Found problems: 4776
Let $\mathbb{N}_0=\{0,1,2 \cdots \}$. Does there exist a function $f: \mathbb{N}__0 \to \mathbb{N}_0$ such that:
\[ f^{2003}(n)=5n, \forall n \in \mathbb{N}_0 \]
where we define: $f^1(n)=f(n)$ and $f^{k+1}(n)=f(f^k(n))$, $\forall k \in \mathbb{N}_0$?
Does there exist a strictly increasing function $f: \mathbb{R} \rightarrow \mathbb{R}$ that takes on only irrational values?
Let $ y_1(x)$ be an arbitrary, continuous, positive function on $ [0,A]$, where $ A$ is an arbitrary positive number. Let \[ y_{n+1}=2 \int_0^x \sqrt{y_n(t)}dt \;(n=1,2,...)\ .\] Prove that the functions $ y_n(x)$ converge to the function $ y=x^2$ uniformly on $ [0,A]$.
Evaluate $ \frac{1}{\displaystyle \int _0^{\frac{\pi}{2}} \cos ^{2006}x \cdot \sin 2008 x\ dx}$
Let $ a, b, c, d \in \mathbb {R}, $ $ b \neq0. $
Find the functions of the $ f: \mathbb{R} \to \mathbb{R} $ such that $ f (x + d \cdot f (y)) = ax + by + c, $ for all $ x, y \in \mathbb{R}. $
Define $x\star y$ to be $x^y+y^x$.Compute $2\star (2\star 2)$.
Each one of $2001$ children chooses a positive integer and writes down his number and names of some of other $2000$ children to his notebook. Let $A_c$ be the sum of the numbers chosen by the children who appeared in the notebook of the child $c$. Let $B_c$ be the sum of the numbers chosen by the children who wrote the name of the child $c$ into their notebooks. The number $N_c = A_c - B_c$ is assigned to the child $c$. Determine whether all of the numbers assigned to the children could be positive.
For all integers $n\geq 2$, let $f(n)$ denote the largest positive integer $m$ such that $\sqrt[m]{n}$ is an integer. Evaluate \[f(2)+f(3)+\cdots+f(100).\]
Let $ r$ be a real number such that the sequence $ (a_{n})_{n\geq 1}$ of positive real numbers satisfies the equation $ a_{1} \plus{} a_{2} \plus{} \cdots \plus{} a_{m \plus{} 1} \leq r a_{m}$ for each positive integer $ m$. Prove that $ r \geq 4$.
Find all $a\in\mathbb R$ for which there exists a function $f\colon\mathbb R\rightarrow\mathbb R$, such that
(i) $f(f(x))=f(x)+x$, for all $x\in\mathbb R$,
(ii) $f(f(x)-x)=f(x)+ax$, for all $x\in\mathbb R$.
Does there exist a function $f\colon \mathbb{R}\to\mathbb{R}$ such that for any $x>y,$ it satisfies $f(x)-f(y)>\sqrt{x-y}.$
Let $G$ be a non-empty set of non-constant functions $f$ such that $f(x)=ax + b$ (where $a$ and $b$ are two reals) and satisfying the following conditions:
1) if $f \in G$ and $g \in G$ then $gof \in G$,
2) if $f \in G$ then $f^ {-1} \in G$,
3) for all $f \in G$ there exists $x_f \in \mathbb{R}$ such that $f(x_f)=x_f$.
Prove that there is a real $k$ such that for all $f \in G$ we have $f(k)=k$
In the tribe of Zimmer, being able to hike long distances and knowing the roads through the forest are both extremely important, so a boy who reaches the age of manhood is not designated as a man by the tribe until he completes an interesting rite of passage. The man must go on a sequence of hikes. The first hike is a $5$ kilometer hike down the main road. The second hike is a $5\frac{1}{4}$ kilometer hike down a secondary road. Each hike goes down a different road and is a quarter kilometer longer than the previous hike. The rite of passage is completed at the end of the hike where the cumulative distance walked by the man on all his hikes exceeds $1000$ kilometers. So in the tribe of Zimmer, how many roads must a man walk down, before you call him a man?
Find all functions $f,g: \mathbb{Q}\to \mathbb{Q}$ such that the following two conditions hold:
$$f(g(x)-g(y))=f(g(x))-y \ \ (1)$$
$$g(f(x)-f(y))=g(f(x))-y\ \ (2)$$
for all $x,y \in \mathbb{Q}$.
Consider a function $f: \mathbb{Z} \rightarrow \mathbb{Z}$. We call an integer $a$ [i]spanning[/i] if for all integers $b \neq a$, there exists a positive integer $k$ with $f^k(a)=b$. Find, with proof, the maximum possible number of [i]spanning[/i] numbers of $f$.
