vector
trigonometry
ratio
analytic geometry
graphing lines
slope

Consider a river of width $ a$ located on the land to a latitude$ \lambda$. In the river flows waters down with a speed$ v_0$ . Demonstrate that the water level to the right side of the river is superior to that of the left side in a quantity:

$ \Delta h$ $ \equal{}$ $ \dfrac{2a \Omega v_0 Sin(\lambda)}{g}$

$ \Omega$ $ \equal{}$ Angular speed of the Earth

Solution.

We must assume that terms of order $ \Omega^2$ are negligible. This means that the centripetal force is negligble in comparison with the coriolis force, and can be ignored.

Let's define a set of orthogonal, unit basis vectors on every point on the Earth as follows:

$ \bf{e_1}$ directed west to east.
$ \bf{e_2}$ directed south to north.
$ \bf{e_3}$ directed in the outwards radial direction.

The river has velocity $ {\bf v_0} = (\, \pm v_0 \, , \, 0 \, , \,0 \,)$. The sign depends on whether the river is flowing west to east, or east to west. In the following argument, the top symbol (in the $ \pm$ or $ \mp$) always represents west to east.

So the angular velocity is $ {\bf \omega} = (\, 0 \, , \, \Omega \cos{\lambda} \, , \, \Omega \sin{\lambda} \,)$. This holds for all latitudes, positive and negative.

The only fictitious force is the coriolis. (The water is not accelerating as measured by an observor corotating with the Earth, thus the "apparent force" term dissapears). We introduce the mass term $ m$, which is the mass assosiated with some elementary volume. But this really isn't important, and serves more importantly as keeping dimensional consistancy.

$ {\bf F} = m(2 {\bf \omega} \times {\bf v_0}) = m(\, 0 \, , \, \pm 2 \Omega v_0 \sin{\lambda} \, , \, \mp 2 \Omega v_0 \cos{\lambda} \, )$

The two forces acting on the water at the surface is a sort of "physical" force, $ {\bf T}$ (rather like tension in a string) from the water underneath, and the gravitational force $ m{\bf g}$. Note that $ {\bf g}=(\, 0 \, , \, 0, \, -g \, )$. It is a well established fact that the equipotential surfaces of a constant (in time and space) vector field, are orthogonal, flat planes. Since the water surface forms and equipotential of $ {\bf T}$, then what we ultimately require is the direction of $ {\bf T}$.

$ {\bf F} = {\bf T} + m{\bf g}$

$ \Rightarrow {\bf T} = m(\, 0 \, , \, \pm 2 \Omega v_0 \sin{\lambda} \, , \, g \mp 2 \Omega v_0 \cos{\lambda} \,)$

We now come to finding the gradient of $ {\bf T}$ which is the ratio of it's components in the positive directions of $ {\bf e_3}$ (up) and $ {\bf e_2}$ (north).

$ m_{T} = \frac{g \mp 2 \Omega v_0 \cos{\lambda}}{ \pm 2 \Omega v_0 \sin{\lambda}}$

So the perpendicular gradient, in this $ ({\bf e_2}, {\bf e_3})$ plane gives the slope of the river surface as height over width. This is because the width is in the north-south direction.

$ m_{surface} = -\frac{1}{m_{T}} = \frac{\mp 2 \Omega v_0 \sin{\lambda}}{g \mp 2 \Omega v_0 \cos{\lambda}}$

The term ${ 2 \Omega v_0 \cos{\lambda}}$ is very small in comparison with $ g$, and the denominator is large in comparison to the numerator so this can be written as:

$ m_{surface} = \frac{\mp 2 \Omega v_0 \sin{\lambda}}{g}$

We take a minus if the river flows west to east, and a plus if the river flows east to west. (by the top sign, bottom sign consistancy used).

Assume that the river flows on the northern hemisphere as to make $ \lambda$ and hence $ \sin{\lambda}$ positive.

In the west to east case the left is the north bank and the right is the south bank. But then the surface gradient is negative, i.e the further north - to the left bank - the more shallow the water. The left bank undermines the right.

In the east to west case the left is the south bank and the right is the north bank. But then the surface gradient is positive, i.e the further north - to the right bank - the more deep the water. The left bank undermines the right.

However, in the southern hemisphere this does NOT hold. This is because $ \lambda < 0$ making $ \sin{\lambda} < 0$. So the water gradients have opposite signs, and thus slope in the opposite directions. Thus lateral rivers flowing in the southern hemisphere undermine the right bank more than the left. These claims also hold when the river flows with a northern or southern component, on each hemisphere.

The last special case is that corresponding to a river flowing on the equator, where $ \lambda = 0$. The river is flat and neither bank is undermined.

Now the difference in the height of the water is clearly given by:

$ \Delta h = a \cdot |m_{surface}| =  \frac{2 a \Omega v_0 \sin{|\lambda|}}{g}$

The case where the river does not flow in the east-west direction can be tackled with similar analysis, replacing the vector $ {\bf v_0}$ appropriately. However, it can be extended from this case. Consider several sets of parallel lines drawn on an inclined plane. Draw lines perpendicular to, and intersecting each pair of parallel lines. Now take a second plane which intersects this inclined plane. How do the components of gradients of these perpendicular lines parallel to this intersecting plane change according to the angles that the parallel lines make with each other on the original plane?


algebra
polynomial
superior algebra
superior algebra solved

I think that grobber won't offend if i use ideas, that he proposed for the same last problems by Moubinool
Denote $P(x)=det(A-xB)$, so it follows(from the condition) that $P(i)P(-i)=0$ and $P(1)=0$, therefore, since $P(x)$-polynomial of degree 4 with real coefficients it must be $P(i)=P(-i)=P(1)$, so $P(x)$ is of the form $P(x)=(x^2+1)(x-1)(xdetB-detA)$  (because the domiant coefficient $detB$ and $P(0)=detA$), so we see that $det(A+B)=P(-1)=4(detB+detA)$
so we are done


Nine copies of a certain pamphlet cost less than $ \$10.00$ while ten copies of the same pamphlet (at the same price) cost more than $ \$11.00$.  How much does one copy of this pamphlet cost?

\[ \textbf{(A)}\ \$1.07 \qquad
\textbf{(B)}\ \$1.08 \qquad
\textbf{(C)}\ \$1.09 \qquad
\textbf{(D)}\ \$1.10 \qquad
\textbf{(E)}\ \$1.11
\]

[hide]e, its the only one that is more than 11 dollars when multiplied by 10[/hide]

[hide="Click for solution"]
Quickly, we note that $ \$1.11 \times 9=\$9.99$, which is close to the upper bound of what we want, so we try multiplying by $ 10$, which yields $ \11.10$, so this works.  We can test every other answer choice to see that just $ \boxed{\textbf{(E)}}$ works.
[/hide]


Asymptote geometry incenter circumcircle

In Asymptote, how do you draw perpendicular bisectors and angle bisectors?

You can find the incenter and draw lines through it and the vertices for the angle bisectors, but I'm not sure how to draw $\perp$ lines, so I can't help you with the $\perp$ bisectors.

The perpendicular bisectors meet at the circumcenter. I assume you can find the midpoint of a line easily enough, so just draw a line that passes through both it and the circumcenter.

Tip: don't actually draw the circumcenter or the midpoint; just don't do draw O; or draw M; or whatever.


trigonometry
inequalities
geometry unsolved
geometry

Given ABCD on the sides AB, BC, CD, DA choosen points M, N, P, Q respectively.
AM=BN=CP=DQ  also given that MNPQ - quadralaterial. Prove that ABCD also quadralaterial

[quote="Alisher"]MNPQ - quadralaterial[/quote]
by "quadralaterial" do you mean cyclic? :maybe:

(we call $ MNPQ$ cyclic iff $ M,N,P,Q$ lie on a circle)

sorry for my english I can't translate it. But
 I mean square. If given MNPQ so MN=NP=PQ=QM  ,MN perpendicular NP, NP perpendicular PQ, PQ perpendicular QM, QM perpendicular MN

Isn't it obvious because triangle AMQ , MBN , NCP ,PQD are equilateral so all side of ABCD is equal and all angle's are 90' If I'm not mistaken

this problem was given in an Iranian olympiad like 12 years ago!
anyway here's the solution:

let $ \angle AQM \equal{} \alpha,\angle NMB \equal{} \beta,\angle PNC \equal{} \gamma,\angle PDQ \equal{} \lambda$,we will show that $ \alpha\equal{}\beta\equal{}\gamma\equal{}\lambda$.

first of all let $ MN \equal{} NP \equal{} PQ \equal{} QM \equal{} a$ and let $ AM \equal{} BN \equal{} CP \equal{} DQ \equal{} r$.

now WLOG assume that $ \alpha$ be the greatest among $ \alpha,\beta,\gamma,\lambda$.

now according to $ \sin$ law in $ \triangle AQM$ we have:

$ \frac {r}{\sin (90 \minus{} \beta)} \equal{} \frac {a}{\sin (90 \minus{} \alpha \plus{} \beta)}$

$ \Rightarrow\frac {r}{\cos\beta} \equal{} \frac {a}{\cos (\alpha \minus{} \beta)}$

$ \Rightarrow\frac {a}{r} \equal{} \frac {\cos (\alpha \minus{} \beta)}{\cos\beta}$

$ \Rightarrow 1 \plus{} \tan\alpha\cdot\tan\beta \equal{} \frac {a}{r\cos\alpha}$    (1)

similary we conclude that:

$ 1 \plus{} \tan\beta\cdot\tan\gamma \equal{} \frac {a}{r\cos\beta}$       (2)

$ 1 \plus{} \tan\gamma\cdot\tan\lambda \equal{} \frac {a}{r\cos\gamma}$      (3)

$ 1 \plus{} \tan\lambda\cdot\tan\alpha \equal{} \frac {a}{r\cos\lambda}$      (4)

now note that $ \alpha\geq \lambda$ thus $ \frac 1{\cos\alpha}\geq\frac 1{\cos\lambda}$ so according to (1),(4) we get that $ \tan\beta\geq\tan\lambda$ hence $ \boxed{\beta\geq\lambda}$.

thus we will have:

$ 1 \plus{} \tan\gamma\cdot\tan\beta\geq 1 \plus{} \tan\gamma\cdot\tan\lambda$

and also by (2),(3) we get that $ \cos\beta\geq\cos\gamma$ hence $ \boxed{\beta\geq\gamma}$.

again we have $ \alpha\geq\gamma$ hence by a similar way we get that $ \boxed{\lambda\geq\gamma}$.

now by the inequalities in the boxes we get that:

$ \boxed{\alpha\geq\beta\geq\lambda\geq\gamma}$ (*)

thus we will have:

$ 1 \plus{} \tan\beta\cdot\tan\gamma\leq 1 \plus{} \tan\alpha\cdot\tan\lambda$ (5)

now using this inequality and (2),(4) we get that $ \beta\leq\lambda$ so according to (*) we'll have:

$ \beta \equal{} \lambda$

now by (5) we get that $ \alpha \equal{} \gamma$ thus:

$ \alpha \equal{} \beta \equal{} \gamma \equal{} \lambda$

now the rest is pretty easy.


Let P be a polynomial that belongs to $R[x]$, $P(x)=x^3+ax^2+bx+c$, a<>0, with the solutions $x_1, x_2, x_3$ which are distinct 2 by 2.
a) $\frac{x_1}{P'(x_1)} + \frac{x_2}{P'(x_2)} +\frac{x_3}{P'(x_3)}$ is equal to ?
b) Let $Q$ be a polynomial of $degre\; 1$ that belongs to $R[x]$, what is the value of $\frac{Q(x_1)}{P'(x_1)} + \frac{Q(x_2)}{P'(x_2)} +\frac{Q(x_3)}{P'(x_3)}$

The first sum is $\frac{x_1} {(x_1-x_2)(x_1-x_3)} +\frac{x_2} {(x_2-x_1)(x_2-x_3)} +\frac{x_3} {(x_3-x_1)(x_3-x_2)} =0$ be clearing the denominators and straightforward checking.

The second is $0$ too. You only have to denote $Q(x) = ax+b$ and use the same way as above (note that the part $ax$ is done by the first sum).

Pierre.


calculus
integration
trigonometry
logarithms
calculus computations

$ I \equal{} \int_{_0}^{^{\frac {\pi}{4}}} x \tan x dx$ ??

Anybody help me ? Thanks !!

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1072333#1072333]this might help[/url]  :)

$ \textbf I \equal{} \int_0^{\dfrac{\pi}{4}}\;x\cdot \tan\,x\;\textbf dx\; \equal{} \;\int_0^1\dfrac{y\cdot\tan^{ \minus{} 1}\,y}{1 \plus{} y^2}\;\textbf dy\qquad (\text{where}\quad y\equal{}\tan\,x\; )$

$ \equal{} \dfrac{1}{2}\int_0^1\tan^{ \minus{} 1}\,y\cdot\textbf d\left(\ln\,(1 \plus{} y^2)\right)$

$ \equal{} \dfrac{1}{2}\left(\tan^{ \minus{} 1}\,y\cdot \ln\,(1 \plus{} y^2)\bigg|_0^1 \minus{} \int_0^1\dfrac{1}{1 \plus{} y^2}\;\ln\,(1 \plus{} y^2)\;\textbf dy\right)$

$ \equal{} \dfrac{1}{2}\left(\dfrac{\pi}{4}\;\ln\,2 \minus{} \left( \dfrac{\pi}{2}\;\ln\,2\; \minus{} \;\textbf K_{Catalan}\right)\right)$

$ \equal{} \dfrac{1}{2}\left(\textbf K_{Catalan} \minus{} \dfrac{\pi}{4}\;\ln\,2\right)$

the   integral $ \int_0^1\dfrac{1}{1 \plus{} y^2}\;\ln\,(1 \plus{} y^2)\;\textbf dy$
has been shown many times on this forum.

