trigonometry
vector
function
inequalities
calculus
calculus computations

A vector is in R3 such that the angles it make with the x,y,z axes are a,b,c which are always between 0 and 90. So we know that cos^2(a)+cos^2(b)+cos^2(c)=1.

Prove or disprove a+b+c<=180.

Consider this as a constrained max-min problem: Find the extreme values of $ a\plus{}b\plus{}c$ for $ a,b,c\in\left[0,\frac{\pi}2\right]$ subject to $ \cos^2a\plus{}\cos^2b\plus{}\cos^2 c\equal{}1.$

First, consider boundary cases:

If $ a\equal{}0,$ then we have $ \cos^2b\plus{}\cos^2c\equal{}0,$ which implies that $ b\equal{}c\equal{}\frac{\pi}2.$ In that case, we have $ a\plus{}b\plus{}c\equal{}\pi.$ There are symmetric cases for $ b\equal{}0$ or $ c\equal{}0.$

If $ a\equal{}\frac{\pi}2,$ then we have $ \cos^2b\plus{}\cos^2c\equal{}1.$ But then $ \cos^2c\equal{}\sin^2b,$ and since we are restricted to the first quadrant, that means that $ \cos c\equal{}\sin b,$ so that $ b$ and $ c$ are complementary angles: $ b\plus{}c\equal{}\frac{\pi}2.$ Thus $ a\plus{}b\plus{}c\equal{}\pi.$ In fact, we have that $ a\plus{}b\plus{}c\equal{}\pi$ in all cases in which any of $ a,b$ or $ c$ are either $ 0$ or $ \frac{\pi}2,$

That leaves only the interior to cope with. Lagrange multipliers yields:
$ 1\equal{}\minus{}2\lambda\cos a\sin a,$  $ 1\equal{}\minus{}2\lambda\cos b\sin b,$ and $ 1\equal{}\minus{}2\lambda\cos c\sin c.$ Hence $ \sin2a\equal{}\sin2b\equal{}\sin2c.$

There a number of cases for this equation. We'll save the obvious $ a\equal{}b\equal{}c$ for last. First consider the possibility $ a\equal{}c,b\equal{}\frac{\pi}2\minus{}a.$ Then
$ \cos^2a\plus{}\cos^2b\plus{}\cos^2c\equal{}\cos^2a\plus{}1\minus{}\cos^2a\plus{}\cos^2a\equal{}1.$
We conclude that $ \cos^2a\equal{}0$ or $ a\equal{}\frac{\pi}2,$ which is a previously explored case. But all cases in which $ a,b,$ and $ c$ are not identical, two of them are the same and one is the complementary angle; these all turn out to be previously explored boundary cases.

Finally we come to the one and only interior critical point, where $ a\equal{}b\equal{}c.$ This leads to $ \cos(a)\equal{}\cos(b)\equal{}\cos(c)\equal{}\frac1{\sqrt{3}}.$ Hence $ a\plus{}b\plus{}c\equal{}3\cos^{\minus{}1}\left(\frac1{\sqrt{3}}\right).$ Since $ \frac1{\sqrt{3}}>\frac12$ and $ \cos^{\minus{}1}$ is a decreasing function, we see that

$ a\plus{}b\plus{}c\equal{}3\cos^{\minus{}1}\left(\frac1{\sqrt{3}}\right)<3\cos^{\minus{}1}\left(\frac12\right)\equal{}
3\cdot\frac{\pi}3\equal{}\pi.$

This proves it. The original statement is true, and we can say the following: that for all such $ a,b,$ and $ c$ (restricted to the first octant, as the problem was stated)
\[ 3\cos^{\minus{}1}\left(\frac1{\sqrt{3}}\right)\le a\plus{}b\plus{}c\le\pi.\]
[color=green][Had this problem been posted in the inequalities section rather than the calculus section, someone else would have solved it by some different method.][/color]


I kinda remember some identity which goes as follows:

If $ a,b,c$ are coplanar, non collinear vectors then

$ \alpha a \plus{} \beta b \plus{} \gamma c \equal{} 0$
=> $ \alpha \plus{} \beta \plus{} \gamma \equal{} 0$

or something like this. Can someone help me remember.

Since we are in two dimensions, the vector equation $ \alpha a \plus{} \beta b \plus{} \gamma c \equal{} 0$ is a system of two linear equations. Non-collinearity ensures that nothing degenerates, so the space of solutions has dimension $ 3\minus{}2\equal{}1$. This means that all such triples $ (\alpha,\beta,\gamma)$ are multiples of each other.
Everything else depends on the vectors $ a,b,c$; indeed, for any given triple $ (\alpha,\beta,\gamma)$ one can find $ a,b,c$ so that $ \alpha a \plus{} \beta b \plus{} \gamma c \equal{} 0$.

Another way of saying this:

What conjurer wrote is simply the [i]definition[/i] of linear independence - the independence of vectors $ a,b,$ and $ c.$

Three vectors that are coplanar must be linearly dependent - which is what mlok said.

If we have vectors $ a,b,$ and $ c$ in $ \mathbb{R}^3$ (or higher dimension) with the property that $ \alpha a\plus{}\beta b\plus{}\gamma c\equal{}0\implies \alpha\equal{}\beta\equal{}\gamma\equal{}0,$ then the vectors, being independent, are not coplanar.

What conjurer wrote was not really linear independence: it had $ \alpha\plus{}\beta\plus{}\gamma\equal{}0$ instead of $ \alpha\equal{}\beta\equal{}\gamma\equal{}0$. I don't know if this was on purpose or not.

I meant $ \alpha \plus{} \beta\plus{} \gamma \equal{} 0$ only .

@mlok: OK I get that a,b,c are multiples of each other but does that mean when we write $ \alpha a\plus{} \beta b\plus{} \gamma c \equal{} 0$ for a coplanar system, then it implies
$ \alpha \plus{} \beta\plus{} \gamma \equal{} 0 ?$

Well here is a question I encountered:

The points represented by [b]a[/b],[b]b[/b],[b]c[/b],[b]d[/b] are coplanar and (sinA)[b]a[/b] + (2sin2B)[b]b[/b] + (3sin3C)[b]c[/b] - 4[b]d[/b] =0 , then the minimum value of
$ sin^2 A \plus{}sin ^22B \plus{}sin^23C$ is?

And here is my ,unfortunately, not so well understood solution:

[hide]Note that if [b]a[/b],[b]b[/b],[b]c[/b],[b]d [/b]are coplanar, then (sinA)[b]a[/b] + (2sin2B)[b]b[/b] + (3sin3C)[b]c[/b] - 4[b]d[/b] will imply that sinA + 2sinB + 3sinC -4 =0

Now by Cauchy Swarz:

$ (sin^2A \plus{}sin^22B \plus{}sin^23C)(1 \plus{} 2^2 \plus{} 3^2) > \equal{} (sinA \plus{} 2sinB \plus{} 3sin3C)^2$

=> $ S(14) > \equal{} (16)$

=> $ S> \equal{} 8/7$

This is the correct answer.But I solved this because I kindof remembered some identity which I quoted above.But I am not sure of that.[/hide]

Being unsure about the identity, you should have tried drawing some triples of vectors on the plane. For example, if $ \mathbf{a}\equal{}\langle 1,0\rangle$, $ \mathbf{b}\equal{}\langle 0,1\rangle$, and $ \mathbf{c}\equal{}\langle 34,\minus{}17.256\rangle$, then $ 34\mathbf{a}\minus{}17.256\mathbf{b}\minus{}\mathbf{c}\equal{}0$. Clearly, the sum of coefficients is not zero.

I don't understand the problem you posted, in particular because it has points (not vectors?) being added. What if $ d\equal{}0$? Then no matter what $ a,b,c$ are, you can take $ A\equal{}B\equal{}C\equal{}0$, thus achieving the minimum of $ 0$.

For points, we want to think in affine terms, where an expression like $ \frac12A$ doesn't define a point. A linear combination with total coefficient 1 (such as $ tA\plus{}(1\minus{}t)B$) is a point and a linear combination with total coefficient zero (such as $ A\minus{}B$) is a vector.

There, we get this:
Three points $ A,B,C$ are collinear if and only if there are some $ \alpha,\beta,\gamma$ not all zero with $ \alpha\plus{}\beta\plus{}\gamma\equal{}0$ and $ \alpha A\plus{}\beta B\plus{}\gamma C\equal{}0$.

This is actually linear dependence of the vectors $ B\minus{}A$ and $ C\minus{}A$, expressed in symmetric form. The question of whether four points are coplanar is similar: there are $ \alpha,\beta,\gamma,\delta$ not all zero such that $ \alpha\plus{}\beta\plus{}\gamma\plus{}\delta\equal{}0$ and $ \alpha A\plus{}\beta B\plus{}\gamma C\plus{}\delta D\equal{}0$.

[quote="jmerry"]

There, we get this:
Three points $ A,B,C$ are collinear if and only if there are some $ \alpha,\beta,\gamma$ not all zero with $ \alpha \plus{} \beta \plus{} \gamma \equal{} 0$ and $ \alpha A \plus{} \beta B \plus{} \gamma C \equal{} 0$.

This is actually linear dependence of the vectors $ B \minus{} A$ and $ C \minus{} A$, expressed in symmetric form. The question of whether four points are coplanar is similar: there are $ \alpha,\beta,\gamma,\delta$ not all zero such that $ \alpha \plus{} \beta \plus{} \gamma \plus{} \delta \equal{} 0$ and $ \alpha A \plus{} \beta B \plus{} \gamma C \plus{} \delta D \equal{} 0$.[/quote]

 Does it necessarily imply that when we write something like $ \alpha A \plus{} \beta B \plus{} \gamma C \plus{} \delta D \equal{}0$ where A,B,C,D are position vectors then we need to have  $ \alpha \plus{} \beta \plus{} \gamma \plus{} \delta \equal{} 0$ such that not all $ \alpha,\beta,\gamma,\delta$ are zero. I mean if the converse is true then it necessarily implies the points are coplanar but what about the original statement ?
[hide]
I am sorry for asking stupid questions, I am new to this topic in particular and mathematics in general.  :oops:  :blush: [/hide]

I don't like the term "position vectors" here, because they don't behave like vectors. While we can make some sense of $ A\plus{}B$, it doesn't represent a point. If we translate everything, points move together, and $ A\plus{}B$ moves twice as far. To make an expression represent a point independent of where we place the origin, it needs total weight exactly 1.
Now, sums with total weight zero do act like vectors- we can add them or multiply by constants, and get back another expression of the same type. That's what the $ \alpha\plus{}\beta\plus{}\gamma\plus{}\delta\equal{}0$ condition is about; without it, $ \alpha A\plus{}\beta B\plus{}\gamma C\plus{}\delta D$ makes no sense, since $ 0$ is a vector and not a point. Making all of the coefficients zero is the trivial solution; we care whether there are any others.


Where can I find amc10b 2006 answers?
By the way,what's the difference between amc10aand AMC10b?

Go up to the toolbar thing on top (not on the side) and click Contests (if AoPS) or Resources (if ML, I think), then click on USA, then AMC10, then 2006, then scroll down, as the first test will be A. Here's a link if you wanna skip that... (but keep in mind how to use AoPS' nice resources section for future use :) )
http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43&year=2006

And starting from 2002 (I think), there were two test dates of the AMC10 and AMC12. Basically, you could take both, giving you 2 chances to do well. So it's two different AMC's that happen to fall on the same year.

Thx. but where is the key?And which is more difficult?AMC10A or AMC10B?

First question:
Use our archive:
http://www.unl.edu/amc/e-exams/e5-amc10/e5-1-10archive/2006-10a/06AMC10AB-answers.html

Second question:  Is answered in many places on this forum.  You can see in the archives that statistics on the two contests are essentially equal.

Steve Dunbar
MAA Director, American Mathematics Competitions


Show that there exists a sequence of polynomials $ P_{0}(x), P_{1}(x),..., P_{K}(x),...,$ such that $ deg P_{k}(x)\equal{}k$, and   $ P_{n}(P_{m}(x))\equal{} P_{m}(P_{n}(x))$,...,           for all pairs of non-negative integers $ m$ and $ n$

$ P_k (x) \equal{} x^k$ seems to work just fine... do you wish to add a further condition?


geometry
perimeter
inequalities proposed
inequalities

$a,b,c$ are the side lengths of a triangle. Prove that
\[\sum{\frac{a}{b+c-a}}\ge \frac{a+b+c}{\sqrt[3]{abc}}\]

if you aren't replying cause i said i have a solution, look at the name of this forum. i'm interested in second opinions. so please post your solution if you have one. or if you want i can post mine. :)

[quote="Albanian Eagle"]$\sum_{cyc}\frac{a}{b+c-a}=\frac{1}{2}\sum_{symm}\frac{p-a}{p-b}\geq$
$\frac{p}{\sqrt[3]{(p-a)(p-b)(p-c)}}$
so we just have to show
$abc\geq (a+b-c)(b+c-a)(c+a-b)$ which is just AM-GM or Schur.

Happy new year every one  :D[/quote]

i dont understand what's $p$ here. is it the perimetre? if it is then how does the expression come? someone please explain. :blush:


[tex]\[
\{ (x,y) \in \mathbb{R}^2 ^2  + y^2  < 4,x \leqslant 3\}
\]
[/tex]

Show if the set is bounded,open,closed or compact.

[hide]Am I missing something or the constraint x <= 3 is useless ??

I would say, obviously bounded (|x|,|y|<= 2), open, not closed (and so not compact).

PiDeltaPhi asked some more details (indeed, less would be difficult   )

S = the set.

* S is bounded : x+y < 4 implies  (|x|,|y|< 2) so the set is bounded.

* S is not closed :

. consider P(n) = (0,2-1/n), P(n) is the set but P(n) -> (0,2) not in the set

or

. T = IR\S, T is not open, B({2,0},r) (open disk, r > 0) is never completly inclosed in S

* S is open :

. f IR -> IR, f(x,y) = x+y. f is continuous and S = f^-1(]-oo,2[), ]-oo,2[ is open

or

. or let (x,y) in S. B({x,y}, (2-sqrt (x+y))/2)) is completly inclosed in S.

