## Strange error message using plot3d...

Asked by:

When trying to perform the following:

p1 := proc (x, y) if x^2+y^2 <= 1 then x*y-y^2 else 0 end if end proc;
plot3d(p1, x = -1 .. 1, y = -1 .. 1);
Error, (in plot3d) expected ranges but received x = -1 .. 1 and y = -1 .. 1

I get this strange error message. To the best of my knowledge x and y ARE provided as ranges. What am I missing/not understanding?

If I omit the ranges in plot3d Maple returns a correct plot, but the default range (-10 .. 10) does not display sufficient details

## Range of function ...

Asked by:

How can i calculate Range (Max. and Min.) of $f(x) = \sqrt{x^2+x+1}-\sqrt{x^2-x+1}$

## Multiplier in a specific range...

Asked by:

Dear all,

I am creating an animation, and I was wondering if I can add a multiplier to my equation in a specific range.  So starting from z:=0.2 add a multiplier (z+1). The code I have so far is added. Does anyone know a code for this?

Kind regards

restart;
with(plots);
a := -1/2; b := 1/2; c := -2; d := 2; n := 20;
g := proc (x) options operator, arrow; value(Int(sigma(t), t = 0 .. x)) end proc;
sigma := proc (z) options operator, arrow; 2*sqrt(2*h^2-4*z^2)*z/h^2 end proc;
h := i/n;
for i to n do
an2[i] := plot(sigma(z), z = -(1/2)*h .. (1/2)*h, view = [a .. b, c .. d], color = AQUAMARINE);
an3[i] := plot(2*g(x), x = 0 .. (1/2)*h, view = [a .. b, c .. d], color = RED)
end do;
p := plots[display]([seq(an2[i], i = 1 .. n)], insequence = true);
q := plots[display]([seq(an3[i], i = 1 .. n)], insequence = true); display(p, q)

## nested ode programs...

Asked by:

> restart;
> with(plots);
> n := [0, .5, 1, 5]; pr := .71; p := 0; l := [1, 2, 3]; b := 0; s := 0; L := [green, blue, black, gold];
[green, blue, black, gold]
> R1 := 2*n/(n+1);
2 [0, 0.5, 1, 5]
------------------
[0, 0.5, 1, 5] + 1
for j from 1 to nops(l) do; for j from 1 to nops(n) do        R1 := 2*n[j]/(1+n[j]);        R2 := 2*p/(1+n[j]); sol1 := dsolve([diff(diff(diff(f(eta),eta),eta),eta)+f(eta)*diff(diff(f(eta),eta),eta)+R1*(1-diff(f(eta),eta)^2) = 0, (1/pr)*diff(diff(theta(eta),eta),eta)+f(eta)*diff(theta(eta),eta)-R2*diff(f(eta),eta)*theta(eta) = 0, f(0) = 0, (D(f))(0) = l+b*((D@@2)(f))(0), (D(f))(-2) =1, theta(0) = 1+s*(D(theta))(0), theta(-2) = 0], numeric, method = bvp); fplt[j]:= plots[odeplot](sol1,[eta,diff(diff(f(eta),eta),eta)],color=["blue","black","orange"]);         tplt[j]:= plots[odeplot](sol1, [eta,theta(eta)],color=L[j]); fplt[j]:= plots[odeplot](sol1,[eta,diff(f(eta),eta)],color=L[j]);      od:od:

Error, (in dsolve/numeric/bvp) unable to store 'Limit([eta, 2*eta, 3*eta]+eta^2*[.250000000000000, .500000000000000, .750000000000000]-.250000000000000*eta^2, eta = -2., left)' when datatype=float[8]
> plots:-display([seq(fplt[j], j = 1 .. nops(n))], color = [green, red], [seq(fplt[j], j = 1 .. nops(l))]);

> sol1(0);

Dear sir

In the  above problem i tried to write a nested program but its not executing and showing the error as Error, (in dsolve/numeric/bvp) unable to store 'Limit([eta, 2*eta, 3*eta]+eta^2*, i want the plot range from -2 to 2 but taking only 0 to -2 ,and -2.5 to 3 but taking only 0 to 1

## How to specify solution range for solve...

Asked by:

Hey,

I want to solve this equation and looking at the plot there are at least 3 solutions. I want the greatest/smallest negative solution. Unfortunately using solve with assumptions produces no results and solve without assumptions only finds two solutions.

Can you please help me?

 > #select greatest negative value from solution
 > restart:
 > expr:= ax*cos(lambda)+ay*sin(lambda)-(a+b*lambda)
 (1)
 > ax:=1:ay:=2:a:=0.5:b:=0.25: #examplanatory values
 > plot(expr)
 >
 > assume(-2*Pi
 >
 > sol_lambda:=[solve(expr=0,lambda, useassumptions)];# returns empty list even though without assumption one solution is found
 (2)
 > sol_lambda:=[solve(expr=0,lambda)]; #returns only two solutions even though looking at the plot 3 are there
 (3)
 > sol_l_v:=evalb~(sol_lambda<~0); #dirty workaraound
 (4)
 > sol_l_add:=[ListTools:-SearchAll(true,sol_l_v)] ; #this seems overly complicated
 >
 (5)
 > lambda:=sol_lambda[sol_l_add[-1]];  #to select the last entry
 >
 (6)
 > expr; #test
 >
 (7)
 >

Download select_solution.mw

Thanks!