I've got the answer. Thank all of you , Madness , Missan  !!  :lol:


if abc=1 prove that 1/a^3(b+c) +1/b^3(c+a) +1/c^3(a+b) is greater than or equal to 3/2

This is well known, it is number 10 on [url]http://www.artofproblemsolving.com/Resources/Papers/MildorfInequalities.pdf[/url], four solutions are given. The third (Cauchy) is the easiest.


AMC
AIME
MATHCOUNTS
AMC 12
USA(J)MO
USAMO
number theory
\/closed

Does the Intermediate Number Theory Class mostly help for aime, usamo.
Because the Introduction to Number theory class problems were much easier and did not cover aime level problems.

Yes, it covers  all the important topics in AIME Number Theory, as well as some easy olympiad problems.
In my opinion, it's a lot harder than the Introduction class, because the Intro only has MAHTCOUNTS, AMC 10, and easy AMC 12 NT problems.

How come there isn't an Intermediate NT textbook?

Remy i have noticed that some problems in the AMC 12  are extremely hard (20- 25) yet they are not covered in intro to nt.

Oh yeah, Intermediate NT also covers hard AMC 12 problems too. I think AMC12 has a big range of difficulty, and the Intermediate class will include the last few problems as well (while the Intro covers the easy one).

@maybach: The staff doesn't have any plans to publish the textbook yet. Writing a book is not an easy task, and it takes a long time to do so. I heard that they will publish the Interm NT textbook [i]someday[/i], but not in the close future.

If you want a supplementary book try [url=http://www.amazon.com/exec/obidos/ASIN/081763245X/artofproblems-20]Number Theory: Structure, Examples, and Problems[/url] by Titu Andreescu and Dorin Andrica.

The book is a bit more difficult than the class.

[quote="ic2149"]i notice the Intermediate NT class only have 8 class, and no book, the class just introduce some concepts? does it offer enough problems? when will the class start?[/quote]
Future classes are not scheduled yet, so you will have to wait. As far as I remember, it introduced concepts, proved them, and had 7-10 problems every week, along with around 14 message board problems each week for homework. However, there's no Challenge Sets like most subject courses do.

Is there going to be a Intermediate NT book in a few years??

[quote="ic2149"]i notice the Intermediate NT class only have 8 class, and no book, the class just introduce some concepts? does it offer enough problems? when will the class start?[/quote]

There are plenty of problems.  We haven't set a date for the course yet.

We are not sure when there will be a corresponding book.

Are there other books that are used as follow up to Intro to NT?

[quote="AlphaBetaTheta"]So, remy did you feel a difference in your number theory skills?
Did it raise your'e amc 10/ 12/ aime/ usamo :| ?[/quote]

Eh I don't know. I took it this summer, so I didn't have a chance to apply them yet. XD But I think I learned important things from the course.


algebra
system of equations
algebra solved

Solve the following system of equations:

x + [y] + {z} =  3.9 (1)
y + [z] + {x} =  3.5 (2)
z + [x] + {y} =  2    (3)

[a] denotes integer part and {a} denotes its fractional part, a = [a] + {a}

You add them all and get :
x+y+z=4,7 (4)
then  you do (4)-(1);(4)-(2);(4)-(3)
and you get x,y,z! :D

x=1,7
y=2,8
z=0,2

simple! cheers! :D :D


geometry
geometric transformation
reflection

i was wondering what are some good jobs that are available to someone who intends to major in mathematics.
i know that i could be an actuary but what else is there.
if you respond, could you please also provide a short description of the job.

 thanks ahead of time.  :)

Here's a page which includes short descriptions:

http://www.cs.xu.edu/math/mathcareers.html

And, here's one with links to other math-related-careers pages (it's not as good as the earlier one, but does have a little bit of additional info):

http://cerebro.cs.xu.edu/~smbelcas/mathcar.html

Your can do quantitative finance job or financial engineering,
here is the introduction:
[url=http://www.quantfinancejob.com]QuantFinanceJob.com[/url]

This is by no means a common career path (and probably doesn't even count as a job per se), but here's what one math major is doing:  http://www.ayria.com

Maybe not completely unique. Phil Alvin, whose rock bands over the years have included the Blasters, has an M.S. in math from my university.

What kind of career options are available to me if I major in pure mathematics for a BSc.?

That is exclusive of teaching and statistics related occupations.

We have a poster here in our office from the American Mathematical Society.  It says, in big letters near the top, "What can I do with a math degree?"

The listed options are:
Actuary
Animator
Appraiser
Benefits Administrator
Climate Analyst
College Professor
Commodities Trader
Cryptanalyst
Epidemiologist
Foreign Exchange Trader
Forensic Analyst
Market Research Analyst
Pollster
Population Ecologist
Public Utilities Analyst
Quantitative Analyst
Research Scientist
Statistician
Teacher
Technical Writer
Urban Designer

See also [url]http://www.ams.org/employment[/url]

I think you have to be careful though - these are jobs that some math major would be able to get, but a BSc in pure math will not automatically qualify you for very much at all.  You should probably reflect a little on what you want to do now and take some electives in a relevant subject or maybe do some volunteer work.  Your math skills will be invaluable, but you need to have displayed an interest in other career paths.  Otherwise people will assume that you are only interested in math and are just settling for something else, and they won't want to hire you.

I should know - I graduated with a BSc in pure math a few years ago and couldn't find any job, so ended up continuing on with my part-time job as a photocopying assistant  and doing other odd unskilled jobs for nearly a year before coming back to university and starting my phd.  I live in fear of the day I have to graduate - don't turn out like me!

Thanks for the advice. yeah i planned to take electives in areas such as physics and programming or some economics.
but thanks.

Here's a recently-updated mathematics careers webpage:

http://www.toroidalsnark.net/mathcareers.html

(Okay, I'll admit it, I made the page)

see http://www.ams.org/employment

ps: I like your malmsteen avatar ;)

[quote]Whenever this question gets asked, it seems like "Consultant" or "Investment Banker" are never listed...is that just because the list is geared towards possible interests of math majors, or is there really a prejudice against them in those two fields (since both fields require a high degree of socialization and schmoozing, which math majors stereotypically do not possess)/?[/quote]

Surprisingly, many, if not most, applied math and statistics folks end up doing "consulting" work.  Their clients might be internal to an organization but virtually all are working with scientists, engineers, and other specialists to solve problems critical to the missions of their organizations.

Interpersonal skills are critical in just about all careers.  Applied math/stat folks are often surprisingly good when communicating with other scientists and professionals in a work environment.  The sterotypes probably originate in the often failed attempts to educate students.

There's a difference between communicating effectively about how your expertise can be utilized to address a problem and "socializing/schmoozing."  You don't necessarily have to have    a wide circle of friends or be the life of the party to succeed in careers that require personal interaction.


find all the answer of:
$2a^2+3b^2-5c^2=1997$

Are you sure that the question is to find [b]all[/b] integer solutions?


1. Let x= a+b+c. Show that $ (a\plus{}b\plus{}c)^2\plus{}(3\minus{}a\minus{}b\minus{}c)^2\equal{} 2(x\minus{}3/2)^2\plus{}9/2$

2. Let $ t\equal{}xy\plus{}yz\plus{}xz$ Find $ (x^3\plus{}y^3\plus{}z^3\minus{}3xyz)^2$ in terms of t.

[quote="bigman"]1. Let x= a+b+c. Show that $ (a \plus{} b \plus{} c)^2 \plus{} (3 \minus{} a \minus{} b \minus{} c)^2 \equal{} 2(x \minus{} 3/2)^2 \plus{} 9/2$

2. Let $ t \equal{} xy \plus{} yz \plus{} xz$ Find $ (x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz)^2$ in terms of t.[/quote]

1. LHS= (a+b+c)^2+(3-(a+b+c))^2

Sub in x, x^2+(3-x)^2

x^2+(9-6x+x^2)

2x^2-6x+9

Complete square:

=2(x - 3/2)^2 + 9/2

2. Similar method but consider the expansion of a^3+b^3+c^3

[quote="bigman"]1. Let x= a+b+c. Show that $ (a + b + c)^2 + (3 - a - b - c)^2 = 2(x - 3/2)^2 + 9/2$

2. Let $ t = xy + yz + xz$ Find $ (x^3 + y^3 + z^3 - 3xyz)^2$ in terms of t.[/quote]

1.  Since $ x=a+b+c$, using that substitution property of inequality.
The $ LHS$ becomes:
$ x^2+(3-x)^2$
Expanding, we get:
$ x^2+x^2-6x+9 \implies \boxed{2x^2-6x+9}$

Expanding the RHS, we get:
$ 2(x^2-3x+9/4)+9/2 \implies \boxed{2x^2-6x+9}$

Therefore, we have proven that $ (a + b + c)^2 + (3 - a - b - c)^2 = 2(x - 3/2)^2 + 9/2.$

2.  Expansion of the expression.  (Not very much progress.)

$ (x^3 + y^3 + z^3 - 3xyz)^2$
$ ([x^3 + y^3] + [z^3 - 3xyz])^2$

$ (x^6+2x^3y^3+y^6)+2(x^3z^3-3x^4yz+y^3z^3-3xy^4z)+(z^6-6xyz^4-9x^2y^2z^2)$

$ x^6+2x^3y^3+y^6+2x^3z^3-6x^4yz+2y^3z^3-6xy^4z+z^6-6xyz^4-9x^2y^2z^2$

I'm stuck.

wow. complex math. WHAT GRADE IS THIS STUFF FOR?!?!?!


Where are the points in a regular triangle which has the following special property: from the distances to the sides (from the points) a triangle can be constructed.

The solution can be seen in the figure attached but I don't know how to prove it. [geogebra]5727740c53fa39d13ead50e74152a9033d0b9418[/geogebra]

Let the distance from the point to AB be p, to AC be q, to BC be r, side length of regular triangle be l.

Obviously, ABp+ACq+BCr=lr=2S, where S is the area of ABC.

since p,q,r need to be able to form a triangle, we need p+q>r, which is equivalent to rl<S. This implies that the point must lie inside BCEG.

Similarly, p+r>q and q+r>p give the point must lie in ACFG and ABFE. Combine all, the point must lie within EFG.


Well by using the well known inequality \[9(a+b)(b+c)(c+a)\geq 8(a+b+c)(ab+bc+ca),\] and $(a+b+c)^{2}\geq 3(ab+bc+ca),$we obtain $(ab+bc+ca)\leq 3$.
And now, we obtain \[(a+b+c)^{2}\geq 3(ab+bc+ca)\geq (ab+bc+ca)^{2},\] thus $a+b+c\geq ab+bc+ca$.

[quote="X-men"]Let a, b, c be positive real numbers such that (a + b)(b + c)(c + a) = 8. Prove that

        ab + bc + ca <= a + b + c.[/quote]
Let $a, b, c $ be positive real numbers such that $ (a + b)(b + c)(c + a) \leq8.$ Prove that

     $  ab + bc + ca \leq a + b + c.$


algebra
polynomial
number theory unsolved
number theory

Let $P(x)$ be a polynomial with integer coeffiences . Let $k$ be a possitive integer . Prove that : There exists $n \in N$ such that : $k \mid P(1)+P(2)+...+P(n)$

[quote="Megus"]$P(1)+...+P(k^2)\equiv k(P(1)+...+P(k)) \equiv 0 \mod k$[/quote]
 Megus , Can you explain clearly ? I think it's wrong .

I think it's right.
$P(m) \mod k$ does only depend on $m \mod k$, thus $P(m) \equiv P(m+k) \equiv P(m+2k) \equiv ... \mod k$.


superior algebra
superior algebra unsolved

Let $ m$ be a positive integer.Call a ring $ R$ a $ m$-ring,if $ x^m\equal{}x$ for all $ x\in R$.

prove that  a necessary and sufficient condition that a $ m$-ring be a Boolean ring is that $ m$

is even and $ m\minus{}1$ has no prime factor of the form $ 2^p\minus{}1$.

Imam Ali (as):

The first fruit of forbearance is that people will sympathize with you and they will go against the man who offended you arrogantly.

this was discussed here:
http://www.mathlinks.ro/viewtopic.php?t=24885


Assuming that $ f: \mathbb{N}^* \rightarrow\mathbb{N}^*$ satisfying that:
for every $ n\in \mathbb{N}^*$, $ f(n\plus{}1)>f(f(n))$
Prove that:
$ f(n)\equal{}n (n\in\mathbb{N}^*)$

This is an old problem. It's originate in IMO 1977. The solution can see in document below(exercise 20):


geometry
trigonometry
inequalities
geometry proposed

Given a triangle with the area $S$, and let $a$, $b$, $c$ be the sidelengths of the triangle.