* S is not compact since it is not closed (a compact is always closed)

Hope it's clear (and correct!).

[/hide]

[quote="belenos"]* S is bounded : x+y < 4 implies  (|x|,|y|< 2) so the set is bounded.[/quote]

Or, alternatively, that's just [tex]||(x, y)||_2 < 2[/tex], so it's bounded by that (since all norms on [tex]\mathbb{R}^2[/tex] are equivalent).


inequalities
geometry
probability
LaTeX
number theory

something says that cauchy's inequality is:

xy :^3: +yz :^3: +zx :^3:  :ge: xyz(x+y+z).

but it confuses me and i dont get the point of it.  so please help!!!

btw, i only know geometry so please dont go into alg ii and all that stuff.

thnx!!!

Cauchy-Schwarz Inequality states that for all reals

[tex]a_1, a_2, \ldots, a_n[/tex] and [tex]b_1, b_2, \ldots, b_n[/tex] (where n is a natural number), the inequality

[tex](a_1^2 + a_2^2 + \ldots + a_n^2)(b_1^2 + b_2^2 + \ldots + b_n^2) \ge (a_1b_1 + a_2b_2 + \ldots + a_nb_n)^2[/tex]

is satisfied; equality is established iff

[tex]{{a_1}\over{b_1}}={{a_2}\over{b_2}}=\ldots={{a_n}\over{b_n}}[/tex].

Could someone post some problems that relate to this because it's new to me and I would like to learn how to use it?  Thanks
P.S. I would prefer you to start with easier ones first.

1. Prove that for reals [tex]a,b,c[/tex],

[tex](a^2 + b^2 + c^2) \ge {{1}\over{3}}(a + b + c)^2[/tex].

2. Prove that for positive reals [tex]x_1,x_2,\ldots,x_n[/tex],

[tex](x_1 + x_2 + \ldots + x_n)(x_1^{-1} + x_2^{-1} + \ldots + x_n^{-1}) \ge n^2[/tex].

3. (a bit harder) Prove that for positive reals [tex]a,b,c,d[/tex] with [tex]a+b+c+d=16[/tex],

[tex](a + c^{-1})^2 + (c + a^{-1})^2 + (b + d^{-1})^2 + (d + b^{-1})^2 \ge 289/4[/tex].

I think this thread belongs in intermediate.

[quote="Osiris"]I think this thread belongs in intermediate.[/quote]

I think so too. I'll move it soon.

Here's a tip on levels: if a problem requires curricular math beyond geometry, it most likely is NOT a Getting Started level problem. Problems with elementary solutions (for example, many number theory problems) fit well on Getting Started, but many problems involve too high a degree of curricular math to really fit on Getting Started.

#3:

[hide]

(1+1+1+1)((a+1/b)^2+(b+1/c)^2+(c+1/d)^2+(d+1/a)^2)  :ge:  (1*(a+1/b)+1*(b+1/c)+1*(c+1/d)+1*(d+1/a))^2

4((a+1/b)^2+(b+1/c)^2+(c+1/d)^2+(d+1/a)^2)  :ge:  (a + b + c + d + 1/a + 1/b + 1/c + 1/d)^2

a+b+c+d = 16, and if min(1/a+1/b+1/c+1/d) = 1, then:

((a+1/b)^2+(b+1/c)^2+(c+1/d)^2+(d+1/a)^2)  :ge:  (16+1)^2/4

((a+1/b)^2+(b+1/c)^2+(c+1/d)^2+(d+1/a)^2)  :ge:  289/4

min(1/a+1/b+1/c+1/d) = 1 is justified:

1/m+1/n=(m+n)/mn.  If m+n is a fixed z, then the minimum occurs when mn is maximized, which is when m=n=z/2 since m+n is fixed (basic algebra shows this).  Thus min(1/m+1/n) = z/(z/2)^2 = 4/z.  Let us partition 16 into x which will further partition into a and b and y which will further partition into c and d.  The minimum value of 1/c+1/d = 4/y.  The minimum value of 1/a+1/b = 4/x.  We want to minimize 4/x+4/y or 1/x+1/y, which as done with m and n, occurs when x = y = 16/2 = 8.  Therefore, minimum value of 1/a+1/b+1/c+1/d is achieved when a=b=c=d.  Now I know there must be a more straightforward way.  Any suggestions?[/hide]

Thanks for the alternative solution.  I don't have to use crude, rudimentary methods anymore :-).  Now I just need find out what AM-HM is...

Probability: you may want to review your #2. Also, see a thread in the advanced forum called "useful inequalities" for more inequalities, including AM-HM.

#3:

Let [tex]a + b + c + d = m = 16[/tex] and [tex]a^{-1} + b^{-1} + c^{-1} + d^{-1} = k[/tex]. (This is done to circumvent a weird LaTeX bug that wouldn't let me post my expression correctly.)

First by Cauchy,
[tex](a + c^{-1})^2 + (c + a^{-1})^2 + (b + d^{-1})^2 + (d + b^{-1})^2 \ge (m + k)^2[/tex]

By ineq. proven in question #2,

[tex]mk \ge 4^2[/tex]

However [tex]m = 16[/tex], so

[tex]k \ge 1[/tex].

Therefore

[tex]\begin{eqnarray}
4[(a + c^{-1})^2 + (c + a^{-1})^2 + (b + d^{-1})^2 + (d + b^{-1})^2] &\ge& (16 + 1)^2\nonumber\\
(a + c^{-1})^2 + (c + a^{-1})^2 + (b + d^{-1})^2 + (d + b^{-1})^2 &\ge& 289/4\nonumber[/tex]

QED.


Hola amigos, propongo el siguiente problema aunque me parece que es algo bien conocido, bueno despues de todo los animo a dar sus soluciones.

Demostrar que existen una potencia de 2 tal que sus primeras cifras sea 2006.

Saludos,

Alexis...

Si mal no recuerdo este problema esta resuelto en el libro " El Ingenio en la Matematicas " de la coleccion La Tortuga de Aquiles

De hecho este problema solo necesita un pequeño lema:

$\{n\alpha\}$ es denso en $(0,1)$ para $\alpha$ irracional fijo y $n$ recorriendo los enteros positivos.

Esto es conocido como el lema de Densidad de Kroenecker, su prueba mas elementeal es con el principio de casillas.

Espero la pista les sirva.


I'm stuck on this construction problem. Can someone help me if possible?
Let ABC be a triangle. Using ruler and compass to construct a line parallel to BC, meeting AB at D, and AC at E, such that DE=DB+EC.

[quote="namelessguy"]
Let ABC be a triangle. Using ruler and compass to construct a line parallel to BC, meeting AB at D, and AC at E, such that DE=DB+EC.[/quote]
Let $ I$ be the incenter of $ \triangle{ABC}$, construct through $ I$ a line parallel to $ BC$.


What is the ratio of the number of space diagonals of a cube to the number of edges? Express your answer as a common fraction.

There are $ 4$ space diagonals and $ 12$ edges, so $ \frac{4}{12}\equal{}\boxed{\frac{1}{3}}$.


Here are some number theory problems!!!!

1) Find out m,n (integers) so that 7^m+1 and 7^(m+n)+2 are both divizible by 5.

2) Let n and m be natural, n odd and S_m=1^n+2^n+..+m^n. Then S_m can be divided with (1+2+..m).

3) Compute n so that n^n has n digits.

thank you!!!
bye bye!!!

You should have posted each problem under a different topic... :D :D

cheers! :D

[b]1)[/b] It looks like 7^m and 7^(m+4) have the same last digit :{7,9,3,1}. Because we need to have the divizibility with 5 we have that 7^m must end with 9. Because 7^2 is the first power of 7 that ends in 9 we have that m=4a+2. And the same we get that m+n=4b+3, from where we get that: n=4c+1 with a,b,c pozitive integers!!!

cheers!! :D :D

[b]2)[/b] Look at this:

we write 2S_m = 1^n+2^n+..+m^n+1^n+2^n+..+m^n=(1^n+m^n)+(2^n+(m-1)^n)+..+(k^n+(m-k+1)^n)+..+(m^n+1).

We can see that every term of this sum k^n+(m-k+1)^n is divizible with k+m-k+1=m+1. So we have that m+1|2S_m
Then 2S_m=2[1^n+..+(m-1)^n]+2m^n=2*S_(m-1)+2m^n.
But m |2S_(m-1) from where we have that m|(2S_(m-1)+2m^n=2S_m. We have that (m,m+1)=1 which gives m(m+1)|2S_m. SO 2S_m=m(m+1)*k, where k is integer, and this gives:
S_m=m(m+1)/2*k=(1+..+m)*k

done!!

cheers! :D :D

[b]3)[/b] hehe we have that 1^1=1 (trivial solution)

we have that 10^(n-1)<=n^n<10^n so we have n<10. Then we have that n^(n-1)*n>=10^(n-1) or n^(n-1)*n>=5^(n-1)*2^(n-1) and so n>5. For 6 and 7 we don't have the required property. For 8^8 and 9^9 we have the required property. (use a computer to see that) :D

cheers! :D :D


search
Diophantine equation
Additive Number Theory

Prove that there exist infinitely many integers $n$ such that $n, n+1, n+2$ are each the sum of the squares of two integers.

For completeness, this problem was [i]A2[/i] at the contest.
We are looking for solutions with $ n \equal{} 4x^2$. Then $ n \equal{} (2x)^2 \plus{} 0^2$, $ n \plus{} 1 \equal{} (2x)^2 \plus{} 1^2$. It remains to find infinitely many $ x$ for which there exist $ y,z\in\mathbb{Z}$ such that $ n \plus{} 2 \equal{} 4x^2 \plus{} 2 \equal{} y^2 \plus{} z^2$, for some $ y,z$. We'll search for solutions having $ y \equal{} z$, that is $ 2x^2 \plus{} 1 \equal{} y^2$. This is a simple Pell Equation and it has infinitely many solutions, all of the form $ (y,x) \equal{} \left(\dfrac{(3 \plus{} 2\sqrt2)^n \plus{} (3 \minus{} 2\sqrt2)^n}2,\dfrac{(3 \plus{} 2\sqrt2)^n \minus{} (3 \minus{} 2\sqrt2)^n}{2\sqrt2}\right)$. It is easy to see that the above numbers are integers and indeed satisfy $ y^2 \minus{} 2x^2 \equal{} 1$.

another (somehow similar) solution is the following:
we know that th equation $ k^2\minus{}2t^2\equal{}1$ has infinite many solutions,now put:
$ n\equal{}2t^2\rightarrow n\equal{}t^2\plus{}t^2$
$ \Rightarrow n\plus{}1\equal{}k^2\rightarrow n\plus{}1\equal{}k^2\plus{}0^2$
$ \Rightarrow n\plus{}2\equal{}k^2\plus{}1\rightarrow n\plus{}2\equal{}k^2\plus{}1^2$

Another solution:
Consider the following sequence:
$ a_{1}\equal{}8$, $ a_{n\plus{}1}\equal{}a_{n}(a_{n}\plus{}2)$
We claim that for each $ n$, $ (a_{n}, a_{n}\plus{}1, a_{n}\plus{}2)$ are three consecutive integers, each being the sum of two squares. We will prove that by induction.
For $ n\equal{}1$, we have $ 8\equal{}4\plus{}4, 9\equal{}9\plus{}0, 10\equal{}9\plus{}1$.
Now assume that $ (a_{n}, a_{n}\plus{}1, a_{n}\plus{}2)$ are sum of two
squares. Then we have

$ a_{n\plus{}1}\equal{}a_{n}(a_{n}\plus{}2)  \longrightarrow$ being the sum of two squares by Lagrange's lemma.
$ a_{n\plus{}1}\plus{}1\equal{}a_{n}(a_{n}\plus{}2)\plus{}1\equal{}(a_{n}\plus{}1)^2\plus{}0$
$ a_{n\plus{}1}\plus{}2\equal{}a_{n}(a_{n}\plus{}2)\plus{}2\equal{}(a_{n}\plus{}1)^2\plus{}1$

The proof is complete.


Putnam
Princeton
college
Harvard
MIT
college contests

is there a restriction on the composition of a team?
for example, does a team need to have one first year exam taker?

The Putnam organization has no rules about that. Each campus organizer invents his own rules. For some, the rules include a local competition or local "qualifying exam." The team must be designated in mid-October. Schools that have a fairly large number of well-qualified students taking the Putnam (which includes your school) may find it particularly difficult to find a rational way of selecting the team.

I have no idea how Princeton selects the team. I think someone in the math department just chooses based on past years' results.

At Harvard, it's just the top 3 from the previous year (non-graduated, of course).

That seems to be how MIT chooses their top 3 also (top 3 from last year).  That seems to be the only reasonable way for the top 3 to be chosen, at MIT at least.

The past three years at UCSB, the Putnam administrator just asked me who should be on the team. I have no idea how they're going to choose next year.

So I guess at Harvard and MIT, there's no way for a hot-shot freshman to make the team (unless they took the Putnam in high school)?

[quote="Ravi B"]...(unless they took the Putnam in high school)?[/quote]
You know who I just had to look up following that remark! In 2001, Gabriel Caroll was indeed on the Harvard team as a freshman, having been a Putnam Fellow in 2000 (taking the exam at UC Berkeley, where he was taking his math courses as a 12th grader in Oakland.) The 2001 Harvard team consisted of two sophomores (George Lee and Sasha Schwartz) and one freshman (Carroll.)

