Complex no

[Saumyojit]

https://math.berkeley.edu/~giventh/complex.pdf

Operation of complex conjugation respects sum and products

means what?

The Result of sum or product of each conjugate term will give me another complex no .
(i.e Closed under complex no with respect to addition and multiplication)
_ _
Z + W ( addition of two Complex nos)
Gives me another complex no
Correct?




Wiki page of exponentiation

Conjugation is commutative under composition with exponentiation to integer powers

Commutative operation means changing the position of operands and getting same result.
this is the expression
( z^n)' = (Z') ^n for all n belongs to integer. ' denotes conjugation

Now if i have a expression like this a+b ; now if i want to express the idea of commutation then i would interchange b and a -> b+a right?
Here i understood and demonstrated "changing the position of operands"

Also in non commutative operation like exponentition i can change the order of operands a^b to this -> b^a .

But if my expression is like this ( z^n)' i want to commute operands z and n wrt to conjugation not exponentiation what should be my approach ?

 

[Dr.Peterson]

Conjugation is commutative under composition with exponentiation to integer powers​

means what?

They tell you what it means:

The operation of complex conjugation respects sums and products: [MATH]\overline{z+w}=\bar{z}+\bar{w}[/MATH] and [MATH]\overline{zw}=\bar{z}\bar{w}[/MATH].​

It means that the conjugate of a sum is the sum of the conjugates, and similarly for products.

Wiki page of exponentiation

Conjugation is commutative under composition with exponentiation to integer powers​

Commutative operation means changing the position of operands and getting same result.
this is the expression
( z^n)' = (Z') ^n for all n belongs to integer. ' denotes conjugation

You lied about where this is quoted from; I had to search to find it is in https://en.wikipedia.org/wiki/Complex_conjugate

Again, they show what they mean, though I can understand that you don't understand, because, as we have repeatedly told you, this is not material you are ready for. STOP READING WIKIPEDIA! Here is what they say:

Conjugation is commutative under composition with exponentiation to integer powers, with the exponential function, and with the natural logarithm for nonzero arguments.​

if z is non-zero​

Here is what that means: Composition means doing one thing after another; when you do any of the three things listed (raising a variable to a power [exponentiation], raising e to a power [exponential function], and natural logarithm), you can interchange the order in which you do that and conjugation. It's the same idea as "respecting" addition and multiplication. Composition "passes through" any of those operations unchanged.

Now if i have a expression like this a+b ; now if i want to express the idea of commutation then i would interchange b and a -> b+a right?
Here i understood and demonstrated "changing the position of operands"

Also in non commutative operation like exponentition i can change the order of operands a^b to this -> b^a .

But if my expression is like this ( z^n)' i want to commute operands z and n wrt to conjugation not exponentiation what should be my approach ?

In the case of composition, the "operands" are the operations, e.g. conjugation and exponentiation! You are changing the order of two functions, not of two numbers. I'm guessing you don't fully understand what composition means, at least at this abstract level.

Composition of two functions f and g produces a new function [MATH]f\circ g[/MATH], defined by [MATH](f\circ g)(x) = f(g(x))[/MATH]. If the two functions are commutative under composition, that means that [MATH]f\circ g = g\circ f[/MATH], i.e. [MATH]f(g(x))=g(f(x))[/MATH]. That's what they're talking about. For example, you could say [MATH]conj(exp(z))=exp(conj(z))[/MATH].

 

[JeffM]

First, the English in this question is quite poor and thus hard to understand. I have no idea what you are even trying to say under your wiki quotation.

Second, you are being quite rude. You give no link to your wiki quotation (the link you do give is to the wrong article) and no indication where in a long article the sentence you are asking about occurs.

Third, here you go again flying off in different directions. One source is talking about complex numbers; the other about abstract algebra.

Fourth, if I understand what you are saying about conjugation of complex numbers, it is correct, but it is not what the statement is getting at.

 

[Saumyojit]

conjugate of a sum is the sum of the conjugates,

ok that's the other way of saying The operation of complex conjugation respects sums . They could have simply said
"conjugate of a sum is the sum of the conjugates"


TheDistributive law has been used or property has been used to be technically correct

STOP READING WIKIPEDIA

I started reading to know about complex conjugation .but did not realize this part is from abstract algebra.

You lied

It was a mistake.

