IMO Shortlist
number theory
Sequence
recurrence relation

Define the sequences $ a_n, b_n, c_n$ as follows. $ a_0 \equal{} k, b_0 \equal{} 4, c_0 \equal{} 1$.
 If $ a_n$ is even then $ a_{n \plus{} 1} \equal{} \frac {a_n}{2}$, $ b_{n \plus{} 1} \equal{} 2b_n$, $ c_{n \plus{} 1} \equal{} c_n$.
If $ a_n$ is odd, then $ a_{n \plus{} 1} \equal{} a_n \minus{} \frac {b_n}{2} \minus{} c_n$, $ b_{n \plus{} 1} \equal{} b_n$, $ c_{n \plus{} 1} \equal{} b_n \plus{} c_n$.
Find the number of positive integers $ k < 1995$ such that some $ a_n \equal{} 0$.

Any solutions to this fine, fine problem?

Does anybody have the official solutions or an own solution???

Can anyone prove that an integer is good(i.e satisfies the given conditions) if and only if it is of the form $k(2k+1)$ where $k$ is a natural number?

**** Incorrect solution due to incorrect reading of problem statement (see my post after the next) ***

Note that the recursive definition of $b_n$ depends only on $b_n$ itself, and the recursive definition of $c_n$ depends only on $b_n$ and $c_n$. One easily gets per induction that $b_{2n-1}=b_{2n}=2^{2+n}$ and $c_{2n}=c_{2n+1}=2^{3+n}-7$.
With this it follows again per indicution (because all $b_n,c_n \ge 0$) that, if $a_{n_0} < 0$ then $a_n < 0$ for all $n\ge n_0$.

However, to solve this problem, one needs only to calculate the $b_n,c_n$ values for $n\le 10$ recursively and use them for the $a_n$ recurrence (sorry, I have no idea how to make a table, any help would be appreciated):

$n$     $b_n$    $c_n$
0  4 1
1  8 1
2  8 9
3 16 9
4 16 25
5 32 25
6 32 57
7 64 57
8 64 121
9 128 121
10 128 249

$n$ $a_n$
0  $k$
1 $\frac{1}{2}k$
2 $\frac{1}{2}k-5$
3 $\frac{1}{4}k-\frac{5}{2}$
4 $\frac{1}{4}k-\frac{39}{2}$
5 $\frac{1}{8}k-\frac{39}{4}$
6 $\frac{1}{8}k-\frac{203}{4}$
7 $\frac{1}{16}k-\frac{203}{8}$
8 $\frac{1}{16}k-\frac{915}{8}$
9 $\frac{1}{32}k-\frac{915}{16}$
10 $\frac{1}{32}k-\frac{3875}{16}$

We see: For $k < 1995$, $a_{10} < 0$ and hence by the above observation $a_n < 0$ for all $n\ge 10$. So $a_n=0$ is only possible for $n<10$ and solving $a_n=0$ for $0 \le n \le 9$ gives as possible solutions $k=10, 78, 406, 1830$ ($k=0$ not possible as $k$ is positive according to problem statement).
So there are 4 such positive integers k as sought in the problem statement.

@hajimbrak: All the solutions are of the stated form: $10=2*5$, $78=6*13$, $406=14*29$ and $1830=30*61$. Note that the smaller factor is each time one bigger than the larger factor if the previous solution. But of course, not all the intergers of the form $m*(2m+1)$ are solutions. I guess it is straightforward to find all the solutions. You have to guess and then prove by induction the general formula for $a_n$, starting at the above given formulas for $b_n$ and $c_n$.

Here is a proof of the fact that all integers of the form $k(2k+1)$ are good.
Let $2^a$ be the largest power of $2$ less than $k$.Hence $k=2^a+b$ where
$0\leq b<2^a$.Now we prove by induction that for some $n$, $a_{n}=2^{a+1}+4b+1,b_{n}=2^{a+2},c_{n}=4b+1$ which also implies that $a_{n+1}=0$.
We can easily check the statement for the few base cases.Now we assume that
the statement is true for all $k$ less than $2^l$.
Let $k=2^l+m$,where $2^l$ is the largest power of two less than $k$.Hence $a_{0}=k(2k+1)=2^{2l+1}+(4m+1)2^{l}+m(2m+1)$
Now $m$ is less than $2^l$ hence the statement is true for it.Hence after sometime the term $m(2m+1)$ will vanish and for some $n$ we will get
 $a_{n}=2^{r+l+1}+(4m+1)2^{r},b_{n}=2^{l-r+2},c_{n}=4m+1$ which implies that $a_{n+r+1}=0$.

@Ingix:I do not think your solution is correct because I have given a proof of the fact that all numbers of the form $k(2k+1)$ are good,hence their are more than four solutions.You can check that $21,36,3,55$etc all are solutions.In fact the correct answer is $31$,you can get a solution in the IMO Compendium.

Sorry, I screwed up. I read "If $a_n$ is even/odd" as "if $n$ is even/odd", thus leading to my incorrect solution above.

For the other part, we refer to the proof given in the IMO compendium where it is proved that if $a_n=0$ then $b_{n-1},c_{n-1}$ uniquely fixes $a_0$.Now let $b_{n-1}=2^m$ and $c_{n-1}=4k+1$.We choose $l=2^{m-2}+k$ and $a_0=l(2l+1)$.This satisfies the conditions according to the proof of the other part and it is unique.Hence it is proved that an integer is good if and only if it is of the form $a(2a+1)$.


In my opinion compounds C and D would not even form the alcohol. I would vote again on structure A.

Ans is C. But what else will C and D form?

Oops!  :cleaning:

Of course! Correct answer is C because this compound does not even form the alcohol: and this because the carbocation it forms is too stable (recall Gomberg's episode).

Obviously the ans is C because the carbocation in C is the most stable.......so it will have a longer life span....and the nucleophile can easily attack from both sides as the leaving group would have departed away in that time!  :maybe:  :maybe:

What is this Gomberg's episode sir?  :maybe:

Sir if C is just too stable to form alcohol , then how will it get racemised?

Answer this question: what's is a racemization?

[hide]equalization of stereoisomers. i.e. in a molecule with one steric center, racemization would result in an equal amount of +/- stereoisomers. Thus, racemization would most likely occur when an organic molecule has a VERY stable carbocation intermediate at the steric center.[/hide]


[quote="sef"]I am searching for good books for the IMC. Do somebody have any
recommendations?

thanks[/quote]problem-solving strategies(Arthur Engel)

In fact, that's more IMO-training than IMC-training. I think solving all IMC problems and all Putnam problems will give a very solid base already. Of course you have to know all the theorems and such, but they are in your 1st year of university courses. If you intend to participate before going to university, that's a different thing.


1.Suppose AB and CD are colinear. Find the locus of the point P such that the angles APB and CPD are equal.

2.Suppose AB and CD are colinear. Find the locus of the point P such that the APB+CPD equals 90 degrees .

For 1:
Suppose that A,B,C,D are arranged in this order
$\widehat{APB}=\widehat{CDP} \leftrightarrow \frac{PA^2}{PD^2}=\frac{AB.AC}{DC.DB}(=const)$ This is Appolonious problems

(Sorry for my bad English)


Let $n$ a positive integer and $x_1,...,x_n$ positive reals such that
$\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}=n$
Find the minimum price of $x_1+\frac{x_2^2}{2}+\frac{x_3^3}{3}+...+\frac{x_n^n}{n}$
When the equality of minimum holds?

[quote="silouan"]Find the minimum price of $x_1+\frac{x_2^2}{2}+\frac{x_3^3}{3}+...+\frac{x_n^n}{n}$[/quote]

0. You get it for free at http://www.mathlinks.ro/Forum/viewtopic.php?t=16636 .  :D

(Actually, you meant "minimum value" rather than "minimum price". Just a language remark.)

  Darij


Ring Theory
quadratics
number theory unsolved
number theory

Let $ p$ be an odd prime.  What can you say about the smallest value of $ k$ such that $ p^2 | F_k$?

[hide="The answer for p | F_k"] We can show that $ k$ divides $ p \minus{} \left( \frac{p}{5} \right)$.  My proof of this fact doesn't extend readily to $ p^2$ in the case that $ \left( \frac{p}{5} \right) \equal{} \minus{}1$, so I'm curious if there is another technique that extends readily to this problem. [/hide]

How do you show that $ k\equal{}p \minus{} \left( \frac {p}{5} \right)$ is smallest such $ k$?

We can only show that $ k$ divides that value, I think.

[hide="Sketch"] If $ \left( \frac{p}{5} \right) \equal{} 1$ then $ \phi, \varphi$ are elements of $ \mathbb{F}_p$ so the Fibonacci sequence $ \frac{\phi^n \minus{} \varphi^n}{\phi \minus{} \varphi}$ is periodic with period dividing $ p\minus{}1$.  In particular, $ \phi^{p\minus{}1} \equiv \varphi^{p\minus{}1} \equiv 1 \bmod p$.

Otherwise, $ \phi, \varphi$ are elements of the splitting field of $ x^2 \equal{} x \plus{} 1$ over $ \mathbb{F}_p$, which is $ \mathbb{F}_{p^2}$.  Then $ \phi^{p\plus{}1}, \varphi^{p\plus{}1}$ both have order dividing $ p \minus{} 1$ and hence are elements of the prime subfield generated by $ 1$.  Moreover, they are conjugates, and conjugation fixes the prime subfield, so in fact $ \phi^{p\plus{}1} \equal{} \varphi^{p\plus{}1}$ and so $ p | F_{p\plus{}1}$. [/hide]

[quote="t0rajir0u"]Otherwise, $ \phi, \varphi$ are elements of the splitting field of $ x^2 \equal{} x \plus{} 1$ over $ \mathbb{F}_p$, which is $ \mathbb{F}_{p^2}$.[/quote]

Could you elaborate on how that splitting field is $ \mathbb{F}_{p^2}$?

Explicitly, the splitting field is $ \mathbb{Z}[\phi]/(p) \equal{} \{ a \plus{} b \phi : a, b \in \mathbb{F}_p \}$.  Because $ p$ is prime over $ \mathbb{Z}[\phi]$ (which is a UFD) in this case, this is indeed a field, and it has $ p^2$ elements.

Sketch of why $ p$ is prime:  if $ \left( \frac{p}{5} \right) \equal{} \left( \frac{5}{p} \right) \equal{} \minus{}1$ then $ p$ is a quadratic non-residue $ \bmod 5$.  But if $ p$ is composite in $ \mathbb{Z}[\phi]$ then $ p \equal{} a^2 \minus{} ab \minus{} b^2 \equiv (a \minus{} 3b)^2 \bmod 5$; contradiction.

Oh, right. I was thinking of the field completely differently. But I am still lost on the rest... I do not completely understand the structure of this field.

I would say that your proof can be extended easily...

If 5 is not a quadratic residue modulo p: (you already solved the other case I suppose)

Consider the ring: $ \mathbb{Z}_{p^2}[\sqrt {5}]$ (where $ \mathbb{Z}_{p^2}$ are the integers modulo $ p^2$ according to my notation - which nobody else uses)

Let $ j$ be the smallest integer such that $ p|F_j$

Let k be the smallest integer s.t. $ p^2|F_k$

Clearly j|k because Fibonacci numbers are a Mersenne sequence

Let's say k=aj

Clearly $ \phi ^j \minus{} \varphi ^j \equal{} mp$ (all the terms of the equality, as well as the equality, are in $ \mathbb{Z}_{p^2}[\sqrt {5}]$)

So $ \phi^k \minus{} \varphi^k \equal{} \phi ^k \minus{} (\phi ^j \minus{} mp)^a \equal{} amp\phi ^j$ (Newton's binomial)

We must have that $ amp\phi ^j \equal{} 0$

Since $ \phi$ is a unit (because $ \phi( \minus{} \varphi) \equal{} 1$) this is equivalent to amp=0.

Case1) If m is a unit, this is equivalent to p|a.

Case2) If m is not a unit we can prove that m=pz for some z in $ \mathbb{Z}_{p^2}[\sqrt {5}]$, so amp=0 for all a.

Clearly  the smallest possible a is a=p (case 1) or a=1 (case 2), so we know that $ k|pj|p(p \plus{} 1)$

I hope I have been clear enough... Let me know if there is something that is not clear..


hey i was wondering if anyone knows where to find the Mu Alpha Theta Log 1 and Ciphering Time Trials contest results. :)

There is a forum on AoPS for just those competitions [url=http://www.artofproblemsolving.com/Forum/index.php?f=209]here[/url].

Or you can go to their official website [url=http://www.natassessment.com/]here[/url].


Let $f: R^{+}\rightarrow R^{+}$ be such that

$f(x)^{2}\geq f(x+y)(f(x)+y)$

Is it true that $f$ must have some fixed point?

[quote="Jumbler"]Let $f: R^{+}\rightarrow R^{+}$ be such [/quote]

Really? Why?

It is easy to get, that $f(x+y)\le f(x) \ \forall y\in R_{+}$.
Therefore if f(x) continiosly it had fixed point.
Contrexample: $f(x)=\frac{1}{x}, 0<x<1, \ f(x)=0, x\ge 1$.


I was looking at the list of future mathjams and I noticed that AOPS is no longer doing the USAMO 2005 Day 1 and 2 math jams that they used to do, as well as the "How to prepare for the USAMO (psychologically)" math jam. What happened? Is there a copyright breach or something?

