I am wondering about this. A well-known class of functions are the "involutive functions" or "involutions", which have that $f(f(x)) = x$, or, equivalently, $f(x) = f^{-1}(x)$ (with $f$ bijective).  Now, consider the "anti-involution" equation $f(f(x)) = -x$. It is possible for a function $f: \mathbb{C} \rightarrow \mathbb{C}$ to have $f(f(z)) = -z$. Take $f(z) = iz$.  But what about this functional equation $f(f(x)) = -x$ for functions $f: \mathbb{R} \rightarrow \mathbb{R}$, instead of $f: \mathbb{C} \rightarrow \mathbb{C}$? Do such functions exist? If so, can any be described or constructed? What about in the more general case of functions on arbitrary groups $G$ where $f(f(x)) = x^{-1}$ (or $-x$ for abelian groups), $f: G \rightarrow G$? Can we always get such $f$? If not, what conditions must there be on the group $G$ for such $f$ to exist?
25227
I've been perusing the internet looking for interesting problems to solve. I found the following problem and have been going at it for the past 30 minutes with no success:     Find a function $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(f(x)) = -x$ for all $x \in \mathbb{R}$.   I am also wondering, can we find $f$ so that is continuous?  I was thinking of letting $f$ be a periodic function, and adding half the period to x each time. I had no success with this, and am now thinking that such a function does not exist.  Source:
123650
The function $f: G→G$ defined by $f(x) =x^2$ is a homomorphism if and only if $G$ is abelian.  Can anyone give me any tips how to work on this question?
29460
Show there does not exist a function $f:\mathbb{Z}\rightarrow \{1, 2, 3\}$ satisfying $f(x)\ne f(y)$ for all integers $x,y$ and $|x-y|\in\{2, 3, 5\}$.   This is taken from "Putnam and Beyond" and I cannot seem to understand why this cannot be true. I let $x$ be fixed and assigned $f(x)=1$, then assumed $f(y)=2, |x-y|=2.$ Then I proceeded to substitute $y-1$ and $y-3$ but did not get anywhere in finding a contradiction.  I am looking for comments about my approach and an explanation of the solution.
84641
I hesitated to post on StackOverflow but I think the problem has little to do with programming and more to do with mathematics. So, here it is:  I wanted to compute the function $ f(n) = 0 \oplus 1 \oplus 2 \oplus \dotsb \oplus n$ in O(1) instead of O(n) so I computed $f(1), f(2),...$ to see the global look of the function and it appears that it has an easily identifiable pattern, which can be used to transform $f$ like so:  $$  f(n) = \left\{  \begin{array}{ll}   n  &amp; \mbox{if } n \equiv 0 \pmod 4 \\                 1  &amp; \mbox{if } n \equiv 1 \pmod 4 \\                 n+1  &amp; \mbox{if } n \equiv 2 \pmod 4 \\                 0 &amp; \mbox{if } n \equiv 3 \pmod 4 \\  \end{array} \right. $$   Does anyone know why the function has such pattern ?
23678
I'm actually in a course of Abstract Algebra, and the teacher asked some interesting questions at the end of a course on the material we had seen. I found the answer to the first question, but the other two leave me perplexed. Could you give me your opinion on the questions (2) and (3)?  (1) Is the function $f : \mathbb{N} \to \mathbb{N} $ such that $$f= \begin{pmatrix} 1 &amp; 2 &amp; 3 &amp; 4 &amp; ...\\ f(1) &amp; f(2) &amp; f(3) &amp; f(4) &amp; ... \end{pmatrix} = \begin{pmatrix} 1 &amp; 2 &amp; 3 &amp; 4 &amp; ...\\ 2 &amp; 3 &amp; 4 &amp; 5 &amp; ... \end{pmatrix}$$ is a permutation? Answer : no, because $1$ isn't an element in the image.  (2) Is there a function $ f \in S_E$  that is factorisable into disjoint cycle of infinite order such that $\mid \{n \in \mathbb{N} : f(n) \not= n\}\mid &lt; \infty$ ?  Definitions :    $S_E$ is the set of all permutation of E.  Let the set $E \not= \emptyset$. W say that a map $f : E \to E$ has an inverse function, if there is a mapping $g : E \to E$ such that both composition $f \circ g$ and $g \circ f$ are the application identity for each $x$ element of $E$. We say that $f$ is a permutation (bijection) of $E$ .    (3) Does the subset of functions $\in S_{\mathbb{N}}$ ($S_E$ with $E=\mathbb{N}$) consistent with $\mid \{n \in \mathbb{N} : f(n) \not= n\}\mid &lt; \infty$ form a group?
108117
Find all functions $g:\mathbb{R}\to\mathbb{R}$ with $g(x+y)+g(x)g(y)=g(xy)+g(x)+g(y)$ for all $x,y$.  I think the solutions are $0, 2, x$. If $g(x)$ is not identically $2$, then $g(0)=0$. I'm trying to show if $g$ is not constant, then $g(1)=1$. I have $g(x+1)=(2-g(1))g(x)+g(1)$. So if $g(1)=1$, we can show inductively that $g(n)=n$ for integer $n$. Maybe then extend to rationals and reals.
