First $\text{f}\left( 1 \right)=1$ beacause  $\text{f}\left( a \right)^{\text{f}\left( 1 \right)}=\text{f}\left( a \cdot 1 \right)$, and $\log_{\text{f}\left( a \right)} \text{f}\left( a \right)^{\text{f}\left( 1 \right)} =\text{f}\left( 1 \right) \log_{\text{f}\left( a \right)} \text{f}\left( a \right) =1$  If $\text{f}\left( 2 \right) = c$ then $\text{f}\left( 2 \right)^{\text{f}\left( 2 \right)}=\text{f}\left( 4 \right)$ so $\text{f}\left( 4 \right)=c^c$ so there should be only one  group of functions as a solution.  So if $\overbrace{x^{\text{...}^{x}}}^{\text{n times}}=2$ then $\overbrace{{\text{f}\left( x \right)}^{\text{...}^{{\text{f}\left( x \right)}}}}^{\text{n times}}={\text{f}\left( 2 \right)}$ then the $n$th super root of $c={\text{f}\left( x \right)}$, $n=\text{slog}_x2$ so ${\text{f}\left( x \right)}$=the $\text{slog}_x2$ root of c.  I am rather incompetent of my answer so if someone could look it over that would be great.  Super roots and slogs
36008
Consider the functional equation problem $$ f : \mathbb R \to \mathbb R \text , $$ $$ f \left( a ^ b \right) = f ( a) ^ { f ( b ) } \text , $$ when $ a , b \in \mathbb R $, $ a , b \ge 0 $. So far the only solution I have is the trivial $$ f ( x ) = x \text . $$ Does there exist any other possible solution? Even if the $ f $ doesn't have a closed form in terms of elementary functions, is there some way I could derive a series or an alternative expression for it? I tried to get a little more creative: $$ f ( w ) = f \left( w ^ { \frac 1 w } \right) ^ { f ( w ) } \text , $$ which nests deep into $$ f ( w ) = f \left( w ^ { \frac 1 w } \right) ^ { f \left( w ^ { \frac 1 w } \right) ^ { f \left( w ^ { \frac 1 w } \right) ^ \vdots } } \text , $$ which is equivalent to $$ \ln \big( f ( w ) \big) = \ln \bigg( f \left( w ^ { \frac 1 w } \right) \bigg) f ( w ) \text . $$ I was hoping to generate some sort of series using this.
8014
I need to prove the  of $\{\text{or},\text{ xor},\text{ xnor}\}$ with the help of $\{\text{not},\text{ or},\text{ and}\}$ (which have been already proven to be functional complete). My attempt is that I only have to show that $\{\text{or},\text{ xor}\}$ is functional complete as $\text{xnor}$ is the negation of $\text{xor}$ while $\text{xor}$ is defined as $(x \wedge ¬y) \vee (¬x \wedge y)$. My attempt is to show that I can use $\{\text{or},\text{ xor}\}$ for $\{\text{not},\text{ or},\text{ and}\}$ but I already fail showing that $¬x$ can be replaced by $\{\text{or},\text{ xor}\}$... $¬x = ¬x+¬x = ¬¬(¬x+¬x)$ at this point I have no clue how to continue any constructive ideas?
76296
Let $k$ be a field and let $K=k(x)$ be the rational function field in one variable over $k$. Let $\DeclareMathOperator{\aut}{Aut}\sigma, \tau \in \aut(K)$ s.t. $$\sigma\left(\frac {f(x)}{ g(x)}\right)=\frac {f(1/x)}{g(1/x)}\qquad \&amp;\qquad \tau\left(\frac {f(x)}{ g(x)}\right)=\frac {f(1-x)}{g(1-x)}.$$ Find the fixed field $F$ of $&lt;\sigma, \tau&gt;$. Determine $\DeclareMathOperator{\gal}{Gal}\gal(k/F)$. Find an $h \in F$ so that $F=k(h)$.   Now for the first automorphism $\sigma$, I think if I concentrate on $k[x]$ it will give me some idea. Now, if $f(x)=a_0+a_1x+a_2x^2+ \cdots + a_nx^n \in k[x]$ then if $x^n \sigma(f(x))=f(x)$ the case will be $a_i=a_{n-i}$ $\forall i=0(1)[n/2]$. So all the funtions in $K$ which will be fixed by $\sigma$ are of the form $\frac {f(x)}{ g(x)}$ where $f(x)=a_0+a_1x+a_2x^2+ \cdots + a_nx^n$ and $g(x)=b_0+b_1x+b_2x^2+ \cdots + b_nx^n$ such that $a_i=a_{n-i}$ $\forall i=0(1)[n/2]$ and $b_i=b_{n-i}$ $\forall i=0(1)[n/2]$.   Please check whether am I right or not?  And any ideas for the rest of the part?