Prove that $a^{2}+4b^{2}+12c^{2}\geq 32\cdot S$.

\[ \boxed {\ \triangle ABC\Longrightarrow a^2+4b^2+12c^2\ge 32S\ }  \]Hello, Flavian ! How are you ? The your proposed problem is nicely and easily.[hide="Here is a short metrical solution."]$D\in (BC)$, $DC=2BD$ and [u]Stewart's relation[/u] $16AD^2+3a^2=4b^2+12c^2\Longrightarrow$
$a^2+4b^2+12c^2=4\left(4AD^2+a^2\right)\ge 16aAD\ge 16aAD\sin \widehat {ADB}=32S$, i.e.
$a^2+4b^2+c^2\ge 32S$, with equality if and only if $AD=\frac a2$ and $AD\perp BC\ .$
[b]Generalization.[/b] $\boxed {\ k>1\Longrightarrow a^2+k^2b^2+k^2\left(k^2-1\right)c^2\ge 4k^3S\ }\ .$
For $k: =2$ obtain the [b]Flavian[/b]'s proposed inequality.
For $k: =\sqrt 2$ obtain the well-known inequality $a^2+2\left(b^2+c^2\right)\ge 8\sqrt 2S\ .$[/hide] I offer you an another inequality: $\triangle ABC,\ k\ge 1\Longrightarrow$ $\boxed {\ 15a^2+5(k^2+k+1)b^2+(k-1)(3k^3+6k^2+4k+2)c^2\ge 15(k+1)(k^2+1)S\ }\ .$

Thank you for the nice solution. I saw another generalisation for this inequality, but I think that is same with your. I am verry good, preparing for olympiad. But how are u? :)

Another generalization:

[b]Generalization[/b] Let $ABC$ be a triangle with side lengths $a, b, c$ and with area $S$. If $m, n, p$ are three non-negative real numbers then
      $ma^2+nb^2+pc^2 \geq 4\sqrt{mn+np+pm}.S$

From this you can lead to the following inequality:

For any positive $m, n, p$ such that $k = 2mn+2np+2pm-m^2+n^2+p^2 \geq 0$ then $\sum ma^2 \geq 4\sqrt{k}S + \sum m(b-c)^2$


I have one example right away and I've seen another one but I cannot recall it now. If you go to the profile of [url=http://www.artofproblemsolving.com/Forum/usercp.php?mode=viewprofile&u=20118]you-know-who[/url], the post count shows 45 posts, but when you click on the "Find all posts by ..." link, you see none. The same happens if you try to use the search function and search by name. To be honest, I haven't seen a single post by this user, so I don't really know whether it is the post counter or the search engine that is malfunctioning. Or, perhaps, there is some other reason behind this. Anyway, I just thought it may be worth reporting.

Those posts the student made are posted in some forums you do not have permission to access (classes forums mainly). There is nothing wrong with the search system.

Thanks for reporting then.


How many common fractions with a value between 0 and 1 have 6
as their denominator?

1/6, 2/6, 3/6, 4/6, and 5/6 are the five fractions, but the middle three all simplify, leaving only [b][u]2[/u][/b], 1/6 and 5/6


Prove that for any positive integer n,

1/4 ((2 + \sqrt 3)2n + 1 + (2 - \sqrt 3)2n + 1)

can be written as the sum of two squares of consecutive integers.

The problem is straightforward:
   Trying to find a such that a^2+(a+1)^2=the number given in the problem, we solve a second degree equation and find that:
  a=1/4{ (1+sqrt(3)*(2+sqrt(3))^n-(sqrt(3)-1)*(2-sqrt(3))^n-2). So, it is enough to prove that the number in the RHS is a positive integer. Let
 (2+sqrt(3))^n=a_n+sqrt(3)*b_n. Then (2-sqrt(3))^n=a_n-sqrt(3)*b_n and making the computations the RHS becomes 1/2(3b_n+a_n-1). So, it remains to prove that a_n+b_n is odd, which is trivial using Newton formula and identifying a_n and b_n from (2+sqrt(3))^n=a_n+sqrt(3)*b_n.


[quote="Scrollphaser"]I can't figure out how to set this one up. Can anyone help me out? It makes my head hurt.
The larger of two numbers is four less than three times the smaller. Also, three times the larger number is six more than seven times the smaller. Find the two numbers.[/quote]
the larger of two numbers : $ x>y$
is four less than three times the smaller: $ x\equal{}3y\minus{}4$
three times the larger number is more than seven times the smaller: $ 3x\equal{}7y\plus{}6$
there ya go :D

[quote="Scrollphaser"]Smaller is 3x=7y+6
Larger is x=3y-4

If so, why wouldn't the smaller be
3(3y-4)=7y+6 [/quote]

It is.  From there, you need to solve for y and plug that into either of your original equations to find x.

[quote="Scrollphaser"]Why did the problem ask for the two numbers if the answers would be expressions?[/quote]

The answers are numbers.  However, you have to use the expressions to find the numbers.

[quote="Scrollphaser"][hide]Thanks. It didn't realize I needed two variables.
You lost me though. Are these the "two numbers" then?
Smaller is 3x=7y+6
Larger is x=3y-4
?
If so, why wouldn't the smaller be
3(3y-4)=7y+6
?
Why did the problem ask for the two numbers if the answers would be expressions?
The instructions at the begining of this problem set say, "Write an equation and use it to solve the problem. Make sure to answer the question asked by the problem."[/hide][/quote]

Finding numbers from expressions: Normally you need $ a\plus{}1$ equations in $ a$ variables to find the numerical solutions, except in the case of Diophantine equations, or when certain restrictions hold.

For simple questions, simply substitute. We know that $ x\equal{}3y\minus{}4$, so $ 3x\equal{}9y\minus{}12$. Then $ 9y\minus{}12\equal{}7y\plus{}6$. $ 2y\minus{}12\equal{}6$, $ 2y\equal{}18$, $ y\equal{}9$.

[quote="ThetaPi"]
Finding numbers from expressions: Normally you need $ a \plus{} 1$ equations in $ a$ variables to find the numerical solutions,...[/quote]

 :huh: $ a\plus{}1$??

[quote="leoxnlin"][hide][quote="ThetaPi"]
Finding numbers from expressions: Normally you need $ a \plus{} 1$ equations in $ a$ variables to find the numerical solutions,...[/quote]

 :huh: $ a \plus{} 1$??[/hide][/quote]

Oh sorry. It should be $ a$ equations for $ a$ variables.


function
limit
calculus
calculus computations

Denote with $ A$ the set of all functions $ f: [\minus{}1,1]\rightarrow\mathbb{R}$, for which the graph is a part of the unit circle $ T\equal{}\{(x,y)\in \mathbb{R}^{2}|x^{2}\plus{}y^{2}\equal{}1\}.$ Find all continuous functions in $ A$.

[hide]One possible solution: Let us study the continuity in the point $ x\equal{}0$. We know that $ f(0)\equal{}\pm 1$. We choose $ \epsilon\equal{}1$. We find the greatest $ \sigma$ $ \left( \sigma \in (0,1] \right)$, so that for all $ x$ satisfying $ |x|<\sigma$, we have $ |f(x)\minus{}f(0)|<1$. If $ \sigma\equal{}1$, then all the points are on the same side of $ x$-axis. So we have $ 2$ continuous solutions. But if $ \sigma<1$, then we study the points $ x\equal{}\pm \sigma$. If $ f(\pm \sigma)$ or on the other side of the $ x$-axis than $ f(0)$, then the function is not continuous in $ x\equal{}\pm \sigma.$ So they are on the same side as $ f(0)$. We again study continuity of the function in the points $ x\equal{}\pm \sigma$ for $ \epsilon\equal{}|f(\sigma)|$. If there do not exist corresponding $ \sigma$-s, then the function is not continuous in points $ x\equal{}\pm \sigma$. But if they exist, then the initial $ \sigma$ (for continuity in $ x\equal{}0$) was not the greatest. Contradiction. [/hide]

Intuitively it is obvious, that the only possible solutions are $ y=\sqrt{x^{2}+1}$ and $ y=-\sqrt{x^{2}+1}$, but sometimes it is tough to write a formal proof. I thought of some possibilities:

[list]2. possibility:
Let us suppose that $ f$ is continuous and that there are two points, which lie in different sides of $ x$-axis. WLOG we can say $ f(x_{0})=-\sqrt{x_{0}^{2}+1}, f(x_{1})=\sqrt{x_{1}^{2}+1}$ and $ x_{0}<x_{1}$.

Now we bisect the interval $ [x_{0},x_{1}]$ and get $ x_{2}=\frac{x_{0}+x_{1}}{2}$. If ${ \textrm{sign}(f(x_{2})})=-\mathrm{sign}({f(x_{0})})$, then we take the interval $ [x_{0},x_{2}]$ and repeat the same procedure, otherwise we take the interval $ [x_{2},x_{1}]$ and repeat the same procedure.

We are repeating this procedure on and on and get the infinite sequence $ \left\{x_{i}\right\}_{i=1}^{\infty}$, which form intervals $ [x_{0},x_{1}]\supset[x_{11},x_{2}]\supset[x_{22},x_{3}]\supset \cdots \supset[x_{ii},x_{i+1}]\supset \cdots$, where $ x_{ii}$ is either $ x_{(i-1)(i-1)}$ or $ x_{i}$ $ \left( \right.$ such that ${ \mathrm{sign}(f(x_{ii})})=-\mathrm{sign}({f(x_{i})} \left. \right)$ .

The length of the $ n$-th interval above is $ \frac{1}{2^{n-1}} (x_{1}-x_{0})$. Since the $ \lim\limits_{n \to \infty}\left(\frac{1}{2^{n-1}} (x_{1}-x_{0})\right)=0$, the sequence $ \left\{x_{i}\right\}_{i=1}^{\infty}$ converges to some $ l$.

Since $ l \in \cdots \subset [x_{ii},x_{i+1}] \subset \cdots  \subset[x_{11},x_{2}] \subset [x_{0},x_{1}]$ and the lengths of the intervals go toward $ 0$, there always exists $ x$, so that ${ \mathrm{sign}(f(l)})=-\mathrm{sign}({f(x)})$ for all $ \delta$-neiborhoods of $ l$. So $ f(x)$ is not continuous in $ x=l$. Contradiction.

Therefore all the points lie on the same side of $ x$-axis. [/list]

3. possibility:
Let us suppose that $ f$ is continuous and that there are two points, which lie in the different side of $ x$-axis. WLOG we can say $ f(x_{0}) \equal{} \minus{} \sqrt {x_{0}^{2} \plus{} 1}, f(x_{1}) \equal{} \sqrt {x_{1}^{2} \plus{} 1}$ and $ x_{0} < x_{1}$.

Let us define the set $ M \equal{} \left\{x; f(x) \equal{} \sqrt {x^{2} \plus{} 1} \wedge x\in[x_{0},x_{1}]\right\}$. Since the set $ M$ is not empty ($ x_{1}\in M$) and has a lower bound, according to Dedekind`s axiom there exists ist infimum $ m$, $ m\in[x_{0},x_{1}]$.

If $ m\neq x_{0},x_{1}$, then the there exist values of $ f$ in any right $ \delta$-neibourhood, which are in the different side of $ x$-axis than the values in any left $ \delta$-neibourhood of $ m$. Hence the function is not continuous in $ m$.

If $ m \equal{} x_{1}$, then the values of $ f$ in any left $ \delta$-neibourhood are in the different side of $ x$-axis than $ f(x_{1})$. Hence the function is not continuous in $ x_{1}$.

If $ m \equal{} x_{0}$, then there exist values of $ f$ in any right $ \delta$-neibourhood, which are in the different side of $ x$-axis than $ f(x_{0})$.  Hence the function is not continuous in $ x_{0}$.

We came into contradiction with the assumption, that there exist two points in the different side of $ x$-axis. Therefore all the points lie on the same side of $ x$-axis.

4.possibility:

We suppose $ f$ is continuous. Let us study the interval $ I\equal{}[\minus{}1\plus{}\frac{1}{n},1\minus{}\frac{1}{n}]$, where $ n \in \mathbb{N}$.  For each $ x^{\ast} \in [\minus{}1\plus{}\frac{1}{n},1\minus{}\frac{1}{n}]$, we have some $ \delta(x^{\ast})$, such that for $ |x|<\delta(x^{\ast})$ we have $ |f(x)\minus{}f(x^{\ast})|<|f(x^{\ast})|$.

Therefore each $ x\in I$ has its neighborhood $ O_{x}$, such that $ O_{x}\equal{}(x\minus{}\delta(x),x\plus{}\delta(x))$. Now according to the Heine-Borel theorem we can choose finitely many $ O$-s, such that they cover $ I$. So $ I\subset O_{x_{1}} \cup O_{x_{2}} \cup \ldots \cup O_{x_{n}}.$

Now we can see from the choice of $ \epsilon$-s ($ |f(x^{\ast})|$), that all the values of $ f$ in $ O_{x_{i}}$ are in the same side of $ x$-axis.

Since $ O$-s are open sets, each adjacent two cover themselves. Hence $ O_{x_{i}} \cap O_{x_{i\plus{}1}} \neq \emptyset$. Therefore values of $ f$ in any adjacent two $ O$-s lie in the same side of $ x$-axis. Inductively we conclude, that the values of $ f$ in all $ O$-s lie in the same side of $ x$-axis. (If we have just one $ O$, we come to the same conclusion).

Therefore all the values of $ f$ from the interval $ [\minus{}1\plus{}\frac{1}{n},1\minus{}\frac{1}{n}]$ lie in the same side of $ x$-axis. Since $ n$ was arbitrary natural number, we can limit the upper interval. Hence all the value of $ f$ from the interval $ \lim_{n \to \infty}[\minus{}1\plus{}\frac{1}{n},1\minus{}\frac{1}{n}]\equal{}(\minus{}1,1)$ lie in the same side of $ x$-axis. Again we conclude, that there are just two solutions $ f(x) \equal{} \minus{} \sqrt {x^{2} \plus{} 1}$ and $ f(x) \equal{} \sqrt {x^{2} \plus{} 1}$.

Note that we could not use this procedure for the interval $ [\minus{}1,1]$. We would have problems in $ x\equal{}\pm 1$.


Let [i]k[/i], [i]m[/i] and [i]n[/i] be integers so that 1 < [i]n[/i] < [i]m[/i] < [i]k[/i]+2.
Determine maximum number of elements of subset S of {1,2,...,k} so that sum of any [i]n[/i] different elements of S:
a) is not equal to [i]m[/i],
b) does not exceed [i]m[/i].

Answering the case (b).

If $ \exists \quad 1\le r< m$ so that
$ r \plus{} (r\plus{}1) \plus{} \dots \plus{} (r\plus{}n\minus{}1)\le m$
and $ (r\plus{}1 ) \plus{} (r\plus{}2) \plus{} \dots \plus{}(r\plus{}n) > m$

Then, $ S\equal{}\{1,2,\dots ,r\plus{}n\minus{}1\}$ is the maximal subset.


calculus
integration
calculus computations

Evaluate:

$\int_{0}^\infty \frac{dx}{\sqrt{x}}$

I'm not exactly sure how to convert this from an integral to a limit problem. Integration is fine for me, but I just need help setting up the problem. Any help?

