Prove that for closed curves $ \gamma_1$, $ \gamma_2$,
$ I(\minus{}\gamma_1,z_0) \equal{} \minus{}I(\gamma_1,z_0)$
and
$ I(\gamma_1 \plus{} \gamma_2,z_0) \equal{} I(\gamma_1,z_0) \plus{} I(\gamma_2,z_0)$
Interpret these results geometrically.

Use the definition

$ I(\gamma,z_0) \equal{} \frac {1}{2\pi i} \underset{\gamma }{\int } \frac {dz}{z \minus{} z_0}$

and theorems

$ \underset{ \minus{} \gamma }{\int } f(z)\, dz \equal{} \minus{} \underset{\gamma}{\int}f(z)\, dz$

$ \underset{ \gamma_1 \plus{} \gamma_2 }{\int } f(z)\, dz \equal{} \underset{ \gamma_1}{\int } f(z)\, dz \plus{} \underset{\gamma_2 }{\int } f(z)\, dz$

For interpreting the results geometrically, I would say that these identities are actually talking about the structure of $ H_1(\mathbb{C} \setminus \{z_0\})$, in particular its linear structure. So, in homology, $ [\gamma_1 \plus{} \gamma_2] \equal{} [\gamma_1] \plus{} [\gamma_2]$, for example.


function
algebra
domain
calculus
real analysis
real analysis unsolved

How would I solve the following 3 problems dealing with functions on Euclidean space:

1. Does the function f : ${\mathbb{R}^{2}}$ \  {${(0, 0)}$} → $\mathbb{R}$, defined by
$f \left( x,y \right) ={\frac{x}{\sqrt{{x}^{2}+{y}^{2}}}}$,
have a limit as (x, y) approaches (0, 0). Justify your answer.

2. Prove that a is a limit point of a set D ⊂ $\mathbb{R}^{p}$ if and only if there is a sequence in D which converges to a.

3. If G : D → $\mathbb{R}^{p}$ is a function with domain D ⊂ $\mathbb{R}^{d}$ is and F : E → $\mathbb{R}^{q}$ is a function
with domain E ⊂ $\mathbb{R}^{p}$, then F(G(x)) is defined as long as x ∈ D and G(x) ∈ E.
Thus,
(F o G)(x) = F(G(x))
defines a function with domain D ∩${G}^{-1}\left( E \right)$ and with values in $\mathbb{R}^{q}$. This is the
composition of the function G with the function F.

[b]Prove the following theorem:[/b]
With F and G as above, if a ∈ D ∩${G}^{-1}\left( E \right)$ , G is continuous at a and F is continuous at G(a), then F o G is continuous at a.

the questions you posted are quite easy, so i will give you hints

for the first one- the limit if it exists has to be unique. Now think about the limit as we approach $(0,0)$ along several curves such as $y=mx$( for suitable choice of m) or $y=x^{2}$ etc.

for the second one, what is your definition of limit point? is it this? a point x is a limit point of the set A if every neighbourhood centred at x contains a point of A? if that is the definition, then consider this for solving your problem. take neighbourhoods around x of radius $1/n$ and for each $n$ choose a point $x_{n}$ belonging to A which lies in that neighbourhood of radius $1/n$

the third one is very common and can be found in any standard calculus book


advanced fields
advanced fields unsolved

Find the number of transitive relations possible in set of cardinality $n$.

I checked multiple sources: no such general formula is known.


Let a,b,c>0 then    abc=1  then
$ \frac{a}{\sqrt{8c^3\plus{}1}}\plus{}\frac{b}{\sqrt{8a^3\plus{}1}}\plus{}\frac{c}{\sqrt{8b^3\plus{}1}} \geq1$

This problem is on M&Y in Vietnam . Lock it please :wink:
And this is problem of can_hang2007


integration
trigonometry
function
calculus
calculus computations

Find the value of :

$ \int\frac{x\cos x\plus{}1}{\sqrt{2x^3e^{\sin x}\plus{}x^2}}\ dx$

Such a random function (I assume you made it for fun  :lol: ) probably hasn't got an elementary antiderivative (checked that with Mathematica-it really doesn't).

[hide="Linguistics"]

By the way, sort of a linguistic question:

"Find the value of"- isn't it more suitable for definite integrals (which "do have a value").

Wouldn't "Integrate" or "Find" be more appropriate for antiderivatives? (This is a question for native English speakers).

P. S. What about calculate? Personally, I consider it to be a phrase more appropriate for definite integrals: as most of calculations end up with a result (i.e. value)[/hide]


Could someone help me with this problem?

Tina randomly selects two distinct numbers from the set {1,2,3,4,5}, and Sergio randomly selects a number from the set {1,2,3,4,5,6,7,8,9,10}. What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina?

Thanks in advance.

[url]http://www.artofproblemsolving.com/Wiki/index.php/2002_AMC_10A_Problems/Problem_24[/url]

Ha I wrote that up yesterday.


geometry
AMC
AIME
algebra
functional equation

It seems like everyone got around the same score on APMO, so it will not be too much of a factor in the selection for IMO team this year.

In my opinion, The Canadian Mathematics Olympiad will be the main factor in the selection for IMO.

What are your predictions and comments on this contest?

My goal is to get a perfect score. :D

which one is harder? APMO?

Good luck  :lol:

well apmo is getting harder and harder from yr to yr so does  the cmo have the same trend?

ya u can tell by the results..I dont think it'll be harder this year

there was a type that turned an nice telescoping equation into a horrendous expression that is impossible to simplify so everyone got owned on this question.

[quote="vincent_five"]but i heard last yr's #1 has some mistakes in the question and it didn't count, that's why the mark is so low[/quote]

ya i remember reading that on the results page...

So did the markers end up omitting question 1 cuz I thought maybe they didnt.

who thinks cmo will be as "biased" as last year, with 3 combinatorics and 2 geo problems, or something of that sort? Or did they do that last year to challenge the penguins (peng, richard peng, etc..) :rotfl:

Do you guys know which countries have the same difficulty level as the CMO ?

Thanks

I don't think i will be biased. Even last year's CMO was not too bad if there was no typo.

i hope diz years #5 is a geometry problem, just like last year...geometry problems are relatively much easier to solve... :P

The last problem is not always the hardest one -- that was the case last yr.
I am sure they will have at least one geometry problem, and you can consider it number 5. :lol:

geometry peoblems r always really straight forward...lots of times i use geometry to solve algebraic problems...that was fun... :D

That's true. If the hardest problem is geometry, that would be like a gift. :lol:
I think last problem will be sequence or combinatorics.

And you will post all problem on this forum when done. Do you agree? :P

I smashed my head over last year's #1 until I realized it was pretty much impossible. That was not a morale booster :(

ya..good luck to all those people writing CMO and the alternate AIME

How'd everyone do?

Crap..... I am so stupid for not getting #3 :wallbash_red:

Stupid...stupid....stupid

I solved 1, 2, 3a, and 4.  My solution to 4 was terribly written (after fumbling around with 1 for 40 minutes, I was really short on time).  23 at the most.   :(

I should have gotten 3b.  I had no time for 5.

I can never sleep well the night before big contests.

sunchips, you got everything other than 3b?  Nice.

I got 1, 2, 3a, 4, 5a completely.
5b i nearly finished i just didn't see how to finish it
3b is bsed. maybe 1 mark if lucky.
So like 24+
hopefully honorable mention

I will post the problems in 1 hr.

yes,

In math class, after the contest, I had nothing to do (since my other choice was to go to sleep, like I usually do), so I took another look at 3b. I got 3b after 2 minutes :wallbash_red: , and swear, I got so mad, I wanted so bad to throw a desk

but ya, I thought #1 was hard .... had to solve in 4 cases, for me.....congrats to you guys, for your performance;

hmm my solution for 1 was nicer... but yeah 1 was harder.

Hey...so what is your nice solution to 3b?  The one I discovered after the contest reduces to showing a quartic equation must have one real root, which is true due to casework + descartes rule of signs.

A friend at school stole my test paper.  :mad:   ...I haven't even read 5b yet.    :oops:

EDIT: nvm...I see 3b now.

[quote="Kileel"]A friend at school stole my test paper.  :mad:   ...I haven't even read 5b yet.   :oops:[/quote]
 :rotfl:

How did he do?

He didn't even take CMO.

He stole my test paper because I owe him something.  I guess he's going to use the paper as collateral. :rotfl:

Oh ya, I never actually knew this, but what's Canada's average, usually, for CMO honorable+?

I think it's something like 28.

I think it'll be lower this year as the contest was harder.

Do you guys think it was harder?

ya, I thought it was harder actaully,

I mean, remember 2003, 2004's paper, and 2005's paper: some easy problems.....

I think we discussed this earlier, but CMO's gettin harder isn't it?

Nope.  Did not solve 5a.   :wallbash_red:  And it is soooo easy.  Wasted so much time on 3b.

1, 2, 3, 4 (all parts). But I had only a diagram for #5  :(
Gah, the geometry is sooo easy (yes, Kileel), but I spent to much time on the other ones.

I found a nice simple solution for 3b though, and I finished it in the last 3 minutes  :D

Oh, and this year's one seemed a _lot_ easier than last years (last year I got between 9 and 13 points; this year I'm hoping for 29 but I'll probably get like 24 if I miss crucial proof bits :( ). I think Winter Camp actually helped with the functional equation -- I probably would never have thought of the nice solution otherwise :)

24!!!  :o Seems like everyone did a great job on CMO.....I guess Olympaids doesn't like me at all.... :(

last year HM was low too but i think thats because they screwed up question 1...i guess by the looks of it it seems that the CMO was harder then last year's

lol, Alex, where did u put the problem discussion page? Anyhow, yea, i got 3b pretty quickly largely due to Winter Camp. But geometry is not my thing. Btw, Cynthia ur solution for 5b is so pro...lol, i cannot say how u can see RF actually bisect that angle..... I　guess no HM for me this year.  In term of difficulty from easiest to hardeest, i would say it is 2, 1, 4, 3, 5b. I'm thinking HM would be like 26ish this year, but it would be close. Looks like everyone get 1,2,4.


algebra
polynomial
function
calculus
integration
complex analysis
complex analysis unsolved

Suppose $f$ is an entire function, and $|f(z)|\leq A+B.|z|^{k}$ for all $z$, where $A,B$, and $k$ are positive numbers. Prove that $f$ must be a polynomial

By Cauchy Integral formula $f^{(k)}(z)=\frac{n!}{2\pi i}\int_{\gamma}\frac{f(\zeta)d\zeta}{(\zeta-z)^{k+1}}$. So $f^{k}$ is bounded. By Liouville's theorem $f^{(k)}(z)$ is constant and $f$ is polynomial.

Or you can use almost the same proof to prove that $f^{[k+1]}(z)$ is zero everywhere, and not bother with quoting Liouville.


[quote="Twiz"]How many positive integer solutions $(x,y,z)$ does $xyz=2004$ have? I know the answer, but I pretty much brute forced it. I'd like a quicker method. I'm sure that this question has been asked before, but I honestly could not find it with a search.[/quote]
[hide="Answer"]2004 factors to 2 ^ 2 * 3 * 167. You add each power by 1 and multiply. Then, you get 12. 12 choose three is [b]220[/b].[/hide]

[hide="12 choose 3 is"]

$\binom{12}{3}= \frac{12 \cdot 11 \cdot 10}{3 \cdot 2 \cdot 1}$[/hide]

That depends-is lets say (1,1,2004) the same as (1,2004,1).

No, different orders mean different solutions in this case.

So can anyone post the solution? please

Its not 12C3, becasue you actually have to choose three whose product is 2004, not just any three factors..

[quote="perfect628"]not correct

[hide="hint"]What is $1\times2\times167$?[/hide][/quote]
Um, I don't really get your hint that much. Can you post the solution?

[quote="Twiz"]How many positive integer solutions $(x,y,z)$ does $xyz=2004$ have? I know the answer, but I pretty much brute forced it. I'd like a quicker method. I'm sure that this question has been asked before, but I honestly could not find it with a search.[/quote]
[hide]$2004 = 2^{2}\cdot 3^{1}\cdot 167^{1}$. So we have $x = 2^{a}\cdot 3^{b}\cdot 167^{c}, ~ y = 2^{d}\cdot 3^{e}\cdot 167^{f},$ and $z = 2^{g}\cdot 3^{h}\cdot 167^{i}$. So we have,
\begin{eqnarray}a+d+g & = & 2\\ b+e+h & = & 1\\ c+f+i & = & 1 \end{eqnarray}
The number of solutions in nonnegative integers for equation (1) is $\binom{4}{2}= 6$, the number of solutions in nonnegative integers for equation (2) is $\binom{3}{2}= 3$ and the number of solutions in nonnegative integers for equation (3) is $\binom{3}{2}= 3$. So the number of ordered triples $(x, y, z)$ such that $xyz = 2004$ is $6 \cdot 3 \cdot 3 = \boxed{54}$. Hope this helps.[/hide]

Wait, then you are implying order matters..what if order doesnt matter

FWIW, it is not too difficult (or take too much time) to list all the factors here.
(If nothing else, one can  check  the  the answer  to see if it is correct, ... that is, the method  you used is  a correct one,  and you have not made a silly mistake)

[hide]I will not list the last factor (last factor  is always greater than or equal to 167 - so essentially I am writing factors of 12 as first 2)
(Also to keep is systematic, second factor is always greater than or equal to first)
$(1,1,\dots)$ **
$(1,2,\dots)$,$(1,3,\dots)$,$(1,4,\dots)$,$(1,6,\dots)$,$(1,12,\dots)$
$(2,2,\dots)$ **
$(2,3,\dots)$ ,$(2,6,\dots)$ ,$(3,4,\dots)$
So there are 10  ways if order does not matter.
If order matters, each (**) above gives 3, and rest gives 6 different orders, so total number is 6+6*8 = 54.
And BTW , according to wording of  the original question: [b]order does matter here.[/b]
[/hide]
If you are really looking for to derive the  formula  for such type:

[hide]Here let us assume $k$  baskets
and we have to put  $n$ number of balls (identical balls) in them.
The number of ways one can put them in is given by:
$\binom{n+k-1}{n}$
In our case, $3$ baskets, and $n$  takes the value of  (2,1,1) (remember there are 2 powers of 2, and 1 each for 3 and 167)

So the answer is $\binom{4}{2}\binom{3}{1}\binom{3}{1}$ [/hide]
One can easily write general formula for general case.

