AMC
AIME
geometry
geometric transformation
rotation

In simple terms, what is the binomial theroem?  Do you have a sample question?

Erm, it's:

$(x+y)^n = \displaystyle\sum_{i=0}^n \binom{n}{i} x^{n-i}r^i$

Here's a fancy AIME:

Two fo the squares of a 7 by 7 checkerboard are painted yellow and the rest green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane of the board. How many inequivalent color schemes are possible?

Would the solution of that problem just be $\displaystyle\frac{\binom{49}{2}}{4}$? Since you can just rotate any pattern into 3 other new ones, so you need to divide the posibilities by four...

That's very nice intuitive step, but what about if the two yellow spots are diametrically opposite? It wouldn't be 4 rotations.

I see...so you can just look at at half of the board and find all the diametrically opposite pairs, and can see that it's just any point on one 7x4 segment except the very middle of the entire board, so $27$ pairs that are diametrically opposite...so you just add back $2\binom{27}{1}$?

Are you sure it is 27?

Think of it as diamtrically opposite, divide by 2.

If it has 4 rotations equivalent divide by four.

[hide]

this took a bit of thinking  :P

$\binom{49}{2}$ total ways, $24$ diametrically opposite, so you have $49\cdot 24 - 24 = 48 \cdot 24$ ways left

out of these, when rotated, 4 of them will be identitical each time, so you divide it by 4 to get $12\cdot 24$. Then you have to add the $24$ diametrically opposite, but since those are repeated twice each you add $24/2 = 12$, so:

$24 \cdot 12 + 12 = 25 \cdot 12 = 300$

[/hide]

Oh yeah, the middle doesn't count, so it'd be $24$ points to choose from that have a diametrically opposite pair.

So $\frac{\binom{49}{2}}{4}+2\binom{24}{1}$


next semester or next fall perhaps, I'll have to select 1 or 2 optional courses. I took my time before taking optional classes, but I'll have to do them sooner or later. I'd like to know your advice on whether it's "better" or opens more doors to take certain options in actuarial sciences, computers or whatever. I'm interested in about every field, computers, physics and so on. Are there courses that you feel will get you to the core of things like I heard analysis of algorithms is really great in computer science, so if I could narrow it down to perhaps one class of computer, one physics and actuary maybe it could fit the bill. Or do you find it's better to concentrate on only of subject at once ? I'm doing a math degree.

Which area interests you the most?  The only one of those choices I can help you with is physics.  I don't know how much experience you have with physics, but you should start with a good full-year course which covers mechanics, and E&M, from a calculus-heavy standpoint.  If you're still interested, continue to take a course on special relativity, waves, and the like.  If you're still interested, then continue on to quantum, advanced E&M, or whatever else interests you.


even though there was a thread about this i want to start fresh because its been some months now and people have different views.

so, how far do you think youll go?

i tried to be pretty specific in the poll, but try to go into more detail in your posts

LOL... i voted for top 10, preferably top 3, and i know what you voted for... unless u didn't vote.

i havent voted yet, but i know what i would vote for, and will get

edit: try to be as realistic as you can. and if you want to stay private its up to you. but if you vote in the polls nobody will know who you are. unless you post about it

another edit: that previous edit was not aimed at Martin, i just had to put it somewhere

work hard and you just may be surprised at your results. but lets say you do get 50th place and they have countdown round from 1-50, i bet you could make your way to 1st! :P

oooh, 4 people for mc champ, this may get interesting

VERY VERY INTERESTING...

hey neal, can u post another one for team?

This should be interesting... 4 national champs at the same time, hmmmm

Last time there were like 4 votes for champion too.

I added the last option, I hope it's OK with you.

I haven't been reading all posts, but who here has made it in the top 50 Nats before? Some of my friends have, but I haven't :( .

im pretty sure that none of the people who have posted so far have gotten top 50 nats. but i am sure that the people who have posted so far will do really well.

Um I haven't and I never will. :( Oh well, I'll try to make  up for it on the AMC.

Yeah... you sound better than qualified for Mathcounts.

Well
a.) I couldn't get my school to register
and
b.) I decided to move fully to 9th thus losing my qualification

I just made it to school team.  The rest is questionable, I hope to get top 10 in chapter.

I voted for top 25 at nationals. I got #69 last year because of a couple of really dum mistakes. I also got 2 easy team round probs wrong for my team, or we wouldve gotten top 10. They beat me up afterwards.


Prove that

$ l_a\plus{}l_b\plus{}l_c\leq\sqrt{3(m_a^2\plus{}m_b^2\plus{}m_c^2)}$

[quote="Ligouras"]Prove that

$ l_a \plus{} l_b \plus{} l_c\leq\sqrt {3(m_a^2 \plus{} m_b^2 \plus{} m_c^2)}$[/quote]

what is $ l_a,l_b$ and $ l_c$? if $ l_a \le m_a,l_b \le m_b,l_c \le m_c$, ineq is clear true by CS...


The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\text{th}}$ term of this sequence.

[hide="Solution"]
In base 3, we are looking at all numbers representible only by $1$s and $0$s.  The 100th term of this sequence is simply 100 represented in base 2, and interpreted in base 3.  This gives us $100 = 64 + 32 + 4 = 1100100_2$, so our answer is $1100100_3 = 9 + 243 + 729 = \boxed{981}$.
[/hide]

Why is the 100th term interpreted in base 2?

So, suppose you have all the numbers from $1,3,3^2,\ldots 3^n$.  Imagine each comma separated number as a "spot" when you're making numbers from powers of three you can either choose to use a number or not use a number from each spot. So you have $2^n$ possible numbers up to $3^n$. Then, you can create a bijection matching the number you create from powers of 2 to powers of three.

Great solution t0rajir0u!  I'll try to explain in greater detail.

[hide]The sequence is $3^{0}, 3^{1}, 3^{1}+3^{0}, 3^{2}, 3^{2}+3^{0}, 3^{2}+3^{1}, 3^{2}+3^{1}+3^{0}\ldots$

Consider the representation of these numbers in base $3$.  The digit places in base $3$ from right to left are $3^{0}, 3^{1}, 3^{2}\ldots$.  If we write the sequence in base $3$, every term will consist only of the digits $0$ and $1$, because we cannot choose the same power of three twice according to the problem.  Also, every positive number in base $3$ that consists only of the digits $0$ and $1$ is in the sequence.  Then the sequence in base $3$ is

$1,10,11,100,101,110,111\ldots$

The sequence in base $3$ consists of the base $2$ representations of the natural numbers.  So the 100th term written in base $3$ will be the base $2$ representation of $100$, which is $1100100$.  $1100100_{3}=3^{6}+3^{5}+3^{2}=\fbox{981}$[/hide]

[quote="chess64"]What? :? I don't really get it.... :([/quote]

You know that the whole concept of bases is that each number can be represented by summing at most $b-1$ of each power of $b$, where $b$ is the base you're working with.  For example, any integer can be represented using at most $1$ of each power of $2$.  For example, $100 = 2^6 + 2^5 + 2^2 = 110100_2$.

Now, this problem asks for a specific subset of numbers - that is, numbers expressible as sums of at most $1$ of each power of $3$.  For example, $93 = 3^1 + 3^2 + 3^4 = 10110_3$.

Notice that because we're only using $1$ of each power of $3$ instead of $2$ (to represent every integer), we're only going to get numbers that are expressed with $1$s and $0$s in base $3$.

$1 = 3^0 = 1_3, 1_2 = 1$
$3 = 3^1 = 10_3, 10_2 = 2$
$4 = 3^0 + 3^1 = 11_3, 11_2 = 3$
$9 = 3^2 = 100_3, 100_2 = 4$
$10 =3^2 + 3^0 = 101_3, 101_2 = 5$
$12 = 3^2 + 3^1 = 110_3, 110_2 = 6$
$13 = 3^2 + 3^1 + 3^0 = 111_3, 111_2 = 7$

Are you seeing the pattern?  The numbers are just strings of $1$s and $0$s in base $3$.

Now, we want the $100^{th}$ term of such a sequence.  Remember that in base $2$, all numbers are representible as $1$s and $0$s, so what we can do is find $100$ in base $2$, and use those $1$s and $0$s as our number in base $3$, which is the $100^{th}$ term.

(In other words, we have made a bijection between the integers base $2$ and the terms of our sequence - that is, for every integer in base $2$ we can find a corresponding term of the sequence just by using the same pattern of $1$s and $0$s.)

(Hopefully the list above made this step clearer.)

Ah, I get it now. When I took the actual test for practice, I had found that each term was numbers in base 3 with only 0's and 1's, but I didn't know how to find the 100th term--thanks! :)

[hide]Our first few terms are $1,3,4,9,10,12,13,\ldots$. Notice that in base $3$, these numbers are $1, 10, 11, 100, 101, 110, 111,\ldots$.

Interestingly, we just found a one-to-one correspondence! The $n^{\text{th}}$ term of this sequence is simply $n$ interpreted in binary, but translated to base 3. To find our answer, we note that $100=1100100_2$, and $1100100_3 =729 +243+9= \boxed{981}$.
[/hide]

Hi, I'm getting an annoying off-by-one error. Please help!
Listing the first few out in base 3, we get
1
10
11
100
101
110
111
1000
1001
1010
1011
1100
1101
1110
1111

1 digit: 1
2 digit: 2
3 digit: 4
4 digit: 8
n digit: 2^(n-1).

Table
1  2  3  4   5   6   7
1  2  4  8  16 32 64

The last 6 digit number is
111111. This is the 63rd number (sum of 1 + 2 + 4 + 8 + 16 + 32)
The 64th number is
1000000.
Now we can add on 31 more be noting that the 31st number is

11111.
So 1011111 is the 64 + 31 - 1 = 94th number.

We can list out now
95: 1100000
1100001
1100010
1100011
1100100
1100101

Thus, 1100101 is the 100th, which is
729  + 243 + 9 + 1 = 982.
But the answer is 981. What did I do wrong?

[quote="hrithikguy"]
The last 6 digit number is
111111. This is the 63rd number (sum of 1 + 2 + 4 + 8 + 16 + 32)
The 64th number is
1000000.
Now we can add on 31 more be noting that the 31st number is

11111.
So 1011111 is the 64 + 31 - 1 = 94th number.

We can list out now
95: 1100000
1100001
1100010
1100011
1100100
1100101

Thus, 1100101 is the 100th, which is
729  + 243 + 9 + 1 = 982.
But the answer is 981. What did I do wrong?[/quote]

When you add $64 + 31$ why do you subtract 1?
Let's look at it this way. We know the first number is $1$, so if we add $1$ in we don't do $64 + 1 - 1$, so that's where your error is.
Otherwise, good approach.

Ahh..... I see. Thank you very much. Do you know how I can prevent such errors in the future (and on the test!)?


Let $x_1,x_2,..,x_n$ be real numbers and $y_1,y_2,..,y_n$ positive real numbers.
Let $n,m,k$ be positive integers such that $n>2$,$m$ is even and $k$ is odd with$m > k$.
Prove the ineq

$\sum_{i=1}^{n}\frac{x_i^m}{y_i^k} \geq \frac{1}{n^{m-k-1}}\frac{( \sum_{i=1}^{n}x_i)^m}{(\sum_{i=1}^{n}y_i)^k}$

I think it is another consequence of the holder's inequality?

Right, it follows from

$\left(\sum_{i=1}^{n} \frac{x_i^m}{y_i^k}\right)\left(\sum_{i=1}^{n} y_i\right)^k \ge \left(\frac{\sum_{i=1}^{n} x_i^{\frac{m}{k+1}}}{n}\right)^{k+1}\cdot n^{k+1}$
$\ge \left(\frac{\sum_{i=1}^{n} x_i}{n}\right)^m \cdot n^{k+1} = \frac{\left(\sum_{i=1}^{n}x_i\right)^m}{n^{m-k-1}}$

where we use the evenness of $m$ to justify the signs going through Holder and Power-mean and the fact that, for integers, $m > k \iff m \ge k+1$.

\[
\left( {\sum\limits_{i = 1}^n {\frac{{x_i^m }}{{y_i^k }}} } \right)\left( {\sum\limits_{i = 1}^n {y_i } } \right)^k \left( {1 + 1 + .. + 1(n{\rm{ }}1s)} \right)^{m - k - 1}  \ge \left( {\sum\limits_{i = 1}^n {x_i } } \right)^m
\]

I think this is more direct..


what do people think of "camp rock"  :!:

ya, i was a disney movie
it was on abc one saturday night (maybe 8-10 weeks ago?)
it also aired on disney channel a few days earlier

it's about a boy that goes to camp rock and makes friends with the people there. they totally rock out with their music at the camp

anyway, about the first post
since i have never seen the movie, i can't say what i thought of it, but i really want to watch it (so that says something)

Vallon, this is the [b]CT[/b] forum, not the MA forum. :wink:

meh
i found out after i posted...

yay
my 1000th post!
go me!!!!!

(i would've wrote this on my last post, but i can't edit it anymore...)


trigonometry
geometry
geometric transformation
rotation

A mass $m$ is hanging from a spring of rest length $L$.
It is then displaced an angle $\theta$ from the vertical.

Find the equations of motion.

(Hint: you can use $\sin \theta \approx \theta$.)

As the hint basically states that $\theta$ is small, does that mean $\theta$ actually is small? If the spring is fixed to some ceiling, such that no rotation of the spring can occur, then how does a spring behave when it is bent? I think that is very hard problem by itself, because the linear restoring force doesn't translate over the stretched length very well, because I don't know what kind of shape the bending will behave like.  The spring would try to be in a minimum energy scenario of course, but it would be very illuminating if someone responded to how a spring would behave if the terminal surfaces were bent away from each other.

On the other hand it is an easier problem interpreting it as if the terminal surfaces are always aligned parallel to one another and not to form a closed loop, as in one could look down the spring. I also assume that the angular displacement from the vertical means that the spring is just rotated by that angle, so that until you remove your hand from the spring to let it go, the length is still some radius, $r$.  In addition, I don't know if this is standard terminology or not, but is the rest length the length of the spring with no load and perpendicular to the force of gravity? I'll assume it is, it only translates to a slight change in variables anyways. The assumption for $\theta$ is still bothering me, does that mean that $\cos{\theta}= 1$ or $1-\theta^{2}$. I t is looking messy if I attempt to solve it without some approximation like that, so I'll get back to you when I complete it with some series of assumptions.

BTW I just had a huge dose of caffeine and haven't been on the physics forum to my knowledge before, so if I am operating in an intolerable mood as evidenced by my writing, just disregard it.