[quote="keta"]as well as the "How to prepare for the USAMO (psychologically)" math jam. [/quote]

I dont remember that math jam... although there should be USAMO Day 1 and 2 Math Jam, as there has alway s bee.n.

I think the USAMO math jams are just not scheduled yet, although I do remember a USAMO prep type math jam last year.


mathleague.com not mathleague.org :ninja:

The contest was today. Does anyone know what day we can start discussing the problems?

Thanks :)

[quote="now a ranger"]i think 3 days from now not sure

If Alamiss is what you are talking about.[/quote]

No, I don't live in Alabama, I live in Missouri.

well, you can most likely discuss it after the due date has passed no doubt about that I think.  :D

I'm pretty sure this exists in every state, and people have already begun to discuss some problems on the intermediate board  :ninja:

Since people can send in solutions up to a week late, I think it would be wise not to discuss the mathleague until a week from Oct. 23.


vector
superior algebra
superior algebra solved

Let $E$ an euclidean vectorial space of dimension $n$ and $A=(x=(x_1,..,x_n)|,\forall i\in [1..n],x_i \ge 0.)$.Let $H$ a subspace of $E$.Show that either $H\cap A\neq \emptyset$ ,either  $H^\bot\cap A\neq \emptyset$,where $H^\bot $ is the set of vectors which are orthogonal on $H$.

Yes.For an euclidian space it is true.Take an orthonormate base of $H$,then complete it with a family of vectors $(e_{r+1},..,e_n)$in order to form an orthonormate base of $E$.$Vect(e_{r+1},..,e_n) \subseteq  H^\bot $.You get..$dim(H^\bot)+dim(H)\ge n$.So their sum must be equal to $n$($H$ and $H^\bot$ they are in direct sum).

I'm not entirely sure my low-dimensional intuition works here as well, but i'll try :).

First of all, it's easy to prove if $V$ (this is your $H$; I'll have some other $H$'s, so I didn't want to get confused) is a hyperplane (or a line, since it's the same thing), so assume we've already done that. We also assume that $V\cap A=\{0\}$ (there was a small slip-up in the statement: $0$ is always contained in both $V$ and $A$, so I guess we're interested in non-trivial intersections).

For each line $d\subset V$, let $\mathcal H_d$ be the hyperplane orthogonal to $d$, and let $V_d=\mathcal H_d\cap V$, i.e. the hyperplane of $V$ orthogonal to $d$. Assume that, for some hyperplane $\mathcal H_d$, $A\setminus\{0\}$ lies strictly on one side of $\mathcal H_d$. This is the same as saying that $A$ does not cut $\mathcal H_d$, so $A\cap d\ne\{0\}$, which is a contradiction, because of the assumption $V\cap A=\{0\}$ and $d\subset V$. This means that every $\mathcal H_d$ cuts $A$ (non-trivially). By projecting everything onto $V$ through the orthogonal projection $p$, we see that every $V_d$ cuts $p(A\setminus\{0\})$.

Now, $p(A\setminus\{0\})$ is a convex subset of $V$ minus one point ($0$), which does not lie strictly in any half-space of $V$. It should be easy to prove now that, in general, such a situation implies the fact that $0$ i a convex combination of non-null elements of $p(A)$ (or, in other words, $0$ is not an extremal point of the subset $p(A)\subset V$), and then the same convex combination of non-null elements of $A$ which are projected onto those which we use to get $0$ must belong to $H^{\perp}$, since it's projected onto $0$ through the orthogonal projection $p$.


What is the least perfect square with 3 different prime factors?


inequalities
calculus
derivative
function
inequalities unsolved

[quote="hydro"][quote="tranthanhnam"]let $x,y,z$ be real numbers such that :$x,y,z \in (0;1] .$
 prove that :$(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})*(x+y+z) \leq 10$[/quote]

I don't know so much about inequalities, but isn't it Chebycheff? cause WLOG $x \ge y \ge z \implies \frac{1}{x}\le \frac{1}{y}\le \frac{1}{z}$
so, $(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \cdot (x+y+z) \leq 9$[/quote]

Your inequality should be reversed.

The Agrady's solution used for student in university, not elementary? Do you think so?

I don't think so, if second derivative of a function is positive, then the function takes the greastest value at the points of $1,2$.


quadratics
number theory
Diophantine equation
number theory proposed
OEIS

Find all positive integer $ m$ ,$ m\ge 2$ such that $ m^2\minus{}1|3^m\plus{}5^m$

I have been trying to solve this problem for a long time, it looks very interesting.
I would really appreciate if someone could post a hint :)

Very nice problem.
Here is my solution.  :)
If $ m$ is even then $ m = 2k$.
 $ ( {\frac {{ - 1}}{{m^2 - 1}}} ) = ( - 1)^{\frac {{m^2 - 2}}{2}} = ( - 1)^{2k^2 + 1} = - 1$ and $ 3^{2k} \equiv - 5^{2k} (\bmod m^2 - 1)$. That gives contracdition.
Thus, $ m$ is odd. Because $ 2^3 \| {(3^m + 5^m )} $, so$ m$ is of the form $ 8k \pm3$. We'll consider two case:
Case 1: $ m = 8k + 3 ( k \in N)$.
$ 1)k = 0,$so $ m = 3.$
$ 2) k > 0$. obvious $ m^2 - 1$isn't divide by$ 5$ and $ m \vdots 3$.Setting $ m = 3l$ then $ m = 24l + 3 ( l \in N)$. Thus, $ m^2 - 1 = 8(6l + 1)(12l + 1).$
If $ m \equiv 2(\bmod 5)$ then $ 6l + 1 = 6(5r + 2) + 1 = 30r + 13$. However,
$ ( {\frac {{ - 15}}{{30r + 13}}} ) = ( {\frac {{ - 1}}{{30r + 13}}} )^2 .( {\frac {{30r + 13}}{3}} )( {\frac {{30r + 13}}{5}} ) = 1.1.( - 1) = - 1.$ Contracdition.
If $ m \equiv 3(\bmod 5)$ then $ 12l + 1 = 12(5r + 3) + 1 = 60r + 37$. Similarly, we have
$ ( {\frac {{ - 15}}{{60r + 37}}} ) = - 1$
If $ m$ is divide by $ 5$ then setting $ m = 5x$. Morever, $ m \vdots 3$ then $ m = 15y (x = 5y)$. So $ m^2 - 1$ will be the form of $ 15z - 1$. Because
$ ( {\frac {{ - 15}}{{15z + 1}}} ) = ( {\frac {{ - 1}}{{15z + 1}}} )^2 .( {\frac {{ - 1}}{3}} ).( {\frac {{ - 1}}{5}} ) = - 1$
we have the contracdition. ( Note that the case $ m\vdots 5$ is true for two cases).
Case 2: $ m = 8k - 3 ( k \in N)$. Similarly, this case doesn't work.
Conclusion:$ n = 3$ .  :)
And what do you mean "Diophantine equation"?. I think it isn't [u]Diophantine Equation[/u]. :wink:

Wow very nice indeed. I had done very similar things, using quadratic reciprocity, legendre's symbol, but I never thought of Jacobi's symbol, that was very clever :)  Thank you

mmm i think thats wrong, your argument only shows $ m\equiv 3(mod\ 15)$.

By the way, do you know some proof of this problem?

Dear Charlydif, can you show me my mistake? I think the case $ m\equal{}8k\minus{}5$, we can do simirlarly.

I think that what he means is:
When you were looking the possible residues $ mod 5$ of $ m$ you mistook $ m$ and $ l$ . If you correct that you will  find that if $ m\equal{}5t\plus{}3$ there is no contradiction.

So I'm afraid it is wrong :(

In that case,
$ \left( {\frac{{ \minus{} 15}}{{60r \plus{} 37}}} \right) \equal{} \left( {\frac{{ \minus{} 1}}{{60r \plus{} 37}}} \right)^2 .\left( {\frac{1}{3}} \right).\left( {\frac{2}{5}} \right) \equal{}  \minus{} 1$.

No, because if $ m \equal{} 5t \plus{} 3 \Longrightarrow l \equal{} 5r$
You put $ l \equal{} 5r \plus{} 3$ instead

Find all integers $ (m,n)$ such that $ n(m^2 \minus{} 1) \equal{} 3^m \plus{} 5^m$. Now it's a diophantine equation!!!! :)

Talking seriously... hxy09 or Harazi, do you know a solution of this problem or at least if it admits one?

Very sorry Charlydif,like many of discussers,I broke down into it and failed to find a solution :blush:
My thoughts are similar to yours(quadratic residues ...),and I tried to use recurrence relation,but none of them works(at least for me)
And I asked harazi about this problem,but he is very busy now and hardly has time to resolve it :(

Perhaps Zetax, or somebody else could help us to solve this one.

This appears to be an open problem. See discussions at
MathOverflow: http://mathoverflow.net/questions/16341/on-polynomials-dividing-exponentials
StackExchange: http://math.stackexchange.com/questions/612346/find-all-positive-integers-n-s-t-3n-5n-is-divisible-by-n2-1
and sequences in the OEIS: http://oeis.org/A234535 and http://oeis.org/A234536


Find $x, y \in \mathbb{Z}$ such that

$2x^{2}+5y^{2}+6y = 21+6xy+2x$

Alternatively, since all the terms are positive, this is a bilinear quadratic form on a euclidean vector space, and its quite an easy algorithm to change it into a basis where this scalar product has martix $I$.

Im not sure it this is what Thales418 had in mind as he was doing this, but it works out the same anyway...

i dont like matrices. heheh

and post your answer here.  dont explain how to. but "do to"

its better

I don't know how to do matricies in latex  :(
Hold on, ill go read up and be back in a bit. (Though maybe not, class in 20 minutes)

hm..

use mathematica or maple to make the matrices
it's simple

you can convert to latex


Consider in the plane a circle $\Gamma$ with centre O and a line l not intersecting the circle. Prove that there is a unique point Q on the perpendicular drawn from O to line l, such that for any point P on the line l, PQ represents the length of the tangent from P to the given circle.

[hide]$\Gamma = (O,R)$
$D \in l, OD \perp l$

$PK$ a tangent segment from $P$ to $\Gamma$

$PK^2 = PQ^2 \Leftrightarrow$
$OP^2-R^2 = PD^2+QD^2 \Leftrightarrow$
$QD^2 = (OP^2-PD^2)-R^2 \Leftrightarrow$
$QD^2 = OD^2 - R^2 = w \ \ (1)$

It is $OD>R$, so we have solution

[color=blue][u]Note[/u][/color]

Actually, there are two solutions for $Q$. They are the intersection points of the circle $B=(D,\sqrt{w})$ with the line $OD$.

Notice that $w=OD^2-R^2$ is the power of $D$ w.r.t. $\Gamma$,
so $DQ$ is the length of the tangent segment from $D$ to $\Gamma$[/hide]

The problem can also be done this way though it is only a regional level olympiad and does not require all this:
Draw the tangents  from P to the circle at L, R. Take Q such that PQ=PR=PL. So consider the circle QLR with center P. Also, let M be the projection of O on $ l$. Obviously, OMPL is cyclic and consider that circle as well. So with the initial circle, we have 3 circles. Now using radical axes, we can find out Q with respect to the radical center. I know this method is very bad but just a solution.


The incircle $C(I,r)$ of the triangle $ABC$ touches its sides in the points $D\in (BC),\ E\in (CA),\ F\in (AB)$. I note the intersections $\Gamma \in BE\cap CF$ (the Gergogne's point), $U\in AB\cap ED,\ V\in AC\cap FD$. Prove that $I\Gamma\perp UV$.
See http://www.mathlinks.ro/Forum/viewtopic.php?t=50592

See http://www.mathlinks.ro/Forum/viewtopic.php?t=5485 .

  darij


Sequence $ (x_n)$ satisfies
                                                                                $ x_0 \equal{} a$;
                                                                 $ x_n^2 \equal{} x_{n \plus{} 1}(2 \minus{} x_n^2)   ,\forall n$;
Find a to $ x_n < 0 \forall n$

Let $ b_1\equal{}\minus{}\sqrt{2}$ and $ b_n\equal{}\minus{}\sqrt{\frac{2b_{n\minus{}1}}{1\plus{}b_{n\minus{}1}}}$.

It is easy to see for natural $ n$,
$ a<b_{2n\minus{}1}$ and $ a>b_{2n}$.

Since $ b_{2n\minus{}1}$ decreasingly converges to $ \minus{}2$, $ b_{2n}$ increasingly converges to $ \minus{}2$.

$ a\equal{}\minus{}2$.


http://acm.sgu.ru/problem.php?contest=0&problem=101

I know the solution is an euler tour. What makes it difficult for me are two things:

1. Multiple edges between the same nodes. Not really a problem, just do ++a[x][y] when adding an edge instead of a[x][y] = 1, but...
2. How do you keep track of which dominos you need to reverse and which not? This is what I can't figure out, because it gets pretty complicated (for me, anyway) if you can have multiple edges between the same nodes.

Any hints?

As you said, the problem is essentially to find a Eulerian path on a graph, but I think you might have the wrong graph. Try a different representation of the input.

EDIT: By the way, the test data for this problem is kind of tricky.


does anyone have suggestions for practicing double stops? i'm working on viotti voilin concerto 22 and the double stops are killing my hand.

Well I play cello but...