[quote="usaha"]umm good solution.  but to follow up on art of owna, is there a formula for order doesnt matter?[/quote]

isn't it just

$\binom{x}{y}$???

i think for when order doesn't matter it's called a combination

im sorry, can you explain that with more detail?  Even with combination, its not 12C3.. with combination- for example, we counted (1,1,2004) 3 times, but with combination we only count it once-also (2,6,167)- we counted taht 6 times but with combination we only count once, so im guessing you need to find the number of ones in which all three terms are the same(which is none)-2 are same and 1 is different-and all three are different

[hide]When you choose one factor, let's say $\varepsilon$, you need to choose 2 factors $\vartheta$ and $\mu$, such that $\vartheta\mu=\frac{2004}{\varepsilon}$.

So there are 12 choices for $\varepsilon$. Let's say that $\varepsilon=1$. Then we need to choose 2  numbers that multiply to 2004. There are 12 more choices for the first number, and then there is only 1 for the third. So that's 12 for $\varepsilon=1$

For $\varepsilon=2$, We need to choose 2 numbers that multiply to 1002. That has 8 factors. So that's 8 more solutions.

For $\varepsilon=3$, we need 2 numbers that multiply to 668. That has 6 factors. So that's 6 more solutions.

Now we need to know all of the factors. There's

1,2,3,4,6,12,167,334,501,668,1002,2004.

But then we now only need the sum of the numbers of the factors of all of those numbers.

Now that's hard to generalize, I think. I'll post this problem in the intermediate forum.[/hide]


geometry
integration
vector
analytic geometry
geometric transformation
rotation
logarithms

Hello everbody. I am new to this website, and I have been trying for hours to solve these couple of questions, yet I seem to be stumped and unsure what exactly to do. If someone could please help me out I would really appreciate it. Thanks.

Problem 1) A small circular loop of area 1.72 cm2 is placed in the plane of, and concentric with, a large circular loop of radius 0.857 m. The current in the large loop is changed at a constant rate from 192 A to -192 A (a change in direction) in a time of 0.802 s, beginning at t = 0. What is the magnitude of the magnetic field at the center of the small loop due to the current in the large loop at (a) t = 0, (b) t = 0.401 s, and (c) t = 0.802 s? (d) Because the inner loop is small, assume the magnetic field due to the outer loop is uniform over its area. What emf is induced in the small loop at t = 0.401 s?

Problem 2) As shown in the diagram, a long rectangular conducting loop, of width L = 22 cm, resistance R = 13 Ω, and mass m = 0.15 kg, is hung in a horizontal, uniform magnetic field of magnitude 1.4 T that is directed into the page and that exists only above line aa. The loop is then dropped; during its fall, it accelerates until it reaches a certain terminal speed vt. Ignoring air drag, find in meter per second that terminal speed.

Problem 3) As shown in the diagram, two straight conducting rails form a right angle. A conducting bar in contact with the rails starts at the vertex at time t = 0 and moves with a constant velocity of 7.11 m/s along them. A magnetic field with B = 0.456 T is directed out of the page. Calculate the magnitude of (a) the flux through the triangle formed by the rails and bar at t = 4.37 s and (b) the emf around the triangle at that time. If the emf is ε = atn, where a and n are constants, what are the values of (c) a and (d) n?

Problem 4) Switch S is closed at time t = 0 s, initiating the buildup of current in the 13.7 mH inductor and the 20.8  resistor. At what time (in ms) is the emf across the inductor equal to the potential difference across the resistor?

[color=blue][b](1)[/b][/color] Magnetic field at the center of the large loop is given by Biot-Savart's law:

$\vec B = \frac{\mu_0 I}{4 \pi} \int_L \frac{\text d\vec l\times \vec R}{R^3}$

Since R = 0.857 m (the large loop radius) is constant and $\vec R$ pointing to the center from any loop element $\text d\vec l$ is perpendicular to this element, the magnetic field has only the component $B_n \neq 0$ perpendicular to the loop (the unit vector $\vec n$ points up when the curent I flows counter-clockwise).

$\vec B = \frac{\mu_0 I \vec n}{4 \pi R^2} \int_L \text dl = \frac{\mu_0 I \vec n}{4 \pi R^2} \cdot 2 \pi R = \frac{\mu_0 I \vec n}{2 R}$

$B_n = \frac{\mu_0 I}{2 R} = \frac{\mu_0 I_0 (1 - \frac{2t}{T})}{2 R}$

where $I_0 = 192\ \text A$ and $T = 0.802\ \text{sec}$. The emf induced in the small loop of area $S = 1.72\ \text{cm}^2$ (convert to $\text m^2$) is constant with time

$\epsilon = -\frac{\text d\Phi}{\text dt} = -\frac{\text dB_n}{\text dt}\ S = \frac{\mu_0 I_0 S}{R T}$

[color=blue][b](2)[/b][/color] Put the z-axis vertical and up, the x-axis horizontal and right (identical with the line aa') and the y-axis into the page (to have a hight-handed coordinate system). Hence, $B_y > 0$. Let z be the height of the top loop segment with the given length L. The emf induced in the loop is

$\epsilon = -\frac{\text d\Phi}{\text dt} = -B_y \frac{\text dS}{\text dt} = -B_y L \frac{\text dz}{\text dt} = -B_y L \dot z$

This emf causes curent $I = \frac \epsilon R$ in the loop, the magnetic field of which opposes the given magnetic field, hence, this current flows clockwise in the figure (from the positive z-axis to the positive x-axis). The magnetic field with the horizontal component $B_y$ exerts Lorentz force on this current. In the left vertical segment, this force is directed left and in the right segment, an equal force is directed right, so they cancel each other. In the upper horizontal segment, this force is directed up and has the magnitude

$F_{Bz} = ILB_y = \frac{\epsilon LB_y}{R} = -\frac{L^2B_y^2 \dot z}{R} > 0$

(because $\dot z < 0$ is downward velocity). The equation of motion of the loop is

$m \ddot z = F_{Gz} + F_{Bz} = -mg - \frac{L^2B_y^2 \dot z}{R}$

$\ddot z + \frac{L^2B_y^2}{mR}\ \dot z + g = 0$

The corresponding differential equation for the vertical velocity $v_z = \dot z$ is

$\dot v_z + \frac{L^2B_y^2}{mR}\ v_z + g = 0$

which has a general solution

$v_z = -\frac{mgR}{L^2 B_y^2} + C \exp \left(-\frac{L^2B_y^2}{mR}\ t\right)$

where C is an arbitrary constant depending on initial conditions. But the right term approaches zero for $t \rightarrow \infty$ regardless of C. Hence, the left term is the terminal velocity (that is, as long as the upper segment of the long-long loop stays above the line aa' during the fall).

Thank you very much yetti. I really appreciate it. I fully understand your work for question 1, yet I have a question regarding number 2.

I understand how the term multiplied by the constant results in zero, so the velocity v will equal:

v = [-(m)(g)(R)]/[(L^2)(By^2)]. By declaring that there is a z-axis, what results in the value for By? Plugging in the numbers, I thought that:

v = [-(.15 kg)(9.81 m/s^2)(13 ohms)]/[(.22 m)^2 x (1.4 T)^2]
v = -201.65 m/s

This answer is incorrect, however. I think I was incorrect in assuming that based on the orientation of the problem, the B value that is provided is the equivalent of the By value in the formula. [b]Having said this, what should I do in order to determine By, and ultimately, determine the velocity?[/b]

[b]Also, if someone can be nice enough to help me with problems (3) and (4), I would really appreciate it.[/b]

Have a good day.

$B_y = 1.4\ \text T$. Please, tell me what is the correct answer, it would make it easier to find an error (if any).

I am unaware of the correct answer. If I knew it, it would make my life easier as well, lol. As soon as I find out how to solve the problem I will deffinately post the solution for you to see. Thank you very much for helping me with question 1 and trying to solve question 2.

[b]In the meanwhile, can someone also help me with questions 3 and 4? I really would appreciate it. Thank you in advance.[/b]

[quote="PhysicsManiac101"]I am unaware of the correct answer...[/quote]

What is then the basis for your previous statement ?

I said that because I was studying and going thru problems located in the text and came across these problems which I found fairly difficult. It is now the weekend, but early next week I will ask someone in school if they could help me solve it.

Can someone help me answer questions 3 and 4 so I know how to solve them? I am a little confused how to go about those questions. Thanks.

If you can guess that a constant terminal speed exists, you can get it from the condition

$0 = F_{Gz} + F_{Bz} = -mg - \frac{L^2B_y^2 v_z}{R}$

$v_z = -\frac{mgR}{L^2 B_y^2}$

Do you agree with that ?

No, I agree. I have a feeling that you are correct and that it is the minus sign that is incorrect. Like I said, I will find out within the next day or two and as soon as I do, I will let you all know what the problem is. But I have a gut feeling that it's the minus sign and they want the magnitude of the velocity (I could be wrong though, but our solution makes sense).

[u]Can someone please help me with problems 3 and 4[/u] as well, this way if I have any questions about those problems I can also ask in class once the weekend is up? Thank you soo much!

The terminal velocity is a vector fully determined by its components in the coordinate system used:

$\vec v = (v_x, v_y, v_z) = \left(0,\ 0,\ -\frac{mgR}{L^2B_y^2}\right)$

Knowing the components in any coordinate system, its magnitude is $v = |\vec v| = \sqrt{v_x^2 + v_y^2 + v_z^2}$. In this case, $v_x = v_y = 0$ and $v = \sqrt{v_z^2} = |v_z| = \frac{mgR}{L^2B_y^2}$.

For problem 3, the magnetic flux is $\phi = BS$, because the magnetic field is perpendicular to the plane of the loop. The area of the loop is the area of an isosceles right angle triangle with the right angle altitude $vt$, hence, $S = v^2t^2$. The emf is $\epsilon = -\frac{\text d\Phi}{\text dt} = -B \frac{\text dS}{\text dt} = -2Bv^2t$ and $|\epsilon| = 2Bv^2t$.

The minus sign is $\epsilon = -\frac{\text d\Phi}{\text dt}$ has this meaning: The area of the loop is a vector $\vec S$ with magnitude S and direction perpendicular to the plane of the loop. But there are 2 such directions, so you choose one arbitrarily, usually to get the magnetic flux $\phi = \vec B \cdot \vec S$ positive. In this problem, parallel to the magnetic field. This makes the loop directed in the direction of rotation of a normal (right-handed) screw traveling in the chosen direction of $\vec S$.  In this problem, the loop becomes directed clockwise when looking on the figure. If $\epsilon$ comes out positive, the induced current flows in the loop direction, if $\epsilon$ comes out negative, the current flows against the loop direction. In this problem, $\epsilon$ came out negative, so the current flows counter-clockwise when looking on the figure.

For problem 4, the current flowing through the circuit is (look at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=80538]emf[/url])

$i(t) = \frac \epsilon R \left[1 - \exp \left(-\frac R L\ t \right)\right]$

$\frac{\text di}{\text dt} = \frac \epsilon L \exp \left(-\frac R L\ t \right)$

At time t,

$R i = \epsilon_R = \epsilon_L = L \frac{\text di}{\text dt}$

Substituting i(t) and $\frac{\text di}{\text dt}$

$\epsilon \left[1 - \exp \left(-\frac R L\ t \right)\right] = \epsilon \exp \left(-\frac R L\ t \right)$

$1 = 2\exp \left(-\frac R L\ t \right)$

$\ln 2 = \frac R L\ t,\ \ t = \frac L R\ \ln 2$

Thank you very very much. Once again, I will check with the troublesome number 2 and report the correct solution. Thanks!

We were right - it was the minus sign that caused us to get the velocity value wrong. Thanks again!


The graph $C_{f}$ of a function $f$ from reals to reals has the following property .
For each points $A,B\in C_{f}$ with $A(a,f(a))$
$B(b,f(b))$ it holds that $\frac{f(a)-f(b)}{a-b}\geq 0$

If $a\in R$ is a given number solve the equation
$a-f(x+f(a))=x$

The condition says $f(a)\geq f(b)$ for all $a>b$. So $f$ is increasing.
Now $x_{0}=a-f(a)$ is obviously a solution of $a-f(x+f(a))=x$,
and for every $x>x_{0}$ we then have $a-f(x+f(a))\leq a-f(x_{0}+f(a))=x_{0}<x$
and for $x<x_{0}$ we have $a-f(x+f(a))\geq a-f(x_{0}+f(a))=x_{0}>x$.
So $x_{0}$ is the unique solution.


I'm supposed to document algorithm for solving determinants of matrices using minors of the first row for matrices bigger than 2x2.

I also need to do the same thing for solveing a system of n linear equations with n independent variables using determinants.

I know how to solve for both... but I'm having lots of trouble for writing the algorithm for it.

Thank you sooo much :D

Do you mean minors?

The appropriate programming technique is recursion: your function calls more copies of itself to handle the subproblems.
This is also an extremely slow and impractical algorithm for large matrices. Gaussian elimination is a much better way to find determinants.


trigonometry
algebra
polynomial
complex numbers

Prove that :

$ cos(\frac{\pi}{5}) \plus{} cos(\frac{3\pi}{5}) \equal{}\frac{1}{2}$ .

[quote="Svejk"]See here. http://www.mathlinks.ro/Forum/viewtopic.php?t=154226
It's not really the same problem but it's the same technique  :)[/quote]
But it is the same problem since $ \cos \theta \equal{}\minus{}\cos (\pi\minus{}\theta)$

I believe the following generalization is also true:

$ \sum_{i = 1}^{k}{\cos(\frac {2i - 1}{2k + 1}}\pi) = \frac {1}{2}$

for all positive integers $ k$.

