Is it true that an orthonormal system $ \{e_i|i \in \mathbb{N}\}$ in separable Hilbert's space H is an orthonormal basis (full orthonormal system)? If true, how to prove it?

If $ (e_n)_{n\in\mathbb{N}}$ is an orthonormal basis, then $ (e_{2n})_{n\in\mathbb{N}}$ is an orthonormal system that is not a basis.  or am i missing something?


The distance from home plate to the fence in dead center in XYZ Field is 420 feet. How far is it from the fence in dead center to third base considering that each base in the infield, which is a square, is 90 feet apart?

If you call $ a$ the distance between home and center, $ b$ the distance from home to third, and $ c$ the distance from third to center, then call $ A$ the angle between home-third and third-center.  Then you can apply the law of cosines.


Let $ a,b,c>0$ such that $ abc\equal{}1$. Prove that
$ (a^2\plus{}b^2\plus{}c^2)^2\ge 2(a^3\plus{}b^3\plus{}c^3)\plus{}3$

[quote="akai"]Let $ a,b,c > 0$ such that $ abc \equal{} 1$. Prove that
$ (a^2 \plus{} b^2 \plus{} c^2)^2\ge 2(a^3 \plus{} b^3 \plus{} c^3) \plus{} 3$[/quote]

Assum $ c\equal{}min${$ a;b;c$}

$ 3.RHS \le\ 2(a^3 \plus{} b^3 \plus{} c^3)(a\plus{}b\plus{}c) \plus{} 3abc(a\plus{}b\plus{}c)$

Need prove $ 3(a^2 \plus{} b^2 \plus{} c^2)^2 \ge 2(a^3 \plus{} b^3 \plus{} c^3)(a\plus{}b\plus{}c) \plus{} 3abc(a\plus{}b\plus{}c)$

$ \leftrightarrow a^4\plus{}b^4\plus{}c^4\plus{}6(a^2b^2\plus{}b^2c^2\plus{}c^2a^2) \ge\ 2\sum ab(a^2\plus{}b^2)\plus{} 3abc(a\plus{}b\plus{}c)$

$ \leftrightarrow M(a\minus{}b)^2\plus{}N(a\minus{}c)(b\minus{}c) \ge\ 0$

With $ M \equal{} (a\plus{}b)^2\plus{}c^2\plus{}6c^2\minus{}2(a\plus{}c)(b\plus{}c)$

           $ \equal{}a^2\plus{}b^2\plus{}4c^2\minus{}2(ac\plus{}bc) \ge\ 0$

        $ N \equal{} ab\plus{}(a\plus{}c)(b\plus{}c)\plus{}6ab\minus{}2(2ab\plus{}bc\plus{}ca)$

           $ \equal{} 4ab\plus{}c^2\minus{}bc\minus{}ca \ge\ 0$ ( Because $ c \equal{} min${$ a;b;c$})

We have done! :)


Can I use the Muirhead in the Math Competiton such as IMO or TSTs?

[quote="Soarer"]why not? :?[/quote]

Because i afraid the examiner don't accept this ineq.(In Macau Mathematic Olympiad)

And did this ineq 'BORN' these years? I think this ineq is universal these years therefore many dunno what this ineq is.


Disclaimer about why this is here:
The reason I post this in intermediate is that the ideas involved are, I think, of an intermediate level.  The questions I ask may not be, but this is really a request for help rather than a challenge problem, and I thought that intermediate-types might also like to see a solution if someone explains it.  If a moderator thinks that this in inappropriate I don't have any objections to its being moved, although I would appreciate it if a shadow were left in the intermediate forum.

Now, on with the problem:

Actually, this tiling problem that came to me while I was in the bathroom.  Let's say we're tiling a plane with two colors of hexagon, as shown.  (The "Bathroom Floor Tiling," if you will.)  It's fairly clear that we can say that "the area of the yellow hexagons is 1/3 of the total area" because we could also tile this plane with a variety of combinations of two white hexagons attached to one yellow hexagon.  (One such possible tile is outlined differently in the diagram.)  However, it's not obviously clear to me that you couldn't tile it with pieces that have, say, 3 white and one yellow hexagon (one such piece is also outlined in the diagram).  I mean, for any such piece, I can show that tiling with in will break down eventually but I can't show that it will in general.  Does anyone have any ideas about this?

Okay, so, second thing:  It's clear that in this tiling, every yellow hexagon is in essentially the same location as every other -- if we shift the plane around to put one on top of another, we can do it without affecting the areas at all.  Thus, I thought of another way of approaching this problem.  Let the fraction of the area represented by yellow hexagons be $p$.  Then we will figure out what $p$ is as follows:
p = (percent of total that is white)(percent of hexagons touching each white hexagon that is yellow) + (percent of total that is yellow)(percent of hexagons touching each yellow hexagon that is yellow)
or
$p=(1-p)\cdot(\frac{1}{2})+(p)\cdot(0)$
which we can solve to get $p=\frac{1}{3}$, luckily the same answer we got before.  So, I wonder -- does this always work?  (Assume we have the condition that, unlike in a [url=http://mathworld.wolfram.com/PenroseTiles.html]Penrose Tiling[/url] or something, every tile of the same color really in interchangable -- lotsa symmetry, in other words.)  What if all of the tiles aren't actually the same shape?  Also, we were lucky in this case that every hexagon bordered only on hexagons along its sides -- there weren't any shapes that just touched its corners.  Would we have to take that into account?  Is it possible to do so?  (Someone might try the method out for a nice tiling with hexagons and triangles, say.)

Anyhow, tell me what you think!

I find it interesting but the picture doesn't work so I don't follow.


:starwars: how to figure out  linear  equiations  :yoda:  :play_ball:  can you tell me

Hi;

Can you post a specific problem? That's how you can get help.

try to be more specific as possible..what is it that you don't understand about linear equations?...I can make it petty simple for you if you want...

Like the others said, more information please. Do you need the basics, usually you could post a problem.

Basically, just use y=mx+b for EVERYTHING. Plug the integers in to see if it works.....hmmmm.......Use standard form (ax+by=c), and also point-slope, graph the points. Your question isn't that specific............... don't really know what you mean by your question.

first you need either a graph or some ordered pairs...then you find the slope and write it in the y=mx+b form. You plug in one of the ordered pairs as x and y.. and m as slope...and you figure out b, guess what! There's ur equation!!! If You have a graph first find the slope and ordered pairs and repeat the process..

The simplest form is y=mx+b where m is the slope of the line and b is the y-intercept. If you can get the equation in this form, you can graph it easily. :first:

do you mean the formula y=ax+b? :maybe: more details plz

_________________
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ummmm guys!! I don't think mr. (or miss) "1365120" really needs our help..i remember, this post dated almost a month ago and that [1365120] person don't even bother to respond to us..so we better stop replying to this post, unless of course, 1365120 would be grateful enough to make clarifications to his/her problems on linear equations..


Can't figure this one ou, could someone provide a formula?

A jet plane traveling at 500 MPH overtakes a propeller plane traveling at 200 MPH that had a 2-hour head start. How far from the starting point are the planes?

[hide]We can write the equation:

$ d \equal{} 200t \equal{} 500(t \minus{} 2)\implies t \equal{} \frac {10}{3}\implies d \equal{} \frac {2000}{3}$[/hide]

don't have a formula for you, was never a math genius. but the answer is 665 miles.

Your answer should be 667 if you rounded up, and 666 if you truncated. How did you get 665?

EDIT: nevermind, you get the same answer

[hide="i think of it."]
I think of it as, the 1st plane started 2 hrs earlier, so it starts ahead by 400 miles, and the faster plane catches up by 500-200=300 miles everyhr, so it would catch up in $ \frac{4} {3}$ hours. then plug it back in to get your answer.[/hide]

I would call the distance (in miles) $ x$, then express times of flight for both planes as a function of $ x$. The jet has been flying for $ x/500$ hours; the propeller plane for $ x/200$ hours. Then
$ x/500\plus{}2\equal{}x/200$
Same result, of course.


Let $S=\{\sigma(n),n\in\mathbb{N}^*\}$.
For all $L\in\mathbb{N}$, can we find $a\in\mathbb{N}$ such that $[a,a+L] \cap S=\emptyset$?

Related question (solved by Myth): [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=26480][u]here[/u][/url]

Here is my advance, not much but is better than nothing:

Let $L$ be a number such that $\{a,a+1,...,a+L-1\}\cap S=\emptyset$ for some $a$. Then we know the following:

$L=1\Rightarrow a_{min}=2$
$L\in\{2,3\}\Rightarrow a_{min}=9$
$L\in\{4,5\}\Rightarrow a_{min}=49$
$L\in\{6,7,8,9\}\Rightarrow a_{min}=423$
$L\in\{10,11\}\Rightarrow a_{min}=1333$
$L\geq 12\Rightarrow a_{min}>1333$

Does somebody want to continue? :)

I have just found an interesting reference related to this problem. Check http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A109322


Following on the bear problem. A hunter walks one mile south, one mile east and one mile north, and is back where he started. Where did he start from? (Ok i know smartass, think of something original)

[hide="I mentioned this problem in the original bear problem topic"]the north pole[/hide]

Because i knew you would say that, see comment after question.

Just to taunt, there is an infinate amount of places he could have started from!

Maybe the hunter is in a projective plane...

Since no one has gotton it immediatly (shame on ye, ye call yourselves mathematicians!) think penguins.

[hide="how about this"]The planet's radius was $\frac1{2\pi}$ miles[/hide]

I will give all possible solutions.

North pole.

one mile north of the ring where going 1 mile east gets you around exactly once around the world.

one mile north of any ring where going 1 mile east gets you x times around the world where x in an integer

plugging in $\infty$ for the previous one gives the last answer: one mile north of the south pole.

[quote="NoSoupForYou"][hide="Or maybe this"]The hunter lied[/hide][/quote]

 :rotfl:  :10:  :clap:  :thumbup:  :wow:

Hmmm, some in there i didnt have! I was thinking the ring thats one mile in an easterly direction, one mile above that. If i re-phrased it, he starts walking south for a mile, then starts going east for a mile, then starts going north for a mile, do we have new solutions?

Yes, but also the ring that's 1/2 mile when travelled east (which you'd walk twice over), and 1/3 mile (which you'd walk 3 times as you walked your mile east), and 1/4 mile, and so on.  Hamster then argued you could be exactly 1 mile above the south pole, as well, but I disagree -- the direction "east" isn't well-defined there, and you couldn't possibly walk a mile east on that point.  (In other words, induction doesn't actually get you to infinity.)

well then let's use an infinitely large number that is not infinity so that for all purposes it is one mile above the south pole :)

I can't tell if that's a joke or not.  We can get [i]arbitrarily close[/i] to one mile north of the south pole and find a position at which the conditions will be satisfied.

Note that in the re-stating, ive introduced even more solutions, he can start near the north pole aswell!


Hello all,
My sister is conducting a survey of people regarding lenses, eyesight and other random stuff like that.

[url=http://tinyurl.com/survey17]Click here to open the survey[/url]

So yeah, take it etc. etc.


modular arithmetic
AMC
AIME
algebra
binomial theorem

The Binomial Expansion is valid for exponents that are not integers.  That is, for all real numbers $ x, y,$ and $ r$ with $ |x| > |y|,$
\[ (x \plus{} y)^r \equal{} x^r \plus{} rx^{r \minus{} 1}y \plus{} \frac {r(r \minus{} 1)}2x^{r \minus{} 2}y^2 \plus{} \frac {r(r \minus{} 1)(r \minus{} 2)}{3!}x^{r \minus{} 3}y^3 \plus{} \cdots
\]
What are the first three digits to the right of the decimal point in the decimal representation of $ \left(10^{2002} \plus{} 1\right)^{10/7}?$

[hide]
Note that $2002 = 7(286)$.

Then clearly the first term has no relevance; it's just $10$ to be power of something big.

The second term looks like it might be important; we'll come back to that.

The third term has $\left(10^{2002}\right)^{-\frac{4}{7}}$, which is $10$ to the power of something very negative, which also will not affect the digits we are looking at... and if we go further the negative only grows bigger so we can stop here.

So the second term is: $\frac{10}{7}\left(10^{2002}\right)^{\frac{3}{7}} = \frac{10}{7} \cdot 10^{858} = \left(1+\frac{3}{7}\right) \cdot 10^{858}$.

But we can ignore the $1$ because it won't affect the digits we want. So we have $\frac{3}{7} \cdot 10^{858}$. Well, now knowledge of the decimal expansion of $\frac{1}{7}$ somes in handy, but you could always long divide. Assuming we know it, $\frac{3}{7}$ repeats this string: $428571$.

So... moving the decimal point $858$ places to the right (or taking it $\pmod{6}$, $0$ places), our desired digits are $428$.[/hide]

[quote="paladin8"]Then clearly the first term has no relevance; it's just $ 10$ to be power of something big.[/quote]

I don't get how this was gotten.  Could someone please explain it to me?  Thanks!

In your binomial formula, x = 10^2002, y = 1, and r = 10/7
So, the first term is x^r. which is 10^(2002*10/7), which is 10^2860, which is 1 with a lot of zeros attached to it.
You are looking for digits to the RIGHT of the decimal point, which means the place values smaller than 1.  Obviously, 10^2860 won't be a part of your answer in any way.  Thus, you can disregard it.
I hope this helps.

[quote]$ (x\plus{}y)^r\equal{}x^r\plus{}rx^{r\minus{}1}\plus{}\frac{r(r\minus{}1)}2x^{r\minus{}2}y^2\plus{}\frac{r(r\minus{}1)(r\minus{}2)}{3!}x^{r\minus{}3}y^3\plus{}\cdots$[/quote]

Shouldn't this be $ (x\plus{}y)^r\equal{}x^r\plus{}rx^{r\minus{}1}y\plus{}\frac{r(r\minus{}1)}2x^{r\minus{}2}y^2\plus{}\frac{r(r\minus{}1)(r\minus{}2)}{3!}x^{r\minus{}3}y^3\plus{}\cdots$ ?

Since y=1, it doesn't affect the problem though.

Sorry for revival if this wasn't necessary.

It definitely should be. Someone should edit the resources section and the wiki!

[hide]

Note that using the Binomial Theorem we get $(10^{2002}+1)^\frac{10}{7}=(10^{2002})^{\frac{10}{7}}+\frac{10}{7}10^{2002*\frac37}+\frac{10}{7}*\frac{3}{7}/2*10^{2002*\frac{-4}{7}}+\cdots$.  Note that all negative exponents are going to give us a neglible situation, we multiply by a value less than $1$ with the beginning part, and then we have a huge value to a large negative expnoent.