What my teacher has me do is:
1) Only play the lower note.
2) Only play the upper note.
3) Only play the lower note but finger the upper note.
4) Only play the upper note but finger the lower note.
5) And voila! Perfect double stops.

well, i play violin, and i will do that piece in about 2 months...

im working on a piece and there are FAST double stops...

one trick is to try to know what interval you play (major 3rd, 4th, etc). it helped me a lot!


We take a triangle $ / ABC$ such that there is a point $ M$ in $ Int[ABC]$ with the property:the sum of the distances from $ M$ to the sides of the triangle is equal to $ (AB\plus{}AC)/2$.Prove that $ M$ is the midpoint of $ (BC)$ and $ m(BAC)\equal{}90$.

Here is a counterexample. Consider points $ A \equal{} (4 \minus{} \sqrt{7}, 0)$, $ B \equal{} ( \minus{} 4 \plus{} \sqrt {7}, 0)$, $ M \equiv C \equal{} (0, 3)$. Then
\[ AB \plus{} AC \equal{} 2 (4 \minus{} \sqrt {7}) \plus{} \sqrt {(4 \minus{} \sqrt {7})^2 \plus{} 3^2} \equal{} 2 (4 \minus{} \sqrt {7}) \plus{} 2 (\sqrt {7} \minus{} 1) \equal{} 6.
\]
The sum of the distances from $ M\equiv C$ to the sides of the triangle equals $ 3 \plus{} 0 \plus{} 0 \equal{} 3$.
[asy]draw((4-sqrt(7), 0)--(-4+sqrt(7), 0)--(0,3)--cycle);
dot((4-sqrt(7), 0),red); dot((-4+sqrt(7), 0) ,red); dot((0, 3) ,red);
label("$A$", (4.1-sqrt(7),-0.1)); label("$B$", (-4.2 + sqrt(7),-0.1));
label("$C \equiv M$", (0,3.2));[/asy]


My biggest problem is learning how to convert word problem data into a math equation leading to the answer by solving for a variable or variables.

This skill separates amateur math people from true mathematicians.  I believe the skill to create an equation from the given information in word problems is unique and wonderful.
Not every student can do this no matter how hard he or she may try.

Here is a sample question:

Peter has nickels, dimes and quarters in his pocket.  The number of nickels is 1 more than twice the number of quarters.
The number of dimes is 1 less than the number of quarters.  If the value of the coins is 85 cents, how many of each coin does Peter have?

MY WORK:

I created the following equation but it does not lead to the correct answer.

I said:

2(0.05x) + 1 + 0.10x - 1 + 0.25x = 0.85

How do I set up the right equation for this question?

What is the right equation?

Again I don't know if this is right

[hide]
another system of equations so
x=nickels
y=dimes
z=quarters

$x=2z+1$
$y=z-1$
$x+y+z=85$

if you have a calculator to solve the equations using rref
[/hide]

[hide]Or simply substitute the two first equations in the second, you are left with only x.  Your third equation is wrong because it is saying that you have 85 COINS total; you must multiply by the VALUE of each coin; that gives you 85.  This would be:

$5x+10y+25z=85$

Now substitute, and get

$5(2z+1)+10(z-1)+25z=85$

Simplify, blah blah blah to get

$45z=90$

Obviously, $z=2, x=5, y=1$

So 5 nickels, 1 dime, and 2 quarters.[/hide]


Find the number of nonnegative integers n for which $(n^2-3n+1)^2+1$ is a prime number.

[hide]There are 2 cases to consider:
$n\equiv0\pmod{2}$
That means that $n^2$ is even, $-3n$ is even, and $(n^2-3n+1)^2$ is odd.  Therefore, the sum is even, and the only even prime is 2.
$n\equiv1\pmod{2}$
That means that $n^2$ is odd, $-3n$ is odd, and $(n^2-3n+1)^2$ is odd.  Therefore, the sum is even, and the only even prime is 2.

$(n^2-3n+1)^2+1=2$
$(n^2-3n+1)=\pm1$
$n^2-3n=0$
Because $n$ is nonnegative, $n=3$ or $n=0$.
$n^2-3n+2=0$
$(n-1)(n-2)=0$
$n=1$ or $n=2$
There are 4 possible nonnegative integral solutions for $n$.[/hide]

[hide]n^2+3n+1 is always odd, so the square is also odd, and when you add one to the square it becomes even.  Hence, the only prime number possible is 2 and the solutions that get you two are 3,2,1,0.
$4$[/hide]


Source: OLYMON, August 2003 (Question 243.)

The inscribed circle, with centre I, of the triangle ABC touches the sides BC, CA and AB at the respective points D, E and F. The line through A parallel to BC meets DE and DF produced at the respective points M and N. The midpoints of Dm and Dn are P and Q respectively. Prove that A, E, F, I, P and Q lie on a common circle.

[img]http://www3.sympatico.ca/yuethomas/images/243.gif[/img]

Solution by T. Yue

[hide]Let AQ produced meet CB at R.  Then AQ=QR and NQ=QD so that RD=AN=AE => CR=CD+DR=CE+AE=CA.  Therefore CAR is isosceles with median CQ.  Hence CQ is perpendicular to AR and Q lies on the angle bisector of <ACR.  Thus I,Q,C are collinear with <IQA=<IFA=90.  Hence AFQIE is concyclic, Also AFPIE is conclyclic and the result follows

[/hide]

You use his own solution to a problem he posed? Come on, you can do better than that.

I posted bc i like his solution  :mrgreen:

Here's my soln:

It's obvious that AEIF is cyclic (*).

It's easy to see (ANF similar to BDF and CDE similar to AME) that FNA and EMA are isosceles, and since FN=EM we get AN=AM, so APQ is the median triangle of DMN (1). It's now easy to see that E and F are the feet of the perpendiculars from N and M respectively in triangle DMN (2). From (1) and (2) we get that A, P, Q, E, F are situated on the Euler circle of triangle DMN (**).

(*) and (**) give what we want.

Inspiration : IMO 1998/5 ?
The situation is very similar but the problem is easier.

Dear,
see also http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=576791942&t=178572
Sincerely
Jean-Louis


function
vector
modular arithmetic
algebra unsolved
algebra

Find all monotone (not necessarily strictly) functions $ f: \mathbb{R}^{\plus{}}_0\rightarrow \mathbb{R}$ such that:

$ f(x\plus{}y)\minus{}f(x)\minus{}f(y)\equal{}f(xy\plus{}1)\minus{}f(xy)\minus{}f(1) \forall x,y \ge 0$;

$ f(3)\plus{}3f(1)\equal{}3f(2)\plus{}f(0).$

[quote="moldovan"]Find all monotone (not necessarily strictly) functions $ f: \mathbb{R}^{ \plus{} }_0\rightarrow \mathbb{R}$ such that:

$ f(x \plus{} y) \minus{} f(x) \minus{} f(y) \equal{} f(xy \plus{} 1) \minus{} f(xy) \minus{} f(1) \forall x,y \ge 0$;

$ f(3) \plus{} 3f(1) \equal{} 3f(2) \plus{} f(0).$[/quote]
1) Let us solve the easier equation $ (E1)$ :
"Find all functions $ g(x)$ from $ \mathbb N\to\mathbb R$ such that : $ g(2x\plus{}y)\minus{}g(2x)\minus{}g(y)\equal{}g(2y\plus{}x)\minus{}g(2y)\minus{}g(x)$ $ \forall x,y\in\mathbb N$"

The set $ \mathbb S$ of solutions is a $ \mathbb R$-vector space.
Setting $ y\equal{}1$, we get $ g(2x\plus{}1)\equal{}g(2x)\plus{}g(1)\plus{}g(x\plus{}2)\minus{}g(2)\minus{}g(x)$
Setting $ y\equal{}2$, we get $ g(2x\plus{}2)\equal{}g(2x)\plus{}g(2)\plus{}g(x\plus{}4)\minus{}g(4)\minus{}g(x)$
From these two equations, we see that knowledge of $ g(1),g(2),g(3),g(4)$ and $ g(6)$ gives knowledge of $ g(x)$ $ \forall x\in\mathbb N$ and so dimension of $ \mathbb S$ is at most $ 5$.
But the $ 5$ functions below are independant solutions :
$ g_1(x)\equal{}1$
$ g_2(x)\equal{}x$
$ g_3(x)\equal{}x^2$
$ g_4(x)\equal{}1$ if $ x\equal{}0\pmod 2$ and $ g_4(x)\equal{}0$ if $ x\neq 0\pmod 2$
$ g_5(x)\equal{}1$ if $ x\equal{}0\pmod 3$ and $ g_5(x)\equal{}0$ if $ x\neq 0\pmod 3$
And the general solution of $ (E1)$ is $ g(x)\equal{}a\cdot x^2\plus{}b\cdot x\plus{}c\plus{}d\cdot g_4(x)\plus{}e\cdot g_5(x)$

2) Solutions of the original equation :
Let $ P(x,y)$ be the assertion $ f(x\plus{}y)\minus{}f(x)\minus{}f(y)\equal{}f(xy\plus{}1)\minus{}f(xy)\minus{}f(1)$
In a first time, let us forget the monotonous constraint and the equation about $ f(0)$

Comparing $ P(xy,z)$ and $ P(xz,y)$, we get $ Q(x,y,z)$ : $ f(xy\plus{}z)\minus{}f(xy)\minus{}f(z)\equal{}f(xz\plus{}y)\minus{}f(xz)\minus{}f(y)$
Let then $ p$ a positive integer.
$ Q(2,\frac mp,\frac np)$ $ \implies$ $ f(\frac{2m\plus{}n}{p})\minus{}f(\frac{2m}{p})\minus{}f(\frac np)\equal{}f(\frac{2n\plus{}m}{p})\minus{}f(\frac{2n}{p})\minus{}f(\frac mp)$

So $ f(\frac xp)$ is a solution of $ (E1)$ and so $ f(\frac xp)\equal{}a_p\cdot x^2\plus{}b_p\cdot x\plus{}c_p\plus{}d_p\cdot g_4(x)\plus{}e_p\cdot g_5(x)$ $ \forall x\in\mathbb N$
Choosing $ x\equal{}kp$, it's easy to see that $ a_p\equal{}\frac{a}{p^2}$, then that $ b_p\equal{}\frac bp$
Choosing $ x\equal{}2kp$, $ x\equal{}3kp$ and $ x\equal{}6kp$, it's easy to see that $ c_p\equal{}c$ and $ d_p\equal{}e_p\equal{}0$

And so $ f(\frac xp)\equal{}a(\frac xp)^2\plus{}b(\frac xp)\plus{}c$ $ \forall x,p\in\mathbb N$
And so $ f(x)\equal{}ax^2\plus{}bx\plus{}c$ $ \forall x\in\mathbb Q^{\plus{}*}$

Now, we must remember that $ f(x)$ is monotonous, and so $ f(x)\equal{}ax^2\plus{}bx\plus{}c$ $ \forall x\in\mathbb R^{\plus{}*}$
The condition "monotonous" implies too that $ ab\geq 0$
The condition $ f(0)\equal{}f(3)\plus{}3f(1)\minus{}3f(2)$ implies $ f(0)\equal{}c$

And so the necessary conditions are :  $ \boxed{f(x)\equal{}ax^2\plus{}bx\plus{}c}$ $ \forall x\in\mathbb R_0^\plus{}$ with $ ab\geq 0$

And it's immediate to check that these necessary conditions are indeed sufficient.


Given are quadirateral $ABCD$ inscribed $(C),AB\cap CD=\{O\}$, $(\alpha)$ is a cricle that touch $OB,OC$ and small arc $BC$, tangent points with $OB,OC$ is $M,N$ prove that line $EF$ is through incenter of triangles $ABC$ and $DBC$. :)

[quote="gemath"]Given are quadirateral $ ABCD$ inscribed $ (C),AB\cap CD \equal{} \{O\}$, $ (\alpha)$ is a cricle that touch $ OB,OC$ and small arc $ BC$, tangent points with $ OB,OC$ is $ M,N$ prove that line $ MN$ is through incenter of triangles $ ABC$ and $ DBC$. :)[/quote]
$ EF$ or $ MN$  is through incenter of triangles $ ABC$ and $ DBC$
If $ MN$ then
Your Problem is Lyness open..... :D


Let $ a(n)$ denote the number of (decimal) digits in $ 2^n$ that are at least $ 5$. Find $ \sum_{i\equal{}0}^{\infty}\frac{a(i)}{2^i}$.

[quote="Peter"]Let $ a(n)$ denote the number of (decimal) digits in $ 2^n$ that are at least $ 5$. Find $ \sum_{i = 0}^{\infty}\frac {a(i)}{2^i}$.[/quote]

Very very nice problem !!!

[hide="My solution"]
I suggest we'll do the count in a "vertical" manner instead an "horizontal" one :

Consider $ x\geq 0$ and $ u(x)=1$ if the unit decimal digit of $ x$ $ \in\{5,6,7,8,9\}$ and $ u(x)=0^$ otherwise.
Consider then $ f(x)=\sum_{k=0}^{+\infty}\frac{u(2^kx)}{2^k}$
$ f(x)$ counts (with weight $ 2^{-k}$) the unit digits over $ 5$ in the sequence $ x,2x,4x,...$.