[quote="turcas_c"]Prove that  $ \cos\frac {\pi}{5} \plus{} \cos\frac {3\pi}{5} \equal{} \frac {1}{2}$ .[/quote]
[b][u]Proof.[/u][/b] $ \left\|\cos\frac {2\pi}{10} \plus{} \cos\frac {6\pi}{10} \equal{} \frac {1}{2}\right\|\ \bigodot\ 2\cos\frac {\pi}{10}\ \ \Longleftrightarrow$

$ \cos\frac {3\pi}{10}\plus{}\cos\frac {\pi}{10}\plus{}\cos\frac {7\pi}{10}\plus{}\cos\frac {5\pi}{10}\equal{}\cos\frac {\pi}{10}$ $ \Longleftrightarrow$

$ \cos\frac {3\pi}{10}\minus{}\cos\frac {3\pi}{10}\plus{}\cos\frac {\pi}{2}\equal{}0$, what is truly.

If you do not like to spend lots of time on these types of problems, then you may use complex numbers (a typical, standard method).  Let $ z \equal{} e^{i\frac {\pi}{5}}$.  It then follows that we are to prove that
\[ \frac {1}{2} \equal{} \frac {z^2 \plus{} 1}{2z} \plus{} \frac {z^6 \plus{} 1}{2z^3}.
\]
But the right hand side is equivalent to $ \frac { \minus{} e^{ \minus{} i\pi}}{2} \equal{} \frac {1}{2}$.

[hide="Alternately"] The roots of the polynomial $ z^4 \minus{} z^3 \plus{} z^2 \minus{} z \plus{} 1 \equal{} 0$ sum to $ 1$, and they are $ \text{cis }\frac{\pi}{5}, \text{cis }\frac{3\pi}{5}, \text{cis }\frac{7\pi}{5}, \text{cis }\frac{9\pi}{5}$.  The real part is therefore

$ \cos \frac{\pi}{5} \plus{} \cos \frac{3\pi}{5} \plus{} \cos \frac{7\pi}{5} \plus{} \cos \frac{9\pi}{5} \equal{} 1$

Now note that $ \cos (2 \pi \minus{} x) \equal{} \cos x$, hence

$ 2 \left( \cos \frac{\pi}{5} \plus{} \cos \frac{3\pi}{5} \right) \equal{} 1$

MellowMelon's generalization follows similarly.[/hide]


I was just wondering about the best way to determine whether molecules with no specific central atom are polar are not. For molecules with a specific central atom, I usually just determine what the molecular geometry of the molecule is and determine whether it is symmetrical or not. Can this opportunity be revised to work for atoms with no specific central atom?

Some specific examples that I am considering are ethane, ethylene, and actylene.

You could split those three molecules up in symmetrical halves, find the overall polarities of the halves, and then combine those polarities to find net polarities of each molecule. Thus, all have a net dipole moment of zero.

At some point, it gets extremely difficult to impossible to judge the vector quantity of polarity. Basically, oxygens, nitrogens, and other non-carbon and non-hydrogen atoms present indicate polar molecules. Hydrocarbons are generally nonpolar.

[quote="Jdlkem"]Hydrocarbons are generally nonpolar.[/quote]

That's right. How about an example of a polar hydrocarbon?

ummm... a polar hydrocarbon could be a nonsymmetrical cis alkene.....

Still, the polarity is not meaningfull. Another example?

azulene!

possible to synthesize from napthalene, a nonpolar aromatic compound.

Carcul, why arent cis alkenes polar? I thought the polarity meant that the compound had a dipole moment...

They have dipole moment which does not mean that they are polar. They can influenced under Electric field.
Whereas in azulene which exists as a zwitterion there are visible polar ends in the same molecule.


[quote="PenguinIntegral"]How many values of x that are integers sanisfy: $(x^2-x-1)^{x+2}=1$?[/quote]

[hide="Solution"]

x^2 - x - 1 can equal either 1 or -1 (provided x + 2 is even).

x^2 - x - 1 = 1
x^2 - x - 2 = 0
(x - 2)(x + 1) = 0
x = 2
x = -1

x^2 - x =0
x(x - 1) = 0
x = 0, because x = 1 doesn't work.

x = 2
x = -1
x = 0

x can also equal -2, because x + 2 = 0.

[b]4[/b]
[/hide]


logarithms
calculus
integration
function
real analysis
real analysis solved

$(a_n)$ a sequence of positive real number such that $\sum \sqrt{a_n} < + \infty$.
$(b_n)$ a sequence of real number.
$E$ is the set of real number $x$ such that $\sum \dfrac{a_n}{|x - b_n|} < + \infty$.
Prove that almost all reals are in $E$.  (i.e. $\mathbb{R}-E$ has null measure)

I think( if I didn't make stupid mistakes ), that one can replace
the condition $\sum \sqrt{a_{n}} < \infty$ by $\sum a_{n} \ln ( \dfrac{1}{a_{n}} ) < \infty$.

For the problem as alekk posted it:

Let $A_n$ be the set of reals $x$ which satisfy $|x-b_n|<\sqrt{a_n}$. That series diverges only if $x$ is in infinitely many $A_n$. However, the series formed by the measures of $A_n$ converges, and it's been shown before on the forum that in this case the set consisting of those $x$ which belong to infinitely many $A_n$ has null measure (we just add up the integrals of the characteristic functions of the $A_n$'s).

Take some sequence $\varepsilon_{1}, \varepsilon_{2}, \ldots \rightarrow 0$
and consider a sequence of functions $f_{n}(x) =$
$\begin{cases} \frac{a_{n}}{| x-b_{n} |} & \varepsilon_{n} \leqslant | x-b_{n}| \leqslant 1 \\ 0 & \text{otherwise}. \end{cases}$

Then $\int_{ - \infty }^{ \infty } f_{n}(x) dx = 2 a_{n} \ln( \frac{1}{\varepsilon_{n}})$. Now if we assume that $\sum a_{n} \ln(\frac{1}{\varepsilon_{n}}) < \infty$, then the series $f_{1}(x) + f_{2}(x) + \ldots$ converges to a measurable function
$f(x)$, such that $\int_{ - \infty }^{ \infty } f(x) dx < \infty$, therefore $f(x) = \infty$ only on a set of measure $0$, so the series $f_{1}(x) + f_{2}(x) + \ldots$ converges for
almost every $x \in \mathbb{R}$, except a set $T$ of measure $0$ . Now, if we define $A : = \bigcup_{n=1}^{\infty} ( b_{n} - \varepsilon_{n} , b_{n} + \varepsilon_{n} )$, then because of $\sum a_{n} < \infty$, if we take $x \notin A$, then convergence
of $f_{1}(x) + f_{2}(x) + \ldots$ is equivalent to a convergence
of $\sum \frac{a_n}{|x - b_n|}$. Therefore the series $\sum \frac{a_n}{|x - b_n|}$ converges if $x \notin A \cup T$. Now,
$m(A) = \sum_{n=1}^{\infty} \varepsilon_{n}$. Now I want to find
such a sequence $\varepsilon_{1}, \varepsilon_{2}, \ldots \rightarrow 0$, that $\sum_{n=1}^{\infty} a_{n} \ln(\frac{1}{\varepsilon_{n}}) , \sum_{n=1}^{\infty} \varepsilon_{n} < \infty$. This is equivalent to $\sum_{n=1}^{\infty} (a_{n} \ln(\frac{1}{\varepsilon_{n}}) + \varepsilon_{n}) < \infty$. Let us find the optimal sequence $\varepsilon_{1}, \varepsilon_{2}, \ldots$. It is easy to see that
the minimum of $a \ln( \frac{1}{\varepsilon} ) + \varepsilon$
attains when $\varepsilon = a$, so if we take $\varepsilon_{n} : = a_{n}$, then we have to demand that $\sum_{n=1}^{\infty} a_{n} \ln(\frac{1}{a_{n}}) , \sum_{n=1}^{\infty} a_{n} < \infty$
and because of $a_{n} \rightarrow 0$, it is enough to have $\sum_{n=1}^{\infty} a_{n} \ln(\frac{1}{a_{n}}) < \infty$. So we
see that if $\sum_{n=1}^{\infty} a_{n} \ln(\frac{1}{a_{n}}) < \infty$, then for all $x \in \mathbb{R}$, may be except a set
of measure $\sum_{n=1}^{\infty} a_{n}$, we have that $\sum_{n=1}^{ \infty } \frac{a_n}{|x - b_n|} < \infty$. Now, for
some $k \in \mathbb{N}$ convergence of $\sum_{n=k}^{ \infty } \frac{a_n}{|x - b_n|}$ implies a convergence of $\sum_{n=1}^{ \infty } \frac{a_n}{|x - b_n|}$, if $x \neq b_{1},b_{2}, \ldots , b_{k-1}$, so we see that  $\sum_{n=1}^{ \infty } \frac{a_n}{|x - b_n|} < \infty$ for every $x$ except a set of measure $\sum_{n=k}^{\infty} a_{n}$ , for every $k$ we take, so we get
that  $\sum_{n=1}^{ \infty } \frac{a_n}{|x - b_n|} < \infty$ for
almost all $x$.

I hope it is correct.


calculus
integration
trigonometry
calculus computations

Let $m,\ n$ be the given distinct positive integers. Answer the following questions.

(1) Find the real number $\alpha \ (|\alpha |<1)$ such that $\int_{-\pi}^{\pi}\sin (m+\alpha )x\ \sin (n+\alpha )x\ dx=0$.

(2) Find the real number $\beta$ satifying the sytem of equation $\int_{-\pi}^{\pi}\sin^{2}(m+\beta )x\ dx=\pi+\frac{2}{4m-1}$, $\int_{-\pi}^{\pi}\sin^{2}(n+\beta )x\ dx=\pi+\frac{2}{4n-1}$.

[quote="bos1234"]There is a big difference between Japan maths and Aus maths  :o  :o[/quote]
 :lol:
Try to find it yourself. some hint:
use $\sin{a}\sin{b}=\frac{1}{2}(\cos{(a-b)}-\cos{(a-b)})$
after you must note something  ( why this problem must be difficult) you must divide interval of integrations in some needed parts.
in problem #2 use
$\sin^{2}x=\frac{1-\cos(2x)}{2}$ and after do analogysli
I believe you solve   it if you want

[quote="bos1234"]There is a big difference between Japan maths and Aus maths  :o  :o[/quote]

 :rotfl:  :D


In the winter sport of curling, players give a 20KG stone a push across a sheet of ice. A curler accelerates a stone to a speed of 3.1 m/s over a time 2.1s.
How much force does the curler exert on the stone?
What average power does the curler use to bring the stone up to speed?

$ F\equal{}ma
\\
\\P \equal{} \frac{\text{Work done}}{\text{time taken}}
\\
\\
\;\text{Work done, W} \equal{} Fs
\\
\\s\equal{}ut\plus{}\frac{1}{2}at^2, \;u\equal{}0\text{ here.}$

I'm sorry but can you be more specific?
[quote="Rajiv"]$ F \equal{} ma \\
 \\
P \equal{} \frac {\text{Work done}}{\text{time taken}} \\
 \\
\;\text{Work done, W} \equal{} Fs \\
 \\
s \equal{} ut \plus{} \frac {1}{2}at^2, \;u \equal{} 0\text{ here.}$[/quote]

But how can you get the ball's acceleration?  I'm sort of lost here.

[quote="Rajiv"][quote="cdymdcool"]
I'm sorry but can you be more specific?

[/quote]

Sure, why not!

$ \text{Force by curler, F} \equal{} \text{[mass of ball (M)]} \times \text{[accleration of ball (a)]}$

$ \text{Average Power (P)} \equal{} \frac {\text{Work done by curler (W)}}{\text{total time for doing the work (t)}}$

$ \text{Where Work done, W} \equal{} Fs$

$ \text{Where s, the displacement is given by }s \equal{} ut \plus{} \frac {1}{2}at^2, \;u \equal{} 0\text{ here, in this case.}$[/quote]

[quote="cdymdcool"]But how can you get the ball's acceleration?  I'm sort of lost here.
[/quote]

$ v \equal{} u \plus{} at \\
 \\
u \equal{} 0 \\
3.1m/s \equal{} a\times(2.1sec) \\
 \\
a \equal{} \frac {3.1}{2.1}m/s^2$

[quote="Immanuel Bonfils"]For the average power, isn't it easier to multiply the force by the average velocity ?[/quote]

That is when velocity is constant, isn't it?
Here the ball is accelerated to $ 3.1m/s$ in $ 2.1s$.

.

[quote="cdymdcool"]But how can you get the ball's acceleration?  I'm sort of lost here.
[/quote]Very simple: $ a_{av} \equal{} \frac {\Delta v}{\Delta t} \equal{} \frac {3.1}{2.1}$

For the average power, just use $ P \equal{} Fv$

[quote="Rajiv"]That is when velocity is constant, isn't it.[/quote]That's why Immanuel said [i]average [/i]velocity (for a constant velocity the word [i]average [/i]would be redundant) ;) After all, only the [i]average [/i]power is asked.

[quote="Rajiv"]$ \text{Where s, the displacement is given by }s \equal{} ut \plus{} \frac {1}{2}at^2, \;u \equal{} 0\text{ here, in this case.}$[/quote]

This is true only if $ a$ is uniform, i.e. constant with respect to time. There is no indication in the problem statement that this is the case.

[quote="Immanuel Bonfils"]For the average power, isn't it easier to multiply the force by the average velocity ?[/quote]

Good thought, but if you do it that way, you'll get a result that is twice the value of the correct answer.

\[ \mbox{Instantaneous power} \equal{} P \equal{}\frac{d W}{d t} \equal{}\mathbf{F}\cdot\mathbf{v}\]

Whereas,

\[ \mbox{Average power} \equal{} \langle P\rangle \equal{}\frac{W}{\Delta t} \equal{}\frac{\Delta KE}{\Delta t}\]

Evidently use of the work-energy theorem, $ W \equal{} \Delta KE$, is intended for this problem.

