Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

 

Is there a way with Maple 18 to place a bracket as shown below to contain two DE's? I am not trying to solve them. Only for note-taking purposes. If there is can you please share how to do it?

 

 

DE

How to solve the equation
2^(sin(x)^4-cos(x)^2)-2^(cos(x)^4-sin(x)^2) = cos(2*x)
symbolically? The solve command produces a weird answer. Evalfing all its values, one sees
0.7853981634, -0.7853981634, 2.356194490, -2.356194490,

1.570796327 - 1.031718534 I, -1.570796327 + 1.031718534 I,

1.570796327 + 1.031718534 I, -1.570796327 - 1.031718534 I,

0.7853981634, -0.7853981634, 2.356194490, -2.356194490,

1.570796327 - 1.031718534 I, -1.570796327 + 1.031718534 I,

1.570796327 + 1.031718534 I, -1.570796327 - 1.031718534 I


The identify command
interprets the real solutions on -Pi..Pi as -3*Pi/4, -Pi/4, Pi/4, 3*Pi/4
(for example,
identify(2.356194490);

3*Pi/4 ).
Is it possible to obtain these with Maple in a simpler way?

PS. Mathematica 10 does the job.

PPS. So does even Mathematica 7.

Hello people in mapleprimes,

 

I have a question of labels of equations.

For example, suppose that there is a following description in a worksheet.

>2*x+y=5;

  2*x+y=5   (1)

>3*x+y=6;

  3*x+y=6   (2)

>solve({(1),(2)},{x,y});

 {x=1,y=3}          (3)

subs((3),100+x+100*y)

    400   

 

And, when I moved (2) to the next to (1) with dragging, and then, clicked !!! to execute it, then the worksheet 

changes as follows, which is wrong as accompanying ??

>2*x+y=5;3*x+y=6;

  2*x+y=5

 3*x+y=6   (1)

>solve({(1),??},{x,y})

{x = x, y = -3*x+6}          (2)

>subs({(1),??},100*x+100*y)

-200*x+600

 

Do you know some good way for ?? not to appear in the above sitiation.

I will appreciate when you tell me some measures about this.

 

Best wishes.

 

taro

 

 

 

 

 

 

Hi 

I'm having a problem with the statements inside a for-loop somehow being read in a different way than outside the loop. 

I've defined some functions earlier, and then I need to perform an integration using these functions, while I change one variable a little for each loop of the for-loop. 

The problem is that IN the for-loop, I get the same value from my integration for all loops. But when I execute the exsact same code OUTSIDE of the loop, I get the correct values, which are changing whenever i change the one variable. 

Here is the loop:

for i from 0 to 42 do

rotorshift := evalf[6](2*((1/180)* *Pi*((1/2)*Ø[gap]))/N[m]-1/2*(tau[p]-tau[s]));

PMmmf_func := x-> proc (x) if type(x-rotorshift, nonnegative) then if type(trunc((x-rotorshift)/tau[p]), odd) = true then -H[c]*l[m] else H[c]*l[m] end if else if type(trunc((x-rotorshift)/tau[p]), odd) = true then H[c]*l[m] else -H[c]*l[m] end if end if end proc

B[g] := proc (x) -> 1000*mu[0]*(PMmmf_func(x)+MMF_func(x))/d[eff, stator](x) end proc;

Flux(i+1) := (int(B[g](x), 0 .. tau[s])+int(B[g](x), 3*tau[s] .. 4*tau[s])+int(B[g](x), 6*tau[s] .. 7*tau[s])+int(B[g](x), 9*tau[s] .. 10*tau[s]))*10^(-3)*L[ro];

end do;

And for all the values in the "Flux" vector, I get the same value. But when I remove the loop, and change the value of manually, I get the correct (changing) values of flux!! 

Any ideas why this may be? This is really dricing me nuts. I've spent the beter part of the day on this, and I just can't seem to find a workaround, much less a reason for this behaviour.

many thanks!

I have generated an 8x8 Jacobian, containing a few variables and several zeros as elements. I would like to translate this to LaTeX code. How can I first simplify what I have, to make it tractable?

