Talk:Inverse function

The main table of inverses.

--JayEB (talk) 02:58, 19 November 2011 (UTC)
I have no issue with the small table of trig inverses, but the main table seems inconsistent, at the least.
For starters, below the table, for f(x)= y = (2x + 8)³, I think f ^-1(x) should be (x^1/3- 8)/2 and f^-1(y) should be (2y + 8)³ .
If not, then specifically why not ?

There's no f^-1(x) to switch to f^-1(y). It says f(x)=(2x + 8)³. So what change are you suggesting exactly? Right now it's consistent in that the variable isolated in the first part is replaced with the inverse function symbol (written as a function of the dependent variable), and if you switched the variables to make both functions functions of x, you're likely to throw people off. ᛭ LokiClock (talk) 23:16, 19 November 2011 (UTC)

--JayEB (talk) 00:18, 20 November 2011 (UTC)
Confused by f^-1(x) ? Then leave it out. I say the f^-1(y) in the article is wrong, and it should be f^-1(y) = (2y + 8)³ .
Specific reasons why not ? Supporting work and references please.

You seem very confused by standard mathematical notation. f(x)= (2x + 8)³ means exactly the same as f(y)= (2y + 8)³, which means exactly the same as f(q)= (2q + 8)³, which means exactly the same as f(β)= (2β + 8)³. You can use any letter you like, it doesn't change the meaning. By the way, please put your signature at the end of your message, not the beginning. -- Dr Greg  talk  02:00, 20 November 2011 (UTC)

Specific reasons why not ?. f ^-1(y)=(y^1/3-8)/2 is reversing the input variable, not a function inverse.
Plots of the inverses mirror across X=Y and compose to X. You seem to be confused.--JayEB (talk) 02:44, 20 November 2011 (UTC)

f⁻¹(y)=(y^1/3−8)/2 means exactly the same as f⁻¹(x)=(x^1/3−8)/2, which means exactly the same as f⁻¹(t)=(t^1/3−8)/2, which means exactly the same as f⁻¹(φ)=(φ^1/3−8)/2. You can use any letter you like, it doesn't change the meaning. All of those are correct answers for f⁻¹ in this case. If you write y=f(x) then you would write x=f⁻¹(y), but if you are simply talking about f(x) with no mention of y, then you are free to use any letter you like for the inverse. -- Dr Greg  talk  14:33, 20 November 2011 (UTC)

Letters are not the point. I have to think that you just don't understand the issue.
Looking for supported comments by those who do. --JayEB (talk) 16:58, 20 November 2011 (UTC)

It's true I don't understand the issue of why you find this so confusing. It's very simple. If f(p)= (2p + 8)³ then f⁻¹(q)=(q^1/3−8)/2 and you can replace either or both of p and q by x, y, or any letters you like. This is very standard mathematical terminology. -- Dr Greg  talk  17:27, 20 November 2011 (UTC)

I don't understand your point, either. Can you write out the original text, and what they'd become after you changed them? f^-1(x) isn't confusing, it doesn't exist. So I don't know why you're saying it should become something. The answer to your question may in fact be that there's no reason why not, or that it's merely a matter of pragmatism. But "why not" isn't an argument, "why" is. But first... what? ᛭ LokiClock (talk) 17:35, 20 November 2011 (UTC)

I need a math notation editor. I uploaded a pdf to avoid the grief of this editor:

Add caption here

--JayEB (talk) 21:30, 20 November 2011 (UTC)

You took f({0, -2}) to get {512,64}, so you have to take f^-1({512,64}), not f^-1({0,64}) to get {0, -2}. That's why it's written f^-1(y), because you're supposed to set y=f(x) to get x back. ᛭ LokiClock (talk) 21:48, 20 November 2011 (UTC)

Are you sure you aren't confusing inverse relation and inverse relationship? Relations are like functions that don't have to pass the vertical line test, so ignore those terms for the purposes of the conversation. You're saying y=e^x is the inverse of x=e^y, but again, it merely plots the inverse of x=e^y, meaning you get the same plot as taking x=ln(x). Saying X and Y instead of x and y is problematic, because then it looks like you're talking about the sets, which are both the real numbers. e^x sends any real number x to another real number, so it sends X to X. Your problems would be solved if you were talking about the pairs of coordinates, the actual objects on the plot, instead of the input and output those objects are employed to graph. Then, truly, if x and y are coordinates, y=e^x produces the coordinates {(x=0,y=1),(x=1,y=e)}, and x=ln(y) produces the same coordinates - {(x=0,y=1),(x=1,y=e)}, except in one case the 2nd coordinate is derived from taking a function of the 1st, and in the other case the 1st coordinate is derived from taking a function of the 2nd. However, if you use x=e^y you {(x=1,y=0),(x=e,y=1)}, the same coordinates you get from the inverse function on the opposite variables, y=ln(x). It's only when the x and y refer to coordinates that you get this illusion that when functions have the same plot then they're the same, and when they have a different plot they're different. It depends on whether x and y have an input-output role, and whether those roles correspond to geometric realities. When we say f(q) instead of f(x), that's like labelling the horizontal axis q instead of x. Instead, you're using the vertical axis for input and the horizontal axis for output, which is a different situation. Taking the inverse function of the variable with the opposite role will produce the same coordinates. That proves that the resulting coordinates have not changed, not that you didn't have to change both the function and input to do so. ᛭ LokiClock (talk) 22:20, 20 November 2011 (UTC)