Now, we find $10^{2002*\frac{10}{7}}+\frac{10}{7}*10^{2002*\frac37}$.  Note that $2002=7*286$ so hence we get $10^{2860}+10^{858}*\frac{10}{7}$.  The first part contribues nothing to our answer.  The seconed part gives us $10^{859}*\frac17$.  Note $\frac17=\overline{.142857}$.  Therefore $10^{859}*\frac17=10^{859\pmod{6}}*\frac17$ in terms of finding the last digits, which is the same as $\frac{10}{7}$, which gives us $1.42857\cdots$.  The rest, as I've said before are neglible since they are so small they don't account for any of the first three digits, so our answer is $\boxed{428}$. [/hide]

Before I get 4 backseat moderators saying "don't revive" this is an old AIME problem.

[quote=m1234567]Why exactly can the negative exponents be ignored?
Thanks
M[/quote]

Because its going to be $10^{-(huge power)}$ and we only care about first $3$ digits of approximation these will not even affect the tota

Sorry for the bump. [url = https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_I_Problems/Problem_7]In the solution for this problem on the wiki[/url], it states that [quote](The remainder after this term is positive by the Remainder Estimation Theorem)[/quote]. What is the Remainder Estimation Theorem? I tried searching it up, but I didn't get an exact definition.

Thanks :)

[hide=Solution]Note that if we have a negative exponent, we are multiplying by something extremely close to zero, so the result is negligible. This means we only need to consider $10^{2860}+\frac{10}{7} \cdot 10^{858}$. The first of these does not have a decimal, so it does not contribute anything. The second of these has $428571$ after the decimal, so the answer is $\boxed{428}$.[/hide]


probability
number theory
least common multiple

Bleh. I'm gonna start introducing two different problems @ once. They shouldn't be hard for you guys though since you guys are so much better than me @ math  :|

1. How many of the first 1000 positive integers are divisible by 2, 3, 4, and 5?

2. If I write down the first 1000 positive integers on paper, what is the probability that I pick a positive integer divisible by 2, 3, 4, and 5?

If you solve number 1 than you can solve number 2
[hide="(1)"]
For the number to be divisible by 2,3,4 and 5 then it has to be divisible by 60.(their lcm)
So $ \frac{1000}{60} \equal{} \boxed{16}$ (ignore the remainder)
[/hide]
[hide="(2)"]
Since we know that 16 numbers fit this condition it is simply $ \frac{16}{1000} \equal{} \boxed{\frac{2}{125}}$
[/hide]

[hide="1)"]
$ \text{LCM}(2,3,4,5)\equal{}60,$ so we need to find how many multiples of $ 60$ there are $ \le1000.$ They are $ 60$ to $ 960$ giving us $ \boxed{16}$ number divisible by $ 2,3,4,5.$[/hide]

[hide="2)"]
Using the first part and the definition of probability, \[ \frac{\text{number of successes}}{\text{total number of possibilities}}\equal{}\frac{16}{1000}\equal{}\boxed{\frac{2}{125}}.\][/hide]


algebra
polynomial
AMC
AIME
function
quadratics
quadratic formula

1.  The equation $2000x^6 + 100x^5 + 10x^3 + x - 2 = 0$ has just two real roots. Find them.

2.  Find the sum of the absolute values of the roots of $x^4-4x^3-4x^2+16x-8$

1) I remember this as an AIME problem.  Since there's no magical method for factorizing a sixth degree polynomial, and since real roots come in conjugate pairs, try and factor it into a fourth and second degree polynomial via the rational root theorem.

is there a way to do it without a lot of casework?  I split it up into $(ax^2+bx+c)(dx^4+ex^3+fx^2+gx+h)$ where a, ..., h are integers, and then can get some equations using viete's formulas but it's a lot of casework or i'm missing something.

Well, the two trailing coefficients multiply to two.  A lot of the coefficients are even, so we probably want to try even coefficients rather than odd ones.  Also, none of the coefficients are too big, so we might want to make $a,d$ as close to each other as possible, e.g, $40,50$.  This doesn't work, so the next one to try is $20,100$.  This ends up working, and we get: $(20x^2 + x - 2)(100x^4 + 10x^2 + 1)$

2. Here's how my thinking went.

[hide]Without the -8 on the end, the other terms factor nicely by grouping:
$x^4-4x^3-4x^2+16x\\
=x^3(x-4)-4x(x-4)\\
x(x^2-4)(x-4)\\
x(x-2)(x+2)(x-4)$.
So this function has roots of -2, 0, 2, and 4, and is therefore symmetric around x=1. This means that the original polynomial is symmetric here also. So shift it over one unit, i.e. replace x with y+1, and it comes out to $y^4-10y^2+1$, which we can easily use quadratic formula to find that $y=\pm\sqrt{5\pm2\sqrt6}$. Hoping that there is a nicer form for this, we get lucky and find that $y=\pm\sqrt2\pm\sqrt3$. This gives us all the $x$ values, and it is easy to finish.[/hide]


Let $ ABCDE$ be a convex pentagon in which $ AB \equal{} 1$, $ \angle BAE \equal{} \angle ABC \equal{} 120^{\circ}$, $ \angle CDE \equal{} 60^{\circ}$ and $ \angle ADB \equal{} 30^{\circ}$. Prove that the area of pentagon $ ABCDE$ is less than $ \sqrt 3$.

It was posted already [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1979048899&t=200045]here[/url] but no solution has given  :wink:


geometry
parallelogram
geometry proposed

$ABCD$ is a convex quadrilateral s.t. $AB$ and $CD$ are not parallel. The circle through $A,B$ touches $CD$ at $X$, and a circle through $C,D$ touches $AB$ at $Y$. These two circles intersect in $U,V$. Show that $AD\|BC\iff$ $UV$ bisects $XY$.

I will prove the ==> part.

Let Z=AB^CD, as AD//BC ==> ZB/ZC=ZA/ZD. From theorem of tangent and secant we have that ZX^2=ZB*ZA and ZY^2=ZC*ZD. Then it holds that (Zx/ZY)^2 = ZB/ZC*ZA/ZD = (Zb/Zc)^2 = (ZA/ZD)^2 which means that AYXD and YBCX are both cyclic.

From theorem of radica axes AX, DY and UV are concurrent, let's say at R. Similarly BX, CY and UV concurr at S.

We have also that <BYC = <BXC = <BAX = <YAX = <YDX, i.e. DY//XB and AX//YC, so RYSX is a parallelogram, then RS or UV bisects XY.

I will try later on to prove the part <==.

only if part.

Let ,G and H be the common chord to the two circles and let M=XY^GH. If U is the (second) point where XY cuts c(DXC) and similarly we define V=YX^c(AYB), then XU=YV. Indeed XM*MU=GM*MH=YM*MV, from which, as XM=YM, if follows that MU=MV which in turn implies the thesis.
From this fact derive what follows.
If Z=AB^CD isnt too hard  to prove that ZX/ZY=r(DXC)/r(AYB). From this one can prove that ZA/ZD=ZB/ZC= r(DXC)/r(AYB).

I will provide proof of this last two statments in a separate lemmaRL151104.

Is it possible to derive from this configuration a very nice construction problem: I will propose that in a separate post.

LemmaRL151104.

Let U,Y,X and V be four collinear points in this order s.t. UY=XV. If R and S are two points on the axes of UX and VY respectively and Z is the point of intersection between the orthogonal lines through X and Y wrt RX and RY respectively, then ZX/ZY = RX/SY.

Dim.

If X' is the fourth vertex of the parallelogram RXX'Y, it is easy to see that X' lies on the axis of YV.  It results that XYZ~X'SY, as their sides are respectively orthogonal and we are done.


ABCD is a rectangle inscribed in a circle

AB, and AD when produced meet the tangent at C in the points  E and F respectively.

Prove that $\frac{BE}{DF}$ = $\frac{AE^{3}}_{AF^{3}}$

[hide]Let $\angle AFE= \delta$, obviously $\angle AFE =\angle BCE =\angle BAC$.
$\frac{BE}{BC}=\frac{DC}{DF}=\tan \delta$ so
$\frac{BE.DC}{DF.BC}=\tan^{2}\delta$ or
$\frac{BE}{DF}= \frac{BC}{DC}\tan^{2}\delta= \frac{BC}{AB}\tan^{2}\delta= \tan^{3}\delta= \frac{AE^{3}}{AF^{3}}$[/hide]


calculus
integration
calculus computations

The forward difference formula can be expressed as
$f'(x_{0}) = \frac{1}{h} [f(x_{0} + h) - f(x_{0})] - \frac{h}{2} f''(x_{0} - \frac{h^2}{6} f'''(x_{0}) + O(h^3)$
use extrapolation to derive an $O(h^3)$ formula for f' (x0)

do not have much of a clue on how to solve this. I know it has something to do with Richardson's extrapolation.

would it have something to do with this formula $N_{j} (h) = N_{j-1} (h) (\frac{h}{2}) + \frac{N_{j-1}(h/2) - N_{j-1} (h)}{4^{j-1} -1}$ ??

Any ideas, hints?
Thank you for any help!

[quote="stunner5000pt"] ...$f'(x_{0}) = \frac{1}{h} [f(x_{0} + h) - f(x_{0})] - \frac{h}{2} f''(x_{0}) - \frac{h^2}{6} f'''(x_{0}) + O(h^3)$... [/quote]
The question is simple  ...but must be solved ! Suppose that   $f\in C^{4}[a,b]$ and  $x_0, x \in [a,b$. Then according to  Taylor formula with
 remainder-term (in integral form)  we have
\[ f(x)=f(x_0)+\sum\limits_{k=1}^{3}\frac{(x-x_0)^k}{k!} f^{(k)}(x_0) +\frac{1}{3!}\int\limits_{x_0}^{x}(x-t)^3f^{(4)}(t)\; dt  \]
Let $x: =x_0+h , h>0 .$ Then above equality is equivalent  to   $(*)\; \; \; \; f(x_0+h)=f(x_0)+\sum\limits_{k=1}^{3}\frac{h^k}{k!} f^{(k)}(x_0) +R[f]$  with $R[f]: = \frac{1}{3!}\int\limits_{x_0}^{x_0+h}(x_0+h -t)^3f^{(4)}(t)\; dt .$ Assume that $|f^{(4)}(t)|\le M \, \; \forall t \in [a,b].$
 Then$|R[f]|\le \frac{M}{3!} |\int\limits_{x_0}^{x_0+h}(x_0+h -t)^3\; dt |= \frac{M}{4! }h^4\;$ or $\frac{1}{h}R[f] ={O}(h^3)$
Therefore from (*)
\[ \frac{f(x_0+h)-f(x_0)}{h}=f^{\prime}(x_0)+\frac{h}{2}f^{\prime\prime}(x_0)+\frac{h^2}{6}f^{(3)}(x_0)+ \frac{1}{h}R[f]  \]
or $f^{\prime}(x_{0}) = \frac{ f(x_{0} + h) - f(x_{0})}{h} - \frac{h}{2} f''(x_{0}) - \frac{h^2}{6} f'''(x_{0}) +\varepsilon[f]$ where  $\varepsilon[f] : =-\; \frac{1}{h}R[f]={O}(h^3)$  for  $f \in C^4[a,b] .$
[size=75] [b] If you are interested in more general methods to approximate $f^{\prime}(x_0)$, please inform me.[/b][/size]


geometry
rectangle
analytic geometry
graphing lines
slope
function
algebra unsolved

[quote="N.T.TUAN"]For which integers $ n>1$ and real numbers $ r$ does the curve $ y=x^{n}+rx$ contain the vertices of a rectangle?[/quote]

Let $ f(x)=x^{n}+rx$

If the four vertices of the rectangle are $ (x_{A},y_{A}), (x_{B},y_{B}), (x_{C},y_{C}), (x_{D},y_{D})$ with $ x_{A}\leq x_{B}\leq x_{C}\leq x_{D}$, then we have obviously $ x_{A}< x_{B}< x_{C}< x_{D}$ and $ f'(x)$ have at least two zeroes in $ [x_{A},x_{D}]$. Then, since $ f'(x)=nx^{n-1}+r$, we must have $ n$ odd and $ r<0$.

Since the two values where $ f'(x)=0$ are in $ [x_{A},x_{D}]$, we have $ x_{A}<0$ and $ x_{D}>0$

Now, we can consider that none of the slopes of the sides of the rectangle are zero (either the other sides would be "vertical", which is impossible). Consider then the slope "$ a$" which is positive (the slope of $ AB$ and $ CD$).

It is rather easy to see that if $ x_{C}<0$, then the length of $ CD$ is greater than the length of $ AB$ (don't forget they are parallel).
It is easy too in the same way to see that if if $ x_{B}>0$, then the length of $ AB$ is greater than the length of $ CD$.

So  $ x_{A}< x_{B}< 0 < x_{C}< x_{D}$

Then if we consider the intersections $ C$ and $ D$ of $ f(x)=x^{n}+r$ ($ n$ odd and $ r<0$) with $ y=ax+b$ ($ a>0$) with $ x>0$, we have at most one couple of points $ C$ and $ D$ for a given length $ CD$.

Hence, since $ f(x)$ is an odd function, since $ AB$ is parallel to $ CD$ and since $ AB=CD$, we need to have $ A$ symetric to $ D$ and $ B$ symetric to $ C$ and the four points are $ A=(-b,-b^{n}-rb)$, $ B=(-a,-a^{n}-ra)$, $ C=(a,a^{n}+ra)$ and $ D=(b,b^{n}+rb)$ for some $ 0< a<b$.

It remains to have $ BD$ perpendiculat to $ CD$ :

Slope of BD is $ \frac{a^{n}+b^{n}+ra+rb}{a+b}$

Slope of CD is $ \frac{a^{n}-b^{n}+ra-rb}{a-b}$

$ BD$ perpendicular to $ CD$ means product of slopes equal to $ -1$, and so $ (a^{n}+ra)^{2}-(b^{n}+rb)^{2}=b^{2}-a^{2}$
And so $ (a^{n}+ra)^{2}+a^{2}=(b^{n}+rb)^{2}+b^{2}$ with $ 0<a<b$

This is possible iff $ g(x)=(x^{n}+rx)^{2}+x^{2}$ has at least one postive root (wth sign changing) for $ g'(x)=0$
$ g'(x)=2(x^{n}+rx)(nx^{n-1}+r)+2x = 2x[(Z+r)(nZ+r)+1]$ with $ Z=x^{n-1}$
The equation $ (Z+r)(nZ+r)+1=0$ has a positive root with sign changing iff $ r^{2}<\frac{4n}{(n-1)^{2}}$, that's to say $ r<-\frac{2\sqrt{n}}{n-1}$ (since we know $ r<0$)

So the curve $ f(x)=x^{n}+rx$ ($ n>1$) contains the four vertices of a rectangle if and only if $ n$is odd and  $ r<-\frac{2\sqrt{n}}{n-1}$ .


