There is a checkerboard (or chessboard for all you chess players) that has an infinite number of squares. There is a positive integer in each equare, and every integer is equal to the average of the four integers located in the north, east, west, and south squares. Prove that all the integers in the squares are equal.
[hide="Hint makes this problem too easy"]Extreme Principle[/hide]

Similar problem:

There is a circle with 2000 points on it, and each point has a number next to it. Each of the numbers is equal to the average of the numbers of the points next to it. Prove that all the numbers are equal.
[hide="Hint makes this problem too easy"]Extreme Principle[/hide]

[hide]locate the smallest number on the board/circle, since it's the smallest number and the numbers around it averages that number, they must equal to that number. Hence, all of the numbers of the borad/circle must be the same.[/hide]

[hide]There is some number that is the smallest. If it is the average of the four neighboring squares, then none of the four squares can be greater than it, since that would imply some square is less than it, which contradicts the fact that it is the smallest number. Thus they are all the same.[/hide]

haha you can also try to locate the largest  :rotfl:

[hide="Way of finding the largest"]You find a largest number. If the numbers next to it are less than the largest number, then some numbers are greater than the original number, which is imopssible. Therefore, all of the numbers are the same.[/hide]

Also, there is the method of finite descent as well, but yes, Extreme Principle does work as well.

I don't think finding the largest works, becuase there are infinite integers.

Is infinite descent when you assume that there is a smallest number, and find a number smaller than that, providing a contradiction?

Could you post some problems that aren't trivial with the Extremal Principle?

[quote="SplashD"][quote="DanK"][quote="vishalarul"]I don't think finding the largest works, becuase there are infinite integers. [/quote]

Why should that matter?  It's the exact same thing as using the smallest number, just in reverse...[/quote]

there's no upper bound[/quote]

He's talking about the Extremal Principle, not Infinite Descent.

Extremal Principle is just contradiction: finding the smallest (or largest) of a set and proving there is a smaller (or larger) number.

[quote="vishalarul"]Is infinite descent when you assume that there is a smallest number, and find a number smaller than that, providing a contradiction?[/quote] No, you described the Extremal Principle.
Infinite Descent is: given a solution, you can find a smaller solution, and from there a smaller solution, etc.


modular arithmetic
combinatorics unsolved
combinatorics

Prove that:

${a\choose b } \equiv {a_0\choose b_0}{a_1\choose b_1} \cdots {a_n \choose  b_n} \pmod{p}$

Where $a_0, a_1 \cdots a_n, b_1, b_2 \cdots b_n$  are the digits of $a, b$ in base $p$,  $p$ prime

I'm pretty sure it's been discussed before. Try searhing for "Lucas Theorem".


function
real analysis
real analysis solved

I just got my hands on a problem and fail to see why the sequences of functions needs to uniformly convergent. This is the problem description:

Prove that every uniformely convergent sequence of bounded functions is uniformely bounded.

Well, uniformely bounded means that $|f_n(x)| < M \; \forall x $ and $n \in \mathbb{N}^{+}$. Every bounded function has a bound which is a real number. Why can't I just create the set of all these numbers and pick the largest one and call in M? It must exist, I don't know how many numbers there is or which it is but still.

Thanks a lot if you can explain how this works to me. I just regard uniformely convergent as that all functions goto some other function with the same speed since there is a single N that limits all of them indepented of my choice of epsilon or x.

Mike

[quote="litton"]Why can't I just create the set of all these numbers and pick the largest one and call in M? It must exist, I don't know how many numbers there is or which it is but still.
[/quote]

This maximum doesn't necessarily exist, if you take the constant functions $f_n = n$, all of them are bounded but not uniformly.

Yo sequence has a lot of functions (infinite set) and u can get an infinite set of their bondaries. But infinite set can have no upper bound. Let's say the that n-th function has upper bound n. Then you can't get the largest, right? For the uniformely convergent sequence it's obvious however, because since some number N  [b]all[/b] the functions lie in the neigbourhood of the limit function (their graphs lie near its graph. Let's say you've got a rectangular paper band with a small height and you put on it a red thread which is yo limit function and a lot of other black threads which are your sequence functions' graphs. Well, it'll be better to draw it, but I don't know how to draw here. If you want I can scan my drawing and attach it here). Since the difference between yo limit function and any of the sequence function (with number more than N) is no more than the band's height and hence:  the limit function < n-th function + height. Note that I don't write here f, not f(x), because uniformly means for each x. So, the limit function is bounded and all other functions lie in its $\epsilon$ neighbourhood, besides some first ones. But there are a finite number of the "first ones"...

Seems to be correct.

Here's what you have:

For each $n$, there exists $M_n$ such that $|f_n(x)|\le M_n$ for all $x$.

There exists a function $f$ such that, given $\epsilon>0,\,\exists N$ such that if $n>N,\,|f_n(x)-f(x)|<\epsilon$ for all $x.$

Here's the proof.

We won't use the limit function $f.$ Instead, we will use that uniform convergent implies uniformly Cauchy. That is, given $\epsilon>0,\,\exists N$ such that if $n,m>N,\,|f_n(x)-f_m(x)|<\epsilon$ for all $x.$

First, take a convenient $\epsilon,$ say $\epsilon=1.$ Find the $N$ that you get from the definition of uniformly Cauchy.

Suppose $m>N.$ Then for all $x,$ $|f_{N+1}(x)-f_m(x)|< 1.$ We can rearrange this to $|f_m(x)|<|f_{N+1}(x)|+1\le M_{N+1}+1.$

Now let $M=\max\{M_1,M_2,\dots,M_N,M_{N+1}+1\}$

We have that for all $n,$ $|f_n(x)|\le M.$

The biggest problem with Stan's proof: he was assuming that the limit function is bounded. But how are you going to prove that? You'll notice that I explicitly ducked this question by relying on uniformly Cauchy. In fact, proving that the limit function is bounded seems to me to about as difficult as the theorem we are trying to prove.

Hi, Kent,

[quote]The biggest problem with Stan's proof: he was assuming that the limit function is bounded[/quote]

Hmmm... Was I? I thought I proved it. :) Let's say $g(x)$ is the limit function, then we have $|g(x) - f_{n}(x)| < \epsilon$ when $n > N$, hence $|g(x)| < |f_{n}| + \epsilon$ and $f_{n}$ is bounded. This is what I wrote but in other words.


limit
calculus
derivative
function
calculus computations

[hide="hint"]I think a good start on this problem would be applying binomial theorem.  I hope that helps.[/hide]

Hey!
  I think it is very nice to solve it by binomial theorem.
  Indeed, i can solve it by the definition of derivative but i am not sure about this. probably you can check it.
   I can't post my sols here because I am in Cupertino library right now and I have only 2 hours. So I will post my idea first. I hope it works.
   We can use substition here for the denominator
    Put $a=1- \sqrt{1-\frac{x}{2}}$ then write the numerator in term of $a$. So we will get something likes $\frac{f(a)}{a}$ then can apply the derivative method.
    Oh, I just have another idea. We can use the rational conjugate to eliminate the denomenator.It will only have a $x$ term in the denomenator. After that, we can apply the derivative method.
     I hope it works. Sorry, I am so busy right now. I will post my sols later.

