MaplePrimes Questions

With Typesetting level: Extended, INTERVAL displays only the first argument. So evalr(1/INTERVAL(-1 .. 1)) is displayed as INTERVAL(-infinity .. -1), and INTERVAL(-1,-1,1,1) is displayed as INTERVAL(-1).

 

 

I attach a file in which i get an error though it performs.

I made a for do statement which performs perfectly for one loop, but gives an error "unexpected ',' , though performs (i cld not find the errors!).
What does this error state then? How do i remove when I use semicolon to end statement? Can any one help me. Thanks.
 

Data made for Data[i,j] [k] where i =1,2; j=1..8. Each Data has 6 values. k = 1..6

For each data i got an output list of 6 values using a module;

I made a for do statement which performs but gives an error "unexpected ','.

Can any onr help to find my mistake. Thanks.

 

restart

R1 := "SE2"; BP__ref := 3.93; BTE__optimum := .334; BP := 2.36; BTE := .271; sfc := .308; Data[1, 1] := [R1, BP__ref, BTE__optimum, BP, BTE, sfc]
NULL

R2 := "SE2"; BP__ref := 3.93; BTE__optimum := .334; BP := 2.75; BTE := .298; sfc := .282; Data[1, 2] := [R1, BP__ref, BTE__optimum, BP, BTE, sfc]

R3 := "SE2"; BP__ref := 3.93; BTE__optimum := .334; BP := 3.14; BTE := .318; sfc := .264; Data[1, 3] := [R1, BP__ref, BTE__optimum, BP, BTE, sfc]

R4 := "SE2"; BP__ref := 3.93; BTE__optimum := .334; BP := 3.53; BTE := .330; sfc := .253; Data[1, 4] := [R1, BP__ref, BTE__optimum, BP, BTE, sfc]

R5 := "SE2"; BP__ref := 3.93; BTE__optimum := .334; BP := 3.88; BTE := .334; sfc := .249; Data[1, 5] := [R1, BP__ref, BTE__optimum, BP, BTE, sfc]

R6 := "SE2"; BP__ref := 3.93; BTE__optimum := .334; BP := 3.93; BTE := .334; sfc := .249; Data[1, 6] := [R1, BP__ref, BTE__optimum, BP, BTE, sfc]

R7 := "SE2"; BP__ref := 3.93; BTE__optimum := .334; BP := 4.32; BTE := .330; sfc := .253; Data[1, 7] := [R1, BP__ref, BTE__optimum, BP, BTE, sfc]

R8 := "SE2"; BP__ref := 3.93; BTE__optimum := .334; BP := 4.71; BTE := .318; sfc := .264; Data[1, 8] := [R1, BP__ref, BTE__optimum, BP, BTE, sfc]

NULL

R1 := "4S"; BP__ref := 3.88; BTE__optimum := .383; BP := 2.36; BTE := .323; sfc := .258; Data[2, 1] := [R1, BP__ref, BTE__optimum, BP, BTE, sfc]

R2 := "4S"; BP__ref := 3.88; BTE__optimum := .383; BP := 2.75; BTE := .350; sfc := .239; Data[2, 2] := [R1, BP__ref, BTE__optimum, BP, BTE, sfc]

R3 := "4S"; BP__ref := 3.88; BTE__optimum := .383; BP := 3.14; BTE := .369; sfc := .229; Data[2, 3] := [R1, BP__ref, BTE__optimum, BP, BTE, sfc]

R4 := "4S"; BP__ref := 3.88; BTE__optimum := .383; BP := 3.53; BTE := .380; sfc := .219; Data[2, 4] := [R1, BP__ref, BTE__optimum, BP, BTE, sfc]

R5 := "4S"; BP__ref := 3.88; BTE__optimum := .383; BP := 3.88; BTE := .383; sfc := .217; Data[2, 5] := [R1, BP__ref, BTE__optimum, BP, BTE, sfc]

R6 := "4S"; BP__ref := 3.88; BTE__optimum := .383; BP := 3.88; BTE := .383; sfc := .217; Data[2, 6] := [R1, BP__ref, BTE__optimum, BP, BTE, sfc]