[quote="1234567890"]I have reall fast cable, but it usually takes 10-15 seconds for me to get to the page after I click. This started last week. Anyone else have the same problem? Before, it took 1-2 secs only to go from link to link.[/quote]Sometimes the server is lagging because of too many visitators. We will solve these problems soon :)

[quote="Valentin Vornicu"]Sometimes the server is lagging because of too many visitators. We will solve these problems soon :)[/quote]

That`s the dream of every forum. :thumbup:

[quote="Valentin Vornicu"]Just to make it clear: we will get rid of the slowdowns, not by decreasing the number of visitators, but by putting better hardware and software behind the site :P  :lol:[/quote]

 :lol:

(btw, isn't it "visitors")

[quote="236factorial"]Only half of the graphics load on each page on this site....[/quote]Can you attach a print screen? I don't experience these issues. Might be something on your side (browser cookies, internet connection, something like that). Try clearing the cache/cookies and restarting the browser (and if you can also the computer - that might help, especially on Windows machines :) ).

[quote="Peter VDD"]Here the site has full speed all around the clock. Doesn't look like a server issue to me...[/quote]It's been fixed partially for now (and in plus you're not actually browsing usually when the peak hour is in the US - it's around 3-4 AM in europe :D).


For how many integers n with 2 is less than or equal to n is less than or equal to 80, is [(n-1) (n) (n+1)]/ 8 equal to an integer?

A) 10
B) 20
C) 59
D) 39
E) 49

[hide="Solution"]We need to find how many numbers with $ 2\leq n \leq 80$ make $ \frac {(n \minus{} 1)(n)(n \plus{} 1)}{8}$ an integer. Since $ 8$ factors as $ 2^{3}$, we need a factor of $ 2$ to appear in the numerator $ 3$ times. At most two of the factors in the numerator can be multiples of $ 2$.

So there are three cases:
(1) If $ n$ is divisible by $ 8$:
There are $ 10$ such values of $ n$.
(2) If $ (n \minus{} 1)$ is divisible by $ 4$:
There are $ 19$ such values of $ n$.
(3) If $ (n \plus{} 1)$ is divisible by $ 4$:
There are $ 20$ such values of $ n$.

Adding up all the cases, we get $ 10 \plus{} 19 \plus{} 20 \equal{} \boxed{49}$ such values of $ n \Rightarrow \boxed{E}$.[/hide]

[quote="123456789"][hide="Solution"]We need to find how many numbers with $ 2\leq n \leq 80$ make $ \frac {(n \minus{} 1)(n)(n \plus{} 1)}{8}$ an integer. Since $ 8$ factors as $ 2^{3}$, we need a factor of $ 2$ to appear in the numerator $ 3$ times. At most two of the factors in the numerator can be multiples of $ 2$.

So there are three cases:
(1) If $ n$ is divisible by $ 8$:
There are $ 10$ such values of $ n$.
(2) If $ (n \minus{} 1)$ is divisible by $ 4$:
There are $ 19$ such values of $ n$.
(3) If $ (n \plus{} 1)$ is divisible by $ 4$:
There are $ 20$ such values of $ n$.

Adding up all the cases, we get $ 10 \plus{} 19 \plus{} 29 \equal{} \boxed{59}$ such values of $ n \Rightarrow \boxed{C}$.[/hide][/quote]

[hide="Correction to your solution"]I think you mean $ 10 \plus{} 19 \plus{} 20 \equal{} \boxed{49}$ such values of $ n \Rightarrow \boxed{E}$.[/hide]

Yes, that's what I meant.

Wow...I can't read my own numbers...haha

I'll edit my post.

Thanks!


How did everyone come out on the Mandelbrot?

I finished with a 60/70 in regionals, missing my goal by 5 points...

umm dude national results are not actually out yet :/

okay dude I think I got like 8 14 8 11 11 or something like that.. yeah... fail

hmm 58 on national... I shouldn't have sucked at round 1 (and 2 and 3 to a lesser extent). Also, Enloe appears to have failed to submit its scores correctly - there could be two people there beating me :(

regional, eighth grader. got 54 (epic...fail...). hopefully, i can get a 60+ in regional or a 45+ in national next year.


How many perfect cubes divide $ 2009^9$?


AMC
AIME
algebra
polynomial
geometric series
binomial theorem

Problem : The [[polynomial]] $ 1\minus{}x\plus{}x^2\minus{}x^3\plus{}\cdots\plus{}x^{16}\minus{}x^{17}$ may be written in the form $ a_0\plus{}a_1y\plus{}a_2y^2\plus{}\cdots \plus{}a_{16}y^{16}\plus{}a_{17}y^{17}$, where $ y\equal{}x\plus{}1$ and the $ a_i$'s are [[constant]]s. Find the value of $ a_2$.

Solution : === Solution 1 ===
[b]Using the geometric series formula, $ 1 \minus{} x \plus{} x^2 \plus{} \cdots \minus{} x^{17} \equal{} \frac {x^{18} \minus{} 1}{x \plus{} 1} \equal{} \frac {x^{18} \minus{} 1}{y}$[/b]. Since $ x \equal{} y \minus{} 1$, this becomes $ \frac {(y \minus{} 1)^{18} \minus{} 1}{y}$, so we want the coefficient of the $ y^3$ term in $ (y \minus{} 1)^{18}$. By the [[binomial theorem]] that is $ {18 \choose 3} \equal{} \boxed{816}$.

it says "using the geometric series formula, $ 1 \minus{} x \plus{} x^2 \plus{} \cdots \minus{} x^{17} \equal{} \frac {x^{18} \minus{} 1}{x \plus{} 1} \equal{} \frac {x^{18} \minus{} 1}{y}$."

but don't you get $ \frac{1\minus{}x^{18}}{x\plus{}1}$ when you use the geometric series formula??

Yes. It should be $ \dfrac{1 \minus{} x^{18}}{x \plus{} 1} \equal{} \dfrac{1 \minus{} (y \minus{} 1)^{18}}{y}$. The coefficient of the $ y^3$ term in $ 1 \minus{} (y \minus{} 1)^{18}$ is $ \minus{} 1\cdot( \minus{} 1)^{15}\cdot\binom{18}{3} \equal{} \binom{18}{3} \equal{} 816$.

I'm guessing that it is a double error, which canceled out, since the $ y^3$ coefficient in $ (y \minus{} 1)^3$ is $ \minus{} 816$ not $ 816$.

Sorry about that  :oops: ; I just fixed it.

By the way, you can also note any mistakes directly on the "discussion" page of the relevant problem (so that you don't have to copy/paste the entire thing over), or even fix them yourself - it is a wiki after all.

why do we want the coefficient of the $ y^3$ term?
Edit: nevermind because we are dividing by y right?


function
ratio
inequalities
geometry solved
geometry

Prove that in a triangle with a>=b>=c we have la/lb+lb/la>=ha/hb+hb/ha.

tschuess!

Maybe squaring works:
we get      la 2 /lb 2  + lb 2 /la 2   \geq ha 2 /hb 2 + hb 2 /ha 2 = a 2 /b 2  + b 2 /a 2
or maybe squaring doesn't work and we should leave the ineq like it was and doing the same we get:
la/lb+lb/la  \geq a/b +b/a
I don't know.... I'm really out of ideas now so, Lagrangia, could you please post your soluton

Cheers :)

If I remember correctly this problem was solved here, but the post was lost (in that info-losing accident a long time ago  :D ). I forgot who solved it, but the soln used sth like this: the function x+1/x is increasing on [1, infinity), so it's enough to take a>=b and I think from here la>=lb (or lb>=la, I'm not sure right now) follows, and we just compare 2 of the ratios (instead of 2 sums of ratios).

I'm not entirely sure this is it, but I definitely remember this problem being solved by smb before right here.

[b]Maverick[/b] said:

"I don't know.... I'm really out of ideas now so, Lagrangia, could you please post your soluton "

hehe .. I wish I had a solution(as you can see I didn't post it at the "Proposed Problem" section).. so I wish that the person who solved it could post the solution again!

cheers! :D :D

I was the one who solved it. The inequality la/lb+lb/la \geq a/b+b/a is equivalent with proving that la/lb  belongs to (-oo,b/a] \cup [a/b,oo). We will prove that la/lb \leq b/a, or (a+c)cos A/2 \leq (b+c)cos B/2, which becomes after calculations (a+b+c)(sin 2 C/2-1/4) \leq 0, true because C \leq  \pi /3.


[hide]Well, if they ate 10 chalupas all together, then it is pretty straightforward from there.... 3 people so the answer is $\boxed{\frac{10}{3}}$[/hide]

[hide]Andy ate 6. Brandon ate 4. Cameron ate 0.
$a=b+2$
$b=c+4$
$a+a-2+a-6=10$
$a=6$
[/hide]

[quote="Arvind_sn"]Andy, Brandon, and Cameron are eating chalupas at taco bell, and they are hungry. Andy ate 2 more chalupas than Brandon. Brandon ate 4 more chalupas than Cameron, and they ate 10 chalupas in total. What is the average of the number of chalupas that they ate[/quote]

[hide="this is 20% luck, 10% skill, and 15% concentrated power of skill"]Cameron: $x$
Brandon: $x+4$
Andy: $x+6$

$3x+10=10$
The only way possible is if $x=0$, so Cameron ate $0$, Brandon ate $4$, and Andy ate $x+6$[/hide]

omg... only one of you can read...

AVERAGE

i saw that... no really, i did see that.
i was simply arriving at the answer by a different method.

[quote]It actually is straightforward, as ragnarok23 said, since there are 3 people, 10 in total, the average is the total divided by the amount, in this case the answer is 10/3

P.S. Yippee!, my 200th post[/quote]

He's right, you know...


Here is the problem:

A unicycle has a wheel with a radius of 1 foot. How many complete revolutions will the wheel make when the unicycle rolls 100 feet? Express your answer to the nearest whole number.

The solution is to find that the circumference is $ C\equal{}2 \pi \cdot r\equal{}2\pi$, and the wheel makes exactly $ \frac{100}{2 \pi} \approx 15.92$.

My question is, [b]Is the answer 15 or 16?[/b] MathCounts says the answer is 16, but it says that "how many [b]complete[/b] revolutions will the wheel make" at first, suggesting the answer should be 15. It contradicts itself, first saying to round down, but then to round to the nearest integer, or up.

What is right and did this cause any problems in 2000?

Well, if it contradicts itself, would there be a "correct" answer?

I agree that the problem is poorly worded.  The word "complete" should be removed.

I think the answer would be 16 because it would always be best to round it up

Complete means FULL revolutions, so the correct answer should have been 15. Did anyone protest that problem?

Guys really, we have noticed the poorly worded question, so u don't need to revive


Each page number of a 488-page book is printed one time in the
book.  The first page is page 1 and the last page is page 488.
When printing all of the page numbers, how many more 4's are
printed than 8's?

The key insight is that there is symmetry in the $ 4$'s and $ 8$'s (in fact, all the nonzero digits greater than $ 3$) from $ 0$ to $ 399$, so we only consider the numbers from $ 400$ to $ 488$! Simple counting yields $ 89 \plus{} 10 \plus{} 9 \equal{} 108$ fours (considering hundreds, tens, and units digits separately) and $ 9 \plus{} 9 \equal{} 18$ eights in this range, for a difference of $ 108 \minus{} 18 \equal{} \boxed{90}$.

After using the symmetry, you don't need to count. Consider MORE symmetry.

Since 4 is coming in hundredth place everytime, and not 8, there are 89 more fours.(400 to 488)
Now consider when tenth digit is 4, 41-81 same..42-82 same.....47-87 same...44-84(1 extra four) 45-85 same..47-87 same..48-88(1 less four to counter the other extra 4)..49-no 89...1 extra 4..

Total 89+1 = 90


algebra
polynomial
inequalities
function
inequalities proposed

hi,
Let $ n \in \mathbb{N}$, prove
\[ \frac{1}{2^{n\minus{}1}} \leq x^{2n} \plus{} (1\minus{}x^{2})^{n} \leq 1,\]
$ \forall x \in [\minus{}1,1]$.

Let $ x \equal{} siny$ , $ \sqrt {1 \minus{} x^2} \equal{} cos y$
Then $ x^{2n} \plus{} (1 \minus{} x^{2})^{n} \equal{} (sin^2 y)^n \plus{} (cos^2 y)^n \ge sin^2 y \plus{} cos^2 y \equal{} 1$

$ 1 \equal{} (sin^2 y \plus{} cos^2 y)^n \le 2^{n \minus{} 1} ( sin^{2n} y \plus{} cos^{2n}y ) \Rightarrow \frac {1}{2^{n \minus{} 1}} \le sin^{2n} y \plus{} cos^{2n}y \equal{} x^{2n} \plus{} (1 \minus{} x^{2})^{n}$ , so done.

[quote="doicu"]hi,
Let $ n \in \mathbb{N}$, prove
\[ \frac {1}{2^{n \minus{} 1}} \leq x^{2n} \plus{} (1 \minus{} x^{2})^{n} \leq 1,
\]
$ \forall x \in [ \minus{} 1,1]$.[/quote]
The function $ f(x) \equal{} x^n$ is convex. Thus $ (x^2)^n \plus{} (1 \minus{} x^2)^n \ge 2 \cdot \left ( \frac {x^2 \plus{} 1 \minus{} x^2}{2} \right ) ^2 \equal{} 2^{ \minus{} n \plus{} 1}$ and $ (x^2)^n \plus{} (1 \minus{} x^2)^n \le 1^n \plus{} 0^n \equal{} 1$.

[quote]Then $ x^{2n} \plus{} (1 \minus{} x^{2})^{n} \equal{} (sin^2 y)^n \plus{} (cos^2 y)^n \ge sin^2 y \plus{} cos^2 y \equal{} 1$[/quote]
What?

