Does there exist an interval (a,b) with 0<a<b and a continuous function
f: (a,b)->(0,1) such that
(i) f is increasing on (a,b)
(ii) for all a<=x<y<=b, we have f(x)*f(y)*(f(y)-f(x))=f(y-x).

Hello test20,

I don't clearly understand :
f is defined in (a,b), not in [a,b]. So f(a) and f(b) are not defined. So why, in (ii), do you write a <= x ?
More curious : in (ii), if y is very near of x, you need  f(y-x) defined, and so f(x) must be defined near 0+ ==> a must be 0. But you wrote a > 0, and not a >= 0.

...
 :?:

--
Patrick


The incircle of a triangle $ABC$ has centre $I$ and touches sides $AB, BC, CA$ in points $C_1, A_1, B_1$ respectively. Denote by $L$ the foot of a bissector of angle $B$, and by $K$ the point of intersecting of lines  $B_1I$ and $A_1C_1$. Prove that $KL\parallel BB_1$.

[b]proposed by L. Emelyanov, S. Berlov[/b]

Let's start with the easy ones :).

Let $T=BL\cap A_1C_1$. $TB_1LK$ is cyclic, so $\frac{IK}{IL}=\frac{IT}{IB_1}\ (*)$. What we want to show is $\frac{IK}{IL}=\frac{IB_1}{IB}$. Combining this with $(*)$, we see that what we need is $IB_1^2=IT\cdot IB$. However, this is clear from the right triangle $IC_1B$. We have $IB_1^2=IC_1^2=IT\cdot IB$.

Excuse my lack of geometry knowledge  :blush:  but when a quadriteral is cyclic and what it gives [I don't understand $(*)$] ?

It's when the four points lie on the same circle. $(*)$ takes place because of the equality of angles of the triangles.

Thanks - drawing proper picture also helps.

Dear Mathlinkers,
and without any calculation? what do you think?
Sincerely
Jean-Louis

An outline:

(1) $ BK $ meets $ AC $ at its midpoint, $ M $.

(2) $ MI $ meets $ BB_1 $ at its midpoint, $ X $.

(3) The pencil $ (KX,KB_1,KL,KM) $ is harmonic. So, $ KL \parallel BB_1 $.

Dear,
yes, this is also my proof...
Sincerely
Jean-Louis


Hi,

I am not quite sure why and when to use expected values. I just finished that chapter from Intro. to Counting and Probability, but I am not confident about it.

I have a couple of questions.

1) The expected value for rolling a die is 3.5. We see that the expected value of the sum of two dice is = 7 which is also equal to 3.5+3.5 = E(one die) + E(one die). Why does this work? Why does the same work with products? I couldn't figure it out.

2) This is a question from the book:

A valuable treasure worth $ \$100,000$ fell out of my airplane and landed in a large swamp. If I don't find it within a day, it will sink to the bottom and never be found again. I can hire one or more helicopters to search the swamp. Each helicopter costs $ \$1000$ to hire for one day, and has a 90% chance of finding the treasure. How many helicopters should I hire?

If we hire one helicopter, we have a 0.9 probability of  recovering the treasure. So our expected gain from finding the treasure is 0.9(100,000) = 90,000 - 1,000 for the helicopter for a total of 89,000 net expected gain.

It doesn't make sense to me what they are doing. Why are they taking 90% off of the money in the treasure.

I am basically unsure when to use expected value and what to do with it, for example taking 90% off the money in the treasure, I don't think I know how to handle those sorts of indirect questions.

Any help would be greatly appreciated.

Thank you very much as usual ;)

Well the first one is simple really. We expect to roll 3.5 (assume 3.5 is a number that is possible to roll) on the first die. We expect to roll a 3.5 on the second also. So the sum is simply 3.5+3.5=7.

You made a Latex error on your second one. Please correct it.

The reason you take 90% of it is because you'll only find it 90% of the time.

For example, let's repeat this situation 10 times with 1 helicopter.  It is expected you will recover it 9 times out of 10, because each helicopter has a 90% chance of finding the treasure.

So, the expected value of your return if you hire one helicopter is $ \underbrace{0.9(100000)}_{\text{expected earning}} \minus{} \underbrace{1000}_{\text{cost}}\equal{}89000$.  Similarly, if you hire two helicopters, the expected value of your return is $ \underbrace{0.99(100000)}_{\text{expected earning}}\minus{}\underbrace{2000}_{\text{cost}}\equal{}99000\minus{}2000\equal{}97000$.

Thank you for the replies pytheagle and AIME15. But what does finding it 90% of the time have to do with the amount that is in it?

This does not have anything to do with the amount of money that is in the treasure. This means the [i]average[/i] amount that you would gain if you repeated this over and over again. Like AIME15 said, if you do the helicopter situation 10 times, you would be expected to get the treasure 9 times and not get it 1 time. The overall amount of treasure you would get is 9*100000 (the successful times) +1*0 (the failure to get the treasure) = 900000, but since you did it 10 times, you have to divide by 10 to get 90000.

Ahh...now that you put it that way aleph0. I didn't think about considering it in terms of that. Thank you, it is all clear to me now :)


I post this question in interme. forum but no one answered it. i was hoping if you can solve the question. (fill free to move the question)
What is the d/dx [log (x)]?

[tex]\displaystyle \frac{d}{dx}\big(\ln x\big)=\frac{1}{x}[/tex]. This is a very common calc formula.

Chris, where do all these problems come from? And why don't you look them up in any calculus book? If you are curious, that's fine; we can try to answer your questions. If you are asking homework questions, we really should not be answering them. There is no way I can tell what the source of your problems is, so I have to be a little bit suspicious.

d/dx (log_10(x) ) is what he wants I think.

[tex]\displaystyle\frac{d}{dx}\log_{10}(x) =\frac{1}{x \cdot \ln{10}}[/tex]

This is derived from the formula that zabelman suggested.

[tex]\displaystyle(\log_{a}(x))'=\left({\frac{\ln(x)}{\ln(a)}\right)' =\frac{1}{\ln{a}} \cdot (\ln{x})' = \frac{1}{x \cdot \ln{a}}[/tex]

Good point Simon, but it's summer - so he is probably just asking b/c he is curious.  I ask a lot of textbook questions just because I don't know who else to ask for help.  Is this okay?  I never know whether to post them in intermediate or in other problem solving topics, or at all...

I did answer this is the intermediate board...

I've asked him where they come from before, and he has yet to answer once.  In fact, I don't think he responds on his own posts, no matter what.  I've personally decided to boycott his messages (that is, to not help) until he explains.

Chris sent me a PM explaining the source of his multitude of questions. He's trying to learn Calc on his own, and the 1 calc book he has isn't very good.


Propose a synthesis of [i]o[/i]-methylacetophenone from toluene.

1. sulphonate to get para product
2. Acylate and treat with acid to get 2 methyl acetophenone

Is it the case with SO3H also?
I thought SO3H is weakly deactivating only so.........  :D

If you introduce -NH2 gorup then if u do F&C then it will introduce new group in the meta postion to -CH3

yes i didnt think of that  :blush:
what are the +R grps weaker than CH3 ?  :D

Find one, but u try a different approach

got it.  :D
Do as i told you till NH2 then make it CN
Use Von Richter Synthesis to get meta methyl benzoic acid
Then it becomes simple  :P

after meta methyl benzoic acid??????????

supposed to prepare the ortho product :rotfl:  :rotfl:

oh!  :blush:
I've done a comedy of errors with this topic :oops:
Okay let me have a try.
1. Oxidise to benzoic acid
2. Nitrate
3. Protect COOH grp using esterification
4. Use Von Richter and do acidic hydrolysis (Get the kinetic product) :) i.e. ortho
5. Use Ruzicka rn  with HCOOH
6. Get back COOH grp
7. Reduce the COOH into CH3 using LAH and P/HI (or only P/HI)

:)

tolune+sulphonate-> para product
now brominate(mono) and remove so3h and then react with mg in ether to get organometallic and react with acetyl chloride

[quote="chemrock"]Does benzoic acid undergo nitration?????????
Also von richter takes place in presecne of -EWG and not with -CH3[/quote]

why cant benzoic acid undergo nitration???  :huh:

Yours is the same as madness but does -SO3H substituted ring undergo F&C???

[quote="VARUNARASU"]tolune+sulphonate-> para product
now brominate(mono) and remove so3h and then react with mg in ether to get organometallic
[/quote]
[b]There you intend to get o- MgBr toluene right? What r u doing after that?[/b]

F and C is Friedel and crafts  :D

the organometallic formed is a good nucleophile and will displace cl

The acyl chloride is very reactive and the end product is a tertiary alcohol. I think you should use organocadmium compounds

suppose i use 1 eq of the organometallic the reaction will stop at nucleophilic subs and addition wont take place(even excess of acetyl chloride)

still it is better to use organocadmium compounds.

[quote="madness"]In that case you can Nitrate to get para isomer
make it NH2 and NHCOCH3 in the para posn to CH3
do what i said
protect the COCH3 grp using acetal
use NaOH and H3PO2 and then use H+  :D[/quote]
you cant carry out f&c with nh2 or nhcoch3 in the ring

And what would be the difference?

Also, Von Richter reaction, in the few cases it [i]may[/i] give yields around 50%, only is good when a para electronwithdrawing group is present.


On the Contests section of the Forum (with the compilation of all the problems), when I use the "Download PDF Version" command, the images don't show up.

For example, on USA AMC12 2005 Day A, on problem 17,

[img]http://www.artofproblemsolving.com/Forum/albumpic.php?picid%20=%20320[/img]

(except some parts are italicized)

shows up instead of the actual image.

I am using Windows XP, IE and Mozilla, but I don't think that really matters.

I am also using Adobe Acrobat Reader

It's a known issue. Have patience please.

Sorry if this has already been answered, but the images still don't show up. When will they show up?


If two triangles have their correspondent sides parallel, then they are similar, and the segments linking correspondent vertices are on the same line or don't intersect. Consider that the correspondet vertices are distinct.

[quote="Sashsiam_2"]the segments linking correspondent vertices are on the same line or don't intersect.[/quote]  This is false -- consider, for example, the case of any triangle and its midpoint triangle.


analytic geometry
geometry
geometry unsolved

Let O,A1,A2,A3,A4 be 3 points on the coordinate plane and S(ij) be the area of the triangle OAiAj. Suppose that S(ij)>=1 for 1<=i<j<=4. Prove that we can find 2 positive integers i',j' such that S(i'j')>=sqrt(2) and 1<=i'<j'<=4.

Actually, if I didn't make another mistake, I think the bound can be brought up to 1.5 instead of root 2... :D


Ok, i got a 138 on the 10B as a freshman...
how does the process work after AIME to make USAMO?
about how many do i need to get right on the AIME?

Since you are a freshman (10th grade and under), you can qualify through the Index or the Floor.

Index: AMC 10/12 Score + 10*AIME Score = Index. If it is above a certain number, you will qualify for the USAMO. Usually this number is around 200 or 210.

They select a bunch of people for the USAMO through the Index. Then they look at the person who qualifies through the index who has the lowest AIME score. This lowest AIME score is the floor. If your AIME score is greater or equal to the floor, and your index is high enough (among people who are qualifying through the floor; i.e. 10th and unders; you could think of it as a "tiebreaker"), then you will qualify for the USAMO. The floor varies from a 6 to an 8 usually, and has reached a 9 before...

Many people hope that somebody gets a 150 on the AMC 12 and a 6 on the AIME.  :wink:

For those that are selected through the "floor" AIME score, you said that the index is a sort of tiebreaker. Approximately what is the cutoff index for these "floor" people?

If I understand correctly, this is different from the 200-210 number you mentioned. Am I right?

Last year it was actually around the same cutoff as the USAMO index for the AMC 12 people.

In the few years before that, it was usually lower.

could some kind person who got a 150 on AMC 12 please purposely get a 6 on AIME???  :P  jk.

6 as a floor score is not necessarily a good thing though. Like last year, since the floor score was so low, the index required was high. So if you did have a 6, you needed a 144 on the AMC 10. But people who had 132/7 or 120/8 didn't make it. So if you have a low amc score, you want don't want such a low floor, because you won't make it either way unless you have at least a 9 or something. If you have a really high amc 10 score, then yeah, you want the floor score to be as low as possible.

[quote="Arcesso"]But people who had 132/7 didn't make it.[/quote]Just like me!

Wait, what? There's no advantage for AMC12 takers? I thought they take the top 350 or so indexes from AMC12, and the top 150 or so AMC10...

That is pretty much how they do it. The floor value only applies to the AMC 10 takers, but if too many have high enough AIME scores then they use the index.


