to whom did i have pleasure of playing frisbee with today? i'm really bad with names...  :P

I love frisbee, where are you guys playing? I would LOVE to play a game of ultimate.

well... i originally wanted to go the gym... then i saw a bunch of ppl that i thought were frosh (although sketical about it) playing frisbee. me, being not able to resist my fav game, asked to play. nice ppl allowed me to play. and i found out it was seminar ppl... this is at uw, between biology and student life centre. i need to get my skills back!! i'll go wonder around see if they're playing again tml

That was a fun game of frisbee, you were really intense! I was the non-Russian white guy with brown hair, I think that identifies me. I might have been wearing my #15 t-shirt. I had no idea you were from AoPS, I thought you were Lino Demoosi's friend.

Hey it was you.
I am the Russian guy on your team.
I think I was wearing trhe umbro t-shirt
You are really good.

And for the seminar, I really liked it :)

[quote]ohh! u were on the other team right? u owned it up! after that game i realized i needed more practice, i've been trying to play everyday after work... how was seminar?[/quote]Thanks, you were really good, if we had played a few more minutes we would have lost.[quote="J-I-M-B-O"]nice nice... glad to hear that u enjoyed it, where u guys live?[/quote]

Seminar was AWESOME. I'm in Vancouver, the only one from BC. I know there were at least two more invited, Farzin from the IMO team, and Jimmy He would have got invited for Fermat. But Danny Shi (gr 9) and Scott Lin (gr 12) didn't get invited even though they got top 5 on Euclid. I have a theory, they invite as many grade 11's as possible, and also they try to invite as many Ontarians as possible, because they want you to come to Waterloo. Unless you're a really strong student and might be traveling across the world to go to a good university, then they will be more likely to invite you.

[quote="J-I-M-B-O"]nice nice... glad to hear that u enjoyed it, where u guys live?[/quote]
I live in Waterloo. I go to waterloo collegiate institute. This year 4 ppl from Waterloo got invited: Ghuyon Lee(I maybe misspelled the name), William Ma, me, and Jennifer Park. The first 3 ppl are from Waterloo Collegiate. Will and Jennifer will go to UW next year.

rem!
how can you spell his name wrong?
after all you've been through together...
 :rotfl:

OMFG! :rotfl:
Please stop now.
Anyways, hey Greg! You were playing frisbee, too, right? :?

yes i was...
practising my ninja skillz  :ninja:
maybe that's why you didn't notice me...
oh yeah
i forgot to return the money i stole from you during that game
oh...
and what's up with your email? you dont use msn do you...
rambler.ru  :rotfl:

were u on my team? wow... how many aopsers played frisbee that day?

I think quite a few AoPsers were playing that day. I didn't know you are on AoPs also, Jim. I am the tall Chinese guy with long hair (Well, compare to others). I was on the other team with Steven (Carpo). Jim, you are really good at frisbee...I wish I have that skills.

thanks... u played really well too! really hard to find a game to join in unless i join intermural team, which i don't have time for... i tried to join in first year, but found out i was overloaded with work... well, if u guys ever want to play, just tell me i'll be there hahaha...

hey Alex, i just live on Hickory, so like steps away from WCI. i'll be here until aug, then i'll be moving to west side of uw...

Startogalaxy, are you by any chance Alan Guo? :?

To otaku, my E-mail is cool :gleam: . I bet nobody on the canadian forum uses a Russian sever! Give me my money back! :rotfl:

just call me Greg
geez, as if you don't know me...
 :mad:

Sorry, Greg. :blush:
How did you like the soccer games at Seminar?

well...
apart from the massive scrape on my arm that still hurts, soccer was fun
at least I scored some goals...

I'm so jelous, there are no frisbee games going on in my area right now.  :(  Any one know if there is a pick up league in the toronto area? What about in waterloo for next september?

man, u gotta start looking
i'm sure there are tons of frisbee leagues in toronto
i could ask some friends for you if you want...

that would be cool, assuming these are actual pick up games, and not a group of friends that I've never met  :D

there r lots of tournaments in toronto. i played in one tournament when i was back in hs... pretty fun, played with really talented ppl... in waterloo, there is a intramural frisbee league u can join, i'm not sure about the details. i played one game during frosh week, it was intense game, like 50 ppl on the field! haha.. glow in the dark too! u can ask around then how to join. there r some really amazing ppl playing. too bad waterloo doesn't have a varsity team.

i don't know how intense you are, but some of my friends play in TUC (Toronto Ultimate Club) leagues - website: http://www.tuc.org

also, i believe university of toronto has pick-up games on wednesday nights at Newman Centre (not as hardcore, i guess  :rotfl: )


Let $ f : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function that satisfies the following conditions:
a) $ f(1000)\equal{}999$, and
b) $ f(x) \cdot f(f(x)) \equal{} 1$ for all $x \in \mathbb{R}$.

Find $ f(500)$.

[quote="prester"]Let $ f : \mathbb{R} \rightarrow \mathbb{R}$ a [i]continuous function[/i] that satisfies the following conditions:

$ a)$ $ f(1000) \equal{} 999$

$ b)$ $ f(x) \cdot f(f(x)) \equal{} 1,$   $ \forall x \in \mathbb{R}$

Find $ f(500).$[/quote]

$ f(x)f(f(x))\equal{}1$ $ \implies$ $ f(x)\ne 0$ and so $ f(f(x))\equal{}\frac 1{f(x)}$ $ \forall x\in\mathbb R$ and so $ f(x)\equal{}\frac 1x$ $ \forall x\in f(\mathbb R)$

$ f(1000)\equal{}999$ $ \implies$ $ 999\in f(\mathbb R)$ $ \implies$ $ f(999)\equal{}\frac 1{999}$ $ \implies$ (using continuity of $ f(x)$) $ [\frac 1{999},999]\subseteq f(\mathbb R)$ $ \implies$ $ 500\in f(\mathbb R)$ $ \implies$ $ \boxed{f(500)\equal{}\frac 1{500}}$

[quote="Mathias_DK"][quote="pco"]$ f(1000) \equal{} 999$ $ \implies$ $ 999\in f(\mathbb R)$ $ \implies$ $ f(999) \equal{} \frac 1{999}$ $ \implies$ (using continuity of $ f(x)$) $ [\frac 1{999},999]\subseteq f(\mathbb R)$ $ \implies$ $ 500\in f(\mathbb R)$ $ \implies$ $ \boxed{f(500) \equal{} \frac 1{500}}$[/quote]
How can one prove $ f(a_1) \equal{} b_1,f(a_2) \equal{} b_2 \Rightarrow [b_1;b_2] \in f(\mathbb{R})$?[/quote]

This is just the intermediate value theorem for continuous functions. see http://en.wikipedia.org/wiki/Intermediate_value_theorem

It was a problem from a context in Italy (Cesenatico). As said above by pco,  the solution of this problem use a theorem called "Intermediate value theorem" that say that for continuous functions in an interval $ [a,b]$ where $ f(a) \ne f(b)$, the function takes all the values between $ f(a)$ and $ f(b)$.
As I know, this theorem was attributed to $ Bolzano$. See wikipedia for more informations.

Thanks a lot for your work.


advanced fields
advanced fields unsolved

Show that $ RP^{3}$ is orientable.

Thank you.

[hide="Hint"]Does an orientation on $ S^3$ descend to $ RP^3$?[/hide]


ratio
inequalities
geometry
Intersection
Triangle
IMO
IMO 1961

Consider triangle $P_1P_2P_3$ and a point $p$ within the triangle. Lines $P_1P, P_2P, P_3P$ intersect the opposite sides in points $Q_1, Q_2, Q_3$ respectively. Prove that, of the numbers \[ \dfrac{P_1P}{PQ_1}, \dfrac{P_2P}{PQ_2}, \dfrac{P_3P}{PQ_3} \]
at least one is $\leq 2$ and at least one is $\geq 2$

A useful identity in all triangles proven at [url]http://www.mathlinks.ro/Wiki/index.php/1988_AIME_Problems/Problem_12[/url] is
$ \dfrac {PQ_1}{P_1P \plus{} PQ_1} \plus{} \dfrac {PQ_2}{P_2P \plus{} PQ_2} \plus{} \dfrac {PQ_3}{P_3P \plus{} PQ_3} \equal{} 1$
So the question is equivalent to
$ x \plus{} y \plus{} z \equal{} 1$, prove out of
$ (1/x \minus{} 1)$,$ (1/y \minus{} 1)$,$ (1/z \minus{} 1)$
one is greater than 2, one is less than 2.
But it is clear that one of x,y,z is $ \le$ 1/3 and one $ \ge$ than 1/3,
which proves the claim. So we are done.

Kalva has an awesome solution to this problem. Take the lines through the centroid parallel to the sides of the triangle. The result is then obvious.

If ABC is the triangle, P - an arbitrary point inside it, D, E and F the intersections of the cevians AP, etc. with the opposite side, there holds the relation: AP/PD = AF/FB + AE/EC and other two similar relations (Van - Aubel). The sum of these three relations is obviously at least 6, hence at least 1 ratio among the ones involved is at least 2.

Note that, if one counts the product of the same ratios will find something at least 8.

Best regards,
sunken rock

[b]Such an easy problem like this will not be seen again, atleast in the IMO[/b]

Let \[[P_1PP_2]=a\],\[[P_2PP_3]=b\],\[[P_1PP_3]=c\].
Then
\[P_1/PQ_1=[P_1PP_3]/[PQ_1P_3]=[P_1PP_2]/[PP_2Q_1]=(a+c)/b\]
Similarly,
\[P_2P/PQ_2=(a+b)/c\] and \[P_3P/PQ_3=(b+c)/a\]
Now if all the ratios are >2 then we have
\[a+c>2b                 ..(1)\]
\[a+b>2c                 ..(2)\]
\[b+c>2a                 ..(3)\]
But (1)+(2) gives
$2a>b+c$ which contradicts with (3).Contradiction.
Hence one of the ratios is <=2.The other proof is similar.

Let:
$\frac{P_1P}{PQ_1}=x,
\frac{P_2P}{PQ_2}=y,
\frac{P_3P}{PQ_3}=z$
We have that $x+y+z+2=xyz$, and we are finished.

Assign masses of $a$, $b$, $c$, $b+c$, $c+a$, $a+b$, and $a+b+c$ to $P_1$, $P_2$, $P_3$, $Q_1$, $Q_2$, $Q_3$, and $P$, respectively, where $P$ is the overall center of mass. Then, we have that $\tfrac{P_1P}{PQ_1}=\tfrac{b+c}{a}$, $\tfrac{P_2P}{PQ_2}=\tfrac{c+a}{b}$, and $\tfrac{P_3P}{PQ_3}=\tfrac{a+b}{c}$ along cevians $P_1Q_1$, $P_2Q_2$, and $P_3Q_3$, in that order. Now, suppose FTSOC that $\tfrac{P_1P}{PQ_1},\tfrac{P_2P}{PQ_2},\tfrac{P_3P}{PQ_3}>2$, i.e.,
$$\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}>2\iff b+c>2a,c+a>2b,a+b>2c.$$
But summing gives $2a+2b+2c>2a+2b+2c$, contradiction, so at least one of the $\tfrac{P_iP}{PQ_i}$ are $\leq2$. Analogously, at least one of the $\tfrac{P_iP}{PQ_i}$ are $\geq2$, hence proven. $\blacksquare$


I just don't get this one and I've tried as hard as I can :( Please help me!

What is the greatest common divisor of all sets of numbers where each member of the set is the sum of three consecutive positive integers?
I know how to find three or what ever number of numbers which add to get a number, but I'm not sure if that'd help in this situation.

Heh... This is actually pretty easy. Let's say the first number is $ x$. The next $ 2$ numbers would be $ x\plus{}1$ and $ x\plus{}2$. So let's add them together. $ x\plus{}x\plus{}1\plus{}x\plus{}2\equal{}3x\plus{}3$ We take $ 3x\plus{}3$ and notice that 3 is the only divisor other than 1, that works. $ 3$ is the answer!!!


calculus
number theory
real analysis
real analysis unsolved

Sorry if this is the wrong section, but I was wondering if anyone had a proof that $e$ and $\pi$ are transcendental numbers.  Thanks.

Doesn't Spivak's [i]Calculus[/i] have these proofs? Try looking there.

He said transcendental, not irrational :)

Probably searching the web is best.

I know he said transcendental; I don't have a Spivak in front of me right now, but I think he has the proof that $e$ is transcendental; I'm less sure about $\pi.$

i just took a look in my spivak, and as kent said and i suspected, there is a whole chapter talking about e beeing transcedental... in fact the chapter's title is "e is Transcedental"

To show that $\pi$ is transcendental, the nicest way I know is to use the Lindemann–Weierstrass theorem which can be found in Alan Baker, Transcendental Number Theory, Cambridge University Press, Chapter 1.
If $\pi$ was algebraic, $2\pi i$ would be algebraic too (since $2i$ is algebraic), and then by the Lindemann-Weierstrass theorem $e^{2\pi i}= 1$ would be transcendental, which is absurd.


quadratics
calculus
integration
symmetry
geometry
3D geometry
algebra

Find all positive integer solutions to $x^3-y^3=xy+61$ Prove that you have found all of them.

