Hurricane Ike is approaching the coast, and according to the National Hurricane Center, my area (Houston) gets a direct hit. We're out of range of the storm surge (thankfully!), but we still will get high winds, so do you think 100 mph winds are very bad?

I've never been in 100mph winds but I can't imagine it being a fun situation.  I don't think such winds are powerful enough to flip cars or anything but it's probably sufficiently strong to take people off their feet.  At least you won't have to deal with the surge.

[color=green]Moderator says: There is never a good reason to quote the entire post immediately preceding yours.[/color]

i really really really doubt that

and the winds gonna be around 70-80 mph...i cant wait... :lol:

The terminal velocity of a skydiver with closed parachute is about 124mph, the predicted wind speed is somewhat short of that so big fat men will stay on their feet if they lean against the wind at an appropriate angle :P, but small children may get dragged though, probably, not airborne (you need a vertical wind for that) and you'll see some dragged cats for sure (they need just 60mph, so unless they are clever enough to find a shelter before the wind rises, they have no chance to stay on their feet). 100mph wind is not a joke, really :(.

Hurricane Local Statement for IKE (from http://www.nhc.noaa.gov/text/WTUS84-KHGX.shtml )
[quote]STRUCTURAL DAMAGE WILL INCLUDE THE MAJORITY OF MOBILE HOMES BEING
SEVERELY DAMAGED. THOSE THAT SURVIVE WILL BE UNINHABITABLE UNTIL
REPAIRED. HOUSES OF POOR TO AVERAGE CONSTRUCTION WILL HAVE MAJOR
DAMAGE...INCLUDING PARTIAL WALL COLLAPSE AND ROOFS BEING LIFTED
OFF. MANY WILL BE UNINHABITABLE. WELL CONSTRUCTED HOUSES WILL
INCUR MINOR DAMAGE TO SHINGLES...SIDING...GUTTERS...AS WELL AS
BLOWN OUT WINDOWS. UP TO ONE QUARTER OF GABLED ROOFS WILL FAIL.

PARTIAL ROOF FAILURE IS EXPECTED AT INDUSTRIAL PARKS...ESPECIALLY
TO THOSE BUILDINGS WITH LIGHT WEIGHT STEEL AND ALUMINUM
COVERINGS. OLDER LOW RISING APARTMENT ROOFS MAY ALSO BE TORN
OFF...AS WELL AS RECEIVING SIDING AND SHINGLE DAMAGE. UP TO ONE
QUARTER OF ALL GLASS IN HIGH RISE OFFICE BUILDINGS WILL BE BLOWN
OUT. AIRBORNE DEBRIS WILL CAUSE DAMAGE...INJURY...AND POSSIBLE
FATALITIES.

NATURAL DAMAGE WILL INCLUDE ALL TREES WITH ROTTING BASES BECOMING
UPROOTED OR SNAPPED. NEARLY ALL LARGE BRANCHES WILL SNAP. BETWEEN
ONE QUARTER AND ONE HALF OF HEALTHY SMALL TO MEDIUM SIZED TREES
WILL BE SNAPPED OR UPROOTED...MOST COMMON WHERE THE GROUND IS
SATURATED. UP TO THREE QUARTERS OF NEWLY PLANTED GROUND CROPS
WILL BE DAMAGED.[/quote]  Mandatory evacuations in several counties, etc.

Hurricanes cause death and destruction in the following ways:

1. Storm surge: the rise of ocean levels as water is pushed ahead of the storm winds.
2. Wind damage: the direct effect of the main circulation winds of the storm on buildings and trees.
3. Freshwater flooding caused by the rains. In rugged areas (like Haiti) this includes landslides and mudslides.
4. Tornados spawned by the hurricane's spreading over land.

The big killers are #1 and #3. It was storm surge that devastated the Mississippi coast during Katrina (and during Camille 30-some years earlier). It was an indirect, inland effect of storm surge that caused such death and misery in New Orleans during Katrina. It was storm surge that killed at least 8000 people in Galveston in 1900.

Among all large metropolitan areas in the U.S., the Houston-Galveston area is very likely the second most vulnerable to storm surge - not downtown Houston, or its west or southwest sides, but the suburbs east and southeast of the city along the Ship Channel and Galveston Bay. (The most vulnerable city remains New Orleans.)

This storm's winds won't be all that high. They'll last a long time, and be damaging for that reason, but wind damage won't be the big story. It is, after all, "only" a category 2 storm in peak wind velocity. But it's a very strange storm - very, very large, and with the wind field spread over a very large stretch of ocean. The result is that it has set a very large amount of water in motion, and the storm surge bearing down on Galveston, Galveston Bay, and (further east) the Beaumont-Port Arthur area, is more like the storm surge you'd get from a category 4/5 storm. There is a great deal to be afraid of there - and it's already too late to get off of Galveston Island.

I don't know where the lotsofmath is from, but to get a sense of what parts of Houston AoPS members might be from, the 2008 USAMO qualifiers were from Bellaire (inner southwest), Sugar Land (outer southwest), and Spring (outer northwest), while in previous years I remember qualifiers from Katy (outer west). Those are all locations that are probably at an elevation of 50-150 feet above sea level and well away from the Gulf and the bay. But if any of our members live in Texas City or Clear Lake City or Baytown or some place like that - I hope you're not home to read this!

It's coming very near my home! How exciting!   :jump:  <== proof of my immaturity *

I remember when it flooded at my cousin's house in China. That was pretty exciting as well. **

*We're near but sufficiently far outside the zone for evacuation for Ike.
**We were on the third floor of an apartment building.

I hear it's fading out a bit, but according to MSN, it is like "a bomb exploded". The Weather Channel also stated that it could be as bad as Katrina, but--more people are prepared this time.

Earlier in the week, Hurricane Ike was predicted to hit near where I live, but I am incredibly fortunate that it turned toward the Galveston area (I live about 150 miles southwest of Galveston).  I do feel really bad for those people affected by Ike, though.  It is estimated that over 4 million people are without electricity--simply incredible.

Just two weeks ago (during Labor Day weekend), my family visited Galveston on vacation.  It's sobering to think that it'll be quite some time before it returns to the way it was when we were there.

I got power back at 8:05 pm last night.

My area got about 9 inches of rain and winds about 80 mph with gusts up to 100 mph. (I was seriously worried the tree in our backyard was going to fall on the house!)

You got power back last night?  That's quick.  I heard on the news this morning that some people could have to wait weeks or months, although I think they were referring to the immediate Galveston area.  Early estimates are 8 billion in damages.  Just for comparison purposes, Katrina was over 80.

The new estimated price of damage....
Around 10 billion, and at this rate it could double.

i got my power back last night around 9, lost it this morning around 7, got it back this afternoon around 3...

katy nvr lost power... :dry:

Just a random question, how is all this about math?

or is that why its in the "other topics" section?

yeah....thats why its called other topics....


A license plate consists of two letters followed by two digits; for
example, $ MP78$. Neither the digits nor the letters may be
repeated, and neither the letter $ O$ nor the digit $ 0$ may be
used. When reading from left to right, the letters must be in
alphabetical order and the digits must be in increasing order.
How many different license plate combinations are possible?

There are $ \binom{26}{2}\equal{}325$ ways to pick the letters and $ \binom{10}{2}\equal{}45$ ways to pick the numbers.  Each set of two letters has exactly one way to be put in alphabetical order; the same is for the digits.  Thus, we just multiply $ 325 \times 45\equal{}\boxed{14625}$.

WAIT A MOMENT. the letter O and the number 0 can be used. So it would be:

25C2=300*36(9C2)=10800 right?

[quote=nav_murali]As per the problem, the letters and numbers have to be in ascending order. How can we just use Choose for this problem without considering the order?[/quote]

It doesn't matter how we choose them because there is only one way to put them because you have to put them in alphabetical or numerical order.

I solved it another way that was a lot less simple but still got it right

[hide=sol] There are $9+8+7+6+5+4+3+2+1 = 45$ ways to pick numbers, since the amount of numbers in increasing order lessen by one every time the number increases. This is the same for the letters, so it would be $25+24+23+...+2+1 = 325$. Multiplying these together, we get $45\times325 = 14625
$

[quote=BanSpeedrun]I don't understand..
Why wouldn't FA32 be okay? What prevents that?[/quote]

[quote=GameBot]A license plate consists of two letters followed by two digits; for
example, $ MP78$. Neither the digits nor the letters may be
repeated, and neither the letter $ O$ nor the digit $ 0$ may be
used. [b]When reading from left to right, the letters must be in
alphabetical order [/b]and the digits must be in increasing order.
How many different license plate combinations are possible?[/quote]

Alternatively, you could note that in half the cases, the letters and in non-alphabetical order and the other half the letters are in alphabetical order. The same goes for the digits. Thus, we could simply calculate $25\cdot24\cdot9\cdot8$ and divide by $2\cdot2=4$ to get $\boxed{10,800}$.


If two regular hexagons are inscribed in a circle of unit radius,then find the minimum area common to the hexagons .

thankyou.

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=327196[/url]


$x^2+y^2=1 \Longrightarrow P=\frac{2(xy+y^2)}{2xy+3x^2+y^2}$ . Put $t=\frac xy$, and you have to study $P(t) = \frac{2(t+1)}{3t^2+2t+1},\forall t\in R$

 :cool:  Finally $1-\sqrt{\frac 32} \leq P \leq 1+\sqrt{\frac 32}$  :cool:

[quote="monster_of_god"]what about $x^2+y^2=1$[/quote]
I think we can try trig substitution.


(a) The complex number $ z=x+iy$, where x and y are real, satisfies the parametric equation $ z=1+2i+t(3-4i)$ where t is a real parameter

(i) Show that the locus P which represents z in an Argand diagram has equation $ 4x+3y-10=0$

(ii) FInd the min. value of |z| and the corresponding value of z in the form a+ib, where a and b are real

----------------------------

(b) The equation z^3 - 1 =0 has a real root 1 and non-real roots $ w,w^{2}$

(i) Show that $ w^{3}=1$ and $ 1+w+w^{2}=0$

(ii) Find the roots of the eqn $ \frac{{1-z}{z})^}{3}= 1$, expressing the two non-real roots in the form $ Aw$ and $ Bw^{2}$, where A and B are real constants whose values are to be determined

[quote="bos1234"](a) The complex number $ z=x+iy$, where x and y are real, satisfies the parametric equation $ z=1+2i+t(3-4i)$ where t is a real parameter

(i) Show that the locus P which represents z in an Argand diagram has equation $ 4x+3y-10=0$

(ii) FInd the min. value of |z| and the corresponding value of z in the form a+ib, where a and b are real

[/quote]
[hide="solution (a)"]
i) $ z$ can be written as $ z=1+3t+i(2-4t)$
So, $ x=1+3t, \ y=2-4t$
Eliminating $ t$ from the two equations we obtain $ 4x+3y-10=0$

ii) $ |z|=\sqrt{(1+3t)^{2}+(2-4t)^{2}}=\sqrt{(5t-1)^{2}+4}$
The minimum is obtained for $ t=\frac{1}{5}$ and is equal to $ 2$.
In this case $ z=\frac{8}{5}+\frac{6}{5}i$[/hide]


Find the sum of the solutions of the equation $ \minus{}32x^2 \plus{} 84x \plus{} 135\equal{}0$.

Using Vieta's, we get $ \minus{}\frac{84}{\minus{}32}\equal{}\frac{21}{8}$.


function
algebra
polynomial
induction
algebra proposed

function $f(x,y):R*R\rightarrow R$,if x is constant then f(x,y) is polynomial of y and for y the property is the same prove or disprove that f(x,y) is the plynomial of x,y.

I happen to know this problem and I don't find it easy at all. It depends where is defined this function. As far as I know, if we work in numerable fields, the answer is no, it doesn't imply that. But if we work on fields like R, then the answer is yes. A very nice and hard problem.

I have invented function f:QxQ->R which is non-polynomial and f(x,.) and f(.,y) are polynomial.

I might be wrong, but I think there's a recent article in Gazeta Matematica about this. I don't know how recent, but it just looks familiar.

Let $q_n$ be sequence of all rationals. We define
\[
f(x)=\sum_{n=1}^{\infty}\prod_{k=1}^n (x-q_k)(y-q_k).
\]

oh! Myth maybe I'm idiot ,you disprove the problem :( but I saw realy incredible solution to the problem :(  :maybe:  :blush:  :P

No, he doesn't disprove it. He confirms what I have said: you can define it for QxQ, as Myth created. But you cannot if it is RxR. And there is no solution yet written here for this part. There is no contradiction!

Can someone please explain Myth's example? I'm having a bit of trouble understanding why it works :?

how much can some be idiot :( I'm the first :blush: excuse me for misunderstanding the example.

Suppose $ p_*(x,y)=p(x,y)-p(x,0)$ $p_*$ has the problem properties.
Now Suppose $p_1(x,y)=p_*(x,y)/y$ Easily $p_1(x,y)$ has problem properties.
And For each $x$ degree of $p_1(x,y)$ is less than degree of $p(x,y)$
With induction it's know obvious(Induction on degree of $p(x,y)$ for constant $x$) :P  :P  :P

[quote="Omid Hatami"]
And For each $x$ degree of $p_1(x,y)$ is less than degree of $p(x,y)$
With induction it's know obvious(Induction on degree of $p(x,y)$ for constant $x$)[/quote]
And we got nothing. Your proof shows that my example for f:QxQ->R doesn't exist!

