Talk:Inverse function/Archive 2

Archive 1

Archive 2

Note on "Note on notation"

Usually and most of the time :

$\sin ^{-1}x=(\sin x)^{-1}.\,\!$

and inverse function of

$f^{-1}(\sin x)=\arcsin x=\mathrm {asin} \,x.\,\!$

If arcsin(x) is used then sin⁻¹(x)=(sin x)⁻¹.

In calculus ƒ⁽ⁿ⁾, where n is a Roman numeral; with or withouth parentheses; that n with equal validity also denotes the nth derivative of a function ƒ. For instance:

$f^{II}(x)={\frac {d^{2}}{dx^{2}}}f(x).$

Finally, in case of derivative nth function could be represented, and often and usually is, with an apostrophe. For instance:

$f^{''}(x)={\frac {d^{2}}{dx^{2}}}f(x).$

All these notations are valid and used.

However, I tried to keep my editing as minimal as possible so I edited only categoric statemant regarding sin(x). Hrvatistan (talk) 02:55, 9 July 2010 (UTC)

I disagree, I have only ever seen sin^{-1}x used to mean arcsin x, even though it is inconsistent with sin^2 x. Where have you seen it used to mean (sin x)^{-1}? I also haven't seen e.g. f^3 without brackets for derivative, presumably because it already has two meanings (multiplicative power and repeated function application). Quietbritishjim (talk) 00:40, 10 July 2010 (UTC)

I agree with Quietbritishjim. -- Dr Greg  talk  08:41, 10 July 2010 (UTC)

Well, that is the reason why I only edited only categoric statemant regarding sin(x) because it is possible to use it both ways. And since Karl Weierstrass mathematic is all about consistency. If You use arcsin(x) then it is general rule that You don't use sin^{-1} x for the same thing. And I said Roman not Arabic numerals. If You haven't seen something it doesn't mean that something does not exist. Or like Dolph Lundgren said it: "There's an old saying: just because you're paranoid doesn't mean they're not out to get you." Hrvatistan (talk) 16:24, 31 July 2010 (UTC)

Bourbaki uses a different notation for this. His notation suppress ambiguity, but is only seldom used. It places the $-1$ exactly on top of the funtion. For a function $f$, it should render something like ${\stackrel {-1}{f}}(x)$, which avoid confusion with the function $f^{-1}$, whose value $f^{-1}(x)$ at $x$ is $f(x)^{-1}$. — Preceding unsigned comment added by 92.104.193.24 (talk) 15:38, 7 November 2011 (UTC)

Technical

I know y'all are going to absolutely hate me for this one ... I perfectly well understand the article and think its actually wonderfully written for me (personally). -- but this is really just because I've messed around with math for long enough that I can actually understand it (with occasional references to old text books ...) but its going to be really hard for most people to read through this without having a lot of trouble. Ideas? Possibly an "Introduction to inverse functions" article? Katanada (talk) 05:03, 28 April 2011 (UTC)

"Introduction to" articles are best created very sparingly. I don't think the problem with the lede was actually a technical language issue, just an instance of confusing language. I attempted to clean that up. If your problem was with the rest of the article, please fill out the section parameter to the template. ᛭ LokiClock (talk) 20:49, 16 June 2011 (UTC)

I just want to question a "transposed form" presented as an "inverse".

--JayEB (talk) 05:13, 2 November 2011 (UTC)
Below the table of inverses, there's a method for finding an inverse which just solves
the equation to isolate the other variable. But the step of trading places of the two variables is not there.
So the result seems to be just an isolation of the other variable.(the plot of the transformed function is identical to the original)
I think any change should be after some talk.(as long as this huge talk page does not continue to grow exponentially)(Cliff's Notes ?)

The process of finding the inverse is, in fact, isolating the other variable. x & y are usually swapped so that x will always represent an input and y an output, but the x input to the function isn't the same as that of the inverse, whether you switch them or not. f(a)=2a is the same function as f(b)=2b. ᛭ LokiClock (talk) 01:48, 8 November 2011 (UTC)

--JayEB (talk) 19:22, 3 November 2011 (UTC)
[The following is derived from Bronshtein, Handbook of Math, 5th, 2007, at 2.1.3.7. ,
but I substitute my personal notation f ^1-(x) as a cure for the tainted shorthand: f ^-1(x) ]
[ f ^[-1](x) is too cumbersome, and f ^-(x) is too close to f '(x) ]
The functions y = f(x) and y = f ^1-(x) are inverse functions of each other, for the domains that are functions.
Starting with y = f(x), to get the explicit inverse, you exchange x and y in the expression,
which gives you x = f(y), and solving for y gives the explicit form y = f ^1-(x) .

The plot of y = f ^1-(x) is a reflection of y = f(x) across the line y = x .
Examples of inverse functions: y = x ² has the inverse y = f ^1-(x) = x ^1/2
y = f(x) = e ^x has the inverse y = f ^1-(x) = ln x
y = f(x) = sin x has the inverse y = f ^1-(x) = arcsin x

Nope, it's irrelevant. Exchanging the variables is purely syntactic and totally unnecessary, because a function f(x)=x^2 is the exact same function as f(y)=y^2. Variable names don't matter here.--greenrd (talk) 23:41, 6 November 2011 (UTC)

--JayEB (talk) 06:48, 7 November 2011 (UTC)
y = f(x) = x^2 is a vertical arch symmetric about the y axis.
x = f(y) = y^2 is a horizontal arch symmetric about the x axis.
The axes are x-horizontal and y-vertical for both.