Well, technically e^x sends X to X⁺, but it still doesn't help you. The line is a separate set from the input and output sets. Y=X is a true fact - Y=R, and X=R. ᛭ LokiClock (talk) 22:54, 20 November 2011 (UTC)

Put a different way, the 1 on the x axis is the same as the 1 on the y axis, but (1,0) is not the same as (0,1). You can certainly take the coordinates (x,f(x)), and then the coordinates (f(y),y) and get different points, but you can't take the function f(x) and the function f(y) and get different values. ᛭ LokiClock (talk) 22:33, 20 November 2011 (UTC)

I did not wade through the last few notes. I give up. In parting, I mixed up p's and q's in that modified example;
I do not agree; the inverse function should be f^-1(q)= (2q + 8)³ as with the actual case f(x)=(2x+8)³.
A plot of f(x) between(-8,-512)and (0, 512), along with line X=Y makes it clear that the inverse runs through (0,-4) with a very small range in Y. --JayEB (talk) 00:34, 21 November 2011 (UTC)

Okay. I "waded through" your PDF and tried to give a thorough explanation. The reflection you keep using only works when you don't change input and ouput axes before you plot the second function. ᛭ LokiClock (talk) 02:30, 21 November 2011 (UTC)

Dr Greg undoes correction of a blunder

Dr Greg: You need to prove that the temperature converse was an inverse. You cannot, so leave my correction alone.--JayEB (talk) 22:53, 29 November 2011 (UTC)

Of course I can prove it, and I've added an extra line to do so (though it is really stating the obvious). In your version you gave two different and contradictory formulas for f, which means you are still very confused about the meaning of the symbols f and f⁻¹ (see discussion above).

As for the second edit you made, please go and read the definition of function, and, in particular, the section Specifying a function. There is no necessity to have a "rule" or "correlation relevance".

By the way, I have a Ph.D. in mathematics, what is your qualification? -- Dr Greg  talk  00:01, 30 November 2011 (UTC)

.......................................
I know several math PhD's who don't understand what an inverse function really means.
You transpose to isolate the other variable, then plug it back into the equation; that will ALWAYS return the original variable!
If you plot your "inverse", you will have the same plot on the same axes; that is NEVER an inverse!
But a transposed converse will plot as the same line. My real inverse plots symmetric to F=C.
You don't provide proof, so what allows you to counter proof ? --JayEB (talk) 00:21, 30 November 2011 (UTC)

Interesting that if several PhDs disagree with you, you assume they are all wrong and you are right. The trouble is, you seem to be making up your own definition of "inverse" instead of using the one stated in the article. If we define f and g by f(C) = 9⁄5C + 32 and g(F) = 5⁄9(F + 32), do you not agree that f and g are both functions and that f(C) = F if and only if g(F) = C? That's precisely the definition given in the article (Inverse function#Definitions). Equivalently g(f(C)) = C for every C, which satisfies the alternative definition given at Inverse function#Inverses and composition.

In your version you had f(C) = C 9⁄5 + 32 and f(F) = 5⁄9 (F – 32), which makes no sense at all as you've used the same letter f with two contradictory meanings. I still think you haven't grasped the concept that the definition f(F) = 5⁄9 (F – 32) means exactly the same as f(C) = 5⁄9 (C – 32) or f(μ) = 5⁄9 (μ – 32). -- Dr Greg  talk  01:41, 30 November 2011 (UTC)

I see one problem is that you use this article to defend this article. Try using serious math sources;
several very serious sources; not your memory; not the palaver here.--JayEB (talk) 03:09, 30 November 2011 (UTC)

I think the "I have a PhD" comment has put this discussion a little off track. Let's concentrate on the facts. The title of this article is very clear: it is talking about "Inverse function[s]". The word function is meant in the mathematical sense: objects which accept a number, and using some rule output another number (this is a slight simplification in a couple of ways not relevant here). They are like little black boxes which have a hole you put numbers in and another hole out of which other numbers fall. To reflect this fact, mathematicians sometimes use notation like f: x → 1 + x² (f takes any number x to the new number 1 + x²), which means precisely the same as f(x) = 1 + x² (and the same as f: y → 1 + y², and the same as f(y) = 1 + y²). Functions know nothing about temperatures, or anything else in the real world: all they can see is the numbers we put into them. They don't even know about graphs (we can use graphs to help understand functions, but that's for our own benefit, and it doesn't affect the meaning of "function"), so this stuff about "plotting on the same axes" isn't relevant.

