I don't know if this exists, but it would make my algebra easier if it did instead of having to use complicated radicals to solve an equation.

If $x \in \mathbb{R}$ and $y \in \mathbb{R}$ and $h(x,y)=x+y$, is there a real or complex function $f$ such that $f(x+y)=x^{2}+y^{2}$? And if so, what is it?

  • 4
    $\begingroup$ If there were, what would $f(4)$ be? Would it be $f(2+2) = 2^2+2^2 = 8$, or $f(3+1) = 3^2+1^2 = 10$? $\endgroup$ – rogerl Oct 26 '18 at 1:52
  • 4
    $\begingroup$ $f$ is a function, so if you feed it a number, it produces one (and only one) number. So if you feed it the number 4, you have to tell me what it produces. But 4 can be written in multiple ways as a sum of two numbers, and thus the value of the function depends on the representation you choose for 4. The function is not well-defined. $\endgroup$ – rogerl Oct 26 '18 at 1:57
  • 1
    $\begingroup$ @user608672 If you show the function is inconsistent, even when restricted to real numbers as input, then it will still be inconsistent if you try to extend it to complex numbers. $\endgroup$ – Cheerful Parsnip Oct 26 '18 at 2:17
  • 2
    $\begingroup$ The point here is that "testing only real numbers" is sufficient because you want it to work for real numbers, and so the fact is that it does not (as @rogerl showed you). $\endgroup$ – hardmath Oct 26 '18 at 2:21
  • 1
    $\begingroup$ "and $h(x,y)=x+y$" - where do you use this? Also, choosing $y=0$ you must have $f(x)=x^2$, but clearly $f(x+y)=(x+y)^2=x^2+2xy+y^2=x^2+y^2$ only if $2xy=0$... $\endgroup$ – Sil Oct 26 '18 at 6:54

It doesn't exist because $f(0)=f(-1 + 1) = 2$ and $f(0)=f(2+-2)=8$.

  • $\begingroup$ Couldn't that simply suggest that this function has multiple branches? Arcsin has multiple output values, and when you feed it a real number, it still outputs a real number over a specific domain. It's comprised of imaginary variables even though its form is unintuitive. So what I would expect is that there is some kind of unusual looking complex/imaginary function where you would have to be selective of the branch you're using to satisfy the x+y formula. $\endgroup$ – user608672 Oct 26 '18 at 2:03
  • $\begingroup$ @user608672: if an input has multiple outputs, then it is not a function. Please look up the definition of function. $\endgroup$ – Cheerful Parsnip Oct 26 '18 at 2:19
  • $\begingroup$ That's still just a cop-out to get out defending your claim. Arcsin has multiple outputs, but we can still adjust it as needed to be a function when we're ready to use it over a specific range, pretending arcsin doesn't exist doesn't solve anything. $\endgroup$ – user608672 Oct 26 '18 at 3:05
  • 2
    $\begingroup$ arcsin does not have multiple outputs. 1) we restrict outputs to only values between -90 and 90 degrees 2) for the few multivalue "functions" that seem to allow multiple outputs. Example $\ln w = x + 2k \pi i$ that output is not actually the infinite values of $x + 2k \pi i$ but the single equivalence class of $\{x + 2k \pi i\}$. The sets $[w] = \{a^2 + b^2| a+b = w\}$ is not an equivalence class. the $[w]$ are not distinct. $\endgroup$ – fleablood Oct 26 '18 at 6:36
  • 4
    $\begingroup$ "That's still just a cop-out to get out defending your claim." That is exceedingly rude! And, as it turns out, fairly ignorant. $\endgroup$ – fleablood Oct 26 '18 at 6:40

It cant be a single value function because if $w = a+ b = c+d$ then $f(w) = a^2 + b^2$ which may not equal $c^2 + d^2$ so it is not well defined.

But in you comments you seem to ask about multivalued "functions" with branches such as $\arcsin$ and $\sqrt{}$ and $\ln$ of complex numbers where if $f(x) = y$ it can actually be argued that there are a set of $\{y_\alpha\}$ where the each of the $y_\alpha$ can be argued is legitimate value for $f(x)$.

This "branching", of course, violates the definition of a function. And for the must part this functions are restricted to only specific outcomes. For example: although $\sin \theta = \sin (\pi - \theta) = \sin (\theta + 2\pi)=x$ it would seem $\arcsin x$ could be all of $\theta, \pi - \theta,$ or $\theta + 2\pi$. But this is not what happens because we restrict, by definition, that $\arcsin x$ must be a value between $-\frac \pi 2$ (exclusive) and $\frac \pi 2$ (inclusive).

Likewis $\sqrt{x}$ is defined to always be non-negative.

This gets iffy with complex numbers that if $e^x = w$ then $e^{x + 2\pi i} = w$ and we don't bother to restrict $\ln w$ to a specific one of the infinite possible $x + 2\pi i$ but instead often say $\ln w = $ all and any of the $x + 2\pi i$ values.

This utterly bollocks the idea of "function" where $f(w)$ really has to be one thing.

Well, the work around is that $\ln$ is not a function of $\ln: \mathbb C \to \mathbb C$ but a function of $\ln:\mathbb C \to \mathbb C/mod 2\pi i$. That is that $\ln$ doesn't map to a complex number but to an equivalence class.

We can say for two different complex $w_i \ne w_j$ that $w_i$ is equivalent to $w_j$ if $w_i - w_j = 2k \pi i$. This relationship is an equivalence relations ship (its reflexive, symmetric and transitive; two elements related in this way might as well be considered for our purposes to be the same). And for a given $w$ then the set of all $\{w_i| w_i - w = 2k\pi i$ for some integer $k\}$ is call an equivalence class.

So $\ln w = w + 2k\pi$ is not a single number, but a single equivalence class.

And $\mathbb C = \cup $ all equivalence classes. And any two equivalence classes are distinct.

So If we want to define $f(w) = a^2 + b^2$ when $w = a+b$ in this way as a multivalued "function" we must consider $a^2 + b^2$ as an equivalence class.

And if we do that, if we want say $w $ is equivalent to $v$ if there is a number $k$ so that $w = (k -e)^2 + e^2$ for some value $a$ and $v = (k-h)^2 + h^2$ for some $h$

But that is not an equivalence relation. And the classes form by this are not distinct.

So this can't make any sense.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.