Nevermind, I did a tad of research and what I said is incorrect.  MIT professors choose the team (Rogers and Stanley usually):

[quote]
Rogers and Stanley have selected the Putnam team for many years. “I’d been involved with it for the last fifteen years,” Rogers said. The first approximation, he said, is “to take the three people who scored the highest the previous year,” but this is not the only criterion. [/quote]
http://www-tech.mit.edu/V124/N14/Putnam04.14n.html

Dan Kane, given as an example in that link, took it as a freshman.  That was a pretty good choice on the professors' part :D


modular arithmetic
number theory unsolved
number theory

A sequence is defined by $x_1=1$, $x_2=4$ and $x_{n+2}=4x_{n+1}-x_n$ for $n\ge 1$. Find all natural numbers $m$ such that the number $2x_n^2+m$ is a perfect square for all natural numbers $n$.

Are you sure the problem's right? Checking for $n=1, 2$ gives that $m+2, m+32$ must be perfect squares, but their difference is $\equiv 2 \pmod{4}$, absurd.


There are five main choices for curriculum in XI and XII : (1) CBSE, (2) ISC, (3) Cambridge (IGCSE/AICE), (4) International Baccalaureate (IB), (5) State.

What are your opinions about the advantages and disadvantages of these different curricula?

I personally am ruling out the "State" curriculum for me since I don't like the quality of the question papers and the curriculum is also not challenging enough. That leaves me with the other four choices.

Cambridge (IGCSE/AICE) and IB curricula are being suggested as good choices if one intends to go abroad for undergraduate studies. However, I noticed that their final exams tend to be in May/June. With many of the Olympiad camps also taking place in May/June, I am wondering whether that is a serious disadvantage. Is there a way to avoid the schedule conflicts? Is it possible to take AICE exams in November of XI and XII grade years to avoid schedule conflicts?

Thanks for any information people can provide regarding the pros and cons of these different curricula.

dude how can u assume that u will be selected for olympiad camps
bade bade logon ki badi badi batein

You missed out APs on the list :P
And I'd love to help you- But you need to give more info. Your background, city, college aims etc.

I intended this theread to be a place for general discussion about the pros and cons of different curricula and not so much about my own situation.

Having said that, in response to nuclear_alchemist's question:

Regarding city : I am willing to relocate to almost any city in India (preferably south) if the attractiveness of the school justifies the move.

Aim & background: My interests are in Computer Science, Mathematics and science in general. My desire is to pursue undergraduate studies in U.S. As a backup plan, I may also study a bit for IITJEE.

I thought AP is not a curriculum choice available in India (Am I wrong?). Instead, my impression was that they are exams one can take. If a school provides training in AP syllabus material, that would be a plus.

Thanks.

Well, I can start a general discussion, but that would probably not serve your purpose. India is the most diverse place on earth. My aim is the same- get through the olympiads, undergrad in the US and IIT as a backup. Yet, our situations are unlikely to be similar. As far as APs are concerned, yes, some schools, especially international schools DO teach it. Though for a guy planning to take on Olympiads, the AP classes can suck with mediocrity. I myself took the AP exams with no formal course. Just random stuff crammed in the last 3 days before the 5 exams. I understand you might be uncomfortable disclosing too much about yourself on a public forum, so you can either try to put up SPECIFIC questions or send me a PM or do something else that might be able to give us a clearer picture.
Remember- the pros and cons of a curriculum are relative to the person and purpose which they may serve.
Good luck!

Better choose a school with less exams and more holidays :rotfl:

As you say that there are five courses for XI and XII. All boards are good for your education, it depends on you how serious you are in your education. If you would like to go abroad for further studies then, it does not give you any disadvantages of any board.


Hi,

I have finished most of trigonometry and geometry at least in my school. But I recently noticed that my problem solving isn't developing as much as it used to. :? So, I'm hoping to find some books that can help me lead the way to more advanced topics. I have AoPS I and Mathematical Circles but I need little bit different kind. More like contest problems but have lectures on some specific topics.

Thank you very much.


find all functions $ f: \mathbb{R^ \plus{} } \rightarrow \mathbb{R}$ which satisfy:

$ f(x) \plus{} f(y)\leq \frac {f(x \plus{} y)}{2}$    ,
$ \frac {f(x)}{x} \plus{} \frac {f(y)}{y}\ge \frac {f(x \plus{} y)}{x \plus{} y}$


We start at 0.

Subtract/Add the number of stars you have.

Join alternating teams. (I imagine that this will be the most fair apportioning system)

[b]5[/b]

[u][b]+ Team[/b][/u]
james4l

[u][b]- Team[/b][/u]

i'll join +team but i won't make much of a difference :)
7

[b]+ Team[/b]
james4l
AceOfDiamonds

[b]- Team[/b]
no one yet

[b]12[/b]

so much for [quote="james4l"](I imagine that this will be the most fair apportioning system) [/quote]
it's like +- already

[quote="james4l"]Join alternating teams.[/quote]

[b]11[/b]

So much for that rule.

Right now 3 people have +, 2 have -. To balance it out the next person should be in the "-" team, and then we restart alternating.

wait... will there ever be a winner??

[b]-5[/b]

Ernie, I give you the honor of finishing this game.

661, W-5.


Hey guys. I think that we should make Wikipedia articles for the winners of MathCounts Nationals. I mean, they do that for a lot of spelling bee kids, why not MathCounts kids?

If people start disputing this, don't get too emotional. Just tell them that this is the premier national mathematics competition, and if this doesn't count then perhaps the spelling bee kids should not get articles either.

I was a Wikipedian (for a while anyways...)

Anyways, if that's the case, then you should my friend Indeed. (He's an admin on the wiki, etc, etc)

Bobby Shen is beast.


induction
series summation
algebra
rational number
IMO Shortlist
IMO Longlist

If $x$ is a positive rational number show that $x$ can be uniquely expressed in the form $x = \sum^n_{k=1} \frac{a_k}{k!}$ where $a_1, a_2, \ldots$ are integers, $0 \leq a_n \leq n - 1$, for $n > 1,$ and the series terminates. Show that $x$ can be expressed as the sum of reciprocals of different integers, each of which is greater than $10^6.$

The first part for now :

Every $x$ can be written $x = f(x)/n!$ with $f(x) \in N$

Let's do it by induction on $n$.

Let's suppose it works for $n$, for all $x$ such as $x = f(x)/n!$

Let $x$ such as $x = f(x)/(n+1)!$.
$f(x) = q(n+1)+r$, for some $q, r$, $r \leq n$
Then $f(x)/(n+1)! = q/n! + r/(n+1)!$ and we can apply hypothesis to $q$.

It works for $n=1$ of course.

If $\dsp x = \sum^n_{k=1} \frac{a_k}{k!} = \sum^n_{k=1} \frac{b_k}{k!}$, with eventually some $a_k$ or $b_k$ null to achieve same number of terms.

Then by multiplying by $n!$ and looking $\mod \ n$ : $a_k = b_k$. We can do it again with $(n-1)!$, ... so decomposition is unique.

Let $ \frac {p}{q} < \frac {1}{N}$ be a rational that we want to express as the sum of inverses of different integers not smaller than $ N$, where $ p,q$ are coprime wlog.  If $ p \equal{} 1$, the sum contains only term $ \frac {1}{q}$, the inverse of integer $ q$.  Otherwise, consider positive integers $ a,b$ such that $ pa \minus{} qb \equal{} 1$, which obviously exist, and $ \frac {p}{q} \minus{} \frac {1}{aq} \equal{} \frac {b}{a}$, where $ aq$ is obviously an integer larger than $ N$ since $ N < \frac {q}{p} < aq$.  Assume now that we pick the solution with least $ a \plus{} b$.  Then, $ a < q$, since otherwise $ a' \equal{} a \minus{} q$, $ b' \equal{} b \plus{} p$ would yield $ a'p \minus{} b'q \equal{} 1$ and $ a' \plus{} b' \equal{} a \plus{} b \minus{} q \plus{} p < a \plus{} b$.  Then, $ b \equal{} \frac {pa \minus{} 1}{q} < p$ is also an integer, so repeating for $ \frac {a}{b}$ instead of $ \frac {p}{q}$, the process cannot continue indefinetely, and  $ \frac {p}{q}$ will eventually be expressed as the sum of inverses of integers, all of them different since at every step $ a\neq1$, or otherwise $ \frac {1}{N} > \frac {p}{q} > b$ absurd!

Assume now that $ \frac {p}{q} > \frac {1}{n}$ must be expressed as the sum of inverses of integers not smaller than $ n$.  Then, choose the least $ N$ such that $ \frac {p}{q} \minus{} \frac {1}{n} \minus{} \frac {1}{n \plus{} 1} \minus{} \frac {1}{n \plus{} 2} \minus{} ... \minus{} \frac {1}{N} < 0$.  Take now $ \frac {p'}{q'} \equal{} \frac {p}{q} \minus{} \frac {1}{n} \minus{} \frac {1}{n \plus{} 1} \minus{} \frac {1}{n \plus{} 2} \minus{} ... \minus{} \frac {1}{N \minus{} 1} < \frac {1}{N}$, which as shown above may be expressed as the sum of inverses of different integers not smaller than $ N$, and we are done.  Note that such an $ N$ always exists since otherwise $ \sum_{k \equal{} n}^\infty\frac {1}{k}$ would be bounded, and this is not true.  Note also that integer $ 10^6$ could be substituted by any arbitrary integer...

Er, the second part,
It is quite well known that any positive rational number can be expressed as a sum of finite number of unit fractions.
Suppose $x=\sum_{n}\frac{1}{\alpha_n}$ is one of them.
Then the consider the representation $x=\sum_{n}\frac{10^6}{10^6\alpha_n}$.

Whenever necessary, replace $\frac{1}{n}$ by $\frac{1}{n+1}+\frac{1}{(n)(n+1)}$.
Iterating this process for finite number of times we get a required representation.

[url=https://aops.com/community/p22672218]ISI 2021/3[/url] copied it.


I don't know if anyone has said this yet, but official AMC results have been emailed out (well, for my school at least). My teacher got the scores today.

wow, now i can laugh at my pathetic score :lol:


Thank You for the paper m8. Nice questions. But what paper is this? /entrance etc?

This is the examination conducted by [url=http://www.nbhm.dae.gov.in/]NBHM[/url] (National board of higher mathematics, India) in order to sample the undergraduate students in India for an interview leading to a [url=http://www.nbhm.dae.gov.in/msc.html]scholarship[/url] while the student pursues his graduate study (called MSc here).

I attended this exam and texed it for the anybody who would be seeking it next year or later. The official copy appears on the NBHM site only by next year.

By the way what is m8?

@Talegari
m8 - mate,I guess
U'll get to know many more such abbreviations over here,I too fail to understand most of them. :oops:


geometry
trapezoid
trigonometry
quadratics
algebra
geometry unsolved

Given a isoceles trapezoid ABCD with the base AD being equal to the Diagonal AC, the Small base $ CD \equal{} \sqrt {2}BC$, find the measure of interior angles of ABCD

hello, let $ a\equal{}\overline{AB}$, $ b\equal{}\overline{DC}\equal{}{AB}$ and $ b\equal{}\sqrt{2}c$, then we get in the triangle $ ADC$ after the law of cosine $ b^2\equal{}a^2\plus{}a^2\minus{}2a^2\cos(\pi\minus{}\alpha)$ or as an equivalent term
$ \frac{c^2}{a^2}\equal{}1\minus{}\cos(\pi\minus{}2\alpha)$. In the triangle $ ACB$ we get after the same theorem
$ a^2\equal{}b^2\plus{}c^2\minus{}2bc\cos(\pi\minus{}\alpha)$ or $ \frac{c^2}{a^2}\equal{}\frac{1}{3\minus{}2\sqrt{2}\cos(\pi\minus{}2\alpha)}$, setting $ t\equal{}\cos(\pi\minus{}2\alpha)$, then we get the quadratic equation
$ 2\sqrt{2}t^2\minus{}t(3\plus{}2\cdot \sqrt{2})\plus{}2\equal{}0$. Here you can derive $ t$ and after this $ \alpha$.
Sonnhard.

Trapezoid ABCD cannot have parallel bases AD || CD. Assume AB is the long base. Let a = AB = AC = BD, c = CD.

$ c \equal{} a \minus{} 2c \sqrt 2 \cos A$ and $ c \equal{} a \cos (\pi \minus{} 2A) \minus{} c \sqrt 2 \cos A$

$ a \minus{} c \sqrt 2 \cos A \equal{} a (1 \minus{} 2 \cos^2 A) \Longrightarrow \cos A \equal{} \frac{c}{a \sqrt 2} \Longrightarrow$

$ a \minus{} \frac{2c^2}{a} \equal{} c \Longrightarrow a \equal{} 2c \Longrightarrow \cos A \equal{} \frac{1}{2 \sqrt 2}$


The medians of a triangle $ABC$ subdivide this triangle into six little triangles (each of these little triangles is formed by one vertex, one side midpoint and the centroid of triangle $ABC$). Prove the circumcenters of these six little triangles lie on one circle.

See
http://www.mathlinks.ro/Forum/viewtopic.php?t=1536 ;
http://groups.google.de/group/geometry.college/browse_thread/thread/6c4569ba3fa00cd3/63cbcbc129fed62c  ;
http://www.math.ust.hk/excalibur/v6_n1.pdf ;
the note "The Lamoen circle" on http://de.geocities.com/darij_grinberg .

Is the theorem really from Neuberg? This would be very surprising, since it is generally attributed to Floor van Lamoen, and I don't recall any related results by Neuberg.

  darij

Dear mathlinkers,

a new synthetical proof of the van Lamoen's circle is presented on my website
http://perso.orange.fr/jl.ayme

Sincerely
Jean-Louis


Is there a way to see what portion of my attachment quota I've used up so far?

Go to your private messages, and then you'll see your quota usage in the top left corner (on the AoPS style) :)


trigonometry vector calculus induction conics derivative AMC

I was thinking about doing that last summer too, but I really don't think it's worth it. Trigonometry is one of the most fundamental parts of math and trying to learn it in 3 months really will not give you a good mastery of it. Ofcourse I'm sure it's *possible* to learn everything and learn it well, but you'll be doing like 4-5 hours of precal everyday, so in my opinion it's a waste of time, just wait for your senior year :)

I took a precalc course over the summer...but it ended up being somewhat useless because it missed out several key things (3-d land, vectors, that whole section).  Anyways, the stuff which I did learn wasn't really connected and I wasn't given any applications or reasons.  Anyways, I'm learning many new things in precalc now.

i skipped pre calc ... if you're going to learn it on your own probably some of the more important things to go over are the trig aspects and polar graphing... a lot of the rest you'll get to see again during calculus.

