Well, I actually have a problem that I need help on. :)
If anyone has any ideas or can give me references, I'd be grateful:

(and don't worry, you're not solving my hw - well, you are, but it's an independent study class and the hw was "ask around about this problem")

We're given objects of n types. We start off with one object of type 1.
We have fixed probabilities p_ij that an object of type i will become an
object of type j. An object of type i can also be "duplicated" into an object
of type i (so one object of a certain type can become two objects of the same type etc), and it can also "disappear," in which case we no longer take it into account. There are fixed probabilities for "duplication" and "disappearance" as well.

At the end of our "process," after a certain number of "rounds," we have different
"realizations," which come from different paths that could have been taken at some
point along the way.
Knowing these "realizations," ie how many objects of types i we have at the end, for
every i between 1 and n, is there some way to approximate all of the transition
probabilities above or at least some of them?

Thank you,
Maria

Well, I am just giving it a quick read, but it seems to be a problem of Marcov Chains. We have the object of class $i$ that has probability $p_{ij}$ to become an object of class $j$. If just this can happen, then, you define $P=(p_{ij})_{n\times n}$ and if you have $x_k$ the objects in the step $k$, then $x_k = Px_{k-1}$.

On the other hand, well, I don't really have a clear idea of how to fix the problem that there is also probability of doubling or vanishing, but look for Finite Marcov Chains, and probably it will be a good point to start.

Best regards,


The length of a rectangle is increased by $ 20\%$ and the width is reduced by $ 20\%$. Write the ratio of the original area to the area of the new rectangle.

Suppose the rectangle was 10 by 10. The new rectangle would have lengths 12 and 8. It is easy to see that the ratio of the areas is $ \frac{100}{96} \equal{} \frac{25}{24}$.

[hide="General"]
Let $ s\equal{}\text{the side length}.$
The new rectangle has sides $ \frac{4}{5}s$ and $ \frac{6}{5}s.$ The area of the new rectangle is thus $ \frac{24}{25}s^2.$
The original area is $ s^2,$ and the ratio is $ \frac{s^2}{\frac{24}{25}s^2}\equal{}\boxed{\frac{25}{24}}.$[/hide]

[quote="mrsaturns"]I just need to ask. The problem does not state to make the ratio into a common fraction. Would you get it wrong if you answered it as a common fraction? Isn't the standard way to do a ratio $ 25: 24$ ?

Anyone know if you have to express this answer as a common fraction or with ":'s"
The problem doesn't say to put it as a common fraction.[/quote]
As a rule, if the problem asks for a ratio, you are supposed to write it as a common fraction.

When I was doing this problem I almost read[quote=FTW]the original area to the area of the new rectangle[/quote]as[quote=not FTW]the area of the new rectangle to the original area[/quote]and almost put in $\frac{24}{25}$ XD


$ p(n)$ is the least prime $ >$ n
$ q(n)$ is the biggest prime$ \leq n$
$ x_n \equal{} \sum _{k \equal{} 2}^n \frac {1}{p(k).q(k)}$
Prove that $ x_n$ converge and find it limit.

More $ h(n)$ is the least prime $ \geq$ n
Let
$ x_n \equal{} \sum_{k \equal{} 2}^n \frac {1}{h(k)q(k)}$
Prove that $ x_n$converge.

Call $ p_i$ the $ i$-th prime.
Now
\[ x_{p_m\minus{}1} \equal{} \sum _{k \equal{} 2}^{p_m\minus{}1} \frac {1}{p(k) \cdot q(k)} \equal{}\\
\equal{} \sum_{i\equal{}1}^{m\minus{}1} \sum_{p_i \leq k < p_{i\plus{}1}} \frac {1}{p(k) \cdot q(k)} \equal{} \\
\equal{} \sum_{i\equal{}1}^{m\minus{}1} \sum_{p_i \leq k < p_{i\plus{}1}} \frac {1}{p_i \cdot p_{i\plus{}1}} \equal{} \\
\equal{} \sum_{i\equal{}1}^{m\minus{}1} \sum_{p_i \leq k < p_{i\plus{}1}} \frac 1{p_{i\plus{}1}\minus{}p_i} \cdot \left( \frac 1{p_i} \minus{} \frac 1{p_{i\plus{}1}} \right) \equal{}\\
\equal{} \sum_{i\equal{}1}^{m\minus{}1} \left( \frac 1{p_i} \minus{} \frac 1{p_{i\plus{}1}} \right) \equal{} \frac 1{p_1} \minus{} \frac 1{p_m}.\]
Thus $ x_n$ converges to $ \frac 1{p_1} \equal{} \frac 12$.

Good solution. :wink:
But I have prove that $ x_n$(in the 2 problem ) converge but now I haven't find out it.
If we can find out it?

If $ x_n \equal{} \sum _{k \equal{} 2}^n \frac {1}{p(k) \cdot q(k)}$ and $ y_n \equal{} \sum_{k \equal{} 2}^n \frac {1}{h(k) \cdot q(k)}$, then by $ \frac {1}{h(k\plus{}1) \cdot q(k\plus{}1)} \equal{} \frac {1}{p(k) \cdot q(k\plus{}1)} \leq \frac {1}{p(k) \cdot q(k)}$ we get that $ y_k \leq x_{k\plus{}1}$, thus $ y_k$ converges as $ x_k$ does.

Yes it easy but which is the ponit converge.?


Can anyone solve:
$ 1) \frac {x}{1 \plus{} x^2} \plus{} \frac {y}{1 \plus{} y^2} \plus{} \frac {z}{1 \plus{} z^2}\le\frac {3\sqrt {3}}{4}, x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$
$ 2) \frac {x}{1 \minus{} x^2} \plus{} \frac {y}{1 \minus{} y^2} \plus{} \frac {y}{1 \minus{} z^2}\ge\frac {3\sqrt {3}}{2}, x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$
$ x,y,z, \ge0$
using only AM-GM?

Let $ x\equal{}\frac{a}{\sqrt{3}}$ , $ y\equal{}\frac{b}{\sqrt{3}}$ , $ z\equal{}\frac{c}{\sqrt{3}}$ . Inequality equivalent to $ \frac{a}{a^2\plus{}3}\plus{}\frac{b}{b^2\plus{}3}\plus{}\frac{c}{c^2\plus{}3}\leq\frac{3}{4}$ . By AM-GM $ a^2\plus{}1\geq2a$ and $ a^2\plus{}3\geq2a\plus{}2$ .We will prove that $ \frac{a}{a\plus{}1}\plus{}\frac{b}{b\plus{}1}\plus{}\frac{c}{c\plus{}1}\leq\frac{3}{2}$ $ \Longleftrightarrow$ $ \frac{1}{a\plus{}1}\plus{}\frac{1}{b\plus{}1}\plus{}\frac{1}{c\plus{}1}\geq\frac{3}{2}$  ,    $ \frac{1}{a\plus{}1}\plus{}\frac{1}{b\plus{}1}\plus{}\frac{1}{c\plus{}1}\geq\frac{9}{a\plus{}b\plus{}c\plus{}3}$   and   $ (a\plus{}b\plus{}c)^2\leq3(a^2\plus{}b^2\plus{}c^2)$ , $ 3(x^2\plus{}y^2\plus{}z^2)\equal{}a^2\plus{}b^2\plus{}c^2\equal{}3$ , $ \frac{1}{a\plus{}1}\plus{}\frac{1}{b\plus{}1}\plus{}\frac{1}{c\plus{}1}\geq\frac{3}{2}$.

[quote="arqady"][quote="Flakky"]Can anyone solve:

$ 2) \frac {x}{1 \minus{} x^2} \plus{} \frac {y}{1 \minus{} y^2} \plus{} \frac {y}{1 \minus{} z^2}\ge\frac {3\sqrt {3}}{2}, x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$
$ x,y,z, \ge0$
using only AM-GM?[/quote]
$ \sum_{cyc}\frac {x}{1 \minus{} x^2}\geq\frac {3\sqrt {3}}{2}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac {x}{1 \minus{} x^2} \minus{} \frac {\sqrt3}{2} \minus{} \frac {3\sqrt {3}}{2}\left(x^2 \minus{} \frac {1}{3}\right)\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {x(3\sqrt3x^3 \minus{} 3\sqrt3x \plus{} 2)}{1 \minus{} x^2}\geq0.$
But by AM-GM we obtain:
 $ 3\sqrt3x^3 \minus{} 3\sqrt3x \plus{} 2 \equal{} 3\sqrt3x^3 \plus{} 1 \plus{} 1 \minus{} 3\sqrt3x\geq3\sqrt [3]{3\sqrt3x^3\cdot1^2} \minus{} 3\sqrt3x \equal{} 0.$ :lol:
Thus, your inequality is proved.[/quote]
The same idea for $ 2)$ but $ \sum_{cyc}\frac {x}{1 \minus{} x^2}\geq\frac {3\sqrt {3}}{2}(x^2 \plus{} y^2 \plus{} z^2)\Leftrightarrow$$ \sum_{cyc}\frac {x}{1 \minus{} x^2} \minus{} \frac {3\sqrt {3}x^2}{2}\geq0\Leftrightarrow\sum_{cyc}\frac {2x \plus{} 3\sqrt {3}x^4 \minus{} 3\sqrt {3}x^2}{2(1 \minus{} x^2)}\geq0$ and so on.

Actually for the second one, we have $$\frac{x}{1-x^2}-\frac{3\sqrt{3}}{2}x^2=\frac{x}{2(1-x^2)}\cdot(\sqrt{3}x+2)(\sqrt{3}x-1)^2\geq 0. $$


Find least value of $ |3z \minus{} 13| \plus{} |3z \minus{} 11|$.
where $ z$ lies in argand plane.

By the triangle inequality, $ |3z \minus{} 13| \plus{} |3z \minus{} 11| \equal{} |3z \minus{} 13| \plus{} |11 \minus{} 3z| \geq |3z \minus{} 13 \plus{} 11 \minus{} 3z| \equal{} 2$. This is attained when $ z \equal{} \frac{13}{3}$.


vector
abstract algebra
superior algebra
superior algebra solved

Let G be a group such that every non-trivial element is of order 2,
  What can you say about the structure of G?
  [Hint: Use the concept of vector spaces, or better, modules.]

I know its very easy to say that G is abelian & it contains $2^n$ elements.But thats trivial ,I want a complete characterisation of the said group.

They can be seen as vector spaces over $\mathbb Z_2$. Every vector space has a basis, so these groups are the direct products of some copies of $\mathbb Z_2$.

Hmm I thought it would it would be isomorphic to $Z_2 X Z_2 X$..
but can you please explain how can they be seen as vector spaces over $Z_2$

Just check that all the axioms of a vector space are satisfied, man. If $a\in\mathbb Z_2$ (so $a$ is either $0$ or $1$), and $x$ is an element of our group, define $ax$ to be $0$ if $a=0$ and $x$ if $a=1$. That's all.


quadratics
inequalities
algebra
polynomial
difference of squares
special factorizations

[quote="SplashD"][quote="junggi"]Find all integer solutions (n,m) to
$n^{4}+2n^{3}+2n^{2}+2n+1=m^{2}$[/quote]

[hide]$n^{4}+2n^{3}+2n^{2}+2n+1=(n+1)^{2}(n^{2}+1)=m^{2}$, from the hint kurt gave us.

The only way that $n^{2}+1$ is a perfect square is if $n=0$, therefore, the solutions set is $(0,\pm1)$[/hide][/quote]

Yes, indeed.

[hide="Or"]$(n^{2}+n)^{2}< n^{4}+2n^{3}+2n^{2}+2n+1 \leq \left(n^{2}+n+1\right)^{2}$ with equality iff $n=0$.[/hide]

[b]Follow-up:[/b]

Find all integer solutions to $n^{4}+n^{3}+n^{2}+n+1 = m^{2}$.

$n^{4}+n^{3}+n^{2}+n+1=\frac{n^{5}-1}{n-1}$

nevermind i didnt really read the question that well.  i didnt want to erase it but i dont feel like writing more, it is very late.

Just look for similar bounds...it has to be a quadratic on both sides of the inequality, with difference less than 1.

$(n^{2}+n+1)^{2}-n(n+1)^{2}$

edit:  the rest of what i wrote was wrong

I don't quite understand how either of your factorizations did anything.

i forgot the n wasn't squared, so the difference of squares didnt work.  the result that followed (although wrong) showed that there were no solutions.

Should we try to find bounds between the polynomial?

Also, one of the bounds would be n^4, right?

Bounds will work, with some caveats.  I think there might actually be a much neater solution, but I have not found it yet.

You will have to be a bit more cunning with the bounds than before, and no, $n^{4}$ is too far away to be a bound ;)

[quote="Elemennop"]it has to be a quadratic on both sides of the inequality, with difference less than 1.[/quote]

hint, most of solution:

[hide]

Some REALLY tight bounds are  on the lower side (holds for n>0) and  on the upper side (holds for  with equality on ). A little bit of casework to show that there is indeed no square between these two, but it the bounds are tight and look nice, I think.

See if you can find some nicer bounds, though.

[/hide]

Denote $f(n) = n^{4}+n^{3}+n^{2}+n+1$.
Lower bound $\left(n^{2}+\frac{n}{2}\right)^{2}$ holds for all real $n$ (equivalent with $\frac{3}{4}n^{2}+n+1 > 0$), no equality possible.
Upper bound $\left(n^{2}+\frac{n}{2}+1\right)^{2}$ holds for all real $n$ (equivalent with $\frac{5}{4}n^{2}\geq 0$), equality iff $n=0$.

If $n=0$, the expression is a perfect square.
If $n$ is even, without being 0, our expression is between two consecutive perfect squares: let $n=2k$, then $\left(4k^{2}+k\right)^{2}< f(n) < \left(4k^{2}+k+1\right)^{2}$.
If $n$ is odd, let $n=2k+1$, then $\left(4k^{2}+k+\frac{1}{2}\right)^{2}< f(n) < \left(4k^{2}+k+\frac{3}{2}\right)^{2}$, so $f(n)$ must be equal to $\left(4k^{2}+k+1\right)^{2}= \left(n^{2}+\frac{n}{2}+\frac{1}{2}\right)^{2}= n^{4}+\frac{n^{2}}{4}+\frac{1}{4}+n^{3}+n^{2}+\frac{n}{2}$, thus $n^{2}-2n-3=0$, i.e. $n=-1,3$.

