Let $ a,b,c>0$ such that $ ab\plus{}bc\plus{}ca\equal{}3abc$ . Show that :

\[ \frac{3a^2b\minus{}a^2}{c\plus{}a}\plus{}\frac{3b^2c\minus{}b^2}{a\plus{}b}\plus{}\frac{3c^2a\minus{}c^2}{b\plus{}c}\ge 3\]

[quote="Pain rinnegan"]Let $ a,b,c > 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 3abc$ . Show that :
\[ \frac {3a^2b \minus{} a^2}{c \plus{} a} \plus{} \frac {3b^2c \minus{} b^2}{a \plus{} b} \plus{} \frac {3c^2a \minus{} c^2}{b \plus{} c}\ge 3\]
[/quote]

let $ x\equal{}\frac{1}{3a},y\equal{}\frac{1}{3b},z\equal{}\frac{1}{3c}$.
ineq <=>
\[ x,y,z > 0,x \plus{} y \plus{} z \equal{} 1,\sum {\frac{{z \minus{} yz}}{{x^2 y \plus{} xyz}}}  \ge 9.\]
\[ \Leftrightarrow \sum {\frac{{\left( {xz \plus{} z^2 } \right)\left( {x \plus{} y \plus{} z} \right)}}{{x^2 y \plus{} xyz}}}  \minus{} 9 \ge 0\]
\[ \Leftrightarrow \sum {x^3 }  \plus{} \sum {xy\left( {x \plus{} y} \right)}  \minus{} 9xyz \ge 0\]
which is true by AM-GM...

By the condition we have $ abc\ge 1$ and $ \frac{b}{ca}\equal{}\frac{3b\minus{}1}{c\plus{}a}$ and similars . So :

\[ LHS\equal{}\sum_{cyc} a^2\cdot \frac{3b\minus{}1}{c\plus{}a}\equal{}\sum_{cyc}\frac{ab}{c}\ge 3\]


pigeonhole principle
arithmetic sequence
number theory unsolved
number theory

prove that for any partition of $x=\{1,2,\dots , 9\}$ into two subsets one of the subsets contains distinct elements $a,b,c$ such dat $2c=a+b$

You can use a color denoting two numbers in the same set, for example red and blue. So by pigenhole you have two numbers x and y that have the same color, say red. Also assume x < y. Now (x+y)/2 must be blue. Then consider x, x + (y-x) = y, y + (y-x) = 2y - x. From this we have that 2y-x must be blue. Also consider y, y + (x-y) = x, x + (x-y) = 2x-y. Because x,y are red, 2x - y must be blue. But now we have (x+y)/2, 2y - x and 2x - y blue. 2y - x - x/2 - y/2 = 3y/2 - 3x/2 and x/2 + y/2 - 2x + y = 3y/2 - 3x/2, so 2x - y, (x+y)/2 and 2y - x are an arithmetic progression and they have same color.

good solution but is the $2x-y$or$2y-x$or$(x+y)/2$always a natural number among
$1$ to$9$ ?

Ahh yeah :(. Forgot to check that heh :S

For example: 9,7,3,8,4; 6,5,3,1 is a set that is not so good :S

Well. Then you can solve it in an ugly way by considering cases. For example you can consider 1,5,9. They cannot be in the same set, and at least two of them have same color. Assume 1,5 are in the same set, A, and 9 in the other, B. Then 3 cannot be in A so it must be in B. Now we have that 3 and 9 are in B. Then 6 cannot be B, so 6 is in A, so 1,5,6 is in A. 7 cannot be in A so it's in B, so 3,7,9 are in B. Now 8 cannot be in B, so 8 is in A and we have 1,5,6,8 in A. We cannot have 2 in A so 2 must be in B and we have 2,3,7,9 in B. Now 4 cannot be in B (because 4+2 = 2*3) so 4 is in A, and we have 1,4,5,6,8 in A. But now 2*5 = 4+6.

Similarly assume 1,9 are in the same set. And do the same argument...

Any other nicer solution?


PEOPLE RUNNING
ernie
AIME15
myyellowducky82

Make your speeches!

ok, i was thinking of using the original presidential thread, but this works just as well  :D

edit: it'll be pointless to make a new post just to say this, so i'll use this one

the post after this is really a pointless post.....

Election will start Saturday, but please post if you wish to run!

[quote="mewto55555"]Actually I intended to do this but w/e. JUst PM your votes to me, not ducky![/quote]
I thought I'd give you a break by doing the thread =]
Oh, and how many people can we vote for?

1 vote at a time, please don't vote yet! And ducky, I can handle it. i have nothing better to do!

Voting starts on Saturday, application to run ends on Friday night! (when I leave AOPS)

I vote for myself.  And as for my speech, I already gave it away.

Vote ernie for your tourneys!  :wink:
____________________________________________________________________________________________

Speech from Another Thread

Hello, all. As we all know, mewto55555 got elected for president of CTC. As many of you know, I am running for VP. Mewto and I were the people who realized that this is too much. So I have the ideas to stop this madness. I have organized tourneys before, so I have the experience. And, if you elect me for VP, your voice will be heard and it may factor into my decision.

Please vote ernie for your tourneys. Thank you!

[quote="myyellowducky82"]i thought we weren't supposed to give speeches yet[/quote]

I already made this speech last week.  Oh well.

EDIT: 700th Post! Yay!

Vote ernie for your tourneys!

AGH, you people! Why did you have to start CTC??
There's really no point! WHY IS EVERYONE JOINING????

If everyone else on AoPS joins CTC (which is likely to happen), I still won't join! Why can't you people just play FTW games and not have to go into all these tournaments???

But most of all, why did you create CTC???? I tell you, mewto55555, that there is NO WAY that you people will be able to maintain CTC if most people can't even maintain a single tournament without it shutting down.

he makes a convincing arguement, however, CTC was started to stop the tournies
because of it, i believe that the amount of tournies being created is decreasing, so apparantly, it's sort of working
after the tourny mania dies down, i believe CTC will die down as well

But if CTC dies down, tourney mania will happen all over again.

and then we can start CTC again!
it's helpful, but if tournies are already under control, CTC just gets annoying

CTC keeps tournies under control [i]ALL THE TIME[/i]


Find all functions $ f:\mathbb{N}\rightarrow\mathbb{N}$ such that $f(m+n)f(m-n) =f(m^{2})$ holds for all integers $m$ and $n$ with $m>n>0$.

$ m\equal{}n$ give $ f(m^{2})\equal{}0$. If $ 0\not\in N$ conradition, else $ m\equal{}1, n\equal{}2,3,...$ give $ f(n)\equal{}9\ \forall n\in N$.

[quote="apollo"]Let ℕ  denote the set of positive integers.Find all functions $ f:\mathbb{N}\rightarrow\mathbb{N}$ such that $ f(m\plus{}n)f(m\minus{}n) \equal{} f(m^{2})$[/quote]

I think that the condition $ f(m\plus{}n)f(m\minus{}n) \equal{} f(m^{2})$ is supposed to be true $ \forall m > n > 0$

Let then $ a > 2$ : $ f(2a\minus{}1)f(1) \equal{} f(a^{2}) \equal{} f(2a\minus{}2)f(2)$ and so $ E1$ : $ \frac{f(2a\minus{}1)}{f(2a\minus{}2)}\equal{}\frac{f(2)}{f(1)}\equal{} q$.
With $ a \equal{} 3$, we have $ \frac{f(5)}{f(4)}\equal{}\frac{f(2)}{f(1)}$.

Let then $ a > 3$ : $ f(2a)f(4) \equal{} f((a\plus{}2)^{2}) \equal{} f(2a\minus{}1)f(5)$ and so $ E2$ : $ \frac{f(2a)}{f(2a\minus{}1)}\equal{}\frac{f(5)}{f(4)}\equal{}\frac{f(2)}{f(1)}\equal{} q$.

And so $ \frac{f(x\plus{}1)}{f(x)}\equal{} q$ $ \forall x > 6$ and $ f(x) \equal{}\alpha q^{x}$ $ \forall x > 6$

Then, for $ m > n\plus{}6$, $ f(m\plus{}n)f(m\minus{}n) \equal{} f(m^{2})$ becomes $ \alpha^{2}q^{2m}\equal{}\alpha q^{m^{2}}$ and $ \alpha \equal{} q \equal{} 1$ and $ f(x) \equal{} 1$ $ \forall x > 6$

Let then $ a > 0$ and $ b > 3$ : $ f(a\plus{}2b)f(a) \equal{} f((a\plus{}b)^{2})$. We have $ a\plus{}2b > 6$ and $ (a\plus{}b)^{2}> 6$ and so $ f(a\plus{}2b) \equal{} f((a\plus{}b)^{2}) \equal{} 1$ and $ f(a) \equal{} 1$ $ \forall a > 0$

And the only solution is $ f(n) \equal{} 1$ $ \forall n\in\mathbb{N}$


How'd you do on the USAMO?
I didn't take it.  :P

We don't need 3 threads on this topic, so let's nip this one in the bud.


Find all continuous functions $f: \mathbb R^{+} \to \mathbb R^{+}$ such that :
\[f\left(\frac{1}{f(xy)}\right)=f(x)f(y)\]
for all $ x,y\in \mathbb R^+$.

[quote="Nbach"]Find all continous function  $ f: R_{\plus{}}\mapsto R_{\plus{}}$ such that :
                             $ f(\frac{1}{f(xy)}) \equal{} f(x)f(y)$ for all $ x,y\in R_{\plus{}}$[/quote]

Let $ g(x)\equal{}f(\frac{1}{x})$
We have $ g(g(xy))\equal{}g(x)g(y)$ and so $ g(g(x))\equal{}g(1)g(x)$ and so $ g(x)\equal{}g(1)x$ $ \forall x\in g(\mathbb{R}^{\plus{}})$
Then, since $ g(xy)\in g(\mathbb{R}^{\plus{}})$, $ g(g(xy))\equal{}g(1)g(xy)$ and so $ g(1)g(xy)\equal{}g(x)g(y)$ $ \forall x\in\mathbb{R}^{\plus{}}$

Let then $ g(x)\equal{}g(1)h(x)$ : we have $ h(xy)\equal{}h(x)h(y)$ and $ h(x)$ continuous. So $ h(x)\equal{}x^{c}$ for some $ c\in\mathbb{R}$

Putting back this expression in the original equation, we find $ f(x)\equal{}1$ or $ f(x)\equal{}\frac{a}{x}$, with $ a>0$

Let $ f(1) \equal{} a$.

Let $ y \equal{} 1$ , $ f(\frac{1}{x}) \equal{} af(x)$.( 1 )
On ( 1 ) move $ x$ by $ xy$ , we get :

$ f(\frac{1}{xy}) \equal{} af(xy) \equal{} f(x).f(y)$.( 2 )

Let $ g(x) \equal{}\frac{f(x)}{a}$ so $ g(x)$ is a continous function .
From ( 2 ), we have $ g(xy) \equal{} g(x)g(y)$.
Then $ g(x) \equal{} x^{t}$ or $ f(x) \equal{} ax^{t}$
And easily to show that $ t \equal{} 0$ or $ t \equal{}\minus{}1$.

-$ t \equal{} 0$, we have $ a \equal{} a^{2}$, then $ a \equal{} 1\rightarrow f(x) \equal{} 1$
-$ t \equal{}\minus{}1$. we have $ f(x) \equal{}\frac{a}{x}$ with $ a\in\mathbb{R^{\plus{}}}$


the vector component of  vector (b ) perpendicular to vec (a) is     [ vec(a) x(vec(b) x vec(a)  )]/magnitude(a)^2

prove it

[b]Hint:[/b] Break up $ \vec{\mathbf{b}}$ into its component parallel to $ \vec{\mathbf{a}}$ and its component orthogonal to $ \vec{\mathbf{a}}$.


Three values below are randomly selected without replacement. What is the probability they are equal? Express your answer as a common fraction.\\
$ \frac{24}{60}$,       $ 40\%$ of 1,     $ \frac25$,      0.4,      $ \frac14$

There are $ \binom{5}{3}\equal{}10$ total ways to choose three values.

The five values, when converted to common fractions, are $ \frac25$, $ \frac25$, $ \frac25$, $ \frac25$, and $ \frac14$.  There are $ \binom{4}{3}\equal{}4$ ways to get three numbers equal to $ \frac25$, and $ \binom{1}{3}\equal{}0$ ways to get three numbers equal to $ \frac14$.

The probability is $ \frac{4\plus{}0}{10}\equal{}\boxed{\frac25}$


vector
linear algebra
matrix
calculus
calculus computations

[quote="KBriggs"]for vectors a,b,c in 3-space, Can someone show me how to prove that $ a\ x (b\ x\ c) \equal{} b(a \cdotp c) \minus{} c(a \cdotp b)$?

I could do it the long and tedious way of expanding it with the determinant notation, but surely there is a simpler way?

This is an exerise posed by three of my textbooks, none of them show how to do it and it' driving me nuts since I can't figure it out.[/quote]
Sorry for my ignorance: What is the definition of $ a x b$? (And is it right that if $ a\equal{}(a_1,a_2,..,a_n)$, $ b\equal{}(b_1,b_2,..,b_n)$ then $ a \cdot b \equal{} a_1b_1 \plus{} ... \plus{} a_nb_n$?)

Sorry, $ axb$ was supposed to be the cross product of a and b - I am not sure how to get the notation right in Latex. These will be vectors in three-space, since the cross product is only valid there, so if a = <a1, a2, a3> and b = <b1, b2, b3> then $ a \cdotp b$ = a1*b1 + a2*b2 + a3*b3. The cross product of a and b can be expressed as the determinant of a 3x3 matrix whose top row is made up of the unit vectors i,j,k and the second row is the compnents of a and the third row the components of b.