[quote]
[quote]
Then $ x^{2n} \plus{} (1 \minus{} x^{2})^{n} \equal{} (sin^2 y)^n \plus{} (cos^2 y)^n \ge sin^2 y \plus{} cos^2 y \equal{} 1$
[/quote]
What?
[/quote]
@Mathias: only the false direction


Prove rigourously:

A line (not necessarily straight) exists in the x-y plane, such that any line parallel to one of the axes does not intersect the line more than once. Prove that the sum of the lengths of the projections of the line onto the x ad y axes is larger than the length of the line.

[quote="JBL"]1. What exactly do you intend to convey by "not straight line"?  A rectifiable curve of finite length?

2. Why do you think this is a combinatorics problem?[/quote]

1. a curve of finite length

2. sorry, I just had to use this in a combinatorial problem  :lol:

i know we can say "wlog, assume the line is piecewise linear", but is there a rigourous proof?


[quote="cgyao15"]these aren't requiring you to use inequalities; just simple subtration questions.[/quote]Right, and consider also that these are [b]Middle School Algebra[/b] questions, which are supposed to be [color=red]easy[/color]. maybach's answers are correct, numerically and format-wise.

[quote="cgyao15"]these aren't requiring you to use inequalities; just simple subtration questions.[/quote]

Sorry, cgayo15, but the problem specifically says for both
[quote="Problem 1"] Write and solve an inequality [/quote]
and
[quote="Problem 2"] Write and solve an inequality to show.........[/quote]

So the answers "6" and 145" wouldn't answer the question and would be incorrect.  :wink:

At this stage, we recommend a solution without using too much calculus but strictly speaking, to find the feasiable region for the given inequalities, we can not forget the significance of gradients involved. Some problems may change theirs questions such as

Given $ 12x\plus{}ky\equal{}$  Profit
and determine the constraint of k so that the profit is maximized thoughout that interval

Of course, a bunch of inequalities are given beforehand.  :yup:

[quote="sri340"]1) Sumiko is allowed to watch no more than 10 hours of television each week. She has watched 4 hours of television already. Write and solve an inequality to    show how many more hours of television Sumiko can watch.

2) A satellite will be released into an orbit of more than 400 miles above theEarth. The rocket carrying it is currently 255 miles above Earth. Write and solve an inequality to show how much higher the rocket must climb before it releases the satellite.[/quote]

400-255=145. Needs at least 145 more miles.

As we already noticed, this is not an inequality. What he asks for is a inequality.
Mods, please lock.

10-4=6 So she can watch 6 more hours of TV :)
Then the rocket needs to go 145 more miles before it can release the satelite :)


i cant find texnic center!  i dowloaded the program and installed it.
I clicked on all programs but there is no texnic center file!
below are the only ones that are inthe folder of miktex
miktex package manager ,
miktex   options,
help web files,
update wizard.

i searched the texmf folder where everything is installed for texcenter, i still couldnt the find texnic center!
please Help!!

texnic center is usually installed under the Program Files folder. Check there, and then on ur start menu as well.

[quote="Tr"]i searched my hard drive for "texnic center" but there was only one result it gave me and that was html file explaining about tex definition etc.
should i delete everything and reinstall all files? ..time consuming though....:([/quote]If you used regular windows search system, you should know that it's extremely bad. I won't find you the files you're looking for that easily. You must either use a specialised program (like Advanced Find and Replace) or you probably searched for the wrong string.

I just installed mictex. the 24 mb one. and I have the same problem cann't find texnicCenter.

You *do* know that TexnicCenter is unrelated to miktex.. and you download it from a completely different place..


The softball team won nine of its twenty first games. How many of the remaining sixteen games does the team have to win to have a 50$ \%$ winning record?

The team will play$ 20 \plus{} 16 \equal{} 36$ games. In order to win half of them, they must win $ 18$ games. Since they have already won $ 9$, $ 18 \minus{} 9 \equal{} \boxed{9}$ more games that they have to win.

There are 16 remaining games, meaning there will be 36 games in total. Since 50% of 36 is 18, they need to win 18 games in total. They've already won 9, so they only need to win 9 more.


how do you make graphs (cubic, quartic, etc) in asymptote?

i'm also having some trouble with tick marks. currently i'm using the olympiad package and pathticks, but i can't find a way to label the tick marks.

I use a case-wise approach for functions. It may be time-consuming and sometimes ineffective, but I get it to work.

Graph a function on your calculator and record a bunch of key points the function passes through (mid-points, vertices, points close to an asymptote, etc...). Then simply copy all these points as pairs and draw the path that connects them.

pair A,B,C,D

A= co-ordinate 1
B = "" 2
C = "" 3
D = "" 4
draw(A..B..C..D);

If you're drawing a circle, you'll want to have something like draw(unitcircle); or draw(dir(30)..dir(60)..dir(120)..dir(220)..dir(330)..cycle);

If you're drawing a periodic function, draw one cycle of the function accurately (as in using lots of points for your curved path), then shift(distance*direction)*path as much as needed to get as many periods of the graph as you want.


Let $ a,\ b,\ c$ be positive integers such that $ a^2 \plus{} b^2 \equal{} c^2$ and $ a,\ b$ are coprime.

Prove the following contexts.

(1) If $ a$ is odd, then $ b$ is even and $ c$ is odd.

(2) If $ a$ is odd, then there exists positive integers $ d$ such that $ a \plus{} c \equal{} 2d^2$.

[hide="2"]Notice that $ (a,b)\equal{}1\Rightarrow (a,c)\equal{}1$ since if $ d|a,d|c$ then $ d|c^2\minus{}a^2\equal{}b^2$.  Returning to the equation, we have $ a^2\plus{}b^2\equal{}c^2\Rightarrow b^2\equal{}(c\minus{}a)(c\plus{}a)$.  We have $ (c\minus{}a,c\plus{}a)\equal{}(2c,c\plus{}a)$.  Thus, their greatest common factor is a factor of $ 2c$.  But, it is obvious that if $ d|c$ then $ d\not| c\plus{}a$ since $ (c,a)\equal{}1$.  Thus, since it is given that $ a$ is odd and from (1) that $ c$ is odd we have $ (c\minus{}a,c\plus{}a)\equal{}2$.  So, because $ (c\minus{}a)(c\plus{}a)\equal{}b^2\Rightarrow \left(\frac{c\minus{}a}{2}\right)\left(\frac{c\plus{}a}{2}\right)\equal{}\frac{b^2}{4}$ and $ (\frac{c\minus{}a}{2},\frac{c\plus{}a}{2})\equal{}1$ we must have $ \frac{c\plus{}a}{2}\equal{}d^2\Rightarrow c\plus{}a\equal{}2d^2$ for some $ d$.[/hide]

[hide="#1"]
the square of an integer must be either 1 mod 4 or 0 mod 4 (even is 0 mod 4 odd is 1 mod 4)
since a is odd, $ a^2$ is 1 mod 4
$ b^2$ is either even (0 mod 4) or odd (1 mod 4)
if b is odd, then $ b^2$ is 1 mod 4 and $ a^2\plus{}b^2$ is 2 mod 4, but there is no positive integer,c,s.t. c^2 is 2 mod 4
so b must be even, and $ a^2\plus{}b^2$ is 1 mod 4 and c^2 is odd
[/hide]


The length of one side of a square is increased by 2 units and its other side is decreased by 2 units. By how many square units do the areas of the original square and the new rectangle differ?

Let the side of the square be x. Then its area is x^2.
The rectangle has one side x+2 and one side x-2. It has area (x+2)(x-2)= x^2-4
The positive difference in area is $ x^2 \minus{} 4 \minus{} x^2 \equal{} \boxed{4}$


Let $M$ be the midpoint of the side $[BC]$ of the triangle $ABC$. The parallel to $BC$ through a point $G \in (AM)$ intersects $AB$ in $X$ and $AC$ in $Y$; $CX$ and $GB$ intersect in $Q$, $CG$ and $BY$ in $P$. Denote $\frac{AG}{AM}$ by $k$.
(a) Prove that $\frac{GP}{PC}= \frac{k}2$.
(b) Express the vector $\overrightarrow{CP}$ as a function of $\overrightarrow{AB}$, $\overrightarrow{AC}$ and the scalar $k$.
(c) Prove that if $G$ is the centroid of $ABC$, then $MPQ$ is similar to $ABC$.

$(a)\quad \frac{GP}{PC}=\frac{GY}{BC}=\frac{1}{2}\cdot\frac{XY}{BC}=\frac{1}{2}\cdot\frac{AG}{AM}=\frac{k}{2}$

$(c)$ Denote: $D=BG\cap AC,\,E=CG\cap AB$
We have: $\frac{BQ}{QG}=\frac{BC}{XG}=\frac{BC}{\frac{2}{3}BM}=\frac{BC}{\frac{1}{3}BC}=3$
$\Longrightarrow BQ=3QG\Longrightarrow BQ=\frac{3}{4}BG=\frac{3}{4}\cdot\frac{2}{3}BD=\frac{1}{2}BN$
$\Longrightarrow MQ\parallel AC.$
Similar: $MP\parallel AB$ and easy to prove that: $PQ\parallel BC$
Hence: $\triangle MPQ\sim \triangle ABC$


A 280 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.1 N/cm. The block becomes attached to the spring and compresses the spring 12 cm before momentarily stopping.

1) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.)

2) If the speed at impact is doubled, what is the maximum compression of the spring?

You can use conservative law of energy.

1) $\frac{1}{2}mv^{2}=mgh\Longleftrightarrow v=\sqrt{2gh}.$

2) $\frac{1}{2}kx^{2}=\frac{1}{2}m(2v^{2})\Longleftrightarrow x=2v\sqrt{\frac{m}{k}}.$

[quote="maokid7"]thanks.
for the first question i got the same answer but for some reason the computer program tells me that it is incorrect.
i think ill talk to my teacher about this flaw. :D[/quote]

Check units. If you forgot that $N=kg*m/s^{2}$ you would get the wrong answer even if you did the formula correctly. That's one possibility. Or the computer program could be wrong.

For the first question, you're ignoring the energy stored in the spring. The proper relationship is $\frac{1}{2}m v^{2}+m g x = \frac{1}{2}k x^{2}$.

why did u include the energr stored in the spring?

The first question asks for the speed of the block [i]before[/i] it hits the spring, so there is no stored energy to consider.
For the second question, the speed is doubled, the velocity is  $2v$, which becomes $4v^{2}$ when squared, not $2v^{2}$.

Exactly. The spring is uncompressed before the block hits the spring, and when the block stops the spring has been compressed 12 cm.


Strongly $ |a_n \minus{} \frac n2 |\le \frac {2(\sqrt 2 \minus{} 1)}{n} \ \forall n > 1$.

thx rust....but can you be more exactly pls ?


Let $ p(x)$ be a polynomial with integer coefficients so that $ p(0) \equal{} p(1) \equal{} 1999$. Show there is no $ x_{0}\in\mathbb{Z}$ so that $ p(x_{0}) \equal{} 0$.

[hide="Hint"]
I realized that $ 1999$ is prime.
[/hide]

Assume $ p(x) \equal{} (x\minus{}x_{0})g(x)$

$ 1999 \equal{} (\minus{}x_{0})g(0) \equal{} (1\minus{}x_{0})g(1)$ so $ \minus{}x_{0}$ and $ 1\minus{}x_{0}$ must be two consecutive integers that are factors of 1999 which does not exist since 1999 is prime. Thus $ p(x)$ has no integer roots

[hide]
Let $ a$ be odd.
$ p(a)\equiv p(1)\equiv 1999\not\equiv 0\mod 2$
So no odd integer can be a root.

Now let $ a$ be even.
$ p(a)\equiv p(0)\equiv 1999\not\equiv 0\mod 2$
So no even integer can be a root.

Thus no integers can be roots.
[/hide]

$ (n,s,p)\in\mathbb{N}\times\mathbb{Z}\times\mathbb{Z}[X]$ we have:
$ \forall i\in\{s,s+1,s+2,...,s+n-1\}:\ n\not|p(i)$ $ \to$ $ p(x) = 0$ has not integrs solution.

$ proof:$
if $ \forall i\in\{s,s+1,s+2,...,s+n-1\}:\ n\not|p(i)$
$ \forall x\in\mathbb{Z},\exists k\in\{0,1,..,n-1\}:\ x\equiv s+k\ (mod\ n)$
then $ p(x)\equiv p(s+k)\ (mod\ n)\not\equiv 0\ (mod\ n)$
so $ \forall x\in\mathbb{Z}:\ p(x)\neq 0$

here we have $ s=0$ and $ n=2$

This is actually an old problem which I solved a long time ago. My idea was basically tjhance's one. Here is my solution:
[hide="My solution"]Let $ p(x)\equal{}a_{n}x^{n}\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}\dots\plus{}a_{1}x\plus{}a_{0}$.
$ p(0)\equal{}1999\implies a_{0}\equal{}1999$, $ p(1)\equal{}1999\implies a_{n}\plus{}a_{n\minus{}1}\plus{}\dots\plus{}a_{1}\equal{}0$. As the sum of an odd number of odd numbers is odd(so can not be zero), we must have $ 2\mid n$. Now let $ r$ be an integer zero of $ p(x)$.

If $ r$ is even, we have $ a_{n}r^{n}\plus{}a_{n\minus{}1}r^{n\minus{}1}\plus{}\dots\plus{}a_{1}r\plus{}a_{0}\equal{}\text{odd}$, which is not possible as $ 0$ is even.

If $ r$ is odd, assume that exactly $ k$ of the $ a_{i}$s are even. Then from $ a_{1}\plus{}a_{2}\plus{}\dots\plus{}a_{n}\equal{}0$, we get $ k$ must be even. Then $ p(r)\equal{}\text{sum of }k\text{ even numbers}\plus{}\text{sum of }n\minus{}k\plus{}1\text{ odd numbers}$ $ \equal{}\text{sum of an even number of even numbers\plus{}sum of an odd number of odd numbers}$ $ \equal{}\text{odd}$ and thus cannot be zero. So there does not exist any integers $ r$ such that $ p(r)\equal{}0$.[/hide]

Well like...you can just say that $ p(r)\equiv p( r\mod 2 )\equiv 1999\equiv 1\mod 2$, but $ p(r)\equal{}0\implies p(r)\equiv 0\mod 2$, which is a contradiction.


Let $ x,y$ be positive integers such that both $ 4x\plus{}7y$ and $ 7x\minus{}4y$ are perfect squares.
Determine the possible least value of the two perfect squares respectively.