This integral doesn't converge anywhere. So, I am not sure you would get anything even if you converted the problem to computing limits.

In fact, $\int_{0}^{\infty}\frac{1}{x^{\alpha}}\,dx$ converges only if $\alpha > 1$. In your problem, $\alpha = \frac12 < 1.$

I don't think $\int_{0}^\infty x^{t}$ ever converges. If $t <-1$, it blows up too much near 0. If $t >-1$, it doesn't decay enough $\to \infty$. If $t =-1$, it has both problems.

If you write it as $\int_{a}^{b}x^{t}$ and evaluate the integral, and then take the limits as $a\to 0$ and $b\to \infty$ you can prove the divergence.


trigonometry ARML calculus derivative integration articles search

Hi all,

I noticed something strange as I went from grade-school mathematics to college mathematics: I used to be horrible at manipulating symbols and equations, esp trig equations, and would frequently have trouble seeing the "tricks" used to manipulate an equation into solvability. Yet, when I started doing higher math proofs came much quicker to me.

Do you guys think that there is an actual difference in "elementary" vs. "higher" mathematics? Could a good grade school mathematician not be very good with proofs and vice versa? Or is it a matter of training?

computational and rigorous mathematics may seem to have a different type of ora about them, but they do both belong to the same beast. in grade school mathematics you basically get a survey of various categories of mathematics but you don't get to justify much. this justification is the actual mathematics and in the higher level math courses i believe you begin to see how everything ticks and tocks and start to answer more of the "why?" and "how?" questions.

[quote="joshuatan"]Do you guys think that there is an actual difference in "elementary" vs. "higher" mathematics? Could a good grade school mathematician not be very good with proofs and vice versa? Or is it a matter of training?[/quote]

Interesting question.  I think a lot of it is training...if you're not taught to ask "why" and "how", I've seen good grade school mathematicians be TERRIBLE at proofs.

More commonly these days, I run into students and most of my professors who only do rigorous proofs, and can't add or subtract.  Like in my last linear algebra class, everytime the prof started writing numbers on the board, we'd all groan and wince (including the professor) because we're so bad with numbers.  We didn't do a single correct numeric example all term, it always took 20 minutes to figure out where the mistakes were hahaha.  I suspect this is from lack of practice though!

Well, I'm now a graduate student, but not so long ago I was a high school student at Stuyvesant, in New York.  I still go back and visit from time to time, and I've found that the math team kids there are often quicker at a variety of tasks now than I am.  Actually, at ARML some of the coaches were joking that we should have a coach vs. team contest, and I was rather pessimistic about the coaches' chances in such a hypothetical matchup.  On the other hand, I find that in writing contests, there are some ideas, some modes of thought, some types of attack on certain problems that are "trivial" to me now, but that I wouldn't have thought of immediately even at my math team prime.

Sorry if this is a bunch of rambling anecdotes, rather than a coherent comment.  Let me add one further: I think that many competent students first run into trouble in mathematics at the level of calculus.  (Some, of course, have run into trouble much earlier; but let's focus on the ones who make it through algebra untroubled.)  I think there are two reasons that calculus is hard: it requires certain technical skills (a mastery of algebraic and trigonometric manipulations, for example) that not all students have, but it also has certain conceptual barriers (the notions of limits, derivatives, antiderivatives, integrals, etc., and what these things all [i]mean[/i] -- how their abstract definitions are related to their content).  To me, at least, these have always seemed to be distinct barriers that must be overcome separately, making me think the answer to the question is that they really are different.

A little bit ago I posted a link to a famous article called [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1296484634&t=192363]Lockhart's Lament[/url]. His basic argument is that the math of grade school is quite different from higher math and that we are losing a bunch of potential students who don't realize how fun math can be. Grade school students think of math as numbers, most people who have been through higher stuff hear the the word math think of tons of things almost none of which involve numbers.

I agree. There are some good comments on the thread about it too.  :)

Of course it's different, but it's silly to think that one can understand higher mathematics without an understanding of elementary mathematics.  To put it plainly, higher mathematics is built on elementary mathematics.  There is no such thing as a competent mathematician who "can't" do elementary mathematics.  They might mess up mechanical manipulations, but that's just mental laziness.  I actually hate it when profs say that they can't get simple computations right at the blackboard.  What they are really saying is, "I don't take my job as a teacher seriously enough to make the necessary effort to do this right."  This is especially true since the prof should have worked out the problem head of time anyway.

Well, when helping people or tutoring I almost always appeal to the stupid computations argument, and I don't believe that it is lack of taking teaching seriously. On the contrary, I think it is of the utmost importance for students to realize that the answer doesn't matter and one shouldn't waste their time fighting for the proper answer if it isn't coming quickly.

I still recall a calc test with four problems on it. I got three answers wrong and one correct but I didn't work out the answer, I just wrote, the answer is the two solutions to this quadratic. I got a 99%. My prof wrote -1 for incorrect answers at the top.

There are always computers and calculators in the real world.

Of course, you shouldn't always get every numerical answer wrong. I'm not advocating that, but I have definitely seen students go into a test unprepared, because they got caught up in trying to make an incorrect answer match the back of the book. If someone was there to tell them that they had the right concept and the error was meaningless, they would have learned more. Showing that that attitude is OK to adopt in those situations is a good teaching strategy. At least it helped me in lots of situations.

To all who took the time to reply: thanks for your thoughts!

I kind of feel that Lockhart is too much of an idealist for my taste, but then again I'm not American so I don't really know whether I would have had the same reaction had I actually been in the K-12 system. I didn't really like how he seemed to come across as (over)emphasising intuition at the expense of rigour, or implying that formalism somehow makes the subject dry (because as you get more and more abstract, I think you need to have formalism to keep your intuition in line). However perhaps he is rebelling against a system that has veered too far to one end, and therefore feels that an extremist(?) seeming argument is necessary to wake people up.

HilbertThm90, you certainly have a point.  In particular, every reasonable grader is lenient with computational mistakes made under timed conditions.  But that argument fades away when it comes to homework or pretty much anything else.  In the real world, students will become doctors or engineers or whatever, and just like in elementary school, the right answers do matter.

More rebuttal:  The techniques that one uses to avoid "stupid" mistakes are good habits that encourage clear thought anyway---writing out your work clearly, not taking huge steps, taking your time, and (my favorite) trying to work the problem in a way that minimizes computation.  It's also important to realize that in the scenario that was being contemplated (math professor teaching first-year college students), there is a huge chasm between what the professor considers to be mindless manipulation and what the students consider to be mindless manipulation.  It's all too easy to start going down a slippery slope from "this is just arithmetic" to "this is just some algebraic manipulation" to "oh, you've seen this a bunch of times and I'm sure you can do it yourself, so it's no biggie that I just royally F'ed this computation."  But I guess when it comes down to it, I concede that it's partly an issue of personal style.

As for the "back of the book" issue, I have to disagree.  Students rightfully want to know if what they are doing is correct.  (The ones that already know that they're essentially right don't need help.)  Checking against the back of the book is a reasonable method of verifying that you are doing things right.  Of course, it doesn't tell you if you are doing things wrong for the reason you mention---maybe it was just a computational mistake.  But that means that if you are careful to avoid making computational errors, you can be more confident that there is a substantive error.  (But I do agree that sometimes students get a little obsessed with the back of the book thing.)

Random remark:  Most high school students take multiple choices tests in school, and even if they don't, then their most important tests (SAT, AP) still have a lot of multiple choice.  So from a practical point of view (at least for high school tutoring), it's important for your students to avoid computation errors.

Okay, I've strayed far from the topic of the thread.  Sorry.


function
calculus
inequalities unsolved
inequalities

Let $ x,y,z $ satisfying : $ x^2 +y^2 =2 $ and $ z^2+2z(x+y)=8 $
   Find max of $ A=z(y-x) $

By a little computation we have $ (y-x)^2=8-\frac{16}{z^2}-\frac{z^2}{4}$
From there, we can go by 2 way:
   1>     $ (z(y-x))^2=8z^2-16-\frac{z^4}{4}$ leads exactly to $ -4\sqrt3 \leq z(y-x) \leq 4\sqrt3.$ ( because RHS is a one variable function)
 or  2>  $ 8= \frac{16}{z^2}+4\frac{z^2}{16}+3\frac{(y-x)^2}{3} \geq 8(\frac{z^6}{16^3}\frac{(y-z)^6}{3^3})^{1/8}$ then get the answer is $ 4\sqrt3$
"TT loves TT" :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D

tot ,tot . May be you are very lucky :P  :P  :D  :D

These seem like they will be simple using Lagrange Multipliers

Can you tell mw what is  this Lagrange method,can we use it to find the min,max point. I have heard about it before but I don't know it.

Using multivariable calculus we find that a necessary condition for an extremum is that, given $f(x,y,z)=0$, with constraints $c_1(x,y,z)=k_1$, $c_2(x,y,z)=k_2$, ..., $c_j(x,y,z) = k_j$ is that:

$\nabla f(x,y,z) = a_1\nabla c_1(x,y,z) + a_2\nabla c_2(x,y,z) + \dotsc + a_j \nabla c_j(x,y,z)$.

where $a_1,a_2,\dotsc a_j$ are constants and where $\nabla$ is the gradient operator.

This leads to a system of $3+j$ equations in $3+j$ variables, if we expand it in scalar form.

Nice solution, darkmaster.  My calculus approach just got me bogged down.

Let $ x,y,z $ be reals such that $ x^2 +y^2 =2 $ and $ z^2+z(x+y)=8 .$ Prove that
   $$ -6\leq z(y-x) \leq 6$$


algebra
polynomial
induction
search
algebra unsolved

P(x) is the 100. degree polynomial and P(k)=2^k  where k= 1,2,3,...,100 then fin P(101 )?

The solution is not 2^101 :)

yes, it is $ 2^{101}\minus{}2$(by induction or Lagrange's Theorem)

[quote="shoki"]yes, it is $ 2^{101} \minus{} 2$(by induction or Lagrange's Theorem)[/quote]

could you send the solution by Lagrange ?

Sorry, that involves a lot of computations but if u know the theorem that's easy(it needs only computations) if not,search it, u will certainly find it :wink:

I think the answer is $ 2^{101}\minus{}1$, not $ 2^{101}\minus{}2$...

Just construct the finite difference table and note that the 100th differences are constant.

there is a problem... epsilon should give the value of $ P(101)$ (because its degree is $ 100$),if not, we can't solve it.
i was wrong by saying that $ P(101) \equal{} 2^{101} \minus{} 2$.in fact there was a problem like this(and i think that epsilon's problem is this in fact):
If $ P(x)$ is a polynomial with $ deg(P(x)) \equal{} n$ such that for $ k \equal{} 1,2,...,n \plus{} 1$ we have $ P(k) \equal{} 2^{k}$. then prove that
$ P(n \plus{} 2) \equal{} 2^{n \plus{} 2} \minus{} 2$.

This is a very well-known problem.  The polynomial is

$ P(x)\equal{}\binom{x}{0}\plus{}\binom{x}{1}\plus{}\cdots\plus{}\binom{x}{100}$


Has anyone here taken AP Statistics? If so, did you take a class, or did you study on your own? I'm taking Calculus right now and I want to get all the math AP tests done this year...any thoughts would be appreciated.

I took a class through my high school. It was quite ridiculous since statistics is heavily dependent upon calculus (all those areas...), but yet we didn't use any calculus. You would probably be much better off getting a good book and learning it on your own.

[quote]You would probably be much better off getting a good book and learning it on your own.[/quote]

Many things are like that...

Yeah, now I regret skipping math...if I'd know AoPS two years ago, then I would've slept through math and studied with AoPS...yet since I learned about AoPS through high school, there's a paradox...

Tare, r u a middle or high skool student?

both, weird, ain't it? :)

Here, you could say I'm in high school since I'm in sophomore math, but usually (well, for all the other classes) I go to middle.

i did that for 2 yrs in middle school too...but ur still considered a middle skooler

Do you know of any good books that would teach me all the skills required?

Me too, this is my second year.
...though this is a [b]math [/b]forum, so we might as well as go with what year math we're in...

I'm not sure about any good books, though.


videos
function
search
complex numbers
complex analysis
AMC

What exactly is Birch and Swinnerton Dyer?

I won't tell you exactly what it is, but it is a statement about the Taylor expansion of a function. The only two of the Millenium Problems I think I could explain to anyone with any coherence whatsoever are the Poincar Conjecture and the Riemann Hypothesis. You can read the CMI official problem statements and/or watch the video lectures. I watched the video for the Riemann Hypothesis, but it doesn't have any particularly deep mathematical content. Still, if you are interested, you should watch the one on Birch Swinnerson-Dyer.

Short statement of the problem:
http://www.claymath.org/Millennium_Prize_Problems/Birch_and_Swinnerton-Dyer_Conjecture/

Link to PDF of the fuller problem statement:
http://www.claymath.org/Millennium_Prize_Problems/Birch_and_Swinnerton-Dyer_Conjecture/_objects/Official_Problem_Description.pdf

Link to a lecture about it:
http://www.claymath.org/Millennium_Prize_Problems/Birch_and_Swinnerton-Dyer_Conjecture/_objects/Lecture_by_Fernando_Rodriguez-Villegas_at_UT.ram

At first I thought this was about Tarepandas :shock:  :lol:

The only thing to add would be that "Birch and Swinnerton Dyer" is just my rank, not my avatar. My avatar is that oddly colored squiglly thing that I made (see if you can figure out the mathematical meaning behind it). For mathfanatic it's the pear, for Syntax Error it's that surfur dude, etc. It's usually a picture.
Birch and Swinnerton Dyer is the longest of the 7 millenium prize problems, thus it was selected to be the highest rank. From experience I assume that the lowest rank (P vs. NP) features the problem with the shortest title and the last (Birch and Swinnerton Dyer) is the longest since the difficulty level of each is not in order, a la ComplexZeta I believe. If it was ordered from easiest to hardest it should start with the Reimann Hypothesis...or something like that.