This seems pretty nasty problem and I need help on finding the answer.  :(

Let $a,b,c,d$ be distinct real numbers such that

\[a+b+c+d = 3, a^2+b^2+c^2+d^2 = 45\]

Find the value of the expression

\[\frac{a^5}{(a-b)(a-c)(a-d)} + \frac{b^5}{(b-a)(b-c)(b-d)} + \frac{c^5}{(c-a)(c-b)(c-d)} + \frac{d^5}{(d-a)(d-b)(d-c)}\]

Consider the determinant

\[
D(a,b,c,d)=\left|
\begin{array}
[c]{cccc}
1 & 1 & 1 & 1\\
a & b & c & d\\
a^{2} & b^{2} & c^{2} & d^{2}\\
a^{5} & b^{5} & c^{5} & d^{5}
\end{array}
\right|
\]
as a polynomial in $a,b,c,d.$ It is not difficult to see that $D$ is a homogenous and skew-symmetric polynomial, with $\deg D=8.$ For $a=b,$ $D=0,$ thus $a-b$ divides $D.$ Similarly, $D$ is divisible by all pairwise differences hence
\[
D=\left(  M\sum a^{2}+N\sum ab\right)  \prod\left(  a-b\right)  .
\]
Taking particular values for $a,b,c,d,$ we deduce that $M=N=1.$

Now, expand $D$ after the last row to obtain
\[
D=\sum a^{5}\left(  b-c\right)  \left(  b-d\right)  \left(  c-d\right)  .
\]
It follows that
\[
\sum\frac{a^{5}}{\left(  a-b\right)  \left(  a-c\right)  \left(  a-d\right)
}=\sum a^{2}+\sum ab.
\]
The latter is easy to compute if $\sum a=3$ and $\sum a^{2}=45.$ The answer is $27$.

Here's another approach (slightly different):

consider the system
\[
\left\{
\begin{array}
[c]{l}
x+y+z+t=0\\
ax+by+cz+dt=0\\
a^{2}x+b^{2}y+c^{2}z+d^{2}t=0\\
a^{3}x+b^{3}y+c^{3}z+d^{3}t=1
\end{array}
\right.
\]
Using Cramer's rule and Vandermonde determinants (I posted this in Algebra Theorems and Formulas) yields
\[
x=\frac{1}{\left(  a-b\right)  \left(  a-c\right)  \left(  a-d\right)  },
\]
and similar equalities. Now, let
\[
S_{n}=\sum\frac{a^{n}}{\left(  a-b\right)  \left(  a-c\right)  \left(
a-d\right)  }.
\]
Replacing in the system, we obtain
\[
S_{0}=S_{1}=S_{2}=0
\]
and
\[
S_{3}=1.
\]
We have to compute $S_{5}.$ Observe that $a,b,c$ and $d$ are the roots of the polynomial
\[
f\left(  x\right)  =x^{4}-\left(  a+b+c+d\right)  x^{3}+\ldots+abcd,
\]
hence
\[
a^{4}-\left(  a+b+c+d\right)  a^{3}+\ldots+abcd=0.
\]
This, together with the other three similar equalities gives
\[
S_{4}-\left(  a+b+c+d\right)  S_{3}+\ldots+abcdS_{0}=0,
\]
therefore $S_{4}=a+b+c+d.$ Similarly, $a,b,c,d$ are roots for
\[
g\left(  x\right)  =x^{5}-\left(  a+b+c+d\right)  x^{4}+\ldots+abcdx,
\]
hence
\[
S_{5}-\left(  a+b+c+d\right)  S_{4}+\left(  \sum ab\right)  S_{3}
+\ldots+abcdS_{1}=0,
\]
which implies that
\[
S_{5}=\left(  a+b+c+d\right)  ^{2}-\sum ab=\sum a^{2}+\sum ab.
\]

Thank you!

I'll look at this tomorrow or day after when I have a lot of time. Thanks VERY MUCH for posting full solutions (TWO!!)!

Let us more generally consider $n$ variables $a_1,...,a_n$ and let

$S_k = \sum_{i=0}^{n} \frac{a_i^k}{\prod_{j\neq i} (a_i-a_j)}$.

Furthermore, denote $C_k(a_1,...,a_n)=\sum_{e_1+...+e_n=k,e_i\geq 0} a_1^{e_1}\cdots a_n^{e_n}$.

Then we have

1) $k<0$: $S_k = (-1)^{n+1} \frac 1{a_1\cdots a_n} C_{-1-k} (a_1^{-1},...,a_n^{-1})$
2) $0\leq k<n-1$: $S_k=0$
3) $k\geq n-1$: $S_k=C_{k+1-n} (a_1,...,a_n)$.

To prove this, first note that the polynomial $P(x)=\sum_{i=1}^{n} \frac{\prod_{j\neq i}(x-a_j)}{\prod_{j\neq i}(a_i-a_j)}$ is of degree at most $n-1$ and for $x=a_1,...,a_n$ we have $P(x)=1$; therefore $P(x)\equiv 1$. Summing up, we know that

$1=\sum_{i=1}^{n} \frac{\prod_{j\neq i}(x-a_j)}{\prod_{j\neq i}(a_i-a_j)}$.

Now, divide by $\prod_{i=1}^{n} (x-a_i)$:

$\frac 1{\prod_{i=1}^{n} (x-a_i)}=\sum_{i=1}^{n} \frac 1{(x-a_i)\prod_{j\neq i}(a_i-a_j)}$.

It is easy to show that $\frac 1{\prod_{i=1}^{n} (x-a_i)}=\sum_{m=0}^{\infty} C_mx^{-m-n}$. On the other hand $\frac 1{x-a_i} = \sum_{m=0}^{\infty} a_i^mx^{-m-1}$. Now, comparing coefficients gives 2) and 3). To get 1), take power series around $0$ instead of power series around $\infty$.


I suck pretty bad at mathcounts, and im really worried now. our mathcounts school sprint competition is next mon, and target is next tues! What should I study and be aware about? I just need a few hints on the problems.

One thing:
DON'T be nervous, that will really raise your score

That's the thing I'm kind of bad at. Thanks for the advice though, it might earn me some extra points.


analytic geometry
trigonometry
geometry open
geometry

Consider the surface of a sphere. Divide it with n equally space meridians and m lines parallel to the equator. These lines divide the surface in grid "squares" - "triangles" near the poles. Consider such a square, S, and any point P. What is the shortest distance between a point in S and P?

Did you try geographic coordinates? By means of spherical trigonometry it will surely work.

Best regards,
sunker rock


I just know the basic one in learned in high school,
its in c++ syntax:

i, j and temp are integers
list is a vector with 20 elements

[code]for(i = 0;  i < 20;  ++i)
     for(j = i;  j < 20;  ++j)
          if(list[i] > list[j])     {
               temp = list[i];
               list[i] = list[j];
               list[j] = temp;
          }[/code]

the algorith puts the elements in accending order.  it puts the elements in order one at a time, that's what the first for is.  the second for is when it puts the element its working on as high up the list that it can.  The three lines in the if statement switch the two values if necessary.
You can make the if statement anything, to order it any way.

when you think about what the computer has to do to sort a list, its incredibly redundant.  people will take the time to put in a switch statement to avoid 5 or 6 unncecessary evalutations, this thing's got to do 210 evalutations, and its only 20 items...

(its been awhile, so if i made any mistakes please correct me)

Any others?

I don't see anything wrong except that it's a very bad algorithm. The algorithm is $O(n^2)$ where the standard is $O(n\log n)$. This is insignificant for sorting 20 items (computers are very fast), but enormous for sorting a million items.

There are a large number of sorting algorithms that work in $n\log n$ time, and they're one of the things you really should have learned when you took computer science

And it's called minsort I think. Most common sorting algorithms use only $N\log N$ comparations while this one needs $N^2$. One such N log N algorithm is quicksort which is already implemented in C standard library stdlib.h so you don't really need to learn it - you just use function qsort. Or if you use stl, there is function sort. But I would advise that you learn at least the basics of how these algorithms work...

Yes, qsort is a very nice function, but I believe it requires that you pass in a pointer to a function... for someone who only knows one sorting algorithem, I recomend that they learn a bubble sort, just to know how it works, followed by quicksort and merge sort.

Also, quicksort doesn't always run in Nlog(N) time... it works best on the most random ordered arrays... in fact it can be as slow as o(N^2). Merge sort is a pretty nice sorting algorithem, and this 1 will always run in time o(Nlog(N)). Depending on what you are sorting and the size, sometimes quicksort will go faster, and other times mergesort will go faster.

[code]
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
vector <int> a;
int main(void)
{
    .... //read a
    sort(a.begin(), a.end());
    return 0;
}
[/code]
 :w00t:

Yeah :), STL is a great improvement over the standard C libraries but not as many people can use it... Sort is actually implemented as quicksort with median of three, or similarly so that it's worst-case time complexity is N log N as opposed to N^2 with standard quicksort.

It's actually intro sort.

[quote]
[2] Earlier versions of sort used the quicksort algorithm (C. A. R. Hoare, Comp. J. 5, 1962), using a pivot chosen by median of three (R. C. Singleton, CACM 12, 1969). Quicksort has O(N log(N)) average complexity, but quadratic worst-case complexity. See section 5.2.2 of Knuth for a discussion. (D. E. Knuth, The Art of Computer Programming. Volume 3: Sorting and Searching. Addison-Wesley, 1975.) The current implementation of sort, however, uses the introsort algorithm (D. R. Musser, "Introspective Sorting and Selection Algorithms", Software Practice and Experience 27(8):983, 1997.) whose worst case complexity is O(N log(N)). Introsort is very similar to median-of-three quicksort, and is at least as fast as quicksort on average.
[/quote]

Does anybody know how introsort works (except for that it is similar to quicksort with median of three :) ) ?


LaTeX
function
real analysis
real analysis unsolved

Hello:

Can anyone, well-versed in infinite series, show me how to arrive at the solution for $\sum_{n=1}^\infty\frac 1{3^n-1}$?
I know it converges. That's easy to test. It converges to 18309/26840. How do we arrive at this answer?. Does anyone know a good book or website on convergent series?.

Thanks,
Cody

[b] Moderator edit: Edited for $\LaTeX$ [/b].

I tired to check it on a computer. I got that the sum of the first 30 terms of the series is about $0.682153502605\dots$ while $\frac{18309}{26840}=0.682153502235\dots$, which is less. For control, I checked the sum $\sum_{n=1}^{30}\frac{1}{3^n}$. It was below $0.5$. So I believe that $\frac{18309}{26840}$ is just a very good approximation but not an exact answer. Actually, I suspect that the sum of the series is irrational but do not have the slightest idea of how to prove it...

Your series is the value at $x=1/3$ of $x p'(x) / p(x) $ where  $p(x)$ is the generating function for partitions (reciprocal of $(1-x)(1-x^2)....$).  $1/p(x)$ is a theta function and can be computed very efficiently (Euler's pentagonal numbers theorem gives the power series expansion).   I don't know about $p'(x)$ and would be surprised if there is a simple formula for the number you posted.

Thank you for your responses.  I just figured it's more trouble than it's worth. I like series, especially when they're convergent, but this sort is more difficult than I realized to do by hand.

Thanks again,
Cody


Which will fall if i drop a half pound of feathers from 1/2 foot high or a pound of gold from 1 and a 1/2 feet high.  [hide="hint"]
[hide="sure you wanna hint?"]
[hide="positive?"]

well, the hint is:
there is no hint!!!!!

[/hide]
[/hide][/hide][/hide][/hide][/hide][/hide][/hide][/hide][/hide][/hide]

nobody waste their time by clicking 120 times to get a stupid hint that doesn't even exist! :mad: :bomb:

it exists...you just have to click 287 times.

[quote="math92"]Which will fall if i drop a half pound of feathers from 1/2 foot high or a pound of gold from 1 and a 1/2 feet high.  [hide=hint]
[hide=sure you wanna hint?]
[hide=positive?]
[hide=absolutley sure?]
[hide=okay]
[hide=if you say so]
[hide=HINT]
[hide=CLICK ME]
[hide=are you still cliking?]
[hide=you must really wanna hint.]
[hide=why do you wanna hint anyways?]
[hide=you can't get the problem?]
[hide=CLIK ME ONE LAST TIME]
[hide=ha i lied]
[hide=you have to clik me 7 more times]
[hide=hahahahaha]
[hide=hahahaha]
[hide=hahaha]
[hide=haha]
[hide=ha]
[hide=is that 7 times yet?]
[hide=oh well]
[hide=HINT!!!]
[hide=clik for money]
[hide=ha, yu get no money!]
[hide=hahahaha]
[hide=hahaha]
[hide=haha]
[hide=are you bored?]
[hide=are you still clicking?]
[hide=hint]
[hide=sure you wanna hint?]
[hide=positive?]
[hide=absolutley sure?]
[hide=okay]
[hide=if you say so]
[hide=HINT]
[hide=CLICK ME]
[hide=are you still cliking?]
[hide=you must really wanna hint.]
[hide=why do you wanna hint anyways?]
[hide=you can't get the problem?]
[hide=CLIK ME ONE LAST TIME]
[hide=ha i lied]
[hide=you have to clik me 7 more times]
[hide=hahahahaha]
[hide=hahahaha]
[hide=hahaha]
[hide=haha]
[hide=ha]
[hide=is that 7 times yet?]
[hide=oh well]
[hide=HINT!!!]
[hide=clik for money]
[hide=ha, yu get no money!]
[hide=hahahaha]
[hide=hahaha]
[hide=haha]
[hide=are you bored?]
[hide=are you still clicking?]
[hide=hint]
[hide=sure you wanna hint?]
[hide=positive?]
[hide=absolutley sure?]
[hide=okay]
[hide=if you say so]
[hide=HINT]
[hide=CLICK ME]
[hide=are you still cliking?]
[hide=you must really wanna hint.]
[hide=why do you wanna hint anyways?]
[hide=you can't get the problem?]
[hide=CLIK ME ONE LAST TIME]
[hide=ha i lied]
[hide=you have to clik me 7 more times]
[hide=hahahahaha]
[hide=hahahaha]
[hide=hahaha]
[hide=haha]
[hide=ha]
[hide=is that 7 times yet?]
[hide=oh well]
[hide=HINT!!!]
[hide=clik for money]
[hide=ha, yu get no money!]
[hide=hahahaha]
[hide=hahaha]
[hide=haha]
[hide=are you bored?]
[hide=are you still clicking?]
[hide=hint]
[hide=sure you wanna hint?]
[hide=positive?]
[hide=absolutley sure?]
[hide=okay]
[hide=if you say so]
[hide=HINT]
[hide=CLICK ME]
[hide=are you still cliking?]
[hide=you must really wanna hint.]
[hide=why do you wanna hint anyways?]
[hide=you can't get the problem?]
[hide=CLIK ME ONE LAST TIME]
[hide=ha i lied]
[hide=you have to clik me 7 more times]
[hide=hahahahaha]
[hide=hahahaha]
[hide=hahaha]
[hide=haha]
[hide=ha]
[hide=is that 7 times yet?]
[hide=oh well]
[hide=HINT!!!]
[hide=clik for money]
[hide=ha, yu get no money!]
[hide=hahahaha]
[hide=hahaha]
[hide=haha]
[hide=are you bored?]
[hide=are you still clicking?]
well, the hint is:
there is no hint!!!!!
[/quote]
this is your quote without the [ /hide]'s
there is no hint and it's pretty anoying to do something like this  :furious:

[quote="LynnelleYe"]Er...they'll both fall. You mean which one will fall first? I suppose the feathers since they're closer to the ground. Right?[/quote]

you havent considered air-resistance

and math92, if we were really desperate to see ur hint, we would as phsyco did, quote your post and avoid unpleasant clicking...............

Well they both will hypothetically "fall".

But, as suggested above, if you meant "Which will fall first?" then you would have to conside a number of factors- how large are the respective objects, is there any wind out, and stuff like that.

Without air resistance, the feathers will hit the ground first, since they're closer.

On Earth, the gold would hit the ground first, as the feathers would float too much, while the gold would drop straight down.

sorry for the big clicking hint incident.  I was just bored.  i'm sorry. :(
colts18 told me to write this problem.

if you use *fake Galilo's idea it should fall at a same time, but you also have to consider the velocity

um...

I say
gold!


function
Support
integration
limit
floor function
calculus
irrational number

Prove or disprove.

If $\alpha$ is an irrational number and $f:\mathbb R\rightarrow\mathbb R$ is a continuous function (it may be assumed to have compact support), such that $f(x)\geq0$ and $\int_{\mathbb R}f(x)dx=1$, then

\[\lim_{\Delta\rightarrow0^+}\left(\frac{\alpha}{\Delta^2}\int_{\mathbb R}f\left(\frac{x}{\Delta}\right)f\left(\frac{\alpha x}{\Delta}\right) \left\{\frac{\alpha x}{\Delta}\right\}\left\{\frac{x}{\Delta}\right\}dx\right)=\frac14\]

where $\left\{t\right\}=t-\lfloor t\rfloor$ is the fractional part of $t$.

If you tried this one before, please note that there was a typo  :blush: .

It should be right now (I think). I appologize  :oops:  :oops:

Don't know what it was, but, as written now, it doesn't seem to make any sense: after the change of variable $y=\frac x\Delta$, you get $\frac \alpha\Delta$ times a fixed integral under the limit sign.... :?
Maybe you mean
\[
\lim_{\Delta\rightarrow0^+}\left(\int_{\mathbb R}f\left({x}\right)\left\{\frac{\alpha x}{\Delta}\right\}\left\{\frac{x}{\Delta}\right\}dx\right)=\frac14\,\,?
\]

[quote="fedja"]\[
\lim_{\Delta\rightarrow0^+}\left(\int_{\mathbb R}f\left({x}\right)\left\{\frac{\alpha x}{\Delta}\right\}\left\{\frac{x}{\Delta}\right\}dx\right)=\frac14\,\,?
\][/quote]

I don't see how is that you eliminate one of the functions in the multiplication.  :?

I'm not saying that this is equivalent to your expression. I'm merely saying that this is something that is true and that  the equality you proposed is clearly false. Sorry if I haven't made it clear :blush:. I just tried to extrapolate from one of your previous posts where you were talking about a random variable $X$ and two associated random variables $Y=\{X\}$ and $Z=\{\alpha X\}$. The equality I wrote is equivalent to asymptotically $0$ correlation between $Y$ and $Z$ when the density of $X$ is $\Delta f(\Delta x)$ and $\Delta\to 0$. I just thought it might be what you really needed, but, perhaps, I'm mistaken.  :? Anyway, the version with two functions and $\Delta$ in the denominator everywhere is clearly false and very strange for the reason I mentioned in my previous post...

:blush: you are right! I made a mistake, but anyway I haven't been able to prove the other post equality!  :blush:

[quote="fedja"]So $\lim_{\Delta\rightarrow0^+}\left(\int_{\mathbb R}f\left({x}\right)\left\{\frac{\alpha x}{\Delta}\right\}\left\{\frac{x}{\Delta}\right\}dx\right)=\frac14
$ is what you need? I just wanted to make it sure before posting a proof, which is not that hard. Note that $\{x\}=\sum_k c_k e^{2\pi i kx}$ and similarly for $\{\alpha x\}$ where the series converge in $L^2_{loc}$. Then, replacing each series by a finite sum and putting $\lambda=\frac 1\Delta$, we get that the integral under consideration is approximately equal $\sum_{|k|,|\ell|\leq N}c_kc_\ell\int_{\mathbb R}f(x)e^{2\pi i\lambda(k+\ell\alpha)x}\,dx$ with error depending on $N$, but not on $\Delta$. Since $k+\ell\alpha\neq 0$ unless $k=\ell=0$, all integrals except the one corresponding to $k=\ell=0$ tend to $0$ as $\lambda\to \infty$. This leaves us with $c_0^2\int_{\mathbb R}f(x)\,dx=1/4$ (because $c_0=1/2$).[/quote]

Fedja,

Thank you very much. I haven't thouht in Fourier Series for this problem. And your solution is quite clever! (more clever that all my preceding attempts :blush: ).

Fedja,

May be I am too purist, but there is somthing, after thinking about that very carefully, that I still don't see, and it is this: I agree that $\displaystyle \sum_{|k|,|\ell|\leq N}c_kc_\ell f(x)e^{2\pi i\lambda(k+\ell\alpha)x}\rightarrow f\left({x}\right)\left\{\lambda\alpha x\right\}\left\{\lambda x\right\}$ as $N\rightarrow\infty$, but I cannot apply dominated convergence. How do you prove that you can interchange both limits, I mean, how do you prove that in the expression,

$\lim_{\lambda\rightarrow\infty}\left(\int_{\mathbb R}f\left({x}\right)\left\{\lambda\alpha x\right\}\left\{x\right\}dx\right) = \lim_{N\rightarrow\infty}\lim_{\lambda\rightarrow\infty}\sum_{|k|,|\ell|\leq N}c_kc_\ell\int_{\mathbb R}f(x)e^{2\pi i\lambda(k+\ell\alpha)x}\,dx$

you can interchange both limits?

This piece of the proof is a bit tricky and it essentially uses that $f$ is bounded and that the support of $f$ is contained in some finite interval $I$. The idea is that the difference between $\{\lambda x\}$ and $S_N(\lambda x)=\sum_{|k|\leq N}c_k e^{2\pi i\lambda kx}$ is small in $L^2(I)$. To prove it, let us note that the integral of the square of the difference over each period of length $1/\lambda$ is $\frac 1\lambda \sum_{|k|>N}|c_k|^2$ and we have not more than $|I|\lambda+2$ periods intersecting $I$, so, if $\lambda>1$, the total integral does not exceed $(|I|+2)\sum_{|k|>N}|c_k|^2$, which can be made as small as we want (say, less than $\delta$) by choosing $N$ large enough. Note that the estimate does not depend on $\lambda$. Analogously, $\{\lambda\alpha x\}$ is close to $S_N(\lambda\alpha x)$ in $L^2(I)$. So, the product is close to the product of the partial sums in $L^1(I)$, which property is preserved after multiplication by a bounded function $f$. Therefore, we can pass to the limit as $\lambda\to+\infty$ keeping the estimate for the difference between the LHS and the RHS small and independent of $\lambda$. The RHS will have the limit $1/4$. As to the LHS, we'll get the conclusion that its upper limit is at most $\frac14+10\max_I|f| \delta$ (or something like that) and its lower limit is at least $\frac14-10\max_I|f|\delta$. Now it remains to recall that $\delta$ was arbitrary. So, you are right, this part should be done "by hand"; a reference to some general theorem wouldn't help much here. Of course, after we know the result for bounded compactly supported functions, we can automatically extend it by continuity to all $L^1$-functions $f$. Hope this helps.  :)

Hi Fedja,

I feel kind of slow :blush: , it took me more than a day to fully understand your solution (or at least to write it down in a way that would convince myself that is correct), but again, thank you very much!, it is not totally obvius, but it works! :first:

Best,

I don't know if someone besides Fedja and myself is interested in this topic, but after a while, I found a nice solution that is probably a lot simpler, and that generalize for all such $f$'s in $L^1(\mathbb R)$. Basically, the more basic concept on approximation theory... Start with step functions!
It works, and the resoult you have is that
\[\lim_{\Delta\rightarrow0^+}\left(\int_{\mathbb R}f\left({x}\right)\left\{\frac{\alpha x}{\Delta}\right\}\left\{\frac{x}{\Delta}\right\}dx\right)= \frac{\|f_+\|_1-\|f_-\|_1}{4}\]
The only misterious part is to apply Weyl distribution theorem, but after that, it is straight forward


[quote="Virgil Nicula"]$ \left\{\begin{array}{c} \triangle\ ABC\ : \ A \equal{} 40^{\circ}\ ;\ B \equal{} 30^{\circ} \\
\ \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \\
\ M\in (AB)\ ,\ m(\widehat {BCM}) \equal{} 30^{\circ} \\
\ N\in (AC)\ ,\ m(\widehat {CBN}) \equal{} 10^{\circ}\end{array}\right\|$ $ \implies$ $ m(\widehat {AMN}) \equal{} \ ?$[/quote]
Nice problem!
[hide="Solution"]
We denote the intersection between $ CM$ and $ BN$ to be $ P$ and we connect $ AP$. We also say that $ \angle AMN \equal{} x$.  Thus, $ \angle MPN \equal{} \angle BPC \equal{} 180 \minus{} \angle PBC \minus{} \angle PCB \equal{} 180 \minus{} 30 \minus{} 10 \equal{} 140$.  Thus, $ \angle NAM \plus{} \angle MPN \equal{} 40 \plus{} 140 \equal{} 180$, so $ ANPM$ is cyclic.  Additionally, we notice that $ \angle AMC \equal{} \angle ABC \plus{} \angle MCB \equal{} 30 \plus{} 30 \equal{} 60\implies 60 \minus{} x \equal{} \angle NMC \equal{} \angle PAC$.  Now, we call $ 60 \minus{} x \equal{} y$.  Thus, $ \angle BAP \equal{} 40 \minus{} y$.  So, by the Sine Form of Ceva's Theorem, we obtain that
\[ \frac {\sin \angle CBN}{\sin \angle ABN}\times \frac {\sin \angle BAP}{\sin \angle PAC}\times \frac {\sin \angle ACM}{\sin \angle MCB} \equal{} 1
\]
\[ \frac {\sin 10}{\sin 20}\times \frac {\sin (40 \minus{} y)}{\sin y}\times \frac {\sin 80}{\sin 30} \equal{} 1
\]
\[ \frac {1}{2\cos 10}\times 2\sin 80\times \frac {\sin (40 \minus{} y)}{\sin y} \equal{} 1
\]
\[ \frac {1}{2\cos 10}\times 2\cos 10\times \frac {\sin (40 \minus{} y)}{\sin y} \equal{} 1
\]
\[ \sin (40 \minus{} y) \equal{} \sin y
\]
\[ y \equal{} 20
\]
\[ x \equal{} \boxed{40}
\]
[/hide]

[quote="Virgil Nicula"]Thank you, [b]The QuattoMaster 6000[/b], the your proof is nicely and quickly. Maybe someone presents us and a synthetical proof for middle-school students.[/quote]
My apologies, dear [b]Virgil Nicula[/b], I did the trignometric proof only so I could find the value of $ \angle AMN$.  Now that I know what it is, we can find a synthetic proof  :blush:
[hide="Alternate Solution"]
Note that $ \angle ACB \equal{} 180 \minus{} \angle BAC \minus{} \angle ABC \equal{} 180 \minus{} 30 \minus{} 40 \equal{} 110$ and $ \angle NBC \equal{} 10\implies \angle BNC \equal{} 180 \minus{} \angle NBC \minus{} \angle NCB \equal{} 60$.  We can also say that $ \angle MBC \equal{} 30$ and $ \angle NCB \equal{} 30\implies \angle CMB \equal{} 120$.  Now, we notice that $ \angle MBC \equal{} \angle MCB\implies MB \equal{} MC$.  This, coupled with the fact with $ \angle BMC \equal{} 120 \equal{} 2(60) \equal{} 2\angle BNC\implies \angle BMC \equal{} 2\angle BNC$ implies that $ M$ is the circumcenter of $ \triangle NBC$, and thus $ MN \equal{} NC$.  With this in mind, we can see that $ \angle MNC \equal{} \angle MCN \equal{} 110 \minus{} 30 \equal{} 80 \implies \angle ANM \equal{} 180 \minus{} \angle MNC \equal{} 100\implies \angle AMN \equal{} 180 \minus{} \angle MAN \minus{} \angle ANM \equal{} 180 \minus{} 100 \minus{} 40 \equal{} \boxed{40}$. [/hide]

Is this a proof without words or I miss something?


