Trying to reconcile function composition problems with sets & formulas

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Homework Help Overview

The discussion revolves around reconciling function composition problems involving sets and formulas. Participants explore the relationships between different methods of function composition, specifically focusing on a set of relations and a pair of functions.

Discussion Character

  • Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants describe their methods for solving function composition problems with sets and formulas. They express confusion about how to reconcile the two approaches, questioning the underlying assumptions and definitions of function composition.

Discussion Status

Some participants have provided insights into the nature of function composition, discussing the relationships between the set notation and the formulaic approach. There is an ongoing exploration of how these methods align or differ, with no explicit consensus reached yet.

Contextual Notes

Participants are working within the constraints of specific homework problems involving function composition with given sets and formulas. There is a noted concern about potential misunderstandings in translating between the two methods.

s3a
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Homework Statement
**Problem involving function composition with sets:**
Consider the set A = {w , x , y, z }, and the relations

S = {(w , x ), (w , y), (x , w ), (x , x ), (z , x )}

T = {(w , w ), (w , y), (x , w ), (x , x ), (x , z ), (y, w ), (y, y), (y, z )}

Find S ◦ T.

**Problem involving function composition with formulas:**
Let f(x) = x + 1 and g(x) = x^2.

Find f ◦ g.
Relevant Equations
The answer for the set one is.:
S ◦T = {(w , w ), (w , x ), (w , z ), (w , y), (x , w ), (x , y), (x , x ), (x , z ), (z , w ), (z , x ), (z , z )}

The answer for the formula one is.:
f( g(x) ) = ( x^2 ) + 1

f( g(x) ) = x^2 + 1
I know how to solve each of those problems. For the set one, I look at the output of the S and try to match it with the input of T and then take the pair (input_of_S, output_of_T), and I do that for each pair.

As for the formula one, I just plug in x = g(y).

My confusion lies in trying to reconcile the two methods as different algorithms that are doing the same thing.

What bothers me is that for the formulaic one x_f = g(x_g), but for the set one, it seems to be to be output_of_S = input_of_T, which to translate that to the formulaic way, would be like saying f(x_f) = x_g instead.

Given that I get similar results, I'm assuming that I'm making a small mistake somewhere, but I'm not sure what it is.

Could someone please help me reconcile the two approaches (to similar problems)?

Any input would be greatly appreciated!

P.S.
Sorry for the weird question. :P
 
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s3a said:
Homework Statement:: **Problem involving function composition with sets:**
Consider the set A = {w , x , y, z }, and the relations

S = {(w , x ), (w , y), (x , w ), (x , x ), (z , x )}

T = {(w , w ), (w , y), (x , w ), (x , x ), (x , z ), (y, w ), (y, y), (y, z )}

Find S ◦ T.

**Problem involving function composition with formulas:**
Let f(x) = x + 1 and g(x) = x^2.

Find f ◦ g.
Relevant Equations:: The answer for the set one is.:
S ◦T = {(w , w ), (w , x ), (w , z ), (w , y), (x , w ), (x , y), (x , x ), (x , z ), (z , w ), (z , x ), (z , z )}
That looks like ##T \circ S## to me.
 
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s3a said:
I know how to solve each of those problems. For the set one, I look at the output of the S and try to match it with the input of T and then take the pair (input_of_S, output_of_T), and I do that for each pair.

As for the formula one, I just plug in x = g(y).

My confusion lies in trying to reconcile the two methods as different algorithms that are doing the same thing.

f \circ g means "do g, then do f to the result". So in set notation with g: X \to Y and f: Y \to Z it would be <br /> \begin{split}<br /> f &amp;= \{ (x, f(x)) : x \in Y \} \subset Y \times Z \\<br /> g &amp;= \{ (x, g(x)) : x \in X \} \subset X \times Y \\<br /> f \circ g &amp;= \{ (x, f(g(x))) : x \in X \} \subset X \times Z \end{split}
 
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Consider that , for one , , while every function is a relation, the converse doesn't hold.
You can consider a set R in your case as a relation r, where (a,b) stands for (a,r(a)), i.e., b:=r(a).

 

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