geometry
3D geometry
Pythagorean Theorem

Points P and Q are diagonally opposite vertices of a cube whose side has length s. Let A be the shortest path from P to Q. Let B be the shortest path an ant can take in crawling from P to Q on the surface of a cube. Find B-A.

[hide]You can get A by using the Pythagorean Theorem twice.

As for B, please see the diagram. The bug is crawling from E to B. "Unfold" the cube, like in the diagram, and you should realize that the shortest distance the ant could crawl is a straight line on that unfolded cube. Use the Pythagorean Theorem to find that distance.

I hope that helps.[/hide]


1)

$ n \in \mathbb{N*}$, find :

$ \lim_{x\rightarrow \frac {\pi}{2n}}\frac {\sqrt {\frac {\sin(2nx)}{1 \plus{} \cos(nx)}}}{4(nx)^2 \minus{} \pi^2}$

2)
find :

$ \lim_{x\rightarrow 0}\frac {\sin(\pi\sqrt {x \plus{} 1} \: )}{x}$

You will get better results posting calculus questions in the Calculus Forum.

[hide="2"]er, so I multiplied top and bottom by $ \pi x\plus{}1$ so I could get use $ \lim_{x\to 0}\frac{sinx}{x}\equal{}1$
Then, we have $ \frac{\pi(x\plus{}1)}{x}$
However, $ \lim_{x\to 0}\frac{x\plus{}1}{x}$ does not exist.

Therefore, the limit does not exist.

[/hide]

Still new to calc
Definitely new to competition calc  :|

P.S.:  Yeah, somebody move this to calculus...

[quote="zapi2007"][hide="2"]er, so I multiplied top and bottom by $ \pi x \plus{} 1$ so I could get use $ \lim_{x\to 0}\frac {sinx}{x} \equal{} 1$
Then, we have $ \frac {\pi(x \plus{} 1)}{x}$
However, $ \lim_{x\to 0}\frac {x \plus{} 1}{x}$ does not exist.

Therefore, the limit does not exist.

[/hide]

Still new to calc
Definitely new to competition calc  :|

P.S.:  Yeah, somebody move this to calculus...[/quote]
I'm new to calc too, but...
[hide]You are attempting to use $ \frac{\sin(\pi\sqrt{x\plus{}1})}{\pi\sqrt{x\plus{}1}}$, but this doesn't work since $ \pi\sqrt{x\plus{}1}$ does not approach $ 0$ as $ x$ approaches $ 0$.[/hide]

This problem has been posted (double posted) in the calculus section.


If 5y + 2x = 23 and x = y+ 1 what is the value of y?

Hint:
2(y+1)

*makes the message a little bit longer.*

solution[hide]
several ways to do this like substituting y+1 to x
making it $5y+2(y+1) = 23$
so $y$ is $3$

[/hide]


Three generations of the Wen family are going to the movies, two from each generation. The two members of the youngest generation receive a $ 50\%$ discount as children. The two members of the oldest generation receive a $ 25\%$ discount as senior citizens. The two members of the middle generation receive no discount. Grandfather Wen, whose senior ticket costs $ \$6.00$, is paying for everyone. How many dollars must he pay?

$ \textbf{(A)}\ 34 \qquad
\textbf{(B)}\ 36 \qquad
\textbf{(C)}\ 42 \qquad
\textbf{(D)}\ 46 \qquad
\textbf{(E)}\ 48$

[hide="Click for solution"]
A senior ticket costs $ \$6.00$, so a regular ticket costs $ \$8.00$.  The children's tickets cost $ \$4.00$, so we have:

\[ 2(6\plus{}8\plus{}4)\equal{}2(18)\equal{}36
\]

So Grandfather Wen pays $ 36$ dollars, or $ \boxed{\textbf{(B)}}$.
[/hide]

[hide]One could also say that since a $ 75\%$ ticket costs $ \$ 6.00$ a full price ticket costs $ \$ 8.00$. Additionally the family is paying for 2 of each $ .5$, $ .75$, and $ 1.0$ priced tickets so $ 8(2)(.5\plus{}.75\plus{}1) \equal{} 36$[/hide]

Here's a quick/clever solution.

[hide]Since the senior price is the mean of the middle/youngest prices, we just multiply grandfather wen's price by 6.

6 * 6 = 36. [/hide]


modular arithmetic
number theory unsolved
number theory

Solve the equation $ 7^{x}\minus{}3^{y}\equal{}4$ in positive integers.

$ 7^{x}\equal{} 3^{y}\plus{}4$.

We consider two cases.

[u]Case 1[/u].$ x \equal{} 1$, then $ y \equal{} 1$.

[u]Case 2[/u].$ x\ge 2$.

-$ y \equal{} 1$, so $ 7^{x}\equal{} 7\Leftrightarrow x \equal{} 1$ contracdiction because $ x\ge 2$
-$ y\ge 2$

We have $ 3^{y}\equiv 3 (mod 7)$.

And easily to show that : $ y \equal{} 6k\plus{}1$, with $ k\in\mathbb{N^{*}}$
Then :

$ 7.(7^{x\minus{}1}\minus{}1) \equal{} 3.(3^{6k}\minus{}1)\equiv 0(mod 12)$.

So , $ x \equal{} 12h\plus{}1$,with $ h\in\mathbb{N^{*}}$.

And now , we have the contracdiction because, $ 7.(7^{12h}\minus{}1)\vdots 9$ but $ 3.(3^{6k}\minus{}1\not\vdots 9$.
So $ (x,y)\equal{}(1,1)$is only pair of root of this euqation . :!:
The original problem is :
Find all trios $ (x,y,z)$ of integers show that :
$ 3^{x}\plus{}4^{y}\equal{} 7^{z}$ :lol:
And the same problem
Find all trios $ (x,y,z)$ of integers show that :$ 28^{x}\equal{} 19^{y}\plus{}87^{z}$ :P

If $ 7^{x\minus{}1}\minus{}1\equiv 0\mod 12$ then more exactly $ x \equal{} 2h\plus{}1$. And we have contradiction only for $ h \equal{} 6\cdot n$.
It is easy to show that necessary $ x\equiv y\equiv 1\mod 2$ and $ \gcd(x,y) \equal{} 1$.

[hide]
We show no solution for $ y\geq 2$. Suppose there were. First, $ 7^{x}\equiv 4\pmod{9}$ requires $ x\equiv 2\pmod{3}$ (since $ 7^{2}\equiv 4\pmod{9}$ and the order of 7 modulo 9 is 3).

Now look modulo 19. Note that $ 7^{3}\equiv 1\pmod{19}$ so that
\[ 3^{y}\equiv 7\pmod{19}\]
We get $ 3^{6}\equiv 7\pmod{19}$ and the order of 3 modulo 19 is 18 (it divides 18 but not 6, and is not equal to 9), so $ y\equiv 6\pmod{18}$. In particular, $ y$ is divisible by 3.

The next mod to look at is 13, since $ 3^{3}\equiv 1\pmod{13}$. We get $ 7^{x}\equiv 5\pmod{13}$. Letting $ x\equal{}3j\plus{}2$ (we know $ x\equiv 2\pmod{3}$), since $ 7^{3}\equiv 5\pmod{13}$, this becomes
\[ 5\equiv 7^{2}\cdot 5^{j}\equiv 10\cdot 5^{j}\pmod{13}\]
\[ 5^{\minus{}j}\equiv 2\pmod{13}\]
However, the powers of 5 modulo 13 are $ \pm 1,\pm 5$. This is a contradiction!
[/hide]


modular arithmetic
greatest common divisor

[quote="ProtestanT"][quote="aufha"]Consider four integers $ {a},{b},{c},{d}$ and a positive integer $ {m}$ such that $ a\equiv b\pmod {m}$.

Prove that $ \frac {a}{e}\equiv \frac {b}{e}\pmod {\frac {m}{\gcd(m,e)}}$, where e is a positive integer that divides $ {a}$ and $ b$.

Could someone explain the proof for this to me and/or direct me to a previous topic that had the same question?[/quote]
what is the point of the integers $ c$ and $ d$?[/quote]
Yeah, c and d aren't necessary.
[hide="Solution"]
Call $ m\equal{}zk$, $ a\equal{}xpk$, and $ b\equal{}ypk$ where $ e\equal{}pk$ and k is the greatest common divisor between m and e.  We have that $ a\minus{}b$ is a multiple of $ m$, so $ (xp\minus{}yp)k$ is a multiple of $ zk$ and thus $ p(x\minus{}y)$ is a multiple of $ z$.  Notice that p is not a factor or multiple of z and thus we have that $ x\minus{}y$ is a multiple of $ z$.  So, $ x\equiv y\bmod z$, thus $ \frac{a}{e}\equiv \frac{b}{e}\bmod {\frac{m}{\gcd(m,e)}}$. [/hide]


abstract algebra
Ring Theory
superior algebra
superior algebra solved

f : P -> P is epimorphism and P is noether ring
Prove that f is isomorphic

you mean: $ f : P \to P$ is a surjective ring homomorphism and $ P$ is a noetherian ring. prove that $ f$ is an isomorphism.

this is straight forward, you have to consider the chain $ ker(f) \subseteq ker(f^2) \subseteq ...$.

For a small generalization, you can also check this out:
[url]http://www.mathlinks.ro/viewtopic.php?p=977545#977545[/url]


hello,
I would like  to solve the following recurrence:
 \[ E_n=\left\lbrace \begin{array}{l} E_0 = 1\\ E_1 = \frac{5}{3}\\ (2n+3)E_{n+1} -(2n+3)E_{n} -2 E_{n-1}=2, \forall n \ge 2 \end{array} \right.  \]

(i.e. To found the explicite formula.)

Thank you

Please look through the forums to determine an appropriate level of difficulty before posting.  Some of them have explicit guidelines.  This is not at all appropriate for the middle school forums.


[quote="anirudh"]I am creating a Mock AMC 8. It has 25 problems, and you are on your honor to do it in 40 minutes.

Please Private Message the username: [b]Mock AMC 8 I Director[/b]

DO NOT PRIVATE MESSAGE THEM TO ME. Please follow these guidelines and it will be easier. These will be graded on Christmas Eve, and the results will be posted on Christmas Day. All Other Guidelines are on the paper. The Paper will be out (hopefully) by the next 2 days.

If you want to sign up, please sign up in this thread.[/quote]
I'm in unless this is like your mock mathcounts...

[quote="anirudh"]No. No answers at the back.[/quote]
It wasn't the answers at the back I was referring to, it was the whole "these questions are copied and pasted from a well known source with a couple numbers changed here and there so I'll call it a mock" that I thought made it kind of useless...

I don't see the point of this as it's 11 months until the next AMC 8.

Exactly. Its not exactly AMC 8 level. Its more like AMC 9 level.

[quote="now a ranger"]Wouldn't just making a mock amc 10 be way easier and encourage more people to join, since the AMC 10 is less than 2 months away?[/quote]

I concur  :)

I don't feel like competing, but I'lll edit/check your answers for you and stuff.

i'll sign up

you should have someone check for difficulty!

-jorian

yes. Actually, its going to, in difficulty, lean toward an amc 10.

so why dont you just make it an amc 10 instead?

[quote="anirudh"]Okay. Its a Mock AMC 10 then.[/quote]

HIP HIP HOORAY!!!!

I'm taking AMC 10 in february   :D

Ok, I have hardly any idea what AMC10's are, and I will never take one, but I'm 10th grade age and I don't have much to do over the next week, so I'll do it.

I will try, but no garuntee I will remeber,  :blush:

~lyra

joining...

got a perfect score when i was in 8th grade.

I'll do this, even if it's like an AMC 10.
Will the answers be graded like an AMC 8 or an AMC 10?
I got a perfect score on the AMC 8 :thumbup: and I want to do the AMC 10.

i think 1337 volunteered

pm him

-jorian

This is Officially Over. Please lock this thread. The other thread is what is now real (The AMC 10 one even though its sort of like AMC 12 level because I didn't know that trig and precalc were not allowed on amc 10)


induction
real analysis
real analysis unsolved

[obvious but sweet, ;)]

$x_n = \frac{1}{ 1 + \frac{1}{2 + ... \frac{}{+ \frac{1}{n}}}}$ is convergent.

[quote="gnosis"]It can be shown by induction that $x_2<x_4<...<x_{2k}<x_{2k+2}<...<x_{2k+1}<x_{2k-1}<...<x_3<x_1$ from which the conclusion follows.[/quote]

please be more specific.

We must use induction to show that $x_{2k}<x_{2k+2}$, $x_{2k+1}<x_{2k-1}$, $x_{2k}<x_{2k+1+2s}$, $x_{2k+2s}<x_{2k-1}$ by starting from the further-most denominator.


algebra
polynomial
function
algebra unsolved

Find all polynomials $p$ with real coefficients, for which
$p(0)=0$ and
$f(f(n))+n = 4f(n)$ for every natural number n, where
$f(n)=[p(n)]$

Due
 $f(f(n))+n = 4f(n),\forall n\in N \Rightarrow f(f(x))+x= 4f(x),\forall x\in R\Rightarrow$
$deg[f(f(x)+x]=degf(x)$

 so
$degf(x)=1, f(0)=0\Rightarrow f(x)=ax.$ But

$a^{2}n+n=4an\Rightarrow a=2\pm\sqrt{3}$ so $f(x)=2\pm\sqrt{3}x$

Well, first of all, you should find all p, not all f, and btw, f isn't a polynomial function, therefore you can't determine degree, and also, f(x) is an integer for all x.


abstract algebra
superior algebra
superior algebra solved

No, $x \mathbb Q[x]+\mathbb Z$ is not noetherian, since $(2^{-n}x)_{n \in \mathbb N}$ is an infinite increasing chain of ideals.

LiMa


We consider the series (xn)n>=0, x0,x1 from R and

x(n+2)=(x(n)*x(n+1))/(4*x(n)-2*x(n+1)), n from N
Prove that the series contain a finite number of integers.

NOTE:
x(n) is read x index n.

I think this works. Assume that eventually we will reach two consecutive rational numbers, or else we will never get infinitely many integers. Therefore, let these two consecutive rational numbers be a/b and c/d, in that order. Then the next term is ac/(2(2ad-bc)), so each time, the denominator gets a new power of 2, but the numerator is the product of the numerators of the previous 2, so it doesn't gain a new power of 2. Therefore, eventually the denominator will contain more powers of 2 than the numerator, and then there can be no more integers.


calculus
integration
real analysis
real analysis unsolved

$y''-5y'+6y'=x{}^2e{}^2{}^x$

I tried but i did not end at an answer . I will show below the method i used and the results i got.

Let LHS = 0 and get the $CF= Y=Ae{}^2{}^x+Be{}^3{}^x$

To find the PI let $y=v(x)e{}^2{}^x$,$y'=(v'+2v)e{}^2{}^x$ and $y''=(v''+4v'+4v)2{}^2{}^x$

Substitute these results into the first  equation and we end up with $v''-v+20v$

This is another second order equation which can be solved by the same method

However CF contains complex values , which we havent done.

You are wrong in the calculation of $y'$. :(

I tried another method as this one gets complicated giving CF as a complex number which we have not done .

I have tried another method with $PI=Cx{}^2e{}^2{}^x$ . Here we get $y'=2Cxe{}^2{}^x+2Cx{}^2e{}^2{}^x$ and $y''=2Ce{}^2{}^x+4Cxe{}^2{}^x+4Cxe{}^2{}^x+4Cx{}^2e{}^2{}^x$ .
Substituting back into equation one i get
$2C-2Cx=x^2$

Here i get constant is 0 and x=0 . So is it possible to get an integral with just the CF?? And if so is it correct or where did i go wrong

Thanks in advance

snake_1987, please correct your differential equation.

For constant number $a$, Let $D=\frac{d}{dx},$ we can find the particular solution of the given differential equation as follows.