Actually it would end with the Riemann Hypothesis since that's the hardest. It's also the most interesting, but that's just personal bias speaking. (Note that my user name is ComplexZeta, for example...)

Oh whoops, sorry.
What does "ComplexZeta" mean anyways?

I thought it was either the complex (a+bi) part of the Riemann Zeta function or somat, or else it's just two math words spliced, diced, and spiced together...but now watch Simon prove me wrong.

mathfanatic is basically right. Complex refers to complex numbers and complex analysis (which is also known by other names, such as complex variables and complex function theory), which is a field I really enjoy. Zeta refers to the Riemann zeta function. Then I just put them together. If you search through the forum memberlist, you'll see a member named RealZeta. He's a friend of mine who made that username because he was inspired by mine.

[quote="ComplexZeta"]...inspired by mine.[/quote]

Oh.  I thought he was trying to be funny.

So now we know over half the namings of the most active members in AoPS (12 I believe)

??? you mean the meaning behind the names??? well.. you can figure mine out really easily. (It wasn't really supposed to have a meaning... lol)

Yup, like mathfanatic's name coming from being a math fanatic and such

mmm i was pondering "mathamaniac" for my username but i went with syntax error because i was thinking of calculators. ah jes

Personally I hate Syntax Errors, they are [b]so[/b] annoying, especially when you're rushing in the competition and it keeps on saying "Error: Syntax" over and over and over...
Doesn't mean you Neal :)

Is there a relationship between the length of the title of the 7 Millenium Prize Problems and how hard they are to understand? (Not necessarily their difficulty level) :lol:


vector
algebra
polynomial
linear algebra
linear algebra unsolved

The following is a Theorem from Friedberg, Insel and Spence.

Theorem 7.1. Let $ T$ be a linear operator on a vector space $ V$, and let $ \lambda$ be an eigenvalue of $ T$. If $ K_\lambda$ represents the generalized eigenspace of $ T$ corresponding to $ \lambda$, then for any scalar $ \mu \neq \lambda$, the restriction of $ T-{\lambda}I$ to $ K_\lambda$ is one-to-one.

Let $ x \in K_\lambda$ and $ (T-{\lambda}I)(x) = 0$. By way of contradiction, suppose that $ x\neq 0$. Let p be the smallest integer for which $ {(T-{\lambda}I)}^p(x)=0$, and let $ y={(T-{\lambda}I)}^{p-1}(x)$. Then $ (T-{\lambda}I)(y)={(T-{\lambda}I)}^p(x)=0$ and hence $ y \in E_\lambda$.

Furthermore, ${{ (T-{\mu}I)(y)=(T-{\mu}I)(T-{\lambda}I})^{p-1}(x)=(T-{\lambda}I})^{p-1}(T-{\mu}I)(x)=0$

At this point, what's the justification for commuting $ (T-{\mu}I)$ and ${ {(T-{\lambda}I})}^{p-1}$? Thanks, I'm having a hard time following that step, but after that, everything is clear.

$ A$ always commutes with any power of $ A.$

For instance: $ A\cdot A^m\equal{}A^{m\plus{}1}\equal{}A^m\cdot A,$ and all I've done there is rearrange some parentheses.

More generally, $ A$ commutes with any polynomial in $ A.$

Ah, that makes sense. I got tripped up on the $ \mu$ and $ \lambda$ constants, but I guess they don't interfere since they are coefficients of $ I$. But if it were some other linear operator $ U\neq I$, the commutativity wouldn't hold.

Thanks, KM.


Given $ a, b, c > 0$. Prove that:

$ \sqrt[3]{\frac{(a\plus{}b)(b\plus{}c)}{ab}} \plus{} \sqrt[3]{\frac{(b\plus{}c)(c\plus{}a)}{bc}} \plus{} \sqrt[3]{\frac{(c\plus{}a)(a\plus{}b)}{ca}} \geq$$ \frac{6}{\sqrt[9]{2}}. \sqrt[9]{\frac{ab^2\plus{}bc^2\plus{}ca^2\minus{}abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}}$
 :)

The hint: by Am-Gm, we can prove the stronger:

$ \frac{3}{4\sqrt[3]{4}}( \sqrt[3]{\frac{(a\plus{}b)(b\plus{}c)}{ab}}\plus{}\sqrt[3]{\frac{(b\plus{}c)(c\plus{}a)}{bc}}\plus{}$$ \sqrt[3]{\frac{(c\plus{}a)(a\plus{}b)}{ca}}) \geq\ 2\plus{} \frac{ab^2\plus{}bc^2\plus{}ca^2\minus{}abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$
 :)


complex numbers
geometry proposed
geometry

[quote="mecrazywong"]Isn't it a bit too trivial? Simply consider the Apollonian Circles...[/quote]

Just love the little things one finds are being claimed trivial from time to time

Daniel

[quote="mecrazywong"]Isn't it a bit too trivial? Simply consider the Apollonian Circles...[/quote]Remember this is for junior team selection, and it is supposed to be an easy problem on the test. It is not so trivial for a 14-year old :)

it's fairly easy with complex numbers also.

let afix of point $X$ be $x$, and $m=0$ . From conditions above we have $|a|=|b+c|, |b|=|a+c|, |c|=|a+b|$, and if we rewrite $a$ as $x+iy$, $b$ as $m+ni$ and $c$ as $p+qi$ we obtain $x^2+y^2=(m+p)^2+(n+q)^2$, $m^2+n^2=(x+p)^2+(y+q)^2$ and $p^2+q^2=(m+x)^2+(y+n)^2$ . After suming this $(x+m+p)^2+(y+n+q)^2=0$, hence $a+b+c=0$ and $M$ is it's centroid.

Here is my solution:$BM=2MB_{1},CM=2MC_{1}$ and $C_{1}B_{1}=\frac{BC}2$ therefore $\frac{C_{1}M}{CM}=\frac{C_{1}B_{1}}{CB}=\frac{B_{1}M}{BC}=2 \iff \triangle C_{1}B_{1}M$ ~ $\triangle CBM$,thus $\angle{CBM}=\angle{C_{1}B_{1}M}$.But $C_{1}B_{1}$ is paralel to  $BC$ therefore $B_{1},M,B$ are colinear.By the same manner we prove that $C_{1},M,C$ are colinear, hence M is the centroid of the triangle and the problem is solved!

We'll find after some computations that MG=0 (we'll use  Leibinz theorem)

[quote=Valentin Vornicu]Let $ABC$ be a triangle and $A_1$, $B_1$, $C_1$ the midpoints of the sides $BC$, $CA$ and $AB$ respectively. Prove that if $M$ is a point in the plane of the triangle such that \[ \frac{MA}{MA_1} = \frac{MB}{MB_1} = \frac{MC}{MC_1} = 2 , \] then $M$ is the centroid of the triangle.[/quote]

Let $G$ be the centroid of triangle $ABC$. Let $A'$ be the reflection of $A$ about $A_1$. See that $(A,A_1;G,A') = -1$ so $M \in \odot (GA')$ (Appolonius, as $\frac{MA}{MA_1} = 2$). Similarly. define $B',C'$. We see that $M$ is the point of concurrence of $\odot (GA'),\odot (GB'),\odot (GC')$. It is easy to see that these three circles cannot be coaxial. So, $M=G$. $\blacksquare$


Happened already? Has this that states which another, is there inexplicable and bizarre? More. Even something by replaced, be (and disappear) -- Instantly will it here. It is why, and for is, universe the what: discovers anyone, ever. If. That states which theory? A is there.

Hitchhiker's Guide to the Galaxy? Answer and Question to Life, the Universe, and Everything.

[hide="Solution"]
Reverse the wording, and reorder the punctuation (it was pretty awkward to read it right?)

There is a theory which states that if ever anyone discovers what the universe is for and why it is here, it will instantly disappear and be replaced by something even more bizarre and inexplicable. There is another which states that this has already happened.

(This is what?)

As brightknight says, it's a quote from HG2G (the radio show actually)

[/hide]


I plan to write a bunch of essays about AMC. The first one - Should You Check Your Answers During Tests? - discusses a mathematicians approach to the question. In my essay I use AMC 10/12 for my sample calculations. The essay is on my blog:
http://blog.tanyakhovanova.com/?p=201

YES You should check your answers during AMCs
I didn't last year and i still feel bad about it

On AMC 10A i got a 114 (cutoff for aime was 120)
when I checked my answers with the online answers i saw that i got 4 problems in the first 20 wrong

Then, on AMC 10B, i got another 114, and when I checked my answers, i saw that i got number 1, 3, 9 and 16 wrong

I think ppl like me tend to check their answrs to only that last 5 problems, assuming that they make no stupid mistakes on the first 20

On the other hand, a lot of people like me need the full 75 minutes to finish all 25 questions.  So I would say that you should finish as many questions as possible, THEN check your work.

in fact, I can't finish the Amc 12 with the full 75 min...though my accuracy in the first 17 problems is usually very good...

There can be different forms of checking. For example, you can check your bubbling (it would be frustrating to get one wrong like this), you can re-read a question to make sure you interpreted the question correctly, you can check your arithmetic, you can check your method, ... You have to be able to balance this as well as time.

It also depends on your own ability. If you're a neat person who rarely makes mistakes, then you don't need to check. However, if you make lots of computational errors, you should check.

do not check ur answers

the AMCs are designed NOT to trick you on the multiple choice
so if ur answer matches one on the multiple choice, then its probably right and if it doesnt match, it means you made a careless mistake

dont guess either, unless you have narrowed it down to 2-3 but i dont really see how thats possible , well actually  it might be but its not too often.

i think you should first try to solve all the problems and then think about checking

iYOA, I would say that isn't entirely true.  That sort of approach also would do no good on the AIME.

AMC 10 tries to trick you, just not as much as they could.  (They would give you a break on some difficult questions, but they still try to trick you a bit.)

I would say check them.

well with AIME its very important to check your work especially if you make lots of careless mistakes
but not so much on AMC. i think in the time it takes for you to find a mistake, you could be solving more problems.

and i did read somewhere that the AMC is not designed to trick a person even though they  knew how to solve the certain problem.

so i think checking should come last, only when you think you have solved all the problems you possibly can. (this only pertains to AMC)

on AIME it all depends on the person i guess.

That I agree with.  I think you should solve them all before checking.

While doing the test, you should read each question carefully.  Read it more than once to make sure that you haven't misread it.  Then make sure that your answer makes sense in the context of the question.  (This probably doesn't count as checking, but its still very important.)  Also, even if you don't have time to do anything else, always check your bubbling at the end.  It only takes half a minute and could save you from missing a question.


calculus
integration
function
calculus computations

Prove the following inequality.

$ \left(\int_0^1 (x^2\plus{}px\plus{}q)\ dx\right)^2\leq \int_0^1 (x^2\plus{}px\plus{}q)^2\ dx\minus{}\frac{1}{180}$

[quote="kenn4000"]iff

$ 0 \le \frac{(p\plus{}1)^2}{12}$

cool one.[/quote]

I also got same by brute force technique

[b][size=150]Is there any elegant way ?[/size]
[/b]

[quote="EulerIntegral"]Interesing.
A Generalization

$\boxed{\boxed{
\left(\int_0^1 \left(\sum _{i=0}^n \alpha _i x^i\right) \, dx\right)^2\leq \int_0^1 \left(\sum _{i=0}^n \alpha _i x^i\right)^2 \,
   dx-\frac{2^{-1-4 n} \pi  (n!)^2}{\Gamma \left(\frac{1}{2}+n\right) \Gamma \left(\frac{3}{2}+n\right)}
}}$

With $\alpha_n = 1, n \ge 1$, and also, the constant is optimal.[/quote]

can u provide some explanation

[quote="EulerIntegral"]Interesing.
A Generalization

$\boxed{\boxed{
\left(\int_0^1 \left(\sum _{i=0}^n \alpha _i x^i\right) \, dx\right)^2\leq \int_0^1 \left(\sum _{i=0}^n \alpha _i x^i\right)^2 \,
   dx-\frac{2^{-1-4 n} \pi  (n!)^2}{\Gamma \left(\frac{1}{2}+n\right) \Gamma \left(\frac{3}{2}+n\right)}
}}$

With $\alpha_n = 1, n \ge 1$, and also, the constant is optimal.[/quote]

Where the hell did this come from?

[quote="EulerIntegral"]Interesing.
A Generalization

$\boxed{\boxed{
\left(\int_0^1 \left(\sum _{i=0}^n \alpha _i x^i\right) \, dx\right)^2\leq \int_0^1 \left(\sum _{i=0}^n \alpha _i x^i\right)^2 \,
   dx-\frac{2^{-1-4 n} \pi  (n!)^2}{\Gamma \left(\frac{1}{2}+n\right) \Gamma \left(\frac{3}{2}+n\right)}
}}$

With $\alpha_n = 1, n \ge 1$, and also, the constant is optimal.[/quote]

I just simplified  the constant

$\boxed{\boxed{
\left(\int_0^1 \left(\sum _{i=0}^n \alpha _i x^i\right) \, dx\right)^2\leq \int_0^1 \left(\sum _{i=0}^n \alpha _i x^i\right)^2 \,
   dx-\frac{ (n!)^4}{(2n)! (2n+1)!}
}}$

With $\alpha_n = 1, n\in\mathbb{N}$

[quote="EulerIntegral"]I made this and proposed as a problem.[/quote]

can u give your solution


I took it in BEIJING,i want to konw my score

Look further down in the forums for the thread.

