Confusion over inverse of a function (involution)

Simonsky

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f(x) = 1/x is given as an example of an 'involution'-that is, the inverse of the function (the function that maps the output to the input)

In the case of the above example, I don't understand why it's the case because if the 'output' is 1/x then couldn't the f^-1 be 1/1/x or (x^-1)^-1 ?

So if x = 3 and the output is 1/3 then to get back to the 3 you have to (f(x))^-1

What am i getting wrong, or am I just doing something tautological without realising it?
 

Dr.Peterson

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f(x) = 1/x is given as an example of an 'involution'-that is, the inverse of the function (the function that maps the output to the input)

In the case of the above example, I don't understand why it's the case because if the 'output' is 1/x then couldn't the f^-1 be 1/1/x or (x^-1)^-1 ?

So if x = 3 and the output is 1/3 then to get back to the 3 you have to (f(x))^-1

What am i getting wrong, or am I just doing something tautological without realising it?
First, to be clear, an involution is not an inverse -- it is a function that is its own inverse.

Also, it's possible that you are confusing f^-1 with x^-1; the notation for the inverse of a function and the reciprocal of a number are similar, but the ideas are very different. In particular, f^-1(x) is not the same as (f(x))^-1, which is 1/f(x). Here, it happens that we are talking about f(x) = x^-1, which confuses things a bit, so you may not be saying anything wrong.

So, how can we see that f is its own inverse? In general, f^-1(f(x)) = x for all x -- that is, the inverse function takes the output of the function back to the input. For an involution, since f^-1 = f, the test is that f(f(x)) = x for all x.

And that is true here: f(f(x)) = f(x^-1) = (x^-1)^-1 = x^(-1*-1) = x, for any x.

Restating that in terms of f(x) = 1/x as you originally stated it, f(f(x)) = f(1/x) = 1/(1/x) = 1 * x/1 = x.

For your specific example, with input x = 3, f(3) = 1/3, and f(f(3)) = f(1/3) = 3.

I'm not sure whether I've answered your question, as there are several things that might be confusing you. Ask more if needed!
 

lev888

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f(x) = 1/x is given as an example of an 'involution'-that is, the inverse of the function (the function that maps the output to the input)

In the case of the above example, I don't understand why it's the case because if the 'output' is 1/x then couldn't the f^-1 be 1/1/x or (x^-1)^-1 ?

So if x = 3 and the output is 1/3 then to get back to the 3 you have to (f(x))^-1

What am i getting wrong, or am I just doing something tautological without realising it?
Involution - a function, transformation, or operator that is equal to its inverse, i.e., which gives the identity when applied to itself.


Second part of your initial statement is not quite right. Involution is not "the inverse of the function", it's a function with an inverse the same as the function itself.

Let's take 1/x: y = 1/x. What is the inverse function? Solve for x: x = 1/y. As you can see the inverse is the same 1/x.
Another way to check is apply it to itself and see if the result is the identity function f(x) = x. So, f(1/x) = 1/(1/x) = x.
 

Simonsky

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Involution - a function, transformation, or operator that is equal to its inverse, i.e., which gives the identity when applied to itself.


Second part of your initial statement is not quite right. Involution is not "the inverse of the function", it's a function with an inverse the same as the function itself.

Let's take 1/x: y = 1/x. What is the inverse function? Solve for x: x = 1/y. As you can see the inverse is the same 1/x.
Another way to check is apply it to itself and see if the result is the identity function f(x) = x. So, f(1/x) = 1/(1/x) = x.
Thanks for replies both Dr. Peterson and Lev888. yes, I was sloppy confusing involution with inverse in general, thanks for clarification. i realise where i went wrong: I was confusing the the operation with the fact that the outputs and inputs are different. It's the fact that the operation works both ways: in the case of f(x) = 1/x , the operation of -1 as a power gives rise to output and return to input.

I came across another example: f(x) = 2-x. so far we could generalise and say: f(x) = 1/x is an involution for all x and f(x) = a-x for all x ( maybe you need to add the sign for real numbers as well? But is there a general approach to formulating involutions that includes these examples?
 

Dr.Peterson

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Thanks for replies both Dr. Peterson and Lev888. yes, I was sloppy confusing involution with inverse in general, thanks for clarification. i realise where i went wrong: I was confusing the operation with the fact that the outputs and inputs are different. It's the fact that the operation works both ways: in the case of f(x) = 1/x , the operation of -1 as a power gives rise to output and return to input.