$ a \times (b \times c)$ is a vector perpendicular to both $ a$ and $ b \times c$.  Once you check that the RHS also has this property, the identity is proven up to a multiplicative factor, and all you have to do to figure out what that factor is is to plug in any particular choice of $ a, b, c$ that makes both sides nonzero.

Hint:
Tray to prove this equality for koordinate vectors e1,e2,e3 (take special cases c=e1,e2,e3) and than use linearity for general case. But, this method is [b]relatively long[/b] and "violent" (in proof You must add and subtract some naisly shosen members). Otherwise, You can find elementary proof of this formula in college books.

Cross and dot product of vector $ \bold a, \bold b$ can be denoted by :
$ (\bold a\times\bold b)_i\equal{}\varepsilon_{ijk}a_jb_k$, $ \bold a\cdot\bold b\equal{}\delta_{ij}a_ib_j$. (Einstein notation)
Where $ \varepsilon_{ijk}$ and $ \delta_{ij}$ denotes Levi-Civita and Kronecker delta symbol respectively.
$ (\bold a\times(\bold b\times\bold c))_i$

$ \equal{}\varepsilon_{ijk}a_j(\bold b\times\bold c)_k\equal{}\varepsilon_{ijk}a_j(\varepsilon_{lmk}b_lc_m)$
$ \equal{}\varepsilon_{ijk}\varepsilon_{lmk}a_jb_lc_m$.
By using the identity $ \varepsilon_{ijk}\varepsilon_{lmk}\equal{}\delta_{il}\delta_{jm}\minus{}\delta_{im}\delta_{jl}$, the form can be rewritten by :
$ (\varepsilon_{ijk}\varepsilon_{lmk}\equal{}\delta_{il}\delta_{jm}\minus{}\delta_{im}\delta_{jl})a_jb_lc_m$
$ \equal{}(b_l\delta_{il})(\delta_{jm}a_jc_m)\minus{}(c_m\delta_{im})(\delta_{jl}a_jb_l)$
$ \equal{}b_i(\delta_{jm}a_jc_m)\minus{}c_i(\delta_{jl}a_jb_l)$.
This means each componat of $ \bold a\times(\bold b\times\bold c)$ and $ \bold b(\bold a\cdot \bold c)\minus{}\bold c(\bold a\cdot\bold b)$ is equal.


geometry
power of a point
radical axis
circumcircle
geometry proposed

[quote="forevercontinue"]Construct the three circles $ (O_1) , (O_2), (O_3)$ each passing through the Gergonne point $ G_e$ and tangent to two sides of triangle $ ABC$, respectively . Prove that $ 6$ points of tangency lie on a circle and $ AG_e$ is the radical axis of $ (O_2),(O_3)$, $ BG_e$ is the radical axis of $ (O_1),(O_3)$, $ CG_e$ is the radical axis of $ (O_2),(O_3)$.[/quote]
[img]http://img521.imageshack.us/img521/204/39521955.png[/img]

[b][color=blue]
$ A$ is intersectionof two tangents $ AC_b$ and $ AB_c$ , it's mean $ A$ lies on the radical of $ (O_2)$ and $ (O_3)$. Similarly for $ B$ , $ C$

We also see that

$ \blacksquare$ midperpendiculars of $ C_aB_a$ ;$ A_bC_b$ ; $ B_cA_c$ are angular bisectors of $ \widehat{A}$ ; $ \widehat{B}$ ; $ \widehat{C}$

$ \blacksquare$ $ M_a$ is midpoint of $ A_bA_c$ , similarly for $ M_b$ ; $ M_c$ $ \Longrightarrow$ midperpendiculars of $ A_bA_c$ ; $ B_aB_a$ ; $ C_bC_a$ are $ IM_a$ ; $ IM_b$ ; $ IM_c$

$ \Longrightarrow$ $ A_b$ , $ A_c$ , $ B_c$ , $ B_a$ , $ C_a$ , $ C_b$ lies on a circle which has center $ I$[/color][/b]

[quote][b]Generalization [/b]:
Given triangle $ ABC$ and an arbitrary point $ O$. Let $ A_1, B_1, C_1$ be arbitrary points on $ AO, BO, CO$, respectively. Then $ (A_1OB_1), (B_1OC_1), (C_1OA_1)$ intersect three sides of triangle $ ABC$ at 6 points and they are concyclic.[/quote]
Nice one!

Let $ C'_1$ and $ C'_2$ respectively be the intersections of $ (O_c)\equiv (OA_1B_1)$ with $ AB$. The same definition holds for $ B'_1,$ $ B'_2$ and $ A'_1,$ $ A'_2$. We have $ AC'_1.AC'_2 \equal{} AB'_2.AB'_1$, yielding that $ C'_1B'_2B'_1C'_2$ is concyclic quadrilateral. Let's call the circumscribed circle of concyclic quadrilateral $ C'_1B'_2B'_1C'_2$ by $ (\mathcal {K})$, also, denote $ D,$ $ E$ by the intersections of $ (\mathcal {K})$ with $ BC$. Now, moreover, we have $ BB_1.BO \equal{} BC'_1.BC'_2$, but the power of $ B$ wrt to $ (O_c)$ is $ BC'_1.BC'_2 \equal{} BB_1.BO$, which implies $ B_1,$ $ O,$ $ D,$ $ E$ are concyclic, denote $ (B_1OED)$ $ \equiv$ $ (O'_a)$. Also with the same argument, $ CB'_1.CB'_2 \equal{} CD.CE$ $ \equal{} CC_1.CO$, implying that $ C_1\in (O'_a)$, which leads to the obvious fact that $ (O'_a)\equiv (O_a)$. Therefore, $ \{D,E\}$ must coincide with $ \{A'_1,A'_2\}$. As the result, $ A'_1,A'_2,B'_1,B'_2,C'_1,C'_2$ are concyclic. Our proof is completed then. $ \square$
___________________________
[i]Remark.[/i] Let $ O\equiv$ with the orthocenter $ H$ of $ \triangle ABC$, then we obtain [url=http://www.mathlinks.ro/viewtopic.php?p=1190553#1190553]Problem 1 of IMO 2008[/url]


vector
superior algebra
superior algebra solved

How can we prove any two basis of a given vector space have the same cardinality??

Look just a few posts below this one. This question was also posted by alekk. Myth and I gave solutions.

[quote="harazi"]Myth and I gave solutions.[/quote]
I have already completely forgotten about this. :blush:


An engineer said he could finish a highway section in $ 3$ days with his present supply of a certain type of machine. However, with $ 3$ more of these machines the job could be done in $ 2$ days. If the machines all work at the same rate, how many days would it take to do the job with one machine?

$ \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 36$

[hide] Let the number of machines be $ x$ and the work done be $ w$. Since the job is done at the same rate, we can write it as $ x\equal{}\frac{1}{3}w$ and $ x\plus{}3\equal{}\frac{1}{2}w$
Solving for $ x$ we get $ \frac{x\plus{}3}{x}\equal{}\frac{3}{2}\Leftrightarrow x\equal{}6$
Now we find $ w$. $ 6\equal{}\frac{1}{3}w \Leftrightarrow w\equal{}18$
The number of days $ d$ needed to do the job with $ 1$ machine is:
$ 1\equal{}\frac{1}{d}18 \Leftrightarrow d\equal{}18$

The answer is D
[/hide]

You can set up an equation:3x=2(x+3), you'll get 3x=2x+6, which means x=6. That means when you'll need 6 machines if you want complete section in three days. one machine is 1/6 of original question, so you do 3*6=18, which is D


Every minute, a snail picks one cardinal direction (either north, south, east, or west) with equal
probability and moves one inch in that direction. What is the probability that after four minutes
the snail is more than three inches away from where it started?

there are 2ways that the snail can go more than 3 inches(3inches first, and then 1inch or going 4inches straight)
1) 3inches-1inches
(1/4)^4*2*4 (because there are 4 directions)

2) 4inches straight
(1/4)^4*4

so, 3/64

you assumed for your second case that the snail HAD to go 3 units in 1 direction then 1 unit in a perpendicular direction.
but, the snail could go 1 unit in 1 direction then 3 units in a perpendicular direction.
your probability for all in the samedirection was right, 1/64

but for 3 and 1, there are 4 ways to choose the primary direction (the direction the snail goes 3 units) and 2 ways to choose the other, secondary direction.

once you have chosen that, there are 4 places for the snail to move in its secondary direction, for a total of
$ 4 \cdot 2 \cdot 4 \cdot (\frac14)^4 \equal{} \frac18$

sp $ \frac {9}{64}$ i think

you forgot about the point obtained by going 2 inches in one direction and then 2 inches in a perpendicular direction  :P

1/64+1/8+6/64=15/64

i'm not sure what you forgot about :D  but i think this should be right.

plot it on a cartesian plane and we see that the snail must end up at $ (0,\pm 4)$ or $ (\pm 4, 0)$ or $ (\pm 1, \pm 3)$ or $ (\pm 3, \pm 1)$.  So count up the number of ways which is $ 4 \plus{} 4\cdot 8 \equal{} 36$

then $ \frac {36}{4^4} \equal{} \frac {9}{64}$

actually never mind someone already got it.  but mikemoneywire, 2 inches up and 2 inches left is $ 2\sqrt{2}\equal{}2.8<3$ from the center  :wink:

:P haha oops..i did a similar problem once, but it involved steps away, not the actual distance.  maybe that's  y i missed it lol


trigonometry
inequalities
calculus
calculus computations

i) Find the closure of the set $ \{(x, \sin(\frac{1}{x})): x > 0 \}$.

My question for finding closure, was how would you find a closure of a graph like $ \sin(\frac{1}{x})$?

ii) Consider $ \{(x,y) \in \mathbb{R} ^{2}: 2x^{2} \minus{} y^{2} \le 1\}$. Is it closed, or bounded, or both?

for this question, by the inequality this set has to be closed, but ...how can I tell if it's bounded or not?

iii) Let $ A \equal{} \{(x,n) \in \mathbb{R}^{2} | x \in \mathbb{Q}, n \in \mathbb{N}\}\cup \{(x,y) \in \mathbb{R}^{2} | 2 < x^{2} \plus{} y^{2} < 3\}$. Find the closure of A, interior points, exterior point, and boundary points.

for the last part, how do i even begin to approach this? Also, this set is open right?

i) Definitely try to graph this with relative accuracy. Notice that at (0, something) weird stuff is going on. Now $ x \equal{} 0$ was left out but you can get infinitely close to $ 0$. What does the graph do there? Does it get infinitely close to $ (0,0)$? Or $ (0, 0.8)$? Or $ (0,2)$? Well, the amplitude of sine isn't changing so everything from $ (0, \minus{} 1)$ to $ (0,1)$ the graph is getting very close to.

If $ S$ is your set, then $ \text{Cl}(S) \equal{} S\cup \{(0,y) : \minus{} 1\leq y \leq 1\}$.

Side note: This is often called The Topologist's Sine Curve.

iii) I would say this is not open. Consider the point $ (42, 42)$ (in commemoration of Douglas Adams). Can you find an open set/open ball around this point that is completely in $ \mathbb{R}^2$?

thanks

for iii) I guess i was looking only at one half of the union, because it was the part i understood the most! My real question would be what's a good way to approach a set as strange as this? By approach I mean, "picturing".

(2) can't be bounded. Pick x large first, and pick a even larger y to make $ 2x^2 \minus{} y^2$ negative, thus smaller than 1.

For (3), you can try to picture the whole thing, it's like you have many "near line" (Q is dense) and an annulus. If you believe that the kind of "near line" can't be open in R^2, the whole thing can't.


limit
trigonometry
function
calculus
calculus computations

This has been puzzling me greatly.  Using this method evaluates one limit correctly but not the other.

$ \displaystyle\lim_{x\to0}1/x\minus{}1/\sin{x}
\\
\\
\displaystyle\lim_{x\to0}1/x\minus{}1/\sin{x}\cdot x/x
\\
\\
\displaystyle\lim_{x\to0}1/x\minus{}x/\sin{x}\cdot 1/x
\\
\\
\displaystyle\lim_{x\to0}1/x\minus{}1\cdot 1/x \equal{} 0$
This is the correct answer, but for this one:
$ \displaystyle\lim_{x\to0}1/x\minus{}1/(e^x\minus{}1)
\\
\\
\displaystyle\lim_{x\to0}1/x\minus{}1/(e^x\minus{}1)\cdot x/x
\\
\\
\displaystyle\lim_{x\to0}1/x\minus{}x/(e^x\minus{}1)\cdot 1/x
\\
\\
\displaystyle\lim_{x\to0}1/x\minus{}1 \cdot 1/x \equal{} 0$
This, however, is not the correct response, as the correct answer is actually 1/2...what mistake am I making here??

In fact, you cannot use these asymptotic estimates on sums and differences of functions, only on products.

In the last line of the second example you are trying to simultaneously use the fact that limits respect products and that limits respect sums.  However, in order to do this you must use those facts [i]sequentially.[/i]  And what happens when you try to do that is that $ \lim_{x \to 0} \frac{x}{e^x \minus{} 1} \cdot \frac{1}{x}$ obviously diverges.

In other words, the product and sum properties of limits only work if those limits [i]exist.[/i]

[quote="Dvagner"]Ok, so is it just a coincidence that it worked for the first? And also, I don't understand how these are "estimates," I am merely manipulating the equation by multiplying by 1.[/quote]

It's what you did after this that I referred to

An explanation for why it worked the first time but not the second.