So it's just plain hard without implementing $\sin{\theta}= \theta$ and $\cos{\theta}= 1-\theta^{2}$ for small enough $\theta$. Thus,
I have two equations: \[\frac{k}{m}L\cos{\theta}-g = \frac{d^{2}L}{dt^{2}}\cos{\theta}-2\frac{dL}{dt}\sin{\theta}\frac{d \theta}{dt}-L\cos{\theta}(\frac{d\theta}{dt})^{2}-L\sin{\theta}\frac{d^{2}\theta}{dt^{2}}\]  \[\frac{k}{m}L\sin{\theta}=\frac{d^{2}L}{dt^{2}}\sin{\theta}+2\frac{dL}{dt}\cos{\theta}\frac{d \theta}{dt}-L\sin{\theta}(\frac{d \theta}{dt})^{2}+L\cos{\theta}\frac{d^{2}L}{dt^{2}}\] With the nice approximation in theta they become: \[\frac{k}{m}L(1-\theta^{2})-g = \frac{d^{2}L}{dt^{2}}(1-\theta^{2})-2\frac{dL}{dt}\theta\frac{d \theta}{dt}-L(1-\theta^{2})(\frac{d\theta}{dt})^{2}-L\theta\frac{d^{2}\theta}{dt^{2}}\]  \[\frac{k}{m}L\theta=\frac{d^{2}L}{dt^{2}}\theta+2\frac{dL}{dt}(1-\theta^{2})\frac{d \theta}{dt}-L\theta(\frac{d \theta}{dt})^{2}+L(1-\theta^{2})\frac{d^{2}L}{dt^{2}}\] Ok from here I get some possibly nice stuff:

by adding sums and differences of the above two equations you get some nice forms ,there is a factor of $\theta$ or $\theta^{2}$ in there, I gotta go.

How can you have $\frac{dL}{dt}$?

$L$ is a constant.  When I did it, I denoted $r$ as the length the spring is stretched, so the radius becomes $L+r$, where downward orientation (stretched away from the attachment point) is positive.

When one pulls a spring, $L$ modifies.
In this particular case, $L$ gets bigger, then smaller, then again bigger etc.

I know, but you need to also know the displacement from the rest length...

Yes I need to use the correct notation, sorry, and I need to fix my errors.

If $L$ is the equilibrium length, then: \[\frac{k}{m}(r-L)\cos{\theta}-g = \frac{d^{2}r}{dt^{2}}\cos{\theta}-2\frac{dr}{dt}\sin{\theta}\frac{d \theta}{dt}-r\cos{\theta}(\frac{d\theta}{dt})^{2}-r\sin{\theta}\frac{d^{2}\theta}{dt^{2}}\]  \[\frac{k}{m}(r-L)\sin{\theta}=\frac{d^{2}r}{dt^{2}}\sin{\theta}+2\frac{dr}{dt}\cos{\theta}\frac{d \theta}{dt}-r\sin{\theta}(\frac{d \theta}{dt})^{2}+r\cos{\theta}\frac{d^{2}\theta}{dt^{2}}\]

Using $\tan{\theta}\frac{d (r \cos{\theta})}{dt^{2}}+g\tan{\theta}= \frac{d(r \sin{\theta})}{dt^{2}}$

We get $g \tan{\theta}= 2 \frac{dr}{dt}\sec{\theta}\frac{d \theta}{dt}+r \sec{\theta}\frac{d^{2}\theta}{dt^{2}}$

The result of that is $\frac{d^{2}r}{dt^{2}}\theta+g\sin{\theta}= \frac{d(r \theta)}{dt^{2}}$

Using $\tan{\theta}\frac{d(r \sin{\theta})}{dt^{2}}+\frac{d(r \cos{\theta})}{dt^{2}}= \frac{k}{m}(r-L)\sec{\theta}-g$

We get $\frac{d^{2}r}{dt^{2}}-r (\frac{d \theta}{dt})^{2}= \frac{k}{m}(r-L)-g \cos{\theta}$

[quote="perfect_radio"]Technically speaking, both notations should give the same result, right?[/quote]

Yes, but you did not specify a constant for the rest length.

Yeah it really doesn't matter, except for the slight error I made in the equation, which does matter.  On that note, how do you solve the differential equations associated with the model? I do not know how, so...help?

[quote="FMako"]Also I should note, that I haven't used the $\sin{\theta}= \theta$ assumption, but I want to hold off using it as long as possible.  I also don't know what $\cos{\theta}$ is in terms of $\theta$ given the previous assumption.  PTynan, did you make up this problem? Mind if I call you Phil?[/quote]

While I'm sure I'm not the first one who thought of it, I did make it up in the sense that I didn't find it in any book.

And yes, Phil is fine.  But how did you know my first name?


A refrigerator is approximately a uniform parallelepiped h = 7 ft tall, w = 3 ft wide, and d = 2 ft deep. It sits upright on a truck with its 3 ft dimension in the direction of travel. Assume that the refrigerator cannot slide on the truck and that its mass is 100 kg.
If the rope  (T) is not there then the maximum acceleration which the truck can move so that the fridge does not tip over is $4.2 \frac{m}{s^{2}}$.

Suppose now that a rope connects the top of the refrigerator with the cab of the truck, which now accelerates at twice the acceleration given above. The refrigerator lifts off slightly at the front but is held in place by the horizontal rope. what is the tension in the rope?

[hide="comment"]
i know that the following two conditions must be satisfied:

1. The net torque about a vertical axis through the CM must be zero.
2. The net force must provide the acceleration.

but i can not find the right equation for findind the friction and the tension[/hide]

Well, the torque by gravity cancels with the first 4.2m/s^2 of acceleration.  So, we know that the torque caused by the tension in the rope must cancel with this additional torque.

So, then disregard the torque caused by gravity and treat it as if it is accelerating at 4.2m/s^2 (where the rope is the only other torsional force acting).  Then, choose the reference frame of the truck.  Since this frame is noninertial, we must account for the fictitious inertial force, -ma, which appears to pull the refrigerator over.

So, let T be the tension, a be acceleration, and m be the mass.

Then, net torque = T*h - ma*h/2 = 0.
Cancelling h, T = ma/2 = 4.2/2 = 2.1m/s^2.

Lol lack of rigor + wrong answer = fruitful

I am going to respond to the comment, because that will yield the most use I hope for you.

The truck is moving with the fridge on it. To make things simple, let's simplify the fridge as a point at the center of mass of the actual fridge. Another important thing to know about this fridge is that it pivots about the edge farthest from the driver. Therefore, about that point there is a torque from the fridge, which is what is keeping it down for the most part with low enough acceleration. Now there is this critical acceleration, where the gravitational fridge torque balances with the torque induced from the vehicle accelerating.  The pivot point of the fridge is tied to the truck itself, so the center of mass of the fridge accelerates at $a$ relative to the pivot. If you draw a diagram properly, you'll see that the competing torques are complementary, for that $3,3.5$ triangle.

Next you know that when you add that new torque from the tension in the rope, it will allow you to accelerate twice as fast,  so the torque from the rope is equal in magnitude to the fridge mass torque and the torque from the truck accelerating( at $4.2 \frac{m}{s^{2}}$), but in the direction of the torque from the truck accelerating and in the opposite direction of the torque from the center of mass differing from the pivot point.

I hope this helps.

Darnit.  I somehow messed up my units and forgot to include mass.

Thus, multiplying the previous result by 100kg gives 210 N for the tension in the rope.


For the following question, schematic graphs are used to prove/find the result. Since there are no variables, the method should be described.

The force between electrostatic charges is described by:

F = [u]q1 x q2[/u]
[color=white].....[/color]4πεor^2

where, F and r are measured variables for known Q1, Q2, and E0. How do you check the inverse square law?

Thanks in advance for the help!

It can be checked by extensive , precise , careful measurement using highly sensitive instrument of the forces between the the 2 charges etc. Remember Cavendish's experiment. Besides, Gauss's Law is an excellent verification of the inverse square law character of many forces!

Direct measurement of the electrostatic force between 2 charges with a torque balance can verify that the power p of the distance r is equal to 2 with the relative uncertainty about $\frac{\triangle p}{2} \doteq \pm 10^{-4}$. One consequence of Coulomb's law is that the electric field everywhere inside the cavity of a charged conductor is $\vec E = 0$, regardless of the conductor and cavity shapes. If the distance power in Coulomb's law was not 2, this would not be the case. The zero electric field inside a big hollow metal conductor can be measured with a great accuracy by observing no oscillation of a little galvanometer through a little double window filled with transparent conductive water solution, while changing the charge distribution on the conductor outer surface with the frequency equal to the characterictic frequency of the galvanometer. In this way, the distance power can be measured with the relative uncertainty about $\frac{\triangle p}{2} \doteq \pm 10^{-10}$.

Thnks for the help.

I figured it out though.

Anyone wanna know the soln...just let me know.

everyone would love to know your 'solution' if its different from the ones I and yetti posted above!


a,b,c -real positive

$ \sqrt{(a^{2}b\plus{}b^{2}c\plus{}c^{2}a)(ab^{2}\plus{}bc^{2}\plus{}ca^{2})} \ge abc\plus{} \sqrt[3]{(a^3 \plus{} abc)(b^3 \plus{} abc)(c^3 \plus{} abc)}$

Divide by $ abc$ and denote $ x \equal{} \frac {a}{b}, y \equal{} \frac {b}{c}, z \equal{} \frac {c}{a}$.

and

$ p \equal{} x \plus{} y \plus{} z, q \equal{} xy \plus{} yz \plus{} zx, r \equal{} xyz \equal{} 1, w \equal{} \sqrt {pq}$.

The inequality becomes :
\[ \sqrt {(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx)} \minus{} 1\geq\sqrt [3]{(x \plus{} y)(y \plus{} z)(z \plus{} x)}\]

\[ \Leftrightarrow\sqrt {pq} \minus{} 1\geq\sqrt [3]{pq \minus{} 1} \Leftrightarrow w(w \minus{} 1)(w \minus{} 3)\geq 0\]
, true because $ w\geq3$.
Equality occurs when $ a \equal{} b \equal{} c$


[quote="RacerX"]Just wondering[/quote]

Correct me if I'm wrong, but doesn't this go into the Questions forum?


function
complex numbers
complex analysis
complex analysis unsolved

Find regular function $ f(x\plus{}iy)$ whose real part looks like $ u(x\plus{}iy)\equal{}y \cos(y) \sinh(x) \plus{} x \sin(y)\cosh(x)$.

Thanks.

Writing everything in terms of exponentials and simplifying we find,
$ \begin{eqnarray*}u(x,y) & = & \frac {i}{4}\left(e^{ - x - iy}x + e^{x - iy}x - e^{ - x + iy}x - e^{x + iy}x + ie^{ - x - iy}y - ie^{x - iy}y + ie^{ - x + iy}y - ie^{x + iy}y\right) \\
& = & \frac {i}{4}\left(ze^{ - z} + \overline{z}e^{\overline{z}} - \overline{z}e^{ - \overline{z}} - ze^{z}\right) \\
& = & \frac {1}{2}\Re\left(ize^{ - z} - ize^{z}\right) \\
& = & \Re \left( - iz\sinh z\right)$


I think this is from an old AMC-12.

sec(x) - tan(x) = 2
Find the value of sec(x) + tan(x)

Hint: [hide]How are sec x and tan x related?  Think about the Pythagorean Theorem, and try drawing a picture.[/hide]

[hide]Think of it as ratios.
a=length of the leg adjacent to angle x
b=length of the leg opposite of angle x
h=length of the hypotenuse.
sec(x)=h/a
tan(x)=b/a
sec(x)-tan(x)=(h-b)/a
sec(x)+tan(x)=(h+b)/a
(h-b)/a*(h+b)/a=(h^2-b^2)/a^2=a^2/a^2=1
Thus (sec(x)-tan(x))*(sec(x)+tan(x))=1
2*(sec(x)+tan(x))=1
sec(x)+tan(x)=1/2[/hide]

This is a good problem, I think I just got a solution.

sec(x)-tan(x)=2
Squaring both sides we get sec^2(x)-2sec(x)tan(x)+tan^2(x)=4, then adding one to both sides and using the identity that tan^2(x)+1=sec^2(x), we get 2sec^2(x)-2sec(x)tan(x)=5, so factoring we get 2sec(x)(sec(x)-tan(x))=5, then using the fact thats given that sec(x)-tan(x)=2, then 4sec(x)=5 so sec(x)=5/4.  I'm too lazy to get my calculator but from there the answer should be able to be calculated.

Ahhh, the divergence of algeabraic and geometric solutions.

$\sec x +\tan x=2$
$\text{Put }\sec x-\tan x=k$
[hide=" Now spoiler-ized"]$\left(\sec x+\tan x\right)\left(\sec x-\tan x\right)=2k$
$\sec^2 x-\tan^2 x=2k$
$1=2k\text{, by the Pythagorean Identity}$
$\frac{1}{2}=k=\sec x-\tan x$[/hide]

[hide]Use the identity $\sec^2 x - \tan^2 x = 1$, and then factor the left side by difference of squares.[/hide]

yeah, i did what ravi did.

sec^2x - tan^2x = 1
(secx+tanx)(secx-tanx) = 1
(secx + tanx)(2) = 1
secx + tanx = 1/2

tah dah.

$sec^2x-tan^2x=1$

$(sec{x}+tanx)(secx-tanx)=1$

$2(secx+tanx)=1$

$secx+tanx=\frac{1}{2}$


trigonometry
trig identities
Law of Cosines
IMO Shortlist

Let $ABC$ be an equilateral triangle and let $P$ be a point in its interior. Let the lines $AP$, $BP$, $CP$ meet the sides $BC$, $CA$, $AB$ at the points $A_1$, $B_1$, $C_1$, respectively. Prove that

$A_1B_1 \cdot B_1C_1 \cdot C_1A_1 \ge A_1B \cdot B_1C \cdot C_1A$.