The "composition" of two functions just means that we apply one function then the other to any given number (like connecting up the "out" hole of one function directly to the "in" hole of the next). An "inverse" of a function f is another function g such that when you compose g with f either way round you always get the number out that you originally put in. Note that an inverse function is still a function: it is still a rule for going from one number to another. As Dr Greg correctly says, f: x → 9⁄5x + 32 has inverse function g: x → 5⁄9(x – 32). This is a mathematical fact. If you do not believe this, then I'm afraid that you have misunderstood how functions work. In that case this article (and presumably the one on functions) has failed you; I'm sorry about that, but the talk page isn't the place to get help. The talk page is for people who already understand the subject to discuss improvements. Quietbritishjim (talk) 01:32, 1 December 2011 (UTC)

Too many have learned that a transposed converse is an inverse. They should not be repeating that mistake here.
References can be found to support that, but check Bronshtein, Wolfram, and the consensus references; and ask engineers who actually work with cancelling inverse functions. Words are just words without proof and citations. Prove it to yourself.--JayEB (talk) 02:49, 1 December 2011 (UTC)

Another problem is that you have now introduced an "x" for a binary function with variables C and F.
The "x" does not belong in this example, and it's deceiving when a converse is composed with a function,
since the result looks very much like a definition of an inverse function.--JayEB (talk) 03:13, 1 December 2011 (UTC)

JayEB, before you can understand what "inverse function" means, you must understand what "function" means. The fact you have said '"x" does not belong in this example' shows that you don't know what a function is. A function is just a rule to go from one number to another; the variable name is not part of the function. f is also equal to "multiply by 9⁄5 then add 32". Are you happy to accept this?

You want references? (I note that you aren't giving any!) Any calculus or early analysis book will do, but here's a good one:

• Calculus By Michael Spivak, p 39, "a function is a rule which assigns, to each of certain real numbers, some other real number." Quietbritishjim (talk) 11:47, 1 December 2011 (UTC)

Some function references: https://picasaweb.google.com/112167498780318401154/ScrapbookPhotos --JayEB (talk) 20:15, 1 December 2011 (UTC)

The ones on your first page are good references, and they agree with what I'm saying, especially the bits you've underlined. As they say, the function is a rule that takes a number to a new number. This is why "multiply by 9⁄5 then add 32" is a function, the one we've been talking about all along. And as the first reference says, "functions can be represented by the points (x,y)". Note that it says that graphs are just ways to represent graphs for our understanding; a graph is not a definition of a function.

The reference on you second page is not authoritative. It is from PlanetMath, which is community-written, like Wikipedia. It still has it pretty much right, but is maybe a bit sloppy in places, and almost sounds like the author had a similar misunderstanding to you. (Actually I think they didn't but were just trying to explain the idea behind the function.)

Stick to the references in your first picture, and you should be fine. Quietbritishjim (talk) 21:31, 1 December 2011 (UTC)

Some folks keep writing “advice” that I’ve known since age 12, and use their ‘news-to-me’ in droning, counter-productive palaver. . WP is not a game show. . . The second page you refer to is from Springer’s European Math Society site; I would not cite from Planet Math for these purposes. . .Why was it thought to be from PM ? . . . The idea of function can be as in set math, where tags on individuals can be a plain correspondence. . . . That is unfortunately called a ‘function’, but that’s just a unique inventory list ( maybe that should be called a “set function”). . . Algebraic functions have a rule involved. That rule can have uncertainty within a correlation, as long as that is explicitly included. . .But the subject here is a simple linear equation of f(C) and f(F), with “x” improperly imposed. . .And why should I let some unknown, and somewhat hostile, person tell me which references to believe ?
Too much wasted time here; need fewer words, more proof and citations.--JayEB (talk) 22:51, 1 December 2011 (UTC)
. . . . Another warning to anyone trying to learn from this article . . . .
In the temperature “example”, f(C) is composed with “f^-1(F)” as a “proof”. But for a real inverse, it would be f(C) composed with f^-1(C)
Furthermore: https://picasaweb.google.com/112167498780318401154/ScrapbookPhotos#5681793568547730738 . . . --JayEB (talk) 16:40, 3 December 2011 (UTC)

Redirects that do not make much sense ....

