This is first type of solving parameter but  i solved some semilar but the shape of them are so different and solving them different too i did my trail but i can't reach out solution also the author solve by another way which i did too but i have to solve in this way too thanks  for any help !

Download F-P-O-W.mw

sometime this code i use make a problem for me and sometime they work very good how i can get something general for use in all equation 

Download seperate_L-NL.mw

my answer is so different but  and i want remove this lambert in my test? can we do something for giving question exactly solve the question by model i have a book of ode which a lot of time i do some trail but have a problem? how i can remove this issue specially for bernoli and  other type?

Download test.mw

A lot of time i finded but i have a dubt about this why this is happen each time number of equation for finding parameter a_12 is 4 but this time is 28 which i thoght some thing must be mistake also the author of paper use  u=2(ln(f))_xx which is wronge and not satisfy but i try to find R which is strange again is not number contain parameter but is satisfy also in equation 14 i don't know each i is 2 or 1 or it can be i remain itself?

thanks for any help ?

t1.mw

In this example by applying the substitution i can get half of paicewise function but how get another  half ? i am looking for B_rs as Piecewise function ?

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why i get error in end and how i can fix this error?

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How do I change the legend position so it doesn't cover the plot label?

Before entertaining non-standard solutions to this problem such as building my own from scratch, is the a standard method for changing the position of the default legend generated along with a standard plot? I couldn't find anything about custom legend positions in the help sheets.

I can't help but wonder why this observed behaviour occurs by default? In what kind of plot would I prefer the plot labels to be obscured by the plot legend?

legend_covers_plot_labels.mw

Can I open Maple 2025 files in Maple 2024? Further, what if the files don't use Maple 2025 features/packages? Does that change the outcome?

Using inttrans package fourier (Dirac(t-T),t,w) gives the correct answer

exp(-I*T*w).

Taking immediatel the inverse transform I get only

invfourier(exp(-I*T*w), w, t),

but the the expression remains unevaluated and cannot be brought to evaluate by any means that I know. Funny thing: Taking just plain Fourier Transform and Inverse in the usual integral form "int" it works flawlessly. Apparently Maple knows how to deal with distributions in this context. What am I doing wrong?

in a lot of my function i have a interval which is make my function singular and i don't know how remove this singularity even when i am change a lot of parameter with explore which explore option for plot is a little bit heavy for more than  7 or 8 parameter for running , and i know the shape of the graph is 2-soliton and 1-breather(zig-zag) but i have to see the shape and make my plot have a best shape  there is any idea for fixing this issue?

singular-interval.mw

I tried the following procedure in a worksheet; Maple did not like it and claimed there was an error. However, I cannot even copy this to a Maple prompt; it jumps to another type of region. Any ideas? If I retype the command there is no problem with an error.

It reminds me of Maple 2 and the letter t which sometimes had to be retyped to get Maple to respond-a very strange bug which was eliminated years ago.

Hey guys, 

I try to solve big systems of polynomial equations and inequalities. Therefore I use the command SemiAlgebraic. In the moment I take those result and want to go on calculating with them. Sadly it turns out, that solve has some problems with RootOf expressions. It doesnt find a solution (althoug the graph shows that there is one) and gives the warning solution may have been lost. So now I though I might just aks SemiAlgebraic to give me solutions without RootOf expressions. For example you can write {x = RootOf(_Z^2 - y)+1, 1 < y, y < 2} as {x=t+1, y=t^2,1<t<2^0.5 . This might be easier to work with for solve. 

So my question is: Is there any way I can tell SemiAlgebraic precisely in what form the solution should be? 
Since the websites are down Im not able to do a first own research on this problem. So thank you in advance. 

Regards

Felix

in some equation i don't have problem but in a lot of them this problem is come up for me and i don't know how fix this issue?

Download critical-point.mw

Hey guys, 

From a former calculation I got a set of points as a implicit RootOf function for an intervall. Now I want to check, if these points are in a certain area. So i thougt I take the RootOf function, the intervall and the inequalities (which describe the target area) and use the solve command. But then I get the warning, solutions may have been lost and no solution. When you draw the implicit function you can see thats in the right area (above y=1 and below y=x). So there should be a clear anwer, giving me back the whole RootOf function in the intervall.

Download QUESTI~2.MW

Since there was an error uploading the picture here the code 

restart;
Sol := {x = RootOf(_Z^2 - y, index = real[2]) + 1, 1 < y, y < 2};
area := {1 < y, y < x};
Sol_area := solve(Sol union area);
print(Sol_area);

So why do I get this warning, the calculation seems quite easy? And is there a workaround? Or a diffrent kind of solve function? SemiAlgebraic is as far as i know only for polynomials. So I got an error as well. Since the websites are down I could start an own reasearch before. So thank you in advance. 

Regards

Felix

Hey guys, 

I have a problem with the solve command. And since the websites are down, I cant help myself. I have a function x(y) and an intervall for y. This function or the set of points described by the function should now be transformed to another area of the plane. So now I can get a(x,y)=1/y and b(x,y)=x/(x+y-1) with just plugging in, then everything depeends from the y Invtervall. But I want to get the form b(a) and an intervall for a. So that I can see the function directly. So i thought I just put everything into the solve command and than ask for a solution for {a,b} and expected to get what I want. (Uploading the script here gives an error)
 

restart;
Sol := solve({a = 1/y, b = x/(x + y - 1), x = (y - 1)^2, 1 < y, y < 5/4});
Sol_ab := solve({a = 1/y, b = x/(x + y - 1), x = (y - 1)^2, 1 < y, y < 5/4}, {a, b});

#expected (or wanted) solution
#with y = 1/a and the inequalities we get 4/5 < a and a < 1 with y
#with y = we get x(a)/a and y(a) and reach b = 1/a - 1
#all together: Sol={b=1/a - 1, 4/4<a, a<1};

THe problem is, that I get an empty set which is obviously wrong. So I somehow make an error when making the variables I want concrete with adding {a,b}. What do I do wrong? Or is there a better command for what I want to achieve? In this case I can solve the problem via hand, but I have more complex tranformations and mor ecomplex functions x(y), so thats why I ask for general help with my problem. 

Thank ypu in advance

Felix

Download QUESTI~1.MW