Therefore, only $f(-1)$, $f(0)$ and $f(3)$ are perfect squares. (If I didn't make any computational mistakes  :roll: )

Edit: fixed a little, quite important, typo.

[quote="Elemennop"]$(n^{2}+\frac{n}{2}+\frac{1}{4})^{2}$[/quote]

The bounds should be integer squares, or else they're meaningless.

[quote="Kurt Gödel"]Therefore, only $f(-1)$, $f(0)$ and $f(3)$ are integers.[/quote]

I think you mean squares, but yes, this is correct :)

[quote="t0rajir0u"][quote="Elemennop"]$(n^{2}+\frac{n}{2}+\frac{1}{4})^{2}$[/quote]

The bounds should be integer squares, or else they're meaningless.
[/quote]

There is nothing wrong with making them non integer squares; it's clear that the lower bound is the square of something RIGHT ABOVE an integer, and the opposite goes for the upper bound. Some casework for n even or odd shows that these work fine and go give the correct answer.


Problem:

$ 1600$ coconuts are distributed over $ 100$ monkeys, allowing some monkeys to be left-handed. Prove that, indepent of the distribution, there always exist four monkeys with the same number of coconuts.

I hope youre interestings solutions.
 :lol:

Assume that only exist $ \leq 3$ monkeys with the same number of coconuts, then total of coconuts $ \geq 3(0\plus{}1\plus{}2\plus{}3\plus{}...\plus{}32)\plus{}33\equal{}1617>1600$, contradiction!


function
advanced fields
advanced fields unsolved

Are there any known asymptotics (or ''exact'' formulas ?) for
\[ \sum_{j = 2}^n \frac{1}{\zeta ( j )}  \]
where $\zeta$ is the Riemman zeta function ?

I moved this to a more "advanced" forum.


geometry
perimeter
rectangle
analytic geometry

1. Determine the area under an archway that was constructed on a straight base AB whose length is 4 feet and with circular arcs AC and AB. (srry i cant draw the pic, it's basically a triangular shape with a flat base of 4 and the other 2 "sides" are actually arcs)

and...

2. The perimeter of an isosceles right triangle is 2k, what is its area? (how can you find this?)

[hide="Hint for 1"]
Break up the archway into two different shapes - a rectangle and a semicircle. Simply find the area of each and add them at the end. This will produce the total area.
[/hide]

Not sure about number 2...

Where's the third question?

WHOOPS i forgot  :blush:
here it is

3. is there any way to find the area of a circle given 3 coordinates that are "on" the circle, all we know is 3 given coordinates.

sorry.. :oops:

That's okay  :P

[quote="DaReaper"]3. is there any way to find the area of a circle given 3 coordinates that are "on" the circle, all we know is 3 given coordinates.[/quote]

Err, if we're allowed to use a calculator, we *might* be able to find the equation of the circle (which ultimately derives us the radius and the area).

Or if you're constructing it, you can always pull out a compass and trace the circle  :D .

the archs that r used as the sides are wayyyy to large to be used as semicirlces (in the pic that is), is u try to make it a triangle, the side will AT MOST become the chord of a larger circle...sigh i have never seen a problem like this

by the way, the answer is
\[ \left(\frac{16\pi}{3}\minus{}4\sqrt{3}\right) ft^2\]

for two, the semi perimeter is defined by $ k$, so
$ a\equal{}b\Rightarrow \frac{2a\plus{}c}{2}\equal{}k\equal{}\frac{2b\plus{}c}{2}$
$ a\equal{}b\Rightarrow k\minus{}a\equal{}k\minus{}b\equal{}\frac{c}{2}$
$ A\equal{}\frac{c\sqrt{4s^2\minus{}c^2}}{4}\equal{}\frac{1}{2}bh$

I don't think I can get it any simpler...

i dont know... :|

it says that the answer is \[ k^2(3 \minus{} 2\sqrt{2})\]

any ideas?

[hide="2"]
Denote the length of either leg as $ l$. The other leg's length is also equal to $ l$, while the hypotenuse is equal to $ l\sqrt2$. Then the perimeter is equal to $ 2l\plus{}l\sqrt2\equal{}(2\plus{}\sqrt2)l$. Set this equal to $ 2k$, and solve for $ l$:
$ (2\plus{}\sqrt2)l\equal{}2k\\
l\equal{}\frac{2k}{2\plus{}\sqrt2}\equal{}\frac{2k(2\minus{}\sqrt2)}{(2\plus{}\sqrt2)(2\minus{}\sqrt2)}\equal{}\frac{2k(2\minus{}\sqrt2)}{2}\equal{}k(2\minus{}\sqrt2)$
The area of the triangle is equal to $ \frac{1}{2}l^2$. Plugging in for $ l$:
$ A\equal{}\frac{1}{2}\left(k(2\minus{}\sqrt2)\right)^2\equal{}\frac{1}{2}k^2(6\minus{}4\sqrt2)\equal{}k^2(3\minus{}\sqrt2)$, which is the answer.
[/hide]


induction
trigonometry
function
complex numbers
real analysis
real analysis unsolved

Greetings, my first post. I have a trimester test in 4 days and im grinding problems.
Ill post two problems i got stuck on. Please keep in mind this is pre-academics. meaning not university level.
So try to avoid using super-duper theorems i probably wont understand.
And no calculators or computers allowed of course.
Thanks in advance.

1. Solve the equation for x.

cos(3x)^2 + cos(4x)^2 + cos(5x)^2 = 1/2 * [ (sin(80)+sin(40)) / sin(70) ]^2

also i would need to change the sin functions to known angles like 15,18,22.5,30,36,45,60,90..
and i need to solve it with identities (or ??). someone told me to use something with e and complex numbers,
but were not at that level yet.

2. Proove with induction, or a different method. ( For all natural numbers of n > 0 )

3/(1*2)^2 + 5/(2*3)^2 + 7/(3*4)^2 + ... + (2n+1)/[n(n+1)]^2  <= 1 - 1/(n+1)^3


USAMTS AMC AIME algebra polynomial function trigonometry

USAMTS Round 4 is up!

It looks challenging.

I don't count, but I thought number 1 was an excellent problem.  Still looking at the others.

Edit:  Actually, I have to say that I think they've collected 5 very interesting problems there.  I give them high marks :)

I'll order the difficulty from my perspective (hardest to easiest):

3 5 4 (1 and 2 tied)

it is difficult to get elegance in #5. and #3 is similar in annoyance as the pebbling number problem.

uve solved all 5? ive only gotten 2 and 4 yet.

Gah.  I've gotten #1 and part of #2, but I still need to find one of those elusive examples...

seems that everyone in the whole world has gotten number 2 except for me.  is it really that easy? obviously i'm not asking for hints..but maybe somebody can reassure me that whatever way i'm chugging through is so much more complicated than it actually is.
beans

I don't want to ruin anything, but I'll give some general advice: keep trying, and in particular, keep trying new things.

the website is down for some reason now... all i get is a bunch of code when i go there.

they must have some errors in their xml code

No, they moved their website.  I got an e-mail with a link to the new site (oddly enough, they didn't tell me it had changed, so it was a good thing I followed the link anyway).

Here we go: [url]http://www.nsa.gov/usamts/[/url].  It's actually simpler than the previous URL, which I had memorized.  (Audience: What a nerd.)

Arg..I hope I'll have time to do some of the problems.

-interesting_move

How do you solve USAMTS problems so quickly?

You need basically a perfect score on the USAMTS to qualify for AIME so isn't it frustrating?

Mathhelp, nobody solves USAMTS all that quickly.  See the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10993]'time spent on USAMTS' poll[/url] and you'll see what I mean.

The main thing is to keep working at the problems.  If you give up because you can't solve them in five or ten minutes, you're not trying hard enough.

I did round 3 basically from 1am-6am the day it was to be sent in, i then fell asleep until 2:30 narrowly getting it into the mailbox in time for the mail. it is doable, but it depends round by round. that is the only round to which i submitted and i got 23/25

It's turn in time!

Note: I was lazy and only did two problems, as I could get (or, am stuck with) a silver award with 5 pts. on this round or all 25.

1. 3 @ n=69
2. Omit
3. Omit
4. -69/64 + 3/32 sin(3x) (not sure; rushed in 15 mins.)
5. Omit

I rushed too. I did only 3 problems in 3 hours (including writing them) last night. tAsDoOm, for some reason I have different solutions compare to yours. For #1, I have this big number starting with 3. for #4 i have -63sin(3x)/64 (I think).

Maybe I'm totally wrong.......

I got all five.  :D

But my answers are very different from tAsDoOm's.  I agree with answerseeker on #1 and #4.

1. smallest possible is 2, with n = (10^14+1)/29 = 344827586somethingorother

2. n=9,15,27,45; one example is 65-156-169

3. a ridiculously long polynomial with 24 terms, each with degree six.  Each term is of the form w*x^2*y^3.

4. -63/64sin(3x) - either use the addition formulas and grind out the algebra, or graph the function and notice that it looks like a sine curve with some period/amplitude changes

5. 2x^3 - 2x^2 - 3x + 2

There'd better not be any mistakes in my solutions, because I spent a while on them.  If I get perfect on this round, I get gold, though.  :D

P.S. After I make sure it's okay, I'll post pictures of my papers (yes, I took pictures of them) on my website.

[quote="zabelman"]I agree with mathfanatic on 4 of the 5 answers... my polynomial was simple: (w-x)(w-y)(w-z)(x-y)(x-z)(y-z)... this might be the same one actually, just factored... anyway, the degree is 6[/quote]

Yes, it's the same one, I think.  Sweet.

I have to say that zabelman's way of doing #4 is very nice.  It makes it clear that switching any two indeterminates will result in the additive inverse of f(w,x,y,z), since the term containing both switched indeterminates will switch from (x-y) to (y-x), for example, whereas all the others will retain their signs.

My way of doing #4 was much less elegant.  Basically, I said "if wx^2y^3 is one term, then switching w and z makes x^2y^3z, which must be a term in -f(w,x,y,z), so -x^2y^3z is a term in f(w,x,y,z).  Continue this reasoning to find all 24 terms.

[quote="Setofallsets"]but I didn't see a nice way to do #4 in general[/quote]

f_n(x) satisfies a simple recursion:
f_(n+3)(x)=3/4 f_(n+1)(x) - 1/4 sin(3x) f_n(x)

[quote="Setofallsets"]I got the same as mathfanatic for 1-4  (#3 generalizes nicely, too :) - but I didn't see a nice way to do #4 in general), but I got something different for #5.  I can't seem to recall it at the moment, tho.  I may have made an arithmetic error.  That would be sad. I want another book.[/quote]

Concerning #5: I got a fourth-degree polynomial, then divided by x-2 because they asked for a third-degree polynomial.  Is that possibly how our answers differ?

I did 1,2,4,5 Friday night from 8pm-1am and mailed them in on Saturday... I was too tired to think by the time I looked at number 3 and I had a 75 going into this round :(

I agree with mathfanatic's answers.

[quote="314159265358979323846"]I agree with mathfanatic's answers.[/quote]

Yay.  Maybe that means I got them right...

Do you really have to discuss the problems so soon? I know of at least one person on this forum who has not submitted solutions yet and plans to, hoping that they will accept them.

Good point.  Although I don't think they'll accept late solutions, I have edited my posts.  I advise others to do the same.

But to anyone who read my answers and plans to submit them late as their own work, please remember the honor code.  Thank you.

[quote="mathnovice"]Do you think that they will take off a point if I said that the polynomial in number 3 was degree 3 instead of degree 6. I thought that the degree of a polynomial was the highest degree of a term in a polynomial not the sum of the degrees for each variable. I had the right answer though, and I have a 71 cumulative score for rounds one to three. : ([/quote]

Depends on whether they notice or not.  :D

I'm in the same boat as you - 71 cumulative right now, need a perfect 25 for gold.  Let's hope we don't lose any points anywhere.

as usual
here are my solutions

http://68.38.137.202/round4.pdf

My Solutions:

1. (10^14 + 1)/29

2. mod 9, 15, 27, 45.

3. (w-x)(x-y)...(w-z), degree 6

4. -63/64 sin (3x)

5. I don't remember the polynomial but I know mine is right.

BTW, I think there are two solutions for #3.  They are (w-x)(w-y)(w-z)(x-y)(x-z)(y-z) or -(w-x)(w-y)(w-z)(x-y)(x-z)(y-z) (since a switch on each one produces the negative version)

there are actually infinitely many solutions to #3... any non-zero multiple of the polynomial will work

[quote="interesting_move"]Does anybody know when we'll get our round 4 results?

-interesting_move[/quote]

If we're lucky, in 2005 or so.  :lol:


How many more vertices does an octagonal prism have than an
octagonal pyramid?

An octagonal prism has 16 vertices, and a octagonal pyramid has 9 vertices, thus there are 7 more vertices.

You just count all of the vertices. Just look it up on Google and you will see the amount of vertices.

Sometimes, it is important to know random geometric shapes :D

A prism always has twice the number of vertices as the base polygon (the other base), and a pyramid always has one more vertex than the base polygon (top vertex). So in general, the difference in the number of vertices for a pyramid and a prism is $2x-(x+1)=x-1$


Your inequality is wrong. :(
 Put for example a=b=c=1

I'm sorryyyyyyyyyyyyyy :( , i dont know why i always make such stupid mistakes :(


In the multiplication problem below,  $ A$, $ B$, $ C$ and $ D$ are
different digits. What is $ A\plus{}B$?

\begin{align*}
ABA \\
\underline{\times\;\;\;CD}\\
CDCD
\end{align*}

[quote="GameBot"]In the multiplication problem below,  $ A$, $ B$, $ C$ and $ D$ are
different digits. What is $ A + B$?

\begin{align*}
ABA \\
\underline{\times\;\;\;CD}\\
CDCD
\end{align*}[/quote]

$ \frac{CDCD}{CD}=101=ABA$.
Thus, $ A+B=1+0=\boxed{1}$.


function
real analysis
real analysis unsolved

$f : (0,+\infty)\to R $ differentable, Prove that function $g(x)=f'(x)+\lambda f(x) $ is non-decreasing iff $ f'(x)e^{\lambda x}$ is non-decreasing

eugene the relation with $g$ and $f$ miss something ?