First, the English in this question is quite poor and thus hard to understand. I have no idea what you are even trying to say

Extremely sorry. I read it and I understood it .
Also dr p understood.
Thanks btw.

 

[Dr.Peterson]

ok that's the other way of saying The operation of complex conjugation respects sums . They could have simply said
"conjugate of a sum is the sum of the conjugates"

They could; but the symbols say that more clearly. They are trying to teach you a new term to use! Learn it for the future ...

The Distributive law has been used or property has been used to be technically correct

Why do you say that?

It was a mistake.

I intentionally used a strong word, along the same lines as JeffM's "rude", to call your attention to the fact that if you are not careful to make yourself understood correctly, you are making things hard for the person you are talking to. Mistakes happen, but if you respect other people you will try to avoid them. (By the way, some of the most knowledgeable helpers here make a lot of typos that I consider just as rude, because it implies they don't respect others enough to proofread. You aren't alone in "lying".)

 

[Saumyojit]

In wiki page of exponentiation

The identities (bc)^x = b^x.c^x and
(b/c)^x = b^x/c^x are valid when b and c are positive real numbers and x is a real number.

The above statement has been said

Now if I take b as a negative integer then

Suppose b=-3 ,c=2, x=2

(bc)^x = b^x.c^x

(-3*2)^2= -3^2 . 2^2
36=36

I took negative real no but the identity still works how

 

[pka]

In wiki page of exponentiation
The identities (bc)^x = b^x.c^x and
(b/c)^x = b^x/c^x are valid when b and c are positive real numbers and x is a real number.
The above statement has been said
Now if I take b as a negative integer then
Suppose b=-3 ,c=2, x=2
(bc)^x = b^x.c^x
(-3*2)^2= -3^2 . 2^2
36=36
I took negative real no but the identity still works how

You are abusing the notation: \((-3)^2\ne-3^2\) because \(-3^2=-9\).

 

[Dr.Peterson]

In wiki page of exponentiation

The identities (bc)^x = b^x.c^x and
(b/c)^x = b^x/c^x are valid when b and c are positive real numbers and x is a real number.

The above statement has been said

Now if I take b as a negative integer then

Suppose b=-3 ,c=2, x=2

(bc)^x = b^x.c^x

(-3*2)^2= (-3)^2 . 2^2
36=36

I took negative real no but the identity still works how

I've corrected what you wrote to express what you presumably mean. But there is something far worse than what pka pointed out.

You are misunderstanding basic logic. When you are told, If A then B, that says nothing about what happens when A is not true! So saying "valid when b and c are positive real numbers" does not mean "NOT valid when b and c are NOT positive real numbers"!!!

They are saying nothing about what happens when b and c are not both positive. They are making a statement about what can be stated for all positive real numbers (when the exponent is not necessarily an integer).

It would have been very nice of you to include a link to the page, to make it as easy as possible for me to check out the context. The quote is from far down in the page where they are talking in the context of complex variables, which is far beyond your knowledge! Specifically, it's from https://en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities, and they then say,

On the other hand, when x is an integer, the identities are valid for all nonzero complex numbers.​

Did you not bother to read a couple lines further???

 

[Saumyojit]

In wiki page of expo under "rational exponent" part there is this paragraph

This sign ambiguity needs to be taken care of when applying the power identities. For instance:
The problem starts already in the first equality by introducing a standard notation for an inherently ambiguous situation –asking for an even root– and simply relying wrongly on only one, the conventional or principal interpretation

the whole para till the end.

The central idea that they are trying to convey is that in this identity
(a^b)^c=a^(b*c)
When the no a =-ve no the identity fails to hold!

Q1 :What is the first equality they are talking about?
-27=(-27){ {2/3 * 3/2} } (this is the first equality)??

Q2: introducing a standard notation .

What is the standard notation.?

They say asking for a even root . In the first line after

For instance:

in the group of expressions that I see even root implies to what I feel is this part "9^3/2"
Right? We are asking the square root of 9^3

[As I know even root will give us two real solns (one +ve and -ve )to the root of a positive no]

What does "Conventional or principal interpretation" mean?
The principal +ve root of the radicand?

 

[JeffM]

You do make us work to try to help you. I puzzled for almost a minute over "no a = -ve." If "a is negative," is English, not gibberish.

No, you are slightly missing the point.