So the result we are looking for is $ f(1) + f(10^{-1}) + f(10^{-2}) + ...$ (consider that $ f(1)$ will count the unit digits of the sequence $ 2^n$, that $ f(10^{-1})$ will count the second digits (power of 10) of the same sequence, $ f(10^{-2})$ will count the third digits (power of 100) of the same sequence, and so on).

Now is the magic result : $ f(x)=\frac{x}{5}$ $ \forall x\in[0,10)$

In order to see this, consider that $ u(2^kx)$ is $ 1$ $ \iff$ $ \{\frac{2^kx}{10}\}\in[\frac{1}{2},1)$ (where $ \{u\}$ is the fractional part of $ x$) and so is just the $ (k+1)^{th}$ binary digit (right of dot, with $ k=0$ being the unit digit) of
$ \frac{x}{10}$

So the required result is $ \frac{1}{5}+\frac{1}{50}+\frac{1}{500}+...$ $ =\frac{2}{9}$
[/hide]


[quote="john lai"]The thing was already due but I still have the paper and it's killing me that this is still blank and I can't figure out the answer or how to do it! Can someone help me please?

A trough is 12 feet long and 3 feet across the top (see figure). Its ends are isoceles triangles with altitudes of 3 feet.

a. If water is being pumped into the trough at 2 cubic feet per minute, how fast is the water level rising when it is 1 foot deep?

b. If the water level is rising at a rate of 3/8 inch per minute when h = 2, determine the rate at which water is being pumped into the trough.

Here is my bad sketch from the book made in paint:[/quote]

Would someone please move this to another forum?


Whos going?  I am.  Do you guys want to make a list?

Indianaman

^Did you go?  I dont actually know who you are, but we were very fortunate in that we did very well this year at ARML.

We got 9th place nationally, which is perhaps the best we've ever done.


ABE is 1/4 of ACD, so BCDE is 3/4 of ACD, or 75%

Hello,

You know that because that the dimensions of ABE are all 1/2 of the larger triangle.

- bartle-doo


number theory
prime numbers
number theory open

Let $p<q$ be prime numbers. Let $0<u<p<q$ be an integer.
It is known that the progressions $u+p\mathbb N$ and $u+q\mathbb N$ contain infinitely many prime numbers.

Do there exist prime numbers $p$ and $q$ such that the smallest primes appearing in each of the above progressions are equal?

Sorry. Except of course for the case $u$ is itself prime.

What I am trying to do with this problem is to construct a graph on the set of prime numbers such that each vertex $p$ is attached to $p-1$ edges and for all $i=\overline{1,p-1}$, there is an edge connecting $p$ to the least prime number $q$ in the arithmetic progresion $i+p\mathbb Z$. I am trying to see if this construction is possible, the problem being that this construction when carried out step by step may give a prime $p$ connected to more than one prime for some $0<i<p$.


rectangle
combinatorics proposed
combinatorics

A $3000\times 3000$ square is tiled by dominoes (i. e. $1\times 2$ rectangles) in an arbitrary way. Show that one can color the dominoes in three colors such that the number of the dominoes of each color is the same, and each dominoe $d$ has at most two neighbours of the same color as $d$. (Two dominoes are said to be [i]neighbours[/i] if a cell of one domino has a common edge with a cell of the other one.)

Let the colors be 0, 1, and 2

if a block is exclusively in column k, then color it k (mod 3).

If a block is in column k and column k+1, then count how many blocks above it are also in column k and k+1. If this number is even, then color it k (mod 3). If it's odd, color it (k+1) (mod 3).

This coloring satisfies all properties in the problem.

No, this doesn't work (it already fails on a $ 6\times 6$ board). And if you claim that this coloring solves the problem, you have to prove that.

Further note: the number of tiles that are in columns $ k$ and $ k \plus{} 1$ is always even (more generally on any board with even height).

Imagine the board as a chessboard. Each domino occupies a white and a black cell. If its white cell is (i,j) then color the domino with color $ (i \minus{} j) \mod3$. And that's all.

This coloring will form a sort of diagonal strips. The only neighbors of a domino which have the same color are on the same diagonal strip, so there are at most two of them.

[quote="ZetaX"]You can have three neighbours in the same diagonal strip:
[code]
XxXxX
xXbAx
XxBaX
cCsSx
XxXxX
[/code]
Here the dominos are aA, bB, cC and sS (where capital letters are white fields), you see that the first three will be coloured the same but all touch sS.[/quote]
But Ss has a different color. the diagonal ABC is different mod 3 from the parallel one that contains S.  :!:

Note that it says the same color [b]as $ d$[/b].

color the table with 1,2,3 diagonally.
1,2,3,1,2,3
2,3,1,2,3,1
3,1,2,3,1,2
1,2,3,1,2,3
now for each domino write the some of two numbers in the cells of it mod 3

and we are done

Color the table with red, green and blue like this, and color each domino with the same colored cell it covers. Done ?:
[code]
ROGOBOROGOBO...
OGOBOROGOBOR...
GOBOROGOBORO...
OBOROGOBOROG...
BOROGOBOROGO...
OROGOBOROGOB...
...............
[/code]

Here is a simple solution:Color the dominos with one part in the first column blue the remaining ones with one part in the second column red the remaining ones with one part in third column green and continue this alnternatingly.It is easy to see this works.

wow amazing
[hide = solution]
Enumerate the cells $(i, j)$ for $1 \leq i, j \leq 3000$. Write a value of $\omega^{i+j}$ on the cell $(i, j)$, where $\omega$ is a primitive third root of unity. Now to each domino, color it with its sum of values. First, observe that each color assigned is one of $-1, -\omega$ or $-\omega^2$ so there are at most three colors. Next, observe that the following picture shows there are at most two dominoes of the same color touching any given one:

[img]https://i.imgur.com/hPdtDZp.png[/img]

(the yellow domino represents the tiled one, whereas the pink ones denote [b]possible[/b] locations for a touching domino of the same color) $\square$.
[/hide]

Label each cell with $x+y \pmod{3}$. Color each domino with its missing label.


An examination has fifteen problems which are numbered 1 through 15. No student answers two consecutive problems correctly. And assume that each question has only two possible answers -- “yes” or “no”; and no student leaves any question unanswered.
If 1600 candidates sit for it, then must at least two of them have identical answer patterns?


analytic geometry
logarithms
function
Support

Can you guys help me SHOWING ALL WORK? I need help with these problems. Solutions before Wednesday would be awesome.

Btw, you MUST show all work, no matter how tedious or insignificant it is, because my math teacher is very picky about this sort of stuff.

Also, when you answer, can you state which file it came from? For example: Doc.1, #4b.

I would really appreciate help on ALL of the problems, even the "easy" ones, because my definition of "easy" may not be the same as yours. :blush:

Thank you. :D

EDIT: oh yeah, and for the graphs, you can just use any program like paint or art pad or whatever and just add that as an attatchment. :D

We don't do homework. EVER. ROFL  :rotfl:

http://mathforum.org/library/drmath/view/66166.html
might help you understand some of concept...i might do your hw for easter :(

woot almost 1 am and here is
[hide="1"]Take the log of both sides and divide by xy
to get
log(x)/x=log(y)/y
clearly, if x & y=a
log(a)/a=log(a)/a[/hide]
and
[hide="2"]since log(x)/x=log(y)/y then log(y)/y=log(x)/x the only integer solution for this is (2,4) and (4,2) other than the (a,a)[/hide]

Why are you doing the problems, he's just asked for you to do his homework... :roll:

dude because if someone gives me free math homework i do it. pathetic, i know.  :( but my algebra 2 barely had this stuff covered so i will do their homework as review making sure they sort of understand.

@funcia: can you be a little more clear on how you got #2?

[quote="kohjhsd"]It does look like homework, but anyway...
I think #2 is actually
[hide]Since we are told to refer to the graph, let's refer to the given graph.

Consider a horizontal line $y=c$, where $0<c<e$. We can say this line has two intersections with the graph of the $f$ since $f$ increases, has a maximum at $e$, and decreases again.

Then for that two point, if the $x$-coordinates of the intersections are $a$ and $b$, we have $(a,c)$ and $(b,c)$ that is on the graph of $f$. This means that $f(a)=f(b)=c$. Therefore, $\frac{\ln a}{a}=\frac{\ln b}{b}$.

Then $a \ln b = b \ln a$.
Then $\ln b^{a}= \ln a^{b}$.
Then since $f(x)=\ln x$ is a one-to-one function, $a^{b}=b^{a}$.

Therefore $(a,b)$ is a solution of $x^{y}=y^{x}$, and clearly there are many of them for different values of $c$.[/hide]

[quote="bigboy82892"]Btw, you MUST show all work, no matter how tedious or insignificant it is, because my math teacher is very picky about this sort of stuff.

Also, when you answer, can you state which file it came from? For example: Doc.1, #4b. [/quote]
Sometimes it is difficult to show some of the work :roll:[/quote]
i sort of see where you are going and it definitely looks to have more rhyme and reason then my post but i have to go :(

for question number 3 there are not other integers wich satisfy $\ x^{y}=y^{x}$ different of (2,4)
the proof:
if we make a horizontal line $\ y=c$
first case c is a negative integer then there isn't two intersection with f(x)
second case c is an integer wich satisfy  $\ e\geq c\geq 0$
then there isn't two intersection apart from $\ c=f(4)$

Ok, I REALLY need that help right now, I have NO IDEA how to do part 2. :blush:

EDIT: and part 1 :oops: . I have no idea about any of this. HELP!

Hello? Can someone PLEASE HELP ME? I CAN'T DO PART 2!

[hide="hint"]#1:  Try rearranging and doing a little bit of algebra
#2:  Consider where the graphs intersect (use a calculator if you have to to get the graphs).
#3:  What if $x=y$?[/hide]

Ok, this is due on Monday (can't be late or no credit :( ) So I need this by Sunday. There are two problems on each that I still can't get, so I'm BEGGING you to PLEASE HELP ME I SUCK! #3(a&b) and 4 on part one (I know, I posted it backwards), and #2&3 on part two. I would REALLY appreciate help with this.

wtf dude do your own work...like it is okay if you are like: "give me a hint so i can figure it out," but instead, you are like: "do my homework so i can just copy what you wrote"

erm, shouldn't we lock this topic? he's going to copy down what someone else did and submit it for credit

I don't mean "give me the answer." I mean "give me the path then let me walk down it alone."

I think that both solutions and hints were thoroughly given for the first file (part 2).

However, instead of saying "I don't know," you might want to say what you don't know :) . Many people do that in the basics forum; getting a "hint" or a key idea. The questions are quite interesting though; maybe it's just because I skipped algebra 2...

[quote="bigboy82892"]Ok. Well, on 3.a. for part one, I know how to take a log of both sides like this: $\ln{x^{2}}=\ln{2^{x}}$. Which can then be written as <skip a few steps>: $\frac{\ln{x}}{x}=\frac{\ln{2}}{2}$. But I'm not sure how you would necessarily graph that. I mean, I guess you could graph $f(x)=\frac{\ln{x}}{x}$ and $g(x)=\frac{\ln{2}}{2}$, plug them into your calculator and find the intercepts. Oh. I guess I can do part a and b, then just do $g(x)=\frac{\ln{whatever}}{whatever}$ for #4. Ha! I CAN do it! Thanks kohjhsd! :P

Now on part two, on #2, I guess I could just make a few graphs then say "by observing these graphs, one can see...", but wouldn't that be kinda empty? I mean, I'd need more of a mathematical explanation than that. Right? And on #3, I could just bring up the table from part 1. Well, thanks. I guess I really haven't tried thinking like this :oops: , so thanks for knocking some sense into me. But I'm still not sure for #2 on part 2, I mean, just saying that since a few cases of graphs support the statement, that it's true is kind of a hollow arguement, it needs more to it. Any thoughts?[/quote]

For 1-3a, I think
[hide]You just need to graph to functions and find the intersections, which means you don't need to modify the equation. Just graph $y=2^{x}$ and $y=x^{2}$ and find the intercepts! This lead to two solutions. Then you can finish the table using the same method. Then you can plot the points and do the rest. (That's what I think; I didn't take a close look at part 1 anyway.)[/hide]

For 2-2, I posted my solution already:
[hide]Consider a horizontal line $y=c$, where $0<c<e$. We can say this line has two intersections with the graph of the $f$ since $f$ increases, has a maximum at $e$, and decreases again. [...]

What can you say about these two points? How many possible values are there for $c$?[/hide]

For 2-3,
[hide]Don't forget you are only finding the integer solutions.[/hide]


function
ratio
geometry
geometric transformation
reflection
inequalities
projective geometry

$AB$ is a diameter of a circle. $C$ and $D$ are points on the tangent at $B$ on opposite sides of $B$. The lines $AC, AD$ meet the circle again at $E, F$ respectively. The lines $CF, DE$ meet the circle again at $G, H$ respectively. Show that $AG = AH$.

We will show that $CDFE$ is an inscribed quadrilateral. Indeed, $\angle ADB=\pi/2-\angle FAB$ and $\angle AEF=\angle ABF=\pi/2-\angle FAB$, since $AB$ is a diametr.

Since $\angle BFD=\angle BEC=\pi/2$ and $\angle CFD=\angle DEC$ we conclude $\angle GFB=\angle HEB$. It follows $BH=BG$, hence $AH=AG$.

Here's what I did :) (even though it's overkill :))

If we keep $C$ fixed and move $D$, the map $AH\mapsto AG$ is projective, so it suffices to show that the conclusion holds in three cases. We can take $D\to\infty,D=B$, and, finally, $BD=BC$.