$\frac{\sqrt[3]{1-\frac{x}{3}}-\sqrt[4]{1+\frac{x}{4}}}{1-\sqrt{1-\frac{x}{2}}}= \frac{-x}{\sqrt{1-\frac x2}-1}\cdot \left(\frac{\sqrt[3]{1-\frac x3}-1}x - \frac{\sqrt[4]{1+\frac x4}-1}x\right)$

Someone wants to use the definition of the derivative? Have at it.

I L'hopital'd it and evaluated the coefficients of the inside parts (inside part of all three mini-functions will always be 1, no matter how much you take the derivative of it) and got a negative number between -0.5 and -1...

Using L Ho'spital's Rule since it is 0/0 form

I got the limit to be -25/36


Solve $(x+y)^n=x^n+y^n$ in $\mathbb{R}$ for all $n\in\mathbb{N}$.

[i](3 points)[/i]

If $n=1$ all $x,y$ works.

When $2|n$ we get $x=0$ or $y=0$, because $(x+y)^n > x^n+y^n$ when $x,y$ have both the same sign and $(x+y)^n < x^n+y^n$ when they have different sign.

When $n$ is odd and bigger than $1$ we have $(x,-x),(x,0),(0,y)$ are all solutions.
To see this, just look to the cases $x,y$ have same sign (trivial).
When they have different sings, let $|x|>|y|>0$ and remark $|(x-y)^n|+|y^n|<|x^n|$.


$ a_{n}$ take value in the range {1,...,k}
$ a_{1}\equal{}1$
$ a_{n}\equal{}a_{n\minus{}1}\plus{}n (mod k)$
Find all k such that the sequence take all k value .

Bad title http://www.mathlinks.ro/Forum/viewtopic.php?t=135915 . Locked!


The area of the cone without the circular base is just the area of the circle after the sector is cut out (see why?)

you mean the surface area of the cone without the circular base is equal to the area of the circle after the sector has been cut off.

[quote="mr. math"]A circle has radius 8. A cone is formed by cutting out a 60 degree wedge and placing together the two ends of the remaining part of the circle. What is the surface area of the cone, rounded to the nearest whole number, excluding the circular base?[/quote]

[hide="solution"]The area of the cone without the base is $ \pi\cdot 8^{2}\cdot\frac{5}{6}=\frac{160\pi}{3}$.

The circumference of the base is $ 2\pi\cdot 8\frac{5}{6}=\frac{80\pi}{3}$.  The radius of the base is $ 2\pi r=\frac{80\pi}{3}=\frac{40}{3}$.

The area of the base is $ \pi\cdot \left(\frac{40}{3}\right)^{2}=\frac{1600\pi}{9}$

Adding them together gives us $ \boxed{\frac{2080\pi}{9}}$[/hide]


Find all primes $p$ for which $17p+1$ is a perfect square

[hide="Solution"] Let $17p+1 = n^{2}$.  Then $17p = (n+1)(n-1)$ which has the obvious solution $n = 18, p = \boxed{ 19 }$.

If $17$ does not take on the value of either term, but only divides it, that would imply $p$ was composite, so this is the only solution. [/hide]

I posted in the wrong forum :(. You aren't missing anything. Sorry.


$ a_1,a_2,..,a_5$ are given different real number.and for every $ i\neq j,a_i \minus{} a_j$ are different
Now you can choose differnent reals $ b_1,b_2,...,b_7$ and let $ N \equal{} \{a_i \plus{} b_j|1\le i \le 5,1 \le j \le 7\}$
Find the smallest value of $ |N|$

(Can $ 5,7$ be generalized?)

The condition that the $ a_i \minus{} a_j$ are all distinct does not determine this value independent of the choice of the $ a_i$ (unless there's something incredibly special about $ (5, 7)$ that I haven't noticed).  A true independence condition would be something like "the $ a_i$ are linearly independent over $ \bf Z$," I think.

For example, $ (5, 7)$ replaced by $ (3, 4)$ we have that choosing $ \{a_i\} \equal{} \{0, 1, 3\}$ allows us to achieve the theoretical minimum value $ |N| \equal{} 7$ (e.g., when $ \{b_i\} \equal{} \{0, 1, 2, 3\}$), but it seems that for $ \{a_i\} \equal{} \{0, 2, 5\}$, the best we can do is $ |N| \equal{} 8$ (e.g., when $ \{b_i\} \equal{} \{0, 2, 4, 7\}$).

Replace $ (5,7)$ by $ (m,n)$
I have solutions for $ m\ge n$ and $ m\equal{}n\minus{}1$
And it is easy to show when $ n$ is much bigger than $ m$ then if replacing $ n$ by $ n\plus{}1$,$ N$ be will be replaced by $ N\plus{}1$
But when $ n$ get close to $ m$ such as $ (5,7)$ I didn't know how to constuct $ a_i,b_j$  :?:


LaTeX
videos
Alcumus
algebra
binomial theorem

So, I was taking a contest and I was stumped by this one problem:

Find the x^8 term in the expansion of            (2 + (1/2)x^2) ^ 6

$ x^8$ term in $ (2 \plus{} \frac {1}{2}x^2)^6$
So it's going to be the term where the $ x^2$ term is raised to the 4th power, so it'll be $ k\cdot2^2(\frac {1}{2}x^2)^4$
Of course, that k can be described by Binomial theorem, the combination of 4 from 6. (don't know how to do latex) which equals 15.
So, the term is $ 15\cdot2^2(\frac {1}{2}x^2)^4\equal{}\boxed{\frac{15}{4}x^8}$

Because we want $ x^8$, we need to look for the 4th term of the expansion, since the fourth term will be in the form of $ ax^2y^4$, and when we do $ y^4$ we get $ \frac{1}{16}x^8$, which is what we are looking for.  WE have already established that the fourth term is $ ax^2y^4$, but we need to find the value of $ a$.  We can do that by doing $ \binom{6}{4} \equal{} 15$ (binomial theorem, I suggest you look it up if you don't know what it is because that's how you solve the problem)).  Plugging in $ a\equal{}15$, we get $ 15x^2y^4$.  We simply plug in $ x\equal{}2$ and $ y\equal{}\frac12x^2$ (from the original equation) and get $ \boxed{\frac{15}{4}x^8}$.

ehh...My solution is kind of confusing, but i highly suggest you look up the [url=http://www.artofproblemsolving.com/Wiki/index.php/Binomial_Theorem]binomial theorem[/url] or watch the video on [url=http://www.artofproblemsolving.com/Alcumus/Videos/ShowVideo.php?video_id=72&m=ea97ea3b667da2e2bd78ec32171f0af7]alcumus[/url].

dang it, i spent to much time doing these hyperlinks.

Perhaps a better explanation than "because of the Binomial Theorem":

You have some coefficients, and then $ x^2$s that are being multiplied 6 times. You need to create an $ x^8$ term, which is $ (x^2)^4$. That is, from the 6 $ x^2$s available from all the stuff being multiplied, you must find all the ways to choose 4 of them, which is obviously $ \binom{6}{4}\equal{}15$.