Here is the Maple code:

eq_1 := Lambda-mu*S-(beta*(H+C+C1+C2)*S+tau*(T+C)*S)/(S+T+H+C+C1+C2+C1M+C2M);
eq_2 := (-beta*(H+C+C1+C2)*T+tau*(T+C)*S)/(S+T+H+C+C1+C2+C1M+C2M)-(mu+mu[T])*T;
eq_3 := (beta*(H+C+C1+C2)*S-tau*(T+C)*H)/(S+T+H+C+C1+C2+C1M+C2M)-(mu+mu[A])*H;
eq_4 := (beta*(H+C+C1+C2)*T+tau*(T+C)*H)/(S+T+H+C+C1+C2+C1M+C2M)-(mu+mu[T]+mu[A]+lambda[T])*C;
eq_5 := lambda[T]*C-(mu+mu[A]+rho[1]+eta[1])*C1;
eq_6 := rho[1]*C1-(mu+mu[A]+rho[2]+eta[2])*C2;
eq_7 := eta[1]*C1-(mu+rho[1]+gamma)*C1M;
eq_8 := eta[2]*C2+rho[1]*C1M-(mu+rho[2]+gamma*rho[1]/(rho[1]+rho[2]))*C2M;
J := VectorCalculus:-Jacobian([eq_1, eq_2, eq_3, eq_4, eq_5, eq_6, eq_7, eq_8], [S, T, H, C, C1, C2, C1M, C2M]);

JQDFE := eval(J, [S = Lambda/(beta-mu[A]), T = 0, H = Lambda*(beta/(mu+mu[A])-1)/(beta-mu[A]), C = 0, C1 = 0, C2 = 0, C1M = 0, C2M = 0]);

Thanks.

Hi,

I am trying to interpolate a set of arbitrary numbers including real ones like Pi and plot them using ArrayInterpolate. This does not work. Is there a simple way around this?

Thank you.

Hi,

I use the VectorCalculus package to calcutate derivative formula for geometric functions, and met difficulity simplifying the result expression.

For example, I define some vectors P, S, V like below:

P:=<Px, Py, Pz>, S:=<Sx, Sy, Sz>, V:=<Vx, Vy, Vz>

then define an intermediate variable Q:=P - S,

then define a function d:= sqrt(DotProduct(Q, Q)-(DotProuct(Q,V))^2)

by calculating the function's derivative w.r.t Px I got a very complex result expression:

dpx:=1/2 * (2Px - 2Sx - 2 ( (Px - Sx) Vx + (Py - Sy) Vy + (Pz - Sz)Vz )Vx ) / (sqrt( (Px-Sx)^2 + (Py-Sy)^2 + (Pz-Sz)^2 - .....)

 

Apparently this expression can be simplified by substituting its sub-expression with pre-defined variables like Q and d.

I know I can use subs, eval, and subsalg to do it manually:

subs(1/(sqrt( (Px-Sx)^2 + (Py-Sy)^2 + (Pz-Sz)^2 - .....) = 1/dv, dfdpx)

subs((Px - Sx) Vx + (Py - Sy) Vy + (Pz - Sz)Vz = dotproduct_q_v, dfdpx)

and I can get a simplified expression like this:

(qx-dotproduct_q_v*vx)/d

 

But it's like my brain does the simplification first, and Maple only does the text substitution for me.

Is there any way to do it automatically?

 

Thanks,

-Kai

 

Hi i 2 questions. all pertaining to solving a systems of equations mod 2

First if i have a large set of equations, 11^3 equations in 11 unknowns and i want maple to give me ALL solutions mod 2 how can i do that? Maples msolve is loosing solutions.

Second suppose i want all unique solutions that say 6 of the variables can have but dont care what the solution to the other variables are as long as it is a solution. 

mini example:

say x=1,y=1,z=1 is a solution as well as x=1,y=1,z=0, i just want to know about x=1,y=1.

 

I have produded a 3D plot where I have used the graphic's lighting->user GUI to set the light color, direction, and ambient lighting to my liking.

 

I would like to save the lighting parameters so that I can reproduce the identical lighting in other plots.  I see no way of reading off the lighting parameters from the GUI.  I tried to "lprint(myplot)" to see if it contains that information but apparently it doesn't.

So my question is: Is there a way to retrieve the lighting parameters from a 3D plot?

 

--

Rouben Rostamian

I experienced strange operation of "union" for sets of vectors.

Mt1:=Matrix(2, 4, [[ 0,1, 0, 0], [ 0,  0,  1, 1]]); Ms := Vector[column](4, [8,4,2,1]); St1 := {}:

St1:= `union`(St1, {Mt1 . Ms});

I am surprised, because each execution of union adds new and the same vector <4 | 3> to set St1:

1

But after copying any set in the clipboard and pasting the set St1 has only one instance of vector <4 | 3>:

2

What does it mean?

How to solve the system
{sqrt((x-1)^2+(y-5)^2)+(1/2)*abs(x+y) = 3*sqrt(2), sqrt(abs(x+2)) = 2-y}
over the reals symbolically? Of course, with Maple. Mathematica does the job.