73580
Let $S = \{1,2,3,4\}$. Let $F$ be the sets of all functions from $S$ to $S$.  Now I think this statement is true:  $\forall f \in F , \exists g\in F$ so that $(g \circ f)(1) =2$  I suppose $f \in F$, and I need to show that $\exists g\in F$ so that $(g \circ f)(1) = 2$  Do I just make an example  of $g$ so that $(g \circ f)(1) = 2$ ?  Like $f(1) = y$  and then  $g (y)=2 ?
115039
Let $f$ be entire and non-constant. Assuming $f$ satisfies the functional equation $f(1-z)=1-f(z)$, can one show that the image of $f$ is $\mathbb{C}$?  The values $f$ takes on the unit disc seems to determine $f$...  Any ideas?
55873
Today I was thinking about composition of functions.  It has nice properties, its always associative, there is an identity, and if we restrict to bijective functions then we have an inverse.  But then I thought about commutativity.  My first intuition was that bijective self maps of a space should commute but then I saw some counter-examples. The symmetric group is only abelian if $n \le 2$ so clearly there need to be more restrictions on functions than bijectivity for them to commute.  The only examples I could think of were boring things like multiplying by a constant or maximal tori of groups like $O(n)$ (maybe less boring).  My question: In a euclidean space, what are (edit) some nice characterizations of sets of functions that commute? What about in a more general space?   Bonus: Is this notion of commutativity important anywhere in analysis?
19698
I have found this exercise.  Let $f, g$ be entire functions.  a) Show that there exists entire $f_{1}, g_{1}$ and $h$ such that $f=hf_{1}$, $g = hg_{1}$ and $f_{1}$ and $g_{1}$ have no common zeros.  b) For $f$, $g$, and $h$ as above, show that there exists entire functions $A$ and $B$ such that $Af+Bg = h$.  Any idea ? How to manage the case ord($f$) = $\infty$ and ord($g$) = $\infty$ ?
100758
Find all functions $f(x)$ such for a given fixed $a\in \mathbb{R}$ such that the following functional equation holds: $$f(x)^{2}=f\left(\frac x a\right)$$  I'm not sure  how to solve this equation other then using the method of power series, any tips?
35538
Find all function $f:\mathbb{R}\rightarrow \mathbb{R}$ so that $f(xf(y))+f(yf(x)=2xy$.  By putting $x=y=0$ we get $f(0)=0$ and by putting $x=y=1$ we get $f(f(1))=1$. Let $y=f(1)\Rightarrow f(x)+f(f(x)f(1))=2x$, which tells us that $f$ is an injective function. The only solutions I came up with so far are $f(x)=x$ and $f(x)=-x$.
4795
Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $f\left( f(x)^2+f(y) \right)=xf(x)+y$ for all real numbers $x$ and $y$.   Clearly $f(x)=x$ is a solution, check by substitution.  I'm at a loss as to how to show this is the only one or find others.  Any help would be very much appreciated.
73022
Find all functions $f: \mathbb{N} \rightarrow \mathbb{N} $ satisfying $\frac{1}{f(a)}+ \frac{1}{f(b)} = \frac{1}{f(c)} $ whenever $\frac{1}{a}+ \frac{1}{b} = \frac{1}{c} \ a,b,c \in \mathbb{N} $   After talking through it with a friend, they believe there is a simple solution only a couple of lines long, but I have yet to spot the method. Any ideas?
98869
How can I determine $ f(x)$ if $f(1-f(x))=x$ for all real $x$?  I have already recognized one problem caused from this: it  follows that $ f(f(x))=1-x $, which is discontinuous. So how can I construct a function $f(x)$?  Best regards and thanks, John
37618
Is there an injective function $f :\mathbb{R}^2\rightarrow\mathbb{R}$?  I approached this problem from the perspective of $f^{-1}$, from which I showed there exists a surjective function $f^{-1}:\mathbb{R}\rightarrow\mathbb{R}^2$. Would this imply that there exists an injective function for the inverse mapping?
72788
Let G be a group. The function $f:G \rightarrow G$ defined by $f(x)=x^2$ is a homomorphism iff G is abelian.  I am having trouble with the 2nd part of the proof.  Proof:  Assume the function $f:G \rightarrow G$ is defined by $f(x)=x^2$ then:   Assume $f$ is a homomorphism then $f$ satisfies the property: $$ f(xy)=f(x)f(y) $$   where $x,y \in G$  Notice how: $$f(xy)=(xy)^2=xyxy$$ $$f(x)f(y)=x^2y^2=xxyy $$  Since $f(xy)=f(x)f(y)$ then by substitution:  $$xyxy=xxyy$$  Multiplying the right by $y^{-1}$ and the left by $x^{-1}$ then:  $$yx=xy$$ which means G is abelian   Assume G is abelian, we want to show $f$ is a homomorphism   For the 2nd part of the proof, would I just work backwards from the first part of the proof? Here is an outline:  then for $x,y \in G$  $$yx=xy$$ by definition of abelian groups.  Multiple the right by y and the left by x, we have:   $$xyxy=xxyy$$  which is equal to: $$(xy)^2=(x)^2(y)^2$$  since we are given $f(x)=x^2$ then translating this to the above statement: $$f(xy)=f(x)f(y)$$  which satisfies the property of homomorphism.  Does that look right?