I have completed Hatami's idea.
It is clear $\exists N\in\mathbb{N} \exists \{x_n\}_{n=1}^\infty\ :\ deg(f(x_n,y))\leq N$.
Now apply Hatami's trasformation N times. As the result we get $f^*(x,y)$ with the same property as $f(x,y)$, but for all $x_n$ we have $f^*(x_n,y)\equiv 0$(!!!).
It follows that for each $y$ polynomial $f^*(x,y)$ has infinitely many roots $x_1,x_2,...$, hence $f^*(.,y)=0$ for each $y$.


topology
real analysis
real analysis unsolved

In the following text overlining indicates the closure of a set.

I have to prove the propostion:

Let $A$ be an open subset of a metric space $E$. Then we have for every set $ B \subseteq E $:

$ A \cap \bar B \subseteq \overline{A \cap B} $.

But what, if I take $ A = \mathbb{Q} \, , \, B = \mathbb{R} \backslash \mathbb{Q} $ and $ E = \mathbb{R} $ ?

Then $ \bar B = \mathbb{R} $ and:

$ A \cap \bar B = \mathbb{Q} $, but $ \overline{A \cap B} = \varnothing .$ ($ \varnothing $ is the empty set)

I think there must be something wrong with my example, but where is the error?

Is $\mathbb Q$ an open subset of $\mathbb R$? :)

Your $A$ is not open.

P.S. Oops, grobber was faster to reply  :) .

As far as I know, it is. Or does your question mean, that it isn't.

So, after being told that $\mathbb{Q}$ is not open in $\mathbb{R}$ I have another question:

Can you find two open sets $ A, B \subseteq \mathbb{R} $, so that $ A \cap \bar B, \bar A \cap B, \bar A \cap \bar B, \overline{A \cap B} $ are all distinct from each other?

Do you consider this last question being too trivial for an answer? For me it is not trivial, so can you help me with this?


function
absolute value
number theory proposed
number theory

${c_{n}\in \mathbb{Z}}$
${\sqrt{u}=c_{1}}$

A number $u$ is a perfect square root, and $k$ numbers after the square root of $u$ is a perfect square root too

${\sqrt{\sqrt{u}+k}=c_{2}}$

$k$ numbers after the square root of $k$ numbers after a perfect square root of $u$ is a perfect square root too.

${\sqrt{\sqrt{\sqrt{u}+k}+k}=c_{2}}$

To simplify, given a function

${f(n)=\sqrt{k+f(n-1)}}$
${f(1)=u-k}$

What is the biggest $n$ such as $f(x)$ is a perfect square root for all $2 \leq x \leq n$ $\land$ $u > 0$ $\land$ $k \neq 0$ $\land$ $k \neq 2$ $\land$ $f(x) \neq 0$?

If $k=2$ the solution is infinity. but. $k \neq 2$

so, what's is the biggest $n$?

It is infinite for any integer $k$.

Let $g(0)=m>|k|$ be positive integer. Define an infinite sequence of positive integers $g(1), g(2), \dots$ by the following recurrent formula:
\[g(j+1)=g(j)^{2}-k \]
Now let's take $n$ as large as we like. And define integers $f(i)=g(n-i)$ for $i=1,\dots,n.$ It is easy to see that $f(i)$ satisfy the required property: $f(i+1)=\sqrt{k+f(i)}.$

i dont understand your point

because i think there is not high solting for this

i try brute force i did not find

and your proof  dont proof $f$ is integer

or i dont understand

can you remake this and proof that $f(n)$ is a integer number for some $n$

and i forgot to tipo $|k| \neq 1$

EDIT: Sorry

you are rigth

now i understand

there is infinitely many solutions

but i cannot say if this is for all k

sorry for the trash of the last posts

just now i realise that g is the inverse process of f

osrry

you are rigth

sorry

and thanks

a god and easy proof

suppose

$f(x) = \frac{1}{k+f(k-1)}$
$f(0) = m$

there is integer solution for this for larges $x$ such as for all $n>x$  $f(x-n)$ is integer too!?

for $k$ different of zero  and the absolute value of $f$ different of 1

for solving this i can find a $g$ that is the inverse of this, isnt it?

and proove or disproove it

[quote="Thales418"]$f(x) = \frac{1}{k+f(k-1)}$[/quote]
Probably, you meant $f(x) = \frac{1}{k+f(x-1)}.$
There is no much freedom here. If from integer $f(x-1)$ we want to obtain integer $f(x)$ by this formula, the denominator $k+f(x-1)$ must be $\pm 1,$ implying that $k=0$ and $f(1)=f(2)=\dots=\pm 1.$

i give a poor example

more general

i have a function $f$ always integer

so, this function $f$ is definied with recurrence

suppose this is a symply recurrence using just $f(n+1)$ and $f(n)$

so, if i solve this for $f(n)$ and  make a substituition

$g(x) = f(n)$
$g(x-1) = f(n+1)$

im not equaling this equations ,just make a substituition

this is what you made

to proof f is integer we just need to proof g is integer

it is true for all f and g?


Im using Internet Explorer and Windows XP . My computer has some pop up blocker or something like that. Even though ive enabled pop ups from mathlinks and aops, they still dont seem to pop up. Any ideas and suggestions on what i should do?

Thanks :)

Sorry, i dont get you. Ctrl + click on what?

I'll say it again. It doesnt even pop up but says pop ups were allowed .

Well I'm guessing that you are clicking on a link? (Sorry if you're not) But if you are, if you hold down the ctrl key while clicking, it automatically allows the popup on PCs.

No, im referring to pms and the same sort. When i click on a link, it comes but pms dont pop up for me :( ive enabled them in my profile too.

Also, im not able to view the latex codes if i click them. They dont pop up but it says pop ups were allowed, the same happens for pms  :)


Find the remainder when 5^99 is divided by 13.

Notice that $ 5^2 \equiv \minus{}1 \mod 13$. We find that $ 5^{99} \equiv \minus{}5^{99 \mod 2} \equiv \minus{}5 \equiv 8 \mod 13$

Therefore, the remainder is $ \boxed{8}$.

EDIT: Thanks Walk.

If you want to use binomial theorem to get the answer, write $ 5^{99}$ as $ (26\minus{}1)^{49}.5$


AMC
USA(J)MO
USAMO
summer program
Mathcamp
Putnam
AIME

Found this in another national form, for any one interested:

1. What town do you live in?
2. What grade are you in?
3. What's your favorite color?(yer kinda random i know)
4. What's your favorite math subject?
5. What's your favorite math problem on AOPS?
6. How did you find this website?
7. What contests (preferrably math-related) have participated in recently?

Feel free to do this free form, you don't have to stick with the questions I posted. I actualy haven't even read all seven, just a cut and paste.

1) WATERLOO!!!!!
2) Grade?  bwahahah.  I'm DONE with high school.  At the time this post was made, I was just finishing up my 1b term at Waterloo.  Maybe by the time you read this, I will have got further.  Or given up, spared myself the pain, and got a job at McDonalds.
3) Colour?  Red.  It's so powerful, passionate, seductive, and well... red.
4) Math subject?  heheh... I like ad hoc, strange, interesting problems, like USAMO 2005 #6... those kind of problems that don't really fit into a genre.  These problems are never bashable, and always require a good amount of insight to solve.  If you forced me to pick my favourite, I'd probably say functional equations, because they tend to have really cool solutions that don't require a lot of knowledge, but rather rely on the brain of the solver to engineer the solution.
5) See above.
6) I don't remember at all... going to all the mathcamps and having tons of math people on my MSN list, SOMEBODY must have first introduced me to this place, but I have no idea who.
7) Ummm... well... Putnam?!?!  LOL.  That was a while ago, but for me, that's my most recent... as for high school stuff... there was the IMO for math and the USAICO for CS.  I also did the bernoulli trials and the Special K, which are Waterloo-only contests.

1. I live in Waterloo.
2. I am going to grade 11.
3. My favourite color is red.
4. My favourite math subject is algebra.(specifically, functional equations)
5. Turkey TST day 1 number 3 is the best problem ever:
If $x,y,z$ are positive real numbers and $xy+yz+zx=1$ prove that
$\frac{27}{4}(x+y)(y+z)(z+x) \geq ( \sqrt{x+y}+\sqrt{ y+z}+\sqrt{z+x})^{2}\geq 6 \sqrt 3$
6. I found this thing when searching for old USAMO and AMC problems.
7. I participated in the Canadian Mathematics Competitions(Euclid, Fermat, COMC, Hypatia) and the olympiads(APMO, CMO, USAMO)

[quote="carpo"]1) Vancouver is a city.
2) Grade 12 next year.
3) Different shades of brown??
4) Number Theory
5) Something I can solve?
6) Google at first, I think. Then C3PO "highly suggested" we all go join at a math camp.
7) Going backwards: USAMO, CMO, AIME, AMC12, COMC. Somewhere in there I wrote CEMC contests starting with E and F and H but they have lowercase letters which is lame.

Did you learn more about me from my responses or the fact that I responded?[/quote]

Euclid, Fryer, Hypatia? hmmm, is that them?

I should answer these questions myself:

1)Currently Toronto, Waterloo Starting in september.
2)See above.
3)Blue
4)My strongest is counting ( not by much ). But algebra/inequalities are damn awesome when you know how to do them.
5)Jessica is studying combinatorics during a 7-week period. She will study a positive integral number of hours every day during the 7 weeks (so, for example, she won't study for 1.5 hours), but she won't study more than 11 hours in any 7-day period. Prove that there must exists some period of consecutive days during which Jessica studies exactly 20 hours.
6) Alex Wice said to me after a computer competition: "You can program better than me. You could likley do math if you tried." He then later suggest the aops books, and I found the site and them at the same time.
7)Euclid, AMC12, COMC, a bit of olymon. Looking forward to the putnam!

[quote="boxedexe"]Fermat is the correction to be made.[/quote]
What do you mean by this post? :maybe:

rzsolt thought that carpo wrote Fryer, but that is not possible.  He wrote Fermat.

Masoud Zargar


The measure of angle B in triangle ABC is 60. altitudes CD and AE intersects at point H. Prove that the circumcenter O is on the angle bisector of angle COE

This one comes together pretty nicely after noticing [hide]since we have a 60 degree angle, some perpendiculars, and some midpoints (created by perpendiculars from the circumcenter), we have a bunch of 30-60-90 right triangles and a bunch of equal lenghts caused by the 1:sqrt3:2 ratios. The rest is simple.[/hide]

[quote="zabelman"]Im pretty sure you mean the angle bisector of CHE.[/quote]

Guevara?

I think different that Zabelman, but at least more fully explained:

[hide]O is the intersection of the perpendicular bisectors of the sides.  Let the midpoint of AB be M, the midpoint of BC be N.  The altitude and perpendicular bisectors to the same side of a triangle are parallel.  Thus, the distance from O to CH is the length of MD and the distance from O to HE is the length of NE.  From 30-60-90 triangle ABE, BE = AB/2 and from 30-60-90 triangle CBD, BD = BC/2.

BM = AB/2 and BN = BC/2.

Thus MD = NE = |AB - BC|/2

Thus O is equidistant from the two lines.

Thus O is on their angle bisector.[/hide]

On second thought, probably just about the same as Z's.

Quote:zabelman wrote:
Im pretty sure you mean the angle bisector of CHE.

Guevara?

Lol.

[quote="JBL"]
Lol.[/quote]

Unfortunately, my geometry skills have been rather neglected, and pithy comments are about all the contribution I can make to geometry problems.

suppose  CD  intersect  circumcircle  at  M , AE intersect  circumcircle  at N ,
Then  ∠MBC=∠MBA+∠ABC=∠ACM+∠HCN=∠ACN  ,gained  AN=CM

AOC is isosceles triangle with [tex]<AOC=2<B=120, <CAO=<ACO=30[/tex].
Also [tex]<AHC=120[/tex], so AHOC is cyclic.Then: [tex]<CHO=<CAO=30=\frac{<CHE}{2} [/tex].Therefore O is on the bisector of [tex]<CHE[/tex].  QED.

Let $CD$ and $AE$ intersect the circumcircle of $\triangle ABC$ at $M$, respectively $N$. Easily $MN=AC$, making $CAMN$ an isosceles trapezoid, i.e. $CM=AN$, thus they are at the same distance from the circumcenter, done.

Best regards,
sunken rock


function
logarithms
algebra unsolved
algebra

What is the coefficient of $ x^{n\minus{}3}$ in the power series expansions of

$ \frac{2}{x\plus{}1}  \minus{} \frac{x\ln x}{x\plus{}1}$

Thanks,
Spai


In the course Precalculus, how many challenge sets are there? Is it one every 3 weeks like in the class I'm currently enrolled in?
Is it the same in general with all subject classes?

i'm not sure, but probably  5 challenge sets
or maybe 2.

Take a look at the course intro document on the My Classes page for precalculus.  It has a schedule of the due dates for the 6=18/3 challenge sets.  They are due the dates of classes 4, 7, 10, 13, 16, and 18+1 (one week past the last class).


Show that if $ 2^n \minus{} 1$ is a prime number then $ n$ must be also prime.
Just below there was a hint: Use that $ \frac{2^{ab} \minus{} 1}{2^a \minus{}1} \equal{} 2^{a(b\minus{}1)} \plus{} 2^{a(b\minus{}2)}\plus{}...\plus{}1$.

Too simple.

Assume on contrary that $ n$ is composite number. Then $ n\equal{}dk$ so:

$ 2^{n}\minus{}1\equal{}(2^{d})^{k}\minus{}1\equal{}(2^{d}\minus{}1)(2^{d(k\minus{}1)}\plus{}2^{d(k\minus{}2)}\plus{}...\plus{}2\plus{}1)\equal{}A.B$ where obviously $ A$ and $ B$ are positive integers greater than 1 which i scontradiction to assumption of problem that $ 2^{n}\minus{}1$ is prime.