It's not unheard of to take pre-calc and calculus simultaneously. Ask if you can take two math courses a year.
I don't remember what exactly was covered in pre-calc, but definitely not all of it was required for a decent understanding of calculus. We did things like mathematical induction, fractals, conic sections and some series. Only the series turned up in calculus and we reviewed them in calculus before actually using them in a calculus context.
You don't need to know about vectors or 3-space for first year calculus...

precalc was actually my favorite math subject in high school. you do alot of little things that may not be necessarily connected but nonetheless interesting. even though they may not be very important for a good understanding of calculus, some things covered in precalc are just plain cool and you don't get to see them in any other context.

nonono Trig is actually pretty easy to learn. If you work hard, you can learn trig in about 1 month and a half. then learn the rest

[quote="probability1.01"]You don't learn much of anything in precalc. Precalc is basically the class where you go like "Hey there are a bunch of interesting concepts in math let's name them and draw pictures and then ignore them." You shouldn't have trouble learning it in a short period of time. Don't worry about the trig; there's really not much there (although apparently there's this rational trig thing or something which looks really cool but utterly unrelated to precalc). In fact, http://www.mathworld.wolfram.com might be enough to teach you most of it.[/quote]

My precalc must be very different...

Get a precalc book, figure out how much of it you need to get through each day to cover the subject, and then do it.
(That's what I'd do anyway)

[quote="Feynmaniac"]I'm interested in learning precalc on my own this summer so that I can take this really amazing postcalc elective my senior year. Any advice on how to cover an honors precalc course in three months?[/quote]

The content of precalculus courses varies wildly from school to school.  (I didn't take precalc, myself; at my school it was a hybrid of honors algebra II and what is now usually the beginning of calculus.)  My advice would be to approach the current honors precalc teacher and ask which sections of the book are covered in the course, and what topics in the course don't appear in the book.  Ask also if you can borrow a copy of the text for a while.  Then you'll have a much better idea of how massive/difficult a task you have ahead of you.

I think that precalc is a complete waste of time.  The only part of it that you actually need is limits, which you can learn in about 5 minutes.  My school has all of the students on the honors math track skip it so we can devote more time to calculus.

[quote="ShoeFactory"][quote="SplashD"]Can someone tell me what exactly is in pre-calc? How much of calculus, how much trig, how much Alg II? I'm not skipping it cause I don't have any time during the summer but I just want to know what's in it so I can be a little ahead.[/quote]

30% TRIG
60% alg 2

10% calculus (limit and that h->0 derivative thing which i dont remember the name of)

just my opinion which is a crude estimation.

The "calculus" you learn will be useless very soon though.[/quote]

Is there calculus in USAMO though?

No but it might help. It has helped me in a few places for math competitions and tests. Calculus is very powerful :O

Precalculus calculus is useless though, except for maybe limits which you may or may not consider calculus.

I just completed "Trigonometry/Math Analysis" (fancy name for precalculus) at my high school. in which i just graduated  :) . Anyways, the course was lame and pretty pointless. I will show you why:

Unit 1: Linear and quadratic functions (covered in algebra 1 and algebra 2)

Unit 2: Inequalities (everything in this unit except polynomial inequalities was covered in previous algebra courses)

Unit 3: Functions (The only new thing we learned was about period ic functions. Everything else such as reflections, translations, and composition of functions etc.. was covered in algebra 2)

Unit 4 through Unit 8 was all trigonometry, which was new. (trigonometric functions, inverse trigonometric functions, the unit circle, trigonometric equations and identities, right triangle trigonometry and other triangle trigonometry such as the law of sines and cosines, the graphs of all the trigonometric equations, inclination of lines, polar coordinates and graphs, and de moivre's theorem)

Unit 9: Polynomial functions (Covered in Algebra 2, except with the addition of Descartes rule of signs)

Unit 10: Exponential and Logarithmic Functions (LAME. COMPLETE ALGEBRA 2 REVIEW)

Unit 11: Vectors (This chapter was new, although i took the basic high school physics class last year so alot of the material in this chapter was review to me)

Unit 12: Sequences and Series (Algebra 2 review with the exception of intervals of convergence, limits of sequences, recursive definitions, and mathematical induction)

Unit 13: Random bits and pieces from the 3 chapters on combinatorics, probability, and statistics (this was also a complete algebra 2 review except for the discussion of normal distributions in the statistics chapter.However i was also concurrently enrolled in AP statistics so i have had lectures on normal distributions up the ass....so yeah. i slept through this unit...as well as most of the other ones)

Unit 14: Matrices (This was a new unit)

Unit 15: Limits, continuity, and rational functions (this was also new)

So yeah....this class was basically 50% new material, and the other 50% was Algebra 2 all over again. So basically i think it would be alright for you to take precalc over the summer. I'm assuming you have had algebra 2, so you can skim through alot of precalc very quickly so you can concentrate most of your time on trig. The trig isn't very hard at all. It's just a pain to memorize all the identities, sum+difference formulas, and the double and half angle formulas. In fact, i already forgot most of that stuff except for the stuff i use regularly D:

dude i skipped trig just like your thinking. I think that taking it on my own I have a better foundation in trig/precalc than anyone else in my school that took the class. you have to take it serious though. for like 3 full chpts. of the book i made sure i did almost all the odd problems (so i could check my answer) it went by pretty fast though. maybe like 2 hrs. a day.

I'm planning to skip precalculus so that I can do calculus in sophmore year. To do so, I must do reasonably well (80 or above) on both the (honors) precalculus midyear and final exams at our school.

In my case, I'm already familiar with most of the concepts in precalculus - it's only a matter of brushing up on a few random details and getting practice. Thus, I'm doing most of the problems in the book we use, which the course follows closely. I don't particularly recommend skipping it unless you already have a general understanding of the concepts involved.

---------------------------------------------------
About the actual course:

The course of our school uses the book "Advanced Mathematics: Precalculus with Discrete Mathematics and Data Analysis, Brown" and is called Adv. Algebra/Trigonometry/Analytic Geo Honors (citing certain topics that the normal American Precalculus course apparently doesn't cover, as per our department head). As for what it contains, it's about 50% review of earlier courses, 25% trigonometry, and 25% other stuff (polar coordinates, vectors, limits, intro to calc).

The exact breakdown of the chapters is:
Semester 1:
Chapter 1 - Linear Equations and Quadratics (basic Algebra 1/2 stuff, including complex numbers)
Chapter 2 - Polynomials (various stuff about polynomials - factors, remainder theorem, etc)
Chapter 3 - Inequalities - linear inequalities, polynomial inequalities where you have to factor, absolute value inequalities, and graphing 2-variable inequalities
Chapter 4 - functions (inverses, symmety, translating graphs, periodic functions)
Chapter 7 - Trigonometry 1
Chapter 8 - Trigonometry 2
Chapter 9 - Trigonometry 3
Chapter 10 - Trigonometry 4
Semester 2:
Chapter 11 - Polar Coordinates and DeMoivre Theorem (new)
Chapter 12 - Vectors and Parametric Equations (new)
Chapter 13 - Sequences and Series
Chapter 14 - Matrices
Chapter 19 - Limits
Chapter 20 - Derivatives/Intro to calc
Chapter 5 - Logarithms
Chapter 6 - conics

So Chapter 1-4 is basic Alg 1/Alg 2, Chap 7-10 is trig, Chapters 11, 12, 19, and 20 are other new things, and Chapters 13, 14, 5, and 6 are pretty much review Alg 2.

If you'd like to see the course description, go to [url]http://ab.mec.edu/abrhs/math/trigonometry331.html[/url]
--------------------------------------------------

i took honors precalc this year and it was rather boring and we didn't learn much; the only new thing was matrices and...some real easy calculus.  the trig was all review from the previous year (everyone has to take trig first, then most people forget everything so we learn it all again in precalc).  still, it might be different/harder in other schools depending on the exact definition of precalc.

Hey Calc Rulz, did you guys use the textbook "Advanced Mathematics: Precalculus with discrete mathematics and data analysis" by Richard G. Brown? lol because we used that textbook and it had the same exact chapters as what you mentioned and in that order as well. My class wasn't the honors class so we didn't cover everything, but we covered most of it. We skipped a section  in chapter 4 (functions of several variables), chapter 6 (conic sections) entirely for some stupid reason, even though we covered it in algebra 2, bits and pieces of chapters 15 through 17 (we did select sections from each of those chapters and crammed it all into one unit), and we skipped the last half of chapter 19 and all of chapter 20 (even though i worked most of chapter 20 on my own).

I haven't taken trig yet.  I learned a lot on my own, but I think it would be beneficial to take the actual class, so I will.

[quote="SplashD"]Would a month to prepare to skip precalc be enough? For me, you have to get a 90 or above to pass and get into calc honors the assessment they give you is the midterm combined with the final.[/quote]

Based on what I've seen  you do, I'd say you already know a bunch of the topics in the precal curriculum already.  I think it's feasible for you to pick up the remaining topics and prepare for the exam in a month.  You'll probably be really bored in pre calculus so I highly suggest you try and do this.  Good luck :)

really precalc is a waste of time.
if you learned algebra 2 well and understood it then basically you dont learn much.  the only really new stuff is the trig, depending on what kind of geometry background you have. and vectors are easy to learn.
so yes i personally think it is possible to "learn" precalc in 3 months.  however if you are planning on using what you learned in math competitions then you might want to take precalc to make sure that you didn't skip over anything important.
also it will be an easy class, somewhere you can study for whatever else it is you need to.

Pre-calc is basically (if you fully know "Algebra 2"):

Algebra II Review
Understand Trig
Memorize Trig Formulas/Identities
Basic knowledge of Vectors
Basic knowledge of Limits
Basic sum/sequence (if you haven't learned from Algebra II)


geometry
3D geometry
tetrahedron
geometric transformation
homothety
geometric series
geometry unsolved

Determine the point M inside a tetraedre ABCD so that MA² + MB² + MC² + MD² is minimal.

Note the following lemma...

Let $ AB$ be a segment with midpoint $ M$, and let $ P$ be an arbitrary point.  Then $ PA^2\plus{}PB^2\equal{}2(PM^2\plus{}MA^2)$.  The proof is by the Law of Cosines.

Let the midpoints of the edges of this tetrahedron be $ P_1, P_2, P_3, P_4, P_5,$ and $ P_6$.  Then, from the lemma, it follows that, given a fixed tetrahedron, the desired quantity relies only on $ \sum\limits_{i\equal{}1}^{6}MP_i^2$.  Now take $ P_1, P_2, P_3$ to be the midpoints of three edges on one face of the tetrahedron, and we notice by again applying the lemma that $ MP_1^2\plus{}MP_2^2\plus{}MP_3^2$ relies only on the sum of the squares of the distances from M to the midpoints of $ P_1P_2, P_2P_3,$ and $ P_3P_1$.  This is because the lengths of $ P_1P_2, P_2P_3,$ and $ P_3P_1$ only depend on the original tetrahedron.

Take $ G_1, G_2, G_3,$ and $ G_4$ be the centroids of the four faces of the tetrahedron.  We can continue the process above, and using a limiting argument, we determine that the value in questions depends only on $ MG_1^2\plus{}MG_2^2\plus{}MG_3^2\plus{}MG_4^2$.  We can use an infinite geometric series summation to determine the exact value in terms of this quantity and the sides of the tetrahedron.

Let $ G$ be the centroid of the tetrahedron.  Note a homothety through $ G$ maps $ ABCD$ to $ G_1G_2G_3G_4$.  So then if $ M$ is the point which minimizes $ MA^2\plus{}MB^2\plus{}MC^2\plus{}MD^2$, then the image of $ M$ under the same homothety through $ G$ gives us the corresponding point in $ G_1G_2G_3G_4$.  But that point must also be $ M$.  Thus, $ M\equal{}G$, the centroid of $ ABCD$.

This method generalizes to a simplex in any number of dimensions.


Given that $ f(x \plus{} y) \equal{} 2f(x)f(y) ,f(0)\not \equal{} 0$  ,  Show that $ f(x)$  is an even function.

[color=red]sorry ,edited[/color]

Not $ f(x)\not \equal{} 0$

there are two trivial solutions, $ f(x) \equal{} 0$ and $ f(x) \equal{} \frac {1}{2}$, which are even. The other solution $ f(x) \equal{} \frac {1}{2}\left(2f(1)\right)^x$ is not even for $ f(1)\not \equal{} \frac{1}{2}, 0$..

[quote="ocha"]there are two trivial solutions, $ f(x) \equal{} 0$ and $ f(x) \equal{} \frac {1}{2}$, which are even. The other solution $ f(x) \equal{} \frac {1}{2}\left(2f(1)\right)^x$ is not even for $ f(1)\not \equal{} \frac {1}{2}, 0$..[/quote]

Since
$ f(0\plus{}0)\equal{}2f(0)f(0)$
$ \implies a\equal{}2a^2 , a\equal{}f(0)$
$ \implies a(1\minus{}2a)\equal{}0$
$ \implies f(0)\equal{} 0$ or $ f(0)\equal{}\frac {1}{2}$
but ,
$ f(0)\not \equal{} 0\implies f(0)\equal{}\frac {1}{2}$

And,

$ f(x\minus{}x)\equal{} 2f(x)f(\minus{}x)\implies f(0)\equal{}2f(x)f(\minus{}x)\implies f(x)f(\minus{}x)\equal{}\frac {1}{4}$

Ocha ,

how can you get  $ f(x) \equal{} 0$?

