Euler
modular arithmetic
AMC
AIME
number theory

One of Euler's conjectures was disproved in then 1960s by three American mathematicians when they showed there was a positive integer $ n$ such that \[133^5 \plus{} 110^5 \plus{} 84^5 \plus{} 27^5 \equal{} n^5.\] Find the value of $ n$.

[hide="Answer"]144. I read about this a while back  :P Don't know how you would get that though.[/hide]

Like the subject of this thread? lol.  :D

[hide="Discover the coolness of Mods"]I'm talking about modular arithmetic...  :P

First off, we know that the units digits of any number $k$ and $k^5$ are the same. Thus, the units digit of $n$ must be equal to $3+0+4+7=4$. Since modular arithmetic seems to be working well for us, let's continue. Notice that $133^5+110^5+84^5+27^5\equiv (-1)^5+1^5+0+0\equiv 0\pmod 3$, which means that $n$ is a multiple of $3$.

With this info, let's find $n$. It's obvious that $n>133$, so the only possibilities are $n=144$ or $n=174$. With some examination, we see that 174 is too large, so the only possible answer is $n=\boxed{144}$.[/hide]

Edit: Guess I was a little late, though I did just go for a jog.  :D

using mod 10 and the to the 5th power thingy we know it ends in 4, using intuition we can guess that it's 144, it's not that hard if you think for around half a minute (134 is obviously too small, 154 just looks too big)

[quote="paladin8"][quote="vvr1590"][quote="4everwise"]Like the subject of this thread? lol.  :D

With this info, let's find $n$. It's obvious that $n>133$, so the only possibilities are $n=144$ or $n=174$. With some examination, we see that 174 is too large, so the only possible answer is $n=\boxed{144}$.[/hide]

Edit: Guess I was a little late, though I did just go for a jog.  :D[/quote]

um, why can't it be greater than 200?[/quote]

Well if you test $174$ and it's too big, that should solve your problem.[/quote]

How would you test 174 without a calculator(since this was an AIME problem)?

See K81o7's post; it has the full solution. Just use some weak bounds and you can see that it's way too big.

This is a really late post, sorry  :(

[hide="Solution"]

Taking the equation with various mods, $n \equiv 0 \bmod{3}, 0\bmod{4}\Rightarrow n \equiv 0 \bmod{12}.$ Thus, we can say that $n = 144, 156,$ etc. We clearly see that $178$ itself is too big, as it greatly overshadows the sum of the left numbers (remember that we are taking 5th powers here, and believe it or not, $145^{5}-144^{5}> 1000000$). Likewise, $156$ is too big, so the answer must be $144.$ Seeing that $156$ wasn't too hard; you just have to see that the answer was really close to $130$ and had to be a multiple of $12,$ which wasn't that hard to arrive at.

[/hide]

@K81o7 - i believe that it should be $\cdots < 27^{5}\left(1+4^{5}+5^{5}+5^{5}\right) < 27^{5}\cdot 7500 < 27^{5}\cdot 6^{5}$  but other than that nice soln.

I believe the conjecture was that for any $n\geq 3$, it is impossible to have a sum of $n-1$ $n$th powers equal to another $n$th power.  It was an extension of Fermat's Last Theorem.

please! why was my previous post rated spam? my rating is decreasing day by day...i just wanted to know if the conjecture was something like "the sum of four 5-th powers is never a fifth power or like "the equation $133^{5}+110^{5}+84^{5}+27^{5}=n^{5}$ has no positive integer solutions" etc.

EDIT: sorry :blush: E^(pi*i)=-1, didn't see your post there.

[quote=nayel]what was the original conjecture? :blush:[/quote]

Double post but [url=https://en.m.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture]see here[/url]

[quote=OlympusHero][hide=Solution]Note that $n \equiv 0 \pmod 2$, $n \equiv 0 \pmod 3$, $n \equiv 4 \pmod 5$, and $n \equiv 4 \pmod 7$. The solution to these congruences is $n \equiv 144 \pmod {210}$. Clearly $354$ and anything above is too big, so the answer is $\boxed{144}$.[/hide][/quote]

Ayo nice 14 year bump


inequalities
function
inequalities proposed

Let $ a,b,c$ be positive real number
Prove that
$ (a\plus{}b\plus{}c)[\frac{a}{(2a\plus{}b\plus{}c)(b\plus{}c)}\plus{}\frac{b}{(2b\plus{}c\plus{}a)(c\plus{}a)}\plus{}\frac{c}{(2c\plus{}a\plus{}b)(a\plus{}b)}] \ge \frac{9}{8}$.

[quote="soruz"]Let $ a,b,c$ be positive real number
Prove that
$ (a \plus{} b \plus{} c)[\frac {a}{(2a \plus{} b \plus{} c)(b \plus{} c)} \plus{} \frac {b}{(2b \plus{} c \plus{} a)(c \plus{} a)} \plus{} \frac {c}{(2c \plus{} a \plus{} b)(a \plus{} b)}] \ge \frac {9}{8}$.[/quote]
Let $ a \plus{} b \equal{} z,$ $ a \plus{} c \equal{} y$ and $ b \plus{} c \equal{} x.$ Hence,
$ (a \plus{} b \plus{} c)[\frac {a}{(2a \plus{} b \plus{} c)(b \plus{} c)} \plus{} \frac {b}{(2b \plus{} c \plus{} a)(c \plus{} a)} \plus{} \frac {c}{(2c \plus{} a \plus{} b)(a \plus{} b)}] \ge \frac {9}{8}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {(x \plus{} y \plus{} z)(y \plus{} z \minus{} x)}{x(y \plus{} z)}\geq\frac {9}{2}\Leftrightarrow\sum_{cyc}(x\plus{}y\plus{}z)\left(\frac{1}{x}\minus{}\frac{1}{y\plus{}z}\right)\geq\frac{9}{2}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x \minus{} y)^2\left(\frac {1}{xy} \minus{} \frac {1}{2(z \plus{} x)(z \plus{} y)}\right)\geq0,$ which obviously true.

Since the inequality is homogenuous I suppose $ a\plus{}b\plus{}c\equal{}1$ and the inequality is equivalent to:

$ \sum \frac{a}{1\minus{}a^2} \geq \frac{9}{8}$
Now considering the function $ f: (0,1)\minus{}> \mathbb R$ , $ f(x)\equal{}\frac{x}{1\minus{}x^2}$ (f is convex) from Jensen's inequality we have :

$ \sum \frac{a}{1\minus{}a^2}\equal{}\sum f(a) \geq 3\cdot f\left( \frac{a\plus{}b\plus{}c}{3}\right)\equal{}3\cdot f \left(\frac{1}{3}\right)\equal{}\frac{9}{8}$

[quote="soruz"]Let $ a,b,c$ be positive real number
Prove that
$ (a \plus{} b \plus{} c)[\frac {a}{(2a \plus{} b \plus{} c)(b \plus{} c)} \plus{} \frac {b}{(2b \plus{} c \plus{} a)(c \plus{} a)} \plus{} \frac {c}{(2c \plus{} a \plus{} b)(a \plus{} b)}] \ge \frac {9}{8}$.[/quote]
Let $ a\plus{}b\plus{}c\equal{}3$
Ineq <=> $ \frac {a}{9\minus{}a^2} \plus{} \frac {b}{9\minus{}b^2} \plus{} \frac {c}{9\minus{}c^2} \ge \frac {3}{8}$
We have
$ \frac{a}{9\minus{}a^2}\ge \frac{5a}{32}\minus{}\frac{1}{32}$
this ineq <=> $ \frac{(a\minus{}1)^2(5a\plus{}9)}{32(9\minus{}a^2)}\ge 0$  It is true
So $ \frac {a}{9\minus{}a^2} \plus{} \frac {b}{9\minus{}b^2} \plus{} \frac {c}{9\minus{}c^2} \ge \frac{5(a\plus{}b\plus{}c)}{32}\minus{}\frac{3}{32}\equal{}\frac{3}{8}$
So ineq was proved  :D

[quote="soruz"]Let $ a,b,c$ be positive real number
Prove that
$ (a \plus{} b \plus{} c)[\frac {a}{(2a \plus{} b \plus{} c)(b \plus{} c)} \plus{} \frac {b}{(2b \plus{} c \plus{} a)(c \plus{} a)} \plus{} \frac {c}{(2c \plus{} a \plus{} b)(a \plus{} b)}] \ge \frac {9}{8}$.[/quote]
See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=61695[/url].
I think the following is stronger:
$ \frac{a}{(2a\plus{}b\plus{}c)(b\plus{}c)}\plus{}\frac{b}{(2b\plus{}c\plus{}a)(c\plus{}a)}\plus{}\frac{c}{(2c\plus{}a\plus{}b)(a\plus{}b)}\ge \frac{9}{8\sqrt{3(ab\plus{}bc\plus{}ca)}}$

[quote="ductrung"]I think the following is stronger:
$ \frac {a}{(2a \plus{} b \plus{} c)(b \plus{} c)} \plus{} \frac {b}{(2b \plus{} c \plus{} a)(c \plus{} a)} \plus{} \frac {c}{(2c \plus{} a \plus{} b)(a \plus{} b)}\ge \frac {9}{8\sqrt {3(ab \plus{} bc \plus{} ca)}}$[/quote]
See here: http://canhang2007.wordpress.com/2009/11/16/inequality-72-p-h-duc/


What is the sum of the reciprocals to the 5 solutions to: $ 5x^5 \plus{} 4x^4 \minus{} 3x^3 \plus{} 2x^2 \plus{} x \minus{} 1$?

[hide]The sum of the reciprocals of the roots $ r_1, r_2, r_3, r_5, r_5$ is equal to
the sum of the products of the roots taken 4 at a time over the product of all the roots.

By Vieta's, this is equal to $ \frac{1/5}{1/5} \equal{} 1$[/hide]

[hide]Let $ f(x) = 5x^5 + 4x^4 - 3x^3 + 2x^2 + x - 1$ and $ g(x) = f\left(\frac{1}{x}\right)$

Then $ g(r) = 0 \iff f\left(\frac{1}{r}\right) = 0$. So this problem becomes "what is the sum of the 5 solutions to g(x)?"

\begin{align*}
g(x) &= f\left(\frac{1}{x}\right) \\
 &=  5\left(\frac{1}{x}\right)^5 + 4\left(\frac{1}{x}\right)^4 - 3\left(\frac{1}{x}\right)^3 + 2\left(\frac{1}{x}\right)^2 + \left(\frac{1}{x}\right) - 1 \\
&= \frac{5+4x-3x^2+2x^3+x^4-x^5}{x^5}
\end{align*}

$ g(x) = 0 \iff
5+4x-3x^2+2x^3+x^4-x^5 = 0$

By vieta's formulas, the sum of the five solutions of $ g(x)$ is $ -\frac{1}{-1}=1$
[/hide]


Sorry if this has been posted before, but when will there be some AMC-8 tests in the Contest Resources area?

[quote="anirudh"]That takes forever to add them. :([/quote]
SIgh, post them in the amc forum, some mod or admin or someone in power will put them in...

[quote="anirudh"]Will I get credit?[/quote]
Does it matter? If you want to share the problems then do so, if you don't then don't.

[quote="anirudh"]That takes forever to add them. :([/quote]

And it's quicker posting them on the boards and putting them in?  I don't buy it.  Maybe for you it is, but then an RManager has to go and enter the problems into the database so really it's not.

Posting them in the wiki is basically the same as posting them on the forum (unless the LaTeX gives you troubles...which is very annoying on the wiki...but still).

[quote="joml88"]Posting them in the wiki is basically the same as posting them on the forum (unless the LaTeX gives you troubles...which is very annoying on the wiki...but still).[/quote]It's by no means the same. The wiki is for information, the forum is for discussion.

So, could you add me to RMManagers, so I can add resources?


Let $ ABC$ be a triangle and let $ I$ be its incenter. Let $ DEF$ and $ MNP$ be its intouch and medial triangles, respectively (i.e. $ D$, $ E$, $ F$ are the tangency points of the incircle with the sidelines; $ M$, $ N$, $ P$ are the midpoints of $ BC$, $ CA$, $ AB$, respectively), and let $ X$, $ Y$, $ Z$ be points on the lines $ AI$, $ BI$, $ CI$, respectively. Prove that triangles $ XYZ$ and $ DEF$ are perspective if and only if $ XYZ$ and $ MNP$ are perspective.

This is a beautiful generalization of the  [url=http://xtec.net/~qcastell/ttw/ttweng/definicions/d_mig_arc2_p.html]second mid-arc point[/url], among other perspectors. Label $EF \equiv \tau_A,$ $FD \equiv \tau_B,$ $DE \equiv \tau_C,$ $NP \equiv \ell_A,$ $PM \equiv \ell_B,$ $MN \equiv \ell_C.$ Notation $\delta(U,\ell)$ stands for the distance from the point $U$ to the line $\ell.$ It's well known that in any scalene triangle, the polar of one vertex with respect to the incircle, the midline referent to a second vertex and the inner angle bisector issuing from the third, concur. Thus, $ED,$ $PM,$ $AI$ concur and $FD,$ $NM,$ $AI$ concur. Consequently, any point $X$ on the line $AI$ will satisfy

$\frac{\delta(X,\tau_C)}{\delta(X,\ell_B)}=\frac{\delta(I,\tau_C)}{\delta(I,\ell_B)} \ \ , \ \ \frac{\delta(X,\tau_B)}{\delta(X,\ell_C)}=\frac{\delta(I,\tau_B)}{\delta(I,\ell_C)} \ \Longrightarrow$

$ \ \frac{\delta(X,\tau_C)}{\delta(X,\tau_B)} \cdot \frac{\delta(X,\ell_C)}{\delta(X,\ell_B)}=\frac{\delta(I,\tau_C)}{\delta(I,\tau_B)} \cdot \frac{\delta(I,\ell_C)}{\delta(I,\ell_B)} \ \ (1)$

Analogously, for points $Y,Z$ on $BI,CI$ we have  the relations

$ \ \frac{\delta(Y,\tau_A)}{\delta(Y,\tau_C)} \cdot \frac{\delta(Y,\ell_A)}{\delta(Y,\ell_C)}=\frac{\delta(I,\tau_A)}{\delta(I,\tau_C)} \cdot \frac{\delta(I,\ell_A)}{\delta(I,\ell_C)} \ \ (2)$

$\frac{\delta(Z,\tau_B)}{\delta(Z,\tau_A)} \cdot \frac{\delta(Z,\ell_B)}{\delta(Z,\ell_A)}=\frac{\delta(I,\tau_B)}{\delta(I,\tau_A)} \cdot \frac{\delta(I,\ell_B)}{\delta(I,\ell_A)} \ \ (3)$

Multiplying the equations $(1),(2)$ and $(3)$ yields

$\left [ \frac{\delta(X,\tau_C)}{\delta(X,\tau_B)} \cdot \frac{\delta(Y,\tau_A)}{\delta(Y,\tau_C)} \cdot  \frac{\delta(Z,\tau_B)}{\delta(Z,\tau_A)} \right] \cdot \left [\frac{\delta(X,\ell_C)}{\delta(X,\ell_B)} \cdot  \frac{\delta(Y,\ell_A)}{\delta(Y,\ell_C)} \cdot \frac{\delta(Z,\ell_B)}{\delta(Z,\ell_A)} \right]=1$

By the converse of Ceva's theorem in the intouch triangle $\triangle (\tau_A,\tau_B,\tau_C)$ and medial triangle $\triangle (\ell_A,\ell_B,\ell_C),$ we conclude that $DX,EY,FZ$ concur $\Longleftrightarrow$ $MX,NY,PZ$ concur.


