@miguelbravo 

The difference LHS-RHS  is very small  (about 10^(-43)) for Digits=40.

It becomes 10^(-80)  for Digits=80  and 10^(-120)  for Digits=120.

So,  RHS-LHS is practically 0. It seems that your constants are not accurate enough or the model is not reliable.
 

@jamunoz 

I think I have explained that it's not possible.
For a given t you cant find a unique pair (t1,t2). E.g. what are t1 and t2  if t = 7?
Your problem is a mathematical nonsense. If it comes from another problem you want to solve then please show that problem and we could try to help.

@Markiyan Hirnyk 

But it's not a bug, it's just a notation (it appears in all the help  examples). Maybe it could be explained better (with that f@phi).
A mathematician would use F(u,v)  for (f@phi)(u,v) but Maple prefers not to use another name; and actually many engineers do the same.

@Markiyan Hirnyk

No, Maple is right. Keep in mind that f(u,v) is actually (f@phi)(u,v)  where phi is the diffeomorphism but Maple uses the same notation.

@Markiyan Hirnyk 

Yes, I also think that `a` should not appear. But if we add params = [a] then it is OK. It is true that in this case the new variables can be omitted. Maybe you are right that "it's not simple" :-)

P.S. I don't understand your PS

@jamunoz 

f must have either one or two variables.

Now, say you have a function f(t) of one variable.
You can define a function g of two variables:

g:=(t1,t2) -> f(t1/(k*m) + t2/(k*m^2));
# Let us compute:
d:=diff(g(t1,t2),t1)*k*m + diff(g(t1,t2),t2)*k*m^2;

#  Observe the "2";  this is the correct result.

# If you want to appear "t" you can use:

eq := t1/(k*m)+t2/(k*m^2) = t;
subs(eq,d);

You don't need and cannot use dchange for this.
 

@jamunoz

It is not a Maple limitation. It's maths!
You will have to explain what you are trying to do.

@Markiyan Hirnyk 

If the change of variables is e.g.  {x=u+v, y=u-v}, then the names of the new vars {u,v} are not necessary.

But for {x=u+v, y=u-v+a}, you must declare {u,v}. It's simple.

A change of variables is mathematicaly a diffeomorphism, so it it acts between spaces of same dimension.
E.g. you cannot have a diffeomorphism from R to R^2.

What result would you expect? f is a function of one variable and it seems that you want to treat it as having two.

@Markiyan Hirnyk 

Yes, the result is correct.

#  Sum( binomial(i,k)*(Sum(1/j!,j=i+1 .. infinity)),i=k..infinity);
#  Maple cannot find the final sum. Change the summation order:
Sum((Sum(  binomial(i,k)*   1/j!,  i=k .. j-1)),j=k+1..infinity);            
value(%);

@Markiyan Hirnyk 

But I have nothing new. I have said that the code executed in 2016.2 gives correct plots; the code was the same in your last worksheet. So, what is the problem? Normal people do not ask the same question twice.

@Markiyan Hirnyk 

Your worksheet is 2017.0 (probably beta).

After deleting the plot and reexecuting in 2016 it's OK.

RectangularWindow, as the help page says multiply the signal by 1, i.e. does nothing.
The reasons to do so seem to be:
- the user can see this as a comment, and knows that no windowing is applied
- the command does some checks and conversions (if needed).

As for the TriangleWindow, listplot was forgotten for Re(u), i.e.
listplot(Re( u ))
must be used.

Define:
n:=4;
dist:=add( (- m*x[i] - q + y[i])^2, i=1..n);
Then solve for m,q:

diff(dist,m)=0, diff(dist,q)=0;