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

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In summary, we are asked to find the linear copying of h o f and h o g, where f, g, and h are linear maps defined for specific input and output values. Using the composition of functions, h o f = h(f(x,y,z)) = (x+y, 2y) and h o g = h(g(x,y,z)) = (x-y, 4z).
  • #1
Physicsissuef
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Homework Statement



Let's say that [tex]f: \mathbb {R}^3 \rightarrow \mathbb {R}^2[/tex], [tex]g: \mathbb {R}^3 \rightarrow \mathbb {R}^2[/tex], and [tex]h: \mathbb {R}^2 \rightarrow \mathbb {R}^2[/tex] are linear copying defined with [tex]f(x,y,z)=(y,x+z)[/tex], [tex]g(x,y,z)=(2z,x-y)[/tex] and [tex]h(x,y)=(y,2x)[/tex]. 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.
 
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  • #2
What does [itex]h \circ f[/itex] mean?
What is its domain and co-domain (e.g. what are A and B in [itex]h: A \to B[/itex])? How is it defined ([itex]h(\cdots)= \cdots[/itex] )?
 
  • #3
CompuChip said:
What does [itex]h \circ f[/itex] mean?
What is its domain and co-domain (e.g. what are A and B in [itex]h: A \to B[/itex])? How is it defined ([itex]h(\cdots)= \cdots[/itex] )?

probably h, f, g are the linear copying
 
  • #4
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 [itex]h \circ f[/itex] 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
[tex]h \circ f: \mathbb{R}^3 \to \mathbb{R}^2[/tex]
is a function defined by
[tex](h \circ f)(x, y, z) = h(f(x, y, z))[/tex].
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 ?
 
  • #5
CompuChip said:
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 [itex]h \circ f[/itex] 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
[tex]h \circ f: \mathbb{R}^3 \to \mathbb{R}^2[/tex]
is a function defined by
[tex](h \circ f)(x, y, z) = h(f(x, y, z))[/tex].
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.
 
  • #6
CompuChip said:
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 [itex]h \circ f[/itex] 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
[tex]h \circ f: \mathbb{R}^3 \to \mathbb{R}^2[/tex]
is a function defined by
[tex](h \circ f)(x, y, z) = h(f(x, y, z))[/tex].
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 said:
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 [itex]h\circle f(x,y,z)[/itex]. What is [itex]f\circle g(u,v)[/itex]?
 
  • #7
HallsofIvy said:
Okay, so do it! If h(u,v)= (v, 2u), and (u,v)= f(x,y,z)= (2z, x- y), what is [itex]h\circle f(x,y,z)[/itex]. What is [itex]f\circle g(u,v)[/itex]?

How do you know that (x,y)=f(x,y,z)=(2z,x-y)? I truly don't know what is [itex]h \circ f(x,y,z)[/tex] and [itex]f \circ g(u,v)[/itex]
 
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  • #8
Because that was what YOU told us:
physicsissuef said:
[tex]f(x,y,z)=(y,x+z)[/tex]
[itex]h\circ f(x,y,z)= h(f(x,y,z))= h(y, x+ z)[/itex]. If h is defined by h(a,b)= (b, 2a), what is h(y, x+z).
 
  • #9
Can you tell me just this, please, there is one line more, I guess. Thanks.
Probably [itex]h(y, x+ z)=(b,2a)[/itex]
But, what next?
 
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  • #10
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) ?
 
  • #11
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.
 
  • #12
HallsofIvy said:
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).
 
  • #13
But it should be h(y, x+z) and (y,2x), so y=y and 2x=x+z. Hmm...
 
  • #14
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. [itex]h\circ f(x,y)= h(f(x,y,z))= h(y, x+z)= (x+ y, 2y) as I said.
 
  • #15
I know that, they are just 'place holders'. But can't understand the whole process.

[tex]h \circ f[/tex] is what we need to find.

[tex](h \circ f)(x,y,z)=h(f(x,y,z))=h(y,x+z)[/tex]

Now if [tex]h(x,y)=(y,2x)[/tex] then [tex]h(y,x+z)=(x+z,2y)[/tex], right?

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

[tex](h \circ g)(x,y,z)=h(g(x,y,z))=h(2z,x-y)[/tex]

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

I think, I understand now.
 
  • #16
And if we get [tex]f \circ h[/tex]?

[tex](f \circ h)(x,y)=f(h(x,y))=f(y,2x)[/tex]

Now if [tex]f(x,y,z)=(y,x+z)[/tex] then [tex]f(y,2x)=??[/tex]

In my book, it says that it is impossible, why?
 
  • #17
What could it be? [itex](y, 2x)[/itex] is an element of [itex]\mathbb{R}^2[/itex], whereas [itex]f[/itex] is a function on [itex]\mathbb{R}^3[/itex]. 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.
 
  • #18
CompuChip said:
What could it be? [itex](y, 2x)[/itex] is an element of [itex]\mathbb{R}^2[/itex], whereas [itex]f[/itex] is a function on [itex]\mathbb{R}^3[/itex]. 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 [tex]h \circ (f+g)[/tex] and [tex]h \circ f + h \circ g[/tex].
First, are they same?
Other thing, how will I solve it?
 
  • #19
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).
 
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FAQ: Solving Linear Copying Problems with Functions: Example using f, g, and h

1. What are linear copying problems?

Linear copying problems involve using a function to copy and transform a given input value to produce an output value. This is commonly seen in math and computer science when working with linear equations or functions.

2. How do you use functions to solve linear copying problems?

To solve linear copying problems, you can use a combination of functions, such as f, g, and h, to manipulate the input value and produce the desired output. This can involve basic operations like addition, subtraction, multiplication, and division, as well as more complex functions like logarithms or exponents.

3. What is the role of f, g, and h in solving linear copying problems?

f, g, and h are commonly used to represent different functions in math and computer science. In the context of solving linear copying problems, these functions can be used to represent the different steps or transformations needed to copy and manipulate the input value to produce the desired output.

4. Can you provide an example of solving a linear copying problem using functions?

Sure! Let's say we have the function f(x) = 2x + 3 and we want to copy and transform the input value of 4. We can use the function f to manipulate the input value as follows: f(4) = 2(4) + 3 = 11. So, the output value is 11, which is the result of copying and transforming the input value of 4 using the function f.

5. How can solving linear copying problems with functions be useful in real-world applications?

Solving linear copying problems with functions can be useful in various real-world applications, such as data analysis, financial modeling, and computer programming. It allows us to easily manipulate and transform data to produce desired outputs, which can be used to solve problems or make predictions.

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