What should the maximum total area of circles $a$, $b$, and $c$ be so that when positioned in a right triangle, they do not overlap and do not cross outside the triangle, but occupy as much area as possible?

[img]http://www.bloatedshrimp.net/temporary.JPG[/img]

are you sure?
i think more info is needed
like how big the triangle is.

This is known as the [url=http://mathworld.wolfram.com/MarbleProblem.html]marble problem[/url].  I don't know if this is intentional, but your diagram is the same as the one in the MathWorld entry for [url=http://mathworld.wolfram.com/MalfattisProblem.html]Malfatti's problem[/url], which is related.


Let $ A$ be the set of all subsets of size three $ S_1,S_2,...,S_{84}$ of $ R \equal{} \{1,2,...,9\}$. Prove that we can partition $ A$ into $ 28$ sets $ B_1,B_2,...,B_{28}$ in such a way that each set $ B_i$ contains three subsets $ S_i$ from $ A$, and the union of the three subsets in each $ B_i$ is $ R$ itself.

i'm interested in this question too
can anyone solve it?

We can construct the $ B_i$s as follows. There are $ 28$ $ S_i$s containing $ 1$; we put one of these in each $ B_i$.
Hence this element is distinct for each $ B_i$.
Now consider the $ 7$ $ B_i$s such that both $ 1$ and $ 2$ belong to the set we put in them.
For such a $ B_i$, we take the second subset as $ f(k), f(k \plus{} 1), f(k \plus{} 3)$ where $ \{1, 2, k\} \in B_i$ and $ f(j) \equal{} (j \minus{} 2) \mod 8 \plus{} 2$. It is clear that the $ 7$ $ B_i$s we get thus are disjoint.
Now we take the $ 6$ $ B_i$s such that for some $ k \geq 4$, $ \{1, 3, k\} \in B_i$. Find a scheme for constructing them so that they do not overlap with the previous ones.
Proceed thus to get the complete structure of $ A$.
Try completing this proof; I'll try to find a better one.


calculus
integration
LaTeX
function
probability
algorithm
calculus computations

I have  a definite integral from a physics problem:

$ \displaystyle\int e^{\minus{}2x^2}\,dx$ with max of 1.2 and min of 0.15 (I can't find
the command to display more than one integer in Latex - anyone?.)

Apparently this integral uses something called the "error function" and "erf" which
I can't find in any of my text books - my question is would this be something you learn in high
school or later on please?

(btw calc101.com tells me it can't solve this problem,
and mathematica won't give me a numerical answer, only a detailed equation
featuring "erf".)

If you want a numerical answer, you should write $ e^{\minus{}2x^2}$ as a power series and integrate that.

This is how to do what you want in LaTeX: $ \int_{0.15}^{1.2} e^{\minus{}2x^2}\,dx$ (click to see the code)

\[ \int_{0.15}^{1.2}{e^{\minus{}2x^2}dx}\equal{}\sqrt{\frac{\pi}{8}}\left(Erf(1.2\sqrt{2})\minus{}Erf(0.15\sqrt{2})\right)\approx 0.468603\] The error function is defined as \[ Erf(y)\equal{}\frac{2}{\sqrt{\pi}}\int_0^y{e^{\minus{}x^2}dx}\] so the relation to your integral is straightforward.  You may encounter this special function in high school if you take a college level course related to probability or statistics.

By the way, Mathematica is certainly capable of performing this integral; use the command
[code]Integrate[Exp[-2*x^2], {x, 0.15, 1.2}][/code]

Thanks for that guys! I'm an adult teaching myself math from
high-school texts, so it does seem like I'd need the higher
level textbooks for erf and power series.

I'm told the result is "roughly 0.47" (as I see you get)
- is there any way to get that number using Mathematica please? (rather than the lengthy
solution it gives with logs and erfs and radicals.)

It should be possible to force the system to give numerical answers. I don't know the specific command in Mathematica.

This function is not an elementary function, and is calculated using power series or numerical integration techniques. The numerical integration methods are generally more accurate than power series estimates for a specific amount of work and a specific integral; the power series is better for looking at the behavior on intervals near zero.

To estimate something like Erf(5) quickly, we want an asymptotic series.
$ \int_M^{\infty}e^{\minus{}x^2}\,dx\equal{}\left[\minus{}\frac1{2x}e^{\minus{}x^2}\right]_M^{\infty}\minus{} \int_M^{\infty}\frac1{2x^2}e^{\minus{}x^2}\,dx$
$ \equal{}\frac1{2M}e^{\minus{}M^2}\minus{}\left[\frac1{4x^2}e^{\minus{}x^2}\right]_M^{\infty}\plus{} \int_M^{\infty}\frac{2}{4x^3}e^{\minus{}x^2}\,dx$
$ \equal{}\frac1{2M}e^{\minus{}M^2}\minus{}\frac1{4M^2}e^{\minus{}M^2}\plus{}\frac{2}{8M^3}e^{\minus{}M^2}\minus{}\cdots$

Sum the terms until they stop getting smaller, and you have a very accurate estimate without too much calculation. Do stop there- the series diverges for all $ x$ if you keep going.
That gives the outside of the integral; use the known integral on $ [0,\infty)$ to complete it.

You can put $ N[\ ]$ around an integral in Mathematica to get a numerical estimate.

Thanks jmerry - would that be something that someone studying intro
physics would be supposed to be able to do though? (The book answer
mentions using a calculator.)

"You can put  N[] around an integral in Mathematica to get a numerical estimate."
Just tried that - same result.  Any other ideas? :)

No, the asymptotic series I mentioned is considerably more advanced. It's also useless close to zero. For your problem, ordinary numerical integration is clearly best. You may remember Simpson's rule; that's the sort of thing I'm talking about. Calculator and computer systems have various algorithms built in to do the same thing without needing to know any more than how to calculate the function at a few points.

[quote="bio223"]"You can put  N[] around an integral in Mathematica to get a numerical estimate."
Just tried that - same result.  Any other ideas? :)[/quote]

JoeBlow's method also works.

Are you sure that you're using the constant $ e$ rather than a variable named e? :)

Oops - that was it, using the wrong 'e' :)
Works fine now, thanks!


Να λυθεί στους ακεραίους η εξίσωση:
$ x^{3}\minus{}y^{2}\equal{}1$

Λιγο βιαστηκά....
$ x^3=y^2+1=>x^3=(y+i)(y-i)=>y+i=(a+bi)^3 \& y-i=(a-bi)^3$ .....απο τις σχεσεις βρισκεις το αποτελεσμα....ΚΑΛΗ ΑΝΑΣΤΑΣΗ ΚΑΙ ΚΑΛΟ ΠΑΣΧΑ!

Oraia, tora na lithei stous akeraious i :
$ x^3\minus{}y^2\equal{}5$

Stergios

Proton den prepi na parasirthoume giati to $ \mathbb{Z}[\sqrt{\minus{}5}]$ den ine UFD $ (6\equal{}2*3\equal{}(1\plus{}i\sqrt{\minus{}5})(1\minus{}i\sqrt{\minus{}5})$. Bebaia exo brei lisi xrisimopoiontas oti o arithmos taxis tou $ \mathbb{Z}[\sqrt{\minus{}5}]$ ine 2(den tin grafo epidi den ine kai toso elementary) .
Einai eukolo na doume oti $ y\equiv1\bmod{4}$. Tote $ (y\minus{}1)(y^2\plus{}y\plus{}1)\equal{}x^2\plus{}2^2$ iparxi p=prime$ \equiv3\bmod{4}$ oste $ p\mid y^2\plus{}y\plus{}1$ giati $ y^2\plus{}y\plus{}1\equiv3\bmod{4}$. Omos tote $ p\mid x^2\plus{}2^2$ ara $ 1\equal{}\left(\frac{\minus{}2^2}{p}\right)\equal{}\left(\frac{\minus{}1}{p}\right)\equal{}\minus{}1$ atopo.
Ara den exei liseis stous akeraious.

Καταρχήν, παρατηρούμε ότι πρέπει $ x>0$, αλλιώς θα είχαμε αρνητικό = θετικό, άρα άτοπο .

Ακόμη, θα υπάρχει $ k\in\mathbf{N}$ , τέτοιος ώστε $ y\equal{}x\minus{}k$, αν έχει λύση στους ακεραίους .

Έτσι γράφεται και :

$ x^3\minus{}(x\minus{}k)^2\equal{}5$

Θεωρώ τη συνάρτηση :

$ G(x)\equal{}x^3\minus{}(x\minus{}k)^2\minus{}5$

Έστω $ k\equal{}1$.

Έχουμε :

$ f(x)\equal{}x^3\minus{}(x\minus{}1)^2\minus{}5$

$ f'(x)\equal{}3x^2\minus{}2(x\minus{}1)\equal{}3x^2\minus{}2x\plus{}1 >0$ (διακρίνουσα αρνητική)

Άρα $ f$ για $ k\equal{}1$ γνησίως αύξουσα .

Ακόμη :

$ f(0)<0,\ f(2)>0$

Άρα έχει μία τουλάχιστον λύση από Bolzano και αυτή οφείλει να 'ναι μοναδική, αφού f γνησίως αύξουσα .

Για $ x\equal{}1$ , $ f(x)\neq 0$, άρα δεν έχει λύση στους ακεραίους. Επομένως, $ k>1$ (1)

Έχουμε ακόμη :

$ G'(x)\equal{}3x^2\minus{}2(x\minus{}k)$

Όμως

$ x\minus{}k < x\minus{}1\implies \minus{}(x\minus{}k)>\minus{}(x\minus{}1)$

Έτσι :

$ G'(x)\equal{}3x^2\minus{}2(x\minus{}k)>3x^2\minus{}2(x\minus{}1)\equal{}3x^2\minus{}2x\plus{}2>0$ (διακρίνουσα αρνητική)

Άρα και η $ G$ γνησίως αύξουσα .

Ακόμη :

$ G(0)\equal{}\minus{}k^2\minus{}5<0$

και $ G(x)\equal{}x^3\minus{}(x\minus{}k)^2\minus{}5> x^3\minus{}(x\minus{}1)^2\minus{}5$

Άρα $ G(2)> 8\minus{}1\minus{}5 \equal{} 2>0$

Άρα $ G(0)\cdot G(2)<0$, άρα από Bolzano έχει τουλ μία λύση στο $ (0,2)$, όμως είναι μοναδική λόγω μονοτονίας της $ G$. Για $ x\equal{}1$, προκύπτει εύκολα ότι δεν έχει λύσεις . Άρα είναι αδύνατη στο $ \mathbf{Z}$.

Στέλιος

Bravo Taso !

O logos pou egrapsa tin askisi einai gia na sas toniso oti oi methodoi tis algebrikis theorias arithmon einai men isxiroi alla theloun prosoxi !
Opos sosta episimanthike i idia apodeiksi opos auti sto post #2 tha itan lathos ( episeis i apodeiksi sto post #2 den einai olokliromeni, sto Z[i] tha prepei na prosexoume oti exoume perisotera units i.e. +1,-1,+i,-i ) dioti sto Z[sqrt(-5)] den exoume monadiki paragontopoiisi se protous. I sosti prossegisi tha  itan na theorisoume to ideal class group tou somatos, to opoio exei taksi 2 kai na xreisimopoiisoume tin  monadiki paragontopoisi se prota ideodi.

Taso : Nomizo oti eprepe na eixa grapsei 20 anti gia 5

Mostel : Askisi : Bres to lathos

Stergios

Ναι, όντως μπάζει η λύση ... Ευχαριστώ τον Σιλουανό για τη διόρθωση  :oops:

Δν ισχύει απαραίτητα $ x>y$

Διχως να εχω ασχοληθει με αλγεβρικη Θεωρια Αριθμων βλεποντας σας να λυνεται ορισμενες μη - γραμμικες διοφαντικες εξισωσεις καταλαβα στο περιπου πως λυνονται μερικες δουλευοντας με μιγαδικους, ετσι εφτιαξα μια διαφορετικη εξισωση που ομως λυνεται περιπου οπως οι παραπανω

Να λυθει στους ακεραιους η εξισωση:

$ y^3 \minus{} x(x \plus{} 6) \equal{} 13$

βαζω τις λυσεις σε hide

[hide]$ (x, y) \equal{} (\minus{}5, 2)$, $ (x, y) \equal{} (\minus{}1, 2)$, $ (x, y) \equal{} (\minus{}14, 5)$, και $ (x, y) \equal{} (8, 5)$[/hide]

H παραπάνω σχέση γράφεται ισοδύναμα $ y^3\equal{}(x\plus{}3)^2\plus{}4$
Όμως η $ y^3 \equal{} x^2 \plus{} 4$ έχει λύσεις τις $ (x,y) \equal{} (2,2), (\minus{}2, 2), (11,5), (\minus{}11, 5)$(η απόδειξη για κάποιον που γνωρίζει αλγεβρική θεωρία αριθμών είναι εύκολη)
Πρέπει να έχει λυθει στο forum γιατί είναι γνωστο problem.
Άρα$ (x\plus{}3,y)\equal{}(2,2), (\minus{}2, 2), (11,5), (\minus{}11, 5)$ άρα $ (x,y) \equal{} (\minus{}5,2), (\minus{}1, 2), (\minus{}14,5), (8, 5)$

γεια σας παιδια.ειμαι καινουριος στο forum.που λετε εχς τρελα με τα μαθηματικα.ειμαι 17 και του χρονου θα δωσω πανελληνιες.λοπον.εχω μια απορια.μια διοφαντικη εξισωση της μορφης χ^3=y^2 +1  λεμε τωρα!λοιπον .μπορειατι να μου πειτε την μεθοδολογια ευρεσησ λυσεων?μπορω να βρω μερικες απο αυτες με το μυαλο,αλλα και παλι δν ειναι σιγουρο οτι θα τις βρω ολες.γι'αυτο ζηταω να μαθω την μεθοδολογια επιλυσης τετοιων εξισωσεων.ευχομαι να μαθω πολλα εδω :) [/hide][/code]


$ x,y,z>0$

prove :

$ 2\sum_{cyc} ab^2 \ge 3abc\plus{}\sum_{cyc} a^2b$

Well, I don't think it is true.
Take $ f(a,b,c)\equal{}LHS\minus{}RHS$
Check that $ \lim_{a\to \infty}f(a,1,\frac{1}{a})\equal{}\minus{}\infty$ :wink:


group theory
abstract algebra
geometry
geometric transformation
superior algebra
superior algebra unsolved

If $ G$ is a group where $ [G: H]\equal{}p$ and $ [G: K]\equal{}q$ where $ (p,q)\equal{}1$, prove that $ G\equal{}HK$. Thank you for the solution before.