$y''-5y'+6y=x^2e^{2x}\Longleftrightarrow (D^2-5D+6)y=x^2e^{2x}$

$\Longleftrightarrow (D-2)(D-3)y=x^2e^{2x},$ thus we have

$y=\frac{1}{(D-2)(D-3)}x^2e^{2x}=\frac{1}{D-2}\cdot \frac{1}{D-3}x^2e^{2x}$

$=\frac{1}{D-2}e^{3x}\int e^{-3x}\cdot x^2e^{2x}dx$

$=\frac{1}{D-2}e^{3x}\int x^2e^{-x}dx$

$=-\frac{1}{D-2}e^{3x}(x^2+2x+2)e^{-x}$

$=-\frac{1}{D-2}(x^2+2x+2)e^{2x}$

$=-e^{2x}\cdot \frac{1}{(D+2)-2}e^{-2x}\cdot (x^2+2x+2)e^{2x}$

$=-e^{2x}\frac{1}{D}(x^2+2x+2)$

$=-e^{2x}\int (x^2+2x+2)dx$

$=-\left(\frac{1}{3}x^3+x^2+2x\right).$

Therefore the desired general solution is $y=C_1e^{2x}+C_2e^{3x}-\left(\frac{1}{3}x^3+x^2+2x\right)e^{2x}.$

kunny

Hi kunny thanks a lot for trying to help though i do not get wat u mean by correct ur differential equation

The equation is in my first post ?

[quote="snake_1987"]$y''-5y'+6y'=x{}^2e{}^2{}^x$ [/quote]

incorrect $6y'\Longleftrightarrow$ correct $6y$

Take a look, #4. :)

snake_1987, sorry for that my answer is wrong, for confusing you.
I will correct this. :blush:

kunny

At last I could solve it!

The general solution is $y=C_1e^{2x}+C_2e^{3x}-\left(\frac{1}{3}x^3+x^2+2x\right)e^{2x}.$ :lol:

Maybe Kent Merryfield will solve it by using Laplace Transformation. ;)

I have just edited #4.

kunny


Is it true that every positive perfect square is a divisor of infinitely many Fibonacci numbers?

Isn't true that every positive integer is a divisor of infinitely many terms of the Fibonacci sequence :?:

[b]The lemma:[/b]
$n, d \in \mathbb{Z^+}$
let $k(d)=min\{n: d|F_n\}$
$d|F_n\iff k(d)|n$

By the lemma we only need to prove, that for any integer $d$ there exist $n$ such that $d|F_n$. If we manage to prove this, then it's obvious, that $d$ divides infinitely many terms of Fibonacci sequence.

This is a particular case (proving that exists at least one), but i is easy to extend it to the original problem...

Take the pairs $(F_i,F_{i+1})$ modulo $n$. There exists two of them, say $(F_i,F_{i+1}),(F_j,F_{j+1})$, which leaves the same remainder modulo $n$: \[ F_i \equiv F_j (mod n) \] \[ F_{i+1} \equiv F_{j+1} \]

But then the pair $(F_{i-1}, F_{j-1})$ also works! In fact $F_{i-1} \equiv F_{i+1}-F_i \equiv F{j+1}-F_j\equiv F_{j-1}$. By induction it is easy to conclude that $(F_0,F_{j-i})$ is a good pair.  And the problem follows from it.


function USA(J)MO USAMO induction algebra

Call a real-valued function $ f$ [i]very convex[/i] if
\[ \frac {f(x) \plus{} f(y)}{2} \ge f\left(\frac {x \plus{} y}{2}\right) \plus{} |x \minus{} y|
\]
holds for all real numbers $ x$ and $ y$. Prove that no very convex function exists.

The source of this problem is USAMO 2000 #1

USAMO2000#1 has much more meaning than unknown . :)

hmm, it rather too easy for USAMO , is it round 1 ?

harazi posted a generalization in the "Real Analysis" section, with $|x-y|$ replaced by $|g(x)-g(y)|$, the conclusion being that $g$ must be constant. Try searching, there's a solution there.

We prove by induction that if a very convex function $f$ exists, then \[\displaystyle \frac{f(x)+f(y)}{2} - f\biggl( \frac{x+y}{2} \biggr) \ge 2^n|x-y|\] for all non negative integers $n$. We are given the base case $n=0$; now, if the inequality holds for a given $n$, then for $a,b$ real, \[\displaystyle \frac{f(a)+f(a+2b)}{2} \ge f(a+b)+2^{n+1}|b|,\] \[f(a+b)+f(a+3b) \ge 2(f(a+2b)+2^{n+1}|b|),\] and \[\displaystyle \frac{f(a+2b)+f(a+4b)}{2} \ge f(a+2b)+2^{n+1}|b|,\]  Summing the three enqualities and setting $x=a, y=a+4b$ completes the induction. Now we only have to notice that the inequality we have proved cannot hold for all arbitrarily large n.

Yup,gorbber,here is your solution:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=g%28x%29-g%28y%29&t=43440[/url]

[hide="Does the following work?"]
Assume that such a function exists.  Let $ \epsilon$ be the number which is "right next to $ 0$ on the number line."  Now, let $ f(m \minus{} \epsilon) \equal{} a \minus{} b\epsilon$, $ f(m) \equal{} a$, and $ f(m \plus{} \epsilon) \equal{} a \plus{} c\epsilon$.  It follows that
\[ \frac {f(m \minus{} \epsilon) \plus{} f(m \plus{} \epsilon)}{2}\ge f(\frac {m \plus{} \epsilon \plus{} m \minus{} \epsilon}{2}) \plus{} |m \plus{} \epsilon \minus{} (m \minus{} \epsilon)| \equal{} f(m) \plus{} 2\epsilon
\]\[ \frac {a \plus{} c\epsilon \plus{} a \minus{} b\epsilon}{2}\ge a \plus{} 2\epsilon\implies \epsilon(c \minus{} b)\ge 4\epsilon\implies c \minus{} b\ge 4
\]
Hence, the slope of the line between $ (m \minus{} \epsilon, f(m \minus{} \epsilon))$ and $ (m, f(m))$ is at least $ 4$ less than the slope between $ (m, f(m))$ and $ (m \plus{} \epsilon, f(m \plus{} \epsilon))$.  Similarly, the slope between $ (m, f(m))$ and $ (m \plus{} \epsilon, f(m \plus{} \epsilon))$ is at least $ 4$ less than the slope between $ (m \plus{} \epsilon, f(m \plus{} \epsilon))$ and $ (m \plus{} 2\epsilon, f(m \plus{} 2\epsilon)$.  This implies that the slope between $ (m \minus{} \epsilon, f(m \minus{} \epsilon))$ and $ (m, f(m))$ is at least $ 8$ less than the slope between $ (m \plus{} \epsilon, f(m \plus{} \epsilon))$ and $ (m \plus{} 2\epsilon, f(m \plus{} 2\epsilon))$.  Continuing in the same manner, we see that the slope between $ (m \minus{} \epsilon, f(m \minus{} \epsilon))$ and $ (m, f(m))$ is at least $ \frac {4}{\epsilon}$ less than the slope between $ (m \plus{} 1, f(m \plus{} 1))$ and $ (m \plus{} 1 \minus{} \epsilon, f(m \plus{} 1 \minus{} \epsilon))$.  Yet, $ \lim_{\epsilon\to 0} \frac {4}{\epsilon} \equal{} \infty$, so the slope between $ (m \plus{} 1, f(m \plus{} 1))$ and $ (m \plus{} 1 \minus{} \epsilon, f(m \plus{} 1 \minus{} \epsilon))$ is infinity, which implies that $ f$ will become a vertical line at $ x \plus{} 1$.  Yet, this means that $ f(x \plus{} 1)$ has infinitely many values, namely those on the vertical line.  Hence, $ f(x \plus{} 1)$ has more than one value and $ f$ is not a function, which is a contradiction.  This means that no very convex function exists. [/hide]

[hide="Solution"]

Note that $ f(x) \ge |2x| \minus{} f(0) \plus{} 2f(\frac {x}{2})$

We prove by induction that
\[ f(x) \ge |2nx| \minus{} (2^n \minus{} 1)f(0) \plus{} 2^n f(\frac {x}{2^n})
\]
The base case $ n \equal{} 1$ is above. Assume it's true for $ n$. Then we have, applying the first inequality to $ \frac {x}{2^n}$
\[ f(\frac {x}{2^n}) \ge |\frac {x}{2^{n \minus{} 1}}| \minus{} f(0) \plus{} 2f(\frac {x}{2^{n \plus{} 1}})
\]
or
\[ 2^n f(\frac {x}{2^n}) \ge |2x| \minus{} 2^n f(0) \plus{} 2^{n \plus{} 1} f(\frac {x}{2^{n \plus{} 1}})
\]
Thus
\[ f(x) \ge |2nx| \minus{} (2^n \minus{} 1)f(0) \plus{} 2^n f(\frac {x}{2^n}) \ge |2nx| \minus{} (2^n \minus{} 1)f(0) \plus{} |2x| \minus{} 2^n f(0) \plus{} 2^{n \plus{} 1} f(\frac {x}{2^{n \plus{} 1}}) \equal{} |2(n \plus{} 1)x| \minus{} (2^{n \plus{} 1} \minus{} 1) f(0) \plus{} 2^{n \plus{} 1} f(\frac {x}{2^{n \plus{} 1}})
\]
and the induction is complete.

Letting $ x \equal{} 1$ and $ x \equal{} \minus{} 1$ and solving for $ f(\frac {x}{2^n})$ we have
\[ f(\frac {1}{2^n}) \le \frac {(2^n \minus{} 1)f(0) \minus{} |2n| \plus{} f(1) } {2^n}
\]
and similarly
\[ f(\frac { \minus{} 1}{2^n}) \le \frac {(2^n \minus{} 1)f(0) \minus{} |2n| \plus{} f( \minus{} 1) } {2^n}
\]
Choose $ n \equal{} N$ sufficiently large that $ |2N| > \max{f(1),f( \minus{} 1)}$. Then $ f(\frac {1}{2^N}),f(\frac { \minus{} 1}{2^N}) < f(0)$. Applying the given inequality to $ x \equal{} \frac {1}{2^N}$ and $ y \equal{} \frac { \minus{} 1}{2^N}$ shows that it is not [i]very convex[/i] and we are done.[/hide]

Notice that we can write $ \frac{d^2 f}{dx^2}$ as
\[ \lim_{h \to 0} \frac{\lim_{h \to 0} \frac{f(x\plus{}2h)\minus{}f(x\plus{}h)}{h} \minus{} \lim_{h \to 0} \frac{f(x\plus{}h)\minus{}f(x)}{h}}{h} \equal{} \lim_{h \to 0} \frac{f(x\plus{}2h)\plus{}f(x) \minus{} 2f(x\plus{}h)}{h^2} \ge \lim_{h \to 0} \frac{|h|}{h^2} \equal{} \infty\] where the operations (i.e. taking the two h's to be the same) can be justified by uniform convergence. This would only work if $ f$ is twice-differentiable, but I wonder whether there is some way to rigorize this idea?

Can somebody rate/give comments on my solution?
[hide="Solution"]Assume, to the contrary, that such a function $ f$ does exist. Let $ f(0)\equal{}a$, and letting $ g(x)\equal{}f(x)\minus{}a$, we have \[ \frac{g(x)\plus{}g(y)}{2}\geq g\left(\frac{x\plus{}y}{2}\right)\plus{}|x\minus{}y|,\] so we can consider the very convex function $ g$ instead, with $ g(0)\equal{}0$. Plugging in $ y\equal{}0$, we have \[ \frac{g(x)}{2} \geq g\left(\frac{x}{2}\right)\plus{}|x|.\quad \left(*\right)\]
We wish to prove that for all positive integers $ n$, $ \frac{g(x)}{2}\geq 2^{n\minus{}1} g\left(\frac{x}{2^{n}}\right)\plus{}|nx|$. For the base case $ n\equal{}1$, we have $ \left(*\right)$. Now, assuming this statement is true for some positive integer $ k$, we have \[ \frac{g(x)}{2}\geq 2^{k\minus{}1} g\left(\frac{x}{2^{k}}\right)\plus{}|kx|. \quad \left(**\right)\] Plugging in $ \frac{x}{2^{k}}$ into $ \left(*\right)$, we obtain $ \frac{g\left(\frac{x}{2^k}\right)}{2}\geq g\left(\frac{x}{2^{k\plus{}1}}\right)\plus{}|k\frac{x}{2^k}|$. Multiplying through by $ 2^k$ gives us $ 2^{k\minus{}1}g\left(\frac{x}{2^k}\right) \geq 2^k g\left(\frac{x}{2^{k\plus{}1}}\right) \plus{} |kx|$, and substituting this into $ \left(**\right)$ gives us \[ \frac{g(x)}{2}\geq 2^{k} g\left(\frac{x}{2^{k\plus{}1}}\right)\plus{}|(k\plus{}1)x|,\] completing the inductive process.
But this it is impossible to satisfy this condition as $ n$ grows arbitrarily large, so we have reached a contradiction. Thus, no very convex function exists.[/hide]

@ Brut: I think your solution is valid.

[hide = outline of a nearly complete geometric solution]
If f(x) is concave, (f(x)+f(y))/2 is always smaller than f((x+y)/2), by the definition of concavity. Because |x-y| is always positive, we reach contradiction in the initial condition of a very convex function. The same reasoning holds for a function which is neither concave nor convex--there is one portion of it that must behave as if it's convex, in which case we use the logic for a concave function.
If f(x) is convex, all that remains to prove is |x-y| is always greater than (f(x)+f(y))/2-f((x+y))/2. Drawing a few pictures, one sees that this is clearly true. However, I'm having trouble rigorizing this last step.
*Note: For my solution i use many actual graphs. My observations come from graphs.
Is it valid?
Sorry for my lack of LaTex[/hide]

[quote="CatalystOfNostalgia"]So your proof is better than a 'proof by example' in that it's a 'proof by lots of examples'?[/quote]

I don't see what you're talking about. If you're getting that impression that I drew specific graphs when coming up with my solution, you are mistaken. Rather, I drew arbitrary concave and convex graphs as I explained above. I was trying to take a more graphical approach.

Otherwise, do you care to do something more constructive than make a sarcastic remark?

[quote="MellowMelon"]"Drawing an arbitrary convex function"? Erm, I don't think that's possible. His comment on your solution has merit, despite being somewhat sarcastic. You need to show that the problem is true for any $ f$, and I don't think a purely graphical approach is going to get you to that point.[/quote]

I understand what you're saying. I guess my underlying reasoning for the graphical approach is as follows: a function can either be convex, concave, or neither, and the graphs of all three categories have particular properties common to all graphs of those types, as given by their definitions (e.g. what does it mean for a graph to be concave/convex?) I was trying to make use of those properties by considering a completely generic concave/convex function (similar to dealing with the graph of the parabola y=x^2 when you are trying to derive general results for parabolas). Then again, I was merely experimenting, so its perfectly reasonable that my approach makes use of skewed logic. I would just like to make sure I see where I go wrong.

[quote="azjps"][quote="mihail911"] convex, concave, or neither, [/quote]

Evidentally, but such a categorization is not going to help: there's no reason to believe $ f$ is smooth, or even anywhere continuous, etc, .. graphing is a first step towards gaining some intuition about the problem, but a statement like "that this is clearly true" clearly [btw, words like "clearly" and "obviously" are almost clearly hand-waving  :wink: ] lacks justification.[/quote]

Thanks, that makes a lot of sense.

Hmm, here is a weird solution.

Let $ x>0$, and assume we have a very convex function $ f(x)$.

Plugging in $ (x,\minus{}x)$ gives $ f(x)\plus{}f(\minus{}x)\ge2f(0)\plus{}4x$.
Plugging in $ (x,0)$ gives $ f(\dfrac{x}{2})\le\dfrac{1}{2}f(x)\plus{}\dfrac{1}{2}f(0)\minus{}x$.
Plugging in $ (\minus{}x,0)$ gives $ f(\minus{}\dfrac{x}{2})\le\dfrac{1}{2}f(x)\plus{}\dfrac{1}{2}f(0)\minus{}x$.
Adding the last two gives $ f(\dfrac{x}{2})\plus{}f(\minus{}\dfrac{x}{2})\le f(x)\plus{}f(0)\minus{}2x$.