I have a very nice solution for the following one:
  If a^2+b^2=c^2+d^2=1, then sqrt(1+a)+sqrt(1+c)+sqrt(ac-bd+1)>=sqrt(2). What about you?

Maybe the trigonometric substition can help?

Namdung

Maybe it works, but my solution is based on something else. Anyway, a inequality is beautiful when it has lots of ways of solving, without being trivial. I will wait a little bit to see if someone finds my solution.

I try to solve it

We have  \sqrt(1+a) + \sqrt(1+c) >= \sqrt(2+a+c)    (1) (trivial!)
And also  \sqrt(1+ac-bd) = \sqrt(a^2+c^2+2ac+b^2+d^2-2bd)/2 = \sqrt((a+c)^2+(b-d)^2)/2 >= |a+c|/\sqrt(2).                  (2)
Now, we will prove that  \sqrt(2+a+c) + |a+c|/\sqrt(2) >= \sqrt(2)  (3)
Indeed, (3) is equivalent to
           \sqrt(4+2(a+c)) >= 2 - |a+c|
<=>            4 + 2(a+c) >= 4 - 4|a+c| + |a+c|^2
<=>        2(|a+c|+a+c) + |a+c|(2-|a+c|) >= 0
last ineq holds due |a+c|<=2.
From (1), (2), (3) we have the desired ineq.

Maybe is yours, Harazi?

Namdung

I must admit your solution is very beautiful. My solution was the following:
  Consider z=a+ib and w=c+id. Then the inequality can be writen |z+1|+|w+1|+|zw+1|>=2 and can be deduced from 2=|zw+1-z(w+1)+z+1|<=|zw+1|+|z||w+1|+|z+1|.

A trigonometric soln does indeed work:

We can put a=cosx, c=cosy, so ac-bd=cos(x+y), so the ineq is:

 \sqrt 1+cosx +  \sqrt 1+cosy +  \sqrt 1+cos(x+y) >=  \sqrt 2 , but we know that 1+cosx = 2(cos(x/2))^2, so th problem becomes
|cosx|+|cosy|+|cosz| >= 1 for x+y+z = 0.

By the transformations x->x+Pi or x->x-Pi or x->Pi-x and so on we can easily take it to the form
|cosA|+|cosB|+|cosC| >= 1, for A, B, C the angles of a triangle. We know that |cosA|+|cosB|+|cosC| >= cosA+cosB+cosC = 1+r/R >= 1, so it's done.

Well, we have already three solutions. So, this one is easy, but nice.


Hello, i need some help with the following exercise :

The joint distribution of X and Y is given by $ f(x, y) \equal{} \frac{exp(\minus{}y)}{y}$, $ 0 < x < y <\infty$ .
Compute $ E(X^2 \plus{} Y^2 |Y \equal{} y)$.

Thanks for any help.

Well, in the first place, that would be $ y^2 \plus{} E(X^2\,|\,Y \equal{} y).$ No need to do any calculations involving $ Y.$

Second: note that if we fix $ Y \equal{} y,$ we have $ X$ ranging from $ 0$ to $ y,$ and the joint density doesn't depend on $ x$ in that range. So our conditional distribution for $ X$ is that it is uniform on $ [0,y],$ from which
\[ E(X^2\,|\,Y \equal{} y) \equal{} \frac1y\int_0^yx^2\,dx \equal{} \frac {y^2}3.
\]
So the final answer is $ y^2 \plus{} \frac {y^2}3 \equal{} \frac {4y^2}3.$


sala/\/\
ye soale tarkibiat didam be nazaram ghashang oomad az ostadan mikhastam rahe halashoono began
yek safhe shatranje namotanahi hame khanehayash sefid ast tedade motanahi az khanehaye in safhe siah shodeand har khane siah ba tedade zoji khane sefid zele moshtarak darad sabet konid khanehaye sefid ra mitavan ghermez va sabz kard tori ke har khane siah ba tedad mosavi khane ghermez va sabz zele moshtarak dashte bashad

rahnami mikonam age kasi rahe hale kamelo khast bege.
aval say konid gerafesh konid masalaro.
dovom in geraf khobe :khone haye sefid ke anhara beham vasl mikonim agar ye siah beyneshan bod L shekl ham vasl konid.
sevom ye taghir kochik to graf bedin
charom  hal sabet konid moalefe hara mishavad ba 2 rang, rang kard betori ke hich 2 rase mojaveri ham rang nabashand(2 bakhshi)


The limit is $0$ because using the $\epsilon$ cryteria it is clearly that $\forall \epsilon >0 \exists x$ such that$\frac{1}{x} <\epsilon$.So taking $\epsilon$ closer to $0$ then $\frac{1}{x}$ is going straight to $0$  :D .

[quote="pkerichang"]plugging infinity into $x$, seeing that any constant over infinity is 0, so the answer is $0$.    You should only worry about the case of infinity over infinity or 0 over 0.[/quote]Remember that phrases like "plugging infinity" only have intuitive meanings because infinity is not a number.

Formally: given $\epsilon>0$, there exists a positive integer $N$ such that for $x>N$, then $\left|\frac1{x}-0\right|<\epsilon$

[quote="ryan22"]solve $\displaystyle\lim_{x\to\infty}\frac{1}{x}$[/quote]

[hide]Limit=$0$[/hide]

[quote="Bruno Bonagura"]Can somebody prove that $\lim_{x\to\infty}\left(\sqrt{x^2+2}-x\right)=0$ ?

Thx in advance...[/quote]

[hide="hint"]you use a very standard limit technique
[hide="a hint that gives it away"] conjugate [/hide]
[/hide]

[hide="solution"]$\sqrt{x^2+2}-x=\frac{2}{\sqrt{x^2+2}+x}$
so the limit is 0 as the numerator goes to infinity[/hide]

..

[quote="Bruno Bonagura"]Can somebody prove that $\lim_{x\to\infty}\left(\sqrt{x^2+2}-x\right)=0$ ?

Thx in advance...[/quote]

[hide="Click to reveal hidden content"]$\lim_{x\to\infty}\sqrt{x^2+2}-x=0$

$\lim_{x\to\infty}\sqrt{x^2+2}=x$

$\lim_{x\to\infty}x^2+2=x^2$

$\lim_{x\to\infty}\frac{x^2+2} {x^2}=\frac{x^2} {x^2}$

$\lim_{x\to\infty}1+\frac{2} {x^2}=1$

Since x approaches infinity, it's just:

$1+0=1$ Which is obviously true.[/hide]

[quote="Zenith Xan"]

[hide="Click to reveal hidden content"]$\lim_{x\to\infty}\sqrt{x^2+2}-x=0$

$\lim_{x\to\infty}\sqrt{x^2+2}=x$

$\lim_{x\to\infty}x^2+2=x^2$

$\lim_{x\to\infty}\frac{x^2+2} {x^2}=\frac{x^2} {x^2}$

$\lim_{x\to\infty}1+\frac{2} {x^2}=1$

Since x approaches infinity, it's just:

$1+0=1$ Which is obviously true.[/hide][/quote]

You cannot put one $x$ to the other side.

EDIT: Sorry for the harsh comment. I have now included a non-rigorous proof:

Let $L=\lim_{x\to\infty}\left(\sqrt{x^2+2}-x\right)$

Then, $L=\lim_{x\to\infty}\left(\sqrt{x^2+2}-x\right)\frac{\sqrt{x^2+2}+x}{\sqrt{x^2+2}+x}$

$L=\lim_{x\to\infty}\frac2{\sqrt{x^2+2}+x}$

As $x\to\infty$, the denominator grows without bounds with a constant numerator so $L\to 0$

Thus, $L=0$.

[quote="Zenith Xan"]

$\lim_{x\to\infty}\sqrt{x^2+2}-x=0$

$\lim_{x\to\infty}\sqrt{x^2+2}=x$

$\lim_{x\to\infty}x^2+2=x^2$

$\lim_{x\to\infty}\frac{x^2+2} {x^2}=\frac{x^2} {x^2}$

$\lim_{x\to\infty}1+\frac{2} {x^2}=1$

Since x approaches infinity, it's just:

$1+0=1$ Which is obviously true.[/quote]

That`s not the way to work with limits...  :noo:


If a cube is inscribed in a sphere of diameter 3, what is the volume of the cube?

[quote="jli"][hide="solution"]The diagonal of the cube is $ 3$ so the side is $ \frac{3}{\sqrt{3}}$.

Cube that to get $ \frac{9}{\sqrt{3}}=\boxed{3\sqrt{3}}$.[/hide][/quote]

why did you erase my post and instead copied my answer which i already posted?

its not like you did much work either..

that wasn't very nice jli. He did have the right the answer even though it was just the answer.

alright its  not likie jli posted anything that was useful...so

the diameter of the sphere is the long space diagonal of the cube (try drawing a cube to visulize), we call a side of the cube $ a$ so

$ \sqrt{a^{2}+a^{2}+a^{2}}=3$, thus $ 3a^{2}=9$, $ a^{2}=3$, meaning $ a=\sqrt{3}$, since the volume is $ a^{3}$ its $ ({\sqrt{3}})^{3}=3\sqrt{3}$

Like i said jli's post is almost just as unuseful as just posting the answer (what I  did) because we have no idea how the side is $ \frac{3}{\sqrt{3}}$ if the diagonal is $ 3$(as he posted), so if you are going to erase a post with the right answer, atleast you SHOULDNT BE A HYPOCRITE..

its because he didnt hide  or show any work and i said that i would erase anything like that

i thought u guys would be smart enough to figure that out

jeez

im tired of hiding people's solutions.  ive probably edited 10 posts in the last 3 days


Hi. I am taking Math Analysis right now and the problem is our professor isn’t really explaining things well. He just kept on reading what he is writing on the board.  Another problem is, I am not really good at this stuff.

One problem is on the Axioms of Real Numbers.  He just gave for axioms without giving the theorems and its consequences (which I have read in other books) and I think we have to solve the problems using the Axioms that he gave us.  I have solved some of the problems on different books, but another problem is, it has no solution, so there is no way to verify if my answer is correct. (My God, our exam is fast approaching).

So that you will know what I am talking about, I’ll write the axioms.

Addition Axioms
A1: Associativity: For all x, y and z in R, (x + y) + z = x + (y + z)
A2: Additive Identity. For all x in R, x + 0 = x and 0 + x = x
A3: Additive Inverse: For all x, y in R, there exists x an element that x + y = 0 and y + x=0
A4: Commutativity: For all x, y in R, x + y = y + x

Multiplication Axioms
M1: Associativity: For all x, y and z in R, (xy)z = x(yz)
M2: Multiplicative Identity. For all x in R, (x)(1) = x and 1(x) = x
M3: Multiplicative Inverse: For all x, y in R, there exists x an element that xy = 1 and yx = 1.
M4: Commutativity: For all x, y in R, xy = yx

Ordering Axioms

01	Trichotomy: For all x, y in R, exactly one of the following holds: (i) x = y, (ii) x < y, (iii) y < x.
02	Transitivity: For all x, y and z in R.  If x <y and y < z, then x < z.
03	If x,y, z in R such that x < y, then x + z , y + z.
04	If x,y and z are in R, and 0 < z and x <y, then xz < yz.

Whew!

Sample1:

Prove that if ac = bc, then a = b.

Proof:
ac = bc
${c^{-1}}ac={c^{-1}}bc$ M3 (existence of a multiplicative inverse)
$a(cc{-1})= b(cc^{-1})$ M1 (associativity of multiplication)
$a(1) = b(1)$ M2 (multiplicative identity)
$a = b.$
Sample2:

Show that if $a > b$, then $a^{2}> b^{2}$

Case 1
$a > b$
$aa > ab (04)$
$a^{2}> ab$
Case 2
$a > b$
$ab >bb (04)$
$ab > b^{2}$

If $a^{2}> ab$ and $ab > b^{2}$, then by (01)Trichotomy, $a^{2}> b^{2}$.

PLEASE HELP ME HERE: Some of the problems from other books which I have not yet solved.

1.	Show that $(a^{-1})^{-1}= a$
2.	Show that $\frac{a}{b}$ = $\frac{1}{\frac{b}{a}}$.
3.	Show that if a and b are positive, then $a^{3}>b^{3}.$

Thanks.

[quote="Kent Merryfield"]Some organization: The "A" and "M" groups of axioms are the axioms for a field. However, two axioms have been left out: the distributive law for multiplication over addition, and the fact that $1\ne0.$ Re-check your notes - the distributive law should be there! Anything based purely on these axioms is a matter of algebra, not analysis.

Your first two problems are such matters of algebra. The key to both of them is to note that $y=x^{-1}$ if and only if $xy=1.$

The "O" group of axioms gives you an ordered field. You don't have the real numbers just yet - there's one more axiom that you may not have been shown yet. Note that the rational numbers satisfy all of the axioms of an ordered field, as do many other things I could construct.

Now, I would warn you that "If $a>b$ then $a^{2}>b^{2}$ is false. Consider $a=1, b=-2.$

A possible path toward proving that $a>b$ implies $a^{3}>b^{3}.$

$a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2}).$

Can you prove that $a^{2}+ab+b^{2}>0?$[/quote]

Thanks for your reply. By the way, you are right, I forgot to mention the axiom of disributivity. And I also forgot that in my sample, $a$ and $b$ are positive. (I tried to make it short, so I actually deleted some part)

Let me try proving $a^{2}+ab+b^{2}>0$  using the axioms:
Since $a$ and $b$ are positive, it follows that $ab > 0$.
Now since $a > b$, therefore $a^{2}> ab$. Multiplying both sides by b, we also know that $ab > b^{2}$. Now, we know that they are all positive, and adding yields a sum greater than 0. Is my proof complete enough? Now, can we consider factoring $a^{3}-b^{3}$ a consequence of distributive property?

