Talk:Inverse function

Percentages non-example

Despite their familiarity, percentage changes do not have a straightforward inverse. That is, an X per cent fall is not the inverse of an X per cent rise.

This could do with some rewriting. There is a straightforward inverse to the function "add a fraction p to x", i.e., f(x) = (1 + p) x, and that inverse is "subtract a fraction p / (p + 1)", i.e., "multiply by 1 / (1 + p)". The current text is confusing to someone who knows this inverse, but not the general concept. QVVERTYVS (hm?) 17:55, 5 October 2014 (UTC)

I see your point. You were thinking something like this?

Despite the familiarity of percentages, some find the inverse of a percentage change to be confusing because an X per cent fall is not the inverse of an X per cent rise. To solve this, percentages may be treated as fractions, p = n / 100, where n is the percentage. The inverse of adding a fraction p to x, i.e., f(x) = (1 + p) x, is subtracting a fraction p / (p + 1), i.e., f(x) = [1 - {p / (1 + p)}] x can be simplified to f(x) = [1 / (1 + p)] x.

Should the heading be changed from "non-example" in this case? —PC-XT+ 03:07, 16 October 2014 (UTC)

I suggest that this "non-example" be removed from the article. I find the phrase "non-example" confusing. It could be anything: "Non-example: James Baldwin." There seems to be a general confusion on this talk page about the word "inverse" in mathematics. The scope of the article should be limited. This is an article about inverse functions. "Percentages" aren't functions, they are ratios and numbers. A brief section like this or something similar could be in the article on percentage.

John (talk) 17:05, 24 January 2015 (UTC)

I agree. I don't think it really adds to the subject, which may actually be the reason for the name "non-example." There are plenty of other functions that have what some would think are strange inverses, but they all follow from laws. This one was probably only added due to familiarity. —PC-XT+ 00:27, 25 January 2015 (UTC)

I removed the section as I too thought the same thing as soon as I saw it Belovedeagle (talk) 00:13, 7 April 2015 (UTC)

Formula for the Inverse

An example which requires taking cube roots is more complicated than the principle it's supposed to illustrate.

A linear equation would be a more appropriate first example. — Preceding unsigned comment added by 66.35.36.132 (talk) 01:09, 18 March 2016 (UTC)

bijectivity would be more sensible

For the sake of generality, the article mainly considers injective functions. Yet, other articles (see for example Bijection#Inverses and Injective function#Injections_can_be_undone) consider that a function is “invertible if and only if it is a bijection”, and they link to the present article whenever mention is made of a complete inverse (i.e., a bijection). To consider injective functions instead of bijective functions, when speaking of inverses, creates much confusion. Even in the present article, in the section #Definitions, in the last paragraph, it is said that “If $f$ and $f^{-1}$ are functions on $X$ and $Y$ respectively, then both are bijections”, even though $f$ and $f^{-1}$ were actually defined above as having $X$ and $Y$ as their respective domains.

In section #Definitions, the article links to Inverse element; this implies that an inverse function is an inverse element with regard to function composition. Yet, as defined in the present article, an inverse function is just a right inverse, not necessarily a (two-sided) inverse. For the sake of both consistency and clarity, then, it seems crucial to make the article center not on injections, but on bijections.--Anareth (talk) 15:11, 28 August 2016 (UTC)

I think the confusion is in the subject mater in that some uses od invertible functions require bijective and some only injective. It would be wrong limit this article to only the bijective case. Yes, this causes confusion, but that's the nature of the subject.David Sherwood (talk) 21:36, 4 October 2016 (UTC)

A function is invertible if and only if it is a bijection. Those who hold that only injectiveness is required are the same set of folk who believe that all functions are onto (that is, they reject the concept of a codomain). Unfortunately, this definitely minority point of view has infected several articles and when not pointed out it certainly creates confusion. The article (especially the lead) needs to be rewritten from the general viewpoint and, if necessary, a section devoted to this minority view should be included much later in the article. I'll try to rewrite the lead with as little jargon as possible and concentrate on the meaning of an inverse function rather than its uniqueness. --Bill Cherowitzo (talk) 04:22, 5 October 2016 (UTC)

What I wrote above reflects my own point of view, but my edits will be much more neutral on this issue. As I started to rewrite I realized that because different editors with different POV's had been editing this article for a while, the result was a confusing mess with references to things that no longer existed and conflicting terminology. I can smooth this out with appropriate references, but it will take a little more time than I originally thought it would. --Bill Cherowitzo (talk) 23:27, 5 October 2016 (UTC)

Multivariate inverse function

Please read 'Multivariate inverse function', Is it easy to be understand? Can you accept it?