$ \sin x \equal{} x \minus{} \frac {x^3}6 \plus{} \cdots$ where "$ \cdots$" is for other terms that are smaller for small $ x.$ Hence

$ \frac1{\sin x} \equal{} \frac {1}{x \minus{} \frac {x^3}6 \plus{} \cdots} \equal{} \frac1x \plus{} \frac {x}6 \plus{} \cdots$

(The last part of that was obtained by long division.)

So $ \frac1x \minus{} \frac1{\sin x} \equal{} \frac1x \minus{} \left(\frac1x \plus{} \frac {x}6 \plus{} \cdots\right) \equal{} \minus{} \frac {x}6 \plus{} \cdots\to0$ as $ x\to0.$

But $ e^x \minus{} 1 \equal{} x \plus{} \frac {x^2}2 \plus{} \frac {x^3}6 \plus{} \cdots,$ so

$ \frac {1}{e^x \minus{} 1} \equal{} \frac {1}{x \plus{} \frac {x^2}2 \plus{} \frac {x^3}6 \plus{} \cdots} \equal{} \frac1x \minus{} \frac {1}2 \plus{} \frac {x}{12} \plus{} \cdots$

(Again, there's some long division involved.)

So $ \frac1x \minus{} \frac {1}{e^x \minus{} 1} \equal{} \frac1x \minus{} \left(\frac1x \minus{} \frac {1}2 \plus{} \frac {x}{12} \plus{} \cdots\right) \equal{} \frac12 \minus{} \frac {x}{12} \plus{} \cdots\to\frac12$ as $ x\to0.$

You got away with it in the $ \sin x$ case because the expansion of $ \sin x$ about zero lacks an $ x^2$ term; you didn't get away with it the second time because of the $ x^2$ term.

Here's another way to explain this: $ \sin x$ is essentially the same size as $ x$ as $ x\to0$ That means that $ \frac1x$ and $ \frac1{\sin x}$ are essentially the same size, which makes $ \frac1x \minus{} \frac1{\sin x}$ essentially .... what?

It makes $ \frac1x \minus{} \frac1{\sin x}$ something that is much smaller than $ \frac1x.$ That's about all we can say for sure without some more work. The problem with this observation is that there's still room in "much smaller than $ \frac1x$ " for the quantity to do nearly anything.


hello, we have $ 7^4\equiv 1 \mod 10$ so we get $ 7^{24}\equiv 1 \mod 10$ and since $ 7^{3}\equiv 3 \mod 10$ we get $ 7^{24\plus{}3}\equiv 3 \mod 10$.
Sonnhard.

hello, Sonnhard has decided to be a troll.
Sonnblowhard

[quote="epiclightningstorm"]I don't really understand. Could you explain what you mean when by the modulos and how they relate to the problem?[/quote]

Well he decided to find the last digit or mod $10$. So he found the number $7^x\equiv 1\mod 10$ which was $4$. So from now on every multiple of $4$ is going to be $1\mod 10$. Therefore since $24$ is a multiple of $4$ it must also be $1\mod 10$. From here we just have to do $7^3$ whose last digit or whose mod $10$ is $3$.

[hide="Alternate Solution"]

Find a pattern.

The units digit of $7^1$ is $[b]7[/b]$

The units digit of $7^2$ is $[b]9[/b]$

The units digit of $7^3$ is $[b]3[/b]$

The units digit of $7^4$ is $[b]1[/b]$

The units digit of $7^5$ is $[b]7[/b]$

The units digit of $7^6$ is $[b]9[/b]$

The units digit of $7^7$ is $[b]3[/b]$

The units digit of $7^8$ is $[b]1[/b]$

The units digit of $7^9$ is $[b]7[/b]$   Notice how the numbers repeat in cycles of 4.

The units digit of $7^3$ is the same as $7^7$ which is the same as $7^{11}$ which is the same as $7^{27}$.

The units digit of $7^3$ is $[b]3[/b]$, therefore, the units digit of $7^{27}$ is $\boxed{3}$

[/hide]

What do you mean by the three lines, is it an equal sign? Also, what is mod? (I'm bad at math).

Thanks!

Three lines means congruency in a certain mod.

You can think of mod as remainder (ehhh), so $5\equiv2\pmod{3}$, because $5\div3$ has the same remainder as $2\div3$.

Pretty bad explanation, Intro to NT has much better one.

What the instructors use:

"Since we are interested in remainders, our first definitions will be in terms of remainders.

As an example, both 37 and 100 leave the same remainder when divided by 7, namely 2. We can also express this by saying that "37 and 100 are congruent modulo 7."

Using mathematical notation, we would write $37\equiv100\pmod{7}$.

More generally, we write $a\equiv b \pmod{m}$ if both $a$ and $b$ leave the same remainder when divided by $m$."


algebra
polynomial
ratio
geometric sequence

1. Let $k$ be an integer. Find the sum of all distinct $k$ such that the equation $x^{3}+2x^{2}+kx-3=0$ has at least one rational root for $x$.

2. If $3$ is divided by $(3+x)$ and $|x|<3$, the result is $a+bx+cx^{2}+dx^{3}+ex^{4}+\ldots$. Find the value of $a+b+c+d$. Express your answer as a common fraction reduced to lowest terms.

1. use rational root theorem... if a root, $r$, is rational, then $r\in{-3,-1,1,3\}}$, use these to find the values of k

2. use a geometric sequence, let the common ratio be $-1<r=\frac{-x}{3}<1$. and the initial term 1, this makes such a sequence


If you make this year's USAMO, will your parents, or yourself, or somebody else, award you something (a gift? money? book?)?
To me, my mother will award me a T-shirt of AoPS(my idea, she spends the money)!  :P

Haha, nope.

edit> and it's not like they don't care about it; my father was on the IMO team and is a mathematician, so he knows what it is...I guess he encourages and expects me to do well

yeah i get get an xbox 360. heh i wish...

No, I'm not getting anything except the great feeling that I qualified for USAMO.  :D

Actually, I think I would get something for all the people that helped me to qualify.

[quote="JSteinhardt"]I get to know that I qualified. And I get the fun of 9 hours of proofs. lol[/quote]

If someone I know had made USAMO, I was going to break the news by coming up to him and saying "you will be happy to know that you have been condemned to lose 9 hours of your treasured vacation time."

[quote="beta"]My parents are giving me 50 lashes[/quote]

award = 25/(points above cut off) lashes

[quote="K81o7"]My mother: "Have fun with the competitions, but don't get too caught up in them.  They're not that important.  Finish your calculus (I'm doing a long distance calculus course)."
But now that I missed it again by one point, she feels really bad...[/quote]

That's one thing I'm happy about, if I didn't think that $\sqrt{27} = 9\sqrt{3}$ on the AIME, I would have get 216.5 and miss by 0.5 point and kill myself...now I don't feel as bad :)

[quote="amitavadhar"]I got a cookie. A REALLY BIG COOKIE.  :w00t:  :laugh:[/quote]

well I get a cake. A REALLY BIG CAKE.

They were happy I qualified, so they didn't yell at me to do chores for a day :P

I tell people I get to take a 9 hour test.
"It's not so bad," I say.  "It's broken up over two days... during spring break."
The real reward is their reactions.

Besides the award I mentioned above, has anyone received some awards from the official or the AMC? like a pin? certificate or something?


logarithms
integration
limit
real analysis
real analysis unsolved

Let $a\geq 1$ be a real number. Put $x_{1}=a,x_{n+1}=1+\ln{(\frac{x_{n}^{2}}{1+\ln{x_{n}}})}(n=1,2,...)$. Prove that the sequence $\{x_{n}\}$ converges and find its limit.

[quote="Yustas"]let $M=\max f'(x),\ x>1$[/quote]
Is there $M$? Are you sure?

When $x>\frac{1}{1-\varepsilon}$, $f'(x)<1-\varepsilon$, so for every fixed $a>1$ $M$ is bounded by $1-\varepsilon$ for some $\varepsilon>0$.

Post concrete your solution, please!  :oops:

The solution i posted before was a bit incorrect, so i'll try to make it full now:
Let $f(x)=1+\ln(\frac{x^{2}}{1+\ln x})$. First, lets check that $0<f'(x)\leq1$ when $x\geq1$
\[f'(x)=\frac{1+2\ln x}{x(1+\ln x)}\]
and $1+2\ln x\leq x(1+\ln x)$ when $x\geq 1$ cause $g'(x)\geq 0, \ g(1)=0$ where
\[g(x)=x(1+\ln x)-(1+2\ln x) \]
Next, $0<x_{k}=f(x_{k-1})=\int_{1}^{x_{k-1}}f'(x)dx+1\leq 1+(x_{k-1}-1)=x_{k-1}$ , so limit exists.
Now it is easy to see that the equation $x=1+\ln\frac{x^{2}}{1+\ln x}$ has only one solution $x=1$.

[quote="Yustas"] $0<f'(x)\leq1$ when $x\geq1$
[/quote]
From here $x_{2}\leq x_{1}$ and $1\leq ...\leq x_{n+1}\leq x_{n}\leq ...\leq x_{1}$, therefore exist $\lim_{n\to+\infty}x_{n}$.


function trigonometry inequalities calculus derivative

If $x+y=5$, $x,y \geq 0$,then find the range of:
\[ f(x,y) = \frac{2x}{2+x} + \frac{3y}{3+y}  \]

Hmm, substituting $y = 5-x$ it becomes $\frac{5(x-6)(x+1)}{(x+2)(8-x)}$, but I don't know a good way to find the range of that...

The function is also $\frac{25 + 5x(y+1)}{16 + x(y+1)}$.  Let $p = x(y+1)$.  It's easy to show that p takes [i]and only takes[/i] the values in $[0,9]$.

Then we want to find the range of $\frac{25 + 5p}{16 + p} = 5 + 5\Bigg( \frac{9}{p+16} \Bigg)$, which ranges from $34/5$ to $125/16$.

[quote="kunny"]The answer is $\frac{5}{2}\leq f(x,y)<5.$

Let $\overrightarrow {a}=(\sqrt{x+2},\ \sqrt{y+3}),\ \overrightarrow {b}=\left(\frac{2}{\sqrt{x+2}},\ \frac{3}{\sqrt{y+3}}\right),$ by $\overrightarrow a\cdot \overrightarrow b=|\overrightarrow a|=|\overrightarrow b|\cos \theta.$ where $\theta\ (0\leq \theta \leq \pi)$ is the angle formed by these two vectors.Thus we have  $(\overrightarrow a\cdot \overrightarrow b)^2=|\overrightarrow a|^2|\overrightarrow b|^2\cos ^2 \theta.$
Since $0\leq \cos ^ 2 \theta \leq 1,$ we have $(2+3)^2=10\left(\frac{4}{x+2}+\frac{9}{y+3}\right)\cos ^ 2 \theta \Longleftrightarrow 0<\frac{2}{x+4}+\frac{9}{y+3}\leq \frac{5}{2}.$
Now we can rewrite $f(x,y)$ as follows.

$f(x,y)=5-\left(\frac{4}{x+2}+\frac{9}{y+3}\right),$ yielding $\frac{5}{2}\leq f(x,y)<5.$

I got it! After all, we can use Cauchy-Schwalz's inequality or A.M-G.M. :lol:

Rewrite $f(x,y)=5-\frac{1}{10}\{(x+2)+(y+3)\}\left(\frac{4}{x+2}+\frac{9}{y+3}\right).$[/quote]

well i check it using differentiation , the range is

$\frac{10}{7}\leq f(x,y)\leq \frac{5}{2}$  i.e when $(x,y)=(5,0)$ ,$f(x,y)$ reaches min ; when $(x,y)=(2,3)$ $f(x,y)$ reaches max.


A circle of radius $3$ is placed at the center of a circle of radius $r > 3$ such that the length of a chord of the larger circle tangent to the smaller circle is $8$.

1.  What is $r$?

2.  What is the area between the larger and the smaller circles?

[hide="1"]
Midpoint of chord meets radius of smaller circle making a right-angled triangle, and the hypotenuse, which is radius of larger circle, is 5 by Pythagoras.

Then I suppose you just use area=$\pi\cdot r^{2}$[/hide]

[hide]Yeah, radius is 5 per above.  Then, just do $(5^{2}-3^{2})(\pi) = 16\pi$[/hide]

[quote="b-flat"][quote="mathgeniuse^ln(x)"][hide]Oh, notice that if the chord tangent to the smaller circle is 8, the area of the region is always $16\pi$[/hide][/quote]

Could you explain what generalization you are saying?  Thanks.[/quote]

Based on my interpretation, he is wrong.

Say r=2 for the small one. r_big = (4+16)^(1/2)

No way that makes a region of 16pi.

neelnal:  The radius of the SMALLER circle is $3$, so the radius of the larger circle must be $>3$ (also it says this in the problem).

So basically, you can't have a radius of 2.

[quote="mathgeniuse^ln(x)"]I will be more specific.

Let there be two concentric circles.  Now let there be a chord of the larger circle that is tangent to the smaller circle.  This has a length 8.  Prove that the area of the annulus is $16\pi$.[/quote]

This is indeed true.

[hide]Let the radius of the smaller circle equal $r$. The area of this circle is $\pi r^{2}$. Half of the tangent segment is 4. By the Pythagorean theorem, this means that the radius of the bigger circle is $\sqrt{r^{2}+16}$. Thus, the area of this circle is $\pi r^{2}+\pi 16$. By subtraction, the area of the annulus is $\pi 16$. Regardless of the radius of the inner circle.[/hide]

This leads me to my further generalization, suppose you have 2 concentric circles. If every chord of the larger circle which is tangent to the smaller circle has radius $c$, the area of the annulus is $\frac{\pi c^{2}}{4}$, regardless of the respective radii.