I came across another example: f(x) = 2-x. so far we could generalise and say: f(x) = 1/x is an involution for all x and f(x) = a-x for all x ( maybe you need to add the sign for real numbers as well? But is there a general approach to formulating involutions that includes these examples?
Your terminology suggests that you still don't quite grasp the concept fully. The phrase "for all x" is out of place; either a function is or is not an involution! It is the function itself, as a single entity, that is the involution, not its value for particular inputs. Properly, you should say that, for any real number a, the function f defined by f(x) = a-x is an involution.

Also, I'm unsure what you are thinking when you say "the operation of -1 as a power gives rise to output and return to input". Can you explain that? It is practically meaningless to me!

There is no way to "formulate" an involution, if you mean some sort of formula that encompasses every involution. But there are ways to think about them more generally. For example, from the definition of involution, can you see what must always be true of its graph? Can you see that this is true for both of your examples? You could invent an involution by sketching a graph that fits this requirement.
 

Simonsky

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Your terminology suggests that you still don't quite grasp the concept fully. The phrase "for all x" is out of place; either a function is or is not an involution! It is the function itself, as a single entity, that is the involution, not its value for particular inputs. Properly, you should say that, for any real number a, the function f defined by f(x) = a-x is an involution.

Also, I'm unsure what you are thinking when you say "the operation of -1 as a power gives rise to output and return to input". Can you explain that? It is practically meaningless to me!

There is no way to "formulate" an involution, if you mean some sort of formula that encompasses every involution. But there are ways to think about them more generally. For example, from the definition of involution, can you see what must always be true of its graph? Can you see that this is true for both of your examples? You could invent an involution by sketching a graph that fits this requirement.

Thanks for help. I'm aware that my language is sloppy, hopefully it will improve in time as my ability to think clearly develops ( I hope!). I think I was just trying to note that I had realised that the operation works both ways, so that, in the case of f(x) = 1/x applying a power of minus one creates the output and applied to the output gives you the input ( for that value of x). I've probably dug myself in deeper with imprecise language-apologies. I've got a long way to go before I can use language with the rigour required by mathematics! I think I now get it even though it might sound like I don't linguistically. I can now grasp that when f(x) = ff(x) then it is an involution.

Thanks for the help-really valuable.
 

Simonsky

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Thanks for help. I'm aware that my language is sloppy, hopefully it will improve in time as my ability to think clearly develops ( I hope!). I think I was just trying to note that I had realised that the operation works both ways, so that, in the case of f(x) = 1/x applying a power of minus one creates the output and applied to the output gives you the input ( for that value of x). I've probably dug myself in deeper with imprecise language-apologies. I've got a long way to go before I can use language with the rigour required by mathematics! I think I now get it even though it might sound like I don't linguistically. I can now grasp that when f(x) = ff(x) then it is an involution.

Thanks for the help-really valuable.
Regarding the graphing function, I tried this. In the case of 1/x we have a discontinuous (?) curve that is symmetrical. Linear functions, I assume will be parallel to the x = y line???
 

JeffM

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Thanks for help. I'm aware that my language is sloppy, hopefully it will improve in time as my ability to think clearly develops ( I hope!). I think I was just trying to note that I had realised that the operation works both ways, so that, in the case of f(x) = 1/x applying a power of minus one creates the output and applied to the output gives you the input ( for that value of x). I've probably dug myself in deeper with imprecise language-apologies. I've got a long way to go before I can use language with the rigour required by mathematics! I think I now get it even though it might sound like I don't linguistically. I can now grasp that when f(x) = ff(x) then it is an involution.

Thanks for the help-really valuable.
One way to avoid the problems of natural language is to formulate your thoughts in mathematical symbols immediately. If you are told that an involution is a function that is its own inverse, put the definition of an inverse down first.

\(\displaystyle g(x) \text { is the inverse of } f(x) \iff g(f(x)) = x = f(g(x)).\)

\(\displaystyle \therefore f(x) \text { is an involution means that the inverse of } f(x) \text { is } f(x).\)

\(\displaystyle f(x) \text { is an involution } \iff x = f(f(x)) \implies f(f(x)) = x.[/f() tex]

\(\displaystyle f(x) = \dfrac{1}{x} \implies f \left ( \dfrax{1}{x} \right ) = \dfrac{1}{\dfrac{1}{x}} = x \implies\)

\(\displaystyle \dfrac{1}{x} \text { is an involution because } f(f(x)) = x.\)