Consider the action of K on the coset space G/H by left translation, the number of elements of the orbit of H under this action is clearly, [K:K \cap H]. Since q[K:K \cap H] = [G:K][K:K \cap H]= [G:K \cap H] = [G:H][H:K \cap H]=p[H:K \cap H], and (p, q)=1, it follows that p|[K:K \cap H]. On the other hand, the number of elements of the orbit of H is less than or equal to p. So we conclude [K:K \cap H] = p. This imply that KH = the union of kH, where k runs over K, is G. :D

[quote="J.S.Siu"]Consider the action of $ K$ on the coset space $ G/H$ by left translation, the number of elements of the orbit of $ H$ under this action is clearly, $ [K: K \cap H]$. Since $ q[K: K \cap H] \equal{} [G: K][K: K \cap H]$$ \equal{} [G: K \cap H] \equal{} [G: H][H: K \cap H]\equal{}p[H: K \cap H]$, and $ (p, q)\equal{}1$, it follows that $ p|[K: K \cap H]$. On the other hand, the number of elements of the orbit of $ H$ is less than or equal to $ p$. So we conclude $ [K: K \cap H] \equal{} p$. This imply that $ KH$ = $ \bigcup kH$, where $ k$ runs over $ K$, is $ G$. :D[/quote]

Doesn't it look better with dollar signs? :)

Thank you for your explanation before :D.

"the number of elements of the orbit of $ H$", what is the definition of "the number of elements of the orbit of $ H$" ?? Please explain to me. Thank you for your explanation before.


$ 1\plus{}10\plus{}101\plus{}1010\plus{}10101\plus{}101010\plus{}\cdots\plus{}$up to $ 2n^{th}$ terms，find the sum in terms of $ n$

 :roll:

[hide]Add pairs of terms to get:
$ 11\plus{}1111\plus{}111111 \plus{} \ldots$, n terms.

Add $ n/9$, or equivalently, add $ 0.1 \ldots$ to each term:
$ 11.1 \ldots  \plus{} 1111.1 \ldots \plus{} \ldots$, n terms

This is a geometric sequence. It's sum is equal to:
$ \frac{10^{2n}/9 \minus{} 1/9}{1 \minus{} \frac{1}{100}} \equal{} \frac{100^{n\plus{}1} \minus{} 100}{891}$

Finally, subtract back out $ n/9$ to get $ \frac{100^{n\plus{}1} \minus{}99n \minus{} 100}{891}$[/hide]


number theory
least common multiple
function
floor function

hehe i know a lot of people who made stupid mistakes on this one.

personally, i didn't see that 2 and 4 share a common factor of 2 XD

I think I would just think that since there are 50 multiples of 2's in 1-100 and only 25 of them are not multiple of four. Then I would think that since there are 33 multiple's of threes are in 1-100 and every other one is a multiple of two so I would divide 32 by two and add one because 99 is odd. So you would have the answer in just a couple of seconds.
I hope this made sense.

25 multiples of 2 that aren't divisible by 4
33 multiples of 3 (yes some are divisible by 4, but I am about to fix that)
16 multiples of 6(which include all the multiples of 3 that were divisible 2 and of course by 4 as well)

25 + 33 - 16 = 42

[quote="l Bob l"]Target 1: How many integers from 1 to 100 are multiples of 2 or 3 but not 4?

My Solution[hide]
List the LCM of 2, 3, and 4:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. I learned how to count! :D
Cross out the numbers that do not follow the criteria:
x, 2, 3, x, x, 6, x, x, 9, 10, xx, xx

5 out of 12 numbers. That is 40 out of 96. 97 to 100 are congruent to 1 to 4 mod 12 respectively, and since 2, 3 work, 98, 99 do too. 40 + 2 = $ [42]$ of them.[/hide]

So how did you do this one? o.O    :maybe:[/quote]

heh, that's the way I used to check.  I found it to be much easier than the PIE method I used (arg, frost pwned me at it this time)

Yay I found a solution(its like jhangil's solution) after like 5 minutes, so I would still fail the target.

So the multiples of 2 that are not multiples of 4 are basically 2*an odd# or else it would have another 2 in it and it would be divisible by 4. So the numbers under this criteria would be 2*1,2*3,2*5.....2*49. We count how many odd number by adding one and dividing. (49+1)/2=25.

Multiples of 3 under 100 that don't have any even factors(we counted them when we count the multiples of 2) so just 3*odd number. Our list is 3*1,3*3,3*5.....3*33, now we count by adding one and dividing by 2. (33+1)/2=17.

Now add 25+17=42. Pretty doable if you think quick.

Yeah, I used the PIE method, but I accidentally put a wrong number into my calculator, since I didn't want to make any mistakes doing it myself.

Remind me never to trust a calculator that I enter stuff into.

I knew how to do this one! I just got it wrong. I was being very stupid.

I got that using the $ \phi$ function, basically the same as the first method.

Lots of good answers, here.  I'm still partial to using Venn Diagrams for 3-variable problems.  With not much practice, you can make a Venn diagram almost as fast as you can write the numbers in the circles.  And I find them to be almost mistake-free.

First, the number of multiples of $ k$ between $ 1$ and $ n$ is given by $ \lfloor\frac{n}{k}\rfloor.$  Second, the number of multiples of $ a$ and $ b$ is the same as the number of multiples of $ \text{LCM}(a,b).$

Draw three circles and fill them in, as shown.  Tally up what the problem asks and you're done.

this is how I did it: 50 multiples of 2 and 33 multiples of 3. Add and you get 83. YOu count the multiples of 6 twice so you must subtract 16 from 83 which gives 67. There are also 25 multiples of 4. Subtract 25 from 67 and you get 42 which is the answer.

I did it very similar to frost.

so there are 50 multiples of two minus 25 which are also 4 so 25

there are 33 multiples of 3 minus 8 which are also 4 so 25

16 multiples of 6. Half of them are divisible by 4, so 8.

50-8=42.

Hehe, I got the wrong answer at first for this problem and was like, wait, let me do this again.  Then I get 42, but b/c I spent the rest of my time checking, I had like 10 sec left before I was like wait, this can't be right and changed my 42 answer for no reason whatsoever.  I just found the total number of multiples of 2 or 3 and subtracted the multiples of 2 and 4 and 3 and 4 (not double-counting the repeats).  But I got that multiples of 2 and 4 were multiples of 8 rather than 4, which totally owned my answer.  Not exactly the best answer ever, but it would have worked for me if I hadn't spazzed out when the reader announced the time left and become paranoid at the last second.  It really made me angry...

Well, its [100/2]+[100/3]-[100/4]-[100/6]
[100/2]+[100/3] is how many multiples of 2 and three
-[100/4] is to remove multiples of 4
-[100/6] is to remove repeats in multiples of 2 & 3 (which are multiples of 6)

It also helps if you know how to program your calculator: use brute force. Here's the program I used.

: 0->T
: For(X,1,100)
: If [X/2]=X/2 or [X/3]=X/3
: T+1->T
: If [X/4]=X/4
: T-1->T
: Disp T
: End

This program essentially tests all the numbers from 1 to 100 and checks if they are multiples of two or three. It then adds one to the total "T" if so. If it is a multiple of 4, it subtracts 1. Run it, and when it stops, the last display is 42.

It really helps to know how to program your calculator as you can get the answer algebraically and then run the program to check your work. While the program is running, you can work on target #2.

Incidentally, the answer to question #1 on Target is the same as the answer to the Ultimate Question.


If $ a,b,c>0$, then

$ \frac{a^2}{b^2}\plus{}\frac{b^2}{c^2}\plus{}\frac{c^2}{a^2}\geq\frac{9}{2}\plus{}(\frac{a}{c}\plus{}\frac{c}{b}\plus{}\frac{b}{a})\minus{}\frac{9}{2}\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}$

The following inequality is also true.
For positives $a$, $b$ and $c$ prove that:
\[\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}-\frac{a}{c}-\frac{b}{a}-\frac{c}{b}\geq\frac{7(a^2+b^2+c^2-ab-bc-ca)}{a^2+b^2+c^2} \]


Consider a sequence x(1),x(2) ... of real numbers s.t. x1=1 and for each n>1:
x(n) = (x(n-1) ^ +- 1) + (x(n-1)^ +- 1) ... + (x(1)^ +-1)
with any choice of the signs + or - in the exponents.
(a) Is it possible that such an infinite sequence is bounded?
(b) Find the least possible value x(10^7).
(c) Given any nine members of the sequence, show at least one is greater than 4.
(d) find the maximal (integer :)) k s.t. there exist infinitely many sequences
with some string of k terms smaller than 4.

I think you mean (x-2) for your second term, instead of repeating (x-1).
Also, it would be bounded if you chose the powers to be -1 for all that are less than 1 and chose it to be 1 for all that are greater than 1.
Do the +- mean that I get to choose whichever I want or that it is both positive and negative 1?


Find the number of permutation $ a_1,a_2,...a_3$ of the set {1,2,...n} such that $ \left| {a_i  \minus{} a_{i \plus{} 1} } \right| \ne 1$ for all i=1,2,...,n-1.

Besides [url=http://www.mathlinks.ro/viewtopic.php?t=192170&search_id=416626497]this[/url], I think

Let $ a_n$ be the number of permutations of $ \{1,2,\dots ,n\}$ satisfying the required property and $ b_n$ be the number of permutations $ \{a_1,a_2,\dots ,a_n\}$ of $ \{1,2,\dots ,n\}$ with the property that there exists exactly one pair such that $ |a_i\minus{}a_{i\plus{}1}|\equal{}1$. Then we have a recursion,

$ a_{n\plus{}1}\equal{}(n\minus{}1)a_n\plus{}b_n\equal{}(n\minus{}1)a_n\plus{}{n\choose 2}(n\minus{}1)a_{n\minus{}2}$

Have you tried to actually generate some terms of the sequence from that recursion and compare them to the right answer ([url=http://www.research.att.com/~njas/sequences/A002464]A002464[/url])?  It is very not right.


Solve for $ x$:
\[ \frac12(8^{2x}) \equal{} 256\]

Express your answer as a common fraction.


I did a search for similar problems...

[hide]$AC^2-(DE)(FB)=(AC)(GB)-(GC)(AB)$
Quadrilateral ABCG is cyclic. Applying Ptolemy's Theorem,
$(AC)(GB)=(GC)(AB)+(GA)(BC)$
$(AC)(GB)-(GC)(AB)=(GA)(BC)$
$AC^2-(DE)(FB)=1001^2$[/hide]

Thanks

But just to clarify, all those segments are congruent right.

[quote="mathgeniuse^ln(x)"]Source:Mathcounts Nationals before 1991 (I think)

A regular heptagon ABCDEFG has side 1001.  What is $AC^2-(DE)(FB)?$

Could you please give a detailed solution?
Thanks[/quote]
that was definitely an ARML problem, I remember it from practice


If the equation 2x :^2:  + 2x = 3a :^2: x + a has two roots which are the additive inverse of each other, find a.

A. 0
B. 2/3
C.  :pm:  :sqrt: (6)/3
D.  :sqrt: (6)/3
E.  :pm: 2/3

[hide]The sum of additive inverses is 0. In a quadratic, the sum of the roots is -b/a. Once you move the equation into standard quadratic form, you see that b=-3a :^2: +2.

So we have (3a :^2: -2)/2=0

3a :^2: -2=0

3a :^2: =2

a :^2: =2/3

a= :pm:  :sqrt: 6/3

Option C is the answer.[/hide]

[hide]

since the two roots are additive inverses of each other, their sum will be 0

collect all terms to one side to put it in quadratic form

2x :^2:  + 2x - 3a :^2: - a = 0

2x :^2: + x(2 - 3a :^2: ) - a = 0

the sum of the roots is 0, so (2-3a :^2: )/2 = 0

2=3a :^2:

solving give a = :pm::sqrt:(6) / 3, where the negative a gives imaginary roots.

so the answer is C, :pm::sqrt:(6) / 3,

[/hide]

You're just missing one little detail. The answer is not C. See if you can find it.

Hint: [hide]The problem is in the constant term of the quadratic equation.[/hide]

Hmm... you're going the right direction, but what you said wasn't true... x * -x does not equal x^2, but if you say it right, it'll lead you to the answer.

sorry, i deleted my post after realizing what i had said wasnt true.. but it was too late. i still have to think about it though

take all the time you need. you're heading the right direction.

This is my solution:

[hide]2x^2 + 2x = 3a^2 x + a

get zero on one side

2x^2+(2-3a^2)x-a=0

Use quadratic formule

(-2+3a^2+ :sqrt: (4-12a^2+9a^4-9a))/4

(-2+3a^2+ :sqrt: (-12a^2+9a^4-9a-4))/4

now since we know there the additive inverses we can do

-(-2+3a^2+ :sqrt: (-12a^2+9a^4-9a-4))/4=(-2+3a^2- :sqrt: (-12a^2+9a^4-9a-4))/4

And that's as far as I got (need to do homework.)[/hide]


function
symmetry
algebra
functional equation
algebra solved

Find all functions $f: \mathbb{R} \mapsto \mathbb{R}$ satisfying $f(x) \cdot f(x+y) = (f(y))^2 \cdot (f(x-y))^2 \cdot 2^{y+4}$.

What is the range of $f$? Is $f$ from $R$ to $R$?

Pierre.