The word "invertible" in the Wikipedia now redirects to "inverse element", which is something in abstract algebra (e.g. group theory). However, "invertible" should redirect to "invertible function", which is the more familiar concept in analysis and mappings between sets. Then "invertible function" gets redirected to "inverse function", which is sometimes not what we are looking for either. Sometimes we just want to know if a function or a mapping is invertible, and we don't need to know what that inverse IS.
98.67.106.59 (talk) 07:17, 4 August 2012 (UTC)

Too technical for most readers to understand

"This article may be too technical for most readers to understand...."
Yes, yes, it is supposed to be. "Too technical for most readers to understand" is necessary and sufficient. Most readers have not taken or completed "upper-level" high school math, and hence they do not know anything about logarithms, trigonometric functions, inverse trigonometric functions, etc. Often the same applies for exponential functions and functions defined by roots (square roots, cube roots, etc.) or absolute values. They certainly do not know anything about calculus. (Note: a fucntion that has a horizontal tangent line fails to have an inverse there.)
This continues for college students on the freshman level. I have tried to teach students on the level of College Algebra and in Precalculus who thought that all functions were straight lines. Hence f(x) = K/x does not make sense to them because they do not know that its graph is a curve, and furthermore, it has two disjoint pieces to it.
One of my colleagues jokingly said that those students are "instinctively prepared" for calculus, in which all differentialble functions, if you look at them through a powerful-enough microscope, look like straight lines!
98.67.106.59 (talk) 07:38, 4 August 2012 (UTC)

This article needs to be split in two in some way

This article needs to be split in two in some way.
Method 1: Split this article into two articles in the Wikipedia.
Method 2: Split this article into two large nonoverlapping sections. Call them Part A and Parr B for the moment.
In either case, Part A will be all about functions of just one real number and their inverse functions. The functions themselves will only have real numbers in their ranges.
In either case, Part B will cover functions and mappings of a lot more generality. The objects in the domains and ranges of the functions can be abstract "points" in the two sets, or to be more specific, pairs of real numbers in both sets, triples of real numbers, complex numbers, quaternions, or functions themselves - whatever you like.
Hence Part A will be strictly on the level of high school students and lower-level college freshmen. Then Part B will be for readers with more background and interest in math, incluing those who have studied multivarible calculus, complex analysis, and whatever.
This ought to cut down on a lot of confusion and arguments - especially concerning people who know something about Part A, but them they want to get argumentative about things in Part B, which is a lot more generalized.
See also the article on the Inverse Function Theorem for some guidance.
98.67.106.59 (talk) 08:25, 4 August 2012 (UTC)

"undoes" another function ?

I think this is really very bad terminology. So "log" undoes "exp"? Then how comes that the function exp still exists, since log undid it? One function cannot undo another function, at best, it can undo (sic) the "action" of the other function.

PS: "putting y into the (...) function g " is roughly as bad. Since when are we putting things into a function? C'mon, there's a difference between being understandable and using baby language, esp. in an article about mathematics...

PS2: "Direct variation function are based on multiplication; y = kx. The opposite operation of multiplication is division and an inverse variation function is y = k/x." --- nonsense! (and bad grammar...) The second function does not divide x by k, as opposed to the first which multiplies x by k. (If you are "putting" kx "into" the 2nd function, you get y=1/x!) Maybe the author should first stick to addition, with y=x+a (aka "translation"), then the inverse operation obviously is y=x-a, and not a-x). — MFH:Talk 04:58, 9 September 2012 (UTC)

The lead paragraph: I agree that the first sentence is not as clear as it could be, although I think you're making a bigger deal out of the difference between a function and its effect than is really warranted. Changing "undoes another function" to "undoes the effect of another function" or "undoes the action of another function" would probably better. Perhaps it would be better to put the rigorous definition first and then put this sentence, making it clear that it's an intuitive point of view. But in that case the rigorous definition should be expressed in words instead of symbols, in my view. Either way, it should be corrected - at the moment it only defines a one-sided inverse! Quietbritishjim (talk) 16:25, 9 September 2012 (UTC)

Inverses in calculus

"Inverses in calculus" should refer to the Fundamental Theorem of the Calculus because the fact that integration and differentiation are the inverse operations of each other is waht the relationship between differential calculus and integral calculus is all about.
If Isaac Newton and Gottfried Wilhelm von Leibnitz had not discovered the Fundamental Theorem of the Calculus (of something simpler than it), then we would be in serious trouble in science, technology, and mathematics. Not even mentioning the Fundamental Theorem of the Calculus is a serious lacking.
This article also has other serious problems. I suggest that it needs to be scrapped and redone from scratch.
98.67.96.230 (talk) 23:37, 19 September 2012 (UTC)