[quote="Rajiv"][quote="Dr. No"]
This is true only if $ a$ is uniform, i.e. constant with respect to time. There is no indication in the problem statement that this is the case.
[/quote]

Yes, I thought the acceleration is assumed to be constant (this being posted in the Introductory Physics forum). Does it matter if the acceleration varies?[/quote]

Sorry for the lateness of my reply. I've been busy for awhile and not able to respond.

The only way to compute the acceleration in this case is to compute the average acceleration:

\[ a \equal{}\frac{3.1\mbox{ m/s } \minus{} 0\mbox{ m/s }}{2.1\mbox{ s}} \equal{} 1.5\; m/s^2\;\mbox{ to two significant figures.}\]

The more important consideration for this problem as far as I'm concerned is the distinction between average and instantaneous power.


Suppose a box with mass $ 2$ kg was moving to the right at $ 6$ m/s on another box with equal mass $ 2$ kg, which was at rest. There is friction between the two boxes, and it was found that $ f_{k}\equal{}\mu_{k}N\equal{}2$ newtons. So, after one second, the velocity of the upper box is reduced to $ v\equal{}u\plus{}at\equal{}6\plus{}(\minus{}1)(1)\equal{}5$ m/s, and the velocity of the lower box becoms $ v\equal{}u\plus{}at\equal{}0\plus{}(1)(1)\equal{}1$ m/s.

Considering the upper box, we find that, using the energy equation, $ \frac{1}{2}mu^{2}\equal{}\frac{1}{2}mv^{2}\plus{}f_{k}S$, where $ S$ is the relative distance between the two boxes. We get that $ \frac{1}{2}(2)(6)^{2}\equal{}\frac{1}{2}(2)(5)^{2}\plus{}f_{k}S$. At the same time, considering the lower box, we find that $ f_{k}S\equal{}\frac{1}{2}(2)(1)^{2}$.

So $ 6^{2}\equal{}5^{2}\plus{}1^{2}$.

What's wrong?

[quote="Radio"]and the velocity of the lower box becoms $ v \equal{} u\plus{}at \equal{} 0\plus{}(1)(1) \equal{} 1$ m/s[/quote]

The acceleration of the lower box is not certainly the same as of the upper one. What are the forces applied at the lower box?

The forces applied are its weight $ mg$, the weight of the upper box, and the friction force. However, the first two are perpendicular to the direction of movement...

[hide="think"]
about the nature of  frame of refrence from which u r using ur force and in particular energy equation.[/hide]

[quote="Radio"]The forces applied are its weight mg, the weight of the upper box, and the friction force. However, the first two are perpendicular to the direction of movement...
[/quote]

There are more forces applied. Have you drawn a scheme for this problem?

[quote="Carcul"]

There are more forces applied. Have you drawn a scheme for this problem?[/quote]

Hmm... what forces are they?  :maybe:

[quote="Carcul"]I am assuming the following picture: we have the "lower box" on an horizontal surface (between which there is friction), and the "upper box" moving on top of the lower box. Then the forces applied to the lower box are: its weight; the normal force exerted by the horizontal surface; the friction force between the horizontal surface and the box; the normal force exerted by the upper box; and the friction force between both boxes.[/quote]

Umm... I think radio assumed there's no friction between the box and the floor.

I think the mistake is somewhere about the distance of friction in your energy equation, but I'm not so sure...

Even so there is friction between both boxes.

I get it  :D

Your distance $ S$ is wrong.


vector
topology
advanced fields
advanced fields solved

funny one: compute the fundamental group of $GL_{2}(\mathbb{R})^{+}$

It's $\mathbb Z$. If $v\in\mathbb R^{2}$ is some non-zero vector, it's not difficult to prove that $A\mapsto\frac{Av}{\|Av\|}$ is a homotopy equivalence of $\mbox{Gl}_{2}^{+}(\mathbb R)$ onto the circle.

yes, I found the same, but my approach was a little longer. The idea was to show that a path in $GL_{2}(\mathbb{R}^{+})$ was homotopic to a path in $SO_{2}(\mathbb{R})$, using [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=homeomorphism&t=64293]this[/url] pb for example.


logarithms
inequalities unsolved
inequalities

[quote="thanhnam2902"][color=blue]Let $ \boxed {0 < a \le b \le c \le d}$. Prove that: $ \boxed {a^b b^c c^d d^a \ge b^ac^bd^ca^d}$.[/color][/quote]
$ <\equal{}> {(\frac{c}{a})}^{d\minus{}b}\ge{(\frac{d}{b})}^{c\minus{}a}$; if  $ ( a \equal{} c)$ or $ ( b \equal{} d )\equal{}>$ equality;
Let $ (a < c)$ and $ (b < d) \equal{}> \frac{\ln{(c)}\minus{}\ln{(a)}}{c\minus{}a}\ge \frac{\ln{(d)}\minus{}\ln{(b)}}{d\minus{}b}$ - which obviously true for $ Ln$ :)


Determine the three first digit of the number:\[\sum^{1000}_{n=1}n^n\]

[hide]Since 1000^1000 is more than 3 digits longer than 999^999, it is 100.[/hide]

[hide]
Look at the # of digits of the biggest numbers, 998^998, 999^999, 1000^1000.
1000^1000 has more than three digits more than 999^999, meaning when we add them all up, the digits in 999^999 will not affect what the first three digits of 1000^1000 is.

It is obvious that the first three digits of 1000^1000 is $\boxed{100}$.
[/hide]

EDIT. two peopel got before me... didnt see it until i scrolled down to see the Topic Review. ;)

[quote="amirhtlusa"]hold on, the first time i read the problem, i tought is asking for the first three digit from the right... ;)[/quote]

it should say then: find the LAST three digits... but it says first, so i assumed it was from the left.

what if it was asking for the last 3 digits from the right?

Then... well... it seems pretty hard. We could use mod 1000 to figure out, but that would be very time consuming. somebody else want to try?

[quote="seamusoboyle"]But your just saying that $1000^{1000}$ has 3 more digits than $999^{999}$ but youll have to prove that $1000^{1000}$ has 3 more digits than $\sum _{k=1} ^ {999} n^n$ All those extra terms could affect the result....[/quote]

You're right,but one can easily prove that the first three digits are 100

[quote="seamusoboyle"]PS last three digits are 700[/quote]

how di d u get it?


hameye maghadire $a$ ra beyabid be sharti ke namosavie zir baraye hameye adade haghighi va mosbate $x,y,z$ sedgh konad.
$2(x^3+y^3+z^3)+3(1+3a)xyz\geq(1+a)(x+y+z)(xy+yz+zx)$

ghashange na?

vaghean yani inghad sakhte?
age fekri be zehnetoon nemirese begam ke ye ideye kolli ine ke be jaye $x,y,z$ adadgozarie hoshmandane(be dard bokhor) bokonin ta baraye $a$ ye hodoodi be dast biyad

Selam alejkum!

Can write this problems in english please!

salamon alaykom va rahmatollahe va barakatoh
find all values of $a$ for which the inequality(top) holds for all positive real numbers $x,y,z$


calculus
integration
LaTeX
real analysis
real analysis unsolved

I(alpha)=integral(0 to 1)(x^(3*alpha-1) + x^(alpha-1))/(x^(4*alpha) +1)
I(alpha)=?

Please use $\LaTeX$ !
[url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]Look here[/url]

In $\LaTeX$:
$I(\alpha)=\int_{0}^{1}\frac{(x^{3\alpha-1}+x^{\alpha-1})}{(x^{4\alpha}+1)}$

$I(\alpha)=$?

yes this is what i meant.i have no time right now to learn latex

I think that the value of your integral is $\frac{\pi}{2\sqrt{2}\alpha}$.

First use the substitution $x^{4\alpha}=y$ and you get that

$I(\alpha)=\frac{1}{4\alpha}\int_{0}^{1}\left(\frac{y^{3/4-1}}{y+1}+\frac{y^{1/4-1}}{y+1}\right)dy$.

Use the substitution $\frac{1}{y}=t$ for calculating the first integral and you get that

$I(\alpha)=\frac{1}{4\alpha}\int_{0}^{\infty}\frac{t^{1/4-1}}{1+t}dt=\frac{1}{4\alpha}\beta(1/4,3/4)=\frac{\pi}{2\sqrt{2}\alpha}$.

I hope I made no computational mistakes.


trigonometry
algebra
system of equations
geometry
trig identities
Law of Sines

[img]http://www.dlstudios.net/uploaded/1tri.gif[/img]

is it even possible to find a, b, c, x, y, and z from the given data?

I've come up with the equations:
a+b+c = 360
x+y+z = 110
a+c+x+y = 300
a+b+y+z = 330
b+c+x+z = 310
a+y = 160
b+z = 170
x+c = 40

but none of them give me any unknowns.
I'm writing this problem for a PreCal (algebra 2 and trig) project, if its impossible, what is the least amount of data I can give the person to make it possible, but keep it as hard as possible?

You have eight equations for six unknowns, but they aren't all distinct, i.e. some provide no new information.  This system can be reduced to four equations for six unknowns, which has an infinite number of solutions, so it's impossible.

To make it possible through a system of equations, you have to provide two new distinct equations involving the angles, but given the triangle in question, I don't think it'd be very difficult to solve, and might seem a little artificial or contrived as a problem.  So instead, I would personally suggest giving one additional angle and two side lengths, and ask for all the angles and the remaining side length.  It's pretty much a simple law of sines exercise, but it might make the problem slightly more interesting (not to say doable).

Wait, to make sure I understand everything, which 4 equations did you reduce the system to?

Well, there's a couple systems that you could reduce it to, but the one I chose was
x+y+z=110
a+y=160
b+z=170
x+c=140
e.g.
x=40, y=50, z=20, a=110, b=150, c=100
and
x=10, y=65, z=35, a=95, b=135, c=130
are two solutions.

[quote="Darklightmw"][img]http://www.dlstudios.net/uploaded/1tri.gif[/img]

is it even possible to find a, b, c, x, y, and z from the given data?

I've come up with the equations:
a+b+c = 360
x+y+z = 110
[color=red]a+c+x+y = 300[/color]
a+b+y+z = 330
b+c+x+z = 310
[color=blue]a+y = 160[/color]
b+z = 170
[color=blue]x+c = 40[/color]

but none of them give me any unknowns.
I'm writing this problem for a PreCal (algebra 2 and trig) project, if its impossible, what is the least amount of data I can give the person to make it possible, but keep it as hard as possible?[/quote]
Wait a second... your equations don't work... add the two blue equations and you get 200, not 300 for the red equation.

I assumed Darklight meant x+c=140, not x+c=40, since x+c+40=180.  If you make the switch, the equations work.


calculus
integration
geometry
3D geometry
sphere
analytic geometry
function

Discuss the convergence of the integral
1/[x^2 + y^2 + z^2 + 1]^2 dxdydz in the whole space.

Here is my work : Since the space is unbounded, the integral is an improper integral so we can consider a sphere with radius N and take the limit as N goes to infinity. I have used spherical coordinates. Theta is between 0 and 2Pi, Phi is between 0 and Pi, and rho is between 0 and N and the integrand becomes
(rho^2)sin(Phi)/[1 + (rho^2)] d(rho) d(phi) d(theta) .
Here again we use substitution : rho = tan x and the integrand becomes ((sin x)^2)d(x). But i can't figure out the upper limit of integration.
For rho = 0, we have tan x = 0 and rho = N, we have tan x = N. So lower limit of x is 0 but what about the upper limit? Is this integral convergent?

Edit : Okay, using the arc tan substitution again, i have found the integral convergent. Excuse me!

Your calculations involving the substitution are the way to calculate the exact value of the integral; if we just want to show it converges, we don't need that.

After expressing it in spherical coordinates, the only part that might cause trouble is $ \int_{0}^{\infty}\frac{\rho^{2}}{\rho^{4}+2\rho^{2}+1}\,d\rho$
(Quote to see how I did that)
This is not improper at zero. At $ \infty$, the integral converges by comparison to $ \int\frac1{\rho^{2}}\,d\rho$.

Suppose you have a function in $ \mathbb{R}^{3}$ that is "tame" near the origin and looks like a constant divided by $ \|\vec{r}\|^{b}$ for points far from the origin, where $ \|\vec{r}\|=\rho$ is the distance from the origin. Then what can we say about convergence of the integral of this function over all of $ \mathbb{R}^{3}?$ Let's say that the function is nonnegative, or that we're asking about absolute convergence. Then the integral converges if $ b>3$ and diverges if $ b\le 3.$ In this case, the function looks like $ \frac1{\|\vec{r}\|^{4}},$ so $ b=4>3$ and the integral converges.

More generally, if we're working over $ \mathbb{R}^{n},$ such an integral would converge for $ b>n$ and diverge for $ b\le n$. Note that this formulation works even for $ n=1.$


Αφιερωμένη στον κ. Πατήλα  :)
Έστω $ a,b,c,d$ θετικοί πραγματικοί αριθμοί τέτοιοι ώστε $ abcd\equal{}1$. Να δείξετε ότι $ \frac{1}{(1\plus{}a)(1\plus{}a^2)}\plus{}\frac{1}{(1\plus{}b)(1\plus{}b^2)}\plus{}\frac{1}{(1\plus{}c)(1\plus{}c^2)}\plus{}\frac{1}{(1\plus{}d)(1\plus{}d^2)}\geq 1$.
Enjoy  :wink:

Ισχύει $ \frac{1}{(1\plus{}a^2)(1\plus{}a^4)}\plus{}\frac{1}{(1\plus{}b^2)(1\plus{}b^4)}\geq\frac{1}{1\plus{}(ab)^3}$ (γνωστη ανισοτητα Vasc η αποδειξη με πραξεις)
Με α το $ a^{1/2}$ παιρνουμε
$ \frac{1}{(1\plus{}a)(1\plus{}a^2)}\plus{}\frac{1}{(1\plus{}b)(1\plus{}b^2)}\geq\frac{1}{1\plus{}(ab)^{3/2}}$ και ομοιως
$ \frac{1}{(1\plus{}c)(1\plus{}c^2)}\plus{}\frac{1}{(1\plus{}d)(1\plus{}d^2)}\geq\frac{1}{1\plus{}(cd)^{3/2}}\equal{}\frac{(ab)^{3/2}}{1\plus{}(ab)^{3/2}}$
Με προσθεση κατα μελη των παραπανω εχουμε το ζητουμενο
Γιωργο αυτη την εποχη εχω πολυ δουλεια
Μη με ξενυχτας......

