ma,A,T1);

                                  mA


                                  ma

        Construction of the median from the point А
>draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mA(color=magenta,thickness=3),ma},printtext=true,axes=NONE);
The setting of the median from the point С in the trianglemedian(mC,C,T1,C1);
> median(mc,C,T1);

                                  mC


                                  mc

        Costruction of the median from the point С
>draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mA(color=magenta,thickness=3),mA,mC(color=maroon,thickness=3)},printtext=true,axes=NONE);




Calculation of the point of  intersection of the median and let's call it "О"

>intersection(O,ma,mb,mC);coordinates(O);

                                  O


                              [7/3, 8/3]

        Construction of medians and marking of the point of  intersection "О" in the triangle
>draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mA(color=magenta,thickness=3),mA,mC(color=violet,thickness=3),O},printtext=true,axes=NONE);
> restart:with(geometry):
> _EnvHorizontalName:=x:_EnvVerticalName=y:       The setting of the heights of the triangle points A, B, C  and let's call it "Т"
> triangle(T,[point(A,7,8),point(B,6,-7),point(C,-6,7)]);

                                  T

       Construction of the triangle
> draw(T,axes=normal,view=[-8..8,-8..8]);


The setting of the altitude in the triangle from the point Сaltitude(hC1,C,T,C1);
> altitude(hC,C,T);

                                 hC1


                                  hC

        Analytical information about the altitude hC from the point С in the triangle
> detail(hC);
   name of the object: hC
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: -99+_x+15*_y = 0

        Construction of the altitude from the point С
> draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hC},printtext=true,axes=NONE);     

  The setting of the altitude in the triangle from the point Аaltitude(hA1,A,T,A1);
> altitude(hA,A,T);

                                 hA1


                                  hA

        Analytical information about the altitude hA from the point А in the triangle
> detail(hA);
   name of the object: hA
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: -28-12*_x+14*_y = 0

        Construction of the altitude from the point А
>draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hA1(color=red,thickness=3),hA1},printtext=true,axes=NONE);       The setting of the altitude from the point В

> altitude(hB1,B,T,B1);
> altitude(hB,B,T);

                                 hB1


                                  hB

        Analytical information about the altitude hB from the point В in the triangle
> detail(hB);
   name of the object: hB
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: -71+13*_x+_y = 0

        Consruction of the altitude from the point В
>draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hA1(color=red,thickness=3),hB1(color=blue,thickness=3),hB1},printtext=true,axes=NONE);     
 Calculation of the point of intersection of altitudes and let's call it "О"

>intersection(O,hB,hA,hC);coordinates(O);

                                  O


                               483  608
                              [---, ---]
                               97   97

        Construction of altitudes and marking of the point of intersection "О" in the triangle
>draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hA1(color=red,thickness=3),hB1(color=blue,thickness=3),hB1,O},printtext=true,axes=NONE);




 

 

 

 

 

 

 

 

 

 

 

 

 

Hi evrey ones in pdsolve we have these commande to use U(x,t) 

> U:= subs(pds:-value(output=listprocedure), u(x,t));

  id like to get du(x,t)/dt

i tried these  

U:= subs(pds:-value(output=listprocedure), du(x,t)/dt);  but is not work 

thank you 

 

Hi

I want to draw  plot y(x) against x but I can't. a, b , _C1 are parameter.

I am unable to solve the attached optimal control problem,please any one who many help  me in guideing .tnx

restart:
unprotect('gamma');

L:=b[1]*c(t)+b[2]*i(t)+w[1]*(u[1])^2/2+w[2]*(u[2])^2/2+w[3]*(u[3])^2/2;
1 2 1 2 1 2
b[1] c(t) + b[2] i(t) + - w[1] u[1] + - w[2] u[2] + - w[3] u[3]
2 2 2
H:=L+lambda[1](t)*((1-p*Psi)*tau+phi* v + delta *r-lambda*(1-u[3])*s-u[1]*varphi*s -mu*s ) +lambda[2](t)*(p*Psi*tau + u[1]*vartheta*s -gamma*lambda* (1-u[3])*v-(mu+phi)*v ) +lambda[3](t)*( (1-u[3])*rho*lambda* (s +gamma*v)+(1-q)* u[2]*eta*i -(mu +beta +chi)*c ) +lambda[4](t)* ((1-rho)*(1-u[3])*lambda*( s +gamma*v) +chi*c - u[2]*eta*i - (mu +alpha )*i) +lambda[5](t)*( beta*c + u[2]*q*eta*i -(mu +delta)*r);
1 2 1 2 1 2
b[1] c(t) + b[2] i(t) + - w[1] u[1] + - w[2] u[2] + - w[3] u[3] + lambda[1](t
2 2 2