[b]Please hide your solutions so that others may answer.[/b]

[hide="Part 1"]Factor the left side: [tex] (x-y)(x^2+xy+y^2)=xy+61[/tex]
Rearrange: [tex](x^2+xy+y^2)-xy=\frac{61}{x-y}[/tex]
Cancel: [tex]x^2+y^2=\frac{61}{x-y}[/tex]
Rearrange: [tex](x-y)(x^2+y^2)=61[/tex][/hide]
[hide="Part 2"]61 is prime, so since [tex]x-y[/tex] and [tex]x^2+y^2[/tex] must be integers, there are 4 cases that make this equation true:
[tex]x-y=1[/tex] and [tex]x^2+y^2=61[/tex]
[tex]x-y=61[/tex] and [tex]x^2+y^2=1[/tex]: No solutions because 2nd equation gives [tex]x=y=1[/tex]
[tex]x-y=-1[/tex] and [tex]x^2+y^2=-61[/tex]: No solutions because sum of squares must be positive
[tex]x-y=-61[/tex] and [tex]x^2+y^2=-1[/tex]: No solutions because sum of squares must be positive
Only the first set of equations holds plausible solutions.[/hide]
[hide="Part 3"]Substitute: [tex]x=y-1[/tex], so [tex](y-1)^2+y^2=61[/tex]
Expand: [tex]2y^2-2y+1=61[/tex]
Make into a quadratic equation: [tex]2y^2-2y-60=0[/tex]
Solving this using the quadratic formula gives [tex]y=6[/tex] or [tex]y=-5[/tex], implying that [tex]x=5[/tex] or [tex]x=-6[/tex].

The only positive integer solution is [tex](x,y)=(5,6)[/tex].[/hide]

Edit: I just realized I screwed up on my algebra early on, so this is not the solution unless I got it through a really weird coincidence.  It should give people some ideas though.

twobomb, you have a problem going from your first line to your second line in part 1.  You didn't divide every single term through.

Here's a bad solution, but it gets the job done.

[hide]First, we check the case where x=y.  This clearly yields no solutions.  Now, because x and y are both greater than 0, we have xy>0, and, xy+61>0.  This means that x^3-y^3 greater than 0, which means that x>y.

We now make the substitution x-y=a, xy=b.  Note that both of these are positive integers.

The equation now becomes $a(a^2+3b)=b+61$, which means that $b(3a-1)=61-a^3$.  We have 3a-1>0, and b>0, which means that 61-a^3>0.  Because a is an integer greater than 0, and 61>a^3, we now have a being an element of the set {1,2,3}.  Of these, only a=1 yields an integral value for b (just take my word on it).  This then yields the only solution, a=1, b=30, or, even better, x=6, y=5.[/hide]

[quote="Magnara"]Here's a bad solution, but it gets the job done.

[hide="Correct Solution1"]First, we check the case where x=y.  This clearly yields no solutions.  Now, because x and y are both greater than 0, we have xy>0, and, xy+61>0.  This means that x^3-y^3 greater than 0, which means that x>y.

We now make the substitution x-y=a, xy=b.  Note that both of these are positive integers.

The equation now becomes $a(a^2+3b)=b+61$, which means that $b(3a-1)=61-a^3$.  We have 3a-1>0, and b>0, which means that 61-a^3>0.  Because a is an integer greater than 0, and 61>a^3, we now have a being an element of the set {1,2,3}.  Of these, only a=1 yields an integral value for b (just take my word on it).  This then yields the only solution, a=1, b=30, or, even better, x=6, y=5.[/hide][/quote]
Yeah, that's it.  :)  I'll post my solution late tomorow.

Here's a somewhat similar solution, though it probably gets us there faster.  It's clear that $x>y$, so we can write $x=y+a$, for $x,y,a\in\mathbb{N}$.

The binomial theorem gives us:

$3y^2a+3ya^2+a^3=y^2+ay+61$

Clearly $y^2a+ya^2\geq y^2+ay$, so we get:

$2y^2a+2ya^2+a^3\leq 61$.

This means that $a<4$, and symmetry gives us $y<5$.  The $\bmod y$ approach gives us $y\vert a^3-61$, so we get $a=1$ and the rest follows easily.

Slightly round about  and tangantial   but the first  thing I noticed  was that  $61=5^3-4^3$ and  the second difference in the table of cubes is given by $6(n-1)$ (That is  $(n^3-(n-1)^3)-((n-1)^3-(n-2)^3) = 6(n-1)$ and  the solution sort of jumped out ..($(6^3-5^3) - (5^3-4^3) = 6\cdot5$)

[quote="diophantus"][quote="Gyan"]Slightly round about  and tangantial   but first  thing I noticed  was that  $61=5^3-4^3$ and  the second difference in the table of cubes is given by $6(n-1)$ (That is  $(n^3-(n-1)^3)-((n-1)^3-(n-2)^3) = 6(n-1)$ and  the solution sort of jumped out ..($(6^3-5^3) - (5^3-4^3) = 6\cdot5$)[/quote]
But does this prove that you found all of the solutions?[/quote]
[i]But does this prove that you found all of the solutions?[/i]

Not if you are looking for a formal proof in the above post, after all this was  just an observation which was reached within a few seconds of reading the original question.

But the “intuition”  (experience ?)  (and may be a few  minutes of writing)  could lead to a solid proof that this the solution is unique.  The idea was to  think of
$n^3-m^3 - (5^3- 4^3)=nm$ as
$k_n + k_{n-1} + k_{n-2} \dots k_{m+1}   -k_5 = nm$
where $k_n = n^3-(n-1)^3$  etc ..
Now since the second difference is $6(n-1) $ even for consecutive terms ..   The values for ‘n’ and ‘m’ can be fixed  pretty soon (If m>5 you can only have n=6 as solution else the LHS would be too large,..smaller values - less than 5 can be ruled out easily)

  and you can form a formal proof (which  is  essentially same as given in a few above posts)

(Point was not to introduce any new "insights" but rather a fun  observation)

Hope this helps.

Ok, here's how I went about solving this problem.
[hide="My Solution"]
If you multiply both sides of the equation by 27 and rearrange, you get this:
$(3x)^3+(-3y)^3+(-1)^3-3(3x)(-3y)(-1)=1646$
Which is equivalent to:
$(3x-3y-1)(9x^2+9y^2+1+9xy+3x-3y)=2\cdot 823$
Note that since 823 is prime and $3x-3y-1\equiv2 \mod{3}$, $3x-3y-1=2$ yielding $x-y=1$ (trust me that $3x-3y-1=2\cdot 823$ yields no solutions)
We have the system:
$x-y=1$
$9x^2+9y^2+1+9xy+3x-3y=823$
Squaring the first and substituting into the second our system simplifies to
$x^2-2xy+y^2=1$
$x^2+xy+y^2=91$
Giving $xy=30$
The positive solution to the system:
$x-y=1$
$xy=30$
which is $(x,y)=(6,5)$ is then proven to be the only positive solution.
QED
[/hide]


logarithms
limit
function
algebra unsolved
algebra

Let $\mathbb{N}^{*}=\mathbb{N}\setminus\{0\}$ (the positive integers) and $n$ be an integer $\ge2$.
Let $U_{n}=\{(x_{1},\cdots,x_{n})|\forall{i}, x_{i}\in{\mathbb{N}^{*}\,and\,x_{1}.x_{2}.\cdots.x_{n}=x_{1}+\cdots+x_{n}\}}$.
Let $\phi(n)=\inf_{U_{n}}(x_{1}.x_{2}.\cdots.x_{n})$.
Is it true that $\forall{n}\ge{2},\,\phi(n)>n$?
Is it true that $\phi(n)\sim{n}$ when $n\rightarrow\infty$?

Apologies for the first question: it's obvious.
Have a look only on the second question.

Hi Rust,
 $(1,\cdots,1,2,n)\in{U_{n}}$ and the product is $2n$which isn't
$\le n+log_{2}(n)+1$.

You interested minimum, when $n=2^{j}-j, \ j=[log_{2}(n)]+1$ we have $s=n+j=n+[\log_{2}(n)]+1$.

Hi Rust,
you have proved that if $n$ can be written $2^{j}-j$ then $\phi(n)\leq{n+[\log_{2}(n)]+1}$ and therefore that $\liminf\frac{\phi(n)}{n}=1$. What about $\lim\frac{\phi(n)}{n}$ ?
Remark:$\phi$ isn't an increasing function; for example:$\phi(210)=240$ and $\phi(211)=224$.


calculus
integration
limit
logarithms
absolute value
calculus computations

I'm having a bit of trouble calculating this integral. I got up to a point where I was pretty sure it was divergent but I just want to make sure:
\[ \int^{4}_{0}{\frac {dx}{x^{2} + x - 6}}
\]
Basically one of the expressions I got was $ \lim_{t\to 2 - }\frac {1}{5}\ln{|x - 2|}$, which grows unbounded in the negative direction as ${ t\to 2 + }$. Does this mean that the entire integral diverges though?

Edit: I'll just post the work that I did so far:

\[ \int^{4}_{0}{\frac{dx}{x^{2}+x-6}}=\lim_{t\to 2-}\int^{t}_{0}{\frac{dx}{x^{2}+x-6}}+\lim_{t\to 2+}\int^{4}_{t}{\frac{dx}{x^{2}+x-6}}\]
and
\[ \int{\frac{dx}{x^{2}+x-6}}=\frac{-1}{5}\ln{|x+3|}+\frac{1}{5}\ln{|x-2|}+C\]
I think what I'm unsure of is the behavior of the logarithms with the absolute value signs. I'm pretty sure I can't dispense with them here, right? And when I go to evaluate the integral, if just one of the limits in the entire expression goes to $ +\infty$ or $ -\infty$, does that mean the entire expression does not exist as a finite number?

Your reasoning seems to be correct, and the integral diverges.

Yes, the integral is divergent.

You can also use $ \frac{1}{x^2\plus{}x\minus{}6} \equal{} \frac{1}{5}\left (\frac{1}{x\minus{}2}\minus{}\frac{1}{x\plus{}3}\right )$


Do you know where i can find AHSME archive problems?

Professor Reiter has some on his page:
http://www.math.uncc.edu/~hbreiter/

If you really want old tests, you can always buy them off from AMC...

There are also a few other places but you'll have to google them.

Here're some 50 problems.

They're rather fun. Some actually involve thinking!  :)

[url=http://www.unl.edu/amc/a-activities/a7-problems/a7-1-archives/1999-pa/anniver.pdf]www.unl.edu/amc/a-activities/a7-problems/a7-1-archives/1999-pa/anniver.pdf[/url]


please ; how can you use the latex help ? I wait your reply

Here's a link:

http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:About

Your questions would be better placed in the $ \text{\LaTeX}$ forum.

There's a nice [url=http://www.artofproblemsolving.com/Forum/weblog.php?ppage=5&w=1157&start=5]$ \text{\LaTeX}$ tutorial[/url] on my friend's blog.


a)Let A and B be closed subspaces of a topological group. Give an example to show that AB need not be closed.

b)let A and B be compact subspces of a topological group. Show that AB is compact.

I guess b) follows directly from the Tychonoff's Theorem. Is it true? But how can I do it if I wanna prove it directly?????

You only need the two-space version of Tychonoff, and that's not hard to prove directly. Products of open sets are a base for the open sets; given an open cover of $A\times B$, we can reduce WLOG to the case of a cover by products $U\times V$. For each point $x\in A$, find a finite subcover of $x\times B$ using compactness of $B$. That finite subcover covers $U\times B$ for such $U$; we then apply compactness of $A$ to cover $A\times B$ by finitely many of these.
In the last step, apply the multiplication map on $X\times X$ to the compact set $A\times B$. The image $AB$ is compact.

The example I'd use for (a):
[hide="Example"]$X=\mathbb{R}$ under addition, $A=\mathbb{Z}$, $B=\sqrt{2}\mathbb{Z}$.[/hide]

Thanks for your help, but I don't get your example for a), why it is not closed? Aren't they just a set of points in X x X, which are closed??? Or my concept is wrong again :?:

$AB$ is the set $\{xy: x\in A\text{ and }y\in B\}$. That's a subset of the group $X$, not $X\times X$.

so [i]xy[/i] in [b][i]AB[/i][/b] should follow the operation of the group? addition in your example?
Sorry for my stupidness, but I still didn't get why it's not closed????? :(


vector
combinatorics proposed
combinatorics

There are 62 non zero vectors, in space.
Prove or disprove: it is possible to find two vectors
such that the angle between two of them is smaller than 30°.


[hide]
assuming that a rotated figure counts as the same figure, is it 18?

*cough* brute force

EDIT: haha i meant 19.
[/hide]

[quote="lizcheang"]How many patterns can you  make using 5 squares making sure that they touch only edges or an edge and no corners

for example here is one

[][][][][] <-- except they would be squares, of course![/quote]
is [][][]
   []   [] same as []   []
                        [][][]?

can you be more clear in the question? you/re saying the squares can only touch on edges yet not on the corners (u mean vertices right?) yet in ur example the vertices do touch................................


inequalities
trigonometry
geometry proposed
geometry

Given triangle $ABC$ centroid $G$ let $\angle BGC=\alpha,\angle CGA=\beta, \angle AGB=\gamma$ prove that
$\frac{abc}{GAGBGC}\ge 8\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}$
 :D

[quote="Albanian Eagle"]the RHS can be improved to $6^{\frac{3}{4}}\sqrt{\sin \frac{\alpha}2 \sin \frac{\beta}2 \sin \frac{\gamma}2}$[/quote]
Can you post solution for it ?


Find the extreme values of

  (a+1)/(ab+a+1) + (b+1)/(bc+b+1) + (c+1)/(ca+c+1)

with abc=k with k real positive number and a,b,c real positive numbers.

It is obvious that for abc=1 the expression is 2.


What is the maximum number of points of intersections when two circles and six lines intersect each other?
Explain/show/prove your answer.

These are all coplanar and are not colinear/concentric, right?
[hide="if so"]Put the two circles together, you get 2
If yopu put the six lines together, you get 0+1+2+3+4+5=15
For every line, it can pass the circles 4 times (if you put all the line intersections inside the area where the circles intersect), so 4*6=24
15+2+24=41
>>41<<[/hide]
<edit by dwx314: ha,ha, i added wrong first time. :blush: >

[hide]I got 41 by drawing a diagram.[/hide]

This is how I did it:

[hide]Two circles can have at most two points of intersection. After you draw the two circles, a line can intersect each circle twice, for a total of four times. The next line can do the same, except that it can also intersect the first line. This process keeps repeating for every extra line (every new line can intersect all lines already drawn and the circles). So the total number of lines is 2+4+5+6+7+8+9=41.