The only solution of $f(x) \cdot f(x+y) = (f(y))^2 \cdot (f(x-y))^2 \cdot 2^{y+4}$ (1) with a root is $f(x) \equiv 0$ which means $f(x_0) = 0$. Suppose $x_0$ is a root of $f(x)$, hence $f(x_0) = 0$. Then for $x = x_0$ we have: $f(x_0) \cdot f(x_0+y) = (f(y))^2 \cdot (f(x_0-y))^2 \cdot 2^{y+4}$. $2^{y+4} \neq 0$, hence
$f(y) = 0$ or $f(x_0-y) = 0$ $\forall y$. This yields $f(y) \equiv 0$ and $f(x) \equiv 0$, respectively.
Plugging $f(x) = -f(x)$ (SYM) in (1), one seems it functional equation is satisfied as well. Due to symmetry we can assume WLOG $f(x) > 0 $. Plug $y = 0$ in (1). $f(x) \cdot f(x) = (f(0))^2 \cdot (f(x))^2 \cdot 2^{4}$ which simplifies too $f(0) = 2^{-2}$ (2). Setting $x=0$ in (1)  makes $f(x) \cdot f(y) = (f(y))^2 \cdot (f(-y))^2 \cdot 2^{y+4}$ which is equivalent to $2^{-y-6} = f(y) \cdot (f(-y))^2$ (3). Replace $y = -y$ in (3) gives $2^{y-6} = f(-y) \cdot (f(y))^2$ (4). (3) and (4) makes $2^{-12} = (f(y) \cdot f(-y))^3$ or $f(-y) = 2^{-4} \cdot \frac{1}{f(y)}$ (5). Substituting (5) in (4) makes $2^{y-6} = 2^{-4} \cdot \frac{1}{f(y)} \cdot (f(y))^2$ which simplifies to $2^{y-2} = f(y)$. Due to (SYM) $f(y) = -2^{y-2}$ this solution also satisfies (1.) Hence the solutions are: $f(x) \equiv = 0$ and $f(y) = \pm 2^{y-2}$.

Well, you had enough time to solve it.  :D

[quote="orl"]..., hence $f(y) = 0$ or $f(x_0-y) = 0$ $\forall y$. This yields $f(y) \equiv  0$[/quote]
Why?

[quote="orl"] Assume WLOG $f(x) > 0 $.[/quote]
Do you mean f is continuous function?


Lol. Sorry, I still don't get it. *sigh* Oh well.

Hey, on a different note, does anyone know how to pronounce "AoPS"? Or do we all just say A-O-P-S?

I also say A-O-P-S.  (Hey, that could be a song ;-) like YMCA).  My boyfriend (whoisn't an active member) says ay-ops.  I'm not sure where he got that from. ;-)

Hmm, I pronounce it as Alison's boyfriend does, as a two-syllable word: A-ops.  Sort of like the word "chaos".

You read this silly topic .. not only that ...you put a post in it....<now ducking for cover>

Ach, Gyan beat me to it!

I also pronounce it A - ops, not A O P S.

We say it the two syllable way.  Long A then "ops" as in cops.

That's the company slang at least.

i just say ops, as in cops... i guess the A is a poor, neglected, letter....  :)

another one here who says A-ops with a long a and short o


Consider this pattern where
the positive, odd integers
are arranged in a triangular
formation. The 1st through
4th rows are shown; each row
has one more entry than the
previous row. What is the sum
of the integers in the 15th row?

[asy]label("1",(0,0),S);
label("3",(-5,-5),S);
label("5",(5,-5),S);
label("7",(-10,-10),S);
label("9",(0,-10),S);
label("11",(10,-10),S);
label("13",(-15,-15),S);
label("15",(-5,-15),S);
label("17",(5,-15),S);
label("19",(15,-15),S);
dot((0,-22));
dot((0,-20));
dot((0,-24));[/asy]

got another way to look at this ...

knowing that sum of odd numbers is a square ... at the end of each row you can compute the sum of all the preceding odd numbers, which is just the square of how many there were:  end of row 2, sum is 3^2, end of row 3, sum is 6^2, etc.

end of row 14 is 105^2, end of row 15 is 120^2, thus, sum of row 15 is (120^2) - ( 105^2) == 3375

easy :-)

Can't u do it by finding the numbers in the row? I did that and must have done something really wrong because I got a totally different number

You could, but the way I did it was to note that the sum of each row was simply a perfect cube corresponding to the row number. Thus, the answer would be $ 15^3\equal{}3375$.


Mr. Norris, State Coordinator, is still working on the compete calendar.  Here comes first event to put on your calendar:

October 27, 2007 SIGMA – Skyview Invitational Games for Mathematical Achievement, at Vancouver, WA.  Please see attached file for details, and contact Carol Hattan, contest director, for information.

Good luck to all who are heading to Tempa Bay next month for MAO 2007 National Convention, the last event to wrap up 2006 - 2007 season.

The eleventh annual Mu Alpha Theta Fall Classic will be held at Thomas Jefferson High School on Saturday, November 3rd, 2007.  The contest is designed to reward excellence in mathematics.  You do NOT have to be a member of Mu Alpha Theta to come to the contest.

Please see registration form.

Here attached is a list of HS Math team contests events and schedule, courtesy of Mr. Norris.
There are other contests that your school may join that are not listed here.  Use this as your planning reference.

Invitation, Schedule, and Signup Form included here

This is a fun event.  Hope to you all there.

Mr. Tosch already posted the result:  http://wamath.net/hs/contests/wastate/state08/WaState08Results.pdf

A hugh "thank you" to Tom N, Tom T, and others for the impressive organization of such a large, successful event.


$\sum_{k \geqslant 0} {n + k \choose 2k} {2k \choose k} \frac{(-1)^k}{k+1}= ? $

Genereal identity:
\[\sum_{k=0}^{n-m} {n + k \choose m+2k} {2k \choose k} \frac{(-1)^k}{k+1}={n-1\choose m-1}.\]

Thank's a lot. What about that sum ?
$\sum_{k \geqslant 0} {n + k \choose 2k} {2k \choose k} \frac{(-1)^k}{m+k+1}$

Are you sure?
It is too difficult to solve such kind of problems if you are not sure that there is closed form...


Find all real coefficient polynomials $ P(x)$ satisfying:For all $ a > 1995$,the number of real roots of $ P(x) \minus{} a$
(including multiplicity of each root ) is greater than $ 1995$ and every roots is greater than $ 1995$ too

[quote="cosinator"]Do you mean $ P(x) \equal{} a$?[/quote]
Sorry,i have some misytkes in my post,i have just edit it
How about the new post


If $ f(x\plus{}1)\equal{}x^{2}\plus{}4x\plus{}5$ and $ f(x\plus{}2)\equal{}x^{2}\plus{}6x\plus{}10$, find $ f(x)$.

[quote="t0rajir0u"]A strange question.  The second piece of information is unnecessary.[/quote]
It was included on the test  :lol:

let $ y\equal{}x\plus{}1$...then $ y\minus{}1\equal{}x$...

$ f(y)\equal{}(y\minus{}1)^2\plus{}4(y\minus{}1)\plus{}5$

[quote="Altheman"]let $ y \equal{} x \plus{} 1$...then $ y \minus{} 1 \equal{} x$...

$ f(y) \equal{} (y \minus{} 1)^2 \plus{} 4(y \minus{} 1) \plus{} 5$[/quote]
$ f(y) \equal{} y^2 \plus{} 2y \plus{} 1$, so $ f(x) \equal{} x^2 \plus{} 2x \plus{} 1$?

[quote="Altheman"]$ f(y) \equal{} (y \minus{} 1)^2 \plus{} 4(y \minus{} 1) \plus{} 5$[/quote]
[quote="Brut3Forc3"] $ f(y) \equal{} y^2 \plus{} 2y \plus{} 1$[/quote]
Well, it's actually $ f(y) \equal{} y^2 \plus{} 2y \plus{} 2$,
$ (y \minus{} 1)^2 \plus{} 4(y \minus{} 1) \plus{} 5 \equal{} y^2 \minus{} 2y \plus{} 1 \plus{} 4y \minus{} 4 \plus{} 5 \iff y^2 \plus{} 2y \plus{} 2$

$ f(x)\not \equal{} x^2 \plus{} 2x \plus{} 1$ because, according to the original equations:
[list]$ f(x \plus{} 1) \equal{} x^2 \plus{} 4x \plus{} 5 \iff f( \minus{} 1 \plus{} 1) \equal{} ( \minus{} 1)^2 \minus{} 4 \plus{} 5$
 $ \equal{} 2$
$ f(x \plus{} 2) \equal{} x^2 \plus{} 6x \plus{} 10 \iff f( \minus{} 2 \plus{} 2) \equal{} ( \minus{} 2)^2 \minus{} 12 \plus{} 10$
$ \equal{} 2$ [/list]
this tells that $ f(0) \equal{} 2$, wheras if $ f(x) \equal{} x^2 \plus{} 2x \plus{} 1 \iff f(0) \equal{} 1$, instead of $ 2$, a contradiction.

$ f(x) \equal{} x^2 \plus{} 2x \plus{} 2$ seems correct based on my approach.

[hide="A Direct Approach"]
(it's late now and I'm not sure I did it correctly so please correct me if needed): [list] Let $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$.
$ f(x \plus{} 1) \equal{} a(x \plus{} 1)^2 \plus{} b(x \plus{} 1) \plus{} c$.
 $ \equal{} ax^2 \plus{} (2a \plus{} b)x \plus{} (a \plus{} b \plus{} c)$.
$ \equal{} x^2 \plus{} 4x \plus{} 5 \iff a \equal{} 1, b \equal{} 2, c \equal{} 2$

$ f(x \plus{} 2) \equal{} a(x \plus{} 2)^2 \plus{} b(x \plus{} 2) \plus{} c$
$ \equal{} ax^2 \plus{} (4a \plus{} b)x \plus{} (4a \plus{} 2b \plus{} c)$
$ \equal{} x^2 \plus{} 6x \plus{} 10 \iff a \equal{} 1, b \equal{} 2, c \equal{} 2$ This also holds with the second equation.
Therefore, we conclude that
$ \boxed{f(x) \equal{} x^2 \plus{} 2x \plus{} 2}$, based on the two given functions. [/list] [/hide]

$ f(x)\equal{}f(x\plus{}2)\circ (x\minus{}2)\equal{}(x\minus{}2)^2\plus{}6(x\minus{}2)\plus{}10\Longrightarrow \boxed{f(x)\equal{}x^2\plus{}2x\plus{}2}$
$ f(x)\equal{}f(x\plus{}1)\circ (x\minus{}1)\equal{}(x\minus{}1)^2\plus{}4(x\minus{}1)\plus{}5\Longrightarrow \boxed{f(x)\equal{}x^2\plus{}2x\plus{}2}$

 :D


algorithm
number theory
greatest common divisor
Euclidean algorithm

Can you explain me step-by-step how to solve this linear congruence with the help of the extended euclidean theorem?

$ 39x\equiv11(mod56)$

Thank you :D

I'm not sure what you mean by extended euclidean theorem; the Euclidean algorithm maybe?  This congruence has a solution if and only if $ \gcd(39,56)$ divides 11.  So let's check for the gcd.
\begin{align*}
&56=1(39)+17\\
&39=2(17)+5\\
&17=3(5)+2\\
&5=2(2)+1\\
&2=2(1)
\end{align*}
so the gcd is 1.  This obviously divides 11, so there is one mutually incongruent solution.  I'll leave you to find it.

I know there is only one solutioon because gcd(56,39)=1

My question is how to find that solution

Use the Chinese Remainder theorem mod 7 and 8.


For the AIME II 2005 #6 (see below), it says in the problem that

"If, after the restacking process, at least one card from [b]each[/b] pile occupies the same position that it occupied in the original stack, the stack is called magical."

In pile A, the 131 remains the same, but how do you address the one in pile B? (the solution doesn't mention how)

Original Problem:
The cards in a stack of $2n$ cards are numbered consecutively from $1$ through $2n$ from top to bottom. The top $n$ cards are removed, kept in order, and form pile $A$. The remaining cards form pile $B$. The cards are then restacked by taking cards alternately from the tops of pile $B$ and $A$, respectively. In this process, card number $(n+1)$ becomes the bottom card of the new stack, card number $1$ is on top of this card, and so on, until piles $A$ and $B$ are exhausted. If, after the restacking process, at least one card from each pile occupies the same position that it occupied in the original stack, the stack is named [i]magical[/i]. Find the number of cards in the magical stack in which card number $131$ retains its original position.

Consider this "mapping" of some sort. By working out a few cases, we see that

$k \rightarrow 2n-2(k-1)-1$ for $k < n$
$n+k \rightarrow 2n-2(k-1)$ for $k < n$

where the number of the bottom card in the new pile is $2n$ and the top is $1$. So we check for $k = 131$, which means $2n-2(131-1)-1 = 131 \Rightarrow 2n = 392$ and $131$ is indeed less than $n = 196$ so it works. Our answer is $392$.

From this solution, we can simply check to find out what the other card is. We solve the other mapping with $n = 196$ to find

$196+k =392-2(k-1) \Rightarrow k = 66$,

so card number $262$ stays in the same position as well (hmm, it's exactly double, which is in fact always the case; just work through the algebra a bit).


One of my students sent this problem from Hampshire summer program:

Winnie the Pooh has 2n+1 jars of honey.  No matter which jar he excludes, he is always able to separate the remaining 2n jars into 2 groups of n jars such that both groups have the same total amount of honey.

Prove each jar has the same amount of honey...

It's from American Mathematical Monthly. It can be proved by using linear algebra in a very appealing way.

But it was stated with the last sentence exchanged to is that always possible or something.

Hmm.. this can be solved if all the jars have a 'rational' amount of honey, but I'm not sure about 'irrational' (just letting the volume be a number). A bit trickier I think..


Compute the number of ten-digit numbers which contain only the digits 1, 2, and 3 with the digit 2 appearing in each number exactly twice.

[hide]Since there are two 2's and every other digit must be either a 1 or a 3, the answer is
$ \binom{10}{2}\cdot2^{8}=\boxed{11520}$[/hide]

Those two answers are equivalent; $ \binom{10}{2}=45$.

Suppose you have 10 blank digits, you have to take 2 positions for the two 2's. THen that is 10 taken two at a time which is $ \binom{10}{2}$. The remaining 8 positions, can only take 1 and 3. That is for each position, you have 2 choices. That is $ 2^{8}$

By multiplication principle, you answer should be
$ \binom{10}{2}$$ (2^{8})$


geometry incenter circumcircle inradius trigonometry ratio inequalities

Given $\triangle ABC$ such that $r$ and $R$ are inradius and circumradius respectively. Let $d$ be the distance between the incenter and circumcenter. Prove that $d=\sqrt{R(R-2r)}$

[hide="But..."]
Without trig...[/hide]

Call incenter is $I$ and circumcenter is $O$, $AI$ cut $(O)$ at $A_{1}\not = A$, $A_{1}O$ cut $(O)$ at $B_{1}\not =A_{1}$, $IC_{1}\perp CA$. We have $OI^{2}-R^{2}=-IA.IA_{1}$ and $IA.IA_{1}=2R.r$ because triangle $B_{1}BA_{1}$ similary to triangle $AC_{1}I$. Done :D

$\left|OI\right|=d$  $M$ and $N$ be the interseciton points with the circle .. Let's say $AI\cap circle = \left\{L\right\}$ $\left(L\neq A \right)$
$\angle BIL = \angle BAI+\angle IBA = \frac{A}{2}+\frac{B}{2}= \frac{A+B}{2}$
$\angle IBL = \angle IBC+\angle LBC = \frac{B}{2}+\frac{A}{2}= \frac{A+B}{2}$
$\Rightarrow \angle BIL = \angle IBL \Rightarrow \left|LB\right|= \left|LI\right|$
from the sinus theorem $\left|LB\right|= 2R.\sin\left(\frac{A}{2}\right)$
$\Rightarrow$ $\left|LI\right|= 2R.\sin\left(\frac{A}{2}\right)$   $-1-$
from the power with respect to point $I$
$\left|MI\right|. \left|IN\right| = \left|AI\right|. \left|IL\right|$ ($MN$ is the diameter through the point $I$)
$\Rightarrow (R-d).(R+d)=\left|AI\right|. \left|IL\right| \Rightarrow R^{2}-d^{2}= \left|AI\right|. \left|IL\right|$   $-2-$
$T$ is the foot of the perpendicular from $I$ to $AB$
$\left|IT\right|=r$ and $\left|AI\right|= \left|IT\right|.\frac{1}{\sin\left(\frac{A}{2}\right)}$  $\Rightarrow \left|AI\right|=r/\sin\left(\frac{A}{2}\right)$  $-3-$
from $-1-$ , $-2-$ and $-3-$
$\Rightarrow R^{2}-d^{2}=2R.\sin\left(\frac{A}{2}\right).r.\frac{1}{\sin\left(\frac{A}{2}\right)}$
$\Rightarrow R^{2}-d^{2}=2Rr$
$\Rightarrow \left|OI\right|=d=\sqrt{R(R-2r)}$

Yes it is.

