Could anyone explain the example below? Why is $F_4 = $ {$0,1,x,x+1$}? (I was learning that it should be $F_4 = $ {$0,1,2,3$}). And how do we get the two tables?
66047
Question: Why is the table of $GF(4)$ look like the one below? I know it has to do with the fact that 4 is composite Let $GF(4) = \{0,1,B,D\}$ Addition: $$ \begin{array}{c|cccc} +  &amp; 0&amp;   1&amp;   B &amp;  D  \\ \hline     0&amp;   0 &amp;  1 &amp;  B &amp;  D \\  1 &amp;  1 &amp;  0 &amp;  D &amp;  B  \\ B &amp;  B &amp;  D &amp;  0 &amp;  1  \\ D &amp;  D &amp;  B &amp;  1 &amp;  0  \end{array} $$ Multiplication: $$\begin{array}{c|cccc} \cdot &amp;   0 &amp;  1 &amp;  B &amp;  D  \\ \hline 0 &amp;  0 &amp;  0 &amp;  0 &amp;  0  \\ 1 &amp;  0 &amp;  1 &amp;  B &amp;  D  \\ B &amp;  0 &amp;  B &amp;  D &amp;  1  \\ D &amp;  0 &amp;  D &amp;  1 &amp;  B     \end{array}$$
110733
$f$ is an entire function with $\operatorname{Im}f \geq 0$. Then which of the followings are true: 1) $f$ is constant. 2) $\operatorname{Re}f$ is constant. 3) $f = 0$. 4) $f'$ is a non-zero constant.  That (3) &amp; (4) are wrong can be shown by using $f(z) = i$. But I'm clueless about the remaining two options.
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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??
27798
Question: Let $F_n$ the sequence of Fibonacci numbers, given by $F_0 = 0, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 2$. Show for $n, m \in \mathbb{N}$: $$F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$$   My (very limited) attempt so far: after creating a small list of the values $F_0=0, F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, F_6=8, F_7=13, F_8=21, F_9=34, F_{10}=55$ i can see that yes it does seem to work for instance $F_{6+3}=F_5 F_3 +F_6 F_4 = 10 +24 = 34 = F_9$. However, I really don't know where to begin as showing that this must hold in general terms. Should I be looking to use limits? Or perhaps induction? What is the best way to solve this?
61009
I'm going through the book Proofs and fundamentals, by Bloch, and it doesn't include a solution manual for it's examples. It doesn't have many examples on notation and proof strategy for certain cases, so I needed a little help.    Although I understand the idea, and can draw it out, I'm not sure how to write it in decent proof notation. Any help would be appreciated!   Edit: apologies, my idea is this: theres some function in $F(C,A\times B)$ such that theres some $x$ within $C$ that means $g(x)$ is within $A\times B$. Similarly, the second two imply that for some function within $F(A,C)$ there exists a value $x$ such that $g(x)=C$. and similarly for the third part from $B\rightarrow C$. It's easy to see with drawing it out that these two are the same because one will have a part within A and the other will lead to a part within $B$, so their cross will be the same. I wanted to go about proving it by setting a function $f$ within $F(C,AxB)$ and then working from there, but I really have no idea where to start or the notation.
26822
Question     Let the function $f:\mathbb R\to\mathbb R$,and such   $$f(xf(y)+x)=xy+f(x)$$      Find all $f(x)$   Let $x=1,y=1$,then $$f(f(1)+1)=1+f(1)$$ let $f(1)=t$,then $$f(t+1)=1+t$$  So I guess $$f(x)=x$$ is such it, and I found $f(x)=-x$ is also such it?   Can you help?  I found this simaler problem：(BMO 1997, 2000) $$f(xf(x)+f(y))=y+(f(x))^2,f:R\to R$$ this answer $f(x)=x$ or $f(x)=-x$,  the full solution can see
47813
Find all functions such that $$ f(m^2 + f(n)) = f(m)^2 + n$$ for all natural numbers (1,2,3,4,..)  I have been stuck on this problem.   I have tried to set f(1) = k, however, did not make any progress.