Honigmelone

## Why are ranges exceeded?...

Asked by:

eq1 := z = y*log(x): eq2 := z = y+x*log(x):

DispIntersecting := implicitplot3d([eq1, eq2], x = 0 .. 10, y = -30 .. 30, z = -40 .. 40, color = [blue, green]):

solve({eq1, eq2}, [x, y, z]);

assign(%):

DispIntersection := spacecurve([x, y, z], x = 0.1e-2 .. 10, color = red, view = [0 .. 10, -30 .. 30, -40 .. 40]):

display(DispIntersecting, DispIntersection, axes = boxed, scaling = constrained);

## Need to restrict range on a graph...

Asked by:

Hello,

I have a procedure which plots a graph. I need the x-axes, which in this case is theta, to range between -3 and +3. However, I am not sure as to how I can create this restricted range. Any help is greatly appreciated! Thank you in advance!

Kind regards,

Gambia Man

 >
 >
 >
 >
 >
 >
 >

Download Poincare_section_Boyd_plot.mw

 >
 >
 >
 >
 >
 >
 >

Download Poincare_section_Boyd_plot.mw

## Solving a pde system...

Asked by:

restart;

with(DETools, diff_table);

kB := 0.138064852e-22;

R := 287.058;

T := 293;

p := 101325;

rho := 0.1e-2*p/(R*T);

vr := diff_table(v_r(r, z));

vz := diff_table(v_z(r, z));

eq_r := 0 = 0;

eq_p := (vr[z]-vz[r])*vr[] = (vr[]*(vr[r, z]-vz[r, r])+vz[]*(vr[z, z]-vz[z, r]))*r;

eq_z := 0 = 0;

eq_m := r*vr[r]+r*vz[z]+vr[] = 0;

pde := {eq_m, eq_p};

IBC := {v_r(1, z) = 0, v_r(r, 0) = 0, v_z(1, z) = 0, v_z(r, 0) = r^2-1};

sol := pdsolve(pde, IBC, numeric, time = z, range = 0 .. 1);

what am I doing wrong?

it's telling me: Error, (in pdsolve/numeric/par_hyp) Incorrect number of boundary conditions, expected 3, got 2
but i did just as in the example :-/

## extract range from solution...

Asked by:

Is there any simple way to extract the feasible range for the variables from solve result ?

For example, when I do solve({x-y = 10, x+y < 100}) I get {x = y + 10, y < 45}

From here, I need -inf<y<45 and -inf<x<55. Is it possible ? (I tried few methods, but it is not working for all cases.)

## Maple error for NonlinearFit...

Asked by:

I have been trying to fit a function to experimental data. To do this i was using

.

The data is of the type

.

When I use initialvalues, I get a result, that fits the data well, but is clearly not the desired minimum. Maple delivers g always bigger than 10000 which is nowhere near -7, where it has to be for physical reasons. When using parameterranges I get the error

Warning, no iterations performed as initial point satisfies first-order conditions.

He stopps computing and simply prints out my initialvalues or the first value that is in the parameterrange with a huge RSS.

How can I use initialvalues and parameterranges together for my data?

## How to adjust color hue...

Asked by:

In a 3d coordinate system I have a circular spacecurve with z-minimum -4 and z-maximum +4. In the same 3d coordinate system I have a 3d surface plot with z-minimum -0.5 and z-maximum +1.3 . When I choose the color option "Z(Hue)" in order to color-code the z-values on the 3d surface and make the topography more clear, I mostly get a totally green 3d surface. It seems that the color scaling is coupled with the spacecurve with z-values of +-4 . How can I uncouple the color scaling from the spacecurve and couple it with the z-range of the surface, while the color-limits shall be at +-1.3 ?

## Plotting a function over a range...

Asked by:

I've got a piece-wise function(for which I've made the procedure) f defined over x<=-1, -1<x<1 and x>=1 which I am trying to plot over the range of (-2,2). I've tried using plot(f,-2..2) but it doesn't show any curve. Should I add a few more parameters to plot()?

## How to fix: 'Error, (in int) integration range' ??...

Asked by:

int(int(y, 0 <= y, x^2+y^2+z^2 <= 1));

Error, (in int) integration range or variable must be specified in the second argument, got 0 <= y

int(int(y, y = 0, x^2+y^2+z^2 = 1));

Error, (in int) integration range or variable must be specified in the second argument, got y = 0

## How to limit a function to an interval...

Asked by:

Hello,

I have a function, lets say g(x,y,...), that depends on many other functions. But I don´t want the results that are inside certain intervals, and I need to receive the results of those functions as something like NULL when asking for a result that is inside any of those intervals. That way, g(x,y,...) would also have to result in something like NULL if any of the lesser functions are NULL fora any given values.

I tryied using the piecewise command, and for the intervals that I wanted, it worked, but for those I wanted to be NULL, they were understood as 0, and so G(x,y,..) continued to exist but with a very different value.

To clarify what I need, I will try an exemple:

Imagine I have the function f(x)=x

I want to disconsider the results for x<2 and x>6, in a way that if I try the command 'f(1)', I will receive something like NULL and know that it is outside the range.

In the same way, I need the plot of this function f(X) to show the function only from 2 to 6, but not existing for the delimited intervals.

Ad if I continue and make g(x)=f(x)+10 , I don´t want g(x) to exist if f(x) doesn´t exist, and same for the g(x) plot, which shouldn´t be shown in the intervals where f(x) don´t exist.

Thank you very much for your atention!

## How to create command...

Asked by:

Hi,

I want to calculate a range.

eg. j=1...n-1 will use this equation

P(j,n):={((m/(2-1))(m/(2))^(j-1))/(2(m/(2))^(n)-1))}

j=n will use other equation.

How to create the command? I keep receive errors.

Thank you.

Lina

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