Note that $ (4x \plus{} 7y)^2 \plus{} (7x \minus{} 4y)^2 \equal{} (x^2 \plus{} y^2)(4^2 \plus{} 7^2)$.  Let $ 4x \plus{} 7y \equal{} a^2$ and $ 7x \minus{} 4y \equal{} b^2$.  Then $ a^4 \plus{} b^4$ is divisible by 65, it is divisible by both 5 and 13.  If one or more of a or b is not divisible by 5, then we have that at least one of $ a^4$ or $ b^4 \equal{} 1 (mod 5)$ and the other is 0 or 1 mod 5 by Fermat's Little Theorem, so 5 does not divide $ a^4 \plus{} b^4$.  Thus a and b are both divisible by 5 and both of 4x+7y and 7x-4y are divisible by 25.  By Fermat's Little Theorem, $ k^{12}$=1 (mod 13) or k is divisible by 13, so if k is not divisible by 13, $ k^4$ is a cube root of 1 mod 13, which are 1, 3, and 9.  The only pair $ a^4,b^4\in\{0,1,3,9\}$ such that $ a^4 \plus{} b^4 \equal{} 0 (mod 13)$ is $ a^4 \equal{} b^4 \equal{} 0 (mod 13)$, which only happens when a=b=0 (mod 13).  Thus both of 4x+7y and 7x-4y are divisible by 5^2 and 13^2, so they are both divisible by $ 65^2$, so the smallest possible value for them is $ 65^2$ which can be attained with x=11*65=715, y=3*65=195.


search
combinatorics proposed
combinatorics

Can the cells of a $2002 \times 2002$ table be filled with the numbers from $1$ to $2002^2$ (one per cell) so that for any cell we can find three numbers $a, b, c$ in the same row or column (or the cell itself) with $a = bc$?

If you are talking about the problem of the russian olympiad, then this is the following.

Can the numbers from $ 1$ to $ 2002^{2}$ be written in the squares of a $ 2002 \times 2002$
board in such a way that, for each square, there exist three numbers in
the union of its row and its column, one of which is the product of the other two?

I rewrite the problem because there is a big difference, in yours, for a fixed cell, the three numbers have to be in the same row (or column) and the problems becomes trivial, it is different if you have to consider the numbers in the union of the row and the column.
I'll post the solution soon.

More general, let's consider an $ n \times n$ board. We say that a board is flap if we can write the numbers from $ 1$ to $ n^2$ in such a way that, for each square, there exist three numbers in the union of its row and its column, one of which is the product of the other two.

[b]Lemma 1: [/b] If a board is flap, another board obtained from this by row permutation is also flap. By the same way, another board obtained from this by column permutation is also flap.
[i]Proof:[/i] Trivial.

Then, suppose that we have written the numbers in such a way that the board is flap. Now, by row and columns permutations, we can put the numbers from $ 1$ to $ (n\minus{}1)$ in the $ (n\minus{}1) \times (n\minus{}1)$ subboard and this is still flap. Consider the union of the $ n^{th}$ row and the $ n^{th}$ column, and in this set we have three numbers $ a$, $ b$, $ c$ such that $ ab \equal{} c$, but $ a,b,c \geq n$, then it must be $ a \equal{} b \equal{} n$ and $ c \equal{} n^{2}$. Contradiction! Then we can not write the numbers in the board in such that way.

if we have a row that contains no numbers such a,b,c which a*b=c
then each column has a number less than 2002 (in this situation we have three a,b,c that a=b*c and the min (b,c) must be less than 2002)
and we have 2002 columns so there must be at least 2002 numbers less than (and not equal) to 2002 and it's wrong !!!!

[quote=alejnaser]

Can the numbers from $ 1$ to $ 2002^{2}$ be written in the squares of a $ 2002 \times 2002$
board in such a way that, for each square, there exist three numbers in
the union of its row and its column, one of which is the product of the other two?

[/quote]
No, the problem states in either the column or in the row. Not in the union of them

The answer is no. For the sake of contradiction, suppose such arrangement is possible. We will cal a row or column good, if it contains three numbers $a,b,c$ such that $a=bc$.

What we have to understand is, either all columns are good, or all rows are good. Why? if at least on column is not good, then its not hard to get that all rows are good. W.L.O.S assume all rows are good. Then in each row $r_i$ there are numbers $a_i,b_i,c_i$ such that $a_i=b_ic_i$. So there are 2002 equations $a_i=b_ic_i$, with all numbers distinct, ranging from $1$ to $2002^2$. So $a_i\leq 2002^2$ meaning at least one of $\{b_i,c_i\}$ is strictly less that 2002. We have only 2001 numbers which can be used as $b_i$ or $c_i$ when there should be 2002. Contradiction. Therefore, such arrangement is not possible


inequalities
linear algebra
matrix
inequalities theorems

Let’s start with an exemple :
For $ j \equal{} 1,\dots,n$ let $ x_j,y_j,z_j\ge0$. Then
\[ \sum x_j^3\sum y_j^3\sum z_j^3\ge \sum x_j^2y_j^{\ }\sum y_j^2z_j^{\ }\sum z_j^2x_j^{\ }\ge \left(\sum x_jy_jz_j\right)^3.\]
The proof is immediat :
For the left one, we have by Hölder $ \sum x_j^3 \sum x_j ^3\sum y_j^3\ge\left(\sum x_j^2y_j^{\ } \right)^3$. Similar for $ x\to y\to z$ cyclically. Multiply them all three.
The right one is directly Hölder.

This inequality chain corresponds to the chain of partitions $ (3,0,0)\succ(2,1,0)\succ(1,1,1)$. It should be clear how to generalize these terms for a partition $ \alpha \equal{} (\alpha_1,...,\alpha_m)\vdash m$, applied to an $ m\times n$-matrix of variables $ x_{ij}\ge0$. Formally the term is
\[ \boxed{p_\alpha: \equal{} \prod_{i \equal{} 1}^m \sum_{j \equal{} 1}^n\prod_{k \equal{} 1}^m x_{kj}^{\alpha_{i \plus{} k}}}\]
with indices mod $ m$. To make this well-defined, we must require $ \alpha_1\ge\cdots\ge\alpha_m$.

Now, [i]the question is if $ p_\alpha\ge p_\beta$ for two partitions of $ m$ whenever they satisfy $ \alpha\succ\beta$. [/i]But I think,
[b]
[i]$ p_\alpha\ge p_\beta$ only holds if $ \alpha\succ\succ\beta$. [/b][/i]

Here the [b]"cyclic majorisation"[/b] $ \succ\succ$ is defined by $ \alpha\succ\succ\beta\Longleftrightarrow (\alpha)\ge(\beta)$, where $ (\alpha)$ and $ (\beta)$ are the cyclical sums (applied to $ m$ variables) as opposed to the symmetrical sums $ [\alpha]$ and $ [\beta]$. See [url=http://www.mathlinks.ro/viewtopic.php?pt=103761]here[/url] for a discussion of this criteria. :)

Suppose there is a decomposition such that $ (\alpha)\ge(\beta)$ can be proved by AM-GM (see [url=http://www.mathlinks.ro/viewtopic.php?p=585871]post#4[/url] in the thread for more details and the $ m \equal{} 3$ case). Using the non-negative coefficients of that, we can prove $ p_\alpha\ge p_\beta$ by Hölder in a similar way as $ p_{(3,0,0)}\ge p_{(2,1,0)}$ above.  :lol:

I just discovered that this isn't new. There is the "Callebaut inequality" which refines Cauchy-Schwarz in exactly the same way by virtue of the interpolation \[ (1,1)\prec\prec(1\plus{}x,1\minus{}x)\prec\prec(1\plus{}y,1\minus{}y)\prec\prec(2,0)\] for $ 0\le x\le y\le 1$.  So we can state it as

$ (\sum a_ib_i)^2\le\sum a_i^{1\plus{}x}b_i^{1\minus{}x}\sum a_i^{1\minus{}x}b_i^{1\plus{}x}$ $ \le\sum a_i^{1\plus{}y}b_i^{1\minus{}y}\sum a_i^{1\minus{}y}b_i^{1\plus{}y}\le\sum a_i^2\sum  b_i^2$. :lol:


The inscribed circle of triangle $ABC$ touches $BC$, $CA$ and $AB$ at $D$, $E$ and $F$ respectively. Let $S$ be the second intersection of $AD$ with the incircle. Prove that line $ES$ bisects segment $AF$ if and only if $AC = BC$.

A hint...

[hide]Let $T$ be the intersection of $ES$ and $AF$.

Prove that $AT = TF \Leftrightarrow \Delta TSA \sim \Delta TAE$.[/hide]


Let n be a positive integer.
Find the number of positive integer solutions x_1, x_2, ... , x_m of the equation:

 a_1 x_1 + a_2 x_2 + ... +a_m x_m = 2^n-1

where a_1, a_2, ..., a_m are positive integers less than 2^n-1 none of which is of the form 2^i-1 for some integer i.


How many different ways can you colour an icosahedron with one of 3 colours on each face?
Umm..what's an icosahedron?  :D (I am just a beginner! lol!) :blush:

Use Burnside's lemma.  http://en.wikipedia.org/wiki/Burnside's_lemma

[quote="mgao"]An icosahedron is a 3D figure with 20 faces.  The faces are triangular.

here is a picture of one:  click link

 :D

by chance, the picture seems to be 3.14 kilobytes!?

Also, try to find the number of ways you can arrange the icosahedron, and find the value of $3^{20}$.  Then divide $3^{20}$ by the number of ways you can arrange the icosahedron (like choose a random face, how many different areas in space can that labeled face be when the icosahedron is rolled).  There is your answer.[/quote]

Hmmmm...thanx! :D


superior algebra
superior algebra unsolved

My definition of a strongly nilpotent element is:

Let a be in ring R the element a is strongly nilpotent if for every sequence a_0,a_1,...,a_i,... such that a_0 =a
and a_{i+1} is in a_iRa_i, there exists an n with a_n =0.

The question is in a theroem I am using that a is not strongly nilpotent, what does it mean

The author is saying

Since a is not strongly nilpotent we have a sequence a_0,a_1,...,a_i,... with a_{n+1} is in a_nRa_n, a_n is different from zero an
and a_1 is in aRa.

I dont understand that why is a_1 is in aRa. how do we get that?

peteryellow: You seem to be learning a serious amount of advanced stuff in a very short time. It occurred to me, in several threads, that you very often don't even know what the author (or you?) is (are) talking about. My advice: do it more slowly. There's nothing to be ashamed of... it's not how much you know, certainly in algebra, it's a way of thinking, and doing it too fast brings you nowhere. Also, don't expect people to answer in complete detail. We often deliberately post incomplete proofs. We want you to be as pissed off as we were when we worked hard to understand the subject ;)

greets


algebra
polynomial
probability
calculus
integration
binomial theorem

How do you prove the binomial theorem?

If there is a way to do it, I would be very happy to see a supporting link or a

description.

Mathematically, with as little English as possible.

The Theorem (in case you didn't know) looks like this:

$\text {For any positive integer n,}$

$(a+b)^n=\displaystyle\sum_{r=0}^{n}{}_nC{}_ra{}^n{}^-{}^rb{}^r$

$= a^n+{}_nC{}_1a{}^n{}^-{}^1b+{}_nC{}_2a{}^n{}^-{}^2b{}^2+......+{}_nC{}_ra{}^n{}^-{}^rb{}^r+......+{}_nC{}_n{}_-{}_1ab{}^n{}^-{}^1+b^n$

(so you don't have to go to Wikipedia)

 :rotfl:  :rotfl:  :rotfl:

Proving is on intermediate section. But since you are new, can somebody prove it?  :rotfl:

[hide="Combinatorial proof"]Write all the binomials next to one another:

$(a+b)(a+b)\dots(a+b)$

By the rule of the polynomial multiplication, every term from every pair of parentheses must be multiplied by every term from every other pair of parentheses. Therefore, from each pair of parentheses pick either $a$ or $b$ and multiply them out. You'll get the term of the form $a^kb^{n-k}$. It remains to be seen in how many ways you can pick exactly $k$ $a$'s and $n-k$ $b$'s, in order to determine the coefficient with $a^kb^{n-k}$: If under each pair of the parentheses you write the picked letter, you'll get something like $aaba\dots bba$, where you have exactly $k$ $a$'a and exactly $n-k$ $b$'s. By the known combinatorial result, these letters can be arranged in $\frac{n!}{k!(n-k)!}={n\choose k}$. Hence the result.[/hide]

[quote="shadysaysurspammed"][quote="Farenhajt"][hide="Combinatorial proof"] If under each pair of the parentheses you write the picked letter, you'll get something like $aaba\dots bba$, where you have exactly $k$ $a$'a and exactly $n-k$ $b$'s. By the known combinatorial result, these letters can be arranged in $\frac{n!}{k!(n-k)!}={n\choose k}$. Hence the result.[/hide][/quote]

Could someone please explain that with a simple example(like $(a+b)^3$?)[/quote]

Write $(a+b)(a+b)(a+b)$. Now you have to multiply "each by each", and all the resulting terms will have the form $a^kb^{3-k}$ where $k=0,1,2,3$. So, start with straight $a$'s - there's only one way to pick three $a$'s, hence the coefficient with $a^3$ will be $1$. Next, there's $a^2b$ - there are three ways to pick $2$ $a$'s and $1$ $b$, namely $aab, aba, baa$ (the first arrangement means that you multiply the $a$ from the first parentheses by the $a$ from the second parentheses by the $b$ from the third parentheses, similarly for the other two) - which means that the coefficient with $a^2b$ will be $3$. Similarly, you can pick $1$ $a$ and $2$ $b$'s in three ways - $abb, bab, bba$ - meaning that the coefficient with $ab^2$ will be $3$, and finally, the coefficient with $b^3$ will be $1$. Writing all that out, you get $(a+b)^3=a^3+3a^2b+3ab^2+b^3$

[quote="mgao"]There [i]is[/i] a second part.

The general term in the expansion of $(a+b+c)^n$ is

$\frac{n!}{p!q!r!}a^pb^qc^r...(p+q+r=n)$

How about this one?[/quote]

The same reasoning applies here, since the general term will have the form $a^pb^qc^r$ where $p+q+r=n$, and the corresponding coefficient will be determined by the number of ways in which $p$ letters $a$, $q$ letters $b$ and $r$ letters $c$ can be ordered - those arrangements are repetitive permutations, and the known formula gives $\frac{(p+q+r)!}{p!q!r!}=\frac{n!}{p!q!r!}$

See also : http://www.mathlinks.ro/Forum/viewtopic.php?t=58166 . ;)

You can also use a probability tree diagram, compare probabilities with an expansion...they are exactly the same.