Anyway, congratulations both of you  :)

[quote="N.T.TUAN"]Nice, But MathTeX speak : Without trig... :D[/quote]
but i saw it later  :D  so nothing to do

You can use inversion with pole the incenter. And use the fact that circumcircle is transformed into cirle with radius $\frac{r}{2}$.

[quote="N.T.TUAN"]Call incenter is $ I$ and circumcenter is $ O$, $ AI$ cut $ (O)$ at $ A_{1}\not \equal{} A$, $ A_{1}O$ cut $ (O)$ at $ B_{1}\not \equal{} A_{1}$, $ IC_{1}\perp CA$. We have $ OI^{2} \minus{} R^{2} \equal{} \minus{} IA.IA_{1}$ and $ IA.IA_{1} \equal{} 2R.r$ because triangle $ B_{1}BA_{1}$ similary to triangle $ AC_{1}I$. Done :D[/quote]
Shouldn't it be $ AI/r\equal{}2R/A'B$ (unless $ A'B\equal{}IA'$)? (let $ A_1\equal{}A'$)
If $ A'B\equal{}IA'$, then $ \angle A'IB\equal{}\angle A'BI$. But $ \angle A'IB\equal{}90\minus{}\angle C'AI\equal{}90\minus{}\angle IAB$, and $ \angle A'IB\equal{}\angle IAB\plus{}90\minus{}A'BI$ Since $ \angle IAB\equal{}90\minus{}\angle A'IB$ we get $ \angle A'IB\equal{}90\minus{}\angle A'IB\plus{}90\minus{}A'BI\implies 2\angle A'IB\equal{}180\minus{}\angle A'BI$.
If it is indeed $ A'B\equal{}IA'$, then wouldn't $ \triangle A'IB$ be equilateral? If so, how to prove it?

Could someone please show why $ IA.IA_{1}\equal{} 2R.r$? If this is true, $ \triangle A_1IB$ should be equilateral according to my previous analysis of angle-chasing...  :?:

This is one of the famous Euler's Theorems in the Geometry of Triangle: if O is circumcenter and I - the incenter of triangle ABC, R - circumradius, r - inradius, then
|OI|² = R² - 2Rr
See the attachment (GIF picture) for details.

Proof: let the line OI intersects the circumcircle in D and D', CI - in C', then, according the property of intersecting chords:
|CI|*|IC'| = |DI|*|ID'| or |CI|*|IC'| = (R - |OI|)(R + |OI|) or
|CI|*|IC'| = R² - |OI|²
Triangle AIC' is isosceles - angles IAC' and AIC' are both equal to (α + γ)/2, so
|CI|*|AC'| = R² - |OI|²
Triangle AIT yields |CI| = r/sin(γ/2),
the Law of Sines, applied to ACC' yields |AC'| = 2Rsin(γ/2), hence
|CI|*|AC'| = (r/sin(γ/2))*(2Rsin(γ/2)) = R² - |OI|², finally
|OI|² = R² - 2Rr as required.

More:
1) Since O ≡ I only for an equilateral triangle we have R ≥ 2r: the equality hold if and only if the triangle is equilateral;
2) if E is a midpoint of the base AI of the isosceles triangle AIC' then
2|AE| = |AI| = r/sin(α/2) or r = 2|AE|sin(α/2) = 2|AC'|sin(α/2)sin(β/2) =
= 4Rsin(α/2)sin(β/2)sin(γ/2), or the ratio r/R is:
r/R = 4sin(α/2)sin(β/2)sin(γ/2)
3) Since r/R ≤ 1/2 and  cosα + cosβ + cosγ = 4sin(α/2)sin(β/2)sin(γ/2) + 1
we obtain two very useful inequalities:
sin(α/2)sin(β/2)sin(γ/2) ≤ 1/8 and cosα + cosβ + cosγ ≤ 3/2
Equalities above <=> equilateral triangle.
4) Generalizations of Euler's Theorem: if I_a, I_b, I_c are the centers if the 3 excircles and their radiuses are r_a, r_b, r_c, then:
|OI_a|² = R² + 2Rr_a; |OI_b|² = R² + 2Rr_b; |OIc|² = R² + 2Rr_c
The proof is similar to above.


Right now I am studying the lymphatic system. My book says that it doesn't have a pump(like the heart),but that when certain muscles contract lymph vessels are squeezed. Unfortunately, I'm not sure which muscles they are. My question is, if someone got paralyzed could their lymphatic system stop working, making them very susceptible to diseases?

Funcia

interesting question, and honestly...i don't know. i know that the lymphatic system works because your movement. buy being paralyzed would make it difficult to function. perhaps the system is slightly controlled by the autonomic nervous sytem, because i guess this same problem would happen while your sleeping.... and something must work, otherwise you would die. but i don't know for sure...

Hmmm..what gr8sk8r said about that...could the movement have to do with the lungs-when we're sleeping we breathe(usually) and you'd be able to breathe when paralyzed so could the movement have to do with the lungs? :huh:

I know that paralyzed patients need to be moved frequently and require extra-soft matressess so that their circulation is not blocked. Maybe there is a similar treatment offered to keep the lymph flowing in such patients. :?:

The problem it's not that simple...
The circulation of the lymphatic fluid is pretty slow.It is reglated by the same factors that intervein in the venous circulation:the contraction of regional musculature,the pulse of arteries.And another very important factor is the hydrostatic pressure in the lymphatic capilaries .it has the value of 1 mm Hg,menwhile at the site of conjunction between the main lymphatic vessels and the main trunks it is 0.The lymphatic vessels have also a inner strangulation (valves) formed by a thin layer of conjunctive tissue ,valves witch have the role of helping the upward lymphatic circulation...also if you consider the structure of the lymphatic vessel  you can see that it is formed of three layers(tunica) of which the medium one has in its structure muscular fibres .these fibers can contract especially in the main vessels helping the fluid to move...

So...as you see there are several colateral ways in which lymph may flow
,so i dont think that paralysis should't affect the lymphatic circulation that much...
 :D
Hope it helped...


algebra
polynomial
function
2003 AIME II problem 15

Let
\[P(x)=24x^{24}+\sum_{j=1}^{23}(24-j)(x^{24-j}+x^{24+j}). \]
Let $z_{1},z_{2},\ldots,z_{r}$ be the distinct zeros of $P(x),$ and let $z_{k}^{2}=a_{k}+b_{k}i$ for $k=1,2,\ldots,r,$ where $i=\sqrt{-1},$ and $a_{k}$ and $b_{k}$ are real numbers. Let
\[\sum_{k=1}^{r}|b_{k}|=m+n\sqrt{p}, \]
where $m,$ $n,$ and $p$ are integers and $p$ is not divisible by the square of any prime. Find $m+n+p.$

I'm not sure if I understand the summation notation, but isn't it just x times $(x^{23}+x^{22}+x^{21}... x+1)^{2}?$

[hide="Solution"] The summation is quite daunting, so let's write out the coefficients neatly.

$P(x) = x+2x^{2}+...+24x^{24}+23x^{25}+...+2x^{46}+x^{47}= x(1+x+...+x^{23})^{2}$

As PenguinIntegral noted.  (The expansion of $(1+x+...+x^{n})^{2}$ gives the generating function for the rolling of two $n+1$-sided die, so it may be distantly familiar to you from the case $n = 5$.)

The distinct zeroes of $P(x)$ are just $0$ and the non-unity $24^{th}$ roots of unity.  We can ignore $0$ as a root, so let

$z_{k}= e^{ \frac{2\pi i k}{24}}, k = 1, 2, ... 23$

Then

$z_{k}^{2}= e^{ \frac{\pi i k }{6}}= \cos \frac{\pi k}{6}+i \sin \frac{\pi k}{6}$

This isn't looking so bad; $b_{k}= \sin \frac{\pi k}{6}, k = 1, 2, ... 23$.  These take on the values

$b_{1}= \frac{1}{2}$
$b_{2}= \frac{ \sqrt{3}}{2}$
$b_{3}= 1$
$b_{4}= \frac{ \sqrt{3}}{2}$
$b_{5}= \frac{1}{2}$
$b_{6}= 0$

And so forth.  When summing up the absolute values, we just repeat this chunk $4$ times, ignoring the last $0$, so our summation is

$m+n \sqrt{p}= 8+4 \sqrt{3}$

Ergo, $m+n+p = \boxed{ 15 }$.

I think. :| [/hide]

[quote="t0rajir0u"]
The expansion of $ (1 \plus{} x \plus{} ... \plus{} x^{n})^{2}$ gives the generating function for the rolling of two $ n \plus{} 1$-sided die, so it may be distantly familiar to you from the case $ n \equal{} 5$.

[/quote]

I'm sorry, but what exactly does this polynomial generate?

The coefficient of $ x^k$ gives the number of ways to roll a sum of $ k\plus{}2$ from two $ n\plus{}1$-sided die with sides $ 1, 2, 3, ... n\plus{}1$.

Factored a bit differently than others, but same ending
[hide=Sol]
We can factor this as $x(x+1)^2(x^{22}+x^{20}+x^{18}+\cdots + x^2+1)^2=0.$ Notice we can substitute $y=x^2$ and get rid of the $x(x+1)^2$ because they give roots that we don't care about. Then this becomes $(y^{11}+y^{10}+y^9+\cdots+y^2+y+1)^2=0$, which becomes $\left( \frac{y^12-1}{y-1} \right)^2=0$ which implies $y^{12}-1=0, y\neq 1$. The imaginary part will be $$2 \sum_{n=1}^{11} \left| \sin(30n) \right| .$$ (We multiplied by 2 because if $y=x^2$ is a root, then $y=(-x)^2$ is also a root.) We can simply calculate this value, and we get $8+4\sqrt3,$ or $8+4+3=\boxed{15}$.


Whenever I work on a math problem, I always save the scratch paper I use for the solution. I don't really know why, I just think it's kind of cool to have a record of my work. Since I started several years ago, I now have a fairly sizeable stack of papers!

Does anyone else do this?

if the problem was worked in my spiral notebook, i stil have it. if not, then it is long gone.

Wow, that sounds like a cool idea!  My dad would never let me do that, he hates it when I leave anything lying around (even a pencil).

Then keep your scratch paper in a binder or something and keep it out of sight.

I don't get your signature... :(

I don't keep my scratch paper for the record; I just forget to throw it away :blush:

I use a spiral
scratch paper usually ends in the trash

i keep a spiral notebook, too
scrap paper gets buried under my other stuff

Yea, I save a whole bunch of stuff.  My dad keeps yelling at me to throw a lot of it away, but its kind of hard.

[size=59][color=white]iwhwiwt[/color][/size]

[quote="mathnerd314"][quote="ZennyK"]I use spiral notebooks, too. And legal pads sometimes . . . and junk mail envelopes . . . and napkins . . . and the margins of history books and just about anything else that happens to be lying around when an idea pops into my head :D[/quote]

But sometimes you have a marvelous 200-page number-theory proof that just won't fit into the margin! :D ;)[/quote]
 :rotfl: rofl! :lol: yeah it's a real bummer when that happens ;)

sometimes i have to clean my room so i have to throw away those papers because i wouldn't have more place left :(

I keep paper under my keyboard, so I always have something to write on.  My desk is [i]covered [/i]with used sheets of scratch paper.

Same here.  I started doing Aops stuff over the summer...so yeah.  I just kinda sat at the computer for somet ime each day and did problems here.  I just used whatever paper was at hand as scratch.  I wrote and drew very small things.  So then after a LOT of paper, my mom got irritated and told me to use this graph paper spiral and gave it tome.  So I have used that or a while.  It's about halfway filled.

i never keep my scrap paper.  I always throw it away. :D

I have notebooks lying around with scratch work in them.  I also have scratch paper all over the place.  I try to throw it away as soon as the sheet is filled to stay neat...but my house is still very messy!

[quote="solafidefarms"]Envelopes, desks, hands, arms, ... anything, especially at competitions. :o Oh, and did I mention windows? :oops:[/quote]

windows? wtf?

[quote="1234567890"]I just randomly write on a sheet of paper that is close to me when I see a cool problem. Unfortunately, that means I waste expensive printer paper.[/quote]
lol that's the only paper I use for scratch work.

My grand mother is a neat freak and she throws it all away.


[quote="fjwxcsl"]The inequality inversely holds.[/quote]
Why?  Try $a=3$ and $b=c=0.$ :wink:

Arqady, I reduced the inequality to proving that $ ab(b+2)+bc(c+2)+ca(a+2)\leq9,$ which looks nice enough :)
But I don`t think it`s allways true..

[quote="andyciup"]Arqady, I reduced the inequality to proving that $ab(b+2)+bc(c+2)+ca(a+2)\leq9,$ which looks nice enough :)
But I don`t think it`s allways true..[/quote]
It's true! $ab(b+2)+bc(c+2)+ca(a+2)\leq9\Leftrightarrow\sum_{cyc}a(a-c)^{2}\geq0.$

[quote="andyciup"]Arqady, I reduced the inequality to proving that $ab(b+2)+bc(c+2)+ca(a+2)\leq9,$ [/quote]

How you reduced it to this one?