Hi Moubi, here is my rephrase.
Suppose $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ is differentiable. Prove that the function $g(x)=f'(x)+\lambda f(x)$ is non-increasing iff $f'(x)e^{\lambda x}$ is non-decreasing.

My question is, are there any conditions on $\lambda$?

you are almost right 3X.lich, except the monotonity condition
Here is the problem:
$f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ is differentiable,$\lambda>0$
Prove that the function $ g(x)=f'(x)+\lambda f(x) $ is non-decreasing iff
$ f'(x)e^{\lambda x} $ is non-decreasing


AoPS users: you can now use the dollar signs to render LaTeX on the message board, as you would when writing a regular LaTeX document.

Examples:

$ x+3 $

\[ x+3 \]

The second is made with the double dollar signs for display mode (leave spaces around the math inside).

Awesome!!!  This will make things a lot easier.

\[ {n\choose k}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!} \]

I don't know what happened to yours nr.  It works for me.

Oh cool!!

\[
\Gamma(t) = \int_{0}^{\infty} x^{t-1} e^{-x} \, dx
\]

$
\forall x \in \mathbb{R}, \exists y \in \mathbb{R}$ such that
\[
x\cdot y = 1
\]

I think this warrants a try. First, the $1,000,000 question: can I use dollar signs normally, and then display some formulas? (Using the preview button, I've discovered that the double-dollar-sign mode will work, but the single will not. Having some coding experience, I doubt that this can be fixed)

\[

\begin{equation}
\oint{\vec{E}\cdot d\vec{A}}=\frac{q_{enc}}{\epsilon_0}
\end{equation}
\begin{equation}
\oint{\vec{B}\cdot d\vec{A}}=0
\end{equation}
\begin{equation}
\oint{\vec{E}\cdot d\vec{s}}=\frac{d\phi_B}{dt}
\end{equation}
\begin{equation}
\oint{\vec{B}\cdot d\vec{s}}=\mu_0\left(\epsilon_0\frac{d\phi_E}{dt}+i_{enc}\right)
\end{equation}
\]

test
\[
\begin{equation}
\oint{\vec{E}\cdot d\vec{A}} = \frac{q_{enc}}{\epsilon_0}
\end{equation}
\]
weird....

[quote="nr1337"][quote="ComplexZeta"][quote="nr1337"]$
\forall x \in \mathbb{R}, \exists y \in \mathbb{R}$ such that
\[
x\cdot y = 1
\][/quote]

Sadly, that's false. x=0 is a counterexample.[/quote]

Zero always causes problems...

$\forall x \in \mathbb{R}, \exists y \in \mathbb{R}$ such that

\[
x \cdot y = 0
\][/quote]
So essentially what you're saying with that statment is $\forall x\neq0 \in\mathbb{R}, y=0 \text{ and when } x=0, y\in\mathbb{R}$

[quote="confuted"][/quote][quote="nr1337"][quote="ComplexZeta"][quote="nr1337"]$
\forall x \in \mathbb{R}, \exists y \in \mathbb{R}$ such that
\[
x\cdot y = 1
\][/quote]

Sadly, that's false. x=0 is a counterexample.[/quote]

Zero always causes problems...

$\forall x \in \mathbb{R}, \exists y \in \mathbb{R}$ such that

\[
x \cdot y = 0
\][/quote][quote="confuted"]
So essentially what you're saying with that statment is $\forall x\neq0 \in\mathbb{R}, y=0 \text{ and when } x=0, y\in\mathbb{R}$[/quote]

Even simpler: either x or y is zero.

[quote="nr1337"][/quote][quote="confuted"][/quote][quote="nr1337"][quote="ComplexZeta"][quote="nr1337"]$
\forall x \in \mathbb{R}, \exists y \in \mathbb{R}$ such that
\[
x\cdot y = 1
\][/quote]

Sadly, that's false. x=0 is a counterexample.[/quote]

Zero always causes problems...

$\forall x \in \mathbb{R}, \exists y \in \mathbb{R}$ such that

\[
x \cdot y = 0
\][/quote][quote="confuted"]
So essentially what you're saying with that statment is $\forall x\neq0 \in\mathbb{R}, y=0 \text{ and when } x=0, y\in\mathbb{R}$[/quote][quote="nr1337"]

Even simpler: either x or y is zero.[/quote]
You don't get to use $\text{\LaTeX}$ if you do that.

A dollarsign:  $\$

This is: (dollar)   \    (dollar)

[quote=Spaced_Out]I can't imagine aops without $\LaTeX$  :huuh:[/quote]

lol same. its really important. [quote=ARay10][verbatim]\[\] and \(\) have been replaced with $$ and $ respectively.(Though you can still use both)[/verbatim][/quote]

y the bump...

[quote=ARay10][verbatim]\[\] and \(\) have been replaced with $$ and $ respectively.(Though you can still use both)[/verbatim][/quote]
why the 16 year bump

Edit: Never mind It was from a LONG time ago :blush:

[quote=Facejo][quote=ARay10][verbatim]\[\] and \(\) have been replaced with $$ and $ respectively.(Though you can still use both)[/verbatim][/quote]

Why did you bump this really old thread?[/quote]

maybe because this is a really important thread...idk

[quote=ArtyA6]This like the oldest thread I’ve seen so far[/quote]

Same. I mean, they just introduced $\LaTeX$ to the site. And some of the users haven't been online since $2005$!
EDIT: also, how did ARay10 find this thread?

you just search by oldest. I have seen several users bump threads from 2003 in Fun Factory

:rotfl: People say I bump. However a person posted, I posted, and I think a mod/admin deleted that person's post.


Gauss
geometry
geometric transformation
homothety
ratio
geometry proposed

Let $ ABC$ be triangle. $ D, E, F$ lie on $ BC, CA, AB$ such that $ AD, BE,CF$ are concurrent. Prove the midpoints of $ BC, DE$ and $ AP$ are collinear :)

This is [b]Gauss[/b]'s line in complete quadrilateral.

From a problem that proposed by [b]Livetolove212[/b], which is:

[quote]Let $ ABC$ be a triangle and $ A_0,B_0,C_0$ respectively be the point such that $ AA_0,BB_0,CC_0$ are concurrent at $ J$. Let $ M_a,M_b,M_c$ respectively be the midpoints of $ BC,CA,AB$; $ P_a,P_b,P_c$ respectively are the midpoints of $ B_0C_0,C_0A_0,A_0B_0$. Prove that $ M_aP_a,M_bP_b,M_cP_c$ are concurrent.[/quote]

[hide="solution of mine by the way"]Consider the homothety with center $ J$, ratio $ k_1 \equal{} 2$. We obtain $ \mathcal {H}(J,k_1): N_a\mapsto A$, $ N_b\mapsto B$, $ N_c\mapsto C$. Therefore $ \mathcal {H}(J,k_1): \triangle N_aN_cN_c\mapsto \triangle ABC$. Now consider the homothety with center $ G$- the centroid of $ \triangle ABC$, ratio $ k_2 \equal{} \minus{} \frac {1}{2}$, we obtain $ \mathcal {H}(G,k_2): A\mapsto M_a$, $ B\mapsto M_b$, $ C\mapsto M_c$. Hence $ \mathcal {H}(G,k_2): \triangle ABC\mapsto \triangle M_aM_bM_c$. Therefore $ \mathcal {H}(G,k_2)$ $ \circ$ $ \mathcal {H}(J,k_1):$ $ \triangle N_aN_bN_c\mapsto M_aM_bM_c$. But $ \mathcal {H}(G,k_2)$ $ \circ$ $ \mathcal {H}(J,k_1)$ $ \equal{} \mathcal {H}(G', \minus{} 1)$. Hence $ M_aN_a,M_bN_b,M_cN_c$ are concurrent at $ G'$ $ \Longrightarrow d_a,d_b,d_c$ are concurrent, which implies to the result of the problem.
[/hide]
I have accidentally reprove the [i]Gauss theorem[/i] about collinear points that you have mentioned with a way that is different to the common way- which using [i]Menelaus theorem[/i]  :D. Here is my proof:

[hide="proof"]Let $ N_a,N_b,N_c$ respectively be the midpoints of $ JA,JB,JC$. Let $ H_1,H_2$ respectively be the orthocenters of $ \triangle C_0BJ$, $ \triangle B_0CJ$. Let $ C_1,J_1,B_1$ respetively be the projections of $ H_1$ onto $ BJ,BC_0,JC_0$. Let $ B_2,C_2,J_2$ be the projections of $ H_2$ onto $ CJ,JB_0,CB_0$. It is followed then $ \overline {H_1C_1}.\overline {H_1C_0} \equal{}$ $ \overline {H_1B_1}.\overline {H_1B} \equal{}$ $ \overline {H_1J_1}.\overline {H_1J}$ and $ \overline {H_2C_2}.\overline {H_1C} \equal{}$ $ \overline {H_2B_2}.\overline {H_2B_0} \equal{}$ $ \overline {H_2J_2}.\overline {H_2J}$, which implies that $ \mathcal {P}_{H_1/(M_a)}$ $ \equal{} \mathcal {P}_{H_1/(N_a)}$ $ \equal{} \mathcal {P}_{H_1/(P_a)}$, where $ (M_a),(N_a),(P_a)$ repsetively are the circles with diameters $ BC,AJ,B_0C_0$. Simliarly for $ H_2$. Thus $ H_1H_2$ are the radical axes of $ (M_a),(N_a),(P_a)$ $ \Longrightarrow M_a,N_a,P_a$ are collinear. With the same argument, we also have $ (M_b,N_b,P_b)$, $ (M_c,N_c,P_c)$ are collinear. Let $ d_a$ be the line which contains $ M_a,N_a,P_a$, define the same for $ d_b,d_c$.[/hide]


Very interesting problem  :lol: ....where can I get this Mandelbrot competition problems ?

[quote="RisingFlame"]At a certain math conference there are 100 people who have met nine of the others before, another 100 who have met eight of the others, and so on to the final 100 who haven't met anybody else before. If two participants are chosen at random, then what is the probability that they have met before?

Is there any way to do this problem using polynomials? :)[/quote]

[hide="Solution"]
The number of pairs of people whom have met each other is half of the sum of the numbers of people that each individual has met (why?). Thus, there are
\[ \frac {100(9 \plus{} 8 \plus{} 7 \plus{} \ldots \plus{} 1)}{2} \equal{} 50 \cdot \frac {9(10)}{2} \equal{} 2250\] pairs of people that have met. The total number of pairs of people is $ {1000 \choose 2} \equal{} 999 \cdot 500$. Thus the probability is \[ \frac {2250}{500 \cdot 999} \equal{} \boxed{\frac {1}{222}}\]
I don't think there is any convenient way of solving this with polynomials, as this is a combinatorics/graph theory problem.
[/hide]

I remember I accidentally put 1/220 :mad: because I did 1/10*45/990  :stretcher:

I remember I did it that way...

Then, I did 100*99 instead of 100C2 :mad:


geometry
projective geometry
geometry theorems

I saw projective geometry being used on this site to solve many geo problems. Can someone tell me what it is and maybe some properties/theorems?

Or perhaps give a link?

Here is a brief outline of some properties and some links/notes for further info:

Some basic axioms...

[b]Plane Properties:[/b]

1. If A and B are distinct points on a plane, there is at least one line containing both A and B.
2. If A and B are distinct points on a plane, there is not more than one line containing both A and B.
3. Any two lines in a plane have at least one point of the plane (which may be the point at infinity) in common.
4. There is at least one line on a plane.
5. Every line contains at least three points of the plane.
6. All the points of the plane do not belong to the same line.

[b]Space Properties:[/b]

1. Three points not all coincident with the same line are coincident with a unique plane.
2. Three planes not all coincident with the same line are coincident with a unique point.
3. For a line and a plane not coincident with it, there's exactly one point that is coincident with both.
4. For a line and a point not coincident with it, there's exactly one plane that is coincident with both.
5. Two distinct planes are coincident with exactly one common line.
6. Two distinct points are coincident with exactly one common line.

There are also a number of resources on the web, and I have provided some here. I think that much more can be found by simply searching for 'Projective Geometry' using Google. Also, Kiran Kedlaya's geometry notes contain an entire section devoted to this subject as well. A copy of his notes can be found on the AMC website.

[b]Links:[/b]

http://mathworld.wolfram.com/ProjectiveGeometry.html
http://www.cs.elte.hu/geometry/csikos/proj/proj.html
http://ai.stanford.edu/~birch/projective/
http://www.kuro5hin.org/story/2003/5/26/92949/1229
http://mathforum.org/library/topics/projective_g/
http://homepages.inf.ed.ac.uk/rbf/CVonline/LOCAL_COPIES/BEARDSLEY/beardsley.html

http://www.imomath.com/index.php?options=mbb|strsljenbfa&p=0
may also be an usefull link :)


[hide][quote="mathnerd314"]Billy Bob wants to build a house. For each of the four walls, he can choose one of six materials: cardboard, paper, gingerbread, glass, diamond, and bananas. However, he doesn't want any two adjacent walls to be made of the same material. How many different arrangements are possible?