As to neelnal...
[quote="neelnal"]
Based on my interpretation, he is wrong.

Say $r=2$ for the small one. r_big = $\sqrt{4+16}$

No way that makes a region of $16\pi$.[/quote]
(LaTeX added in quote)

neelnal did indeed set up a correct case. $r=2$ and r_big=$\sqrt{20}$. The areas of these circles are $4 \pi$ and $20 \pi$, respectively, and thus the annulus [i]is[/i] $16 \pi$, so the [i]is[/i] a way that makes a region of $16\pi$ :lol:.

There was a problem like the ones stated before (chord tangent) at nationals on the team round. My team got it :). It was way too easy for a nats team round...

but on any test there has to be a range of questions that are easy and hard... would you rather they were [i]all [/i]like number 7 or number 10?

[quote="besttate"](LaTeX added in quote)

neelnal did indeed set up a correct case. $r=2$ and $r_{b}ig=\sqrt{20}$. The areas of these circles are $4 \pi$ and $20 \pi$, respectively, and thus the annulus is $16 \pi$, so the is a way that makes a region of $16\pi$ :Very Happy:.[/quote]

But the problem says the SMALLER radius is 3, and the larger circle has a radius of $r>3$...

[quote="b-flat"]
But the problem says the SMALLER radius is 3, and the larger circle has a radius of $r>3$...[/quote]

No, neelnal was commenting not on the original problem, but on a specification of the generalization made by mathgeniuse^ln(x).


geometry
geometric transformation
homothety
ratio
geometry proposed

Let $ \triangle{ABC}$ and points $ D,E,F$ on sides $ BC,AC,AB$ respectively such that $ AD,BE,CF$ concur. Let $ D_1,E_1,F_1$ points on $ BC,AC,AB$ such that: $ BD\equal{}CD_1,CE\equal{}AE_1,AF\equal{}BF_1$, and let $ L_1,L_2,L_3$ the parallel lines to $ AD,BE,CF$ through $ D_1,E_1,F_1$ respectively. Prove $ L_1,L_2,L_3$ concur.

AD,BE,CF concur at P. Call $ M_A$, $ M_B$, $ M_C$ the midpoint of BC,CA,AB. Call $ T_1$, $ T_2$, $ T_3$ the line from $ M_A$, $ M_B$, $ M_C$ parallel to AD,BE,CF. Then $ T_1$, $ T_2$, $ T_3$ concur on Q that is the complement of P. Clearly $ L_1$, $ L_2$, $ L_3$ concur on the simmetric of P w.r.t. Q that is the anticomplement of P.

Clearly it's also true this: Let D',E',F' the result of D,E,F by an homothety with center respectively $ M_A$, $ M_B$, $ M_C$ and ratio $ \minus{}k$ and let $ S_1$, $ S_2$, $ S_3$ the line from D',E',F' parallel to AD,BE,CF. Then $ S_1$, $ S_2$, $ S_3$ concur (and we can obtain the point of concurrence with an homothety with center the centroid and ratio $ \minus{}\frac{3k\plus{}1}{2})$.


modular arithmetic
number theory unsolved
number theory

Let $ p_n$ denote the $ n$-th smallest prime ($ p_1 \equal{} 2, p_2 \equal{} 3, p_3 \equal{} 5$), etc.

For a given $ n \geq 10$, let $ r$ be the smallest integer such that $ p_r > n \minus{} 3$, and let $ N_s \equal{} sp_1p_2... p_{r\minus{}1} \minus{} 1$ for $ s \equal{} 1, 2,... , p_r$. Show that there
exists $ j$, $ 1 \leq j \leq p_r$, such that none of $ p_1, p_2, ... , p_n$ divides $ N_j$.

Is this right?

If $ N_s\equiv0\pmod{p_i}$ then $ p_1p_2\cdot p_{r\minus{}1}s\equiv1\pmod{p_i}$.  Since $ p_i$ is a prime, this gives at most one unique solution for $ s\in\mathbb{Z}_{p_i}$.  It is obvious that $ p_i\not|N_s$ for $ i\equal{}1,2,...,r\minus{}1$.  But, for each $ p_i,i\equal{}r,r\plus{}1,...,n$, at most one number out of $ 1,2,...,p_r$ cannot be used (since $ p_r\le p_i$ and so there are no two numbers such that $ x\equiv y\pmod{p_i}$ for $ x,y\equal{}1,2,...,p_r$).  The problem will be solved if it is shown that $ n\minus{}r\plus{}1<p_r$.  We demonstrate this: since $ n\ge 10$, we have $ r>4$ and so $ n\minus{}r\plus{}1<n\minus{}3<p_r$, as desired.


I need some help with the following problem:

Show that $ \displaystyle\frac { \minus{} 1 \plus{} i\sqrt {3}}{2}$ is a cube root of 1.

I would just cube $ \displaystyle\frac { \minus{} 1 \plus{} i\sqrt {3}}{2}$.

This one is pretty easy if you can find the trick. The trick is...
[hide] Cube both sides. You get (((i*/sqrt3)-1)^3)/8=1
We then get ((i*/sqrt3)-1)^3=8. This simplifies to 8, if you do it on a calculator, and we get 8=8[/hide]
Q.E.D
P.S. Could someone put this in LaTeX for me (I am new and just learning about the many features of AOPS).

Seems easier to me if you do

$ x^{3} \equal{} 1$
$ x^{3} \minus{} 1 \equal{} 0$
$ (x \minus{} 1)(x^{2} \plus{} x \plus{} 1) \equal{} 0$

and then just perform the quadratic, rather than expand.

Thanks for the replies. I didn't realize that it was as simple as cubing it.

:) The conclusion can be gotten with DeMoivre's.

Latex is done by putting dollar signs around the code that you want Texed.

I.e. $ \sqrt{5}$ Check by mousing over the code to see what's done.

[code]$ LaTeX $[/code]


Find root of this system: $ \begin {cases} x^2 \minus{} y^2 \plus{}x\minus{} y \equal{}5 \\
x^3\minus{} x^2y\minus{}xy^2 \plus{}y^3\equal{}6\end {cases}$

[quote="azjps"]Let $ u \equal{} x \minus{} y,\ v \equal{} x \plus{} y \Longleftrightarrow x \equal{} \frac {u \plus{} v}{2},\ y \equal{} \frac {v \minus{} u}{2}$.
\[ \begin {cases} x^2 \minus{} y^2 \plus{} x \minus{} y \equal{} (x \minus{} y)(x \plus{} y \plus{} 1) \equal{} u(v \plus{} 1) \equal{} 5 \\
x^3 \minus{} x^2y \minus{} xy^2 \plus{} y^3 \equal{} (x \minus{} y)^2(x \plus{} y) \equal{} u^2v \equal{} 6\end {cases}
\]
The first equation yields $ v \equal{} \frac {5}{u} \minus{} 1$, and substituting we find $ 5u \minus{} u^2 \equal{} 6 \Longrightarrow u \equal{} 2,3$. Then $ v \equal{} \frac {3}{2}, \frac {2}{3}$, respectively, and $ (x,y) \equal{} \boxed{\left(\frac 74, \minus{} \frac 14\right)\left(\frac {11}6, \minus{} \frac 76\right)}$.[/quote]

Thank you very much. Your solution is nice. :lol:


I'm just wondering if somebody wouldn't mind checking this.  I supposed to rank these according to their relative strength as reducing agents, from strongest to weakest.

H−   H2 + 2e- = 2H-          EV= -2.25V
Br−   Br2 + 2e- = 2Br-          EV= 1.07V
OH−    O2 (g) + 2 H2O (l) + 4 e- = 4OH- (aq)    EV= 0.401V
Ag+   Ag+ + e- = Ag                                      EV = 0.80   (but this is switched to -0.80V)
Fe   Fe2+ + 2e- = Fe                EV = -0.44V
Ca   Ca2+ + 2e- = Ca              EV = -2.87V

 it will be:  Ca > H- > Ag+ > Fe > OH- > Br-

[quote="Sacha"]but this is switched to -0.80V[/quote]

The correct value is +0.80 V. As you have written the half-reactions as reduction half-reactions, then the species with greatest reducing power will correspond to the more negative potential in your list: so the order is Ca > H- > Fe > HO- > Ag+ > Br-.

Carcul


Let $x_n=\sqrt{ n + \sqrt{n^2-1} }$ for all $n \geq 1$. Express $\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_49}$ as $a+b\sqrt{2}$ for some integers $a$ and $b$.

[hide]
$\sqrt{n+\sqrt{n^2-1}}=\sqrt{\frac{n+1}{2}+\frac{n-1}{2}+2\sqrt{\frac{n+1}{2}\cdot \frac{n-1}{2}}}=\sqrt{\frac{n+1}{2}}+\sqrt{\frac{n-1}{2}}$.

Rationalizing the numerator, $\frac{1}{x_n}=\sqrt{\frac{n+1}{2}}-\sqrt{\frac{n-1}{2}}$. Every two terms telescope, so the sum is $\sqrt{\frac{50}{2}}+\sqrt{\frac{49}{2}}-\sqrt{\frac 12}-\sqrt{\frac 02}$ which simplifies to $\boxed{5+3\sqrt{2}}$.
[/hide]


modular arithmetic
Diophantine equation
number theory proposed
number theory

Solve the equation $ x^2\plus{}y^2\plus{}xy\minus{}3y\equal{}2009$ in the set of rational numbers.

Hint: [hide]To write $ x^2\plus{}y^2\plus{}xy\minus{}3y$ as $ X^2\plus{}XY\plus{}Y^2$ migh be helpfull.[/hide]

[quote="ozgurkircak"]Solve x²+y²+xy-3y=2009 over Q.[/quote]
We have (2x+y)²+3(y-2)²=4².503, but (-3/503)=-1, contradiction (just note that it works over Q). []

Excuse me, my math is little bad. What is this?
[quote="bboypa"] but (-3/503)=-1[/quote]
Pell's equation?

No,it's not pell equation, it is legendre symbol

${ (\frac { - 3}{503}} )= - 1$ ,Because $ 503\equiv - 1\pmod{3}$


What is the total sum of all the numbers from 1 to 2x (inclusive)?

Could you explain your process, Bluoctober

I think he did this $ \frac{(n)(n\plus{}1)}{2}$ which is a nice way to add all integers from 1 to $ n$.

let $ n\equal{}2x$

then the sum of $ 1$ to $ n$ is $ \frac{n^{2}\plus{}n}{2}$ as hunter34 stated.

so the total sum of numbers from $ 1$ to $ 2x$ is $ \frac{(2x)^{2}\plus{}2x}{2}$

$ \sum^{2x}_{i\equal{}1}1\plus{}\cdots\plus{}2x\equal{}\frac{(2x)^{2}\plus{}2x}{2}\equal{}2x^{2}\plus{}x$

A basic demonstration may serve our purpose better binomial_4eva. Complexity is the true form of simplicity, but don't forget what it might really look like.

THE SUM OF THE INTEGERS FROM 1-100 IS????
[hide="4321"][b]WRONG![/b][/hide]
[hide="5555"][b]WRONG![/b][/hide]
[hide="5050"]$ 1\plus{}100\equal{}101$;
$ 2\plus{}99\equal{}101$;
$ 3\plus{}98\equal{}101$; ....;
$ 50\plus{}51\equal{}101$

Apparently, $ n\plus{}1$, $ \frac{n}{2}$ times, is the sum.[/hide]
[hide="11000"][b]WRONG![/b][/hide]

Ok, I understand that but then what would be the sum of all integers from x+1 to 2x.

There are two methods that I might consider when solving the problem you just posted (the sum from x+1 to 2x)
[hide="Method 1"]$ (x\plus{}1)\plus{}(x\plus{}2)\plus{}\cdots\plus{}(2x)\equal{}(1\plus{}2\plus{}\cdots\plus{}2x)\minus{}(1\plus{}2\plus{}\cdots\plus{}x)$
$ \equal{}\frac{2x(2x\plus{}1)}{2}\minus{}\frac{x(x\plus{}1)}{2}$
$ \equal{}\frac{4x^{2}\plus{}2x\minus{}x^{2}\minus{}x}{2}$
$ \equal{}\boxed{\frac{3x^{2}\plus{}x}{2}}$[/hide]
[hide="Method 2... easier"]Rearrange to get:
$ (x\plus{}x\plus{}...\plus{}x)\plus{}(1\plus{}2\plus{}\cdots\plus{}x)\equal{}x^{2}\plus{}\frac{x(x\plus{}1)}{2}$
$ \equal{}\boxed{\frac{3x^{2}\plus{}x}{2}}$[/hide]

[hide]$ \sum_{k \equal{} 1}^{x}(x\plus{}k) \equal{} x^{2}\plus{}\sum_{k \equal{} 1}^{x}k \equal{} x^{2}\plus{}\frac{x(x\plus{}1)}{2}\equal{} x^{2}\plus{}\frac{x^{2}\plus{}x}{2}\equal{}\frac{3x^{2}\plus{}x}{2}$, as desired.[/hide]

[quote="Lazarus"]This question is totally not bumming off USAMTS 19/1.[/quote]

 :rotfl:  :rotfl:

Well, if he didn't even know the formula, he probably won't be able to do it...

I am merely looking for inspiration to do it. What I want to see is how you would think about approaching it.


geometry
parallelogram
similar triangles
geometry unsolved

We have a quadrilateral. We mark the midpoint of each side. Then we choose a point P inside the quad. We join the point P with the midpoints. In that way the quad will be divided into four parts. The problem is to find such a point P so that the original quad will be divided into 4 equal parts.