By the Law of Cosines,

$ A_1B_1^2 \equal{} A_1C^2 \plus{} B_1C^2 \minus{} A_1C\cdot B_1C \geq A_1C\cdot B_1C$ etc so multiply together

and by Ceva, $ A_1B \cdot B_1C \cdot C_1A \equal{} AB_1\cdot BC_1\cdot CA_1$

[hide=Solution]From Ceva's Theorem, we obtain $$\frac{AC_1}{C_1B}\cdot\frac{BA_1}{A_1C}\cdot\frac{CB_1}{B_1A}=1\Leftrightarrow AC_1\cdot BA_1\cdot CB_1=C_1B\cdot A_1C \cdot B_1A.$$ Now, from the Law of Cosines, $$A_1C_1^2=C_1C^2+A_1B^2-C_1BA_1B.$$ From the trivial inequality, $$A_1C_1^2=C_1C^2+A_1B^2-C_1B\cdot A_1B\geq C_1B\cdot A_1B.$$ Similarly, we find that $$A_1B_1^2=B_1C^2+A_1C^2-B_1C\cdot A_1C\geq B_1C\cdot A_1C \text{ and }$$$$ B_1C_1^2=B_1A^2+C_1A^2-B_1A\cdot C_1A\geq B_1A\cdot C_1A.$$ Multiplying these inequalities together, we obtain $$(A_1B_1 \cdot B_1C_1 \cdot C_1A_1)^2 \geq A_1B \cdot C_1B \cdot B_1C \cdot A_1C\cdot C_1A \cdot B_1A=$$$$(A_1B\cdot B_1C \cdot C_1A)\cdot(C_1B \cdot A_1C \cdot B_1A)=(A_1B\cdot B_1C \cdot C_1A)^2\Leftrightarrow$$$$A_1B_1 \cdot B_1C_1 \cdot C_1A_1 \ge A_1B \cdot B_1C \cdot C_1A,$$ as desired.[/hide]

By Ceva it suffices to show that $$A_1B_1^2 \cdot A_1C_1^2 \cdot B_1C_1^2 \geq B_1C \cdot CA_1 \cdot A_1B \cdot BC_1 \cdot C_1A \cdot AB_1.$$ But $$A_1B_1^2 \geq CB_1 \cdot CA_1 \iff \sin \angle CB_1A_1 \cdot \sin \angle CA_1B_1 \leq \frac 34,$$ but this is evident as the angles sum to $120^\circ$.


We recently received an order from Korea in which I believe the buyer used a Korean character that was rendered by our software as a text string in standard English characters (ampersands, numbers, number sign, and semi-colon).  Here's what I am guessing the address looks like with the Korean character in it (I'm copy-pasting the address here, and the string renders back as its Korean character I hope):

GOJAN 1DONG 628-4 302호
ANSAN 425-804
 Korea, Republic of (South)

Does this look like a reasonable address?

Yes, it looks reasonable.
The Korean character usually stands for the apartment/house/etc number in a building.

Thanks!  And I'm sure the person in Korea thanks you as well. . .

[quote="Spencer"]the korean character just states that it is a suite number or office number or apartment number.  In this case apartment number.[/quote]

(that was my mom!!!!!) :diablo:  :starwars:
I like Star Wars!! :roll:

It's not old.. it's just kind of.. like a book behind a bookshelf.


calculus
integration
function
calculus computations

Hi,

I think that this problem is interesting:

Let $a\in[0,1]$. Evaluate in closed form:

$\int_{0}^{\infty}\frac{e^{x}}{\sqrt{(e^{x}-a)^{3}}}dx$.

Because $a\in [0,1]$, the function is continous on $(0, \infty)$.  Make the substitution $u=e^{x}-a$, then we have

$\int_{0}^\infty \frac{e^{x}}{\sqrt{(e^{x}-a)^{3}}}\ dx = \int_{1-a}^{\infty}\frac{du}{u^{3/2}}= \frac{2}{\sqrt{1-a}}$


Stanford
college
calculus
derivative
limit
real analysis
real analysis unsolved

This problem appears in Stanford Mathematics Tournament as a part of the official problem but I don't get the official solution!
Here's the problem:Prove that: lim(h goes to 0) [f(x+2x)-2f(x+h)+f(x)]/(h^2)=f''(x)(the 2nd derivative of f(x))
The solution says that: according to the definition of f''(x), it infers that lim(h goes to 0) [f(x+2x)-2f(x+h)+f(x)]/(h^2)=f''(x)
But I know that the "true" definition of f''(x) is: lim(h goes to 0) [f(a+h)-2f(a)+f(a-h)]/h^2=f''(a).If we follow this formula, the LHS can't be derived to be equal to: lim(h goes to 0) [f(x+2x)-2f(x+h)+f(x)]/(h^2).
So is lim(h goes to 0) [f(x+2x)-2f(x+h)+f(x)]/(h^2)=f''(x) wrong? If it is, the official solution must be wrong! :o

I'm going to guess that that's a typo in the problem and $ f(x\plus{}2x)$ was supposed to have been $ f(x\plus{}2h),$ and that the solution was written under the assumption that the problem said $ f(x\plus{}2h).$

[quote="ghjk"]
Here's the problem:Prove that: lim(h goes to 0) [f(x+2x)-2f(x+h)+f(x)]/(h^2)=f''(x)(the 2nd derivative of f(x))[/quote]

that is not possible at all. you typed the question [b]wrong.[/b] period. stanford cannot make that kind of a mistake.

it has to be
$ f^{(2)}(x) \equal{} \lim_{h\to 0} \; \dfrac{f(x \plus{} 2h) \minus{} 2f(x \plus{} h) \plus{} f(x)}{h^2}$

see eq 10 of
http://mathworld.wolfram.com/Derivative.html

[quote="ghjk"]
But I know that the "true" definition of f''(x) is: lim(h goes to 0) [f(a+h)-2f(a)+f(a-h)]/h^2=f''(a).[/quote]

 that is   [b]"one"[/b] of the   definitions for  $ f^{(2)}(x)$. there is no one and only one true definition. what you have   happens to be one of the [i]symmetric[/i] definitions. it all depends on how the secant eventually becomes a tangent.

just like

$ f'(x) \equal{} \lim_{h\to 0} \; \dfrac{f(x \plus{} h) \minus{} f(x)}{h} \equal{} \lim_{h\to 0} \; \dfrac{f(x) \minus{} f(x \minus{} h)}{h} \equal{} \lim_{h\to 0} \; \dfrac{f(x \plus{} h) \minus{} f(x \minus{} h)}{2h}$

you can also have

$ f^{(2)}(x) \equal{} \lim_{h\to 0} \; \dfrac{f'(x \plus{} h) \minus{} f'(x)}{h} \equal{} \lim_{h\to 0} \; \dfrac{f'(x) \minus{} f'(x \minus{} h)}{h} \equal{} \lim_{h\to 0} \; \dfrac{f'(x \plus{} h) \minus{} f'(x \minus{} h)}{2h} \equal{} \lim_{h\to 0} \; \dfrac{f(x \plus{} 2h) \minus{} 2f(x \plus{} h) \plus{} f(x)}{h^2} \equal{} \lim_{h\to 0} \; \dfrac{f(x) \minus{} 2f(x \minus{} h) \plus{} f(x \minus{} 2h)}{h^2} \equal{} \lim_{h\to 0} \; \dfrac{f(x \plus{} 2h) \minus{} 2f(x) \plus{} f(x \minus{} 2h)}{4h^2}$

[quote="ghjk"]   If it is, the official solution must be wrong! :o[/quote]

i don't think so. it is you who wrote down the question   wrong.

The problem is posted on the SMT website with $ f(x \plus{} 2x),$ so it's not a matter of ghjk miscopying that. I still assume it's a misprint. I've PM'd Sly Si - see his response below.

It should in fact be $ f(x\plus{}2h)$; I seem to recall having to announce that during the tournament.

ghjk, the definition you've given is equivalent to the one that this problem requires.  Take yours and replace $ a$ with $ x\plus{}h$, and you get the other one.  Since h goes to zero, they give the same answer as long as the second derivative is continuous.


Assume there is a test for a terminal disease that is deemed to be 98% accurate(meaning that is someone has the disease, they will test positive 98% of the time and if someone doesn't have the disease they will test negative 98% of the time).  Further assume that only 0.5%(1 in 200) actually have the disease.

Your doctor has informed you that you have tested positive for this deadly disease.  What are the chances that you actually do have the disease?(expressed as a decimal to the nearest tenth)

Should you be worried?

not sure
[hide]If you test positive then I don't really think it should matter that 1/200 of the people have it.. isn't it 1.0? heheh[/hide]

haha. Nope, not that simple. The likelyhood of you having the disease definitely plays a part.

That is an interesting new decimal you have discovered dwx :D

[hide]It says 1 in 200, and the chances that he is right is 98%, so you have 49/50*1/200=49/10000, which is 0.49% or 0.0049, so no, you shouldn't be worried.[/hide]

.49% is an important part of the answer, but not the answer

This is like the "What's the Verdict" problem you posted a few weeks ago.

Is it .2?
I'm doing this really quick at 10:00 pm, so I'm probabily wrong. I tried to use Bayes' Theorem.

[hide][quote="dwx314"]Evil conditional prob.
[hide]Table!
test says +, he's wrong: 199/200*1/50
test says -, he's right: 199/200*49/50
test says +, he's right: 1/200*49/50
test says -, he's wrong: 1/200*1/50

we only need the 1st and 3rd, because they you have the desiese.
So its really:
Good: 49/10,000
All: 49/10,1000+199/10,000

>>49/248<<[/hide]I think...[/quote]

I agree with dwx314. This is an example of conditional probability.
The answer is the probability that you have the disease and the test shows positive over the total probability that the test will show positive (this can happen in two ways) :
(49/50 x 1/200) / ((49/50 x 1/200) + (1/50 x 199/200))
= (49/10000) / (49/10000 + 199/10000)
= (49/10000) / (248/10000)
= (49/10000) x (10000/248)
= 49/248[/hide]

Magix, please hide you answer (yse the "Hide" button on top of the text area)

Sorry frost13. ;)

[hide]49/248~19.8%
>>19.8<<[/hide]
My bad

[quote="frost13"]It is juts like the "What's the verdict?"
I was just going to say you were wrong, but it might just be your form that is wrong. A percent to one decimal place.[/quote]
Sorry! I think it is my form that's wrong. [hide]I got [b]19.8%[/b].  I know I shouldn't be worried, but if there is a 1/5 chance that I have a deadly disease, I would be.[/hide]

Ahhh,
So the next time I see a list of figures like that, remind me to do the math.
~1/5 is a all less accurate than any of us expected.

we have enough right answers, but a lot of issues with the form of answers :(

Be careful with that at competitions everyone.

19.8 % is correct


Euler
inequalities
geometry theorems
geometry

in the triangle $ ABC$,we have the sides $ a,b,c$,and $ R,r,S,s$,which $ 2s \equal{} a\plus{}b\plus{}c$
we have $ R\geq 2r$and $ a^{2}\plus{}b^{2}\plus{}c^{2}\geq 4sqrt{3}S$
$ sinA\plus{}sinB\plus{}sinC \equal{}\frac{s}{R}$
....
who can post all the identity and  inequality?

Thanks.

You can find here http://www.math.dartmouth.edu/~euler/ , or use Google with good keywords .

because my English study is not very good.so can someone copy the theorem here and let everyone saw it rapidly?

thx..


inequalities
function
algebra
polynomial
algebra unsolved
Functional Equations

Find all nondecreasing functions $ f:\mathbb{R}^{\plus{}}\rightarrow\mathbb{R}^{\plus{}}$ such that for all $ x\in\mathbb{R}^{\plus{}}$, $ f(x\plus{}3)\plus{}f(x)\ge 2\cdot f(x\plus{}2)$

Can anyone solve it? It's been over a week. Any ideas or comments would be welcome.

I have a comment that if $ f$ is bounded from above then $ f$ must be constant.

Oh mods please delete my previous post. In fact I mean $ f$ is period with period $ 1$, ie... $ f(x\plus{}1)\equal{}f(x)$ for all $ x$.

Take $ f(x) \equal{} e^{x}$.  Obviously, $ f(x\plus{}3)\plus{}f(x) \equal{}\left(e^{3}\plus{}1\right)e^{x}> e^{3}e^{x}> 2e^{2}e^{x}\equal{} 2f(x\plus{}2)$.  In fact, for any non-negative $ a$ such that $ e^{3a}\plus{}1\geq2e^{2a}$, and for any real $ b$, then $ f(x) \equal{} e^{ax\plus{}b}$ is a solution.  In a similar way, it is also possible to show that $ f(x) \equal{} cosh(x)$ is also a solution.  Once this is established, any non-decreasing function of $ x$ multiplied by either $ e^{x}$ or $ cosh(x)$ will also satisfy the condition.  Or, we may expect an infinite number of solutions.  However, one issue is clear: no finite-degree polynomial is able to satisfy the given condition, since if $ f(x) \equal{} ax^{n}\plus{}bx^{n\minus{}1}\plus{}...$, then the $ n\minus{}1$-th degree coefficient in $ 2f(x\plus{}2)\minus{}f(x\plus{}3)\minus{}f(x)$ is $ an$, while the $ n$-th degree coefficient would be 0, or for $ x$ sufficiently large the inequality would not hold.  So all functions that grow at least as fast as $ cosh(x)$ or $ e^{x}$ are good answers.  This gives a lot of room...

[quote="daniel73"]one issue is clear: no finite-degree polynomial is able to satisfy the given condition[/quote]

How about $ f(x) \equal{} c$?  ;)

The equality case $ f(x\plus{}3)\plus{}f(x) \equal{} 2 f(x\plus{}2)$ corresponds to a third-order linear recursion with characteristic equation $ t^{3}\minus{}2t^{2}\plus{}1 \equal{} (t\minus{}1)(t^{2}\minus{}t\minus{}1)$, so it therefore has solutions of the form

$ f(n) \equal{} A\plus{}B\phi^{n}\plus{}C\varphi^{n}$

where $ \phi,\varphi$ are the roots of $ t^{2}\minus{}t\minus{}1 \equal{} 0$.  As daniel73 noted, other exponential functions that grow faster than $ \phi^{n}$ should also be solutions.

Edit:  pluricomplex:  a nondecreasing function with period $ 1$ is constant, no?

Edit 2:  Note that only values of $ B, C$ that produce non-decreasing functions $ f(n)$ are valid.  I believe it is enough that $ f(0)\le f(1)$.

[quote="t0rajir0u"][quote="daniel73"]one issue is clear: no finite-degree polynomial is able to satisfy the given condition[/quote]

How about $ f(x) \equal{} c$?  ;)

[/quote]

Uh-oh, got me   :blush:
Obviously "non-constant" should have been added right next to "finite-degree polynomial".   :oops:
Torajiru's answer is definitely much more complete than mine, since it does not only show why my answers work, but also bounds the set of functions that satisfy the inequality.  Good job Torajiru!


How do I do this type of problem:

$x+y+z=1$
What is the largest possible value of $x^3 y^2 z?$

Thanks.

The problem's kind of vague. Is this what you mean?

Find the maximum value of $x^{3}y^{2}z$ if $x+y+z=1$.

I think it's too hard for Getting Started.

[quote="draconis"]My apologies I don't know how to use LaTeX in this forum. What's the tag? I assumed it was [math] then when I realized it was wrong the system won't let me change it saying it's my first post and I have to wait another 10 min.

No I meant x+y+z=1, calculate maximum of (x^3)(y^2)(z).