Answer: 41 [/hide]

With my drawing a diagram, I also found that pattern, then I didn't finish my diagram.

[quote="solafidefarms"]I typed up a great explanation and then... I closed the tab. :(  :(  :(  :(  :( And my undo close tab didn't work.

I will see if I can recreate it.
[hide]
The circles can intersect a maximum of 2 times. Each line can intersect each circle 2 times and it can intersect every other line once. Therefore, the formula is $2+4n+\frac{n(n-1)}{2}$. So 2 circles and 6 lines gives 2+24+15=41.[/hide]

BIlly[/quote]

Please explain your formula so that it can be applied to any problem such as this!  It looks like a great formula!

[quote="mathgeek2006"][hide]I believe that his formula goes more like this (Hope I'm right!)

$x+2x(n)+\frac{n(n-1)}{x}$

Where $x$=#circles
and $n$=#lines[/hide][/quote]

Can we apply this to lines only (no  circles) as well?  How about lines and triangles?  Lines and squares?  etc.

I'm working on those other formulas, but right now I'm not getting them. The formula I supplied is only for circles and lines (I believe).

[hide]I Drew a diagram where each line crossed every other line and got 41[/hide] :)

Eewgh.?

LOL.

I can never draw right, so I always stick to formulas. Or course, this one's not hard to deduce from the facts given, thankfully.

My teacher always used to say, "diagrams are a mathematician's killer." With [i]that[/i] killer, I now know why.

[quote="probability1.01"]Circles and lines formula:

For x circles, the circles intersect one another first.  There are x(x-1)/2 pairs of circles, and each intersect at 2 points.  So we first have x(x-1) as a term.  Then for lines between lines, if there are n lines, there are n(n-1)/2 pairs of lines, each intersecting once.  Between lines and circles, each circle intersects with each line twice.  That gives us 2xn.  The sum of these is x(x-1) + 2xn + n(n-1)/2.[/quote]

Could I modify this to work for just lines?

Yes. Probability points out the formula just for lines. It is $\frac{n(n-1)}{2}$. Notice for the $n$th line you add, it intersects optimally with the previous $n-1$ lines. Thus forming the formula similar to this one that you will recognize.

$1 + 2 + 3 + \ldots + n = \frac{n(n-1)}{2}$  :D


Hi all,

It's been a long time... I've just completed permutations and combinations but I just cannot get around this problem :(  Hope one of you can enlighten me...

Question:

You have a Chess board (8*8). Starting from the bottom left square, your objective is to get to the top right. You can only travel up and across lines (not diagonally) and cannot move in the direction away from the top right square, that is to say, you can only move towards the top right. So how many ways are there?
The answer is [u]12870[/u] but I do not know how to get there - any help would be appreciated.

thanks once again,

Kesh

Say you make a list of the moves you make with u being up and r being right. You could do:

uuuuuuuurrrrrrrr

With 8 of each. Now think of just that list. There are 16 slots for each letter. For the u's you can put them in $\binom{16}{8}$ ways and the r's are determined from that. Thus, the answer is $\binom{16}{8} = \frac{16\cdot15\cdot14\cdot13\cdot12\cdot11\cdot10\cdot9}{8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2} = 13\cdot11\cdot10\cdot9 = \boxed{12870}$

It's basically how many ways you can rearrange the word UUUUUUUURRRRRRRR. And that is $\frac{16!}{8!8!}$. The formula for an m by n grid is $\frac{(m+n)!}{m!n!}$. :D

[quote="lotrgreengrapes7926"]Wait, wouldn't it be only $\frac{14!}{7!7!}$? Since we start on the chess board and we only need to move 7 up and 7 over. :?[/quote]

Maybe we start on a corner and go along lines?


Let be 2 systems of linear equations: AX = 0; BX = 0 (A, B are mxn matrix). Prove that:
Two systems of linear equations are equivalent [i]if and only if[/i] there is exist the mxm matrix C such that: A = CB


I just received three hours ago email notification of how the young people in my administration of the AMC 8 did. That's exceptionally timely service, and very convenient. I appreciate the innovations the AMC 8 management is making this year.  :D

One pleasant surprise in the results is that the top-scoring young person in our group actually got one point more than he recalled getting. Two people in our group repeated from last year, and they both did very well. The group average was quite good too--more to the credit of recruiting "ringers" than to the credit of my coaching.

Now I'll have to have the top scorers try our first-ever AMC 10. This looks like a good year in the AMC programs.

I'm glad you appreciate the email notification of results on the AMC 8.
We've actually been doing that for several years, and although I wish I could claim it as an innovation, it is actually due to our hard-working AMC staff, in particular, our programmer-system manager-and-everything-else-computer-related person Marsha.

The follow-up postal mail notification with printed results and reports, pins, certificates, etc. is now being packaged and going out daily in the mail.  Watch for it in your mailbox soon!

I should note that we have now scanned all the answer sheets received so far (some were still arriving in the mail as late as yesterday!) and we are in the process of editing bad answer sheets and simultaneously preparing reports.  This was all in spite of having our largest scanner down for (expensive) repairs and being the last stages of preparing to send the orders for the AMC 10 and AMC 12.  Hats off to the AMC staff!

Steve Dunbar, AMC Director

It would be interesting to know how many perfect scorers are there so far on AMC8.  We have one from our school.  This is our first year for AMC8.  Students, teachers and parents were all excited.

Good question, but even I don't know the answer yet!

Last night, we were putting together some of the statistical reports on the AMC 8  but they are not finished yet nor are they double-checked, and so I can't say with certainty.  We are striving to get them done very soon, and will put them on our Web-page.  That should answer your question, and many others!  Please be patient for a few more days!

Steve Dunbar, AMC Director

It's very kind of the AMC Director to have joined this forum.  And it's even nicer of him to take the time to read the messages and post his replies.

Thanks for the dedication to Math, Steve.  We do appreciate it.

The statistical summaries for the AMC 8 are now up and on the AMC website at
http://www.unl.edu/amc

The global statistics are available as well as summaries for each state.

The last of the printed school summaries with the certificates, pins, etc, went out in this morning's mail, so schools should be receiving the printed summaries in no later than a week.

Steve Dunbar, AMC Director

[quote="AMCDirector"]The statistical summaries for the AMC 8 are now up and on the AMC website at
http://www.unl.edu/amc

The global statistics are available as well as summaries for each state.[/quote]

Thanks for letting us know. That was fast work covering a huge amount of data. I continue to be impressed by the performance of students at Chiang Tsui Junior High in Taiwan (the neighborhood school of the neighborhood I lived in from December 1998 to July 2001), which my niece used to attend. There are more perfect scores in that one school (on a test given in what is, after all, a foreign language!) than in whole states of the United States.

The top scorer in my administration was, for the second year in a row, the top scorer in his grade in the whole state of Minnesota. He has two years of eligibility left, and has already learned most of the curriculum assumed by the AMC 10, which he will take for the first time in February 2004. The other repeat test-taker in our group did very well too. I seem to have figured out how to recruit "ringers," now I just have to figure out how to help them learn more math so they can go on beyond the AMC 8 to higher-level contests.

At least I have the preliminary makings of a reasonably competitive MATHCOUNTS team . . .

When you click on "nevada" on the state statistics page, it takes you to the nebraska page.  Would you please fix the link to the nevada page?  (I know the page is on the web - it's at [url]http://www.unl.edu/amc/e-exams/e4-amc08/e4-1-8archive/2003-8a/03statestats/nv.html[/url]), but some people may be confused because the link is wrong.)  Thanks.

The same goes for Kentucky (which takes you to Manitoba) and Pennsylvania (which takes you to Puerto Rico).


inequalities
integration
real analysis
real analysis unsolved

If $ \mu$ is $ \sigma$-finite and $ 1\leq p \leq \plus{} \infty$, $ 1 \leq p^{\prime} \leq \plus{} \infty$ and $ \frac {1}{p} \plus{} \frac {1}{p'} \equal{} 1$, then show that
$ \parallel{}f\parallel{}_p \equal{} \sup{|\int fg d\mu |}$ where supremum is taken over all bounded $ g$ which vanish outside sets of finite measure and for which $ \parallel{}g\parallel{}_{p'}\leq 1$ and $ \int fg d\mu$ exists.

for $ 1<p<\plus{}\infty$ I think it follows from Hahn-Banach theorem.

But only after you know that the dual of $ L^p$ is isometric to $ L^q$, for which you need the fact quoted by the original poster.

Consider $ g\equal{}|f|^{p\minus{}2}f$, with appropriate interpretation.


number theory
greatest common divisor
modular arithmetic
Vieta
algebra
polynomial
quadratics

1. Is there any $ a\in\Re$ so that $ P(X) \equal{} X^5 \minus{} ((10*a^2)/3)*X^3 \plus{} 5*a^4*X \minus{} 8*a^3$ has a zero of multiplicity exactly 3?

2. Prove that P(X)= 1 + X + ... + X^(p-1) is prime over $ Q$ (rational numbers) for every $ p$ prime.

3. $ m , n \in N$ , $ P(x) \equal{} 1 \plus{} (n \plus{} 1)*X \plus{} 2(m \plus{} 1)*X^2 \plus{} n*X^3 \plus{} x^4 \in Z[X]$. Prove that P(x) is prime over $ Q$ (rational numbers).

4.Prove that if $ (n \minus{} 1)! \plus{} 1 \equiv 0 mod n$ then n is a prime number.

5. Prove that if p is prime (and not 3) then (p-2)*2^(p-2) + 1 is not prime.

6. Prove that if $ gcd (a, n) \equal{} 1$ and  $ gcd (b, n) \equal{} 1$, the equation $ ax \plus{} by \equal{} c$ has exactly n solutions in $ Z/n$ (Z mod n)

#4 is wrong. It should be either one of the following:

4. If $ (n\minus{}1)!\equiv 0\pmod n$ then $ n$ is composite.

4. If $ (n\minus{}1)!\not\equiv 0\pmod n$ then $ n$ is prime.

Yeah thanks, you're right, I forgot to write "+1". Corrected.

[quote="jaden"]Yeah thanks, you're right, I forgot write "-1". Corrected.[/quote]Counterexample: let $ n \equal{} 5$, $ (5 \minus{} 1)! \minus{} 1 \equal{} 4! \minus{} 1 \equal{} 23\not\equiv0\pmod{5}$. I think you meant "+1", which is simply Wilson's Theorem.

T_____T I'm sorry, it's already corrected, I hope the rest is OK.

[hide="1"]Suppose that $ r$ is a zero with multiplicity exactly 3. If $ r \equal{} 0$, $ p(0) \equal{} \minus{} 8a^3 \equal{} 0\Rightarrow a \equal{} 0$ and this implies that $ r$ will have a multiplicity of 5>3, which is a contradiction to our assumption. So $ r\ne0$.

Let $ p(x) \equal{} (x \minus{} r)^3(x^2 \plus{} bx \plus{} c)$. Since the sum and product of the zeroes of $ p$ are $ 0$ and $ \minus{} 8a^3$, respectively, by Vieta's equations we have $ 3r \minus{} b \equal{} 0$ and $ \minus{} r^3c \equal{} \minus{} 8a^3$. So, $ p(x) \equal{} (x \minus{} r)^3\left(x^2 \plus{} 3r \plus{} \frac {8a^3}{r^3}\right)$.

Expand $ p(x)$ to get

$ p(x) \equal{} x^5 \plus{} \left( \minus{} 6r^2 \plus{} \frac {8a^3}{r^3}\right)x^3 \plus{} \left(8r^3 \minus{} \frac {24a^3}{r^2}\right)x^2 \plus{} \left( \minus{} 3r^4 \plus{} \frac {24a^3}{r}\right)x \minus{} 8a^3$

Comparing the coefficients with $ p(x) \equal{} x^5 \minus{} \frac {10a^2}{3}x^3 \plus{} 5a^4x \minus{} 8a^3$ we have the equations:

$ \minus{} 6r^2 \plus{} \frac {8a^3}{r^3} \equal{} \minus{} \frac {10a^2}{3}$ (I), $ 8r^3 \minus{} \frac {24a^3}{r^2} \equal{} 0$ (II), $ \minus{} 3r^4 \plus{} \frac {24a^3}{r} \equal{} 5a^4$ (III)

Equation (II) yields $ r^5 \equal{} 3a^3$. The rest is using algebra on (I) and (III). Since $ a\in\mathbb{R}$, $ a \equal{} \pm\sqrt {3}$[/hide]

[hide="2"]Use Eisenstein's Criterion on $ p(y\plus{}1)$.[/hide]

[hide="4"]This is also known as Wilson's Theorem. I think nayel already suggested how to prove this easily.[/hide]

I'll post 5,6 if no one else tries them....

EDIT#1: Nevermind, 2,3. Maybe t0r can help you with these two :D

EDIT#2: Got #2. Still trying to find a way to do #3...

Formatting advice:

\cdot instead of * for multiplication
\frac{}{} for fractions
Don't switch between $ x$ and $ X$ in the same formula
\gcd for $ \gcd$
\mathbb or \bf for blackboard bold or just regular bold: $ \mathbb{Z}$ $ {\bf Z}$

In the future, please don't post multiple problems in the same thread.  Discussions are formatted more nicely when separate problems are discussed in separate threads.

[hide="2 (discussion)"] $ 1 \plus{} x \plus{} ... \plus{} x^{p \minus{} 1}$ is the [url=http://mathworld.wolfram.com/CyclotomicPolynomial.html]cyclotomic polynomial[/url] $ \Phi_p(x)$, where $ p$ is prime.  They are all irreducible (although the proof is not elementary). [/hide]
[hide="3"] Suppose that $ P(x)$ is reducible.