R7 := "4S"; BP__ref := 3.88; BTE__optimum := .383; BP := 4.32; BTE := .378; sfc := .221; Data[2, 7] := [R1, BP__ref, BTE__optimum, BP, BTE, sfc]

R8 := "4S"; BP__ref := 3.88; BTE__optimum := .383; BP := 4.71; BTE := .365; sfc := .229; Data[2, 8] := [R1, BP__ref, BTE__optimum, BP, BTE, sfc]

 

``

Data[1, 8][4]

4.71

(1)

``

``

for i to 2 do for j to 8 do y[i, j] := action1:-Mproc(Data[i, j]); print(Data[i, j], y[i, j]) end do end do

["4S", 3.88, .383, 4.71, .365, .229], ["4S", .4311313490, 0.5411807293e-1, 0.3357509744e-1, .4235561571, 0.6183000001e-1]

(2)

NULL

N1 := NULL; for j to 8 do N1 := N1, Data[1, j][4] end do; print(N1)

2.36, 2.75, 3.14, 3.53, 3.88, 3.93, 4.32, 4.71

(3)

for i to 2 do N[i] := NULL; for j to 8 do N[i] := N[i], Data[i, j][4] end do end do; print(N[2])

Error, `,` unexpected

 

N[2]

2.36, 2.75, 3.14, 3.53, 3.88, 3.88, 4.32, 4.71

(4)

N[1]

2.36, 2.75, 3.14, 3.53, 3.88, 3.93, 4.32, 4.71

(5)

``


 

Download Doubt_A_ReactantValues.mw

Doubt_A_ReactantValues.mw

How does one delete an empty cell. I can't see to do that easily, please see the attached for what I am trying to explain.

eq1 := 3*x^2+x+2

``

if 6*x+1 = diff(eq1(x), x) then print(yes) else print(no) end if


 

Download 2.mw

 

 

 

How do you indent your code neatly ? There should be a feature for automatic indentation, the code structure closesly follows python coding style and having it organized would be very helpful.

 

Any advise is appreciated.

Hello.

I have a problem for determining a and b?

please help me.

thanks

PADE.mw



Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/PADE.mw .
 

Download PADE.mw



 

 

where i'm wrong
 

> with(plottools);
> with(plots);
> m := PLOT(polygon([[2, 2], [2, 3], [4, 3], [4, 2]], color = blue), polygon([[1, 1], [1, 4], [5, 4], [5, 1]], color = red), polygon([[1, 4], [3, 6], [5, 4]], thickness = 3, color = red), curve([[4.5, 5.5], [5, 6]]), polygon([[3.5, 5], [4, 5.5], [4.5, 5], [4, 4.5]], color = red));
> n := PLOT(polygon([[2, 2], [2, 3], [4, 3], [4, 2]], color = green), polygon([[1, 1], [1, 4], [5, 4], [5, 1]], color = red), polygon([[1, 4], [3, 6], [5, 4]], thickness = 3, color = red), curve([[5.5, 6], [6, 6.5]]), polygon([[3.5, 5], [4, 5.5], [4.5, 5], [4, 4.5]], color = red));
> display([m, n], insequence = true);
Error, (in plots:-display) expecting plot structures but received: [m, n]
>

 

I was to define a function of two variable in terms of two other functions.  This other function depends on another two functions which are unknown.  For example

g(u,v)=alpha(u,v)*beta(u,v)

h(u,v)=alpha(u,v)+beta(u,v)

f(u,v) = D_1g(u,v)*h(u,v).

 

I tried by writtinge this line 

g:=(u,v)->alpha(u,v)*beta(u,v)

h:=(u,v)->alpha(u,v)+beta(u,v)

f:=(u,v)->D[1](g)(u,v)*h(u,v).

Is this correct?   

If this is correct, then how I make maple show f in tems of the unkown functions alpha and beta?

 

Thanks,

 

Sergio

i want to use by part integration for triple integral but parts dont work properly, how i can fix it?