Although, coffeym, a probability tree diagram doesn't provide proof for all n, although it can for fixed, finite n (with some effort).

[quote="José"][quote="kokosai"]You can also use induction.[/quote]

Know how?[/quote]We want to show that $(a+b)^n=\sum_{k=0}^n\binom{n}{k}a^kb^{n-k}$ is true for all nonnegative integral $n$.  The statement is true for $n=0$, so assume that it is true for some nonnegative integer $k$.  But $(a+b)^{k+1}=(a+b)(a+b)^k=a(a+b)^k+b(a+b)^k$, from which, in combination with the identity $\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}$, the statement for $k+1$ quickly proves true.


What is the sum of all positive integer values of $ n$ for which $ \frac{n\plus{}8}{n}$ is an integer?

We can see that $ n$ will have to be a factor of $ 8$. Thus, the answer is $ \frac{1\plus{}8}{1}\plus{}\frac{2\plus{}8}{2}\plus{}\frac{4\plus{}8}{4}\plus{}\frac{8\plus{}8}{8}$, which simplifies to $ 9\plus{}5\plus{}3\plus{}2\equal{}\fbox{19}$.

They asked for the sum of the values of n, so it is $1+2+4+8=\boxed{15}$


LaTeX
inequalities proposed
inequalities

Let a, b, c, A, B, C, D be reals such that (Ax + B)(Cx + D) = ax^2 + bx + c for all real x.

Show that at least one of a, b, c is greater than or equal to 4(A + B)(C + D)/9.

This is a little bit unusual - I like it though. My proof is a bit messy, so I'll just give an outline.

As a first step, I divide through by $\frac{a+b+c}{3}$ (just to tidy things up later - it's easy to check that the result is unaffected).

Obviously $(A+B)(C+D)=a+b+c=3$, so we want to show that at least one of $a,b$ and $c$ is at least $\frac{4}{3}$.  If this is not the case, then we can show that $b^2-4ac<0$, contradicting the requirement that $A,B,C,D$ be real.

More particularly, given $a+c=k$, $ac$ is minimised when $|a-c|$ is as large as possible (assuming $a,c>0)$.  Given $a,c<\frac{4}{3}$, it follows that $ac>\left(\frac{4}{3}\right)\left(k-\frac{4}{3}\right)$.  But now, $k=3-b$, so we get $b^2-4ac<b^2-4\left(\frac{4}{3}\right)\left(b+\frac{5}{3}\right)<0$ for $0<b<\frac{4}{3}$.

I've changed notation in going from my notes to Latex, so I may have made some horrenduous typos, but the main idea should be clear enough...

[hide=*][quote=Arne]Let a, b, c, A, B, C, D be reals such that (Ax + B)(Cx + D) = ax^2 + bx + c for all real x.Show that at least one of a, b, c is greater than or equal to 4(A + B)(C + D)/9.[/quote][/hide]
Let $a, b, c, A, B, C, D$ be reals such that $(Ax + B)(Cx + D) = ax^2 + bx + c $
for all real $x$.Show that at least one of $a, b, c$
is greater than or equal to $\frac{4(A + B)(C + D)}{9}$.


calculus
integration
number theory unsolved
number theory

For a positive n, show that the number of integral solutions (x,y) of x^2 + xy + y^2 = n is finite and a multiple of 6.

By unique prime decomposition in $\mathbb{Z} [ \zeta _3]$ one gets that there are no solutions when not all prime divisors of $n$ that are $\equiv 2 \mod 3$ occure an even number of times and otherwise there are exactly
\[ 6 \cdot \prod_{\substack{p \in \mathbb{P} \\ p\equiv 1 \mod 3}} (v_p(n)+1)  \]
of them.
One then easily concludes that there are
\[ 6 \left( \cdot \sum\limits_{\substack{ d \in \mathbb{N} \\ d|n \\ d \equiv 1 \mod 3}} 1 - \sum\limits_{\substack{ d \in \mathbb{N} \\ d|n \\ d \equiv 2 \mod 3}} 1 \right)  \]
solutions, independent of the prime divisors of $n$.
So the problem get's trivial ;)

I can remember something like this: If there exist $(x,y) \in Q^2$ such that $x^2+xy+y^2=n$ So we have $(x,y) \in Z^2$ too.

Yes, this follows from Davenport-Cassels-Theorem.

A solution based on [b]Minkowski's theorem[/b] :)


p^2-p+1=q^3 where p and q are prime
Find p and q.

$ p^{2} \minus{} p \equal{} q^{3} \minus{} 1$ $ \implies$
$ p(p \minus{} 1) \equal{} (q \minus{} 1)(q^{2} \plus{} q \plus{} 1)$
$ (q \minus{} 1,q^{2} \plus{} q \plus{} 1) \equal{} 1,3$
if $ p > 3$
$ i)q \minus{} 1 \equal{} p.u ,q^{2} \plus{} q \plus{} 1 \equal{} \frac {p \minus{} 1} {u}$   $ \implies$
$ q^{2} \plus{} q \plus{} 1 < q \minus{} 1$ $ \implies$  $ q^2 < \minus{} 2$  Contradiction!
$ ii)q^{2} \plus{} q \plus{} 1 \equal{} p.u ,q \minus{} 1 \equal{} \frac {(p \minus{} 1)} {u}$  $ \implies$
$ p \equal{} uk \plus{} 1,q \equal{} k \plus{} 1$ $ \implies$
$ q^{2} \plus{} q \plus{} 1 \equal{} (uk \plus{} 1).u \equal{} (k \plus{} 1)^{2} \plus{} k \plus{} 2 \equal{} u^{2}.k \plus{} u$  $ \implies$
$ k^{2} \minus{} k(u^{2} \minus{} 3) \plus{} 3 \minus{} u \equal{} 0$ $ \implies$ $ (u^{2} \minus{} 3)^{2} \plus{} 4(u \minus{} 3)$ must be perfect square
$ u^{4} \minus{} 6u^{2} \plus{} 4u \minus{} 3 \equal{} c^{2}$ $ \implies$
$ (u^{2} \minus{} 3)^{2} < u^{4} \minus{} 6u^{2} \plus{} 4u \minus{} 3 < (u^{2} \minus{} 2)^{2}$  $ \implies$  we get any solution.
$ u^{4} \minus{} 6u^{2} \plus{} 4u \minus{} 3\leq u^{4} \minus{} 6u^{2} \plus{} 9$ $ \implies$  $ 4u\leq 12,u\leq 3$
$ u \equal{} 1$ $ \implies$ $ q \equal{} p,p^2 \plus{} p \plus{} 1 \equal{} p$ Contradiction!
$ u \equal{} 2$  $ \implies$ $ 2^{4} \minus{} 6.2^{2} \plus{} 4.2 \minus{} 3 \equal{} \minus{} 3 \equal{} c^{2}$ Contradiction!
$ u \equal{} 3$  $ \implies$  $ 3^{4} \minus{} 6.3^{2} \plus{} 4.3 \minus{} 3 \equal{} 6^{2}$  $ \implies$ $ u \equal{} 3,q^{2} \plus{} q \plus{} 1 \equal{} 3p,3q \minus{} 2 \equal{} p$
$ q^{2} \plus{} q \plus{} 1 \equal{} 9q \minus{} 6 ,q^{2} \minus{} 8q \plus{} 7 \equal{} 0$ $ \implies$ $ (q \minus{} 1)(q \minus{} 7) \equal{} 0$ $ \implies$ $ q \equal{} 7,p \equal{} 19$
$ (u^{2} \minus{} 2)^{2}\leq u^{4} \minus{} 6u^{2} \plus{} 4u \minus{} 3$ $ \implies$ $ 2u^2 \plus{} 4u \minus{} 7\leq0$ $ \implies$ $ u\leq1$
$ u \equal{} 1$  $ \implies$ $ q \equal{} p,p^{2} \plus{} p \plus{} 1 \equal{} p$ Contradiction!
$ p\equal{}3$ $ \implies$  $ 3^{2}\minus{}3\plus{}1\equal{}7\equal{}q^{3}$ Contradiction!
unique solution is $ (19,7)$ :wink:

http://www.mathlinks.ro/viewtopic.php?p=224998#224998

BMO 2005/2

There's a $ \text{\LaTeX}$ forum under ML/AoPS. And, if you know how to work in $ \text{\LaTeX}$, just put dollar signs around the equation. You can see another tutorial under LaTeX section on AoPS site.


Ring Theory
superior algebra
superior algebra unsolved

Could you please describe the structure of the spectrum of $ \mathbb Z[x]$? And what would be the structure of $ \mathbb Z_p[x]$ for prime p?

if $ F$ is a field, $ F[X]$ is a PID, so that $ Spec F[X]$ is easy to describe.

if $ R$ is a PID, $ Spec R[X]$ can be determined with the fibres of $ Spec R[X] \to Spec R$. try it!

this is in Atiyah-Mcdonald

look here http://www.math.uchicago.edu/~allanaa/atiyah.pdf

problem 1.16 part e)

the description doesn't seem to be complete though...


Literally. i who have posted just 12 articles( :| ) was suspicious of moderators. could u tell me about ways to choose mods.

Danielsinx wants to be mod.(visionary scheme?)

Suspicious of what¿

And of what do you want to become a moderator¿ They are chosen by the admins. This is no democracy, but adminocracy.
But note: Almost no moderator got moderator by proposing himself  :wink:

[quote="ZetaX"]But note: Almost no moderator got moderator by proposing himself  :wink:[/quote]I did. :D

[quote="james4l"]Technically that's an oligarchy (nearly all power is represented by a small elite group).[/quote]

Ha! I just learned about that in history class! Anyway, lots of people have posted about this subject. Just DONT ASK IF YOU CAN BE A MOD


The median of the set $ \{n, n \plus{} 5, n \plus{} 6, n \plus{} 9, n \plus{} 15\}$ is 9. What is the mean?

We know the median is n+6, and also 9, so we solve for n.

$ n \plus{} 6 \equal{} 9$

$ n \equal{} 3$

Our mean is the five terms added and then divided by 5. In terms of n, this is

$ \frac{n \plus{} n \plus{} 5 \plus{} n \plus{} 6 \plus{} n \plus{} 9 \plus{} n \plus{} 15}{5}$

$ \frac{5n \plus{} 35}{5}$

$ n \plus{} 7$

Plugging in n=3, we get

$ 3 \plus{} 7 \equal{} \boxed{10}$


Find all $ x,y\in\mathbb{Z}$, which satisfying equation $ \frac{x^7\minus{}1}{x\minus{}1}\equal{}y^5\minus{}1.$


group theory
abstract algebra
superior algebra
superior algebra unsolved

H is a sub-group of G such that whenever $ Ha \ne Hb$, we have $ aH \ne bH$. Prove that for all $ g \in G$ we have $ gHg^{\minus{}1} \subset H$

we have:
            If $ b^{\minus{}1}a\in H$ $ \longrightarrow$ $ ab^{\minus{}1}\in H$
Let $ b\equal{}gh^{\minus{}1}$ and $ a\equal{}g$ $ \longrightarrow$ $ ghg^{\minus{}1}\in H$.


linear algebra
matrix
vector
complex numbers
linear algebra unsolved

Show that two orthogonal projection operators $ P_A$ and $ P_B$ are orthogonal ( $ P_AP_B = 0$ ) if and only if their ranges are orthogonal.
[hide="Sth. is not right"]
Let $ \textbf{x,y}$ arbitrary nonzero vectors.
Range of $ P_A$ is $ P_A\textbf{x}$,
Range of $ P_B$ is $ P_B\textbf{y}$

Then, for orthogonality, the inner product $ < P_A\textbf{x}, P_B\textbf{y} >$ is zero.

$ < P_A\textbf{x}, P_B\textbf{y} > = (P_B\textbf{y})^HP_A\textbf{x} = \textbf{y}^HP_B^HP_A\textbf{x} = 0$

If any $ \textbf{x,y}$ vectors satisfy this condition, $ P_B^HP_A = 0$ ( $ P_A^HP_B = 0$ is equivalent )
But it says $ P_AP_B = 0$
Is any orthogonal operator $ P = P^H$ ?? (How??)
[/hide]

Another question:
About any $ P_A$ and $ P_B$ (they're not necessarily orthogonal in this question)
Is $ P_AP_B = P_BP_A$ ?

Yes, any orthogonal projection operator $ P$ satisfies $ P\equal{}P^H$. The easiest way to see this is to show that $ \left<v,Pw\right>\equal{}\left<Pv,w\right>$ for any two vectors $ v$ and $ w$. Why? Try writing $ v\equal{}Pv\plus{}\left(v\minus{}Pv\right)$ on the left hand side and $ w\equal{}Pw\plus{}\left(w\minus{}Pw\right)$ on the right hand side. Time to use the orthogonality, then.

  darij

[quote="gunes"]I suppose, what you mean by [quote]Time to use the orthogonality, then.[/quote]
is that $ < v \minus{} Pv,Pw > \equal{} 0$
Also, what about the second question?
That, is it $ P_AP_B \equal{} P_BP_A$ for any projections?(Just a logical guess)[/quote]

Oh, I was sure I answered to this, but now I see I deleted my answer before I submitted it. No, $ P_AP_B\equal{}P_BP_A$ doesn't even hold for orthogonal projections in general. Draw two lines (non-perpendicular) in plane, and a point. Project this point on one line and then project the projection on the other one. Change the order of the lines. You don't get the same image, do you?

  darij

Hmm, I was trying to show that if $ P_A \plus{} P_B$ is a projection operator ($ P_A$ and $ P_B$ are orthogonal projection operators), $ P_A$ and $ P_B$ are orthogonal.
$ P_A \plus{} P_B$ is a proj. op.

So, it should satisfy
$ (P_A \plus{} P_B)^2 \equal{} P_A \plus{} P_B$
$ P_A^2 \plus{} P_B^2 \plus{} P_AP_B \plus{} P_BP_A \equal{} P_A \plus{} P_B$
$ P_A \plus{} P_B \plus{} P_AP_B \plus{} P_BP_A \equal{} P_A \plus{} P_B$
$ P_AP_B \plus{} P_BP_A \equal{} 0$

So, I simply assumed $ P_BP_A \equal{} P_AP_B$  :lol:
How can I proceed to show that $ P_AP_B \equal{} 0$ ??

By the way, $ < v \minus{} Pv,Pw > \equal{} 0$ is OK, right?