Actually, Yes!
Here is how I reduced the problem:
\[ \frac{5a}{b^{2}+5}=a-\frac{ab^{2}}{b^{2}+5},\]so it suffices to show that \[ \sum \frac{ab^{2}}{b^{2}+5}\leq\frac{1}{2}.\]But $\frac{ab^{2}}{b^{2}+5}\leq\frac{ab(b+2)}{18}$ :)

Very nice .  :)  . Thank you for this great solution and Arqady thank you for the great problem .  :lol:

[quote="andyciup"]
But $\frac{ab^{2}}{b^{2}+5}\leq\frac{ab(b+2)}{18}$ :)[/quote]
I think, it's wrong. Try $a=1$ and $b=1.5.$ :wink:
$\frac{ab^{2}}{b^{2}+5}\leq\frac{ab(2b+1)}{18}$ is true
but $ab(2b+1)+bc(2c+1)+ca(2a+1)\leq9$ is wrong. :(

i was thinking of using cauchy:

$(\sum\frac{ab^{2}}{b^{2}+5})^{2}\leq (\sum\frac{a^{2}}{b^{2}+5})(\sum\frac{b^{4}}{b^{2}+5})$

even these sums look pretty nice, they seemed easier to deal with before i actually started to..  :(

[quote="arqady"]Let $a,$ $b$ and $c$ are non-negative numbers such that $a+b+c=3.$
Prove that $\frac{a}{b^{2}+5}+\frac{b}{c^{2}+5}+\frac{c}{a^{2}+5}\ge \frac{1}{2}.$[/quote]

Let $a=3x$, $b=3y$, $c=3z$. Thus the inequality is equivalent with:

$\sum{\frac{x}{9y^{2}+5}}\geq \frac{1}{6}$, with $x+y+z=1$.

So by weighted Jensen appyled for the convex function $f(t) = \frac{1}{9t^{2}+5}$, we have that:

$\sum{\frac{x}{9y^{2}+5}}= xf(y)+yf(z)+zf(x) \geq f(xy+yz+zx) = \frac{1}{9(\sum{xy})^{2}+5}\geq \frac{1}{6}$, the last one beeing true because $3(xy+yz+zx) \leq 1$.

[quote="pohoatza"]
   $f: [0,1]\to \mathbb R$
 convex function $f(t) = \frac{1}{9t^{2}+5}$,

[/quote]

Are you sure ? :maybe:.Because $f''$ tells me somthing else.($f''(x)\geq 0$ is not true for all $x\in [0,1]$)


trigonometry
real analysis
real analysis unsolved


Hi everybody,

If I did not make an error of miscalculation, here is a statement about primality. Is it new? I do not know.
How to prove it?
I expect that it will not be easier to do it.
Maybe it is just another way to write Wilson's theorem.
Thank you for any comment or explanation.

Ps : I'm sorry I forgotten to say that p>=7

It's Wilson's theorem.  I'll leave it as an exercise for you.

Why does there appear to be a "rediscovered Wilson's theorem" trend on this board lately?   :|

There is another statement to come where a better expression will be put in place.
I will post it when finished.
This statement is hard to prove I think. I'm not 100% sure anyway.


I can't see how to solve this one...

Might it be intermediate?

[quote="chess64"]Prove that the tens digit of $x^y$ repeat every 100 for all $x$.[/quote][quote="chess64"]In other words, prove that $x^{89808098} \equiv x^{98} \pmod {100}$--they have the same units and tens digits.[/quote]I'm not getting this at all.. the original post is unclear to me and I don't see how the second relates to the first. Can someone explain what the question is?

i think i get it, but it is very badly worded.


logarithms
integration
inequalities
calculus
derivative
calculus computations

If $ x,y$ are positive numbers, then

$ |x\ln{x} \minus{} y\ln{y}|\leq \sqrt [3]{|x^2 \minus{} y^2|} \plus{} \sqrt [3]{|x^2 \minus{} y^2|^2},$

with equality if and only if $ x \equal{} y.$

[quote="ye109"][quote="Ji Chen"]If $ x,y$ are positive numbers, then

$ |x\ln{x} - y\ln{y}|\leq \sqrt [3]{|x^2 - y^2|} + \sqrt [3]{|x^2 - y^2|^2},$

with equality if and only if $ x = y.$[/quote]
$ |x\ln{x} - y\ln{y}|\leq \sqrt [3]{|x^2 - y^2|} + \sqrt [3]{|x^2 - y^2|^2},$
let $ x\geq y$,and $ t = x^2 - y^2 \geq 0,b = y^2 > 0$
it's equal to$ |{\int_{b}^{b + t}\frac {\ln(x) + 2}{4\sqrt {x}}}dx |\leq \sqrt [3]{t} + \sqrt [3]{t^2}$
just need to prove when $ b \geq \frac {1}{e^2}$ or $ b + t \leq \frac {1}{e^2}$
when $ b\geq \frac {1}{e^2}$
$ {\int_{b}^{b + t}\frac {\ln(x) + 2}{4\sqrt {x}}}dx \leq \sqrt [3]{t} + \sqrt [3]{t^2}$
$ \Leftarrow{\int_{b}^{b + t}\frac {\ln(x) + 2}{4\sqrt {x}}}dx \leq \sqrt [3]{t + b} + \sqrt [3]{(t + b)^2} - (\sqrt [3]{b} + \sqrt [3]{b^2})$
$ = \int_{b}^{b + t}(\frac {1}{3}x^{\frac { - 2}{3}} + \frac {2}{3}x^{\frac {1}{3}})dx ($when $ b\geq \frac {1}{e^2}$)
when $ b + t\leq \frac {1}{e^2}$
$ { - \int_{b}^{b + t}\frac {\ln(x) + 2}{4\sqrt {x}}}dx \leq \sqrt [3]{t} + \sqrt [3]{t^2}$
$ \Leftarrow{ - \int_{b}^{b + t}\frac {\ln(x) + 2}{4\sqrt {x}}}dx \leq \sqrt [3]{t + b} + \sqrt [3]{(t + b)^2} - (\sqrt [3]{b} + \sqrt [3]{b^2})$
$ = \int_{b}^{b + t}(\frac {1}{3}x^{\frac { - 2}{3}} + \frac {2}{3}x^{\frac {1}{3}})dx$ (when $ b + t\leq \frac {1}{e^2}$)
so we only want to show that
${ \frac {\ln(x) + 2}{4\sqrt {x}}}\leq \frac {1}{3}x^{\frac { - 2}{3}} + \frac {2}{3}x^{\frac {1}{3}}$when $ b\geq \frac {1}{e^2}$
and${ \frac { - \ln(x) - 2}{4\sqrt {x}}}\leq \frac {1}{3}x^{\frac { - 2}{3}} + \frac {2}{3}x^{\frac {1}{3}}$when $ b + t \leq \frac {1}{e^2}$
it's easy to solve that
when t=0 ,the inequality holds[/quote]

Perfect job,it has puzzled me for several days


As an 8th grader who scored honor roll on the AMC 8, I've found myself in a bit of a conundrum: I've been invited to take the AMC 10 and some of the questions are over my head. Would someone mind going over some of the harder problems from older (specifically, last year's B) AMC 10s?

Thanks.

There were three "Math Jams" on this site that covered the harder questions on recent AMC 10 (and 12) tests.  Here are the links:

[url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=48]2005 AMC 10/12 A Math Jam[/url]

[url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=28]2004 AMC 10/12 B Math Jam[/url]

[url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=25]2004 AMC 10/12 A Math Jam[/url]

If there's any specific questions you'd like to ask about, I'm sure I and many others would not mind helping you with the ideas.


Sequence $ \{a_n\}$ set condition $ a_{n}\equal{}a_{n\minus{}1}\minus{}\frac{2}{n}a_{n\minus{}2}$ for $ n\geq 2$ ($ a_{0}$ and $ a_{1}$ are any numbers). Prove that $ \sum_{n\equal{}0}^{\infty}{|a_n|}<\infty$.


I have yet again another question. Usally i don't have any problems doing functions, but when i work with sin- & cos-functions i get problems. Here's my task:

f(x)=2*cos(2x)+1, x€[0;phi]

Can anyone explain to me how do draw this equation? I have the TI-84 plus calculator, and here's a couple of spots i can use to draw from:

[u]x=0[/u]
3

[u]phi[/u]
2,9880

[u]0,5phi[/u]
2,9970

Are these results right or wrong?

[quote="Sporally"]I have yet again another question. Usally i don't have any problems doing functions, but when i work with sin- & cos-functions i get problems. Here's my task:

f(x)=2*cos(2x)+1, x€[0;phi]

[/quote]

well if you let f(x)=cosx
then we have to draw 2f(2x)+1
so translate the normal graph of y=cosx first by moving it 1 up the y axis
then double the y coordinates and half the x coordinates to get the graph function which you need.

I have a TI-84 plus. Do you know if i can make it show me the function? And what should i write in the 'y=' tag?

Can anyone plz tell me how to make my TI-84 show the graph?


Find all real-valued functions $ f$ on the reals which have at most finitely many zeros and satisfy $ f(x^4 \plus{} y) \equal{} x^3f(x) \plus{} f(f(y))$ for all $ x, y$.

It is APMO contest .
The condition $ f(x)$ have finite zeros is not necessary .
[u]Solution [/u]
1. Prove $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$
2. Consider $ f(x\plus{}1)$ in two ways. Use properties :
If $ P(x)\equal{}g(x)$ for infinite $ x\in R$ then $ f(x)\equal{}g(x)$
It follow that :
$ f(x)\equal{}0,f(x)\equal{}x$ is solution.[/code]


function
logarithms
number theory
Diophantine equation
algebra
linear equation
algebra proposed

the function $ f(x) \equal{} x^{\frac{1}{x}}$ is strictly monotonic in the intervals $ (\minus{}\infty, e]$ and $ [e, \infty)$.

So if we rewrite the equation as, $ x^{\frac{1}{x}} \equal{} 2^{\frac{1}{2}}$.

Obviously x=2 is a solution. There can be only one another solution. Which is x=4.

No other solutions exist.

[quote="hsbhatt"]Obviously x=2 is a solution. There can be only one another solution. Which is x=4.

No other solutions exist.[/quote]
I haven't read your post totally. But I have a feeling there is a negative solution.

$ f(x) \equal{} x^2 \minus{} 2^x$ is continuous.

$ f(0) \equal{} \minus{} 1$
$ f( \minus{} 1) \equal{} \frac {1}{2}$
And mean value theorem.

Here is the graph.solutions  2 and other two are -0.784 and 4
the negative is in approx .
Correct if any one can get more approx value.

The negative solution can be written in the form of:

${ x_0 = - \frac{2}{\ln 2} W(\frac{\ln 2}{2}})$

where $ W(\cdot)$ is the Lambert W-function which can be approximated to any desired accuracy with:

$ W(x) = \sum^{\infty}_{n = 1} \frac{(-1)^n n^{n-2}}{(n-1)!} x^n$

This solution can be verified knowing that

$ W(x) e^{W(x)} =x$

Now:

${{ (- \frac{2}{\ln 2} W(\frac{\ln 2}{2}}))^2 = 2 ^{- \frac{2}{\ln 2} W(\frac{\ln 2}{2}})}$

${{ (- e^{-W(\frac{\ln 2}{2}})})^2 = 2 ^{- \frac{2}{\ln 2} W(\frac{\ln 2}{2}})}$

${{ (- e^{-W(\frac{\ln 2}{2}})})^2 = e ^{- \ln 2 \frac{2}{\ln 2} W(\frac{\ln 2}{2}})}$

${{ e^{-2 W(\frac{\ln 2}{2}})} = e ^{- 2 W(\frac{\ln 2}{2}})}$

$ Q.E.D$

thanx. i had read about this, only now its sinking into my consciousness  :)

$ x \equal{} 2$ or $ x \equal{} 4$ or $ x \simeq \minus{} 0,766664696...$
 We can solve with $ x > 0$. I use my computer to slove $ x \simeq \minus{} 0,766664696...$

[quote="hsbhatt"]the function $ f(x) \equal{} x^{\frac {1}{x}}$ is strictly monotonic in the intervals $ ( \minus{} \infty, e]$ and $ [e, \infty)$.

So if we rewrite the equation as, $ x^{\frac {1}{x}} \equal{} 2^{\frac {1}{2}}$.

Obviously x=2 is a solution. There can be only one another solution. Which is x=4.

No other solutions exist.[/quote]
 :D  :D  :blush:  :D

[quote="milin"]The negative solution can be written in the form of:

${ x_0 = - \frac {2}{\ln 2} W(\frac {\ln 2}{2}})$

where $ W(\cdot)$ is the Lambert W-function which can be approximated to any desired accuracy with:

$ W(x) = \sum^{\infty}_{n = 1} \frac {( - 1)^n n^{n - 2}}{(n - 1)!} x^n$

This solution can be verified knowing that

$ W(x) e^{W(x)} = x$

Now:

${{ ( - \frac {2}{\ln 2} W(\frac {\ln 2}{2}}))^2 = 2 ^{ - \frac {2}{\ln 2} W(\frac {\ln 2}{2}})}$

${{ ( - e^{ - W(\frac {\ln 2}{2}})})^2 = 2 ^{ - \frac {2}{\ln 2} W(\frac {\ln 2}{2}})}$

${{ ( - e^{ - W(\frac {\ln 2}{2}})})^2 = e ^{ - \ln 2 \frac {2}{\ln 2} W(\frac {\ln 2}{2}})}$

${{ e^{ - 2 W(\frac {\ln 2}{2}})} = e ^{ - 2 W(\frac {\ln 2}{2}})}$

$ Q.E.D$[/quote]

Huh! It's so hard!  :oops:

Hard? No! I can give you complicated, at best.. ;)

Basically, you assumed this is a number theory problem-a Diophantine equation-your first assumption clearly shows that. Being posted in the Algebra subforum, it's assumed on the other hand by the other solvers that the set in which the equation is being solved is $ \mathbb{R}$ and not $ \mathbb{N}$ or $ \mathbb{Z}$.

[quote="tim26270746"][quote="hungctqb"]Solve: $ x^2 \equal{} 2^x$[/quote]

If I am correct,
This problem is easy.
Because all divisors of 2^x are powers of 2,
thus x=2^t for some t.

