Solving Linear Copying Problems with Functions: Example using f, g, and h

[Physicsissuef]

Homework Statement

Let's say that f: \mathbb {R}^3 \rightarrow \mathbb {R}^2, g: \mathbb {R}^3 \rightarrow \mathbb {R}^2, and h: \mathbb {R}^2 \rightarrow \mathbb {R}^2 are linear copying defined with f(x,y,z)=(y,x+z), g(x,y,z)=(2z,x-y) and h(x,y)=(y,2x). Find the linear copying:

h o f and h o g

Homework Equations

The Attempt at a Solution

Sorry, but again, I don't know the principle of solving this task. My book is not good at all. Thanks from the start.

 

[CompuChip]

What does h \circ f mean?
What is its domain and co-domain (e.g. what are A and B in h: A \to B)? How is it defined (h(\cdots)= \cdots )?

 

[Physicsissuef]

What does h \circ f mean?
What is its domain and co-domain (e.g. what are A and B in h: A \to B)? How is it defined (h(\cdots)= \cdots )?

probably h, f, g are the linear copying

 

[CompuChip]

I don't know what language you translated it from, but I think you mean "linear map" instead of "linear copying".
Also I think you are a little confused as to what h \circ f means. f is a function that takes something in R³ and gives you something in R². Into h you put something in R² and get something (else) in R² back. Now what you can do, is use the output of f as input for h. So
h \circ f: \mathbb{R}^3 \to \mathbb{R}^2
is a function defined by
(h \circ f)(x, y, z) = h(f(x, y, z)).
So to calculate (h o f)(x, y, z) you first calculate (u, v) = f(x, y, z) and then h(u, v).
Does that clarify ?

 

[Physicsissuef]

I don't know what language you translated it from, but I think you mean "linear map" instead of "linear copying".
Also I think you are a little confused as to what h \circ f means. f is a function that takes something in R³ and gives you something in R². Into h you put something in R² and get something (else) in R² back. Now what you can do, is use the output of f as input for h. So
h \circ f: \mathbb{R}^3 \to \mathbb{R}^2
is a function defined by
(h \circ f)(x, y, z) = h(f(x, y, z)).
So to calculate (h o f)(x, y, z) you first calculate (u, v) = f(x, y, z) and then h(u, v).
Does that clarify ?

Yes, sir, that's what it means. Sorry for my mistranslation.

 

[HallsofIvy]

I don't know what language you translated it from, but I think you mean "linear map" instead of "linear copying".
Also I think you are a little confused as to what h \circ f means. f is a function that takes something in R³ and gives you something in R². Into h you put something in R² and get something (else) in R² back. Now what you can do, is use the output of f as input for h. So
h \circ f: \mathbb{R}^3 \to \mathbb{R}^2
is a function defined by
(h \circ f)(x, y, z) = h(f(x, y, z)).
So to calculate (h o f)(x, y, z) you first calculate (u, v) = f(x, y, z) and then h(u, v).
Does that clarify ?

Yes, sir, that's what it means. Sorry for my mistranslation.

Okay, so do it! If h(u,v)= (v, 2u), and (u,v)= f(x,y,z)= (2z, x- y), what is h\circle f(x,y,z). What is f\circle g(u,v)?

 

[Physicsissuef]

Okay, so do it! If h(u,v)= (v, 2u), and (u,v)= f(x,y,z)= (2z, x- y), what is h\circle f(x,y,z). What is f\circle g(u,v)?

How do you know that (x,y)=f(x,y,z)=(2z,x-y)? I truly don't know what is h \circ f(x,y,z)[/tex] and f \circ g(u,v)

 

[HallsofIvy]

Because that was what YOU told us:

f(x,y,z)=(y,x+z)

h\circ f(x,y,z)= h(f(x,y,z))= h(y, x+ z). If h is defined by h(a,b)= (b, 2a), what is h(y, x+z).

 

[Physicsissuef]

Can you tell me just this, please, there is one line more, I guess. Thanks.
Probably h(y, x+ z)=(b,2a)
But, what next?

 

[CompuChip]

I think you are having some troubles with basic calculus concepts.
Suppose we have a function f(x) = 2 x + 3.
Then can you tell me what are f(3), f(a) and f(y + 1) ?

 

[Physicsissuef]

f(3)=2*3+3
f(a)=2a + 3
f(y+1)=2(y+1)+3

No, I don't have troubles with basic calculations.

 

[CompuChip]

If h is defined by h(a,b)= (b, 2a), what is h(y, x+z).

Well, this is exactly the same, but there are just two values to substitute. So
h(y, x + z) is just (b, 2a) with a = y, b = x + z, i.e. h(y, x + z) = (x + z, 2y).

 

[Physicsissuef]

But it should be h(y, x+z) and (y,2x), so y=y and 2x=x+z. Hmm...

 

[HallsofIvy]

No, you should have learned long ago that the letters used as variables in defining functions are just "place holders"- you can replace them with anything you want. If f is defined by f(x, y)= 3x+ y then f(a, b)= 3a+ b, f(y, x)= 3y+ b, etc. h\circ f(x,y)= h(f(x,y,z))= h(y, x+z)= (x+ y, 2y) as I said.

 

[Physicsissuef]

I know that, they are just 'place holders'. But can't understand the whole process.

h \circ f is what we need to find.

(h \circ f)(x,y,z)=h(f(x,y,z))=h(y,x+z)

Now if h(x,y)=(y,2x) then h(y,x+z)=(x+z,2y), right?

I think I understand now. Now for h \circ g is the 2nd think that we need to find.

(h \circ g)(x,y,z)=h(g(x,y,z))=h(2z,x-y)

Now if h(x,y)=(y,2x) then h(2z,x-y)=(x-y,4z), right?

I think, I understand now.

 

[Physicsissuef]

And if we get f \circ h?

(f \circ h)(x,y)=f(h(x,y))=f(y,2x)

Now if f(x,y,z)=(y,x+z) then f(y,2x)=??

In my book, it says that it is impossible, why?

 

[CompuChip]

What could it be? (y, 2x) is an element of \mathbb{R}^2, whereas f is a function on \mathbb{R}^3. In other words, h produces a pair of numbers, whereas f expects a triple as input. So you cannot plug the output of h into f, in other words: you can't compose them in that order.

 

[Physicsissuef]

What could it be? (y, 2x) is an element of \mathbb{R}^2, whereas f is a function on \mathbb{R}^3. In other words, h produces a pair of numbers, whereas f expects a triple as input. So you cannot plug the output of h into f, in other words: you can't compose them in that order.

Ok, and what for h \circ (f+g) and h \circ f + h \circ g.
First, are they same?
Other thing, how will I solve it?

 

[CompuChip]

Can you write down the definitions of (f + g) and h o (f + g) and (h o f) + (h o g)?
In particular, write down from what space into what space they map, like I did in my last post. Then you can already see if there will be a problem in the composition (like: if the space (f + g) maps to is different than the space h maps from, you cannot compose them, as you have just seen).