In lawn-tennis the player who scores at least four points, while his opponent scores at least two points less, wins a game. The player who wins at least six games, while his opponent wins at least two games less, wins a set. What minimum percentage of all points does the winner have to score in a set?

Let $ a_i$ be the points win by $ A$ in the game $ i$
Let $ A\equal{}\sum_{i}{a_i}$, idem for $ B$ and $ b_i$

WLOG $ A$ won the game, Let $ P\equal{}\frac{A}{A\plus{}B}$
The results in games can be: $ 6\minus{}0, 6\minus{}1,6\minus{}2,6\minus{}3,6\minus{}4$ clearly the last result gives a minimum percentage, just add a $ 0\minus{}4$ game you have the same numerator in $ P$ but a bigger denominator.
WLOG, and for a more emotional game,  $ B$ won the first four games and $ A$ the last six games then:
$ b_1\equal{}b_2\equal{}b_3\equal{}b_4\equal{}4$ and $ a_5\equal{}a_6\equal{}\cdots\equal{}a_{10}\equal{}4$
$ P\equal{}\frac{a_1\plus{}a_2\plus{}a_3\plus{}a_4\plus{}24}{16\plus{}b_5\plus{}b_6\plus{}\cdots\plus{}b_{10}\plus{}24\plus{}a_1\plus{}a_2\plus{}a_3\plus{}a_4}$
then $ b_5\plus{}b_6\plus{}\cdots\plus{}b_{10}$ should be maximize then $ P\equal{}\frac{a_1\plus{}a_2\plus{}a_3\plus{}a_4\plus{}24}{16\plus{}2\plus{}2\plus{}\cdots\plus{}2\plus{}24\plus{}a_1\plus{}a_2\plus{}a_3\plus{}a_4}\equal{}\frac{24\plus{}a_1\plus{}a_2\plus{}a_3\plus{}a_4}{52\plus{}a_1\plus{}a_2\plus{}a_3\plus{}a_4}$ notice that $ \frac{24\plus{}x}{52\plus{}x} \ge \frac{24}{52}$ for nonnegative $ x$ then
$ P\equal{}\frac{24\plus{}0\plus{}0\plus{}0\plus{}0}{52\plus{}0\plus{}0\plus{}0\plus{}0}\equal{}\frac{6}{13}$


Hey guys, I am confused about this topic. I have a question regarding this topic:

A computer store is advertising. A survey of 200 customers says 60 people learned this from the paper, 50 from TV, and 30 from both the paper and the TV.
What is the probability of each event?

a) A randomly selected customer saw the ad in at lease one of the two media.
b) A randomly selected customer saw the ad in exactly one form.

This is what I have:

P(Paper) = 60/200
P(TV) = 50/200
P(Paper and TV) = 30/200

For question a, I'm thinking that a customer can hear about it from the paper or the TV and it is okay for them to hear about it from both sources:

P(Paper) + P(TV)
= 6/20 + 5/20
= 11/20

For question b, I'm thinking that a customer can hear about it from the paper or the TV and it is not okay for them to hear about it from both sources:

P(Paper) + P(TV) - P(Paper and TV)
= 6/20 + 5/20 - 3/20
= 11/20 - 3/20
= 8/20
= 2/5

The answer does not match mine for either question.

Please help! Thanks!

The easiest way (at least for me) to approach problems of this sort is to draw out Venn diagrams.  Then finding probabilities is just a matter of adding and subtracting different categories, and then dividing at the end.  :)

Edit:  But if you want to know why your current way doesn't work, in question 1 you count the people who learned from both TV and the Paper twice by just adding.  Similarly for question 2, you just need to subtract P(TV and Paper) again.

Oh my gosh, Venn diagrams really work :!:.  I never liked them but now I will :blush: .

Thanks, t0rajir0u. But could you or anyone else please explain why my way doesn't work. I got the answer now from the Venn diagram  :)  but I really wanna know why my previous way does not work. Thanks

[quote="Numbnut247"][quote="236factorial"][hide]In the first question, your step of adding P(Paper) and P(TV) are correct, but when you do that, you count P(Paper and TV) twice. Thus you should subtract P(Paper and TV) from P(Paper) + P(TV), and the desired answer would result.

In the second question, you had everything correct except you made the same mistake as in the first one. Since P(Paper and TV) is counted twice, to eliminate this category completely (leaving only exclusively TV or paper advertising), we have to subtract this probability [b]twice[/b] from P(Paper) + P(TV). [/hide][/quote]

Come again? Sorry :blush:

How did I count P(Paper and TV) twice? Is it because it could be Paper and TV or TV and Paper?
Same thing goes with the second question. Thanks[/quote]

By counting 6/20 and 5/20 [for the first question], you are not taking account of the people who are counted twice, because they saw it in both TV and paper. So, it's ....
[hide="answer"]$(60/200-30/200)+(50/200-30/200)+(30/200)=80/200=\boxed {2/5}$. I hope that's the right answer.[/hide]

use this method to solve the second question.

[quote="Numbnut247"]

How did I count P(Paper and TV) twice? Is it because it could be Paper and TV or TV and Paper?
Same thing goes with the second question. Thanks[/quote]

Your notation is slightly unclear - mentally, you didn't distinguish between P(Paper but not TV) and P(Paper) - they're different probabilities.  In the first question, you want P(Paper but not TV) + P(TV but not Paper) + P(Paper and TV), but instead you summed P(Paper) + P(TV), which isn't the same quantity, because

P(Paper) = P(Paper but not TV) + P(Paper and TV)
P(TV) = P(TV but not Paper) + P(Paper and TV)

So you counted P(Paper and TV) twice, you see?

[hide]Well, actually, this is just a poorly written question.  It could just as easily mean 60 people saw it only in the paper, 50 only on TV, and 30 from both, which is what I thought it meant at first.  Then, you could just add them, giving you 140, and divide by 200, giving you $\boxed{\frac{7}{10}}$ for question a..., and then you would just need to do $\frac{60+50}{200}$ for question b, giving $\boxed{\frac{11}{20}}$.  So the problem isn't in people's logic, it's in the way the problem is written...[/hide]

Well... it does not explicitly say that 60 people only learned from paper and did not see it from TV.  Sincie it does not explicitly say that, you would have to assume it.  Usually when worded like that, you should approach with the venn diagram method... as annoying as it is, it helps.


Is it normal for the computer to take about 15 seconds to compile a 3 to 5 line document?

[quote="rrusczyk"]I've not had that happen...

If you're using \usepackage to include something huge, this might happen.  Or if you have simply way too many programs running on your machine.[/quote]
Or a very slow computer.

Also PDF and PS files tend to compile slower than DVI files.

I am using the tutorial in the LaTeX section of this site.

I am also compiling to PDF.

Sorry for asking so many questions (I started doing the tutorial today so I need some help to get started), but every time I turn on my computer a window shows up telling me to register for the Virtual Printer. What should I do?

Compiling to PDF and DVI take basically the same amount of time, 10 to 15 seconds. This is referring to:

\documentclass[11pt]{article}
\begin{document}
Hello, world!
\end{document}

which is only 4 lines long. How long does it take everyone else?

It took me 3 seconds. I don't know what could be slowing your computer that much except for, uh, your computer. But then again, I'm not much of a tech guy, so I wouldn't trust anything computery that I say!

Also, when I do DVI and press F5, I get a series of errors, which I have to keep clicking cancel or OK, before I get to the actual document.

What errors are they?

These all sound rather weird. I've never had any hiccups with my MiKTeX or TeXnicCenter.

Hmm... I got it to work on my other computer in about that time. This is sort of strange though because Mathematica also does not work on this computer. Apparently this computer abhors math : (


Find the inverse function of $ f(x) \equal{} \frac {1}{3}\ln{(x \plus{} \sqrt {x^2 \plus{} 1})}$.

[hide="Mehhh"]Let $ u=f^{-1}(x)$.

\begin{eqnarray*}
f(u)=x&=&\frac{1}{3}\ln u+\sqrt{u^2+1}\\
3x&=&\ln u+3\sqrt{u^2+1}\\
&=&\ln [u\cdot e^{3\sqrt{u^2+1}}]\\
e^{3x}&=& u\cdot e^{3\sqrt{u^2+1}}
\end{eqnarray*}

This looks really ugly. The inverse function of $ f(x)=xe^x$ is $ W(x)$, the Lambert W function, which cannot be expressed in elementary functions. That considered, it seems like there isn't a nice way to solve this for $ u$.[/hide]

I'll keep trying...

Sorry, it's my bad. I forgot the parenthesis.

Hm..

[url=http://en.wikipedia.org/wiki/Hyperbolic_function]Hyperbolic functions[/url]

With that in mind, we can see that the inverse of $ f(x) \equal{} \ln(x \plus{} \sqrt {x^2 \plus{} 1})$ is $ f^{ \minus{} 1}(x) \equal{} \sinh x$, so the inverse of $ \frac 13 f(x)$ is $ f^{ \minus{} 1}(3x) \equal{} \sinh 3x$.

Wow, missing parentheses  :|

[hide="Solution"]
Let $ u=f^{-1}(x)$.

\begin{eqnarray*}
e^{3x}&=&u+\sqrt{u^2+1}\\
e^{6x}-2ue^{3x}+u^2&=&u^2+1\\
e^{6x}-1&=&2ue^{3x}\\
\boxed{\frac{e^{3x}-e^{-3x}}{2}}&=&u
\end{eqnarray*}

Slightly cleaner result: $ \boxed{u=\sinh 3x}$ by definition.
[/hide]


Hey there,

I have the following equation and need to solve for X. Please help me out.

145 = 150(X) + 150(1-X)*0.6

Thanks alot!

Hint: Distribute and combine like terms.

This belongs in HSB or below.

Edit: Detnvvp -- Parentheses are usually used to denote multiplication.


Find all functions $ f: R\longrightarrow R$ such that
$ f(x\cdot y)\leq x^{2n}\cdot f(y)$
holds for all $ x,y\in R$ and $ n\in N^*.$

P.S.
$ N^*=\{n\in N|n\geq 1\}$

i'm sorry, but what does the notation usually $ \mathbb{N}^*$ mean?

:$ y \equal{} 0\Rightarrow f(0) \le x^{2n}.f(0) \Rightarrow f(0) \equal{} 0$
Suppose :$ x^{2n}.g(x) \equal{} f(x)\forall x \ne 0$
$ \rightarrow g(xy) \le g(y) \forall x,y \ne 0$
$ \Rightarrow g(x) \equal{} const$
$ \rightarrow f(x) \equal{} Cx^{2n} \forall x\in R$


geometry
circumcircle
cyclic quadrilateral
projective geometry
geometry proposed

Let a cyclic quadrilateral $ ABCD$ with circumcenter $ O$  and let $ G$ the point of intersection of the lines $ AC,\ BD$ , $ E$ the point of intersection of the lines $ AB,\ CD$ and $ F$ the point of intersection of the lines $ AD,\ BC$

Finally let $ N$ the point of intersection of the lines $ EG,\ OF$
Prove that the quadrilateral $ AGNB$ is cyclic .

Let $ w=(O,R)$ be the circumcircle of $ ABCD$

[color=blue][size=150]Lemma[/size]
Each pair of the points $ G,E,F$ defines the polar line of the third point, wrt the circle $ w$[/color]
(see [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=316994#p316994]here[/url])

Namely
$ GE$ is the polar line of $ F$
$ EF$ is the polar line of $ G$
$ FG$ is the polar line of $ E$
$ \hline$

$ EN$ is the polar line of $ F$ and $ N\in FO\Rightarrow N$ is the inverse of $ F$

Then $ ON \perp NE \Rightarrow N$ belongs to the circle $ t$ with diameter $ EO$

Let $ r^{2}$ be the power of the point $ E$ w.r.t. $ w$
We draw the circle $ e=(E,r)$. Let $ E_{1},E_{2}\in w \cap e$

The inversion wrt the circle $ e$ transforms the circle $ t$ to the line $ E_{1}E_{2}$,

Hence, $ N$ and $ G$ are inverse wrt $ e$, or else $ EN\cdot EG=r^{2}$
But $ EA\cdot EB = EC \cdot ED = r^{2}$ and this is enough to conclude that both of $ AGNB,DGNC$ are cyclic


calculus
integration
function
derivative
calculus computations

Let F(x)= (infinite sum) 2x/[(n+(x^2))^2]. First find the integral (sum from 0 to 1) F(x)dx. Then compute the exact value of F'(0).