The midpoints form a parallelogram.  Draw the diagonals of that parallelogram, and P is their intersection, the center of the parallelogram. The four small triangles outside the parallelogram are each 1/8 of the area of the quadrilateral [similar triangles if you draw diagonals of the quadrilateral.]  The diagonals of the parallelogram, which is half the quadrilateral also divide that into four equal parts, each 1/8 of the parallelogram.


Hello, does anyone know where I can find AMC 10 printable worksheets that are good? I already know agmath and unl.edu, so are there any other sites? Please do not say "Google it." Please be as specific as to give a link. Thanks!


let $a,b,c >0 , ab+bc+ca = 1$ find the greatest value of permutation :
$\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c}$

ok , please delete this topic !!!!!!!!!!!  :blush:   :blush:

Why you want to delete this topic. The problem is wrong ?thank you

1)It is wrong :rotfl:
2)black_puma send it agan in other topic :|


The seqence given by:$ x_0\equal{}a,x_1\equal{}b$ and $ x_{n\plus{}1}\equal{}\frac{1}{2}(x_{n\minus{}1}\plus{}\frac1{x_n})$   is periodic.Prove that: $ ab\equal{}1$

[quote="math10"]The seqence given by:$ x_0 \equal{} a,x_1 \equal{} b$ and $ x_{n \plus{} 1} \equal{} \frac {1}{2}(x_{n \minus{} 1} \plus{} \frac1{x_n})$   is periodic.Prove that: $ ab \equal{} 1$[/quote]

Let $ u_n\equal{}x_nx_{n\plus{}1}$
We have $ u_0\equal{}ab$ and $ u_n\equal{}\frac 12(1\plus{}u_{n\minus{}1})$ and so $ u_n\equal{}1\plus{}\frac{ab\minus{}1}{2^n}$

If $ x_n$ is periodic, so is $ u_n$ and so we have $ u_j\equal{}u_k$ for some $ j\neq k$ and so $ \frac{ab\minus{}1}{2^j}\equal{}\frac{ab\minus{}1}{2^k}$ and so $ ab\equal{}1$

Q.E.D.

nice solution :lol:
This is basic problem  of algebra.


complex analysis
real analysis
real analysis theorems

I have to do complex analysis in self-study (due to overlapping courses in my shedule) but I appear to have trouble just understanding the basic concepts... (eg. main theorem, cauchy formula, ... I'd like some way of showing this more intuitively, more graphically, since I don't understand the proof for it either :blush: it is based on a theorem I don't know.)

It may be due to our papers it may be due to me but anyhow, I'd need some other source to learn the basics of complex analysis... does anyone know a good free online document/site on it?

Thanks a lot!! :)

I honestly recomend the book [i]Complex Analysis[/i] by [b]L.V. Ahlfors[/b]. It is, for me, the best text for basic complex analysis I have ever seen. It should be in your collegue library, I suppose, it is a very well known text.

Other that is a little more advance but very good indeed is the book of the same name by [b]Stein & Shakarchi[/b].

I hope it helps. Best regards,

Hi,

You can also check [url]http://www.geocities.com/alex_stef/mylist[/url] out. There you can find som texts on Complex Analysis.

Regards,


logarithms
function
calculus
calculus computations

why: 0×infinity indeterminate form
why 1^infinity indeterminate form not 1

For example, relative to $ 0 \times \infty$, $ \frac1n \cdot n \to 1$ while $ \frac{1}{\ln n} \cdot n \to \plus{} \infty$.

Regarding the case $ 1^{\infty}$, we have $ \left(1 \plus{} \frac1n\right)^n \to e$ while $ \left(1 \minus{} \frac1n\right)^n \to \frac1e$.

These forms (and the others you should know) are indeterminate because the behaviour depends on the particular functions at hand.


Well, in order of preference, I'd want Paul (you forgot him!), Romney, McCain. I consider Huck ........confuseded and flipflopping. I consider Guiliani dropping out a relief, since now he can't be president and take even more power for the executive branch, like he did as mayor.

Meh My choices
If president was Democratic: Obama.
If president was Republican: Mccain.

[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?p=1001819#1001819]Reason[/url]

I haven't followed the Republican race all that closely, so I don't have a very informed opinion to give. All I can say is that if I were wandering around and happened into a polling station and someone said "ok you are voting in the Republican primary, go!" I would probably vote for McCain. This is based only on the fact that he doesn't believe in torture.

If I seriously wanted to vote in the Republican primary I would have to study the social and economic plans of the candidates more closely. I am most concerned by how the next president will treat our civil rights and the civil rights of foreigners, and by how the next president will try to shape our government's fiscal policies. I don't want the rights of the people to be trampled upon the way they have been for the past eight years (didn't Franklin say that someone who will sacrifice freedom in exchange for security deserves neither? What happened to that philosophy? And how about the pursuit of happiness bit?) and I don't want the government spending money on ridiculous stuff (both parties tend to do this; I don't care if it's blowing up middle easterners or giving stuff to poor people, I think we already do both of those things too much and it's got to get under control).

But anyway I don't know whose fiscal policy is the closest to what I would want and I don't know which candidate is least likely to (say) amend the Constitution to ban gay marriage and/or abortion, so for now I'm just going to have to make my call according to what I do know.

Ron Paul, because of his consistent opposition to the war against Iraq. :!:

[quote="kimmystar94"]for your information, the war is against terrorism. not Iraq.[/quote] :rotfl:

  :|

 :rotfl:

..

Let's just keep it at "the war is at the expense of Iraq"

[quote="GoBraves"]You forgot Ron Paul  :mad:[/quote]
I'm sorry. My mistake.

it's interesting how Romney got no votes here, and the media tries to portray it as a two-man race between Romney and McCain...

I think Romney should stop splitting the Conservative vote with Huckabee. Notice that the amount that McCain is ahead of Huckabee on RCP average is exactly equal to Romney's support. However, Romney probably has too much resourses put in this election to drop out.

Another interesting thing is that 9 people (37%) don't even want to participate in Republican primaries....

My order

McCain, Thompson (out) <gap> Huckabee, <gap> Romney (out), Giuliani, <gap> Paul

What happens to the elected delegates of candidates who have dropped out by now? Suppose Romney (for instance) says he will support McCain from now on, are his delegates legally obliged to do the same then? :maybe:

Well since Romney endorsed McCain, all of his delegates will go to him.

Delegates are not _legally_ obliged to do anything, including vote; certainly they are not obligated to vote for whoever their favorite candidate says.  And no, even if Romney endorsed McCain, his delegates are not obliged to vote for McCain. For that matter, McCain's delegates aren't obliged to vote for McCain. (But they will if they want to be delegates again.)

[quote="solafidefarms"]And no, even if Romney endorsed McCain, his delegates are not obliged to vote for McCain. [/quote]
And are there any precedents like that? :?:

[quote="miyomiyo"]Why is this topic reserved for republicans?  I feel very strongly that Hillary would be the best president out of the current options, but I think that Obama would more easily beat McCain, which is more important at this point.[/quote]

Cough. There is another thread in the Polls forum dedicated to Democrats. Cough.

Ron Paul was added. All these moderators here and none of them added him... shame. I voted for "I'm a democrat" anyways, though...


binomial coefficients
number theory unsolved
number theory

In IMO 1990 no. 3, the solution for n where $n | 2^{n}+1$ is $n = 1$ or $n = 3$. (Will be proved if necesarry)
Thus, if $n^{2}| n^{3}| 2^{n^{2}}+1$, we obtain $n = 1 \Rightarrow m = 3$
Hence, since $27 | 2^{9}+1$, it follows that $m^{3}| 2^{m^{2}}+1$.  :D

I don't think that $n|2^{n}+1$ implies only $n=1$ or $n=3$

the actual IMO question is if $n^{2}|2^{n}+1$ where $n$ is a positive integer then $n=1 or 3$, so it doesn't quite apply.


limit
real analysis
real analysis unsolved

[quote="Moubinool"]$x_n$ is decreasing positive it converge to 0[/quote]

come on man give me more detailes,what you've written was at the end of my book, I need the entire demonstration

All right, more explanation.

We can show that for $y>0,\,\arctan(y)<y.$

Therefore, for $x>0,\,x^2\arctan\left(\frac1x\right)<x.$

This means that $x_{n+1}<x_n.$

A monotone bounded sequence must converge. We have shown that this sequence is monotone, and it is certainly bounded below by $0.$

Let $L=\lim_{n\to\infty}x_n.$ Either $L=0$ or $L>0.$

Suppose $L>0.$ Then $L=L^2\arctan\left(\frac1L\right).$ But the right hand side of this expression is always less than $L$, so this equation has no solution.

The only possibility left is that $L=0.$

Once you get close to zero, the convergence is extremely rapid - what the numerical analysts call second order. But if $x_0$ is large, the sequence changes very slowly for a very long time.

[quote="spx2"]$x_0>0$, $x{}_n{}_(n\in N)$ a sequence defined by:
$x_{n+1}=(x_n )^2arctan(\frac{1}{x_n})$
proove that
$x_n$ converges(is convergent),calculate its limit to $\infty$[/quote]

This is another one with $x \to x - ax^k + ...$.  We may need a thread for the general case.

it would be of real help if you could make a thread just for this and also
explain what that $x \to x - ax^k + ... $ means actually.
also i didnt understood that $O(....)$ stuff meant.
is it about order of increase or what ?

please explain stuff about $O(...)$ and also separate theory from some practical examples.
give also problems proposed for solving.
i will look at them but for now i have an exam at romanian language in a few days.
the storm is coming  :ninja:


Rules are here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213311

[hide="DPatrick"]
DPatrick - 2
dragon96 - 1
[/hide]

Maintain the scoreboard. Keep it in order.

1 point for guessing
2 points for being guessed
5 points for being correct

Guess: isabella2296

DPatrick - 2
Isabella2296-2
dragon96 - 1
Jongao-1

[hide="isabella2296"]isabella2296 - 4
DPatrick - 2
dragon96 - 1
Jongao
r15s11z55y89w21 - 1[/hide]

[hide="isabella2296"]
isabella2296 - 6
DPatrick - 2
dragon96 - 2
Jongao - 1
r15s11z55y89w21 - 1[/hide]

vRusczyk
isabella2296 - 6
DPatrick - 2
dragon96 - 2
Jongao - 2
vRusczyk-2
r15s11z55y89w21 - 1

[hide="isabella2296"]isabella2296 - 8
dragon96 - 3
DPatrick - 2
Jongao - 2
vRusczyk-2
r15s11z55y89w21 - 1[/hide]

@ Jongao quote the user above

[hide="dragon96"]
isabella2296 - 8
dragon96 - 9
DPatrick - 2
Jongao - 2
vRusczyk-2
r15s11z55y89w21 - 1[/hide]

[hide="DPatrick"]
dragon96 - 11
isabella2296 - 10
DPatrick - 4
Jongao - 2
vRusczyk-2
r15s11z55y89w21 - 1[/hide]

[hide="Jongao"]dragon96 - 11
isabella2296 - 10
DPatrick - 4
Jongao - 4
vRusczyk-2
r15s11z55y89w21 - 2[/hide]

[hide]
dragon96
dragon96 - 13
isabella2296 - 10
Jongao-5
dPatrick-4
vRusczyk-2
r15s11z55y89w21 - 2[/hide]

[hide="Jongao"]
dragon96 - 13
isabella2296 - 10
Jongao-7
dPatrick-4
vRusczyk-2
r15s11z55y89w21 - 2
Bugi - 1
[/hide]

[hide="dragon96"]
dragon96 - 13
isabella2296 - 10
Jongao-7
dPatrick-4
vRusczyk-2
r15s11z55y89w21 - 2
Bugi - 1
jjfun1-1
[/hide]

@Jongao: I was supposed to get 5 points for that.

[hide="james4l"]dragon96 - 13
isabella2296 - 10
r15s11z55y89w21 - 8
Jongao - 7
dPatrick - 4
vRusczyk - 2
james4l - 2
Bugi - 1
jjfun1 - 1 [/hide]

[hide="r15s11z55y89w21"]
dragon96 - 16
isabella2296 - 10
r15s11z55y89w21 - 10
Jongao - 7
dPatrick - 4
vRusczyk - 2
james4l - 2
Bugi - 1
jjfun1 - 1
[/hide]

Guys, remember to give the person you guess 2 points. And check with the person above you so they get their 5 points.

