Given the ode   y''(x)*y'(x)=0, clearly it has solutions for y''=0 and y'=0.

These are y=c1+x*c2 and y=c1. But Maple gives 3 solutions

Download why_3_solutions_may_24_2024.mw

Where did the third solution come from? The 3 solutions are correct ofcourse, but why 3? There should only be two. 

FYI, I am using 

Physics:-Version()

The "Physics Updates" version in the MapleCloud is 1746. The 

   version installed in this computer is 1745 created 2024, May 

I have to check earlier Maple versions to see if same thing happens there.

This ode can be solved by just looking at it

ode:=(x+y(x))*diff(y(x),x)=0;

We see the solution is y=-x and y=c__1 because either (x+y)=0 or y'=0

But for some reason ODESteps(ode) says it cannot compute integral.

Any idea why?

Download odestep_quadrature_unable_to_solve_maple_2024.mw

update:

Here is another simpler example that also confused it
 

Download odestep_quadrature_unable_to_solve_v2_maple_2024.mw

update

Here is another one which it gets wrong. 
 

The correct solution is given by dsolve, which is y=0 and y=constant (I do not know why dsolve put minus sign in front of the constant, but it is still correct).

Download odestep_quadrature_unable_to_solve_v3_maple_2024.mw

I am stuck this command works seemlessly in Maple:

ThermophysicalData:-CoolProp:-Property(D, T = 20*Unit('degC'), P = 760*Unit('mmHg'), water)

but it does not work in Maple Flow. Does anyone knows why? Thank you so much for your help in the matter.

Why does 'simplify' not work when calculating Eigenvectors? Further, how can we express (2) in a more simplified form by using 'simplify'?simplify.mw

Is there a shortcut in Maple that will comment out a block of highlighted code? I tried searching on Mapleprimes but everytime I search I get a page generation error.

Hello :) 

I have a math problem, where I first need to use Linear regression to find the equation based on a set of data. I did that, no problem. 

However, in the next part of the problem I need to check if the residuals are under "normal distribution". Usually, I check if a dataset is normally distributed via "QQ-plot", and there will be no problems. But this time, because I need to check the residuals, I need to use the "residualQQplot(data,LinReg)" command to make it happen. But when I read the mean-value, mu, it says "-0," and nothing else? I know it should be "-3,2752*10^-15. 

The standard deviation is correct.

How do I fix this, so the residualQQplot shows me the right result? 

I have attached the worksheet here. worksheet_-_linear_reg_and_residuals_for_normal_distribution.mw

Thank you! 

what is the mathematics behind isolve? How can one selct special solutions?

Hello :) 

I don't know how to further simplify a complex equation as a solution to a differential equation. I have already tried "simplify(%)" etc. but it only repeats the same equation. I also know that it can be simplified to a much more "nice" equation in WordMat and TI-Nspire. I have attached a screenshot of my work. 

Thank you! 

Download worksheet_.mw

I can't understand this behavior. Any idea why it happens?

Solve is able to solve equation   f(y)=x+A for y, but can't solve   f(y)=x for y.

This is unexpected for me. I do not see why it can solve it when RHS is x+A but not when RHS is just x.

I can trick it to solve  f(y)=x for y  by asking it to solve f(y)=x+A for y and then set A=0 in the solution. But one should not have to do this. Is this a bug or Am I missing something?

Download why_solve_when_adding_term_only_may_22_2024.mw

Dear Colleague. 

I am trying to improve the results of abs(res[i] - exy) in the following codes.

restart;
Digits := 30:

# Define the function
f := proc(n)
    -0.5*y[n] + 0.5*sin(x[n] - Pi)
end proc:

# Define equations
e1 := y[n+2] = 2*h*delta[n] + y[n] - h^2*(-2*sin(u)*f(n)*u^2 - 2*sin(u)*f(n+2)*u^2 + 2*sin(2*u)*f(n+1)*u^2 + 2*cos(u)*f(n)*u - 2*cos(u)*f(n+2)*u + 2*cos(2*u)*f(n+1)*u - 2*cos(2*u)*f(n)*u - 2*sin(u)*f(n) + 2*sin(u)*f(n+2) + sin(2*u)*f(n) - sin(2*u)*f(n+2) - 2*f(n+1)*u + 2*f(n+2)*u)/((2*sin(u) - sin(2*u))*u^2):
e2 := y[n+1] = h*delta[n] + y[n] - (1/2)*h^2*(-sin(u)*f(n)*u^2 - sin(u)*f(n+2)*u^2 + sin(2*u)*f(n+1)*u^2 + 2*cos(u)*f(n)*u - 2*cos(u)*f(n+2)*u + 2*cos(2*u)*f(n+1)*u - 2*cos(2*u)*f(n)*u + 4*sin(u)*f(n+1) - 4*sin(u)*f(n) - 2*sin(2*u)*f(n+1) + 2*sin(2*u)*f(n) - 2*f(n+1)*u + 2*f(n+2)*u)/((2*sin(u) - sin(2*u))*u^2):
e3 := h*delta[n+2] = h*delta[n] + h^2*(2*sin(u)*f(n)*u + 2*sin(u)*f(n+2)*u - 2*sin(2*u)*f(n+1)*u - 2*cos(2*u)*f(n+1) + cos(2*u)*f(n) + cos(2*u)*f(n+2) + 2*f(n+1) - f(n) - f(n+2))/(u*(2*sin(u) - sin(2*u))):