I want to solve maximize of equation,but the maximize failed to solve it,who can help me.thanks.

c[1] := (1/8)*w*{(1/((x+y+z)^2+1))^(3/2)+(1/((x+y)^2+1))^(3/2)+(1/((x+z)^2+1))^(3/2)+(1/((y+z)^2+1))^(3/2)+(1/(x^2+1))^(3/2)+(1/(y^2+1))^(3/2)+(1/(z^2+1))^(3/2)+1}+(1/8)*{x/((x+y+z)^2+1)+x/((x+y)^2+1)+x/((x+z)^2+1)+x/(x^2+1)}:

c[2] := (1/8)*w*{(1/((x+y+z)^2+1))^(3/2)+(1/((x+y)^2+1))^(3/2)+(1/((x+z)^2+1))^(3/2)+(1/((y+z)^2+1))^(3/2)+(1/(x^2+1))^(3/2)+(1/(y^2+1))^(3/2)+[1/(z^2+1)]^(3/2)+1}+(1/8)*{y/((x+y+z)^2+1)+y/((x+y)^2+1)+y/((y+z)^2+1)+y/(y^2+1)}:

t[1] := diff(c[1], x);

(1/8)*w*{-(3/2)*(1/((x+y+z)^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/((x+y)^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/((x+z)^2+1))^(1/2)*(2*x+2*z)/((x+z)^2+1)^2-3*(1/(x^2+1))^(1/2)*x/(x^2+1)^2}+(1/8)*{1/((x+y+z)^2+1)-x*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-x*(2*x+2*y)/((x+y)^2+1)^2+1/((x+z)^2+1)-x*(2*x+2*z)/((x+z)^2+1)^2+1/(x^2+1)-2*x^2/(x^2+1)^2}

(1)

t[2] := diff(c[2], y);

(1/8)*w*{-(3/2)*(1/((x+y+z)^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/((x+y)^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/((y+z)^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2}+(1/8)*{1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2}

(2)

eliminate({t[1], t[2]}, w);

[{w = -{1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2}/{-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2}}, {{1/((x+y+z)^2+1)-x*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-x*(2*x+2*y)/((x+y)^2+1)^2+1/((x+z)^2+1)-x*(2*x+2*z)/((x+z)^2+1)^2+1/(x^2+1)-2*x^2/(x^2+1)^2}*{-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2}-{1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2}*{-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(x^2+2*x*z+z^2+1))^(1/2)*(2*x+2*z)/((x+z)^2+1)^2-3*(1/(x^2+1))^(1/2)*x/(x^2+1)^2}}]

(3)

w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*sqrt(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*sqrt(1/(x^2+2*x*y+y^2+1))*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*sqrt(1/(y^2+2*y*z+z^2+1))*(2*y+2*z)/((y+z)^2+1)^2-3*sqrt(1/(y^2+1))*y/(y^2+1)^2);

w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2)

(4)

sub(w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2), c[1]);

sub(w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2), (1/8)*w*{(1/((x+y+z)^2+1))^(3/2)+(1/((x+y)^2+1))^(3/2)+(1/((x+z)^2+1))^(3/2)+(1/((y+z)^2+1))^(3/2)+(1/(x^2+1))^(3/2)+(1/(y^2+1))^(3/2)+(1/(z^2+1))^(3/2)+1}+(1/8)*{x/((x+y+z)^2+1)+x/((x+y)^2+1)+x/((x+z)^2+1)+x/(x^2+1)})

(5)

subs(w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2), c[2]);

-(1/8)*(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)*{(1/((x+y+z)^2+1))^(3/2)+(1/((x+y)^2+1))^(3/2)+(1/((x+z)^2+1))^(3/2)+(1/((y+z)^2+1))^(3/2)+(1/(x^2+1))^(3/2)+(1/(y^2+1))^(3/2)+[1/(z^2+1)]^(3/2)+1}/(-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2)+(1/8)*{y/((x+y+z)^2+1)+y/((x+y)^2+1)+y/((y+z)^2+1)+y/(y^2+1)}

(6)

"#"Iwant to maximize the equation (5)and (6),under the conditon of x,y,z are negative or positive at the same time.

 

NULL

 

Download maximize.mw

Does anyone know how to incorporate the tetrad with the directional derivative? I tried using the SumOverIndices, but get crazy results. I know Maple can find the answer easily because I have done the same thing by hand. What am I missing?

The directional derivative should take the form f,1 = eaμ df/dxμ . The answer is Y,1 = dY/dζ – Ybar dY/du.  I obviously do not get this result.