And regarding the first two problems, we all know that its too easy. But can I manipulate both sides of the equation? I mean in proving, I know its valid, but our professor just kept manipulating one side of the equation.

If I would solve number 1, I would solve it like this:

Since $(a^{-1})^{-1}= a$ then I can multiply $a^{-1}$ on both sides which will yield:

$(a^{-1})^{-1}) (a^{-1}) = a(a^{-1})$
$1 = 1.$ Do you think its a valid proof? My professor told us, that in a proof, the places of the given shoud be reversed. For example in the given $(a^{-1})^{-1}= a$ , our final equation should be $a = (a^{-1})^{-1}$. Can you show me how to do the reverse technique? (I think it isnt right to restrict the solution for proofs)

Your M3 is incorrect -- 0 has no multiplicative inverse.  As a result, your "sample 1" is also false (because c could be 0).  Also, depending how detailed you really want to be, in the proof of sample 1 you used commutativity and associativity at least twice each.

In addition to the problems Kent noted in Sample 2, your final step is by Transitivity (O2), not trichotomy.

For proving $a > b$ implies $a^{3}> b^{3}$, I would probably use the method you used in your sample 2, not the method Kent suggests.  Then it's just repeated applications of O4 and O2.

[quote]If I would solve number 1, I would solve it like this:
Since $(a^{-1})^{-1}= a$ then I can multiply $a^{-1}$ on both sides which will yield:[/quote]

If one says, "Since blahblah," that implies that one has already proven "blahblah."  In this case, you are trying to prove "blahblah," so you certainly don't want to start out like that.  Instead, try starting with the expressions $a\cdot a^{-1}$ and $a^{-1}\cdot (a^{-1})^{-1}$.  You can evaluate these expressions.  Then you can apply transitivity of equality.  Then you can apply your Sample 1.

What your professor does is the following: he is trying to prove something like $X = Y$.  He starts with the expression X and subjects it to certain transformations (probably one axiom at a time) until he arrives at expression Y.  What I just suggested you do is more complicated, but equally correct: we manipulate two different expressions until we have them equal to the same expression, and then we say that they are equal to each other.  (Of course, anything we can do my way we can do your professor's way.)
What you did is different: you started with the thing you wanted to prove (you also called it a given, which is wrong -- it's not given, it's what you're trying to get!) and then manipulated it until you came to something true.  [i]That is not a proof![/i]  Any equation can be manipulated to yield something true.  What you need to be doing is starting with true things and arriving at the thing you're trying to prove.
It is possible to turn your proof upside-down and make it an actual proof.  In fact, it will end up looking just like my suggested proof.

[quote="guile"](I tried to make it short, so I actually deleted some part)

Let me try proving $a^{2}+ab+b^{2}>0$  using the axioms:
Since $a$ and $b$ are positive, it follows that $ab > 0$.
Now since $a > b$, therefore $a^{2}> ab$. Multiplying both sides by b, we also know that $ab > b^{2}$. Now, we know that they are all positive, and adding yields a sum greater than 0. Is my proof complete enough? Now, can we consider factoring $a^{3}-b^{3}$ a consequence of distributive property?[/quote]
Re: making it short.  The point of what you're doing is to reconstruct "things you know" with a rigorous basis.  "Making it short" is in some ways exactly the opposite of what you are 'supposed' to be doing.
The first part of this proof is good: you get $a^{2}> 0, b^{2}> 0$ and $ab > 0$.  Then you said that because they are all positive, adding them together gives something positive.  That's not an axiom.  You may have proven it in class (you actually probably would have proven that if $w > x$ and $y > z$ then $w+y > x+z$) in which case you can use it, but if we're proving things from the axioms, you'll have to show this fact as well.  (It follows from applications of O2 and O3, I believe.)  Finally, the factorization of $a^{3}-b^{3}$ is the result of repeated applications of distributivity (on both sides), associativity and commutativity (which is why I would avoid doing this problem in this way).

Can you recommend a site that has a detailed explanation of these stuff? It's really making me dizzy. Anybody wants to try number 2?


I am a senior in high school right now. I am not super-gifted at anything. I am just an average kid who works hard. I am not super-competitive when it comes to college apps so i was hoping i could apply to Cal Poly Pomona because i hear they have a good math/science/engineering program in which i hope to be a part of. I am aware of their MINIMUM enrollment criterion but the minimum requirements for schools nowadays aren't really a good reference point. So i was wondering how my chances are at getting into that school. Here are my math/science grades.

*key---> grade number) math course (1st semester grade, second semester grade) science course (1st semester grade, 2nd semester grade) 1st semester GPA, 2nd semester GPA

9) Algebra I (B, B-) Biology (B, C+) 3.57, 3.57
10) Geometry (B, B+) Chemistry (A, A) 3.86. 3.86
11) Algebra II (A, A) Physics (B, A-) 3.67, 4.00
12) Currently taking Trigonometry/Math Analysis, AP Statistics, and AP Physics B

SAT Score: (1580 something....equivalent to about an 1100 or 1200 on the old scoring system...so not great but not horrible)

assuming they use sophmore and junior year grades to calculate GPA then my GPA is:

(3.86(2) + 3.67 + 4.00)/ 4 = 3.85

So are my chances good as long as i don't screw up this year (lol)?

A rising trend in grades looks good. I'm not really sure how competitive Cal Poly Pomona is these days, so that's the part of your questions I'll leave to the Californians to answer.


geometry
rectangle
function
combinatorics open
combinatorics

Notice that $ f(R,C)\equal{}1\plus{}f(R\minus{}C,C)$ for $ R \ge C$
or $ f(R,C)\equal{}1\plus{}f(R,C\minus{}R)$ for $ R < C$

I'm sorry, could you explain a little more? I'm still vague. what's f(R,C) ? Is it the function of R and C ? But what kind of function is it ?
I understand in your given function : F(R,C) = 1 + F(R-C, C) for R >= C
"1" means is a square. But this is just special case of a "complete cut".
But the thing I want to prove is "There will be a "complete cut" which will produce a minimum squares eventually ? or I might misunderstand your hint.
Sorry for my poor English!

I try to explain...
let $ f(R,C)$ be the amount of squares for a complete cut.
the recursion is based on a complete cut dividing the rectangle in a big square and a smaller rectangle.
For example: for a rectangle of $ 5 \cdot 7$ you cut $ 1$ square of $ 5 \cdot 5$ and leaves a rectangle of $ 5 \cdot (7\minus{}5)$ then $ f(5,7)\equal{}1\plus{}f(5,2)$

Oh, sorry. I see what you mean. But can you prove it more general. The prime thing I want to prove is I suppose there will be one complete cut dividing the rectangle into [b]2 sub-rectangle[/b] which is the best complete cut. Anyway thanks for your help. I appreciate it.


Prove that the sequence $ x_1 \equal{} 2, x_2 \equal{} 4$ with $ x_{n \plus{} 1} \equal{} \sqrt {x_nx_{n \minus{} 1}}$ is convergent and find  $ \lim_{n \to \infty}{x_n}$.

Let $ x_n \equal{} 2^{y_n}$, we can rewrite the reccurence as $ y_{n \plus{} 1} \equal{} \frac {1}{2}(y_n \plus{} y_{n \minus{} 1})$ with $ y_1 \equal{} 1,\ y_2 \equal{} 2$.

$ y_{n \plus{} 1} \minus{} y_n \equal{} \minus{} \frac {1}{2}(y_n \minus{} y_{n \minus{} 1})$

$ y_{n \plus{} 1} \plus{} \frac {1}{2}y_n \equal{} y_n \plus{} \frac {1}{2}y_{n \minus{} 1}$

$ \therefore y_{n \plus{} 1} \minus{} y_n \equal{} \left( \minus{} \frac {1}{2}\right)^{n \minus{} 1}(y_2\minus{}y_1) \equal{} \left( \minus{} \frac {1}{2}\right)^{n \minus{} 1}$

$ y_{n \plus{} 1} \plus{} \frac {1}{2}y_n \equal{} y_2 \plus{} \frac {1}{2}y_1 \equal{} \frac {5}{2}$.

Eliminating $ y_{n \plus{} 1}$ gives $ y_n \equal{} \frac {5}{3} \minus{} \frac {2}{3}\left( \minus{} \frac {1}{2}\right)^{n \minus{} 1}$.

$ \therefore x_n \equal{} 2^{\frac {5}{3} \minus{} \frac {2}{3}\left( \minus{} \frac {1}{2}\right)^{n \minus{} 1}}$, yielding $ \lim_{n\to\infty} x_n \equal{} 2\sqrt [3]{4}$.


Question- Where is the line of "frivolousness" drawn between the Games and Fun Factory forums? Thanks in advance. :)  :?:

It is not a well-defined line. In this case, common sense should help guide you as to where you should post. One question you might want to ask is, "Is this of any possible importance in contribution to problem solving?" If the answer is "absolutely not," well then, into the Fun Factory it should definitely go. Also, if you're in doubt, most likely it would go into the Fun Factory; afterward, you could always ask a moderator or an administrator for advice if you wish.

Okay. That seems like a great question to ask oneself. Thanks! :)

One further thing about "frivolousness": I made this one post game weird thing in the Fun Factory-
PostIt
frequent aops user? please join

--------------------------------------------------------------------------------

This is quite a pointless game. In this game, there will be 5 rounds. In each round, I will post "Start!". After this, please post here as soon as you can. The first one to post will recieve 10 points, 2nd 7, 3rd 5, 4th-6th 2, 7th-10th 1 points. Whoever has the most points at the end is the champion. After 24 hours, from when I make the post to start a round, all other competitiors who have not posted will recieve 0 points. The time between rounds varies, but is usually reasonably soon after the last round ended. Hope you play! Sign up, there's ten ppl who can play. I will post the points standings often.

http://www.artofproblemsolving.comForumviewtopic.phpt=271400

I was not sure why it was locked; I thought the Fun Factory was for that type of pointless thing. Thanks!

Because even FunFactory doesn't allow arbirary high amounts of stupidity for games  :wink:

Well, it's not locked anymore.  I am posting a reply right now.


probability
function
expected value
probability and stats

I'm casually reviewing introductory statistics after taking it several years ago as AP and never really understanding.

My book says that the mathematics behind the n - 1 in the standard deviation formula is very complicated, but gives a vague explanation--that the sample will never have as much variability as the population. My old text was even worse with its "intuitive" explanation--something about one number being taken away to stand for the mean--it left me with the impression of a point being consumed in a cloud of smoke in the numerator. But by the time I was reviewed standard deviation, I had my own intuitive, visual explanation, explained below.

The forum for the median point (n + 1)/2 was really bugging me because I could think of concrete examples for why it worked mathematically but didn't have a good visualization of it. Then I realized that, arithmetically, you're counting the space between 0 and 1 as if 0 were a point in the set leading up to n.

It looks something like this:

0---1________n

If you divided n by 2, you'd end up shifted 1/2 units to the left, which incidentally works out well if you redistribute the formula as n/2 + 1/2.

Visually, though, to balance it you go:

0---1________n---n + 1

This set's larger but has the same center.

Intuitively, when you recognize that the problem with using just n is that it includes too much, you want to subtract 1, but that would leave you with a set of the right size but shifted one unit to the left.

In this model, each point is an even 1 unit apart from each other point so that they're (a) ranked in order of ascending value and (b) breaking the "line" creates portions proportionate to the size of the subsets thus created, like the top 50% vs. the bottom 50%, as long as you remember to adjust for the length between 0 and 1.

The standard deviation is also a measure of spread, so it uses this same type of mathematical model. This is different from the mean, which is the sum over just plain n, because there's no ranking in the mean and thus no line is created, just a bunch of discrete values put into a pool and then broken into the original number of pieces but evened out.

When you're trying to find standard spread, you divide by the total units of spread, which as shown above is n - 1.

My question is: is this conceptualization at least as close as the ones I was offered? To me, intuitive means a visualization or a mathematically explanation from which a visualization can be derived.

I'd also like to know more about the actual mathematics that go into the standard deviation formula, and any glaring problems with my visualization.

Edit: Ouch, I just came up with one. The median works with the larger set w/ same center but any other percentile points would be off. Hmm... Doesn't immediately affect the spread, though, I don't think.

The context in which that $ n\minus{}1$ makes sense: You have some unknown probability distribution. Given $ n$ samples from this distribution, you want to estimate the variance (or standard deviation) of the original distribution.