I applied the M-Test to the function on the interval [0,1] and found (by equating the derivative to 0) that maximum occurs at x = sqrt(n/3). Inserting this into the function, i obtained the series 3 sqrt(3) (sum)  1/(n^(3/2)), which is convergent by p-test. So f_n is uniformly convergent on this interval and we can apply termwise integration. Then i used the transformation u = n + x^2 and found that  (sum from 0 to 1) F(x)dx = 1.

As for the second part, first, by using M-test i tried to prove(but couldn't finish) that f_n '(x) is convergent on [0,1]. If so, we can use term-by-term differentiation and get   F'(0) = Sum (2n/n^3) = ((Pi)^2)/3.

So how can we prove that f_n '(x) is convergent on [0,1]? Are the other parts of the solution true? Lastly, is there another efficient way other than M-test to prove uniform convergence on an interval? Thanks!

[quote]maximum occurs at x = sqrt(n/3).[/quote]
Which is irrelevant, because this point lies outside of $ [0,1]$ for all but finitely many values of $ n$. The M-test is appropriate, but there is no need to find the exact maximum of $ |f_n|$ on $ [0,1]$: a good upper estimate will do. For example, $ f_n(x) \equal{} \frac {2x}{(n \plus{} x^2)^2}\le \frac {2}{n^2}$ as long as $ x\in [0,1]$. And the series $ \frac {2}{n^2}$ is convergent.

Similarly, $ |f_n'(x)| \equal{} \frac {2|n \minus{} 3x^2|}{(n \plus{} x^2)^3}\le \frac {2n}{n^3} \equal{} \frac {2}{n^2}$ as long as $ x\in [0,1]$ and $ n\ge 3$ (when $ n\le 2$, this estimate may fail because $ n \minus{} 3x^2$ may be negative, but who cares?). Again, the series $ \frac {2}{n^2}$ converges and you can differentiate term-by-term.


[quote="chien than"]Let $a;b;c$ be real numbers.Prove that:
$a^{2}+b^{2}+c^{2}-ab-bc-ca \geq 3(a-b)(a-c)$[/quote]
Let a=10000,b=1,c=1 => the ineq is wrong  :)

with more conditions , $c \geq a \geq b$ and with a few changes , the RHS is $2(a-b)(a-c)$ .... so we can prove it !:D

[quote="vietnam2007"]In fact it was proposed in spain   1996
the correct form is :
$a^{2}+b^{2}+c^{2}-ab-bc-ca \geq 3(a-b)(b-c)$

which is proved simply by using $a^{2}+b^{2}\geq 2ab$ for every reals $a,b$[/quote]
Ineq <=>$(a-2b+c)^{2}\geq 0$
Done  :oops:

the inequality becomes
$\sum(a^{2}-ab-ac+bc) \geq 3(a-b)(b-c)$
$\sum(a-b)(a-c) \geq 3(a-b)(b-c)$
$(a-b)(a-c)+(c-a)(c-b) \geq 4(a-b)(b-c)$
$(a-c)(a-b-c+b) \geq 4(a-b)(b-c)$
$(a-b+b-c)^{2}\geq 4(a-b)(b-c)$
it's quivalent to $(x+y)^{2}\geq 4xy$ which is true for all reals number x;y :D

[hide]what about the problem with my conditions :D , i think it's nice,too :D[/hide]


A dog weighs 40 pounds plus $ \frac{4}{5}$ of its own weight. How many pounds does the dog weigh?

Let's say the dog's weight is $ \alpha$. We can write an equation that looks like this:

$ 40 \plus{} \frac{4}{5}\alpha \equal{} \alpha$

We have to bring the $ \alpha$ terms together and leave the constant on the other side.

$ 40 \equal{} \alpha \minus{} \frac{4}{5}\alpha$

Remember, the coefficient of any variable by itself is 1, so this becomes:

$ 40 \equal{} \frac{1}{5}\alpha$

Divide both sides by $ \frac{1}{5}$.

$ 40 \div \frac{1}{5} \equal{} \alpha$

$ 40 \cdot 5 \equal{} \alpha$

$ \boxed{200} \equal{} \alpha$

That's one heavy dog!

Another way. 40 ponds is (1-4/5) of its weight so 40 is 1/5 of its weight. 400x5=200


geometry perimeter induction LaTeX trigonometry circumcircle inradius

Simply put guys (and gals?) im just posting this to try to make a master list of knowledge for this forum, so that people can look through and see what types of formulas exist.  I'm only taking AP AB calc at my school and have very little opportunity for other courses.  Post any formulas that you guys know that are beyond [i][b]standard[/b][/i] high school curriculum, not to mention an example or proof if you can and what the formula does.  Thanks guys.

Ill start off with Hero's Formula for finding the area of a triangle:
Area = sqrt(s(s-a)(s-b)(s-c)).

S = semi perimeter of the triangle
a, b, and c are the lengths of the sides of the triangle.

The sum of a series $1+2+3+\dots+n=\frac{n(n+1)}{2}$.

[quote="tim1234133"]Continuing the apparent theme:

$1^{3}+2^{3}+\dots+n^{3}=\frac14n^{2}(n+1)^{2}$.[/quote]
I think this is better known as and slightly easier to use as $1^{3}+2^{3}+3^{3}+...+n^{3}=(1+2+3+...+n)^{2}$

[quote="nebula42"][quote="anirudh"]do we have to show proofs. it takes a long time to use indduction[/quote]
Induction is not the only way to prove formulas.  :wink:  For example, some of the "Proofs Without Words" (on this site!) are very good.  :D
Also, it would be nice to try to give something along the lines of an explanation (as long as it is elementary enough for High School Basics though).[/quote]
I'd like to see you prove heron's with induction ;).

[quote="anirudh"]Here is a proof of heron's (the one i know)

[hide]
\begin{eqnarray*}\text{Area}&=&\frac{1}{2}(\mbox{base}) (\mbox{altitude})\\ &=& \frac{1}{2}ab\sin(C)\\ &=& \frac{1}{4}\sqrt{4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}\\ &=& \frac{1}{4}\sqrt{(2a b-(a^{2}+b^{2}-c^{2}))(2a b+(a^{2}+b^{2}-c^{2}))}\\ &=& \frac{1}{4}\sqrt{(c^{2}-(a-b)^{2})((a+b)^{2}-c^{2})}\\ &=& \frac{1}{4}\sqrt{(c-(a-b))((c+(a-b))((a+b)-c))((a+b)+c)}\\ &=& \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}. \end{eqnarray*}

P.S. It took forever to type this up in latex :P
[/hide][/quote]
I never like deriving things with trig. This proof is straight from AoPS vol. 2 yes.
I derived it once without trig. It turns out that my method is very similar to that in the intro to geometry...

[u]Some useful geometry (in no apparent order)[/u]:

[b]Power of a Point Theorem[/b]:
From any point $P$, draw two lines that intersect a given circle at $A$ and $B$, and $C$ and $D$, respectively. ($P \in \ \stackrel{\rightarrow}{AB}\text{ and }P \in \ \stackrel{\rightarrow}{CD}$). Then, $PA \cdot PB =PC \cdot PD$.

         --This can be interpreted (depending on where the point is in relation to the circle) as four different theorems combined into one: the Chord-Chord, Secant-Secant, Tangent-Secant, and Tangent-Secant Theorems. For the theorems involving tangents, the two points of one (or both) of the lines are the same point (the tangency point).

[b]Extended Law of Sines[/b]:
In a triangle $ABC$ with side lengths $a, \ b, \c$ and circumradius $R$, we have:

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}= 2R$.

[b]Circumradius and Inradius Formulas[/b]:
For a triangle $ABC$, with circumradius $R$ and inradius $r$, we have:

$R=\frac{abc}{4 \triangle}\ \ \text{ and }\ \ r=\frac{\triangle}{s}$
where $\triangle$ is the area of triangle $ABC$, and $s=\frac{a+b+c}{2}$ is the semiperimeter.

[b]Ptolemy's Theorem[/b]:

In a cyclic quadrilateral (one that can be inscribed within a circle) $ABCD$, we have:

$AB \cdot CD+BC \cdot AD = AC \cdot BD$

     --In other words, if you take the two pairs of opposite sides, multiply their lengths, and add the two products it will equal the product of the diagonals' lengths.

[b]Brahmagupta's Formula (also known as Bretschneider's Formula)[/b]:

In a cyclic quadrilateral $ABCD$ with side lengths $a, \ b, \ c, \ d$, we have:

$\text{ Area }=\sqrt{(s-a)(s-b)(s-c)(s-d)}$,

where $s=\frac{a+b+c+d}{2}$ is the semiperimeter.

     ---Actually, Hero's Formula can be taken directly from this: just have $d=0$, and a triangle is cyclic (it has to be, as three non-collinear points determine a circle), so from this we get Hero's $A=\sqrt{s(s-a)(s-b)(s-c)}$.

I'll post more later on another area of the mathematics that I know well enough-- I just figured that since this thread started with geometry, I might as well continue on that theme.

Um... D.P.L., making a spammy reply to an admittedly rather spammy post is hardly much better than making an initial spam post.

Some Number Theory:

$a-b|a^{n}-b^{n}$ for all positive integers $a,b,n$
$a+b|a^{n}+b^{n}$ for all positive integers $a,b,n$ with $n$ [b]odd[/b].
$a^{p}\equiv a \mod p$ for all primes $p$ and integers $a$. (Fermat's Little theorem)
If $m$ and $n$ are coprime integers and $a$ and $b$ integers, then the system of equations $x \equiv a \mod m$ and $x \equiv b \mod n$ has one solution $x \mod mn$. (Chinese Remainder Theorem)

[quote="D.P.L"][quote="neelnal"]1. a=a
2. if a=b, then b=a
3. if a=b and c=d, a+c = b +d.
4. a x 1 = a
5. a + 0 = a[/quote]

 :spam: $\Rightarrow$ :mad: $\Rightarrow$  :furious:[/quote]

Those are...axioms... :roll: :mad:

Law of Sines: $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

Law of Cosines: $c^{2}=a^{2}+b^{2}-2ab \cos c$

Oops... these have already been posted. Hmmm...

[quote="mathnerd314"]Law of Sines: $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$[/quote]
More helpfully: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$
EDIT: Sorry, already posted.

$A=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^{2}\left(\frac{A+B}{2}\right)}$ for all quadrilaterals.

${n \choose 1}+{n\choose2}+{n\choose3}+\dots+{n\choose n}=2^{n}$

[quote="nutz_for2.718281828"]Log laws:

$\log_{a}b+\log_{a}c=\log_{a}bc$

$(\log_{a}b)(\log_{c}d)=(\log_{a}d)(\log_{c}b)$

$\frac{(\log_{a}c)}{(\log_{a}b)}=\log_{b}c$

$n\log_{a}b=\log_{a}b^{n}$

$\log_{a^{n}}b^{n}=\log_{a}b$

$\log_{a}b-\log_{a}c=\log_{a}\frac{b}{c}$

This law can be used to find logs on your calculator. Just plug in 10 for a.

$\frac{(\log_{a}c)}{(\log_{a}b)}=\log_{b}c$[/quote]
The second one is called the lazy jim property, (or just lazy jim). It is a generalization of the chain rule of logs.
Here is a proof of it.
It is somewhat obvious if you think about it.
$(\log_{a}b)(\log_{c}d)=$
$(\frac{\log_{x}b}{\log_{x}a})(\frac{\log_{x}d}{\log_{x}c})=$
$(\frac{\log_{x}d}{\log_{x}a})(\frac{\log_{x}b}{\log_{x}c})=$
$(\log_{a}d)(\log_{c}b)$
P.S. It doesn't matter what x is. X is any nonzero number.


Prove that   $n*(p+1)^{n+m} - (n+m)*(p+1)^{n} + m$ is divisible by $p^2$  for all integers $ p > 0 , n \ge 0 , m \ge -n .$
 :cool:

standard proof using binomial expansion
1> for m+n=0 the expression reduces to 0
2>for m+n=1 the expression reduces to

S= n(p+1)-(p+1)^n - n +1= np+1 - (p+1)^n

for n=0,1 the expression is zero ,for n>1
S= (np+1) - ( p^n  +  nC1 * p^(n-1)    + ........... + nC1 *p + 1)

so S/p^2  gives remainder = (np+1) - (nC1 *p +1)= 0
3> for   n+m>1

S=  n[ p^(n+m) +........+(n+m)C1 * p +1] - (n+m)[ p^n +....+nC1 *p+1]+m
so remainder = n[ (n+m)C1 * p +1] - (n+m)[ nC1 *p+1]+m
=n(n+m)p +n -n(n+m)p -n -m +m =0
proved


trigonometry
complex numbers
algebra proposed
algebra

Another problem:

Find all interpretations of the question above.

[hide="This?"]$1^{\pi}=e^{i2n\pi^2}=\cos(2n\pi^2)+i\sin(2n\pi^2)$, and there's infinitely many of those, since $\pi$ is irrational, and so are the periods of the trig functions.

