Invert a triple composite function p(q(r(x)))

In summary: A knockdown-drag-out proof could begin: "Consider the two functions A(x) = h_0^{-1}f( h_0^{-1}(x)) and B(x) = g(h_0(h_0^{-1}(x))) ."If A(x) = y then show that x = h_0 f^{-1} (x) . Use that expression for x to show that B(x) = y also. In a similar manner show that if B(x) = y then A(x) = y . Argue that A(x) and B(x) have the same domain and range.
  • #1
infk
21
0
Hey,
Let ##(f,g) \in B^A## where ##A## and ##B## are non-empty sets, ##B^A## denotes the set of bijective functions between ##A## and ##B##.
We assume that there exists ##h_0: A \rightarrow A## and ##h_1: B \rightarrow B## such that ##f = h_1 \circ g \circ h_0 ##.
This implies that ##g = h^{-1}_1 \circ f \circ h^{-1}_0##, according to my teacher, but why is that?
We have that ##h^{-1}_1 \circ f = g \circ h^{-1}_0 ##, but how do I proceed from that?

I have drawn a graph with sets and arrows representing functions such that going from ##A## to ##B## via ##h_0##, ##g## and ##h_1## is the same as going from ##A## to ##B## via ##f##. I then manipulated this graph a little while maintaining the proper relations between the sets, which showed that going from ##A## to ##B## via ##g## is the same as going through (in order) ##h^{-1}_0##, ##f## and ##h^{-1}_1##, but this is hardly a proof at all.

Thanks
 
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  • #2
infk said:
We have that ##h^{-1}_1 \circ f = g \circ h^{-1}_0 ##, but how do I proceed from that?

Using magic and superstition, if we begin with the equation [itex] h(x) = h(x) [/itex] and apply "equal functions" to both sides of that equation then we can get to [itex] h^{-1}( f ( h(x)) = g(h^{-1}(h(x)) = g(x) [/itex]. The problem is how to justify that procedure.


Your materials may have proven that procedure as a theorem. We'd need to see those materials in order to find an easy proof. Are you asking for some "slick" way of proving the result?

A knockdown-drag-out roof could begin: "Consider the two functions [itex] A(x) = h^{-1}f( h(x)) [/itex] and [itex] B(x) = g(h^{-1}(h(x)) [/itex] ."

If [itex] A(x) = y [/itex] then show that [itex] x = h((f^{-1}(h^{-1}(x)) [/itex]. Use that expression for [itex] x [/itex] to show that [itex] B(x) = y [/itex] also. In a similar manner show that if [itex] B(x) = y [/itex] then [itex] A(x) = y [/itex]. Argue that [itex] A(x) [/itex] and [itex] B(x) [/itex] have the same domain and range. Thus they are identical functions.

.
 
  • #3
Stephen Tashi said:
Using magic and superstition, if we begin with the equation [itex] h(x) = h(x) [/itex] and apply "equal functions" to both sides of that equation then we can get to [itex] h^{-1}( f ( h(x)) = g(h^{-1}(h(x)) = g(x) [/itex]. The problem is how to justify that procedure.Your materials may have proven that procedure as a theorem. We'd need to see those materials in order to find an easy proof. Are you asking for some "slick" way of proving the result?

A knockdown-drag-out roof could begin: "Consider the two functions [itex] A(x) = h^{-1}f( h(x)) [/itex] and [itex] B(x) = g(h^{-1}(h(x)) [/itex] ."

If [itex] A(x) = y [/itex] then show that [itex] x = h((f^{-1}(h^{-1}(x)) [/itex]. Use that expression for [itex] x [/itex] to show that [itex] B(x) = y [/itex] also. In a similar manner show that if [itex] B(x) = y [/itex] then [itex] A(x) = y [/itex]. Argue that [itex] A(x) [/itex] and [itex] B(x) [/itex] have the same domain and range. Thus they are identical functions.

.
Hi and thanks for the response. What I meant was of course that ##h^{-1}_1 \circ f = g \circ h_0##, it seems though that you did not notice my typo. Also, could you use subscript notation, it is not clear which of ##h_1## and ##h_2## you mean. Cheers
 
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  • #4
Using magic and superstition, if we begin with the equation [itex] h_0^{-1}(x) = h_0^{-1}(x) [/itex] and apply "equal functions" to both sides of that equation then we can get to [itex] h_1^{-1}( f ( h_0^{-1}(x)) = g(h_0(h_0^{-1}(x)) = g(x) [/itex]. The problem is how to justify that procedure.

Your materials may have proven that procedure as a theorem. We'd need to see those materials in order to find an easy proof. Are you asking for some "slick" way of proving the result?
 
  • #5
for your question! Let's work through this step by step.
First, let's define the functions we are working with:
- p: A -> B
- q: B -> C
- r: C -> D

Now, we are given that p(q(r(x))) is a triple composite function. This means that we can break it down into three separate functions:
- f: A -> D (defined as p(q(r(x))) )
- g: B -> D (defined as q(r(x)) )
- h: C -> D (defined as r(x) )

Now, let's consider the function g. We know that g = h^-1 * f. This is because g is the composition of two functions, h and f. And since h and f are both invertible, we can use the property of inverse functions to write g as the composition of their inverses, h^-1 and f^-1.

So, we can rewrite g as h^-1 * f. Now, let's plug in our definitions for f and h:
g = h^-1 * p(q(r(x)))

Next, let's consider the function p. We know that p = f * q^-1. This is because p is the composition of two functions, f and q. And since f and q are both invertible, we can use the property of inverse functions to write p as the composition of their inverses, f^-1 and q^-1.

So, we can rewrite p as f * q^-1. Now, let's plug in our definitions for f and q:
p = p * q^-1 * q(r(x))

Finally, let's substitute this into our previous equation for g:
g = h^-1 * p(q(r(x)))
g = h^-1 * p * q^-1 * q(r(x))

Now, we can see that q^-1 * q cancels out, leaving us with:
g = h^-1 * p

And since we know that p = f * q^-1, we can substitute that in:
g = h^-1 * f * q^-1

And there we have it! We have shown that g = h^-1 * f, just as your teacher had mentioned. I hope this helps clarify the process for you.
 

1. What is the definition of a triple composite function?

A triple composite function is a mathematical function that is composed of three individual functions, where the output of one function becomes the input of the next function, and so on. It can be expressed as p(q(r(x))).

2. How do you invert a triple composite function?

To invert a triple composite function, first solve for the innermost function, then work your way outwards. So for p(q(r(x))), you would first solve for r(x), then q(r(x)), and finally p(q(r(x))). The resulting inverted function would be r(q(p(x))).

3. Can a triple composite function always be inverted?

No, not all triple composite functions can be inverted. The innermost function must be one-to-one for the entire function to be invertible. If the innermost function is not one-to-one, the resulting inverted function would not be a true inverse.

4. What is the purpose of inverting a triple composite function?

Inverting a triple composite function can be useful in solving equations or simplifying complex expressions. It can also help in understanding the relationships between the individual functions within the composite function.

5. Are there any rules or guidelines for inverting a triple composite function?

Yes, there are a few rules to keep in mind when inverting a triple composite function. First, the order of the functions must remain the same in the inverted function. Second, the domain and range of the original function must match the range and domain of the inverted function. Lastly, the inverse of a triple composite function is not always the same as the composition of the inverses of the individual functions.

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