[hide="dragon96"]

dragon96 - 16
isabella2296 - 10
r15s11z55y89w21 - 10
Jongao - 7
dPatrick - 4
vRusczyk - 2
james4l - 2
jjfun1 - 2
Bugi - 1 [/hide]

[hide="spaceguy524"]Blazer123 - 705
isabella2296 - 450
r15s11z55y89w21 - 402
jjx1- 384
jjfun1 - 374
spaceguy524 - 190
dragon96 - 186
jxl28 - 152
DPatrick - 138
evilhamster - 116
PowerOfPi - 84
myyellowducky82 - 67
LaTeX - 61
meewhee009 - 45
vahalla - 42
Jongao - 35
stormfirejulian - 34
levans - 25
AIME15USAMO - 25
game.guy - 24
james4l - 22
AceOfDiamonds - 22
ernie - 20
Hopkinsmathclub - 18
cgyao15 - 20
vRusczyk - 16
Falcon1 - 11
Mr. Happy-10
pacman2812 - 7
dajoe - 4
winner2009 - 4
Bugi - 3
algebra1337 - 3
Valentin Vornicu - 2
rrusczyk - 2
nsato - 2
Chompy - 2
GameBot - 2
AIME15 - 2
SonyWii - 1
xsquare - 1
Sabien5 - 1
[/hide]

[hide="spaceguy524"]

Blazer123 - 706
isabella2296 - 450
r15s11z55y89w21 - 402
jjx1- 384
jjfun1 - 374
spaceguy524 - 192
dragon96 - 186
jxl28 - 152
DPatrick - 138
evilhamster - 116
PowerOfPi - 84
myyellowducky82 - 67
LaTeX - 61
meewhee009 - 45
vahalla - 42
Jongao - 35
stormfirejulian - 34
levans - 25
AIME15USAMO - 25
game.guy - 24
james4l - 22
AceOfDiamonds - 22
ernie - 20
Hopkinsmathclub - 18
cgyao15 - 20
vRusczyk - 16
Falcon1 - 11
Mr. Happy-10
pacman2812 - 7
dajoe - 4
winner2009 - 4
Bugi - 3
algebra1337 - 3
Valentin Vornicu - 2
rrusczyk - 2
nsato - 2
Chompy - 2
GameBot - 2
AIME15 - 2
SonyWii - 1
xsquare - 1
Sabien5 - 1
[/hide]

[hide="Blazer123"]
Blazer123 - 713
isabella2296 - 450
r15s11z55y89w21 - 402
jjx1- 384
jjfun1 - 374
spaceguy524 - 193
dragon96 - 186
jxl28 - 152
DPatrick - 138
evilhamster - 116
PowerOfPi - 84
myyellowducky82 - 67
LaTeX - 61
meewhee009 - 45
vahalla - 42
Jongao - 35
stormfirejulian - 34
levans - 25
AIME15USAMO - 25
game.guy - 24
james4l - 22
AceOfDiamonds - 22
ernie - 20
Hopkinsmathclub - 18
cgyao15 - 20
vRusczyk - 16
Falcon1 - 11
Mr. Happy-10
pacman2812 - 7
dajoe - 4
winner2009 - 4
Bugi - 3
algebra1337 - 3
Valentin Vornicu - 2
rrusczyk - 2
nsato - 2
Chompy - 2
GameBot - 2
AIME15 - 2
SonyWii - 1
xsquare - 1
Sabien5 - 1
[/hide]

[hide="Blazer123"]Blazer123 - 715
isabella2296 - 450
r15s11z55y89w21 - 402
jjx1- 384
jjfun1 - 374
spaceguy524 - 193
dragon96 - 186
jxl28 - 152
DPatrick - 138
evilhamster - 117
PowerOfPi - 84
myyellowducky82 - 67
LaTeX - 61
meewhee009 - 45
vahalla - 42
Jongao - 35
stormfirejulian - 34
levans - 25
AIME15USAMO - 25
game.guy - 24
james4l - 22
AceOfDiamonds - 22
ernie - 20
Hopkinsmathclub - 18
cgyao15 - 20
vRusczyk - 16
Falcon1 - 11
Mr. Happy-10
pacman2812 - 7
dajoe - 4
winner2009 - 4
Bugi - 3
algebra1337 - 3
Valentin Vornicu - 2
rrusczyk - 2
nsato - 2
Chompy - 2
GameBot - 2
AIME15 - 2
SonyWii - 1
xsquare - 1
Sabien5 - 1 [/hide]

[hide="evilhamster"]

Blazer123 - 716
isabella2296 - 450
r15s11z55y89w21 - 402
jjx1- 384
jjfun1 - 374
spaceguy524 - 193
dragon96 - 186
jxl28 - 152
DPatrick - 138
evilhamster - 124
PowerOfPi - 84
myyellowducky82 - 67
LaTeX - 61
meewhee009 - 45
vahalla - 42
Jongao - 35
stormfirejulian - 34
levans - 25
AIME15USAMO - 25
game.guy - 24
james4l - 22
AceOfDiamonds - 22
ernie - 20
Hopkinsmathclub - 18
cgyao15 - 20
vRusczyk - 16
Falcon1 - 11
Mr. Happy-10
pacman2812 - 7
dajoe - 4
winner2009 - 4
Bugi - 3
algebra1337 - 3
Valentin Vornicu - 2
rrusczyk - 2
nsato - 2
Chompy - 2
GameBot - 2
AIME15 - 2
SonyWii - 1
xsquare - 1
Sabien5 - 1[/hide]

[hide="Blazer123"]
Blazer123 - 718
isabella2296 - 450
r15s11z55y89w21 - 402
jjx1- 384
jjfun1 - 374
spaceguy524 - 194
dragon96 - 186
jxl28 - 152
DPatrick - 138
evilhamster - 124
PowerOfPi - 84
myyellowducky82 - 67
LaTeX - 61
meewhee009 - 45
vahalla - 42
Jongao - 35
stormfirejulian - 34
levans - 25
AIME15USAMO - 25
game.guy - 24
james4l - 22
AceOfDiamonds - 22
ernie - 20
Hopkinsmathclub - 18
cgyao15 - 20
vRusczyk - 16
Falcon1 - 11
Mr. Happy-10
pacman2812 - 7
dajoe - 4
winner2009 - 4
Bugi - 3
algebra1337 - 3
Valentin Vornicu - 2
rrusczyk - 2
nsato - 2
Chompy - 2
GameBot - 2
AIME15 - 2
SonyWii - 1
xsquare - 1
Sabien5 - 1
[/hide]

[hide="Blazer123"]Blazer123 - 720
isabella2296 - 450
r15s11z55y89w21 - 402
jjx1- 384
jjfun1 - 374
spaceguy524 - 194
dragon96 - 186
jxl28 - 152
DPatrick - 138
evilhamster - 125
PowerOfPi - 84
myyellowducky82 - 67
LaTeX - 61
meewhee009 - 45
vahalla - 42
Jongao - 35
stormfirejulian - 34
levans - 25
AIME15USAMO - 25
game.guy - 24
james4l - 22
AceOfDiamonds - 22
ernie - 20
Hopkinsmathclub - 18
cgyao15 - 20
vRusczyk - 16
Falcon1 - 11
Mr. Happy-10
pacman2812 - 7
dajoe - 4
winner2009 - 4
Bugi - 3
algebra1337 - 3
Valentin Vornicu - 2
rrusczyk - 2
nsato - 2
Chompy - 2
GameBot - 2
AIME15 - 2
SonyWii - 1
xsquare - 1
Sabien5 - 1[/hide]

[hide="evilhamster"]
Blazer123 - 720
isabella2296 - 450
r15s11z55y89w21 - 402
jjx1- 384
jjfun1 - 374
spaceguy524 - 195
dragon96 - 186
jxl28 - 152
DPatrick - 138
evilhamster - 127
PowerOfPi - 84
myyellowducky82 - 67
LaTeX - 61
meewhee009 - 45
vahalla - 42
Jongao - 35
stormfirejulian - 34
levans - 25
AIME15USAMO - 25
game.guy - 24
james4l - 22
AceOfDiamonds - 22
ernie - 20
Hopkinsmathclub - 18
cgyao15 - 20
vRusczyk - 16
Falcon1 - 11
Mr. Happy-10
pacman2812 - 7
dajoe - 4
winner2009 - 4
Bugi - 3
algebra1337 - 3
Valentin Vornicu - 2
rrusczyk - 2
nsato - 2
Chompy - 2
GameBot - 2
AIME15 - 2
SonyWii - 1
xsquare - 1
Sabien5 - 1
[/hide]

[hide="spaceguy524"]

Blazer123 - 721
isabella2296 - 450
r15s11z55y89w21 - 402
jjx1- 384
jjfun1 - 374
spaceguy524 - 197
dragon96 - 186
jxl28 - 152
DPatrick - 138
evilhamster - 127
PowerOfPi - 84
myyellowducky82 - 67
LaTeX - 61
meewhee009 - 45
vahalla - 42
Jongao - 35
stormfirejulian - 34
levans - 25
AIME15USAMO - 25
game.guy - 24
james4l - 22
AceOfDiamonds - 22
ernie - 20
Hopkinsmathclub - 18
cgyao15 - 20
vRusczyk - 16
Falcon1 - 11
Mr. Happy-10
pacman2812 - 7
dajoe - 4
winner2009 - 4
Bugi - 3
algebra1337 - 3
Valentin Vornicu - 2
rrusczyk - 2
nsato - 2
Chompy - 2
GameBot - 2
AIME15 - 2
SonyWii - 1
xsquare - 1
Sabien5 - 1 [/hide]

This game is dying...

Fine I'll revive it.

[hide="Blazer123"]Blazer123 - 723
isabella2296 - 450
r15s11z55y89w21 - 402
jjx1- 384
jjfun1 - 374
spaceguy524 - 197
dragon96 - 186
jxl28 - 152
DPatrick - 138
evilhamster - 129
PowerOfPi - 84
myyellowducky82 - 67
LaTeX - 61
meewhee009 - 45
vahalla - 42
Jongao - 35
stormfirejulian - 34
levans - 25
AIME15USAMO - 25
game.guy - 24
james4l - 22
AceOfDiamonds - 22
ernie - 20
Hopkinsmathclub - 18
cgyao15 - 20
vRusczyk - 16
Falcon1 - 11
Mr. Happy-10
pacman2812 - 7
dajoe - 4
winner2009 - 4
Bugi - 3
algebra1337 - 3
Valentin Vornicu - 2
rrusczyk - 2
nsato - 2
Chompy - 2
GameBot - 2
AIME15 - 2
SonyWii - 1
xsquare - 1
Sabien5 - 1 [/hide]

[hide="evilhamster"]

Blazer123 - 724
isabella2296 - 450
r15s11z55y89w21 - 402
jjx1- 384
jjfun1 - 374
spaceguy524 - 197
dragon96 - 186
jxl28 - 152
DPatrick - 138
evilhamster - 136
PowerOfPi - 84
myyellowducky82 - 67
LaTeX - 61
meewhee009 - 45
vahalla - 42
Jongao - 35
stormfirejulian - 34
levans - 25
AIME15USAMO - 25
game.guy - 24
james4l - 22
AceOfDiamonds - 22
ernie - 20
Hopkinsmathclub - 18
cgyao15 - 20
vRusczyk - 16
Falcon1 - 11
Mr. Happy-10
pacman2812 - 7
dajoe - 4
winner2009 - 4
Bugi - 3
algebra1337 - 3
Valentin Vornicu - 2
rrusczyk - 2
nsato - 2
Chompy - 2
GameBot - 2
AIME15 - 2
SonyWii - 1
xsquare - 1
Sabien5 - 1
[/hide]

[hide="evilhamster"]
Blazer123 - 724
isabella2296 - 450
r15s11z55y89w21 - 402
jjx1- 384
jjfun1 - 374
spaceguy524 - 197
dragon96 - 186
jxl28 - 152
DPatrick - 138
evilhamster - 138
PowerOfPi - 84
myyellowducky82 - 67
LaTeX - 61
meewhee009 - 45
vahalla - 42
Jongao - 35
stormfirejulian - 34
levans - 25
AIME15USAMO - 25
game.guy - 24
james4l - 22
AceOfDiamonds - 22
ernie - 20
Hopkinsmathclub - 18
cgyao15 - 20
vRusczyk - 16
Falcon1 - 11
Mr. Happy-10
pacman2812 - 7
dajoe - 4
winner2009 - 4
Bugi - 3
algebra1337 - 3
Valentin Vornicu - 2
rrusczyk - 2
nsato - 2
Chompy - 2
GameBot - 2
AIME15 - 2
SonyWii - 1
xsquare - 1
Sabien5 - 1
Louiethechewy123 - 1[/hide]

What? How did GameBot get 2 points?????

I'm guessing he posted twice.

[hide="Blazer123"]Blazer123 - 726
isabella2296 - 450
r15s11z55y89w21 - 402
jjx1- 384
jjfun1 - 374
spaceguy524 - 197
dragon96 - 186
jxl28 - 152
evilhamster - 139
DPatrick - 138
PowerOfPi - 84
myyellowducky82 - 67
LaTeX - 61
meewhee009 - 45
vahalla - 42
Jongao - 35
stormfirejulian - 34
levans - 25
AIME15USAMO - 25
game.guy - 24
james4l - 22
AceOfDiamonds - 22
ernie - 20
Hopkinsmathclub - 18
cgyao15 - 20
vRusczyk - 16
Falcon1 - 11
Mr. Happy-10
pacman2812 - 7
dajoe - 4
winner2009 - 4
Bugi - 3
algebra1337 - 3
Valentin Vornicu - 2
rrusczyk - 2
nsato - 2
Chompy - 2
GameBot - 2
AIME15 - 2
SonyWii - 1
xsquare - 1
Sabien5 - 1
Louiethechewy123 - 6
[/hide]

Blazer123 - 726
isabella2296 - 450
r15s11z55y89w21 - 402
jjx1- 384
jjfun1 - 374
spaceguy524 - 197
dragon96 - 186
jxl28 - 152
evilhamster - 139
DPatrick - 138
PowerOfPi - 84
myyellowducky82 - 67
LaTeX - 61
meewhee009 - 45
vahalla - 42
Jongao - 35
stormfirejulian - 34
levans - 25
AIME15USAMO - 25
game.guy - 24
james4l - 22
AceOfDiamonds - 22
ernie - 20
Hopkinsmathclub - 18
cgyao15 - 20
vRusczyk - 16
Falcon1 - 11
Mr. Happy-10
pacman2812 - 7
dajoe - 4
winner2009 - 4
Bugi - 3
algebra1337 - 3
Valentin Vornicu - 2
rrusczyk - 2
nsato - 2
Chompy - 2
GameBot - 2
AIME15 - 2
SonyWii - 1
xsquare - 1
Sabien5 - 1
Louiethechewy123 - 6
mathelete - 1
mewto - 2


algorithm
function
geometry
geometric transformation
rotation

Hello all,

I am working on a personal project (not for school or work), and am having trouble creating an algorithm. The program consists of a container of specified size (x by y by z) loaded with many different packages (p by q by r ). The container could change size and not all of the packages will be the same size (p1,q1,r1 may not be equal to p2,q2,r2). With this information is there an algorithm or theory that I could use to find the most efficient loading order of the packages into the container?