\(\displaystyle f(x) = 2 - x \implies f(f(x)) = f(2 - x) = 2 - (2 - x) = x \implies 2 - x \text { is an involution.}\)\)
 

Dr.Peterson

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Thanks for help. I'm aware that my language is sloppy, hopefully it will improve in time as my ability to think clearly develops ( I hope!). I think I was just trying to note that I had realised that the operation works both ways, so that, in the case of f(x) = 1/x applying a power of minus one creates the output and applied to the output gives you the input ( for that value of x). I've probably dug myself in deeper with imprecise language-apologies. I've got a long way to go before I can use language with the rigour required by mathematics! I think I now get it even though it might sound like I don't linguistically. I can now grasp that when f(x) = ff(x) then it is an involution.

Thanks for the help-really valuable.
I think when you say "the operation", you mean "the function"; and you are saying that the same thing you do to get the output, you can do to the output to get the input back. That's right.

When you said "f(x) = ff(x)", you meant either "ff(x) = x" or "f^-1(x) = f(x)". Right?

Yes, I am pointing out incorrect language because precision is important in math, and the way to learn it, unfortunately, is to be caught saying something you don't mean, and gradually learn to catch it yourself so other don't have to. It's more or less the same idea as proofreading what you write before hitting Send, rather than after.

Regarding the graphing function, I tried this. In the case of 1/x we have a discontinuous (?) curve that is symmetrical. Linear functions, I assume will be parallel to the x = y line???
Yes, the graph of y = 1/x is discontinuous (it consists of two "branches"), which is not specifically relevant here, and it is symmetrical with respect to the line y=x, which is exactly the point. Your other example, y = a - x, is in fact perpendicular to that line; can you see how that makes it symmetrical, and therefore an involution?

A line that is parallel to y=x will not be symmetrical, unless it is the line y=x. The Wikipedia article I referenced in my first response gives some other examples. Another, more subtle, is y = (2x + 1)/(3x - 2).
 

Simonsky

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I think when you say "the operation", you mean "the function"; and you are saying that the same thing you do to get the output, you can do to the output to get the input back. That's right.

When you said "f(x) = ff(x)", you meant either "ff(x) = x" or "f^-1(x) = f(x)". Right?

Yes, I am pointing out incorrect language because precision is important in math, and the way to learn it, unfortunately, is to be caught saying something you don't mean, and gradually learn to catch it yourself so other don't have to. It's more or less the same idea as proofreading what you write before hitting Send, rather than after.



Yes, the graph of y = 1/x is discontinuous (it consists of two "branches"), which is not specifically relevant here, and it is symmetrical with respect to the line y=x, which is exactly the point. Your other example, y = a - x, is in fact perpendicular to that line; can you see how that makes it symmetrical, and therefore an involution?

A line that is parallel to y=x will not be symmetrical, unless it is the line y=x. The Wikipedia article I referenced in my first response gives some other examples. Another, more subtle, is y = (2x + 1)/(3x - 2).
Yes...(coughs politely)..I did , in fact mean "f^-1(x) = f(x)" ( another attack of mental lethargy which assails me frequently). But, in the case of involution, doesn't ff(x) = f^-1(x) or am I confusing input/output values with the function by saying that? Time for bed here in the UK-I'll sleep on it! Thanks so much for your help and time.
 

Dr.Peterson

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Yes...(coughs politely)..I did , in fact mean "f^-1(x) = f(x)" ( another attack of mental lethargy which assails me frequently). But, in the case of involution, doesn't ff(x) = f^-1(x) or am I confusing input/output values with the function by saying that? Time for bed here in the UK-I'll sleep on it! Thanks so much for your help and time.
Sleep well!

But, no, ff, that is, f composed with f, in the case of an involution, is the identity function: its value is always the same as its input, f(f(x)) = x. That is not what f^-1(x) is; its value is the input of f, for a given output x.
 

Simonsky

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Sleep well!

But, no, ff, that is, f composed with f, in the case of an involution, is the identity function: its value is always the same as its input, f(f(x)) = x. That is not what f^-1(x) is; its value is the input of f, for a given output x.
I think I get it ( I find I get a flash of understanding but my brain doesn't seem to hold it, so next time I come across the concept it has to be reworked - a bit like I have to keep re-inventing the wheel, or the Myth of Sisyphus-could be a cognitive issue there but that's a subject for another type of thread!).