) ((1 - p Psi) tau + phi v + delta r - lambda (1 - u[3]) s - u[1] varphi s

- mu s) + lambda[2](t) (p Psi tau + u[1] vartheta s

- gamma lambda (1 - u[3]) v - (mu + phi) v) + lambda[3](t) ((1 - u[3]) rho

lambda (s + gamma v) + (1 - q) u[2] eta i - (mu + beta + chi) c) + lambda[4](t

) ((1 - rho) (1 - u[3]) lambda (s + gamma v) + chi c - u[2] eta i

- (mu + alpha) i) + lambda[5](t) (beta c + u[2] q eta i - (mu + delta) r)
du1:=diff(H,u[1]);

w[1] u[1] - lambda[1](t) varphi s + lambda[2](t) vartheta s
du2:=diff(H,u[2]);du3:=diff(H,u[3]);
w[2] u[2] + lambda[3](t) (1 - q) eta i - lambda[4](t) eta i

+ lambda[5](t) q eta i
w[3] u[3] + lambda[1](t) lambda s + lambda[2](t) gamma lambda v

- lambda[3](t) rho lambda (s + gamma v)

- lambda[4](t) (1 - rho) lambda (s + gamma v)

ddu1 := -A[1] u[1] + psi[1](t) beta x[1] x[3] - psi[2](t) beta x[1] x[3]

ddu2 := -A[2] u[2] - psi[3](t) k x[2]
sol_u1 := solve(du1, u[1]);
s(t) (lambda[1](t) varphi - lambda[2](t) vartheta)
--------------------------------------------------
w[1]
sol_u2 := solve(du2, u[2]);sol_u3 := solve(du3, u[3]);
eta i (-lambda[3](t) + lambda[3](t) q + lambda[4](t) - lambda[5](t) q)
----------------------------------------------------------------------
w[2]
1
---- (lambda (-lambda[1](t) s - lambda[2](t) gamma v + lambda[3](t) rho s
w[3]

+ lambda[3](t) rho gamma v + lambda[4](t) s + lambda[4](t) gamma v

- lambda[4](t) rho s - lambda[4](t) rho gamma v))
Dx2:=subs(u[1]= s*(lambda[1](t)*varphi-lambda[2](t)*vartheta)/w[1] ,u[2]= eta*i*(-lambda[3](t)+lambda[3](t)*q+lambda[4](t)-lambda[5](t)*q)/w[2], u[3]=-lambda*(lambda[1](t)*s+lambda[2](t)*gamma*v-lambda[3](t)*rho*s-lambda[3](t)*rho*gamma*v-lambda[4](t)*s-lambda[4](t)*gamma*v+lambda[4](t)*rho*s+lambda[4](t)*rho*gamma*v)/w[3] ,H );
2 2
s (lambda[1](t) varphi - lambda[2](t) vartheta)
b[1] c(t) + b[2] i(t) + -------------------------------------------------
2 w[1]

2 2 2
eta i (-lambda[3](t) + lambda[3](t) q + lambda[4](t) - lambda[5](t) q)
+ ------------------------------------------------------------------------- +
2 w[2]

1 / 2
------ \lambda (lambda[1](t) s + lambda[2](t) gamma v - lambda[3](t) rho s
2 w[3]

- lambda[3](t) rho gamma v - lambda[4](t) s - lambda[4](t) gamma v

/
\ |
+ lambda[4](t) rho s + lambda[4](t) rho gamma v)^2/ + lambda[1](t) |(1
\

/ 1
- p Psi) tau + phi v + delta r - lambda |1 + ---- (lambda (lambda[1](t) s
\ w[3]

+ lambda[2](t) gamma v - lambda[3](t) rho s - lambda[3](t) rho gamma v

- lambda[4](t) s - lambda[4](t) gamma v + lambda[4](t) rho s

\
+ lambda[4](t) rho gamma v))| s
/

2 \
s (lambda[1](t) varphi - lambda[2](t) vartheta) varphi |
- ------------------------------------------------------- - mu s| +
w[1] /