[b]Case:[/b]  $ 1$ is a rational root.  Then $ n \plus{} 2 \plus{} 2m \plus{} 2 \plus{} n \plus{} 1 \equal{} 2(n \plus{} m) \plus{} 5 \equal{} 0$, contradiction.

[b]Case:[/b]  $ \minus{} 1$ is a rational root.  Then $ \minus{} n \plus{} 2m \plus{} 2 \minus{} n \plus{} 1 \equal{} 2(m \minus{} n) \plus{} 3 \equal{} 0$, contradiction.

[b]Case:[/b]  $ P(x)$ is the product of two irreducible quadratic factors.  Then

$ x^4 \plus{} nx^3 \plus{} 2(m \plus{} 1)x^2 \plus{} (n \plus{} 1)x \plus{} 1 \equal{} (x^2 \plus{} ax \plus{} 1)(x^2 \plus{} bx \plus{} 1) \equal{} x^4 \plus{} (a \plus{} b)x^3 \plus{} (ab \plus{} 2)x^2 \plus{} (a \plus{} b)x \plus{} 1$

but clearly $ n \neq n \plus{} 1$; contradiction.  (The constant term of the quadratics can't be $ \minus{} 1$ because then the $ x$-coefficient of their product is negative.) [/hide]

[quote="t0rajir0u"]They are all irreducible (although the proof is not elementary)[/quote]Eisenstein criterion works and I believe such topic was used in IMO like [url=http://www.mathlinks.ro/viewtopic.php?t=62187]1993 IMO Problem 1[/url]... however I do agree with you but there are many instances where the borderline between elementary and non-elementary  are sometimes confusing.

Eisenstein can't prove the irreducibility of the cyclotomic polynomials in general.  In any case, the problem you linked to is a trivial consequence of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=628409297&t=150]Perron's criterion[/url].

See http://www.mathlinks.ro/Forum/viewtopic.php?t=150951 for irreducibility.

[hide="5"]Take $ \mod p$: $ (p\minus{}2)2^{p\minus{}2}\plus{}1\equal{}p2^{p\minus{}2}\minus{}2^{p\minus{}1}\plus{}1\equiv0\pmod p$ by Fermat's Little Theorem.[/hide]

:)  wow, nice, thank you very much
I like the solution to number 1  :P
next time I'll try to post every problem on a different post (and to learn a bit more of LATEX), sorry


[quote="pardesi"]have alooka t the question i posted [url=http://www.physicsforums.com/showthread.php?t=227829]here[/url][/quote]
hey if i click on the url it comes page cant be displayed nways i am member of physics forum where have you posted your question cant you copy paste it here for everyone to see here??? :D

OOPS SORRY da got it now will see and reply!!!
here it is there is no concept of conservative field here bcoz however the motion of the particle whether circular or helical or cycloidal it doesnt matter coz the work done by magnetic field is zero it is always ZERO and is both PATH AND STATE INDEPENDENT!!!!! :o

a conservative field is so if it depends on only the end points but here it is not dependent even on the end points!!!!

got it????

 :D

yes that's what i am telling it's independent of the end points so can u call that conservative or not...
i know the answer is no :)  but the reason is n't this only :)

Hmm..Interesting... Intuitively the answer is no.. But then what exactly is the definition of a conservative field?? Is it only that the work done should be path independent.

And usually a conservative field has a potential associated with it? Am I right?

yes and more importantly a scalar potential and not a vector one.
A magnetic field has a Vector potential associated with it
It's a very peculiar situation since the force law of the magnetic field unlike others makes the force perpendicular to field and speed
so evn thoug the line integral of the magnetic field is not 0 around a closed loop but the net work done by the field in moving a charge is 0

Can you explain a Vector field?

And even if a body enters a magnetic field with a velocity in all three directions with the field being in only 1. F.v is always zero.

[quote="Euclidean Geometer"]Isnt it non conservative because work done in a round trip is non zero :?:
For conservative forces like spring force , work done in a round trip is zero.[/quote]

Oi. Work done by magnetic field is always zero?? So it doesn't matter whether it goes in a straight line, a circular path or a helix.

Oh i was talking abt induced electric fields. :D


they did half amount of work using 4 people and it took 15 days

now...

if it took 8 people, it would have taken 7.5 days
(shows it's linear )
using this

since

10 / 4 = 2.5

15/2.5 = ?

[hide]If takes a person 60 days to work half the job. That means it will take 10 people 60/10=6 days. [/hide]

[quote="polarpan523"]OK

A work team of four people completes half of a job in 15 days. How many days will it take a team of ten people to complete the remaining half of the job? (they all work at the same speed)[/b][/quote]
this problem is on eof the basic problems and it is nice. :D
and here is nice solutions but i think we can solve it more atractive than now.

Let number of people be $x$, the time it takes to do the job $y$, and the part of the job $z$.

so, since more days mean more parts of the job completed,

$\frac{y}{z}$.

Also, since more people mean more parts of the job completed,

$\frac{x}{z}$.

Thus, the time it takes to do the job and the number of people are [b]directly proportional[/b] to the part of the job. Thus,

$\frac{xy}{z}$.

Substituting the numbers,

$\frac{60}{\frac{1}{2}} = 120$.

$120$ is the constant.

since we have to find the number of days, the equation becomes:

$120 = \frac{4y}{\frac{1}{2}}$
$120 = 8y$
$y = 15$

Thus, it takes 15 days for 10 people to do half a job.

[quote="paradox102"]Let number of people be $x$, the time it takes to do the job $y$, and the part of the job $z$.

.....
since we have to find the number of days, the equation becomes:

$120 = \frac{4y}{\frac{1}{2}}$
$120 = 8y$
$y = 15$

Thus, it takes 15 days for 10 people to do half a job.[/quote]

$x$ is 10 here so $y=6$ instead

nice problem and cool solvers.i think everyday lots of mathematicans are born in every where like you and me. :D

shyong, your right.  my bad.  I solved the givens instead!  :D


function
calculus
derivative
calculus computations

In a graph of the function f(x) has all the following properties.
f(x) is defferentiable and concave up over the interval [0.4]
f'(1)=1
f(1)=2
f(3)=5
This condition would have been impossible if the last condition were given as f(3)=3.9 instead of f(3)=5. Explain why it would have been impossible, using the mean value theorem and the fact that the first derivative is increasing.

From mean value theorem for some $\xi \in (1,3)$ we have

$\\ \frac{f(3)-f(1)}{3-1}= f'( \xi) \Leftrightarrow \frac{3.9-2}{2}= f'( \xi) \Rightarrow f'( \xi) <1 =f'(1)$ .

But $f'$ is increasing...


sorry, sort of unrelated, but whose school does both A&B tests?

Mine does.

I split the topics because the two things are really unrelated...and added a poll for fun.

thanks.  i thought it would be waste of space, but this is probablly better.

That's because it takes me a minute or two to add it.  There.  See it?

in the past we've only taken the A, but im planning on asking my teacher if we could do both, but i'm doubtful that they'd say yes... i'd really like it though if we could take both.

Both last year.

Hopefully we don't get a snow day on any of the competition dates though.

Yeah, that would be ironic, wouldn't it?  Normally snow days are good, because snow day => full day to study for math contests.

how can you find out what date your school is taking. do you ask your maths teacher or something?

Yes.  Most people do the A date if they're only going to do one.  (Although I'm sure there are exceptions...)

Our homeschool group schedules two dates, in large part because we have a test-taker or two who want to take both the AMC10 and the AMC12 in the same year. I hope we can keep that going for several more years.

We do both here also.  And I'm taking the AMC 10 and the AMC 12.


inequalities
linear algebra
matrix
linear algebra unsolved

If $A, B \in M_n (R)$, such that $r (B) = 1$, prove that
\[ \det (A - B)(A + B) \leq \det A^2 \]

alexilic

At first let's suppose that $A$ is invertible then to generalise we'll use the density of invertible matrices.
Thus your question is equivalent to proving that for a matrix $C$ such that $r(C)=1$ :
                                            $\det(I_n-C) \det(I_n+C) \leq  1$
we have $dim(Ker(C))=n-1$ so let's consider the base where the matrix$D$ similar to $C$ in that base has only one non null column let's consider that it's the first column and let's consider $\lambda$ the first element of that column.
we have  $\det(I_n-C)= \det(I_n-D)$  and  $\det(I_n+C)= \det(I_n+D)$
And $\det(I_n-D)=1-\lambda$ ,$\det(I_n+D)=1+\lambda$
so   $\det(I_n-C) \det(I_n+C)=1-\lambda^2 \leq  1$  :).


What is the smallest real finite value possible (that is, a value furthest to the left on the number line) using every one of the following symbols from the number of times listed, BUT not using any other symbols different from this list?

3 -> 0 [zeroes]
3 -> !  [exclamation points for factorial signs]
1 -> ( ) [1 pair of parentheses for grouping and/or multiplication]
1 -> - [used for subtraction and/or negation]
1 -> ^ [used for exponentiation]

Rule:  0^0  is indeterminate.  It and any equivalent
forms of it, if possible, do not count here.

0! = 1 is standard

That solution doesn't count.

This isn't about equivalent answers being allowed; it is about following the instructions:  "...using every one of the following symbols from the number of times listed..."

You may resubmit with the intended symbols and their respective times they occur.

Here is a negative one (pun intended):

[hide]-(0!0!)^0![/hide]


Two grade seven students were allowed to enter a chess tournament otherwise composed of grade eight students. Each contestant played once with each other contestant and received one point for a win, one half point for a tie and zero for a loss. The two grade seven students together gained a total of eight points and each grade eight student scored the same number of points as his classmates. How many students for grade eight participated in the chess tournament? Is the solution unique?

[hide]The answer is 7 or 14, the solution is not unique.

Suppose there are $ n$ grade 8 students, each scored $ k$ points. Hence we have
\[ 8\plus{}kn\equal{}\frac{(n\plus{}1)(n\plus{}2)}2,
\] i.e.,
\[ n^2\minus{}(2k\minus{}3)n\minus{}14\equal{}0.
\] Hence $ n$ is a divisor of 14, namely $ 1,2,7$ or $ 14$. If $ n\equal{}1$ or $ n\equal{}2$, then $ k$ is negative. It is easy to see that $ n\equal{}7$ and $ n\equal{}14$ are both possible.[/hide]


It last happened 68 years ago.
It will happen again in 21 years.
What is it?

[hide="Answer"]When a year is a square[/hide]

Nice riddle. [hide]How about when a year is square?[/hide]


geometry
3D geometry
sphere
induction
calculus
calculus computations

There is a spherical bubble of radius 1. a hemispherical bubble could be placed on the sphere bubble such that the radius of the hemispherical bubble is smaller than the radius of the spherical bubble. we could add more hemispherical bubbles this way to the previous system in a way such that the radius of next hemisphere is smaller than the previous hemisphere  .prove that the maximum height of the column is $1+n^{0.5}$, where n is the number of partitions..

can someone give me a hint on how to solve it? thanks in advance..

p.s: i rephrased the question myself, and i'm not sure if it is clear to the reader. can someone post the actual question? the book is currently not with me :p..

Here is the original question:

A hemispherical bubble is placed on a spherical bubble of radius 1. A smaller hemispherical bubble is then placed on the first one. This process is continued until chambers, including the sphere, are formed. (The figure below shows the case $n=4$.)
Use mathematical induction to prove that the maximum height of any bubble tower wer with $n$ chambers is $1+\sqrt{n}$.


Where can I buy a pdf of "An Imaginary Tale: The Story of i"

if I can't find any I'll just order one from amazon, but if anyone knows where I can get a pdf version, let me know

I hope that I have given you assistance

http://www.ams.org/notices/199910/rev-blank.pdf
http://mathdl.maa.org/mathDL/46/?pa=content&sa=viewDocument&nodeId=152


Oh, sorry you have already posted first, [b]blahblahblah[/b].

kunny

Of course, it should read $\sum x^2$ instead of $x^2$  :)


Find all positive integer solution of the equation
$ (a\plus{}b)^{b}\equal{} a^{b}\plus{}b^{a}$

[hide="Not sure if this is correct..."]Note that the right side of the equality is symmetric in $ a,b$. So if $ (a,b)$ is a solution then so must be $ (b,a)$, which implies $ (a\plus{}b)^{b}\equal{} a^{b}\plus{}b^{a}\equal{} (a\plus{}b)^{a}\implies a \equal{} b$. Thus $ 2^{a}a^{a}\equal{} 2a^{a}\implies a \equal{} 1$. Hence the only solution is $ (a,b) \equal{} (1,1)$.[/hide]

EDIT: Okay thank you test20. :oops:

[quote="nayel"]Note that the right side of the equality is symmetric in $ a,b$.
So if $ (a,b)$ is a solution then so must be $ (b,a)$.[/quote]
This is not a valid conclusion...
Consider the equation $ 5a\equal{}a\plus{}b$: The right hand side is symmetric in $ a,b$,
and $ (a,b)\equal{}(1,4)$ is a solution. This does NOT imply that $ (4,1)$ is a solution.

[hide]
Obviously, if $ b\equal{}1$, then all positive integers $ a$ satisfy the equation.

If $ b\equal{}2$, then we have $ (a\plus{}2)^{2}\equal{}a^{2}\plus{}2^{a}\Rightarrow4(a\plus{}1)\equal{}2^{a}$. Thus $ a$ must be one less than a power of two. If $ a\leq3$, then we have $ 4(a\plus{}1)>2^{a}$, and if $ a\geq7$ then $ 4(a\plus{}1)<2^{a}$. Thus no solutions exist for $ b\equal{}2$.

Let us consider the case $ b\geq3$.