with(LinearAlgebra):
epsilon := 10^(-10):
inx := 0:
ind := 1:
iny := 0:
h := 0.01:
n := 0:
omega := 1:
u := omega * h:
tol := 1e-4:
N := solve(h * p = 8 * Pi, p):

err := Vector(round(N)):
exy_lst := Vector(round(N)):

c := 1:
for j from 0 to 2 do
    t[j] := inx + j * h:
end do:

vars := y[n+1], y[n+2], delta[n+2]:

step := [seq(eval(x, x = c * h), c = 1 .. N)]:
printf("%6s%15s%15s%16s%15s%15s%15s\n", "h", "Num.y", "Num.z", "Ex.y", "Ex.z", "Error y", "Error z");

st := time():
for k from 1 to N / 2 do
    par1 := x[0] = t[0], x[1] = t[1], x[2] = t[2]:
    par2 := y[n] = iny, delta[n] = ind:    
    
    res := eval(<vars>, fsolve(eval({e1, e2, e3}, [par1, par2]), {vars}));

    for i from 1 to 2 do
        exy := eval(sin(c * h)):
        exz := eval(cos(c * h)):
        printf("%6.5f%17.9f%15.9f%15.9f%15.9f%13.5g%15.5g\n", h * c, res[i], res[i+1], exy, exz, abs(res[i] - exy), abs(res[i+1] - exz));
        
        err[c] := abs(evalf(res[i] - exy));
        if Norm(err) <= tol then 
            h := 0.1 * h * (c + 1) * (tol/Norm(err))^(0.2);
        else 
            break
        end if;
        exy_lst[c] := exy;
        numerical_y1[c] := res[i];
        c := c + 1;
    end do;
    iny := res[2];
    ind := res[3];
    inx := t[2];
    for j from 0 to 2 do
        t[j] := inx + j * h;
    end do;
end do:
v := time() - st;
v / 4;
printf("Maximum error is %.13g\n", max(err));
NFE = evalf((N / 4 * 3) + 1);

# Get array of numerical and exact solutions for y1
numerical_array_y1 := [seq(numerical_y1[i], i = 1 .. N)]:
exact_array_y1 := [seq(exy_lst[i], i = 1 .. N)]:

# Get array of time steps
time_t := [seq(step[i], i = 1 .. N)]:

# Display graphs for y1
with(plots):
numerical_plot_y1 := plot(time_t, numerical_array_y1, style = point, symbol = asterisk, color = blue, symbolsize = 20, legend = ["TFIBF"]);
exact_plot_y1 := plot(time_t, exact_array_y1, style = point, symbol = box, color = red, symbolsize = 20, legend = ["EXACT"]);

display({numerical_plot_y1, exact_plot_y1});
Error_plot_y1 := plot(time_t, err, style = line, symbol = box, tickmarks = [piticks, decimalticks], color = navy, labels = [`h=Pi/8`, typeset(`Absolute Errors`)]);

I am suspecting that I didnt update the new h properly (I may be wrong, though). Please kindly help modify the code to allow the values of abs(res[i] - exy) to about 10^(-11). Thank you and best regards.

I am trying to solve the following recursion for any n, given a constant c. Here is my code for it:

c := 2:

A[i] := rsolve({a(0) = 1/n, a(i) = ((n - i + 1)/(n - i) + 1/(c*(n - i)))*a(i - 1)}, a);

total := evala(Simplify(sum(eval(A[i], i=k), k=0..n-1)));
evalf(eval(total, n = 6));

For c = 1, I get a valid (and correct) output, however for c = 2 for example, rsolve is returning , which does not make sense when n is an integer. Is there something I am doing wrong here? Not sure why this is happening. Thanks!

Include print level in procedure
In the procedure code printlevel is not accepted
-enviroment variable
-interface variable 
error message : Error, (in interface) unknown interface variable, printlevel

More convenient in my opinion is to include on the printlevel depth in the procedure call?

Download MP_vraag_printlevel_in_procedure_-lukt_niet.mw

Hello.

I am very new to Maple. Many this are great, but I do not understand how maple deals with, especially, radiological units. In particular regarding joules [J] wich Maple seems to have alt least three types:

1) J - in relation to work

2) J(radiation) in relation to Gy (Gray) J/kg

3) J(dose_equivalent_index) in relation to Sv also J/kg

Why does Maple distinguish between these "joules"? As a phycisist they are all (well maby not entirely for Sv) equal to me. How can I make Maple treat them all at the "same joule"?

I have tried the following first:

with(Units[Standard]) and

with(Units[Natural])

Best,

Carsten

Let S1: x^2 + (y - 2)^2 + (z + 1)^2 = 29 be a sphere and two points A(0, 0, 4), B(6, -2, 6); the line d passing through point C(4, -8, 4) and have direction v=(1, -1, 2). Find the point M such that M lies on the sphere S1, the angle AMB equals to 90 degree and distance from M to the line d is minimum. 

If I understand correctly, both of 

int(RETURN(is(y::positive)), y = 0 .. x) assuming 0 <= x, x < 1;
int(RETURN(coulditbe(y = 1)), y = 0 .. x) assuming 0 < x, x < 1;

should output . However, Maple simply returns  for the second one. 
Isn't this result incorrect? Or am I missing something?