The calculation that this comes from isn't too hard; the key is the concept of an [url=http://en.wikipedia.org/wiki/Unbiased_estimator]unbiased estimator[/url].
Start with some random variable $ X$, with unknown mean $ a$ and unknown variance $ b^2$. Sample it $ n$ times ($ x_1,x_2,\dots,x_n$), and compute the variance $ V\equal{}\frac1n\sum_{i\equal{}1}^n x_i^2\minus{}\left(\frac1n\sum_{i\equal{}1}^n x_i\right)^2$ of the sample. What is the expected value of $ V$? It's easy to see that each $ x_i^2$ has an expected value of $ a^2\plus{}b^2$, so the average of all $ n$ has expected value $ a^2\plus{}b^2$. For the second sum, we expand it out: it's $ \frac1{n^2}\sum_{i\equal{}1}^n\sum_{j\equal{}1}^n x_ix_j$. If $ i\neq j$, $ x_i$ and $ x_j$ are independent and thus uncorrelated; the expected value of the product $ x_ix_j$ is thus $ a^2$. If $ i\equal{}j$, of course, the expected value of $ x_i^2$ is $ a^2\plus{}b^2$. There are $ n$ of the second and $ n^2\minus{}n$ of the first; adding these up and dividing by $ n^2$, the expected value of $ \left(\frac1n\sum_{i\equal{}1}^n x_i\right)^2$ is $ a^2\plus{}\frac{1}{n}b^2$.
This makes the expected value of $ V$ equal to $ \frac{n\minus{}1}{n}b^2$. The modified estimate $ \frac{n}{n\minus{}1}V$ then has expected value equal to the variance of $ X$, making it an unbiased estimator.

This estimate is in fact the best possible given this level of information. If our distribution has a known mean $ a$, we can estimate the variance more accurately by $ \frac1n\sum_{i\equal{}1}^n (x_i\minus{}a)^2$.

You can see from the calculation that the difference does indeed come from the variation in the mean of our sample. Another idea that might help with intuition: look at what happens for very small values of $ n$. If I told you that my one random number was 42, what could you say about the variance of my version of "random"? Nothing. You need at least two points to do anything.

I just wanted to add two small things, the first of which is a clarification in case there was any confusion on how the expected value of $ x_i^2 \equal{} a^2 \plus{} b^2$ was obtained.  This is simply a rearrangement of

$ {\rm Var}[x_i] \equal{} {\rm E}[x_i^2] \minus{} {\rm E}[x_i]^2$,

or

$ b^2 \equal{} {\rm E}[x_i^2] \minus{} a^2$.

Second, the takeaway from jmerry's excellent exposition is that the sample variance is an estimator, which is a function of the observations made; whereas the population variance is a fixed property of a given probability distribution.  We are free to choose the construction of our estimator--however, in the case of the canonical formula for sample variance, we have chosen the particular function that makes the estimator unbiased when the samples are drawn from a population of unknown fixed mean and variance.

Your Explanation check falls below my Resistance roll...

Okay, I was following you up to the part where j comes in, at which point I'm completely lost. I can't even tell what you mean by "for the second sum". Additionally, shouldn't there be another pair of brackets in the variance formula? And the wiki article you linked says, "For example the square root of the unbiased estimator of the population variance is not an unbiased estimator of the population standard deviation." So I can't see that the standard deviation formula is unbiased from a demonstration that the variance formula is. But I would like to understand that demonstration!

(And, thanks, the concept of biased vs. unbiased estimator enriches my life and certainly my general vague idea of what this whole thing called "statistics" is.)

There are several ways to rearrange the definition of variance. I was using the one atomicwedgie mentioned.

What do I mean by "the second sum"? I mean the second term on the RHS if the equation $ V\equal{}\frac1n\sum_{i\equal{}1}^n x_i^2\minus{}\left(\frac1n\sum_{i\equal{}1}^n x_i\right)^2$, and I did refer back to the original form of that sum a few lines later. The sum with both $ i$ and $ j$ in it is simply what you get by writing $ \left(\frac1n\sum_{i\equal{}1}^n x_i\right)\cdot\left(\frac1n\sum_{j\equal{}1}^n x_i\right)$ as a double sum. That double sum needs two indices.

The square root of this population variance is not an unbiased estimator of the standard deviation, because square roots and expected values don't commute.
A simple example: let $ X$ be the probability distribution which is $ 1$ or $ \minus{}1$ with equal probabilities $ \frac12$. It has mean zero and variance $ 1$. Sample it twice. Half the time, our samples are equal, and the variance estimate we get is zero. The other half of the time, our samples differ by $ 2$, and the variance estimate we get is $ 2$. The average of those estimates is $ 1$, confirming the fact that it's an unbiased estimator.
Now, what if we take the square root of this estimated variance and call it an estimate for the standard deviation? We get zero half the time and $ \sqrt{2}$ the other half, for an expected value of $ \frac{\sqrt{2}}{2}$, not equal to the original standard deviation.

This sort of thing is guaranteed to happen. The concavity of the square root means that the expected value of this standard deviation estimate will always be too low. Worse, we can't simply scale our way out of the problem, because the details of how low it is depend on the distribution. Try the same two-point estimate on a uniform distribution on $ [0,1]$; the average length between our two points is $ \frac13$, so the estimated standard deviation is $ \frac1{3\sqrt{2}}$. The real standard deviation is $ \frac1{2\sqrt{3}}$, a multiple of $ \frac{\sqrt{3}}{\sqrt{2}}$ larger.
There is no generic unbiased estimator for the standard deviation. We have to make do with a biased one, and the square root of our unbiased variance estimator is a good choice.


geometry parallelogram algebra polynomial

[color=darkred]Denote $ X(x)$- the point $ X$ with the affix $ x\in\mathbb C$. Prove (and [b]memorize[/b]) $ A\in BC$ $ \Longleftrightarrow$ $ a\cdot\overline b+b\cdot\overline c+c\cdot\overline a\in\mathbb R$ $ \Longleftrightarrow$ $ |\begin{array}{ccc}a &\overline a & 1\\ \ b &\overline b & 1\\ \ c &\overline c & 1\end{array}| = 0$.

[b]Preliminary.[/b] The conjugate $ \overline z$ of the complex number $ z = x+y\cdot i$ is the complex number $ x-y\cdot i$, i.e. $ \overline{x+y\cdot i}= x-y\cdot i$. Observe that :

[b]$ z\in\mathbb R\Longleftrightarrow z =\overline z$ ; $ P\in\mathbb R[X]\implies$ for any $ z\in\mathbb C$, $ \overline{P(z)}= P(\overline z)$.

$ |z|^{2}= z\cdot\overline z$ ; $ |z| = 1\Longleftrightarrow\overline z =\frac{1}{z}$ ; $ AB = |a-b|$.[/b]

[b]Some exercises.[/b] $ \{a,b\}\subset\mathbb C$ $ \implies$ $ \{\begin{array}{cccc}1\blacktriangleright &\mathrm{the\ rule\ of\ the\ parallelogram}& : & |a+b|^{2}+|a-b|^{2}= 2\cdot(|a|^{2}+|b|^{2})\\ \\ 2\blacktriangleright &\mathrm{the\ rule\ of\ the\ triangle}& : & |a+b|\le |a|+|b|\\ \\ 3\blacktriangleright & |a| = |b| = |a+b| &\implies & a^{2}+ab+b^{2}= 0\\ \\ 4\blacktriangleright &\frac{a^{2}-a+1}{a^{2}+a+1}\in\mathbb R &\Longleftrightarrow & a\in\mathbb R\ \ \vee\ \ |a| = 1\\ \\ 5\blacktriangleright & |a^{2}+1| = |4z+3| &\implies & |a|\le 9+5\sqrt 5\end{array}$.[/color]

0. Obviously, $ z\in\mathbb{R}\iff z \equal{}\bar z$. Hence the equivalent statement is

$ A\in BC\iff\bar ab\plus{}\bar bc\plus{}\bar ca \equal{} a\bar b\plus{}b\bar c\plus{}c\bar a$

Substitute $ a \equal{} b\plus{}x$. After elementary rearranging, that yields

$ (\bar b\minus{}\bar c)x \equal{} (b\minus{}c)\bar x$

and that in turn means

$ (a\minus{}b)(\bar b\minus{}\bar c)\in\mathbb{R}$

Divide the left side by real number $ |b\minus{}c|^{2}$ to get

$ {a\minus{}b\over b\minus{}c}\in\mathbb{R}$

which obviously means that $ AB$ and $ BC$ are collinear. All the steps are equivalences, hence both directions are proven. QED

[quote="Virgil Nicula"]$ 5\blacktriangleright |a^{2} \plus{} 1| \equal{} |4a\plus{} 3|\implies |a|\le 9 \plus{} 5\sqrt 5$.[/quote]
Very nice until now. And the upper exercise ?!

[quote="Virgil Nicula"][quote="Virgil Nicula"]$ 5\blacktriangleright |a^{2} \plus{} 1| \equal{} |4a \plus{} 3|\implies |a|\le 9 \plus{} 5\sqrt 5$.[/quote]
Very nice until now. And the upper exercise ?![/quote]
$ |a^2\plus{}1|^2\equal{}|4a\plus{}3|^2$
$ \Rightarrow (a^2\plus{}1)(\overline a^2\plus{}1)\equal{}(4a\plus{}3)(4\overline a\plus{}3)$
$ \Rightarrow |a|^4\plus{}2\textrm{Re}(a^2)\plus{}1\equal{}16|a|^2\plus{}24\textrm{Re}(a)\plus{}9$
$ \Rightarrow |a|^4\minus{}2|a|^2\plus{}1\leq16|a|^2\plus{}24|a|\plus{}9$
$ \Rightarrow |a|^4\minus{}18|a|^2\minus{}24|a|\minus{}8\leq0$
The lefthandside polynomial has 4 real roots: $ \minus{}2\minus{}\sqrt2,2\minus{}2\sqrt2,\sqrt2\minus{}2$ and $ 2\plus{}2\sqrt2$, of which only the latter is positive.
Whence $ |a|\leq2\plus{}2\sqrt2$

[quote="Virgil Nicula"]$ 5\blacktriangleright |a^{2} + 1| = |4a + 3|\implies |a|\le \sqrt {9 + 5\sqrt 5}$.[/quote]
[b][u]Proof.[/u][/b] Denote $ |a|=r$ . Thus, $ |a^{2} + 1| = |4a + 3|$ $ \ \Longleftrightarrow$ $ (a^2+1)(\overline a^2+1)=(4a+3)(4\overline a+3)$ $ \ \Longleftrightarrow$

$ r^4+(a^2+\overline a^2)+1=16r^2+12(a+\overline a)+9$ $ \ \Longleftrightarrow$ $ r^4+[(a+\overline a)^2-2r^2]+1=16r^2+12(a+\overline a)+9$ $ \ \Longleftrightarrow$

$ (r^4-18r^2)+[(a+\overline a)^2-12(a+\overline a)]=8$ $ \ \Longleftrightarrow$ $ (r^2-9)^2+(a+\overline a-6)^2=125$ . Therefore,

$ \{\begin{array}{c}
|r^2-9|\le 5\sqrt 5\\\\
a+\overline a-6\le 5\sqrt 5\end{array}\|$ $ \implies$ $ \{\begin{array}{c}
|a|\le \sqrt {9+5\sqrt 5}\\\\
\sqrt {6-5\sqrt 5}\le a+\overline a\le\sqrt { 6+ 5\sqrt 5}\end{array}$ .


probability
geometry
3D geometry
sphere
LaTeX
ARML
MATHCOUNTS

If you didn't make MOP, which I don't believe any middle schooler did, the summer brings no math competitions with it. So, instead of waiting until the school year to do anything, I created a monthly tournament called the Failure Cup.

The format goes as follows. From June to September there will be a monthly tournament and in October there will be a Failure Cup Championship among the top 10 (and ties).

On the first of each month, I will post a 10 question Failure Test. The first 5 are worth 1 point each and the last 5 are worth 2 points each. Anyone participating should pm me their answers by the 13th. [b]Submit only once, no changing answers.[/b] Also, please only give me your answers. I don't need solutions.

The top 15 and ties from this first part will participate in part 2 beginning on the 16th and ending on the 28th. This second section will consist of 5 questions with increasing point values from 1 to 5. Each previous answer will be necescery to solve the next question (like a relay.)

By the end of the month, I will determine the top 7 based on cumulative scores from parts 1 and 2, and announce these. First place receives 7 Failure Cup standings points, second receives 6, third receives 5, etc. If there is a tie, I will average the points. For example, if there is a 2-way tie for first, each competitor receives 6.5 points. If there is a a 3 way tie for third each receives 4 points. This way, the total number of standings points will be the same each month.

On September 30th, along with the September winners and season standings, I will announce the top 10 (and ties) and their total number of points. These people will be allowed to participate in the Failure Cup Championship. On October 3rd, I will post 10 questions with answers due on October 16th.

The point value of a question will vary based on how many people get it right.
If everyone or nobody gets a question, it is worth 0 points. Other scenarios are illustrated below:

Number of people correct  /      If correct    /    If incorrect
1 /                                      +4   /                  0
2 /                                      +3.5  /                -0.5
3 /                                      +3  /                   -1
4  /                                     +2.5  /                -1.5
5  /                                     +2 /                    -2
6 /                                      +1.5 /                 -2.5
7 /                                      +1 /                    -3
8 /                                      +0.5  /                -3.5
9  /                                     0    /                   -4

The points earned (or lost) in the final round will be added (or subtracted) to each finalist's season total of standings points. On October 20th, all top 5 ties will be announced with a 4 question tiebreaker conducted from October 22nd to October 28th, where total score will break the tie, followed by performance on question 4, followed by performance on question 3, followed by performance on question 2, followed by submission time. The top 5 will be announced on October 31st. If there are no ties, they will be announced on October 20th.

The questions for June will be posted tomorrow. Best of luck and thank you for actually reading this whole post.  :)

Wow there sure are a lot a mock contests being hosted, ever since jackdillon's.

so pretty much anyone can sign up, all you have to do is pm your answers?

Sorry for such a long post, I just wanted people to know what this was leading up to.

To neurosurgeon: everyone can participate in Part One.