If we keep $C$ fixed and move $D$ around, $H$ and $G$ move. We can then define the function taking $AH$ to $AG$, and it's easy to see that this is a projective transformation from the pencil of lines through $A$ to itself. We want to show that this projective transformation coincides with the one obtained by reflecting $AH$ in $AB$. Since it's projective (conserves cross-ratios of quadruples of lines through $A$), it's determined by the images of three different lines through $A$, or, in other words, by three distinct positions of $D$. I took the three positions $D\to\infty$ (the infinity point), $D=B$, and $D$ s.t. $BD=BC$.

The two transformations which we are trying to show coincide (our transformation and the reflection in $AB$) are the same for these three positions of $D$, so they must be identical.

[quote="grobber"]...it's easy to see that this is a projective transformation...[/quote]

I can't see why this map preserves cross-ratio!

Could you explain this? Thanks.

$AB^2=AE.AC=AF.AD$therefore $\angle FCD=\angle FED$
$FHGE$ is cyclic so $\angle FED=\angle FGH$
therefore $\angle FCD=\angle FGH$ it means that $GH$ is perpendicular to $AB$
We R done.

[color=indigo]If we prove that [b]EFDC[/b] is an inscribed quadrilateral the problem is solved.(because then QC*QF=QE*QD => [b]QF:QE=QD:QC[/b] (1). and [b]EGHF [/b]is an inscribed quadrilateral => [b]QE*QH=QF*QG [/b]=> QF:QE=QH:QG =>(1) =>QH:QG=QD:QC => [b]GH[/b] is parallel with [b]CD[/b]  => AG=AH (?)) so lets prove it:
DF.AD=DB^2=DA^2 - AB^2 & CE.CA=CB^2=AC^2 - AB^2
=>DF.AD - CE.CA=AD^2 - AC^2 => AD(DF-AD)=AC(CE-AC) =>
AD(-AF)=AC(-AE) => [b]AD.AF=AC.AE[/b] =>[b]EFDC[/b] is an inscribed quadrilateral and problem is solved.[/color] :roll:

What do you exactly mean?I cant understand you?can you expalin?

dear Jensen
Number 1 didn't understand Grobber solution.

thank you mathx.i think i understand now.

Yes, I don't understand it. Now, can you Help me?

Well, actualy i understand perfectly grobbner's solution.
I just can' t see why this map $AH\mapsto AG$ preserves cross-ratio.

If I undestand properly, we have to prof: $(AH_1, AH_2,AH_3,AH_4) = (AG_1, AG_2,AG_3,AG_4)$, true?

number1 i think my solution is easier then grobber's.i think you can understand it well why don't you enjoy that? :D

I think you are nice person jensen. Thank you trying help me.
However I stil want to have my ansver. I think perspective and projective transformations are very very poverfull toys, as inversion is.

indeed, projective geometry is a very powerful tool (especially in the hands of grobber)

as for your question, i'm not very sure whether what you're trying to prove is right, but my instinct tells me it is.

if you prove that, please share it with us  ;)

lineAG cut the line DC at point w .
by using inversion we can find that  AEWD is a cyclic quadrilateral.
and it means that the angles AWD and AEF are equal.
look at the angle AGFplz.
AGF=GWB-GCW.
its done .because it means that the angles FGH and FCD must be equal.
and then again it means that the lines HG andCD must be parallel.
and if they are parallel the lins AG andAH will be equal.

i want an easier solution to this problem :D

i know nobody can solve it as easier as i solved :D

may be you are right but i think all of solutions are beautiful and atractive and all of them are the easier solution in solver mind!  :D

I have enought of this! I' am mad!
I think grobber actualy can NOT prove projectivity of map $AH\mapsto AG$!!!

[quote="Number1"]I think grobber actualy can NOT prove projectivity of map $AH\mapsto AG$!!![/quote]

Do you, now? :) The only reason why I did not answer was that I don't like being dragged into old, dead (at least to me) topics only because some dude whom I do not know decides he did not understand my solution. Normally, I would have continued to ignore the thread, but since you're challenging me, what choice do I have? :)

$F\mapsto G$ is projective (on the circle), because it's an involution with center $C$. $EH\mapsto AF$ is projective (from the pencil through $E$ to the pencil through $A$) because the lines $EH,AF$ intersect on $CD$; this means that $H\mapsto F$ is projective. By composing the two projective transformations $H\mapsto F,F\mapsto G$ we find $H\mapsto G$ to be a projective self-map of the circle, and this is just another way of saying that $AH\mapsto AG$ is projective.

I do hope you're satisfied, Number1, since I am going to ignore any subsequent messages which might appear in this thread. ;)

thank you so much grobber i think nobody can help number1 but you and thank you so much. :lol:

number 1 .why do u say inversion isnt power full :mad:
i think its because of that u dont know what is inversion.

:?
[quote="ashegh"]number 1 .why do u say inversion isnt power full
i think its because of that u dont know what is inversion.[/quote]

:?:  :?:  :?:

When did I say that?

Come on, it's enough of spam. Number1 said that inversion [i]is[/i] powerful, but even if he would have said the opposite, it would be just an unsubstantial assertion not worth replying. And, Ashegh, please [b]do not post 4 subsequent messages[/b], especially if 3 of them are trash ("shoma ridi" - would anybody please translate this; "i want an easier solution to this problem" - then look at post #2 by Myth; "i know nobody can solve it as easier as i solved" - are you just trying to flame or do you think that somebody will reply).

Here is my solution of the problem (essentially the same as Myth's one):

[color=blue][b]Problem.[/b] Let A and B be two points, and k the circle with diameter AB. Let C and D be two points on the tangent to the circle k at the point B. The lines AC and AD meet the circle k at the points E and F (apart from A), and the lines CF and DE meet the circle k at the points G and H (apart from F and E). Prove that AG = AH.[/color]

[i]Solution.[/i] Since the point E lies on the circle k with diameter AB, we have < AEB = 90°. On the other hand, the tangent to the circle k at the point B is perpendicular to the line AB (since the tangent to a circle at a point is perpendicular to the diameter of the circle through this point, and AB is the diameter of the circle k through the point B). Since the point C lies on this tangent, we thus have < ABC = 90°. Hence, < AEB = < ABC. Also, trivially, < EAB = < BAC. Thus, the triangles AEB and ABC are similar, so that AE : AB = AB : AC. Hence, $AE\cdot AC=AB^2$. Similarly, $AF\cdot AD=AB^2$. Thus, $AE\cdot AC=AF\cdot AD$, so that, by the converse of the intersecting chords theorem, the points E, C, F and D lie on one circle. Thus, < CFD = < CED, so that < AFG = 180° - < CFD = 180° - < CED = < AEH. Now, consider the chords AG and AH of the circle k; their chordal angles < AFG and < AEH are equal (< AFG = < AEH), and thus they must be equal themselves, i. e. we have AG = AH. Problem solved.

  darij


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Suppose $A_1,A_2,\dots ,A_n\ (n\geq 3)$ are $n\times n$ matrices which $\det A_1=\det A_2=\dots =\det A_n=1$ and $A_1A_2\dots A_n=I$. Suppose $\forall \ 2\leq k\leq n$, $A_1-A_k$ are different nonzero scalar matrices. Prove that $A_1+(-1)^nA_n$ is also scalar matrix.

($X$ is scalar matrix iff there exist $c\in \mathbb F$ which $X=cI$)
[hide="Hint"]
You must use Cayley-Hamilton.
[/hide]

I imagine that $A_{i}$ are all $n$ x $n$ matricies?

[quote="Omid Hatami"] scalar matrices. [/quote]

what is scalar matrix ?

I believe he meant scalar multiples of idendity  :?

Well, I forgot some parts of question. Now everything is clear.

Hey Omid,
     I might have royally messed up, but it would seems to me that I have a counter example   :?

Suppose we are working in $\mathbb C$, and we take $n = 4$. And define $A_k$ for $k = 1,2,3,4$ as follows,

\[ A_1 = \left( \begin{array}{*{20}c} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right),  \]
and $A_2 = A_1 + (i - 1)I_4$ , $A_3 = A_1 - 2I_4$, $A_4 = A_1 + (-i - 1)I_4$, where $I_4$ is the $4$ x $4$ identity matrix, and $i = sqrt(-1)$. It can be checked that $det(A_i) = 1$, and that $A_1 - A_2, A_1 - A_3,$ and $A_1 - A_4$ are all distinct, non-zero scalar matricies. Since $A_1$  isn't scalar, it follows that $A_1 + A_4$ isn't scalar.

Did I mess up? (I very much could have)

The problem is certainly correct. You must  check your counterexample. :)  :)  Also I could help you with the solution. Consider the equation $\det(A_1-cI)=1$

Edit:

 Nonsense !!!  :(  :D  Thanks jmerry

The flaw: $p$ is not the characteristic polynomial of $A$; it's shifted by 1. This is fatal.

[quote="Ramanujan"][quote="Soarer"]To be frank, I think his counterexample works.
In general, write A_1 in his jordan form, and set A_2,...,A_n to be matrices, shifting only diagonal entries. If A_1 is not diagonalizable, then A_1+A_n is definitely not a scalar matrix.  :?[/quote]

The matrices sould satisfy the condition

[quote="Omid Hatami"] $A_1-A_k$ are different nonzero scalar matrices[/quote][/quote]

What's the problem with realzeta's example then.  :?

I checked my solution, and I did not catch any error.

Excuse me  :oops:  :oops: I added something to the problem. I think now it's correct.

I think there is a problem again . I get $A_1A_2\dots A_n=-I$

Τhe polynomial $p(x)=\det(A_1-xI)-1$ has $n-1$ distinct solutions $a_{k}$ where $A_{1}-A_{k}=a_{k}I$ and $\deg p =n$.
Therefore $p(x) =(x-c)(x-a_{2})(x-a_{3})...(x-a_{n}) = \det(A_1-xI)-1$. But $p(0) = (-1)^n c \cdot a_{2}\cdot a_{3}...a_{n} = \det(A_1)-1 =0 \Rightarrow c=0$
hence $p(x) =x(x-a_{2})(x-a_{3})...(x-a_{n}) = \det(A_1-xI)-1$. Then $p(A_{1})=A_1A_2\dots A_n=-I$

Am i wrong again ???  :?

Well, consider the equation $\det(A_1-cI)=1$ then you could see easily that it has $n$ solutions in the field $\mathbb F$. If $c_1,\dots,c_n$ are solutions then, you could see $(A_1-c_1I)(A_1-c_2I)\dots (A_1-c_nI)=p(A)$ which $p(x)=\det (xI-A_1)+(-1)^{n+1}$ and because of Cayley-Hamilton $p(A)=(-1)^{n+1}I$. So $A_1-c_nI=(-1){n+1}A_n$. Therefore $A_1+(-1)^nA_n=c_nI$ :)  :)

what's wrong with ramanujuan's argument?!


A particle moves through the first quadrant as follows: during the first minute, it moves from the origin to (1, 0). It them moves up 1 unit, left 1 unit, up 1 unit, right 2 units, down 2 units, right 1 unit, up 3 units, left 3 units, up 1 unit, right 4 units, down 4 units, right 1 unit, up 5, and so on as suggested by the diagram, at a rate of 1 unit per minute. At which point will the particle be after exactly 1989 minutes?

$ \textbf{(A) } (35,44) \qquad \textbf{(B) } (36,45) \qquad \textbf{(C) } (37,45) \qquad \textbf{(D) } (44,35) \qquad \textbf{(E) } (45,36)$

[hide="Solution"]At $ 2^2\minus{}1$ minutes, it's at $ (0,1)$.

At $ 3^2\minus{}1$ minutes, it's at $ (2,0)$.

At $ 4^2\minus{}1$ minutes, it's at $ (0,3)$.

And so on...

So at $ 44^2\minus{}1\equal{}1935$ minutes, it's at $ (0,43)$ and has $ 54$ minutes left. One minute is spent moving to $ (0,44)$. Now it has to move $ 44$ spaces to the right until it reaches $ (44,44)$ with $ 9$ minutes to spare. Then it moves down for $ 9$ minutes to $ \boxed{\textbf{(D)}\ (44,35)}$[/hide]

That would have been a LOT easier if the diagram the problem alluded to were attached...
But for anyone trying to visualize it, it's a sort of square spiral pattern that spreads out in the first quadrant (plot a few points and it's clear).

[quote="grn_trtle"]That would have been a LOT easier if the diagram the problem alluded to were attached...[/quote]

The source from which this problem came from didn't have an asy pic attached, sorry. :(


Predict the most likely mode of radioactivity for sulfur-35, sodium-22, and polonium-210.

Ok, I actually figured these out.

The expected mode of decay for sulfur-35 is beta decay because it has too many neutrons for stability (high N/Z ratio).

Sodium-22 is expected to undergo electron capture because it lies below the band of stability (too few neutrons).

And polonium-210 is expected to undergo alpha decay. All nuclides with Z>83 are unstable and generally undergo alpha decay. Also, the polonium-210 would be transformed into lead-206, which has N=82. This is a [i]magic number[/i].


what TI calcs do you guys have?
i have a TI84 (so does drunken math)

I use an 83+ and a 89... like on tests, i use both calcs at the same time  :D

well, i own a 83+ but i've traded it with someone for an 89 just for this year. i also have a school issued 85 for ap calc.