Πολύ ωραία η απόδειξή σας! Από την άλλη δεν είχα σκοπό να σας κουράσω πόσο μάλλον να σας ξενυχτισω :( .
Δείτε και εδώ http://gbas2010.wordpress.com/2010/01/21/inequality-32vasile-cirtoaje/


AMC AIME geometry trapezoid ratio quadratics algebra

A trapezoid ABCD has AB parallel to DC, and DC = AB + 100. The line joining the midpoints of AD and BC divides the trapezoid into two regions with areas in the ratio 2 : 3. Find the length of the segment parallel to DC that joins AD and BC and divides the trapezoid into two regions of equal area.

This is #6 on AIME2 in 2000.
Kalva's solution goes:

Midpoint line has length AB+50. So (AB+75/(AB+25) = 3/2. Hence AB = 75. Suppose distance between AB and CD is h and between requd line and AB is kh. Then length = AB + 100k. So k(AB +50k) = (1/2)(AB+50) or 2k(75+50k) = 125, or 4k2+6k-5=0, k=(√29-3)/4, so length = 75+25√29-75 = 25√29.

When taking the test, I got to the point where AB=75, however I got stuck after that, and I do not understand Kalva's solution.  Can someone clarify how to finish off this problem for me?  Thanks!

[hide="Answer"]$DC=AB+100$, so the length of the segment joining the midpoints is $\frac{AB+(AB+100)}{2}=AB+50$. Further, since that is the mid-segment, the height of $ABCD$ is split in half, so the ratio between the two regions is $\frac{AB+(AB+50)}{2}: \frac{(AB+50)+(AB+100)}{2}=2: 3$, so $\frac{AB+25}{AB+75}=\frac{2}{3}\Rightarrow AB=75$.

Now, to divide the regions into equal area is a little trickier. We know that the area of the entire trapezoid is $\frac{75+175}{2}\cdot h=125h$, where $h$ is the height. Therefore, we must have the area of each smaller trapezoid equal to $\frac{125h}{2}$. Therefore, we have $\frac{75+(75+x)}{2}\cdot (h-k)=\frac{(75+x)+175}{2}\cdot k=\frac{125h}{2}$, where $k$ is the height of the trapezoid with base $DC$. The important thing to notice now is that $\frac{100-x}{100}=\frac{k}{h}\Rightarrow xh=100h-100k$. Manipulating the first two parts of the equation for the areas of the trapezoids, we have $150h+xh=400k+2xk$. Substituting in, we now have $250h-100k=400k+2xk\Rightarrow 250h=500k+2xk=k(500+2x)$. Next, we rearrange the ratio equation to get $k=h\left(\frac{100-x}{100}\right)$, so we have $250h=h\left(\frac{100-x}{100}\right)(500+2x)\Rightarrow 250=500-5x+2x-\frac{x^2}{50}$, and from there we find the quadratic $x^2+150x-12500=0$. Using quadratic formula, we find that $x=-75+25\sqrt{29}$, so $75+x=\boxed{25\sqrt{29}}$.[/hide]

I think kalva's solution is better, but I came up with this on my own, so nyah. :P


algebra
polynomial
quadratics
algebra solved

Is this some known result, where p is a prime number? I've tried up to 7919 with Maple, so it seems to be true.

Yes it's true. More general, every cyclotomic polynomial is irreducible.
See http://www.mathlinks.ro/Forum/viewtopic.php?t=15750 for the prove.

Put $x=y+1$ and apply Eisenstein's criterion.

Is it enough just to substitute x=1, and remark that the value of the given polynomial is the prime p, and so cannot be reduced over Z except in the case where x is a factor (since we have set x=1), but clearly this doesn't hold?

That doesn't suffice, take a look at $(x-5)(x-7)$ and set $x=8$.

Ah thanks.

Are there any nice ways of using the fact that a whole polynomial can be prime?

[quote="ZetaX"]If $p(x)$ is an integer polynomial with degree $n$ and there are $2n+1$ different places $k_{1}, ..., k_{2n+1}\in\mathbb{Z}$ for that $p(k_{i})$ is always $\pm 1$ or $\pm$prime, then $p(x)$ is irreducible.

A bit stronger:
If $p(x)$ is an integer polynomial with degree $n$ and there are $n+1$ different places $k_{1}, ..., k_{n+1}\in\mathbb{Z}$ with $|k_{i}-k_{j}|\geq 2$ for $i \neq j$ such that $p(k_{i})$ is always $\pm 1$ or $\pm$prime, then $p(x)$ is irreducible.[/quote]

Can anyone give more explanation?
thx..

Here's an example of a fourth-degree polynomial which factors into a product of two quadratics, and is prime in 6 places. Seeing how the example works should help clarify why there can't be many more.

$p(x)=(x^{2}-x-1)(x^{2}+x-5)$

This is $\pm$prime at $-2,-1,0,1,2,3$.

The issue is all of those places where a factor is $\pm 1$; how many can there be before that factor must be constant?


I will compete in the AMC 8 for the first time soon. Could you guys give me some questions from the amc 8 to give me an idea of what I'm in for?

Here is a link to the [url=http://www.artofproblemsolving.com/Forum/viewforum.php?f=133]Art of Problem Solving AMC Forum[/url]. There are various mock AMC test questions there.


geometry
3D geometry
number theory proposed
number theory

Solve in $ \mathbb{Z}$ this equation :

$ x^6\plus{}x^3y^3\plus{}y^6\equal{}3^n$

For x=0 infinitely many y-s and vice versa. Now assume $ x,y\ne0$.
Denote $ x^3 \equal{} a$ and $ y^3 \equal{} b$. We have $ a^2 \plus{} ab \plus{} b^2 \equal{} 3^n$. Examining towards a we obtain $ D \equal{} 4.3^n \minus{} 3b^2$. Let $ b \equal{} 3^m.r$, where $ m\in\mathbb{N}$, $ r\in\mathbb{Z}$ and $ \left(r,3\right) \equal{} 1$. So $ D \equal{} 4.3^n \minus{} 3^{2m \plus{} 1}.r^2$. If $ r^2 \ge9$ then $ D > 0$ and $ n > 2m \plus{} 1$ so $ 3^{2m \plus{} 1}|D$ but $ 3^{2m \plus{} 2}$ doesn't so D is not a perfect square. Otherwise $ r^2\le 4$. If $ r \equal{} \pm 2$ then $ b \equal{} \pm 2.3^m$ which is obviously not a cube. Thus it remains $ r \equal{} \pm 1$. So $ D \equal{} 4.3^n \minus{} 3^{2m \plus{} 1}$. If $ n > 2m \plus{} 1$ as above D is not a perfect square. So $ n\le 2m \plus{} 1$. Since $ D\ge 0$ we have $ 2m \plus{} 1 \equal{} n$ or $ 2m \plus{} 1 \equal{} n \plus{} 1$. If $ 2m \plus{} 1 \equal{} n$ we get $ a \equal{} 3^m$ and $ a \equal{} \minus{} 2.3^m$ and since the last can't be a cube it remains $ a \equal{} 3^m$. If $ 2m \plus{} 1 \equal{} n \plus{} 1$ then $ a \equal{} 0$ or $ a \equal{} \minus{} 3^m$. Both of them could be cubes but if $ a \equal{} 0$ then $ x \equal{} 0$. So $ a \equal{} 3^m$ and $ a \equal{} \minus{} 3^m$ and since $ a \equal{} x^3$ we take m divisible by 3. The same for b and y.


$Q$ is  the sum of the digits of a natural number.Prove that $Q(Q(Q(2005^{2005})))=7$.

Can exactly be done like http://www.mathlinks.ro/Forum/viewtopic.php?highlight=4444&t=31409


superior algebra
superior algebra theorems

prove that  a group G   that order(g)=sqr(p) ,p is a prime number    ia solution

You probably mean: show that a group of order $ p^2$, $ p$ prime, is solvable.

In fact, all of them are abelian. Or you prove that every group of order $ p^k$ is solvable.


A bullet of mass 7.00 gis fired horizontally into a wooden block of mass 1.30 kg resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.230. The bullet remains embedded in the block, which is observed to slide a distance 0.300 m along the surface before stopping.

What was the initial speed of the bullet?

1) Momentum Conservation to find the final velocity of Bullet+Block System
$ mv \equal{} (M \plus{} m)v'$

2) Work Done by friction = $ \mu (M \plus{} m)gd$

3) Work Done by Friction = $ \frac {1}{2}(m\plus{}M)v'^2$

Hi I'm sorry. My mistake. I gave a wrong formula for W.D By Friction. :( Just calculated, the answer turns out to be around 218.


In triangle $ABC$, $AB:AC=4:3$ and $M$ is the midpoint of $BC$. $E$ is a point on $AB$ and $F$ is a point on $AC$ such that $AE:AF=2:1$. It is also given that $EF$ and $AM$ intersect at $G$ with $GF=72$. Find the length of $GE$

[hide]
$\frac{GE}{GF}=\frac{\triangle{AFG}}{\triangle{AGE}}=\frac{\triangle{AFG}}{\triangle{AMC}}\cdot \frac{\triangle{ABM}}{\triangle{AGE}}=\frac{AF\cdot AG}{AM\cdot AC}\cdot \frac{AB\cdot AM}{AE \cdot AC}=\frac{2}{3}$

Hence $GE=48$.

Note for this problem: There is an arithmetic sequence formed by the ratio of the segments. Can you find it? :P [/hide]


geometry
circumcircle
area of a triangle

Points $A, B,$ and $C$ are on a circle such that $\triangle ABC$ is an acute triangle. $X, Y ,$ and $Z$ are on the circle such that $AX$ is perpendicular to $BC$ at $D$, $BY$ is perpendicular to $AC$ at $E$, and $CZ$ is perpendicular to $AB$ at $F$. Find the value of \[ \frac{AX}{AD}+\frac{BY}{BE}+\frac{CZ}{CF},  \] and prove that this value is the same for all possible $A, B, C$ on the circle such that $\triangle ABC$ is acute.

[asy]
pathpen = linewidth(0.7);
pair B = (0,0), C = (10,0), A = (2.5,8); path cir = circumcircle(A,B,C);
pair D = foot(A,B,C), E = foot(B,A,C), F = foot(C,A,B), X = IP(D--2*D-A,cir), Y = IP(E--2*E-B,cir), Z = IP(F--2*F-C,cir); D(MP("A",A,N)--MP("B",B,SW)--MP("C",C,SE)--cycle); D(cir); D(A--MP("X",X)); D(B--MP("Y",Y,NE)); D(C--MP("Z",Z,NW)); D(rightanglemark(B,F,C,12)); D(rightanglemark(A,D,B,12)); D(rightanglemark(B,E,C,12));[/asy]

[quote="Altheman"]i was lazy and did not type up my solutions...but i have a [b]really[/b] simple solution for #3
Lemma 1:
in a nonright triangle ABC, $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ (easy to show with a+b+c=180 and expanding)

Proof:
note that $D$ lies on segment $BC$ because the triangle is acute, then by power of a point, $BD*DC=AD*DX \implies \frac{DX}{AD}=\frac{BD*DC}{AD^2}=\cot B*\cot C$, then adding 1, we get $\frac{DX}{AD}+1=\frac{AX}{AD}= \cot B*\cot C+1$ using since there is nothing special about the ratio I chose, the same logic follows for the other ones, adding the relations up,

$\frac{AX}{AD}+\frac{BY}{BE}+\frac{CZ}{CF}=\cot A \cot B+\cot A \cot C+\cot B\cot C+3$
$=\frac{\tan A +\tan B+\tan C}{\tan A\tan B\tan C}+3=\boxed{4}$...i wonder if i could have gotten a commended solution for this?[/quote]

Altheman means  [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=409638#p409638]this[/url]

This is the solution I sent in.  It received a 5.

[hide]Since AX=AD+DX, we have

|

$\displaystyle\frac{AX}{AD}=\displaystyle\frac{AD}{AD}+\displaystyle\frac{DX}{AD}=1+\displaystyle\frac{DX}{AD}$

|

Similarly,

|

$\displaystyle\frac{BY}{BE}=1+\displaystyle\frac{EY}{BE}, \displaystyle\frac{CZ}{CF}=1+\displaystyle\frac{FZ}{CF}$

|

Adding these equations, we get

|

$\displaystyle\frac{AX}{AD}+\displaystyle\frac{BY}{BE}+\displaystyle\frac{CZ}{CF}=3+\displaystyle\frac{DX}{AD}+\displaystyle\frac{EY}{BE}+\displaystyle\frac{FZ}{CF}$

|

$\angle FCB$ and $\angle BAD$ are both complements of $\angle ABC$, therefore they are equal.  $\angle BCX$ subtends the same arc as $\angle BAD$, so they are equal. This means that $\angle BCX$ is equal to $\angle FCB$. By ASA congruency, $CHD\cong CDX$, where H is the orthocenter.