Along with those 4 $ x^2$s, you'll get $ 4$ $ \frac{1}{2}$s being multiplied together, which gives us a $ \frac{1}{16}$.

If we're multiplying 4 of the terms involving the variables, we have our $ x^8$, so we don't need any more $ x$s. For that reason, we choose $ 2$s for the last 2 terms to get our 6 multiplications. That gives us a $ 4$ being multiplied.

Put it all together and we get $ 15\cdot \frac{1}{16}\cdot 4\cdot x^8\equal{}\boxed{\frac{15}{4}x^8}$.


[quote="GameBot"]Two friends each have five coins consisting of pennies, nickels, dimes or quarters, with neither friend having more than two of any particular coin. What is the greatest possible difference in cents between the sum of the values of their coins?[/quote]

One friend: Q Q D D N =  75 cents

Other friend: P P N N D = 22 cents

75-22=53 cents


induction
combinatorics proposed
combinatorics

A black pawn and a white pawn are placed on the first square and the last square of a $ 1\times n$ chessboard, respectively. Wiwit and Siti move alternatingly. Wiwit has the white pawn, and Siti has the black pawn. The white pawn moves first. In every move, the player moves her pawn one or two squares to the right or to the left, without passing the opponent's pawn. The player who cannot move anymore loses the game. Which player has the winning strategy? Explain the strategy.

Well, I think it goes inductively very easy.
For $ n\equal{}3,4$ check that white wins and that for $ n\equal{}5$ black wins.
Claim that for every $ n\equal{}3k\plus{}2$ black wins, and loses for other n.
Proof:
We know (from induction) that white wins for $ n\equal{}3k,3k\plus{}1$ and black wins for $ n\equal{}3k\plus{}2$
Observe:
a)$ n\equal{}3k\plus{}3$, white makes a one-step move which wins (same position for $ n\equal{}3k\plus{}2$ with switched colours)
b)$ n\equal{}3k\plus{}4$, white makes a double-step which wins (the same argument as before)
c)$ n\equal{}3k\plus{}5$, whatever move white makes white will land in position a) or b) with switched colours which loses by force which finishes our inductive proof :wink:


Consider the set ${1,2,3,...,100}$ partitioned into 20 "subsets" of consecutive integers, notice that $2$ elements, $a$ and $b$, in the same "subset" satisfy $-4\leq a-b \leq4$. Now to ensure there will be always possible to find such elements, by pidgeonhole principle, it is easy to see the minimun of $k$ is $21$.

 :)  :lol:  :!:

(Is "subset" the most apropiated word in the expresions above?)

Yup the original question uses the word subset.. I got careless on this one and put 20 instead.  :blush:

[quote="aidan"]SMO2007 Open, #1

Let $A$ be any $k$-element subset of the set ${1,2,3,4,\cdots,100}$. Determine the minimum value of $k$ such that we can always guarantee the existence of two numbers $a$ and $b$ in $A$ such that $|a-b|\leq 4$.[/quote]
$S_{k}=\{M\subset \{1,2,...,n\}|\ |M|=k\}$
$max_{M\in S_{k}}(min\{|a-b|\ |\ (a,b)\in M^{2}\})=[n/k]$
if $k$ is  the minimum value such that we can always guarantee the existence of two numbers $a$ and $b$ in any $k$-element subset of the set ${1,2,3,4,\cdots,n}$ such that $|a-b|\leq p$.[/quote]
$[n/k]=p$ and $[n/(k+1)]<p$
then $k=[n/(p+1)]+1$

for exemple, if $p=4,n=100$
$k=[100/4+1]+1=21$


Let $ ABC$ is a right triangle at angle $ A$. Let bisector of angle $ A$ meet $ BC$ at $ D$, let $ DE \perp AC$ and $ DF \perp AC$. Prove that: $ \frac{1}{DB^2}\plus{}\frac{1}{CD^2}\equal{}\frac{1}{DE^2}$

You can compute $ DB,CD$ and $ DE$ via $ b\equal{}AB,c\equal{}AC$.

AEDF is a square with side length s. $ \triangle BED$ and $ \triangle CFD$ are similar, so $ \frac{BE}{s}\equal{}\frac{s}{CF}$, or $ s^2\equal{}BE \cdot CF$
We have to show that
$ \frac{1}{BE^2\plus{}s^2}\plus{}\frac{1}{CF^2\plus{}s^2}\equal{}\frac{1}{s^2}$
Which is equivalent to
$ s^2\equal{}BE \cdot CF$ !

call DB = x, DC = y and DE = h, we have $ \frac{1}{DB^2} \plus{} \frac{1}{CD^2} \equal{} \frac{\sin^2 B}{h^2} \plus{} \frac{\sin^2 C}{h^2} \equal{} \frac{1}{h^2}$ so we have also that the thesis is true iff $ A\equal{}90^\circ$.

Can you solve this problem not use $ sin$, $ cosin$?


if 0<[i]a[/i]<  (:pi:/4)

[i]sin a[/i] ln[i](sin a)[/i]<[i]cos a[/i] ln[i](cos a)[/i]
It is NOT very Easy.
It is NOT very Hard.
(you can use AG-GM)

check inequality because I think there's no $b$ while probably there ought to be - anyway it should be in "unsolved" section

I've made a mistake.In fact there is no "b".Forgive me please.
I've solve this inequality by AG-GM after working on it for 3 hours.
In order to let you believe it look at the following graph.
Do you want to have a try?

Thanks for your advice. :D Forgive me please?

It's ok - but correct it - I'd like to try this one


Which is bigger, the infinity of positive numbers, or that of all integers?

Which is bigger, the infinity of integers, or that of all rational numbers?

Which is bigger, the infinity of rational numbers, or that of all real numbers?

 ;)

For the first one, the infinity of all positive numbers, because if it was all numbers, then everything would cancel out, thus becomming $0$.

EDIT: I thought you meant sum, well, if you mean the number of integers and positive numbers, then shouldn't it be the positive numbers, because fractions are considered to be positive numbers and there are more fractions between consecutive integers, so...positive numbers are more.

I'm not sure that these are right, but oh well...

[hide]
1)  They're equal (if 'positive numbers' refers to integers.  If it's reals, then the set of positive numbers is larger)
2)  They're equal
3)  The number of reals is larger
[/hide]

The sizes of the set of positive integers and the set of all integers is the same, since they can be compared as follows:
Positive integers: $1\qquad2\qquad3\qquad4\qquad5\qquad6\qquad7$
All integers: $0\qquad1\qquad-1\qquad2\qquad-2\qquad3\qquad-3$

Cantor's diagonal principle for (3). The infinity of reals is much larger. In fact, we can find a 1 to 1 correspondence b/w all rational numbers and all reals between 0 and 1.