89..kicks 84, 83 because it has a home screen and pretty font!  :D  :wow:

... and symbolic calculations.  I have an 89 (titanium) and an 86, but I lent my 86 to a friend for this school year.

Why aren't the 92(+) and Voyage 200 listed?

I have a 81 that I stole from OSU (I don't think they'll want it back though  :D ) and an 83+.

You left out the TI-82!!!!! That's discrimination!!! :furious:

True, I have a TI-84+ Silver Edition, but I persuaded my parents to give me a Ti-89 Titanium next year when I take Pre-Calc.

Is TI-89 useful for calculus?

i <3 my TI-83 silver. it got me through Algebra II, trigonometry, AP Calculus BC, AP Chemistry, and is currently working it's way through discrete math, AP Physics C, and Statistics.
There will be no other calculator for me. I love my TI-83.
Ever since my friend's Casio died on the SAT,
There will be no other calculator for me.
Nothing more will you ever need
For Calculus, even if it's AP
But to be an engineer, you may see
That you will need more than your trusty 83.

 :blush:

I have an 83+, but after getting my sleek looking 89 titanium, I use it whenever possible. The only thing that I've got to resort to my 83+ for is stats and tables and stuff.. still don't get how to work those, and my user manual was a disc and it doesn't quite work.

[quote="aznmonkey1992"]*sigh*
well, a TI 89 is good AP calculus, but as education.ti.org puts it, the voyage 200 is a "must-have" for calc[/quote]

There is no difference between the 89 and the v200, except in shape.  One is as good as the other for calculus, or any subject for that matter.  In addition, the 89 (Ti) is allowed on standardized tests, while the v200, with its QWERTY keyboard, is not.

The Ti-89 owns all other non-QWERTY calculators. :P How good is the 89 Titanium compared to the regular 89?

BTW, If you're going to get a QWERTY-calculator, get a Ti-92+ -- it's a lot better than the Voyage 200.

compared to the 89, the titanium has lots more memory..lots more. it also comes with lots of cool apps, like eepro for engineering, etc.etc.

[quote="mathnerd314"]I'm stuck with a TI-82... *sigh*

I WANT AN 89
I WANT AN 89
I WANT AN 89.......[/quote]

i got an 89 titanium

by the way, thanks for the cubic equation program mathnerd

I have an OLD 82 - from Christmas '02 - they've updated the style since; and I have a 84+Silver I got for being the top 8th grader at a university math meet last year and a 84+Silver I got for going to National Mathcounts. :) I use all of them on contests. The 82 is faster than the 84s.

I have an 82 that might not have any batteries at the moment, come to think of it. My main calculator is an 86 that got me from precal all the way to ... now, pretty much. I occasionally use my 89 but distance myself as much as possible as it is, after all, a representative of that satanic organization known as mathcounts.

[quote="solafidefarms"]The 82 is faster than the 84s.[/quote]

[i]Real[/i]ly? How so?

Why is there no Voyage 200 option on this poll? I only take standardized tests maybe three times a year, and my teachers let me use it, so I do for pretty much everything. It does have a bigger screen than the 89...but I put 83 because I still have a couple of those.

About the 82 being faster than an 84

Different operating systems run at different speeds, the 82 might have a good OS but the 84 does not.  For example, I have a 84+ SE, but I wrote a test program, and an 83+ solved it faster.  The reason, I was low on RAM.  Other than available RAM (If you have lots more programs installed like the 84+ SE allows, you won't have as much RAM), or different OS's, I don't know.

Test program:
:For(I,1,500)
:End
:Disp I
:Pause

Just see which displays 500 fastest.

When you use

:For(I,1,500)
:Disp I
:End

you are mainly seeing which can generate output fastest.  Different calculators might be able to generate output faster, but are slower on computation.  I guess that shows that 82s are better for generating output.

If you want, you could try some problem that would normally require the Chinese Remainder Theorem, but make them guess and check instead.  Then you could find out if everything is faster, or just output is faster.  I would be interested to know what the outcome is.

If you don't know what the Chinese Remainder Theorem is, it is when you have a number that leaves a remainder of 5 when divided by 7, 2 when divided by 6,  and a remainder of 11 when divided by 13.  NOTE:  This is just an example.

I don't own a calculator. It's a bit of a problem when it comes to tests in my AP Chem class.

[quote="solafidefarms"]I ran this program on both my TI-82 with 10 BASIC programs on it and my out-of-the-box TI-84+SE with no BASIC programs and the normal array of apps.

For(I,1,500)
Disp I
End

The TI-82 did it in 28 seconds; the 84+SE in 34.

My question is when will we get Linux calculators?[/quote]
[url=http://focus.ti.com/general/docs/wtbu/wtbusplashcontent.tsp?templateId=6123&path=templatedata/cm/splash/data/linux_com_portal]Everything runs linux someday.[/url]

Google finds everything.


calculus
function
Princeton
college
geometry
linear algebra
complex analysis

I know a lot of you are taking math past single-variable calculus - are you able to skip college classes with them or do you have to retake a lot of them in college? For example, if you've been able to take diffy equations at a community college. A lot of you seem to be taking math far beyond single-variable calculus but is that really a good idea to skip college math courses because you've done so? Colleges will probably cover the same topics in much more depth although acceleration is probably the only choice for high school students because no pre-calculus classes cover the material in depth.

I will be heading to the University of Washington next year and will have taken the BC Calculus exam (with some selfstudy), although I have taken the equivalent of AP Calculus AB. However, it is suggested to me that I start with the first year Honors Calculus sequence.

Depth and plug-and-chug calculus courses are mutually exclusive.  If they recommend that math majors take an analysis (or analysis-like sequence), don't skip it.  Otherwise, do skip it.

It's okay to skip lots of college classes. I did plenty of that in my first year. During that first year, I had some trouble with the material, but now (in my second year) I find it not too challenging, and I managed to pick up most of the intermediate material on the way.

What college courses did you skip, ComplexZeta?

I know that for some places, like University of Michigan, where honors classes are definetely worth taking for single variable analysis, so it depends.....

[quote]Every class between calculus and grad classes except complex analysis. Admittedly, I had read a fair amount about the stuff I skipped, but normally I would have had to take those classes (and I would have learned a bunch in doing so).[/quote]

Wow, I see. Did your college even have Honors courses?

I'm thinking about not even skipping single-variable analysis at the local university because I've heard that the Honors courses were pretty rigorous and proof-based. Also that most who are taking the course already took Calc in HS.

What are the courses called?  I tend to think that there's some benefit to taking an honors linear algebra course depending on how much background you have, but 'honors' single variable calculus courses will probably be more of the same with a few epsilons thrown in.  If you're comfortable with single variable calculus, or even multivariable calculus, its not a bad idea at all to skip straight to analysis courses.

[quote="ComplexZeta"]Every class between calculus and grad classes except complex analysis. Admittedly, I had read a fair amount about the stuff I skipped, but normally I would have had to take those classes (and I would have learned a bunch in doing so).[/quote]

geez.... so you basically skipped your undergraduate math education, and taught yourself through books? i imagine that had to take alot of time, and quite a bit of mathematical maturity.

To some extent I knew the undergraduate math curriculum before I went to college. I had read something on nearly every topic, but I hadn't learned the material nearly thoroughly enough. As a result, I had to learn the undergraduate material while learning the graduate material, so it was a somewhat difficult task. But it can be done with a bit of hard work. For me it was definitely a good solution since being bored in classes is terrible for me.

You have to be careful about that though, because what worked for ComplexZ at his school may not work best for someone else at some other institution.

In general, I think it's a bad idea to skip introductory undergraduate classes. Yes, it is true that if you know 40% or even 60% of the material for an introductory class you may be bored by sitting through it again. But if you have a good enough teacher, you may find some new points that you didn't see before, or you may even be able to extend the results or see more deeply into them the second time around.

Also you may think you know the material for a certain course - but you may not know what you don't know (sic). Prerequisites are there for a reason. In most cases they come from years of experiences of students skipping them and then being insufficiently prepared for later material.

One good way to see if you're prepared to skip a class is to view some of the past final exams for the course, or if the department has some kind of general exams, to view past questions from that subject area. If you can answer a fairly significant number of the questions then you should skip the class.

Keep in mind though that math is not a race to see who can get to the most advanced material the fastest. You should go at a pace that works best for you and your overall understanding.

By the way, past senior comprehensive exams in Algebra, Complex Analysis, and Real Analysis for Princeton's Math Department can be found here: http://www.math.princeton.edu/undergraduate/seniorinfo.html

On a dissenting note, I can say that I regret not skipping more courses than I did.  Because I didn't feel like fighting the math department for credit/permission before I had ever taken a single credit hour at my university, I ended up taking some courses which were tremendously dull, and missed out on some opportunities to take some interesting courses that, because of scheduling constraints, won't be offered for another two years (Galois Theory, Calculus of Variations, etc,).  The courses I enjoyed most were generally the ones I took without prereqs.

The only course I think that is 'unskippable', unless you've done a significant amount of background work at a significant level, is real analysis.  For example, I knew a lot of 'stuff' by the time I entered university.  I knew what, for example, a group was.  I knew what a generating function was.  I knew what a linear transformation was.  More significantly, I could do some relatively non-trivial things with the aforementioned 'objects'.  Conversely, I didn't know what a compact set was, I didn't know what a connected set was, I didn't know what a uniformly continuous function was, and I certainly had no clue about Riemann-Stieltjes integrals.  Topics from analysis tend to be introduced only in analysis, and you could miss out altogether if you skip such a course.

As an addendum to Gauss's post, those Princeton exams (I believe) represent the expected learning outcomes for undergraduates at the time that they graduate (that's worded awkwardly, but it should suffice).  In particular, don't use it to decide whether you can skip courses in real analysis, complex analysis, or analysis, because few, if any high school students, know that much math (well, actually the complex analysis exams were pretty easy).

If you've had the class, don't take it again. You can probably show your college some information about what you took (definitely include the relevant catalog passages) and convince them it's equivalent. I got out of the core real analysis requirement at Berkeley this way, by showing them that I had already taken an essentially equivalent analysis course, and then doing more analysis after that.
Of course, I recklessly skipped prerequisites while I was in high school, going from geometry one year to calculus the next, and then taking college courses without their direct prerequisites- that analysis class was actually the second semester of a sequence. I essentially stopped that in college- I was two years ahead anyway, and I didn't feel like rushing through everything. There was still time to take the graduate algebra and analysis, and to get myself breadth.

The one class I should have skipped and didn't was the introductory linear algebra class- I should have started with the second linear algebra class.

Calculus classes are essentially equivalent everywhere, and that makes AP calculus probably the most useful AP. If you have a 4 or 5 on it, don't take calculus again. Multivariable calculus and linear algebra are also pretty universal. After that, the curriculum is too fragmented to worry about matching credit.

many of the top notch colleges don't accept 4 on most (or any) of the AP exams in order to get credit.


geometry
geometric transformation
number theory unsolved
number theory

Find all finite arithmetic sequences $ a_1,a_2,..a_n$ with all elements of form $ \frac{1}{x}$ where $ x$ is a positive integer, such that $ a_1 \plus{} a_2... \plus{} a_n \equal{} 1$

Thanks in advance

Is this problem from Tournament of the towns 2007?

Yes. unleast that's how i understood our translation which was terrible :lol:

For any n we have solution $ a_i\equal{}\frac 1n,i\equal{}1,2,...,n$ with $ h\equal{}0$.
If $ h<0$, then for sequenct $ b_1\equal{}a_n,b_2\equal{}a_{n\minus{}1},...,b_n\equal{}a_1$ we get sequence with $ h>0$. Therefore we can consider only case, suth that $ h>0$. Then $ a_n>\frac 1n$ and $ a_1\plus{}a_n\equal{}\frac 2n .$
Because $ a_k\equal{}\frac{1}{x_k}$ and $ \frac 1n \le a_{[(n\plus{}1)/2]}<...<a_n\le \frac{2}{n\plus{}1}$ we get, that
if n is odd $ a_{(n\plus{}1)/2}\equal{}\frac 1n, a_{(n\plus{}1)/2 \plus{}k}\equal{}\frac{1}{(n\plus{}1)/2 \minus{}k},k\equal{}1,2,...(n\minus{}1)/2$.
From $ a_n\minus{}a_{n\minus{}1}\equal{}a_{n\minus{}1}\minus{}a_{n\minus{}2}$ we get $ n\equal{}3,a_1\equal{}\frac 16, a_2\equal{}\frac 13, a_3\equal{}\frac 12$.
If n=2k is even $ a_{2k}\le \frac{1}{k\plus{}1}\Longrightarrow a_{2k\minus{}l}\equal{}\frac{1}{k\plus{}1\plus{}l},l\equal{}1,2,...k\minus{}1$. It give $ a_{k\plus{}1}\equal{}\frac {1}{2k}$ contadition with $ a_{k\plus{}1}>\frac{1}{2k}$ ($ h>0$).


It is similar to my recent one. In my book, this problem is already proposed and proved in a stronger appearance (also, there are a lot of similar problem).

http://www.mathlinks.ro/Forum/viewtopic.php?t=148360

If $a,b,c,d$ are non-negative with sum $4$ then
\[a^{2}(b+c)+b^{2}(c+d)+c^{2}(d+a)+d^{2}(a+b)+4abcd \le 16. \]
This is a super nice and hard inequality.