This means that HD=DX, so that we can replace DX by HD in the above equation. Exactly the same argument shows that HF=FZ and HE=EY. We get

|

$\displaystyle\frac{AX}{AD}+\displaystyle\frac{BY}{BE}+\displaystyle\frac{CZ}{CF}=3+\displaystyle\frac{HD}{AD}+\displaystyle\frac{HE}{BE}+\displaystyle\frac{HF}{CF}$

|

Since $[ABC]=[AHB]+[BHC]+[CHA]$, using the formula for the area of a triangle we get

|

$HF*c+HD*a+HE*b=CF*c=AD*a=BE*b$

|

Using this equation, we can write $HF=rCF, HD=sAD, HE=tBE$, where $r+s+t=1$. It is thus found that

|

$\displaystyle\frac{HD}{AD}=s, \displaystyle\frac{HE}{BE}=t, \displaystyle\frac{HF}{CF}=r\Rightarrow 3+\displaystyle\frac{HD}{AD}+\displaystyle\frac{HE}{BE}+\displaystyle\frac{HF}{CF}=3+r+s+t=3+1=4$[/hide]


videos
geometry
geometric transformation
reflection
Support
blogs

The description pretty much says it. People make recordings of their playing and provide sheet music in some way, and others analyze the music, then give input. I think this would be a good way for everyone to improve their playing. Anyone interested?

hmmm, this sounds very interesting.  Unfortunately, making the recording would be very time-consuming for me, but I would love to discuss piano music and talk with other piano players.

Same as Quickster.
Maybe we can also give each other some recommendations
for good pieces?

The minimum quality need for the recording is for some sort of difference in volume to be heard. Crappy laptop, some mp3 players will do. With a crappy laptop I just plug it in and warm up while it's starting. It takes me about 30 minutes total to do one piece after editing, which isn't that bad.

Like istos said, maybe we could recommend some music that you thought was really fun, challenging, etc.
    I don't know how to get sheet music on here or anything, but it would be really cool if we could give each other advice on some of our pieces.
      Does anyone on here happen to be doing a piano competition (I can't remember the name of it at the moment) this November that goes on from states to regionals and nationals?

well, I guess I'll post the pieces I'm working on:

Just started Haydn Piano Concerto in D Major
will be doing a Chopin Nocturne (my first)
Bach Prelude+Fugue number 21 in the WTC

The pieces that I'm working on right now are:

     1. Franz Schubert, March Militaire
     2. Dmitri Kabelevsky, Sonatina No.2, Op.13 movements 1, 2, and 3
     3. Haydn Sonata
     4. Bach Inventions 13 and 14

And pieces that I'm done working on but can still play:

     1. Felix Mendelssohn, Hunting Song
     2. Bela Bartok, Bear Dance
     3. Bach Inventions 1 and... 8?
     4. Chopin, Minute Waltz

All of these songs (I think) are really fun, but I would especially recommend Hunting Song and the Minute Waltz.

And just wondering, who is everyone's all-time favorite composer? I don't really have a favorite, but Mendellsohn wrote some neat songs.

I don't really have a fav composer, I like and dislike songs from most of the composers i've played.

What are people's favorite style? baroque/classical/romantic/20th Century?

hmmmm...i'm actively working on

mendelssohn-variations serieuses
bach-partita no. 5: praeambulum (probably the other movements at some point as well)
shostakovich-concerto no. 2: I (maybe the other movements if i ever find time)
saint-saens-concerto no. 2: I (just decided to start it yesterday  :P )

then i eventually have to bring back premiere comunion de la vierge from vingt regards sur l'enfant jesus by MESSIAEN and the op. 10, no. 3 beethoven sonata for later competitions. i think.

I just switched teachers, so not much:
Beethoven Sonata Op.10 No.2
Chopin Nocturne Op.9 No.1
Actually I'm not working on them, I finished them both in under 5 days so :P
I recently played:
Chopin Waltz Op.34 No.3 (this one is fun :))

I've played all the Bach Inventions, and nearly every Mozart Sonata 1st movement so yea...

How many Bach Inventions are there? I've only played four of them, but I like them cuz Bach didn't put any markings in, so you get to play them however you want.
    And how long has everyone played piano, anyway? And how long do you practice a day?

There's 14 or 15, something like that. I don't remember how many sinfonia.
I've been playing 7 years, now I practice 3 hours a day.

I've played for 10.5 years, and how much I practice really depends on how much I want to. I've gone strings of three days while practicing like 4 hours total in those three days, and I've done 5-6 hours every day of a week a few times (that's how I learned the Sonnetto 104 del Petrarca in like 2 weeks).

I've only played 5 years... and I practice 1-2 hours a day. How old and what grade are you guys in? You seem a lot older than me.

gonna be a freshman nxt yr, played since I was 5-6ish, had some unproductive years in the middle with an easy teacher, I have a very good 1 now.  I usually practice 1 hr a day

(Just saying...Lionblaze does not play piano)

im going to New York this weekend to play in a Concert at Carnegie and Dicapo Opera Theatre
^^

...WHOA WOW THAT IS AMAZING. How did you get to do that?

Ohh, answering Wolfheart's question... i got there by auditioning for the American Fine Art Music Festival. They basically go to almost every state in the US and audition kids. If your good, they send you to Carnegie. If you almost as good, they send you to Dicapo, and if you slightly under that, they put you into a concert hall in your area or somewhere else.

does anyone know much about liszts 8th rhapsody and Ligeti's musica ricercata.
Im playing them and 5 other things for a contest.

[quote="mrperson"]THis song is crazy!! it's for piano. mostly...... well it's written for piano but shouldn't be played by one, you'd brake it.
[youtube]9xbkovCQMmc[/youtube][/quote]

I can play that with one hand.

[quote="jasonzhang"]does anyone know much about liszts 8th rhapsody and Ligeti's musica ricercata.[/quote]
Great to see someone else playing Musica Ricercata!  My piano teacher's private students each learned one a couple years back as part of one of his professional performances honoring Ligeti.  Unfortunately I only learned I, III, and half of VII (it was turned into a duet).  Are you playing all 11 pieces?

[quote="NuncChaos"]Whoo!!

Just got Islamey! Mostly to fool around with, maybe play at small recitals...

Anyways I am now also playing Bach's c minor partita and Rachmaninoff's prelude in g minor... Once I figure out how to record with decent sound I guess I'll post an audio thingy.[/quote]

Islamey!  Practicing that too, but also just for kicks.  It's such a...cool piece.

Having a major major piano relapse in college right now...
Learned Scriabin 5 and Bach WTC I no.12 over winter break, Chopin Ballade 1 over spring break, I need to get around to recording Chopin Ballade 4 and Scriabin 2 too and post them someday.

Think I will try to do Mazeppa and Liszt B minor over the summer.

I'm doing Liszt's Hungarian Rhapsody No.12 which is really time consuming, but I also have Bach's Italian Concerto mvt 1+2, Chopin's Revolutionary Etude, Chopin's Grande Valse Brilliante in e flat major, Debussy's prelude no.8 (i think), Debussy's THe Little Shephard (from the Children's Corner), Staircases by Tom Peterson (contemporary piece i was forced to learn for a competition), Beethoven's Sonata in C minor (not the Pathetique, the other one), and Rachmaninoff's Serenade (dunno what no/key, forgot).
Altogether: 10 pieces for June's piano test. :)

Well i finished up the ricercare :P.
Anyways,  I am working on finishing all 27 chopin etudes by the end of this year, the totentanz by liszt, Les jeux d'eaux à la Villa d'Este by liszt, bach's 6th french suite in e major, the dante sonata by liszt, and La fuente y la campana by mompou.

Chopin Scherzo No. 1 in B Minor.

Also doing Arensky Piano Trio, and soon to start Schumann Carnaval Op. 9 and, if I have time, Chopin Scherzo No. 2.


Julio had four markers and three pencils. He bought the markers for an average of $ \$0.75$ each and the pencils for an average of $ \$0.50$ each. How much money, in dollars, did he spend to purchase all seven of these items?

Note: The fact that it's an average price does not affect the outcome of the problem. Do you see why?

Textangle, I entered 4.5 and I got it wrong. When I went to review, the answer was 4.50. I was like, are you kidding me. So I guess the 0 counts.

ANSWER: 4.50


Do blogs that are marked for deletion ever get deleted? Because I deleted mine a while back and it's still not deleted... can I get it back somehow? (Yes, I am aware that I've deleted my blog like 4 times, so I won't delete it again after this :blush:)


Let a,b,c>0 such that   $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.

Prove that:

   $ (2 \minus{} a)(2 \minus{} b)(2 \minus{} c) \geq \frac {25}{27}$

Let  be $ c\equal{} Min(a,b,c)$ then $ c \in [0,1]$
We have:$ 2(2\minus{}a)(2\minus{}b) \equal{}8\minus{}4(a\plus{}b)\plus{}2ab \equal{} 8\minus{}4(a\plus{}b) \plus{}(a\plus{}b)^2\minus{}a^2\minus{}b^2 \equal{}(a\plus{}b\minus{}2)^2 \plus{}4\minus{}a^2\minus{}b^2 \geq c^2\plus{}1$
Thus, $ f(c) \equal{}(2\minus{}c)\left(\frac{c^2\plus{}1}{2}\right)$
$ f'(c)\equal{}0 \to c\equal{}\frac{1}{3} , c\equal{}1$ Hence, $ Min f(c) \equal{} Min \left( f(0),f(1),f\left(\frac{1}{3} \right) \right) \equal{}\frac{25}{27}$
The equality holds if and only if $ a\equal{}\frac{5}{3},b\equal{}\frac{1}{3},c\equal{}\frac{1}{3}$ or any cyclic permutations.


geometry
3D geometry
sphere
parameterization
rotation
geometric transformation
trigonometry

I have a problem involving Stokes' Theorem where I need to find a parametrization of the intersection of $x^{2}+y^{2}+z^{2}=1$ and $x+y+z=1$.  I have two questions regarding this:

1. How do I find the parametrization?

2. If I were to project this intersection onto the plane $\mathbb{R}^{2}$, how would I find the equation of the projection?  I am assuming it would be an offset circle.

These may be easy questions, but I'm drawing a blank right now so any help would be greatly appreciated :).

hmm, a possible idea is to rotate the plane so that it still has the same cut of the sphere, but it's projection onto $\mathbb{R}^2$ is a regular circle (then you can go with the regular polar parametrization).

Do you know anything about rotation matrices?

Yeah, but I was really just looking for the cylinder which when intersected with the sphere would give the same intersection as the plane.

I think a paramaterization would be:

$(x(t),y(t),z(t)) =$
$\frac{1}{3}(1 + \sin t + \sqrt 3 \cos t, 1 + \sin t - \sqrt 3 \cos t, 1 - 2\sin t)$

To project this onto the x-y plane you would just set the z-coordinate of the above equal to zero.

OK, well that seems to work with what I am doing.  Could you please show how you arrived at that though?  Thanks a lot :).

Edit:  Did you just rotate $x^{2}+y^{2}+\left(1-x-y\right)^{2}=1$ in order to eliminate the $xy$ term?

I rotated the parametric curve/family of vectors $\frac{\sqrt 6}{3}(\cos t, \sin t, 0)$ by $\cos^{-1} (\sqrt 3/3)$ degrees around the x-axis, and then $45^o$ around the z-axis. Then I added the vector $(1/3, 1/3, 1/3)$ to shift the center of the circle to the right position.

[I did this rotation because it sends the vector (0,0,1) to the vector direction (1,1,1), i.e. it sends the normal vector of the circle in the x-y plane to the normal vector of the circle that you want to parameterize. The rigid rotations ensure that it stays a circle with the proper size].

Yes of course.  All I need are two orthogonal vectors to find the projection, which is an ellipse with radii of $\sqrt{\frac{2}{3}}$ and $\frac{\sqrt{2}}{3}$ respectively.

Thanks! :)


I was browsing wikipedia, and this (see image) randomly came up, around 6:30 PM...

[quote="pianoforte"]Seriously. It was weird.

Oh, and if anyone could translate that (assuming it has any meaning), thanx.[/quote]

O_O exactly what were u searching???

its just a random configuration of the words: Fire, Water, Wind, Cow, Gold, Darkness, Fur, and Earth. ASSUMING its Chinese

OMG that is like so scary even for comp hackers to do

mz: Some stuff pertaining to Avatar the Last Airbender  :oops: . Some random page, I accidentally clicked back and couldn't get to the thing again.

The wikipedia template was weird, you could see the Wikipedia logo and search bar on the side (this particular pic was scrolled down, so you can't see the Wikipedia logo - I have another one with the Wikipedia logo, but you can't read what it says). The rest of the page was static.

I just realized that "Zodiac" refers to a serial murderer, whose symbol was the one in the pic I showed. The symbol (it's celtic) linked to the wikipedia article on it (I didn't bother to click it though).

I should've checked the source code.

Edit: http://piccu.livejournal.com/32430.html

for someone else who saw the same. Also Avatar related.

Edit 2: similar event, a few months ago, and today: http://forums.xkcd.com/viewtopic.php?f=2&p=771274

Edit 3: http://en.wikipedia.org/w/index.php?title=Template:Contains_Chinese_text&oldid=226745000

lmao i should do that to a site people surf on alot but has a low security.....thast creepy though, so far AoPS is ok  :P

Um, you do know who the Zodiac killer is, right?  Some crazy murderer dude in California who has yet to be caught, and the circle with the cross is his symbol, so...yeah....creepy.  PRobably just some guy who thought it was funny to creep you out with a bunch of chinese characters.  I think... :maybe:

Oh, so it wasn't the entire website being hacked. Just a single page.

weird . . .

probably just a hack, nothing that wikipedia can't take care of

but, i read thru the whole conversation from that xkcd link that pianoforte posted up there, and it's pretty scaaaary...

hm. maybe... a STOLEN computer?!?!?! just a thought, tho.

sorry, if im talking on an old topic, but dreambold, I think that if he had a stolen computer he'd have to connect it to the internet, and that would make him have to have a IP address away from his home if he wanted to stay anonymous. That would take either walking into a cafe or taking the time to set up an internet connection out in some hideaway in the woods.

In conclusion: This trick was probably created by some random guy who looked completely normal, not the Zodiac.

wow that was one of my longest posts.

You haven't seen the news lately, have you?
Becuase if you did, you would know who "the Zodiac" was

i dont know who the "zodiac" is. but he's some serial killer dude, and thats all i need to know to go hide under my covers... :(

Ah-ha! After searching through for 3 minutes, I found this page:

http://en.wikipedia.org/w/index.php?title=User_talk:125.76.226.12&diff=prev&oldid=226747092

Apparantly, it just seems to be vandalism, nothing more.