[quote="towersfreak2006"]I don't think that's true. The set of all real numbers is the same set of real numbers in any finite range, so then the set of all real numbers would be the same size as the set of rational numbers.[/quote]
what?  :what?:  :huuh:
the last deduction that you did, are you sure? $\mathbb{R}\neq\mathbb{Q}$

No, you are correct about the set of reals in a finite range having a 1 to 1 correspondence with the set of reals in general. But the set of reals is much larger than the set of rationals, much much much larger. It is in fact called uncoutable rather than infinite.

Actually I was wrong about the 1-1 correspondence on the interval 0-1 between reals and all rationals. In reality for any interval with length $p>0$ the reals in that interval will be much more than all the rationals, the rationals will have measure 0.

Consider an interval $\phi>0$. We line up all the rational numbers. We match the first rational with the reals in an interval of length $\frac{\phi}{2}$, the second with the reals in an interval of length $\frac{\phi}{4}$, the third with the reals in an interval of length $\frac{\phi}{8}$, ..., the $n$th with the reals in an interval of length $\frac{\phi}{2^n}$. We easily compute that the total length of all these intervals is $\phi$ and we have proved that the the infinity of rationals is of measure zero compared to reals.

Interesting problem: Let $m$ denote the number of rationals in set $S$ and $d$ denote the number of reals. Prove that the number of subsets of $S$ is equal to the number of reals, in other words prove that $2^m=d$

Heh, Palytoxin, I meant that there is not a 1 to 1 correspondence between the elements in the set of all rational numbers and all elements in the set of real numbers between 0 and 1. Sorry for the confusion.

I see what you meant, my comment about the 1-1 correspondence would bring about a contradiction.  :blush:

[quote="DPopov"]... the rationals will have measure 0...[/quote]
measure 0? in intermediate topics? i think it's enough to say that the set of rationals can have 1-1 with the set of positive integers. i don't know what measure means, i've heard of it, and i know it is not intermediate topics. for the original problem, i think it suffices to show that the real numbers $\mathbb{R}$ has a higher cardinality than $\mathbb{Q}$ by the diagonal process.

i like that interesting question but working with an infinite number of reals and an infinite number of rationals and show $2^m=d$... i mean $m$ and $d$ are no longer numbers...

Here's one mapping that shows the set of all real numbers has the same size as the set of real numbers in the range $-1<x<1$:
$x\in(-1,1)$ goes to $\tan\left(\frac{\pi*x}{2}\right)$
Note that we can manipulate the argument of the tangent function to fit different intervals.

There's also the oh so cool way of bending the real line into a semicircle...


These are direct applications of Holder. I hate to write all of them. Maybe I just prove one "$\geq$" and the rest are similar. So now, I am closing my eye and randomly pick one....
.
.
.
.
.
.
The lucky one is ..........

$(a^7x+b^7y+c^7z+d^7w)(ax^7 +by^7+cz^7+dw^7) $

$\geq (a^6x^2+b^6y^2 +c^6z^2+d^6w^2)(a^2x^6+b^2y^6+c^2z^6+d^2w^6)$.

Here is the proof:

$a^6x^2+\cdots = (a^7x)^{5/6} (ax^7)^{1/6} + \cdots \leq (a^7x + \cdots)^{5/6} (ax^7 + \cdots)^{1/6}$

Similarly,

$a^2x^6+\cdots = (a^7x)^{1/6} (ax^7)^{5/6} + \cdots \leq (a^7x + \cdots)^{1/6} (ax^7 + \cdots)^{5/6}$

Multiply together, we get desired result.


In a group of twelve people, four are sox fans, and eight are not sox fans. If two people are randomly selected from this group, what is the probability that one is a sox fan and the other is not a sox fan? Express your answer as a common fraction.

There are two cases:

The first is a Sox fan and the second isn't
OR
The first isn't a Sox fan and the second is.

The cases give probabilitites of $ \frac{1}{3} \cdot \frac{8}{11}\equal{}\frac{8}{33}$ and $ \frac{2}{3} \cdot \frac{4}{11}\equal{}\frac{8}{33}$.

Summing these gives $ \frac{16}{33}$, which is the answer


How do I prove this?

Let V be a R inner product space. Show that the set of all self-adjoint applications A : V --> V is a vectorial space.

In English, one usually says "operator" rather than "application."

You need merely to show that the sum of two self-adjoint operators is self-adjoint and that the scalar multiple of a self-adjoint operator is self-adjoint. This is one of those "just follow the definition" proofs. Write out what it means for an operator to be self-adjoint, and do what you have to do. (The fact that the inner product is bilinear will be used.)


Express the sum $S_m(n)=1^m+2^m+\ldots +(n-1)^m$ with Bernolli numbers.

Let $ S_m(n) = \sum_{k=1}^{n}k^m$ where $ m$ is a positive integer.

Now let

$ \begin{equation*}
\begin{alingned}
\left({v + u}\right)^m &= v^m + m v^{n-1} u + \cdots \\
\left({v + u - 1}\right)^m &= v^m + m v^{n-1} \left({u-1}\right) + \dfrac {m-1} {2} v^{m-2} \left({u-1}\right)^2 + \cdots
\end{alingned}$

Subtracting we have

$ \left(v+u\right)^m-\left(v+u-1\right)^m =mv^{m-1}+ \dfrac{m-1}{2}v^{m-2}[u^2-(u-1)^2]+...$

Now let $ u^2-\left(u-1\right)^2=0;u^3-(u-1)^3=0;...$

then we have that $ \left(v+u\right)^m-\left(v+u-1\right)^m =m v^{n-1}$

Taking $ v = 1,2,3, \ldots, n$ we have

$ \begin{equation*}
\begin{alingned}
\left(u+1\right)^m-\left(u\right)^m &= m1^{m-1} \\
\left(u+2\right)^m-\left(u+1\right)^m &= m2^{m-1} \\
\vdots \\
\left(u+n\right)^m-\left(u+n-1\right)^m &=mn^{m-1}
\end{alingned}$

Then we sum all of the equations and we obtain:

$ \begin{equation*}
\begin{alingned}
(u+n)^m-u^m &= m(1^{m-1}+2^{m-1}+...+n^{m-1}) \\
\dfrac{(u+n)^m-u^m}{m} &= 1^{m-1}+2^{m-1}+...+n^{m-1}
\end{alingned}$

where $ u$ is a Bernoulli number


Tidy Painters will paint Taylor's house for a charge of $ \$30$ per hour.  It takes 40 hours for their crew to complete the job.  How many dollars does the paint job cost?

This is simply $ 30 \times 40$, or $ \boxed{1200}$.


trigonometry
function
algebra unsolved
algebra

Find all functions such that:
\[f(x)+f(y)=2f(\frac{x+y}{2})f(\frac{x-y}{2})\]
$x, y \in R$

I think that I should add something to it...
\[f(0) \not = 0\]
And I've proved that $f(x)=f(-x)$ and $f(x)=f(x+4)$, but can i find sth more about it??

The condition $f(0)  \neq 0$ isn't really necessary -- the only thing it changes is that the solution $f(x) \equiv 0$ no longer exists.