No, you are talking about a 2-dimensional plot of a function, which is not the essence of a function, it's just a drawing of it. All you've demonstrated is that the same function can be plotted in more than one way. Conventionally, the dependent variable is usually on the vertical axis and the independent variable on the horizontal. If you plot a function with a log-scale vertical axis, that doesn't make it a different function.--greenrd (talk) 19:38, 7 November 2011 (UTC)

You can also turn the piece of paper sideways, and then the x axis runs up & down. You haven't inverted the function by doing that unless you cross out x & y and switch them. ᛭ LokiClock (talk) 01:48, 8 November 2011 (UTC)

--JayEB (talk) 06:25, 8 November 2011 (UTC)I will assume you are not putting us on. A function in a real-number cartesian X-Y field is properly
represented by a plot of the function on X-Y coordinates. And its inverse, whether it's also a function or not
is properly represented by a plot of that inverse. One problem in math is that folks don't plot things early on.
Your statement carries a burden to prove that the "essence" of such a function is not properly represented in a plot of the function.
We are not dealing with imaginary numbers here. That's here: http://en.wikipedia.org/wiki/Imaginary_number#Geometric_interpretation

The real numbers aren't an X-Y field. X and Y are labels for the 1st & 2nd dimensions, simple conventions. Cartesian coordinates can use any labels, or none. The Z and Y axis switch roles frequently - either can be used for depth or height. Properly, x, y, and z are just variables. What they usually represent is "a number in the real numbers," so that when the solution to an equation is x, the solution means "any number in the real numbers." When you say f(x)=x, you can interpret f as something with an infinite number of values. It has a solution in terms of x, which will give f as many solutions as their are real numbers. Since the x is the same on both sides of the equation, the solution corresponds to the value in R. When a function depends on two variables, both defined as "any number in R," the symbol being the same on both sides of the equation is what distinguishes the two "any numbers." But you don't label the numbers by what order you pick them in for the equation, nor does picking labeling them give them any sign of input or output - the output of the function is strictly evaluation of f(x). It's the convention on paper that makes y=f(x) for whatever f you happen to be plotting. ᛭ LokiClock (talk) 12:15, 9 November 2011 (UTC)

"All you've demonstrated is that the same function can be plotted in more than one way."
These are not the "same function"; they are inverses of each other. And X = Y² is not even a function;
its a relation, since there are two Y's for each positive X.
Consider the Exponential function Y =e^X, and its inverse X=e^Y (the Log function).

That's a mistake. It's easy to misinterpret the vertical line test in that way. What it actually says is that the true inverse of x², ±sqrt(x) which is the same as ±sqrt(y) and ±sqrt(a), isn't a function, because it maps multiple outputs (±sqrt) to one input (x, y, or a). x=y² isn't the inverse of y=x², it's the same function on different variables. When you plot x=y² without relabeling the axes, you're switching the input/output ROLE of x and y, and the function is unchanged. Then you actually perform the "vertical line test" using a horizontal line - so take the name with a grain of salt. ᛭ LokiClock (talk) 12:21, 9 November 2011 (UTC)

"If you plot a function with a log-scale vertical axis, that doesn't make it a different function."
We are in full agreement with the World on that.

"You can also turn the piece of paper sideways, and then the X axis runs up & down.
You haven't inverted the function by doing that unless you cross out X & Y and switch them."
We are in full agreement with the World on the first part; but re-labeling the axes would not change
the nature or operation of any planar real-number function, nor its inverse.

It does, because by changing the graph in that manner you're actually changing the function you're graphing. On your graph, you've chosen x for input and y for output. So when you relabel the axes on the graph of y=x², it becomes a graph of y=±sqrt(x). Geometrically, reflection about the line y=x is equivalent to taking the inverse through syntactic isolation. Try it with a logarithmic & exponential function of the same base. ᛭ LokiClock (talk) 11:53, 9 November 2011 (UTC)

--JayEB (talk) 00:45, 10 November 2011 (UTC)
The proof I asked for ? After reviewing things, I stand by what I wrote.
And re: x = y² isn't the inverse of y = x²
The inverse of f(x)= x² is f ^-1(x)= x^1/2.
The check is: f [f ^-1(x)] = x (the identity line Y = X )
I've had enough of this thread. The previous article section needs work.

I think you need to ask more questions before you decide. You deny that f(x) and f(y) are the same thing, but you haven't explained why they should be different functions. The "identity line" picture also requires consistency of input/output conventions - x must be input to f, and y must be its output, so to reverse it the inverse must take the output as an input - so it must be f^-1(y), not f^-1(x), when testing f^-1(f(x))=x. That means to change the order to f(f^-1(x))=x, f could only be f(y) and f^-1 could only be f^-1(x). By the logic that f(x) is different from f(y), satisfying the opposite version of the identity statement changes the function and its inverse. In other words, if f(x)≠f(y), f(f^-1(x))≠f^-1(f(x)), and therefore x≠x. ᛭ LokiClock (talk) 10:35, 10 November 2011 (UTC)