We will prove by induction that $ f(x)\plus{}f(\minus{}x)\ge2f(0)\plus{}4kx$ for all $ k\in\mathbb{N}$. The base case, $ k\equal{}1$, is clear from the first line. Assume it's proven for $ k$.

Plugging in $ \dfrac{x}{2}$ into the inductive hypothesis gives $ f(\dfrac{x}{2})\plus{}f(\minus{}\dfrac{x}{2})\ge2f(0)\plus{}2kx$. But combining this with the fourth line, we have

$ 2f(0)\plus{}2kx\le f(\dfrac{x}{2})\plus{}f(\minus{}\dfrac{x}{2})\le \dfrac{1}{2}(f(x)\plus{}f(\minus{}x)\plus{}f(0)\minus{}2x$,

and ignoring the expression in the middle we get that $ f(x)\plus{}f(\minus{}x)\ge2f(0)\plus{}2(2k\plus{}2)x\equal{}2f(0)\plus{}4(k\plus{}1)x$, completing the induction. To finish, plug in $ x\equal{}1$ and we see that $ f(1)\plus{}f(\minus{}1)\ge2f(0)\plus{}4kx$, and since $ f(0),f(\minus{}1),f(1)$ are constants (given a very convex $ f(x)$), we can take an arbitrarily large $ k$ such that the inequality becomes reversed, a contradiction. Thus a very convex function cannot exist.

Here is my proof (a bit lacking in the detail...)

Let $ x_k\equal{}\frac{k}{2N}(b\minus{}a)\plus{}a$ where $ b\ge a$. Then $ x_{k\plus{}1}\equal{}2x_k\minus{}x_{k\minus{}1}$. Let $ a_k\equal{}f(x_k)$. Let $ d\equal{}\frac{2(b\minus{}a)}{N}$. Easily, we get

$ (a_2\minus{}2a_1\plus{}a_0)\ge d$
$ 2(a_3\minus{}2a_2\plus{}a_1)\ge 2d$
...
$ N(a_{N\plus{}1}\minus{}2a_N\plus{}a_{N\minus{}1})\ge Nd$
$ N(a_{N\plus{}2}\minus{}2a_{N\plus{}1}\minus{}a_{N})\ge Nd$
...
$ a_{2N}\minus{}2a_{2N\minus{}1}\plus{}a_{2N\minus{}2}\ge d$

Summing, we get $ a_{2N}\minus{}2a_{N}\plus{}a_0\ge (N)(N\plus{}1)d$ or
$ f(a)\minus{}2f\left(\frac{a\plus{}b}{2}\right)\plus{}f(b)\ge 2(N\plus{}1)(b\minus{}a)$

but this obviously cannot hold if we fix $ a,b$ and increase $ N$.

There's a cool solution that uses tiny differences, such as in calculus when you calculate areas or derivatives by the limit. It results in the same inequality as catalyst's as you'll see.

Suppose for all $ i$ that $ \Delta i\equal{}f\left(\frac{i}{n}\right)\minus{}f\left(\frac{i\minus{}1}{n}\right)$

We also have that $ f\left(\frac{i}{n}\right)\plus{}f\left(\frac{i\plus{}2}{n}\right)\ge 2f\left(\frac{i\plus{}1}{n}\right)\plus{}\frac{4}{n}$

Which gives us $ f\left(\frac{i\plus{}2}{n}\right)\minus{}f\left(\frac{i\plus{}1}{n}\right) \ge f\left(\frac{i\plus{}1}{n}\right)\minus{}f\left(\frac{i}{n}\right)\plus{}\frac{4}{n}$.
In other terms, $ \Delta (i\plus{}1)\ge \Delta i \plus{}\frac{4}{n}$

If we use this inequality for $ n$ consecutive values of $ i$, we have that $ \Delta (i\plus{}n) \ge \Delta i \plus{}4$

Now sum this for $ i\equal{}[1,n]$ to obtain $ f(2)\minus{}f(1)\ge f(1)\minus{}f(0)\plus{}4n$.

Clearly this cannot hold true for all $ n$ and thus we are done.

A solution possibly identical to one of the users above (edit: seems similar to calc rulz's)
[hide]Suppose a very convex function $ f$ exists.

Lemma: for a non-zero $ k$ and sufficiently large $ n$, $ f(x \plus{} \frac {k}{2^n}) < f(x)$.
Proof: Plugging in $ x, x \plus{} \frac {k}{2^n}$,
$ \frac {f(x) \plus{} f(x \plus{} \frac {k}{2^n})}{2} \geq f(x \plus{} \frac {k}{2^{n \plus{} 1}}) \plus{} \frac {|k|}{2^n}$
$ \iff f(x \plus{} \frac {k}{2^{n \plus{} 1}}) \leq \frac {f(x)}{2} \minus{} \frac {|k|}{2^n} \plus{} \frac {f(x \plus{} \frac {k}{2^n})}{2}$ (*)
Applying (*) on $ n \minus{} 1, n \minus{} 2, ..., 1$ we obtain
$ f(x \plus{} \frac {k}{2^{n \plus{} 1}}) \leq f(x)\cdot \sum_{i \equal{} 1}^{n \plus{} 1} \frac {1}{2^i} \minus{} (n \plus{} 1)\cdot \frac {|k|}{2^n} \plus{} \frac {1}{2^{n \plus{} 1}}f(x \plus{} k)$
$ \equal{} (1 \minus{} \frac {1}{2^{n \plus{} 1}})\cdot f(x) \minus{} (n \plus{} 1)\cdot \frac {|k|}{2^n} \plus{} \frac {1}{2^{n \plus{} 1}}f(x \plus{} k)$
$ \equal{} f(x) \minus{} \frac {1}{2^{n \plus{} 1}}(2(n \plus{} 1)|k| \minus{} (f(x \plus{} k) \minus{} f(x)))$
For sufficiently large $ n$, $ 2(n \plus{} 1)|k| \minus{} (f(x \plus{} k) \minus{} f(x)) > 0$, which implies $ f(x \plus{} \frac {k}{2^{n \plus{} 1}}) < f(x)$. $ \blacksquare$

By lemma, for sufficiently large $ n$, we have $ f(x \plus{} \frac {k}{2^n}) < f(x)$ and $ f(x \minus{} \frac {k}{2^n}) < f(x)$. Thus $ \frac {f(x \plus{} \frac {k}{2^n}) \plus{} f(x \minus{} \frac {k}{2^n}) }{2} < f(x)$. But $ \frac {f(x \plus{} \frac {k}{2^n}) \plus{} f(x \minus{} \frac {k}{2^n}) }{2} \geq f(x) \plus{} 2\cdot \frac {|k|}{2^n}$, contradiction. Therefore no very convex function exists.
[/hide]

I found this solution by graphing...so graphing still has its merit?

[quote="Brut3Forc3"]Can somebody rate/give comments on my solution?
[hide="Solution"]Assume, to the contrary, that such a function $ f$ does exist. Let $ f(0) \equal{} a$, and letting $ g(x) \equal{} f(x) \minus{} a$, we have
\[ \frac {g(x) \plus{} g(y)}{2}\geq g\left(\frac {x \plus{} y}{2}\right) \plus{} |x \minus{} y|,\]
so we can consider the very convex function $ g$ instead, with $ g(0) \equal{} 0$. Plugging in $ y \equal{} 0$, we have
\[ \frac {g(x)}{2} \geq g\left(\frac {x}{2}\right) \plus{} |x|.\quad \left(*\right)\]
We wish to prove that for all positive integers $ n$, $ \frac {g(x)}{2}\geq 2^{n \minus{} 1} g\left(\frac {x}{2^{n}}\right) \plus{} |nx|$. For the base case $ n \equal{} 1$, we have $ \left(*\right)$. Now, assuming this statement is true for some positive integer $ k$, we have
\[ \frac {g(x)}{2}\geq 2^{k \minus{} 1} g\left(\frac {x}{2^{k}}\right) \plus{} |kx|. \quad \left(**\right)\]
Plugging in $ \frac {x}{2^{k}}$ into $ \left(*\right)$, we obtain $ \frac {g\left(\frac {x}{2^k}\right)}{2}\geq g\left(\frac {x}{2^{k \plus{} 1}}\right) \plus{} |k\frac {x}{2^k}|$. Multiplying through by $ 2^k$ gives us $ 2^{k \minus{} 1}g\left(\frac {x}{2^k}\right) \geq 2^k g\left(\frac {x}{2^{k \plus{} 1}}\right) \plus{} |kx|$, and substituting this into $ \left(**\right)$ gives us
\[ \frac {g(x)}{2}\geq 2^{k} g\left(\frac {x}{2^{k \plus{} 1}}\right) \plus{} |(k \plus{} 1)x|,\]
completing the inductive process.
But this it is impossible to satisfy this condition as $ n$ grows arbitrarily large, so we have reached a contradiction. Thus, no very convex function exists.[/hide][/quote]

Apologies for reviving, but I'm not sure if the last statement is necessarily true, because if n grows arbitrarily large, then $ g(\frac {x}{2^n})$ can also grow arbitrarily "small", so that the inequality is not necessarily violated. Or am I missing something?

I don't think you're missing anything; I think a few more details should have been given more details as to how the problem is finished from there.

Anyway, here's how I'd finish it (hopefully I didn't mess up somewhere):
From $ \frac {g(x)}{2} \geq 2^k g(\frac {x}{2^{k + 1}}) + |(k + 1)x|$, we have that $ g(\frac {1}{2^{k + 1}})$ must be negative for sufficiently small $ k$, or else $ g(1) = 2^k g(\frac {1}{2^{k + 1}}) + |(k + 1)| \geq |(k + 1)|$, which is unbouded. Setting $ x = - 1$ tells us that $ g(\frac { - 1}{2^{k + 1}})$ must be negative for sufficiently large $ k$. Hence, there is some $ y$ such that $ g(y)$ and $ g( - y)$ are both negative (pick some sufficiently large $ k$ such that $ g(\frac { - 1}{2^{k + 1}}) < 0$ and $ g(\frac {1}{2^{k + 1}) < 0}$.) But then, $ 0 > \frac {g(y) + g( - y)}\geq g(0) + |y - ( - y)| = |2y| \geq 0$, a contradiction.

[quote="CatalystOfNostalgia"]Plugging in $ (\minus{}x,0)$ gives $ f(\minus{}\dfrac{x}{2})\le\dfrac{1}{2}f(x)\plus{}\dfrac{1}{2}f(0)\minus{}x$.
Adding the last two gives $ f(\dfrac{x}{2})\plus{}f(\minus{}\dfrac{x}{2})\le f(x)\plus{}f(0)\minus{}2x$.[/quote]Flipped some signs here, but it seems to just be a typo (the induction is still correct).

This took me far longer than it should have D:

[hide="Solution"]Suppose FTSOC that such an $f$ exists, and WLOG let $f(0)=0$. Then taking $y=0$ yields $f(x)\ge 2 f\left(\frac{x}{2}\right)+2x$ for $x\in\mathbb{R}^{+}$. By taking $x=\frac{1}{2^k}$ for nonnegative $k$, straightforward induction yields $f\left(\frac{1}{2^n}\right)\le\frac{f(1)-2n}{2^n}$; hence $f\left(\frac{1}{2^n}\right)<0$ for sufficiently large $n$. Similarly we can find sufficiently large $n$ such that $f\left(-\frac{1}{2^n}\right)<0$, whence $f\left(\frac{1}{2^n}\right)+f\left(-\frac{1}{2^n}\right)<0$. But taking $x=\frac{1}{2^n}$, $y=-\frac{1}{2^n}$ in the given condition yields $f\left(\frac{1}{2^n}\right)+f\left(-\frac{1}{2^n}\right)\ge 4\cdot\frac{1}{2^n}>0$, contradiction. So we're done! $\blacksquare$[/hide]

[quote=mishai]Does the following solution work?
[hide=Tentative]Let $f(c)$ be the minimum value of $f$ such that $f'(c)=0$. Then define $g(x)=f(x)-f(c)$ such that $g(c)=0$. Next substitute $x=c+a$ and $y=c-a$ for some positive a. Then $\dfrac{g(c+a)+g(c-a}{2}\geq g(c)+2a=2a$. Therefore, we must have $\dfrac{g(c+a)+g(c-a)}{4a}\geq1$.
Now note that $\lim_{a\rightarrow0}\dfrac{g(c+a)+g(c-a)}{4a}$ is of the form $\dfrac{0}{0}$ , so we can use L'Hopital's to achieve $\lim_{a\rightarrow0}\dfrac{g'(c+a)+g'(c-a)}{4}=0\le 1$ The last step follows since we defined $f'(c)=g'(c)=0$. Therefore there cannot be a "very convex" function.[/hide][/quote]

It's not valid --- you can't take the derivative of $f$ without first having shown its differentiable. Unfortunate this is probably about as hard as the original problem. :(

Asa rule of thumb, for 99% of Olympiad functional equations, derivatives/continuity type arguments won't help unless they're given conditions.

Very similar to some of the above.
[hide=Solution]
Let $g(x)=f(x)-f(0)$; we have $g(0)=0$ and the problem is equivalent to proving there doesn't exist a $g$ such that \[\frac{g(x)+g(y)}{2}\geq g\left(\frac{x+y}{2}\right)+\vert x-y\vert.\] Let $P(x, y)$ denote this assertion. Then $P(x, 0)$ gives \[g(x)\geq 2g\left(\frac{x}{2}\right)+2\vert x\vert.\quad \quad (1)\]

We'll now prove by induction that $g(1)\geq 2^k\cdot g\left(\frac{1}{2^k}\right)+2k$ for all positive integers $k$. The base case is obvious by plugging in $x=1$ into $(1)$; now assume the result holds for $k-1$; we have \[g(1)\geq 2^{k-1}\cdot g\left(\frac{1}{2^{k-1}}\right)+2(k-1)\geq 2^{k-1}\cdot \left(2g\left(\frac{1}{2^k}\right)+\frac{2}{2^{k-1}}\right)+2(k-1)\geq g(1)\geq 2^k\cdot g\left(\frac{1}{2^k}\right)+2k,\] so the result holds for $k$ as well. Combining the base case and the inductive hypothesis, the induction is proven. Similarly, we can prove that \[g(-1)\geq 2^k\cdot g\left(-\frac{1}{2^k}\right)+2k.\] Notice that \[P(x, -x): g(x)+g(-x)\geq 4\vert x\vert.\] Adding the inequalities for $g(1)$ and $g(-1)$ gives \[g(1)+g(-1)\geq 2^k\left(g\left(\frac{1}{2^k}\right)+g\left(-\frac{1}{2^k}\right)\right)+4k\geq 4+4k.\] However, because $g(1)$ and $g(-1)$ are constants, we can take a sufficiently large $k$ such that the inequality above reverses direction; thus, there exists no function $g$ that satisfies the condition, and the result follows.
[/hide]

We wish to prove by induction that,
$$\frac{f(x)+f(y)}{2} \geq f(\frac{x+y}{2})+2^n|x-y| \quad \forall n \in \mathbb{N}_0$$
The base case $n=0$ is our question statement.

Inductively, $\frac{f(x)+f(y)}{2} \geq f(\frac{x+y}{2})+2^n|x-y|$. Let's call this assertion $P(x,y)$.