(I don't remember how many materials there were in the original version.)[/quote]

We'll start at an arbitrary wall which can be made of 6 materials, and we'll take both cases:

Case 1: 2 adjacent walls are same material.
Here, we have 6 for the first wall, and 5 for the 2 walls, and 5 for the last wall (since it has to be different from the 2 adj. walls which are identical). 6*5*5 = 150

Case 2: 2 adjacent walls are different material.
Here, we also have 6 for the first wall, 5 for one of the adj. walls, 4 for the other adj. wall (since it can't be identical to adj. wall #1), and 4 for the last wall (since it has to be diff from the two walls). 6*5*4*4=480

480+150=630[/hide]
btw, 4 materials in orig. version

[hide]Choose a single wall and stick stuff on it.  There are 6 ways to do that.

Now, there are 2 cases:  The two adjacent wall are the same, and the two adjacent walls are different.

Case 1:
There are 5 choices to stick stuff on these walls.  Then, there are 5 more choices for the last remaining wall, making $5\times5=25$ total ways to build the walls.

Case 2:
There are 5 choices for the first wall, and 4 choices for the other wall.  Order matters here because they are different walls.  The last wall can be built in only 4 ways, for a total of $5\times4\times4=80$ ways.

Add up the cases and multiply by 6 to get:

$\boxed{6(80+25)=\boxed{630}}$[/hide]

In the original question it was 4 the answer was 84 in the real problem.

You need to consider the four cases, just like the previous poster.


I have posted a new competition in the Mathcounts forum. I am in the registration stage, and it is really good practice for everyone, although challenging. The url is in my signature.

It's not a big deal this time, but please don't spam forums with messages that point to other forums.


The sequence (c1,c2,....,cn) of positive integers is called admisble if each integer k with
1 <= k <= 2c(1)+.......+2c(n) can be represented in the form k=a(1)c(1)+.....+a(n)c(n) with a(i)in {-2,-1,0,1,2}.For each n ,find max c(1)+......+c(n)

http://www.om.edu.pl/index2.php?poziom=9&main=zadania.php
if you know polish ;]

Thanks, but Idonnot know polish.Can you translate from polish into English about Poland   olympiad for us?the first is this problem and inequaliti Poland 45 .Thankks you very much

[tex]C_1+C_2+...+C_n \le  \frac{ 5^n-1}{4}[/tex].
Note:
 The sequence [tex](c_1,c_2,....,c_n)[/tex] of positive integers is called admisble if each integer k with
[tex]k=1;2;...;1002c_1+.......+1002c_n[/tex] can be represented in the form[tex] k=a_1c_1+.....+a_nc_n[/tex] with [tex]a_i[/tex] in {-1002,-1001;...;1001;1002}.For each n ,find max [tex]c_1+......+c_n[/tex] :lol:


The Ackermann function is

$A(m,n)=$

$n+1$ if m=0

$A(m-1,1)$ if m>0 and n=0

$A(m-1,A(m,n-1))$ if m>0 and n>0

Compute $A(4,2)$.

Is there an easiier way than brote force, or should I put this in the olympiad forum?

You can prove a number of identities about the Ackermann function for small fixed $m$.  Consult

http://en.wikipedia.org/wiki/Ackermann_function#Table_of_values


Two astronaut are travelling in a space ship each of mass 120kg

The astronauts do a slimming  exercise and reduce their mass to

90 kgs.Initially the space ship is travelling with a speed 15km/sec.

Initially the spaceship and astronaut's total mass is 660kg

After doing the slimming exercise find the velocity of the space ship

[space ship is out of the earth's surface]

[hide]

momentum initial = final momentum

m * v = m * v

both astronauts lose 30 kgs each.. making the new mass 600kg

660 * 15000 =  600 * v

Final velocity = 660 * 15000 / 600 = 660 * 25 =  16500 m/s

[hide]


[hide]In 1 hour, they are 520 miles apart. They will be 1560 miles apart in 3 hours.[/hide]

Use the physics formula for a more elegant solution
[hide]$d=t*v\\v=v_{1}+v_{2}\\v=560\frac{miles}{hour}\\ 1560miles=560\frac{miles}{hour}*t\\t=3hours$[/hide]

[hide]3 hrs. (I used vectors :D )[/hide]

[hide]In one hour, the westbound plane would have traveled 270 miles.
In one hour, the eastbound plane would have traveled 250 miles.
In one hour, they would be 250+270 = 520 miles apart.
Therefore, in x hours, they would be 520x miles apart.

520x = 1560
x = 3 , [b]3 hours[/b][/hide]

[hide]In an hour, the planes would be 520 miles apart.(250+270=520) Therefore it would take $3$ $hours$ for them to be 1560 miles apart.(1560/520=-3)[/hide]

[hide][quote="math92"]Two airplanes leave Baltimore at the same time, one traveling east and one traveling west.  The westbound plane travels at 270 mph and the other eastbound at 250 mph.  In how many hours will they be 1560 Miles apart?[/quote]

1560/(250+270)
1560/520
3[/hide]

Well one hour is $ 250+270 = 520$
So $ 1560$ divided by $ 520 = 3$
So in three hours

I think so :oops:

[quote="math92"]Two airplanes leave Baltimore at the same time, one traveling east and one traveling west.  The westbound plane travels at 270 mph and the other eastbound at 250 mph.  In how many hours will they be 1560 Miles apart?[/quote]

[hide]Assuming they are going in totally opposite directions...

Every hour they get 520m apart. 1560/520=3. So 3.[/hide]

there are 9 people that posted the same answer.

[hide] 3 hours[/hide]

this makes it 10

I learned this by this topic..

Avoid yourself from reviving old topics that has been already solved.
(I used to get this kind of message,too..but I learned that it can be kind of annoying.. so let's not do it..)


Fie $ABCDA'B'C'D'$ un paralelipiped dreptunghic. Notăm cu $M, N, P$ centrele feţelor $ABCD$, $BCC'B'$, respectiv $ABB'A'$. De asemenea, notăm cu $s$ aria triunghiului $MNP$ şi cu $S$ aria totală a paralelipipedului $ABCDA'B'C'D'$.
Dacă $\frac{S}{s}=16\sqrt{3}$, demonstraţi că paralelipipedul $ABCDA'B'C'D'$ este cub.

$ABCD$, $ABB'A'$ si $BCC'B'$ sunt dreptunghiuri, deci $MP$, $PN$ si $MN$ repr. $1/2$ din $A'C'$, $A'D$ si $DC'$, $\Rightarrow $triunghiurile $MPN$ si $A'DC'$ sunt asemenea, iar raportul ariilor este $1/4$. Cu teorema celor trei perpendiculare determinam inaltimea $DQ$, unde $Q$ este proiectia lui $D'$ pe $A'C'$.
Se ajunge la $A[A'DC']=\sqrt{a^{2}c^{2}+a^{2}b^{2}+b^{2}c^{2}}/2$ si de aici calcule.

[quote="mychrom"]$ABCD$, $ABB'A'$ si $BCC'B'$ sunt dreptunghiuri, deci $MP$, $PN$ si $MN$ repr. $1/2$ din $A'C'$, $A'D$ si $DC'$, $\Rightarrow$triunghiurile $MPN$ si $A'DC'$ sunt asemenea, iar raportul ariilor este $1/4$. Cu teorema celor trei perpendiculare determinam inaltimea $DQ$, unde $Q$ este proiectia lui $D'$ pe $A'C'$.
Se ajunge la $A[A'DC']=\sqrt{a^{2}c^{2}+a^{2}b^{2}+b^{2}c^{2}}/2$ si de aici calcule.[/quote]
Pe aceeasi idee am mers si eu ... pana la "calcule"...  :)

in calcule am inclus si restrangerile . daca e nevoie le postez. aceasta a fost cea mai grea problema a concursului?

Se poate face si exprimarea conditiei din ipoteza prin inegalitatea AM-PM.
Probabil ca a fost cea mai dificila problema din concurs. Este totusi olimpiada locala.


Suppose $ f$ is a bounded nonnegative function on $ (X,m)$ with $ m(X) = \infty$.

Show that $ f$ is integrable if and only if
${ \sum_{n=0}^{\infty}} \frac{1}{2^n} m\{x \in X : f(x) > \frac{1}{2^n} \} < \infty$

Variation #1: remove the hypothesis that $ f$ is bounded. Then

$ f$ is integrable if and only if $ \sum_{n\equal{}\minus{}\infty}^{\infty}2^nm\{x\in X: f(x)>2^n\}<\infty.$

Variation #2:

$ f\in L^p,$ $ 0<p<\infty,$ if and only if $ \sum_{n\equal{}\minus{}\infty}^{\infty}2^{np}m\{x\in X: f(x)>2^n\}<\infty.$

Thank you for the response Kent.  Can you give me some hints on how
to go about solving these problems.  I have a qual in two days and really
need some help.

Let $ A_n\equal{}\{x: f(x)>2^{\minus{}n}\}$ and let $ \mathbf{1}_{A_n}$ be the function that is $ 1$ on $ A_n$ and zero off of it.

How are $ g(x)\equal{}\sum_n2^{\minus{}n}\mathbf{1}_{A_n}(x)$ and $ f(x)$ related?


What number is 17 less than its negative? Express your answer as a decimal to the nearest tenth.

Let the number be $ n$.

\[ n \minus{} (\minus{}n) \equal{} 17 \]

\[ 2n \equal{} 17 \]

\[ n \equal{} 8.5 \]

However, we're looking for the smaller number, so our answer is $ \boxed{\minus{}8.5}$.

Or we can solve directly for the number. It's confusing to have your number be different than the variable you solve for.

n=-n-17


That's what my teacher calls it, but it has nothing really to do with math. Can ne1 help me? Here's and example; 26=L. of the A.         translates to 26 Letters of the Alphabet.

1) 435= M. of the H. of R.
2) 120= Y. in a F.F.       (It's 120 yards in a ________ field, not football)
3) 8= T. that E.T. has been M.
4) 1= R. A. in E.B.
5) 18 1/2= M.E. from the W.T.                            (has to do with history)
6) 10= D. in a T.N. (includint the A.C.)

There's some for starters. I would really appreciate your help!

interesting math class. sleeping is more productive, try your hand at that

I asked my mother, who likes playing word games, about these, and she said she thinks the third one is times <something> has been moved, but she doesn't know what it is.

oh, I was kinda thinking that it was times that <someone> has been mayor, but I don't know.

[quote]4) 1= R. A. in E.B. [/quote]

1 = Rippledance Admirer in Etc Board

I'm being serious fierytycoon. These are extra credit problems I was given in Algebra and I don't think my teacher would know about your intense crush on rippledance...

:) That won't be true, though.
It would be >

I have no clue about these, though.

...wait, MathFiend, isn't that kinda cheating?

Well, if a math teacher has a right to give these sorts of questions that have nothing to do with math as extra credit, then the students probably have the right to ask on this forum.

Here's some more, let's see if you can get them instead.

7) 270= E.V.N. to W. a P.E.
8) 4= O. in a M.Q.P. (B.C.)
9) 46= H.C.
10) 9= S.W. by B.
11) 500= S. of P. in a R.

Hehe, ok...

1 = Rotten Apple in Every Barrel

You guys really gotta loosen up though...  :wink:

i seem to be missing why shes wasting your time. the teacher i mean

my teacher's a he, but oh well. Hey, you should know that neal, there's no reason to make that mistake

[quote="MathFiend"]9) 46= H.C.[/quote]

46=Human Chromosomes.

oh, sorry. ive only had 1 male teacher before (for core classes), and i shouldve remembered. but my rememory aint workin

[quote]These are extra credit problems I was given in Algebra and I don't think my teacher would know about your intense crush on rippledance...[/quote]

By the way...I just checked and all the answers were online. Just google each one, if you're really desperate for answers by tomorrow.

These are really fun :)
All I have is 9=Sub Ways by Brookline :) , but I'm pretty positive that's wrong :)

and Fiery, I think you're a bit obssessed about Ripply...It's as if she's controling your mind...I'm getting too affected by Astral :P

I just found all the answers [hide]

([/hide]435 members of the house of representatives, 8 times that elizabeth taylor has been married, 18.5 minutes erased from the Watergate tapes, 24 karats in pure gold, 10 digits in a telephone number (including the area code), 270 electoral votes needed to win a presidential election, 4 ounces in a McDonald's quarter pounder (before cooking), 9 symphonies by Beethoven, and 500 sheets of paper in a ream. thanks for all your help, especially fierytycoon for the google advice)

thats kind of nuts... i cant imagine having gotten many of those right without looking it up.

what was the 120 yards in FF?

Wow...no idea [b]at all[/b] :)

Here's really easy ones from me:
299792458=S of L in V (EV, U =300000k)
PB>10000000/y
ESB>10000000Bs
First 4 moves of C=318979564000
J=0.01s
123000000=C on Hs
1525000000=PL in US
1=C.G. :)
And here's the one I thought was most interesting:2500 l-h people d from using products for r-h people

OK, so maybe they're not [b]that[/b] easy, but they were easy to [b]make[/b], anyways :)
And they're all 100% original (or so I think...), so you can't use Google to cheat on these. Good luck! You'll need it :)

[color=cyan]How about "52 = H. V."?[/color]

120 yards in a frisbee field
57 heinz varieties
How about..........0=a S. C. in H.

ComplexZeta wrote:299792458 = speed of light in a vacuum.

I don't kow about the others.

You're right! (EV, U =300000k) is easy if you know that. Just think about the =300000k part...the rest'll come to you eventually...

Hints: 2.) eyes

3.) NYC

4.) Go

5.) I think there's a car repair/maintainance company called something-lube

6.) Eisenhower

7.) ring ring!

7.) Internet

8.) a joke about me

9.) it's black humor...think about  [hide]oven mitts[/hide]

I love the guessing game!  Especially that feeling I get when I solve one. It's exhilarating! It's like proving the Reimann Hypothesis! It's like eating 100 packs of altoids at once!!!

OK, I'm a bit sugar-high, even though I'm hungry


group theory
abstract algebra
superior algebra
superior algebra unsolved

I was going through my old notes, when I ran into this simple problem: " Let $G$ be a finite group, with $H$ as its normal cyclic subgroup. Prove that every subgroup of $H$ is normal in $G$. "
I remember discussing the problem for an infinite group $G$. Now I want your idea for the infinite case please.