Expanding the left side of the equation, we have
$ (a\plus{}b)^{b}\equal{}a^{b}\plus{}ba^{b\minus{}1}b\plus{}{b\choose2}a^{b\minus{}2}b^{2}\plus{}...\plus{}b^{b}\equal{}a^{b}\plus{}b^{a}$

We can subtract an $ a^{b}$ from both sides and rearrange to get
$ a^{b\minus{}1}b^{2}\plus{}{b\choose2}a^{b\minus{}2}b^{2}\plus{}...\plus{}b^{b}\equal{}b^{a}$

Obviously, we must have $ a>{b}$ or the left side of the equation will be greater than the right side. Let us divide the equation by $ b^{2}$. Then, for some positive integer $ n$,

$ a^{b\minus{}1}\plus{}b(n)\equal{}b^{a\minus{}2}$

This means we must have $ b|a^{b\minus{}1}$

Reinserting this into the original equation gives

$ (a\plus{}b)^{b}\equal{}a^{b}\plus{}(a^{b\minus{}1}k)^{a}$ for some integer $ k$.
Because $ a>b\geq2$, we can assure that $ b<ab\minus{}a$.
This means $ (a\plus{}b)^{b}\equal{}(a^{b})(m)$ for some integer $ m$.

Thus $ a^{b}|(a\plus{}b)^{b}\Rightarrow{a|a\plus{}b}\Rightarrow{a|b}$.

However, this contradicts the statement that $ a>b$. Thus there are no solutions $ (a,b)$ to the original equation that satisfy $ b\geq3$.

This means the only solutions $ (a,b)$ to the original equation are $ (x,1)$, for all positive integers $ x$.[/hide]

[quote="FightTheTide"][hide]
Obviously, if $ b \equal{} 1$, then all positive integers $ a$ satisfy the equation.

If $ b \equal{} 2$, then we have $ (a\plus{}2)^{2}\equal{} a^{2}\plus{}2^{a}\Rightarrow4(a\plus{}1) \equal{} 2^{a}$. Thus $ a$ must be one less than a power of two. If $ a\leq3$, then we have $ 4(a\plus{}1) > 2^{a}$, and if $ a\geq7$ then $ 4(a\plus{}1) < 2^{a}$. Thus no solutions exist for $ b \equal{} 2$.

Let us consider the case $ b\geq3$.

Expanding the left side of the equation, we have
$ (a\plus{}b)^{b}\equal{} a^{b}\plus{}ba^{b\minus{}1}b\plus{}{b\choose2}a^{b\minus{}2}b^{2}\plus{}...\plus{}b^{b}\equal{} a^{b}\plus{}b^{a}$

We can subtract an $ a^{b}$ from both sides and rearrange to get
$ a^{b\minus{}1}b^{2}\plus{}{b\choose2}a^{b\minus{}2}b^{2}\plus{}...\plus{}b^{b}\equal{} b^{a}$

Obviously, we must have $ a >{b}$ or the left side of the equation will be greater than the right side. Let us divide the equation by $ b^{2}$. Then, for some positive integer $ n$,

$ a^{b\minus{}1}\plus{}b(n) \equal{} b^{a\minus{}2}$

This means we must have $ b|a^{b\minus{}1}$

Reinserting this into the original equation gives

$ (a\plus{}b)^{b}\equal{} a^{b}\plus{}(a^{b\minus{}1}k)^{a}$ for some integer $ k$.
Because $ a > b\geq2$, we can assure that $ b < ab\minus{}a$.
This means $ (a\plus{}b)^{b}\equal{} (a^{b})(m)$ for some integer $ m$.

Thus $ a^{b}|(a\plus{}b)^{b}\Rightarrow{a|a\plus{}b}\Rightarrow{a|b}$.

However, this contradicts the statement that $ a > b$. Thus there are no solutions $ (a,b)$ to the original equation that satisfy $ b\geq3$.

This means the only solutions $ (a,b)$ to the original equation are $ (x,1)$, for all positive integers $ x$.[/hide][/quote]
Good
Are there any method
For example
Let $ d\equal{}gcd(a,b)$
Then use same method.


farz konid az har raese mosalas N khate sewaee rasm shodeh ast ke hich 3taee anha
yekdigar ra dar dakhele mosalas ghate nemikonand.in 3N khat daroon mosalas ra be chand nahieh mojaza taghsim mikonand???


I've got some kind of problem in understanding the idea of problem 2 from round 6. Are you sure there's no typo in it???

Oooops  :blush: I combined two different problems to get something trivial. I have restated the problem now :)

[quote="New Version of Problem 2"]A $n\times n$ matrix is filled with non-negative real numbers such that on each line and column the sum of the elements is 1. Prove that one can choose $n$ positive entries from the matrix, such that each of them lies on a different line and different column.[/quote].

[quote="perfect_radio"]What about the solutions to the problems?

I solved the circles one by computations (btw how do you prove sintetically that there's only one inner circle and one outer circle which tangent to all three?).[/quote]
Also see this[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=407992#p407992]www.mathlinks.ro/Forum/viewtopic.php?p=407992#p407992[/url] ;)


This is a special case (p = 2) of the following inequality:

For any four real numbers a, b, c, p, we have

$\left(\frac{pa-b}{a-b}\right)^2+\left(\frac{pb-c}{b-c}\right)^2+\left(\frac{pc-a}{c-a}\right)^2\geq p^2+1$.

[i]Proof.[/i]

$\left(\frac{pa-b}{a-b}\right)^2+\left(\frac{pb-c}{b-c}\right)^2+\left(\frac{pc-a}{c-a}\right)^2-\left(p^2+1\right)$
$=\left(\frac{ab^2+bc^2+ca^2-ba^2p-cb^2p-ac^2p+3abcp-3abc}{\left(b-c\right)\left(c-a\right)\left(a-b\right)}\right)^2\geq 0$.

  Darij

how can you come up with such a beautiful identity? :?

Is there really something to say? I searched for the minimum of the term $\left(\frac{pa-b}{a-b}\right)^2+\left(\frac{pb-c}{b-c}\right)^2+\left(\frac{pc-a}{c-a}\right)^2$ for a given p. Denoting this minimum by M(p), some experiments with computer algebra software revealed that M(-2) = 5, M(-1) = 2, M(0) = 1, M(1) = 3, M(2) = 5 (this was the initial problem), M(3) = 10. Fortunately, what made everything much easier was that, once the right value for M(p) was found, proving that it was indeed the minimum was easy: $\left(\frac{pa-b}{a-b}\right)^2+\left(\frac{pb-c}{b-c}\right)^2+\left(\frac{pc-a}{c-a}\right)^2-M\left(p\right)$ always turned out to be the square of some rational function. Note that the value M(1) = 3 is a kind of trap, but ignoring this value and collecting data about the other values, I came up with the conjecture $M\left(p\right)=p^2+1$ which actually holds for every $p\neq 1$. So this is how I found the value $p^2+1$. The identity now followed by straightforward factorization.

  darij

How did you factor

$\left(\frac{pa-b}{a-b}\right)^2+\left(\frac{pb-c}{b-c}\right)^2+\left(\frac{pc-a}{c-a}\right)^2-\left(p^2+1\right)$

because to me, it seems like a very heavy calculations (like IMO 2004 #2  :D) , although the end result is nice...


Let $a_i>0$ be real positive numbers, $i=\overline{1,n}$, $n\in\mathbb N$, such that
\[ \frac1{1+a_1}+\frac1{1+a_2}+\cdots+\frac1{1+a_n}=1.  \]
Put $s=a_1+a_2+\cdots+a_n$. Prove that
\[ \frac1{a_1}+\frac1{a_2}+\cdots+\frac1{a_n}\geq(n-1)\left\{\frac1{s-a_1}+\frac1{s-a_2}+\cdots+\frac1{s-a_n}\right\}.  \]

[quote="zhaobin"]I don't where I am wrong,But I think the condition
\[ \frac1{1+a_1}+\frac1{1+a_2}+\cdots+\frac1{1+a_n}=1.  \]is not necessary.[/quote]

Yes, Zhao Bin. You're right. The condition is redundant. Let's solve and generalize the following for positive numbers:
\[ \frac1{1+a}+\frac1{1+b}+\frac1{1+c}+\frac1{1+d}=1.  \]
Prove that
\[ a+b+c+d\geq 9\Bigl(\frac1a+\frac1b+\frac1c+\frac1d\Bigl).  \]


algebra
polynomial
modular arithmetic
number theory unsolved
number theory

Suppose an integer has exponent 10 modulo an odd prime p. Prove that

$ \sum^{10}_{i\equal{}1} a^{i^2} \equiv 0$ $ (mod p)$

Since the exponent of $ a$ modulo $ p$ is 10, $ a$ represents a zero of the cyclotomic polynomial $ \Phi_{10}(x) \equal{} x^4 \minus{} x^3 \plus{} x^2 \minus{} x \plus{} 1$ modulo $ p$.
We have
\[ \sum^{10}_{i \equal{} 1} a^{i^2} \equiv \sum^{10}_{i \equal{} 1} a^{i^2\bmod 10} \equiv 2(a^9 \plus{} a^6 \plus{} a^4 \plus{} a) \plus{} a^5 \plus{} 1 \equiv 0\pmod{p}\]
since $ a^5\equiv \minus{} 1\pmod{p}$ and $ x^9 \plus{} x^6 \plus{} x^4 \plus{} x \equal{} (x^5 \plus{} x^4 \plus{} x^2 \plus{} x)\cdot \Phi_{10}(x).$


Find the $ 3$ lowest multiples of $ 100$ that can be obtained by taking out any amount of plus signs from the sum and joining the two numbers that were next to the plus sign:

$ 1\plus{}2\plus{}3\plus{}\cdots \plus{} 13$

Note: you cannot move the numbers from their places.

Example, take out the second plus sign:

$ 1\plus{}23\plus{}4\plus{}5\plus{}\cdots\plus{}13 \equal{} 109$

[hide="My answer"]
By noticing that every time you take out a plus sign you add a multiple of 9 to the total sum, you have to find multiples of $ 100$ such that when you subtract $ 91\equal{}1\plus{}2\plus{}\cdots\plus{}13$ you get a multiple of $ 9$. The 3 smallest are  $ 100,1000,1900$.

When I took the test I had written down the ways to get to those numbers, but they took my papers away  :(
[/hide]

[quote="Thinkertoys"]I have a little doubt about this problem. I solved it but I didn't write down the ways to get those numbers. Was it really necessary to write these ways? And if so, how many points we will lose if we didn't write them?

I also would like to know how can we find the ways to get those numbers without use trial and error.

Thanks in advance.

Thinkertoys[/quote]

I guess it wasn't necessary to write down the ways to make those numbers, but it would make the solution better to show an example that works, instead of stating that it can be made.

You don't need to use Trial and Error on this problem, you can approach it like this:

Notice that if you take out a $ \plus{}$ sign then the sum is increased by a multiple of $ 9$, therefore we need to find multiples of $ 100$ such that when you subtract $ 91$ you get a multiple of $ 9$. In other words,

$ 100\cdot k\equiv 91\pmod{9}\implies k\equiv 1\pmod{9}$ and the three smallest natural $ k$s that work are $ 1,10,19$, and there are your three answers.


inequalities
trigonometry
algebra unsolved
algebra

$ x \equal{} 5$ is root of inequation?
$ sin(x \plus{} 1)\sqrt [3]{cosx} \minus{} sinx\sqrt [3]{cos(x \plus{} 1)} < \sqrt [3]{cosxcos(x \plus{} 1)}$

I'm not sure what your question is, but I believe the inequality is in reverse.  It should be: $ \sin (x\plus{}1)\sqrt[3]{\cos x}\minus{}\sin x \sqrt [3]{\cos (x\plus{}1)}>\sqrt[3]{\cos x \cos (x\plus{}1)}$ (wherever both sides are real).


inequalities calculus derivative algebra polynomial

Let $p(x)$ be a polynomial with real coefficients. Prove that if\[ p(x)-p'(x)-p''(x)+p'''(x)\geq0 \]for every real $x$, then $p(x)\geq0$ for every real $x$

Let $f(x)=\Big[p(x)-p''(x)\Big]e^{-x}$, $g(x)=\Big[p(x)-p'(x)\Big]e^{x}$, $h(x)=p(x)e^{x}$
\begin{align*}
f'(x)&=-\Big[p(x)-p'(x)-p''(x)+p'''(x)\Big]e^{-x}\leqslant0\\
f(x)&=\Big[p(x)-p''(x)\Big]e^{-x}\\
&\geqslant\lim\limits_{x\to+\infty}\Big[p(x)-p''(x)\Big]e^{-x}=0\\
g'(x)&=\Big[p(x)-p''(x)\Big]e^{x}\geqslant0\\
g(x)&=\Big[p(x)-p'(x)\Big]e^{x}\\
&\geqslant\lim\limits_{x\to-\infty}\Big[p(x)-p'(x)\Big]e^{x}=0\\
h'(x)&=-\Big[p(x)-p'(x)\Big]e^{-x}\leqslant0\\
h(x)&=p(x)e^{-x}\geqslant\lim\limits_{x\to+\infty}p(x)e^{-x}=0
\end{align*}
Thus $p(x)\geqslant0$

Similar below.
Let $p(x)$ be a polynomial with real coefficients. Prove that if\[ p(x)-2p’(x)+p''(x)\geq0 \]for every real $x$, then $p(x)\geq0$ for every real $x$.

[i]Source: University of Warsaw and Editorial of Delta Math League, February 2018

Both problems follow from two lemmas for polynomials.

[b][u]Lemma 1:[/u][/b]
$p(x) \ge p'(x)$ for all real $x$, then $p(x) \ge 0$ for all real $x$

[u]Proof:[/u]
Obviously p must have even degree with positive leading coefficient, thus p must have a global Minimum at some point c.
Then $p(x) \ge p(c) \ge p'(c) = 0$

[b][u]Lemma 2:[/u][/b]
$p(x) \ge p''(x)$ for all real $x$, then $p(x) \ge 0$ for all real $x$

[u]Proof:[/u]
Obviously p must have even degree with positive leading coefficient, thus p must have a global Minimum at some point c.
Then $p(x) \ge p(c) \ge p''(c) \ge 0$

[u]Comment 1[/u]
For solving the original problem look at $p(x)-p''(x)$
For the other one take $p(x)-p'(x)$

[u]Comment 2[/u]
A similar statement does not hold for the third derivate, try e.g. $x^6+120x^3+720$


algebra
polynomial
algorithm
calculus
derivative
trigonometry
calculus computations

I'm having trouble simplifying:
$\sum_{n=0}^{\infty}\frac{x^{3n}}{(3n)!}$
I see that it solves $f'''(x)=f(x)$ and $f(x)+f'(x)+f''(x)=e^{x}$...