Actually, what's the locus of all such points? It's obviously a subset of the circle $|z|=1$, but what points are missing?[/hide]

Farenhajt: And what about quarternions, then ? ;) Why do you restrict yourself to complex numbers ?
Anyway I don't get the point of the problem by seamusoboyle.


but, how to prove the l'hopital rule if: $ x\rightarrow\infty$

if:$ \left(\underset{x\rightarrow\infty}{\lim}f\left( x\right) \equal{}\underset{x\rightarrow\infty}{\lim}g\left( x\right) \equal{}\infty\right)\wedge\left(\underset{x\rightarrow\infty}{\lim}\frac{f^{\prime}\left( x\right) }{g^{\prime}\left( x\right) }\equal{}\infty\right)$,then:$ \left(\underset{x\rightarrow\infty}{\lim}\frac{f\left( x\right) }{g\left( x\right) }\equal{}\underset{x\rightarrow\infty}{\lim}\frac{f^{\prime}\left( x\right) }{g^{\prime}\left( x\right) }\right)$

ps. i can prove  this ,if:$ \left(\underset{x\rightarrow a}{\lim}f\left( x\right) \equal{}\underset{x\rightarrow a}{\lim}g\left( x\right) \equal{} 0\right)\wedge\left(\underset{x\rightarrow a}{\lim}\frac{f^{\prime}\left( x\right) }{g^{\prime }\left( x\right) }\equal{} c\right)$,then:$ \left(\underset{x\rightarrow a}{\lim }\frac{f\left( x\right) }{g\left( x\right) }\equal{}\underset{x\rightarrow a}{\lim}\frac{f^{\prime}\left( x\right) }{g^{\prime}\left( x\right) }\right)$

One version of the argument: substitute $ t\equal{}\frac{1}x$, and see what the limits look like in terms of $ F(t)\equal{}f(\frac{1}t)$ and $ G(t)\equal{}g(\frac{1}t)$.

simple proof:

$ if: \underset{x\rightarrow\infty}{\lim}f\left(  x\right)  \equal{}\underset
{x\rightarrow\infty}{\lim}g\left(  x\right)  \equal{}\infty,\underset{x\rightarrow
\infty}{\lim}\frac{f^{\prime}\left(  x\right)  }{g^{\prime}\left(  x\right)
}\equal{}\infty$

use: $ \underset{x\rightarrow\infty}{\lim}f\left(  x\right)  \equal{}\underset
{x\rightarrow\infty}{\lim}g\left(  x\right)  \equal{}\infty,\underset{x\rightarrow
\infty}{\lim}\frac{f^{\prime}\left(  x\right)  }{g^{\prime}\left(  x\right)
}\equal{}l,then: \underset{x\rightarrow\infty}{\lim}\frac{f\left(  x\right)
}{g\left(  x\right)  }\equal{}l$

$ \Longrightarrow \underset{x\rightarrow \infty }{\lim }\frac{g\left(
x\right) }{f\left( x\right) }\equal{}0$

$ \Longrightarrow \underset{x\rightarrow \infty }{\lim }\left\vert
\frac{f\left( x\right) }{g\left( x\right) }\right\vert \equal{}\infty$

$ \Longrightarrow \underset{x\rightarrow \infty }{\lim }\frac{f\left(
x\right) }{g\left( x\right) }\equal{}\infty$


[quote="ifai"]Do there exist functions $f$ and $g$ with either of the following properties?
\[f(x)+g(y)=xy,\ \forall x,y\in\mathbb{R}\]

\[f(x)g(y)=x+y,\ \forall x,y\in\mathbb{R}\]
[/quote][hide]Add the two equations: \[f(x)+g(y)+f(x)g(y)=x+y+xy.\] Add one to both sides and factor: \[[f(x)+1][g(y)+1]=(x+1)(y+1).\] Set $y$ constant. Then \[{f(x)+1=\frac{(x+1)(y+1)}{g(y)+1}}=C(x+1).\] Hence, \[g(y)+1=\frac{y+1}{C}.\] Then \[f(x)+g(y)=C(x+1)+\frac{y+1}{C}-2=xy.\] Take $y=-1$ to get \[C(x+1)=-x+2,\] which has no solution in $C$. Thus, there are no such functions.[/hide]

Nice proof, what leads to that thinking? Is there a clear way that this contradiction idea forms by examining the function? It seems quite tough to narrow in on such a specific conclusion!
Thanks in advance for your response

Whenever you see $x+y$ and $xy$ in the same problem, a prime candidate to consider is the factorization of the form $(x+1)(y+1)$.

Yes, up to that point it is the same as I had done (I tried the problem before looking at answer)
it is just the setting constant and "estimating" kind of method, it loses a lot of generality and most of the time too much (the solution reminds me of randomly zooming into a picture while trying to focus on something tiny, and randomly finding it in view)

We can simplify the solution significantly:

From the first equation, putting $y=1$, we get $f(x)=x-g(1)$. Then the second equation yields $[x-g(1)]g(1)=x+1$ (again by putting $y=1$). From there $g(1)x-g(1)^{2}=x+1\iff g(1)=1\land g(1)^{2}=-1$, which is clearly impossible.

Or we can just analyze the first equation. For $y=1$ we get $f(x)=x-g(1)\neq\text{const.}$, and for $y=0$ we get $f(x)=-g(0)=\text{const.}$. Obviously, a function can't be non-constant and constant at the same time, hence no such functions exist.


principal value of trignometry lies between 0 to 2pie but principal value of complex lies between -pie to pie why.

i am talking about two type of solution one is principal solution and other is general solution .in books it is wrriten that principal solution lies between 0 to 2 pie  for trig.=ns.and for complex polar form it is  -pie to pie

why

according to conventions,
measurement of complex angles starts from -ve x axis (3rd quadrant)

why it is so i can't answer as this is convention

[url]http://en.wikipedia.org/wiki/Complex_number[/url]

its even better
[url]http://mathworld.wolfram.com/ComplexNumber.html[/url]

u'll read abt it very quickly

mathworld.wolfram is the best site 4 maths -it gives elementary info on all topics and quickly(infact that is the 1 which generated my interest n MAths)

but for detailed explanation try"sparknotes"--its the US curriculum
don't remember the exact link.try searching in google


Hi everyone, I am a new member to this site, and I would just like to introduce myself. I have just discovered an interest in math, and would like to eventually learn as much as many of you on here. I would like to at least get to the level of AMC 12, and eventually attempt some IMO problems. I am currently a college student, 2nd year, and majoring in Chemistry. My goal is to compete in Putnam competitions, after I have some more experience under my belt of course. I wish I had learned about this site, when I was 13, maybe I would have appreciated math a lot more.

I have a couple of questions:

What math classes did some of the IMO contestants take? I heard of Arnav Tripathy taking graduate math classes while in high school, is this generally the norm?

How can I reach my goal of competing in Putnam? At this point in time, I am finishing up multivariable calculus. I am not a math major, so I will probably have to teach myself the more upper division courses.

What textbooks should I read or what classes should prepare me for this tests?

Thank you!Hi everyone, I am a new member to this site, and I would just like to introduce myself. I have just discovered an interest in math, and would like to eventually learn as much as many of you on here. I would like to at least get to the level of AMC 12, and eventually attempt some IMO problems. I am currently a college student, 2nd year, and majoring in Chemistry. My goal is to compete in Putnam competitions, after I have some more experience under my belt of course. I wish I had learned about this site, when I was 13, maybe I would have appreciated math a lot more.

I have a couple of questions:

What math classes did some of the IMO contestants take? I heard of Arnav Tripathy taking graduate math classes while in high school, is this generally the norm?

How can I reach my goal of competing in Putnam? At this point in time, I am finishing up multivariable calculus. I am not a math major, so I will probably have to teach myself the more upper division courses.

What textbooks should I read or what classes should prepare me for this tests?

Thank you!

The Putnam exam is open to everyone, and I don't see any particular reason to wait to take it -- the experience will be useful for when you take it again later with more math classes under your belt.  Unfortunately, you've missed the signup date for this year's exam.  I suggest you contact someone in the undergraduate math office (or whatever it's called at your university) and ask which professor is in charge of administering it, then contact that professor and ask if you can take it unofficially.  (The exam is this coming Saturday.)


We can all agree that nothing is better than math.  Now we also know that a ham sandwich is better than nothing.  Thus, a ham sandwich is better than math.

  ...the results speak for itself...

[quote="G-UNIT"]We can all agree that nothing is better than math. [/quote]

That is a false statement.

What in the world is better than math?  :P

  Its funny how two different interpretation of "nothing" can have such a drastic effect on a statement. :D

[quote="MithsApprentice"]No offense, but a lot of stuff is. *points to cable television*[/quote]and would you have a cable television without math? ;)

[quote="Peter VDD"]and would you have a cable television without math? ;)[/quote]

Doesn't matter. It's still better. :P

and if math>math, then math/math>math/math, so 1>1  :D

[quote="white_horse_king88"]Cable television is the result of cleverly applied mathematics. Thus, math is better than itself in almost every form...

I think that's the conclusion here...[/quote]
nope I disagree so you say that what is original is much better than upgraded???? according to u an old car is much better than a new one?

[quote="236factorial"][quote="MithsApprentice"][quote="G-UNIT"]What in the world is better than math?  :P[/quote]

No offense, but a lot of stuff is. *points to cable television*[/quote]

I hate television.  :P

Nothing can beat math forums!  :D[/quote]

Not even mithapprentice's avatars?Wooot,no offence. :)

But if it's self-contradicting, then it's not very logical, so logic is safe, and we can eat our ham / bologna / nothing sandwiches in peace.

[quote="G-UNIT"][quote="nr1337"]There exists a person, place, thing, or idea named "something" such that nothing is better than something.  However a BOLAGNA sandwich is better than nothing, so a bolagna sandwich is better than something.  However, nothing is better than something, so logic contradicts itself QED[/quote]

  But your proof fails when we use Lewis Carol's definition that "nothing" is "something". ( Im doing "nothing", thus I'm doing "something").  :D[/quote]

That's the point :D

How do we know that there exists something that nothing is better than? What if there is an infinite sequence of things, each better than the last? :?

There are some logical fallacies in your arguement

First off your reasoning is incorrect although you were making a joke(sorry this is what i get for taking debate and learning logical arguements)

You used to different definitions of nothing

According to this line of reasoning i could also claim that(i am sorry if this offenses anyone but i am not christian either)

God is love
love is blind
ray charles is blind
therefore
God is ray charles

There are two different interpretations of blind and two different interpretations of love

There is also a problem because the circles dont match up

Also another flaw (irrelevant to g unit) is that the converse is not true where blind is not ray charles so they would not really cancel out you would end up with

God is love
love is blind
~blind is ray charles
so you would not reach the right conclusion

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

[quote="G-UNIT"]Its funny how two different interpretation of "nothing" can have such a drastic effect on a statement.  :D  [/quote]


Maria paid $ \$600$ for her computer.  One day she sold it to one of her friends for $ \$700$.  A month later she bought back her computer from her friend for $ \$800$.  A week later she sold it fro $ \$900$.  What was Maria's total profit?

She made $ \$100$ the first time she sold it, and she made $ \$100$ the second time too. Thus her total profit was $ \boxed{\$200}$.


If the probability of it snowing tomorrow is 99%, and the probability of me owning the competition if it occurs is 99.9999%, then what is the probility that I will not own?

BTW= Own means to beat badly.

[hide]
$1 - .99*.999999$, whatever that is[/hide]

The probability that you do "own" the competition is the probability that it does not snow and you "own" the competition.

$(1-0.99)\cdot0.999999$
$0.01\cdot0.999999$
$0.00999999$

The probability you do NOT "own" the competition is $1-\text{P(You win the competition)}$

$1-0.00999999=0.99000001$

Basically, because it might snow, you have almost no chance of "owning" the competition.


Let $ n$ be a positive integer. Prove that there exists an integer $ N$ such that for any prime $ p$ satisfying $ p>N$, there exist positive integers $ x$, $ y$, $ z$ satisfying $ x\not\equiv 0\mod p$, $ y\not\equiv 0\mod p$, $ z\not\equiv 0\mod p$ and $ x^n \plus{} y^n\equiv z^n\mod p$.

a more exact form of the problem is:

let $ p$  be a prime number and $ n$ a natural number such that  $ p > 3n!$. prove that: the equation
$ x^n \plus{} y^n \equiv z^n$ $ (\mod p)$
has roots $ x$, $ y$, $ z$ such that   $ p \not | xyz$

We can replace 3 by e=2.718281828....

Use this theorem :

Let {1,2,3,....,n} be partitioned into k subsets A(1),A(2),....,A(k) . Prove that if n>= e * k! then there exist an integer

i such that  1<= i <= n and  there are x,y,z  (not necessary distinct) such that x-y=z and x,y,z are members of A(i).

_________
Now suppose an integer number g such that  g^1 , g^2 , ... , g^(p-1) be different mod p and continue ...! :D

This result is not due to Dickson, but to Schur. And, it is probably the first one in Ramsey theory, even before Ramsey theorem itself...
I will post a solution later, if nobody do it.