[quote="ali"]Clearly f(x)=cosx is a solution. but cos(x+4) is not cosx :?
[quote]
shobber wrote:
f(x)=f(x+4) , but can i find sth more about it??
[/quote][/quote]
My mistake...
I was taking that $f(1)=0$ and get the results...
And is there any other solutions?? How can we find them out??

how do you get:
g(x+y) has to be equal to f(x)g(y)+g(x)f(y).
[/code]


Is there any perfect scorer on the IMO 2005 from Mathlinks... I think no ...  and i am asking my self everytime this question ... so maybe i am wrong . And as i remember there were 16 perfect scorers or 13...

Davron

come on, immediately after i read your question i could remember 3 perfect scorers pretty active on ML (Peter Scholze, Iura, Ghavr), and in am sure there are way more...

  darij


inequalities calculus quadratics function

If $ a,b,c$ are positive such that $ a\plus{}b\plus{}c \equal{} 1$ then
$ \frac{1}{1\plus{}a}\plus{}\frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c} \plus{}6(ab\plus{}bc\plus{}ca) \leq \frac{17}{4}$

A third class solution to your inequality:
[hide="solution"]
hint: assume that 0<a[u]<[/u]b[u]<[/u]c[u]<[/u]1. (as the inequality required to prove is symmetric). Also, a[u]<[/u]1/3. Then we see that
 [b]           1/(1+a) + 1/(1+b) + 1/(1+c)+6(ab+bc+ca)[u]<[/u]3/(1+a) + 6.3.a^2
                                                                      [u]<[/u]9/4 + 2
                                                                        =17/4
[/b]
QED

[/hide]

You might laugh at me for giving such a wild and "irrational sol" :(

[quote="Potla"]A third class solution to your inequality:

hint: assume that 0<a[u]<[/u]b[u]<[/u]c[u]<[/u]1. (as the inequality required to prove is symmetric). Also, a[u]<[/u]1/3. Then we see that
 [b]           1/(1+a) + 1/(1+b) + 1/(1+c)+6(ab+bc+ca)[u]<[/u]3/(1+a) + 6.3.a^2
                                                                      [u]<[/u]9/4 + 2
                                                                        =17/4
[/b]
QED

[/quote]
I maybe wrong but how come $ a<b<c$ then $ ab\plus{}bc\plus{}ca \leq 3a^2$ ?

@ Gibbenergy's second question:
I think you have the inequality sign pointed the wrong way...it's > instead of <
because if a < b < c,
then a.a < a.b;
       a.a < a.c;
 and a.a < b.c
so the sum of these three inequalities leads to my assumption
[edited]

No the equality sign is correct, but the logic is wrong...

By cauchy, or rearrangement, or muirhead, or AM-GM..

$ a^2\plus{}b^2\plus{}c^2 \ge ab\plus{}bc\plus{}ac$ with equality when $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$

$ \Rightarrow ab\plus{}bc\plus{}ca \le \frac{1}{3}$

The real problem is the first part...
By cauchy;

$ \frac{1}{a\plus{}1}\plus{}\frac{1}{b\plus{}1}\plus{}\frac{1}{c\plus{}1} \ge \frac{(3)^2}{3\plus{}a\plus{}b\plus{}c} \equal{} \frac{9}{4}$

So $ \frac{9}{4}$ is the minimum, and the inequality doesn't follow...

Just use Mixing variable.
$ f(a,b,c)\equal{}\frac{1}{1\plus{}a}\plus{}\frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c} \plus{}6(ab\plus{}bc\plus{}ca)$
Let $ a\leq\ b\leq\ c$ so
$ f(a,\frac{b\plus{}c}{2},\frac{b\plus{}c}{2})\geq\ f(a,b,c)$
THe rest is obvious

$ \frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {1}{1 \plus{} c} \plus{} 6(ab \plus{} bc \plus{} ca) \leq \frac {17}{4}$

$ \frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {1}{1 \plus{} c} \plus{} 3(a \plus{} b \plus{} c)^2 \minus{} 3a^2 \minus{} 3b^2 \minus{} 3c^3 \le \frac {17}{4}$

Let $ f(x) \equal{} \frac {1}{1 \plus{} x} \minus{} 3x^2 \plus{} 1$. Then $ f(\frac {1}{3}) \equal{} \frac {1}{1 \plus{} \frac {1}{3}} \minus{} 3(\frac {1}{3})^2 \plus{} 1$
$ \equal{} \frac {3}{4} \minus{} \frac {1}{3} \plus{} 1 \equal{} \frac {5}{12} \plus{} 1 \equal{} \frac {17}{12}$.

Hence we need to show that $ f(a) \plus{} f(b) \plus{} f(c) \le 3f(\frac {a \plus{} b \plus{} c}{3})$. If we show that $ f(x)$ is concave, then we are done by jensen's inequality.

$ f'(x) \equal{} \minus{} (\frac {1}{1 \plus{} x})^2 \minus{} 6x$

$ f''(x) \equal{} 2(\frac {1}{1 \plus{} x})^3 \minus{} 6\le 2 \minus{} 6 \equal{} \minus{} 4 < 0$.

Q.E.D.

Rofler's solution is correct.
I just want to comment that if we change the inequality a little bit, we will have a new inequality which is much difficult to prove, and Rofler's approach is not valid. See Can_Hang's inequality http://www.mathlinks.ro/viewtopic.php?t=246634

Can_Hang inequality
Let $ a \plus{} b \plus{} c \equal{} 1$ and positive. Prove

 $ \frac {1}{1 \minus{} a} \plus{} \frac {1}{1 \minus{} b} \plus{} \frac {1}{1 \minus{} c} \plus{} 6(ab \plus{} bc \plus{} ca) \geq \frac {13}{2}$

[quote="Gibbenergy"]If $ a,b,c$ are positive such that $ a \plus{} b \plus{} c \equal{} 1$ then
$ \frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {1}{1 \plus{} c} \plus{} 6(ab \plus{} bc \plus{} ca) \leq \frac {17}{4}$[/quote]
If we dont want to use calculus, the Cauchy Schwarz Inequality or AM-GM will be a good choice! For example, we can prove it by this way:
Rewrite the inequality as
\[ \frac {a}{1 \plus{} a} \plus{} \frac {b}{1 \plus{} b} \plus{} \frac {c}{1 \plus{} c} \plus{} \frac {5}{4} \ge 6(ab \plus{} bc \plus{} ca)
\]
Now, by AM-GM Inequality, we have
\[ \frac {1}{1 \plus{} a} \plus{} \frac {9(a \plus{} 1)}{16} \ge \frac {3}{2}
\]
It follows that
\[ \sum \frac {a}{1 \plus{} a} \ge \sum a\left( \frac {3}{2} \minus{} \frac {9(a \plus{} 1)}{16}\right) \equal{} \sum a \left(\frac{15}{16} \minus{} \frac {9}{16} a\right) \equal{} \frac {15}{16} \minus{} \frac {9}{16}(a^2 \plus{} b^2 \plus{} c^2)
\]
And we can deduce the inequality to
\[ \frac {9}{16} (a^2 \plus{} b^2 \plus{} c^2) \plus{} 6(ab \plus{} bc \plus{} ca) \le \frac {35}{16}
\]
which is obvious since it is equivalent to
\[ a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} ca
\]
:)