Note: $\mathbb{Z}$ represents the set of all integers, so $f$ is a function from the set of integers to itself. $f^k(a)$ is defined as $f$ applied $k$ times to $a$.
Define two functions $g(x),\ f(x)\ (x\geq 0)$ by $g(x)=\int_0^x e^{-t^2}dt,\ f(x)=\int_0^1 \frac{e^{-(1+s^2)x}}{1+s^2}ds.$
Now we know that $f'(x)=-\int_0^1 e^{-(1+s^2)x}ds.$
(1) Find $f(0).$
(2) Show that $f(x)\leq \frac{\pi}{4}e^{-x}\ (x\geq 0).$
(3) Let $h(x)=\{g(\sqrt{x})\}^2$. Show that $f'(x)=-h'(x).$
(4) Find $\lim_{x\rightarrow +\infty} g(x)$
Please solve the problem without using Double Integral or Jacobian for those Japanese High School Students who don't study them.
For a positive integer $n$ consider any partition of the set $\{ 1,2,\ldots ,2n \}$ into $n$ two-element subsets $P_1,P_2\ldots,P_n$. In each subset $P_i$, let $p_i$ be the product of the two numbers in $P_i$. Prove that
\[\frac{1}{p_1}+\frac{1}{p_2}+\ldots + \frac{1}{p_n}<1 \]
$d(n)$ shows the number of positive integer divisors of positive integer $n$. For which positive integers $n$ one cannot find a positive integer $k$ such that $\underbrace{d(\dots d(d}_{k\ \text{times}} (n) \dots )$ is a perfect square.
For a subset $ S$ of vertices of graph $ G$, let $ \Lambda(S)$ be the subset of all edges of $ G$ such that at least one of their ends is in $ S$. Suppose that $ G$ is a graph with $ m$ edges. Let $ d^*: V(G)\longrightarrow\mathbb N\cup\{0\}$ be a function such that
a) $ \sum_{u}d^*(u)\equal{}m$.
b) For each subset $ S$ of $ V(G)$: \[ \sum_{u\in S}d^*(u)\leq|\Lambda(S)|\]
Prove that we can give directions to edges of $ G$ such that for each edge $ e$, $ d^\plus{}(e)\equal{}d^*(e)$.
Consider a directed graph $G$ with $n$ vertices, where $1$-cycles and $2$-cycles are permitted. For any set $S$ of vertices, let $N^{+}(S)$ denote the out-neighborhood of $S$ (i.e. set of successors of $S$), and define $(N^{+})^k(S)=N^{+}((N^{+})^{k-1}(S))$ for $k\ge2$.
For fixed $n$, let $f(n)$ denote the maximum possible number of distinct sets of vertices in $\{(N^{+})^k(X)\}_{k=1}^{\infty}$, where $X$ is some subset of $V(G)$. Show that there exists $n>2012$ such that $f(n)<1.0001^n$.
[i]Linus Hamilton.[/i]
For how many positive integers $x$ is $\log_{10}{(x-40)} + \log_{10}{(60-x)} < 2$?
${ \textbf{(A)}\ 10\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}}\ 20\qquad\textbf{(E)}\ \text{infinitely many} $
For any integer $k\ge1$, let $p(k)$ be the smallest prime which does not divide $k$. Define the integer function $X(k)$ to be the product of all primes less than $p(k)$ if $p(k)>2$, and $X(k)=1$ if $p(k)=2$. Let $\{x_n\}$ be the sequence defined by $x_0=1$, and $x_{n+1}X(x_n)=x_np(x_n)$ for $n\ge0$. Find the smallest positive integer, $t$ such that $x_t=2090$.
A function $f(S)$ assigns to each nine-element subset of $S$ of the set $\{1,2,\ldots, 20\}$ a whole number from $1$ to $20$. Prove that regardless of how the function $f$ is chosen, there will be a ten-element subset $T\subset\{1,2,\ldots, 20\}$ such that $f(T - \{k\})\neq k$ for all $k\in T$.
The positive integer-valued function $f(n)$ satisfies $f(f(n)) = 4n$ and $f(n + 1) > f(n) > 0$ for all positive integers $n$. Compute the number of possible 16-tuples $(f(1), f(2), f(3), \dots, f(16))$.
[i]Proposed by Lewis Chen[/i]
Determine all functions $f:(0,\infty)\to\mathbb{R}$ satisfying $$\left(x+\frac{1}{x}\right)f(y)=f(xy)+f\left(\frac{y}{x}\right)$$ for all $x,y>0$.