 


restart; with(Physics); with(Tetrads)

0, "%1 is not a command in the %2 package", Tetrads, Physics

(1)

`#msup(mi("ds",mathcolor = "#af00af"),mn("2",mathcolor = "#af00af"))` := Physics:-`*`(Physics:-`*`(2, dzeta), dzetabar)+Physics:-`*`(Physics:-`*`(2, du), dv)+Physics:-`*`(Physics:-`*`(2, H(zetabar, zeta, v, u)), (du+Physics:-`*`(Ybar(zetabar, zeta, v, u), dzeta)+Physics:-`*`(Y(zetabar, zeta, v, u), dzetabar)-Physics:-`*`(Physics:-`*`(Y(zetabar, zeta, v, u), Ybar(zetabar, zeta, v, u)), dv))^2)

2*dzeta*dzetabar+2*du*dv+2*H(zetabar, zeta, v, u)*(du+Ybar(zetabar, zeta, v, u)*dzeta+Y(zetabar, zeta, v, u)*dzetabar-Y(zetabar, zeta, v, u)*Ybar(zetabar, zeta, v, u)*dv)^2

(2)

X = [zetabar, zeta, v, u]

X = [zetabar, zeta, v, u]

(3)

PDEtools:-declare(`#msup(mi("ds",mathcolor = "#af00af"),mn("2",mathcolor = "#af00af"))`)

Ybar(zetabar, zeta, v, u)*`will now be displayed as`*Ybar

(4)

Setup(automaticsimplification = true, coordinatesystems = (X = [zetabar, zeta, v, u]), metric = 2*dzeta*dzetabar+2*du*dv+2*H(zetabar, zeta, v, u)*(du+Ybar(zetabar, zeta, v, u)*dzeta+Y(zetabar, zeta, v, u)*dzetabar-Y(zetabar, zeta, v, u)*Ybar(zetabar, zeta, v, u)*dv)^2)

[automaticsimplification = true, coordinatesystems = {X}, metric = {(1, 1) = 2*H(X)*Y(X)^2, (1, 2) = 1+2*H(X)*Y(X)*Ybar(X), (1, 3) = -2*H(X)*Y(X)^2*Ybar(X), (1, 4) = 2*H(X)*Y(X), (2, 2) = 2*H(X)*Ybar(X)^2, (2, 3) = -2*H(X)*Ybar(X)^2*Y(X), (2, 4) = 2*H(X)*Ybar(X), (3, 3) = 2*H(X)*Y(X)^2*Ybar(X)^2, (3, 4) = 1-2*H(X)*Y(X)*Ybar(X), (4, 4) = 2*H(X)}]

(5)

g_[]

g_[mu, nu] = (Matrix(4, 4, {(1, 1) = 2*H(X)*Y(X)^2, (1, 2) = 1+2*H(X)*Y(X)*Ybar(X), (1, 3) = -2*H(X)*Y(X)^2*Ybar(X), (1, 4) = 2*H(X)*Y(X), (2, 1) = 1+2*H(X)*Y(X)*Ybar(X), (2, 2) = 2*H(X)*Ybar(X)^2, (2, 3) = -2*H(X)*Ybar(X)^2*Y(X), (2, 4) = 2*H(X)*Ybar(X), (3, 1) = -2*H(X)*Y(X)^2*Ybar(X), (3, 2) = -2*H(X)*Ybar(X)^2*Y(X), (3, 3) = 2*H(X)*Y(X)^2*Ybar(X)^2, (3, 4) = 1-2*H(X)*Y(X)*Ybar(X), (4, 1) = 2*H(X)*Y(X), (4, 2) = 2*H(X)*Ybar(X), (4, 3) = 1-2*H(X)*Y(X)*Ybar(X), (4, 4) = 2*H(X)}))

(6)

``

NULL

NULL

eqn3 := SumOverRepeatedIndices(Physics:-`*`(d_[mu](Y(X)), e_[1, `~mu`]))

((Y(X)*Ybar(X)-1)*(diff(Y(X), zetabar))+(Y(X)*Ybar(X)-1)*(diff(Y(X), zeta))+(diff(Y(X), u)+diff(Y(X), v))*(Y(X)+Ybar(X)))*2^(1/2)/((-(Ybar(X)^2+1)*(Y(X)^2+1)/(Y(X)+Ybar(X))^2)^(1/2)*(2*Y(X)+2*Ybar(X)))

(7)

NULL

``

NULL


Download Directional_Derivative.mw

Dear All,

I'm trying to solve the following in Maple.

minimize(int(0.1e-3+.5*t+0.2e-2*t^2-b*t-a, t = 0 .. 300), location = true)

But Maple told me that the answer is

Float(-infinity), {[{a = Float(infinity), b = Float(infinity)}, Float(-infinity)]}.

I really need to get a kind of numerical answer. Would it be possible? Please Help me!!

First 1285 1286 1287 1288 1289 1290 1291 Last Page 1287 of 2224