9546
Let's say $f$ is an isomorphism $f:G  \rightarrow G'$, where $G$ and $G'$ are multiplicative groups.  Then for $x\in G$, if $f(x) = x' \in G'$. Do we have always have $\text{ord}(x) = \text{ord}(x')$? Why?
83599
Prove that $\left|\dfrac{z-w}{1-\bar{z}w}\right| = 1$ where $\bar{z}$ is conjugate of $z$ and $\bar{z}w\ne 1$ if either $|z| = 1$ or $|w| = 1$.  I used $|c_1/c_2| = |c_1|/|c_2|$ and multiply out with $z = x + iy$ and $ = a+ib$ but I am getting stuck near finish.
39394
Suppose $cf(x)=f(cx)$ and $f:\mathbb{R}\to\mathbb{R}$. I believe it follows that $f(x+y)=f(x)+f(y)$.   Proof: There is some $c$ such that $y=cx$. Then  $$f(x+y)=f\left((1+c)x\right)=(1+c)f(x)=f(x)+cf(x)=f(x)+f(cx)=f(x)+f(y)$$  QED.  I wonder if the same thing holds for when $f:\mathbb{R}^n\to\mathbb{R}^n$? I can't use the same trick, because all vectors are not scalar multiples.  I tried thinking of it in terms of basis units, but didn't get anywhere.
69949
I need to show log $Fib_{n}$ is $\theta(n)$ by the Fibonacci numbers defined as   $$ F_n=F_{n-1}+F_{n-2}$$  for  $$ n \geq 2 $$ $ F_{0} = 0 $ and $ F_{1} = 1 $  I'm not sure how to approach this.   I can see it grows exponentially as I've shown a basecase for $F_{6}$.  Basecase for $F_{6}$:  $$ F_{2} = F_{1} + F_{0} = 1 + 0 $$ $$ F_{3} = F_{2} + F_{1} = 1 + 1 $$ $$ F_{4} = F_{3} + F_{2} = 2 + 1 $$ $$ F_{5} = F_{4} + F_{3} = 3 + 2 $$ $$ F_{6} = F_{5} + F_{4} = 5 + 3 $$  But how I prove it's true for log $Fib_{n}$ is $\theta(n)$ I don't know. Hope someone can help!
95028
What can be said about$\ f^{-1}(f(S))$ and$\ S$, where$\ f(S):=\{f(x):x \in S\}$?   I feel as if$\ f^{-1}(f(S))$ will be smaller than or equal in size (number of elements) to$\ S$.  For example, if$\ f(x)=x$, then$\ f(\{-1, 0, 1, 2\})=\{-1, 0, 1, 2\}$and$\ f^{-1}(f\{-1, 0, 1, 2\})=\{-1, 0, 1, 2\}$ (same size).  But if$\ f(x)=x^2$, then$\ f(\{-1, 0, 1, 2\})=\{0, 1, 4\}$ and$\ f^{-1}(f\{0, 1, 4\})=\{0, 1, 2\}$, which is a little bit smaller than$\ S$.   What can be said about$\ f(f^{-1}(U))$ and$\ U$, where$\ f^{-1}(U):=\{x \in X:f(x) \in U\}$?   The same size as each other? Always??
9389
This came up in proving non-regularity of a certain language (powers of 2 over the ternary alphabet). Any clue to the above equation could help me move forward.  Edit: Of course, $x = 1, y = 1$ is a solution. I am looking for non-trivial solutions.
23938
So this is a piecewise defined function I have that I need to talk about:     f(a,b) = \left\{      \begin{array}{lr}        \text{open} &amp; : \text{RMSD}_\text{s-open}\ge6, \text{RMSD}_\text{closed}\ge6\\        \text{closed} &amp; : \text{RMSD}_\text{closed}\le2 \\        \text{semiopen} &amp; : \text{RMSD}_\text{s-open}\le2\\        \text{transition} &amp; :f(a,b)\notin\{\text{open}, \text{closed}, \text{semiopen}\} \\       \end{array}    \right.   So what's the way to align up the colons (conditions) so that it looks all nice and pretty?   Also, if I want to describe transition to be the value of the function when none of the conditions of the previous 3 are satisfied, is the way I wrote it out with f(a,b) not belonging to the set {open,closed,semiopen} a good way of writing it? seems really unprofessional...
