Talk:Inverse function

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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)[reply]

Integration and differentiation are inverse operations; this is, however, the article about inverse functions. I fail to see what the history of the calculus has to do with a discussion about inverse functions. A function is a one-to-one mapping from a set A (the domain) to a set B (the range or image). If the mapping is one-to-one, then the inverse exists. If it is one-to-one and onto, the inverse relation is a function.

If you see a "serious problem" then spell it out specifically.

Some calculus may be appropriate in a discussion of inverse functions, but it is not at all necessary to use calculus to define the inverse of a function.

John (talk) 03:44, 24 January 2015 (UTC)[reply]

simple English

some of it could be simpler, is this the simple English Wikipedia. For example I have just learnt that a way to think of the inverse of a number is 1 divided by that number. Rather than that number divided by 1. So this clearly says what is inverted. The mathematical explanations do not say that in written English. This inappropriate writing style (as found here) give the problems in understanding mathematics. — Preceding unsigned comment added by 141.244.80.133 (talk) 14:20, 15 January 2013 (UTC)[reply]

This is not the article about the multiplicative inverse in the field of real numbers $\mathbb {R}$ , it is about inverse functions. The operation of multiplication in a number field is not particularly relevant here, and an off-topic digression is not a simplification.

John (talk) 00:05, 26 January 2015 (UTC)[reply]

Inverse operation

Definition. * is an operation on A, i.e. a*x =b where a, x, b are at A, then x = a^-1*b, if * is conmutative, we can write x = b*a^-1; else x = bºa; at this case º is named inverse operation of *.--201.240.86.150 (talk) 17:42, 26 April 2013 (UTC)[reply]

non-invertible functions

I would like to have here explicit examples ( and discussion) of functions that do not have inverse function. Regards.--Adam majewski (talk) 09:54, 4 January 2014 (UTC)[reply]

There is a detailed discussion in inverse function #Partial inverses. The last paragraph of inverse function #Definitions also consider this. Note that section levels currently appear to be broken, with “examples” sections following “Definitions” as ===s (subsections) for no reason; I refer to the last paragraph of “Definitions” proper. Incnis Mrsi (talk) 11:35, 4 January 2014 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

invertibility implies bijectivity (?)

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 Failed to parse (syntax error): {\displaystyle f ^{−1}} are functions on $X$ and $Y$ respectively, then both are bijections”, even though $f$ and Failed to parse (syntax error): {\displaystyle f ^{−1}} were actually defined above as having $X$ and $Y$ as their respective domains.