97092
In  thread, the question was to find a $f: \mathbb{R} \to \mathbb{R}$ such that  $$f(f(x)) = f(x) + x$$  (which was revealed in the comments to be solved by $f(x) = \varphi x$ where $\varphi$ is the  $\frac{1+\sqrt{5}}{2}$).    Having read about  shortly before though, I came up with this train of thought:  $$f(f(x)) = f(x) + x$$ $$\Leftrightarrow f^2 = f^1 + f^0$$ $$f^2 - f - f^0 = 0$$  where $f^n$ denotes the $n$'th iterate of $f$.  Now I solved the resulting quadratic equation much as I did with plain numbers  $$f = \frac{1}{2} \pm \sqrt{\frac{1}{4} + 1}$$ $$f = \frac{1 \pm \sqrt{1+4}}{2} = \frac{1 \pm \sqrt{5}}{2}\cdot f^0$$  And finally the solution $$f(x) = \frac{1 \pm \sqrt{5}}{2} x .$$  Now my question is: *Is it somehow allowed to work with functions in that way?** I know that in the above, there are denotational ambiguities as $1$ is actually treated as $f^0 = id$ ... But since the result is correct, there seems to be some correct thing in this approach.  So can I actually solve certain functional equations like this? And if true, how would the correct notation of the above be?
55097
A trick question about combinators (Stars and Bars)- Need some explanations.     Let $S=\{1,2,3,4,\dotsc,N\}$ and $X=\{ f: S \rightarrow S\mid x &lt; y  \Rightarrow f(x)\leqslant f(y)\}$. Then $|X|$ is equal to what?   How to use Stars and Bars here? Thank you very much.
2933
Solve the recurrence relation  $$f(n) = 2f(n - 1) + f(n - 2)$$ with initial conditions $f(0) = a, f(1) = b$. (here a and b are fixed, arbitrary integers).  Can anyone show me how to solve this? a and b are fixed, arbitrary integers. Does it mean I can put any value for a and b?
111724
The Problem:   Let $f : \mathbb R \to \mathbb R$ be a real-valued function defined on the set of real numbers that satisfies $$f(x + y) \leq yf(x) + f(f(x))$$ for all real numbers $x$ and $y$.  Prove that $f(x) = 0$ for all $x ≤ 0$.  (IMO $2011$ , Pr: $3$)  Here is the solution:  Before we prove that $f (x) = 0$ for all $x ≤ 0$, we will prove that $f (0) = 0.$ $\color{black}{\large\text{Case} \thinspace  1:}$ $f(0)&gt;0$ Let, $\varepsilon &gt;\dfrac{f(f(0))}{f(0)}$ and $x&lt;- \dfrac{\varepsilon+f(f(0))}{f(0)}$  we have, from $f(x)\leq xf(0)+f(f(0))$,  $f(x)&lt;-\varepsilon.$ Then from, $f(f(x))\leq f(x)f(0)+f(f(0))$ we have $f(f(x))&lt;0$. Finally,  since $0&lt;f(0)\leq f(f(0))$ ,we get $f(0)\leq-xf(x)+f(f(x))&lt;0$ which imply $f(0)&lt;0$, which gives a contradiction. $\color{black}{\large\text{Case} \thinspace  2:}$ $-\alpha \leq f(0)&lt;0$, where $\alpha\in\mathbb{R^+}.$ From $f(x + y) \leq yf(x) + f(f(x))$ we have: $\begin{cases} f(x)\leq f(f(x)) \\ f(x) \leq xf(0)+ f(f(0)) \end{cases} \Longrightarrow f(x)\leq f(x)f(0)+f(f(0)) \Longrightarrow f(x)(1-f(0))\leq f(f(0))$. Then applying $x=f(0)$, we get $f(f(0))\leq 0$, which imply $f(x)\leq 0$, which gives $f(f(x))\leq 0$. In this case, we have $f(x)&lt;0.$ Because, if $f(x)=0$ at least for some $x$, from $f(x)\leq f(f(x))$, we get $f(0)\geq 0$, which gives a contradiction. Then let, $x&lt;-\sqrt {\alpha}$,  from  $f(x + y) \leq yf(x) + f(f(x))$  we have: $$\begin{align} f(z)\leq(z-x)f(x)+f(f(x)) \Longrightarrow f(f(x)) \leq (f(x)-x)f(x)+f(f(x))\Longrightarrow f(x)(f(x)-x)\geq 0 \Longrightarrow f(x) \leq x &lt;-\sqrt {\alpha} \Longrightarrow f(x)&lt; -\sqrt{\alpha}.