Alternatively, following very nice one are in my book, too

If $a,b,c,d$ are non-negative with sum $4$ then
\[ab(a+b+c)+bc(b+c+d)+cd(c+d+a)+da(d+a+b) \le 8. \]

[quote="hungkhtn"]If $ a,b,c,d$ are non-negative with sum $ 4$ then

$ a^{2}(b \plus{} c) \plus{} b^{2}(c \plus{} d) \plus{} c^{2}(d \plus{} a) \plus{} d^{2}(a \plus{} b) \plus{} 4abcd \le 16$.[/quote]
Actually, the following stronger inequality holds :

$ a^{2}(2 \minus{} a) \plus{} b^{2}(2 \minus{} b) \plus{} c^{2}(2 \minus{} c) \plus{} d^{2}(2 \minus{} d) \plus{} 8abcd\leq 2$

for $ a,b,c,d\geq0$ and $ a \plus{} b \plus{} c \plus{} d \equal{} 2$.

[hide="Proof."]$ \frac {(a \plus{} b \plus{} c \plus{} d)^4}{16}$

$ \minus{} \frac {a \plus{} b \plus{} c \plus{} d}{4}[a^2(b \plus{} c \plus{} d) \plus{} b^2(c \plus{} d \plus{} a) \plus{} c^2(d \plus{} a \plus{} b) \plus{} d^2(a \plus{} b \plus{} c)] \minus{} 4abcd$

$ \equiv F(a,b,c,d) \equal{} F(a,a \plus{} 2x,a \plus{} 2x \plus{} y,a \plus{} 2x \plus{} y \plus{} 2z)$

$ \equal{} (3x^2 \plus{} y^2 \plus{} 3z^2 \plus{} 2yz \plus{} 2zx \plus{} 2xy)a^2$

$ \plus{} (10x^3 \plus{} y^3 \plus{} 2z^3 \plus{} 10x^2y \plus{} 10x^2z \plus{} 3y^2z \plus{} 5y^2x \plus{} 10z^2x \plus{} 4z^2y \plus{} 10xyz)a$

$ \plus{} (3x \plus{} y \plus{} z)(3x^3 \plus{} z^3 \plus{} 3x^2y \plus{} 3x^2z \plus{} y^2x \plus{} z^2x \plus{} z^2y \plus{} 2xyz)\geq 0$,

which is clearly true for $ a\leq b\leq c \leq d$.[/hide]


geometry
parallelogram
trigonometry
cyclic quadrilateral
geometry unsolved

1.Let ABC be an isosceles triangle with AB=AC . Point D and P lie on  the sides of AB and AC

,respectively . The line passing through B and parallel to AC meets the line DE at F . The line

passing through C and parallel to AB meets the line DE at G . Prove that  Area[DBCG]/Area[FBCE]=AD/AE

2. Let ABCD be a cyclic quadrilateral and let P and Q be points on the sides AB and AD respectively such

that AP=CD and AQ=BC let M be the point of intersection of and PQ . Show that M is the midpoint of PQ

Thankyou

1.
Let $ h_b, \ h_c$ be the heights from B and C in the triangle ABC. Then $ h_b\equal{}h_c$

$ Area[DBCG]\equal{}\frac{(DB\plus{}CG)h_c}{2}, \ Area[FBCE]\equal{}\frac{(BF\plus{}CE)h_c}{2}$

Then $ \frac{Area[DBCG]}{Area[FBCE]}\equal{}\frac{DB\plus{}CG}{BF\plus{}CE}$

$ \frac{DB}{AD}\equal{}\frac{BF}{AE}\Rightarrow DB\equal{}\frac{BF\cdot AD}{AE}$

$ \frac{CG}{AD}\equal{}\frac{CE}{AE}\Rightarrow CG\equal{}\frac{CE\cdot AD}{AE}$

Then $ DB\plus{}CG\equal{}\frac{AD}{AE}(BF\plus{}CE)\Rightarrow\frac{DB\plus{}CG}{BF\plus{}CE}\equal{}\frac{AD}{AE}$

[quote="poonprang"]

2. Let ABCD be a cyclic quadrilateral and let P and Q be points on the sides AB and AD respectively such

that AP=CD and AQ=BC let M be the point of intersection of and PQ . Show that M is the midpoint of PQ

Thankyou[/quote]

[color=darkblue]You see again your post at "let M be the point of intersection of and PQ". You tell me M be the point of intersection of [size=200]?[/size] and PQ.[/color]

Sorry Let M be the point of intersection of AC and PQ Show that M is the midpoint of PQ

2. Let $X$ on $AB$ and $Y$ on $AD$ such that $CX\parallel DA$ and $CY\parallel BA$. Then $AXCY$ is a parallelogram, so $CA$ bisects $XY$. Since $\frac{CX}{CB}=\frac{\sin\angle XBC}{\sin\angle BXC}=\frac{\sin\angle YDC}{\sin\angle DYC}=\frac{CY}{CD}$. Thus $\frac{AP}{AQ}=\frac{CD}{CB}=\frac{AX}{AY}$, so $XY\parallel PQ$ so $AC$ also bisects $PQ$.


algebra
polynomial
function
calculus
derivative
complex analysis
domain

Prove that: if $f\in C^\infty(\mathbb{R})$, st for all $x$, $f^{(n)}(x)=0$ for some $n$, then $f$ must be a polynomial.

That's a very beautiful and hard problem. I saw a long solution using Baire's theorem, but there is a great solution using an interesting lemma in a french book for preparation for agregation. I even posted this lemma as proposed problem, but nobody answered. Too bad! I will remind the lemma:
    Let $f$ be a $C^{\infty}$ function defined on an interval $I$, which is not polynomial. Then there is a compact sub-interval of $I$, call it $J$ such that $f$ is not polynomial on $J$ and has no zeroes there.
    The proof is completely elementary and very nice. Next, it suffices to apply this lemma several times for different derivatives of the function and then to use the method of closed containing intervals.

Well I did a copuple of days ago and as u say using baire's theorem, well thats what i did:

Since for each closed interval $I$, \[\displaystyle\bigcup_{n\in\mathbb{N}}(Z(f^{n})\cap I)=I\]
using baire's theorem we can coclude that  $P=\{x\in\mathbb{R}/f$ is a polinomial in a neighborhood of $x\}$ is dense and clearly open.
Now let be $A=\{x\in\mathbb{R}/f $ is analytic in a neighborhood of $x\}$  , using the density of $P$ is easy to conclude that $P=A$.

Well if $P\neq \mathbb{R}$ lets take a conex component of $P$, $(a,b)$. Then $b\notin P$  so $f$ cannot be analytic at $b$ so
\[\limsup\sqrt[n]{\left|\frac{f^{(n)}(b)}{n!}\right| }=\infty\]
(just the convergence radio formula). So there is a $N$ st for $n\geq N$,  $|f^{(n)}(b)|>n!$ but since $f$ is a polynomial in $(a,b)$ for $n$ enoughtly big $f^{(n)}(b)=0$ which is a contradiction  :) .

i was wondering if this fact depends on baire's theorem or in other words  if it  would be possible to prove it without using it..... in the complex analysis forum i posted an holomorphic form of this fact hoping(wich is quite easier) that somebody  could prove it without using baire's theorem.... well it's just curious of myself :P.

PD. I'll try your lemma............

So what if $P$ is the whole of reals? Though I guess it is straightforward but I still cannot see how one concludes there is one polynomial for the whole domain, not many... Thanks!

Taylor's theorem: $f(x)=\sum_{j=0}^{n}\frac{f^{(j)}(0)}{j!}x^{j}+\frac{f^{(n+1)}(t)}{(n+1)!}x^{n+1}$, where $t$ is some number between zero and $x$. If $f^{(n+1)}$ is identically zero, this error term is identically zero and $f$ is equal everywhere to its $n$th Taylor expansion at zero.

[quote] If $f^{(n+1)}$ is identically zero ... $f$ is equal everywhere to its $n$th Taylor expansion at zero.[/quote]

Surely the intention was that $n$ depends on $x$.  For fixed $n$, that is, to show that functions whose derivative of order $n+1$ vanishes are polynomials, one needs only the case $n=0$ (constant functions) and induction.

Hi,I am sorry I want to ask two stupid questions on $hcast's$ proof?
1.what is defination of $f$ is analytic(I just know what is analytic if $f$ is defined on complex numbers,not on real numbers)  and why $P=A$?
2.If we proved $A=R$,then how we conclude that $f$ is a polynomial?

Thanks for answer :)

[quote="zhaobin"]Hi,I am sorry I want to ask two stupid questions on $hcast's$ proof?
1.what is defination of $f$ is analytic(I just know what is analytic if $f$ is defined on complex numbers,not on real numbers)  and why $P=A$?
[/quote]
The function $f$ is analytic at point $x$ if its Taylor series converges to $f$ in some neighborhood of $x$. An equivalent definition is, the function is analytic at point $x$ if it can be extended to a complex analytic function in some (complex) neighborhood of $x$.

[quote]
2.If we proved $A=R$,then how we conclude that $f$ is a polynomial?
Thanks for answer :)[/quote]
If two analytic functions are equal on an open set then they are equal everywhere (we assume that they are defined on the same connected set). Since $f$ is equal to a polynomial on an open set, it is equal to this polynomial everywhere.

Thanks very much.
But can you tell me how to show $P=A$?

[quote="zhaobin"]But can you tell me how to show $P=A$?[/quote]
First of all $P\subset A$, since every polynomial is analytic.

Now consider a connected component $A_{0}$ of $A$. Since the set $P$ is everywhere dense, it has a non empty intersection with $A_{0}$. Since the set $P$ is open, this intersection is open. Hence $f(x)$ is equal to a polynomial on the inersection, and therefore, it is equal to this polynomial on $A_{0}$. This means $A_{0}\subset P$.

[quote="Yury"]First of all $P\subset A$, since every polynomial is analytic.

Now consider a connected component $A_{0}$ of $A$. Since the set $P$ is everywhere dense, it has a non empty intersection with $A_{0}$. Since the set $P$ is open, this intersection is open. Hence $f(x)$ is equal to a polynomial on the inersection, and therefore, it is equal to this polynomial on $A_{0}$. This means $A_{0}\subset P$.[/quote]
Thanks agian. :)

I was looking at hcast's proof, and it seems to me that it needs some repair. How can we conclude from the non-analiticity at b that the Taylor series expansion at that point has zero convergence radio? Consider the following function:

f(x)=e^(-1/x^2) unless x=0, f(0)=0

That function has derivatives of all orders, and they all equal zero at zero. Hence, the Taylor series at zero converges to zero everywhere, unlike f. f is therefore non-analytic at zero. Indeed, we could replace the left part of the function with the zero function, and then suinto the whole domain a polynomial function, obtaining a smooth (C inf) function, that would serve as a counterexample to the proof. Of course, it is not a counter example of the theorem! the function obtained in this way should have real, positive values of x for which no derivative vanishes. Well, maybe there's something I'm not seeing, but I'll post this.

I got the function at Wikipedia. Search for non-analytic smooth function.

marcosaedro

it's jmerry proof complete??? i think yes.. it's very simple!

I see there's no solution so far. I'd be happy to see the solution in the end.. can anyone help?

Yet another try to ask for help..

My own attempt: let $ deg(x) = min \{n \in \mathbb{Z}| f^{(n)}(x) = 0\}$. Then the problem reduces to showing that the following cannot occur:

$ f$, as in the problem statement, is a polynomial on $ [a,b]$ and for any $ \epsilon > 0, max\{ deg(x) | x \in [a, b+\epsilon]\}= \infty$


geometry
parallelogram
incenter
perpendicular bisector
geometry proposed

Consider a circle $S$, and a point $P$ outside it. The tangent lines from $P$ meet $S$ at $A$ and $B$, respectively. Let $M$ be the midpoint of $AB$. The perpendicular bisector of $AM$ meets $S$ in a point $C$ lying inside the triangle $ABP$. $AC$ intersects $PM$ at $G$, and $PM$ meets $S$ in a point $D$ lying outside the triangle $ABP$. If $BD$ is parallel to $AC$, show that $G$ is the centroid of the triangle $ABP$.