[quote="gunes"]Hmm, I was trying to show that if $ P_A \plus{} P_B$ is a projection operator ($ P_A$ and $ P_B$ are orthogonal projection operators), $ P_A$ and $ P_B$ are orthogonal.
$ P_A \plus{} P_B$ is a proj. op.

So, it should satisfy
$ (P_A \plus{} P_B)^2 \equal{} P_A \plus{} P_B$
$ P_A^2 \plus{} P_B^2 \plus{} P_AP_B \plus{} P_BP_A \equal{} P_A \plus{} P_B$
$ P_A \plus{} P_B \plus{} P_AP_B \plus{} P_BP_A \equal{} P_A \plus{} P_B$
$ P_AP_B \plus{} P_BP_A \equal{} 0$

So, I simply assumed $ P_BP_A \equal{} P_AP_B$  :lol:
How can I proceed to show that $ P_AP_B \equal{} 0$ ??[/quote]

You don't have $ P_BP_A \equal{} P_AP_B$, and you don't need it either.

Just think the following way: Every element of $ \mathrm{Im}P_A\cap\mathrm{Im}P_B$ must be $ 0$ due to $ P_AP_B \plus{} P_BP_A \equal{} 0$ (since $ P_A$ and $ P_B$ fix such elements), so $ \mathrm{Im}P_A\cap\mathrm{Im}P_B\equal{}0$, but on the other hand, $ \mathrm{Im}\left(P_AP_B\right)\subseteq\mathrm{Im}P_A\cap\mathrm{Im}P_B$ (again, this uses $ P_AP_B \plus{} P_BP_A \equal{} 0$), so that $ \mathrm{Im}\left(P_AP_B\right)\equal{}0$ and thus $ P_AP_B\equal{}0$.

[quote="gunes"]By the way, $ < v \minus{} Pv,Pw > \equal{} 0$ is OK, right?[/quote]

Yes.

  darij

I'm really sorry but I couldn't follow the below thing. For ex. In $ \mathrm{Im}P_A\cap\mathrm{Im}P_B$, are we talking about the set that imaginary parts of all elements of matrix $ P_A$ are in, or does Im refer to sth. else??
[quote="darij grinberg"]
Just think the following way: Every element of $ \mathrm{Im}P_A\cap\mathrm{Im}P_B$ must be $ 0$ due to $ P_AP_B \plus{} P_BP_A \equal{} 0$ (since $ P_A$ and $ P_B$ fix such elements), so $ \mathrm{Im}P_A\cap\mathrm{Im}P_B \equal{} 0$, but on the other hand, $ \mathrm{Im}\left(P_AP_B\right)\subseteq\mathrm{Im}P_A\cap\mathrm{Im}P_B$ (again, this uses $ P_AP_B \plus{} P_BP_A \equal{} 0$), so that $ \mathrm{Im}\left(P_AP_B\right) \equal{} 0$ and thus $ P_AP_B \equal{} 0$[/quote]

Thanks for all help, BTW.

Ouch. I didn't even expect this to be ambiguous...

What I called $ \mathrm{Im}A$ (for a linear map $ A$) is the image of $ A$, also called the range of $ A$, and also denoted by $ A\left(V\right)$ (where $ V$ is the vector space on which $ A$ is defined). It has nothing to do with the imaginary part of a complex number, which is called $ \mathrm{Im}$ as well. Sorry.

Note that I identify a matrix with the linear transformation which it defines. I hope you have no problems with this (as you seem to be doing this, too ;) ).

To be honest, I tend to work with the $ A\left(V\right)$ notation, but since you never introduced a name for the vector space, I used $ \mathrm{Im}A$ instead. I hope this makes things clearer, finally.

  darij

It's been 2 years since I took linear algebra course, sorry, I now recall what $ Im$ means :)
Also, you said
[quote="darij grinberg"]Every element of $ \mathrm{Im}P_A\cap\mathrm{Im}P_B$ must be $ 0$ due to $ P_AP_B \plus{} P_BP_A \equal{} 0$ (since $ P_A$ and $ P_B$ fix such elements) [/quote].

From $ P_AP_B\plus{}P_BP_A\equal{}0$ I can only reach the following:
$ P_Ab_i\plus{}P_Ba_i\equal{}0$ for $ P_A\equal{}[a_1 \quad a_2\quad ... \quad a_n]$ and  $ P_B\equal{}[b_1\quad b_2\quad ... \quad b_n]$
where $ a_i$ and $ b_i$ are $ n$ by 1 vectors. From this equation, I can say that some of the elements of $ Im(P_A)$ is associated with $ \minus{}Im(P_B)$ But I couldn't reach a conclusion about the whole range space of these matrices.

$ \mathrm{Im}\left(P_AP_B\right)\subseteq\mathrm{Im}P_A\cap\mathrm{Im}P_B$
I'm also wondering this part, too.  :blush:

I'm stuck here. If anyone's got an idea about what darij explained in the end, please reply  :(

Sorry for forgetting this topic (should egosearch more often).

[quote="darij grinberg"]Every element of $ \mathrm{Im}P_A\cap\mathrm{Im}P_B$ must be $ 0$ due to $ P_AP_B \plus{} P_BP_A \equal{} 0$ (since $ P_A$ and $ P_B$ fix such elements) [/quote]

More details: Every element $ t$ of $ \mathrm{Im}P_A\cap\mathrm{Im}P_B$ is fixed by $ P_A$ (since it lies in $ \mathrm{Im}P_A$, and $ P_A$ is a projection) and by $ P_B$ (similarly), thus also by $ P_AP_B$ and by $ P_BP_A$. Hence, $ \left(P_AP_B \plus{} P_BP_A\right)t \equal{} 2t$. But this is $ 0$...

[quote="gunes"]$ \mathrm{Im}\left(P_AP_B\right)\subseteq\mathrm{Im}P_A\cap\mathrm{Im}P_B$
I'm also wondering this part, too.  :blush:[/quote]

$ \mathrm{Im}\left(P_AP_B\right)\subseteq\mathrm{Im}P_A$, and $ P_AP_B \plus{} P_BP_A \equal{} 0$ yields $ \mathrm{Im}\left(P_AP_B\right) \equal{} \mathrm{Im}\left( \minus{} P_BP_A\right) \equal{} \mathrm{Im}\left(P_BP_A\right)\subseteq\mathrm{Im}P_A$.

Note that this proof (that if $ P_A$, $ P_B$ and $ P_A \plus{} P_B$ are projections, then $ P_AP_B \equal{} P_BP_A \equal{} 0$) works over any field of characteristic $ \neq 2$ and does not require these projections to be orthogonal projections!

A kind of generalization (not completely, because of the characteristic): Over a field of characteristic $ 0$, let $ P_1$, $ P_2$, ..., $ P_n$ be $ n$ projections such that $ P_1\plus{}P_2\plus{}...\plus{}P_n$ is also a projection. Prove that $ P_iP_j\equal{}0$ for all $ i\neq j$.
(Warning: this is much harder to prove.)

  darij

[quote="gunes"]Hmm, I was trying to show that if $ P_A \plus{} P_B$ is a projection operator ($ P_A$ and $ P_B$ are orthogonal projection operators), $ P_A$ and $ P_B$ are orthogonal.
$ P_A \plus{} P_B$ is a proj. op.

So, it should satisfy
$ (P_A \plus{} P_B)^2 \equal{} P_A \plus{} P_B$
$ P_A^2 \plus{} P_B^2 \plus{} P_AP_B \plus{} P_BP_A \equal{} P_A \plus{} P_B$
$ P_A \plus{} P_B \plus{} P_AP_B \plus{} P_BP_A \equal{} P_A \plus{} P_B$
$ P_AP_B \plus{} P_BP_A \equal{} 0$[/quote]
In fact, you have in general that $ x^2\equal{}x,y^2\equal{}y,xy\equal{}\minus{}yx$ implies that $ xy\equal{}0$ in any (not necessarily commutative) ring (doesn't need a unit, either) in which $ 2$ isn't a zero-divisor. The proof is rather short:
$ xy \equal{} \minus{}yx$ implies (square it) $ xyxy\equal{}yxyx$ implies (replace $ yx$ by $ \minus{}xy$ and vice versa) $ xxyy\equal{}yyxx$ implies (as $ x^2\equal{}x,y^2\equal{}y$) $ xy\equal{}yx$. Now just add $ xy\equal{}\minus{}yx$ to get $ 2xy\equal{}0$, giving the result.


When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed in a c.p. calorimeter, the temp. of the mixture increases from 22.30°C to 23.11°C. Calculate D H for this reaction, assuming that the combined solution has a mass of 100.0 g and a S.H. = 4.18 J/g °C.

I figured out that the answer is -68000J

AgNO3 + HCl ---> AgCl + HNO3

Now does +68000J go to the right side of the equation, with the products? This is probably a really stupid question but I am paranoid. (test tomorrow)

Yes, for exothermic reactions the energy released always goes on the right side, with the products. In exothermic reactions, energy (in the form of heat and/or light) always leaves the system.


Let $ABC$ be an isosceles triangle at angle $A$.Suppose that $M$ is a point on $BC$,$N$ is a point on $AC$ such that $\hat{MAC}=\hat{NBA}=30^{o}$.Caculate $\hat{MNC}$

Are you sure with your problem? The Geometer's Sketchpad shows that $\angle{MNC}$ changes when we change the measure of $\angle{A}$.


At tz's website you can get national sprints, but where do you get the answers to them. Could someone tell me where the answer keys are?
Thanks

Yeah, I noticed this too.  Also, can anyone get the correct state answers from 1999?  Thanks.

Why don't we just start posting problems in the same format that RManagers do for the Resources section?  Then I (or someone else) can compile it all into not just an answer key, but a solution document as well.

[quote="Zmastr"]What is that?  I don't know what the first sentence was, but if anyone can help out that knows more about computers than I do (probably most people) I would appreciate it.[/quote]

Umm... OK.  This is kinda hard to explain.  You see the tab that says contests up on the AoPS toolbar?  Those contests are compiled into a database from individual AoPS post that were formatted a special way.  Perhaps ask [b]joml88[/b] (Joe Laurendi); he knows more about it than I do.

AoPS was denied that by MATHCOUNTS.
That is why

For the national answer keys, i assume you are referring to the sprints that I posted. I don't have the answers, but maybe someone can make them.

[quote="biffanddoc"]i assume you are referring to the sprints that I posted[/quote]
yes I was

[quote="goldendomer"]Why did MathCounts deny AoPS?[/quote]

Maybe because it's bad for the foundation?  Think about it...  Once those problems get circulated you can't pull them back, so if they wanted to compile a list of problems, they'd be competing with a free database.

You'd have to ask whoever requested it on that.

[quote="goldendomer"]Then why do they allow us to scan problems and post them on websites?[/quote]

LOL.  That is actually illegal.  I have a feeling that as soon as MathCounts finds out about tz's website, it'll be shut down for good.

Do you guys have the answers to the 1999 state sprint and target rounds?  My answers are wrong, or are for the wrong competitions.

I do.
If I had access to the scanner today...

I have a scanner, but I need someone to teach me how to use it.  Seriously.  PM me if you have any idea.

1999 state? I'll post those. I have them.
Wait, where should I post them?

[quote="Zmastr"]Does anyone that has a scanner and the will and knowledge to scan it that can post the 1999 state answers?[/quote]

I have all but the last.  lol.

Klebian, if you have a link, post it in a Forum, or post the file name on a website and give the link for the website where you posted it.  I have no idea how to use LaTeX.

http://www.4shared.com/dir/257226/be0f945c/Mathcounts_States.html

That's my folder with Mathcounts states, if you can read, you [u]should[/u] be able to find the 1999 state test with answers.

I went over my formula sheet the day before States and I was fine.

But then again, when I do cramp the day before, the stuff I learn is usually never on the test  :lol:

hey Zmastr, if you go back to klebian's 4shared onlline space, he updated the 1996 sprint/target so that they do have the answers now.

[quote="pgpatel"][quote="Zmastr"]Thanks a lot! States in 4 days. Must.....study..... :wow:  :wow:[/quote]
DON'T study for 2 days before. If you do it is suicide. Trust me, I learned this the hard way. :([/quote]
Idk, I did 2 tests the day before and 3 tests 2 days before and I made nationals...
I found it better, when I was starting practice with nats tests, to just do the test with no worry about time to focus more on the problems themselves. But with 4 days, I don't know... XD

Thanks.  I needed a lot of those States.  I'll save them for next year!   :D

Klebian's s 1996 states, or nats?  Cause I already have the answers to the 1996 states.

I believe it was states. I didn't have 1996 national sprint, so there can't be answers...

[quote="pgpatel]DON'T study for 2 days before. If you do it is suicide. Trust me, I learned this the hard way. :([/quote] Dude, I've been studying for the past 5 weeks.  It's not like I'm cramming.  It's just I would give up a lung and a kidney to go the nats.  :D

I would give my right arm (heck why not my left too) for making countdown at nats.

To make Countdown?  Not just my lung and kidney, how about all motor skills?  I believe it would be worth it.

I wouldn't
1. you can't hit the buzzer
2. you can't go to any more competitions because you can't write

:|  You make a point. let's make it my left hand instead  :D

[quote="xxm0rg07hxx"]You could hit the buzzer with your head.  But that would be uncomfortable.  Plus you'd be in pain in a hospital probably at the time of countdown.[/quote] :rotfl: . THREE DAYS TO GOOOOOOOOOOOOOOOO

[quote="pgpatel"][quote="Zmastr"]Thanks a lot! States in 4 days. Must.....study..... :wow:  :wow:[/quote]
DON'T study for 2 days before. If you do it is suicide. Trust me, I learned this the hard way. :([/quote]

I did a Sprint round the day before the competition and I got second.

I did tests for about 2.5 hours straight before chapter and I got 6th, I played gamecube and got a good night's sleep before state and made it to Nationals

[quote="ehehheehee"]I played gamecube and got a good night's sleep before state and made it to Nationals[/quote]Nice.  Reaaaaaaaaaaal nice.  :rotfl:  :rotfl:  :rotfl:

Back on topic: so can anyone get nats answers to tz's website?

Thx

This is slightly off topic, but I can post nats here from 1999 and on including answers. Unfortunately I have written in some :blush: . I'll put up any that biff is missing and answers. Can someone convert it for me?

Yes. I can convert to pdf and upload on the website.

I just found by 2000 set so can you do 2005-2004-1999?

Um... I've written in some of them. Let me go check  :ninja: 23 hours....