Here are the Part One questions for June. 1 thru 5 are one point and 6 thru 10 are 2 points. PM me your answers by June 13th.

1. If x/(x-2)=(y-2)/y, Find y in terms of x.

2. The first letter in a list is a Z, followed by 2 Ys, 3 Xs, 4 Ws, 5 Vs, etc. What is the 189th letter on the list?

3. If a#b= [a(a+1)+b]/2 and f(a)=a-6, Find 6#f(4)

4. Horace and Joe are playing a coin flip game. Horace flips the coin twice, then Joe flips it once. This continues until someone flips a tail and loses. What is the probability that Joe wins? (Assume that if one person loses, the other wins.)

5. Find y if y is the smallest multiple of 8 such that abs(y)<13pi

6. The numbers 10 through 25 are arranged in a 4x4 grid such that every column has the same sum. What is this sum?

7. Jason has an infinite supply of post-it notes that read 3 and 5. The greatest integer he can't make by adding post-its is A. Frank has an infinite supply of post-it notes reading 6 and 7. The greatest integer he can't make by adding post-its is B. Find A+B.

8. The space diagonal of a cube is 12 \sqrt{3} units. Find the volume of the sphere inscribed inside the cube, in cubic units. Express your answer in terms of pi.

9. Chester rolls two die with faces numbered 1 to 6 and Jerry rolls two die with faces numbered 1 to 4. What is the probability that the sum of sum of Chester's die is greater than the sum of Jerry's die?

10. Jay's favorite ice cream flavors in no particular order are VANILLA, CHOCOLATE, MINT, STRAWBERRY and RASPBERRY. He likes Chocolate more than Raspberry, but less than Mint. He likes Strawberry more than Vanilla. Mint is not his favorite. How many permutations can be made in the letters of Jay's favorite flavor?

Good luck. The top 15 will advance to Part Two.

can someone latexify some of the problems so they are easier to read?
(I'm too lazy")

Since I'm bored--you are in luck. :P

1. If $ \frac {x}{x \minus{} 2} \equal{} \frac {y \minus{} 2}{y}$, find $ y$ in terms of $ x$.

2. The first letter in a list is a $ Z$, followed by $ 2$ $ Y$s, $ 3$ $ X$s, $ 4$ $ W$s, $ 5$ $ V$s, etc. What is the $ 189$th letter on the list?

3. If $ a\#b \equal{} \frac {a(a \plus{} 1) \plus{} b}{2}$ and $ f(a) \equal{} a \minus{} 6$, find $ 6\#f(4)$.

4. Horace and Joe are playing a coin flip game. Horace flips the coin twice, then Joe flips it once. This continues until someone flips a tail and loses. What is the probability that Joe wins? (Assume that if one person loses, the other wins.)

5. Find $ y$ if $ y$ is the smallest multiple of $ 8$ such that $ |y| < 13\pi$.

6. The numbers $ 10$ through $ 25$ are arranged in a $ 4$x$ 4$ grid such that every column has the same sum. What is this sum?

7. Jason has an infinite supply of post-it notes that read $ 3$ and $ 5$. The greatest integer he can't make by adding post-its is $ A$. Frank has an infinite supply of post-it notes reading $ 6$ and $ 7$. The greatest integer he can't make by adding post-its is $ B$. Find $ A \plus{} B$.

8. The space diagonal of a cube is $ 12\sqrt {3}$ units. Find the volume of the sphere inscribed inside the cube, in cubic units. Express your answer in terms of pi.

9. Chester rolls two die with faces numbered $ 1$ to $ 6$ and Jerry rolls two die with faces numbered $ 1$ to $ 4$. What is the probability that the sum of sum of Chester's die is greater than the sum of Jerry's die?

10. Jay's favorite ice cream flavors in no particular order are VANILLA, CHOCOLATE, MINT, STRAWBERRY and RASPBERRY. He likes Chocolate more than Raspberry, but less than Mint. He likes Strawberry more than Vanilla. Mint is not his favorite. How many permutations can be made in the letters of Jay's favorite flavor?

edit: if you can, you should probably NOT go by my stuff, cause I might've typoed. No blaming me for anything. :D

Beaten to it! oh well i made a PDF version
EDIT: I too take no responsibility and am not liable to typos in my latex!

Um...can you let some high schoolers or people check the problems and say what they think the answers are? Because some of these are a little ambiguous.

I've read through briefly and I haven't spotted any gaping errors; however, I haven't actually tried to solve anything, so I could have missed some possibilities.

math154, do you have any specific problem numbers you think are ambiguous?

(I'm currently a freshman)

Well, if you use common sense,there aren't any typoes, but I mean, if you look at it rigorously...w/e I mainly meant like let other non-participants check the answers.

um i have a question about number ten and the definition of permutation, for instance since "all" has two l's are you supposed to count that as one of the possibilities when scrambling it around?
all x2
lal x2
lla x2
here so basically instead of three would their be six?

I don't think so.

the number of ways that arranging 'all' would be 3.

l and l counts as same.

Due to ambiguity, I will accept 2 answers to number 10. I'll get my dad to check the answers.

Thanks to those who latexed it.

Um...can you post the correct answers here after each deadline so they're open for discussion etc.?

What are you talking about, cognos?

Edit: Does weight for number 4 mean the chance that it is rolled or the chances that the number across from it is rolled?

I think he meant to post in yongyi's mock test thread?

Never mind, I got a different answer for 3 now, and it doesn't work for 4...

By the way, I will be gone 8/4-8/16 for a EPGY summer camp. Dunno if I will be able to do Round 1.

I am pretty sure that number 3 gives a very nice answer. Only 3 people sent me their answers.

did you mean for the answer to number 3 to be left in unsimplified form?  b/c if so, the answer would work for 4, but I think that's a little ambiguous...

I simplified, then I sent it in... did I get number 3 right? Because it looks way too nice for me.

Can you post round 1 questions? Or delay the end of round 1 a bit? Since I'll be gone 8/4-8/16...

Well, if he meant it to be unsimplified, i think i would like to request an extension for the last set, b/c it didn't seem clear to me, and other ppl as well (if they had they same problem)

If there is an extention, I can't do it, because I am leaving in 3 hours :(

I didn't get enough people to send in answers. I'll give until the end of August for July's questions, and then give new questions in September. The regular season will be condensed from 4 months to 3 months. The Failure Cup Championship will still be in October.

It would help if you answered our questions...

Whether or not our questions on the prev set of questions get answered, I unfortunately will have to drop out, WOOT will be taking up too much of my time when it starts

Even if you don't participate, you will qualify for the championship.

So here are our limited results for July/August and our problems for September.

In seventh place,
[hide]7+0=7 echen[/hide]

In sixth place,
[hide]11+0=11 ghjk[/hide]

In fifth place,
[hide]12+0=12 vamathletes[/hide]

In fourth place,
[hide]13+0=13 math154[/hide]

In second place,
[hide]14+1=15 chenhsi and 15+0=15 Quickster94[/hide]

In first place,
[hide]13+6=19 Bacteria
Good job for getting first again![/hide]

I don't think there was a problem with the questions, because the people who thought there were plug-in problems with 3 and 4 had gotten other questions incorrect, and also Bacteria got number 3 correct.

That aside, our updated standings are

1. Bacteria: 13.5
2. Quickster94: 12
3. math154: 8
4. vamathletes: 7
5. chenhsi: 6.5
6. xpmath: 4
7. ghjk: 2
7. abacadaea: 2
9. echen: 1

7 more people will earn points this month (possibly already on the list) and than the top 10 will qualify for the Failure Cup Championship.

Here are September's round one problems. The first 5 are each one and the next 5 are each two.

1. Find the sum of the infinite geometric series, 9,-3,1, -1/3...

2. John, Barack, and Write-in are playing the election game. There is a 1/2 chance that Barack wins outright, a 2/5 chance that John wins outright, a 1/30 chance that they tie and flip a fair coin to determine the winner, there is a 1/20 chance that Barack and Write-in tie and need to flip a coin, and a 1/60 chance that John and Write-in tie and need to flip a coin. What is the probability that John does not win?

3. If a+b=c, a+c=b, and b+c=a, find a+b+c.

4. a=10b and b>a. What must true be true about b?

5. In a group of 36 students, 19 went to MathCounts Nationals, while 12 went to USAMO. If  8 went to neither, how many went to both?

6. Express 24 in base (-2).

7. An example of an up-and-down number is 1537241. How many 4 digit up-and-down numbers can be made using the numbers 1,2,3, and 4 each once?

8. Quentin Quaffledorf gets 17 points for every "Q" in his name and trades each point for an 8x8 chessboard. Quentin counts all the squares (not just the 1x1s) on all the chessboards, but counts wrong, and is off by 3. What are the 2 possible numbers Quentin counted?

9. A unit equilateral triangle and a unit square are both inscribed in circles. Find the positive difference between the areas of the circles.

10. If c^3-d^3=a and c^2-d^2=b and c=6, find a in terms of b and d.

Good luck, answers are due on September 14th.

Well, could you post the correct answers to the last round???


function
algebra
polynomial
limit
algebra unsolved

find all functions f: ${{\{R}^{+}to \{R}^{+}$ satisfying the condition
f(f(x))=6x-f(x)  for all positive number x

If f(x) consists of a finite sum of powers of x then f(x) = ax + b since if there is a term cx^n with n not 1 then the power x^(2n) appears in f(f(x)) and hence in f(x).  But then the power x^(4n) appears in f(f(x)) and so on indefinitely so that f(x) does not have a finite number of terms.

Then f(f(x)) = a(ax + b) + b = a^2x + b(a + 1) = 6x - (ax + b).    (*)

Therefore a^2 + a = 6, a = -3 or +2.  f(x) = -3x + b does not map from positive reals to positive reals for all x.  Therefore f(x) = 2x + b only.

By (*), b(a + 1) = -b, and since a = 2, b = 0.

Therefore f(x) = 2x is the only solution with finitely many terms.

This is the solution that works not only for polinomials:
First fix arbitrary [b]x>0[/b].Now Consider the sequence [b]a[1]=x,a[n+1]=f(a[n]).[/b]
From the equation we get:
[b]a[n+2]=-a[n+1]+6a[n][/b].The roots of the characteristic equation are [b]-3[/b] and [b]2[/b].
This means [b]a[n]=c1(-3)^n+c2.(2)^n[/b].
But now if [b]c1>0[/b],for sufficiently large and odd[b] n[/b] we get[b] a[n]<0[/b] which is a contradiction.Similarly for[b] c1<0[/b]  we get [b]a[n]<0[/b] for sufficiently large and even[b] n[/b].
All this means that [b]c1=0 [/b]so [b]a[2]=2a[1][/b] whch means [b]f(x)=2x[/b],since x was arbitrary the problem is solved.

This method is new to me but very clever.  Many thanks for posting.  I presume this generalises so that if
A(n + k) = (b1)A(n + k - 1) + (b2)A(n + k - 2) + ... ... (bk)A(n) where the b1 are real coefficeints then letting A(n + k) = x^k, the resulting polynomial equation is called the characteristic of the sequence?  And then A(n) = sum(1 = 1 to k)[(ci)(Ri)^n] where the Ri are the roots of this equation and the ci are constants to be determined by solving for k appropriate terms in the sequence?

What happens if the characteristic equation does not have k distinct real roots?

If your root has multiplicity $n$, then you have to construct a polynomial of degree $n-1$.

[quote="Hawk Tiger"]Have been posted many times.
It not hard to find that the solution of $f(f(x))+af(x)=bx$
where $a,b>0,f: R^{+}-\to R^{+}$[/quote]
Yes, see here http://www.mathlinks.ro/Forum/viewtopic.php?t=128763  :wink:

[quote="Hawk Tiger"]Have been posted many times.
It not hard to find that the solution of $f(f(x))+af(x)=bx$
where $a,b>0,f: R^{+}\to R^{+}$[/quote]
$x^{2}+ax-b=0$ has $2$ real solutions $\lambda>\mu$, as $a^{2}+4b>0$,
with $\lambda\mu=-b<0$, so $\lambda>0>\mu$, and $\lambda+\mu=-a<0$, so $0<\lambda<-\mu=|\mu|$.

Then $f^{(n)}(x)=\underbrace{f(f(\cdots f(x)\cdots))}_{n\text{ times}}={f(x)-\mu x\over\lambda-\mu}\lambda^{n}+{f(x)-\lambda x\over\mu-\lambda}\mu^{n}$ for all $x\in\mathbb{R}_{+}$ and $n\in\mathbb{N}_{0}$ by induction.
So in case $f(x)\not =\lambda x$ we have $\lim_{n\to \infty}{f^{(n)}(x)\over \mu^{n}}={f(x)-\lambda x\over\mu-\lambda}\not =0$,
and ${f^{(n)}(x)\over \mu^{n}}$ have all the same signs for all $n\in\mathbb{N}$ sufficiently large, which contradicts $\mu<0$ and $f^{(n)}(x)>0$ for all $n\in\mathbb{N}$.
So $f(x)=\lambda x$ for all $x\in\mathbb{R}_{+}$ is the only solution.


Let $ ABC$ is a triangle. Let $ BC\equal{}a, AC \equal{} b, CA\equal{}b$ and let $ a\plus{}b\plus{}c\equal{}2$.

Prove that $ a^2\plus{}b^2\plus{}c^2\plus{}2abc<2$.

[quote="thanhnam2902"]Let $ ABC$ is a triangle. Let $ BC \equal{} a, AC \equal{} b, CA \equal{} b$ and let $ a \plus{} b \plus{} c \equal{} 2$.