$P(a,a+2b) \implies \frac{f(a)+f(a+2b)}{2} \geq f(a+b)+2^n|2b| =  f(a+b)+2^{n+1}|b| $

$P(a+2b,a+4b) \implies \frac{f(a+2b)+f(a+4b)}{2} \geq f(a+3b)+2^n|2b| =  f(a+3b)+2^{n+1}|b| $

$P(a+b,a+3b) \implies f(a+b)+f(a+3b) \geq 2f(a+2b)+2^{n+1}|2b| =  2f(a+2b)+2^{n+2}|b| $

Summing up this three inequalities, we get

$\frac{f(a)+f(a+4b)}{2} \geq f(a+2b)+|b|(2^{n+1}+2^{n+1}+2^{n+2}) =  f(a+2b)+2^{n+3}|b| = f(a+2b)+2^{n+1}|4b|$

Taking $x=a,y=a+4b$ completes our induction.

Now $n$ can be arbitrarily large. So no such function exists.

I saw many calculus attempts in the above posts. As mentioned, they all have a flaw - the function may not be smooth, it may not have a derivative. Strictly speaking, it's true of course. But, you konow, this function is not a random one, it's convex. Those functions are very good ones, they have derivatives almost everywhere (moreover they have derivatives everywhere except at countable many points) and can be represented as
	$$f(x)=f(a)+\int_a^x f'(t)\,dt$$
	as being absolutely continuous. $f'(t)$ is defined almost everywhere and is non decreasing. This property is enough to conclude
	$$f(x+2h)+f(x)-2f(x+h)=o(h)$$
	as it was (non rigorously) pointed out in #11. Let's see why it holds.
	\begin{align*}
	\displaystyle f(x+2h)+f(x)-2f(x+h)&=f(x)+\int_x^{x+2h} f'(t)\,dt +f(x)-2f(x)-2\int_x^{x+h} f'(t)\, dt\\
	&=\int_x^{x+2h} f'(t)\,dt - 2\int_x^{x+h} f'(t)\, dt\\
	&=\int_{x+h}^{x+2h} f'(t)\,dt - \int_x^{x+h} f'(t)\, dt\\
	&=\int_{x}^{x+h} f'(t+h)-f'(t) \,dt
	\end{align*}
		Since $ f'(t)$ is non decreasing, for any $t\in [x,x+h]$ we have:
	$$\displaystyle 0\le f'(t+h)-f'(t)\le \max_{u\in[x,x+2h]} f'(u)-\min_{u\in[x,x+2h]} f'(u)\le |f'(x+2h)-f(x)|$$
	Hence
	$$\displaystyle |f(x+2h)+f(x)-2f(x+h)|\le h|f'(x+2h)-f'(x)|$$
	Now, since $f'(x)$ in non decreasing it's continuous everywhere except on a countable set. It's enough to take a point $x$ where $f'$ is continuous and get
	$$\displaystyle \lim_{h\to 0}\frac{|f(x+2h)+f(x)-2f(x+h)|}{h}=0$$
	which contradicts to the initial condition imposed on $f$.

[b]EDIT[/b]. There is an issue with this solution. I somehow presumed the function was convex. But nothing is stated about $f$ except that so called very convex condition. Of course, $f$ is mid-point convex, but it's not enough to conclude it's convex. There are pathological counterexamples, for example via Hamel bases we could construct a "bad" solution of the Cauchy functional equation and the same function would be mid-convex but not convex. If additionally, we know $f$ is continuous (or even bounded, or even measurable), the convexity follows.
Fortunately, in our case it can be fixed. Note that mid-point convexity implies $f$ is convex in the set $D$ of dyadic rational points, i.e. the points $\frac{k}{2^{\ell}}, k,\ell\in\mathbb{Z}.$ It's not difficult to obtain that the restiction of $f$ over $D$ can be extended to a continuous and convex function $\tilde{f}$ over $\mathbb{R}$ with $f(x)=\tilde{f}(x),\forall x\in D$. Since $D$ is dense in $\mathbb{R},$ the extra convex condition over $D$ implies $\tilde{f}$ is extra convex too. That's why, we can already apply the above argument to $\tilde{f}$ and get contradiction.
I edited it [url=https://dgrozev.wordpress.com/2020/08/26/very-convex-function-usamo-2000-problem-1/]here[/url]

Here is another solution.
Fix some $h>0$ and consider the points $x_j=x_0+j\cdot h, j=0,1,\dots,n+1$ where $x_0\in\mathbb{R}, n\in\mathbb{N}$. Applying the given condition to $x=x_0+jh, y=x_0+(j+2)h$ yields
$$\Delta^2_h f(x_j):=f(x_j)+f(x_{j+2})-2f(x_{j+1})\ge 4h \quad (1)$$
The above notation follows the notation of corresponding [url=https://dgrozev.wordpress.com/2019/09/15/finite-differences-in-olympiads/]finite difference[/url] of 2nd order. Note that
$$\Delta^2_h f(x_j)=\Delta_h f(x_{j+1})-\Delta_h f(x_j)=(f(x_{j+2})-f(x_{j+1}))-(f(x_{j+1})-f(x_{j})).$$
Hence, we can sum up $(1)$ for $j=0,1,\dots,n-1$ and telescoping it to obtain
$$4nh\le \sum_{j=0}^{n-1}\Delta_h^2 f(x_j)=\Delta_hf(x_{n})-\Delta_hf(x_0)$$
Putting $x_0=0,h=1/n$ for any $n\in\mathbb{N}$ it yields
$$ 4\le \Delta_h f(1)-\Delta_h f(0) \quad (2)$$
In case $f$ is continuous it brings us to a straightforward contradiction, since the RHS converges to $0$ as $n\to \infty.$ But, we don't know it a priori, so a few more moves are needed.
Note that $f$ is convex on the set $D$ of all dyadic rational numbers $D:=\left\{\frac{k}{2^{\ell}}: k\in\mathbb{Z}, \ell\in \mathbb{N}\right\}.$ It follows (as in the case of the convex functions) the slopes, taken at these points are increasing, i.e. for any $d_1<d_2\le d_3<d_4$ in $D$ we have
$$\frac{f(d_2)-f(d_1)}{d_2-d_1}\le \frac{f(d_4)-f(d_3)}{d_4-d_3}$$
Thus, it yields
$$\frac{f(0)-f(-1)}{1}\le \frac{\Delta_h f(0)}{h}\le \frac{f(2)-f(1)}{1}$$
for any $h=2^{-n},n\in\mathbb{N}$. The same is true for $\Delta_hf(1)$ in the place of $\Delta_hf(0)$.Hence
$$\max\Big(|\Delta_hf(0)|,|\Delta_hf(1)| \Big)\le C|h|$$
for some constant $C>0.$ Using it back in $(2)$ we get contradiction when $h\to 0$.

[hide=Solution (with less variables)]
If such an $f(x)$ exists, then one can verify that an equivalent $f(x)$ such that $f(0)=0$ also satisfies the inequality constraint.
Unlike in previous solutions, we can now use actual constants to describe the phenomenon that is happening.
In particular, $$f(1) \ge 2f(1/2)+2 \ge 4f(1/4)+4 \ge 8f(1/8)+6 \ge 16f(1/16)+8 \ge 32f(1/32) + 10 \ge ...$$ and $$f(-1) \ge 2f(-1/2)+2 \ge 4f(-1/4)+4 \ge 8f(-1/8)+6 \ge 16f(-1/16)+8 \ge 32f(-1/32) + 10 \ge ...$$
Adding them up, we get $$f(1)+f(-1) \ge 2^n[f(1/2^n)+f(1/2^{-n})]+4n$$ for all positive integers $n$.
The inequality constraint gives us $$f(1/2^n)+f(1/2^{-n}) \ge 1/2^{n-2}$$
so that $f(1) + f(-1) \ge 4+4n$ for all positive integers $n$; clearly impossible.
[/hide]

Note that after shifting the function, we wlog assume that $f(0)=0$.
Firstly taking $x\rightarrow 2x$ and $y=0$, then $$f(2x)\geq 2f(x)+4\vert x\vert.\quad (1)$$

[b]Claim.[/b] If $f(x)+f(-x)\geq a\vert x\vert$, then we also conclude that $f(x)+f(-x)\geq (a+4)\vert x\vert $.
[i]Proof.[/i] Now using $(1)$ and the given condition, we obtain that
$$f(2x)+f(-2x)\geq 2(f(x)+f(-x))+8\vert x\vert \geq (2a+8)\vert x\vert ,$$
hence $f(x)+f(-x)\geq (a+4)\vert x\vert$.

By the claim we can take $a$ incredibly large in $f(x)+f(-x)\geq a\vert x\vert$. Fixing $x$, we obtain contradiction.

Let $P(x,y)$ denote the assertion $\frac{f(x)+f(y)}{2} \geq f(\frac{x+y}{2})+|x-y|.$ We claim that there is no such function. WLOG $f(0)=0$ by shifting.

$P(x,-x)\implies \frac{f(x)+f(-x)}{2} \geq 2x$
In particular $f(x/2)+f(-x/2)\geq 2x$ by setting $x=x/2.$
$P(x,0)+P(-x,0)\implies \frac{f(x)+f(-x)}{2} \geq f(x/2)+f(-x/2)+2x \geq 4x$
Iterating this $\frac 12 (f(x)+f(-x))\geq 2nx$ for positive integer $n.$
A clear contradiction after fixing $x$ and making $n$ indefinitely large.

For $b\in \mathbb R^+$ define $P(b)$ to be the statement: $(f(x+a)-f(x))-(f(x)-f(x-a))\ge ab$ for all $x\in \mathbb R, a\in \mathbb R^+$ Note that since $(f(x+a)-f(x))-(f(x)-f(x-a))$ is a real number, there exists a maximum value $b$ for which $P(b)$ is true. Now, summing these four expressions up:
\begin{align*}
2((f(x+a)-f(x))-(f(x)-f(x-a)))&\ge 2ab \\
(f(x+2a)-f(x+a))-(f(x+a)-f(x))&\ge ab \\
(f(x)-f(x-a))-(f(x-a)-f(x-2a))&\ge ab \\
\end{align*}
we get that $(f(x+2a)-f(x))+(f(x)-f(x-2a))\ge 4ab$, or $P(2b)$ to be exact. Since $P(b)\implies P(2b)$, $P(4)$ implies that $P(b)$ is true for infinitely large $b$, contradiction.

Since "very convex"ness does not care about shifting, WLOG $f(0)=0.$ Then, let $f(1)=c$.\\

Claim: $$f(\frac{1}{2^n})\leq \frac{c}{2^n}-\frac{n}{2^{n-1}}$$ for nonnegative integers $n$. We will use induction. Clearly this is true for $n=0$. We will use induction. Letting $x=\frac{1}{2^n}$ and $y=0$ we have $$f(\frac{1}{2^{n+1}})\leq \frac{f(\frac{1}{2^{n}})}{2}-\frac{1}{2^{n}}\leq  \frac{c}{2^{n+1}}-\frac{(n+1)}{2^n},$$ which completes the induction. Similarly, if we let $d=f(-1)$, $$f(-\frac{1}{2^n})\leq \frac{d}{2^n}-\frac{n}{2^{n-1}}.$$ Now, if we plug in $x=\frac{1}{2^n}$ and $y=-\frac{1}{2^n}$, we get $$\frac{f(1/2^n)+f(-1/2^n)}{2}\geq 0+\frac{1}{2^{n-1}}.$$ Thus, from the two above claims, we have $$\frac{c}{2^{n+1}}+\frac{d}{2^{n+1}}-\frac{n}{2^{n-1}}\geq \frac{1}{2^{n-1}},$$ which is just $$c+d-4n\geq 4,$$ which is clearly a contradiction for sufficiently large $n$, QED.

Consider the second derivative of $f$. We claim that $f''(x) \geq C$ for some positive constant $C$. This proves that the function is impossible because we would have it define two values.

We can see that we have \[\frac{f(x+n) + f(x)}{2} \geq f \left( \frac{2x+n}{2} \right) + n\]. Thus we can see that over each infinitessimaly small segment of length $n$ (we choose $n$ so that $f$ is effectively linear). Notice that we must pass through the point that is $n$ units down from the midpoint of this segment (it's also perpendicular). This means that the second derivative is always greater than that constant.

Note: in fact if we were to try to draw this function according to the second derivative definition, it would be a dot. $\blacksquare$

I think this should work but I am not sure

Second derivative? You even don't know if the first derivative exists!? And it may not, at least at lot of points, see #34.

[hide=Solution]
Assume for the sake of contradiction that there existed a very convex function $f$. Let $P(x, y)$ denote the assertion. Note that for any $a, b$, we have that $$P\left(b, \frac{a+b}{2}\right) \implies f\left(\frac{a+3b}{4}\right) \leq \frac{f(b) + f\left(\frac{a+b}{2}\right)}{2} - \frac{|a-b|}{2},$$ and $$P\left(a, \frac{a+b}{2}\right) \implies f\left(\frac{3a+b}{4}\right) \leq \frac{f(a) + f\left(\frac{a+b}{2}\right)}{2} - \frac{|a-b|}{2}.$$ Therefore, we have that $$f\left(\frac{a+3b}{4}\right) + f\left(\frac{3a+b}{4}\right) \leq \frac{f(a)+f(b)}{2} + f\left(\frac{a+b}{2}\right) - |a-b|.$$ Now, note that $$P\left(\frac{a+3b}{4}, \frac{3a+b}{4}\right) \implies f\left(\frac{a+b}{2}\right) \leq \frac{f\left(\frac{a+3b}{4}\right) + f\left(\frac{3a+b}{4}\right)}{2} - \frac{|a-b|}{2} = \frac{\frac{f(a)+f(b)}{2} + f\left(\frac{a+b}{2}\right) - |a-b|}{2} - \frac{|a-b|}{2},$$ so $$2f\left(\frac{a+b}{2}\right) \leq \frac{f(a)+f(b)}{2} + f\left(\frac{a+b}{2}\right) - 2|a-b|,$$ so $$f\left(\frac{a+b}{2}\right) \leq \frac{f(a)+f(b)}{2} - 2|a-b|,$$ i.e. $$\frac{f(a)+f(b)}{2} \geq f\left(\frac{a+b}{2}\right) + 2|a-b|.$$ Then, it follows that $$\frac{(f/2)(a) + (f/2)(b)}{2} \geq (f/2)\left(\frac{a+b}{2}\right) + |a-b|,$$ so $f/2$ is very convex. However, by the above, this would imply that $$\frac{(f/2)(a) + (f/2)(b)}{2} \geq (f/2)\left(\frac{a+b}{2}\right) + 2|a-b|.$$ However, then we can deduce that $f/4$ is convex by dividing everything by $2$, and then applying this logic over and over again, it follows that $\frac{f}{2^n}$ is convex for every nonnegative integer $n\geq 0$, i.e. $$\frac{(f/2^n)(a) + (f/2^n)(b)}{2} \geq (f/2^n)\left(\frac{a+b}{2}\right) + |a-b|.$$ Then, $$\frac{f(a) + f(b)}{2} \geq f\left(\frac{a+b}{2}\right) + 2^n|a-b|.$$ Plugging in $a = 1$, $b = 0$, we have that $$2^n \leq \frac{f(1)+f(0)}{2} - f\left(\frac{1}{2}\right)$$ for all nonnegative integers $n\geq 0$, absurd. So, we have a contradiction and so no very convex function exists. $\blacksquare$


Find all solutions to:
$ \frac {y \minus{} 3x}{2x \minus{} y} \equal{} n$

with x,y,n natural numbers

[hide]
First, note that $ 2x < y < 3x$, otherwise, the $ LHS$ is negative. Let's manipulate the equation to obtain:

$ \frac {y \minus{} 3x}{y \minus{} 2x} \equal{} \minus{} n$

$ \frac { \minus{} x}{y \minus{} 2x} \equal{} \minus{} n \minus{} 1$

$ \frac {x}{y \minus{} 2x} \equal{} n \plus{} 1$

Therefore, $ y \minus{} 2x$ must divide $ x$. Let $ k$ be a proper divisor of $ x$. Then all solutions are of the form $ (x, 2x \plus{} k, \frac{x}{k} \minus{} 1)$.
[/hide]


geometry
circumcircle
Euler
projective geometry
cyclic quadrilateral
geometry unsolved

A convex quadrilateral $ ABCD$ stisfy $ \angle B \equal{} \angle C \equal{} \angle D$. Let $ E,F$ be the projectives of $ A,C$ to $ BC,DA$ and $ M$ be midpoint of $ BD$.  prove that E,M,F are collinear.