/
|
lambda[2](t) |p Psi tau
\

2
s (lambda[1](t) varphi - lambda[2](t) vartheta) vartheta /
+ --------------------------------------------------------- - gamma lambda |1 +
w[1] \

1
---- (lambda (lambda[1](t) s + lambda[2](t) gamma v - lambda[3](t) rho s
w[3]

- lambda[3](t) rho gamma v - lambda[4](t) s - lambda[4](t) gamma v

\
\ |
+ lambda[4](t) rho s + lambda[4](t) rho gamma v))| v - (mu + phi) v| +
/ /

// 1
lambda[3](t) ||1 + ---- (lambda (lambda[1](t) s + lambda[2](t) gamma v
\\ w[3]

- lambda[3](t) rho s - lambda[3](t) rho gamma v - lambda[4](t) s

\
- lambda[4](t) gamma v + lambda[4](t) rho s + lambda[4](t) rho gamma v))|
/

1 / 2 2
rho lambda (s + gamma v) + ---- \(1 - q) eta i (-lambda[3](t)
w[2]

\ \
+ lambda[3](t) q + lambda[4](t) - lambda[5](t) q)/ - (mu + beta + chi) c| +
/

/
| / 1
lambda[4](t) |(1 - rho) |1 + ---- (lambda (lambda[1](t) s
\ \ w[3]

+ lambda[2](t) gamma v - lambda[3](t) rho s - lambda[3](t) rho gamma v

- lambda[4](t) s - lambda[4](t) gamma v + lambda[4](t) rho s

\
+ lambda[4](t) rho gamma v))| lambda (s + gamma v) + chi c
/

2 2
eta i (-lambda[3](t) + lambda[3](t) q + lambda[4](t) - lambda[5](t) q)
- ------------------------------------------------------------------------
w[2]

\ /
| |
- (mu + alpha) i| + lambda[5](t) |beta c
/ \

+

2 2
eta i (-lambda[3](t) + lambda[3](t) q + lambda[4](t) - lambda[5](t) q) q
--------------------------------------------------------------------------
w[2]

\
|
- (mu + delta) r|
/
ode1:=diff(lambda[1](t),t)=-diff(H,s);ode2:=diff(lambda[2](t),t)=-diff(H,v);ode3:=diff(psi[3](t),t)=-diff(H,c);ode4:=diff(lambda[4](t),t)=-diff(H,i);ode5:=diff(lambda[5](t),t)=-diff(H,r);
d
--- lambda[1](t) = -lambda[1](t) (-lambda (1 - u[3]) - u[1] varphi - mu)
dt

- lambda[2](t) u[1] vartheta - lambda[3](t) (1 - u[3]) rho lambda

- lambda[4](t) (1 - rho) (1 - u[3]) lambda
d
--- lambda[2](t) = -lambda[1](t) phi
dt

- lambda[2](t) (-gamma lambda (1 - u[3]) - mu - phi)

- lambda[3](t) (1 - u[3]) rho lambda gamma

- lambda[4](t) (1 - rho) (1 - u[3]) lambda gamma
d
--- psi[3](t) = -lambda[3](t) (-mu - beta - chi) - lambda[4](t) chi
dt

- lambda[5](t) beta
d
--- lambda[4](t) = -lambda[3](t) (1 - q) u[2] eta
dt

- lambda[4](t) (-u[2] eta - mu - alpha) - lambda[5](t) u[2] q eta
d
--- lambda[5](t) = -lambda[1](t) delta - lambda[5](t) (-mu - delta)
dt
restart:
#Digits:=10:


unprotect('gamma');
lambda:=0.51:
mu:=0.002:
beta:=0.0115:
delta:=0.003:
alpha:=0.33:
chi:=0.00274:
k:=6.24:
gamma:=0.4:
rho:=0.338:;tau=1000:;Psi:=0.1:;p:=0.6:;phi:=0.001:;eta:=0.001124:q:=0.6:varphi:=0.9:;vatheta:=0.9:
b[1]:=2:;b[2]:=3:;w[1]:=4:;w[2]:=5:;w[3]:=6:
#u[1]:=s(t)*(lambda[1](t)*varphi-lambda[2](t)*vartheta)/w[1]:
#u[2]:=eta*i*(-lambda[3](t)+lambda[3](t)*q+lambda[4](t)-lambda[5](t)*q)/w[2]:;u[3]:=lambda*(-lambda[1](t)*s-lambda[2](t)*gamma*v+lambda[3](t)*rho*s+lambda[3](t)*rho*gamma*v+lambda[4](t)*s+lambda[4](t)*gamma*v-lambda[4](t)*rho*s-lambda[4](t)*rho*gamma*v)/w[3]:
ics := s(0)=8200, v(0)=2800,c(0)=1100,i(0)=1500,r(0)=200,lambda[1](20)=0,lambda[2](20)=0,lambda[3](20)=0,lambda[4](20)=0,lambda[5](20)=0:
ode1:=diff(s(t),t)=(1-p*Psi)*tau+phi* v(t) + delta *r(t)-lambda*(1-u[3])*s(t)-u[1]*varphi*s(t) -mu*s(t),
diff(v(t), t) =p*Psi*tau + u[1]*vartheta*s(t) -gamma*lambda* (1-u[3])*v(t)-(mu+phi)*v(t) ,
diff(c(t), t) =(1-u[3])*rho*lambda* (s(t) +gamma*v(t))+(1-q)* u[2]*eta*i(t) -(mu +beta +chi)*c(t),
diff(i(t), t) =(1-rho)*(1-u[3])*lambda*( s(t) +gamma*v(t)) +chi*c(t) - u[2]*eta*i(t) - (mu +alpha )*i(t),
diff(r(t), t) = beta*c(t) + u[2]*q*eta*i(t) -(mu +delta)*r(t),
diff(lambda[1](t), t) = -lambda[1](t)*(-lambda*(1-u[3])-u[1]*varphi-mu)-lambda[2](t)*u[1]*vartheta-lambda[3](t)*(1-u[3])*rho*lambda-lambda[4](t)*(1-rho)*(1-u[3])*lambda,diff(lambda[2](t),t)=-lambda[1](t)*phi-lambda[2](t)*(-gamma*lambda*(1-u[3])-mu-phi)-lambda[3](t)*(1-u[3])*rho*lambda*gamma-lambda[4](t)*(1-rho)*(1-u[3])*lambda*gamma,diff(lambda[3](t),t)=-lambda[3](t)*(-mu-beta-chi)-lambda[4](t)*chi-lambda[5](t)*beta,diff(lambda[4](t),t)=-lambda[3](t)*(1-q)*u[2]*eta-lambda[4](t)*(-u[2]*eta-mu-alpha)-lambda[5](t)*u[2]*q*eta,diff(lambda[5](t),t)=-lambda[1](t)*delta-lambda[5](t)*(-mu-delta);
d
--- s(t) = (1 - p Psi) tau + phi v(t) + delta r(t) - lambda (1 - u[3]) s(t)
dt

d
- u[1] varphi s(t) - mu s(t), --- v(t) = p Psi tau + u[1] vartheta s(t)
dt

d
- gamma lambda (1 - u[3]) v(t) - (mu + phi) v(t), --- c(t) = (1 - u[3]) rho lambda
dt

(s(t) + gamma v(t)) + (1 - q) u[2] eta - (mu + beta + chi) c(t), 0 = (1

- rho) (1 - u[3]) lambda (s(t) + gamma v(t)) + chi c(t) - u[2] eta - mu

d d
- alpha, --- r(t) = beta c(t) + u[2] q eta - (mu + delta) r(t), ---
dt dt

lambda[1](t) = -lambda[1](t) (-lambda (1 - u[3]) - u[1] varphi - mu)

- lambda[2](t) u[1] vartheta - lambda[3](t) (1 - u[3]) rho lambda

d
- lambda[4](t) (1 - rho) (1 - u[3]) lambda, --- lambda[2](t) =
dt
-lambda[1](t) phi - lambda[2](t) (-gamma lambda (1 - u[3]) - mu - phi)