[hide="PS"]One will soon find out that it is not $G$ but $H$ whose cardinal matters. [/hide]

If $H$ is cyclic (infinite or not) then for every positive integer $n,\ H$ has at most one subgroup of index $n$. This means that all subgroups of $H$ are characteristic subgroups of $H$. If $H$ is normal in $G$ and $K$ is characteristic in $H$, then $K$ is normal in $G$. So the assertion holds no matter what the cardinalities are: if $H$ is a normal cyclic subgroup of $G$, then all subgroups of $H$ are normal in $G$.

Characteristic subgroups. Exactly the way I thought of this problem recently. Thank you grobber. Any other approach guys?(In my opinion this is the best one, though.)


Hey Friends Solve this one. its an excellent one:

We name a sequence of points A1, A2,, An   ε-linear if for all i  , the angle between the segments AiA(i+1) and A(i+1)A(i+2)  is between π and (1-ε)π. Prove that every sequence of
kn +1  points contains a  1/n- linear sequence of k+1 points.
 :)

I saw another post of you you wrote k^n+1.
Now it's kn+1?
Then it must be really hard.
can you post the solution?

this problem was given as quiz in summer and we are allowed just 10 minute to solve it :(  any way I remember we should use $\chi(G\bigcup H)\leq \chi (G).\chi (H)$also I think $K^n +1 $ is right for kn+1 u can make counter example.

$\chi (G)$ of graph G is minimum number of color needed that we can color the verteces of graph that any 2 neighbors have different colors

Also called 'the chromatic number of the graph'....

Pierre.


Allowing each soldier five ounces of protein each day, an army of 600 soldiers has enough provisions for five weeks. With the same total amount of provisions, how many soldiers could be fed for 10 weeks if each soldier receives only three ounces of protein per day?

[hide]
Right now there are $5\cdot600\cdot7\cdot5=105000$ ounces of protein.
That means that under the new conditions, they have $10\cdot70$ days.
$105000/70=1500$ per day.
Divide by 3 because each soldier gets 3 ounces.
$500$[/hide]

[hide]5*600*5=15000 is the ounces of protein total. 15000/10/3=500 soldiers.[/hide]


limit
inequalities
logarithms
calculus
calculus computations

Compute:
I)
\[
\displaystyle \lim_{n\to+\infty}\left(\left(1+\frac 1 n\right)^{1/2}+\left(1+\frac 1 n\right)^{2/3}+\cdots+\left(1+\frac 1 n\right)^{n/(n+1)}-n\right)
\]
II)
\[
\displaystyle \lim_{n\to+\infty}\left(\sqrt[n]n\cdot\sqrt[n+1]{(n+1)!}-\sqrt[n+1]{(n+1)}\cdot\sqrt[n]{n!}\right)
\]

I) $1$
 Denote the expression by $a_n$
By AM-GM we have :
 $\displaystyle x_n=n\big((1+{1/n})^{\frac {{1/2}+{2/3}+...+{n/(n+1)}} n }-1\big)\leq a_n\leq y_n=n\big((1+{1/n})^{n/(n+1)}-1\big)$
Note that $\lim_{x\to 0} \frac {e^x-1} {x} =1$,so:
 $\displaystyle \lim x_n=\lim nln(1+{1/n})(\frac { {1/2}+...+{n/(n+1)} } {n})=1.1=1$
 $\lim y_n=\lim nln(1+1/n)(n/(n+1))=1$.So $\lim a_n=1$.
II) I think the result is $e^{-1}$ but i need to fix something in my solution.

Ok,I've completed it.The main idea is using Stirling formula $n!\approx{(\frac n e)^n.\sqrt {2\pi n}}$ and $\lim_{x\to 0} \frac {e^x-1} {x}=1$.
My solution is quite monstrous although the idea is quite simple.

I followed this way for the first limit. From the same inequality
\[
n\left[\left(1+\frac 1 n\right)^{\displaystyle\frac{\displaystyle
\sum_{k=1}^n \frac k {k+1}} {n}}-1\right]\le a_n\le n\left[\left
(1+\frac 1 n\right)^{\frac n{n+1}}-1\right]
\]
The left term can be written as:
\[
\displaystyle \frac{\left(1+\frac 1 n\right)^{n\cdot\displaystyle\frac{\displaystyle
\sum_{k=1}^n \frac k {k+1}} {n^2}}-1}}{\displaystyle\frac{\displaystyle
\sum_{k=1}^n \frac k {k+1}} {n^2}}\cdot \frac n{\displaystyle
\sum_{k=1}^n \frac k {k+1}} }
\]
which $\to$ 1 when n $\to +\infty$
Also the right term
\[
\displaystyle\frac{\left(1+\displaystyle\frac 1 n\right)^
{n\cdot\displaystyle\frac 1{n+1}}-1}}{\displaystyle\frac 1{n+1}
\cdot\displaystyle\frac{n+1}n}}
\]
which $\to$ 1 when n $\to +\infty$
O.K for the second limit. The Sterling formula works very well and easily gives us $ e^{-1}$

[quote="nalpaction"]Moubinool,I had the same thought with you.But your arguments should be completed by proving the existence of $\lim x_n$ before apply Cesaro theorem.You can't always say imply $\lim (x_{n+1}-x_n)=l$ if $\lim \frac {x_n} {n}=l$.For example,$x_n=n+(-1)^n$ ,$\lim \frac {x_n} n=1$ but $\lim (x_{n+1}-x_n$ does not exist.
I've managed to fix it,by straightforward using Stirling formula.[/quote]
I have a thougt.we have to calculate ${L=\lim{n \to +\infty}}(\sqrt[n+1]{n!}-\sqrt[n]{(n-1)!}) $
use caughy's theorem,we get $L=\frac{\sqrt[n+1]{n!}}{n}$
hope it is right :)


Okay. A month ago I got rejected from RSI, and thats understandable..

...but governor's school of math and sci? This is ludicrous, especially considering I have one of the highest math reputations in the state of PA...high AMC scores, decent AIME scores...top three in almost every local competition.

This puts so much pressure on me to make MOP...

What is the governor's school of math and science?

Yeah. What are the requirements for governer's school?

Governor's school is basically a free camp for the top 90 students in PA, funded by the commonwealth. The requirements are just "having a strong interest in science or math."

I think one reason may actually have been my counselor's reccomendation. Both my teachers took two days to write it, but I know my counselor just half-a**ed it in about half an hour...

So...

Apparently I'm not in top 90?

I think not.


inequalities
trigonometry
inequalities proposed

What an unexpected week of median! I 'd like to join too. :D  :D  :D  :D
     Let ABC be a triangle. Prove that
    $2\sum \frac{m_a}{h_a} \leq \sum \frac{1}{sin{\frac{A}{2}}} $

I don't know if this news is good but the median week is not over. :D.So please try this one. :D
   PS I intend to make a median month :D (just kidding)

[b]treegoner[/b] posted the following very interesting chain of inequalities at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16365

[quote="treegoner"]Here is a series of inequality :
  $2\sum \frac{m_a}{l_a} \leq \sum\frac{a}{b} + \frac{b}{a} \leq 2\sum \frac{m_a}{h_a} \leq \sum \frac{1}{sin \frac{A}{2}} \leq 2(1+\frac{R}{r})$ .
   The leftmost inequality holds for all acute triangles. For obtuse triangles :P , I 'm not sure. The rests holds for all triangles.[/quote]

I know the proofs of three "$\leq$"s in the chain except $2\sum \frac{m_a}{h_a} \leq \sum \frac{1}{\sin \frac{A}{2}}$ ------(*)

I have tried to transform it into the equivalent forms:

(1) $2(m_aa+m_bb+m_cc) \leq (a+b+c)(AI + BI + CI)$ where I is the incenter.

(2) $2(m_aa+m_bb+m_cc) \leq l_a(b+c) + l_b(c+a) + l_c (a+b)$

(3) $2(m_aa+m_bb+m_cc)  \leq a(l_b+l_c) + b(l_c+l_a) +c (l_a+l_b)$

Don't you think these forms are better looking than (*)?  :D  :D

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

A couple of my wild guesses:

(G1) Is $AG+BG+CG \leq AI + BI + CI$ true? where G is the centroid.

(G2) Is $2m_a \leq AI + BI + CI$ true?, etc.

If either (G1) or (G2) is true, then (1) can be proved. But I have no idea.

Any help is appreciated!

(G2) is wrong, even for acute-angled triangles.

(G1) seems to be true with reversed sign! (Would be a nice problem, BTW.)

Sorry for the laconic reply, I have much to do now...

  Darij


limit trigonometry geometry inequalities calculus function derivative

Can anyone prove to me why $\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$ without using L'Hopital's rule or Taylor series?

See [url]http://web.mit.edu/wwmath/calculus/limits/trig.html[/url]
or [url]http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/squeezedirectory/SqueezePrinciple.html#PROBLEM%204[/url] - (see problem 9.)

Will the Squeezing theorem do?.

I'm going to use h instead of x. No particular reason, just used to it that way.

Assume that h satisfies $0<h<\frac{\pi}{2}$ and construct the angle h(as in the diagrams):

[img]http://img85.imageshack.us/img85/6356/squeezetheorem0io.gif[/img]

The terminal side of h intersects the unit circle at P(cos(h),sin(h)) and intersects the vertical line through

B(1,0) at Q(1,tan(h)).

Look at the figures:

$0<\text{area of triangle OBP}<\text{area of sector OBP}<\text{area of triangle OBQ}$

But,

$\text{area of triangle OBP}=\frac{1}{2}\text{(base)(altitude)}=\frac{1}{2}(1)sin(h)=\frac{1}{2}sin(h)$

$\text{area off sector OBP}=\frac{1}{2}(1)^{2}(h)=\frac{1}{2}h$

$\text{area of triangle OBQ}=\frac{1}{2}(base)(altitude)=\frac{1}{2}(1)tan(h)=\frac{1}{2}tan(h)$

Therefore,

$0<\frac{1}{2}sin(h)<\frac{1}{2}h<\frac{1}{2}tan(h)$

Multiply through by $\frac{2}{sin(h)}$:

$1<\frac{h}{sin(h)}<\frac{1}{cos(h)}$

Take reciprocals:

$cos(h)<\frac{sin(h)}{h}<1$.............[1]

We got here under the assumption that $0<h,\frac{\pi}{2}$

The inequalities in [1] are also valid if $\frac{-\pi}{2}<h<0$

So, [1] holds for all h in the interval $(\frac{-\pi}{2},\frac{\pi}{2})$, except h=0.

But,

$\lim_{h\rightarrow{0}}cos(h)=1$ and $\lim_{h\rightarrow{0}}1=1$

so $\lim_{h\rightarrow{0}}\frac{sin(h)}{h}=1$ by the Squeezing theorem.

Thank you very much fredoniahead and seesaw. I never knew that this limit can be evaluated geometrically in such a beautiful way.  :cool:

Strange. I think that the above proof in any Romanian high-school calculus textbook. But then again, it all depends on what you mean by $\sin$.

That proof is also in several American calculus textbooks. If you're really picky, you'll note that this makes the tacit assumption that the cosine function is continuous at zero, something that really ought to be proved.

I would point out that you can't "prove" this limit by L'Hôpital's rule. This limit is the derivative of the sine at zero; you must know this limit in order to be able to compute the derivatives of sine and cosine in general. So any method that uses knowledge of the derivatives of trigonometric functions cannot be used to prove this limit, becuase you had to know this limit before you had the derivative formulas.

The only other path I know is Newton's: use an area argument in a circle and the binomial series to find a power series for the arcsine. Invert that series to find a series for the sine. Manipulate this series (squares, square roots) to find a series for the cosine.

It's history. It's also an enormous amount of work.

(concerning the $\cos$ continuity problem posed by Kent Merryfield)

That's why I think it is best that we define $\sin x = \frac{e^{ix} - e^{-ix}}{2i}, \, \cos x = \textrm{blah}$ and $e^x$ by its power series, like I saw in the free preview of Rudin's book.

After all, what on Earth is the measure of an angle? I found that it is [i]the length of the arc cut out by the angle, divided by the circle's radius[/i]. But what's the length of an arc? Can't really compute that without integrals, I think. These definitions really can confuse a person.

Maybe one can define the measure of (directed) angles through the above $\sin \& \cos$ and then check the addition formula, or maybe one cannot.

I think only the length of a beautiful curve can be found by integral. For a rectifiable parametrized curve $\phi: [a,b] \Rightarrow \mathbb{R}^2$ you just take a partition of $[a,b]: a = t_0 < t_1 < \cdots < t_n = b$ and consider $\miu (\phi) = \sum ||phi(t_0) - phi(t_1)||$ where $||v||$ is a norm. When the limit exists the curve is defined to be rectifiable and the limit is the length. When the curve is beautiful enough the sum is similar to a riemann sum thus computable by integral.


calculus integration derivative function probability LaTeX

In my calculus textbook, I found an exercise where I need to derive Stein's identity. The identity is of course  the improper integral from negative infinity to positive infinity of derivative of any differentiable functions wiht bounds with respect to x times e^(-x^2/2) dx is same as an improper integral from negative infinity to positive infinity of x times any differentiable function with bounds times e^(-x^2/2) dx. The fist thing I tried was to see if the derivative of xg(x)e^(-x^2/2) would in fact, be the same function as g'(x)e^(-x^2/2)  (g(x) denotes any differentiable function. But it did not work, so I discarded that method, and went on to try substitution with w(it can be any variable) = -x^2/2. But it also did not bear successful fruit... so how am I supposed to be able to derive the identity?

i appreciate any help i can gather from you.[/i][/hide]

See [url=http://www.cs.ucla.edu/~falaki/pub/biased.pdf]Lemma 3.1[/url].  This uses some probability notation but they're the same thing.

I nearly cannot comprehend what the article talks about. It would be darn helpful if you could post the derivation of it in this forum with just normal signs.