[quote="10000th User"]Perhaps he meant

$ f(x) \equal{} \begin{cases}\frac12 & \text{if }x \equal{} 0 \\
0 & \text{otherwise}\end{cases}$

which also satisfies the conditions of the problem. :)
He's also pointing out a solution that is not necessarily an even function...  :ninja:[/quote]

Sorry, I don't think so.

If $ f(x) \equal{} \begin{cases}\frac12 & \text{if }x \equal{} 0 \\
0 & \text{otherwise}\end{cases}$ works,

How can you exclude $ f(x)f(\minus{}x)\equal{}\frac{1}{4}$ in general case?

[quote="vinskman"]
How can you exclude $ f(x)f( \minus{} x) \equal{} \frac {1}{4}$ in general case?[/quote]

What do you mean?
Note that if $ f(x)$ is even then $ f(x)f( \minus{} x) \equal{} \frac {1}{4}$ implies $ f(x)\equal{}\frac{1}{2}$, so there is clearly something wrong with the question, because i have already pointed out a non-constant solution

I meant

the idea of 10000th user : $ f(x)\equal{}0$ otherwise

is impossible,

because $ f(x)f(\minus{}x)\equal{}0$ ,it contradicts with $ f(x)f(\minus{}x)\equal{}\frac{1}{4}$ in general case.

[quote="ocha"]Note that if $ f(x)$ is even then $ f(x)f( \minus{} x) \equal{} \frac {1}{4}$ implies $ f(x) \equal{} \frac {1}{2}$[/quote]

How did you make that inference? Note that your own non-constant solution satisfies $ f(x) f(\minus{}x) \equal{} 1/4$

Viskman, you're right $ f(x)\equal{}0$ is not a solution, sorry

[quote="TZF"][quote="ocha"]Note that if $ f(x)$ is even then $ f(x)f( \minus{} x) \equal{} \frac {1}{4}$ implies $ f(x) \equal{} \frac {1}{2}$[/quote]
How did you make that inference? Note that your own non-constant solution satisfies $ f(x) f( \minus{} x) \equal{} 1/4$[/quote]

If $ f(x)$ is even $ f(x)\equal{}f(\minus{}x)$ so $ f(x) f( \minus{} x) \equal{}f^2(x) \equal{} \frac{1}{4} \Rightarrow f(x)\equal{}\frac{\pm 1}{2}$, but we discard the negative because $ f(0)\equal{}\frac{1}{2}$.
and anyway, the non-constant solution isn't even.

[quote="vinskman"]I meant

the idea of 10000th user : $ f(x) \equal{} 0$ otherwise

is impossible,

because $ f(x)f( \minus{} x) \equal{} 0$ ,it contradicts with $ f(x)f( \minus{} x) \equal{} \frac {1}{4}$ in general case.[/quote] :ewpu: Oops, I misunderstood the question. Still, given that functional equation, there is that non-constant function that isn't even... I'm wondering where PKMathew got this problem from...


By definition, $ 5!\equal{}5\cdot4\cdot3\cdot2\cdot1\equal{}120$. What is the sum of $ 6!$ and $ 7!$?


linear algebra
matrix
vector
superior algebra
superior algebra solved

I found in a mathematical magazine (I forgot it's name) the following beautiful problem:
  Prove that there exists a {0,1} nxn matrix (which means an nxn matrix with 0 or 1 elements) whose square is the matrix all of whose elements are 1 if and only if n is a perfect square. I have proved the neccesity, but I'm not able to prove sufficiency.

So, if I understand correctly, you can prove that n must be a perfect square, but you haven't proved that this works for all perfect squares?

I think I have an example which works for any perfect square n=k^2:

I hope I can explain it correctly. You divide the nxn square in n smaller squares of dimension kxk. You put 1 on the first column of every such square (dim kxk) which is situated on the first line of the initial matrix (the smaller squares are arranged in a kxk matrix of squares of dimension k); you put 1 on the second column of every smaller square which is situated on the second line and so on. I hope I made myself understood. I'll let you figure out why this works. Here are examples for k=2 and k=3:

k=2; n=k^2=4
[10] [10]
[10] [10]

[01] [01]
[01] [01]

k=3; n=k^2=9
[100] [100] [100]
[100] [100] [100]
[100] [100] [100]

[010] [010] [010] This is all just one big matrix. I've divided it like this for
[010] [010] [010] you to see the smaller kxk squares
[010] [010] [010]

[001] [001] [001]
[001] [001] [001]
[001] [001] [001]

Oh, I forgot to mention: I'm still working on the other part, proving that n must be a perfect square.

Hope you don't mind if I give a solution to the necessity. Maybe it's different from yours.

By [x1,x2,...,xn] I shall dentote the vertical nx1 vector with entries x1, x2,...,xn. Suppose that there is a matrix A with entries a i,j such that A 2 =J, where J is the all 1 matrix.

Let Li be the sum ai,1+ai,2+...+ai,n, the sum of elements of the line i. We have, for every k, the following identity:
ak,1L1+ak,2L2+...+ak,nLn=n

Of course, to prove this, it is sufficient to consider only those ak,i's that are non-zero. Suppose they are ak,1,...,ak,j (I supposed this only to make my notations easier). Then, consider the jxn matrix B formed only by the lines 1 to j. Each column i in B will have exactly one entry 1 (because in the matrix A, the product of the line k and the column i must be 1 since A 2 =J). That means that B has n entries so n=L1+...+Lj, and in the general case, ak,1L1+ak,2L2+...+ak,nLn=n.

So let's write down all these equalities for all k and we will get AxL=[n,n,...,n], where L is the vector [L1,L2,...,Ln]. Multimply with A and get JxL=Ax[n,n,...,n], so
[L1+L2+...+Ln, L1+L2+...+Ln,...,L1+L2+...+Ln]=[nxL1, nxL2,...,nxLn]. Thus all the Li's are equal and they will equal a p. Return to ak,1L1+ak,2L2+...+ak,nLn=n, which yields p 2 =n

I have another approach for the necessity. Just consider B the matrix which contains only 1 and you will find very easy that B has the proper values 0,0,...,0 and n. Thus, if x_1,...,x_n are the proper values of A , we must have x_1=0,...,x_n-1=0 and (x_n)^2=n, from where |Tr(A)|=|x_n|=sqrt(n). Since Tr(A) is an integer, n is a perfect square.


calculus integration logarithms trigonometry limit function derivative

Consider making it a closed curve by adding the segment of the $x$-axis from $-1$ to $1.$ The integral along that segment is easily calculated. Having a closed curve gives us a reason to look at Green's Theorem.

[quote="color"]Calculate
\[\int_\mathcal{C}\sqrt{x^{2}+y^{2}}dx+y(xy+\ln{(x+\sqrt{x^{2}+y^{2}})})dy \]
where $\mathcal{C}$ is the upper half of the unit circle oriented anticlockwise.[/quote]

you have,
$r\;=\;\sqrt{x^{2}+y^{2}}\;=\;1,\;\qquad\;0\leq\theta\leq\pi\;\quad\; ( \;upper\;\;half\; ) \;$

$x\;=\;\cos\theta,\;\qquad\;y\;=\;\sin\theta$

$dx\;=\;-\sin\theta\;d\theta,\;\qquad\;dy\;=\;\cos\theta\;d\theta$

so,
$I\;=\;\int_{0}^{\pi}\; ( \;-\sin\theta+\sin^{2}\theta\cos^{2}\theta+\sin\theta\cos\theta\; ( \ln{(1+\cos\theta)}) \; ) \;d\theta$

i am sure you can do the 1st one.

for the 2nd one, use
$\;\quad \sin^{2}\theta\cos^{2}\theta=\frac{1}{4}\sin^{2}( 2\theta )\;=\;\frac{1}{4}\;\cdot\;\frac{1}{2}\; ( 1-\cos{( 4\theta ) ) }$

and as for the 3rd one, let
$u=\cos\theta,\;\quad\;du\;=\;-\sin\theta\;d\theta$

we get
$\int_{-1}^{1}\;\ln{(1+u)}\;\cdot \;u\;du\;=\;\left( \ln{(1+u)}\;\cdot\;\frac{u^{2}}{2}\right)_{-1}^{1}\;-\;\int_{-1}^{1}\;\left( \;\frac{1}{1+u}\;\cdot\;\frac{u^{2}}{2}\;\right) \;du\;\quad\;\;\;...\;(1)$
(using integration by parts)

so, you can do the rest now?  :rotfl:

PS: write
$\frac{1}{2}\;\ln{0}\;\quad\;as\;\quad\;\frac{1}{2}\;\lim_{\eta\rightarrow{0}}\;\ln{\eta}$

don't worry, it eventually cancels out after you evaluate the RHS in $(1)$

Call the line integral we're trying to compute $I.$

The integral along the $x$-axis needed to complete the closed curve is

$\int_{-1}^{1}\sqrt{x^{2}}\,dx=\int_{-1}^{1}|x|\,dx=1.$

Then, $\frac{d}{dx}\,y\left(xy+\ln(x+\sqrt{x^{2}+y^{2}}\right)-\frac{d}{dy}\sqrt{x^{2}+y^{2}}=y^{2}.$

Thus, by Green's theorem, if $E$ is the region above the $x$-axis and below the semicircle in question,

$I+1=\iint_{E}y^{2}\,dx\,dy$

$=\int_{0}^{1}\int_{0}^{\pi}r^{3}\sin^{2}\theta\,d\theta=\frac{\pi}8$

From this, the integral we want is $I=\frac{\pi}8-1.$

We do need to acknowledge that our functions are not well-behaved at the origin. I should apply Green's theorem not to all of $E$ but rather to $E$ less a small semicircle around the origin. As it turns out, our differential form and the $y^{2}$ from the derivatives of that differential form tend to zero at the origin, so the contrubutions of that cut-out piece tend to zero, and our original computation is justified.

Aha! I knew that the expressions had to turn out easy, but I forgot an extra $2$ when differentiating the first differential form so I got an extra $\frac{1}{\sqrt{x^{2}+y^{2}}}$ with the $y^{2}$ term.

So thank you very much for clarifying the hole thing  :D .
Also thanks for your considerations on the validity on applying green.

My Idea was the same as yours, nameley to consider the segment $\gamma : x\in[0,1]$
and the given curve $\mathcal{C}$, then Greens theorem is applied.
So we have
\[\int_\mathcal{C}Pdx+Qdy=\iint_{(\mathcal{C}\cup \gamma)=\partial D}(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})-\int_\gamma P(x,0)dx \]
So the RHS becomes easy

Again thanks  :)


Let $ S_n$ represent $ a_1 \plus{} a_2 .....\plus{} a_n$ where $ a_n \equal{} (n\plus{}1)2^n$. Find the value of $ \frac{S_{2008}}{2008}$.

[hide]
Is the answer $ 2^{2009}$?

$ S_{2008} \equal{} \sum_{i\equal{}1}^{2008}(i\plus{}1)2^i$
            $ \equal{}\sum_{i\equal{}1}^{2008}i*2^i   \plus{} \sum_{i\equal{}1}^{2008}2^i$

For $ \sum_{i\equal{}1}^{2008}i*2^i$,use AGP method,taking it to be S,evaluating 2S and subtracting both to get S.

[/hide]

hello, yes, the correct answer is $ 2^{2009}$.
Sonnhard.


[u][b]Problem[/b][/u]:f(1) < f(2) < f(3) < ... is a sequence of positive integers. S(n) = f(1)+f(2)+...+ f(n). For n > 2001, f(n) divides S(n-1). Show that for some N, f(n) = S(n-1) for all n > N.
[u][b]Note:  [/b][/u]can f(n) is an increasing sequence but not strictly so that the conlusion is true?

[quote="dreammath"][u][b]Note:  [/b][/u]can f(n) is an increasing sequence but not strictly so that the conlusion is true?[/quote]
But what if $f(n)=1$ for all $n$?

I'm pretty sure I've seen this before around here.

yes, i posted and i think you solved it!


Could someone please show me the working for this problem-

Two fisherman each had a certain number of trophies. The first said to the second "If you give me 7 of your trophies, I will have twice as many trophies as you." The second fisherman, however, said, "If you give me 7 of your trophies I will have as many as you"
How many trophies does each have?

The answers are supposed to be First = 49, Second = 35

So far I have-

1/ x + 7 = 2(y - 7)
2/ y + 7 = x - 7

1/ x = 2y - 21

substitute into 2/

y + 7 = 2y - 21 - 7
      y = 2y - 28
     -y = -28
Wrong

$ x \plus{} 7 \equal{} 2(y \minus{} 7)$

and

$ y \plus{} 7 \equal{} x \minus{} 7$

so $ y \equal{} x \minus{} 14$

$ x \plus{} 7 \equal{} 2(x \minus{} 14 \minus{} 7)$

$ 7\plus{}4*7\equal{}x$

$ x\equal{}35$

$ y\equal{}35\plus{}14\equal{}49$

That's great!! Thanks for clearly explaining it


here are  some question  as  well  as  dobts  of  mine-

1-does an air bubble act as a concave lens and why?(too easy i guess)
2-if a bird is directly overhead a fish in water and starts flying towards it,will the fish see the bird coming faster than the actual speed or slower?
3-a concave mirror is placed over a beaker with water upto a height of 33.25cm(ref index=1.33)and image of the object placed at the bottom of the beaker is formed at a dist of 25 cm from th etop,then the focal of the mirror is...?(fought with tut sir one this one too,but i think i'm wrong)

hey comeon, aren't these repeat from your previous forum. Just beacuse nobody answered there as yet, doesn't mean you can spam !! :spam:

Rushil,I thought no one saw it over there cause it was stuffed inbeteen the comments.To make it more readable,i posted em again.Idont think i'm spamming.These are doubts of mine and i need the solutions :lol: .

Wwhy dont ya forget whether u have seen it before and just answer them?Let the moderators decide if its a spam.these are just my doubts and if the moderators think its a spam,they are welcome to delete this post of mine.

concerning 2), see my post for solution of problem no. 5 [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=46586]here[/url], especially its last sentence.