[i]Arnoldo Aguilar (El Salvador)[/i]

Be $Q$ the midpoint of $AM,$ $T$ the intersection between the perpendicular bisector of $AM$ and $AP,$ and $R$ the foot of the perpendicular drawn from $B$ to $AG.$ $QT\parallel PM$ and $Q$ is midpoint of $AM$ $\to$ $T$ is midpoint of $AP.$ and $C$ is midpoint of $AG$ $\to$ $AC=CG$ $\to$ $MC\parallel BG$ $\to$ $BG=AD$ (because $AM=MB$) $\to$ $ADBG$ is a parallelogram and $AG=GB=BD=DA$ $\to$ $\angle{ADP}=\angle{PDB}=\angle{BGD}=\angle{DGA}=\alpha$ and $\angle{BAG}=\angle{BAD}=\angle{ABG}=\angle{ABD}=90-\alpha$ as $AP$ and $PB$ are tangents to $S$ $\to$ $\angle{BAP}=\angle{ABP}=2\alpha.$ $\angle{ABR}=90-\angle{BAR}=90-(90-\alpha)=\alpha$ Be $X$ the intersection point of $PM$ and $S$ $\angle{ABX}=\angle{ADX}=\alpha$ then $R,$ $X,$ and $B$ are colineal. $AMXR$ is cyclic, so $\angle{AMR}=\angle{AXR}=\angle{ADB}=2\alpha$ be $T1$ intersection between $MR$ and $AP$ $\to$ $AMT1$ is isoceles $\to$ $T1$ lies on the perpendicular bisector of $AM$ $\to$ $T1=T.$ Also, $\angle{MRA}=\angle{MXA}=90-\angle{MAX}=90-\alpha$ $\to$ $MR=AM.$ $QCRB$ is cyclic $\to$ $\angle{BQR}=\angle{BCR}=2\alpha$ $\to$ $QR\parallel AT$ ($\angle{BQR}=\angle{BAT}=2\alpha$) $\to$ as $QM=QA,$ $MR=RT$ $\to$ $AT=AB.$ Be $T'$ the point where $BG$ touches $AP$ by a little angle chasing we get: $\angle{ABT'}=\angle{AT'B}=90-\alpha$ $\to$ $T'=T.$ Then $G$ is the intersection of $PM$ and $BT$ ($M$ and $T$ being midpoints of $AB$ and $AP$ respectively) so it's the gravicenter.

[quote="Aldo Pacchiano"]Be $ T'$ the point where $ BG$ touches $ AP$ by a little angle chasing we get: $ \angle{ABT'} \equal{} \angle{AT'B} \equal{} 90 \minus{} \alpha$ $ \to$ $ T' \equal{} T.$ [/quote]

I can't see how this implies $ T'\equal{}T$ (we still don't know $ \angle{ATB}$, do we?)

Let $ I$ be the second intersection of $ PD$ with $ \mathcal{S}.$ Since $ \angle PBI \equal{} \angle IDB \equal{} \angle IBM,$ it follows that $ I$ is the incenter of $ \triangle PAB$ $ \Longrightarrow$ $ BD$ is the external bisector of $ \angle PBA$ $\Longrightarrow$ Parallel $ AC$ to $ BD$ cuts $ PB$ at $ N$ such that $ \triangle BNA$ is isosceles with apex $ B.$ Let $ \angle ABD \equal{} \theta.$ Since $ BNCM$ and $ ABIC$ are both cyclic quadrilaterals, we have $ \angle ACB \equal{} \angle AIB \equal{} 2\theta$ and

$ \angle BCM \equal{} \angle BNM \equal{} \angle ACB \minus{} \angle ACM \equal{} 2\theta \minus{} (\pi \minus{} 2\theta) \equal{} 4\theta \minus{} \pi.$

On the other hand, if $ O$ is the center of $ \mathcal{S},$ we obtain

$ \angle PAB \equal{} \angle AOI \equal{}\pi \minus{} 2\theta \Longrightarrow  \angle APB \equal{} \pi \minus{} 2(\pi \minus{} 2\theta) \equal{} 4\theta \minus{} \pi \equal{} \angle BNM$

Thus, $ NM \parallel PA$ $\Longrightarrow$ $ N$ is the midpoint of segment $ PB,$ which implies that $ G \equiv PM \cap AC$ is the centroid of $ \triangle PAB.$


inequalities
vector
inequalities unsolved

Prove that for every Euclidian vector space we have :
1) $|a| + |b| + |c| + |a+b+c| \geq |a+b| + |b+c| + |c+a|$ for all vectors $a,b,c$ in the space.
2) $|a|\cdot |b-c| + |b|\cdot |a-c| \geq |c|\cdot |a-b|$ for all vectors $a,b,c$ in the space.

I think that 1) isn't trivial even for the case of the reals.
But 2) is easier for sure.

Do you know where can i found the proof of this ?

I can prove Hlawka for C(complex numbers) but I think it works in euclidian vector space.

Would you mind posting your proof, please ?

Second one is trivial. just use |x|+|y|>=|x+y|.

Can you please write some more details, i can't see how you use this inequality direct to prove it.
(but be carefull, because $|a|\cdot |b|$ does not equal $|a\cdot b|$ in general. )

I'm not considering the space to be C, but a randon euclidean vector space.

I don't think that this topic should be in the inequalities section, because i am not sure if the background needed for the proof of these inequalities is elementary. ( I seemed to me to be more "college- algebra" than that ,but i can also be mistaken.)

Here's my proof for  Hlawka inequality in $ \mathbb{C}$.

 (H): $ |a\plus{}b|\plus{}|b\plus{}c|\plus{}|c\plus{}a|\leq |a|\plus{}|b|\plus{}|c|\plus{}|a\plus{}b\plus{}c|.$

We have $ 2|a\plus{}b|\cdot |b\plus{}c| \equal{} 2|b(a\plus{}b\plus{}c)\plus{}ca|\leq 2|b|\cdot|a\plus{}b\plus{}c|\plus{}2|a|\cdot|c|$.
       We write the other two inequalities and by suming with the well known identity:

                $ |a\plus{}b|^{2}\plus{}|b\plus{}c|^{2}\plus{}|c\plus{}a|^{2}\equal{} |a|^{2}\plus{}|b|^{2}\plus{}|c|^{2}\plus{}|a\plus{}b\plus{}c|^{2}$ the conclusion follows
        imediately.

As a reference,
http://www.mathlinks.ro/Forum/viewtopic.php?t=16600


A triangle ABC is given such that <B>120 and on AC is chosen a point H such that AH=BH, BH is perpendicular to the line BC. Points D and E are the midpoints of sides AB and AC, respectively. The line passing through the point H is parrallel to the side AB and intersect with DE at F. Prove that <ACD=<BCF

$ \angle A\equal{}\angle ABH\equal{}\angle ABC\minus{}\angle HBC\equal{}30^o\Rightarrow \angle ACB\equal{}30^o$
So $ AE\equal{}EC\equal{}\frac{AC}{2}$, HE=HB/2=HA/2=AE/3=AC/6
$ EF\equal{}HE\sqrt{3}\equal{}\frac{AC\sqrt{3}}{6}$, $ ED\equal{}\frac{BC}{2}\equal{}\equal{}\frac{AC\sqrt{3}}{2}$
$ \Rightarrow EC^2\equal{}EF.ED\Rightarrow \triangle EFC\sim\triangle ECD\Rightarrow ECD\equal{}\angle EFC\equal{}\angle BCF$


ugly, but probably not too difficult. Try expanding the LHS.By the way maki so vikas za moeto resenie za romanskata od JBMO?

Yes, I guess it's weak, but for $a=b=c=d$ we have $LHS=22,3338\ldots$ so it's [i]almost[/i] the best constant.

So it still isn't solved...  :(    Makaveli, could you post your solution with the expanding of the LHS?


A long room has $ 100$ lightbulbs numbered $ 1$ to $ 100$. Outside is a queue of people, also numbered $ 1$ to $ 100$.

Person $ 1$ walks into the room and flicks the switches of all the lightbulbs which are multiples of $ 1$.

Person $ 2$ walks into the room and flicks the switches of all the lightbulbs which are multiples of $ 2$.

Person $ 3$ walks into the room and flicks the switches of all the lightbulbs which are multiples of $ 3$.

...   ...   ...   ...

Person $ 100$ walks into the room and flicks the switches of all the lightbulbs which are multiples of $ 100$.

Which lightbulbs are on after all $ 100$ people have walked throught the room and which ones are off?

Each lightbulb is on if and only if it has an odd number of divisors, which occurs if and only if it is a perfect square.


Try to solv :D
P1: Let x,y,z be nonnegative real numbers satisfy: x+y+z+xyz=4. Prove that x+y+z>= xy+yz+zx.

the solution can be found here:

http://www.mathlinks.ro/viewtopic.php?t=1027

cheers!:D


geometry
topology
calculus
probability
search
number theory
set theory

Do most mathematicians take math from all areas, or just one or two?

Because I've realized in the college I'm probably going to in two years, Michigan, will have math offered in all these pure areas:
Analysis
Algebra
Topology
Number Theory
Set Theory/ Logic, etc
Combinatorical Theory.

So I counted, there are 33 grad level courses in these areas. And yes, I want to take them all. :lol: . And I'll have Calc III, Linear Algebra, Differential Equations, and Real Analysis under my belt before entering college (2 a year), so I can take grad courses beginning of junior year. I can probably get through 8 out of 33 grad courses I want to take by undergrad, but then I realized it will take me 3 Ph.D's to get through the rest! I really am interested in number theory/ combinatorics , but I fear that in order to become a good mathematician, you need to take only all things.  Does some mathematicians only take a few courses outside their specializations?

You will get advice at your university about which courses to take to get a broad undergraduate education in math. Then you will be advised to go to a DIFFERENT grad school, possibly bringing along graduate-level credits from your undergraduate school, to start a Ph.D. course from the bottom, but with a good background. Read

[url=http://www.amazon.com/exec/obidos/tg/detail/-/082183455X/learninfreed]A Mathematician's Survival Guide: Graduate School and Early Career Development[/url]

by Steven G. Krantz for more information about how to plan undergraduate courses to lead on to further study.

So, those 33 courses aren't all offered all the time.  (Some of them are offered rarely.)  And whether you'll want to take some of them will depend on who is teaching them at the time you're ready to take them.

Probably the most important things to do when you get to UMich are (1) make sure to take the honors frosh sequence (IIRC, one version is like calculus, with proofs, on speed; the other is advanced combinatorics) and (2) repeat, whenever speaking to an advisor-type person, that you want to take the grad-school-track courses.  Otherwise they tend to forget that any undergrads want to do that (even those who take the honors courses!).  If you can get to know a permanent faculty member a bit, that's the ticket.  He (or she, there are a few permanent women in the department now) can help you navigate the bureaucracy and make sure you get into appropriate classes.

I'm been researching many, many colleges since summer started and I'm having difficulty in choosing colleges that I want to apply to. I've read many discussions on this website and I've learned alot in terms of what a "good" math department is. BUt i still have a bit more question.

I've been on many websites for colleges and the first thing that i look at are the courses that the math department offers. Whenever i see courses such as number thoery,set thoery, mathematical logic, etc I immediately put that college on my "list." But in truth I only look at these several courses becuase i really do not know what all the other courses are about. Courses such as analysis I and II, discrete mathematics, linear algebra, and much much much more. On the other I've been exposed to what set thoery, mathematical logic, number thoery are and when I got to experience these courses that are not taught at my High School...i immediately fell in love with it. um..It's amazing..and this is one of the major reasons why i only look at whether colleges have this courses. (which i found is not much help beucase many many colleges do have these courses).

so i was wondering whether someone can sort of explain what alot of these math courses are. Then from that maybe i can see which courses I may be interested in and then that may help me limit my college chooses a bit more?

 I want to major in pure mathematics in college for undergrad and then go into grad school to get my masters in education and credentials to teach math in the middle school and high school levels. (Stanford's STEP program caught my attention in terms of a good education program. One of the things that got my attention was the fact that i can get my masters and my credentials in 12 months) I think that as a future teacher I need and feel obligated to get a good math education for my students. And i just like learning math. So I know for sure that I'm not that much into applied mathematics but i also realized that to get a general mathematics major i have to be  exposed all kinds of mathematics course.

thanks alot...

[quote="SLeon"]. . . . I only look at these several courses becuase i really do not know what all the other courses are about. Courses such as analysis I and II, discrete mathematics, linear algebra, and much much much more. . . .

so i was wondering whether someone can sort of explain what alot of these math courses are. Then from that maybe i can see which courses I may be interested in and then that may help me limit my college chooses a bit more?[/quote]

The courses called analysis I and analysis II are courses in what is called "real analysis," the explanation of how calculus really works. You may also see undergraduate courses in "complex analysis," which is like real analysis except applied to complex numbers--which makes for very different results sometimes. Discrete mathematics means MOSTLY mathematics of integers as those appear in combinatorics and perhaps probability, as contrasted with the mathematics of real numbers (which is what one studies in analysis). Linear algebra is chock full of applications to applied subjects of all kinds and also has strong links to "pure" math subjects. The high-school-level topic that most linear algebra courses start out with is matrices and solving systems of linear equations. A lot of teacher training programs now emphasize at least one course in abstract algebra, which is a very standard undergraduate course for math majors.

[quote="SLeon"]I want to major in pure mathematics in college for undergrad and then go into grad school to get my masters in education and credentials to teach math in the middle school and high school levels. . . . I think that as a future teacher I need and feel obligated to get a good math education for my students. And i just like learning math. So I know for sure that I'm not that much into applied mathematics but i also realized that to get a general mathematics major i have to be  exposed all kinds of mathematics course.[/quote]

You have a very worthy goal, which I admire. One Web resource I recommend to you right now, even though you won't understand it all right away, is

[url=http://gallery.carnegiefoundation.org/cbennett/Assignments/BKPD.pdf]Advanced Mathematics for Secondary Teachers: A Capstone Experience[/url]

which is a pretty good overview of fundamental university-level mathematics for anyone. More generally, you can refer to a [url=http://www.google.com/search?q=Mathematics+Teachers+Capstone]Google search[/url] for similar materials and browse around to see what teachers are expected to know after earning a math major.

A very useful book to read is [url=http://www.amazon.com/exec/obidos/tg/detail/-/0192832026/learninfreed]From Here to Infinity[/url] by Ian Stewart, which you should be able to get by interlibrary loan.

You have my best wishes.