It's hard to understand what you're trying to say.  Please use $ \text{\TeX}$.  The link I provided is straightforward if you ignore the notation.  It seems like you want to prove that $ \int_{-\infty}^{\infty}f'(x)e^{-x^{2}/2}\,dx=\int_{-\infty}^{\infty}xf(x)e^{-x^{2}/2}\,dx$.  Just use integration by parts: set $ u=e^{-x^{2}/2}$, $ du=-xe^{-x^{2}/2}\,dx$, $ dv=f'(x)\,dx$, $ v=f(x)$ (or if you want to go right to left, set $ u=f(x)$ and $ v=xe^{-x^{2}/2}\,dx$ (this integration by parts is particularly easy because you know $ e^{-x^{2}/2}$ doesn't have an elementary antiderivative, so you can't set it to be $ v$).

I clicked on the link you provided which is TEX but still did not help as I did not how it helps me write down the notation. The integration by parts method is apparently a useful method, but it will also yield f(x)e^(-x^2/2) as a by product aside from the resulting integral if you know what I mean.

Thanks for the help, and I hope you could aid me in understanding the mechanism of TEX (though probably there is not much that obstructs in the way of understanding it or I don't know)[/u]

Art of Problem Solving has a useful tutorial on how to learn $ \text{\LaTeX}$-just click on the bar on the left side of the screen.

Yes, there is an $ f(x)e^{-x^{2}/2}\bigg|_{-\infty}^{\infty}$ term, but this is really easy to deal with (the same thing happens as $ x\to\pm\infty$-what is it?).

I guess it will converge to zero, but I think that it depends on what the function f(x) looks like. If it looks something like X^X^X^X function, I believe that e^(-x^2/2) can't dominate that function. So Stein's identity might be false for some instances or would it? So in that case, that term will not be zero, and we can't eliminate f(x)e^(-x^2/2) term.

How do you go about that JRav?


function
superior algebra
superior algebra unsolved

Okay this problem has been annoying me for a while now. I am sure most of you have seen it before and if you could offer some hints on how to proceed that would be great.

Let $ K$ be a field. $ K(x)$ the corresponding field of rational functions in $ x$. Prove that $ Aut_K K(x) \equal{} \{ \frac{aX\plus{}b}{cX\plus{}d}  | ad\minus{}bc \neq 0\}$.

Containment is clear to me, since those functions are easily seen to be invertible. Just not sure how to prove reverse inclusion. Thanks for the help.


[quote="ragnarok23"]A group of 9 people are nominated for a 6 person committee. From this committee, a 3 person subcommittee will be formed. How many ways are there to organize the committee and subcommittee?[/quote]
[hide]There are $\binom{9}{6}$ ways for the first, and $\binom{6}{3}$ for the second. Multiplying these, we get $\binom{9}{6}\binom{6}{3}=\frac{9!6!}{6!3!3!3!}=\frac{8!}{2!2!3!}=280$[/hide]
Assuming I didn't make careless errors...

[quote="ragnarok23"]A group of 9 people are nominated for a 6 person committee. From this committee, a 3 person subcommittee will be formed. How many ways are there to organize the committee and subcommittee?[/quote]

[hide]For the committee, there are ${9\choose 6}$ ways to choose. For the subcommittee, there are ${6\choose 3}$ ways to choose. Therefore, you multiply them toghether. ${9\choose 6}\cdot{6\choose 3}=1680$.[/hide]

[quote="b-flat"][hide]First, there are $\binom96$ ways to choose the committee.  Then, there are $\binom63$ ways to choose the subcommittee.

Thus, there are $\binom96\binom63=\frac{(9\times8\times7)(6\times5\times4)}{(3!)(3!)}=3\times4\times7\times5\times4=6\times280=\boxed{1680}$
[/hide]

[hide="bpms"]Your answer doesn't quite make sense - $8!\approx40,000$, and $2!2!3!=24$...[/hide][/quote]
Heh, I cancelled wrong. You are right...


Prove that all odd polynomials have at least 1 real root.

More : every odd-degree polynomial (with real coeficients) has at least 1 real root.


probability
calculus
derivative
geometry
3D geometry
pyramid
integration

Now, that we are finally back up and all the most recent discussions are lost I think I'll post a few challenge problems to get things going again.  As always, leave these problems to other people if the problem is absolutely trivial for you.

I would also like to thank the administrators for their hard work and dedication to this site.  The forum could still not be up if they did not care as much as they do!  I would also like to say that the improved forum is really cool, thanks!

1.  Three real numbers are chosen between 0 and 10.  What is the probability that their sum is less than 20?

2.  Evaluate: \[\sum_{k=1}^n{(-1)^{k+1}k\binom{n}{k}}=\binom{n}{1}-2\binom{n}{2}+3\binom{n}{3}-\cdots+(-1)^{n+1}n\binom{n}{n}\]

3.  Find the terminating digit of $\left(\frac 58\right)^5$

For number two:

Differentiation is fun!

For 1:

I did it two ways, so I'm pretty sure it's right (assuming random distribution)  Just draw the cube and visualize its intersection with the plane $z=20-x-y$.  It cuts off a pyramid with

$\text{base} = \frac {s^2}{2}$

$\text {height}={s}$

Alternately, triple integrate

$\displaystyle \frac {\int^{10}_{0} \int^{10}_{10-y}\int^{10}_{20-x-y} dV}{\int_{[0,10]^3} dV}$

The answer is [hide]$\frac {1}{6}$[/hide]
Here's an addendum:

Given the same conditions of the problem, what is the probability that $x^2+y^2+z^2\leq 10$?

[quote="blahblahblah"]For 1:

I did it two ways, so I'm pretty sure it's right[/quote]

Are you sure tht you counted the right thing?  :D

Oh and I forgot to mention that number 3 should be done without a calculator.

Yes, if by 'less than' you really meant 'greater than'

#3 can be brute forced quite easily....

Answer
[hide]$6$[/hide]

Nice try :P ...less than means less than.

That's not right for number 3.

Well, then, it's $\frac {5}{6}$ .

For 3, it's well known ? that $\frac {5}{8}=0.625$

Now what happens when you raise all those 5s to arbitrary (integral) powers?

I was wondering...if we had 4 numbers instead of 3 how would you solve that?


function
integration
limit
calculus
real analysis
real analysis unsolved

Let $ \{f_n(x) \}$ is a sequence continuous function defined on $ [\minus{}1,1]$ such tha:
(1)$ \int_{\minus{}1}^1 f_n(x)dx\equal{}1$
(2)For any $ \delta>0$,$ f_n(x)$ uniformly converges to $ 0$ for $ x \in  [\minus{}1,\minus{}\delta]\cup [\delta,1].$
Then can we prove for any continuous function $ g(x)$ defined on $ [\minus{}1,1]$,then:$ \lim_{n \to \plus{}\infty}\int_{\minus{}1}^1f_n(x)g(x)dx\equal{}g(0)$

I don't know even if the conclusion is true.

You should ask for just a little bit more: in addition to your conditions, we need $ \sup_n\int_{\minus{}1}^1 |f_n(x)|\,dx<\plus{}\infty$. If this extra condition holds, it is true. Otherwise, it is false.

Hi,Thank you.And I know the proof if give the additional condition.
But can you give an counterexample otherwise?

I believe [url=http://en.wikipedia.org/wiki/Convergence_of_Fourier_series]this[/url] is the counterexample fedja was thinking of (see the "convergence at a given point" section). Doing it constructively is hard, of course.

For a constructive example, how about an unbalanced dipole?
Let $ h(x)$ be the piecewise linear function with vertices $ (0,0)$, $ (\frac12,2)$, and $ (1,0)$; $ h$ has integral $ 1$.
Let $ f_n(x) \equal{} \begin{cases}n(n \plus{} 1)h(nx) & 0\le x\le \frac1n \\
n^2h( \minus{} nx) & \minus{} \frac1n\le x < 0 \\
0 & \text{otherwise}\end{cases}$

Let $ g(x) \equal{} x$. We have $ \int_0^1 g(x)h(nx)\,dx \equal{} \int_0^n \frac {t}{n}h(t)\cdot\frac1n\,dt \equal{} \frac1{n^2}\int_0^1 t h(t)\,dt \equal{} \frac1{2n^2}$. Similarly, $ \int_{ \minus{} 1}^0 g(x)h( \minus{} nx)\,dx \equal{} \frac1{2n^2}$. Combining these, $ \int_{ \minus{} 1}^1 f_n(x)g(x)\,dx \equal{} \frac {2n^2 \plus{} n}{2n^2} \equal{} 1 \plus{} \frac1{2n},$ which goes to $ 1.$

In the language of distributions, my $ f_n$ converge to $ \delta \minus{} \delta'$.

Thanks,Jmerry.Your counterexample is nice,

Hi jmerry,in your example
$ f_n(x)\equal{}\minus{}n^2 h(\minus{}nx)\quad \minus{}\frac{1}{n}\leq x<0$.

Hi zhaobin,can you give the detail of the proof.

[quote="water2011"]Hi jmerry,in your example
$ f_n(x) \equal{} \minus{} n^2 h( \minus{} nx)\quad \minus{} \frac {1}{n}\leq x < 0$.

Hi zhaobin,can you give the detail of the proof.[/quote]

proof for what?

[quote="zhaobin"][quote="water2011"][quote="zhaobin"]Let $ \{f_n(x) \}$ is a sequence continuous function defined on $ [ \minus{} 1,1]$ such tha:
(1)$ \int_{ \minus{} 1}^1 f_n(x)dx \equal{} 1$
(2)For any $ \delta > 0$,$ f_n(x)$ uniformly converges to $ 0$ for $ x \in [ \minus{} 1, \minus{} \delta]\cup [\delta,1].$
Then can we prove for any continuous function $ g(x)$ defined on $ [ \minus{} 1,1]$,then:$ \lim_{n \to \plus{} \infty}\int_{ \minus{} 1}^1f_n(x)g(x)dx \equal{} g(0)$

I don't know even if the conclusion is true.[/quote]

Prove it with that additional condition.[/quote]

$ |\int_{ \minus{} 1}^1f_n(x)g(x)dx \minus{} g(0)| \equal{} |\int_{ \minus{} 1}^1 f_n(x)(g(x) \minus{} g(0))dx| \le 2M\left(\int_{ \minus{} 1}^{ \minus{} \delta} \plus{} \int_{\delta}^1 \right)|f_n(x)|dx \plus{} \int_{ \minus{} \delta}^{\delta} |f_n(x)\parallel{}(g(x) \minus{} g(0))|dx$

Where we assume $ |g(x)| \le M$
Choose $ \minus{} \delta$ small enough,and then choose $ n$ large enough.
I hope you can understand what I wrote.If not,I will write more :)[/quote]

Please give more details. Thank you

[quote="water2011"]

Please give more details. Thank you[/quote]
OK,First we assume $ |g(x)| \le M, \int_{ \minus{} 1}^{1} |f_n(x)|dx \le M$ by the condition.
For any $ \epsilon > 0$,we can choose $ 0 < \delta < 1$,such that for any $ x \in ( \minus{}\delta,\delta)$
we have:$ |g(x) \minus{} g(0)| < \frac {\epsilon}{2M}$
Then because $ f_n(x)$ uniformly converges to $ 0$ for $ [ \minus{} 1, \minus{} \delta] \cup [\delta,1]$,We can choose $ N > 0$
such that for any $ n > N$ we have:$ |f_n(x)| < \frac {\epsilon}{8M}$,when $ x \in [ \minus{} 1, \minus{} \delta] \cup [\delta,1]$.
Thus:
$ |\int_{ \minus{} 1}^1f_n(x)g(x)dx \minus{} g(0)| \equal{} |\int_{ \minus{} 1}^1 f_n(x)(g(x) \minus{} g(0))dx| \le 2M\left(\int_{ \minus{} 1}^{ \minus{} \delta} \plus{} \int_{\delta}^1 \right)|f_n(x)|dx \plus{} \int_{ \minus{} \delta}^{\delta} |f_n(x)\parallel{}(g(x) \minus{} g(0))|dx$
$ \le 2M\cdot \frac {2\epsilon}{8M} \plus{} \frac {\epsilon}{2M}\cdot M \equal{} \epsilon$
Then,If you still don't understand I will write nothing :P


Prove or disprove:

For every collection $\{ O_{i}\}$ of open subsets in $\mathbb{R}$:
\[\mathbb{Q}\ne \bigcap_{i = 1}^\infty O_{i}\,, \]
where $\mathbb{Q}$ is the set of rational numbers.

Apply the Baire category theorem to $\bigcap O_{i}\cap\bigcap_{q\in\mathbb{Q}}(\mathbb{R}\setminus\{q\})$.


function
real analysis
integration
trigonometry
real analysis unsolved

Hi, could you help me to explain why the closed ball of

\[ C([0,1])\equal{}\{f: [0,1]\rightarrow \mathbb{R}, f \mbox{ continuous }\}\]

of center  $ 0$ and radius $ 1$ is not compact.

Thank you.

Find a sequence with no convergent subsequences.  Is this with respect to the supremum norm?

Ya, the distance between $ f$ and $ g$ is defined to be

\[ \max\{d(f(p),g(p))\}: p\in E\equal{}C([0,1])\equal{}\{f: [0,1]\rightarrow \mathbb{R}, f \mbox{ continuous }\}.\]

The sequence in interest is $ f_n(x)\equal{}x^n$. But is the following reasoning correct:

\[ f_n(x)\rightarrow 0 \mbox{ for } 0\leq x<1, \qquad f_n(x)\rightarrow 1\mbox{ for } x\equal{}1.\]

But, if we choose $ x_n\equal{}1\minus{}\frac{1}{n}$, then

\[ f_n(x_n)\equal{}(1\minus{}\frac{1}{n})^n\rightarrow \frac{1}{e}\].

Therefore, $ f_n(x)$ is not convergent. Hence the closed ball of

\[ C([0,1])\equal{}\{f: [0,1]\rightarrow \mathbb{R}, f \mbox{ continuous }\}\]

is not compact.