 

The following procedure describes the path of a projectile fired from the origin with a launch angle of alpha and a launch velocity of u.

z := proc (x) options operator, arrow; tan(alpha)*x-(1/2)*g*x^2/(u^2*cos(alpha)^2) end proc

The projectile's path must pass through the point [a, h], both values greater than zero.

Can an expression be found for the minimum value of u and its related value for alpha?

the infint  summation  maple cant evaluated it  when my variables  are more than one decimal number 

;

I have a problem that seems rather simple in theory, yet if implemented in maple there arise problems. I want to solve systems of linear equations that involve roots of unity.

This problem arises when trying to find explicit equations of multivariate polynomials of fixed degree through given points (together with conditions on the derivatives of the curve at these points). In the following case, the points are dented by P[i] and I'm looking for the explicit equation of the degree 9 polynomial through these points. Actually, I know there is a two parameter family of such polynomials, so the solution will be of dimension two.

Sorry for the rather crude implementation.

with(Groebner): with(algcurves): with(LinearAlgebra):

t := exp((1/9)*(2*Pi*I)):
s := exp((1/3)*(2*Pi*I)):
a_1 := -1*t:
a_2 := -s*t:
a_3 := -s^2*t:
P[1] := a_1, 0, 1: P[2] := a_2, 0, 1: P[3] := a_3, 0, 1:
P[4] := 1, t*a_1, 0: P[5] := 1, t*a_2, 0: P[6] := 1, t*a_3, 0:
P[7] := 0, 1, a_1: P[8] := 0, 1, a_2: P[9] := 0, 1, a_3:

G := 0: var1 := 0:
for n from 0 to 9 do
G := G+add(c[k, n-k]*x^k*y^(n-k), k = 0 .. n):
for k from 0 to n do var1 := var1, c[k, n-k]:
end do: 
end do:
F := Homogenize(G, z):
Fx := diff(F, x):
Fy := diff(F, y):
Fz := diff(F, z):
Fxx := diff(F, x, x):
Fxy := diff(F, x, y):
Fyx := Fxy:
Fyy := diff(F, y, y):
Fxz := diff(F, x, z):
Fzx := Fxz:
Fzz := diff(F, z, z):
Fyz := diff(F, y, z):
Fzy := diff(F, z, y):

f := unapply(F, [x, y, z]): fx := unapply(Fx, [x, y, z]): fy := unapply(Fy, [x, y, z]): fz := unapply(Fz, [x, y, z]): fxx := unapply(Fxx, [x, y, z]): fxy := unapply(Fxy, [x, y, z]): fyx := unapply(Fyx, [x, y, z]): fyy := unapply(Fyy, [x, y, z]): fxz := unapply(Fxz, [x, y, z]): fzx := unapply(Fzx, [x, y, z]): fzz := unapply(Fzz, [x, y, z]): fyz := unapply(Fyz, [x, y, z]): fzy := unapply(Fzy, [x, y, z]):