Any help would be appreciated

Thanks

~ tZf

Thanks. Can anyone show me how that can be derived (preferably by hand)?

How would you solve this differential equation?

$y'''-y=0; y(0)=1,y'(0)=0,y''(0)=0$

I can't figure out how to solve that  :(

It's a linear, constant coefficient, homogeneous differential equations.The standard procedure for solving such equations goes like this:

1. First, make the structured guess that there are solutions of the form $y=e^{rx}.$ Put that into the differential equation, and divide the resulting equation by $e^{rx}.$ (Safe to divide by because it's never zero.) You get a polynomial equation for $r$ of degree $n,$ where $n$ is the order of the original differential equation. In this case, $n=3.$

2. Solve the resulting polynomial equation, keeping all $n$ of the roots. (This is the weak link of the procedure, and why it falls short of being a foolproof algorithm - see Abel-Galois unsolvability results. But that's not a problem here - we know how to solve this equation.)

3. We will represent the general solution as a linear combination of $n$ terms, each of which is a solution. Each distinct real root $r=r_{j}$ of the equation contributes a $c_{j}e^{r_{j}x}$ term.

For a double root, use two terms: $c_{1}e^{rx}+c_{2}xe^{rx}.$ For a triple root, use three terms: $(c_{1}+c_{2}x+c_{3}x^{2})e^{rx}.$ And so on.

For a compex conjugate pair $r=a\pm bi,$ we can recombine the complex exponentials into $c_{1}e^{ax}\cos(bx)+c_{2}e^{ax}\sin(bx).$

(For repeated complex conjuage roots - you figure that one out.)

4. Finally, assuming you have initial conditions, take the derivative, second derivative, and so on, for your general form and put in the value of $x$ as which the intial conditions occur. The result should be $n$ linear equations for the $n$ unknown coefficients.

For the differential equation $y'''-y=0$ I'll take you through step 3: The characteristic equation is $r^{3}-1=0.$ That has roots $r=1$ and $r=-\frac12\pm\frac{i\sqrt{3}}2.$ The general solution of the differential equation is

$y=c_{1}e^{x}+c_{2}e^{-x/2}\cos\left(\frac{\sqrt{3}}2\,x\right)+c_{3}e^{-x/2}\sin\left(\frac{\sqrt{3}}2\,x\right).$

I'll leave step 4 for you.

I'm getting:
$\frac{e^{x}}{3}+\frac{2}{3}e^{-x/2}cos\left(\frac{\sqrt{3}}{2}x\right)$
and it seems to be working.

Thanks a lot for your help  :)


algebra
polynomial
function
geometry
3D geometry
calculus
calculus computations

With eatch polinomial $ p(x)$ we associate a number $ D(p)$ by following rules:

$ \quad D(\alpha_1 p_1\plus{} \alpha_2 p_2)\equal{}\alpha_1D(p_1)\plus{}\alpha_2D(p_2)$
$ \quad D(p_1p_2)\equal{}D(p_1)\ p_2(1/2)\plus{}p_1(1/2)D(p_2)$,  $ \alpha_1,\alpha_2\in {R}$

Prove that
$ D(p)\equal{}c\ p'(1/2)$, $ c\equal{}$  const.
If we replace polinomial $ p$ in definition of number $ D(p)$ with arbitrarry function $ f\in {C[0,1]}$, prove that then is $ D(f)\equal{}0$, for every such $ f$

the first property shows that $ D$ is uniquely determined by its values at $ x^n$ for $ n\in \mathbb{R}$ and at the constant polynomials. together with the second property, it follows that $ D$ is uniquely determined by its values at $ x$ and at the constants. by property 2, $ D(x) \equal{} D(1x) \equal{} 1\cdot D(x) \plus{} \frac {1}{2}D(1) \equal{} D(x) \plus{} \frac {1}{2}D(1)$, from which it follows that $ D(1) \equal{} 0$. by property 1, $ D(r) \equal{} D(r\cdot 1) \equal{} rD(1) \equal{} 0$ for every constant polynomial $ r$. note that the definition $ D(x) \equal{} c$ and $ D(r) \equal{} 0$ for every constant $ r$ coincides with the definition $ D(p) \equal{} cp^{\prime}(\frac {1}{2})$ at $ p \equal{} x$ and at all constants, so they coincide at all $ p$. checking that $ D(p)\equal{}cp^{\prime}(\frac {1}{2})$ satisfies the two relations completes the proof.

@pleurestique
Very nice, bat this is only the part of solution.
What about
D(f)=0 for every continuos f?

What does it do to something like a cube root? From that product rule, we can generate a power rule $ D(p^n)\equal{}D(p^{n\minus{}1})p(\frac12)\plus{}(p(\frac12))^{n\minus{}1}D(p)\equal{}\cdots \equal{} n\left(p(\frac12)\right)^{n\minus{}1}D(p)$.

Thus, if $ f(\frac12)\equal{}0$, $ Df\equal{}3\left(f^{\frac13}(\frac12)\right)^2D(f)\equal{}0$. Then, if $ g(x)\equal{}f(x)\plus{}c$, $ Dg\equal{}Df\plus{}Dc\equal{}0\plus{}0\equal{}0$. That's everything.

@jmerry

Yes, by your hint, general case may be:

For each $ f$, let $ f_0(x) \equal{} f(x) \minus{} f(1/2)$. Then $ f_0(1/2) \equal{} 0$
Then, $ f_0 \equal{} h_1 \minus{} h_2, h_1 \equal{} \max \{f_0,0\}, h_2 \equal{} \minus{} \min \{f_0,0\}$
Then  $ h_i\geq 0$,  and both are continuos functions.
Now,  $ h_1(1/2) \equal{} 0$ and so
$ D(h_1) \equal{} D(\sqrt {h_1}\sqrt {h_1}) \equal{} 2\sqrt {h_1(1/2)}D(\sqrt {h_1})\Rightarrow D(h_1) \equal{} 0$
Similary for  $ h_2$,  and then linearity.

[b]Problem is solved![/b]


[hide="Generalization"] Given two sequences of integers $x_{i}, y_{i}$ such that $\sum x_{i}= \sum y_{i}$,

$\frac{ \prod x_{i}! }{ \prod y_{i}! }$

Is an integer if $x_{i}\succ y_{i}$ (see: http://mathworld.wolfram.com/Majorization.html ).  A hint to the proof:

http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=141316

In particular, the multinomial $\frac{(\sum a_{i})! }{ \prod a_{i}! }$ is always an integer. [/hide]

does $\frac{ \binom{2a}{a}*\binom{2b}{b}}{\binom{a+b}{a}}$ give a workable method of attack?


$ y \equal{}\sqrt{x}$ gives $ y^{2}\minus{}5y\plus{}6 \equal{} (y\minus{}3)(y\minus{}2) \equal{} 0$, so $ y \equal{} 2$ or $ y \equal{} 3$, implying that $ x \equal{} 4$ or $ x \equal{} 9$.

another way is this
[hide] the equation given is $ x\minus{}5\sqrt{x}\plus{}6\equal{}0$. move $ x\plus{}6$ to the right to get $ \minus{}5\sqrt{x}\equal{}\minus{}x\minus{}6$. square both sides to get $ 25x\equal{}x^{2}\plus{}12x\plus{}36$ now move $ 25x$ over and factor to get $ 0\equal{}(x\minus{}4)(x\minus{}9)$ using the zero factor property we get what mellowmellon said. [/hide]


[quote="most-wanted"]how to solve this ?

I    $3x-5y+4z = 5$
II.  $7y+2y-3z = 2$
III. $4x+3y-7y =-11$[/quote]
There is a method with matrices, but for those uncomfortable with matrices, you can solve it by eliminating one variable at a time or by isolating one variable with equation I, then plugging that into the others and so on.

[quote="most-wanted"]how to solve this ?

I    $3x-5y+4z = 5$
II.  $7y+2y-3z = 2$
III. $4x+3y-7y =-11$[/quote]

I think in the second equation, the first term on LHS is 7x not 7y.

Anyways, a short way to go this is to notice that when you add I and III, you get a similar equation to II other than the fact that the LHS will have a -2y instead of +2y and the RHS will have a -6 instead of 2.

So, to solve for y, do (I+III)-II.
Then, the rest is easy using simple elimination of one of the two remaining variables.

You could also use Gaussian Elimination, as described below

http://mathworld.wolfram.com/GaussianElimination.html


inequalities
function
inequalities proposed

[color=darkred]Given are the real numbers  $ x_k\in[a,b]\ ,\ k\in\overline {1,n}\ ,$

where $ 0 < a < b\ ,\ n\in \mathcal N^*$ and $ 4nab > (a + b)^2\ .$

 Prove that $ (a+b)^2\cdot(\sum_{k = 1}^nx_k)^2\ge 4nab\cdot\sum_{k = 1}^nx^2_k\ .$

[b]Remark.[/b] The particular case  $ \{\begin{array}{c}
n=4\\\
a = 1\\\
b = 3\end{array}$ belongs to [b]V. Cartoaje & M.Lascu .[/b][/color]

[color=darkblue]Where is the your curiosity ?!
Falsely or truly ?! Easy or hard ?! Ugly or nice ?!
Please, at least an opinion ...[/color]

see RHS-LHS as a function on x_i
It is convex, so obtains maximum on endpoints.
Therefore it is enough to consider the case when i of them are a and n-i of them are b.
This does not seem too hard :lol:

Thanks, [b]Ywl[/b] ! On the your idea is and the my proof. In conclusion, it is easy, with the standard technics ....


i dont really know how to explain it but i'll try...
how long does a topic have to remain idle for it should not be replied to anymore?
like if a topic has been dead for a month you shouldnt reply to it right??

Depends what topic it is, and how you reply to it. It's usually safe to reply as long as you're not spamming.

Different forums have different rules. For example, take the High School Basics forum. If you reply to an old topic, your only reason should be either you have a question, or you present a completely different solution for arriving at the answer. However, in the Games and Fun Factory, usually new topics arise to replace the old threads. If you have a question, you can always ask the mod who's in charge of that forum (providing that he/she is still active that is...)

General Rule of Thumb:

Don't reply to an old thread unless you have some new insight to the problem, or have noticed a mistake someone made.

Once, someone asked a question, and asked how to solve it. I think about 3 people responded, me and two others. We offered different solutions, and no one else replied...until, about a month later, someone just said, "Yeah, 32's the right answer."

That was totally unnecessary, and it's spam. That's what you DON'T want to do. But if that someone had thought of a different solution, and posted it, I would think it's fine.


I've been doing practice ARMLs, and I keep getting 3s and 4s  :blush:

Does anyone have any advice?

Do you know how to do them, but are just not getting them, or do you not know how to do them?

If you know how to do them, then I would try to write out a decent solutions (one you could show to people), as your work for each problem.  Also, read Stop Making Stupid Mistakes (forgot where it is, but somewhere), if you make stupid mistakes and do what they say.

If you don't know the information, I would suggest learning some simple things that you need a lot.
For example:
Basic Trig is easy to learn, as are Logs.  Then, you can always use memorize formulas for areas of geometric figures, and stuff like that.

If you already know most of the simple things, then you can't do much in a short amount of time to improve your knowledge.

AoPS books are good to read through.


LaTeX
arithmetic sequence
geometric sequence

In a twenty element arithmetic sequence with first term 8 and common difference 3 what is the largest element that can be part of a three-term geometric sequence using elements of the arithmetic sequence?

Let A, B, C be a part of the geometric sequence, we know that B=AR and C=ARR

Notice that all elements of the arithmetic sequence are congruent to 8 mod 3. In otherwords n is an element of the Set If  $ n \equiv 8 \mod(3)$. (How do you do the congruence symbol? IE how do you input LaTeX in Aops?) (This also implies $ n \equiv 2 \mod(3)$ )

Also we know A can be expressed as 8+3k where k is an integer.

So B can be expressed as 8r+3kr. And since B is an element of the arithmetic sequence,  $ 8r \plus{} 3kr \equiv 2 \mod(3)$. (Notice that $ 8r \plus{} 3kr \equiv 2r \mod(3)$) This also implies $ 8r^2 \plus{} 3kr^2 \equiv 2 \mod(3)$  (and $ 8r^2 \plus{} 3kr^2 \equiv 2r^2 \mod(3)$).

I believe the answer then follows by seeing when 2r is congruent to 2 mod 3.

[hide="solution"] r has to be equal to 1, 4, 7, 10, 13...
Notice that if r=4 then Even if k=0, C becomes 'too large' for the set
So r has to be equal to one.

Assuming that the geometric sequence has distinct elements, there therefore is no solution, otherwise A=B=C so 8+19*3 is your answer, or 65.

Possibly there is a problem with this solution (non-integer solutions for r?)*, but it would seem to make more sense if the total elements of the series was 41+
*r could be 1/5? But then this would mean that since 50 is an element 15 is an element... which is not

[/hide]

For LaTeX, you use "$ \$$" symbols surrounding the LaTeX code.


geometry
3D geometry
sphere
Locus problems
Locus
IMO
IMO 1978

We consider a fixed point $P$ in the interior of a fixed sphere$.$ We construct three segments $PA, PB,PC$, perpendicular two by two$,$ with the vertexes $A, B, C$ on the sphere$.$ We consider the vertex $Q$ which is opposite to $P$ in the parallelepiped (with right angles) with $PA, PB, PC$ as edges$.$ Find the locus of the point $Q$ when $A, B, C$ take all the positions compatible with our problem.