After a little thought, it is clear that if the bird is in height $H$, the fish sees him roughly in height $H' = nH$. Since the observed velocity $v' = \frac{dH'}{dt} = n\frac{dH}{dt} = nv$, where $v$ is the real velocity, the fish sees the bird to come $n$ times faster. However, this doesn't mean that the fish sees the bird come earlier, because he also comes from larger height.


Solve the following equation in complex numbers:

$1.\left(2x+3i\right)^{n}-\left(2i-x\right)^{n}=0$

$2.\left(1+i\right)z^{4}+\left(1+\sqrt{3}\right)-\left(1-\sqrt{3}\right)i=0$

Complex numbers $\neq$ complex analysis.  These questions probably belong in the Pre-Olympiad forum.

[hide=" Partial soln. for (a)"]
Factor, we get $(3x+i)\left(\sum_{k=0}^{n-1}_{n-1}C_{k}(2x+3i)^{k}(2i-x)^{n-1-k}\right)=0 \Rightarrow x= \frac{-1}{3}i$, but i don't know what to do with the other junk.[/hide]

[hide="b"]
$z^{4}= \frac{ (1-\sqrt{3}) i-(1+\sqrt{3})}{1+i}= \frac{ 2 \sqrt{2}\text{ cis}15^\circ}{\sqrt{2}\text{ cis}45^\circ}= 2 \text{ cis}(-30^\circ)$, obvious from there.[/hide]

for a you just divide...and you are like woa...
2i is not a solution for any n\ne 0 (where the solution set is R), so we can divide
$\frac{2x+3i}{2i-x}=\zeta_{n}$ where $\zeta_{n}$ is a nth root of unity, the rest is just isolating x,

is n a complex number or positive integer?


calculus
trigonometry
induction
integration
function
calculus computations

I am creating a small collection of problems that use the number 2009.
The attached file contains three problems.

Please either solve the problems posted or post your own additions.

[b]4:[/b]

Evaluate: $ \left ( \int_0^{14} \sin^2(\pi x) dx \right )^2 \left [7 \left( \sum_{k \equal{} 0}^{\infty} k \frac {1}{7^k} \right)^{ \minus{} 1} \plus{} 5 \right ]$

Now that looks pretty. (the answer can also be guessed)

I asked this on a 2007 contest - but we can change the number easily enough.

Find the integer part of
\[ \sum_{k\equal{}1}^{2009^2} \frac{1}{\sqrt{k}}\]

A slight variation on problem 1166 of the Pi-Mu-Epsilon Journal:  :D

Suppose that the functions, $ f$, $ g$, $ f^{\prime}$, and $ g^{\prime}$ are continuous over $ [0,1]$, $ g(x) \neq 0$ for $ x \in [0,1], f(0)\equal{}0, g(0)\equal{}\pi$, $ f(1)\equal{}1004\frac{1}{2}$, and $ g(1)\equal{}1$. Find the value of

$ \int_{0}^{1} \frac{f(x)g^{\prime}(x)\{g^{2}(x)\minus{}1\} \plus{} f^{\prime}(x)g(x)\{g^{2}(x) \plus{} 1\}}{g^{2}(x)} dx.$

Regards.

[quote="coquitao"]A slight variation on problem 1166 of the Pi-Mu-Epsilon Journal:  :D

Suppose that the functions, $ f$, $ g$, $ f^{\prime}$, and $ g^{\prime}$ are continuous over $ [0,1]$, $ g(x) \neq 0$ for $ x \in [0,1], f(0) \equal{} 0, g(0) \equal{} \pi$, $ f(1) \equal{} 1004\frac {1}{2}$, and $ g(1) \equal{} 1$. Find the value of

$ \int_{0}^{1} \frac {f(x)g^{\prime}(x)\{g^{2}(x) \minus{} 1\} \plus{} f^{\prime}(x)g(x)\{g^{2}(x) \plus{} 1\}}{g^{2}(x)} dx.$

Regards.[/quote]

the answer is

(f/g)(1) - (f/g)(0) + (fg)(1) - (fg)(0).

= 2009/2 + 2009/2

= 2009.

[quote="Attila the Hun"]The answer is

(f/g)(1) - (f/g)(0) + (fg)(1) - (fg)(0).

= 2009/2 + 2009/2

= 2009.[/quote]

Nice going, Attila!  :wink:

Rajiv wrote [quote]There is an error occuring while downloading.............. RE-upload the file please.
[/quote]

The PDF file on the top of this thread is working fine, try it again.


someone have proved a,b.c>=0, :prod: (a 2 +2)>=9(ab+bc+ca)
and now,the ineq is a,b,c>=0,( :Sigma: a^2/3) :^3: >=9(ab+bc+ca)

oh I'm sorry it's wrong,but i think a=b=c=1,it's right
but a=b=1,c=0,it's wrong
------------------
I'm sorry

Hey, guys, don't we have other better things to do except for insisting on proving that the proposed inequalities are wrong? I simply do not understand the case which has been created on this problem. If it is correct, then solve it. If it isn't, then throw it!


[hide="3)"]Be $\arctan x+\arctan \frac{{1-x}}{{1+x}}= p$

and
$\arctan x = \alpha \to \tan \alpha = x$

$\arctan \frac{{1-x}}{{1+x}}= \beta \to \tan \beta = \frac{{1-x}}{{1+x}}$

$\tan (\alpha+\beta ) = \tan p$

so

$\frac{{x+\frac{{1-x}}{{1+x}}}}{{1-x\frac{{1-x}}{{1+x}}}}= \tan p$

simplifying
$1=tanp$

then
$p = \frac{\pi }{4}+k\pi$
for

$k=0$

$p=\frac{\pi }{4}$

$\boxed{\arctan x+\arctan \frac{{1-x}}{{1+x}}= \frac{\pi }{4}}$[/hide]

[hide="1"]
Change of base:
\[a^{ \log_{b}c}= a^\frac{ \log_{a}c }{ \log_{a}b}= c^\frac{1 }{ \log_{a}b}= c^{ \log_{b}a}\]
We apply this to the first three terms, and voila! they match the second three terms. Thus the expression is equal to $\boxed{0}$.
[/hide]

[quote="felipesa"]Be $\arctan x+\arctan \frac{{1-x}}{{1+x}}= p$

and
$\arctan x = \alpha \to \tan \alpha = x$

$\arctan \frac{{1-x}}{{1+x}}= \beta \to \tan \beta = \frac{{1-x}}{{1+x}}$

$\tan (\alpha+\beta ) = \tan p$

so

$\frac{{x+\frac{{1-x}}{{1+x}}}}{{1-x\frac{{1-x}}{{1+x}}}}= \tan p$

simplifying
$1=tanp$

then
$p = \frac{\pi }{4}+k\pi$
for

$k=0$

$p=\frac{\pi }{4}$

$\boxed{\arctan x+\arctan \frac{{1-x}}{{1+x}}= \frac{\pi }{4}}$[/quote]
[quote="fai"]Felipesa, I think your answer is only for x>-1...[/quote]
And $-\frac34\pi$ for $x<-1$.


advanced fields
advanced fields unsolved

If in a poset every contraction has fixed points, is it true that the poset has minimal elements?

What do we know about the relation between the metric and partial order?

Mh, nothing because I don't consider the metric order.. Anyway, I was told that there is a counterexample.

Of course there is, if order and metric are not related. Just combine the order structure of $ \mathbb R$ with the metric of $ [0,1]$, using some bijection between the sets.


Να βρεθούν όλοι οι θετικοί ακέραιοι x,y τέτοιοι ώστε $ 3^x\equal{}2^{x}y\plus{}1$

Για $ x\ge5$ θεωρούμε $ d$ τον ελάχιστο ώστε $ 3^d\equiv1\bmod{2^x}$. Όμως $ 3^{2^{x\minus{}3}}\minus{}1\equiv(3\minus{}1)(3\plus{}1)(3^{2^1}\plus{}1)...(3^{2^{x\minus{}4}}\plus{}1)\not\equiv0\bmod{2^x}$ και $ 3^{2^{x\minus{}2}}\minus{}1\equiv(3\minus{}1)(3\plus{}1)(3^{2^1}\plus{}1)...(3^{2^{x\minus{}3}}\plus{}1)\equiv0\bmod{2^x}$. Άρα $ d\nmid2^{x\minus{}3}$ και $ d\mid2^{x\minus{}2}$ άρα $ d\equal{}2^{x\minus{}2}$ άρα $ 2^{x\minus{}2}\mid x$. Όμως $ 2^{x\minus{}2}>x$ για κάθε $ x\ge5$(αποδεικνύεται εύκολα επαγωγικά) άρα ισχύει $ 3^x\not\equiv1\bmod{2^x}$ για κάθε $ x\ge5$.
Για $ 1\le x\le4$ δοκιμάζοντας παίρνουμε τις λύσεις $ (x,y)\equal{}(1,1),(2,2),(4,5)$.


geometry
incenter
inradius
circumcircle
angle bisector
geometry proposed

Let $ABC$ be a triangle. The incircle of triangle $ABC$ touches the sides $BC$, $CA$, $AB$ at the points $A_{1}$, $B_{1}$, $C_{1}$ respectively. The perpendicular to the line $AA_{1}$ through the point $A_{1}$ intersects the line $B_{1}C_{1}$ at a point $X$.
Prove that the line $BC$ bisects the segment $AX$.

Let $T$ be the second point where $AA_1$ cuts the incircle, and let $S$ be $B_1C_1\cap BC$. Also, let $P=TS\cap A_1X, R=B_1C_1\cap AA_1$. It's easy to see that all we need to show is that $PT\|AX$ (because $S$ is the midpt of $TP$).

We apply Menelaus in $PTA_1$ with transversal $RSX$ to get $\frac{XP}{XA_1}=\frac{RT}{RA_1}=\frac{AT}{AA_1}$ (the last equality holds because $A_1B_1TC_1$ is a harminic quadrilateral), and this is all we need to show that $TP\|AX$.

Here is my solution
Let $P,R$ be the interesction of line $A_1I$ with $B_1C_1$ and the incircle respectively. Also denote $M$ and $N$ to be the intersection of lines $AP$ and $AR$ with side $BC$ . It is easy to see that $N$ is the point where excircle of the triangle (of side $BC$) touches side $BC$. It is a bit harder to prove that $M$ is the midpoint of side $BC$ ( By the way, this fact was a problem in Moldovian Olimpiad in 8-th grade a few years ago). Now let $T=B_1C_1\cap BC$ and $V=AX\cap BC$. Let $U=AA_1\cap MI$ Obviously $AU=UA_1$. Using Menelaos Theorem in triangle $AA_1P$ with transversal $MU$ we have that: $\frac{AU}{UA_1}\frac{A_1I}{IP}\frac{MP}{MA}=1$ and since $AU=UA_1$ we have $\frac{A_1I}{IP}\frac{MP}{MA}=1$ (1). Using the Polar is easy to prove that $IT\perp AA_1$ and thus $IT \parallel A_1X$ and so $\frac{A_1I}{IP}=\frac{XT}{TP}$ (2). Using Menelaos in triangle $AXP$ with transversal $MV$ we get $\frac{AV}{VX}\frac{XT}{TP}\frac{MP}{MA}=1$ (3). Combining the results from (1), (2)and (3) we get the desired result $AV=VX$  q.e.d.

[quote="Andrei"]Here is my solution
Let $P,R$ be the interesction of line $A_1I$ with $B_1C_1$ and the incircle respectively. Also denote $M$ and $N$ to be the intersection of lines $AP$ and $AR$ with side $BC$ . It is easy to see that $N$ is the point where excircle of the triangle (of side $BC$) touches side $BC$. It is a bit harder to prove that $M$ is the midpoint of side $BC$ ( By the way, this fact was a problem in Moldovian Olimpiad in 8-th grade a few years ago).[/quote]

Yes, I have posted a proof of this on http://www.mathlinks.ro/Forum/viewtopic.php?t=5830 .

[quote="Andrei"]Now let $T=B_1C_1\cap BC$ and $V=AX\cap BC$. Let $U=AA_1\cap MI$ Obviously $AU=UA_1$. Using Menelaos Theorem in triangle $AA_1P$ with transversal $MU$ we have that: $\frac{AU}{UA_1}\frac{A_1I}{IP}\frac{MP}{MA}=1$ and since $AU=UA_1$ we have $\frac{A_1I}{IP}\frac{MP}{MA}=1$ (1). Using the Polar is easy to prove that $IT\perp AA_1$ and thus $IT \parallel A_1X$ and so $\frac{A_1I}{IP}=\frac{XT}{TP}$ (2). Using Menelaos in triangle $AXP$ with transversal $MV$ we get $\frac{AV}{VX}\frac{XT}{TP}\frac{MP}{MA}=1$ (3). Combining the results from (1), (2)and (3) we get the desired result $AV=VX$  q.e.d.[/quote]

Both of you have very nice proofs, but the proposed solution is really beautiful. I'm attaching a PDF with a slightly rewritten form of that solution. Enjoy.

[b]PS.[/b] Is it right that now the whole Moldavian IMO team is on MathLinks?  :)

  Darij

Let $I$ be the triangle incenter, $r$ the inradius and $P$ the intersection of the line $B_1C_1$ with the angle bisector $AI$. From similarity of the right angle triangles $\triangle AIC_1 \sim \triangle C_1IP$,

$\frac{r}{IA} = \frac{IC_1}{IA} = \frac{IC_1}{IP} = \frac{IP}{r}$

Consequently, the triangles $\triangle AIA_1 \sim \triangle A_1IP$ are similar by SAS: the angle $\measuredangle AIA_1 \equiv \measuredangle A_1IP$ is common and

$\frac{r}{IA} = \frac{IA_1}{IA} = \frac{IP}{IC_1} = \frac{IP}{r}$

Thus the angles $\measuredangle IAA_1 = \measuredangle IA_1P$ are equal. The lines $AA_1 \perp A_1X$ are perpendicular by definition and the lines $AI \perp B_1C_1$ are perpendicular, because the line $B_1C_1$ is a polar of the vertex $A$ with respect to the incircle $(I)$ perpendicular to the line connecting its pole $A$ with the center of the polar circle $(I)$. Since the point $X \equiv B_1C_1 \cap A_1X$ is on the line $B_1C_1$, the right angle triangles $\triangle APX, \triangle AA_1X$ have the common hypotenuse $AX$. The midpoint $M$ of this hypotenuse is the common circumcenter of the triangles $\triangle APX, \triangle AA_1X$.  The quadrilateral $APA_1X$ is cyclic with the circumcircle $(M)$. The angles $\measuredangle PAA_1 = \measuredangle PXA_1$ are equal because they span the same arc $PA_1$ of the circle $M$. But then the angles $\measuredangle IAA_1 \equiv \measuredangle PAA_1 = \measuredangle PXA_1$ are also equal (see above). This means that the line $IA_1$ is a tangent to the circle $(M)$ at the tangency point $A_1$. Since $r = IA_1$ is the inradius, the lines $IA_1 \perp BC$ are perpendicular. As a result, the line $BC$ passes through the center $M$ of the circle $(M)$, identical with the midpoint of the segment $AX$. Q.E.D.