Pierre.


It is known that $ a_1 \equal{} 2004$ and $ a_{n\plus{}1} \equal{}\Bigg[\frac{a_n \plus{} [\frac{2004}{a_n}]}{2}\Bigg], 1\leq n\leq 2003$. Find the minimum of $ a_1, a_2, . . . , a_{2004}$.

min([(a+[2004/a])/2]) => a+[2004/a] is min
Let S=a+[2004/a] => S+1 >= a+2004/a
By AM-GM => S+1 >= 2(sqrt(2004))=2(44.76) => S > 88

The minimum value = [S/2]=[89/2]=44
I checked this is a part of sequence, but still need to prove it. :(

[hide="My idea"]Prove that $ a_{n \plus{} 1}\leq a_n$ and when does equal occur?[/hide]

My solution:
If $ a_n\geq 45$ then $ \frac {2004}{a_n} < a_n\Rightarrow \Big [\frac {2004}{a_n}\Big ] < a_n\Rightarrow \frac {a_n \plus{} [\frac {2004}{a_n}]}{2} < a_n\Rightarrow a_{n \plus{} 1} < a_n$
Then exits $ a_k \equal{} 44$


function
integration
real analysis
calculus
derivative
topology
real analysis unsolved

Suppose $F$ is absolutely continuous on $[a,b]$ and nondecreasing. Let $A = F(a)$ and $B = F(b)$.

(a) Prove that there exists such an $F$ that is in addition strictly increasing, but such that $F'(x) = 0$ on a set of positive measure.

(b) The $F$ in (a) can be chosen so that there is a measurable subset $E \subset [a,b]$, $m(E) = 0$, so that $F^{-1}(E)$ is not measurable.

(c) Prove, however, that for any increasing absolutely continuous function $F$, and $E$ a measurable subset of $[A, B]$, the set $F^{-1}(E) \cup \{F'(x) > 0\}$ is measurable.

[Hint: For (c), prove first that $m(O) = \int_{F^{-1}(O)}F'(x)dx$ for any open set $O$.]

---------------------

(a) is a really cool concept, I think, and it isn't too hard to think of an example. Hint: consider generalized Cantor sets with positive measure.

(b) is an extension of a previous problem from this text, where we showed that the inverse image of a set of measure 0 can be nonmeasurable. This uses the Cantor-Lebesgue function.

(c), however, we have no idea on. I'm not even sure how to prove the hint, let alone how to use the hint on the problem. Any ideas?

No matter what I do, I can't see what's wrong with the following, which is meant to be a counterexample to (c):

Let $A$ be a dense open subset of $I=[a,b]$ whose complement has positive measure, and let $\chi_A$ be its characteristic function. We'll take $f(x)=\int_a^x\chi_A\ \mbox dm=m(A\cap[a,x])$. Then $f$ is clearly strictly increasing ("strictly" because $A$ has positive measure in every interval) and absolutely continuous, being the integral of an $L^1$ function. Moreover, its derivative is equal to $\chi_A$ almost everywhere (it's a fact that absolutely continuous functions are almost everywhere differentiable, which you must have assumed since you're talking about the derivative of $f$, with $L^1$ derivatives, and they are equal to the integrals of their derivatives, up to an addition of a constant). Notice that this also provides an example for (a).

$f$ takes $I$ onto an interval $J$ of measure $m(A)$, and it takes $A$ onto a set of the same measure (it takes each interval contained in $A$ onto a set of the same measure as the interval, and then we just add the countably many disjoint intervals which make up $A$). This means that it takes the complement of $A$ onto a set of measure $0$.

Since they only differ in a set of measure zero, we can replace that $\{f'(x)>0\}$ with $A$, i.e. what we want to prove becomes: $f^{-1}(E)\cup A$ is measurable for every measurable $E\subset J$. Let $F$ be a non-measurable subset of $I\setminus A$ (there are such sets, because $I\setminus A$ has positive measure), and let $E=f(F)$. $E$ is, of course, measurable simply because it has measure zero. Since $f$ is a bijection, we have $f^{-1}(E)=F$, and since $A$ is measurable, $F$ is not, and they are disjoint, it's [i]not[/i] true that $f^{-1}(E)\cup A$ is measurable.

Now, what am I doing wrong? :?

P.S.

It's also clear from the above how to get an example for (b) (in fact, it's constructed there: that $E=f(F)$ thingie).

I really have no idea why. We spent something like two hours discussing this the other day, and we couldn't figure anything out. I suppose it is possible that Stein was wrong, or that there's some sort of typo in the problem...

It's my typo. The union in part (c) should be an intersection. I think we have a solution, also, but it's somewhat lengthy. I'll post it in a few days, if I remember -- I have to get through some end-of-semester stuff first.

Ok, thanks for clarifying.

I guess this just leaves (c) (see the P.S. added to my first message). Since $f^{-1}(E)\cap\{f'>0\}=\bigcup_{n\ge 1}\left(f^{-1}(E)\cap\left\{f'>\frac 1n\right\}\right)$, it suffices to prove that given some $t>0$, the set $f^{-1}(E)\cap\{f'>t\}$ is measurable whenever $E\subset f(I)$ is measurable.

$E$ is a borel set $B$ plus a set of measure zero $E\setminus B$. $f^{-1}(B)$ is Borel and hence measurable, and so is $\{f'>t\}$, so it suffices to consider the case when the measure of $E$ is zero. In this situation, we can show that the measure of $M=f^{-1}(E)\cap\{f'>t\}$ is actually zero, and the conclusion will follow.

Let $\varepsilon>0$, and let $D$ be an open set containing $E$ of measure $<t\varepsilon$. Then $f^{-1}(D)$ will be an open subset of $I$ containing $f^{-1}(E)$, which means that $f^{-1}(D)\cap\{f'>t\}$ contains $M$. For each subinterval $(u,v)$ of $f^{-1}(D)$ we have $m(f((u,v)))=f(v)-f(u)=\int_u^vf'\ \mbox dm$, and if we sum this after all (countably many) maximal subintervals $(u,v)$ contained in $f^{-1}(D)$, we get something $<t\varepsilon$. This means that the measure of $f^{-1}(D)\cap\{f'>t\}$ is $<\varepsilon$, which, in turn, implies that the external measure of $M$ is $<\varepsilon$. Since $\varepsilon>0$ was arbitrary, we’re done.

That's pretty much the same approach we used, except we went from the other side. We proved Stein's hint for open sets, and then extended that to Borel sets, and said that since $F'(x) > 0$ on the set we're considering, if $m(B) > 0$, then necessarily $m(F(B)) > 0$. Or maybe that should be $F^{-1}$, I can't remember right now.

Anyway, thanks -- I think your approach is more direct, as it seemed like at every opportunity we turned around and said "let's prove the contrapositive!" and in the end the argument was a mess to straighten out.


logarithms
algorithm
function
complex analysis
number theory
puzzles

Can anyone help me with this problem?

(x2-5x+5)x2-9x+20=1

I think you're supposed to start with logarithms, but I'm not sure. I'd appreciate any assistance.

Well, if take the logarithm on both sides, you get :
$(x^2-9x+20)ln(x^2-5x+5)=ln(1)$
But, ln(1) = 0, so :
$(x^2-9x+20)ln(x^2-5x+5)=0$

Which means :
$x^2 - 9x + 20 = 0$ or $ln(x^2-5x+5) = 0$
I guess you can solve the first one pretty easily, to get : x = 4 or x = 5
And, the second one : x = 1 or x = 4
So..

;)

(I hope I didn't make any silly mistake)

No, it's far easier than that. We have that the LHS is equal to 1 if the exponent is 0 or if the base is 1 or -1, the latter only if the exponent is even. $x^2-9x+20=0$ for $x=4,5$, $x^2-5x+5=1\Rightarrow x^2-5x+4=0$ for $x=1,4$, and $x^2-5x+5=-1\Rightarrow x^2-5x+6=0$ for $x=2,3$. Testing the exponent for these values, we see that it comes out even in both cases, so both work. We have $4$ twice, so we can discount that, and all possible values are $x={1,2,3,4,5}$.

Wouldnt this be better in the Intermediate Topics forum rather than the Puzzles???

:lol: Thanks, JesusFreak. The only answers I got were 1, 4, and 5. I would have never figured out that -12y rule.

[quote="pleaseread"]use a TI-82 or Ti-82 (or plus).  If you type in the left hand side
into y1 and "1" into y2 and then use the "intersect" feature, I do not see how you can come up with the x=2 and x=3 values.[/quote]
This is a ridiculous idea. Calculators can calculate, but they can't solve any problems. Your TI calculator does not give 2 and 3 because there are some values near the roots that are too big or small to be handled by the calculator(I tried the graphing and looked at the table). 2 and 3 do work.

1)  Yes, certain calculators CAN solve problems when they're filled with algorithms or such, but what does THAT have to do with here?!

2)  Don't bring up the use of the table.  If it can't be done by graphing with adjustments to the window and the use of the "intersect" mode, then that's a flaw that I was addressing.

3)  "2 and 3 do work."  I NEVER said they didn't.  Why state it?  Hmm?

4)   Now, you don't get on here and misquote me or enter your own statements again to put yourself in a diferent light.  STICK to the topic I presented in MY post at hand.

His point was that there is no reason to use a calculator, as they will not always properly calculate things. For example, my TI-89 won't calculate some infinite sums that do converge, it won't factor some equations, etc. And if you didn't say that 2 and 3 don't work, then it sure sounded like it, and you should think more about how you say something before you actually say it, as that's the impression that I got, as well.

As for 2), there's no flaw in the problem, there's just a flaw in the calculator, as has already been stated. The problem was not meant to be solved with a calculator, ESPECIALLY not the way you were trying to solve it. The only time you might use a calculator here is to factor the equations, but they're simple enough that you don't even need to do that.

Also, you have no reason to get upset at him like you did. First of all, he did not misquote you. Second, he did stick to the topic that you presented, as far as I can see. If you think he didn't, then you need to actually clarify what you're addressing in your original post.

Is this from Queens College Math Competition written by John Kennedy?

Sorry, pleaseread. I didn't mean to offend you. My point was that posting a solution that uses calculator did not seem right. We solve problems to improve our problem-solving ability, so killing a problem with a calculator is pointless. And it doesn't always work anyway(like this problem).

[quote="pleaseread"]attention:

You don't pretend to know to speak for lightrhee; let it come from himself!  If it sounded as if "2" and "3" wouldn't work from what I said, then that's YOUR PROBLEM. Understand?  "The problem was not
meant to be solved with a calculator..."  You can only state that AFTER the fact.  An exponential curve and a constant curve fit well into the y1, y2, "intersect" mode for solution IN GENERAL.  This one does not fit that.  Yes, he did misquote me.  No, he didn't stick to the specific topic I was talking about.

JesusFreak197, stay out of and off from my posts unless they're direct with you.  DO NOT speak for other posters who had issues against me.[/quote]

He did speak for himself already, and you didn't listen to him. And it's not just "my problem" if it sounded like that to everyone.

Wow, enough fighting already. It's amazing how something as simple as saying the problem can't be solved using a calculator can escalate into a conflict.

And wow, jesusfreak, you're only 15? That's insanity. I wish I was that smart when I was 15.

[quote="pleaseread"]attention:

You don't pretend to know to speak for lightrhee; let it come from himself!  If it sounded as if "2" and "3" wouldn't work from what I said, then that's YOUR PROBLEM. Understand?  "The problem was not
meant to be solved with a calculator..."  You can only state that AFTER the fact.  An exponential curve and a constant curve fit well into the y1, y2, "intersect" mode for solution IN GENERAL.  This one does not fit that.  Yes, he did misquote me.  No, he didn't stick to the specific topic I was talking about.

JesusFreak197, stay out of and off from my posts unless they're direct with you.  DO NOT speak for other posters who had issues against me.[/quote]

You have a pretty bad attitude for someone who's been posting here for less than 10 days.

First of all, your original statement was cryptic at best.  It's pretty clear that the ti-82 does not pick up the roots at 2,3 because $(x^2-5x+5)^{x^2-9x+20}$ is not well defined for 'many' values in the interval $[2,3]$ and hence a numerical method for finding roots there like bisection will inevitably fail.  If that was all you were trying to get across, then it was tangential at best and completely irrelevant at worst.  Why would it be particularly significant as to why an obsolete calculator failed to find the roots of a particularly pathological function?

Furthermore, even if your statement was both relevant and clear, misunderstandings inevitably happen, and are generally not best responded to by saying 'stay out of and off from my posts unless they're direct [sic] with you'.

blahblahblah (nondescript name),

I have a forceful attitude that your mind picks up as "bad" because
you are threatened and weak.

It doesn't matter whether I had posted on here for one day or 1000 days, I'm going to set you ignoramuses in your places.  Got it?
If not, hurry up and understand that fact.

That calculator in question is not obsolete; it is in use regularly.  It is the problem itself that is out of the mainstream.

That is THAT user's probem if they don't know how to deal with my telling them that; they should NOT be speaking for someone else.
blahblahblah, YOU should not pretend to be speaking for another user.