Thank you,
Chris

I would move this to the High School section, not here.

put the bigger packages first and then the smaller ones

Well the problem about this question is that you didn't define "most efficient."
That could either mean putting them in to use the least space, or to somehow put them in most easily.
Since the second one is subjective, I assume you are talking about the first one.
To use the least space, it would be similar to the "packrec" problem in the USACO training pages, except you now need to find a general algorithm.
The general algorithm could be a recursive function that repeatedly tests different 90* rotations on the box, placing it in different corner positions.
I'm really not sure if that is the most efficient algorithm, but it could work.


Suppose $ A: \mathbb{R}^{n} \rightarrow \mathbb{R}$ is linear. Show there exists $ y\in \mathbb{R}^{n}$ such that $ Ax \equal{} \langle y,x \rangle$.

Do I just show for a $ y\in A$ that $ yx \equal{} \parallel{}y\parallel{}\parallel{}x\parallel{}$? I ' don't really know where to go with this...

What does $ A$ do to each element of the standard basis for $ \mathbb{R}^n?$

Ok. So if I take the standard basis of $ \mathbb{R}^{n}$ as my $ y$ and map it with $ A$, then:

$ A ( \left[ \begin{array}{c} e_{1}\\ \vdots \\e_{n} \end{array} \right]  ) \equal{} A( \left[ \begin{array}{ccc}1& \ldots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \ldots & 1 \end{array} \right] ) \equal{} \left[ e_{1}, \ldots , e_{n} \right]$ which has n columns and 1 row, because it's in $ \mathbb{R}$ which is one-dimensional.

Then, taking $ Ax \equal{} \left[e_{1}, \ldots , e_{n} \right] \left[ \begin{array}{c} x_{i} \\ \vdots \\x_{n} \end{array} \right]$

By basic rules of matrix multiplication, we have:

$ Ax \equal{} \sum_{i \equal{} 0}^{n} e_{i} x_{i} \equal{} < y,x>$

Is this correct?

Ok, I made it weird...

I should've just expressed $ A$ as a 1 x n matrix: $ \left[ a_{1}, \ldots , a_{n} \right]$ , with $ x$ as a one-dimensional vector. Then the rest would've followed...


calculus
trigonometry
algebra unsolved
algebra

$ x,y\not\equal{} 0 (1\minus{}x^2)(1\minus{}y^2)\equal{}1$   calculate the $ A\equal{}(1\plus{}|x|)(1\plus{}|y|)$

[quote="pluristiq"]$ A$ is not uniquely determined. what are you really asking?[/quote]
just find the the min value of A  !

[quote="KapioPulsar"] just find the the min value of A  ![/quote]
Was not very clear in the original request.

Obviously $ 1\minus{}x^2>0$ would imply $ 1\minus{}y^2>0$ and so $ 1>1\minus{}x^2>0$ and $ 1>1\minus{}y^2>0$ and so $ 1>(1\minus{}x^2)(1\minus{}y^2)$

So $ x^2>1$ and $ y^2>1$ and we can write $ x^2\equal{}1\plus{}\left(\frac{2t}{1\minus{}t^2}\right)^2$ and $ y^2\equal{}1\plus{}\left(\frac{1\minus{}t^2}{2t}\right)^2$ for some $ t\in(0,1)$

Then $ A\equal{}\frac{1\plus{}t}{t\minus{}t^2}$ and we just need to find the minimum of $ f(t)\equal{}\frac{1\plus{}t}{t\minus{}t^2}$ when $ t\in(0,1)$ which is easy to find :

$ t\equal{}\sqrt 2\minus{}1$ and $ \boxed{A_{\text{min}}\equal{}3\plus{}2\sqrt 2}$ (which is trivially obtained when $ |x|\equal{}|y|\equal{}\sqrt 2$)

[quote="pco"][quote="KapioPulsar"] just find the the min value of A  ![/quote]
Was not very clear in the original request.

Obviously $ 1 \minus{} x^2 > 0$ would imply $ 1 \minus{} y^2 > 0$ and so $ 1 > 1 \minus{} x^2 > 0$ and $ 1 > 1 \minus{} y^2 > 0$ and so $ 1 > (1 \minus{} x^2)(1 \minus{} y^2)$

So $ x^2 > 1$ and $ y^2 > 1$ and we can write $ x^2 \equal{} 1 \plus{} \left(\frac {2t}{1 \minus{} t^2}\right)^2$ and $ y^2 \equal{} 1 \plus{} \left(\frac {1 \minus{} t^2}{2t}\right)^2$ for some $ t\in(0,1)$

Then $ A \equal{} \frac {1 \plus{} t}{t \minus{} t^2}$ and we just need to find the minimum of $ f(t) \equal{} \frac {1 \plus{} t}{t \minus{} t^2}$ when $ t\in(0,1)$ which is easy to find :

$ t \equal{} \sqrt 2 \minus{} 1$ and $ \boxed{A_{\text{min}} \equal{} 3 \plus{} 2\sqrt 2}$ (which is trivially obtained when $ |x| \equal{} |y| \equal{} \sqrt 2$)[/quote]  Nice just see my solution $ (1\minus{}x^2)(1\minus{}y^2)\equal{}1$ <=> $ 1\minus{}x^2\minus{}y^2\plus{}(x^2)(y^2)\equal{}1$ so 1/x^2+1/y^2=1 and so they exist a corner b (o,π/2) => 1/|x|=sin(b) 1/|y|=cos(b) so A=(1+1/sin(b))(1+1/cos(b))= 1+(sin(b)+cos(b)+1)/(sin(b)cos(b))=1+2(2^(1/2)*sin(π/4+b)+1)/(sin(2b))=1+2^(3/2)*(sin(π/4+b))/sin(2b))+2/sin(2b) the min value is when sin2b einai is max , when b=π/4 so the min value is $ A\equal{}3\plus{}2\sqrt{2}$

[quote="pco"][quote="KapioPulsar"] just find the the min value of A  ![/quote]
Was not very clear in the original request.

Obviously $ 1 \minus{} x^2 > 0$ would imply $ 1 \minus{} y^2 > 0$ and so $ 1 > 1 \minus{} x^2 > 0$ and $ 1 > 1 \minus{} y^2 > 0$ and so $ 1 > (1 \minus{} x^2)(1 \minus{} y^2)$

So $ x^2 > 1$ and $ y^2 > 1$ and we can write $ x^2 \equal{} 1 \plus{} \left(\frac {2t}{1 \minus{} t^2}\right)^2$ and $ y^2 \equal{} 1 \plus{} \left(\frac {1 \minus{} t^2}{2t}\right)^2$ for some $ t\in(0,1)$

Then $ A \equal{} \frac {1 \plus{} t}{t \minus{} t^2}$ and we just need to find the minimum of $ f(t) \equal{} \frac {1 \plus{} t}{t \minus{} t^2}$ when $ t\in(0,1)$ which is easy to find :

$ t \equal{} \sqrt 2 \minus{} 1$ and $ \boxed{A_{\text{min}} \equal{} 3 \plus{} 2\sqrt 2}$ (which is trivially obtained when $ |x| \equal{} |y| \equal{} \sqrt 2$)[/quote]

Another one great solution.....
Can you please tell us how did you think the ''whole $ t$ thing''????? :)

[quote="Dimitris X"]
Another one great solution.....
Can you please tell us how did you think the ''whole $ t$ thing''????? :)[/quote]

In fact I wrote $ x^2\equal{}1\plus{}a$ and then $ y^2\equal{}1\plus{}\frac 1a$ and $ A\equal{}(1\plus{}\sqrt{1\plus{}a})(1\plus{}\sqrt{1\plus{}\frac 1a})$

In order to simplify the parenthesis, I wrote then $ a\equal{}\tan(b)^2$ and, some calculus later, I used $ t\equal{}\tan(\frac b2)$ hence the direct expression :

$ x^2\equal{}1\plus{}\tan(b)^2\equal{}1\plus{}\left(\frac {2\tan(\frac b2)}{1\minus{}\tan(\frac b2)^2}\right)^2$
...

And I skipped these working preliminary steps in my answer  :blush:


If $ p\equal{}3$, then $ 2^p\plus{}p^2\equal{}17$. So, $ p\equal{}3$ works. If $ p>3$, then  $ 2^p\plus{}p^2\equiv (\minus{}1)^p\plus{}1\equiv 0\, \mod 3$, since $ p$ is odd. Thus, the only solution is for $ p\equal{}3$


linear algebra
matrix
linear algebra unsolved

I don't know wherether the following statement is true or false.I think it's true,but I don't have any effective approach:
If real symmetric square-matrix $A$ has all its sub-determinant with the same index of rows and columns is nonnegative then it has expression:$A=C^{T}C$ there $C$ is also real square matrix.
 Is this statement false?

A necessary condition is that $A$ is symmetric. You didn't mention this :?.

Yes,ofcourse I've forgot it.
Do you mean it's true?

yes it's true
the condition for symmetric matrices is just the one you need
for a matrix over R to be positive definite or semipositive definite

then every symmetric matrix over a field (with characteristic non two) is congruent with a diagonal matrix
so C^t*A*C=D
since it is (semi)positive definite this is just the condition you need for all the diagonal elements to be zero or more
so you can take a diagonalmatrixsquare out of D

so C^t*A*C=P*P
or C^(-t)*P*P*C^(-1)

does anyone know a faster solution perhaps? :ninja:


[quote="TheHeinrich"]How come kalva doesn't work?[/quote]
I noticed this, too. Very strange,indeed. :maybe:

Interesting... Here is the link I use for Kalva... it's much better, and has some older olympiads that were not there before.

[url]http://web.archive.org/web/20040405065644/http://www.kalva.demon.co.uk/index.html[/url]
 :)


floor function
combinatorics unsolved
combinatorics

Why $(a+b)^{n}+(a-b)^{n}=2\sum_{k=0}^{n}{n \choose 2k}a^{2k}b^{n-2k}$ ? I don't see how summing the two binomial expansions I get this result.

And if I'd want to represent in terms of $(a+b)^{n}$ a sum like this $\sum{n \choose 3k}$?

I was interested in the general computation.

Anyway, is it possibile to represent in terms of $(a+b)^{n}$ and $(a-b)^{n}$ a sum involving ${n \choose 3k}$?

[quote="Stewie"]I was interested in the general computation.[/quote]

It's often the case that a small, easily-handled case can shed a light on the general one.  In this case, N.T.TUAN was merely attempting to show how terms cancel in the general case by giving a small, concrete example.  Let me attempt to shed some additional light on the general computation.  We know that

$(a+b)^{n}= \sum_{k=0}^{n}{n \choose k}a^{k}b^{n-k}$
$(a-b)^{n}= \sum_{k=0}^{n}{n \choose k}a^{k}(-b)^{n-k}$

Let's rewrite the second expansion by separating out its positive and negative terms:

$(a-b)^{n}= (-1)^{n}\sum_{k=0}^{ \lfloor \frac{n}{2}\rfloor}{n \choose 2k}a^{2k}b^{n-2k}+(-1)^{n+1}\sum_{k=0}^{ \lfloor \frac{n}{2}\rfloor}{n \choose 2k+1}a^{2k+1}b^{n-(2k+1)}$

Then of course we can make the same revision to the first expansion:

$(a+b)^{n}= \sum_{k=0}^{ \lfloor \frac{n}{2}\rfloor}{n \choose 2k}a^{2k}b^{n-2k}+\sum_{k=0}^{ \lfloor \frac{n}{2}\rfloor}{n \choose 2k+1}a^{2k+1}b^{n-(2k+1)}$

Depending on the sign of $n$, one set of terms or the other will cancel for the desired result.

[quote="Stewie"]Anyway, is it possibile to represent in terms of $(a+b)^{n}$ and $(a-b)^{n}$ a sum involving ${n \choose 3k}$?[/quote]

The summation you want is

$(a+b)^{n}+(a+\omega b)^{n}+(a+\omega^{2}b)^{n}$

Where $\omega = e^{i \frac{2\pi}{3}}$.

Thank you very much. I tried myself to compute the general result, since I've noticed how the terms canceled, but I failed.

So if I'd want a sum like this $\sum_{k=0}^{n}(-1)^{k}{n \choose 3k}$, it would have to be $a=1$ and $b=-1$?

Something like that.  Usually it's more straightforward to consider one variable $x$ rather than two variables $a, b$, by which I mean if we have

$P(x) = (1+x)^{n}= \sum_{k=0}^{n}{n \choose k}x^{k}$

We could find $\sum{n \choose 3k}$ easily by computing

$\frac{P(1)+P(\omega)+P(\omega^{2})}{3}$

As for $\sum (-1)^{k}{n \choose 3k}$, we merely have to negate $x$; in other words, we are finding

$\frac{P(-1)+P(-\omega)+P(-\omega^{2})}{3}$


algebra
polynomial
abstract algebra
geometry
topology

Έστω $T: U\rightarrow U$ ένας γραμμικός μετασχηματισμός.Αν $m_{T}(\lambda)$ το ελάχιστο πολυώνυμο του $T$ και $\chi_{T}(\lambda)$ το χαρακτηστιστικό του πολυώνυμο.Τότε ισχύει $ker(m_{T}(\lambda))=ker(\chi_{T}(\lambda))$;;;

Νομίζω πως έχω μια απόδειξη αλλά δεν ξέρω αν είναι σωστή.....