- lambda[3](t) (1 - u[3]) rho lambda gamma

d
- lambda[4](t) (1 - rho) (1 - u[3]) lambda gamma, --- lambda[3](t) =
dt
d
-lambda[3](t) (-mu - beta - chi) - lambda[4](t) chi - lambda[5](t) beta, ---
dt

lambda[4](t) = -lambda[3](t) (1 - q) u[2] eta

- lambda[4](t) (-u[2] eta - mu - alpha) - lambda[5](t) u[2] q eta,

d
--- lambda[5](t) = -lambda[1](t) delta - lambda[5](t) (-mu - delta)
dt

sol := dsolve({c(0) = 0, i(0) = 0, r(0) = .1, s(0) = 0, v(0) = 0, diff(c(t), t) = (1-u[3])*rho*lambda*(s(t)+gamma*v(t))+(1-q)*u[2]*eta*i(t)-(mu+beta+chi)*c(t), diff(i(t), t) = (1-rho)*(1-u[3])*lambda*(s(t)+gamma*v(t))+chi*c(t)-u[2]*eta*i(t)-(mu+alpha)*i(t), diff(r(t), t) = beta*c(t)+u[2]*q*eta*i(t)-(mu+delta)*r(t), diff(s(t), t) = (1-p*Psi)*tau+phi*v(t)+delta*r(t)-lambda*(1-u[3])*s(t)-u[1]*varphi*s(t)-mu*s(t), diff(v(t), t) = p*Psi*tau+u[1]*vartheta*s(t)-gamma*lambda*(1-u[3])*v(t)-(mu+phi)*v(t), diff(lambda[1](t), t) = -lambda[1](t)*(-lambda*(1-u[3])-u[1]*varphi-mu)-lambda[2](t)*u[1]*vartheta-lambda[3](t)*(1-u[3])*rho*lambda-lambda[4](t)*(1-rho)*(1-u[3])*lambda, diff(lambda[2](t), t) = -lambda[1](t)*phi-lambda[2](t)*(-gamma*lambda*(1-u[3])-mu-phi)-lambda[3](t)*(1-u[3])*rho*lambda*gamma-lambda[4](t)*(1-rho)*(1-u[3])*lambda*gamma, diff(lambda[3](t), t) = -lambda[3](t)*(-mu-beta-chi)-lambda[4](t)*chi-lambda[5](t)*beta, diff(lambda[4](t), t) = -lambda[3](t)*(1-q)*u[2]*eta-lambda[4](t)*(-u[2]*eta-mu-alpha)-lambda[5](t)*u[2]*q*eta, diff(lambda[5](t), t) = -lambda[1](t)*delta-lambda[5](t)*(-mu-delta), lambda[1](20) = 0, lambda[2](20) = 0, lambda[3](20) = 0, lambda[4](20) = 0, lambda[5](20) = 0}, type = numeric);
Error, (in dsolve/numeric/process_input) invalid specification of initial conditions, got 1 = 0

sol:=dsolve([ode1,ics],numeric, method = bvp[midrich],maxmesh=500);

Error, (in dsolve/numeric/process_input) system must be entered as a set/list of expressions/equations

dsolve[':-interactive']({});
Error, `:=` unexpected
sol:=dsolve([ode1,ics],numeric, method = bvp[midrich],maxmesh=500);
Error, (in dsolve/numeric/process_input) system must be entered as a set/list of expressions/equations

eq1:=diff(s(t), t)=(1-p*Psi)*tau+phi* v(t) + delta *r(t)-lambda*(1-u[3])*s(t)-u[1]*varphi*s(t) -mu*s(t);
eq2:diff(v(t), t) =p*Psi*tau + u[1]*vartheta*s(t) -gamma*lambda* (1-u[3])*v(t)-(mu+phi)*v(t);
eq3:=diff(c(t), t) =(1-u[3])*rho*lambda* (s(t) +gamma*v(t))+(1-q)* u[2]*eta*i(t) -(mu +beta +chi)*c(t);
eq4:=diff(i(t), t) =(1-rho)*(1-u[3])*lambda*( s(t) +gamma*v(t)) +chi*c(t) - u[2]*eta*i(t) - (mu +alpha )*i(t);
eq5:=diff(r(t), t) = beta*c(t) + u[2]*q*eta*i(t) -(mu +delta)*r(t);

d
--- s(t) = (1 - p Psi) tau + phi v(t) + delta r(t) - lambda (1 - u[3]) s(t)
dt

- u[1] varphi s(t) - mu s(t)
d
--- v(t) = p Psi tau + u[1] vartheta s(t) - gamma lambda (1 - u[3]) v(t)
dt