My brain balked at this question:

The functions of f and g are defined as f(x) = 2x+1 and g(x) = 3x-2 find g^-1f^-1(x)

I had a go and did this: first worked out f^-1(x): (y-1)/2 then g^-1(x): (x+2)/3

then applied the inverse of g to the inverse of f: ( (x+2)/3 -1)/2 (hope I've notated that correctly-can't use Latex yet)

Well, that might be nonsense. The answer in the book put it like this: (fg)^-1(x) = g^-1f^-1(x) does that correspond to what I got above or have I 'lost the plot'?
 

Dr.Peterson

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I think I get it ( I find I get a flash of understanding but my brain doesn't seem to hold it, so next time I come across the concept it has to be reworked - a bit like I have to keep re-inventing the wheel, or the Myth of Sisyphus-could be a cognitive issue there but that's a subject for another type of thread!).

My brain balked at this question:

The functions of f and g are defined as f(x) = 2x+1 and g(x) = 3x-2 find g^-1f^-1(x)

I had a go and did this: first worked out f^-1(x): (y-1)/2 then g^-1(x): (x+2)/3 <-- you meant x

then applied the inverse of g to the inverse of f: ( (x+2)/3 -1)/2 (hope I've notated that correctly-can't use Latex yet)

Well, that might be nonsense. The answer in the book put it like this: (fg)^-1(x) = g^-1f^-1(x) does that correspond to what I got above or have I 'lost the plot'?
Your error here is that you plugged the result of g^-1 into f^-1, rather than vice versa, even though you said you were doing the right thing. You found f^-1(g^-1(x)). People often do this; you have to pay close attention.

I like to write out the steps:

g^-1(f^-1(x)) = g^-1((x-1)/2) = (((x-1)/2)+2)/3 and then simplify.

Incidentally, I'm not sure whether your book uses slightly non-standard notation, or you are just not trying to write everything just right, but what you write as fg should really be f○g, using the symbol for composition of functions; and what you write as g^-1f^-1(x) should be either (g-1○f-1)(x) or g^-1(f^-1(x)). I didn't use Latex there but it still took more effort than I like; it's fine to write (g^-1 o f^-1)(x) to indicate composition, using the letter o. Here it is using Latex: \(\displaystyle \left(g^{-1}\circ f^{-1}\right)(x)\)
 

Simonsky

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Your error here is that you plugged the result of g^-1 into f^-1, rather than vice versa, even though you said you were doing the right thing. You found f^-1(g^-1(x)). People often do this; you have to pay close attention.

I like to write out the steps:

g^-1(f^-1(x)) = g^-1((x-1)/2) = (((x-1)/2)+2)/3 and then simplify.

Incidentally, I'm not sure whether your book uses slightly non-standard notation, or you are just not trying to write everything just right, but what you write as fg should really be f○g, using the symbol for composition of functions; and what you write as g^-1f^-1(x) should be either (g-1○f-1)(x) or g^-1(f^-1(x)). I didn't use Latex there but it still took more effort than I like; it's fine to write (g^-1 o f^-1)(x) to indicate composition, using the letter o. Here it is using Latex: \(\displaystyle \left(g^{-1}\circ f^{-1}\right)(x)\)

'...even though you said you were doing the right thing' - I certainly wasn't that confident by a long chalk! (Is that a British Idiom?) Anyway, so I fell into the trap of getting the order wrong, i was trying to avoid that but must have reversed it in my head ..zzzzzz..

By the way, I have indeed reproduced the notation used in the book. One needs a lot of patience to not be put off by one's own cognitive deficiencies when learning maths- many thanks.
 

JeffM

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I think I get it ( I find I get a flash of understanding but my brain doesn't seem to hold it, so next time I come across the concept it has to be reworked - a bit like I have to keep re-inventing the wheel, or the Myth of Sisyphus-could be a cognitive issue there but that's a subject for another type of thread!).

My brain balked at this question:

The functions of f and g are defined as f(x) = 2x+1 and g(x) = 3x-2 find g^-1f^-1(x)

I had a go and did this: first worked out f^-1(x): (y-1)/2 then g^-1(x): (x+2)/3

then applied the inverse of g to the inverse of f: ( (x+2)/3 -1)/2 (hope I've notated that correctly-can't use Latex yet)

Well, that might be nonsense. The answer in the book put it like this: (fg)^-1(x) = g^-1f^-1(x) does that correspond to what I got above or have I 'lost the plot'?
\(\displaystyle f(x) = 2x + 1 \implies f^{-1}(x) = \dfrac{x - 1}{2} \ \because\)

\(\displaystyle f(f^{-1}(x)) = f \left ( \dfrac{x - 1}{2} \right ) = 2 * \dfrac{x - 1}{2} + 1 = x - 1 + 1 = x.\)

\(\displaystyle g(x) = 3x - 2 \implies g^{-1}(x) = \dfrac{x + 2}{3}\ \because\)

\(\displaystyle g(g^{-1}(x)) = g \left ( \dfrac{x + 2}{3} \right ) = 3 * \dfrac{x + 2}{3} - 2 = x + 2 - 2 = x.\)

So all of this is well done.