What is being got at is if the base is negative and exponents are rational numbers rather than integers, there may be no answer in the real numbers. This CREATES an ambiguity in certain cases

[MATH](- 27)^{\{(2/3) * (3/2)\}}= (- 27)^1 = - 27.[/MATH]
[MATH](- 27)^{\{(2/3) * (3/2)\}} = \{(- 27)^2\}^{\{(1/3) * (3/2)\}}= (729)^{1/2} = 27.[/MATH]
[MATH]\therefore 27 = - 27.[/MATH]
Rather obviously, something is wrong with that conclusion, which we reach

IF WE ASSUME THAT WHEN WE TAKE AN EVEN ROOT OF A POSITIVE NUMBER, WE MUST TAKE THE POSITIVE ROOT.

In short, we cannot decide which even roots of a negative number are relevant.

 

[Dr.Peterson]

Q1 :What is the first equality they are talking about?
-27=(-27){ {2/3 * 3/2} } (this is the first equality)??

Yes, that's what they are calling "the first equality" in that line.

Q2: introducing a standard notation .

What is the standard notation.?

I'm not sure exactly what they mean by that; I see no actual error at that point, but later, where the 3/2 power is really a two-valued "function". I wouldn't be surprised if this paragraph is the result of several people in a tug-of-war editing it. (As we've discussed before, not everything in Wikipedia should be assumed to make sense at a particular point in its editing history, so you shouldn't try too hard to make sense of every detail.)

in the group of expressions that I see even root implies to what I feel is this part "9^3/2"
Right? We are asking the square root of 9^3

[As I know even root will give us two real solns (one +ve and -ve )to the root of a positive no]

What does "Conventional or principal interpretation" mean?
The principal +ve root of the radicand?

Yes, the even root refers to the 3/2 power, which as you say really has two values; the "conventional or principal interpretation" refers to treating that as the [principal] root (the positive one), rather than retaining both signs. I think you're understanding them correctly at this point.

Ultimately, the problem is simply that your (a^b)^c=a^(b*c) does not apply when you use only the principal root.

 

[Saumyojit]

I will come back to post 11 in a short but before that can you tell me ;
In Wiki page of Exponential function

In mathematics, an exponential function is a function of the form
f(X) =a*b^x

Why did they give 'a' for?
They could have simply equated to b^x

 

[Otis]

Why did they give 'a' for?

It's to make a general definition, Saumyojit, because most exponential functions have a coefficient that is not +1. Without the symbolic coefficient (symbol 'a'), the Wiki definition would ignore a lot of exponential functions.

Here's another example, for the same reason. We don't define quadratic polynomials as x^2+bx+c. Most quadratic polynomials have a leading coefficient other than +1; therefore, the general definition is ax^2+bx+c.

?

 

[Dr.Peterson]

can you tell me ;
In Wiki page of Exponential function

Why did they give 'a' for?
They could have simply equated to b^x

The quick answer, as Otis said, is that this is a definition, which means it can be whatever someone chooses to use the word to mean, if others accept it. And not all functions of the form ab^x can be written as b^x, since the latter is always equal to 1 when x = 0, so the two definitions would not be equivalent. That definition is used (for example, in typical precalculus textbooks) because it is useful; there are many important problems in which such exponential functions appear.

But it is indeed odd that they start with that definition, and then, as far as I can see, they never mention the coefficient "a" again after the first couple paragraphs. So the article titled Exponential function is not really about what they say it is about at the start! The rest of the article is about b^x. In fact, most of it is specifically about e^x only, "the exponential function" (including the section "Formal definition", which you would expect merely to formalize what they say in the first paragraph).

That only confirms what we have told you, the Wikipedia is not perfect, and you should not assume that if you don't understand something there it is your fault.

 

[Saumyojit]

coefficient that is not +1.

f(X)=2*3^x can be a exponential function then

the two definitions would not be equivalent.

If y=b^x then if (X=0) y=1
But y=ab^x then (X=0) y = a
So the definition are not equivalent right?

 

[Dr.Peterson]

f(X)=2*3^x can be a exponential function then


If y=b^x then if (X=0) y=1
But y=ab^x then (X=0) y = a
So the definition are not equivalent right?

Correct on both. In your example, f(0) = 2, so it can't be equivalent to b^x for any value of b.

 

[JeffM]

f(X)=2*3^x can be a exponential function then


If y=b^x then if (X=0) y=1
But y=ab^x then (X=0) y = a
So the definition are not equivalent right?

