I said
You claim that j is an operator. Then j^j is raising an operator to a power
like i exampled in my previous post.
It is a Complex Number which's square is -1, named specially imaginary number

as an allusion.

Because you are saying that j is an operator and j is the imaginary number.
Then I can ask a question is multiplication a number ?
Or are there any operators which are numbers ?

Regards
Posted on 2003-05-16 15:06:20 by inFinie

I said

as an allusion.

Because you are saying that j is an operator and j is the imaginary number.
Then I can ask a question is multiplication a number ?
Or are there any operators which are numbers ?

Regards

I think you have to abort the thinking that j is like the real numbers. Its different because its not included in the subset of the complex numbers called the real numbers.

Therefore j has properties over and above those in the real number set.

One of those properties is that it acts as an operator based on the sign of other operators for other terms in an expression or for the magnitude of the power its raised to.

Complex

j^2 = -1

j^3 = -j

j ^4 = 1

j ^ 5 = j

Can you see how the numbers above are acting as operators on j (or vis versa) giving a result in four different quadrants.

There is your proof that numbers can be operators (at least complex numbers).

At the same time j maintains the properties of a real number.

Association...etc..etc
Posted on 2003-05-16 15:46:36 by IwasTitan
InFinie,

I just want to ask at this point: Have you recieved enough understanding to answer your origional question (how it relates to electronics)?

:NaN:
Posted on 2003-05-16 18:00:18 by NaN

There is your proof that numbers can be operators (at least complex numbers).

By that logic, the number 2 can be concidered an operator, because any number I multiply by it, I get back twice the orgional.

Have we all wasted enough time on this point yet?
Posted on 2003-05-16 18:03:30 by Ernie
inFinie,

Because you are saying that j is an operator and j is the imaginary number.
Then I can ask a question is multiplication a number ?
Or are there any operators which are numbers ?

I said that j is an operator. I never said that j is an "imaginary number". However, j can transform a real number in to a quadrature number by rotating it. But j by itself should be thought of as an operator. Look back at my previous posts. An operator is not a number, and I said that before. I also said before that most of the world writes down j alone and assumes everyone understands it means j(1) or 1 rotated by 90?. An example is 1+j. This notation gives j an ambiguous meaning as a value in the last example, or as a operator in a term like j30. To be precise at the expense of being a little verbose, it should be written like 1+j(1) or j(30), where j is considered strictly as an operator. Ratch
Posted on 2003-05-16 18:52:21 by Ratch
OK

Nan and Ernie are correct.

The subject has gotten way off base.

Anyways from my schooling i learned that complex numbers were used in electrical theory...not exclusively electronics.

As NaN mentioned...reactance and similar electrical quantities.

I'm done on the math thing.

:alright:
Posted on 2003-05-16 22:48:56 by IwasTitan
In algebra squareroot( sqrt(n) ) is defined in n>=0
If n is less than zero, sqrt is undefined. But consider the equation:
x^2+1=0
It is an equation an must have roots, indeed it has two roots cause of degree of 'x'.
x^2=-1

x=sqrt(-1) {this is a congruent root}

And scientists said (IMO) let i=j=x

If you have any incomprehansions i advice you to read and understand:
http://mathworld.wolfram.com/j.html
http://mathworld.wolfram.com/ImaginaryNumber.html

And in here, Turkey, a proverb says that "Talking constitutes talking", therefore nothing is off-topic, cause of relating w/former posts.

In z = a + bi form; a and b are real numbers, z is the complex number. I don't need to say what is i anymore.

I think the links are sufficient to end this discussion.
Posted on 2003-05-17 05:13:26 by inFinie
inFinie,

In algebra squareroot( sqrt(n) ) is defined in n>=0
If n is less than zero, sqrt is undefined.

Above statement is false, inFinie. Sqrt(n) can only be real if n>=0. If n<0, then root is CALCULATED to be ?j(sqrt(|n|)) or ?sqrt(|n|)/_90? in polar form, which I like to call a quadrature number and you would call an "imaginary" number. Note that ?j(sqrt(|n|)) or its equivalent ?sqrt(|n|)/_90? is NOT undefined when n<0.

In the links you gave, the author treats the j-operator as a constant and confuses you. Instead of writing j=sqrt(-1), he should be clearer and write j(1) = sqrt(-1), where j(1) means 1 rotated by 90?. Ratch
Posted on 2003-05-17 06:09:55 by Ratch
"I'm sorry Dave, this conversation can serve no further purpose."
Posted on 2003-05-17 21:42:09 by Ernie