But now I am not sure what the problem even is.

\(\displaystyle g^{-1} f^{-1}(x)\) is meaningless.

Does it mean \(\displaystyle g^{-1}(x) * f^{-1}(x)\) or \(\displaystyle g^{-1}(f^{-1}(x))\)?

And is \(\displaystyle (fg)^{-1}(x)\) supposed to represent a composition or a product of functions?

EDIT: I see Dr. Peterson has beat me to it, but it does look as though there are some notation differences between your text and what I am used to.
 
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Dr.Peterson

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'...even though you said you were doing the right thing' - I certainly wasn't that confident by a long chalk! (Is that a British Idiom?) Anyway, so I fell into the trap of getting the order wrong, i was trying to avoid that but must have reversed it in my head ..zzzzzz..

By the way, I have indeed reproduced the notation used in the book. One needs a lot of patience to not be put off by one's own cognitive deficiencies when learning maths- many thanks.
Let's rephrase that: "What you said you were doing was the right thing", namely, "applied the inverse of g to the inverse of f".

As to notation, this must be another of many areas where notation varies, though I hadn't seen this one. That's why I started with "I'm not sure whether your book uses slightly non-standard notation"; I've learned not to say someone's notation, or even grammar, is "wrong", because it may be what they were taught.

And I've never heard of a long chalk. We're deprived over here.
 

Simonsky

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Let's rephrase that: "What you said you were doing was the right thing", namely, "applied the inverse of g to the inverse of f".

As to notation, this must be another of many areas where notation varies, though I hadn't seen this one. That's why I started with "I'm not sure whether your book uses slightly non-standard notation"; I've learned not to say someone's notation, or even grammar, is "wrong", because it may be what they were taught.

And I've never heard of a long chalk. We're deprived over here.


We're deprived over here. Perhaps not, as American accents have preserved the earlier forms of English 'we've' lost (http://www.bbc.com/culture/story/20180207-how-americans-preserved-british-english)

Anyway, back to the question, if I may. So we are left with (((x - 1)/2) +2)/3. Not sure what to do with that.... perhaps: ((x-1)/2) + 2 = ((x-1)/2) +4/2 = ((x-1) + 4)/2 = (x+3)/2

I've probably messed that up and I'm still not sure what we are supposed to be left with...oh dear!
 

Simonsky

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We're deprived over here. Perhaps not, as American accents have preserved the earlier forms of English 'we've' lost (http://www.bbc.com/culture/story/20180207-how-americans-preserved-british-english)

Anyway, back to the question, if I may. So we are left with (((x - 1)/2) +2)/3. Not sure what to do with that.... perhaps: ((x-1)/2) + 2 = ((x-1)/2) +4/2 = ((x-1) + 4)/2 = (x+3)/2

I've probably messed that up and I'm still not sure what we are supposed to be left with...oh dear!
Sorry, should be((x+3)/2))/3.....'bad day at the office' (or should that be 'worse day than usual at the office').....glad you folks are so patient and helpful!
 

Dr.Peterson

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Sorry, should be((x+3)/2))/3.....'bad day at the office' (or should that be 'worse day than usual at the office').....glad you folks are so patient and helpful!
Very good so far. How about one more step, since complex fractions (fractions containing fractions) are not very popular? One way to take that step is to see that division by 3 is the same as multiplication by 1/3.

There are other ways to "simplify" an expression like this ("simple" is in the eye of the beholder), but you've made a good choice.
 

Simonsky

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Very good so far. How about one more step, since complex fractions (fractions containing fractions) are not very popular? One way to take that step is to see that division by 3 is the same as multiplication by 1/3.

There are other ways to "simplify" an expression like this ("simple" is in the eye of the beholder), but you've made a good choice.
I thought about that but then made the mistake of multiplying the whole thing by 1/3 when it should be (x+3)/2 * 1/3 ( I'm stepping on all the landmines today!) so doing that we're left with (x+3)/6 but comparing that to the original functions and inverses I'm not sure what to conclude.
 
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