It is a nice solution Can Hang.
Here is my solution for the original problem
The inequality is based on this estimation
\[ \frac {1}{1 \plus{} a} \leq 3a^2 \minus{} \frac {41}{16}a \plus{} \frac {61}{48} \]

It is because

\[ \frac {1}{1 \plus{} a} \minus{} \left ( 3a^2 \minus{} \frac {41}{16}a \plus{} \frac {61}{48} \right) \equal{} \minus{} \frac {1}{48} \left (\frac {(16a \plus{} 13)(3a \minus{} 1)^2}{1 \plus{} a}\right ) \leq 0
\]
Hence
\[ \sum \frac {1}{1 \plus{} a} \plus{} 6(ab \plus{} bc \plus{} ca) \leq 3\sum a^2 \minus{} \frac {41}{16} \sum a \plus{} \frac {61}{16} \plus{} 6 (ab \plus{} bc \plus{} ca)
\]
\[ \equal{} 3\sum a^2 \plus{} 6(ab \plus{} bc \plus{} ca) \plus{} \frac {5}{4}
\equal{} 3 \left ( a \plus{} b \plus{} c \right )^2 \plus{} \frac {5}{4}
\equal{} \frac {17}{4}
\]

Whoops!! I had made a casting mistake in my post #2! Of course I neglect it now!
 :blush:
But however, I think that Gibbenergy's solution is the most perfect!

Thanks,

A stronger version for Gibbenergy's problem:
\[ \frac{1}{1\plus{}a}\plus{}\frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c}\plus{}ab\plus{}bc\plus{}ca \le \frac{31}{12}\]
With equality holds when $ (a,b,c)\equal{}(1/3,1/3,1/3)$ and $ (a,b,c)\equal{}(1/2,1/2,0).$
Will your way work here, Gibbenergy? :)

[quote="Can_Hang2007"]
\[ \frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {1}{1 \plus{} c} \plus{} ab \plus{} bc \plus{} ca \le \frac {31}{12}
\]
[/quote]
Dear Hung,

Your proof is true and it can solve  the general case as well.
I just add some details to your proof
Assume that $ f(a,b,c) \equal{} \frac {1}{{1 \plus{} a}} \plus{} \frac {1}{{1 \plus{} b}} \plus{} \frac {1}{{1 \plus{} c}} \plus{} ab \plus{} bc \plus{} ac$
with condition $ a\leq b \leq c$ so let $ 2t \equal{} b \plus{} c \geq \frac {2}{3}$
Hence
$ f(a,t,t) \minus{} f(a,b,c) \equal{} \frac {(b \minus{} c)^2}{4} \left( \frac {b^2 \plus{} c^2b \plus{} b^2 \plus{} c^2 \plus{} 4bc \plus{} 3b \plus{} 3c \minus{} 2}{{(2 \plus{} b \plus{} c)(1 \plus{} b \plus{} c \plus{} bc)}} \right)\geq0$
And lead to inequality with one single variable.

Can_Hang asks me if my technique works for his case, I would say it may work but it will be more complicated
For example, in my solution, I make a quadratic approximation
\[ \frac {1}{1 \plus{} a} \leq 3a^2 \minus{} \frac {41}{16}a \plus{} \frac {61}{48}
\]
This is a weak approximation. We can enhance the approximation by mixing with other functions. For instance, I mix it with $ f \equal{} \frac {1}{ax \plus{} b}$ and got a stronger approximation
\[ \frac {1}{1 \plus{} a} \leq \left(\frac {3}{4} a^2 \minus{} \frac {1}{2}a \plus{} \frac {7}{3} \right) \minus{} \frac {4}{3 \minus{} a}
\]
It is true because
\[ LHS \minus{} RHS \equal{} \minus{}\frac {1}{12} \frac {a(a \minus{} 2)(3a \minus{} 1)^2}{(a \plus{} 1)(a \minus{} 3)} \leq 0
\]
Note: the inequality holds when $ a \equal{} 0$ or $ a \equal{} \frac {1}{3}$.

Using the second approximation, one can find a new solution to my original inequality. However, this approximation does not work for the case of Can_Hang inequality. But I believe that if we mix more functions , we can find the solution.
But it is a boring task of doing so.

Now, let's look back our function.
\[ f(a,b,c) \equal{} \frac {1}{{1 \plus{} a}} \plus{} \frac {1}{{1 \plus{} b}} \plus{} \frac {1}{{1 \plus{} c}} \plus{} ab \plus{} bc \plus{} ac
\]
which has a "nice properties"
\[ f''(a) \equal{} \frac {2}{(1 \plus{} a)^3} > 0
\]
It is also true if we differentiate with respect to $ b$, and $ c$. So the function is convex, which means that we only need to check the value at the boundary of $ a$, $ b$, and $ c$.
Or basically, we only need to check $ max \equal{} {f(0,0,1) ; f(0, \frac {1}{2} , \frac {1}{2}) , f(\frac {1}{3}, \frac {1}{3}, \frac {1}{3}})$

In the same manner, one can find that Can_Hang inequality
\[ \frac {1}{1 \minus{} a} \plus{} \frac {1}{1 \minus{} b} \plus{} \frac {1}{1 \minus{} c} \plus{} 6(ab \plus{} bc \plus{} ca) \geq \frac {13}{2}
\]
also have the same properties
\[ f''(a) \equal{} \frac {2}{(1 \minus{} a)^3}
\]
and we can solve the problem in the same manner.


Excellent solution.  When I did it, I immediately plugged in 2, being fast at arithmetic.  But your way is much nicer!


geometry
trapezoid
parallelogram
rhombus
inequalities
geometry solved

Can anyone please tell me what are the conditions of the sides and angles in a trapezoid so that a circle can be inscribet inside it?

[quote="shobber"][quote="StRyKeR"]A quadrilateral ABCD can inscribe a circle if and only if AB + CD = BC + DA.[/quote]
Is that the only one?  :?[/quote]

Sure. Let $AB + CD = BC + DA$. If $ABCD$ is a parallelogram, $AB = CD, BC = DA$, then $2AB = 2CD = AB + CD = BC + DA = 2BC = 2DA$, i.e., $ABCD$ is a rhombus and every rhombus is tangential. Assume that the sides $BC, DA$ are not parallel, i.e., they intersect at a point $P$. WLOG, assume that the point $P$ is on the rays $BC, AD$. Let $(I)$ be the incircle of the triangle $\triangle ABP$. Assume that the quadrilateral $ABCD$ is not tangential, i.e., $CD$ is not a tangent of this incircle, for example, it intersects it. Let $t$ be a tangent of the circle $(I)$ parallel to $CD$, intersecting the rays $BC, AD$ at points $C', D'$. The quadrilateral $ABC'D'$ is tangential, hence $AB + C'D' = BC' + D'A$. But $C'D' < CD$ and $BC' > BC, D'A > DA$. Consequently, $AB + CD > AB + C'D' = BC' + D'A > BC + DA$, which contradicts the assumption $AB + CD = BC + DA$. The possibility that $CD$ does not have a common point with the circle $(I)$ is disproved similarly, just the direction of all inequalities is reversed.