Is that still too hard for getting started?[/quote]

let y be a really big negative number.

Ummm, unless there's some restriction on the variables, like that they must be positive, it can be infinity by letting $y$ be an infinitely large negative number and $x$ an infinitely large positive number, with $z=0$.

Otherwise, I'm thinking it's $\left(\frac{1}{3}\right)^{3}\left(\frac{1}{3}\right)^{2}\frac{1}{3}=\frac{1}{3^{6}}=\frac{1}{729}$, but you can correct me if I'm wrong.

Btw draconis, the tags for $\LaTeX$ are dollar signs on either side of the stuff you want to use it with...

[b]Generalization.[/b] Let $a_{1},a_{2},\cdots,a_{n}$ be a sequence of positive real numbers such that their sum equals $A\in\mathbb{R}^{+}$. If $b_{1},b_{2},\cdots,b_{n}$ are positive integers with sum $B$, then \[\max\left(\prod_{i=1}^{n}a_{i}^{b_{i}}\right) = \left(\frac{A}{B}\right)^{B}\left(\prod_{i=1}^{n}b_{i}^{b_{i}}\right).\] Equality is achieved when $\frac{a_{i}}{b_{i}}$ is constant.

[quote="DanK"]Ummm, unless there's some restriction on the variables, like that they must be positive, it can be infinity by letting $y$ be an infinitely large negative number and $x$ an infinitely large positive number, with $z=0$.[/quote]

There is no such thing in the conventional real number system as an "infinitely large" anything.  The way mathematicians say what you're trying to say is to use the words "arbitrarily large," meaning "as large as you happen to want to make it at any given instant."  Then the given product can "go to infinity."  But it can't actually [i]be at[/i] infinity (or be infinitely large).

Also, you probably don't want to have $z = 0$, because then your product is going to sit around at 0 no matter how you choose $x$ and $y$.  $z = 1$ is much happier.


If angles $ A$ and $ B$ are complementary and equal in measure, what is the degree measure of each angle?

Complementary angles add up to 90 degrees, so if they have equal measures, they must both be equal to $ 90/2 \equal{} 45^\circ$.


Let $ f : R^\plus{} \to R^\plus{}$ be an increasing function. For each $ u \in R^\plus{}$ , we denote $ g(u) \equal{}$ inf$ \{ f(t)\plus{}u/t|t>0\}$. Prove that
   (a) If $ x\le g(xy)$ , then $ x\le 2f(2y)$
   (b) If $ x\le f(y)$ , then $ x\le g(xy)$.

We assume y>0 as f(2y) is defined.
Now,
(a)
$ x \le inf \{ f(t) \plus{} u/t|t > 0\} \\
\implies x \le f(2y) \plus{} x/2 \\
\implies x \le 2f(2y)$

(b)
Suppose
$ x > g(xy) \equal{} inf \{ f(t) \plus{} (xy)/t|t > 0\} \\
\implies f(y) \ge x > f(t_0) \plus{} \frac {xy}{t_0}>f(t_0)$for some $ t_0 > 0$.
then by increasing ness $ y > t_0 \implies xy/t_o > x \implies f(t_0) < 0$ which is impossible.


Given the set of $ 2n$ integers {$ \pm \mathop a\nolimits_1,\pm \mathop a\nolimits_2,...\pm \mathop a\nolimits_n$} and a positive integer $ m<2^n$ , show that it is always possible to select a subset $ S$ such that

1 $ \pm \mathop a\nolimits_i$ are not both contained in $ S$
2 the sum of the elements of $ S$ is divisible by $ m$

Consider $ 2^n$ sums with the form $ S(i_1,i_2,...,i_k)\equal{}\sum_{j\equal{}1}^k a_{ij}$ where $ 1 \le i_1 < i_2 < ... < i_k \le n$ then since $ m < 2^n$, by Pigeon hole theorem, there must be some sums $ S(a_1,a_2,...,a_u)\equal{}S(b_1,b_2,...,b_v)$, the difference of them yields a set which satisfies the two conditions. :)

[quote="Mashimaru"]Consider $ 2^n$ sums with the form $ S(i_1,i_2,...,i_k) \equal{} \sum_{j \equal{} 1}^k a_{ij}$ where $ 1 \le i_1 < i_2 < ... < i_k \le n$ then since $ m < 2^n$, by Pigeon hole theorem, there must be some sums $ S(a_1,a_2,...,a_u) \equal{} S(b_1,b_2,...,b_v)$, the difference of them yields a set which satisfies the two conditions. :)[/quote]A small flaw: you included the empty subset. :wink:  This can be trivially fixed though.


inequalities
geometry
inequalities proposed

Let $a, b, c$ be the sides of a triangle, $m_a,m_b,m_c$ be the corresponding medians. Prove that

$m_aa^2+m_bb^2 + m_cc^2 \geq 4m_am_bm_c$.

From the formulas $4m_a^2 = 2b^2 + 2c^2 - a^2$, $4m_b^2 = 2c^2 + 2a^2 - b^2$, $4m_c^2 = 2a^2 + 2b^2 - c^2$, we can derive $\displaystyle a^2 = \frac49 \left(2m_b^2 + 2m_c^2 - m_a^2\right)$, $\displaystyle b^2 = \frac49 \left(2m_c^2 + 2m_a^2 - m_b^2\right)$, $\displaystyle c^2 = \frac49 \left(2m_a^2 + 2m_b^2 - m_c^2\right)$. Using these identities, your inequality rewrites as

$\displaystyle m_{a}\cdot \frac49 \left( 2m_{b}^{2}+2m_{c}^{2}-m_{a}^{2}\right) +m_{b}\cdot \frac49 \left( 2m_{c}^{2}+2m_{a}^{2}-m_{b}^{2}\right) +m_{c}\cdot \frac49 \left( 2m_{a}^{2}+2m_{b}^{2}-m_{c}^{2}\right) \geq 4m_{a}m_{b}m_{c}$,

i. e. as

$\displaystyle m_{a}\cdot \frac49 \left( 2m_{b}^{2}+2m_{c}^{2}-m_{a}^{2}\right) +m_{b}\cdot \frac49 \left( 2m_{c}^{2}+2m_{a}^{2}-m_{b}^{2}\right) +m_{c}\cdot \frac49 \left( 2m_{a}^{2}+2m_{b}^{2}-m_{c}^{2}\right) - 4m_{a}m_{b}m_{c} \geq 0$.

Now, since the medians of every triangle are the sidelengths of another triangle, the numbers $m_a$, $m_b$, $m_c$ are the sidelengths of a triangle, so we can substitute $m_a = y+z$, $m_b = z+x$, $m_c = x+y$ and get

$\displaystyle m_{a}\cdot \frac{4}{9}\left( 2m_{b}^{2}+2m_{c}^{2}-m_{a}^{2}\right) +m_{b}\cdot \frac{4}{9}\left( 2m_{c}^{2}+2m_{a}^{2}-m_{b}^{2}\right) +m_{c}\cdot \frac{4}{9}\left( 2m_{a}^{2}+2m_{b}^{2}-m_{c}^{2}\right) -4m_{a}m_{b}m_{c}=\frac{8}{9}\cdot \left( x^{3}+y^{3}+z^{3}+3xyz-yz\left( y+z\right) -zx\left( z+x\right) -xy\left( x+y\right) \right) $

what is $\geq 0$ after Schur. Hence, the inequality is proven.

  Darij

Nice proof!. I have another hlaf of the inequality:

$\displaystyle \frac{4}{9} (m_a+m_b+m_c)(m_am_b+m_bm_c + m_cm_a) \geq m_aa^2+m_bb^2 + m_cc^2 \geq 4m_am_bm_c$.

which could be proved using your method and Schurs too.

I derived these inequalities with different routes. Again, it's nice to see someone came up with a different solution. Thanks.

By the way, this can be derived from:

http://www.mathlinks.ro/Forum/viewtopic.php?t=15943

by taking $x=m_a, y=m_b, z=m_c$.

This is closely related to a very interesting geometry problem found at:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6109

Let me try to recap if I understand correctly:

Darij translated the problem into the following inequality:

$\left( a^{2}+b^{2}+c^{2}\right) \left( am_{a}+bm_{b}+cm_{c}\right) \geq 4m_{a}m_{b}m_{c}\left( a+b+c\right) $ ------------- (*)

Hxtung proved it in two steps:

Step 1. $\displaystyle (a^2+b^2+c^2)(am_a+bm_b+cm_c) \geq (a+b+c)(a^2m_a+b^2m_b+c^2m_c)$

Step 2. $\displaystyle  a^2m_a+b^2m_b+c^2m_c \geq 4m_{a}m_{b}m_{c}$

As noted by harazi, (*) is also equivalent to the problem posted at:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14693


Find all the real numbers $ a<b<1$ such that :

$ (\forall)x\in [a,b]\Rightarrow \frac{x}{x\minus{}1}\in [a,b]$

There are none, as $ {x\over x\minus{}1}$ is negative.

[quote="Farenhajt"]There are none, as $ {x\over x \minus{} 1}$ is negative.[/quote]

Are you sure ? Who said a,b are positives ?

[hide="Solution"]$ a\leq{x\over x\minus{}1}\leq b$

Since $ x\minus{}1<0$, we get

$ b(x\minus{}1)\leq x\leq a(x\minus{}1)$

1. $ bx\minus{}b\leq x\implies (b\minus{}1)x\leq b\implies x\geq{b\over b\minus{}1}$, since $ b<1$

2. $ x\leq ax\minus{}a\implies a\leq(a\minus{}1)x\implies{a\over a\minus{}1}\geq x$, since $ a<1$

Therefore the condition is equivalent to

$ {b\over b\minus{}1}\leq x\leq {a\over a\minus{}1}$

Segment $ [a,b]$ must be a subset of this, hence $ {b\over b\minus{}1}\leq a\land{a\over a\minus{}1}\geq b$

The inequations give $ b\geq a(b\minus{}1)\land a\leq b(a\minus{}1)$, since both $ a$ and $ b$ are less than $ 1$.

Rearranging gives $ a\plus{}b\geq ab\land a\plus{}b\leq ab$

Therefore $ a\plus{}b\equal{}ab$, which gives $ b\equal{}{a\over a\minus{}1}$

Since $ a<b<1$, we get $ a<{a\over a\minus{}1}<1$, which after solving gives $ a<0\land a>a\minus{}1$.

Therefore, all such pairs are given by $ a<0\land b\equal{}{a\over a\minus{}1}$[/hide]

We have $ \frac{d}{dx} (\frac{x}{x\minus{}1})\equal{}\minus{}\frac{1}{(x\minus{}1)^2}$, so $ \frac{x}{x\minus{}1}$ is strictly decreasing in $ [a,b]$

The question becomes $ \begin{cases} \frac{a}{a\minus{}1} \le b \\
\frac{b}{b\minus{}1} \ge a \end{cases}$ $ \Leftrightarrow \begin{cases} a\plus{}b \ge ab \\
a\plus{}b \le ab \end{cases}$, $ \Leftrightarrow b\equal{}\frac{a}{a\minus{}1}$

With the condition $ a<b<1$ gives $ a<0 \land b\equal{}\frac{a}{a\minus{}1}$ as the solutions.


[hide]Scale the inequality so that $a^{2}+b^{2}+c^{2}=1$

A simple check verifies that the function $f(u)=\frac{u^{3}}{1-u}$ is convex on (0,1)

Thus, by Jensen's Inequality:

$\sum_{cyc}\frac{a^{6}}{b^{2}+c^{2}}=\sum_{cyc}\frac{(a^{2})^{3}}{1-a^{2}}\ge 3 \frac{(1/3)^{3}}{1-(1/3)}=\frac{1}{6}$

We must show that $\frac{abc(a+b+c)}{2}\le \frac{1}{6}$

Recalling that $a^{2}+b^{2}+c^{2}=1$, this is equivalent to:

$3abc(a+b+c) \le (a^{2}+b^{2}+c^{2})^{2}$

After multiplying everything out, the result follows by Muirhead[/hide]


symmetry
function
number theory unsolved
number theory

Let S is union intervals $ (0,1),(2,\infty)$ and $ f: S\to S$:
\[ f(x) \equal{} \sqrt {\frac {x^2 \minus{} 4}{x^2 \minus{} 1}}.\]
Prove that 1)$ f(f(x)) \equal{} x$
2) If $ x,y\in S$ and $ xy \equal{} 2$, then $ f(x)f(y) \equal{} 2$, if $ f(x)f(y) \equal{} 2$, then $ xy \equal{} 2$.
3) If x- rational, then $ f(x)$ is irrational.
4) If $ x,y,f(x)f(y)$ are rational, and $ xy\equal{}n$ n- natural,  then $ xy \equal{} f(x)f(y) \equal{} 2$.

The function isn't real valued on all of $ S$ and, even where it is, none of the assertions true.  I assume there's a typo?


What does the words below our names mean? For example P versus NP on my name?

It stands for some famous theory or equation, in this forum it represents how many posts you have done.
I read about PvsNP some day.

[quote="Iversonfan2005"][quote="math92"]i just noticed that this question has been asked so many times...I even asked it when i just joined :)[/quote]

Math92, have you posted anything that relates to math on this forum? ;)[/quote]

i don't know, he's posted in 40 topics  :lol:

What's after Navier Stokes Equations? Birch and something dyer?

How many posts does it take to get you to Birch and Swinnerton Dyer?  Like 5000?

i think its 5000
they should make something after Birch and Swinnerton Dyer

But are there any more millenium problems left?  Also, are they in order of difficulty?  Because it would be kind of hard to rank the difficulty of unsolved problems.

New Member: 0 posts
P vs. NP: 20 posts
Hodge Conjecture: 50 posts
Poincare Conjecture: 100 posts
Riemann Hypothesis: 250 posts
Yang-Mills Theory: 500 posts
Navier-Stokes Equations: 1000 posts
Birch and Swinnerton Dyer: 2500 posts

I think that's what it is.


Find the number of subsets of $\{1, 2, \cdots, 2000 \}$, the sum of whose elements is divisible by $5$.