Then we can have equation 2t=x=2^t
then we have t=2^(t-1)
Thus, this problem becomes very easy and the answers are t=1 or t=2
which means x=2 or x=4[/quote]
I don't think so:$ x \in R$

[quote="milin"]The negative solution can be written in the form of:

${ x_0 = - \frac {2}{\ln 2} W(\frac {\ln 2}{2}})$

where $ W(\cdot)$ is the Lambert W-function which can be approximated to any desired accuracy with:

$ W(x) = \sum^{\infty}_{n = 1} \frac {( - 1)^n n^{n - 2}}{(n - 1)!} x^n$

This solution can be verified knowing that

$ W(x) e^{W(x)} = x$

Now:

${{ ( - \frac {2}{\ln 2} W(\frac {\ln 2}{2}}))^2 = 2 ^{ - \frac {2}{\ln 2} W(\frac {\ln 2}{2}})}$

${{ ( - e^{ - W(\frac {\ln 2}{2}})})^2 = 2 ^{ - \frac {2}{\ln 2} W(\frac {\ln 2}{2}})}$

${{ ( - e^{ - W(\frac {\ln 2}{2}})})^2 = e ^{ - \ln 2 \frac {2}{\ln 2} W(\frac {\ln 2}{2}})}$

${{ e^{ - 2 W(\frac {\ln 2}{2}})} = e ^{ - 2 W(\frac {\ln 2}{2}})}$

$ Q.E.D$[/quote]

I thing that's correct! :wink:

$ x^{2}\equal{}2^{x}$. If $ x\equal{}2k$,
\[ (2k)^{2}\equal{}2^{2k}\] \[ 2k\equal{}2^{k}\] \[ k\equal{}2^{k\minus{}1}\] \[ 2\equal{}k^{\frac{1}{k\minus{}1}}\]
Obviously $ k$ needs to be a power of $ 2$. Therefore, if $ k\equal{}2^{m}$,
\[ 2\equal{}(2^{m})^{\frac{1}{2^{m}\minus{}1}}\]
\[ 2\equal{}2^{\frac{m}{2^{m}\minus{}1}}\] and then $ \frac{m}{2^{m}\minus{}1}\equal{}1$. This means \[ m\equal{}2^{m}\minus{}1\] \[ m\plus{}1\equal{}2^{m}\] Since $ m\plus{}1$ is a linear equation and $ 2^{m}$ is an exponential curve, they can intersect at a maximum of $ 2$ points. This means there are only $ 2$ possible values for $ k$ and therefore only $ 2$ possible values for $ 2k\equal{}x$. Since the equation works for $ x\equal{}2,4$ these are the two possible values.

what's wrong with my solution that made me not get the negative solution?

[quote="t0rajir0u"]Setting $ k \equal{} 2^m$ restricts $ k$ to positive values if $ m$ is taken to be real.[/quote]

Oh right... didn't notice that


Let $ S(n)$ be the number of digits of $ n^2$. Prove that \[ \frac{1}{1\cdot S(1)}\plus{}\frac{1}{2\cdot S(2)}\plus{}\frac{1}{3\cdot S(3)}\plus{}\ldots \plus{}\frac{1}{n\cdot S(n)}<6\]

 :(

It's false -- the sum is strictly larger than ${ \frac {\ln{10}}{2}\sum \frac {1}{n \ln n}}$, and this expression exceeds 6 when $ n$ is on the order of $ 4 \cdot 10^{79}$ (that is, $ \exp(\exp(\frac {12}{\ln 10}))$).  That it diverges follows from the integral test.


What is the ratio of width to length of a folded sheet of paper if the rectangle formed when the unfolded sheet is folded in half so the new longest side is the shortest side of the unfolded paper and the shortest side is half the unfoled longest side, when the the folded paper is similar to the unfolded rectangle.

Behind the messy wording, the thing is simple:

[hide]$\frac{b}{a/2}=\frac{a}{b}\implies b^2=\frac{a^2}{2}\implies \frac{a^2}{b^2}=2\implies \frac{a}{b}=\sqrt{2}$

Note: That dimension ratio is actually used in European paper size standards $A0-A6$[/hide]

I thought so, but I saw interpertations in two different ways. I just wanted to make sure someone else got the same answer as I:(.


$\text{Basic ones}: -$

1)If $cos(\alpha-\beta)+cos(\beta-\gamma)+cos(\gamma-\alpha)= \frac{-3}{2} ,prove%Error. "that" is a bad command.
$

$cos\alpha + cos\beta + cos\gamma= sin\alpha + sin\beta + sin\gamma = 0$

2)If $tanx+tan(x+\frac{\pi}{3}) + tan(x+\frac{2\pi}{3}) = 3,prove%Error. "that" is a bad command.
$

$\frac{3tanx-tan^3x}{1-3tan^2x}=1$

[hide="1"]
$\cos(a-b) + \cos (b-c) + \cos (c-a) = -\displaystyle \frac{3}{2}$

$\cos a \cos b + \sin a \sin b + \cos b \cos c + \sin b \sin c + \cos a \cos c + \sin a \sin c = -\displaystyle \frac{3}{2}$

Adding $(\cos a + \cos b + \cos c)^2$ and $(\sin a+\sin b+\sin c)^2$ together, substituting the above expression and using the identity $\sin^2 \theta + \cos^2 \theta = 1$ yields a sum of zero.

Since $(\cos a + \cos b + \cos c)^2 + (\sin a+\sin b+\sin c)^2=0$, both $(\cos a + \cos b + \cos c)^2$ and $(\sin a+\sin b+\sin c)^2$ must equal zero.

[/hide]

[hide="2"] Use the identity $\tan(a+b)=\frac{\tan a + \tan b}{1-\tan a \tan b}$ to get $\tan x + \frac{\tan x + \sqrt{3}}{1-\sqrt{3} \tan x} + \frac{\tan x - \sqrt{3}\tanh}{1+\sqrt{3}}=3$. Then, $9\tanh - 3\tan^3 x = 3 - 9 \tan^2 x \implies \frac{3\tan x - \tan^3 x}{1-3\tan^2 x}=1$. [/hide]
Kinda easy... maybe things like this should go into Beginning forum?


[quote="N.T.TUAN"]We examine the following two sequences: The Fibonacci sequence: $F_{0}= 0, F_{1}= 1, F_{n}= F_{n-1}+F_{n-2 }$ for $n \geq 2$; The Lucas sequence: $L_{0}= 2, L_{1}= 1, L_{n}= L_{n-1}+L_{n-2}$ for $n \geq 2$. It is known that for all $n \geq 0$
\[F_{n}=\frac{\alpha^{n}-\beta^{n}}{\sqrt{5}},L_{n}=\alpha^{n}+\beta^{n}, \]
where $\alpha=\frac{1+\sqrt{5}}{2},\beta=\frac{1-\sqrt{5}}{2}$. These formulae can be used without proof.

Show that $L_{2n+1}+(-1)^{n+1}(n \geq 1)$ can be written as a product of three (not necessarily distinct) Fibonacci numbers.[/quote]
Obviosly $L_{n}L_{n+1}=\alpha^{2n+1}+\beta^{2n+1}+(-1)^{n}$.

[quote="Rust"]
Obviosly $L_{n}L_{n+1}=\alpha^{2n+1}+\beta^{2n+1}+(-1)^{n}$.[/quote]
[quote="Rust"]$5*F_{n}F_{n+1}=\alpha^{2n+1}+\beta^{2n+1}-\alpha^{n}\beta^{n}(\alpha+\beta)=L_{2n+1}+(-1)^{n+1}$,
[/quote]

that's why one had problems in understanding your answer ;-)


function
integration
inequalities
advanced fields
advanced fields unsolved

I have two questions on weak solutions. Consider the equation \[\ddot x=f(t,x).\] with $x\colon[a,b]\to\mathbf R^{n}$, $\Omega\subset\mathbf R^{n}$ open and $f\in L^{p}([a,b]\times\Omega)$.

In this case $x\in L^{p}(a,b)$ is a weak solution if there exists sequence $(x_{n})$ of $C^\infty(a,b)$ functions such that

(1) $x_{n}\to x$ in $L^{p}(a,b)$
(2) $\dot x_{n}$ converges in $L^{p}(a,b)$
(3) $\ddot x_{n}\to f(\cdot,x)$ in $L^{p}(a,b)$.

But it seems natural to avoid (2).
Here're the questions:

1. Do we really need (2)?
2. It is possible that (1) and (2) are fulfilled while (2) not?

what does $x^{''}_{n}\to f$ in $L^{p}$ mean here ?

A more general question would be:  If $x_{n}$ and $x_{n}''$ both converge in $L^{p}$, must it be true that a subsequence of $x_{n}'$ converges in $L^{p}$?  I think the answer is yes.

Proof:  Bounds on the $L^{p}$ norms of $x_{n}$ and $x_{n}''$ give you a bound on the $L^{p}$ norm of $x_{n}'$.  (This is called interpolation.)  And now that we have that all three are bounded, the result follows by the Rellich Theorem.  (Or if you prefer, Arzela-Ascoli plus addition argument.)

I'm not sure if you really have to take a subsequence here (that is, I don't know whether $x_{n}'$ must itself converge with these hypotheses), but I don't immediately see an obvious proof or counterexample.  Maybe someone else does.

could you developpe a little bit more the interpolation part ? I'm interested. Thank you  :)

I posted this topic because I don't understand the following thing. The Sobolev space $W^{k,p}(\Omega)$ is used for solving the equation of the form \[P_{k}u=f\text{ in }\Omega\] with $(P_{k}u)(x)=\sum_{|\alpha|\le k}a_\alpha(x)\partial\,^\alpha u(x)$ and $f\in L^{p}(\Omega)$.

But if for some $0<|\alpha_{0}|< k$ $a_\alpha=0$ it seems natural not to demand the existence of $\partial\,^{\alpha_{0}}u$ while $u\in W^{k,p}(\Omega)$ automatically implies that $\partial\,^{\alpha_{0}}u\in L^{p}(\Omega)$ exists.

In the post $k=2$ and $\alpha_{0}=1$.

alekk:  Here's a proof for a function $f$ on the interval $[0,3]$.  Choose any point $x'\in[0,1]$ and any point $x''\in[2,3]$.  By the MVT, there is some point $y$ with $|f'(y)|\leq|f(x'')-f(x')|$.  Using this point and Fund Thm Calc, we see that for ANY $x\in[0,3]$, $|f'(x)|\leq |f(x')|+|f(x'')|+\int_{0}^{3}|f''|$.  Now integrate over the $x'$ and $x''$ variables to obtain $|f'(x)|\leq \int_{0}^{3}|f|+\int_{0}^{3}|f''|$.    Now use Holder and integrate the $x$-variable, and voila.

Vladimir:  The philosophical answer is that the point of Sobolev spaces is NOT that we're looking for the minimum regularity needed to interpret the equation.  Indeed, that is what distributions are for.  The point of Sobolev spaces is that they have properties that are very useful for us.  Specifically, they behave a bit like the familiar $C^{k}$ spaces, except that they have some nicer properties wrt PDEs.  In particular, they ought to be nested.

A different sort of answer to your question:  It is true that the Poincare inequality tells us that the norm on $W^{k,p}_{0}(\Omega)$ IS equivalent to the norm you get by just looking at the $k$-th derivatives.  This is NOT true for  $W^{k,p}(\Omega)$, but it is reasonable to ask whether the norm here is equivalent to the norm you get by just looking at the $k$-th AND $0$-th order deriviatives.  A similar statement is true for the $C^{k}$ spaces (and Holder too), so I wouldn't be surprised if it were true here.  But unfortunately, I don't quite see a proof.  Philosophically speaking again, even if it is true, I doubt it would be very useful, and that's why it won't show up in your textbook.  (Indeed, I only know the interpolation proof I gave above, because it's a less refined version of the usual sort of interpolation theorem, which IS important.)


Dear mathlinkers
Would you tell me where can i find the solutions of problems of In the world of mathematics specially year 2000-2008.

Its not that nobody [i]wants[/i] to help, its that nobody [i]can[/i] help!


Solve for x,y,z
$ \left\{ \begin{array}{l}
 x \equal{} y^3  \minus{} 3y \\
 y \equal{} z^3  \minus{} 3z \\
 z \equal{} x^3  \minus{} x \\
 \end{array} \right.$

[hide="Partial Solution"]Let $ f(t) \equal{} t^3 \minus{} 3t$. Notice that since $ f(x) \equal{} z$, $ f(z) \equal{} y$, and $ f(y) \equal{} x$, we have that $ f(f(f(x))) \equal{} x$. In the intervals $ ( \minus{} \infty, 2]$ and $ [  2, \infty)$, $ f(t)$ is increasing, so we may assume that $ f(f(f(x))) \equal{} f(x) \equal{} x$ (in those intervals). So,

$ x^3 \minus{} 3x \equal{} x$

$ x^3 \minus{} 4x \equal{} 0$

$ x(x \minus{} 2)(x \plus{} 2) \equal{} 0$

Hence, we have solutions $ (0,0,0), (2,2,2),$ and $ ( \minus{} 2, \minus{} 2, \minus{} 2)$. However, I don't know find solutions in the interval $ ( \minus{} 2,2)$. [/hide]

[quote="mrmohamed"][quote="JakiroVn"]Solve for x,y,z
$ \left\{\begin{array}{l} x \equal{} y^3 \minus{} 3y \\
y \equal{} z^3 \minus{} 3z \\
z \equal{} x^3 \minus{} x \\
\end{array} \right.$[/quote]
hi
is th third equation
z= x^3 - 3x ??[/quote]

In either case, there are quite a few ugly (real) solutions. Maybe we need another condition on $ x,y,z$...?

[b]Edit:[/b] Heh, I didn't think it through.

[quote="math154"]

In either case, there are quite a few ugly (real) solutions. Maybe we need another condition on $ x,y,z$...?[/quote]

After plotting the two curves at http://www.quickmath.com,
I counted 27 real solutions.

At http://www.quickmath.com

Click to these in succession:
"Equations" -> "Solve" -> "Advanced"

Setting:
Entered as these horizontal forms:
x = y^3 - 3y
y = (x^3 - 3x)^3 - 3(x^3 - 3x) . . . after substituting the z's expression into y,
and using the supposed $ z \equal{} x^3 \minus{} 3x$ version instead of what was given.

Suggest using two different colors.

Suggest:
x runs from -4 to 4
y runs from -4 to 4

Click "Tick marks" and "Grid lines."

Leave it on (or click on "Automatic origin.")

I suggest it be at "One to one."

You can leave it at the text output size of 20.

Then click "Plot" in the bottom right-hand corner.

It's already been shown that outside $ [ \minus{} 2, 2]$ that there are no solutions.

Let $ x \equal{} 2\cos a$, $ y \equal{} 2\cos b$, and $ z \equal{} 2\cos c$.  When plugged in, we get:

$ 2\cos a \equal{} (2\cos b)^3 \minus{} 3(2\cos b) \equal{} 2(4\cos^3 b \minus{} 3\cos b) \equal{} 2\cos 3b$

By the same token, $ 2\cos b \equal{} 2\cos 3c$ and $ 2\cos c \equal{} 2\cos 3a$.

If we use the triple angle formula on $ \cos a \equal{} \cos 3b$, we get that $ \cos c \equal{} \cos 3a \equal{} \cos 9b$.  Doing it again, we get $ \cos b \equal{} \cos 3c \equal{} \cos 27b$.  Then $ b \equal{} \pm 27b \plus{} 2\pi k$ or:

$ 26b \equal{} 2\pi k\Rightarrow b \equal{} \frac {\pi k}{13}$.
$ 28b \equal{} 2\pi k\Rightarrow b \equal{} \frac {\pi k}{14}$.