A tangential isosceles trapezoid with $BC = DA$ has a useful property that its midline $MN = \frac{AB + CD}{2}$ parallel to the trapezoid bases $AB \parallel CD$ is equal to the shoulder sides: $MN = \frac{BC + DA}{2} = BC = DA$.


[quote="Gyan"]My [img]http://www.artofproblemsolving.com/Forum/images/avatars/45961890041e4a6671adbd.jpg[/img]was counting cats at a fair,
This number  was,  exactly  third of a square.
But  quarter of the cats were frightened  away.
What remained was exactly a cube by the way,
At least how many cats were there?

Please Hide the answer, and NO  the solution  need  not rhyme! :)[/quote]

[hide="Wrote this in 30 seconds.......solution is very rough!"]Beginning:
$\dfrac{c^{2}}{3}$ cats

End:
$\dfrac{c^{2}}{3}* \dfrac{3}{4} =\dfrac{c^{2}}{4}$ cats $= d^3$, for some d.

$\dfrac{c^2}{3}$ must be of the form $4*2^{3n}*3^{3m}*p^3$, where p is not divisible by 2 or 3, $n>-1, m>0$ by given conditions.
So to minimize this, let$n = 0, m =1$

So $\dfrac{c^2}{3} = 4 * 2^{3*0} * 3^{3*1} = 4*27 = \fbox{108}$   [/hide]

[quote="1234567890"]
[hide="Wrote this in 30 seconds.......solution is very rough!"]Beginning:
$\dfrac{c^{2}}{3}$ cats

End:
$\dfrac{c^{2}}{3}* \dfrac{3}{4} =\dfrac{c^{2}}{4}$ cats $= d^3$, for some d.

$\dfrac{c^2}{3}$ must be of the form $4*2^{3n}*3^{3m}*p^3$, where p is not divisible by 2 or 3, $n>-1, m>0$ by given conditions.
So to minimize this, let$n = 0, m =1$

So $\dfrac{c^2}{3} = 4 * 2^{3*0} * 3^{3*1} = 4*27 = \fbox{108}$   [/hide][/quote]

It is certainly 3 times a square, but  number of  cats who remained (81) is,   alas not a cube!

[hide]1234567890 made a very slight mistake...
The number of cats which then ran free
Would have to be a multiple of three
However, the number which remained at the fair
Would also be a perfect square
They would number as the sixth power of three
The number of cats which didn't run free
For the result multiplication will do
It spits out the number nine seventy two
In other words...
$\dfrac{c^2}{3}$ cats originally.  After the cats' being scared away, you get $\dfrac{c^2}{4}=d^3$.  Thus, $c^2=2^2d^3$, and d must at least be a mutiple of 3.  However, d must be a perfect square.  So the smallest possible value for d is $3^2$, which means the smallest possible such number of cats was $2^23^5=972$[/hide]

K81o7 :lol:  :lol: Sheer class  :lol:  :lol:


induction
geometry
circumcircle
analytic geometry
AMC
AIME
symmetry

Suppose I have $n$ circles,  $n\geq1$ . Find the maximum number of pts. of intersection in terms of n.

[hide="Hint"]
Let $a_n$ be the maximum number of intersections created by $n$ circles.
Can you find the recursive formula for $a_n$?
(Writing out the first several terms might help.)
[/hide]

[hide="solution"]
$\text{Visually we see that for each circle you add (let's say you }$
$\text{just added the nth circle), there are n new possible points of intersection formed.}$
$\text{This reduces to the question: what is the sum of the first }n\text{ numbers?}$
$\text{(This is version of a strategy called bijection)}$
$\text{The general formula is }n(n+1)\text{.}$
[/hide]

That solution is incorrect, but close. For example, plugging in $n=1$ would give $2$ pts. of intersection.
When I approahced the problem, I noticed that the number of pts. of intersection for $n$ cirlces was 2 times the sum of the first $n-1$ integers. This can be proven with induction, using the fact that two distinct cirlces can intercept in atmost 2 places (someone should prove that assertion).

Thm: The maximum number of pts. of intersection of $n$ distinct circles in 2-space is $n(n-1)$

[hide]Proof: The case where $n=1$ is easily verified since $1(1-1)=0$ gives the correct number of pts. of intersection for a single circle. Now suppose we have k distinct circles with a maximum of $k(k-1)$ pts. of intersection. Adding another distinct circle we have atmost another $2k$ pts. of intersection, since this $k+1st$ circle can intersect with each of the previous $k$ circles in atmost 2 pts. Thus the maximum number of pts. of intersection is given by $k(k-1)+2k=k^{2}+k=k(k+1)=(k+1)((k+1)-1)$, as desired.

Alternate Proof: For each pair of circles, we have atmost 2 distinct pts. of intersection. Thus there are atmost $2*\binom{n}{2}=2\frac{n(n-1)}{2}=n(n-1)$ pts. of intersection.[/hide]

Here is a lemma used in the proof

Lemma: Two distinct circles intersect in atmost 2 pts.

Proof: Assume that two distinct circles intersect in atleast 3 pts. These three pts. form a triangle whose circumcircle is unique.

To see that the circumcircle is unique, first note that the location and radius uniquely determine a circle. The center of the circumcircle, or the circumcenter of the triangle, is the unique pt. of concurrency of the three perpendicular bisectors of the triangle's sides. Further, the radius is uniquely determined to be $\frac{1}{4}$ the product of the length of the sides of the triangle, divided by the triangles area.

Since both circles pass through the three vertices of this triangle, they are both circumcircles, contradicting the fact that the circumcircle is uniquely determined.

oops my mistake.... I meant $n(n-1)$... when I'm in a rush, I always get it mixed up...

I have a good proof I came up with for the other day (of perp. bisectors), I will post it tomorrow when I have time

Here's an outline of my proof (of the perpendicular bisectors thing).

For a given perpendicular bisector, it is the set of points equidistant from two vertices.  So draw in two perpendicular bisectors, the first equidistant from $A$ and $B$ and the second from $B$ and $C$ (the vertices of the triangle are $A$, $B$ and $C$).  The intersection point is therefore equidistant from $A$ and $B$, and from $B$ and $C$.  Thus, it is equidistant from all three vertices, and thus, being equidistance from $A$ and $C$, the third perpendicular bisector must pass through it.

[quote="drunner2007"]How would you prove the fact that the perpendicular bisectors of a triangle are concurrent, w/o resorting to coordinate geom.?[/quote]

Ceva's should be fine.

About Ceva (In spanish): http://garciacapitan.auna.com/bella/htm/ceva.htm

About Ceva, now in english: http://mathworld.wolfram.com/CevasTheorem.html

I put a link in spanish because of the proof.

[quote="drunner2007"]Nice proof.
I don't know Ceva's yet, because I wasn't able to get through that section of AOPS before the AIME, but I'll be scanning it today for the AIME tomorrow.[/quote]

Might as well learn Menelaus with that.