Υ.Γ:Τι είναι τα κουτάκια πάω από κάθε post???

Basically, i askisi auti einai pure theoria tis Linear Algebra. Einai ameso apo ti theoria oti isxuei auto kai malista oti kathe riza-kai pexnidi ginetai me tin pollaplotita autis- tou minimal polynomial (kai tou characteristic fisika) einai kai idiotimi tou pinaka tou T. Fisika ipothetoume ta panta pano apo soma (iparxei, opos ksereis kai theoria ton R-modules pou diapragmateuetai pio eidika i Algebraic Geometry kai Topology, alla den einai tou parontos). Iparxei kai kapos geometriki eksigisi autou, alli epopteia mporei na erthei meta apo arketi douleia se poli proxorimeni Diaforiki Geometria, opote kai pali den einai tou parontos. Genika, polla pragmata tis Grammikis Algebras eksigountai fisika kai epoptika se proxorimeno epipedo, eno sto proto i deutero etos ta parousiazoun mono os prakseis...(ie Cayley-Hamilton, Jordan...etc.)
Hope that was helpful...if not. let me know...
Kali sinexeia...


1. Let $ a,b,c,d$ be positive real numbers with $ a\plus{}b\plus{}c\plus{}d\equal{}4$.
Prove that
\[ a^{2}bc\plus{}b^{2}cd\plus{}c^{2}da\plus{}d^{2}ab\leq 4.\]

2. A set of balls contains $ n$ balls which are labeled with numbers $ 1,2,3,\ldots,n.$ We are given $ k>1$ such sets. We want to colour the balls with two colours, black and white in such a way, that

(a) the balls labeled with the same number are of the same colour,

(b) any subset of $ k\plus{}1$ balls with (not necessarily different) labels $ a_{1},a_{2},\ldots,a_{k\plus{}1}$ satisfying the condition $ a_{1}\plus{}a_{2}\plus{}\ldots\plus{}a_{k}\equal{}a_{k\plus{}1}$, contains at least one ball of each colour.

Find, depending on $ k$ the greatest possible number $ n$ which admits such a colouring.

3. Let $ k$ be a circle and $ k_{1},k_{2},k_{3},k_{4}$ four smaller circles with their centres $ O_{1},O_{2},O_{3},O_{4}$ respectively, on $ k$. For $ i\equal{}1,2,3,4$ and $ k_{5}\equal{}k_{1}$ the circles $ k_{i}$ and $ k_{i\plus{}1}$ meet at $ A_{i}$ and $ B_{i}$ such that $ A_{i}$ lies on $ k$. The points $ O_{1},A_{1},O_{2},A_{2},O_{3},A_{3},O_{4},A_{4}$ lie in that order on $ k$ and are pairwise different.

Prove that $ B_{1}B_{2}B_{3}B_{4}$ is a rectangle.

4. Determine all pairs $ (x,y)$ of positive integers satisfying the equation
\[ x!\plus{}y!\equal{}x^{y}.\]

Thanks for posting the Individual Competition of the first MEMO, I posted the problems into seperate topics:

1) http://www.mathlinks.ro/Forum/viewtopic.php?p=933236#933236
2) http://www.mathlinks.ro/Forum/viewtopic.php?p=933238#933238
3) http://www.mathlinks.ro/Forum/viewtopic.php?p=933239#933239
4) http://www.mathlinks.ro/Forum/viewtopic.php?p=933240#933240


geometry
3D geometry
number theory
prime factorization
algebra
difference of squares
special factorizations

What is the sum of the prime factors of $ 2^{16}\minus{}1$?

Use difference of squares; it'll make the prime factorization process easier.

$ 2^{16}\minus{}1\equal{}(2^8\plus{}1)(2^8\minus{}1)\equal{}(257)(2^4\minus{}1)(2^4\plus{}1)\equal{}(257)(15)(17)\equal{}(257)(3)(5)(17)$

This ends up to be our prime factorization. The sum of the prime factors is thus $ 257\plus{}17\plus{}5\plus{}3\equal{}\boxed{282}$.

But what would you do in a case like $ 2^{15}\minus{}1$?

[quote="Maybach"]You'd do
$ (2^{15/2} \plus{} 1)(2^{15/2} \minus{} 1)$[/quote]
As good as that sounds, I don't think that would work.
How about using the fact that $ a^3 \minus{} b^3 \equal{} (a \minus{} b)(a^2 \plus{} ab \plus{} b^2)$
Only that in this case $ a \equal{} 2^5$ and $ b \equal{} 1$,
\[ 2^{15} \minus{} 1 \equal{} (2^5 \minus{} 1)(2^{10} \plus{} 2^5 \plus{} 1) \equal{} (31)(1024 \plus{} 32 \plus{} 1) \equal{} (31)(1057) \equal{} 31\cdot 7\cdot 151\]
Thus, the sum of the prime factors is,
\[ 31 \plus{} 7 \plus{} 151 \equal{} \boxed{189}\]
Edit: Whoops, wow, but yea the same idea still applies.(fixed)

Does the factoring technique work with any form of $ 2^x\minus{}1$?

Difference of squares only works when $ x$ is even. Similarly, difference of cubes only works when $ x$ is divisible by 3.

What jxl used was just difference of cubes - $ a\minus{}b \equal{} (\sqrt[3]{a} \minus{} \sqrt[3]{b})(\sqrt[3]{a})^2 \plus{} (\sqrt[3]{a})(\sqrt[3]{b} \plus{} (\sqrt[3]{b})^2)$.

[quote="isabella2296"]What jxl used was just difference of cubes - $ a \minus{} b \equal{} (\sqrt [3]{a} \minus{} \sqrt [3]{b})(\sqrt [3]{a})^2 \plus{} (\sqrt [3]{a})(\sqrt [3]{b} \plus{} (\sqrt [3]{b})^2)$.[/quote]

I don't think that's correct. Put $ a\equal{}b\equal{}1$ into that thing and you'll get $ 0\equal{}2$, which is a contradiction unless the world has switched to $ \mod 2$ and I was not informed :D

The more common form is $ a^3\minus{}b^3\equal{}(a\minus{}b)(a^2\plus{}ab\plus{}b^2)$. Plugging in $ b\equal{}\minus{}b$ gets you the sum of cubes formula, $ a^3\plus{}b^3\equal{}(a\plus{}b)(a^2\minus{}ab\plus{}b^2)$.

[quote="Mewto55555"][quote="grn_trtle"]Plugging in $ b \equal{} \minus{} b$ [/quote]

Oh sure. You're one to criticize people saying 0=2  :rotfl:[/quote]

You know what I mean  :P
Fine, $ b\equal{}\minus{}c$ or something, doesn't really matter.

[quote="isabella2296"]What jxl used was just difference of cubes - $ a \minus{} b \equal{} (\sqrt [3]{a} \minus{} \sqrt [3]{b})(\sqrt [3]{a})^2 \plus{} (\sqrt [3]{a})(\sqrt [3]{b} \plus{} (\sqrt [3]{b})^2)$.[/quote]

lol, I don't know what I was thinking when I wrote that.


Sa se arate ca multimea punctelor in care o functie convexa $f:I\to R$ nu este derivabila,este cel mult
  numarabila. ;)

Solutie.

    Pentru a arata ca multimea punctelor in care $f$ nu este derivabila este cel mult numarabila,von nota pentru
  fiecare punct $a$ din $I$ in care $f$ nu este derivabila,intervalul deschis $I_a=(f'_s (a), f'_d (a)) $.Aceste
  intervale sunt disjuncte,deoarece fiecare interval contine cel putin un numar rational.Concluzia decurge
  imediat. ;)

Multumesc, Cezar. Sa stii ca inteleg destul de bine si engleza, deci de acum poti posta solutiile direct in forumurile respective. ;)  Si vezi ca esential este ca intervalele alea sunt disjuncte (ce decurge din convexitate) si nu faptul ca ele contin numere rationale. Oricum, in esenta este aceeasi solutie ca si a lui grobber. ;)

Era de datoria mea sa sa postez o solutie,din moment ce tu mi-ai adresat aceasta intrebare cu toate ca grobber
postase deja o solutie. ;)


How many 5 digit numbers satisfy the condition that each digit is greater than the digit to the right of it?

[hide="hide nextime treething"]
it number where all the digits are distinct - 10 * 9 * 8 * 7 * 6. This will get all five-digit numbers where the numbers are distinct including numbers such as 05612. There is one way out of 120 to order them in greatest-to-least order, and $\frac{10*9*8*7*6}{120}$ = $\boxed{252}$.

I think that looks right...[/hide]

[hide]just choose 5 digits from 10 to get 252.

notice that the answer is different if it was from smallest to largest[/hide]


induction
algebra
polynomial
algebra proposed

Consider the set $ S_n$ of all the $ 2^n$ numbers of the type $ 2\pm \sqrt{2 \pm \sqrt {2 \pm ...}},$ where number $ 2$ appears $ n\plus{}1$ times.
$ (a)$ Show that all members of $ S_n$ are real.
$ (b)$ Find the product $ P_n$ of the elements of $ S_n$.

[quote="moldovan"]Consider the set $ S_n$ of all the $ 2^n$ numbers of the type $ 2\pm \sqrt {2 \pm \sqrt {2 \pm ...}},$ where number $ 2$ appears $ n \plus{} 1$ times.
$ (a)$ Show that all members of $ S_n$ are real.
$ (b)$ Find the product $ P_n$ of the elements of $ S_n$.[/quote]

[b](a)[/b] Let $ T(n)$ represent any of the members of $ S(n)$

Clearly $ 0<T(1)<4$ is real.
Assume that $ 0<T(k)<4$ is real, then $ T(k\plus{}1)\equal{}2 \pm \sqrt{T(k)} \Rightarrow 0<T(k\plus{}1)<4$ is real.

So by principle of mathematical induction we have $ 0<T(n)<4$ is real for all $ n$, done.

[b](b)[/b] $ 2\pm \sqrt {2 \pm \sqrt {2 \pm ...}}$ are the roots of the polynomial $ ((...((x\minus{}2)^2\minus{}2)^2\minus{}2)^2\minus{}2)...))^2\minus{}2\equal{}0$, where $ \minus{}2$ appears $ n\plus{}1$ times.

So $ P_n\equal{}$ product of the roots $ \equal{}((...((\minus{}2)^2\minus{}2)^2\minus{}2)^2\minus{}2)...))^2\minus{}2\equal{}2$

All numbers are $ 2sin\frac{k\pi}{2^{n\plus{}2}}$, were k is odd $ 0<k<2^{n\plus{}1}.$


search
real analysis
real analysis unsolved

Can Anyone help me with my problem set?

http://isites.harvard.edu/fs/docs/icb.topic67955.files/ps2.pdf

Though the problem set has math exercises, it contains a lot of terms from the field of Economics that I think, if I am not wrong, not many people here understand fully. Also, I am not sure if forum members would do a complete homework set for anyone on this forum.

I know this isn't very helpful!

oh yea, sorry, not the whole thing, just #3, the rockerfeller one

I think you just need to use Lagrange multiplier. Do you know this method? If not, you can search for it online (wikipedia!), or you can post here and we will explain it and/or give references.


Find the [b]exact solutions[/b] to the following equation in $x$:
\[
8x^3-6x+1=0
\]

So, this is a neat trick.  (Sorry it's in Latex, so much of it shows through.)

[hide]Let .  Then  implies  and  and, using a fund trig identity, .  Thus , so , or  and the roots are the cosines of these values.[/hide]


complex numbers
real analysis
real analysis unsolved

1. The set of all complex numbers (that is, R2)

Closed : yes;   Open: yes;   Perfect: Yes;    Bounded: No

How can the set be both closed and open? or if it's either closed or open, can some1 pls explain y? thxx

2. The segment (a,b)

Closed: no;    Open :blank;   Perfect :No;   Bounded: Yes

The reason is that the segment (a,b) is not open if we regard it as a subset of R2, but it is an open subset of R1.

I got y it's open for R1, but not for R2. :huh:

#1: The full space and the empty set are always both closed and open. It's a fundamental property of topological spaces.
For spaces like $ \mathbb{R}^n$, there are no other sets which are both closed and open. This property has a name: the space is connected.

#2: As you noted, the question isn't stated properly. The questions "Is $ S$ open?" and "Is $ S$ closed?" have no independent meaning; we need instead to ask "Is $ S$ open in the space $ X$?" and "Is $ S$ closed in the space $ X$?

The set $ \{x\in\mathbb{Q}: x^2<2\}$ is both open and closed in $ \mathbb{Q}$. It is neither open nor closed in $ \mathbb{R}$.

#1.What do you mean by a full space?? :huh:
#2.I get why the set is neither open or closed in R cuz the set consists only of isolated dots. But why would the set be both open and closed in Q? can u pls explain it?

Thxx a lot 4 the reply!

No, it doesn't consist of isolated points. Given any of those points, any ball I draw around it contains other points.

If I'm talking about whether it is open or closed in $ \mathbb{Q}$, there are no numbers outside the rational universe. In particular, there's no square root of $ 2$, which is what lets it be closed.

As for your question about #1, look at what I said for #2. We can only ask such questions about subsets of a topological space (and metric spaces are a special case of that). The full space is the set of "everything".