sys := [f(P[1]) = 0, fx(P[1]) = 0, fy(P[1]) = 0, fz(P[1]) = 0, fxy(P[1]) = 0, fxz(P[1]) = 0, fxx(P[1]) = 0, fyy(P[1]) = 0, fzz(P[1]) = 0, fyz(P[1]) = 0, f(P[2]) = 0, fx(P[2]) = 0, fy(P[2]) = 0, fz(P[2]) = 0, fxy(P[2]) = 0, fxz(P[2]) = 0, fxx(P[2]) = 0, fyy(P[2]) = 0, fzz(P[2]) = 0, fyz(P[2]) = 0, f(P[3]) = 0, fx(P[3]) = 0, fy(P[3]) = 0, fz(P[3]) = 0, fxy(P[3]) = 0, fxz(P[3]) = 0, fxx(P[3]) = 0, fyy(P[3]) = 0, fzz(P[3]) = 0, fyz(P[3]) = 0, f(P[4]) = 0, fx(P[4]) = 0, fy(P[4]) = 0, fz(P[4]) = 0, fxy(P[4]) = 0, fxz(P[4]) = 0, fxx(P[4]) = 0, fyy(P[4]) = 0, fzz(P[4]) = 0, fyz(P[4]) = 0, f(P[5]) = 0, fx(P[5]) = 0, fy(P[5]) = 0, fz(P[5]) = 0, fxy(P[5]) = 0, fxz(P[5]) = 0, fxx(P[5]) = 0, fyy(P[5]) = 0, fzz(P[5]) = 0, fyz(P[5]) = 0, f(P[6]) = 0, fx(P[6]) = 0, fy(P[6]) = 0, fz(P[6]) = 0, fxy(P[6]) = 0, fxz(P[6]) = 0, fxx(P[6]) = 0, fyy(P[6]) = 0, fzz(P[6]) = 0, fyz(P[6]) = 0, f(P[7]) = 0, fx(P[7]) = 0, fy(P[7]) = 0, fz(P[7]) = 0, fxy(P[7]) = 0, fxz(P[7]) = 0, fxx(P[7]) = 0, fyy(P[7]) = 0, fzz(P[7]) = 0, fyz(P[7]) = 0, f(P[8]) = 0, fx(P[8]) = 0, fy(P[8]) = 0, fz(P[8]) = 0, fxy(P[8]) = 0, fxz(P[8]) = 0, fxx(P[8]) = 0, fyy(P[8]) = 0, fzz(P[8]) = 0, fyz(P[8]) = 0, f(P[9]) = 0, fx(P[9]) = 0, fy(P[9]) = 0, fz(P[9]) = 0, fxy(P[9]) = 0, fxz(P[9]) = 0, fxx(P[9]) = 0, fyy(P[9]) = 0, fzz(P[9]) = 0, fyz(P[9]) = 0]:
var := [c[0, 0], c[0, 1], c[1, 0], c[0, 2], c[1, 1], c[2, 0], c[0, 3], c[1, 2], c[2, 1], c[3, 0], c[0, 4], c[1, 3], c[2, 2], c[3, 1], c[4, 0], c[0, 5], c[1, 4], c[2, 3], c[3, 2], c[4, 1], c[5, 0], c[0, 6], c[1, 5], c[2, 4], c[3, 3], c[4, 2], c[5, 1], c[6, 0], c[0, 7], c[1, 6], c[2, 5], c[3, 4], c[4, 3], c[5, 2], c[6, 1], c[7, 0], c[0, 8], c[1, 7], c[2, 6], c[3, 5], c[4, 4], c[5, 3], c[6, 2], c[7, 1], c[8, 0], c[0, 9], c[1, 8], c[2, 7], c[3, 6], c[4, 5], c[5, 4], c[6, 3], c[7, 2], c[8, 1], c[9, 0]]:
A, b := GenerateMatrix(sys, var):

Solving this with

simplify(LinearSolve(A, b))

Gives a rather nicely looking solution. But the calculation strongly depends on the form of s, t and the a_i.

For example if we instead define 

s:=t^3:

The computation takes much longer and the result doesn't take a particularly nice form. Similarly if we instead take

a_1 := -t: a_2 := -t^4: a_3 := -t^7:

the computation takes hours / days and the result looks rather ugly (huge coefficients). Yet both of these inputs are mathematically equivalent to the original one.

The thing is that I have similar systems of linear equations that I know must have solutions, but the solution vector that maple puts out has coefficients with thousands of digits (which after the above I suspect to be wrong). I know the systems are vastly overdetermined, but even removing all redundant equations does not solve the problen.

Is there a way to get the correct solutions with maple? Does anyone know whether a different program would handle this problem better?

How can I add legend to a multicolored histogram using display?

 

I want to make some equation and convert to LaTeX to have some Exercises "Solve the equation" 
My code:
restart; A := [1, 2, 3/2, 4, 5/9]; B := [2, 5, -7, 1, 8]; C := [-1, -2, -3, -4, -5/3]; for i to nops(A) do 'solve*the*equation'*A[i]*x^2+B[i]*x+C[i] = 0 end do
How can I get code in LaTeX, it's mean, I get "Solve the equation $x^2+2x-1=0$" and get roots (if have)

how to approximate this number to The nearest decimal number by maple

0.6364562590

a:=[1,2,3,4,5]:

if I want b to have all elements less than 4 why can't I just write

b:=a<4;

?

 

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