[quote="crazychinesemaniac"]I didnt feel like reading the entire chinese stuff. But the first line (reading like how chinese people do) says "All the cities shall be doused in flames." Sounds a little extremist, doubt its the Zodiac Killer though. It would be pretty stupid of him seeing as they could track his IP assuming he did this from his home. But he could hardly walk into a cafe.[/quote]

they could use proxies...

I'm going to revive this topic because something very similar happened to me. Here's the video.

[youtube]VExNOjztgVA[/youtube][/url]


Find $ x; y; z \in R$ of this system: $ \begin {cases} x\plus{}\frac{1}{y}\equal{}2\minus{}(y\minus{}z)^2\\ y\plus{}\frac{1}{z}\equal{}2\minus{}(z\minus{}x)^2 \\z\plus{}\frac{1}{x}\equal{}2\minus{}(x\minus{}y)^2 \end {cases}$

hello, the only real solution i have found is $ x\equal{}y\equal{}z\equal{}1$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, the only real solution i have found is $ x \equal{} y \equal{} z \equal{} 1$.
Sonnhard.[/quote]
How do you prove it?

$ \Rightarrow  (x\plus{}\frac{1}{y})\plus{}(y\plus{}\frac{1}{z})\plus{}(z\plus{}\frac{1}{x})\equal{}6\minus{}[(y\minus{}z)^2\plus{}(z\minus{}x)^2\plus{}(x\minus{}y)^2]$ and: $ 6\minus{}[(y\minus{}z)^2\plus{}(z\minus{}x)^2\plus{}(x\minus{}y)^2]\le 6.$   (1)
[b]1).[/b] If $ x;y;z>0$, then:
$ x\plus{}\frac{1}{x}\ge 2; \ y\plus{}\frac{1}{y}\ge 2; \ z\plus{}\frac{1}{z}\ge 2\Rightarrow \\
\Rightarrow (x\plus{}\frac{1}{y})\plus{}(y\plus{}\frac{1}{z})\plus{}(z\plus{}\frac{1}{x})\equal{}(x\plus{}\frac{1}{x})\plus{}(y\plus{}\frac{1}{y})\plus{}(z\plus{}\frac{1}{z})\ge 6.$   (2)
And (1), (2) $ \Rightarrow (x\plus{}\frac{1}{y})\plus{}(y\plus{}\frac{1}{z})\plus{}(z\plus{}\frac{1}{x})\equal{}6\minus{}[(y\minus{}z)^2\plus{}(z\minus{}x)^2\plus{}(x\minus{}y)^2]\equal{}6$   (3)
 $ \Leftrightarrow (y\minus{}z)^2\plus{}(z\minus{}x)^2\plus{}(x\minus{}y)^2\equal{}0  \Leftrightarrow x\equal{}y\equal{}z.$ (4)
(3) and (4) $ \Rightarrow 3.(x\plus{}\frac{1}{x})\equal{}6\Rightarrow x\plus{}\frac{1}{x}\equal{}2\Rightarrow x\equal{}1.$ (5)
(4) and (5)$ \Rightarrow x\equal{}y\equal{}z\equal{}1.$

With Maple:
We obtain $ p(z)(z\minus{}1)\equal{}0$ where $ p$ is an irreducible polynomial of degree 21.
$ p$ has only 3 real roots: $ a_1\approx{\minus{}4.166},a_2\equal{}\approx{\minus{}2.641},a_3\approx{\minus{}0.08316}$. $ x,y$ are also roots of $ p$.
Conclusion: there exists an unique (up to a circular permutation) negative solution:
$ x\equal{}a_1,y\equal{}a_2,z\equal{}a_3$.
Beware $ x\equal{}a_2,y\equal{}a_1,z\equal{}a_3$ is not a solution.

I think the system is easier than you suppose.
I'll give hints here.

Take look at the cases some two of the variables are equal.
Take a look at the case all three of the variables are equal.

Now we shoult take a look at the remaining case - all variables are
different.

A) substract equation 1) and 2) - we get 1')
B) substract equation 2) and 3) - we get 2')
C) substract equation 3) and 1) - we get 3')

D) substract 1') and 2') - we get 4')

E) add 1'), 2'), 3') - we get 5')

F) add 1), 2), 3) - we get 6')

Using 4'), 5'), 6') we can get a new system for:
x+y+z
xyz
xy+yz+zx

[quote="Dr Sonnhard Graubner"]hello, the only real solution i have found is $ x \equal{} y \equal{} z \equal{} 1$.
Sonnhard.[/quote]

Hi Dr Sonnhard Graubner. Can you post your solution of this system?


[url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php#coursedetails3]Here[/url], it says that future dates haven't been scheduled yet.

[quote="SonyWii"]When is AoPS next offering the Interm C&P class? It's not on the "Enroll" page...  :([/quote]

We'll offer it in summer, 2010.


inequalities
function
inequalities proposed

\[ \ x,y,z\geq 0 ; x \neq y \neq z\]
Prove that : \[ \frac{x^2\plus{}2yz}{(y\minus{}z)^2}\plus{}\frac{y^2\plus{}2zx}{(z\minus{}x)^2}\plus{}\frac{z^2\plus{}2xy}{(x\minus{}y)^2}\geq 4\]
 :rotfl:  :rotfl:

[quote="mai quoc thang"]
\[ \ x,y,z\geq 0 ; x \neq y \neq z
\]
Prove that :
\[ \frac {x^2 \plus{} 2yz}{(y \minus{} z)^2} \plus{} \frac {y^2 \plus{} 2zx}{(z \minus{} x)^2} \plus{} \frac {z^2 \plus{} 2xy}{(x \minus{} y)^2}\geq 4
\]
:rotfl:  :rotfl:[/quote]

By the Cauchy Schwarz we get
\[ \left( {\sum {\frac {{x^2 \plus{} 2yz}}{{\left( {y \minus{} z} \right)^2 }}} } \right)\left( {\sum {\left( {x^2 \plus{} 2yz} \right)\left( {y \minus{} z} \right)^2 } } \right) \ge \left( {\sum {x^2 \plus{} 2yz} } \right)^2
\]
It suffices to show that
\[ \left( {\sum {x^2 \plus{} 2yz} } \right)^2 \ge 4\left( {\sum {\left( {x^2 \plus{} 2yz} \right)\left( {y \minus{} z} \right)^2 } } \right)
\]
We have
\[ \sum {\left( {x^2 \plus{} 2yz} \right)\left( {y \minus{} z} \right)^2 } \equal{} 2\sum {xy\left( {x^2 \plus{} y^2 } \right)} \minus{} 2\sum {x^2 y^2 } \minus{} 2xyz\sum x
\]
Assume $ x\plus{}y\plus{}z\equal{} 3$ then it becomes
\[ 81 \ge 4\left( {8r \minus{} 6q^2 \plus{} 18q} \right)
\]
Using the well-known $ r \le \frac {{q^2 }}{9}$ we only need to show that
\[ 81 \ge 4\left( {\frac {{8q^2 }}{9} \minus{} 6q^2 \plus{} 18q} \right)
\]

\[ \Leftrightarrow 184q^2 \minus{} 648q \plus{} 729 \ge 0
\]
Then it's done. :)

Equality does not hold.

\[ \ x,y,z\geq 0 ; x \neq y \neq z
\]
k is the max constant such that:

\[ \frac {x^2 \plus{} 2yz}{(y \minus{} z)^2} \plus{} \frac {y^2 \plus{} 2zx}{(z \minus{} x)^2} \plus{} \frac {z^2 \plus{} 2xy}{(x \minus{} y)^2}\geq k
\]
k=???
 :rotfl:  :rotfl:

[quote="arqady"][quote="mai quoc thang"]
\[ \ x,y,z\geq 0 ; x \neq y \neq z
\]
Prove that :
\[ \frac {x^2 \plus{} 2yz}{(y \minus{} z)^2} \plus{} \frac {y^2 \plus{} 2zx}{(z \minus{} x)^2} \plus{} \frac {z^2 \plus{} 2xy}{(x \minus{} y)^2}\geq 4
\]
:rotfl:  :rotfl:[/quote]
I think it's wrong. Try $ x \equal{} 1,$ $ y \equal{} 2$ and $ z \equal{} 1.$ :wink:  :rotfl:[/quote]
hey ..... \[ \ x \neq y \neq z \]

[quote="fjwxcsl"]$ kmax \equal{} \frac {5(1 \plus{} \sqrt5)}{2}$.[/quote]

$ k_{\max }$ is not $ \frac{{5\left( {1 \plus{} \sqrt 5 } \right)}}{2}$. You can try with $ x\equal{}1,y\equal{}2,z\equal{}0$.

Let

                  $ x \equal{} 0,y \equal{} 1,z \equal{} \frac {3 \plus{} \sqrt5}{4} \plus{} \frac {\sqrt { \minus{} 2 \plus{} 6\sqrt5}}{4}$,

then
                        $ k\leq\frac {5(1 \plus{} \sqrt5)}{2}$;

and if $ k \equal{} \frac {5(1 \plus{} \sqrt5)}{2}$,  we  have

     $ \frac {x^2 \plus{} 2yz}{(y \minus{} z)^2} \plus{} \frac {y^2 \plus{} 2zx}{(z \minus{} x)^2} \plus{} \frac {z^2 \plus{} 2xy}{(x \minus{} y)^2}\geq k$

     <=>

     $ 2G61 \minus{} 2G62 \plus{} 6G63 \minus{} 4G64 \plus{} 4G65 \minus{} 5(1 \plus{} \sqrt5)G66$

       $ \equal{} q(x,y,z)(x \minus{} y)(x \minus{} z) \plus{} q(y,z,x)(y \minus{} z)(y \minus{} x) \plus{} q(z,x,y)(z \minus{} x)(z \minus{} y)\geq0$,

 in which

     $ q(x,y,z) \equal{} 2x^4 \minus{} 2x^3(y \plus{} z) \plus{} 6x^2yz \minus{} 4xyz(y \plus{} z) \plus{} 4y^2z^2$

     $ \minus{} 5(1 \plus{} \sqrt5)x^2(y \minus{} z)^2$.

    Suppose

          $ x > y,y > z,x \equal{} a \plus{} b \plus{} c,y \equal{} a \plus{} b,z \equal{} a\geq0, b,c > 0$,
then

     $ q(x,y,z) \minus{} q(y,z,x) \plus{} q(z,x,y)$

     $ \equal{} (a^2bc \plus{} ab^2c \plus{} abc^2)\sqrt {105} \plus{} 10b^2c^2 \plus{} 6bc^3 \plus{} 4b^4 \plus{} 2c^4 \plus{} 6a^2b^2 \plus{} 6a^2c^2$

        $ \plus{} 8ab^3 \plus{} 4ac^3 \plus{} 12b^3c \plus{} 28a^2bc \plus{} 38ab^2c \plus{} 30abc^2\geq0$,

  threfore

             $ q(x,y,z)(x \minus{} y)(x \minus{} z) \plus{} q(y,z,x)(y \minus{} z)(y \minus{} x) \plus{} q(z,x,y)(z \minus{} x)(z \minus{} y)\geq0$,

and  we  have

              $ k_{max} \equal{} \frac {5(1 \plus{} \sqrt5)}{2}$.

The first slight, I have same opinion with fjwxcsl but after I read the post of Honey_S, everything in my mind was changed!? Because we have:
\[ f(a,b,c)\equal{}LHS\]
So if we try:
\[ f(1,2,0)\equal{}8.25> \frac{5(1\plus{}\sqrt{5})}{2}\equal{}8.090169944\]
It means $ \frac{5(1\plus{}\sqrt{5})}{2}$ is not the best constant such that above inequality holds
But if we prove that the inequality holds when at least one variable equal to 0, without loss of generally we choose $ c\equal{}0$ we will find the minimum value of the following function:
\[ f(a,b,0)\equal{}\frac{a^2}{b^2}\plus{}\frac{b^2}{a^2}\plus{}\frac{2ab}{(a\minus{}b)^2}\]
\[ \equal{}f\left(\frac{a}{b}\right)\equal{}f(x)\equal{}x^2\plus{}\frac{1}{x^2}\plus{}\frac{2x}{(1\minus{}x)^2}\]

Easily to check it out the minimum value of $ f(x)\equal{}\frac{5(1\plus{}\sqrt{5})}{2}$.
I guess that the equaly occurs when 3 variables are different from with each other and don't have any variable equal to 0 !!! Am I wrong?!

I think honey_S's mistake is we don't have:
\[ f(a,b,c)\ge 8.25,\forall a,b,c\ge 0\]
I think we can find some example to define above inequality is not correct ;)

[quote="mai quoc thang"]
hey .....
\[ \ x \neq y \neq z
\]
[/quote]
$ 1\neq2\neq1$ is true!