- (mu + phi) v(t)
d
--- c(t) = (1 - u[3]) rho lambda (s(t) + gamma v(t)) + (1 - q) u[2] eta i(t)
dt

- (mu + beta + chi) c(t)
d
--- i(t) = (1 - rho) (1 - u[3]) lambda (s(t) + gamma v(t)) + chi c(t)
dt

- u[2] eta i(t) - (mu + alpha) i(t)
d
--- r(t) = beta c(t) + u[2] q eta i(t) - (mu + delta) r(t)
dt
eq6:=diff(Q(t),t)=b[1]*c(t)+b[2]*i(t)+w[1]*(u[1])^2/2+w[2]*(u[2])^2/2+w[3]*(u[3])^2/2;
d 1 2 1 2 1 2
--- Q(t) = b[1] c(t) + b[2] i(t) + - w[1] u[1] + - w[2] u[2] + - w[3] u[3]
dt 2 2 2
ics:=s(0)=8200, v(0)=2800,c(0)=1100,i(0)=1500,r(0)=200,Q(0)=6700;
s(0) = 8200, v(0) = 2800, c(0) = 1100, i(0) = 1500, r(0) = 200, Q(0) = 6700
sol0:=dsolve({eq1,eq2,eq3,eq4,eq5,eq6,ics},type=numeric,stiff=true,'parameters'=[u[1],u[2],u[3]],abserr=1e-15,relerr=1e-12,maxfun=0,range=0..50):
Error, (in dsolve/numeric/process_input) system must be entered as a set/list of expressions/equations
with(plots):
Q0:=6700;
6700
obj:=proc(u)
global sol0,Q0;
local ob1;
try
sol0('parameters'=[u[1],u[2],u[3]]):
ob1:=subs(sol0(20.),Q(t)):
catch :
ob1:=0;
end try;
#ob1:=subs(sol0(20.),Q(t));
if ob1>Q0 then Q0:=ob1;print(Q0,u);end;
ob1;
end proc;
proc(u) ... end;
obj([1,1,1]);
0
obj([3,2.5],4);
0
u0:=Vector(3,[0.,0.,0.],datatype=float[8]);
Vector[column](%id = 85973880)

Q0:=0;
Q0 := 0
with(Optimization);
[ImportMPS, Interactive, LPSolve, LSSolve, Maximize, Minimize, NLPSolve,

QPSolve]
sol2:=NLPSolve(3,obj,initialpoint=u0,method=nonlinearsimplex,maximize,evaluationlimit=100):
sol0('parameters'=[3.18125786060723, 2.36800986932868]);
sol0(parameters = [3.18125786060723, 2.36800986932868])
for i from 1 to 3 do odeplot(sol0,[t,x[i](t)],0..20,thickness=3,axes=boxed);od;
Error, (in plots/odeplot) input is not a valid dsolve/numeric solution

 

  Elena, Liya

  "Researching turkish song: the selection of the main element and its graphic transformations",