Thank you.

The reasoning is promising, but not complete. You have hinted - but not really shown - that the sequence diverges, and  you haven't shown what you really need to show, which is that it has no convergent subsequences.

I personally would probably have used a piecewise linear example - I find such functions slightly easier to work with. Your example will work, too, but I don't feel like spending that much effort on the details.

For instance: let $ f_n(x)$ be the piecewise linear continuous function whose graph connects the point $ (0,1),\left(\frac1n,0\right),$ and $ (1,0)$ with line segments. That is, $ f(x) \equal{} \max(1 \minus{} nx,0).$ What we need to do is to find $ d(f_n,f_m).$

Assume WLOG that $ m > n.$ The point at which the distance between the two curves is greatest happens at $ x \equal{} \frac1m.$ At that point, $ f_m(x) \equal{} 0$ and $ f_n(x) \equal{} 1 \minus{} \frac nm.$  From this calculation, we see that $ d(f_n,f_m) \equal{} \frac {|n \minus{} m|}{\max(n,m)}.$

Suppose we had a convergent, hence Cauchy, subsequence $ f_{n_k}.$ If it were Cauchy, then given $ \epsilon > 0,$ we could find $ n_k$ such that $ \sup_{j > k}d(f_{n_k} \minus{} f_{n_j}).$ But we have that this sup must be $ \sup_{j > k}\frac {n_j \minus{} n_k}{n_j},$ and for any $ n_j\to\infty,$ this sup is $ 1.$ We can't make it arbitrarily small.

We have shown that no subsequence of our sequence can be Cauchy, so no subsequence can converge. Since we have an sequence with no convergent subsequence, the set we chose this sequence from cannot be compact.

[quote="vnca1"]But, if we choose $ x_n \equal{} 1 \minus{} \frac {1}{n}$, then
\[ f_n(x_n) \equal{} (1 \minus{} \frac {1}{n})^n\rightarrow \frac {1}{e}
\]
.

Therefore, $ f_n(x)$ is not convergent.[/quote]
You need to show the following:  if $ f_n \to f$ uniformly, then $ f_n(x_n) \to f(x)$ for any sequence $ x_n \to x$.  I believe this is an exercise in Rudin, so perhaps that's what you're going off of.

Thank you.

To Kent Merryfield: I'll need to read carefully what you wrote.

To t0rajir0u: could you be more specific? Clearly I forgot to say that if $ f_n\rightarrow f$ uniformly,
then $ f_n(x_n)\rightarrow f(x)$ for any $ x_n\rightarrow x$.
But how is that related to showing that
$ f_n(x)\equal{}x^n$ has no convergent subsequence?

Thank you.

Sorry; what I meant to say was that the same reasoning applies to any convergent subsequence.  Let $ n_k$ be a sequence such that $ f_{n_k} \to g$ uniformly.  Then $ f_{n_k}(x_{n_k}) \to g(x)$ for any $ x_n \to x$.  But setting $ x_{n_k} \equal{} 1$ and then setting $ x_{n_k} \equal{} 1 \minus{} \frac{1}{n_k}$ we obtain a contradiction.

Here's an alternative sequence in the unit ball of $ C[0,1].$

Let $ g_n(x) \equal{} \sin(n\pi x).$ We can see that $ \|g_n\|_{\infty} \equal{} 1$ (That's the sup norm, the norm you are talking about.

But these are also square integrable functions, and we can compute their 2-norms:

$ \|g_n\|_2^2 \equal{} \int_0^1\sin^2(n\pi x)\,dx \equal{} \frac12.$

And the inner products: if $ n\ne m,$ then

$ \langle g_n,g_n\rangle \equal{} \int_0^1\sin(n\pi x)\sin(m\pi x)\,dx \equal{} 0.$

That means that $ \|g_n \minus{} g_m\|_2^2 \equal{} \|g_n\|_2^2 \plus{} \|g_m\|_2^2 \minus{} 2\langle g_n,g_n\rangle \equal{} 1.$

So we have a sequence of functions $ (g_n)$ in the closed unit ball of $ C[0,1]$ with the property that $ d(g_n,g_m)\ge 1$ whenever $ n\ne m.$ It's abundantly clear that this sequence can have no convergent subsequence.

But $ \|g_n \minus{} g_m\|_2^2 \equal{} \int_0^1(g_n(x) \minus{} g_m(x))^2\,dx\le\int_0^1\sup(|g_n(x) \minus{} g_m(x)|^2\,dx$

$ \equal{} \|g_n \minus{} g_m\|_{\infty}^2.$


algebra
polynomial
function
linear algebra
matrix
Functional Analysis
complex numbers

Two real matrices which are unitarily similar are actually orthogonally similar.

I'll use the following result on operators on a Hilbert space: given normal operators $M,N$ and an invertible $T$ such that $T^{-1}MT=N$, if $T=UP$ is the polar decomposition of $T$ (i.e. $U$ is unitary and $P$ is the unique positive square root of $T^*T$), then $U^*MU=N$.

Let $A=M_AP_A,\ B=M_BP_B$ be the polar decompositions of $A,B$, which we may assume are invertible, by eventually adding some $\lambda\in\mathbb R$ to both ($M_A,M_B$ are orthogonal, since they are unitary and real in this case, and $P_A,P_B$ are real symmetric positive definite matrices). Also let $U$ be unitary such that $U^*AU=B\ (0)$.

Apply the $^*$ operation to $(0)$ to get $U^*A^tU=B^t$, which, upon multiplication with $(0)$ yields $U^*A^tAU=B^tB\Rightarrow P_A^2U=UP_B^2$. We thus get $\mathcal P(P_A^2)U=U\mathcal P(P_B^2)$ for all real polynomials $\mathcal P$, and by considering a sequence of polynomials which approximate the $\sqrt$ function arbitrarily well on the union of the spectra of $P_A^2,P_B^2$, we get $P_AU=UP_B\ (1)$. It also follows from this and $(0)$ that $M_AU=UM_B\ (2)$.

If $X,Y$ are the real and imaginary part of $U$, respectively (i.e. $U=X+iY$), then $(1),(2)$ translate to $P_A(X+tY)=(X+tY)P_B,\ \forall t\in\mathbb R\ (3)$ and $M_A(X+tY)=(X+tY)M_B,\ \forall t\in\mathbb R\ (4)$. Let $P(t)=\det(X+tY)$. Since $P(i)=\det U\ne 0,\ P$ is not identically null, so there is a real $t$ such that $P(t)\ne 0$, i.e. $X+tY$ is a real invertible matrix. We now apply the result mentioned in the first paragraph to $(3),(4)$ with $T=X+tY$ to get an orthogonal matrix $M$ such that $M^tM_AM=M_B,\ M^tP_AM=P_B$. Multiply these together (in this order) to get precisely $M^tAM=B$, as desired.


If the minimum number of Friday the 13ths taht can occur in a calendar year is $ V$ and the maximum number of Friday the 13ths that can occur in a calendar year is $ H$, find $ V\plus{}H$.

I know that the answer is 4, (1+3), but is there a mathematical way to do this?

Consider the days of the months $ \bmod7$. if one 13th is a Friday the 13th, then the next 13th will be the corresponding residue left (for example if March 13 is a Friday, then April 13 will be a Monday because $ 31\equiv3\bmod7$.). The maximum of $ 3$ is achieved when one of the fridays is in February, since then March and November will be Friday the 13ths. In the same way, the minimum will be achieved when June has a Friday the 13th.

if we started on Feb, then wouldn't Mar's 13th be on a monday? (so, each month's 13th is 3 days later, since 3 is our residue)?

ohh..not all months have 31 days. is that the reason?

also, how did you get the minimum to be june?

[quote="DaReaper"] ohh..not all months have 31 days. is that the reason? [/quote]
Yes. There are $ 28$ days in February, and $ 28\equiv0\pmod7$.
[quote="DaReaper"]also, how did you get the minimum to be june?[/quote]
At first, simply by thinking that the minimum must be near the middle of the year, and finding that June worked. But now I find that you should consider the sums $ \bmod7$, then simply find which residues are congruent and not. May will also yield a minimum, and it can also be maximized with January, April and July (assuming its a leap year).


In a plane, two congruent squares share a common vertex but have no other points in common. Connect pairs of the remaining six vertices to get 3 parallel segments. If two segments have lengths 6 and 8, what are all possible lengths of the third segment?

[hide]
Consider two sqraues $ ABCD$ and $ AB'C'D'$ that meet at $ A$ and have side length $ l$
Make three triangles that join $ BB', CC'$ and $ DD'$ to point $ A$
$ \angle BAB' = \theta$, $ \angle DAD' = 180-\theta$ and $ \angle CAC' = 90 + \theta$

So using the Cosine law
$ (BB')^2=2l^2 - 2l^2 \cos{\theta}$
$ (DD')^2=2l^2 + 2l^2 \cos{\theta}$
$ (CC')^2=4l^2 + 4l^2 \sin{\theta}$

Notice that length $ BB'$ and $ DD'$ cover the same range as the square rotates around point $ A$ so they will both give the same solution. Therefore we only need to evaluate $ [(BB')=6$ and $ (DD')=8 ]$ and $ [(BB')=6$ and $ (CC')=8  ]$
So we find that the only other lengths could be $ 14$ or $ 4$
[/hide]


calculus
derivative
vector
analytic geometry
parameterization
calculus computations

Hi everyone. I was wondering how would you evalute the following?: $\frac{\partial\hat{e}_{r}}{\partial\theta}, \frac{\partial\hat{e}_{\theta}}{\partial\theta}$
I know that you can evaluate it straightforwardly by converting $\hat{e}_{r}$ and $\hat{e}_{\theta}$ into rectangular coordinates and then taking the derivative. But is there a more intuitive way to see that the answer $\hat{e}_{\theta}$ and $-\hat{e}_{r}$?

It would be helpful to know how exactly $\hat e_{r}$ and $\hat e_{\theta}$ depend on $\theta$. But a partial answer canbe obtained as follows: since $\hat e_{r}\cdot \hat e_{r}=1$ for all $\theta$, the $\theta$-derivative of this dot product is $0$. The dot product obeys the product rule for derivatives, because it is bilinear (remember?  :) ) Therefore, \[\hat e_{r}\cdot \frac{\partial\hat{e}_{r}}{\partial\theta}=0\] Similarly, \[\hat e_{\theta}\cdot \frac{\partial\hat{e}_{\theta}}{\partial\theta}=0\] Finally, differentiate the relation $\hat e_{r}\cdot \hat e_{\theta}=0$ to get \[\hat e_{r}\cdot \frac{\partial\hat{e}_{\theta}}{\partial\theta}+\hat e_{\theta}\cdot \frac{\partial\hat{e}_{r}}{\partial\theta}=0\] All of this combined implies that $\frac{\partial\hat{e}_{r}}{\partial\theta}=c\hat e_{\theta}$ for some number $c$ (possibly depending on $\theta$), and $\frac{\partial\hat{e}_{\theta}}{\partial\theta}=-c\hat e_{r}$ with the same $c$.

[quote="mlok"]All of this combined implies that $\frac{\partial\hat{e}_{r}}{\partial\theta}=c\hat e_{\theta}$ for some number $c$ (possibly depending on $\theta$), and $\frac{\partial\hat{e}_{\theta}}{\partial\theta}=-c\hat e_{r}$ with the same $c$.[/quote]
How? Please Explain.

$\frac{\partial\hat{e}_{r}}{\partial\theta}$ can be expanded in the orthonormal basis $\hat{e}_{r}$, $\hat e_{\theta}$. The coefficients of such expansion are dot products of $\frac{\partial\hat{e}_{r}}{\partial\theta}$ with the basis vectors. We know that one of these products is zero, but don't know the other (call it $c$). Same for the other derivative.

I am sorry, but I still don't know what you mean.  :huh:

For any vector $\hat{v}$ in the plane we have \[\hat{v}=(\hat{v}\cdot \hat{e}_{r})\hat{e}_{r}+(\hat{v}\cdot \hat{e}_{\theta})\hat{e}_{\theta}\] This is a key property of an orthonormal basis. Use this with $\hat{v}=\frac{\partial\hat{e}_{r}}{\partial\theta}$ and with $\hat{v}=\frac{\partial\hat{e}_{\theta}}{\partial\theta}$.

I see  :lol: . So,
$\hat{v}=(\hat{v}\cdot \hat{e}_{r})\hat{e}_{r}+(\hat{v}\cdot \hat{e}_{\theta})\hat{e}_{\theta}$
For $\hat{v}=\frac{\partial\hat{e}_{r}}{\partial\theta},$

$\frac{\partial\hat{e}_{r}}{\partial\theta}= (\frac{\partial\hat{e}_{r}}{\partial\theta}\cdot \hat{e}_{r})\hat{e}_{r}+(\frac{\partial\hat{e}_{r}}{\partial\theta}\cdot \hat{e}_{\theta})\hat{e}_{\theta}=-(\hat{e}_{r}\cdot \frac{\partial\hat{e}_{\theta}}{\partial\theta})\hat{e}_{\theta}$
$\Rightarrow \frac{\partial\hat{e}_{r}}{\partial\theta}= c\hat{e}_{\theta}$
For $\hat{v}=\frac{\partial\hat{e}_{\theta}}{\partial\theta}$,

$\frac{\partial\hat{e}_{\theta}}{\partial\theta}=(\frac{\partial\hat{e}_{\theta}}{\partial\theta}\cdot \hat{e}_{r})\hat{e}_{r}+(\frac{\partial\hat{e}_{\theta}}{\partial\theta}\cdot \hat{e}_{\theta})\hat{e}_{\theta}=-(\hat{e}_{\theta}\cdot \frac{\partial\hat{e}_{r}}{\partial\theta})\hat{e}_{r}$
$\Rightarrow \frac{\partial\hat{e}_{\theta}}{\partial\theta}= c\hat{e}_{r}$

Don't know how you decide if you have $+c$ or $-c$...