But it is still difficult to find $ f(2000)$. Can you find a closed-form formula of $ f(n)$?

an alternative approach:
Let $ \epsilon ^{5} \equal{} 1$ and $ x_i$ be the amount of subsets whose sum = $ i$ modulo $ 5$
Then $ \sum _{i \equal{} 1}^{5}\epsilon^{i}x_i \equal{} \sum_{k \equal{} 0}^{2000}\sum_{1\le a_1 < ... < a_k\le 2000}\epsilon^{a_1 \plus{} ... \plus{} a_k}$  $ (*)$
but $ f(x) \equal{} (x \plus{} \epsilon)(x \plus{} \epsilon^2)...(x \plus{} \epsilon^{2000}) \equal{} (x^5 \plus{} 1)^{400}$
the $ RHS$ of $ *$ is obviouly $ f(1)$
We obtain $ \sum _{i \equal{} 1}^{5}\epsilon^{i}x_i \equal{} 2^{400}$
combining with $ \sum _{i \equal{} 1}^{5}x_i \equal{} 2^{2000}$
We conclude $ x_5 \equal{} \frac {2^{2000} \plus{} 2^{402}}{5}$
QED  :)

[quote="Albanian Eagle"]$ f(5m) \equal{}\frac{2^{5m} \plus{} 2^{m \plus{} 2}}{5}$

[/quote]

Why is it true, [b]Albanian Eagle[/b]?


For $ a,b,c$ positive reals prove that

$ \sum \sqrt{\frac{2a(b\plus{}c)}{(2b\plus{}c)(b\plus{}2c)}} \geq 2$

[quote="manlio"]For $ a,b,c$ positive reals prove that

$ \sum \sqrt {\frac {2a(b \plus{} c)}{(2b \plus{} c)(b \plus{} 2c)}} \geq 2$[/quote]
By Holder we obtain:
$ \left(\sum_{cyc}\sqrt {\frac {2a(b \plus{} c)}{(2b \plus{} c)(b \plus{} 2c)}}\right)^2\cdot\sum_{cyc}\frac{a^2(2b\plus{}c)(2c\plus{}b)}{2(b\plus{}c)}\geq(a\plus{}b\plus{}c)^3.$
Thus, it remains to prove that
$ (a\plus{}b\plus{}c)^3\geq2\sum_{cyc}\frac{a^2(2b\plus{}c)(2c\plus{}b)}{b\plus{}c}.$ But $ (a\plus{}b\plus{}c)^3\geq2\sum_{cyc}\frac{a^2(2b\plus{}c)(2c\plus{}b)}{b\plus{}c}\Leftrightarrow$
$ \Leftrightarrow(a\plus{}b\plus{}c)^3(a\plus{}b)(a\plus{}c)(b\plus{}c)\geq2\sum_{cyc}a^2(a\plus{}b)(a\plus{}c)(2b^2\plus{}5bc\plus{}2c^2)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^5b\plus{}a^5c\minus{}2a^3b^3\minus{}2a^4bc\plus{}a^3b^2c\plus{}a^3c^2b)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a\minus{}b)^2(a^3b\plus{}2a^2b^2\plus{}ab^3\minus{}a^2bc\minus{}b^2ac)\geq0,$ which true because
if $ a\geq b\geq c$ then $ S_c\geq0,$ $ S_b\geq0$ and $ S_a\plus{}S_b\geq0.$


linear algebra
matrix
trigonometry
calculus
integration

Fie $A \in M_n(C)$ cu proprietatea ca $\exists a \in[-2,2]$ ,astfel incat $A^2-aA+I_n=O_n$.
Sa se arate ca $\forall m \in N,\exists$ si este unic $a_m \in[-2,2]$ pentru care:
\[
A^{2m}  - a_m A + I_n  = O_n .
\]
(Aici nu este eficienta semnatura mea)

[quote="silviu"]
A^{2m}  - a_m A + I_n  = O_n .
[/quote]

It seems to me the coefficient of $I_n$ must depends with $m$

$A^{2m}-a_mA+b_mI_n=0$

e.g. if we compute $A^4$ in  term of $A,I_n$ from $A^2=aA-I_n$ we have

$A^4=(a^3-2a)A+(1-a^2)I_n$

Consider the Eucildean division of $x^{2m}$ by $x^2-2ax+1$

there exist $a_m,b_m$ such that

$x^{2m}=(x^2-2ax+1)q(x)+a_mx+b_m$ substitute $x$ by $A$ we get

$A^{2m}-a_mA-b_mI_n=0$
----------------------------------------------------------------
Another thing the matrix $A$ must be $\neq 0$

$A(A-aI_n)=-I_n$
Avem astfel ca A este inversabila si astfel inmultind cu inversa sa relatia data obtinem $A+A^{-1}=2aI_n$
Cum $a \in [-2,2] $ exista x real  astfel incat $a=2\cos x$
Prin ridicare la putere si inductie obtinem $A^m+A^{-m}=2\cos (mx)I_n$
Prin inmultire cu $A^m$ obtinem relatia ceruta unde $a_m=2\cos (mx) \in  [-2,2]$
Mai folosim ca $\cos (nx) =2^{n-1}\cos^n x-C_n^1\cos^{n-2}x-.........$ (relatia se poate gasi in cartea "Calcul diferential si integral", autor Gh. Siretchi)
Si iata ca a iesit :P


Demuestra que si se construyen cuadrados exteriormente sobre los lados de un trapecio isosceles, entonces el cuadrilatero formado por los cuatro centros de los cuadrados tiene diagonales de igual longitud y perpendiculares.

[quote="cuenca"]es mas general: para todo cuadrilatero(no necesariamente convexo)[/quote]

para ser exacto ese es el teorema de van aubel: [url]http://mathworld.wolfram.com/vanAubelsTheorem.html[/url]

Sugerencia al problema propuesto por fabycv:
Invertir con centro O y radio OA. Un problema muy bonito!!!
 :lol:

[quote="fabycv"]considere  un semicirculo de centro  $O$ y  de diametro $AB$ . Una linea interseca a $AB$  en  $M$ y al semicirculo en $C$ y $D$  tal que $MB < MA$ y $MD < MC$ . Los  circuncirculos   de los triangulos  de $AOC$ y $DOB$  se intersecan  en el segundo punto $k$ . Demuestre que $MK$ y $KO$  son perpendiculares.

  resuelvanlo  :D
faby[/quote]
Considerando al semi circulo como circunferencia de inversion (Circuferencia de magnus), las rectas BD y AC se invierten en las circunferencias que pasan por BDO y ACO  respectivamente , donde el segundo punto de interseccion es K, entonces las rectas AC Y BD deben de cortarse pues este punto sera el inverso de K, llamemosle K', luego sea T el punto de interseccion de AD con BC, R1 y R2 los puntos de tangencia de las rectas trasadas desde K' a la semiscircunferencia, entonces la polar de M es TK', , La polar de R1 es K'R1, de R2 es K'R2, como concurren las polares entonces M, K1, K2, son colineales. Por otra parte como K es el inverso de K', entonces R1, R2, K soncolineales, entonces M, K, R1, R2, COLINEALES, entonces <NKK' =90.

De hecho, existe una generalización para n-gonos cuya demostración es algo dificil, la solución que pude observar usaba números complejos. Adicional al teorema de Van Aubel, si se hace esta construcción interiormente entonces también resulta otro cuadrado. El área de ABCD es entonces igual a la diferencia del área de estos dos cuadrados y a este resultado se le llama (por analogía al triángulo) teorema extendido de Napoleón.

[quote="fabycv"]Considere  un semicírculo de centro  $O$ y  de diámetro $AB.$ Una linea interseca a $AB$  en  $M$ y al semicírculo en $C$ y $D$  tal que $MB < MA$ y $MD < MC.$ Los  circuncírculos   de los triángulos  de $AOC$ y $DOB$  se intersecan  en el segundo punto $K.$ Demuestre que $MK$ y $KO$  son perpendiculares.[/quote]
La inversión en el círculo con centro $O$ y radio $OA$ transforma a las circunferencias $\odot(OCA)$ y $\odot(ODB)$ en las rectas $AC,DB$ y a la recta $DC$ en la circunferencia $\mathcal{N} \equiv \odot(ODC).$  $P \equiv AC \cap BD$ y $E \equiv \mathcal{N} \cap AB$ diferente de $O$ son pues los inversos de $K,M,$ pero  por ser $D,C$ las proyecciones ortogonales de $A,B$ en $PB,PA$ y $O$ el punto medio de $AB,$  se sigue que $\mathcal{N}$ es el círculo de los 9 puntos de $\triangle PAB$ $\Longrightarrow$ $PE \perp BA.$ Luego como $E,M$ y $K,P$ son pares inversos, entonces los puntos $P,K,E,M$ son concíclicos $\Longrightarrow$ $MK \perp OK.$

El problema anterior propuesto, puede ser facilmente generalizado:

$A,B,C,D$ son cuatro puntos en una circunferencia $(O).$ Los circuncírculos de $\triangle ODA$ y $\triangle OCB$ se cortan en $O$ y $K.$ Si $M \equiv DC \cap AB,$ entonces probar que $MK \perp OK.$

La inversión en $(O,R)$ transforma a los círculos $\odot(ODA),$ $\odot(OCB$) en las rectas $DA,CB$ y por tanto $P \equiv AD \cap BC$ es el inverso de $K$ $\Longrightarrow$ $P$ yace en $OK.$ Si $OK$ corta a $(O)$ en $L,N,$ resulta que $R^2=OL^2=ON^2=OK \cdot OP$ $\Longrightarrow$ $(L,N,O,P)=-1,$  por tanto $MK$ es la polar de $P$ con respecto a $(O)$ $\Longrightarrow$ $MK \perp OK.$


calculus
derivative
special factorizations

Let b > 0. What is the least value of bx2 + cx + a?

[quote="wigianto"]Let $b > 0.$ What is the least value of $bx^{2}+cx+a$?[/quote]
Elementary calculus is enough; however, it isn't necessary. Here is a standard technique.

$bx^{2}+cx+a$
$= b\left( x^{2}+2\cdot\frac{c}{2b}\cdot x+\frac{a}{b}\right)$
$= b\left( x^{2}+2\cdot\frac{c}{2b}\cdot x+{(\frac{c}{2b})}^{2}+\frac{a}{b}-{(\frac{c}{2b})}^{2}\right)$
$= b\left({(x+\frac{c}{2b})}^{2}+\frac{4ab-c^{2}}{4b^{2}}\right)$

We note that the minimum of the above expression occurs when $x=-\frac{c}{2b}$ and the minimum value is $\frac{4ab-c^{2}}{4b}.$

elementary calc. method:
bx^2 + cx + a;
taking derivative and setting it equal to zero, to find critical points:
2bx +c=0;
minimum value occurs at: x=-c/2b
and we plug that in.

[quote="shake9991"]elementary calc. method:
$bx^{2}+cx+a$;
taking derivative and setting it equal to zero, to find critical points:
$2bx+c=0$;
minimum value occurs at: $x=-\frac{c}{2b}$
and we plug that in.[/quote]
$x=-\frac{c}{2b}$ is a point of local minimum or maximum. You still need to show that it is a point of absolute minimum by taking the second derivative and arguing from there on.

[quote="lotrgreengrapes7926"]Can't we just use that the minimum of $ax^{2}+bx+c$ is at $x=-\frac{b}{2a}$? Is it necessary to prove that?[/quote]
Not really, but it's safer to write an extra few lines down.
I think this is not a pre-olympiad problem, because students in grade 10 learn the technique of completing the square, and this problem is totally trivial. :!:

[color=blue]Moved from Pre-Olympiad to Intermediate.[/color]

Calculus is NOT needed for this. Completing the square is sufficient. lotrgreengrapes' method is also good.

you say this was a Texas A&M problem? what level?


Can the following reaction occur?
Cu(s) + 2H+(aq) --> Cu2+ (aq) + H2(g)

I calculated the following:
2H+  + 2e-  ---> H2 (g)
Cu(s) + 2e- ---> Cu2+(aq)

So, yes the reaction can occur. Is this correct, or am I missing something?

I just worked it out:

Answer is no. Cu(s) reduces by 2 so Cu2+ can't occur.

This is actually a typical experiment done in school. Reacting acids with metals of varying reactivity. So we put a piece of copper into hydrochloric acid and nothing happens. However, if we put copper into nitric acid a reaction takes place, can you think why?


There are 22 points placed inside of the interval [0,1]. At each iteration, one can remove two of the remaining points (e.g, A and B) and replace them by their midpoint, (A+B)/2. Show that one can always choose the points to be replaced in such a way that after 20 such iterations the distance between two remaining points is no greater than 0.001.

[quote="ghjk"]There are 22 points placed inside of the interval [0,1]. At each iteration, one can remove two of the remaining points (e.g, A and B) and replace them by their midpoint, (A+B)/2. Show that one can always choose the points to be replaced in such a way that [b]after 20 such iterations [/b]the distance between two remaining points is no greater than 0.001.[/quote]

Hi, ghjk.
Did you mean "any sequence of iterations" or "there exist a sequence of iterations" so that the distance $ \leq 0.001$ ?
 :)

[quote="ffao"]"there exists", as the "any" version is clearly false. Was this problem posted before? I can't seem to find a link, although I think it wasn't solved the last time.[/quote]
Yeah, I posted it before in the Pre-Olympiad box but nobody solves it. Does anybody mind helping me with this problem? I'm stuck with it for a long time though I think of it for around 2 hours(and have no idea where to start)

Well... Probably we can get a distance $ \le 2^{ \minus{} 10}$, since $ 10\times2 \equal{} 20$, the number of iterations (maybe this is obvious, but you said you have no idea...).


function calculus derivative limit vector

Corner pointz, cusps and vertical tangentz are ALL formed by the graph of continuous functionz .. !!

But i am facing difficulty in understanding them .. like what type of functionz will have these sort of graphz .. ????   :?:

[b]Would anyone be kind be enough to help me by explaining each of these with at least 2 examplez !![/b]  :|

NOTE : I WANT TO KNW THT IF THEREZ ANY WAY WE CAN IDENTIFY THAT A FUNCTION HAS A CUSP, VERTICAL TANGENT OR CONTINUOUS FUNCTION BY SIMPLY LOOKING AT IT NAD NOT DRAWING THE GRAPH .. ????  :idea:

For all of these, you will need to look at the first derivative function, $ f'(x)$.

A [b]vertical tangent[/b] occurs at $ x = c$ $ f'(x)$ either increases or decreases without bound as $ x$ approaches $ c$, but $ f(c) \in \mathbb{R}$.

An example of this is $ f(x) = x ^{\frac {1}{3}}$. $ f'(x) = \frac {1}{3} x^{ - \frac {2}{3}} = \frac {1}{3 \sqrt [3]{x^2}}$.

Hence, $ f(0) = 0$ but $ f'(0) = \frac {1}{0}$, which is undefined (in this case, it approaches $ + \infty$).

A [b]cusp[/b] occurs in a reversal of the direction of the graph. If we have a function, by nature, a cusp must be at a point where the tangent is vertical. Here, the left and right hand limits of the derivative are unbounded, but have opposite signs. Consider $ f(x) = \sqrt { |x|}$ at $ x = 0$.

