Queries on Operators

  • Thread starter Longmarch
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Can anyone help with the 2 questions below:


1) Suppose f is a mathematical quantity that can take on the "states", or "values", f1, f2,...fi,...fn., where n can be finite or infinite.

So, F is the Set { f1, f2,... fi,.. fn } or
F = { f1, f2, fi,... fn }
= { all possible f's}

Now, suppose A is a operator such that when it operates on fi, it gives fj, where i is not equal to j, i.e.
Afi = fj, i =/= j
Afj = fk, j =/= k, and so on

Then can one say anything about A operating on f ?
Af = ?

Note that A is therefore a Mapping that maps F into itself. It maps one element, fi, into a different element fj .
Would it be correct to say that
Af = Mf, where M is some form of modifying factor, that is NOT a function of f ?



2) If Af = Mf is indeed true, then is it necessarily true that A has an Inverse, B, such that
BAf = f ?

If not, then under what conditions will A have such an Inverse?
 

Answers and Replies

  • #2
Fredrik
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Can anyone help with the 2 questions below:


1) Suppose f is a mathematical quantity that can take on the "states", or "values", f1, f2,...fi,...fn., where n can be finite or infinite.

So, F is the Set { f1, f2,... fi,.. fn } or
F = { f1, f2, fi,... fn }
= { all possible f's}
f is a sequence, and F is its range.

[itex]f:Z\rightarrow\mathbb R[/itex], where Z is either a set of the form {1,2,...,n} for some n, or the set of positive integers.
f(Z)=F.

Now, suppose A is a operator such that when it operates on fi, it gives fj, where i is not equal to j, i.e.
Afi = fj, i =/= j
Afj = fk, j =/= k, and so on
So A changes the order of the terms of the sequence, leaving no term in its original place?

Then can one say anything about A operating on f ?
Af = ?
You could do something like this: For each permutation [itex]\sigma:Z\rightarrow Z[/itex] such that [itex]\sigma(n)\neq n[/itex] for all n in Z, define [itex]A_\sigma[/itex] by [tex](A_\sigma(f))(n)=f(\sigma(n))[/tex] for all n in Z. Note that [itex]A_\sigma[/itex] is a function that takes sequences to sequences, and that a sequence is a function that takes positive integers to real numbers.

But I'm not 100% sure that this is what you had in mind.

Note that A is therefore a Mapping that maps F into itself. It maps one element, fi, into a different element fj .
If A acts on f, then that's not true since f is not a member of F.
 
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Thanks for your replies.

Can you tell me how to make quotes like you do? I can't find the icon that enable me to do so.
 
  • #4
Fredrik
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Just use the QUOTE button next to the post you want to quote. If you want to quote smaller parts of the post, as I did, you will have to insert more quote tags manually, and delete the text that you don't want to include. To insert more quote tags, just copy and paste the ones that were typed automatically when you hit the QUOTE button.

(I recommend that you always use the preview feature to make sure that it looks OK before you post).
 
  • #5
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I can see the NEW REPLY button but not the QUOTE button. And the PREVIEW doesn't seems to work either.
 
  • #6
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Originally Posted by Longmarch
<1) Suppose f is a mathematical quantity that can take on the "states", or "values", f1, f2,...fi,...fn., where n can be finite or infinite.

So, F is the Set { f1, f2,... fi,.. fn } or
F = { f1, f2, fi,... fn }
= { all possible f's} >

[f is a sequence, and F is its range.]

Let us call f1, f2,... fi,.. fn Events. So, f is a sequence of events.



[f:Z→R, where Z is either a set of the form {1,2,...,n} for some n, or the set of positive integers.
f(Z)=F.]

What if f1, f2,... fi,.. fn are functions instead of numbers?



Originally Posted by Longmarch
<Now, suppose A is a operator such that when it operates on fi, it gives fj, where i is not equal to j, i.e.
Afi = fj, i =/= j
Afj = fk, j =/= k, and so on >

[So A changes the order of the terms of the sequence, leaving no term in its original place?]

A changes Event fi to an event other than itself. So, I guess A should leave no term (event) in its orignal place.



Originally Posted by Longmarch
<Then can one say anything about A operating on f ?
Af = ?

[You could do something like this: For each permutation σ:Z→Z such that σ(n)≠n for all n in Z, define Aσ by (Aσ(f))(n)=f(σ(n)) for all n in Z. Note that Aσ is a function that takes sequences to sequences, and that a sequence is a function that takes positive integers to real numbers.]

This is a beyond me.

What are σ(f) & σ(n)?

Can you further explain "(Aσ(f))(n)=f(σ(n))"?


[But I'm not 100% sure that this is what you had in mind.]

Actually, I am trying to see if the above could lead to some equation similar in form to an Eigenvalue Equation, Af = af.

Can (Aσ(f))(n)=f(σ(n)) be made into the same form, at least?



Originally Posted by Longmarch
<Note that A is therefore a Mapping that maps F into itself. It maps one element, fi, into a different element fj . >

[If A acts on f, then that's not true since f is not a member of F.]

So, if A does not act on f, it is a Mapping then?
 
  • #7
Fredrik
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I can see the NEW REPLY button but not the QUOTE button. And the PREVIEW doesn't seems to work either.
There are quote buttons all over the place. One in each post. I am unaware of any problems with the preview feature. It's working fine when I'm typing this post.

What if f1, f2,... fi,.. fn are functions instead of numbers?
As long as they all have the same domain and codomain, that's not a problem. [itex]Y^X[/itex] is a standard notation for "the set of all functions from X into Y". We could say that f is a sequence in [itex]Y^X[/itex], and you can choose X and Y to be anything you want.

f:Z→YX, where Z is either a set of the form {1,2,...,n} for some n, or the set of positive integers.