You have two different $c$s in your formulas, but they are different only by the sign. It's better to write $+c$ in one formula and $-c$ in the other, so that the meaning of $c$ stays the same.

I see. I have one more quick question though. In this case our $c$ is 1. But I was wondering, in any orthogonal coordinate sytem, if we take the derivative of one of the unit vectors with respect to some other coordiante, are we always gonna some other unit vector scaled by either $0$,$1$ or $-1$?

I'll answer no, even though I don't fully understand the question. I don't even know what $\hat e_{r}$ and $\hat e_{\theta}$ are. You never defined them. I presumed that they are elements of an orthonormal basis that smoothly depends on a parameter $\theta$, and this was all I used.

In general, if $\hat{e}_{1}(t),\dots,\hat{e}_{n}(t)$ is an orthonormal basis smoothly depending on $t$, then \[e_{i}\cdot \frac{\partial \hat e_{i}}{\partial t}=0,\quad i=1,\dots,n\]  \[e_{i}\cdot \frac{\partial \hat e_{j}}{\partial t}+e_{j}\cdot \frac{\partial \hat e_{i}}{\partial t}=0, \quad 1\le i<j\le n\] for same reason.


AMC USA(J)MO USAMO Euler trigonometry inequalities geometry

If someone qualifies for the USAMO, does he or she have to memorize a lot of fomulas that we don't study in our regular class?
For example, Menelaus' Theorem, Lagrange's Interpolation Formula, Fermat Theorem, Euler (Totient) Function....
There are huge amounts of formulas in one of my Olympic books. And I hardly know any of them!
I know there are a lot of people on this website qualified the USAMO before. Do you guys think it necessary to memorize them?
Does the USAMO really need those formulas? :?:

i think the key is to not call it memorizing

just like many more elementary formulas (ie. law of cosines); those olympiad level theorems go into your head after practicing many olympiad level problems

in your olympiad book, there are probably specific questions geared at each type of problem. running thorugh a few for example menelaus collinear problems should get the theorem into your head

There are a lot of things you need to know for the USAMO. The important thing is to know which things are going to be more useful, for example, you definately want to know AM-GM because it is the basis of almost every single inequality problem in existance.

However, the important thing is not memorization, but understanding how they came to be. The USAMO gives you plenty of time if you need to rederive some theorems that you don't remember perfectly. And also, knowing how to derive the theorems also makes them easier to remember.

Well I wouldn't call Menelaus' Theorem or Fermat's Theorem "a formula". You need to know some basic theorems to tackle successfully problems on the USAMO. When discussing about geometry and number theory on the usamo, Thales, Menelaus, Ceva, Sine Theorems are basic tools which need to be memorized (respectively Fermat's Theorem, Euler Totient formula, Euler's Theorem). Of course this can only be done with a lot of practice and guidance :)


function
inequalities
real analysis
real analysis solved

If $K$- is metric compact space and $f$is surjective function from $K$ to $K$ such that $d(f(x),f(y))\leq d(x,y)$ for any $x,y\in K$ then
$f$-isometry

You need more. Maybe $f$ is surjective? :? Or maybe the sign of the inequality is reversed?

I haven't checked all the details, but wouldn't the argument in post number $12$ in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=isometry&t=18269]this topic[/url] work by substituting $f(x)$ with an arbitrary point from the set $f^{-1}(x)$?


The forum lock ends tommorow so I'd probably wait until then. You might get away with posting them in HSB or Intermediate for now and the mods will just move them back to AMC when the lock's over.

EDIT: The forum lock does not end on the 18th, I got the dates mixed up. Sorry.

[quote="aufha"]So... we're allowed to ask questions about older AIMES just as long as their not the current one, right?[/quote]There's no ban, but we'd prefer you didn't.  One way that we make sure nobody is discussing the AIME is to do a global search for the word "AIME", and discussing old questions just creates false positives for us to investigate.

[quote="7h3.D3m0n.117"][quote="the future"]Are we allowed to talk about it in PMs to people that we know 100% took the AIME?[/quote]

Unfortunately there is no way in preventing this from happening but the answer is still no.[/quote]

Alright, i probably can talk to them in school tomorrow anyways.

Please don't look for loopholes in our policy.  It's very simple: do not discuss the AIME until Sunday.


function
number theory unsolved
number theory

Let $ a_0$ be a natural number. The sequence $ (a_n)$ is deﬁned by $ a_{n\plus{}1}\equal{}\frac{a_n}{5}$ if $ a_n$ is divisible by $ 5$
and $ a_{n\plus{}1}\equal{}[a_n \sqrt{5}]$ otherwise . Show that the sequence $ a_n$ is increasing starting from some term.

any ideas? I can not have a useful idea for this problem  :(

[quote="pco"][quote="math10"]Let $ a_0$ be a natural number. The sequence $ (a_n)$ is deﬁned by $ a_{n \plus{} 1} \equal{} \frac {a_n}{5}$ if $ a_n$ is divisible by $ 5$
and $ a_{n \plus{} 1} \equal{} [a_n \sqrt {5}]$ otherwise . Show that the sequence $ a_n$ is increasing starting from some term.[/quote]

The sequence is such that $ a_{n \plus{} 1}$ is derived from $ a_n$ with one of the two transformations $ f(x) \equal{} \frac x5$ or $ g(x) \equal{} [x\sqrt 5]$

Since $ f(x) \equal{} \frac x5$ is decreasing and $ f(g(x)) \equal{} \frac {[x\sqrt 5]}5$ is decreasing too, we must have at some point two successive usages of $ g(x)$ then :

$ x\sqrt5 \minus{} 1 < g(x) < x\sqrt 5$ (the second $ <$ instead of $ \le$ is because $ \sqrt 5$ is irrational)
So $ g(x)\sqrt5 \minus{} 1 < g(g(x)) < g(x)\sqrt 5$ and  $ 5x \minus{} 4 < 5x \minus{} \sqrt 5 \minus{} 1 < g(g(x)) < 5x$
So $ g(g(x))$ can not be divisible by $ 5$ and the next function to apply is again $ g(x)$

So, as soon as two successive applications of $ g(x)$ occur, only $ g(x)$ will be applied from this point and the sequence become increasing. And since we first showed that we need to have two successive application of $ g(x)$ at some point, the required result is obtained.[/quote]
Nice ,thank you very much  :lol:

[b]Dear pco[/b]
How do you can prove it? Thank you for explain!

"And since we first showed that we need to have two successive application of $ g(x)$ at some point, the required result is obtained."


[quote]There are 2 registered users, of which 14 (or 700.00%) are administrators (see Project:Administrators).[/quote]
 :roll:
http://www.artofproblemsolving.com/Wiki/index.php/Special:Statistics
I was alerted to this by [url=http://www.artofproblemsolving.com/Wiki/index.php/User:Azjps]Azjps[/url],and did not seem to see a thread about it. Why did this happen? [/url]

[quote="Temperal"] I was alerted to this by Azjps,and did not seem to see a thread about it. [/quote]
See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100298[/url].


geometry
geometric transformation
rotation
analytic geometry
trigonometry
function
conics

We're not doing this in our mathbook in school right now, but is there any way to figure out how to rotate an object in a coordinate plane a certain number of degrees (like 42 degrees or something, just anything random) WITHOUT using a protractor? (right now in school, we only do rotations in blocks of 90 degrees. ex: rotate 90, 180, or 270 degrees.that stuff is easy.)

If it is not in blocks of 90 degrees, then it would require polar coordinates and trigonometry, and that would probably go in high school basics (and it wouldn't come out to a nice number anyways).

So yes, it is possible, but not with classroom math methods.

Do you mean like rotating a function and finding the new graph?

like 1/x is a hyperbola rotated 45 degrees.

The rotation of a point $ (x, y)$ through an angle of $ \theta$ counterclockwise is given by

$ x' \equal{} x \cos \theta \minus{} y \sin \theta$
$ y' \equal{} x \sin \theta \plus{} y \cos \theta$.

This can be proven by writing $ (x, y)$ in [url=http://en.wikipedia.org/wiki/Polar_coordinates]polar coordinates[/url] and using the sine and cosine addition formulas.  Verify that this gives familiar formulas when $ \theta \equal{} 90^{\circ}, 180^{\circ}, 270^{\circ}$.

If you want to rotate a set of points, rotate each point individually.  If you want to rotate a graph, such as the parabola $ y \equal{} x^2$, substitute the new coordinates for the old.  andersonw is correct that this is not really "classroom" math.

I got beat to the answer  :( .

But, yes this is definitely Alg II material. Mods, please move to High School Forum.

cmon, i didn't know it was THAT complicated... but there is a way??? (not that i really want to ever use it.)

actually some teachers teach it in school so........... :lol:


From the set of integers $ \{1,2,3,\ldots,2009\}$, choose $ k$ pairs $ \{a_i,b_i\}$ with $ a_i<b_i$ so that no two pairs have a common element. Suppose that all the sums $ a_i\plus{}b_i$ are distinct and less than or equal to $ 2009$. Find the maximum possible value of $ k$.

[quote="Ihatepie"]I just did a: I hope this problem is really simple with a 669.[/quote]
How do you even get 669? Many people put that down at my school but even if you paired up 1 and 2008, 2 and 2006, ..., 668 and 673, 669 and 671, and 670 and 672, you get 670, not 669.

Note the for the solution to be valid, you have to find a configuration that works after proving the bound. This reduces to finding permutations of $ 1,2,\ldots,803$ and $ 804,805,\ldots,1606$ such that the sums are all distinct.

[quote="Smartguy"]I got 669. Is that the right answer?
ignite: if you pair them up, its 1, 2008 ....... 669, 672; 670, 670[/quote]
The way he paired them up, it allowed for 1 extra pair by doing 668 and 673 instead of 668 and 674

zlc's post has the right answer, but does not have a rigorous complete solution fwiw.

Ignite's modification at the end actually motivates the solution for the full problem. You can extend a pairing between two sequences of three consecutive numbers that leaves all the sums distinct to a pairing between two sequences of 803 consecutive numbers.

[quote="flamefoxx99"][quote="zlc"]For [b][i]k[/i][/b] pairs of number, the minimum sum of those [b][i]2k[/i][/b] numbers are 1+2+3... +2k = 2k(2k+1)/2 = k(2k+1), since the sum of each pair has to be distinct and less than 2010, the maximum sum of the sums are 2009+2008+... + (2010-k) = (4019-k)k/2.

In order for k to be a valid one, we must have k(2k+1) <= (4019-k)k/2, which gives the max value of k=803[/quote]

What is the motivation for considering the bounds on the sum of the $k$ numbers?[/quote]

Getting quick bounds is surprisingly helpful; you can always try it on problems like "find the maximum value of", "find all $k$ such that", etc. It takes like 2 seconds and it is worth to try getting bounds on any problem you can.

[quote=smartguy888][quote=Smartguy]I got 669. Is that the right answer?
ignite: if you pair them up, its 1, 2008 ....... 669, 672; 670, 670[/quote]

you're impersonating me :(
reported.
[quote=Alex-131]Nice, does anyone have a solution.[size=50]and yes, I bumped a 12 year old thread for no reason[/size][/quote]

use aops wiki
https://artofproblemsolving.com/wiki/index.php?title=Main_Page[/quote]

[s]they created their acc first

By [url=https://aops.com/community/p3160560]ISL 2012 C2[/url] (see my solution in that thread) the answer is $\left\lfloor2\cdot\frac{2009-1}5\right\rfloor=\boxed{803}$ (the factor of $2$ is for the ordering of the pairs $\{a_i,b_i\}$).

For general $n$, I claim that the answer is
\[k\leq \left\lfloor\frac{2n-1}{5} \right\rfloor.\]
First, I claim that the number of pairs $k$ can be at most $\left\lfloor\frac{2n-1}{5} \right\rfloor$. Note that if we have $k$ pairs, the $2k$ numbers, since distinct, have a sum of at least
\[1+2+\dots+2k=\frac{2k(2k+1)}{2}=k(2k+1),\]
and since they must form $k$ disjoint pairs with distinct sums less than $n$, we have that they have a sum of at most
\[n+(n-1)+\dots+(n-k+1)=\frac{k(2n+1-k)}{2},\]
and taking these two sums and comparing, we have that in order for $k$ pairs to exist satisfying the problem conditions, we must have
\[k(2k+1)\leq \frac{k(2n+1-k)}{2},\]
\[\iff 2k+1\leq {2n+1-k}{2} \iff 5k+2\leq 2n+1 \iff k\leq \frac{2n-1}{5},\]
and since $k$ is an integer, we must have that $k\leq \left\lfloor\frac{2n-1}{5} \right\rfloor$. Now I claim that such a number can be achieved. I'll take this in two cases. \\\\
First, if $k$ is odd, let $k=2m+1$. Then for all integers $0\leq x\leq m$, let $2x+1$ be paired with $3m+2-x$. These pairings give the sums $3m+3$, $3m+4$, $\dots$, $4m+3$. Finally, for all integers $0\leq x\leq m-1$, let $2x+2$ be paired with $4m+2-x$. These pairings give the sums $4m+4$, $4m+5$, $\dots$, $5m+3$. It can be shown that $5m+3$ is always less than $n$ through casework mod $5$ on the condition that $k\leq \left\lfloor\frac{2n-1}{5} \right\rfloor$. \\\\
Finally, if $k$ is even, let $k=2m$. For all integers $0\leq x\leq m-1$, let $2x+1$ be paired with $3m+2-x$. These pairings give the sums $3m+3$, $3m+4$, $\dots$, $4m+2$. Finally, for all integers $0\leq x\leq m-1$, let $2x+2$ be paired with $4m+2-x$. These pairings give the sums $4m+4$, $4m+5$, $\dots$, $5m+3$. It can be shown that $5m+3$ is always less than $n$ through casework mod $5$ on the condition that $k\leq \left\lfloor\frac{2n-1}{5} \right\rfloor$. This shows that $k= \left\lfloor\frac{2n-1}{5} \right\rfloor$ can be achieved, finishing the problem.