:D  .... \[ \ x \neq y  ;  y \neq z ;  z \neq x \]   .... :rotfl:  :rotfl:

[quote="mai quoc thang"]
\[ \ x,y,z\geq 0 ; x \neq y \neq z
\]
Prove that :
\[ \frac {x^2 \plus{} 2yz}{(y \minus{} z)^2} \plus{} \frac {y^2 \plus{} 2zx}{(z \minus{} x)^2} \plus{} \frac {z^2 \plus{} 2xy}{(x \minus{} y)^2}\geq 4
\]
:rotfl:  :rotfl:[/quote]
Assume $ z \equal{} \min\{x,y,z\}$, then we have
\[ 0 < y \minus{} z \le z \Rightarrow \frac {x^2 \plus{} 2yz}{(y \minus{} z)^2} \ge \frac {x^2}{(y \minus{} z)^2} \ge \frac {x^2}{y^2}
\]
Similarly, we have
\[ \frac {y^2 \plus{} 2zx}{(z \minus{} x)^2} \ge \frac {y^2}{x^2}
\]
Thus
\[ LHS \ge \frac {x^2}{y^2} \plus{} \frac {y^2}{x^2} \plus{} \frac {2xy}{(x \minus{} y)^2}
\]
Setting $ t \equal{} \frac {x}{y} \plus{} \frac {y}{x} > 2$, then
\[ LHS \ge t^2 \minus{} 2 \plus{} \frac {2}{t \minus{} 2} \equal{} f(t)
\]
We have
\[ f'(t) \equal{} 2t \minus{} \frac {2}{(t \minus{} 2)^2}
\]

\[ f'(t) \equal{} 0 \Leftrightarrow t \equal{} \frac {3 \plus{} \sqrt {5}}{2}
\]
From this, we can easily check that
\[ f(t) \ge f\left( \frac {3 \plus{} \sqrt {5}}{2}\right) \equal{} \frac {5(\sqrt {5} \plus{} 1)}{2}
\]
Hence
\[ LHS \ge \frac {5(\sqrt {5} \plus{} 1)}{2}
\]
It is easy to check the equality.
We have done. :)

Let $x,y,z $ be three non-negative distinct real numbers. Prove that$$ \frac{x^2+yz}{(y-z)^2}+\frac{y^2+2zx}{(z-x)^2}+\frac{z^2+2xy}{(x-y)^2}\geq \frac{5(1+\sqrt5)}{2}$$


LaTeX
inequalities
inequalities proposed

[quote="hxy09"]I added an attachment for I am not good at Latex :oops:
Try to solve it Mathlinkers![/quote]
Try $ a \rightarrow 0$.

Sorry  :oops:
run through it
change >= to <=
sorry again
Nak could you please help me to make a Latex for the correct problem?thank you!

Let $ a,b,c \in \mathbb R^{\plus{}}$, $ a \plus{} b \plus{} c \equal{} 3$ Then prove:

$ \frac {a}{2b \plus{} 1} \plus{} \frac {b}{2c \plus{} 1} \plus{} \frac {c}{2a \plus{} 1} \leq \frac {1}{abc}$

[quote="Nak"]Let $ a,b,c \in \mathbb R^{ \plus{} }$, $ a \plus{} b \plus{} c \equal{} 3$ Then prove:

$ \frac {a}{2b \plus{} 1} \plus{} \frac {b}{2c \plus{} 1} \plus{} \frac {c}{2a \plus{} 1} \leq \frac {1}{abc}$[/quote]
It's obviously trues because: $ 2(ab^2 \plus{} bc^2 \plus{} ca^2) \plus{} 3abc \leq\ 9$
 :wink:

:oops:
You are right.This ineq is not hard(I have to admit it)
I use sum(a^2b)+abc<=4 then pqr with SCHUR and (sum(ab))^2>=3abcsum(a).It gose trivial.
but I think my solution is not ''Rapid''
Thank you!

[quote="hxy09"]:oops:
You are right.This ineq is not hard(I have to admit it)
I use sum(a^2b)+abc<=4 then pqr with SCHUR and (sum(ab))^2>=3abcsum(a).It gose trivial.
but I think my solution is not ''Rapid''
[/quote]
My full proof: Using Cauchy Schwarts:
$ \sum\ \frac {a}{2b \plus{} 1} \leq\ \sum\ \frac {a(\frac {2}{b} \plus{} 1)}{9} \leq\ \frac {1}{abc}$
<=> $ 2(ab^2 \plus{} bc^2 \plus{} ca^2) \plus{} 3abc \leq\ 9$
We have done  :)
Sorry, I had a mistakes in type  :oops:


calculus
integration
limit
logarithms
college contests

Let $ 0 < a < b $.
a) Find integral  $ I(\lambda) = \int^1_0[bx+a(1-x)]^\lambda dx $
b) Find  $lim_{\lambda -> 0} [I(\lambda)]^{\frac 1{\lambda}}$.

$I(\lambda)=\int_0^1{[(b-a)x+a]^\lambda}dx=\frac 1{b-a}\int_a^bx^{\lambda}dx=\frac 1{b-a}\frac{b^{\lambda+1}-a^{\lambda+1}}{\lambda+1}$.

We only have to find $\lim_{\lambda\to 0}(\frac{b^{\lambda+1}-a^{\lambda+1}}{b-a})^{\frac 1\lambda}$, since $\lim_{\lambda\to 0}(\frac 1{\lambda+1})^{\frac 1\lambda}=\frac 1e$. This limit is equal to $e^{\frac{b\ln b-a\ln a}{b-a}}$, so the limit we're asked to calculate is $e^{\frac{b\ln b-a\ln a}{b-a}-1}$.

There might be some computation mistakes though..

To grobber: [b]a[/b] :a bit mistake: $\lambda =-1$ and $\lambda \neq -1$


analytic geometry
probability
geometry
graphing lines
slope
geometric transformation
reflection

[size=200]Sign-ups still allowed![/size]
[b]
[size=200]Intermediate Chess Marathon[/size][/b]
[b][size=150]Rules[/size][/b]
The chess game will proceed as follows. Black will propose a question ( screened by an admin or mod for intermidiate difficulty) to white. The first team member on white to answer correctly and with a good proof( if a proof question) will be allowed to make the first move on a chess board I will photoshop when the game begins. White will then propose a question to black, with the team member on black that answers correctly being allowed to make the move for that turn. If the problem is not solved within 72 hours of being proposed, the team trying to solve it forfits its turn. All othe rules are exactly the same as regular chess.

[b][size=150]Sign up[/size][/b]
Post a message to sign up.

[b]Teams so far[/b]
White
-------
1)PenguinIntegral
2)vulgarfraction
3)solafidefarms
4)Altheman
5)Jimhu

Black
--------
1)Chess64
2)krustyteklown
3)Boy Soprano II
4)Melissa
5)4everwise

yes, nothing better to spur people to do math problems than more competition
lol, i'll get in, altho i'll prob only do the problem solving part, and not the chess planning : |

[quote="4everwise"][quote="PenguinIntegral"][size=200]Intermidiate Chess Marathon![/size][/quote]I'm sorry, but I don't think that we can have a Chess Marathon in the Intermediate Math forum... ;)[/quote]
Chess is only the hook. It will make problem solving fun(er).

[quote="4everwise"][quote="PenguinIntegral"]Chess is only the hook. It will make problem solving fun(er).[/quote]All right, I'll let it slide.  :)

I just hope this doesn't start a bunch of new marathons that eventually get less and less math related.  ;)[/quote]
I'm planning to start a 2 team competition that is pure math if this goes well, but otherwise, no.

[quote="probability1.01"]OK, but I don't think this is in the "Intermidiate" forum :D.[/quote]
Do you want to sign up? I can put you on Chess's team.

I'm in. I don't mind which team.

However, what did you mean [i]'forfeit'[/i] their turn? Because you can't really do that, and the team answering the question can manipulate that to avoid that condition with a German name where you are forced to play to your own disadvantage because those are the only legal moves, in the later stages of a game. I suppose you can make it so that the team posing the question could move for them, but that would be a bit unfair- that's usually gaining a very good piece or even forcing a checkmate.

If you can think of some other way to make a time control that is better, I will edit the rules.

Eh, why not? Seems fun. Penguin, you can toss a coin to determine which team I'm on.

[quote="Boy Soprano II"]I'll join.[/quote]
You're in.

our move is g1-f3 (i.e. the king side knight to f3)

$\begin{tabular}[t]{cc} White&Black\\ 1.e2-e4 &2.c7-c5\\ 3.g1-f3 &4.\\ 5.&6.\\ \end{tabular}$

(no reason why we cannot use fully algebraic notation)

[quote="Altheman"]$\begin{tabular}[t]{cc} White&Black\\ 1.e2-e4 &2.c7-c5\\ 3.g1-f3 &4.\\ 5.&6.\\ \end{tabular}$[/quote]

It should be

\[ \begin{tabular}[t]{lr} White&Black\\ 1.e2-e4 & c7-c5\\ 2.g1-f3 & \\ 3.&\\ \end{tabular} \]

[quote="Altheman"][quote="chess64"]

[size=150]Problem 2 for White[/size]

Prove that the sum of the medians of a triangle lie between $\frac{3}{4}p$ and $p$, where $p$ is the perimeter of the triangle.

(Thanks to Boy Soprano)[/quote]

[hide="solution"]Let a triangle be vectors $2u$, $2v$ and $2u-2v$, the sum of the sides is $|2u|+|2v|+|2u-2v|=P$. Then the midpoints are $u$,$v$,$u-v$. Thus the medians (by subtracting the vectors) are $2u-v$, $2v-u$ and $u+v$ and the sum of the medians is $|2u-v|+|2v-u|+|u+v|=S_m$.
By the triangle inequality,
$|2u-v|+|u+v| \ge |3u|$, $|2v-u|+|u+v| \ge |3v|$, and $|2u-v|+|u-2v| \ge |3u-3v|$ add them together to get:
$2(|2u-v|+|2v-u|+|u+v|) \ge \frac{3}{2}*(|2u|+|2v|+|2u-2v|) \implies S_m\ge \frac{3}{4}*P$
again, by the triangle inequality, $|2u|+|2v| \ge |2u+2v|$, $|2v|+|2u-2v| \ge |4v-2u|$, and $|2u|+|2u-2v| \ge |4u-2v|$ add them together to get:
$2( |2u|+|2v|+|2u-2v|) \ge 2( |2u-v|+|2v-u|+|u+v| ) \implies P \ge S_m$, notice that the equality condition is when the vectors have the same direction, but that would make the triangle degenerate, so we have

$P > S_m > \frac{3}{4}*P$[/hide][/quote]
[hide="Alternate Solution"]  Let the triangle be $ABC$, with medians $AA'$, $BB'$, and $CC'$, with centroid $M$.

To prove that the sum of the medians is greater than $\frac{3}{4}p$, we apply the triangle inequality to $AMB$ with respect to $AB$.  Adding three such inequalities gives us

$\frac{4}{3}(m_{a}+m_{b}+m_{c})>p$.

To prove that the sum is less than $p$, we extend $ABC$ to parallelogram $ABA''C$, and apply the triangle inequality with respect to $AA''=2m_{a}$.  Adding three such inequalities gives us

$2(m_{a}+m_{b}+m_{c})<2p$.

Q.E.D.[/hide]

Um...
If a, b, c, d.... are integers, prove that (x-a)(x-b)(x-c).....+1 can not be writen as the product of polynomials with integer coefficients.

[hide="Start of a solution"]
The last term of the product is a*b*c*d*... + 1. It can't be divisible by any of a, b, c, d, ... , because it always leaves a remainder of 1. Notice also that |a*b*c*d*... + 1| is the absolute value of the product of the roots, so the roots cannot be a, b, c, d, ... . In other words, the polynomial cannot have the factors a, b, c, d, ... , which are the integers.

Someone else can finish, I have to go :(

It seems to be pretty much the same as Euler's proof for infinity primes though.
[/hide]

Clearly we are talking about polynomials with degree greater than one, but you also need to stipulate that the polynomials are distinct :[hide]With the quadratic case, it is easy to prove that it is necessary and sufficient for $a$ to be odd and $b=a\pm 2$ if it is to factor as the product of two binomials.  E.g., $a=1,b=3$ yields the quadratic $x^2-4x+3\Longrightarrow x^2-4x+4=(x-2)(x-2)$.[/hide]

hmm looks like a contradiction to me : |

I'm thinking black forfits their move...

I DON'T KNOW CHESS BUT I'M PRETTY GOOD AT MATH

For the incredibly massive necro...

This is dead and has been for a while.


calculus
integration
Ring Theory
superior algebra
superior algebra unsolved

In a ring R x^n =x for all x for some n>1.Show that every prime ideal is maximal.

This is an exercise in Atiyah-MacDonald. But they assumes that rings are commutative and with unit. I can show how the proof in this case follows:

Let $ p$ be a prime ideal of $ A$. Form $ A/p$, which will be an integral domain. Given any non-zero $ x\in A/p$, there will be a suitable $ n\in\mathbb{N}\setminus \{ 1\}$, such that $ x^n \equal{} x$ or equivalently $ x(x^{n\minus{}1} \minus{} 1) \equal{} 0$. This implies that $ x^{n\minus{}1} \equal{} 1$, hence that $ x$ is invertible. Therefore, $ A/p$ is a field and thus $ p$ is a maximal ideal.


Find integer solutions of the system
$4x^{3}+y^{2}= 16$
$z^{2}+yz = 3.$

[hide="Hint"]Notice that $z$ is divisible by 3 so $z\in \{\pm 1;\pm 3\}$
Each value of $z$ give us a value of $x$ and a value of $x$. And all of them must be integers.
[/hide]
[hide="But"]There is no roots[/hide]


Suppose $ m$ and $ n$ are positive odd integers. Which of the following must also be an odd integer? \\ \\
$ \text{(A)}\ m\plus{}3n \qquad \text{(B)}\ 3m\minus{}n \qquad \text{(C)}\ 3m^2\plus{}3n^2 \\ \text{(D)}\ (mn\plus{}3)^2 \qquad \text{(E)}\ 3mn$ \\ \\
Select the most appropriate letter.

To be more general and clear, recall a few rules:
[list][*]An odd number times an odd number equals an even number.
[*]An odd number minus an odd number equals an even number.
[*]An odd number plus an odd number equals an even number.
[*]An odd number squared results in an odd number[/list]
So...
[hide="$A$ is Even"]$m+3n=odd+3\times{odd}=odd+odd=\boxed{even}$[/hide]

[hide="$B$ is Even"]$3m-n=3\times{odd}-odd=odd-odd=\boxed{even}$[/hide]

[hide="$C$ is Even"]$3m^2+3n^2=3\times{odd^2}+3\times{odd^2}=3\times{odd}+3\times{odd}=odd+odd=\boxed{even}$[/hide]

[hide="$D$ is Even"]$(mn+3)^2=(odd\times{odd}+3)^2=(odd+3)^2=(even)^2=\boxed{even}$[/hide]

[hide="$E$ is Odd"]$3mn=3\times{odd}\times{odd}=3\times{odd}=\boxed{odd}$[/hide]

Therefore, $\boxed{E}$ is the correct answer.