Edit: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=132366[/url]


inequalities
geometry
circumcircle
inradius
function
trigonometry
inequalities proposed

Let $ ABC$ be an acute triangle with three sides $ a,b,c$ and $ R , r$ is the circumradius and the inradius of the triangle.Prove that:
$ \sqrt{\frac{b+c-a}{a}}+\sqrt{\frac{c+a-b}{b}}+\sqrt{\frac{a+b-c}{c}} \ge \sqrt{9+\frac{1}{2}(1-\frac{2r}{R})^{2}}$

[quote="quykhtn-qa1"]Let $ ABC$ be an acute triangle with three sides $ a,b,c$ and $ R , r$ is the circumradius and the inradius of the triangle.Prove that:
$ \sqrt {\frac {b + c - a}{a}} + \sqrt {\frac {c + a - b}{b}} + \sqrt {\frac {a + b - c}{c}} \ge \sqrt {9 + \frac {1}{2}(1 - \frac {2r}{R})^{2}}$[/quote]
Since this inequality,we have:
$ \sqrt{\frac{b+c-a}{a}}+\sqrt{\frac{c+a-b}{b}}+\sqrt{\frac{a+b-c}{c}} \ge 3$

 :)

[quote="quykhtn-qa1"]Let $ ABC$ be an acute triangle with three sides $ a,b,c$ and $ R , r$ is the circumradius and the inradius of the triangle.Prove that:
$ \sqrt {\frac {b + c - a}{a}} + \sqrt {\frac {c + a - b}{b}} + \sqrt {\frac {a + b - c}{c}} \ge \sqrt {9 + \frac {1}{2}(1 - \frac {2r}{R})^{2}}$[/quote]
This inequality very nice and not easy.No one like it?  :)

[quote="quykhtn-qa1"]Let $ ABC$ be an acute triangle with circumradius $ R$ and inradius $ r$ .Prove that

$ \sqrt {\frac {b + c - a}{a}} + \sqrt {\frac {c + a - b}{b}} + \sqrt {\frac {a + b - c}{c}} \ge \sqrt {9 + \boxed {\frac {1}{2}}\left(1 - \frac {2r}{R}\right)^{2}}$ .[/quote]
[color=darkblue][b]I think that is better  $ 2$ than $ \frac 12$ in above proposed inequality.

I dedicate my proof to my friend, the student[/b][/color] [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=61896][color=darkred][b][u]Mateescu Constantin, Pitesti, RO[/u].[/b][/color][/url]

[quote="Virgil Nicula"][color=darkred][b][u]Lemma 1[/u].[/b] In any [u]acute triangle[/u] $ ABC$ there is the inequality $ \boxed {\ \frac {a^3 + b^3 + c^3}{abc}\ \ge\ 1 + \frac Rr\ }\ \ (1)$ .[/color][/quote]
[color=darkblue][b]Proof.[/b] Apply the [u][b]Walker[/b]'s inequality[/u], i.e. $ s^2\ge 2R^2 + 8Rr + 3r^2\ (*)$ . Indeed,

$ \frac {a^3 + b^3 + c^3}{abc} = 3 + \frac {a^3 + b^3 + c^3 - 3abc}{abc} =$ $ 3 + \frac {(a + b + c)\left[\left(a^2 + b^2 + c^2\right) - (ab + bc + ca)\right]}{abc} =$

$ 3 + \frac {s^2 - 3r^2 - 12Rr}{2Rr} =$ $ \frac {s^2 - 3r^2 - 6Rr}{2Rr}\ \stackrel{(*)}{\ge}\ \frac {\left(2R^2 + 8Rr + 3r^2\right) - 3r^2 - 6Rr}{2Rr} =$ $ 1 + \frac Rr$ .[/color]

[quote][color=darkred][b][u]Lemma 2[/u].[/b] In any [u]acute triangle[/u] $ ABC$ there is the inequality $ \boxed {\ \sin\frac A2 + \sin\frac B2 + \sin\frac C2\ \ge\ \frac 54 + \frac {r}{2R}\ }\ \ (2)$ .[/color][/quote]
[color=darkblue][b][u]Proof[/u].[/b] Apply the [u][b]Popoviciu[/b]'s inequality[/u] to the concave function  $ \mathrm{COS}$ on the interval $ \left (0\ ,\ \frac {\pi}{2}\right )$  :

$ \cos A + \cos B + \cos C +$ $ 3\cos\frac {A + B + C}{3}\ \le\ 2\cdot\left(\cos\frac {A + B}{2} + \cos\frac {B + C}{2} + \cos\frac {C + A}{2}\right)\ \Longleftrightarrow$

$ 1 + \frac rR + \frac 32\ \le\ 2\cdot\left(\sin\frac A2 + \sin\frac B2 + \sin\frac C2\right)\ \Longleftrightarrow\ \sin\frac A2\ +$ $ \ \sin\frac B2\ + \ \sin\frac C2\ \ge\ \frac 54\ + \ \frac r{2R}$ .[/color]

[quote][color=darkred]Prove that in any [u]acute triangle[/u] $ ABC$ there is the inequality $ \boxed {\ \sum\sqrt {\frac {b + c - a}{a}}\ \ge\ \sqrt {5 + \frac Rr + \frac {4r}{R}}\ \ge\ \sqrt {9 + 2\cdot\left(1 - \frac {2r}{R}\right)^2}\ \ge\ 3\ }$ .[/color][/quote]
[color=darkblue][b][u]Proof[/u].[/b] $ \left(\sum\sqrt {\frac {b + c - a}{a}}\right)^2 =$ $ \sum\frac {b + c - a}{a} + 2\cdot\sum\sqrt {\frac {(a + b - c)(a + c - b)}{bc}} =$ $ \sum\left(\frac ab + \frac ba\right) - 3 + 4\cdot\sum\sqrt {\frac {(s - b)(s - c)}{bc}} =$

$ \sum\frac {c^2 + 2ab\cdot\cos C}{ab} - 3 + 4\cdot\sum\sin\frac A2 =$ $ \sum\frac {c^2}{ab} + 2\sum\cos C - 3 + 4\cdot\sum\sin \frac A2 =$ $ \frac {a^3 + b^3 + c^3}{abc} + 2\cdot\left(1 + \frac rR\right) - 3 + 4\cdot\sum\sin\frac A2\ \stackrel{(1)\wedge (2)}{\ge}$

$ 1 + \frac Rr + 2\left(1 + \frac rR\right) - 3 + 5 + \frac {2r}{R} =$ $ 5 + \frac Rr + \frac {4r}{R}\ \Longrightarrow\ \sum\sqrt {\frac {b + c - a}{a}}\ \ge\ \sqrt {5 + \frac Rr + \frac {4r}{R}}$ . Prove easily that and [u]right inequality[/u] is truly.

Indeed, $ \sqrt {5 + \frac Rr + \frac {4r}{R}}\ge\sqrt {9 + 2\cdot\left(1 - \frac {2r}{R}\right)^2}$ $ \Longleftrightarrow 5 + \frac Rr + \frac {4r}{R}\ge 9 + 2\cdot\left(1 - \frac {2r}{R}\right)^2\Longleftrightarrow$ $ \frac {(R - 2r)^2}{Rr}\ge 2\cdot\left(1 - \frac {2r}{R}\right)^2\Longleftrightarrow R\ge 2r$ .[/color]

We can also prove that:
$ \sqrt{\frac{b+c-a}{a}}+\sqrt{\frac{c+a-b}{b}}+\sqrt{\frac{a+b-c}{c}} \ge \sqrt{9+3(1-\frac{2r}{R})^2} $
For any acute  triangle $ ABC $ ,we have:
$ sin\frac{A}{2}+sin\frac{B}{2}+sin\frac{C}{2} \ge m+\frac{(3-2m)r}{R} $ ,where $ m\le k=\frac{\sqrt{4-2\sqrt{2}}+3\sqrt{2}-5}{3\sqrt{2}-4} $
We can check that: $ k>\frac{4}{3}>\frac{5}{4} $.
With $ m=\frac{4}{3} $, we have Jackgarfunkel inequality.

$ \sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2} \ge m+\frac{(3-2m)r}{R} $ ,where $ m\le k=\frac{\sqrt{4-2\sqrt{2}}+3\sqrt{2}-5}{3\sqrt{2}-4}. $

a hard result.

you why
\ssin\frac{A}{2},ssin\frac{A}{2}.

[quote="quykhtn-qa1"]
For any acute  triangle $ ABC $ ,we have:
$ sin\frac{A}{2}+sin\frac{B}{2}+sin\frac{C}{2} \ge m+\frac{(3-2m)r}{R} $ ,where $ m\le k=\frac{\sqrt{4-2\sqrt{2}}+3\sqrt{2}-5}{3\sqrt{2}-4} $
We can check that: $ k>\frac{4}{3}>\frac{5}{4} $.
With $ m=\frac{4}{3} $, we have Jackgarfunkel inequality.[/quote]

This is indeed quite a strong inequality and the main difficulty in proving it by the use of some classic methods is due to the special equality case which occurs when $\triangle\, ABC$ is either [b]equilateral[/b] or [b]right-isosceles[/b]. However, we can overcome this "inconvenience" by making use of the following [b]VERY STRONG[/b] inequality, the [u][b]equality case[/b] of which is attained for any [b]isosceles triangle[/b][/u]:
\[\boxed{\ \sum\, \sin\frac A2\, \ge\, \sqrt {\frac {9R-2r}{4R}+\sqrt {\frac {s^2+r^2+2Rr}{2R^2}}}-\frac 12\ }\ \ \ \ .\]
I derived this amazing (although not aesthetically pleasing) inequality by using the identity
\[\boxed{\ \sum\, \cos\frac {B-C}2=\left(\frac 12+\sum\, \sin\frac A2\right)^2+\frac r{2R}-\frac 54\ }\]
and the following beautiful inequality, which I proved it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2315537#p2315537][b]here[/b][/url], that is:
\[\boxed{\ \sum\, \cos\frac {B-C}2\, \ge\, 1+\sqrt {\frac {s^2+r^2+2Rr}{2R^2}}\ }\ \ \ \ .\]
Thus, it remains to show that:
\[\sqrt {\frac {9R-2r}{4R}+\sqrt {\frac {s^2+r^2+2Rr}{2R^2}}}-\frac 12\, \ge\, k+\frac {\left(3-2k\right)r}R\]
which further simplifies to:
\[\boxed{\ s^2\, \ge\, \frac {\left[\left(4k^2+4k-8\right)R^2+\left(-16k^2+16k+14\right)Rr+\left(6-4k\right)^2r^2\right]^2-8R^2\left(2Rr+r^2\right)}{8R^2}\ }\ \ \ \ ,\]
where $k=\tfrac{\sqrt{4-2\sqrt 2}+3\sqrt 2-5}{3\sqrt 2-4}$ is the best possible constant for the type of inequality in question. Finally, the latter inequality can be proven by using the well known results which are valid in [b]non-obtuse[/b] triangles, viz. $\boxed{\, s^2\, \ge\, 2R^2+8Rr+3r^2\, }$ ([u][b]Walker[/b][/u]) and $\boxed{\, s\, \ge\, 2R+r\, }$ . Specifically, it can be shown - at least by the use of a computer programme - that:
\[\boxed{\ \begin{array}{rrrr}
2R^2+8Rr+3r^2\ge\tfrac {\left[\left(4k^2+4k-8\right)R^2+\left(-16k^2+16k+14\right)Rr+\left(6-4k\right)^2r^2\right]^2-8R^2\left(2Rr+r^2\right)}{8R^2} & \text{holds for} & 2\le\tfrac Rr\le 1+\sqrt 2 \\ \\
\left(2R+r\right)^2\ge\tfrac {\left[\left(4k^2+4k-8\right)R^2+\left(-16k^2+16k+14\right)Rr+\left(6-4k\right)^2r^2\right]^2-8R^2\left(2Rr+r^2\right)}{8R^2} & \text{holds for} & \frac Rr\ge 1+\sqrt 2\end{array}\ } \ \ \ \ \ .\]
We are done. :)

[quote=quykhtn-qa1]
\[\left\{\begin{array}{c}\triangle ABC \\\\  \max\{A, B, C\} \le 90^{\circ} \\\\ k=\frac{\sqrt{4-2\sqrt{2}}+3\sqrt{2}-5}{3\sqrt{2}-4} \end{array}\right\|\ \ \implies\ \ \boxed{\boxed{\sin\frac A2 + \sin\frac B2 + \sin\frac C2 \ge k + \left(3 - 2k\right)\cdot\frac rR}}\]
[/quote]

This slightly sharper version of Jack Garfunkel's inequality can also be proven using the following STRONG results:

\[
\boxed{
\begin{array}{lll}
\sin\frac A2 + \sin\frac B2 + \sin\frac C2 & \ge & \frac 12 + \sqrt{1 - \sqrt{1 - \frac {2r}R}} + \frac 12 \sqrt{1 - \frac {2r}R} \\\\
\sin\frac A2 + \sin\frac B2 + \sin\frac C2 & \ge & \frac{\sqrt{2}}2 + \sqrt{\frac 12 + \frac {\sqrt{2} - 1}2\cdot\frac rR}
\end{array}
}\ .
\]

[url=http://www.artofproblemsolving.com/community/c6h1221278p6108147][b]The first inequality[/b][/url] holds in any triangle and [u]attains equality for [b]most isosceles triangles[/b][/u]; the second one holds in non-obtuse-angled triangles, [u]attains equality for any [b]right-angled triangle[/b][/u] and it is a simple consequence of identity [b]141.2[/b] from [url=http://www.artofproblemsolving.com/community/c6h412623p2481722][b]here[/b][/url] (just replace $\phi$ by $\pi / 2$). Now we can prove the conclusion in two steps as in my previous post, namely:

\[
\boxed{
\begin{array}{lllcr}
\frac 12 + \sqrt{1 - \sqrt{1 - \frac {2r}R}} + \frac 12 \sqrt{1 - \frac {2r}R} & \ge & k + \left(3 - 2k\right)\cdot\frac rR & \text{holds for} & 2\le\frac Rr\le 1 + \sqrt{2} \\\\
\frac{\sqrt{2}}2 + \sqrt{\frac 12 + \frac {\sqrt{2} - 1}2\cdot\frac rR}         & \ge & k + \left(3 - 2k\right)\cdot\frac rR & \text{holds for} & \frac Rr \ge 1 + \sqrt{2}
\end{array}
}\ .
\]