What is the smallest positive integer $ x$ such that $ x\sqrt{115}$ will be greater than 115?

hello, we have $ x>\sqrt{115}$ and the smallest positive integer is $ x\equal{}11$ because $ x^2>115$ and $ 11^2\equal{}121$.
Sonnhard.


Using the acronym package, is it possible to make long words hyphenate properly? Currently, I have an acronym which has a long definition broken only by hyphens - the code looks like this:
[code]\acrodef{METAC}{2-methacryloyl-oxyethyl-trimethylammonium chloride}.[/code]
When this acronym is expanded, the definition is not broken but sticks out into the right margin. It seems to be inside a \mbox. Is there any way of making sure the definition is hyphenated properly.

Could you give a minimal example, a small document that shows the problem? I tried your acronym and it split the line correctly giving [quote]2-methacryloyl-oxyethyl-trimethylammonium
chloride[/quote] and in another case[quote]2-methacryloyloxyethyl-
trimethylammonium chloride[/quote]Perhaps something else is interfering, hence the request for a minimal example.

Hmm... it now seems to work perfectly, as long as I define hyphenation points using \- rather than the \hyphenation command. How odd.


I didn't get what you did after you factored X can you explain more?

[quote="ch1n353ch3s54a1l"][quote="Mathalete"]Can anyone help me on this problem?

Solve for $x$:$y$=$\displaystyle\frac{mx}{p-x}$[/quote]
[hide]
$y=\frac{mx}{p-x}$
$yp-xy=mx$
$yp=xy+mx$
$yp=x(m+y)$
$\boxed{x=\frac{yp}{m+y}}$[/hide][/quote]

well i just took $x$ out of $xy+mx$ by distribution which leads to $x(m+y)$ and then i divided both sides by $(m+y)$ to get $\boxed{x=\frac{yp}{m+y}}$


function
complex analysis
complex analysis unsolved

Why a signal can not be time-limited and band-limited at the same time?

May I Ask to recommend me a related source?

You need an emergency? Look up "Uncertainty principle" and you'll have it.

Because a signal with bounded spectrum is an analytic function (even entire).
And an analytic function (of one variable) cannot be zero on any interval without being identical zero.
Literature: any advanced textbook on functions of complex variable. For example,
E. Stein and Sharkachi, Complex Analysis.
Keywords: Wiener-Paley theorem, Indeterminacy Principle.


algebra
difference of squares
special factorizations

Awesome can you post some stuff on vedic math?

I am looking for materials on vedic math. also.
Any help in this topic will be much appreicated! :)

The difference of squares formula can be used for 96*94=(95+1)(95-1). For 95*95, 9*10 is 90, then append 90 to 5*5=25, so 96*94=9025-1=9024

You got the first trick from the Introduction to Number Theory book, right? Or did you think of it yourself?

Oh?
I learnt that trick in secondary school.
Some of the  tricks are found by myself.
After all, I realize that these shortcuts are only application of some algebraic identities in a smart way.
**********************************************
Another example ( I got this in a book.):
$ A^3 \equal{} (A \minus{} d)A(A \plus{} d) \plus{} d^2A$.
Then, $ 49^3 \equal{} 48*49*50 \plus{} 1^2*49$
48*49=48(50-1)=2400-48=2352
48*49*50=2,352*50=235,200/2=117,600
Therefore, $ 49^3 \equal{} 117,600 \plus{} 49 \equal{} 117,649$.

$ 98^3 \equal{} 96*98*100 \plus{} 2^2*98 \equal{} 940,800 \plus{} 392 \equal{} 941,192$.

[quote="luimichael"]Oh?
I learnt that trick in secondary school.
Some of the  tricks are found by myself.
After all, I realize that these shortcuts are only application of some algebraic identities in a smart way.[/quote]

Wow! you must be a hardworking person to figure out and teach yourself those tricks!
They are cool :lol:

Manipulating the square of a two digit number.
When calculating the square of a two digit number, say 38, I will first generate a squence
09,48,64.
Reconsider it as
[size=150]0[color=violet]94[/color][color=green]86[/color]4[/size], then I calculate the sum of digits in the same colour, and get [size=150]1444[/size].

Resoning behind:
Making use of $ (a \plus{} b)^2 \equal{} a^2 \plus{} 2ab \plus{} b^2$
Take a = 30, b = 8.
$ 38^2 \equal{} 0900 \plus{} 480 \plus{} 64$
That explains why we have to consider
9+4---->13
8+6---->14
then 3 + 1 = 4
and we have 1444
***********************************************************************
$ 28^2$:
04,32,64------>784

$ 29^2$:
04,36,81------>841

$ 73^2$:
49,42,09------>5329

$ 84^2$:
64,64,16------->7056

This method may be useful when at least one of the digits is less than 5.
Do you know why?


trigonometry
limit
algebra unsolved
algebra

The sequence $ (x_n)^{\infty}_{n\equal{}1}$ is defined by $ x_1 \equal{} 1$ and $ x_{n\plus{}1} \equal{}\frac{(2 \plus{} \cos 2\alpha)x_n \minus{} \cos^2\alpha}{(2 \minus{} 2 \cos 2\alpha)x_n \plus{} 2 \minus{} \cos 2\alpha}$, for all $ n \in\mathbb{N}$, where $ \alpha$ is a given real parameter. Find all values of $ \alpha$ for which the sequence $ (y_n)$ given by $ y_n \equal{} \sum_{k\equal{}1}^{n}\frac{1}{2x_k\plus{}1}$ has a finite limit when $ n \to \plus{}\infty$ and find that limit.

A. The sequence $ (x_{n}) $ will be a constant sequence, for $ n $ big enough, if $ x_{n+1} = x_{n} $:
	\[\frac{(2+\cos 2\alpha)x_{n}-\cos^{2}\alpha}{(2-2\cos 2\alpha)x_{n}+2-\cos 2\alpha} = x_{n}\]
	\[x_{n}^{2}(1-\cos(2\alpha))-\cos(2\alpha).x_{n}+\frac{(\cos \alpha)^{2}}{2}=0\]
$ x_{n} $ exists and is real if $ \Delta \geq 0 $, if $ 1 - 8 (\sin \alpha)^{2}(\cos \alpha)^{2} \geq 0 $
	\[(\sin(2\alpha))^{2}\leq 0 \]
	\[\frac{-\pi}{8}+ k.\pi \leq \alpha \leq \frac{\pi}{8}+ k.\pi \quad \quad \quad \frac{-3\pi}{8}+ k.\pi \leq \alpha \leq \frac{3\pi}{8}+ k.\pi\]
	\[\left[\frac{-\pi}{8},\frac{\pi}{8}\right] \quad \quad \quad \left[\frac{3\pi}{8},\frac{5\pi}{8}\right] \ldots \ldots\]
$ x_{n} $ goes to a constant number $ \neq 0$ and $ (y_{n}) $  has no finite limit.
B. Exception: if $\alpha = k . \pi , k \in Z $, the sequence $ x_{n} $ is increasing and the sequence $ y_{n} $ is decreasing. $ x_{n}: 1,2,5,14, \ldots $; $ x_{n+1} = 3x_{n} -1$.
	\[\lim_{n\rightarrow+\infty}\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{11}+\frac{1}{29}+ \ldots + \frac{1}{2x_{n}+1}\right)\]
	\[\lim_{n\rightarrow+\infty}\left(\frac{1}{3^{0}+2}+\frac{1}{3^{1}+2}+\frac{1}{3^{2}+2}+\frac{1}{3^{3}+2}+ \ldots + \frac{1}{3^{n}+2}\right)\]
	\[\lim_{n\rightarrow+\infty} \sum_{i=0}^{n}\frac{1}{3^{i}+2} = 0.676908\ldots\]
C.The sequence $ x_{n} $ doesn't go to a constant number if $ \alpha \in \left[\frac{\pi}{8},\frac{3\pi}{8}\right]$
What does $ x_{n} $ if, e.g. $ \alpha = \frac{\pi}{4}$?
Then
	\[x_{n+1}=\frac{2x_{n}-0.5}{2x_{n}+2} \]
$x_{n}$ is decreasing (even alternating), because $ x_{n+1}< x_{n}$, $ \frac{2x_{n}-0.5}{2x_{n}+2} < x_{n}$, $ 2x_{n}-0.5 < 2(x_{n})^{2}+2x_{n}$, OK
Therefore $	\frac{1}{2x_{k}+1} $ is increasing and \[ \lim_{n\rightarrow+\infty} \sum_{k=1}^{n}\frac{1}{2x_{k}+1}\] is divergent.


Putnam
induction
number theory proposed
number theory

Define a sequence ${u_n, n \geq 0}$ by $u_0=u_1=u_2=1$ and thereafter by the condition
$u_nu_{n+3}-u_{n+1}u_{n+2} = n!$
 $n\geq 0$. Show that $u_n$ is an integer for all $n$ (By condition, $0!=1$).
(William Lowell Putnam Math. Competition)

Nice problem !
 We can prove by induction : $\frac{u_{n+2}}{u_n}=n+1$


floor function
limit
real analysis
real analysis unsolved

Show that $\sum_{n=1}^{\infty}\frac{\left(-1\right)^{\left\lfloor \sqrt{n}\right\rfloor }}{\sqrt n}$ diverges.

Hint: [hide]$\lim_{n \to \infty}\sqrt{n+\sqrt{n}}-\sqrt{n}= \frac{1}{2}$[/hide]

Let's Consider more general case:
for which $ p$ the series converge?
$ \sum\frac{(-1)^{[\sqrt{n}]}}{n^{p}}$
We consider something like this sum
$ \sum_{m^{2}\leq n <(m+1)^{2}}(-1)^{m}n^{-p}$
for $ 1/2<p \leq 1$

Indeed, these sums are the "clusters" which compose the series in question. It seems that they should diverge for the $ p$ in question, but the hint supplied with the question should perhaps suggest a different approach.


calculus
integration
calculus computations

$ \displaystyle\int\displaystyle\int  x^2y^2\,$
[b]     R[/b]

R is the bounded portion of the first quadrant lying between xy = 1
and xy = 2 and the two straight lines y = x and y = 4x.
Can I know how to sketch the region of integration and evaluate the double integral?

Sure:
Let's get the expressions in terms of $ y$ first:
$ y\equal{}\frac{1}{x}$ $ y\equal{}\frac{2}{x}$ $ y\equal{}x$ $ y\equal{}4x$
Now this is an easy job for a graphing calculator.  You'll notice that $ y\equal{}\frac{2}{x}$ is the upper boundary and $ y\equal{}\frac{1}{x}$ is a lower boundary, and then of course, $ y\equal{}x$ is a lower boundary, and $ y\equal{}4x$ an upper.  Now you'll have to find the limits of integration and once you do that, the integral is easy.


What's the explicit form for $ a_n$ of the sequence $ a_1\equal{}2, a_2\equal{}3, a_3\equal{}4, a_4\equal{}6, \cdots, a_{n\plus{}1}\equal{}a_n\plus{}a_{n\minus{}3}$?

[quote="nealth"]What's the explicit form for $ a_n$ of the sequence $ a_1 \equal{} 2, a_2 \equal{} 3, a_3 \equal{} 4, a_4 \equal{} 6, \cdots, a_{n \plus{} 1} \equal{} a_n \plus{} a_{n \minus{} 3}$?[/quote]

What about the following: let's define $ f(x)$ as

\[ f(x) \equal{} \frac{2\plus{}t\plus{}t^2\plus{}2\,t^3}{1\minus{}t\minus{}t^4}\]

Then $ f(x)$ has the following [url=http://en.wikipedia.org/wiki/Taylor_series]Taylor series[/url]:
\[ f(x) \equal{} 2\plus{}3\,t\plus{}4\,{t}^{2}\plus{}6\,{t}^{3}\plus{}8\,{t}^{4}\plus{}11\,{t}^{5}\plus{}15\,{t}^{6}\plus{}21\,{t}^{7}\plus{}29\,{t}^{8}\plus{}40\,{t}^{9}\plus{}...\]

The coefficients list is exactly your sequence.

Sergey Markelov
markelov@mccme.ru


Whenever I go to the classroom on my Mac, many parts of it are completely black, and when I try to post something, the classroom crashes.  Does anyone know what the problem is?  It works fine on my other computers.

Does anyone else have this problem on a Mac?

Do you know which Mac operating system you are using?

I'm using Mac OS X 10.3.1.  I called Apple and they told me to try a different browser because some browsers (like Safari, the one I always use) do Java differently then Internet Explorer, so I'm going to use IE in tonight's class and see if it works.

Please let us know if that fixes the problem.

Yes, now it works.  I was able to get through the class.  It was a very good class.


geometry
rectangle
modular arithmetic
combinatorics proposed
combinatorics

Let $k$ and $n$ be positive integers. Consider an array of $2\left(2^n-1\right)$ rows by $k$ columns. A $2$-coloring of the elements of the array is said to be [i]acceptable[/i] if any two columns agree on less than $2^n-1$ entries on the same row.