\end{align}$$ Finally, from $f(0)\leq-xf(x)+f(f(x))$ we get $f(0)&lt;-\alpha$, which gives a contradiction. So, we deduce that $f(0)=0$. Then, applying $f(0)=0$, from  $f(0)\leq-xf(x)+f(f(x))$ and $f(x) \leq xf(0)+ f(f(0))$ we have, $$\begin{cases} xf(x)-f(f(x)) \leq 0 \\ f(x) \leq 0 \\ f(f(x)) \leq 0 \end{cases} \Longrightarrow  \begin{cases} xf(x)\leq0 \\ f(x)\leq 0 \end{cases} \Longrightarrow \begin{cases} xf(x)\leq0, x\in\mathbb {R} \\ xf(x)\geq 0, x\leq 0 \end{cases} \Longrightarrow xf(x)=0, x\leq 0 \Longrightarrow \color{blue} {\boxed{ f(x)=0, \text{for all} \thinspace x\leq 0.}}$$ End of the Proof.  Can you verify this solution? Are there any mistakes, gaps, etc.? Thank you very much!
35445
I have a feeling this question is a duplicate, but it's not coming up in "Questions that may already have your answer."  We all know very well that $F(0) = 0$, $F(1) = 1$ and $F(n) = F(n - 2) + F(n - 1)$ for all $n &gt; 1$.  I'm wondering about $n &lt; 0$. My first thought was $F(-n) = -F(n)$, which is appealing from a multiplicative point of view, as it seems to preserve certain identities, like $F(n)^2 = F(n - 1) F(n + 1) - (-1)^n$.  But it doesn't quite make sense from an additive point of view, it doesn't seem to work both "forwards" and "backwards." For example, it would give us $F(-1) + F(0) = -1 \neq F(1)$.  How do we extend $F(n)$ to negative $n$ so as to maintain both the related identities and the basic defining identity?
91883
I want to find a function $f:[0,1] \to [0,1]$ such that $f$ takes on each value in $[0,1]$ exactly twice. I think this means there are an infinite number of discontinuities. Can anyone help me figure this one out?  Anyone have any pointers?
4291
Given: $f_1 = f_2 = 1$ and for $n \in\mathbb{N}$, $f_{n+2} =f_{n+1} + f_n$. Prove that $f_2 + f_4 + \dots   + f_{2n} = f_{2n+1}- 1$.  Would you start with setting $f_2 + f_4 + \dots   + f_{2n}= a_n$? Then for the base case let $a_1=1$ LHS$=1$ and RHS$=2-1=1$ so base case holds. Then the inductive hypothesis: Assume $f_2 + f_4 + \dots   + f_{2n} = f_{2n+1}- 1$ $\textbf{NTS}$: $f_2 + f_4 + \dots   + f_{2n} +f_{2n+2} = f_{2n+3}- 1$ Inductive step: By inductive hypothesis $f_2 + f_4 + \dots   + f_{2n}=f_{2n+1}- 1$ So $f_{2n+1}- 1+f_{2n+1}$=$f_{2n+2}- 1$. As was to be shown. Is this correct or did I need to show more algebra in my inductive step ?
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Prove: $\binom{n}{0}F_0+\binom{n}{1}F_1+\binom{n}{2}F_2+\cdots+\binom{n}{n}F_n=F_{2n}$;  I was stuck with this question for a while... Help me please!!! Thanks!!!
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Determine all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for all $x, y \in \mathbb{R}$, $$f((x + 1) f(y)) = y (f(x) + 1).$$    Source: Vuong Lam Huy, on a Facebook group.
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"" (number $5$).  I'm looking at this, and i'm not sure what the $\equiv$ means in this instance? I know the rest of the values, but what does $\equiv$ mean right now?  Thanks in advance!