In general, for a parametric function ${ \mathbf{f}(t)}: \mathbb{R} \to \mathbb{R}^n$, we have a cusp at $ t = c$ if $ \lim_{t \to c^ + } \text{D}\mathbf{f}(t) + \lim_{t \to c^ - } \text{D}\mathbf{f}(t) = \mathbf{0}$

Consider $ f(t) = (t^2, t^3)$ at $ t = 0$: $ \text{D}f = (2t,3t^2)$.
If $ t \to 0 ^ -$, then the vector points left (albeit with infinitesimal magnitude) and if $ t \to 0^ +$, then the vector points right.

A [b]corner[/b] is a more general type of point - it is a point where the function is continuous but derivative is non-removably discontinuous - that is, $ \lim _{x \to c^ - } f'(x) \not = \lim_{x \to c^ + } f'(x)$.

Easy example: $ f(x) = | x|$. From the left side as $ x \to 0$, $ f'(x) = - 1$ but from the right side, $ f'(x) = 1$.

hey !!

thankz a lot .. !!

but can u give me any other example of [b]corner pointz[/b] .. :S!!


integration
trigonometry
real analysis
real analysis unsolved

[quote="thanhnam2902"]$ I \equal{} \int_{0}^{\frac {\pi}{2}}\frac {4\sin{x}}{(\sin{x} \plus{} \cos{x})^3}dx$[/quote]

i thought i already helped you with one of these before.  :|

$ I \equal{} 4 \int_{0}^{\frac {\pi}{2}}\frac {\sin{x}}{(\sin{x} \plus{} \cos{x})^3}\; dx$

$ \Rightarrow 2I \equal{}  4 \int_{0}^{\frac {\pi}{2}}\frac {\sin{x}}{(\sin{x} \plus{} \cos{x})^3}\; dx \plus{}  4 \int_{0}^{\frac {\pi}{2}}\frac {\cos{x}}{(\cos{x} \plus{} \sin{x})^3}\; dx$

$ \Rightarrow I\equal{}2 \int_{0}^{\frac {\pi}{2}}\frac {1}{(\sin{x} \plus{} \cos{x})^2}\; dx\equal{} 2 \int_{0}^{\frac {\pi}{2}}\frac {1}{1\plus{}\sin\; 2x}\; dx$

$ \equal{} 2\; \frac {1}{1\plus{}\cot\, x} \bigg| _{0}^{\frac {\pi}{2}} \equal{} 2$

[hide][color=darkblue]Thank you very much.[/color] :P [/hide]


A swimmer is standing at a corner of a square swimming pool. She swims at a fixed speed and runs at a fixed speed (possibly different). No time is taken entering or leaving the pool. What path should she follow to reach the opposite corner of the pool in the shortest possible time?

There's no information about how fast the swimmer can run/walk? This seems like a basic optimization problem, but without that info I'm stuck.


calculate the (Δx)(Δp) for the gaussian wavefruction $ \psi (x)=\sqrt[4]{\frac{ \lambda}{\pi}} e^{- \frac{ \lambda x^2}{2}}$ and prove that is (Δx)(Δp)$ \geq \frac{ \hbar}{2}$

This belongs in the Advanced Physics forum.


Penniless Pete’s piggy bank has no pennies in it, but it has $ 100$ coins, all nickels, dimes, and quarters, whose total value is $ \$8.35$. It does not necessarily contain coins of all three types. What is the difference between the largest and smallest number of dimes that could be in the bank?

$ \textbf{(A)}\ 0 \qquad
\textbf{(B)}\ 13 \qquad
\textbf{(C)}\ 37 \qquad
\textbf{(D)}\ 64 \qquad
\textbf{(E)}\ 83$

[hide="Solution"]
Let $ n$, $ d$, and $ q$ represent the number of nickels, dimes, and quarters in the bank, respectively.  Then $ n\plus{}d\plus{}q\equal{}100$ and $ 5n\plus{}10d\plus{}25q\equal{}835$.  Also, $ n$, $ d$, and $ q$ must be non-negative integers.  Dividing the second equation by $ 5$ yields $ n\plus{}2d\plus{}5q\equal{}167$, and subtracting the first equation from this gives $ d\plus{}4q\equal{}67$.  Because $ q$ cannot be negative, $ d$ is at most $ 67$, and we check that $ 67$ dimes and $ 33$ nickels produces $ \$8.35$.  Also, $ d$ cannot be $ 0$, $ 1$, or $ 2$ because then $ q$ would not be an integer.  Thus, the smallest $ d$ can be is $ 3$, so $ q\equal{}16$.  We can check that $ 16$ quarters, $ 3$ dimes, and $ 81$ nickels works, so the largest $ d$ can be is $ 67$, and the smallest it can be is $ 3$, so the difference is $ 64$.  This corresponds to answer choice $ \textbf{D}$.
[/hide]

Hope I helped!


I need to prove that $ {n}^2\plus{}n\ge 42$ when $ n\ge 6$ and $ n\le \minus{}7$

Do I take the cases separately?

[hide="Hint"]
Move to one side, factor, and test points.
[/hide]
[hide="Solution"]
$ n^2 \plus{} n \minus{} 42 \geq 0$
$ (n \plus{} 7) (n \minus{} 6) \geq 0$
Changes signs at $ \minus{}7, \ 6,$ and is less than 0 when $ n \equal{} 0$, so we may reason that $ n^2 \plus{} n \geq 42$ when $ n \geq 6$ and $ n \leq \minus{}7$
[/hide]

can this be answered by induction please?

Induction? That's not useful, seeing as $ n$ is not necessarily an integer.

Okay N IS AN INTEGER AND ALL THE OTHER PREREQUISITES NEEDED FOR INDUCTION. I WOULD LIKE THIS DONE BY INDUCTION AND AM GETTING SICK TO THE BACK TEETH OF EVERYONE TELLING ME TO DO IT A DIFFERENT WAY. I KNOW IT IS TRUE; I CAN SEE WITHOUT THINKING IT IS TRUE- THE ANSWER IS NOT THE GOAL HERE, FIGURING OUT HOW TO DO INDUCTION IS. I MAY NOT BE AS GOOD AT MATHS AS A LOT OF THE PEOPLE ON THIS BOARD BUT SERIOUSLY IF YOU ARE GOING TO HELP OUT CAN YOU PLEASE DO JUST THAT AND NOT TELL ME THE QUESTION YOU WOULD LIKE ME HAVE POSTED. I'm sorry I'm just sick of it.

No need to be snippy; you never specified that $ n$ was an integer.  (This is also not a number theory question.)

To prove the result by induction for integers $ n$, you do indeed need to take the two cases separately.  The case $ n \ge 6$ can be handled as follows:  equality occurs when $ n \equal{} 6$, and this is your base case.  If the inequality holds for some $ k \ge 6$, then

$ (k\plus{}1)^2 \plus{} (k\plus{}1) \minus{} (k^2 \plus{} k) \equal{} 2k \plus{} 2 \ge 14 > 0$

so the LHS is strictly increasing and the inequality holds for $ k \plus{} 1$.

The case $ n \le \minus{}7$ is similar.

Number Theory method

n^2 + n >=  42 for n >= 6 OR n <= -7

Let e stand for epsilon and be a real value greater than 0,

so e > 0.

For n >= 6
------------

Note that 6^2 + 6 = 42.

Let n = 6 + e

Substitute that into the LHS:

(6 + e)^2 + (6 + e) >= 42?
36 + 12e + e^2 + 6 + e >= 42?
42 +13e + e^2 >= 42?
13e + e^2 >= 0  . . . True, because both terms are positive on the LHS.

For n <= -7
-------------

Note that
(-7)^2 + (-7) =
49 - 7 =
42

Let n = -7 - e

Substitute that into the LHS:

(-7 - e)^2 + (-7 - e) >= 42?
49 + 14e + e^2 - 7 - e >= 42?
42 + 13e + e^2 >= 42?
13e + e^2 >= 0 . . . True, because both terms are positive on the LHS.


I am an idiot trying to cheat on the Polish MO. Whenever one of my threads gets locked, I start a new one about exactly the same problems. Life is fun, isn't it?

[original contents of the message:

a^5=5*b^3-4*c
b^5=5*c^3-4*a
c^5=5*a^3-4*b
resolve in real
I know that a=b=c=0,-1,-2,1,2 but I don't know how show that are all resolves.]

This is a problem form an ongoing contest  :!:  :!:  :!:  (this year's Polish Mathematical Olympiad)


function
algebra
functional equation
real analysis
real analysis unsolved

Find the function $f(x)>0$ which is differentiable and satisfies the following functional equation  for any real numbers $x,y$ with initial condition $f(0)=2,f'(0)=4$,

\[f\left(\frac{x+y}{2}\right)=\sqrt{f(x)f(y)}\]

Put $h(x)=lnf(x)$. Then the original equation becomes $h\BIG(\frac{x+y}{2}\BIG)=\frac{h(x)+h(y)}{2}$.And $h(x)$ is continious.
Well this is the Cauchy's functional equation with solution $h(x)=ax+b$ so $f(x)=e^{ax+b}$ put $f(0)=2,f'(0)=4$ to get $e^b=2\Rightarrow b=ln2,ae^b=4\Rightarrow a=2$. So $f(x)=2e^{2x}$. :)

Yes,you are right, hardsoul.

Please edit , $f(x)=2e^{2x}$. :)


Does anyone play defend your castle here?
I got to level 165.
I have 856 wizards, 901 archers, and 200 repairmen.
Try it, if you haven't played it.
Heres the link:
http://www.xgenstudios.com/play/castle

OH MAN I LOVE THAT GAME

the only thing is that saying which level you're at is sort of silly since once you get reasonably advanced, you can just go eat a banana or something and your archers and repairmen take care of stuff all by themselves

i usually watch tv while it goes, in fact i have a game going right now. :D

Hmm for some reason my mouse dies by like Level 19  :(

Yeah I used to play and made it to about level 70 or so...had like 500 wizards, 400 repairmen, and 1500 archers.  The guys start to come more and faster, so i decided to turn the quality down.  Of course, I could just walk away and leave my guys to do the stuff.  But, of course, it just became a waste of CPU.  And somebody deleted cookies...computer probably crashed or got rebooted or something...

[quote="fireforce"]Yeah, i used to love playing that game untill i figured out the glitch.

[hide]Are u sure you want to know?[hide]Are you REALLY sure??[hide]By viewing this glitch you accept any damage or harm in the proccess of using this glich is your own FAULT (i.e. sore wrists, tired hands, bloodshot eyes, etc.) The trick works with old micromedia flash players, u can pause the game (by right clicking) it right after you let go of the left mouse to drop the attacker. Then you move the mouse back down and click the right button again when your right in front of a ne person. Plus, this offers sometime for your sore hand :P

Note: the glitch itself requires lots of practise and extremely percise timing I only recomend u use this to save yourself out of sticky situations it is impossible or at least EXTREMELY tiring to win using the glitch alone[/hide][/hide][/hide]

It was really fun in the first 50 rounds because u can really buy and experiment with new things. My personal favorite item is the temple since, I think its pretty cool to convert at least a few of those relentless attackers on your side. I even bought that wrist band they had for sore wrists  :D . What's your favorite item?[/quote]

um...you sort of have to get that before any of the others

my favorite item is the wizards becuase all i do is convert like mad when i have them
 :)

a good game,but my mouse does not work well.

If you like such games, check out http://www.sinn-frei.com/exklusiv.php?bid=2147 . However, the best game for multitasking is still RS...

  Darij


[hide]
Since the radius of the water tank is 30, the area of the base would be $30^2\pi = 900\pi$
The formula for the volume of a cylinder is $r^2h\pi$, so you can divide $2700 \pi$ by $900\pi$ to get 3.
[/hide]

[quote="MCrawford"]A water tank in the shape of a right circular cylinder has a base radius of 30 meters and a volume of $2700 \pi$ cubic meters.  What is the number of meters in the height of the tank?[/quote]

[hide]$V=\pi r^2 h=900\pi h=2700\pi \Rightarrow h=3$[/hide]


If you wish, post your prediction for the winning team and winning individual at the 2008 MA states

Eh, maybe Kevin, but idk. one of the lexington schools will almost definitely win.

Winning team is obviously one of the Lexington teams.

Individual ... Idk, I guess either Kevin or me. If I lose really badly, just forget I said that.

I hope there's gonna be some new genius next year, otherwise MA doesn't have much of a chance at Nats.

1. Clarke
2. Diamond

woo

then

1. Alan Chou
2. don't care
3. don't care
4. don't care
5. Kevin Wen

I wonder how many brilliant 6th graders there will be next year at MA states.

[quote="alkjash"][quote="alanchou"]Who are you anyways, alkjash?[/quote]

Oh, I'm the 8th place person this year. I'm from Acton. (That's where Elizabeth Synge is from!!!! Except she is home schooled.)

I don't really put much faith into new 6th graders. Probably at least half of elementary schools go up to 6th grade, so only half of the 6th graders will be going to MC.

Why is Alan Chou first on your list, Carl? Are the Lexington coaches that good, they get first place every year? I'm itching to prove you wrong, just you wait and see.

Really need to get AoPS vol. 1...[/quote]

dude first of all
elizabeth is from lexington

second of all
you only beat him by like two points
third of all
i believe he is much smarter than you right now
fourth
he picks up stuff in a non-human speed
i would not be surprised if he finished AoPS 1 in under ten days

fifth of all
I will probably come in second.

[quote="alkjash"] Why is Alan Chou first on your list, Carl? Are the Lexington coaches that good, they get first place every year? I'm itching to prove you wrong, just you wait and see.
[/quote]

Wow, next year might get a tad interesting after all.

It really all depends on how much and how you practice.
At least our coach (not telling right now, most people probably know anyways) has a lot of old warm-ups and workouts.

[quote="13375P34K43V312"]
I will probably come in second.[/quote]

...How is that possible? Are 9th graders allowed to participate?

Are you asking about my real name?

If so, it's pretty much my username.

-Alan Chou

[quote="perfect628"]i do! but i'm not gonna tell...[/quote]

eh think again
i do not concur

4th at states from diamond. u don't know who he is? interesting...

Well, he didn't go to states, so that could explain part of it.  But I mentioned it somewhere before, so yeah.

[quote="13375P34K43V312"]I was in 7th grade last year
or maybe I'm lying?
maybe I'm staying back?
who knows?[/quote]

I do since I go to the same school.

[quote="illusiat"]Acton's team is going to be pretty good, and will get 3rd place. Acton will probably have a top 4 individual (alkjash), 2 top 20s, and a top 30.[/quote]

Well, that's pretty cool

[quote="illusiat"]Acton's team is going to be pretty good, and will get 3rd place. Acton will probably have a top 4 individual (alkjash), 2 top 20s, and a top 30.[/quote]

What the heck? Stop making claims you can't keep.
You really think you can make top 20?