Define vectorOP=p, vectorPA=a, vectorPB=b, vectorPC=c
As (p+a)^2=r^2, (p+b)^2=r^2, (p+c)^2=r^2, a.b=0, b.c=0, c.a=0
Then we will have this equality: (p+a+b+c)^2=3r^2-2p^2
So the locus is a sphere, radius R=sqrt(3r^2-2p^2), the centre of it is O.

For this problem, do you have to prove that given any $Q$ on the circle you can always find $A,B,C$?


function
calculus
integration
inequalities
real analysis
inequalities solved

Let $f$ be a convex function on an interval $\left[a;\ b\right]$. Prove that

$f\left(\frac{a+b}{2}\right)\leq\frac{1}{b-a}\int_{a}^{b}f\left(x\right)dx\leq\frac{f\left(a\right)+f\left(b\right)}{2}$.

I am adding a condition to your problem:

[color=blue]Let f be a continuous convex function on an interval [a; b]. Prove that

$f\left(\frac{a+b}{2}\right)\leq\frac{1}{b-a}\int_{a}^{b}f\left(x\right)dx\leq\frac{f\left(a\right)+f\left(b\right)}{2}$.[/color]

Hereby, I have required the function f to be continuous in order to prevent from pathologic cases. Probably a much weaker condition would be enough.

[i]Solution.[/i] Any number $x\in\left[a;\;b\right]$ can be written in the form $x=ta+\left(1-t\right)b$, with some real t satisfying $0\leq t\leq 1$. Thus, the number a + b - x, which also lies in the interval [a; b] (in fact, since $x\geq a$, we have $a+b-x\leq b$, and since $x\leq b$, we have $a+b-x\geq a$, so that $a\leq a+b-x\leq b$, and thus $a+b-x\in\left[a;\;b\right]$), can be written as $a+b-x=a+b-\left(ta+\left(1-t\right)b\right)=\left(1-t\right)a+tb$.

On the other hand, by the convexity of the function f, we have

$f\left(ta+\left(1-t\right)b\right)\leq tf\left(a\right)+\left(1-t\right)f\left(b\right)$        and
$f\left(\left(1-t\right)a+tb\right)\leq \left(1-t\right)f\left(a\right)+tf\left(b\right)$.

Thus,

$f\left(x\right)+f\left(a+b-x\right)=f\left(ta+\left(1-t\right)b\right)+f\left(\left(1-t\right)a+tb\right)$
$\leq \left(tf\left(a\right)+\left(1-t\right)f\left(b\right)\right)+\left(\left(1-t\right)f\left(a\right)+tf\left(b\right)\right)=f\left(a\right)+f\left(b\right)$,

so that $\frac{f\left(x\right)+f\left(a+b-x\right)}{2}\leq\frac{f\left(a\right)+f\left(b\right)}{2}$.

On the other hand, since the function f is convex on the interval [a; b], we have

$\frac{f\left(x\right)+f\left(a+b-x\right)}{2}\geq f\left(\frac{x+\left(a+b-x\right)}{2}\right)=f\left(\frac{a+b}{2}\right)$.

Thus, we get the chain of inequalities

$f\left(\frac{a+b}{2}\right)\leq\frac{f\left(x\right)+f\left(a+b-x\right)}{2}\leq\frac{f\left(a\right)+f\left(b\right)}{2}$.

Taking the mean value over all numbers x in the interval [a; b], we get

$f\left(\frac{a+b}{2}\right)\leq\frac{1}{b-a}\int_{a}^{b}\frac{f\left(x\right)+f\left(a+b-x\right)}{2}dx\leq\frac{f\left(a\right)+f\left(b\right)}{2}$.

But

$\int_{a}^{b}\frac{f\left(x\right)+f\left(a+b-x\right)}{2}dx=\int_{a}^{b}\frac{f\left(x\right)}{2}dx+\int_{a}^{b}\frac{f\left(a+b-x\right)}{2}dx$
$=\int_{a}^{b}\frac{f\left(x\right)}{2}dx+\int_{a}^{b}\frac{f\left(x\right)}{2}dx$         (linear substitution $x\to a+b-x$ in the second integral)
$=\int_{a}^{b}f\left(x\right)dx$.

Thus, we get

$f\left(\frac{a+b}{2}\right)\leq\frac{1}{b-a}\int_{a}^{b}f\left(x\right)dx\leq\frac{f\left(a\right)+f\left(b\right)}{2}$,

and the problem is solved.

  Darij

[quote="darij grinberg"]Hereby, I have required the function f to be continuous in order to prevent from pathologic cases. Probably a much weaker condition would be enough.
[/quote]

In fact, convex functions on open intervals are continuous.  So $f$ is certainly continuous on $(a,b)$ and hence continuous almost everywhere on $[a,b]$, that is, except possibly on a set of measure zero and hence Riemann integrable.  This gives your result without the stipulation that $f$ be continuous.

By the way, I believe this is Hadamard's inequality.


If the radius of the quadrant is $ R$, find the colored area.

[hide="Answer"]

$ \dfrac{\pi R^2}{16}$[/hide]

I must be blind, because i can't find why $ r \equal{} \dfrac{R}{4}$, where $ r$ is the radius of the small circle. Somebody can tell me why?

You are missing one important line AO''E (the 3 points are colinear). with this hint I assume you'll make it.

Thanks delta. That solve everything. I almost calculate all that segments to go nowhere. It's obvious that 3 points are collinear because the small circle and the quadrant are tangent.  :wallbash_red:

we can do an inversion about the large circle...

[quote="DraX"]Sorry but i dont see why it is collinear, can anyone explain it to me, the tangent thing of the small circle and the quadrant.. i mean what are the properties that u use? thanks[/quote]
AE and O"E are both perpendicular to a tangent at E

thanks for the explanation!
but i still cant solve why r =R /4 anyone can show the solution?

[hide="One Solution"]
$ \Delta O''O'D: \overline{O''D}^2 \equal{} \left(\dfrac{R}{2} \plus{} r\right)^2 \minus{} \left(\dfrac{R}{2} \minus{} r\right)^2 \Longrightarrow \overline{O''D} \equal{} \sqrt {2Rr}$

$ \Delta O''AD: (R \minus{} r)^2 \equal{} (\sqrt {2Rr})^2 \plus{} r^2 \Longrightarrow r \equal{} \dfrac{R}{4}$

Hence,

$ S \equal{} \dfrac{\pi\cdot R^2}{4} \minus{} \dfrac{\pi\cdot(R/2)^2}{2} \minus{} \pi\cdot\left(\dfrac{R}{4}\right)^2 \equal{} \dfrac{\pi\cdot R^2}{16}$
[/hide]


trigonometry
geometry
integration
calculus
calculus computations

Hey, I have to do some basic calc II questions, I'm not sure on the last few and was wondering if any of you could hel me out, I have most of them already done, just need to know if they are correct. I created a pdf here: http://armor4life.com/help.pdf

please lemme know how I did, and any explainations to why I am wrong
Thank you very much

on Number 4, the fourth expression should be: $(u + 5)\sqrt{u+1}du$, the next expression didnt distribute right, try a trig substitution
[hide="hint"]$1 = sin^2x + cos^2x$[/hide]

i havent done a problem like number 8 in a while, but where ever you define temperature t, you want the same amount of area above and below the line, if i remember correctly $\frac{1}{b - a} \int_a^bf(x)dx$ serves that purpose, so check to see if thats it


vector
linear algebra
matrix
analytic geometry

Find the dimension of the vector space V spanned by
$ v_1\equal{}(1,1,1,1,1),
v_2\equal{}(1,0,1,1,1),
v_3\equal{}(1,2,1,1,1),
v_4\equal{}(0,\minus{}1,0,0,0).$

Show that vectors $ v_2,v_3$ form a basis.

Then find the matrix Q which expresses the basis $ v_1,v_3$ in terms of the basis $ v_2,v_3$. Write the coordinates of the vector $ v_4$ in terms of these two bases.


geometry
circumcircle
cyclic quadrilateral
geometry solved

How do we prove that a fermat point exists for a triangle?? i.e. how do we prove (without actual construction) that there exist a point s.t. all sides of the triangle subtend angles of $120^{\circ}$ at that point. In the general case, how do we prove or disprove that for an $n$ sided polygon ( both even and odd) there exists a point such that each side subtends an angle of \[ \dfrac{360}{n} \] at that point. My calculations seem to suggest that we can prove it for $n$ odd but my conjecture is false for even --- though I may be wrong!!! Assume a 'normal' case -- convex etc....

[quote="Rushil"]How do we prove that a fermat point exists for a triangle?? i.e. how do we prove (without actual construction) that there exist a point s.t. all sides of the triangle subtend angles of $120^{\circ}$ at that point.[/quote]

Well, here is a proof that automatically yileds a construction:

Let ABC be our triangle. Also, assume that all angles of triangle ABC are <= 120° (this is necessary for some arrangement considerations; if an angle of triangle ABC is > 120°, then some of the 120° angles must be replaced by 60° angles). Now, you want to prove that there exists a point F such that < BFC = < CFA = < AFB = 120°.

Well, erect equilateral triangles BA'C, CB'A, AC'B outwardly on the sides BC, CA, AB of triangle ABC. Then, of course, since the triangles BA'C, CB'A, AC'B are equilateral, < BA'C = < CB'A = < AC'B = 60°. Now, let F be the point of intersection of the circumcircles of triangles BA'C and CB'A (apart from C). Then, the quadrilateral BA'CF is cyclic, so that < BFC = 180° - < BA'C = 180° - 60° = 120°, and similarly the cyclic quadrilateral CB'AF yields < CFA = 120°. Now, < AFB = 360° - < BFC - < CFA = 360° - 120° - 120° = 120°. Hence, our point F satisfies < BFC = < CFA = < AFB = 120°. This completes the proof.

Also, we can find a nice collateral result: Since < AFB = 120° = 180° - 60° = 180° - < AC'B, the quadrilateral AC'BF is cyclic, i. e. the point F lies on the circumcircle of triangle AC'B. So we have proved the existence of a point F such that < BFC = < CFA = < AFB = 120° and found a construction of this point F as the point of intersection of the circumcircles of triangles BA'C, CB'A and AC'B.

All I can say about the n-points problem is that [i]in general[/i], for n points A1, A2, ..., An in the plane, the point P which minimizes PA1 + PA2 + ... + PAn cannot be constructed with compass and ruler. It can be found by solving algebraic equations (whose degree depends on n) or by approximation algorithms. I don't think this will change that much if the points lie in space.

[quote="Rushil"]In the general case, how do we prove or disprove that for an $n$ sided polygon ( both even and odd) there exists a point such that each side subtends an angle of $\dfrac{360}{n}$ at that point.[/quote]

This is wrong for every $n\geq 3$. This should also answer your problems with http://www.mathlinks.ro/Forum/viewtopic.php?t=472 ...

  darij

OK!! Thanks!!

So you mean to say, we CANNOT precisely determine the properties of the required point which minimies $\sum\limits_ {i=1}^{n} PA_{i}$ ??? Actually, if you consider just the $n =$ odd case, you get that IF there exists a point such that every side subtends an angle = $\dfrac{360}{n}$ at that point, then that point $P$ is the general fermat point. The only 'catch' is the exstence of that point $P$. Am I going wrong somewhere???

Please help!


x is congruent to 3 (mod 5)

x is congruent to 2 (mod 13)

find all values of x?

im in summer school holidays right now in australia, flipping through maths textbooks, and ive come across a simultaneous congruency problem.. and i'm lost as how to attack this type of problem...

because 5 and 13 are relatively prime,
x can be exactly determined modulo 5*13.

search the first values:

x = a*5 + 3 => 3, 8, 13, 18, 23, 28, 33
x = b*13 + 2 => 2, 15, 28, 41

so x is congruent to 28 (mod 65)

wish u nice holidays

If you know chinese remainder theorem , that is exactly the one of its application where a number when divided by 5 left 3 , when divided by 13 left 2 .

exactly; I didn't know the name of the theorem, I forgot it. But we discussed it under that name.


I want to solve

0.62243x + 0.51824y = 0.70524
0.71497x + 0.59496y = 0.80996

but different calculators and mathematical programs give me different arithmetic solutions.

Why does this happen?
Which is the exact solution?
Which solving method is the best and most accurate?
And can I have an exact solution with 5‐digit precision?

Thank you!

I'm assuming that the decimal approximations are a little off, but these appear to be linearly dependent, i.e. they are the same line. So any $x$ and $y$ that work for the first one will work for the second. And this doesn't belong in this forum.


Just getting things straight for Tuesday's AMC, it's 6 points for correct, 2.5 for each unanswered, and 0 for each wrong? I thought each unanswered was 1 point, so all of my scores on practices are in the 80's, and if it's 2.5 for each, I think I might make an awesome score this year  :D


complex analysis
complex analysis unsolved

Show that if f(z) has a pole of order m at z(sub 0), we define g(z) = f'(z)/f(z), then g(z) has a pole with order 1 at z(sub 0).

just use that if f has a pole of order m at $a$ then we can write $f(z)=(z-a)^{-m}\times h(z)$ where h is holomorphic and non vanishing in a neighbourhood of $a$.

Note that if you follow through this, you discover not only that $g$ has a simple pole at $z_{0}$ but also that its residue there is $-m.$ And if $f$ had had a zero of order $m$ at $z_{0},$ then $f(z)=\frac{f'(z)}{f(z)}$ would have had a simple pole at $z_{0}$ with residue $m.$ That's all there if you continue what kouboy started.


Quadrilateral ABCD is inscribed in a circle.  BD is a diameter, angle CAB is 45 degrees, and CD=2.  Find the area of the circle.