Let $ O$ be the intersection of $ AD$ and $ BC$.
$ E,M,F$ are collinear iff $ \frac{FO}{FD}.\frac{MD}{MB}.\frac{EB}{EO}\equal{}1$ (Menelaus's theorem)
$ \Leftrightarrow \frac{FO}{FD}\equal{}\frac{EO}{EB}$
$ \Leftrightarrow \frac{FO}{EO}\equal{}\frac{FD}{EB}\equal{}\frac{CF}{AE} (\Delta DFC\sim \Delta BEA)$
But $ CFAE$ is cyclic quadrilateral then $ \frac{FO}{EO}\equal{}\frac{CF}{AE}$
So $ (*)$ is true.
We are done!

This problem only need $ \angle B \equal{} \angle D$. But if $ \angle B \equal{} \angle C \equal{} \angle D$ then we have another collinear:
Let $ H, I$ be the orthocenter and circumcenter of triangle $ BCD$. Prove that $ A,H,I$ are collinear.
[hide]Can use Tri-Ceva!  :wink: [/hide]

Let $ M' \equiv EF \cap BD,$ $ X \equiv BC \cap AD.$ By Menelaus' theorem for $ \triangle XBD$ cut by the transversal $ \overline{EM'F},$ we have

$ \frac {EB}{EX} \cdot \frac {XF}{FD} \cdot \frac {DM'}{M'B} \equal{} 1$

But since $ \triangle EAB \sim \triangle FCD$ and $ CF,EA$ are antiparallel WRT the rays $ XA,XB$

$ \frac {EB}{FD} \equal{} \frac {EA}{FC} \equal{} \frac {EX}{XF} \Longrightarrow M'D \equal{} M'B \Longrightarrow M \equiv M'.$

As Linh pointed out, in this case we only need  $ \angle B \equal{} \angle D.$ If $ \angle B \equal{} \angle C \equal{} \angle D,$ then $ A$ lies on the Euler line of  $ \triangle CBD.$ Let $ Y \equiv DC \cap AB$ and $ N,L$ the midpoints of $ CD, CB.$ $ XN$ and $ YL$ are perpendicular  bisectors of $ CD,CB.$ $ XD,XN,XC$ meet at $ X$ and $ YD,YL,YB$ meet at $ Y$  $\Longrightarrow$ Intersections $ XN \cap YM$ (circumcenter of CBD),  $ BN \cap DL$ (centroid of CBD) and $ A \equiv XD \cap YB$ are collinear, due to Pappus theorem.


inequalities
trigonometry
inequalities unsolved

[quote="Moonmathpi496"]Show that for every acute angled triangle,
\[ \sum_{cyc} \frac {a}{\cos A}\geq 2(a \plus{} b \plus{} c)
\]
[/quote]

By Chebyshev : $ LHS \ge \frac{1}{3}(a\plus{}b\plus{}c)\sum_{cyc} \frac {1}{\cos A} \ge 2(a\plus{}b\plus{}c)$

[quote="Sergic Primazon"][quote="Moonmathpi496"]Show that for every acute angled triangle,
\[ \sum_{cyc} \frac {a}{\cos A}\geq 2(a + b + c)
\]
[/quote]
By Chebyshev : $ LHS \ge \frac {1}{3}(a + b + c)\sum_{cyc} \frac {1}{\cos A} \ge 2(a + b + c)$[/quote]

Very nice solution! I believe that you have used $ \sum_{cyc} \sec A \geq 6$

My solution is a bit different (and longer than yours)
\[ \sum_{cyc} \frac {a}{\cos A} = \frac {a^2}{a\cos A} \geq \frac {\left (\sum a \right)^2}{%Error. "suma" is a bad command.
\cos A} = \frac {\left (\sum a \right)^2}{R \left ( \sum \sin 2A \right )}
\]
Using Cauchy–Schwarz inequality.

We know that, $ \sum \sin 2A = 4\prod \sin A$ and $ (\triangle ABC) = sr = 2R^2 \prod \sin A$
Now it is enough to prove that,
\begin{align*} \frac {\left (\sum a \right)^2}{R \left ( \sum \sin 2A \right )} \geq 2 \left (\sum a \right ) & \iff & \sum a \geq 2R \left ( \sum \sin 2A \right ) \\
& \iff & R \sum a \geq 4 \cdot 2R^2 \prod \sin A = 4 (\triangle ABC) \\
& \iff & R \left( \sum a \right ) \geq 2 r \left( \sum a \right ) \\
& \iff & R \geq 2r \end{align*}

[quote="Moonmathpi496"]
\[ \sum_{cyc} \frac {a}{\cos A} = \frac {a^2}{a\cos A} \geq \frac {\left (\sum a \right)^2}{%Error. "suma" is a bad command.
\cos A} = \frac {\left (\sum a \right)^2}{R \left ( \sum \sin 2A \right )}
\]
Using Cauchy–Schwarz inequality.
[/quote]
Unfortunately here I made a number of typos. :(
It should be
\[ \sum_{cyc} \frac {a}{\cos A} = \sum_{cyc}\frac {a^2}{a\cos A} \geq \frac {\left (\sum a \right)^2}{\sum a \cos A} = \frac {\left (\sum a \right)^2}{R \left ( \sum \sin 2A \right )}
\]


modular arithmetic
Euler
number theory unsolved
number theory

1.3

Let $a_0, a_1, \ldots, a_n, x_0, x_1, \ldots, x_n$ ($n\geq 2$) be integers, and $r \geq 2$ an integer. It is known that $\sum_{j=0}^n a_j x_j^k = 0$ for $k=1, 2, \ldots, r$.
Prove that $\sum_{j=0}^n a_j x_j^m \equiv 0 \pmod{m}$ for all $m \in \{ r+1, r+2, \ldots, 2r+1 \}$

$Lemma$:For each integer $m\geq 2$ there exists $1\leq t(m)\leq \frac m 2$ s.t
 $x^m\equiv x^{t(m)}(mod m) \forall x\in\mathbb Z$
$Proof$:
   Assume that $m={p_1^{a_1}}...{p_k^{a_k}}$ then $\varphi (m)=\prod p_i^{a_i-1}{(p_i-1)}$.Use modulo $4$ we see that $\varphi (m)|m$  $\iff$  $m=2^k$ or $m={2^k}{3^t}$ where $k,t \in\mathbb N$
   If $m=2^k$ choose $t(m)=2^{k-1}$
   If $m={2^k}3^t$ choose $t(m)={2^k}3^{t-1}$
   Now if $\varphi (m)\nmid m$ we choose $t(m)=m-{\varphi (m)}{\left[\frac m {\varphi (m)}\right]}$.This is because of the fact $t(m)\geq max\{a_1,...,a_k\}$
 and Euler theorem${x^{\varphi (m)}\equiv1(mod m)}$ if $gcd(x,m)=1$(q.e.d)
 Return to our problem ,note that $m\in\{r+1,...,2r+1\}$ then $t(m)\in\{1,2,...,r\}$.
  I'm really sorry but i hadn't enough time to write more clearly.


Bob is 13 years old.  Bobella is 12 years old.  Yoda's mom is 1673 years old.  Joe is 56 years old.  What is the average age of everyone?

did you use a calculator or do it in your head?

the adding or the division? lol.. adding in my head but paper for division.. too lazy to do the division in my head.

[hide]1754/4=438.5[/hide]
These ages are a little varied...

[hide]438.5 I used a calculator  :P  Lololol, Yoda's mom.... lolol Very creative.[/hide]


integration
calculus
derivative
real analysis
real analysis unsolved

Let $ D\equal{}\{x\in\mathbb{R}^n: |x|>1\}, u(x)\equal{}|x|^t$. Let $ \minus{}\frac{n}{q}<t<\minus{}\frac{n}{p}$. Then $ u\in W^{1,p}(D)$. But $ u\notin L^q(D), (p<q)$.

I hope you mean $ 1\leq q < p$ or something make sense in the followings
1) Why $ u\not\in L^q(D)$ ?
[hide="Hint for this"]$ \int_D|u(x)|^qdx \equal{} \int_{|x| > 1}|x|^{qt}dx$
$ \equal{} \int_{1}^{ \plus{} \infty}r^{qt}\int_{x\in\partial B(0;r)}dx \equal{} n\cdot\alpha(n)\int_1^{ \plus{} \infty}r^{tq \plus{} n \minus{} 1}dr \equal{} \infty$
(here note that $ tq \plus{} n > 0$ and $ \alpha(n)$ is volume of unit ball $ B(0;1)$.)[/hide]
2) and why $ u\in W^{1,p}(D)$ ?
[hide="Hint for this"]Use the same way above to abtain $ u\in L^p(D)$. Now its $ 1$-st gennerilized derivatives of $ u$. It's easy to see $ u_{x_i}(x) \equal{} t\cdot x_i\cdot |x|^{t \minus{} 2}$ a.e in $ D$. To prove them belong to $ L^p(D)$ it sufficient to show $ \int_D|x_1|^p|x|^{p(t \minus{} 2)}dx < \plus{} \infty$. But this not hard because $ |x_1|\leq |x|$ and $ t \minus{} 1 < t < \minus{} \frac {n}{p}$ which permite us apply the above result or the same arguments.[/hide]

Yes, I had a mistake , with $ q\in [1,p)$. And I had a solution,thanks a lot.


Putnam
logarithms
integration
floor function
Putnam calculus

Define $ f: \mathbb{R}\to\mathbb{R}$ by
\[ f(x)\equal{}\begin{cases}x&\text{if }x\le e\\ xf(\ln x)&\text{if }x>e\end{cases}\]
Does $ \displaystyle\sum_{n\equal{}1}^{\infty}\frac1{f(n)}$ converge?

We prove this by the integral test. Once we know that the integral test applies (more on that below,) we have that $ \sum_{n\equal{}1}^{\infty}\frac1{f(n)}$ converges if and only if $ \int_1^{\infty}\frac1{f(x)}\,dx$ converges. Then,
\[ \int_1^{\infty}\frac1{f(x)}\,dx\equal{} \int_1^e\frac1x\,dx\plus{}\int_e^{\infty}\frac1{xf(\ln x)}\,dx\]
Make the substitution $ u\equal{}\ln x$ to turn the second integral into $ \int_1^{\infty}\frac1{f(u)}\,du.$ Hence:
\[ \int_1^{\infty}\frac1{f(x)}\,dx\equal{}1\plus{}\int_1^{\infty}\frac1{f(x)}\,dx\]
which implies that that integral is infinite and hence the sum diverges.

We have to work a little harder than that; for one thing, we do have to show that $ f(x)$ is increasing so that we can have $ \frac1{f(x)}$ decreasing in order to justify the integral test. Define $ I_1\equal{}(1,e],I_2\equal{}(e,e^e],I_3\equal{}(e^e,e^{e^e}],$ and so on, with $ x\in I_{k\plus{}1}$ if and only if $ \ln x\in I_k.$

$ f$ is differentiable on the interior of each $ I_k.$ We will show that $ f'(x)>0$ for all such points, so that $ f$ is increasing within each $ I_k.$ Note that $ f'(x)\equal{}1>0$ in the interior of $ I_1.$ Now suppose that $ f'(x)>0$ for $ x$ in the interior of $ I_k.$ Consider $ x$ in the interior of $ I_{k\plus{}1}.$ Then $ f'(x)\equal{}f(\ln x)\plus{}f'(\ln x).$ We have $ f(\ln x)>0$ since the function is constructed to be positive and and $ f'(\ln x)>0$ since $ \ln x\in I_k.$ Hence $ f'(x)>0$ for $ x$ in the interior of $ I_{k\plus{}1}.$ Next, we show that if $ x\in I_k$ and $ y\in I_{k\plus{}1},$ then $ f(x)<f(y).$ If $ k\equal{}1,$ then we have $ 1<x\le e$ and $ e<y\le e^e.$ $ f(x)\equal{}x\le e,$ while $ f(y)\equal{}yf(\ln y)\equal{}y\ln y>y\ln e\equal{}y>e.$ Hence $ f(x)<f(y).$ A further induction then shows that if $ x\in I_j,y\in I_k$ with $ j<k,$ then $ f(x)<f(y).$ Hence $ f(x)$ is increasing and $ \frac1{f(x)}$ is decreasing, justifying the integral test. As for that integral test, consider this:
\[ \int_1^{\infty}\frac1{f(x)}\,dx\equal{}\sum_{k\equal{}1}^{\infty}\int_{I_k}\frac1{f(x)}\,dx\]
But then we claim by induction that $ \int_{I_k}\frac1{f(x)}\,dx\equal{}1$ for all $ k.$ We have that that integral over $ I_1$ is 1 by direct calculation. Assume that $ \int_{I_k}\frac1{f(x)}\,dx\equal{}1$ for some $ k\ge 1.$ Then

$ \int_{I_{k\plus{}1}}\frac1{f(x)}\,dx\equal{}\int_{I_{k\plus{}1}}\frac1{xf(\ln x)}\,dx\equal{}\int_{I_k}\frac{1}{f(u)}\,du\equal{}1.$

Hence $ \int_1^{\infty}\frac1{f(x)}\,dx\equal{}\sum_{k\equal{}1}^{\infty}\int_{I_k}\frac1{f(x)}\,dx\equal{}\sum_{k\equal{}1}^{\infty}1\equal{}\infty$ and the integral (hence also the sum) diverges.

I went and bounded it below instead:

First define $ ^kx$ as the power-tower (that is, $ ^kx = (^{k - 1}x)^x, ^1x = x$)

Then for $ ^ke < n\leq^{k + 1}e$ we have $ f(n) = \prod_{i = 0}^k\ln^i(n)$ where the power is a functional iterate, $ \ln^2(n) = \ln(\ln(n))$.

By monotonicity and convexity of the log function we can bound that above by $ f(n)\leq\prod_{i = 0}^k\ln^i(^{k + 1}e)$ and it's easy to see $ \ln(^{k + 1}e) = ^ke$, so the bound is $ \prod_{i = 0}{}^k{}^ie$

Now the series uses the reciprocal, so $ \frac {1}{f(n)}\geq(\prod_{i = 0}{}^k{}^ie)^{ - 1}$ for $ k = \ln^*(n)$, the iterated logarithm.

The number of terms between $ ^ke$ and $ ^{k + 1}e$ is simply $ \lfloor^{k + 1}e\rfloor - \lfloor^ke\rfloor$, and we can see $ \lfloor a\rfloor - \lfloor b\rfloor = a - [a] - (b - [b]) = a - b + [b] - [a] > a - b - 1$

Collecting like terms, $ \sum_{n = 1}^\infty\frac {1}{f(n)} > \sum_{k = 1}^\infty(^{k + 1}e - ^ke - 1)(\prod_{i = 0}{}^k{}^ie)^{ - 1}$
$ = \sum_{k = 1}^\infty^{k + 1}e(\prod_{i = 0}{}^k{}^ie)^{ - 1} - \sum_{k = 1}^\infty^ke(\prod_{i = 0}{}^k{}^ie)^{ - 1} - \sum_{k = 1}^\infty(\prod_{i = 0}{}^k{}^ie)^{ - 1}$

The last term goes down as an inverse of a tetration, and can be easily shown to converge. The second last term cancels one of the tetration products but the others remain, and similarly converges. We're left with

$ = \sum_{k = 1}^\infty^{k + 1}e(\prod_{i = 0}{}^k{}^ie)^{ - 1} - const$

and invoke the ratio test: $ \left|\frac {a_{k + 1}}{a_k}\right|=\frac {^{k + 2}e}{^{k + 1}e}\frac{\prod_{i = 0}{}^k{}^ie}{\prod_{i = 0}^{k + 1}^ie}=\frac {(^{k + 1}e)^e}{(^{k + 1}e)^2}=(^{k + 1}e)^{e - 2} > 1$

And we have bounded the series from below with a rather quickly diverging series (collected from $ ^ke$ terms though)

I correctly got that each I_k integral was equal to 1 by induction, but I forgot to actually justify using the integral test (just used it without justification)...how many points do you think that's worth?