Wow, I had no idea that anyone could be so stupid. You register, immediately begin insulting those whom you don't like simply because they disagreed with you, make completely faulty arguments that at least three people have given good reasons to you why they shouldn't work in the first place, then you come back and call us all idiots, including a mod.

*watches pleaseread get banned*

[b]EDIT:[/b] Out of curiosity, why do you think you're so much better than everyone else here? :huh:

I have been called many things before, but 'threatened and weak' is a new one.  Threatened by what?  Your elite TI-82 skills, perhaps?  I admit that I called it obsolete just to prove a point (even though it's true - it was replaced by the TI-83, which was replaced by the TI-83+, which was replaced by the TI-84...) - even though infinitely more advanced software such as Maple 10 does not pick up those other roots either.  Nevertheless, I gave you a cogent explanation as to why the roots are not picked up.

However, on a more general note, unlike, say, the sci.math forums, the general thrust of this site is problem solving, and while it is perfectly acceptable to use calculators, it is generally frowned upon in instances like this, where an analytic solution is easy to arrive at.  Solving problems that are out of the mainstream is in fact the main focus here, and certainly there is little pedagogical value to be found in plugging something into a graphing calculator and relying blindly (or on the algorithmic skill of some Texas Instruments engineers) on the answer produced.

And I stand by what I said about you having a bad attitude.  You have lectured individuals on using the word easy in their solutions of problems.  While I agree that this can be overused, these forums are sorted to some degree by difficulty, and there is absolutely nothing wrong with saying something like 'this is a trivial application of Rouche's theorem' in a problem in the complex analysis forum, or 'this is easy using Euler's theorem' in the number theory section.  If you are offended by such language, then it's your prerogative to 'catch up', and in particular, it's not your place to lecture others who in fact have presented solutions on what language they should be using.

Moreover, your little manifesto on your mental stature and strength of will might be gratifying to you, but it really means nothing when no one has ever seen you solve a problem here :!:

I'm sorry for my post, blahblahblah is doing a much better job of expressing my P.O.V. than I am.

all you do is blahblahblah back to me because you won't listen to reasons I've spelled out!  Getting mighty tiresome!  Hmm?

Both of you must be one of the same.  (Odds are.)

JesusFreak197, that is YOUR insecurity and YOUR problem that YOU feel that I am better than everyone else here.  It is time that YOU
had people stop leading YOU around by the hand protecting YOUR
feelings and YOUR ego about competitive people.  Deal with it!
Hurry up and understand this fact!  I'm not going to be YOUR babysitter.

[quote="pleaseread"]all you do is blahblahblah back to me because you won't listen to reasons I've spelled out!  Getting mighty tiresome!  Hmm?[/quote]

He has already completely addressed any arguments you gave. You are the one who won't listen to him.

[quote="pleaseread"]JesusFreak197, that is YOUR insecurity and YOUR problem that YOU feel that I am better than everyone else here.[/quote]

No, you misunderstood. I don't think you're better than everyone else here; [b]you[/b] do.


ratio
geometry
parallelogram
geometry unsolved

One sides $BC$ , $CA$ ,$AB$ of triangle $ABC$ lie points $A_1$,$B_1$,$C_1$ such that the medians $A_{1}A_{2}$,$B_{1}B_{2}$,$C_{1}C_{2}$ of triangle $A_{1}B_{1}C_{1}$ are parallel to $AB$,$BC$,$CA$ respectively.Determine in what ratios the points $A_{1}B_{1}C_{1}$ divide the sides of $ABC$.

Solution(so you can draw a nice picture :) ) is
$\frac{BA_1}{A_1C}=\frac{CB_1}{B_1A}=\frac{AC_1}{C_1B}=\frac12$.

Let $M, P, N$ be midpoints of segments $AB_1,CA_1, BC_1$ in that
order. $A_2M||C_1A$ so $A_2M||AB$ so points $A_1, A_2, M$ are
collinear, similarly points $B_1, B_2, N$ and $C_1, C_2, P$ are
collinear. Let $T$ be the centroid of $\triangle A_1B_1C_1$ (in other words,
intersection point of $A_1A_2, B_1B_2, C_1C_2$), then
$NBA_1T$ is parallelogram so $NT=BA_1$. In other hand
$NT||BP$ and $N$ is a midpoint of $BC_1$ so $NT=\frac12 BP$
 so $BA_1=\frac12 BP$, from that $\frac{BA_1}{A_1C}=\frac12$.
For the others, proof is the same.


Kaleb defined a $ \emph{clever integer}$ as an even integer that is greater than 20, less than 120, and such that the sum of its digits is 9. What fraction of all clever integers is divisible by 27? Express your answer as a common fraction.

It is divisible by 9 and is even.
36, 54, 72, 90, 108 work.

There are 2 that are divisible by 27, so 2/5

\[ \frac {126 \minus{} 27}{9} \plus{} 1\Rightarrow12
\]

\[ \frac {success}{possible}\Rightarrow\frac {\frac {108 \minus{} 27}{27} \plus{} 1}{12}\Rightarrow\boxed{\frac {1}{3}}
\]

EDIT
Ignore mine and use AIME15's because I didn't read the EVEN part.


Does anyone know of any good website where I can find AMC 10 and 12/AHSME questions?

Thanx.

Thanx. Those old tests can be good practice (plus I'm going to start practicing for next year really early.)


calculus
derivative
linear algebra
matrix
algebra
polynomial
function

Let $M_{n\times n}$ be the space of complex $n\times n$ matrices and let $det: M_{n\times n}\to \mathbb C$ denote the determinant. Show that if $I$ is the indentity matrix then $D det(I)B=TrB$, ie... $\frac{d}{dt}det (I+tB)|_{t=0}=TrB$

Thanks~

Can you prove it for trigonal matrices ?

General case: $\det(I+tB)$ is a polynomial in $t$, and we are looking for its linear term. By definition $\det(I+tB)$ is the sum of $n!$ products with $n$ terms in each. If more than one entry in a product is off-diagonal, we don't get any linear terms. On the other hand, if $n-1$ entries are on the main diagonal, then all of them are. Thus, the linear term comes only from $\prod_{i=1}^{n}(1+tb_{ii})=1+t\sum_{i=1}^{n}b_{ii}+O(t^{2})$.

Following up on alekk's hint:

By Schur's Lemma, we can find unitary $U$ and upper triangular $T$ such that $B=U^{*}TU.$ It then follows that $I+tB=U^{*}(I+tT)U$ and $\det(I+tB)=\det(I+tT).$

If $T$ has diagonal elements $\lambda_{j},$ then

$\det(I+tT)=\prod_{j=1}^{n}(1+t\lambda_{j})$

$\frac{d}{dt}\det(I+tT)=\left[\prod_{j=1}^{n}(1+t\lambda_{j}) \right]\sum_{j=1}^{n}\frac{\lambda_{j}}{1+t\lambda_{j}}$

Evaluate that at $t=0$ to get $\sum_{j=1}^{n}\lambda_{j}= \text{Tr }T=\text{Tr }B.$

This is closely related to the way I would prove that $\det(e^{A})=e^{\text{Tr }A}$ for $A\in M_{n}(\mathbb{C}).$ Indeed, the result feels somehow related.

thanks Kent Merryfield and all. Your solution is very nice, and the left question is not difficult to me :). My teacher said he has a sol by using some calculating. I will wait him tomorow. :wink:

Here's a very simple proof. Let $(x_{k})_{1 \le k \le n}$ be the columns of $B$, then $f(x_{1},\ldots,x_{n}) = \det B$ is $n-$linear functional on $\mathbf R^{n}$. Our columns are functions of $t$: $x_{k}(t)=e_{k}+tx_{k}$ where $e_{k}$ is vector of which all the coordinates equal to 0 except for $k-$th one which equals to $1$. We have to evaluate $\frac d{dt}f(x_{1}(t),\ldots,x_{n}(t))$ at the origin.

It equals to
\[\sum_{k=1}^{n}f(x_{1}(0),\ldots,x_{k}'(0),\ldots,x_{n}(0)) = \sum_{k=1}^{n}f(e_{1},\ldots,x_{k},\ldots,e_{n}) \]

\[= \sum_{k=1}^{n}x_{k}^{k}= \mathrm{Tr}\,B. \]

A language note: the English for "colon" is "column".
The proof is fine of course.


For those of you in high school, how has homecoming been? were you guys involved in any skits/airbands? :) who won, what were the themes, etc.

What are you talking about?  All we ever had for homecoming was a "spirit week" (each day there was a theme around which you were supposed to dress), coronation, the football game, and the dance.

I'm not much for school dances, our homecoming theme was "Hollywood" that's all I know.
Also our football team hasn't done well at all these past few years; this season didn't start off well, they always beat us 35ish to 0, but on the homecoming game that was very much reversed. Somewhere around 50-0.
Which didn't make the game any more exciting.

hm...do they have homecoming for freshmen?

i have no idea what the "cool" people do at my school...

[quote="furious"]our homecoming bonfire and parade were canceled due to rain.[/quote]

That sucks.  :(

I guess Homecoming  is different for every school then. at our school each class has a color (frosh-blue, soph-yellow, junior-black, senior- red) and we have a homecoming week. Each day one class puts on a skit that promotes school unity and there are airbands and a lot of cheering and screaming.

Awww...I wish our schools did homecoming, sounds like a blast.  *sigh*  Canadians!!!

My high school didn't even have a football team, so I was totally clueless when I heard the homecoming game is tomorrow. (I'm at Brown now.)
We weren't that big on school spirit.

Well, my school has 4000 students and football is extremely important (in fact, our football stadium costed 80 million dollars) so homecoming is quiet big. But I don't particularly like football, so I just went to the dance the day after. Traditionally, homecoming is the first home football game played.

Football is where it's at at my school. We have one of the best teams in the nation, so that could explain why... HC game is this coming Friday, the dance the day after. It's a pretty big deal, anyway, all the talk in the last month has been pretty much about HC. I get to work the concession stand at the game for Math Club (tons of money, yay!) and I'm going to the dance, too. Fun stuff.


let $ABCD$ be a square.$F$ and $E$ are on $BC$ and $AB$ in which,$AB=2AE$ , $CF=2BC$.

fine the :$\angle EDF$

[hide]$\angle EDF={\pi\over 2}-2\arctan{1\over 2}={\pi\over 2}-\arctan{{2\cdot{1\over 2}}\over{1-\left({1\over 2}\right)^2}}={\pi\over 2}-\arctan{4\over 3}=\arctan{3\over 4}$[/hide]

do you mean ${F}$ is on ${BC}$ and ${E}$ is on ${AB}$?

if so

how can ${CF} = 2{BC}$?

do you mean that $2{CF} = {BC}$

if so the answer is - $\angle EDF = 30^{\circ}$


A group of students decided to buy a radio from Rs170 to Rs195 but at the last moment,2 students backed out of the decisino so that thr remaining students had to pay Re 1 more than they planned.Find price and no. of students.

[quote="chess64"]I don't think Americans will be familiar with that form of currency--or for that matter, even know that it [i]is[/i] currency :P[/quote]

[i]ohh[/i]... that was [i]currency[/i].  :D I thought that that was like a model of radio, (like TI83 calc.).  :oops:

[hide="hint1"]
 Number Of Students have to be a whole no.
[hide="hint2"]
19
[hide="hint3"]
Roots
[/hide]
[/hide]
[/hide]


Suppose a,b,c be side of triangle such $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}= 3$ .Find the minimum of: P= $\sum_{cyc}{\sqrt{a+b-c}}$
I thing your slon in you book not true!!!! Can you soln it for me Flove

Here is solution in Flove'book or hungkhtn'book :
       Applying inq $\sum_{cyc}\frac{1}{a^{2}+b^{2}}\geq \frac{10}{(a+b+c)^{2}}$ .(*)
then if a,b,c >0 such:$\sum_{cyc}\frac{1}{a^{2}+b^{2}}=1$so $a+b+c \geq \sqrt{10}$ or if m,n,p >0 satisfy :$\sum_{cyc}\frac{1}{m}=1$ thus $\sum_{cyc}\sqrt{m+n-p}\geq  2\sqrt{5}$
 Therefor if $\sum_{cyc}\frac{1}{x}=3$ we have$\sum_{cyc}\sqrt{x+y-z}\geq  \frac{2\sqrt{5}}{\sqrt{3}}$
But i thng equality hold when a=0 and b=c or ....
but with a,b,c be side of triangle equality  not hold .What do you thing about it??????

You say it's wrong?  :roll: No, I dont think so.

you can find when equality hold???????????
I thinks it can be equality!!!

I think you should go here for a look,

http://www.mathlinks.ro/Forum/viewtopic.php?t=76881

it will explain for u.  :wink:

[quote="hunter_cvp_ic"]       Applying inq $\sum_{cyc}\frac{1}{a^{2}+b^{2}}\geq \frac{10}{(a+b+c)^{2}}$ .(*)[/quote]
I cannot find that one on ML. I guess it is easy with the EV method. But are there other elegant proofs of it?  :roll:

In Algebraic Inequality, page 341, we can find the inequality

$\sum\frac 1{b^{2}+rbc+c^{2}}\geq \frac{4(2r+5)}{(r+2)(a+b+c)^{2}}$,

which is true for $\frac{-3}{4}\leq r \leq \frac 7{4}$.