When $ 0\leq b\leq\pi$ there are new solutions for $ y$.  Thus if $ b\in$
$ \{0, \frac {\pi}{14}, \frac {\pi}{13}, \frac {\pi}{7}, \frac {2\pi}{13}, \frac {3\pi}{14}, \frac {3\pi}{13}, \frac {2\pi}{7}, \frac {4\pi}{13},$
$ \frac {5\pi}{14}, \frac {5\pi}{13}, \frac {3\pi}{7}, \frac {6\pi}{13}, \frac {\pi}{2}, \frac {7\pi}{13}, \frac {4\pi}{7}, \frac {8\pi}{13}, \frac {9\pi}{14},$
$ \frac {9\pi}{13}, \frac {5\pi}{7}, \frac {10\pi}{13}, \frac {11\pi}{14}, \frac {11\pi}{13}, \frac {6\pi}{7}, \frac {12\pi}{13}, \frac {13\pi}{14}, \pi\}$, then $ 2\cos b \equal{} y$, $ 2\cos 3b \equal{} x$, and $ 2\cos 9b \equal{} c$ works.

This is 27 solutions, and if you take the algebraic approach and plug in the equations 3 times, you get a degree $ 3\cdot3\cdot3$ or degree 27 equation, so there are at most 27 solutions.  We have found them all.


function
floor function
irrational number
real analysis
real analysis unsolved

if function f is Riemann integrable on [a, b]
is [f] also integrable, where [] means the integer part of f

No, this is false.

On $ [0,1],$ let $ f(x)\equal{}\begin{cases}1\minus{}\frac1q&\text{ if }x\equal{}\frac{p}{q}\in\mathbb{Q}\\1&\text{ otherwise}\end{cases}.$

Then $ f,$ which is continuous at every irrational number, is Riemann integrable. But $ \lfloor f(x)\rfloor$ is $ 0$ for $ x$ rational and $ 1$ for $ x$ irrational, and hence not Riemann integrable.

thanks
quite elegant
I have once considered negative Riemann function, but not this one!


I am interested in the solution to problem 1, I didnt solve it.

#1 is actually the first [url=http://mathworld.wolfram.com/Rogers-RamanujanIdentities.html]Rogers-Ramanujan Identity[/url].

A weird problem to be put on a contest.

[quote="billzhao"][quote="Fiachra"]I couldn't get it either. Billzhao, do you have a proof (since there isn't one on your link).[/quote]

No.

Though according to Hardy and Wright, there're no "easy" proofs of this identity.

And Schur found a 50-page combinatorial proof of it.[/quote]The proof is at most 2 pages long. I think you are trying to kill a mosquito with a laser-canon Billzhao :)

[quote="Valentin Vornicu"]The proof is at most 2 pages long. I think you are trying to kill a mosquito with a laser-canon Billzhao :)[/quote]

I'd be very interested to see that proof.

well, then lets discuss problems 2 and 3, Fiahcraand Billzhao, how did you do?

#2 is a nice problem in my opinion. I used Menalaus and then Power of a Point.

For #3, after finding the explicit formula for $x_n$, the problem just falls apart.

For #2 I used pespectivity: if we construct a point $F$ on $AD$ so that $FL || CD$, then we get homothetic triangles and the result follows (I hope) fairly easily.

For #3... same as Billzhao. It may have taken me slightly longer though.

ok, thanks! in number 3 i did the same then but it was a cool problem!
for number 2 i used menelaus and sine law, but i think it was a cool problem!

For 2 I just constructed some point and used parallel lines and con-cyclic points.

For 3 I found the explicit formula, and used wolstenholme.

By the way, a solution of number 1 can be found here, if anyone is interested.

[url]http://arxiv.org/PS_cache/math/pdf/0411/0411072.pdf[/url][/url]


Let
$f(x) = \frac{|x|-1}{\sqrt{1-x}}$, when $x<1$
and
$f(x) = \frac{|1-x|}{1-\sqrt{x}}$, when $x>1$

Find $\lim_{x\rightarrow 1}f(x)$


modular arithmetic
quadratics
number theory proposed
number theory

Prove that there are no positive integers $ x,y$ such that $ 3x(x\plus{}1)\plus{}7\equal{}y^3$

[quote="SUPERMAN2"]Prove that there are no positive integers $ x,y$ such that $ 3x(x \plus{} 1) \plus{} 7 \equal{} y^3$[/quote]

$ 3|y^3\minus{}7 \Rightarrow y\equal{}3k\plus{}1$ for some integer $ k$.

So the equation becomes :

$ x(x\plus{}1)\equal{}9k^3\plus{}9k^2\plus{}3k\minus{}2 \equiv 1 \pmod{3}$

[b]Case 1: [/b]$ x \equiv 0 \pmod{3} \Rightarrow x(x\plus{}1) \equiv 0(0\plus{}1) \equiv 0 \pmod{3}$, impossible.
[b]Case 2: [/b]$ x \equiv 1 \pmod{3} \Rightarrow x(x\plus{}1) \equiv 1(1\plus{}1) \equiv 2 \pmod{3}$, impossible.
[b]Case 3: [/b]$ x \equiv 2 \pmod{3} \Rightarrow x(x\plus{}1) \equiv 2(2\plus{}1) \equiv 0 \pmod{3}$, impossible.

So $ 3x(x \plus{} 1) \plus{} 7 \equal{} y^3$ has no integer solutions.

We have that $ p \mid x^p \minus{} x$ for all $ x \in \mathbb{Z} \text{ and }p \in \mathbb{P}$, so $ y^3 \equiv y \pmod 3 \implies y \equiv 7 \equiv 1 \pmod 3$. Substituting (as Stephencheg did) $ y \equal{} 3k \plus{} 1$ we obtain that $ 3 \mid x^2 \plus{} x \minus{} 1 \implies 3 \mid 4(x^2 \plus{} x \minus{} 1) \implies 3 \mid (2x \plus{} 1)^2 \minus{} 5 \cdot 1^2$, but 5 is a quadratic residue of p iff $ 10 \mid p^2 \minus{} 1$, that is not our case.


geometry
circumcircle
trigonometry
symmetry
incenter
angle bisector
power of a point

Here is the "official" text:

Let ABC be a triangle, D be a point on the side BC and omega be the circumcircle of the traingle ABC. Let beta and tetha be the circles tangent to omega, AD,BD and omega,AD,DC respectively. Show that beta and tetha are tangent if and only if : m(BAD)=m(CAD).

Solution: Let DX, DY be the bisector lines of the angles ADB, ADC respectively. We are looking for the centers K,L of the circles beta and tetha such that beta and tetha be tangent to the circle omega. Then M, N be the projections of the point on the segments BD,AD respectively and let k=KM=KN be the radius of the circle beta. Then beta and omega are tangent if and only if OK=R-k(1). Lets denote ADB=2a and m(KDO)=b. Considerinf the lenght x=DN=DM as a variable length of the problem we get DK= x/cosa and k=xtan(a). The the equality (1) becomes OK=R-xtan(a) (1'). Aplying the cosine theorem in the triangle ODK we get: OK^2=DK^2+Do^2-2DK*DO cos b  which is equivalent to:
(R-x tan(a))^2=(x/cos(a))^2+DO^2-2x/cosa * DO * cosb. The last equality can be written under the form:
x^2+2*lambda*x+DO^2-R^2=0 (2)
where lambda=(Rsina-DO cos(b))/cosa (3)

Using a simmilar technique for the point L, we get the equality:
y^2+2*upsilon*y+DO^2-R^2=0 (2') where
upsilon =)R cosa-Do sinb)/sin a and DP=y.
we observe that in this case the angles a and b must be replaced by 90-a and 90-b, respectively.

Now, the condition for the circles beta and tetha to be tanget is: x=y. It means that the equations (2) and (2') have a common root. But this condition is equivalent to lambda=upsilon. In this way, the following equality is obtained: Rcos(2a)=DO sin(b-a). Let observe that b-a=m(ADO). Hence, by the sine theorem in the triangle ADO we obtain:
Rcos(2a)=Rsin(DAO).
Let H denote the foot of the altitude from A in the triangle ABC. Then cos(2a)=sin (HAD). It follows that m(HAD)=m(DAO). Taking into account that the lines AH and AO are isogonal conjugates in the vertex A it follows  that AD is the bisector line of the angle BAC.

all done!!
cheers!! :D

I remember I tried really hard to solve this problem (not at the competition, I was too young! :D ) and eventually I came up with something using an inversion. But that was a year ago. There's no possible way I could retrace the steps of the proof tonight..

There is also a solution using [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=146125]Casey's theorem[/url]!!! And it is much more beautiful!! Smile

Let E be the intersection point of the line AD and the circle omega. Like in the 1st solution, let M,N be the tangent points of the circle beta with BD,AD respectively and let P,Q be the points of the circle tetha with DC and AD. We apply Casey's Theorem fo the circles B,C,E and betha, where the point B,C, E are considered as degenerate circles. SO we get: BE*CM+CE*BM=BC*EN. By applying once again Casey for the circles B,C,E and tetha we get: BE*CP+CE*BP=BC*EQ. the circles beta and tetha are tangent iff N and Q coincide. By the above relations, this condition is equivalent to: BE(CM-CN)=CE(BN-BM) which gives (BE-CE)MN=0.
It follows BE=CE and this is equivalent to BAE=CAE.

cheers! :D

Does anyone have the proof of the converse of Casey's theorem?

I know there is a dutch site with it. If anyone knows it please post it!!!

cheers! :D

You know, my mother tongue is dutch so if this is important for you I could translate the main parts of the proof you're speaking about ... but I certainly won't do this tonight (bedtime ...), maybe tomorrow ...

well I just need the name of the site.. I understand quite well dutch!! :D
no need to bother translating that! :D

cheers!! thank you  anywayz!!

O sorry,

I thought you meant you found the proof on an dutch site and you didn't understand it. You know, I didn't expect to meet another dutch - speaking person on this site ... what a suprise !

The site is

http://www.pandd.demon.nl/casey.htm

PS. Who are you in fact ? Are you a native dutch - speaking person ?

Bye
Tot ziens  :)

Well I am Romanian, but I understand quite good some foreign languages!
I love foreign languages(English, German, Italian, Dutch) :D and I know but I don't like (French,Spanish and Portuguese)! :?

No problem about the misunderstanding!!!

thank you for the name of  the site!!

cheers! :D

I have another problem that is solved using Casey! The problem was given
at the 20th IMO which was held in Bucharest, ROmania.

In the triangle ABC, AB = AC. A circle is tangent internally to the circumcircle of the triangle and also to AB, AC at P, Q respectively. Prove that the midpoint of PQ is the center of the incircle of the triangle.

There is a solution on http://www.kalva.demon.co.uk but  I remember the solution was not using Casey!

I just looked now on that site and the solution is like so:

By symmetry the midpoint, M, is already on the angle bisector of A, so it is sufficient to show it is on the angle bisector of B. Let the angle bisector of A meet the circumcircle again at R. AP is a tangent to the circle touching AB at P, so angle PRQ = angle APQ = angle ABC. Now the quadrilateral PBRM is cyclic because the angles PBR, PMR are both 90. Hence angle PBM = angle PRM = (angle PRQ)/2, so BM does indeed bisect angle B as claimed.

The solution with Casey is also nice!

cheers!! :D

hey Lagrangia, why don't you like french ??
It's a so beautiful langage ;-)

hey ya
french is really beautiful (i studied it for over a year but had to give up for some reasons) but the grammar is goddamn tough :D

no offence but I don't like french either - it's just a personal opinion - i just don't like it for some unknown by-me reasons :D

[quote="galois"]Let $ABC$ be a triangle, and $D$ a point on the side $BC$. Let $w$ be the circumcircle of triangle $ABC$. Show that the circle tangent to $w$, $AD$ and $BD$ and the circle tangent to $w$, $AD$ and $DC$ are tangent to each other if and only if $\measuredangle BAD = \measuredangle CAD$.[/quote]

Can you solute this problem by inversion? Please, post it.

[quote="phanthanhtom"]Can you solute this problem by inversion? Please, post it.[/quote]

From what MysticTerminator post at this page:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=romania+1997&t=46342#
then we have done. :D

Those two circles are tangent to each other iff AD is their radical axis. It's not hard to verify that the midpoint of arc BC is on their radical axis, so the result basically follows. If you want to use inversion, invert about the midpoint of arc BC.

The inversion about the point of tangency of these two circles solves the one direction automatically. I am not sure about the second implication but probably it can be deduced from the first one with some contradiction argument.

Do you have a proof of this?  It's very nice.

[quote="wojto111"]Let N and M be points of tangent on DC and AD. Try to prove lemma that line NM pass through incenter of ABC. If you prove it will be easy.[/quote]


search
number theory proposed
number theory

Let $ n$ be a positive integer, p is a prime satisfying $ p \equiv 7 (mod 8)$. Prove that:
\[ \sum_{k\equal{}1}^{p\minus{}1} \{\frac{k^{2^n}}{p} \minus{} \frac{1}{2} \}\equal{}\frac{p\minus{}1}{2}\]
 :roll:  :P

It's also a problem of Romanian team selection test 2005, u can search on forum


least common multiple
superior algebra
superior algebra unsolved

1. if p is a positive prime integer, prove any field with p elements is isomorphic to Z of p.

2. Suppose that R and S are rings with positive characteristics m and n respectively. If k is the least common multiple of m and n, prove the direct sum of R and S has characteristic k.

Please help thanks!!

These are both pretty simple. Can you show what you've done so far?

[quote="akg9r"]nothing yet[/quote]
So why do you expect us to do anything?  :roll:


function
complex analysis
complex analysis unsolved

Let $ D \equal{} \{z \in \mathbb{C}: |z| < 1\}$ and $ B \equal{} \{z \in \mathbb{C}: |z| < 1, z \neq 0\}$.

I would like to know if there is a surjective conform map $ f: D \rightarrow B$.  ( $ f$ is analitic and $ f'(z) \neq 0$ for all $ z \in D$).

Thanks in advance!

Map $ D$ onto the left half-plane $ \{\text{Re\,}z<0\}$ by some standard fractional linear transformation. Then apply $ e^z$.


linear algebra
matrix
combinatorics proposed
combinatorics

You are given a matrix of real numbers. You choose a row or column of the matrix, and multiply all the numbers in that row (column) by some nonzero real number x (you multiply all the numbers in the row/column by the same number x). next you choose again a row or column of the matrix, and multiply all the numbers in that row (column) by some nonzero real number y. And so on. Prove: it is always possible to make the sum of each row and each column non-negative in a final number of steps.

It´s enough to multiply by $-1$ some rows or columns. There are a finite number of boards you can reach (at most $2^{mn}$, or even better $2^{m+n}$). Take one with maximal sum and this one will satisfied that each row and each column has non-negative sum because if the sum is negative we multiply by $-1$ and the sum increases.

This actually gives us a way to find which rows/columns to multiply: at each turn, select any row or column which has a negative sum and multiply it by -1. This process is guaranteed to stop, and when it ends the matrix will have no row/column with negative sum.