For multivariate function $x_{0}=f(x_{1},x_{2},\cdots ,x_{i},\cdots ,x_{n})$,$f_{i}:x_{i}\mapsto f(x_{1},x_{2},\cdots ,x_{i},\cdots ,x_{n})$.

If $f_{i}$ is bijection for any $x_{j}(j=1,2,\cdots ,n,j\neq i)$ we call $f_{i}^{-1}$ an multivariate inverse function about $x_{i}$. Introduce unary operator $I_{i}$ and denote :

$f_{i}^{-1}=I_{i}(f)$.

For example,$f(x_{1},x_{2})=x_{1}^{3}+x_{2}^{2}$ is invertible about variable $x_{1}$ and is not invertible about variable $x_{2}$.

Partial inverses can be extend to multivariate functions too. We can define multivariate inverse function for an irreversible function if we can divide it into r partial functions $f_{(k)},k=1,\cdots ,r$ and denote its inverses as :

$f_{i,(k)}^{-1}=I_{i}(f_{(k)}),k=1,\cdots ,r$.

For example,$f(x_{1},x_{2},x_{3})=x_{1}^{2}+x_{2}^{3}+x_{3}^{4}$

$x_{0}=f_{(1)}(x_{1},x_{2},x_{3})=x_{1}^{2}+x_{2}^{3}+x_{3}^{4},x_{1}\geq 0,x_{3}\geq 0$

$x_{0}=f_{(2)}(x_{1},x_{2},x_{3})=x_{1}^{2}+x_{2}^{3}+x_{3}^{4},x_{1}\geq 0,x_{3}<0$

$x_{0}=f_{(3)}(x_{1},x_{2},x_{3})=x_{1}^{2}+x_{2}^{3}+x_{3}^{4},x_{1}<0,x_{3}\geq 0$

$x_{0}=f_{(4)}(x_{1},x_{2},x_{3})=x_{1}^{2}+x_{2}^{3}+x_{3}^{4},x_{1}<0,x_{3}<0$

$x_{1}=f_{1,(1)}^{-1}(x_{0},x_{2},x_{3})=I_{i}(f_{(1)})(x_{0},x_{2},x_{3})=[x_{0}-x_{2}^{3}-x_{3}^{4}]^{1/2},x_{0}\geq 0,x_{3}\geq 0$

$x_{3}=f_{3,(3)}^{-1}(x_{0},x_{1},x_{2})=I_{3}(f_{(3)})(x_{0},x_{1},x_{2})=-[x_{0}-x_{1}^{2}-x_{2}^{3}]^{1/4},x_{0}\geq 0,x_{0}\geq 0$

The concept of multivariate inverse function is useful to express the solution of a transcendental equation. — Preceding unsigned comment added by Woodschain175 (talk • contribs) 22:37, 25 June 2017 (UTC)

Notation

Hello all! I believe I read somewhere that the inverse of a function f(x) can be denoted as inv(f(x)). I think it was in a computer science context, rather than pure math. Does this sound familiar to anyone? I'm going to look around to see if I can find any references for this notation eventually before putting it in the article. If anyone knows a reference, I'd appreciate you putting it in! Thanks for reading. :) JonathanHopeThisIsUnique (talk) 04:33, 30 November 2017 (UTC)

This is computer code, not mathematical notation. For instance Mathematica uses Inversefunction[f][x] but translates back into the conventional mathematical expression when asked to present in "TraditionalForm".Limit-theorem (talk) 11:18, 30 November 2017 (UTC)