Kent
Is it necessary to verify that f(x) is continuous

It's necessary to verify that it's monotone increasing- which means that showing that the limits from below and from above at the key points are the same, in this case. It's actually differentiable at all of those points except $ 1$ (we could define $ f(x) \equal{} 1$ on $ [0,1]$ and use the functional equation everywhere above that).

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244073]Putnam 2002 A6[/url]. I was instantly reminded when I saw this problem.

I found that it diverged using the integral test...but i'm not sure my proof was wordy enough.  I think I said something along the lines of the Lim t->infinity of ln(f(t)) approached infinity because the terms would always be the log of t*ln...of a lot of stuff.  I really wanted to get the calc related question as I have no experience in anything but calc/diff eqs/a little discrete math.

[quote="jmerry"]It's necessary to verify that it's monotone increasing- which means that showing that the limits from below and from above at the key points are the same, in this case. [/quote]
One can do a little worse; the limit from the left just needs to be at most the limit from the right.

I ridiculusly did the problem wrong.....
I used integral test and came to the conclution that it converges... :blush:

[quote="jmerry"]It's necessary to verify that it's monotone increasing[/quote]
Hm, this seemed obvious to me so I didn't explain it. I mean, just note that $ f$ is increasing on $ [1, e]$ and that $ f$ is greater than 1 when $ x > e$ so if $ x > e$ then $ xf(x)$ is a product of increasing functions and thus still increasing, and also doesn't drop at $ x\equal{}e$. This works for the first interval ($ [e, e^e]$) and you can just bootstrap up with the previous result to get the next one and cover everything. I hope they don't kill me for just claiming that it's increasing...

Could someone tell me what I did wrong?

$ \sum_{n\equal{}1}^{\infty} \frac{1}{f(n)} \leq \int_1^{\infty} \frac{1}{f(x)} dx \equal{} \int_1^e \frac{1}{x} dx \plus{} \int_e^{e^e} \frac{1}{f(x)} dx \plus{} \int_{e^e}^{e^{e^e}} \frac{1}{f(x)} dx \plus{} \cdots$

$ \int_1^e \frac{1}{x} dx \equal{} 1$

Let $ x$ be from $ 1$ to $ e$.

Then define $ f_n(x) \equal{} f(e^{e^{e^{\cdots^x}}})$ where there are $ n$ $ e$'s and $ f_0(x) \equal{} f(x)$.  Thus $ f_n(x) \equal{} e^{e^{\cdots^x}} f_{n\minus{}1}(x)$.

$ f_n(x) \geq e^{e^{\cdots^x}}f_0(x) \geq e^{e^{\cdots^1}}x$.

$ \int_{e^{\cdots^e}}^{e^{e^{\cdots^e}}} \frac{1}{f(x)} dx \equal{} \int_{1}^{e} \frac{1}{f_n(x)} dx$ where the first tower has $ n$ $ e$'s and the second has $ n \plus{} 1$ $ e$'s.

$ \frac{1}{f_n(x)} \leq \frac{1}{e^{\cdots^e}x}$ and so $ \int_{1}^{e} \frac{1}{f_n(x)} dx \leq \int_1^e \frac{1}{e^{\cdots^e}x} dx \equal{} \frac{1}{e^{\cdots^e}}$.

Thus $ \sum_{n\equal{}1}^{\infty} \frac{1}{f(n)} \leq 1 \plus{} \frac{1}{e} \plus{} \frac{1}{e^e} \plus{} \frac{1}{e^{e^e}} \plus{} \cdots$ which converges.

[quote="peter_"]$ \int_{e^{\cdots^e}}^{e^{e^{\cdots^e}}} \frac {1}{f(x)} dx \equal{} \int_{1}^{e} \frac {1}{f_n(x)} dx$[/quote]
This step is not correct; you're changing variables, so you need to multiply by the appropriate conversion factor.  I'm not actually sure if your recursion is set up properly; you should be getting that this integral is equal to $ \int_1^x \frac{1}{f(x)} \, dx$.

Claim: $f$ is increasing over $\mathbb{R}_{\ge 1}$.

Proof: Fix $x>y$. Let $y_0=y, y_i=\log(y_{i-1})$ if $y_{i-1}\ge e$. Define $x_i$ similarly. Let $k$ be the maximum $k$ such that $y_k$ is defined. Since $x_j>y_j$ for all $0\le j\le k$ it follows that $$f(y) = \prod\limits_{j=0}^k y_j < \prod\limits_{j=0}^k x_j \le f(x)$$

This means we can apply the integral test, and only need to worry about whether $$\int_1^{\infty} \frac{1}{f(x)} dx$$ diverges or not.

To make $f(x)$ easier to describe, we find the intervals when $f(x)=x, x\log x, x\log x \log \log x$ and so on, so on.

Let $L$ be the iterated logarithm (i.e. $L_k(x) = L_{k-1}(\log x)$)

Let $x_0=1, x_j=e^{x_{j-1}}$ for all $j\in \mathbb{N}$. Then $f(x) = \prod\limits_{j=0}^k L_j(x)$ if and only if $x\in [x_k, x_{k+1})$.

Thus,

$$\int_1^{\infty} \frac{1}{f(x)} dx = \sum\limits_{n\ge 0} \int_{x_n}^{x_{n+1}} \frac{1}{x L_1(x) L_2(x) \cdots L_n(x)} dx $$

To compute $$\int_{x_n}^{x_{n+1}} \frac{1}{x L_1(x) L_2(x) \cdots L_n(x)} dx $$

Let $x=e^y$ then $dx=e^y dy$. Thus,

$$\int_{x_n}^{x_{n+1}} \frac{1}{x L_1(x) L_2(x) \cdots L_n(x)} dx  = \int_{y=x^{n-1}}^{x_n} \frac{1}{e^y y L_1(y) \cdots L_{n-1}(y)} e^y dy = \int_{x_{n-1}}^{x_n} \frac{1}{x L_1(x) L_2(x) \cdots L_n(x)} dx$$

Thus, all terms in the sum are equal, so this integral diverges.


geometry
circumcircle
trigonometry
geometry proposed

$\triangle  ABC$ is equilateral with side-length $a.$ Let $(O,r)$ and $(O,R)$ be the incircle and cirumcircle of $\triangle ABC.$ $P$ is a point on $(O,r)$ and $X,Y,Z$ are the projections of $ P$ onto $ BC,$ $CA,$ $AB.$ Circles $ \mathcal{T}_1, \mathcal{T}_2$ and $ \mathcal{T}_3$ are tangent to $BC,CA,AB$ through $X,Y,Z$ and tangent to $(O,R)$ internally (their centers lie on different sides of BC,CA,AB WRT A,B,C). Prove that the sum of the lengths of the common external tangents of  $ \mathcal{T}_1, \mathcal{T}_2$ and $ \mathcal{T}_3$ is a constant value equal to  $ \frac{_{35}}{^{16}}a.$

$ R$ is circumradius, $ r = R/2$ inradius. $ X, Y, Z$ are feet of perpendiculars from $ P \in (I)$ to $ BC, CA, AB.$ Area of the pedal triangle $ \triangle XYZ$ is
${{ [\triangle XYX] = \frac {p(P, (O))}{4R^2} [\triangle ABC] = \frac {R^2 - r^2}{4R^2} }}[\triangle ABC] = \frac {_3}{^{16}}[\triangle ABC].$ Then
$ AY \cdot AZ + BZ \cdot BX + CX \cdot CY = \frac {2}{\sin 60^\circ} \left([\triangle ABC] - [\triangle XYZ]\right) = \frac {_{13}}{^{16}} a^2.$
Let $ t_{23}, t_{31}, t_{12}$ be the external tangent lengths of the pairs $ \mathcal T_2, \mathcal T_3$; $ \mathcal T_3, \mathcal T_1$; $ \mathcal T_1, \mathcal T_2$. Inversion with center $ A$ and power $ a^2$ takes $ BC$ to $ (O)$ and the other way around, therefore it takes $ \mathcal T_1$ to itself $ \Longrightarrow$  $ \mathcal T_1$ is perpendicular to the inversion circle $ \Longrightarrow$ tangent length from $ A$ to $ \mathcal T_1$ is $ a.$ By Casey theorem for points $ B, C$ and circles $ \mathcal T_2,\mathcal T_3,$ all tangent to $ (O),$

$ a^2 = BC \cdot t_{23} + CY \cdot BZ = a \cdot t_{23} + (a - AY)(a - AZ).$

Summing this and two other equation obtained by cyclic exchange,

$ 3a^2 = a (t_{23} + t_{31} + t_{12})$ $ + \ 3a^2 - a(AY + AZ + BZ + BX + CX + CY) + AY \cdot AZ + BZ \cdot BX + CX \cdot CY,$

$ t_{23} + t_{31} + t_{12} = 3a - \frac {_{13}}{^{16}}a = \frac {_{35}}{^{16}} a$

Thanks for your nice solution. Mine is almost the same  :P

Denote $\delta_1$ the tangent segment form $A$ to $\omega_1.$ By Casey's theorem for $(A,B,C,\omega_1)$ we have $L \cdot BA'+L \cdot CA'=\delta_1 \cdot L$ $\Longrightarrow$ $\delta_1=L.$ Similarly, we have $\delta_2=\delta_3=L.$ By Euler's theorem for the pedal triangle $\triangle XYZ$ of $P$ we get:

${{ [XYZ] = \frac {p(P, (O))}{4R^2} [ABC] = \frac {R^2 - r^2}{4R^2} }}[ABC] = \frac {_3}{^{16}}[ABC] \ .$ Therefore

$ AY \cdot AZ + BZ \cdot BX + CX \cdot CY = \frac {2}{\sin 60^\circ} \left([ ABC] - [XYZ]\right) = \frac {_{13}}{^{16}} L^2. \ (*)$

By Casey's theorem for $(B,C,\omega_2,\omega_3),$ we get

$ \delta_2 \cdot \delta_3 = L^2 = BC \cdot \delta_{23} + CY \cdot BZ = L \cdot \delta_{23} + (L - AX)(L - AZ). $

By cyclic exchange we have the expressions

$L^2= L \cdot \delta_{31}+(L-BZ)(L-BX) \ , \ L^2=L \cdot \delta_{12} +(L-CX)(L-CY)$

Adding the three latter equations yields

$ 3L^2 = L (\delta_{23} + \delta_{31} + \delta_{12}) + \ 3L^2 - 3L^2 + AZ \cdot AY + BZ \cdot BX + CX \cdot CY$

Combining with $(*)$ we obtain $ \delta_{23} + \delta_{31} + \delta_{12} = 3L - \frac {_{13}}{^{16}}L = \frac {_{35}}{^{16}} L.$


geometry
incenter
geometric transformation
homothety
inradius
ratio
geometry proposed

On $BC$ side of $ABC$ triangle, take $D,E$ such that the three incircles of non overlap triangles $ABD,DAE,EAC$ are equals. Prove that: $\frac{AD}{AE}=\frac{BE}{DC}$

[b]Lemma:[/b] Let D be an arbitrary point on the side (segment) BC of a $ \triangle ABC.$ Let I, J be the incenters of $ \triangle ABD,\ \triangle ADC$ and E, F the excenters of these 2 triangles. The the center lines IJ, EF meet at a point X on the line BC.

Let IJ meet BC at X. X is the external homothety center of the circles (I), (J) and A is the external homothety center of the circles (I), (E). The external homothety center Y of the circles (J), (E) lies of the external homothety axis AX of (I), (J), (E). A is also the external homothety center of the circles (J), (F), so that the external homothety center of the circles (E), (F), the intersection of their center line EF with their common external tangent BC, lies on the external homothety axis AY of (J), (E), (F). But $ AY \equiv AX$ meets BC at X. QED.

In particular, the excircles (E), (F) are congruent iff the incircles (I), (J) are congruent. The common external homothety center $ X \in BC$ of these 2 circle pairs is then at infinity.

Let $ \rho$ be the equal inradii of $ \triangle ABD,\ \triangle AEC$ and $ \rho'$ inradius of $ \triangle ADE.$ (For now, we do not need $ \rho' = \rho.$) Let $ r_{e},\ r_{d}$ be inradii of $ \triangle ABE,\ \triangle ADC,$ let $ r_{ae},\ r_{ad}$ be their exradii against A, and let $ h_{a}$ be the common A-altitude of these 5 triangles. Since the inradii of $ \triangle ABD,\ \triangle AEC$ are equal, we have

$ \frac {1}{\rho} + \frac {1}{\rho'} - \frac {r_{e}}{\rho \rho'} = \frac {2}{h_{a}} = \frac {1}{\rho} + \frac {1}{\rho'} - \frac {r_{d}}{\rho \rho'}$

$ \frac {1}{\rho}\left(1 - \frac {r_{e}}{\rho'}\right) = \frac {1}{\rho}\left(1 - \frac {r_{d}}{\rho'}\right),\ \ \ \rho' - r_{e} = \rho' - r_{d}$

Thus the inradii $ r_{d} = r_{e} = r$ are equal and according to the above lemma, the exradii $ r_{ad} = r_{ae} = r_{a}$ are also equal. Assume now that the inradii $ \rho' = \rho$ are equal. Let $ s_{e},\ s_{d}$ be semiperimeters of $ \triangle ABE,\ \triangle ADC.$

$ s_{e}r = |\triangle ABE| = |\triangle ABD| + |\triangle ADE| = (s_{e} + AD) \rho$

$ \frac {r}{\rho} = \frac {s_{e} + AD}{s_{e}}$

Substituting this to the inradii relation for $ \triangle ABE$ with $ D \in BE$ (multiplied by r and then by $ s_{e}^{2}$),

$ \frac {2r}{h} = \frac {2r}{\rho} - \frac {r^{2}}{\rho^{2}}$

$ \frac {BE}{s_{e}} = 2\ \frac {s_{e} + AD}{s_{e}} - \frac {(s_{e} + AD)^{2}}{s_{e}^{2}}$

$ s_{e}\cdot BE = 2s_{e}(s_{e} + AD) - (s_{e} + AD)^{2} = s_{e}^{2} - AD^{2}$

$ AD = \sqrt {s_{e}(s_{e} - BE)}$

In exactly the same way, we get $ AE = \sqrt {s_{d}(s_{d} - CD)}$ and

$ \frac {AD}{AE} = \sqrt {\frac {s_{e}(s_{e} - BE)}{s_{d}(s_{d} - CD)}}$

Substituting the incircle/excircle ratios $ \frac {r}{r_{a}} = \frac {s_{e} - BE}{s_{e}} = \frac {s_{d} - CD}{s_{d}}$ leads to

$ \frac {AD}{AE} = \sqrt {\frac {s_{e}^{2}}{s_{d}^{2}}\cdot \frac {r}{r_{a}}\cdot \frac {r_{a}}{r}} = \frac {s_{e}}{s_{d}}$

But $ \frac {s_{e}}{s_{d}} = \frac {rs_{e}}{rs_{d}} = \frac {|\triangle ABE|}{|\triangle ADC|} = \frac {BE}{CD}$

so that $ \frac {AD}{AE} = \frac {BE}{CD}.$

[quote="yetti"]

$ \frac {1}{\rho} \plus{} \frac {1}{\rho'} \minus{} \frac {r_{e}}{\rho \rho'} \equal{} \frac {2}{h_{a}} \equal{} \frac {1}{\rho} \plus{} \frac {1}{\rho'} \minus{} \frac {r_{d}}{\rho \rho'}$

[/quote]
If you don't mind, could you please explain how you got this?  :oops:

Hint for another solution :
If A incircle and excircle tangent to CA at S and T then AT*SA = EA*AE
You can prove this by using homotety


$ (X_n)$ are independent and uniform on $ (0,1)$ random variables.
Does the series $ \sum_{n\equal{}1}^{\plus{}\infty}Y_n$ converge almost surely?
Where $ Y_n\equal{}\frac{X_1^2\plus{}...\plus{}X_n^2}{X_1\plus{}...\plus{}X_n}$.