   Russia, Kazan, school #57

The setting and visualization of the melodic line of the song
> restart:
> with(plots):with(plottools):
> p0:=plot([[0.5,9],[1,7],[2,9],[4,11],[6,9],[7,11],[8,7],[10,9],[12,9],[14,9],[16,7],[16.5,9],[17,7],[18,9]],color=magenta):p1:=plot([[18,9],[20,11],[22,9],[23,11],[24,9],[26,11],[28,11],[29.5,8],[30,11],[32,9],[33.5,8],[34,9],[36,7],[37.5,5],[38,9],[40,7],[42,5],[44,5],[46,4],[47,5],[48,2],[50,4],[51,5],[51.5,4],[52,2],[54,4],[56,4],[56.5,5],[57,4],[58,5],[60,7],[62,5],[64,7],[66,5]],color=cyan):
> p2:=plot([[66,5],[68,5],[69,5],[70,4],[71,5],[71.5,4],[72,2],[73,4],[74,5],[75,7],[76,5],[78,4],[78.5,7],[80,5],[82.5,4],[83.5,4],[84,2],[86,4],[88,4],[90.5,4],[91.5,4]],color=red):
> p3:=plot([[91.5,4],[92,2],[94,4],[96,4],[96.5,9],[97,7],[98,9],[100,11],[100.5,9],[101,11],[102,9],[104,11],[106,9],[108,9],[109,9],[109.5,9],[110,7],[111,9],[112,7],[113,7],[114,9],[116,11],[116.5,9],[117,11],[118,9],[119.5,11],[120,9],[122.5,9],[124,9],[124.5,9],[125,11],[125.5,9],[126,11],[128,9],[129,7],[130,9],[132,11],[132.5,9],[133,11],[134,9],[136,11],[136.5,9],[138.5,9],[140,9],[140.5,9],[141,11],[141.5,9],[142,11],[143,7],[143.5,7],[144,9],[144.5,9],[145,7],[146,9],[148,11],[148.5,9],[149,11],[150,9],[151.5,11],[152,9],[154.5,9],[156,9],[156.5,9],[157,11],[157.5,9],[158,11],[160,9],[161,7],[162,9],[164,11],[164.5,9],[165,11],[166,9],[168,11],[168.5,9],[171.5,9],[172,9],[172.5,9],[173.5,11],[174,9],[174.5,11],[175,7],[175.5,7],[176,9],[176.5,9],[177,7],[178,9],[180,11],[180.5,9],[181,11],[182,9],[183.5,11],[184,9],[186.5,9],[188,9],[188.5,9],[189,11],[189.5,9],[190,11],[192,9],[192.5,9],[193,7],[194,9],[196,11],[196.5,9],[197,11],[198,9],[200,11],[201.5,9],[202,11],[203,9],[203.5,8],[204,9],[205,7],[205.5,9],[206,11],[207,9],[208,7],[209,8],[209.5,7],[210,9],[211,7],[212,5],[213,5],[213.5,5],[214,9],[215,7],[216,5],[217,5],[217.5,5],[218,7],[219,5],[220,4],[221,4],[221.5,4],[222,7],[223,5],[224,4],[225,4],[227,4],[227.5,4],[228,2],[230,4]],color=blue):
> p4:=plot([[230,4],[232,4],[232.5,5],[233,4],[234,5],[236,7],[236.5,5],[237,5],[238,9],[240,7],[242.5,5],[244,5],[245,5],[246,4],[246.5,5],[247,4],[248,2],[250,4],[250.5,7],[251,5],[252,4],[254,4],[254.5,7],[255,5],[256,4],[258,4]],color=brown):
> p5:=plot([[258,4],[259,4],[260,2]],color=green):
> plots[display](p0,p1,p2,p3,p4,p5,thickness=2);

 

 

The selection of the main melodic element in graph of whole song. The whole song is divided into separate elements - results of transformationss0:=plot([[7,11],[8,7],[10,9],[12,9],[14,9],[16,7],[16.5,9]],color=blue):
> s1:=plot([[118,9],[119.5,11],[120,9],[122.5,9],[124,9],[124.5,9],[125,11],[125.5,9]],color=blue):
> s2:=plot([[134,9],[136,11],[136.5,9],[138.5,9],[140,9],[140.5,9],[141,11],[141.5,9]],color=blue):
> s3:=plot([[150,9],[151.5,11],[152,9],[154.5,9],[156,9],[156.5,9],[157,11],[157.5,9]],color=blue):
> s4:=plot([[166,9],[168,11],[168.5,9],[171.5,9],[172,9],[172.5,9],[173.5,11],[174,9]],color=blue):
> s5:=plot([[182,9],[183.5,11],[184,9],[186.5,9],[188,9],[188.5,9],[189,11],[189.5,9]],color=blue):
> s6:=plot([[250,4],[250.5,7],[251,5],[252,4],[254,4],[254.5,7],[255,5],[256,4]],color=blue):
> plots[display](s0,s1,s2,s3,s4,s5,s6);
> s:=plots[display](s0,s1,s2,s3,s4,s5,s6):

 

Animated display of grafical transformation of the basic element (to click on the picture - on the panel of instruments appears player - to play may step by step).m0:=plot([[7,11],[8,7],[10,9],[12,9],[14,9],[16,7],[16.5,9]],color=blue):
> pm:=plot([[118,9],[119.5,11],[120,9],[122.5,9],[124,9],[124.5,9],[125,11],[125.5,9]],color=red,style=line,thickness=4):
> iop:=plots[display](m0,pm,insequence=true):
> plots[display](iop,s0);

> m0_t:=translate(m0,110,0):
> m0_r:=reflect(m0_t,[[0,9],[24,9]]):
> plots[display](m0,m0_r,insequence=true);
> m0r:=plots[display](m0,m0_r,insequence=true):