Prove that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc < 2$.[/quote]

We can prove the general inequality:
Let $ ABC$ is a triangle, $ BC\equal{}a,CA\equal{}b,AB\equal{}c$ and $ p\equal{}\frac{a\plus{}b\plus{}c}{2}$
Prove that
 $ 13(\frac{2p}{3})^3 \leq 2p(a^2\plus{}b^2\plus{}c^2)\plus{}4abc<4p^3$

for the first one,
put $ a\equal{}x\plus{}y ...$ and $ q\equal{}xy\plus{}yz\plus{}zx$ and $ r\equal{}xyz$
then we must prove that,
$ 2\sum x^2 \plus{}2q\plus{}2r< 2$ with $ x\plus{}y\plus{}z\equal{}1$
$ \Leftrightarrow 2(1\minus{}q)\plus{}2r< 2$
$ \Leftrightarrow q> r$
wich is true because by AM-GM, $ q\equal{}pq\geq 9r>\equal{}r$ equality hold if $ r\equal{}0$ and $ x\equal{}y\equal{}z$ wich is imppossible!
then $ q>r$

Yes! Thank you very much. This is a solution similar the [b]peine[/b]'solution.

Let $ a \equal{} x \plus{} y, b \equal{} y \plus{} z, c \equal{} z \plus{} x$. Easy we get $ x \equal{} \frac {a \plus{} c \minus{} b}{2}, y \equal{} \frac {a \plus{} b \minus{} c}{2}, z \equal{} \frac {b \plus{} c \minus{} a}{2} \Longrightarrow x,y,z > 0$.

Because $ a \plus{} b \plus{} c \equal{} 2 \Longrightarrow x \plus{} y \plus{} z \equal{} 1$.

Hence the inequality $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc < 2$ become:

$ (x \plus{} y)^2 \plus{} (y \plus{} z)^2 \plus{} (z \plus{} x)^2 \plus{} 2(x \plus{} y)(y \plus{} z)(z \plus{} x) < 2$

$ \Longleftrightarrow 2(x^2 \plus{} y^2 \plus{} z^2) \plus{} 2(xy \plus{} yz \plus{} zx) \plus{} 2(1 \minus{} z)(1 \minus{} x)(1 \minus{} y) < 2$

$ \Longleftrightarrow 2\left [ (x \plus{} y \plus{} z)^2 \minus{} 2(xy \plus{} yz \plus{} zx) \right ] \plus{} 2(xy \plus{} yz \plus{} zx)$ $ \plus{} 2\left [1 \minus{} (x \plus{} y \plus{} z) \plus{} (xy \plus{} yz \plus{} zx) \minus{} xyz\right ] < 2$

$ \Longleftrightarrow 2\left [ 1 \minus{} 2(xy \plus{} yz \plus{} zx) \right ] \plus{} 2(xy \plus{} yz \plus{} zx)$ $ \plus{} 2\left [1 \minus{} 1 \plus{} (xy \plus{} yz \plus{} zx) \minus{} xyz\right ] < 2$

$ \Longleftrightarrow \minus{} xyz < 0$, it's true because $ x,y,z > 0$.

Hence problem have done.


You can use the definition of combination: $ _{n}C_{r}\equal{}\frac{n!}{r!(n\minus{}r)!}$.

Remark $ 0!\equal{}1$.

Or, you could also think of $ \binom{n}0$ as choosing 0 things, which you can only do in 1 way; take nothing.

In general, $ \binom{n}{k}\equal{}\binom{n}{n\minus{}k}$.  By definition, $ \binom{n}{k}\equal{}\frac{n!}{k!(n\minus{}k)!}$.
Then $ \binom{n}{n\minus{}k}\equal{}\frac{n!}{(n\minus{}k)!(n\minus{}(n\minus{}k))!}\equal{}\frac{n!}{k!(n\minus{}k)!}\equal{}\binom{n}{k}$.


For what values of $ m$ does the equation $ x^2 \plus{} (2m \plus{} 6)x \plus{} 4m \plus{} 12 \equal{} 0$ has two real roots, both of them greater than $ \minus{}1$.

Consider the condition such that the quadratic equation $ x^2\plus{}2ax\plus{}2a\equal{}0$ has 2real roots $ x\geq \minus{}1$.

Rewrite thse equation as $ \minus{}x^2\equal{}2a(x\plus{}1)$, by the graphs of $ y\equal{}\minus{}x^2$ and $ y\equal{}2a(x\plus{}1)$ we have the range of $ a$ is $ 2a\leq 0\Longleftrightarrow a\leq 0$.

Then the given problem is for $ a\equal{}2(m\plus{}3)$, yielding the answer is $ m\leq \minus{}3$.


trigonometry
limit
real analysis
integration
calculus
absolute value
real analysis unsolved

[quote="swag"]Show that:

$lim_{n \rightarrow \infty} \sin^{2}(nx) \neq 0$

in $D'(R)$.

Note that $lim_{n \rightarrow \infty} \sin(nx) = 0$[/quote]
$\sin^2(nx) = \frac{1}{2}\left(1- \cos(2nx) \right)$.

[quote="blahblahblah"]This isn't an answer, but $\lim_{n \rightarrow \infty} \sin(nx) = 0$ in $\mathcal{D}'(\mathbb{R})$ is just a consequence of the Riemann-Lebesgue lemma, right?[/quote]
You're working too hard. Nowhere near the full strength of the Riemann-Lebesgue lemma, more like a simple version of a preliminary lemma within that argument.

Let $\varphi\in\mathcal{D}$ be a test function. Then,

$\lim_{n\to\infty}\int_{-\infty}^{\infty}\sin(nx)\varphi(x)\,dx =$

$\lim_{n\to\infty} \left.-\frac1n \cos(nx)\varphi(x)\right|_{-\infty}^{\infty}+\frac1n\int_{-\infty}^ {\infty}\cos(nx)\varphi'(x)\,dx.$

The boundary terms are zero, and the absolute value of the integral is clearly bounded by

$\frac1n\int_{-\infty}^{\infty}|\varphi'(x)|\,dx$, which goes to zero.

And, of course, hpe just showed that $\sin^2(nx)\to\frac12$ in $\mathcal{D}'.$


geometry
parallelogram
vector
Locus problems
Locus
IMO Longlist
IMO Shortlist

Let $ABCD$ and $A^{\prime }B^{\prime}C^{\prime }D^{\prime }$ be two arbitrary parallelograms in the space, and let $M,$ $N,$ $P,$ $Q$ be points dividing the segments $AA^{\prime },$ $BB^{\prime },$ $CC^{\prime },$ $DD^{\prime }$ in equal ratios.

[b]a.)[/b] Prove that the quadrilateral $MNPQ$ is a parallelogram.

[b]b.)[/b] What is the locus of the center of the parallelogram $MNPQ,$ when the point $M$ moves on the segment $AA^{\prime }$ ?

(Consecutive vertices of the parallelograms are labelled in alphabetical order.

We use vectors, anf from relations of the type $MN=kAB+(1-k)A'B', PQ=kCD+(1-k)C'D'$ we find $MN+PQ=0$ (two-letter combinations are all vectors). This shows that $MNPQ$ is a parallelogram.

We use similar methods to show that $OP=kPO'$, where $O,O',P$ are the centers of $ABCD,A'B'C'D',MNPQ$ respectively. The locus is the segment $OO'$.

[Edit: it's not a solution, just an approach; this is to make orl happy :D]

Please modify your post. Look [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=113358#113358]here[/url].  :)

Here is my solution (sorry, Grobber, my notations conflict with yours, but actually I find your notations a bit confusing, since you have two different points labeled with P  :) ).

Since the points M, N, P, Q divide the segments AA', BB', CC', DD' in equal ratios, we can find a real number u such that u = AM : AA' = BN : BB' = CP : CC' = DQ : DD'. Then, if O is an arbitrary point in space, we have the vectorial identity

$\overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{AM}=\overrightarrow{OA}+u\cdot \overrightarrow{AA^{\prime }}=\overrightarrow{OA}+u\cdot \left( \overrightarrow{OA^{\prime }}-\overrightarrow{OA}\right) $
$=\left( 1-u\right) \cdot \overrightarrow{OA}+u\cdot \overrightarrow{OA^{\prime }}$,

and similarly

$\overrightarrow{ON}=\left( 1-u\right) \cdot \overrightarrow{OB}+u\cdot \overrightarrow{OB^{\prime }}$;
$\overrightarrow{OP}=\left( 1-u\right) \cdot \overrightarrow{OC}+u\cdot \overrightarrow{OC^{\prime }}$;
$\overrightarrow{OQ}=\left( 1-u\right) \cdot \overrightarrow{OD}+u\cdot \overrightarrow{OD^{\prime }}$.

Since the quadrilaterals ABCD and A'B'C'D' are parallelograms, we have $\overrightarrow{AB}=\overrightarrow{DC}$ and $\overrightarrow{A^{\prime }B^{\prime }}=\overrightarrow{D^{\prime }C^{\prime }}$. But

$\overrightarrow{MN}=\overrightarrow{ON}-\overrightarrow{OM}=\left( \left( 1-u\right) \cdot \overrightarrow{OB}+u\cdot \overrightarrow{OB^{\prime }}\right) -\left( \left( 1-u\right) \cdot \overrightarrow{OA}+u\cdot \overrightarrow{OA^{\prime }}\right) $
$=\left( 1-u\right) \cdot \left( \overrightarrow{OB}-\overrightarrow{OA}\right) +u\cdot \left( \overrightarrow{OB^{\prime }}-\overrightarrow{OA^{\prime }}\right) $
$=\left( 1-u\right) \cdot \overrightarrow{AB}+u\cdot \overrightarrow{A^{\prime }B^{\prime }}$,

and similarly

$\overrightarrow{QP}=\left( 1-u\right) \cdot \overrightarrow{DC}+u\cdot \overrightarrow{D^{\prime }C^{\prime }}$.

Since $\overrightarrow{AB}=\overrightarrow{DC}$ and $\overrightarrow{A^{\prime }B^{\prime }}=\overrightarrow{D^{\prime }C^{\prime }}$, we thus get $\overrightarrow{MN}=\overrightarrow{QP}$, and it follows that the quadrilateral MNPQ is a parallelogram. This completes the solution of part [b]a.)[/b]. Now, if U, V and W are the centers of the parallelograms ABCD, A'B'C'D' and MNPQ, respectively, then these points U, V, W are the midpoints of the segments AC, A'C', MP (since the center of a parallelogram is the midpoint of any of its diagonals), and thus we have

$\overrightarrow{OU}=\frac{1}{2}\cdot \left( \overrightarrow{OA}+\overrightarrow{OC}\right) $;
$\overrightarrow{OV}=\frac{1}{2}\cdot \left( \overrightarrow{OA^{\prime }}+\overrightarrow{OC^{\prime }}\right) $;
$\overrightarrow{OW}=\frac{1}{2}\cdot \left( \overrightarrow{OM}+\overrightarrow{OP}\right) $.

The last equation yields

$\overrightarrow{OW}=\frac{1}{2}\cdot \left( \overrightarrow{OM}+\overrightarrow{OP}\right) $
$=\frac{1}{2}\cdot \left( \left( \left( 1-u\right) \cdot \overrightarrow{OA}+u\cdot \overrightarrow{OA^{\prime }}\right) +\left( \left( 1-u\right) \cdot \overrightarrow{OC}+u\cdot \overrightarrow{OC^{\prime }}\right) \right) $
$=\frac{1}{2}\cdot \left( \left( 1-u\right) \cdot \left( \overrightarrow{OA}+\overrightarrow{OC}\right) +u\cdot \left( \overrightarrow{OA^{\prime }}+\overrightarrow{OC^{\prime }}\right) \right) $
$=\left( 1-u\right) \cdot \frac{1}{2}\cdot \left( \overrightarrow{OA}+\overrightarrow{OC}\right) +u\cdot \frac{1}{2}\cdot \left( \overrightarrow{OA^{\prime }}+\overrightarrow{OC^{\prime }}\right) $
$=\left( 1-u\right) \cdot \overrightarrow{OU}+u\cdot \overrightarrow{OV}$.

This shows that the point W lies on the segment UV and that we have UW : UV = u. When the point M moves on the segment AA', the ratio u = AM : AA' varies from 0 to 1 (in fact, we have u = 0 for M = A, and u = 1 for M = A'). Thus, while the point M moves on the segment AA', the locus of the point W is the segment UV (hereby, every point on this segment can be considered as the point W for some value of u). Thus we have shown that the locus of the center W of the parallelogram MNPQ is the segment UV. Note that this locus degenerates to a single point if and only if U = V, i. e. if and only if the parallelograms ABCD and A'B'C'D' have the same center. Thus the solution of part [b]b.)[/b] is complete.

  Darij


linear algebra
matrix
algebra
polynomial
Gauss
geometry
3D geometry

Lets consider the 3rd problem(it's my favourite) :D

Lets consider C=U-In
and we have that U^2=nU
C^(-1)=aU+bIn =>
C*C^(-1)=aU^2-aU+bU-bIn=In =>
[(n-1)a+b]U-(b+1)In=0
b=-1
a=-b/(n-1)=1/(n-1)
so C^(-1)=1/(n-1)*U-In

cheers! :D:D

For the 4th one I computed D^2 and D^4 hope that the answers are correct :)

D^2=4*P.
where
P=
1 0 0 0
0 0 0 1
0 0 1 0
0 1 0 0

D^4=4^2I4
 for the inverse if det(D)<>0 then I have thought to do this D*X=In and solve this matrcial ecuation!
also for the 1st problem the same way AX=In and solve the ecuation.
They are not very inteleigent solutions, if someone has the fastest solution please post! :D :D

cheers!