Since $ Y_n\ge X_1^2/n$ and $ X_1>0$ a.s., the convergence of the series as written is highly unlikely. You certainly meant something else to ask. What was it? :?

Sorry, the random series is
$ \sum_{n \equal{} 1}^{ \plus{} \infty}\left(Y_n \minus{} \frac {1}{3}\right)$.

No, the exercise is easy.
If the random series is $ \sum_{n \equal{} 1}^{ \plus{} \infty}\left(Y_n\minus{}2/3 \right)$, i think it is too hard to prove the convergence.


topology
calculus
floor function
LaTeX
set theory
combinatorics unsolved
combinatorics

Let $ A_1, A_2, A_3, \ldots$ be sets such that $ A_1 \cap A_2 \cap \ldots \cap A_n \neq \emptyset$ holds for all $ n$. Must it be that $ \bigcap_{n\equal{}1}^{\infty} A_n \neq \emptyset$?

Not sure this is really a combinatorics question -- the intervals $ [n, \infty)$ are a simple counter-example.

Apologies, I wasn't too sure which topic set theory fell under. By the intervals $ [n, \infty )$ you mean $ A_n \cap A_{n\plus{}1} \cap \ldots \cap A_{\infty}$?

No, I mean let $ A_n \equal{} [n, \infty)$.  Or if you prefer, $ A_n \equal{} (0, \frac {1}{n})$ is essentially the same example and works (with a small tweak) in every metric space.

I'm actually not sure where I would have put this topic, either -- I feel like it's the sort of question most likely to show up in an intro topology or analysis class, or at least it has that flavor to it, so I probably would have put it in one of the calculus fora.  But it's not really calculus, either.  And despite being "set theoretic," it's not about foundations in the sense that word usually means.  Ah, well, such is (mathematical) life  :wink:

Follow-up question: the two examples I gave have the property that every finite subcollection has nonempty intersection while every infinite subcollection has empty intersection.  This appears at first sight to be stronger than the condition that you gave (in which we ask only that the intersections of the first $ n$ be nonempty and that the intersection of all of them be empty).  Is it actually stronger (in which case, can you find an example that satisfies one condition but not the other) or are they equivalent?

[quote="JBL"]This appears at first sight to be stronger than the condition that you gave (in which we ask only that the intersections of the first $ n$ be nonempty and that the intersection of all of them be empty).  Is it actually stronger (in which case, can you find an example that satisfies one condition but not the other) or are they equivalent?[/quote]

[hide]I feel that these conditions are equivalent.
con 1: Every finite collection has a non empty intersection.
con 2: For every finite $ n$,
\[ \bigcap_{i\equal{}1}^{n} A_i\ne \emptyset\]

(a) con 1 $ \Rightarrow$ con2 due to finiteness.
(b) Let the finite collection be $ \{A_{k_1},A_{k_2},\dots A_{k_n}\}$ where $ k_1<k_2<\dots k_m$ (assuming $ A$'s don't repeat, if they do we may remove the duplicate copies as they do not affect intersection) and $ \{k_1,k_2,\dots,k_m\}\subset \{1,2,\dots ,k_m\}$. Since $ \bigcap_{i\equal{}1}^{k_n}\ne \emptyset$, we have $ \bigcap_{i\equal{}1}^{m} A_{k_m}\ne \emptyset$[/hide]

For infinite, it's stronger: you can take something like $ A_n\equal{}(0,1/n)$ for even $ n$ and $ A_n\equal{}(0,1/n)\cup[1,2]$ for odd $ n$.

Why is the previous example stronger? They seem (to me) to be equivalent to the example $ \left(0, \frac{1}{n} \right)$. Please correct me there!

This concept of infinite collections is a little confusing to me since I was just introduced to it in a first-year uni course! If you were thinking of a set of objects instead of the real number line, the example probably wouldn't have hit you.  :maybe:

Well, the first example I gave works just as well for integers as for real numbers.  The key idea is just that you need $ \bigcup_i A_i$ to be infinite and to exclude "not too many" elements of this union from each of the $ A_i$.

The point of zhoraster's example is that if we add the extra interval $ [1, 2]$ to the odd cases, we can have an infinite collection of sets in which
1) every finite intersection is nonempty,
2) the intersection of all the sets is empty, and
3) there exists an infinite subcollection of the sets whose intersection is nonempty (in this case, the collection $ \{A_{2n \plus{} 1} \mid n \in \textbf{N}\}$ which has intersection $ [1, 2]$).

Ah, I see the light. Thanks lot people  :) This, however, seems to say that half of infinite (i.e. all odd sets) is the same as infinite (all sets)!

Sure, and there's a nice bijection between $ \textbf{N}$ and $ \{2n \plus{} 1 \mid n \in \textbf{N}\}$: send $ n \mapsto 2n \plus{} 1$ to go one way and $ m \mapsto \left\lfloor \frac {m}{2}\right\rfloor$.

One characterization of infinite sets is that they are precisely the sets that can be put in bijection with one of their proper subsets.  Another way of putting this is that infinite sets are precisely those sets for which "is a proper subset" does [i]not[/i] imply "is of smaller cardinality."

[Edit: fixed minor LaTeX error]

Nice - This is precisely why I just borrowed a book called "Numbers, Sets and Axioms" - time to clear my doubts about this!


function
calculus
derivative
geometry
geometric transformation
reflection
real analysis

Determine all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ which have an antiderivative on $\mathbb{R}$ and fulfill the following condition:
\[ f(f(x)) = x , (\forall) x \in \mathbb{R}\]
It's easy to see that $f$ is either strictly increasing or decreasing. The first case is easy to work out, but what about the second one?

Manifestly $f$ is bijective, and because of $f$ is the derivative of a function then $f$ must be monotonic.
1) $f$ is increasing .
$x =< f(x) \Longrightarrow f(x) =< f(f(x))=x \Longrightarrow f(x) = x $
$x >= f(x) \Longrightarrow f(x) >= f(f(x))=x \Longrightarrow f(x) = x $
So $f(x) = x , \forall x $
2) $f$ is decreasing .
$x =< f(x) \Longrightarrow f(x) >= f(f(x))=x \Longrightarrow f(x) = x $
$x >= f(x) \Longrightarrow f(x) =< f(f(a))=x \Longrightarrow f(x) = x $
So $f(x) = x , \forall x $ , but that is not a decreasing function , so the case with$f$ decreasing  does'nt exist !

:cool:  Do you accept " tertium non datum" ?  :cool:

What about the obvious $f(x)=-x$?

Your argument is simply $x\ge f(x)\Rightarrow f(x)\le x$, and then you say this is a contradiction, but it certainly isn't :).

:blush:  Manifestly I'm mind absent  :blush:

Actually, $f$ is continuous (follows immediately from its Darboux property and it's bijectivity), so the problem is, in fact, finding all continuous functions from $\mathbb R$ to $\mathbb R$ s.t. $f\circ f=1_{\mathbb R}$. Are you sure this is do-able?

We should find all continuous functions which have a graph symmetric wrt the line $x=y$. I, for one, can imagine lots and lots of continuous curves which are symmetric wrt that line and constitute the graph of some function.

[quote="grobber"]What about the obvious $ f(x)=-x $?[/quote]
And, more generally, $f(x)=a-x$ for any specific real $a$.

The set of solutions is even broader than that. Let $g$ be any continuous strictly decreasing function on the whole line. By the Intermediate Value Theorem, there exists $a$ such that $g(a)=a$. Now define $f$ so that for $x\le a, f(x)=g(x)$ and for $x>a, f(x)=g^{-1}(x)$. That works. (Intuitively, just reflect the part of the graph that lies above $y=x$ around that line and glue the pieces together.)

I'm not guaranteeing that that's all, having not explored what might or might not happen if you relax continuity.

[quote]Actually, $ f $ is continuous (follows immediately from its Darboux property and it's bijectivity)[/quote]
Thanks. I wasn't thinking that quickly, and I hadn't read your post when I put mine up. The set of solutions is, as we both suggest, quite large.

[quote="Kent Merryfield"]The set of solutions is even broader than that. Let $g$ be any continuous strictly decreasing function on the whole line. By the Intermediate Value Theorem, there exists $a$ such that $g(a)=a$. Now define $f$ so that for $x\le a, f(x)=g(x)$ and for $x>a, f(x)=g^{-1}(x)$. That works. (Intuitively, just reflect the part of the graph that lies above $y=x$ around that line and glue the pieces together.)[/quote]

Actually, this seems like a reasonable solution to the problem (a very neat way to classify all such functions). Maybe that's what wally's teacher wanted for a solution.


integration
calculus
inequalities proposed
inequalities

I'd like to know how hard this one is, it's in the middle of some easy excercices but it took me really long before I found the sol. (although the solution is easy once you got it)

Let $Q_n = 1 + \frac14 + \frac19 + \cdots + \frac1{n^2}$. Then prove that for $n\geq3$ we have: \[\frac{19}{12} - \frac1{n+1} < Q_n < \frac74 - \frac1n\]

I see. I thought of that method but I could not justify the use of an integral for these distinct values... well I'm not quite familiar with this kind of problem, I'll keep it in mind for next time.

Thanks! :)

I think an elementary way to prove this is:

$\displaystyle Q_n > \left(1 + \frac14\right) + \frac{1}{3\cdot4}+ \cdots + \frac1{n(n+1)} = \frac{19}{12} - \frac1{n+1}$.

and

$\displaystyle Q_n <\left(1 + \frac14\right) + \frac{1}{2\cdot3}+ \cdots + \frac1{(n-1)n} = \frac74 - \frac1n$.


hallo,
prove that the equation $x^{2}+y^{2}+z^{2}=x^{3}+y^{3}+z^{3}$ has infinitely many integer solutions.  :)

hallo,
is there any generalization possible?

Use permutations from $S_{3}$ form $x\to y\to \to z \to x$.

hallo,
[quote]
Use permutations from S_{3} form x\to y\to \to z \to x. [/quote]
i dont think i understand this. what do you mean ? how does your generalization look like ?


algebra
polynomial
linear algebra
matrix
inequalities
algebra unsolved

[quote]Hmm... Maybe you should look through the official solution?[/quote]
And...where would I find the official solution?

I remember I saw it, but now I can't find  :(

[quote="cpu886"]Let a sequence of polynomials be defined by P_1(x) = x^2 - 1,  P_2(x) = 2x^3 - 2x, and (P_i(x))^2 - (x^2 - 1)^2 = P_i-1 (x) * P_i+1 (x). Find the highest power of 2 dividing the sum of the absolute values of the coefficients of P_k(x) for each k.[/quote]

Well, the equation is a 2x2 determinant.  If there is a nice 2x2 matrix $M$ (depending on $x$ only)
with $v_{i+1} = M v_i$,  where $v_i  = (P_{i+1}, P_i)$  then the rest should be easy.  In other words, you can probably solve for $P_i$ explicitly, perhaps in terms of Chebyshev polynomials.


So Mathcounts chapters is almost here........any ideas.... of who will win (school) or how hard it might be compared to last year?

I'm [b]not[/b] sure, but didn't Jordan come in 2nd last year? They probably will get 1st this year then, but there are also lots of other good schools I know of.

[quote="bookaholic"]Yeah, Jordan is good (1st, not 2nd last year--JLS was 2nd). They have two people who regularly come to Paly to beat us at contests. I'm rooting for JLS, obviously, and I still think we have a chance, depending on how the other schools are. Ms. Beddoes is terrified of Nueva but they haven't been that great the past two years.[/quote]

Yeah the two people from Jordan placed first and second at the Miller Math Marathon. They aren't in our chapter I believe, so thats good.


modular arithmetic
number theory proposed
number theory

WLOG, we may assume $a_i > b_i$ or we can just switch $a_i$ and $b_i$.

So the sum just becomes $\sum a_i - \sum b_i$.

But we know $\sum a_i + \sum b_i = \frac{1998 \cdot 1999}{2} = 999 \cdot 1999$ is odd.

So $\sum a_i - \sum b_i$ is odd as well.

Since $\sum a_i - \sum b_i$ is the sum of $999$ $1$'s or $6$'s and is odd, we know there must be an even number of $6$'s. Let $2x$ be the number of $6$'s.

Then $\sum a_i - \sum b_i = 6(2x)+1(999-2x) = 10x + 999 \equiv 9 \pmod{10}$ as desired. QED.

Assume WLOG that $6=b_1-a_1=\cdots=b_k-a_k$ and $1=b_{k+1}-a_{k+1}=\cdots=b_n-a_n$. For $i,j\le k$, we say that $\{a_j,b_j\}$ [b]links onto[/b] $\{a_i,b_i\}$ if $\{a_i,b_i\}$ if $a_i<a_j<b_i<b_j$ or $a_j<a_i<b_j<b_i$.

Since pair $\{a_i,b_i\}$ links onto$\{a_j,b_j\}$ if and only if $\{a_j,b_j\}$ links onto $\{a_i,b_i\}$, there are an even number of ordered pairs of integers $(i,j)$ such that pair $j$ links onto pair $i$.

For a given $i\le k$, there are an odd number of integers between $a_i$ and $b_i$, so the number of pairs that link onto$\{a_i,b_i\}$ is odd. Thus, the total number of ordered pairs of integers $(i,j)$ such that pair $j$ links onto pair $i$ is congruent to
\[ \sum_{i=1}^k(\text{number of pairs that link onto pair \textit{i}}) \equiv \sum_{i=1}^k 1\equiv k \bmod 2.  \] Therefore, since we already know that this number is even, we find that $k$ is even, and the value of the given sum is
\[ 6k+(999-k) = 999+5k\equiv 9\bmod10.  \]

Take the equation modulo 5. Then the sum is $ 999\cdot 1\equiv4\pmod{5}$. As shown above, the equation is odd, so the sum must be 9 mod 10.

Here is one proof. I do not think it's very well written, especially the lemma (did I even need a lemma??) but here it is. (Sorry, I also didn't know how to make the congruence sign, and I was in a hurry, so I just put equal signs.)

[hide="lemma"]
Define $ x$ as the number of pairs where $ \[ |a_{i}\minus{}b_{i}| \equal{} 6$, and similarly define $ y$ as the number of pairs where $ \[ |a_{i}\minus{}b_{i}| \equal{} 1$.

We can prove that $ a \equal{} 0 (mod 2)$.

Assume a was not even. Then, we would have some interval, $ (m, m \plus{} 2n)$, where $ m$ and $ n$ are positive integers, where the middle numbers weren't any $ a_i$ or $ b_i$. We could not assign pairs $ |a_{i}\minus{}b_{i}| \equal{} 1$ to these integers.

So, a must be even.
[/hide]

[hide="main proof"]
Using definitions of $ a$ and $ b$ from the lemma, we know that
$ a \plus{} b \equal{} 999$.
Using our results from the lemma,

$ a \equal{} 0 (mod 2)$
$ \Leftrightarrow 5a \equal{} 0 (mod 10)$
$ \Leftrightarrow 5a \plus{} 999 \equal{} 999 (mod 10)$
$ \Leftrightarrow 5a \plus{} 999 \equal{} 9 (mod 10)$

Using the equation $ a \plus{} b \equal{} 999$, we substitute:

$ 5a \plus{} a \plus{} b \equal{} 9 (mod 10)$
$ \Leftrightarrow 6a \plus{} b \equal{} 9 (mod 10)$

We notice that 6a + b represents the total sum, and it is congruente to 9 modulo 10, i.e. its last digit is nine.