I can do 04 but not 05. I've written in 99 so maybe I'll consider erasing and posting...

my coach probably has 2005, so I can do them.
As for 1999, if I can find it...


geometry
3D geometry
algebra unsolved
algebra

Let $ a>1$ be positive integer. $ \frac{a^2\plus{}1}{2}$ ($ a$ odd)  or  $ \frac{a^2\plus{}4}{4}$ ($ a$ even) can be perfect cube?

[quote="mayhem"]Let $ a > 1$ be positive integer. $ \frac {a^2 \plus{} 1}{2}$ ($ a$ odd)  or  $ \frac {a^2 \plus{} 4}{4}$ ($ a$ even) can be perfect cube?[/quote]
I really don't understand reza1370's solution...

We have to cases:
$ a$ even:
$ a \equal{} 2k, k \in \mathbb{N}$: $ \frac {a^2 \plus{} 4}{4} \equal{} k^2 \plus{} 1 \equal{} c^3, c \in \mathbb{N}_0$
Then we have $ k^2 \equal{} (c\minus{}1)(c^2 \plus{} c \plus{} 1)$ but $ (c^2 \plus{} c \plus{} 1, c\minus{}1) \equal{} (2c \plus{} 1, c \minus{} 1) \equal{} (3, c\minus{}1)$.
So we have to cases:
A) $ c\minus{}1 \equal{} m^2, m \in \mathbb{N}_0$ and $ c^2 \plus{} c \plus{} 1 \equal{} n^2, n \in \mathbb{N}_0$ $ \iff$ $ m^4 \plus{} 3m^2 \plus{} 3 \equal{} n^2$, but $ m^4 \plus{} 3m^2 \plus{} 3 \equiv 3 \bmod 4$ and thus there is no solution.
B) $ c\minus{}1 \equal{} 3m^2, m \in \mathbb{N}_0$ and $ c^2 \plus{} c \plus{} 1 \equal{} 3n^2, n \in \mathbb{N}_0$ $ \iff$ $ 9m^4 \plus{} 9m^2 \plus{} 3 \equal{} 3n^2$ $ \iff$ $ 36m^4 \plus{} 36m \plus{} 12 \equal{} 12n^2$ $ \iff$ $ (6m^2 \plus{} 3)^2 \plus{} 3 \equal{} 3(2n)^2$ $ \iff$ $ 3(2m^2 \plus{} 1)^2 \equal{} (2n\minus{}1)(2n\plus{}1)$. Since $ (2n\minus{}1,2n\plus{}1) \equal{} (2, 2n\plus{}1) \equal{} 1$ we have to cases again:
B1) $ 2n \minus{} 1 \equal{} 3p^2, p \in \mathbb{N}$, $ 2n \plus{} 1 \equal{} q^2, q \in \mathbb{N}$. But then $ q^2 \minus{} 3p^2 \equal{} 2$ $ \Rightarrow$ $ q^2 \equiv 2 \bmod 3$. Which is impossible.
B2) $ 2n \minus{} 1 \equal{} p^2, p \in \mathbb{N}$, $ 2n \plus{} 1 \equal{} 3q^2, q \in \mathbb{N}$. Thus we have $ 3q^2 \minus{} p^2 \equal{} 2$. This somewhat reminds me of pells equation. Can anyone finish this?

$ a$ odd:
$ a \equal{} 2k \plus{} 1, k \in \mathbb{N}_0$:  $ \frac{a^2 \plus{} 1}{2} \equal{} 2k^2\plus{}2k\plus{}1 \equal{} c^3, c \in \mathbb{N}_0$ $ \iff$ $ (2k\plus{}1)^2 \plus{} 1 \equal{} c^3$ Which is pretty much the same as $ a$ even above.


Find all polinomials with complex coefficients in modulus at most 1, which have real roots.

There are some problems with the question; we could just scale any polynomial with real roots down. Is there some other condition?

Yes, the polynomial ( :D ) must have the coefficient of the greatest power of $X$ equal to 1.(the polynomial must be monic). Sorry  :blush:


You can find that article [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration#Showing_Asymptote_error_messages_in_TeXnicCenter]here[/url]. This method also incorporates giving you error messages for errors in asymptote.

Except starting with 1.57 version you have to call asy.bat instead of asy.exe to make Asymptote work correctly on Windows, which has two effects: firstly, the DOS window doesn't close if there are errors and, secondly, the postprocessor thingy doesn't seem to work (for some reason, the batch files do not get executed as postprocessors in TeXnic Center).

Oh darn :\. I didn't know a new version came out already.


probability
induction
probability and stats

I have to prove this -

If $ X_1,...X_n$  are independent random variables so $ X_k\sim\Gamma(\alpha_k,\beta), k\equal{}1,2,\ldots,n$ .
Prove that $ Y: \equal{} X_1\plus{}X_2\plus{}\ldots \plus{}X_n \sim \Gamma(\alpha_1\plus{}\ldots\plus{}\alpha_n,\beta)$

Next, calculate density of $ \chi^2$-distibution with m degrees of freedom.

Thnx!!

Can be found in every normal book in Probability Theory.
The main idea is to use mathematical induction, Proof for $ n\equal{}2$ is based on fact that p.d.f. of sum of independent r.v. is a сonvolution of p.d.f of addendums.
Chi-square distribution with n degress of freedom is  a particulate form of gamma-distribution - $ \Gamma(\frac n 2,1/2)$.


At x approaches 0 , all of these functions 1/x , 1/x^2 , 1/ root x  becomes infinite. Which of these inc most rapidly?

The function $ f(x)\equal{}\frac{1}{x^2}$ will increase most rapidly followed by $ f(x)\equal{}\frac{1}{x}$ and then $ f(x)\equal{}\frac{1}{\sqrt x}$.

consider $ f/g$ and  use the L'hospital rule to find the limit when $ x\to\infty$

Consider $ f/g,$ yes, but [i]don't[/i] use L'Hôpital's rule. Be more elementary than that!


are you in a private or public school?
me:public

I am in university.

I'm a big kid now! :blush:

[quote="solafidefarms"]:P :)  :)  :) HOMESCHOOL FOREVER!  :)  :)  :)  :P[/quote]

Agreed.  Homeschool is the ultimate school.  No question about it.  Too bad the system owns me... :(

Just wondering.  How would you qualify for Mathcounts and other stuff while being homeschooled?

We find a parent who can be the coach, we raise money for the math club by asking for contributions from the members of Math Club, and then we tell everyone in the homeschool group about it. If we have enough money for a team and enough people(generally the people part is the harder part - homeschoolers don't generally do math stuff), we have a team. If we don't have enough people, we have some individuals.

This year our math club is having a full team and several individuals - we had several people more then last year eligible. :) Last year, when I was 8th grade, we had 5 people try out. This year we are expecting about 10. Last year our team [i]won[/i] chapter - we have high hopes for this year!

Homeschoolers don't usually compete in math though. :( They don't know about AoPS yet. Maybe if AoPS advertised in the homeschool magazines... I'm telling everyone in our Math Club about the AoPS forum!

Billy

public (forgot boarding school) and there are home schooling magazines and if i was homeschooled i would do a lot of math

Public..

My school is brand new, just opened this year..so there's like nothing here.no math club, nothing :wallbash_red:  :wallbash_red:  :mad:

really big public school, this year we had exactly 3999 students 9th-12th.

private if you mean by that a statal school...it sucks :ninja:

I ve been to many school (11) even though Im 14 I dont really see any difference in public or private. You still have to study yourself :P

Im in CAVA its a virtual acadamy im taking geometry in 8th grade and i get to talk to my teacher through elluminate and cool stuff like that.  You could call it public or homeschool, i call it homeschool.

I go to a public community college (BCC). I was homeschooled through freshman year in high school, though, and did only part-time public school as a sophomore.

Public Junior High and partial homeschool for me (I got extremely bored with the school math curriculum and now I'm homeschooling myself  :lol: )

We have about 900 students grades 7-9.

BTW, ZennyK, what does your BCC stand for?

Private School
really expensive
at least they let me skip grades in math, science, and spanish


topology
limit
calculus
calculus computations

$ x_n \equal{} \{1, \frac{1}{2}, \frac{1}{3} ... \frac{1}{n} , 0, 0, 0 ... \}$
Does this sequence have a limit as $ n 	\to \infty$ ? If it has, is it $ 0$?

Are you talking about the sequence of terms or about the sequence of sequences

$ x_1 \equal{} \{ 1, 0, 0, ... \}$
$ x_2 \equal{} \{ 1, \frac{1}{2}, 0, 0, ... \}$
...
$ x_n \equal{} \{ 1, \frac{1}{2}, ... \frac{1}{n}, 0, 0, ... \}$?

If the latter, then the answer depends on the topology; its [url=http://en.wikipedia.org/wiki/Uniform_convergence]uniform limit[/url] is $ \{ 1, \frac{1}{2}, ... \frac{1}{n}, \frac{1}{n\plus{}1}, ... \}$, which is the same as its $ L^p$ limit, $ p > 1$, but it does not have an $ L^1$ limit.

Sorry for confusion, I'm asking for sequence of terms

Then the notation you want is just $ x_n \equal{} \frac{1}{n}$.  The limit of this sequence is $ 0$; do you know the epsilon-delta definition of a limit?

Yeah, I know. Actually, I was curious about the following example.
If a finitely nonzero sequence $ x_n \equal{} {1,2...,n,0,0,0...}$ has a limit or not?

Again the ambiguity.

I suppose you mean the sequence $ (x_n)$ with

$ x_1 \equal{} 1$, $ x_2 \equal{} 2$, i.e. $ x_n \equal{} n$ for $ n \in \mathbb{N}$.

This sequence tends to infinity, so no.

But if you mean what you wrote (the sequence $ (x_i)_{i \in {\bf N}}$ for which $ x_i \equal{} i$, $ 1 \leq i \leq n$, $ x_i \equal{} 0$ for $ i > n$), then of course $ \lim_{i \to \infty} x_i \equal{} 0$.  (For any $ \varepsilon > 0$, take $ N \equal{} n \plus{} 1$ in the definition.)


calculus
integration
trigonometry
real analysis
real analysis unsolved

Can you prove the following identities?

$ \int_{\minus{}\infty}^\infty {a\choose b\plus{}x}\, dx\equal{}\sum_{k\in\Bbb{Z}} {a\choose b\plus{}k}\equal{}2^a$

$ \int_{\minus{}\infty}^\infty {a\choose b\plus{}x} {c\choose x}\, dx\equal{}\sum_{k\in\Bbb{Z}}  {a\choose b\plus{}k}  {c\choose k}\equal{} {a\plus{}c\choose b\plus{}c}$

$ \int_{\minus{}\infty}^\infty {a\choose b\minus{}x} {c\choose x}\, dx\equal{}\sum_{k\in\Bbb{Z}}  {a\choose b\minus{}k}  {c\choose k}\equal{} {a\plus{}c\choose b}$

Or can you show that for $ p\in\{2,3\}$

$ 2\int_{\Bbb{R}} \cos\pi x\; {2a\choose b\plus{}x}^p\, dx\equal{}\frac{(pa)!}{a!^p}\cos(a\minus{}b)\pi\equal{}\sum_{n\in\Bbb{Z}} (\minus{}1)^n {2a\choose b\plus{}n}^p$

$ 2\int_{\Bbb{R}} \sin\pi x\; {2a\choose b\plus{}x}^p\, dx\equal{}\frac{(pa)!}{a!^p}\sin(a\minus{}b)\pi$

After Kummer's Theorem $ \,_2 F_1(\alpha,\beta;1\plus{}\alpha\minus{}\beta;\minus{}1)\equal{}\frac{\Gamma(1\plus{}\alpha\minus{}\beta)\, \Gamma\left(1\plus{}\frac{\alpha}{2}\right)}{\Gamma(1\plus{}\alpha)\, \Gamma\left(1\plus{}\frac{\alpha}{2}\minus{}\beta\right)}$ .

This is equivalent to $ \sum_{n\equal{}0}^\infty (\minus{}1)^n \, \frac{\Gamma(\alpha\plus{}n)\, \Gamma(\beta\plus{}n)}{n!\, \Gamma(1\plus{}\alpha\minus{}\beta\plus{}n)}\equal{}\frac{\Gamma\left(\frac{\alpha}{2}\right)\, \Gamma(\beta)}{2\cdot \Gamma\left(1\plus{}\frac{\alpha}{2}\minus{}\beta \right)}$ .

Since $ {a\choose n}^2\equal{}\frac{a!^2}{n!^2\, (a\minus{}n)!^2}\equal{}\frac{a^2}{\Gamma^2(1\minus{}a)}\, \frac{\Gamma^2(\minus{}a\plus{}n)}{n!^2}$ we can write

$ \sum_{n\equal{}0}^\infty (\minus{}1)^n \, {a\choose n}^2\equal{}\frac{a^2}{\Gamma^2(1\minus{}a)}\sum_{n\equal{}0}^\infty (\minus{}1)^n \frac{\Gamma(\minus{}a\plus{}n)\, \Gamma(\minus{}a\plus{}n)}{n!\, \Gamma(1\minus{}a\plus{}a\plus{}n)}$

$ \equal{}\frac{a^2}{\Gamma^2(1\minus{}a)} \, \frac{\Gamma\left(\minus{}\frac{a}{2}\right)\, \Gamma(\minus{}a)}{2\, \Gamma\left(1\plus{}\frac{a}{2}\right)}\equal{}\frac{\Gamma\left(1\minus{}\frac{a}{2}\right)}{\Gamma(1\minus{}a)\, \Gamma\left(1\plus{}\frac{a}{2}\right)}\equal{}\cos\frac{a\pi}{2}\; \frac{a!}{\left(\frac{a}{2}\right)!^2}$ .

Hence $ \frac{(2a)!}{a!^2}\, \cos a\pi\equal{}\sum_{n\in\Bbb{Z}} (\minus{}1)^n {2a\choose n}^2$ .

Particularly $ \sum_{n\equal{}0}^\infty (\minus{}1)^n {\minus{}\frac12 \choose n}^2\equal{}\frac{1}{\sqrt{2}}\,\frac{\left(\minus{}\frac12\right)!}{\left(\minus{}\frac14\right)!^2}\equal{}\frac{\sqrt{2\pi}}{2\,\Gamma^2\left(\frac34\right)}$ .

This also confirms the second hypergeometric series in that [url=http://www.mathlinks.ro/viewtopic.php?p=1638610#1638610]topic[/url].