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Hello Everyone.
e^x - (e^x*1i)^-1i - (e^pi)*sin( x + [pi/2] ) and symbolic solve fully ?
Thanks in advance for your time and help.
Regards.
Many thanks, F.M.
Regards.
Open video with windows media player.
The video shows that, in the first graph, by increasing the angle θ of the point z at each turn around the origin, the function of z, in the second graph, remains on the same branch. There isn't the jump on a new branch. Then the origin is not a critical branch point aand the function is single valued.
Hi, F.M. . I think my question is not clear, (sorry) :
Best Regards.
Hi Loi Le,
Hi, F.M.
I greatly appreciate your time and help. They are very, very helpful to me.
Best Regards.
Loi
Hi Loi Le,
I would solve the problem as follows:
Greetings
FM
I have lost track of which problem should be solved, but Mupads simplification
is wrong! Those two expressions are not equivalent.
Sometimes Mupad even has a slight idea of this problem and refuses further simplifications (see below) but ever so often it simplifies too simple and wrong, or lets say, incomplete, as it chooses arbitrary one specific value out of an infinite number of possibilities, ignoring all the others.
Here is a situation where MuPad at least is somewhat cautious:#
and from a mathematical point of view we have to consider that
even though "arg" is often limited to a certain range (like 0..2pi or -pi .. pi) for convenience reasons (to make it a unique function, like we do with arcsin, or the square root in the real domain). But in complex calculations of expressions we have to consider the many multiples of 2pi.
So we get
and therefore
Zero is just one of an infinite number of possible outcomes.
Hello Werner,
I consider the results I obtained accurate. Moreover, mathcad, for me, is not in error.
Bye
FM
To say, that the solution of x^2=4 is x=2 is also accurate and not in error. nevertheless its incomplete - thats my point.
I disagree.
1) Its not a function at all because its not unique valued.
2) Hope you agree that e^(j*z)=e^(j*(z+k 2pi))=e^(j*z)*e^(j k 2 pi)
And now (....)^(-j)
Its similar to z^(1/2)=...2 values or 1?
Its a question of definition of exponentiation which is different in R and C.
Mathcad's results are no proof for anything.
That was the point all the time! Mathcad does not respect the multitude of possible result but only chooses one.
The representation of a complex number with Real- and Imaginary part is unique, but the representation with magnitude and phase is not, unless you arbitrary limit the phase (to -pi to pi or to 0 to 2pi). Thats the reason for the multi-valueness of exponentiated expressions.
We have a similar case with the third root (or any other), which as you probably know, has three results in C.
Mathcad will only give you the so called "main value" (and this makes sense in many situations, even if we are missing the two other results)
This often is confusing for people as they would at least expect -2 as result. But the main value is defined as the one with the smallest phase.
Thats the reason we have
in Mathcad as sort of a compromise. But thats another story.
Mathcad will only present you all three results if you rewrite it as an equation
Nevertheless the same three results are valid for (-8)^(1/3) but Mathcad is not showing all of them.
We have a similar situation here:
I am sure why Mathcad does not simplify the constraint to
and I doubt that its a necessary constraint.
Best Regards
Loi
assume, x = real
Regards.
It follows my attempt to solve this equation:
Four days ago you wrote here:
and now you write
Bravo! I guess now finally you got it that this expression has more than one single "result"! At least here on earth - I'm not sure about heaven 😉
But there is no need to distinguish between k and n here. You may simplify 2 pi k + 2 pi n to just k'*2pi with k' in Z.
....Offenbar Wunder geschehen im Himmel und auf der Erde!
Doch alles hat eine logische und matematische Erklärung.
assume, x = real
Regards.
assume, x = real
Regards.
assume, x = real
Regards.