Wikipedia:Reference desk/Archives/Mathematics/2019 February 20

= February 20 =

Inverse of function
I need to know how I calculate a inverse of this function if $$y=f(x)$$

$$x^a+x^b=y$$--192.161.6.16 (talk) 00:20, 20 February 2019 (UTC)


 * It depends on what $a$ and $b$ are. Can they be negative?  Do they have to be integers?  Even assuming positive integers, it may not have an inverse, and if it does, the inverse may not be expressible in terms of elementary functions.  For example, $$y = x^2 + 1$$ doesn't have an inverse, while $$y = x^5 + x^3$$ does, although I think to express it, you'd have to use something like the Bring radical (not 100% sure though).  –Deacon Vorbis (carbon &bull; videos) 00:56, 20 February 2019 (UTC)


 * The inverse of x^2 + 1 is:
 * x = y^2 + 1
 * y^2 = x - 1
 * y = sqrt(x-1)
 * What's wrong with the above?? (You're saying that the function has no inverse.) Georgia guy (talk) 01:00, 20 February 2019 (UTC)
 * y = sqrt(x-1)
 * What's wrong with the above?? (You're saying that the function has no inverse.) Georgia guy (talk) 01:00, 20 February 2019 (UTC)
 * What's wrong with the above?? (You're saying that the function has no inverse.) Georgia guy (talk) 01:00, 20 February 2019 (UTC)
 * What's wrong with the above?? (You're saying that the function has no inverse.) Georgia guy (talk) 01:00, 20 February 2019 (UTC)


 * $$f(x) = x^2 + 1$$ is not an injective function (aka not one-to-one), so it doesn't have an inverse. What you've done is found the inverse of the function which is the restriction of $f$ to the non-negative reals.  But without such qualification, $f$ would generally be understood to have a domain equal to the reals.  –Deacon Vorbis (carbon &bull; videos) 01:07, 20 February 2019 (UTC)
 * What you mean is, it's a function whose inverse is not a function. Georgia guy (talk) 01:08, 20 February 2019 (UTC)
 * Er, no. I mean that it doesn't have an inverse – period.  I'm not sure what you're getting at, but an inverse of a function (if it exists)...is a function, by definition.  –Deacon Vorbis (carbon &bull; videos) 02:51, 20 February 2019 (UTC)
 * If you think of a function as a special type of relation, you can look at the converse relation (or inverse relation) and check whether that is also a function. —Kusma (t·c) 09:34, 20 February 2019 (UTC)


 * That doesn't matter. Every binary relation has an inverse. And a function can be considered a special type of a binary relation. Then you can say you have a binary relation $$R\subset X\times Y$$ which is a function $$R:X\to Y$$ and its inverse relation $$R'\subset Y\times X$$ which is not a function from $$Y$$ to $$X$$. But this way you explicitly start from a notion of relation and classify it as being a function or not. If the inverse relation is a function, too, then that function is an inverse for the former function.
 * However, when you start from the notion of a function, then the Inverse function is defined as a function which... and so on. So, if the inverse relation is not a function, then the orginal function does not have an inverse. Its relation has an inverse, but the function itself does not. That's all, that's how definition states it. --CiaPan (talk) 10:30, 20 February 2019 (UTC)
 * God this place is awful. CiaPan, perhaps you should reserve your pedantry about the phrase "inverse function" for a situation in which that phrase has been used?  It does not appear in the OP or in Georgia Guy's comments, and what Kusma said is 100% correct. --JBL (talk) 12:27, 20 February 2019 (UTC)
 * @JBL: To keep the thing short I'll just quote this: “how I calculate a inverse of this function”. --CiaPan (talk) 12:44, 20 February 2019 (UTC)
 * Yes: this is a different phrase from "inverse function", and in particular does not imply that the inverse object must also be a function object. As Kusma has very clearly and correctly explained.  --JBL (talk) 13:41, 20 February 2019 (UTC)
 * I may say the following:
 * If by "inverse" the OP means an inverse function of the given function, then the function $$x^a+x^b=y$$ doesn't always have an inverse function, e.g. when $$a=b=2$$ (because both a non-positive $$x$$ and a non-negative $$x$$ may solve the equation).
 * However, if by "inverse" the OP means an inverse relation of the given relation, then the answer to the OP's question depends on what they mean by "how to calculate": If they are looking for a formula, using elementary operations only (addition, subtraction, multiplication, division, and extracting roots, whether square roots or cubic ones and the like), then it's always possible to calculate the inverse relation, provided that $$a$$ and $$b$$ are natural numbers smaller than five (and that complex unreal numbers are allowed as well, unless both $$a$$ - not being smaller than $$b$$ - is odd, and trigonometric functions - as well as their inverse functions - are allowed to be used in the formula, in which case the domain of discourse is allowed to be the real numbers).
 * However, if by "how to calculate" the OP allows numerical approximation as well, then there are many numerical methods for calculating the inverse relation, even for $$a$$ and $$b$$ unlimited. HOTmag (talk) 10:44, 20 February 2019 (UTC)


 * To elaborate a bit on HOTmag’s comments: with rare exceptions, if a,b are positive integers with a≥b and if the inverse function or relation of $$f(x)=x^a+x^b$$ is defined over the real numbers, then a necessary condition for the inverse relation to be a function is that a be odd, with one category of exceptions. This dismissing of cases of even a is because nonreal complex solutions of $$x^a+x^b-y=0$$ for x given y come in pairs, so the number of real solutions is the value of the exponent a minus an even number (the even number of complex nonreal solutions). So if a = 2 or 4 or 6 etc., the number of real solutions is even and hence is not equal to 1 and hence the inverse relation is not a function. The one category of exceptions to the dismissing of cases with a even arises if a is even and y is such that the polynomial equation has a duplicate solution, thus giving an odd number of distinct solutions to $$x^a+x^b-y=0$$ at this value of y and hence the possibility that there is exactly one distinct real root, making the inverse relation a function at that y value. Loraof (talk) 16:19, 27 February 2019 (UTC)