This is a beyond me.

What are σ(f) & σ(n)?
σ is a permutation of Z. That means that it's a bijective function from Z onto Z. Bijective means that the following two conditions are satisfied:

1. For all n,m in Z, σ(n)=σ(m) implies n=m.
2. For all m in Z, there's an n in Z such that σ(n)=m.

I never wrote σ(f).

Can you further explain "(Aσ(f))(n)=f(σ(n))"?
It was [itex](A_\sigma(f))(n)=f(\sigma(n))[/itex]. The nth term of a sequence g can be written as g(n) or gn. I was using the former notation. So [itex]f(\sigma(n))[/itex] denoted the σ(n)th term of the sequence f. I'll use the latter notation in this post, so now [itex]f_{\sigma(n)}[/itex] denotes the σ(n)th term of the sequence f.

[itex]A_\sigma[/itex] is a function that takes sequences to sequences. So [itex]A_\sigma(f)[/itex] is a sequence. I denoted its nth term by [itex](A_\sigma(f))(n)[/itex]. In the other notation, the formula looks like this: [tex](A_\sigma(f))_n=f_{\sigma(n)}.[/tex] The reason that σ appears as an index on A is that the function we're trying to define is a different function for each choice of σ, so I prefer to denote it by [itex]A_\sigma[/itex] instead of just A. However, it's OK to ignore how the function depends on the permutation and just say something like this: Given a permutation σ, define the function A (that takes sequences defined on Z to sequences defined on Z) by [tex](Af)_{n}=f_{\sigma(n)}[/tex] for all n in Z. (Here I chose to write Af rather than A(f), because more parentheses don't make the formula easier to understand).

Actually, I am trying to see if the above could lead to some equation similar in form to an Eigenvalue Equation, Af = af.
What would af mean? Is it the sequence with nth term afn? In that case, it doesn't seem possible. A rearrangement of the terms being equivalent to simply multiplying each term by the same number? I would guess that it only works for constant sequences.

So, if A does not act on f, it is a Mapping then?
Depends on how you define your A.
 
  • #8
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Well, it is getting deeper into some mathematics that I do not understand and drifting away from the question I am asking. Let me try to put the question with an example to make my question clearer:




Suppose f is a mathematical quantity that can take on the "states", or "values", f1, f2,...fi,...fn., where n can be finite or infinite.

So, F is the Set { f1, f2,... fi,.. fn } or
F = { f1, f2, fi,... fn }
= { all possible f's}

Now, suppose A is a operator such that when it operates on fi, it gives fj, where i is not equal to j, i.e.
Afi = fj, i =/= j
Afj = fk, j =/= k, and so on

Then can one say anything about A operating on f ?
Af = ?




----------------





And consider the following example:



Let there be a group of boys: John, Will, Charles, Dick, Phillip, .....

So, G = {J, W, C, D, P,....}



Each of John, Will, Charles, Dick, Phillip, ..... is a Boy

So, each of J, W, C, D, P,.... is a B



Now, a Teacher will decide the way one boy passes a relay-stick to another:

Say, TW = D

TD = J

TJ = P and so on...



Knowing the above, my question is "What is TB", TB = ?



We know that TB must be a boy but not the boy who is currently holding the relay-stick. So, is it right to say

TB = mB, where m is a "mathematical" quantity or function that reflects the way the Teacher chooses the boy to receive the relay-stick?



If so, Can we then say

Af = Mf, where M is some form of modifying factor, that is NOT a function of f ?
 
  • #9
Fredrik
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Let there be a group of boys: John, Will, Charles, Dick, Phillip, .....

So, G = {J, W, C, D, P,....}
OK.

Each of John, Will, Charles, Dick, Phillip, ..... is a Boy

So, each of J, W, C, D, P,.... is a B
I'm guessing that this is just your way of saying that you will use the word "boy" to refer to members of G. I don't know what else you could mean.

Now, a Teacher will decide the way one boy passes a relay-stick to another:

Say, TW = D

TD = J

TJ = P and so on...
You are describing a permutation T:G→G.

Knowing the above, my question is "What is TB", TB = ?
What makes you think that the notation "TB" has any meaning until you have defined it?

I could guess that it means what's usually denoted by T(G), i.e. the set of all T(x) such that x is in G. Since T is a permutation, this would mean that TB=T(G)=G.

We know that TB must be a boy but not the boy who is currently holding the relay-stick.
Is this your way of defining the notation? Are you in fact saying "Let TB be a member of G that's not holding the stick."? Then why did you ask what TB is?

Wait, did you use "B" to denote an arbitrary member of G? Then you were asking for the value of T at B. It's of course impossible to answer that without knowing both

a) the function T, i.e. how the teacher has decided to order the boys, and
b) what boy the symbol B represents.
 
  • #10
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[What makes you think that the notation "TB" has any meaning until you have defined it?

I could guess that it means what's usually denoted by T(G), i.e. the set of all T(x) such that x is in G. Since T is a permutation, this would mean that TB=T(G)=G.]

I am not a mathemtician. So, I am not familiar with the the accepted notations that mathematician use.

[Wait, did you use "B" to denote an arbitrary member of G? Then you were asking for the value of T at B. It's of course impossible to answer that without knowing both

a) the function T, i.e. how the teacher has decided to order the boys, and
b) what boy the symbol B represents.]

If the function T is not yet known and B is an arbitrary member of G, can we write TB = mB, where m is some modifying function yet to be determined?
 

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