> pm0:=plots[display](pm,m0):
> plots[display](pm0,m0r);

> m0:=plot([[7,11],[8,7],[10,9],[12,9],[14,9],[16,7],[16.5,9]],color=blue):
> pn:=plot([[134,9],[136,11],[136.5,9],[138.5,9],[140,9],[140.5,9],[141,11],[141.5,9]],color=blue,thickness=3):
> iop:=plots[display](m0,pn,insequence=true):
> plots[display](iop,s0);

> m0_t1:=translate(m0,126,0):
> m0_r1:=reflect(m0_t1,[[0,9],[24,9]]):
>
> plots[display](m0,m0_r1,insequence=true);
> m0r1:=plots[display](m0,m0_r1,insequence=true):

> pm01:=plots[display](pn,m0):
> plots[display](pm01,m0r1);

 

> pm2:=plots[display](pn,pm,m0):
> plots[display](pm0,m0r,pm01,m0r1);

> pt_i_1:=seq(translate(pm,5*11*i,0),i=0..4):
> plots[display](pt_i_1);

> pm_i:=seq(translate(pm,5*11*i,0),i=0..4):
> plots[display](pm_i);
> iop1:=plots[display](pm_i,insequence=true):
> plots[display](iop1,s0);

 

> pm_i_0:=seq(translate(m0_r,5*11*i,0),i=0..4):
> plots[display](pm_i_0);
> iop2:=plots[display](pm_i_0,insequence=true):
> plots[display](iop2,s0);

 

 

 

 

 

 

Hi,

I am using the solve command to solve an equation of the form "linear over quadratic is equal to a constant" where the constant is assumed to be nonzero. This is easily solved by hand, of course, but I to use the solution in other computations. So I asked maple to solve it for me. But when I check maple's solution (i.e. just plug the two solutions in on the left hand side and simplify) maple does not return the original constant. Can anyone help me understand what is going wrong?

Construction of arabesques of melodic line BACH

Elena, Liya "Construction of arabesques of melodic line BACH", Kazan, Russia, school#57
       
> restart:
> with(plots):with(plottools):

      The setting and visualization of line BACH: B - note b-flat, A - note la, C - note do, H - note si.
> p0:=plot([[0,1],[2,0],[4,1.5],[6,1]],thickness=4,color=cyan,scaling=constrained);
>
>   p0 := PLOT(
>
>         CURVES([[0, 1.], [2., 0], [4., 1.500000000000000], [6., 1.]])
>
>         , SCALING(CONSTRAINED), THICKNESS(4), AXESLABELS( ,  ),
>
>         COLOUR(RGB, 0, 1.00000000, 1.00000000),
>
>         VIEW(DEFAULT, DEFAULT))
>
> plots[display](p0);
> r_i:=seq(rotate(p0,i*Pi/4),i=1..8):
> p1:=display(r_i,p0):plots[display](p1,scaling=constrained);

> c1:=circle([0,0],6,color=blue,thickness=2):
> plots[display](c1,p1,scaling=constrained);
> p_c:=plots[display](c1,p1,scaling=constrained):

> pt_i_2:=seq(translate(p1,0,2*6*i),i=0..4):
> plots[display](pt_i_2,scaling=constrained);
> pt_i_22:=seq(translate(p1,0,6*i),i=0..4):
> plots[display](pt_i_22,scaling=constrained);
> pt_i_222:=seq(translate(p1,0,1/2*6*i),i=0..4):
> plots[display](pt_i_222,scaling=constrained);

> pr:=rotate(p1,Pi/8):
> plots[display](pr,scaling=constrained);
> plots[display](p1,pr,scaling=constrained);
> pr_i:=seq(rotate(p1,Pi/16*i),i=0..8):
> plots[display](pr_i,scaling=constrained);


> pt_1:=translate(p1,0,2*6):
> pr_1_i:=seq(rotate(pt_1,Pi/3.5*i),i=0..6):
> plots[display](pr_1_i,scaling=constrained);
> pr_11_i:=seq(rotate(pt_1,Pi/5*i),i=0..10):
> plots[display](pr_11_i,scaling=constrained);
> pr_111_i:=seq(rotate(pt_1,Pi/6.5*i),i=0..12):
> plots[display](pr_111_i,scaling=constrained);

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