I was just wondering if there is any formula or quick method to calculate the sum of the factors of a number. I have seen problems involving this calculations a number of times.  :)

first prime factorize the number
you will get
$ p_1^{e_1}p_2^{e_2}p_3^{e_3} \ldots p_n^{e_n}$
the sum of the number of prime factors is
$ \prod^n_{k\equal{}1} \sum^{e_k}_{l\equal{}0}p^l$
if that is kinda complicated, i will demonstrate this technique for the number 45360
$ 45360\equal{}2^4 \cdot 3^4 \cdot 5^1 \cdot 7^1$
sum of factors= $ (1\plus{}2\plus{}4\plus{}8\plus{}16)(1\plus{}3\plus{}9\plus{}27\plus{}81)(1\plus{}5)(1\plus{}7)$
equals some big number
the reasoning behind this formula is quite complicated but also quite elegant

[quote="stevenmeow"]first prime factorize the number
you will get
$ p_1^{e_1}p_2^{e_2}p_3^{e_3} \ldots p_n^{e_n}$
the sum of the number of prime factors is
$ \prod^n_{k \equal{} 1} \sum^{e_k}_{l \equal{} 0}p^l$
if that is kinda complicated, i will demonstrate this technique for the number 45360
$ 45360 \equal{} 2^4 \cdot 3^4 \cdot 5^1 \cdot 7^1$
[/quote]
From this step let me u tell the method in which I solved this question.
define a func f(n,k)=(kn-1)/(k-1). where k is the prime factor and n is the pth value of k.
if we take this example only.
f(4,2)=(8-1)(2-1)=7
similarly find for others and multiply them.
 If you want to find the sum of factors of 720
f(16)=(2*16-1)/(2-1) = 31
f(9) = (3*9-1)/(3-1) = 13
f(5) = (5*5-1)/(5-1) = 6
ur answer is 31*13*6 = 2418

Note, this is a function called sum-of-divisors function (Denoted $ \sigma(x)$).

This method intuitively makes sense. To take a small example, $ 24\equal{}2^3\cdot3$ we get $ (1\plus{}2^1\plus{}2^2\plus{}2^3)(1\plus{}3)$ Expanding this we get $ 1\plus{}2^1\plus{}2^2\plus{}2^3\plus{}3\plus{}2^13^1\plus{}2^23^1\plus{}2^33^1$ What we just listed out are all of the factors of 24.

Wikipedia also gives a nice trigonometric series:
$ \sigma(n)\equal{}\sum_{k\equal{}1}^{n}\sum_{j\equal{}1}^{k}\cos\left(\frac{2\pi jn}{k}\right)$.

Perhaps not the easiest thing to evaluate on paper, but it's interesting nonetheless.

See http://en.wikipedia.org/wiki/Divisor_function


Do a normalization: $ a\plus{}b\plus{}c\equal{}1$, the inequality becomes:

$ \sum_{cyc}{\sqrt{\frac{a^3}{1\minus{}3a\plus{}3a^2}}} \geq 1$

We choose appropriate constants $ \alpha, \beta$ such that

$ \sqrt{\frac{x^3}{1\minus{}3x\plus{}3x^2}} \geq \alpha x \plus{} \beta$ for all $ x\in [0,1]$

and that $ \alpha \plus{} 3 \beta \equal{} 1$.

Done!

$ \sqrt {\frac {a^3}{a^3 \plus{} (b \plus{} c)^3}}\geq\frac {a^2}{\sum a^2}$
$ \Leftrightarrow a^3(\sum a^2)^2\geq a^4[a^3 \plus{} (b \plus{} c)^3]$
$ \Leftrightarrow a^3(b^4 \plus{} c^4) \plus{} 2a^3\sum a^2b^2 \minus{} a^4(b^3 \plus{} c^3) \minus{} 3a^4(b^2c \plus{} bc^2)\geq 0$
$ \Leftrightarrow (b^4 \plus{} c^4) \plus{} 2\sum a^2b^2 \minus{} a(b^3 \plus{} c^3) \minus{} 3a(b^2c \plus{} bc^2)\geq 0$
$ \Leftrightarrow (b^2 \plus{} c^2)^2 \plus{} 2a^2(b^2 \plus{} c^2)\geq a(b \plus{} c)^3$
that $ LHS\geq \frac {(b \plus{} c)^4}{4} \plus{} a^2(b \plus{} c)^2 \equal{} (b \plus{} c)^2[\frac {(b \plus{} c)^2}{4} \plus{} a^2]\geq a(b \plus{} c)^3$
$ \Rightarrow \sum\sqrt {\frac {a^3}{a^3 \plus{} (b \plus{} c)^3}}\geq\sum\frac {a^2}{\sum a^2} \equal{} 1$

this inequality is very easy :lol:  . THANKS VERY MUCH

[quote="lia_h1"]Let $ a,b,c > 0$

Prove that:
$ \sqrt \frac {a^3}{a^3 \plus{} (b \plus{} c)^3} \plus{} \sqrt \frac {b^3}{b^3 \plus{} (b \plus{} c)^3} \plus{} \sqrt \frac {c^3}{c^3 \plus{} (a \plus{} b)^3} \ge 1$

I didn't prove that, so I'll get a bad mark :(

P/s: Sorry, I'm a newbie, so if I posted it in wrong box, please del it.  :blush:[/quote]
$ a\plus{}b\plus{}c\equal{}1$ then Jensen :wink:

[quote="lia_h1"]Let $ a,b,c > 0$

Prove that:
$ \sqrt \frac {a^3}{a^3 \plus{} (b \plus{} c)^3} \plus{} \sqrt \frac {b^3}{b^3 \plus{} (b \plus{} c)^3} \plus{} \sqrt \frac {c^3}{c^3 \plus{} (a \plus{} b)^3} \ge 1$

I didn't prove that, so I'll get a bad mark :(

P/s: Sorry, I'm a newbie, so if I posted it in wrong box, please del it.  :blush:[/quote]

See   http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62256


vector
inequalities
topology
triangle inequality
real analysis
real analysis unsolved

a) Let $ \parallel{} \parallel{}$ be any norm on $ \mathbb{R}^{m}$ and let $ B \equal{} \{x \in \mathbb{R}^{m} | \parallel{}x\parallel{} \le 1\}$. Prove that $ B$ is compact.

b) Show that any 2 norms on $ \mathbb{R}^{m}$ are equivalent (i.e. for $ \parallel{} \parallel{}_{1}$ and $ \parallel{} \parallel{}_{2} \exists c > 0$ s.t. $ \parallel{}x\parallel{}_{1} \le c\parallel{}x_{2}\parallel{}$  and $ \parallel{}x\parallel{}_{2} \le c\parallel{}x_{1}\parallel{}, \forall x \in \mathbb{R}^{m}$)

My biggest challenge is with the second part of this question, because how is it that for a norm which is a vector with distance and magnitude be equal to some other vector due to a specified $ c$?

If you have trouble in the abstract setting, try something more concrete first.
For example, in $ \mathbb{R}^2$ consider the two norms $ \|(x,y)\|_1 \equal{} \sqrt{x^2\plus{}y^2}$ and $ \|(x,y)\|_2 \equal{} |x|\plus{}|y|$.

For a), can I use the fact that $ \mathbb{R}^{m}$ is a metric space and try to show that it contains the limits of all its Cauchy sequences to prove completeness. And then from there....I could use some help or at least how I could use the inequality $ \parallel{} x \parallel{} \le 1$

a) is a straightforward consequence of the Heine-Borel theorem.  For b) try picking a basis and using the triangle inequality along with the list of norms of each basis vector.

could you explain what you mean by list of norms for each basis vector is?  Thanks

I mean pick a basis $ e_1, e_2, ... e_m$, look at the list of norms $ \parallel{}e_1\parallel{}, \parallel{}e_2\parallel{}, ... \parallel{}e_m\parallel{}$, and use it to bound the norm of an arbitrary vector.

So was the idea to use the linear independence of a basis and triangle inequality , as suggested, to come up with the constant $ c$?

for a) is there any way to form a closed ball in $ B$ so that I can show that it is closed? I can't think of any other way to show that it might be closed unless there are other ideas?

The question is slightly ambiguous; until you know that two norms on $ \mathbb{R}^m$ generate the same topology it's not clear whether "compact" means "compact with respect to the given norm" or "compact with respect to the usual norm."  In the former case, the ball is automatically closed with respect to the given norm, since the norm is continuous with respect to itself.  In the latter case, part b) implies that any two topologies generated by a norm are equivalent.


Is there such a number that if you increase and decrease it by 15, the digits are still the same as the original number (just arranged)?

probably, but I haven't figured out what it is  ;) .

To reverse the digits of a two-digit number, you have to increase/decrease it by a multiple of $9$.
To reverse the digits of a three-digit number, you have to increase/decrease it by a multiple of $99$.
To reverse the digits of a four-digit number, you have to increase/decrease it by a multiple of $999$.
AND SO ON...

Thus the problem has no solution.

You do not have to reverse the digits, just rearrange them.

Oh yeah, but either way my point is that no matter how large the number is, to rearrange the digits in any way you have to increase/decrease the number by a multiple of $9$.

When I first did the problem, I didn't think that there was a solution either.

[quote="tarquin"]Well if you increase it and decrease it by 15 you still have the same number ... right?[/quote]

I don't get it. I think tarquin's right.

I can prove that there is no possible solution.

First, no number under 100 will work, because to rearrange the digits of one of these numbers, you must reverse them.  Second, any number between 15 and 85 mod 100 will not work, because the digits beyond the hundreds digits will not change, and it will follow the same rules as numbers under one hundred.  Finally, numbers between 85 and 99 and 0 to 15 mod 100 will also not work, because, if there is a number, it will work for both the positive and negative parts.  Therefore, you will have at least one number of the resulting 3, that is between 15 and 85 mod 100, that must work for either the positive or the negative.  Since none of these numbers works for either, we can conclude that no numbers at all will work.


ratio
trigonometry
geometry proposed
geometry

Let $ ABC$ be a triangle and $ P$ is a point lie on bisector of $ A$, let $ A'B'C'$ be pedal triangle of $ P,PA\cap B'C' \equal{} I,PA'\cap B'C' \equal{} M$ prove that $ \frac {B'C'}{IM}\sin\frac {|B \minus{} C|}{2} \equal{} 2\frac {b \plus{} c}{a}\cos\frac {A}{2}$.


[quote="mdk"]Find the positive reals $ x_{1},...,x_{10}$ satisfying the sytem of equations $ (x_{1},...,x_{k})(x_{k}+...,x_{10})=1$ $ k=1,...,10$[/quote]
Please edit your problem.

Maybe he means the shortage of $ +$ signs.

wow, I really didn't see that, my fault. Since I couldn't edit my post, here it goes again

Find the positive reals $ x_{1},...,x_{10}$ satisfying the sytem of equations $ (x_{1},...,x_{k})(x_{k}+...+x_{10})=1$ $ k=1,...,10$


function
calculus
integration
limit
real analysis
real analysis unsolved

Let $ J_{n}(x)$ be a Bessel function.

Show that:

1) $ \int_{0}^{\infty}J_{n}(x) dx = 1$

2) $ \int_{0}^{\infty}\frac{J_{n}(x)}{x}dx = \frac{1}{n}$

Perhaps use the integral form definition for the Bessel functions to get a double integral that may be easier to work with?

Number 2) can be solved as follows:

$ \int_{0}^{\infty}\frac{f(t)}{t}dt=\int_{0}^{\infty}F(s)ds$, where

$ F$ is the Laplace transform of $ f$.

Now, we know that the Laplace transform of the Bessel function $ J_{n}$ is given by:

$ F(J_{n})(p)=\frac{(-p+\sqrt{p^{2}+1})^{n}}{\sqrt{p^{2}+1}}$, and hence

$ \int_{0}^{\infty}\frac{J_{n}(x)}{x}dx=\int_{0}^{\infty}\frac{1}{(p+\sqrt{p^{2}+1})^{n}\sqrt{p^{2}+1}}dp$

$ =\frac{-1}{n(p+\sqrt{p^{2}+1})^{n}}\biggm|_{0}^{\infty}=\frac{1}{n}$.


linear algebra
matrix
Gauss
algebra
system of equations
linear algebra unsolved

I have to solve system of equations using Gauss-Zeidel method, this method is only appliable to systems whose matrix are diagonally dominant (http://en.wikipedia.org/wiki/Diagonally_dominant_matrix) . My system of equations is:

2.7*x1+3.3*x2+1.3*x3 = 2.1;
3.5*x1-1.7*x2+2.8*x3 = 1.7;
4.1*x1+5.8*x2-1.7*x3 = 0.8;

As you see it's matrix is not diagonally dominant, how can i transform it, that i could solve using Gauss-Zeidel method wit it. Thank you.


Does there exist a continuos $f: R \rightarrow R$ such that:
$f(x)$ is rational iff $x$ is rational?

Yup! $f(x)=x$ :D. Maybe you meant $f(x)$ is [b]irrational[/b] iff $x$ is rational? In that case the answer is no:

$f$ must take two different irrational values, so it takes all the irrational values in an interval because it's continuous so it has Darboux property. This means that it takes $\mathbb Q$ to $\mathbb {(R\setminus Q)}\cap (a.b)$, and this can't be done because the second set has continuum cardinality and the first set is countable.

You see f(x)+x is continious and also it can not get rational numbers.Contradiction

Sorry for revising this topic after an year or so, but I think there is an easier solution to this problem (please let me know if I am repeating what someone else has already written in another topic):

Does there exist a continuous function $f: \mathbb{R}\to\mathbb{R}$ such that $f(x) \in \mathbb{Q}$ if and only if $x \in \mathbb{R}\setminus \mathbb{Q}$?