Exactly.

By the way, I do not fully agree with what Dr. Peterson said about the wikipedia article on exponential function. The logic of the article is that they start with a general definition

[MATH]f(x) = ab^x.[/MATH]
Almost everyone who knows any mathematics will agree that such a beast is called AN exponential function in English. So they start by simply defining a term in complete generality.

The article, as is typical of wikipedia, gives a general definition suitable for anyone who knows a bit of mathemtics (algebra in this case) and then, with little preamble, goes off toward the fact that there is something else normally called THE exponential function in English. (The article makes clear that this normally used term is an abbreviation for the technically correct term of “the natural exponential function.”) That function is

[MATH]g(x) = e^x[/MATH] where e is the irrational number that is the base of the natural logarithms.

Notice that a = 1 and b = e for that particular case of an exponential function.

Moreover, the article assumes that anyone who knows about derivatives knows that a mere coefficient just changes the scale of things, not more fundamental characteristics. So in going from the general, which may and usually does have a non-zero coefficient other than 1, to the specific e^x, which has a coefficient of 1, they discuss the general characteristic of all exponential functions where the coefficient is 1. They go from completely general to less general to specific.

It makes sense if you see the overall picture, but it provides no guideposts along the way. Wikipedia is not a textbook, nor does it pretend to be. I read Wikipedia until I realize that they have reached a depth of knowledge that I do not have and then stop. In this case, they start talking about calculus in paragraph 3. Time for you to stop until you understand calculus.

 

[Saumyojit]

because −3^2

why does -3^2 translate to -9 .
I thought -3^2 meant (-3)^2 which gives me 9 . Is there any connection of Bedmas with the evaluation of this expression.

You are misunderstanding basic logic. When you are told, If A then B, that says nothing about what happens when A is not true! So saying "valid when b and c are positive real numbers" does not mean "NOT valid when b and c are NOT positive real numbers"!!!

They are saying nothing about what happens when b and c are not both positive. They are making a statement about what can be stated for all positive real numbers (when the

Ok got the main point.
If A then b , but if A fails that certainly doesn't mean b is invalid!

Suppose A is my proposition "c and b are positive real no's and X is a real no" and B is another proposition "the identities are valid "

Then If A is true I see then B has to be true for any combination of no's I take; with keeping in mind the values are taken matching with the constraint in proposition A.

But there is no A =true B= False Case according to my experiment .

Now if A is false then B can be true still.
I checked.(this was the misconception.i had)

And another case can be both A and B are false which they showed to prove the failure of the law.


Ok the law is valid for all Integer exponent and b and c must be non zero Complex no's .

 

[pka]

You are abusing the notation: \((-3)^2\ne-3^2\) because \(-3^2=-9\).

why does -3^2 translate to -9 .
I thought -3^2 meant (-3)^2 which gives me 9 . Is there any connection of Bedmas with the evaluation of this expression.

In my original reply I did say that you are abusing the notation.
1) \(-3^2\) is read as "the negative of three squared."
2) \((-3)^2\) is read as "the square of negative three."
Surely you do not argue that those two say the same thing.

 

[Dr.Peterson]

why does -3^2 translate to -9 .
I thought -3^2 meant (-3)^2 which gives me 9 . Is there any connection of Bedmas with the evaluation of this expression.

The form "BEDMAS" does not explicitly mention negation, but that can be thought of as a form of multiplication (by -1); negation comes after exponentiation, so -3^2 means -(3^2). Think "BENDMAS", if you like.

Were you never taught that??

Ok got the main point.
If A then b , but if A fails that certainly doesn't mean b is invalid!

Suppose A is my proposition "c and b are positive real no's and X is a real no" and B is another proposition "the identities are valid "

Then If A is true I see then B has to be true for any combination of no's I take; with keeping in mind the values are taken matching with the constraint in proposition A.

But there is no A =true B= False Case according to my experiment .

Now if A is false then B can be true still.
I checked.(this was the misconception.i had)

And another case can be both A and B are false which they showed to prove the failure of the law.

Right. Your mistake was a classic fallacy, called the "inverse error" or "denying the antecedent": https://en.wikipedia.org/wiki/Denying_the_antecedent

Logic can be studied at a high level, like abstract algebra, but the basics (as in that Wikipedia page) are something everyone should know.