Given $n$, determine the maximum value of $k$ for an acceptable $2$-coloring to exist.

this problem follows from problem 2 of IMO 98.  :)

http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16&year=1998&p=124458

Yes, it does mostly, but we need some extra things like an example:

First, to show that $k \leq 2^n$, note that the conditions require each pair of columns to have at least $2^n$ disagreements. This is at least $k(k-1)2^{n-1}$ disagreements in total. If we look at rows instead of columns, a row with $a$ white squares and $b$ black squares has $ab$ disagreements, where $a+b=k$. So there are exactly $\sum ab$ disagreements; then $k(k-1)2^{n-1} \leq \sum ab$. But $ab \leq (a+b)^2/4=k^2/4$, so $k(k-1)2^{n-1} \leq 2(2^n-1)k^2/4$. Rearranging, $k \leq 2^n$.

We make the example by induction. We will make it with $2^n-1$ rows instead of $2(2^n-1)$ and $2^{n-1}$ disagreements instead of $2^n$ (because then it suffices to duplicate the rows). For $n=1$, take one row of length $2^1$ with a $0$ and a $1$ (here we code black squares as $0$'s and white squares as $1$'s). If we have an example for $n$, we construct the one for $n+1$ as follows: take the top row and write $0$s in the first $2^n$ squares, then $1$s in the other $2^n$. We are left with a $2^{n+1}-2 \times 2^{n+1}$ board. Divide it into four $2^n-1 \times 2^n$ rectangles. Fill in the two rectangles under the $0$s, and one of the two under the $1$s, with the $n$-example; fill in the remaining rectangle with the $n$-example $+1 \pmod{2}$.


function vector algebra domain calculus derivative integration

let [tex] f:u \rightarrow R^n [/tex] be a differentiable function with a differentiable inverse [tex] f^{-1}: f(u) \rightarrow R^n [/tex].  if every closed form on u is exact, show that the same is true for f(u).

Hint: if dw=0 and [tex]f^{\star}w = d\eta, [/tex] consider [tex](f^{-1})^{\star}\eta. [/tex]

i don't know where to start with the problem.  what is a closed form?  what does it mean that "every closed form on u is exact"?

A differential 1-form is an object of the form

$\omega=f_1dx_1+f_2dx_2+\cdots+f_ndx_n$

for which one may compute a line integral.

Such a 1-form is closed iff $\frac{\partial f_j}{dx_k}=\frac{\partial f_k}{dx_j}$ for all $j,k.$

(In 3 dimensions, equating a 1-form with a vector field, this would be a vector field with zero curl.)

The 1-form is exact if it can be integrated. That is, if there exists $g$ differentiable on the whole domain $U$ such that $dg=\omega.$ This is the same as saying that $\frac{\partial g}{dx_j}=f_j.$

Every exact form must be closed; this is simply the equality of mixed partial derivatives (for smooth enough functions.) Here's the classic simplest example of a closed form that is not exact:

On $\mathbb{R}^2\setminus \{(0,0)\}$ consider the form $\frac{-y\,dx+x\,dy}{x^2+y^2}.$ This is closed, but not exact. (Its integral should be the $\theta$ of polar coordinates, but that cannot be continuously defined on the whole domain.)

Hi,

thanks for your response, Kent.  I understand closed and exact forms now but i'm still stumped on the question.  I'm not quite sure what the hint means or how to use it.  is it true that [tex]f^{\star} [/tex] is basically another way of writing a differential - eg. df?  I didn't really get a clear definition of it in my text.

anyway here's my thoughts so far:

consider [tex]w[/tex] as a form on U.  suppose [tex]w[/tex] is closed.  then dw = 0.  since every closed form on U is exact, then there exists a [tex] \eta [/tex] on U such that [tex]w = d\eta [/tex].

but how is it that [tex] f^{\star}w = d\eta [/tex] (given in the hint)?  like, how is this relationship derived?  if [tex]w = d\eta [/tex]  and also [tex] f^{\star}w = d\eta [/tex], then we have [tex]w = f^{\star}w [/tex]?  I find that really confusing, and  I'm not sure how to continue the problem.  Any help is greatly appreciated.  thanks in advance!


If the seven-digit integer $ 132ABCD$ is a palindrome, what is the greatest possible value of $ A \times B \times C \times D$?

We immediately get that $ B\equal{}3$, $ C\equal{}2$, and $ D\equal{}1$. The maximum of $ A$ is $ 9$, so $ 9(3)(2)(1)\equal{}\boxed{54}$.


induction
combinatorics unsolved
combinatorics

In a $ 15*15$ board initially all numbers in the squares are 0, in a operation, a row or column can change its numbers to a permutation of $ 1, 2, 3, ..., 15$. In a finite number of operations, find the maximal sum of the numbers in the squares.

Sorry about the sloppy reply But I am going to suggest an answer without proof. I Believe that I have a proof.

(15+14)(15)+(14+13)(14)+...+(1)(1)

Which I believe is not that tedious to calculate.
Please message me if you want a proof

Yes, i want  a proof. I have a proof too, but not much time, now. My idea is the following:
- if $ x_i$ is the number of i in the board, then after n plays 1*15 and n plays 15*1(n<15), one can choose n rows and n columns such that there are at least $ n^2$ numbers between 1, 2, ..., n(by induction), then $ x_1\plus{}x_2\plus{}....\plus{}x_j>\equal{}j^2$ and conclusion follows.

im sorry I do not understand your sketch at all.

My idea is to show that the maximum number of 15 on the board is 29.

Note: 1.each column permutation reduces the number of 15's in that column to 1.
        2.each row permutation reduces the number of 15's in that row to 1.
        3.each rearrangement can only add one 15 to a row or column
        4.if a row or column already has more than one 15 in it, permuting the row or column cannot increase the total number of 15's in that row.

=> just row operations can increase the number of 15's to 15 (one in each row).

note that any further row operations will either decrease the number 15's that are in the row (if some column operations are performed to increase the number of 15's per row) or leave it at one.

=> now, a column operation can add a 15 to each column that has no 15.
(this is at most 14 columns by the assumption that the row operations all added a 15 to the same column)

=> the maximum number of 15's on the board is at most 29.

by a similar argument for 14's (leaving the 15's unmoved) and so on, you get the same answer as I had posted before.

I had a proof similar to that too showing that there can only be 29 15s. What I couldn't do was show that the maximum number of 14s was 27, etc. It sounds easy to say "just fix the 15s", but how do you prove you can't get more 14s if you set up your 15s like this?

[code]15 -- -- -- ... --
15 15 -- -- ... --
-- 15 15 -- ... --
-- -- 15 15 ... --
..................
-- -- -- -- ... 15[/code]

Good question!

well, If you leave out the 15's you will end up with a 14*14 square! QED

If you leave out the 15s, it's not a 14 by 14 square that results. It's a 15 by 15 one with a big crack in the middle, and in terms of the problem those don't necessarily have the same properties. I still think you need to strengthen your proof.

no I do not need to strengthen it.
Yes I realized you are right. it is not a 14 by 14 square, but in order to maximize the sum, we need to minimize the number of 15's in as many rows and columns as possible. your arrangement possibly maximizes it.

when I say that I mean if you were to do any operations on the center 13*13 square, you will reduce the number of 15's in that row or column. Hence the number of 15's is not 29 anyway.


linear algebra
matrix
linear algebra unsolved

Let $A$ be an $n\times n$ matrix with its entries from a field $\mathbb{F}$. Suppose $\lambda\in \mathbb{F}$ is an eigenvalue of $A^{n}$ with $X$ as its eigenvector such that for all $a_{0}, a_{1},\dots , a_{n-1}\in \mathbb{F}$, if $\left(\sum_{i=0}^{n-1}a_{i}A^{i}\right)X=0$ then $a_{0}=a_{1}=\dots =a_{n-1}=0$. Prove that $(A^{n}-\lambda I)^{n}=0$.

[quote="alekk"]this is really strange: the condition says that $e_{k}= A^{k}X$, $0 \leq k \leq n-1$ is a basis. And it is clear that $(A^{n}-\lambda I) A^{k}X = A^{k}(A^{n}-\lambda I) X = 0$. Where is my mistake ?[/quote] I can't get where the strange part is, alekk. :|  But I guess that you haven't noticed the powers of $A$.

[quote="alekk"]if $(A^{n}-\lambda I_{d}) A^{k}X = 0$ for all $k$ and $\{A^{k}X\}_{0 \leq k \leq n-1}$ is a basis, this means that $A^{n}-\lambda Id = 0$  :maybe:[/quote] Let me call them linearly independent rather than a basis. And for your conclusion I can merely say that If the coefficient you mentioned was a scalar rather than a matrix it would be true but now?!

alekk means that the linear map that multiplies $A^{n}-\lambda I$ on the left is a zero map because it takes 0 value on a basis of $\mathbb{F}^{n}$


Let be $ a,b,c\in (0,1)$ such that $ a\plus{}b\plus{}c\equal{}2$ . Prove that :

$ abc\ge 8(1\minus{}a)(1\minus{}b)(1\minus{}c)$

[hide]
let $ a\equal{}1\minus{}x, b\equal{}1\minus{}y, c\minus{}1\minus{}z$ then $ x\plus{}y\plus{}z \equal{} 1$ and;

$ (1\minus{}x)(1\minus{}y)(1\minus{}z) \ge 8xyz$

$ \therefore (y\plus{}z)(z\plus{}x)(x\plus{}y) \ge 8xyz$

This is tru by AM-GM[/hide]

[quote="alex2008"]Let be $ a,b,c\in (0,1)$ such that $ a \plus{} b \plus{} c \equal{} 2$ . Prove that :

$ abc\ge 8(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)$[/quote]

$ abc \geq (2\minus{}2a)(2\minus{}2b)(2\minus{}2c) \Leftrightarrow abc \geq (\minus{}a\plus{}b\plus{}c)(a\minus{}b\plus{}c)(a\plus{}b\minus{}c)$
Which is classic and it is left to the reader :P ...

[quote="alex2008"]Let be $ a,b,c\in (0,1)$ such that $ a \plus{} b \plus{} c \equal{} 2$ . Prove that :

$ abc\ge 8(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)$[/quote]
It's an RMO problem  in India, which is very easy to prove.
Apart from Ocha's solution, heree is my solution:
$ a\plus{}b\plus{}c\equal{}2\implies$:
$ (1) 1\minus{}a\plus{}1\minus{}b\equal{}c\implies \frac{c}{2}\geq \sqrt{(1\minus{}a)(1\minus{}b)}$
$ (2) 1\minus{}b\plus{}1\minus{}c\equal{}a\implies \frac{a}{2}\geq \sqrt{(1\minus{}b)(1\minus{}c)}$
$ (3) 1\minus{}a\plus{}1\minus{}c\equal{}b\implies \frac{c}{2}\geq \sqrt{(1\minus{}a)(1\minus{}c)}$

Hence $ (1)\times (2)\times (3)$ yields the result.
I hope you like this solution. :lol:


The sum of all numbers of the form $ 2k \plus{} 1$, where $ k$ takes on integral values from $ 1$ to $ n$ is:

$ \textbf{(A)}\ n^2 \qquad\textbf{(B)}\ n(n \plus{} 1) \qquad\textbf{(C)}\ n(n \plus{} 2) \qquad\textbf{(D)}\ (n \plus{} 1)^2 \qquad\textbf{(E)}\ (n \plus{} 1)(n \plus{} 2)$

[hide="Solution"]Strategy: pick a small value for $ n$, manually find the sum, and determine which formula(s) yield(s) the same result.

For this solution, let $ n \equal{} 2$. The terms are $ 2\cdot1 \plus{} 1 \equal{} 3$ and $ 2\cdot2 \plus{} 1 \equal{} 5$, so the sum is $ 3 \plus{} 5 \equal{} 8$.

Taking each expression and substituting $ 2$ for $ n$ gives the following:

$ \mathrm{(A)}\quad n^2 \equal{} 2^2 \equal{} 4\ne8$

$ \mathrm{(B)}\quad n(n \plus{} 1) \equal{} 2(2 \plus{} 1) \equal{} 6\ne8$

$ \mathrm{(C)}\quad n(n \plus{} 2) \equal{} 2(2 \plus{} 2) \equal{} 8$

$ \mathrm{(D)}\quad(n \plus{} 1)^2 \equal{} (2 \plus{} 1)^2 \equal{} 9\ne8$

$ \mathrm{(E)}\quad(n \plus{} 1)(n \plus{} 2) \equal{} (2 \plus{} 1)(2 \plus{} 2) \equal{} 12\ne8$

Since the only expression of the five that works for $ n \equal{} 2$ is $ n(n \plus{} 2)$, the correct answer is $ \mathrm{(C)}$.[/hide]

[hide="Or else"]realize that it is an arithmetic sequence/series, and that the sum is given by
$ \frac{n[3\plus{}(2n\plus{}1)]}{2}$ where $ 3$ is the first odd number added, $ (2n\plus{}1)$ is the last one, and there are $ n$ of them.[/hide]

[hide="Another Solution"]Note that the general formula for the sum of the first $ n$ odd numbers is $ n^2$.

In this case, the sum starts at $ 3$ instead of $ 1$, so $ n$ becomes $ n \plus{} 1$, and $ 1$ must be subtracted from the sum. Thus, the formula becomes $ (n \plus{} 1)^2 \minus{} 1$. When expanded, this becomes $ n^2 \plus{} 2n \plus{} 1 \minus{} 1$, or $ n^2 \plus{} 2n$ when simplified. Finally, factoring gives $ n(n \plus{} 2)\Rightarrow\mathrm{(C)}$.[/hide]

[hide="Solution"]$ \sum_{k\equal{}1}^{n}2k\plus{}1\equal{}2\cdot\frac{n(n\plus{}1)}{2}\plus{}n\equal{}n(n\plus{}2)$[/hide]

Just in case anyone was wondering how you'd derive this without knowing the $ n^2$ closed form.