Btw, your poll is messed up.
Alan Chiao is definitely not gonna win...
I seriously doubt he's even going to make it to state.

Well, Carl thinks the poll is messed up even more in saying it's biased.

I somewhat agree in a way.

Hello

I am Carl

I believe the poll is biased

Furthermore I will be grading at States this year so don't piss me off

I will misgrade your papers

And the lack of a cohesive appeal system at States will keep you from reaching your goal, whatever it is

Mark my words

Aren't you only allowed to have one account?
Stop messing with my mind, people.

And there are 2 scorers for each paper, right?
So they would find your "mistakes" and fire you.

Yes, two people score each paper, but Carl would not do anything intentionally wrong anyway!!  Although he will do his best to mess with your head :lol:

Good luck and see you at state,

mr Frost

[quote="alkjash"]Aren't you only allowed to have one account?
[/quote]

Uhh, as long as you keep making new email addresses, you can have as many as you want, I think


What is the greatest common factor of 154 and 252?

hello, it is $ 154\equal{}2\cdot7\cdot11$ and $ 252\equal{}2^2\cdot3^2\cdot7$ and so the greatest common factor is $ 2\cdot7\equal{}14$.
Sonnhard.

[quote="isabella2296"]Another way is to use the Euclidean Algorithm. If you don't know what it is, either read the Into to NT book or private message me for an explanation.

$ 252 \minus{} 154 \equal{} 98$

$ 154 \minus{} 98 \equal{} 56$

$ 98 \minus{} 56 \equal{} 42$

$ 56 \minus{} 42 \equal{} \boxed{14}$[/quote]

Um why did u stop at $14$? Just wondering.

I think it's because it's relatively easy to see that 14 is the GCF of 14 and 42, assuming you know 42 = 14 * 3.


real analysis
inequalities
logarithms
advanced fields
advanced fields unsolved

We define the $s$-dimensional upper Minkowski contents of a set $A \subset R^{n}$ by

$M^{*}^{s}(A) = \limsup_{\epsilon \to 0}\; (2\epsilon)^{s-n}\lambda_{n}(A(\epsilon))$

where $\lambda_{n}$ is the Lebesgue measure on $R^{n}$ and $A(\epsilon)$ is the set of all points $x \in R^{n}$ such that $d(x, A) \le \epsilon$.

Then we define the upper Minkowski dimension as

$\overline{\text{dim}}_{M}A = \inf\{s : M^{*}^{s}(A) = 0\}= \sup\{s : M^{*}^{s}(A) > 0\}$

The problem: find two bounded sets $A \subset R^{n}$ and $B \subset R^{m}$ so that $\overline{\text{dim}}_{M}(A \times B) < \overline{\text{dim}}_{M}A+\overline{\text{dim}}_{M}B$ (the inequality is strict).

This is easier with an equivalent definition of $\overline{\text{dim}}_{M}A$, namely $\overline{\text{dim}}_{M}A=\limsup_{\epsilon\to0}\frac{\log N(A,\epsilon)}{\log(1/\epsilon)}$, where $N(A,\epsilon)$ is the smallest number of balls or radius $\epsilon$ needed to cover $A$. Note that $N(A,\epsilon)N(B,\epsilon)\ge N(A\times B,C(n)\epsilon)$. It remains to construct $A$ and $B$ so that \[\limsup_{\epsilon\to0}\frac{\log (N(A,\epsilon)N(B,\epsilon))}{\log(1/\epsilon)}<\limsup_{\epsilon\to0}\frac{\log N(A,\epsilon)}{\log(1/\epsilon)}+\limsup_{\epsilon\to0}\frac{\log N(B,\epsilon)}{\log(1/\epsilon)}.\] Think of two Cantor-type sets that are fat on some scales and thin on others.


inequalities
function
quadratics
triangle inequality
real analysis
real analysis unsolved

Hello, I would like to find the constant k wich verify:
\[\sup |f+f'| \geq k\times (\sup |f|+ \sup |f'|)\]
where $f$ is $C^1$ on $[0;1]$ and $\sup$ is taken on $[0;1]$ and $f(0)=0$

Thanks !

Moreover than    $k\leq \frac{\ sup |f+f'|}{ \sup |f|+ \sup |f'|}\leq 1$ , I can't get anything else.  $k$ will depends  of every $f$. But the  condition $f(0)=0$ confused me

I think the question is to find the largest fixed $k$ that would make the inequality true regardless of what function is chosen as $f$.

Some sort of constraint on $f(0)$ is required to avoid letting us get away with taking $f(x)=e^{-x}$, which otherwise would immediately give that 0 is the best possible bound.

With the constraint, it seems like the statement is true iff $k \leq 1/3$

To see that $k=1/3$ is okay, note that if $q=\sup(|f(x)+f'(x)|)$, then

\[-q e^x \leq (f(x)+f'(x))e^x \leq q e^x.  \]

The antiderivative of the function in the middle is $f(x)e^x$.  Integrating from 0 to a now gives $- q e^a \leq f(a) e^a \leq q e^a $ (here is where we need $f(0)=0$), so $\sup|f| \leq q$.  It follows from the triangle inequality that $\sup |f'| \leq 2q$, which gives the 1/3 bound (and hence a bound for any $k<1/3$.

To see that nothing bigger than 1/3 works we construct a (semi-explicit) counterexample.

$f(x)= x-x^2, 0 \leq x \leq \frac{1}{2}$
$f(x)=1-a(x-1/2)^2, 1/2 \leq x \leq \frac{1}{2}+\frac{1}{4a}$
For $x>\frac{1}{2}+\frac{1}{4a}$ we take $f$ to be a quadratic function with the right value of $f$ and $f'$ at $\frac{1}{2}+\frac{1}{4a}$ and an exceedingly large value of $f''(\frac{1}{2}+\frac{1}{4a})$.  Once $f'$ becomes 0 we make $f$ a constant function.

One can check that this function is piecewise continuously differentiable, and (assuming $f''$ is large enough for our quadratic) has  $\sup |f| = 1/4$, $\sup |f'|=-1/2$, and $\sup |f+f'|$ approaching 1/4 as $a \rightarrow \infty$.  Therefore by making $a$ sufficiently large we can make $\sup |f+f'| < k (\sup |f|+\sup |f'|)$ for any $k>1/3$.


Was there really a simple 4 in the middle of the first equation?

No there wasn't.  He probably accidentaly wrote the subscript of the $x_4$ as a $+4$.

[hide]Find a way to express a perfect square as the sum of some combinations of the sum of the three consecutive lower perfect squares.

So we want to solve for $a$, $b$, and $c$ in $ax^2+b(x-1)^2+c(x+1)^2=(x+2)^2$. Expanding and matching like terms gives us:
$a+b+c=1$
$-2b+2c=4$
$b+c=4$, from which we solve find the solutions $a=-3$, $b=1$, and $c=3$.

Therefore, when we are given three consecutive perfect squares $(x-1)$, $x$, and $x+1$, then next perfect square is:
$(x-1)^2-3x^2+3(x+1)^2$.

So we know:
$16x_1=3\cdot 9x_1-3\cdot 4x_1+x_1$
$25x_2=3\cdot 16x_2-3\cdot 9x_2+4x_2$.
etc.

Summing these up we have:

$16x_1+25x_2+...+100x_7=3(9x_1+16x_2+...+81x_7)-3(4x_1+9x_2+...64x_7)+(x_1+4x_2+...+49x_7)$

Since all these values are given, our answer is $3(123)-3(12)+(1)=\boxed{334}$[/hide]

[quote="chess64"]Ah, I missed the last step in your solution, frt. :( I got up to (C).[/quote]What I meant by the last step was adding $7x_1+9x_2+11x_3+13x_4+15x_5+17x_6+19x_7=211$ (D) and $9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7=123$ (one of the given equations).

That gives
$16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7=334$.

Is that clear?

No, I meant that I had everything up to the last step when I worked them. In other words, my solution was the same as yours except I didn't get the last step. :)


Find all positive integer $ n$ satisfying the following condition: There exist positive integers $ m$, $ a_1$, $ a_2$, $ \cdots$, $ a_{m\minus{}1}$, such that $ \displaystyle n \equal{} \sum_{i\equal{}1}^{m\minus{}1} a_i(m\minus{}a_i)$, where $ a_1$, $ a_2$, $ \cdots$, $ a_{m\minus{}1}$ may not distinct and $ 1 \leq a_i \leq m\minus{}1$.

After having solved this problem today, I thought exactly the same xD But IMHO, this problem isn't even hard, just much to much to compute :(

[quote="hxy09"]I have to admit that it is indeed an awful and ugly problem.The answer is $ 2,3,5,6,7,8,13,14,15,17,19,21,23,26,27,30,47,51,53,55,61$
The solution was extremely complicated and required much calculation. :wink:
But we can easily prove there only exist finitely $ n$ satisfy the condition by Lagrange 4-square Theorem.[/quote]
$a_i =1$ then every square sastify!!!!!


integration
logarithms
limit
calculus
calculus computations

YES ! i [i]finally[/i] did it !! cool.  :wink:  :D

But your post contains no solution, then how did you solve it.

It does exist an elementary way, besides it's beautiful.

[quote="|-.-|"]But your post contains no solution, then how did you solve it.[/quote]

i solved it   in $ 4$ different ways. in how many ways can [b]you[/b] do it ?

[quote]
It does exist an elementary way, besides it's beautiful.[/quote]

OK ,  let's see what you got....  :|

[quote="misan"]i solved it   in $ 4$ different ways. in how many ways can [b]you[/b] do it ?[/quote]
In post #3 you just said that you solved it, but that does not have a solution, that's why I said "your post contains no solution."

The beautiful solution is [url=http://www.mathhelpforum.com/math-help/calculus/41550-challenging-improper-integral.html]here[/url].


http://www.artofproblemsolving.com/Forum/viewtopic.php?t=266455

I found this one rather interesting.

rrusczyk
DPatrick

literal modkill of whoever chose those names.

dragon96
nsun48

[quote]dragon96
nsun48[/quote]

[quote]"dragon96 over nsun84"[/quote]

2.

Chompy
pacman2812

A yellow circle with eyes and a mouth that chomps on dots instead of neatly swallowing them.

dragon96
isabella2296

love

Aww This message is too small. Please make the message longer before submitting.

Those aren't names. Please do not post if you are not contributing to the thread :mad:

AAKZGORG-AIRCSG-KLAZT
starwar

What?
A sudden event in the force...
They have considered renaming R2-D2...
AAKZGORG-AIRCSG-KLAZT
 :o  :o
How can you say that??

Haha...

Maybach
Spaceguy524

Mee am wheeeeeeing having funnnnnnnnn runnnnnning in the sunnnnnnnn.

dragon90
dragon96
mathdragon96
DragonLover996
eragon97
98Dragons

Something indescribable, considering I don't even know all those people...

[quote="AAKZGORG-AIRCSG-KLAZT"]dragon90
dragon96
mathdragon96
DragonLover996
eragon97
98Dragons[/quote]

Are those all real users?

BTW:

james4l
Chompy

The most chaotic event has happen in james4lville;
GIANT CHOMPY'S HAVE TAKEN OVER!!!
There is now a war in the town...
tsk tsk.

lightbluemathangel
KING1

[quote="Chompy"]dragon245.5[/quote]

I would think it was more along the line of matheragonlovers245.5.

Oh well, you forgot me in that list. But then again...

[quote="meewhee009"]lightbluemathangel
KING1[/quote]

The light-blue math angel is the first king of... hmm... what should I say? The country of 1?

How about this: A king that has a famous saying: "There are two kinds of logical discrepancies. One kind has short legs. The other kind has a long nose. Your kind just happens to be that which has a long nose."

AAKZGORG-AIRCSG-KLAZT
dragon96

a person who made a wrong conjecture.

maybach
Chompy

A car that eats everthing in its way.

spaceguy524
starwar

An exploding Saturn V rocket.

knockout
batteredbutnotdefeated

[quote="bluecarneal"]knockout
batteredbutnotdefeated[/quote]

A weird boxing match where both players have infinite time to get back up.

bluecarneal
westiepaw

a bluepaw

sdkudrgn88
isabella2296
AIME15
Phantom.

2 ghosts who like sudoku and dragons

jjfun1
jjx1

jjxun1

dragon96
jjx1
monkeygirl12
isabella2296
KitKat.LoL

Love with a monkey eating a kit-kat named jjx. :)

AIME15
Mewto55555
Dojo

A mewto that pwns AIME by getting a 15, and plays Super Smash Bros. Dojo all day... :P

meewhee09
levans
 :P

A mod with funny rules

meewheeo9
jjfun1

[quote="bluecarneal"]meewheeo9
jjfun1[/quote]

I divided by zero. oh SH-- [those two hyphens are a sign of an abrupt stop, not a bleep.]

dragon96
me

Oh look.
Math,
$ Sudoku \plus{} Dragon^2 \plus{} 187$

Ha,

it's nice to know you think I'm funny... well... uhm, EH! :P

worthawholebean
MellowMelon
(Yes those are real users :rotfl:)

Melon that's worth a whole bean

meewhee009
Chompy
lightbluemathangel

[quote="spaceguy524"]a person that chomps on angels

jjfun1
jjx1
meewhee09
Chompy
AIME15
dragon96
me[/quote]

Okay, how do I say this...

jjspacewheechompydragonguy647

AAKZGORG-AIRCSG-KLAZT
r15s...I forget the numbers.

-complex array of numbers and letters-

spaceguy524
jjfun1

funspaceguy

sdkudrgn88
dragon96
jjfun1
AIME15
AIME15USAMO
levans
Jongy

levans doing usamo and aime on sdkudragn92 with jongy

ilovepink
pinkmuskrat
lightbluemathangel

a muskrat with a house in pink that is blue.

AceOfDiamonds
meewhee09
maybach
jjfun1

meejjacebach

spaceguy524
lightbluemathangel
meewhee009

a flying angel in space that says mee whee.

lightbluemathangel
AAKZGORG-AIRCSG-KLAZT
jjfun1

@sdkudrgn88: r15 is AAKZGORG-AIRCSG-KLAZT. He just changed his username.

an angel who has fun memorizing AAKZGORG-AIRCSG-KLAZT's name

spaceguy524
skdudrgn88
dragon96
isabella2296
AIME15
jjx1
Dojo
levans

the new version of ftw

AwesomeToad
spaceguy524

[quote="biblioman"]Sorry to revive.

biblioman (ME!)
levans
Thunder365
AIME15USAMO[/quote]

Please don't revive in FF, next time.