[hide="I think I got it right..."]Since $\angle CAB = 45^{\circ}$ then $\angle CAD = 45^{\circ}$ since $\widehat{BCD}=180^{\circ}$ and $\angle BCD = 90^{\circ}$ because $BD$ is a diameter. So... there's an isoceles right triangle $\triangle CBD$ to be used ($CD = CB$ because $C$ is in the middle of $\widehat{BCD}$), and $CD = 2$ therefore $BD = 2 \sqrt2$ and so the radius of the circle is $\sqrt2$. This makes the area $\boxed{2\pi}$.[/hide]

I got the same answer. My reasoning was that the quadrilateral is a square with sidelength 2.

:!: It's not necessarily a square :!:
Try to draw a picture that satisfies the givens and makes the quadrilateral not a square, and then you'll see why.
But yeah, that will still get you the right answer  :P

hmm, if CAB = 45 degrees, doesn't that make CAD, ACB and ACD 45 degrees, making a square?

Yes, the answer is true for all cuadrilaterals, not only for a square..........

[quote="Hokkage"]Here's a diagram showing a contradictory case. All that is required is that A is anywhere on arc $\widehat{BD}$ without coinciding with B or D.[/quote]

That clears it up thanks :)

Well, technically, it's only true for [i]cyclic[/i] quadrilaterals.


Express $ \sqrt{1296a^{18}b^{10}}$ in simplest form.

Just pluck out perfect squares from under the radical to get $ \boxed{36a^9b^5}$.


Diophantine equation
number theory unsolved
number theory

Let p be a prime number. Prove that there is only a finite number of $ (x,y)\in \mathbb{N}^2$ such that:

1) $ x^2\minus{}py^2\equal{}1$

2) $ x\plus{}y$ is a power of p

The first point is not true. In fact in general the equation $ x^{2}\minus{}Dy^{2}\equal{}1$ is called Pell's equation and it is well-known that
I)the equation has infinitely many solutions in naturals if and only if $ D\neq l^{2}$
II)if $ D\equal{}l^{2}$ than $ x^{2}\minus{}y^{2}l^{2}\equal{}(x\minus{}ly)(x\plus{}ly)\equal{}1$ which has no solutions in naturals :wink:

I'm sorry, but I am afraid that you misunderstood the problem (probably because of my terrible english :blush: )...

You have to prove that there is only a finite number of positive integers such that both 1) and 2) are true, that is to say "x,y is a solution to the pell equation $ x^2 \minus{} py^2 \equal{} 1$ [i]and[/i] x+ y is a power of p".


limit
logarithms
calculus
calculus computations

Find
$\lim_{x\rightarrow \infty}(\frac{2\sqrt{x}}{2\sqrt{x}-\sqrt{2}}\ln (x-\sqrt{2x}+1)-\frac{2\sqrt{x}}{2\sqrt{x}+\sqrt{2}}\ln (x+\sqrt{2x}+1))$


Simplify this ugly summation without summation symbols.
sum of (k,a) * (k-a,a) * 4^(k-a), as a ranges from 0 to largest integer less than k/2, and (n,k) is n!/k!(n-k)!

So we can see that this is the value of the constant term in

$2^k(x+2+\frac{1}{x})^k$

Rewrite as

$2^k((x+\frac{1}{x})+2)^k$

If $(x+\frac{1}{x})$ is raised to an odd power you won't have a constant term.

If it is raised to an even power you will have a constant term of $\binom{a}{\frac{a}{2}}$.

So we have

$2^k(2^k + \binom{2}{1}\binom{k}{2}2^{k-2} + \binom{4}{2}\binom{k}{4}2^{k-4} + ... + \binom{2a}{a}\binom{k}{2a}2^{k-2a} + ...)$

Bah! This is what we started with. Well w/e, use my $2^k(x+2+\frac{1}{x})^k$
if you can.

[quote="epitomy01"]neat idea.
so you finish it off by noticing that (y+1/y+2) = (x+1/x)^2 where x^2 = y; so that the constant term in (y+1/y+2)^k is the constant term in (x+1/x)^(2k); so you end up with (2k,k)*2^k.[/quote]

Oh dang how did I not see that? Thanks. lol


[quote="Dr Sonnhard Graubner"]hello, wolfram alpha says no result founds in terms of standard mathematical functions.
Sonnhard.[/quote]

Since you mentioned wolfram alpha, I checked for [url=http://www.wolframalpha.com/input/?i=Integrate[Cos[x]^2%2Cx]]Integrate[Cos[x]^2,x][/url]

and
[url=http://www.wolframalpha.com/input/?i=Integrate%5BCos%5Bx%5D%5E3%2Cx%5D]Integrate[Cos[x]^3,x][/url]

As you said, [url=http://www.wolframalpha.com/input/?i=Integrate%5BCos%5Bx%5D%5Ex%2Cx%5D]Integrate[Cos[x]^x,x][/url], there's no result.

What can I do, now?

Is there a connection with finding the antiderivative of x^x over [0,1] ?


algebra
polynomial
modular arithmetic
calculus
integration
algorithm
number theory unsolved

Let $ F(X)\equal{}X^D\plus{}A_1X^{D\minus{}1}\plus{}A_2X^{D\minus{}2}\plus{}...\plus{}A_{D\minus{}1}X\plus{}A_D$ be a polynomial whose coefficients $ A_1,  A_2,..., A_D$ are integers. Let $ p$ be a prime number. Show that the congruence $ F(X)\equiv 0\pmod{p}$ has at most $ D$ solutions with $ 0\leq X<p$.

In fact, any polynomial of degree $ n$ has at most $ n$ roots in any integral domain. Here you even have a field.
You can use division algorithm to write $ f(x) \equal{} (x\minus{}a) g(x)$ if $ a$ is a root of $ f$. Now iterate this, until $ g$ gets constant.


Simion conjecture's the following:

"Given the Tribonacci sequence defined by $x_1 = 1, x_2 = 2, x_3 = 3$, and $T_n = T_{n-1}+T_{n-2}+T_{n-3}$ for $n\geq 4$, the only perfect square in the sequence is $x_1$."

This is NOT the solution!

We have that

$X_1=1$, $x_2=2$, $x_3=3$ and $x_n=x_{n-1}+x_{n-2}+x_{n-3}$ for all $n\geq 4$.

Each $x_n$ of the succession can be calculated in terms of $x_1$, $x_2$ and $x_3$.

For example $x_6=2+3+1+2+3+1+2+3+3=20$. So here we have $x_6=2x_1+3x_2+4x_3=1 \cdot 2+2 \cdot 3+3 \cdot 4$.
In general $x_n=U_n+2D_n+3T_n$ (for $n=6$, $U_n=2$, $D_n=3$ and $T_n=4$).
Now it is quite easy to obtain these equations:

$T_n=D_{n+1}-U_n$
$U_n=T_{n-1}$
$D_n=U_n+U_{n-1}$

so

$x_n=U_n+2D_n+3T_n=U_n+2(U_n+U_{n-1})+3(D_{n+1}-U_n)= 3U_n+2U_{n-1}+3(U_{n+1}+U_n-U_n)=3U_{n+1}+3U_n+2U_{n-1}$

Now it is to be shown that a number of the form $3n+3k+2m$ with $n,k,m$ integers
and $n>k>m$ is never a square :D

Well the last one does not work: $n=2 , k=1 , m=0$.

I have also edited your post: please use {} not () for the indizes.

[quote="ZetaX"]Well the last one does not work: $n=2 , k=1 , m=0$.
[/quote]
Yes, I realized that few minutes ago :oops:

From $U_4$, the sequence is this: http://mathworld.wolfram.com/TribonacciNumber.html

So we have an explicit formula for $U_n$, right? This can help us to find the formula for $x_n$, since we know that $x_n=3U_{n+1}+3U_n+2U_{n-1}$.


function
induction
algebra proposed
algebra

Find all continuous functions $f:\mathbb{R}\rightarrow\mathbb{R}$ satisfying: \[f(x+y+f(xy))=xy+f(x+y)\]

Hmm it's a very nice problem. I can't solve it though, and I'm not sure if my progress is on the right track or not!

For $k \notin (0,4)$ we can let $x+y=xy=k$, then we have
$f(k+f(k))=k+f(k)$
so $k+f(k)$ is a fixed point. This means that the range of $k+f(k)$ for $k \notin (0,4)$ are all fixed points, so if you showed that $k+f(k)$ can't have a range $(a,infinity), (infinity,b), (a,b)$  and dealt with the interval $(0,4)$ you'd be in pretty good shape, it would imply $f(x)=x$ or $f(x)=-x$ which I think are the only solutions.

Yes, I like it too (and I don't usually enjoy solving functional equations :)).

First of all, putting $y=0$ gives $f(x+f(0))=f(x),\ \forall x$. Now put $y=-x$ to get $f(f(k))=k+f(0),\ \forall k\le 0$ (you get a $-x^2$ which I named $k$). Take $k<0,\ k<-f(0)$. We get $f(f(k+f(0)))=k+2f(0)$ on the one hand, and also $k+f(0)$ on the other hand, so $f(0)=0$. This also shows that $f(f(k))=k,\ \forall k\le 0$.

We have $f(a+f(b))=b+f(a),\ \forall a,b$ s.t. $a^2\ge 4b$. If we take $b<0$, then we can prove by induction on $n$ that $f(a+nf(b))=nb+f(a)\ (*)$, and this holds for all $a$, all $n$, and all $b\le 0$.

I'll assume there is a sequence $b_k\to 0,\ b_k<0$ s.t. $f(b_k)>0$ (we proceed similarly in the other case). Put $a=0$ in $(*)$, choose a $t>0$, and construct the sequence $n_k=n_k(t)$ s.t. $n_kf(b_k)\to t$ as $k\to\infty$. $(*)$ shows that $\frac{b_k}{f(b_k)}\to\frac{f(t)}t$, and this holds for all $t>0$. We can do something similar for $t<0$ (but the construction is a bit different: we take $a=t$ in $(*)$ and $n_k=n_k(t)$ s.t. $t+n_kf(b_k)\to 0$) to conclude that $\frac{f(t)}t$ is constant, and it folows immediately that it must be $-1$ (remember, we assumed that there is a sequence $b_k\to 0,\ b_k<0,\ f(b_k)>0$). The other solution ($f(x)=x,\ \forall x$) can be obtained in a similar manner, by considering the other case: if $b_k\to 0,\ b_k<0$, then $f(b_k)<0$ for large enough $k$.

I hope it's correct.

let $f(0) = k$. Plug in $y = 0$, get $f(x + k) = f(x)$, i.e. $f$ is periodic with period $k$. Plug in $y = -x$, get $f(f(-x^2)) = -x^2 + k \Rightarrow f(f(-x^2 + k)) = f(f(-x^2)) = -x^2 + k$, i.e. for all $x \le k$, we have $f(f(x)) = x$. But then take $x, y \le k$ such that $x = y + k$; we have $f(f(x)) = x, f(f(y)) = y$, but $f(x) = f(y+k) = f(y) \Rightarrow f(f(x)) = f(f(y)) \Rightarrow x = y \Rightarrow k = 0$. Thus $f(0) = 0$ and $f(f(x)) = x$ for $x \le 0$. Now, as $f$ is continuous, we know the image of the nonpositive reals is going to be some interval (it can't be a single point, as f is injective on the nonpositive reals), and it has to contain $0$ (as $f(0) = 0$); thus, we know that this interval contains some closed interval $[0, \varepsilon]$ (although we don't know if $\varepsilon$ is positive or negative). Now, for any nonnegative $t, k$, plug in $y = -t, x = t + k$. We get $f(k + f(-t(t+k))) = -t(t+k) + f(k)$. Now plug in $y = t, x = -t - k$. We get $f(-k + f(-t(t+k))) = -t(t+k) + f(-k)$. Thus, we have for any nonnegative $t$ and for all $k$, $f(k + f(-t(t+k))) = -t(t+k) + f(-k)$. Now, by varying $t$, we can let $-t(t+k)$ take on any value in $[0, -\infty)$, so writing $r = f(-t(t+k))$, we rewrite what we have obtained as $f(k + r) = f(r) + f(k)$ (as we know $f(r) = -t(t+k)$), for $r \in [0, \varepsilon]$. Now, by induction, we have that for all natural numbers $n$, $f(nr) = nf(r)$, and as plugging in $k = -r$ yields $f(-r) = -f(r)$, we can easily extend the above result to all integers $n$. Thus, as we can do this $\forall r \in [0, \varepsilon]$, we have that for any rational $q$, $f(q\varepsilon) = qf(\varepsilon)$ (this is because, if $q = \frac{a}{b}, a \in \mathbb{Z}, b \in \mathbb{Z}^+$, we choose $r = \frac{\varepsilon}{b}, n = b$ and then $r = \frac{\varepsilon}{b}, n = a$). Finally, as $f$ is continuous, we have that for any real $t$, $f(t\varepsilon) = tf(\varepsilon)$, so choosing $t = \frac{1}{\varepsilon}$ and substituting back in, we have $\forall t \in \mathbb{R}, f(t) = tf(1)$. Substituting this into $f(f(-x^2)) = -x^2$, we find $t = \pm 1$, and indeed, $f(x) = \pm x$ both satisfy the functional equation.


geometry
geometric transformation
rotation

Show that there doesnt exist an equilateral triangle whose vertices are lattice points, by using [b]complex[/b] numbers.

Let $ OAB$ be the equlateral tringle such that ${ O(0,\ 0),\ A(a,\ b),\ B(c,\ d)\ (a,\ b,\ c,\ d\in Z})$. $ \triangle{OAB}=\frac{\sqrt{3}}{4}(a^{2}+b^{2})=\frac{1}{2}|ad-bc|$

$ \Longleftrightarrow \sqrt{3}=\frac{2|ad-bc|}{a^{2}+b^{2}}$, which contradicts.

Or

$ a+bi=({1\over 2}\pm i{\sqrt{3}\over 2})(c+di)$ (the rotation), yielding $ a={c\mp d\sqrt{3}\over 2}$, and the final argument is similar, after expressing $ \sqrt{3}$.