Hello Arne,

I dont know this technic Gauss-Jordan , you did not write the colomns
6->10,

If a invertible matrix is triangular superior with diagonal (a,b,c,d,e),
its inverse has on its diagonal (1/a,1/b,1/c,1/d,1/e).

For triangular matrix with size not to big we can write a system from AX=Y as Lagrangia said, and by combination express x_i in terms y_i

x_1+2x_2+3x_+4x_4+5x_5=y_1
x_2+2x_3+3x_4+4x_5=y_2
x_3+2x_4+3x_5=y_3
x_4+2x_5=y_4
x_5=y_5

gives after computation
x_5=y_5
x_4=y_4-2y_5
x_3=y_3-2y_4+y_5
x_2=y_2-2y_3+y_4
x_1=y_1-2y_2+y_3

A^(-1)=
1  -2  1  0  0
0  1  -2  1  0
0  0   1 -2  1
0  0   0  1 -2
0  0   0  0  1

I would like to see the Gauss-Jordan technic, if you don't mind
can you write it Arne ?

cheers
oubinool

[quote="liyi"]
by the way, what does M_n(R) mean? I've never seen this sign before.[/quote]

it means the set of matrices n times n with real elements. It's a Romanian standard notation. Also French, by the looks of it :D

Well, I found this technique to calculate an inverse matrix in a schoolbook of mine.

It says :

Given a matrix A in Rnxn, if you want to find A-1 then :

* write the matrix B in Rnx2n with A in the first n columns and In in the last n columns ;
* reduce B to its canonical form ;
* then you get In in the first n columns and A-1 in the last n columns.

Correct ?

This method can be applied with a TI 83 or so. (That is the "rref" - function.)

It is very difficult to type all calculations (which are straightforward !).
I tried it but it becomes incomprehensible :(

Good work liyi  :)  for the 2th problem

6) Find all matrix A=
[a b]
[c d] \in M_2(R) such that
for any k \in N\{0}
A^k=
[a^k b^k]
[c^k d^k]

Arne i remember  :)  my teacher learned me this technic Gauss-Jordan elimination
http://mathworld.wolfram.com/Gauss-JordanElimination.html
in 1985 when i was in grade 12th in Reunion Island when he explained
i did not pay attention because a chameleon arrived in the classroom by the window, so i was playing :D . In Paris i learned more technics for inverse matrix here some

*solve the system AX=Y express x_i in terms of y_i and a(ij)
Exercice: find the inverse of problem 1) with the system technic.

**find a polynome Q(x)=x^p+c_(p-1)x^(p-1)+...+c_1x+c_0  whith c_0=/=0 such that Q(A)=0
A^p+c_(p-1)A^(p-1)+...+c_1A+c_0.Id=0 then
A( A^(p-1)+c_(p-1)A^(p-2)+...+c_1.Id)/(-c_0) = Id by unicity of the inverse
A^(-1)= ( A^(p-1)+c_(p-1)A^(p-2)+...+c_1.Id)/(-c_0)
 7) Use this technic for the inverse of
A=
[0 1 1]
[1 0 1]
[1 1 0]

T=
[0 0 ......................................1]
[0 0....................................1  0]
[0 0.................................1  0  0]
..............................................
..............................................
[1 0 .................................0  0 0]

*** we have the formula A.A*= detA.Id
where A* is the transpose of the comatrice of A
http://mathworld.wolfram.com/MatrixInverse.html
8) use this technic fo finding the inverse of A,B
A=
[1 2002 ]
[2003 2004]

B=
[1 1 2]
[3 4 5]
[6 7 8]

**** 9) By using formal power serie find the inverse of
A=
[1 2 ......................................n]
[0 1 2 ..............................(n-1)]
..................................................
..................................................
[0 0 0 .......................................1] only 1 on its diagonal

consider (1-x)(\sum_{k>=0}k.x^(k-1))

***** Using polynom
10) Find the inverse of A=(a(i,j))=( C(j-1,i-1)) whith j \geq i
Find a linear map f:R_(n-1)[X]-->R_(n-1)[X] such that the
matrix of f in the canonical basis (1,X,...,X^(n-1)) is A and give the
inverse of f and A^(-1)

****** Find the inverse by product of bloc matrix

cheese heu sorry cheers  :)
oubinool

Sorry the guest was oubinool i don't know why sometimes internet
conection cut
chers
oubi

6)

Lets do the following thing:
A^k=A^(k-1)*A
so we have that:
B=CD=E
where
B=
a^k b^k
c^k d^k

C=
a^(k-1) b^(k-1)
c^(k-1) d^(k-1)

D=
a b
c d

E=
a^k+cb^(k-1)  a^(k-1)+b^(k-1)d
c^(k-1)a+cd^(k-1)   c^(k-1)b+d^k

from where we get that:
a^k =a^k+cb^(k-1)
b^k = a^(k-1)+b^(k-1)d
c^k  = c^(k-1)a+cd^(k-1)
d^k  =c^(k-1)b+d^k

so we get that:
cb^(k-1)=0
c(c^(k-1)-d^(k-1))=ac^(k-1)
b(b^(k-1)-a^(k-1))=db^(k-1)
bc^(k-1)=0

we have now the cases:
1) c=0,b=0 and we get that
A=
a 0
0 d
2)c=0, b<>0 we get :
a)d=0,a=b
A=
a a
0 0
b) d<>0
A=
0 b
0 b
3)c<>0,b=0
a)a=0
A=
0 0
c c
b)a<>0
A=
a 0
0 a
these are all cases cheers! :D :D

Yes Lagrangia good work for6)  :) , for 7) the T compute T
oubi

Hi Lagrangia,

For the 6th my solution is in the same spirit of your, just work with
A, A^3 here

A=
(a b)
(c d)
=
(a+bc  ab+bd)
(ca+dc  cb+d)

we have the system

bc=0
(a+d-b)b=0
(a+d-c)c=0

if c=/=0
then b=0 and a+d=c

A^3=
(a^3    0)
(c^3 d^3)

we get with the cube of A
c(a+ad+d)=c^3
ad=0
So A is of the form
(0 0)
(c c)

(a 0)
(a 0)

If b=/=0
A is of the form
(0 b)
(0 b)

or

(a a)
(0 0)

if b=c=0
A=
(a 0)
(0 d)

chers
oubi

11) you probably know that every real number can be the limit of a sequence of rational numbers, Q is dense in R. We have something like that in Mn(R)

Prove that any A \in Mn(R) is a limit of  a sequence W_p invertible
matrix.
------------------------------------------------------------------
11 bis) application of 11)
let A,B _in Mn(R) prove that det(A.B -Xid) = det (B.A-XId) by the way
det(M-Xid) is called characteristic polynomial
http://mathworld.wolfram.com/CharacteristicPolynomial.html
---------------------------------------------------------------
12) let A,B \in Mn(R) and C= A+i.B prove that if A,C are invertible
then  T = A + B.A^(-1) is invertible and give the expression of C^(-1)
in terms of A,B,M
----------------------------------------------------------------------
 :D  \cup \beta  \int  \cap  :D  :D  \lambda
http://volcano.und.nodak.edu/vwdocs/volc_images/africa/Piton_Fournaise9.jpg

In grade 8th our teacher of biology/geology went the clasrom for
an excursions on this volcano Piton de la Fournaise on Runion Island.

we know that similar matrices have the same characteristic polynomial.
if A ~ B, there exists a inverse matrix X, such that X-1AX=B. hence,
det (B- \lambda I) = det(X-1AX - \lambda I) = det(X-1(A-\lambda I)X) = det(X-1)det(A-\lambda I)det(X) = det(A-\lambda I).

Are A.B and B.A similar?

[quote="Moubinool"]No AB, BA are not similar[/quote]

ehm... if there is a non-singular matrix in A and B, then AB ~ BA.
If A and B are both singular, AB and BA are not similar.

First consider the first  case A is invertible, and use properties on
determinant for 11 bis


\[128=2^{7}\]
We can rewrite the first equation to
\[2^{x^{2}+y-7}+2^{x+y{2}-7}=1 \]
The 2nd equation tells us that
\[x,y\geqq0 \]
.

It is clear that
\[x^{2}+y<7 \]
and
\[x+y^{2}<7 \]
, so it is easy to see that
\[2\leqq{x,y}<3 \]
. If you look at the 2nd equation we have to have that
\[x=y=2 \]

[quote="Meghrabi"], so it is easy to see that
\[2\leqq{x,y}<3\]
[/quote]

Could you elaborate? The only thing I'm obtaining is that $\sqrt{7}\geq \max(x,y) \geq 2 \geq \min(x,y) \geq 0$...

[quote="Kurt Gödel"][quote="Meghrabi"], so it is easy to see that
\[2\leqq{x,y}<3 \]
[/quote]

Could you elaborate? The only thing I'm obtaining is that $\sqrt{7}\geq \max(x,y) \geq 2 \geq \min(x,y) \geq 0$...[/quote]

You can change $3$ by $\sqrt{7}$, but there is not equality because then you would have 1+something (>0)=1.

But I was to quick, from the 2 ineqs (if we subtract them in 2 ways (the ineq stays because the leftsides are both pos)): $(x-y)(x+y) < (x-y)$ , $(y-x)(y+x) < (y-x)$
In the first case we say x>y and we see that it doesn't work, in the 2nd case we see that y>x also doesnt work, we conclude x=y ...

[quote="Meghrabi"]But I was to quick, from the 2 ineqs (if we subtract them in 2 ways (the ineq stays because the leftsides are both pos)): $(x-y)(x+y) < (x-y)$ , $(y-x)(y+x) < (y-x)$
In the first case we say x>y and we see that it doesn't work, in the 2nd case we see that y>x also doesnt work, we conclude x=y ...[/quote]

This is wrong. You substracted the inequalities in a wrong way. If $a < b$ and $c < d$, then $a-d < b-c$, and not necessarily $a-c < b-d$.


trigonometry
Euler
calculus
derivative
real analysis
real analysis unsolved

Discuss the convergence (absolutely and conditionally) of the following series for $ a \in \Bbb N$:
\[ \sum_{n \equal{} 1}^\infty (\sin(\pi n! e))^a\]
, where $ e$ is the Euler constant.

$ n!e\equal{}N_n\plus{}b_n,$ where $ b_n\equal{}\frac1{n\plus{}1}\plus{}\frac1{(n\plus{}1)(n\plus{}2)}\plus{}\cdots.$

Exploring this for small values of $ n,$ we see that $ N_n$ is even when $ n$ is odd and odd when $ n$ is even. That means that the derivative of $ \sin{\pi x}$ is equal to $ (\minus{}1)^{n\plus{}1}$ when $ x\equal{}N_n.$

As for $ b_n,$ we see that $ \frac1{n\plus{}1}<b_n<\frac1n.$

Hence, $ \sin(\pi n!e)\equal{}(\minus{}1)^{n\plus{}1}\sin(\pi b_n),$ which is comparable in size to $ \frac1n.$

For $ a\ge 2,$ the series converges absolutely by comparison to $ \sum\frac1{n^a}.$

For $ a\equal{}1,$ the series does not converge absolutely, but the alternating series test applies and the series converges conditionally. (Note that $ b_n,$ hence $ \sin(\pi b_n),$ is strictly decreasing.)


geometry
cyclic quadrilateral
perpendicular bisector

Let ABCD be a cyclic quadrilateral and M be the mid-point of BD. Suppose $\angle AMB=\angle CMB$. Prove or disprove $\angle ADC=\angle AMB$.

I can tell you that your assertion is true. Anyway, DO NOT post any topic related to the problem set given at our training again before the expiration date. Such kinds of posts are strictly forbidden in mathlinks.

This topic is unlocked now, so anyone who wants to have a go at this nice problem is welcome to do so!

[hide="Eventually, I can solve it myself..."]$\angle ADC=\angle AMB$
$\Leftrightarrow \triangle ADC\sim\triangle AMB$
$\Leftrightarrow \angle BAM=\angle CAD$
$\Leftrightarrow \angle BAM=\angle CBM$
$\Leftrightarrow \triangle ABM\sim\triangle BCM$
$\Leftrightarrow AM\cdot CM=BM^2=BM\cdot MD$
$\Leftrightarrow$ (Let the extension of AM intersect the circle at P) $MC=MP$
This can be proved easily by reflecting $\triangle BMC$ with respect to the perpendicular bisector of BD...[/hide]
PS Sorry for posting this problem before the date due. I didn't know that this is forbidden. I won't do it anymore. :(


Show that 2^(3^k) + 1 is divisible by 3^k for all positive integers k.

[hide]

2 = -1 mod 3

so we have -1^3^k + 1 = 0 mod 3

but 3^k is always odd, so -1^odd power = -1, so

-1+1 = 0 mod 3

[/hide]

Yep that looks good to me Belenos.. TheSouthpaw, I think you only showed its divisible by 3..


algebra
polynomial
function
geometry
3D geometry
number theory unsolved
number theory

Let $ f$ be a third degree polynomial function with integer coefficients such that for each nonnegative integer $ n$ there is a nonnegative number $ k_n$ for which $ f(n) \equal{} k_n^3$.
Prove that there are integers $ a, b$ such that $ f(x) \equal{} (ax \plus{} b)^3$.

[ [b]Kvant 1823[/b] ]

see:[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=326560#326560[/url]
posts 8 , 10


Prove that the set of integers of the form 2^k-3(k=2,3,......) contains an infinite subset in which every two members are relatively prime.