QED

 [/hide]

[hide="Solution"]WLOG $a_i>b_i\forall i$. The sum clearly leaves a remainder of $4$ upon division by $5$, so it suffices to show that it's odd. But this follows from the fact that it has the same parity as the sum of the elements of the original set (since they differ by $2\sum b_i$). $\blacksquare$[/hide]

[hide="Solution by codyj and ithinksomuch"]
Let the sum be $S$. Since $|a_i-b_i|$ is either $1$ or $6$, we have $|a_i-b_i|\equiv1\pmod5$. Clearly, $S\equiv999\equiv4\pmod5$. Also, using the fact that $|a_i-b_i|\equiv a_i+b_i\pmod{2}$, we have that $S\equiv1+2+\dots+1998\equiv1\pmod{2}$. From the Chinese Remainder Theorem, we have $S\equiv9\pmod{10}$, as desired. $\blacksquare$
[/hide]

[hide=" Solution"]

[b]Claim:[/b] $\displaystyle \sum_{i=1}^{999} |a_i-b_i|$ is odd.
[i]Proof:[/i] Note that $|a_j-b_j| \equiv a_j-b_j \pmod{2}$ for all $1 \leq j \leq 999$ since $x \equiv -x \pmod{2}$ for all integers $x$. Hence, it suffices to show that $\displaystyle \sum_{i=1}^{999} a_i-b_i$ is odd, or equivalently, $\displaystyle \sum_{i=1}^{999} a_i$ and $\displaystyle \sum_{i=1}^{999} b_i$ have different parities. Assume otherwise. This implies $\displaystyle \sum_{i=1}^{999} a_i + \displaystyle \sum_{i=1}^{999} b_i$ is even. However,
\[\displaystyle \sum_{i=1}^{999} a_i + \displaystyle \sum_{i=1}^{999} b_i = \displaystyle \sum_{i=1}^{1998} i =  999 \times 1999\]
is odd, which is a contradiction. $\blacksquare$

By the claim, the number of odd terms in the set $\{|a_i-b_i|\}_{i=1}^{n}$ is odd. Since the only possible values $|a_i-b_i|$ can take are 1 and 6, the number of $i$ satisfying $|a_i-b_i|=1$ is odd. This implies the number of $i$ satisfying $|a_i-b_i|=6$ is even. Let the number of integers $i$ satisfying $|a_i-b_i|=6$ be $x$. Then, our desired sum is simply $6x + 999-x= 5x + 999.$ We wish to show that $5x+999 \equiv 9 \pmod{10} \iff 5x \equiv 0 \pmod{10},$ which is true since we have already shown that $x$ is even. $\blacksquare$ [/hide]

[hide=Solution]Note that the set $\{1,2,\cdots, 1998\}$ contains $999$ even numbers and $999$ odd numbers. If $|a_i-b_i|=6$ then clearly $a_i\equiv b_i\pmod2$.Otherwise, if $|a_i-b_i|=1$, then we must have $a_i\equiv b_i+1\pmod 2$.Since the amount of each parity is odd, there will be an odd number of pairs $(a_i,b_i)$ such that $|a_i-b_i|=1$. Suppose there are $k$ pairs $(a_i,b_i)$ such that $|a_i-b_i|=1$. Then \[\sum_{i=1}^{999} |a_i-b_i| \equiv k + (999-k)\cdot 6 \equiv 4-5k\pmod {10}.\] However, since $k$ is odd, this is essentially $9 \pmod {10}$, as desired. $\blacksquare$[/hide]

Ok my sol was overkill:
Here are the two lemmas. First has been proved by many of above solutions, second is true by construction. (just need to describe it nicely.")

[hide= proof sketch]

Lemma 1: The statement is equivalent to proving that the number of i for which $\mid a_i - b_i \mid$ is even.

Lemma 2: Consider the interval of numbers from 1 to 1998 on the number line and a valid distribution of 2-spanning segments and 7-spanning segments, with all numbers being associated with exactly one vertex. Define a n-chain of 7-spanning segments to be a set of n overlapping segments in a valid configuration. Only even n are valid.

Since n is even, summing over all n give that the number of i for which $\mid a_i - b_i \mid$ = 6 is even, hence proving lemma 1 and the problem statement.
[\hide]

Let $k$ of the differences be six, and let the rest be one. Let $A$, $B$, and $C$ denote the number of even-even, even-odd, and odd-odd pairs. Clearly, there are $B+2A=999$ evens, hence $B$ is odd. It follows that $k$ is even, so $5k+999\equiv 9 \mod 10$ as desired. $\square$

[hide=Solution]
Taking this mod 2, it is basically the same as $1+2+...+1998$ mod 2 which is 1 mod 2. And taking it mod 5 is obvious that it will be 4 mod 5. Thus, it must end in the digit 9 since it is 9 mod 10.
[/hide]

Let $S$ be the sum. Modulo $2$, \[ S = \sum |a_i-b_i| \equiv \sum (a_i+b_i) 	= 1 + 2 + \dots + 1998 \equiv 1 \pmod 2. \] Modulo $5$, \[ S = \sum |a_i-b_i| = 1 \cdot 999 \equiv 4 \pmod 5. \] So $S \equiv 9 \pmod{10}$.

I had the same solution as @above, but I couldn't get the $1$ modulo $2$ part. Why can we add the numbers like that? (I don't really get the $1+2+...+1998$ part).

@above It's because $x\equiv -x\pmod{2}$ for all integers $x$, so $|a_i-b_i|\equiv a_i-b_i\pmod{2}$ regardless of whether $a_i\ge b_i$.

Mod 5, the expression is obviously 4 mod 5. Now observe that $$|a_1-b_1| \equiv a_1+b_1 \pmod 2,$$ implying the desired sum is congruent to $$\sum_{i=1}^{1998} i \equiv 1 \pmod 2.$$ Combining gives the result.

WLOG let $a_i>b_i$ and let $a_i-b_i=d_i$. Then $d_i$ equals $1$ or $6$, and we only need to show that $$\sum_{i=1}^{999}d_i\equiv 1\pmod 2.$$
Notice that $$\sum_{i=1}^{999}d_i+2\sum_{i=1}^{999}b_i=\sum_{i=1}^{999}a_i+\sum_{i=1}^{999}b_i\equiv 1\pmod 2,$$ and we are done. $\blacksquare$

The number $6$ comes 'even-even' or 'odd-odd'. We claim that the number of $6$'s must be even. If odd then we get greater than $999$ odd or even number, this is a contradiction because $\{1,2,\cdots 1998\}$ have $999$ even and $999$ odd. So
\begin{eqnarray*}
\sum _{i=1}^{999} |a_i-b_i|=6\times 2k+1\times(999-2k)\equiv 10k+999\equiv 9 \hspace{6mm} \text{(mod 10)}
\end{eqnarray*}

By CRT, we can break this into $\pmod{2}$ and $\pmod{5}.$ For $\pmod{2},$ note that $$|a_1-b_1| \equiv a_1+b_1 \pmod{2},$$ since absolute value doesn't matter in $\pmod{2}.$ So our sum is just $1+2+...+1998 \equiv 1 \pmod{2}.$ For $\pmod{5},$ just notice that its always $1\pmod{5}$ for all of $|a_1-b_1|, |a_2-b_2|, ...$ so our answer is just $4\pmod{5}.$ Using CRT, we get an answer of $9\pmod{10}.$ $\blacksquare$

Note that $|a - b| \equiv a - b \equiv a + b \pmod 2$ for all integers $a, b$.

The original sum then becomes $$|a_1 - b_1| + \cdots + |a_{999} - b_{999}| \equiv a_1 + b_1 + \cdots + a_{999} + b_{999} = 1 + \cdots + 1998 \equiv 1 \pmod 2. $$

Now we have that the sum $$|a_1 - b_1| + \cdots + |a_{999} - b_{999}| \equiv \underbrace{1 + \cdots + 1}_{\text{999 times}} = 999 \equiv 4 \pmod 5. $$

Hence this sum is $1 \pmod 2$ and $4 \pmod 5$, so by CRT we have that the sum is $9 \pmod {10}$, as desired. $\blacksquare$

Let the sum be $S$. Note that $S\equiv 999\cdot 1\mod 5\Longleftrightarrow S\equiv 4\mod 5$. Also note that $S$ has the same parity as $a_1+b_1+a_2+b_2+\ldots+a_{999}+b_{999}=1+\ldots+1998$ which clearly odd, so $S\equiv 1\mod 2$. Hence the remainder of $S\mod 10$ is unique by the Chinese Remainder Theorem, and it is easy to check that it is $9$ indeed.

[quote=Alcumusgrinder07]We will consider the sum mod 2 and mod 5 we see that since $6\equiv 1\pmod 2$ the sum of 999 terms has a remainder of 1 when divided by 2 similarly we see that $6\equiv1\pmod 5$ so the sum of 999 terms has a remainder of 4 when divided by 5 so by CRT we see that the sum has a remainder of 9 when divided by 10 which implies that the units digit is 9[/quote]

$6\equiv 1\mod 2$???

Note that the expression is $4 \pmod 5$, so it suffices to show that it is odd. But notice that we have $$\left|a_i-b_i\right| \equiv a_i-b_i \pmod 2,$$ and we have $1+2+3+\dots+1998 \equiv 1 \pmod 2$, which finishes.

Since $1$ and $6$ differ by $5$, the ending digit is $9$ whenever there is an odd number of $i$ such that $|a_i-b_i|=1$ and is $4$ otherwise. There are $999$ odds and $999$ evens so there can't be an even number of $i$ such that $|a_i-b_i|=1$, because that would imply an even number of odds and an even number of evens. Therefore, the ending digit must be $9$.

[quote=MithsApprentice]Suppose that the set $\{1,2,\cdots, 1998\}$ has been partitioned into disjoint pairs $\{a_i,b_i\}$ ($1\leq i\leq 999$) so that for all $i$, $|a_i-b_i|$ equals $1$ or $6$. Prove that the sum \[ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|  \] ends in the digit $9$.[/quote]

my teach gimme that as homework
[hide]
$[ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| \] \equiv \sum^{999}_{i=1}a_i+b_i = \sum^{1998}_{i = 1}i \equiv 1\pmod 2$
$[ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| \] \equiv \sum^{999}_{i=1}1 = 999 \equiv 4 \pmod 5$
so as CRT we got the desired answer.
[/hide]

Suppose there are $x$ pairs with difference $1$. Then our sum is \[(999-x)+6x = 999+5x,\] which has units digit either $4$ or $9$.

Note that this sum also has the same parity as $1 + 2 + \ldots + 1998$, which is odd, since \[\lvert a_i - b_i \rvert \equiv a_i - b_i \equiv a_i + b_i \mod 2.\]

Thus our sum has units digit $9$. $\blacksquare$

Clearly, the desired sum is $4\pmod 5$, so it remains to show that it’s $1\pmod 2$.

Because $|a-b|\equiv a-b\equiv a+b\pmod 2$ for any integers $a$ and $b$, this means
\[|a_1-b_1|+|a_2-b_2|+\dots+|a_{999}-b_{999}|\equiv 1+2+\dots + 1998\equiv 1\pmod 2\]
and we are done. $\square$

We note that this sum is $\equiv 4\pmod{5},$ we need to prove that this is $1 \pmod{2},$ note that $|a-b|,$ has the same parity as $a+b,$ so we need to prove $a+b \equiv 1\pmod{2},$ so $|a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|  \equiv 1+2+3...+1998 \equiv 1\pmod{2},$ hence proved $\blacksquare$

WLOG assume $a_i>b_i$ for each $i$.

Notice that

\[\sum a_i + \sum b_i = 999 \cdot 1999,\]

which is odd. Thus, in order to have integer values, we must have that $\sum a_i - \sum b_i$ is odd. Thus, the number of $6$'s, denote as $2k$, is even. We have

\[\sum a_i - \sum b_1 = 6 \cdot 2k + (999-2k) = 10k+999 \equiv 9 \pmod{10}. \ \square\]

WLOG $a_i>b_i$ for all $i$.

Now note that,
\[ \displaystyle\sum |a_i-b_i| = \displaystyle\sum a_i - b_i \equiv \displaystyle\sum a_i + b_i = \dfrac{1998\cdot 1999}{2} \equiv 1 \pmod{2}. \]

Also,
\[ \displaystyle\sum a_i - b_i \equiv \displaystyle\sum 1 = 999 \equiv 4 \pmod{5}. \]

Thus combining these by CRT, we get that $\displaystyle\sum |a_i-b_i| \equiv 9 \pmod{10}$. :yoda:

Cute.
[hide=Solution]
Note that each term $|a_i - b_i| \equiv 1 \pmod 5$, so $\sum_{i = 1}^{999} |a_i - b_i| \equiv 4 \pmod 5$. Which means the sum must end with either $4$ or $9$. We show its the latter.
Now let $S = \sum_{i = 1}^{999} (a_i + b_i) = \frac{(1998)(1999)}{2} = 999(1999)$.
Clearly $S$ is odd, and note that to get the desired sum we change each $b_i$ to $-b_i$, but at each step, that only changes $S$ to $(S - 2b_i)$, keeping the parity of $S$ unchanged, thus our desired sum must also be odd, so it can't end with $4$.
Thus, it must end with $9$. [/hide]

$|a_i - b_i| = 1$ or $6$. Say $|a_i - b_i| = 1$ for $k$ of these $999$ pairs. The sum would then be equal to $k + 6(999 - k) = 5994 - 5k$. Note that a pair with difference $6$ has same parity and a pair with difference $1$ does not. All pairs of difference $6$ would give us an even number of even numbers and an even number of odd numbers. If $k$ was even, that suggests there are even numbers of odd numbers and even numbers of even numbers. However, since there is a total of $999$ even numbers and $999$ odd numbers, this cannot be true. So thus, $k$ is odd, and the sum would end with digit $9$.

Overkill solution, but proud of it  :play_ball:

For the sake of simplicity, we will write all pairs $\{a_i, b_i\}$ such that $a_i < b_i$.

Let there be $x$ six-pairs $(a_i, b_i)$ such that $|a_i - b_i|=6$. Then, there are $999-x$ one-pairs such that $|a_i - b_i| = 1$. The given expression evaluates to $$6x + (999-x) = 5x + 999$$
For this to be $9 \pmod{10}$, it suffices to show that $x$ is even.

Note that along with any six-pair $(n, n+6)$, in order to make successful pairings, we must have either exactly one of the six-pairs $$(n+1, n+7), (n+3, n+9), (n+5, n+11),$$or one of the following sets of three six-pairs $$\{(n+1, n+7), (n+2, n+8), (n+3, n+9)\}$$ $$\{(n+1, n+7), (n+2, n+8), (n+5, n+11)\}$$ $$\{(n+1, n+7), (n+4, n+10), (n+5, n+11)\}$$

Each option generates an even number of six-pairs, so $x$ is even, as desired $\blacksquare$


limit
trigonometry
calculus
calculus computations

Suppose that $ f$ is continuous and$ \underset{n\rightarrow\infty }{\lim}f\left( n\right) \equal{} l$,then$ \underset{x\rightarrow\infty}{\lim }f\left( x\right) \equal{} l$

1. isnt true

$ f(x)\equal{}\sin {\pi x}$ is a counterexample

[quote="kenn4000"]$ f(x) \equal{} \sin {\pi x}$ is a counterexample[/quote]
this is not a counterexample!
$ f(x\plus{}1)\minus{}f(x) \equal{}\sin(\pi\,{x}\plus{}\pi)\minus{}\sin(\pi{x})\equal{}\minus{}\sin(\pi\,{x})\minus{}\sin(\pi{x})\equal{}\minus{}2\,\sin(\pi{x})$
and $ \mathop{\lim}\limits_{x\to\infty}\minus{}2\,\sin(\pi{x})$  does not exists!

[quote="§outh§tar"]grigkost, read what kenn4000 wrote again.[/quote]
Thanks SouthStar!
I thought that it is a counterexample for 2 ( Sorry kenn4000 )!
By the way I think the proposition 2 it is true.