Suppose such a function $f(x)$ exists. Consider the function $f(f(x))$. We have that $f(f(x))\in \mathbb{Q}$ if and only if $x \in \mathbb{Q}$. Therefore the functions $g(x): = f(x) - f(f(x))$ and $h(x) = f(x) + f(f(x))$ enjoy the property $g(x), h(x) \in \mathbb{R}\setminus\mathbb{Q}$ for all $x\in\mathbb{R}$. And since $g(x)$ and $h(x)$ are continuous functions, they must be identically equal to (irrational) constants, which means that $f(x) = \frac{1}{2}(g(x) + h(x))$ must be identically equal to a constant, which is the desired contradiction.

Cheers,


[quote="Beat"]$a,b,c,d,e$ are positive real numbers, such that $a^2+b^2+c^2+d^2+e^2\le 1$. Prove that $\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+cd}+\frac{1}{1+de}+\frac{1}{1+ea}\ge \frac{25}{6}$[/quote]
Very easy.
We have $1\ge a^2+b^2+c^2+d^2+e^2\ge ab+bc+cd+de+ea$
And $\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+cd}+\frac{1}{1+de}+\frac{1}{1+ea}\ge\frac{25}{5+ab+bc+cd+de+ea}\ge\frac{25}{6}$ :lol:


Let $ S(n)$ denote the sum of the digits of a natural number $ n$ (in base $ 10$). Prove that for every $ n$, $ S(2n) \le 2S(n) \le 10S(2n)$. Prove also that there is a positive integer $ n$ with $ S(n)\equal{}1996S(3n)$.

[quote=bboypa]Already posted.[/quote]

Where

[quote=mistakesinsolutions][quote=bboypa]2S(n)=2S(10n)[/quote]
wrong)[/quote]

Why is it wrong?
This is correct because $S(n) = S(10n)$ obviously


12Xcubed - 61Xsquared+2x + 30 = 0

Root = -2/3

Please Help!

hello, do you want to solve the equation $ 12x^3\minus{}61x^2\plus{}2x\plus{}60\equal{}0$?
Sonnhard.


Find all positive integer solutions to $ 7^x\minus{}3^y\equal{}4$.

x=y=1 is a solution, I claim there are no others.

Consider the equation mod 8. $ ( \minus{} 1)^x \minus{} 3^y\equiv 4$ gives x,y odd.

Now, if $ y\ge 2$ consider the equation mod 9. $ 7^x\equiv 4$ so a quick check shows that 7^x cycles 7,4,1,7,.. so x is congruent to 0 mod 3. Therefore x is congruent to 3 mod 6. Let x=3+6k with $ k\ge 0$.

Mod 13, we have $ 7^{3 \plus{} 6k} \equiv 5,8$, and $ 3^y \equiv 1,3,9$ The only pair that subtracts to 4 mod 13 is (5,1), so $ 3^y \equiv 1$ mod 13. A quick check shows that 3^y cycles 1,3,9,1,3,9,... so x is congruent to 0 mod 3.

Let x=3m with $ m\ge 1$.

$ (7^{1 \plus{} 2k})^3 \minus{} (3^m)^3 \equal{} p^3 \minus{} q^3 \equal{} (p \minus{} q)(p^2 \plus{} pq \plus{} q^2) \equal{} 4$ so $ p^2 \plus{} pq \plus{} q^2|4$ but
$ p^2 \plus{} pq \plus{} q^2 \ge q^2 \ge 9$ so we have a contradiction!

You can use \equiv   $ \equiv$   instead of \cong   $ \cong$   for congruences.

Another tip: $ 7 \equiv 3 \pmod{4}$ is a good way to include the modulus in the $ \text{\LaTeX}$ code.


Does anybody here play DotA (Defense of the Ancients)? Discuss.

ya i play a wicked shadow shaman, witch doctor, and axe in league play

Oh... so thats what Dota is. I thought it was Dork_of_the_ages or something

I play DotA. My favorite hero is Rikimaru.

[quote="echen"]I've heard the song... I may sound like an idiot but their is a song called DotA by Basshunter and the music video is all of them playing DotA.[/quote]

DotA is the most popular song in FFR. (I wonder if the reason for that is the game... the song isn't that good)

Hm, I really suck and I'm not really experienced, but I kinda like silencer. Glaives is awesome for harassing and for dps, and global silence and pretty awesome. Although I have to say, the other two skills are incredibly useful (for last word, a lot of spells are out of 700 range, and even if they aren't they can just animation cancel it).

I play a wide variety of heroes. My favorite and the best has to be anti-mage though. I also like broodmother for mad pushing, rooftrellen and pandaren brewmaster for tanking, mortred for late-game carrying. (Mortred got really better with stifling dagger whereas the old shadow strike was virtually useless)

I'm playing regulary with some friends since half a year, but about 2 till 6 human players only. I dislike some stressful heros (especially those that have lot of clones/minions one needs to take too much care of, worst one being Geomancer), but there is no hero I like most (Ursa Warrior was once my favorite, but others get better at time).

DotA is extremely popular with the Asians primarily at my high school; I believe one crafty one placed DotA on the school network (hidden among teacher files) when he hacked some of the higher level accounts. (Along with Counter Strike  :P ).

I've never played Warcraft, actually - because my dad never buys computer games, but mostly it's because FPS's are better?  :D

[quote="AznLeeBoy"]I play a different hero every time.
but my favorites have been
-Traxex(Drow Ranger)
-Clinks(Bone Fletcher)
-Puck(idk.)
-Ursalis(Ursa)
-Axe(idk)

I also have seperate builds for each kind of hero

^*-Agility-*^
=-Ursa
=-Troll
=-Drow
=-B. Fletcher
-Bracer.
-Wraith Band.
-Healing Salve
-Perseverance
-sell wraith band
-power treads
-Helm of Dominator(or Vladmir's Offering)
-Yasha
-Sell Bracer
-Sange
-SnY
-Butterfly
-Radiance.
i pretty much destroy everyone after that :D

^*-Strength-*^
=-Axe
=-Crixalis
-Vanguard
-Boots of Travel
-Vladmir's Offering(or Helm of Dominator.)
-BattleFury
-Monkey King Bar
-SnY
this is a very professional build,
you're going to need early kills
and first bloods to be able to afford.
but if you manage to get all of these
you will dominate the game.

^*-Intelligence-*^
-Pugna
-Invoker(SO HARD TO USE.)
I don't use intelligence that much.
but i know a little bit about the
early build.

-Helm of Rejuvination
-Bracer
-Perseverance
-Oblivion Staff
-Mystic Wand
-i don't really know.
haven't got far enough as a int hero.

i'll post up whatever i can once i get even MORE experienced.

RATE ME HIGH IF THIS HELPED YOU IN ANY WAY! :]

thanks.[/quote]

I don't recommend Sange n Yasha for any heroes except a select few (like the Juggernaut). Also, you should never build a perserverance unless you are planning to use it for some higher level item.

Pugna can be built in varieties of ways. One of my friend loved to stick a Dagon on him.

As for Invoker, here is three items I always use with him:

-Guinsoo: A must for Kael
-Linken: For defense
-Boots of Travel: mobility


Let
\[ \left(a_nx^n\plus{}\dots \plus{}a_1x\plus{}a_0\right)(b_kx^k\plus{}\dots \plus{}b_1x\plus{}b_0)
\equal{} 1\plus{}x\plus{}x^2\plus{}\dots\plus{}x^{n\plus{}k}\]
where the coefficients $a_i, b_i$ are non-negative reals and $ a_0\equal{}b_0\equal{}1$.

Prove every $ a_i$ and every $ b_i$ is either $0$ or $1$.

Try some complex number substitution perhaps.

It is sufficient to show that the $a_i$ and $b_i$ are integers. :D (Easy to see why, I'll leave that to the reader.)

Now, all the roots of both LHS polynomials $a(x)$ and $b(x)$ are $(n+k+1)$th primitive roots. Those roots have certain  [url=http://en.wikipedia.org/wiki/Cyclotomic_polynomial]cyclotomic polynomials[/url] as minimal polynomials. As those divide either $a(x)$ or $b(x)$, we can write $a(x)$ and $b(x)$ as products of cyclotomic polynomials, possibly with a constant factor. But as all absolute terms are $1$ and $a_0=b_0=1$, the factor is $1$. And so $a_i,b_i\in\mathbb Z$ qed.


group theory
superior algebra
superior algebra solved

Let $ G$ be a group with this property that for every two elements $ a,b\in G$, $ (ab)^{n}=a^{n}b^{n}$, which $ n$ is a positive integer. Prove that for every elements like $ a,b\in G$ the following holds :

                                                                \[ a^{n}b^{n-1}=b^{n-1}a^{n}\]

$ (ab)^{n}\equal{} a^{n}b^{n}\Rightarrow a(ba)^{n\minus{}1}b \equal{} a^{n}b^{n}\Rightarrow (ba)^{n\minus{}1}\equal{} a^{n\minus{}1}b^{n\minus{}1}\Rightarrow$
$ (ba)^{n}\equal{} (ba)\cdot(a^{n\minus{}1}b^{n\minus{}1})$. But $ (ba)^{n}\equal{} b^{n}a^{n}\Rightarrow b^{n}a^{n}\equal{} ba^{n}b^{n\minus{}1}\Rightarrow b^{n\minus{}1}a^{n}\equal{} a^{n}b^{n\minus{}1}$.

A more interesting problem with $ (ab)^{n}\equal{} a^{n}b^{n}$ is to show that if $ n$ does not divide $ |G|$ (finite) then all $ n\minus{}1$ degree elements commute with everything else.

This one follows from the fact that $ a\to a^n$ is group homomorphism and since $ n$ does not divide $ |G|$ it follows that it is isomorphism and hence for any $ a\in G$ there is some $ u\in G$ so that $ u^n\equal{}a$ but $ \alpha^{n\minus{}1}$ commutes with $ u^n$ an so commutes with $ a$. But $ a$ was arbitrary chosen so $ \{\alpha^{n\minus{}1}|\alpha\in G\}\subset Z(G)$


Let $a,b$ integer positive numbers...

Prove that:

$a^3 b - b^3a = 0 \mod (6).$
for all $a,b$

Carlos Bravo :)
Lma -PERU

$a^3b-b^3a=b(a-1)a(a+1)-a(b-1)b(b+1)$
The product of the three consecutive numbers is the multiple of $6$, we are done.

$a^3b-b^3a=ab(a-b)(a+b)$ first we prove 2 divides it. if one of the numers is even, we are done, otherwise $a-b$ is even so we are equally done. Now we prove 3 divides it. If 3 divides one of the numbers we are done, otherwise one between $a+b$ or $a-b$ is a multiple of $3$ since they are different and they cannot be 1 and 2 mod 3 because it would imply  one of $a$ or $b$ to be $0$ mod 3


topology
abstract algebra
advanced fields
advanced fields unsolved

Let $G$ be a path-connected and locally path-connected topological group. $\mu: G \times G \to G$ is its composition law.
1/$\phi: \pi_{1}(G,1) \times \pi_{1}(G,1)$ sends a couple of (class of) paths $(\gamma_{1}, \gamma_{2})$ to a path in $\pi_{1}(G,1)$ defined by
$\phi(\gamma_{1}, \gamma_{2})(t)=\mu(\gamma_{1}(t), \gamma_{2}(t))$. Show that $\phi$ is a well-defined morphism from $\pi_{1}(G,1) \times \pi_{1}(G,1)$ towards $\pi_{1}(G,1)$ (is $\phi$ sends $(\gamma_{1}, \gamma_{2})$ towards $\gamma_{1}. \gamma_{2}$ where $.$ is the usual composition law of $\pi_{1}(G,1)$).

2/ show that $\pi_{1}(G,1)$ is abelian.

is it [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=abelian+group+topological&t=120436]this[/url] one?

I admit I don't really understand what you have tried to show in this post cited above ?

well, i admit i didn't spend much time giving explanations, in that topic..

then: i think the first point is pretty straightforward (i didn't even try it).
as regarding the second one.. just consider an homotopy between two loops $\alpha, \beta$ as a map from $S(I)$* to your space.
now $S(I)$ is homeomorphic to the square $I\times I$, and you can choose this homeomorphism to send $[(0,0)]$ (the equivalence class of the point $(0,0)$, which is a point in $S(I)$) to $[0,0]$, and $[(1,1)]$ to $(1,1)$.
finally, consider the map $H: I\times I \to G$, $H: (s,t) \mapsto \alpha(s)\beta(t)$.
using the homeomorphism defined before, what you get is an omotopy from $\alpha * \beta$ (which is the low-right border of your square) and $\beta * \alpha$ (which is its left-high border).
by the way, the diagonal is your loop $\alpha\cdot \beta$, where $\cdot$ is the group operation.

* $I = [0,1]$, and $S(X)$ is the suspension of $X$, which is $X\times I$ where points of $X\times\{0\}$ and points of $X\times\{1\}$ are identified (and are two distinct points, not to one single point).


ratio
number theory unsolved
number theory

Find all [tex]a[/tex], [tex]b[/tex] integers, if:
[tex](a^2+ab+b^2)/(ab-1)[/tex]
is also an integer.

If I am not wrong then the equation:
[tex] \frac{a^2+b^2+1}{ab-1}=p [/tex]
have infinitely number solutions.And the least solutions are:
For [tex] p=3;a=4,b=2 [/tex].
For [tex] p=6;a=2,b=1 [/tex]. :)

I think it has already been posted. The topic was started by Valentin, if I remember correctly.

If you will look more carefully you will see that:
- [tex]a,b[/tex] any integers
(case when they have opposite signs is lost: [tex]\frac{a^2-ab+b^2}{ab+1}=p[/tex] )
- i asked for a general explicit solution (not only the possible values of that [tex]p[/tex]), but i doubt that it is possible   :maybe: