Wikipedia:Reference desk/Archives/Mathematics/2024 March 20

From Wikipedia, the free encyclopedia
Mathematics desk
< March 19 << Feb | March | Apr >> Current desk >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.


March 20[edit]

Whether it coincides with a simpler function[edit]

Is y = sin (arcsin x) (1) the same function as y=x, if we consider all branches of logarithm (of any real number) and all branches of inverse sine function? Or does (1) remain meaningless for any argument outside the range [-1;1] when we restrict it to real value for both the domain and the image, and (1) will coincide with the identity function only when we regard it as a function that map complex numbers to complex number? Does the logarithm of negative numbers lead to the presence of removable singularities for (1)? (In contrast, the function y=x obviously does not contain any singularity). I was able to prove that y = arcsin (sin x) and y = sin (arcsin x) are not always the same, but I still can't settle the aforementioned problems. 2402:800:63AD:81DB:105D:F4F:3B26:74C5 (talk) 14:36, 20 March 2024 (UTC)[reply]

A univalued function and a multivalued function possibly partial, can be represented by a relation The total identity function corresponds to the identity relation Function composition corresponds to relation composition: The multivalued function inverse correspond to relation converse:
Just like the multivalued complex logarithm is the multivalued inverse of the exponential function , the complex including all branches is the multivalued inverse of function So
Generalizing this from the sine function to an arbitrary (univalued) function , we have:
Clearly, this implies so the composed relation is the identity relation on the range of representing the identity function on that range.  --Lambiam 18:21, 20 March 2024 (UTC)[reply]