A world that is verry nasty.

rruscyzk
nsato
Valentin Vornicu
copeland
levans

modland

Scamper
Musicnmath
Richard rusyck(my first teacher)
Wobster109

[color=red][size=84]Hi, this was locked because it's a revival. Start a revival thread if you wish. [/size][/color]


[quote="AoPS: Volume 2"]Another root location method is finding upper and lower bounds.  Suppose we use synthetic division to find $ f(x)/(x - c)$ where $ f(x)$ has a positive leading coefficient and $ c\geq 0$ as below:

$ %Error. "polyhornerscheme" is a bad command.
[x = 3]{x^4 - x^3 + 2x^2 + 6x}$

If all the resulting coefficients in the quotient are positive (including the remainder), as in the example above, then no roots are greater than $ c$.  (Why?)  This $ c$ is called an [b]upper bound[/b] on the solution since no roots can be higher.  Similarly, if $ c < 0$ and the coefficients of the quotient and remainder alternate in sign, then there is no root smaller than $ c$ (which we then call a [b]lower bound[/b] for the roots).  Locating upper and lower bounds will often help you shorten your search for roots.[/quote]

Above is an excerpt from AoPS: Volume 2.  However, I don't get the part after the synthetic division is carried out. (Sorry, the $ \text{\LaTeX}$ coding didn't work out.)  Could someone please explain it to me?

Thanks in advance!

Suppose $ f(x) \equal{} (x \minus{} c)g(x)$ for some $ c\ge 0$ and all the coefficients of $ g(x)$ are positive. For some $ x > c$, $ (x \minus{} c) > 0$, and $ g(x) > 0$, since $ x > 0$ and all of the coefficients are positive. Thus $ f(x) \equal{} (x \minus{} c)g(x) > 0$ for all $ x > c$, so $ f(x)$ can't have any roots greater than $ c$.

Suppose $ f(x) \equal{} (x \minus{} c)g(x)$ for some $ c < 0$ and the coefficients of $ g(x)$ alternate signs. For some $ x < c$, $ (x \minus{} c) < 0$. Since $ x < 0$, $ x^m$ is positive for even $ m$ and negative for odd $ m$. Since the coefficients also alternate signs, either all the terms are positive or all the terms are negative. So either $ g(x) > 0 \Rightarrow f(x) \equal{} (x \minus{} c)g(x) < 0$ for all $ x < c$ or $ g(x) < 0 \Rightarrow f(x) \equal{} (x \minus{} c)g(x) > 0$ for all $ x < c$, so $ f(x)$ can't have any roots less than $ c$.

[quote="314.math"]What is $ g(x)$?[/quote]$ g(x)$ is the polynomial that you would get, along with a remainder $ R$, if you divide your original polynomial, let's say $ P(x)$ by $ (x\minus{}c)$. In other words
$ P(x)\equal{}(x\minus{}c)g(x)\plus{}R$.

This exact question has been asked [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1176842966&t=165188]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=939276842&t=99548]here[/url].


Let $ A\subset \mathbb{R}^n$ such that $ |A|\leq \plus{}\infty$. Is there any space between $ L^{\infty}(A)$ and $ L^p(A),p>0$?

There are plenty of spaces. One interesting example is [url=http://en.wikipedia.org/wiki/Bounded_mean_oscillation]BMO[/url].


calculus
integration
logarithms
trigonometry
calculus computations

Just out of curiosity, are there any integrals that are made easier to solve by converting them back into a Riemann sum?

One example is the integral $ \displaystyle\int_0^{\frac{\pi}{2}}\ln\sin xdx$.

You can use the identity $ \displaystyle\sin\frac{\pi}{m}\sin\frac{2\pi}{m}\sin\frac{3\pi}{m}...\sin\frac{(m\minus{}1)\pi}{m}\equal{}\frac{m}{2^{m\minus{}1}}$ for $ m\equal{}2,3,4,...$, and from there construct a Riemann sum.

[quote="Carcul"]Mitochondria, aren't you complicating something that is already very easy to solve?[/quote]
You are right Carcul, so maybe this is a bit more suitable.

First use $ \displaystyle\sin\frac{\pi}{m}\sin\frac{2\pi}{m}\sin\frac{3\pi}{m}...\sin\frac{(m\minus{}1)\pi}{m}\equal{}\frac{m}{2^{m\minus{}1}}$ to show $ \displaystyle\Gamma\left(\frac{1}{m}\right)\Gamma\left(\frac{2}{m}\right)\Gamma\left(\frac{3}{m}\right)\cdots\Gamma\left(\frac{m\minus{}1}{m}\right)\equal{}\frac{(2\pi)^{(m\minus{}1)/2}}{\sqrt{m}}$.
Then use the latter to evaluate $ \displaystyle\int_0^1\ln\Gamma(x) dx$.

Or is that too easy, also?  :maybe:


geometry
circumcircle
trigonometry
cyclic quadrilateral
geometry proposed

Let $ AB, AC$ the tangents of a point $ A$ ouside of a circle and let  $ D$ a point at the extesion of the segment $ AB$ (to the side of $ B$ ). The cicrumcircle of the triangle $ ADC$ intersects the first circle at the point $ P$. Let $ Q$ the projection of  $ B$ to the line $ CD$ .Prove that  $ \angle DPQ \equal{} 2\angle ADC$

this is  IRAN 1995 problem $ 2$ I think,which is a difficult problem,it has a beautiful solution using inversion...
anyway here is two other solutions:
let us change the notations a bit:
[b]let $ KL,KN$ be tangent to circle $ C$.$ M$ is a point on $ KN$ s.t $ N$ lies between $ K,M$.and let $ P$ be the intersection point of $ C$ and the cirumcircle of $ \triangle KLM$,other than $ L$.now let $ Q$ be the projection of $ N$ on $ ML$.prove that:
$ \angle MPQ = 2\angle KML$[/b]

[b]solution [u]1[/u]:[/b]
let $ E$ be the intersection of $ ML$ with $ C$ and let $ PE$ intersect $ MK$ at point $ F$.now in the cyclic quadrilateral $ MKLP$ we have:

$ \angle PMK = 180-\angle PLK (1)$

now $ KL$ is tangent to $ C$ so $ 180-\angle PLK =\frac{\over frown{LXP}}{2}$ in the other hand $ \angle LEP =\frac{LXP}{2}$ thus we have:

$ 180-\angle PLK =\angle LEP (2)$

now according to $ (1),(2)$ we get that:

$ \angle PMK =\angle LEP =\angle FEM$

so two triangle $ MEF,MPF$ are similar,therefor:

$ \angle EPM =\angle EMF,\frac{MF}{EF}=\frac{FP}{MF}\Rightarrow MF^{2}= EF.FP (3)$

now in the other hand the equation form $ (3)$ is the power of point $ F$ wrt circle $ C$ and also $ FN$ is tangent to the circle,so we can say that:

$ FN^{2}= FE.FP (4)$

now by $ (3),(4)$ we get that:

$ FN^{2}= MF^{2}\Rightarrow FN = MF$

so $ QF$ is the median of the right-angled triangle $ NMQ$ so:

$ QF = MF\Rightarrow\angle FME =\angle MQF$

and also we had $ \angle FME =\angle EPM$ so:

$ \angle MQF =\angle EPM (I)$

so $ FMPQ$ is cylci,thus we get that:

$ \angle EMF =\angle EPQ (II)$

now by $ (I),(II)$ we get that:

$ 2\angle EMF =\angle EPQ+\angle EPM\Rightarrow 2\angle EMF =\angle QPM$

aswanted...
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[b]solution [u]2[/u]:[/b]
draw the lines $ NE,NL,KP$,now let $ KP$ intersect $ C$ at $ H$,now connect $ H$ to $ N,L$.now assume that $ \angle KML =\alpha,\angle NLH =\beta$

so we have:

$ \angle KML =\angle KPL =\angle HNL =\angle HLK =\alpha (1)$

$ \Rightarrow\over frown{HL}= 2\alpha$

also:

$ \angle NLH =\angle HNK =\beta (2)$

$ \Rightarrow\over frown{NH}= 2\beta$

in the other hand:

$ \angle NEL =\frac{\over frown{NH}+\over frown{HL}}{2}=\frac{2\alpha+2\beta}{2}=\alpha+\beta$

$ \Rightarrow\angle NEL =\angle ENM+\angle NME =\alpha+\beta\Rightarrow\angle ENM =\beta$

also we know that:

$ \angle ENM =\frac{\over forwn{NE}}{2}=\beta$

$ \angle NLE =\frac{\over frown{NE}}{2}$

now from the last two equalities we get that:

$ \angle NLE =\beta (3)$

now from $ (1),(2),(3)$ we get that:

$ \angle KPM =\angle KLM = 2\beta+\alpha$

$ \Rightarrow\angle LPM =\angle LPK+\angle KPM = 2\alpha+2\beta$

now we know that $ \angle NQM = 90$ so we can say that:

$ MQ = MN.\cos\alpha$

$ LQ = NL.\cos\beta$

$ \Rightarrow\frac{MQ}{LQ}=\frac{MN}{NL}.\frac{\cos\alpha}{\cos\beta}(4)$

now in the triangle $ NML$ we know that:

$ \frac{MN}{NL}=\frac{\sin\beta}{\sin\alpha}$

so we can write $ (4)$ in this way:

$ \frac{MQ}{LQ}=\frac{\sin\beta}{\sin\alpha}.\frac{\cos\alpha}{\cos\beta}(5)$

now let $ Q'$ be a point on $ ML$ s.t $ \angle Q'PL = 2\beta$ now we had $ \angle LPM = 2\alpha+2\beta$ thus $ \angle Q'PM = 2\alpha$ so $ \angle Q'PM = 2\angle NMQ$
so its sufficient to show that $ Q\equiv Q'$...
now let $ \angle PKL =\theta_{1},\angle PKM =\theta_{2}$.
now by $ \sin$ law in triangles $ MPQ',LPQ'$ we get that:

$ \frac{MQ'}{PQ'}=\frac{\sin 2\alpha}{\sin\theta_{1}}$

$ \frac{LQ'}{PQ'}=\frac{\sin 2\beta}{\sin\theta_{2}}$

$ \Rightarrow\frac{MQ'}{LQ'}=\frac{\sin 2\alpha}{\sin 2\beta}.\frac{\sin\theta_{2}}{\sin\theta_{1}}$

$ \Rightarrow\frac{MQ'}{LQ'}=\frac{\sin\alpha.\cos\alpha}{\sin\beta.\cos\beta}.\frac{\sin\theta_{2}}{\sin\theta_{1}}(I)$

now according to the $ \sin$ from of ceva in $ \triangle KNL$ we get that:

$ \frac{\sin\alpha}{\sin\beta}.\frac{\sin\alpha}{\sin\beta}.\frac{\sin\theta_{2}}{\sin\theta_{1}}= 1$

$ \Rightarrow\frac{\sin^{2}\beta}{\sin^{2}\alpha}=\frac{\sin\theta_{2}}{\sin\theta_{1}}(II)$

now using $ (II)$ we can rewrite $ (I)$ as:

$ \frac{MQ'}{LQ'}=\frac{\cos\alpha.\sin\beta}{\cos\beta.\sin\alpha}(6)$

now from $ (5),(6)$ we get that:

$ \frac{MQ'}{LQ'}=\frac{MQ}{LQ}\Rightarrow\frac{MQ'}{MQ'+LQ'}=\frac{MQ}{MQ+LQ}$

$ \Rightarrow\frac{MQ'}{ML}=\frac{MQ}{ML}\Rightarrow MQ' = MQ$

now we know that $ M,Q,Q'$ are on a line,so $ Q\equiv Q'$,so:

$ \angle QPM = 2\alpha,\angle NMQ =\alpha\Rightarrow\angle QPM = 2\angle NMQ$

as wanted...

Denote $ S\equal{}CD\cap (O)$ and $ T\equal{}PS\cap AB$. We have $ \angle TDP\equal{}180^{\circ}\minus{}\angle ACP$ (because of the cyclic quadrilateral $ ACPD$)
On the other hand, $ AC$ is tangent to the circle $ (O)$, therefore $ 180^{\circ}\minus{}\angle ACP\equal{}\angle CSP\equal{}\angle TSD$.
Hence we conclude that $ \triangle TDP\sim\triangle TSD$, so $ \angle TPD\equal{}\angle TDS\equal{}\angle TQD$ (because of the right-angled triangle $ BQD$)
It follows that the quadrilateral $ DPQT$ is cyclic. Notice that $ TD\equal{}TQ$, which follows that $ \angle TPD\equal{}\angle TPQ$.

Hence $ \angle DPQ\equal{}2\cdot\angle TPQ\equal{}2\cdot\angle ADC$ as desired.

April isn't your solution the same as mine (solution 1)...? :huh:

Your solutions are very nice .Thank you Babak and April .
Here is a solution made by Michael Savvas (member of the Greek IMO team 2007 and bronze medalist)
Here is the sketch of that
Let $ K$ the midpoint of BC . Then PQKC is a cyclic quadrilateral and PA is the symmedian the triangle PBC .
So $ \angle{APB}\equal{}\angle{KPC}$ . The rest is an easy angle chasing .


you know tha calcium carbonate dissolves in HCL. Write a balanced chemical equation that represents this reaction which i already did. then this says calculate the equillibrium constant for this rection. (HINT use KSP and KA of carbonic acid and add a few equations. Calculate the mass of calcium carbonate that dissolves in 1 L of PH 4 HCL.
the KSP is 8.7 *10^-9
the Ka is 4.2 * 10 ^ -7

then i divided 4.2 * 10 ^ -7/8.7 *10^-9= 48. 2 M
is this right if im wrong can u show me how i can do it

Let K1 and K2 designate the two acidity constants of carbonic acid. Then:

$CaCO_{3}(s) \rightleftharpoons Ca^{2+}(aq)+CO_{3}^{2-}(aq), k_{sp}$
$CO_{3}^{2-}(aq)+H_{3}O^{+}(aq) \rightleftharpoons H_{2}O(l)+HCO_{3}^{-}(aq), \frac{1}{k_{2}}$
$HCO_{3}^{-}(aq)+H_{3}O^{+}(aq) \rightleftharpoons H_{2}CO_{3}(aq)+H_{2}O(l), \frac{1}{k_{1}}$

Now sum these equations and we get

$CaCO_{3}(s)+2H_{3}O^{+}(aq) \longrightarrow Ca^{2+}(aq)+H_{2}CO_{3}(aq)+2H_{2}O(l), k = \frac{k_{sp}}{k_{1}\cdot k_{2}}$.

(Naturally, the carbonic acid decomposes to carbon dioxide and water.)

thanks man, so all i do is divide ksp of carbonic acid and divide it over its k1 and k2, thanks man