52061
Does $f(x)=ax^b$ grow faster than $g(x)=\ln x$ for all $a, b &gt; 0$? Can I say that $f(x) &gt; g(x)$ as $x$ approaches infinity?  I thought the answer is yes, but this graph appears to be telling a different story.    Is the polynomial (the green curve) going to cross the log function (the red curve) and exceed in value for some large value of x?  If the answer is yes, does it mean that if I subtract the two functions and set it to zero, the resulting equation will have two roots? What are the roots?
108100
I would like to find all functions $f:\mathbb{R}\backslash\{0,1\}\rightarrow\mathbb{R}$ such that   $$f(x)+f\left( \frac{1}{1-x}\right)=x.$$  I do not know how to solve the problem. Can someone explain how to solve it?   In one of my attempts I did the following, which is confusing to me: By the substitution $y=1-\frac{1}{x}$ one gets  $f(y)+f\left( \frac{1}{1-y}\right)=\frac{1}{1-y}$. So with $x=y$ it follows that  $0=x-\frac{1}{1-x}$. So it would follow that there is no solution. Is that possible or is there a mistake?  Best regards
102322
Question: let $f$ be an entire function on $\mathbb{C}$ and let $g(z)=\overline{f(\bar{z})}$ then which of the following is/are correct?  (a) if $f(z) ∈\mathbb{R}$ for all $z∈\mathbb{R}$ then $f=g$  (b) if $f(z)∈\mathbb{R}$  for all $z∈\{z : Im\text{ $z$ }=0\}∪\{z : Im\text{ $z$ }=a\}$ for some $a&gt;0$ then $f(z+ ia)=f(z-ia)$ for all $z∈\mathbb{C}$  (c) if $f(z)∈\mathbb{R}$  for all $z∈\{z : Im\text{ $z$ }=0\}∪\{z : Im\text{ $z$ }=a\}$ for some $a&gt;0$ then $f(z+ 2ia)=f(z)$ for all $z∈\mathbb{C}$  (d) if $f(z)∈\mathbb{R}$  for all $z∈\{z : Im\text{ $z$ }=0\}∪\{z : Im\text{ $z$ }=a\}$ for some $a&gt;0$ then $f(z+ ia)=f(z)$ for all $z∈\mathbb{C}$  My attempt: clearly, if $z∈\mathbb{R}$ then $\bar{z}=z∈\mathbb{R}$. Hence if $f(z)∈\mathbb{R}$ then $f(\bar{z})=f(z)∈\mathbb{R}$ so that $g(z)=\overline{f(\bar{z})}=f(\bar{z})=f(z)$ for $z$ in $\mathbb{R}$ hence (a) is correct.  But as beginner i am unable to prove or discard (b), (c), (d). I am stuck on this from hours, Please help me..
26175
Let $f:X\to Y$ be a function, and let $\{S_{i}:i\in I\}$ be a family of subsets of $X$. Then, $$f\left(\bigcup_{i \in I}S_i\right) = \bigcup_{i \in I}f(S_i).$$  The case where $f(A\cup B)= f(A)\cup f(B)$ is trivial and I've proved this many times in other classes. However, I believe that the problem I am running into is with notation. That is, I don't understand what the set $I$ is. Will my proof method be just the same?  Also, I would like a little help on one more problem. If $S_{1}$ and $S_{2}$ are subsets of a set $X$, and if $f:X\to Y$ is an injection, then $f(S_{1}\cap S_{2})=f(S_{1})\cap f(S_2)$. Now, I know how to prove $f(S_{1}\cap S_{2})\subseteq f(S_{1})\cap f(S_2)$ if our function is not injective, and I know counterexamples of why it isn't equal our function isn't injective. Unfortunately, I am not sure how to use the fact that our function is injective to prove $f(S_{1})\cap f(S_2) \subseteq f(S_{1}\cap S_{2})$. Any help would be much appreciated. Thank you very much!  Note: These questions are coming from Rotman's Intro to Abstract Algebra Chapter 2.
34259
I think I have found the final answer $\left(4 \left[\frac{2t \cosh(4t) - \sinh(4t)}{t^2}\right]\right)$ but need to verify it.
108570
When a $\overbrace{11\ldots1}^{123 \text{ times}}$ is divided by $271$, what is the remainder?   I don't know how could I proceed because the number is too big to divide and I can't even factor it out. Please someone let me know how to solve these kind of problems, also if there is any shortcut related to these kinda problems please let me know.
90080
$f$ is entire without any zeros then there is an entire function $g$ such that $f=e^g$  What I think is since $f$ do not have any zero for some bounded domain, I can define a branch of logarithm $(\log f)$ on that domain which will gives my desired result $f =e^{\log f}$. I don't know if I am doing it right? If this is right I don't know how do I argue $(\log f)$ is entire. Hint please.