Here's the proof I had come up with:
Start with triangle ABC. Draw perp. bisectors to AB and AC. Mark their intersection D. By congruent triangles (SAS) we show that AD and BD are the same. By symmetry we can show that AD and CD are the same. From the transitive property it follows that BD and CD are the same. Thus we can drop the angle bisector of angle BCD to where it meets BC. By SAS we prove that the two triangles formed by subdividing triangle BCD are congruent, thus the angle bisector we dropped from the intersection of the other two perp. bisectors was in fact the last perp. bisector.$\text{Q.E.D.}$


The root(s) of $ \frac {15}{x^2 \minus{} 4} \minus{} \frac {2}{x \minus{} 2} \equal{} 1$ is (are):

$ \textbf{(A)}\ \minus{} 5\text{ and }3 \qquad\textbf{(B)}\ \pm 2 \qquad\textbf{(C)}\ 2\text{ only} \qquad\textbf{(D)}\ \minus{} 3\text{ and }5 \qquad\textbf{(E)}\ 3\text{ only}$

[hide]

$ \frac{15}{(x\plus{}2)(x\minus{}2)}\minus{}\frac{2}{x\minus{}2}\equal{}1$
$ \frac{15\minus{}2(x\plus{}2)}{(x\plus{}2)(x\minus{}2)}\equal{}1$
$ 11\minus{}2x\equal{}X^2\minus{}4$
$ 0\equal{}x^2\plus{}2x\minus{}15$
$ 0\equal{}(x\plus{}5)(x\minus{}3)$

The roots are $ \boxed{A}$
[/hide]


Can anyone please help me to solve the following problem:

[color=green]Let f0, f1, f2, . . . be the Fibonacci sequence defined in class. Let L0, L1, L2, . . .
be the Lucas sequence defined recursively by
L0 = 2, L1 = 1 and Ln+1 = Ln + Ln−1 for n  1.

(a) Make a table that shows the first ten entries in the Fibonacci sequence and
the first ten entries in the Lucas sequence.

(b) Using the table in (a), guess a formula for Ln+1 as a sum of two terms from
the sequence f0, f1, f2, . . . .

(c) Use induction to prove the formula that you guessed in (b) for all n >= 0.[/color]

I've done part a and b......the answer for b should be Ln+1 = fn + fn+2
but I've no idea how to do part c.

THANKS FOR HELPING!!! :lol:


[quote="April"]Find all integers $ a$, $ b$, $ n$ greater than $ 1$ which satisfy
\[ \left(a^3 \plus{} b^3\right)^n \equal{} 4(ab)^{1995}
\]
[/quote]
[b]Solution:[/b]
Suppose that $ (a,b)\equal{}d$ and $ a\equal{}x\cdot d,b\equal{}y\cdot d$.Therefore,we can rewrite our equation as follows:

\[ d^3n\left(x^3\plus{}y^3\right)^{n}\equal{}4d^{3990}\cdot\left(xy\right)^{1995}.
\]

It is easy to understand that $ 4\left(ab\right)^{1995}\equal{}\left(a^3\plus{}b^3\right)^{n}\geq\left(2\sqrt{a^3b^3}\right)^{n}$,
and as we know $ a,b\geq 2$,therefore:
\[ \boxed{\frac{3n}{2}\leq 1995}.
\]

Now let's return to the first equation:

\[ \left(x^3\plus{}y^3\right)^{n}\equal{}4d^{3990\minus{}3n}\cdot\left(xy\right)^{1995}.
\]

As we know $ (x,y)\equal{}1$ and $ y|(x^3\plus{}y^3)^{n}$,therefore $ y\equal{}1$ and from the same reason we conclude that $ x\equal{}1$.
So $ a\equal{}b\equal{}c$,and hence
\[ \left(2c^3\right)^{n}\equal{}4\left(c\right)^{3990}
\]

Our goal is to find all $ n$,such that $ \boxed{3990\minus{}3n|n\minus{}2}$.

Let $ n\minus{}2\equal{}k\cdot(3990\minus{}3n)$,where $ k\in\mathbb{N}$,or

$ n(3k\plus{}1)\equal{}k\cdot 3990\plus{}2$.
Therefore,it is enough to find all $ k$,such that

$ k\cdot 3990\plus{}2$ is divisible by $ 3k\plus{}1$.

As we know $ k\cdot 3990\plus{}2\equal{}1330(3k\plus{}1)\minus{}1328$.

We need to find all $ k$,such that $ 3k\plus{}1$ divides $ 1328$.

Solving last relation,we finally conclude that all solutions are:
$ \{a,b,n\}$:
\[ \{1,1,2\},
\{2,2,998\},
\{32,32,1247\},
\{2^{55},2^{55},1322\},
\{2^{221},2^{221},1328\}.
\]

[quote="arshakus"][quote="Erken"][quote="April"]Find all integers $ a$, $ b$, $ n$ greater than $ 1$ which satisfy
\[ \left(a^3 \plus{} b^3\right)^n \equal{} 4(ab)^{1995}
\]
[/quote]
[b]Solution:[/b]
Suppose that $ (a,b)\equal{}d$ and $ a\equal{}x\cdot d,b\equal{}y\cdot d$.Therefore,we can rewrite our equation as follows:

\[ d^3n\left(x^3\plus{}y^3\right)^{n}\equal{}4d^{3990}\cdot\left(xy\right)^{1995}.
\]

It is easy to understand that $ 4\left(ab\right)^{1995}\equal{}\left(a^3\plus{}b^3\right)^{n}\geq\left(2\sqrt{a^3b^3}\right)^{n}$,
and as we know $ a,b\geq 2$,therefore:
\[ \boxed{\frac{3n}{2}\leq 1995}.
\]
nice solution
Now let's return to the first equation:

\[ \left(x^3\plus{}y^3\right)^{n}\equal{}4d^{3990\minus{}3n}\cdot\left(xy\right)^{1995}.
\]

As we know $ (x,y)\equal{}1$ and $ y|(x^3\plus{}y^3)^{n}$,therefore $ y\equal{}1$ and from the same reason we conclude that $ x\equal{}1$.
So $ a\equal{}b\equal{}c$,and hence
\[ \left(2c^3\right)^{n}\equal{}4\left(c\right)^{3990}
\]

Our goal is to find all $ n$,such that $ \boxed{3990\minus{}3n|n\minus{}2}$.

Let $ n\minus{}2\equal{}k\cdot(3990\minus{}3n)$,where $ k\in\mathbb{N}$,or

$ n(3k\plus{}1)\equal{}k\cdot 3990\plus{}2$.
Therefore,it is enough to find all $ k$,such that

$ k\cdot 3990\plus{}2$ is divisible by $ 3k\plus{}1$.

As we know $ k\cdot 3990\plus{}2\equal{}1330(3k\plus{}1)\minus{}1328$.

We need to find all $ k$,such that $ 3k\plus{}1$ divides $ 1328$.

Solving last relation,we finally conclude that all solutions are:
$ \{a,b,n\}$:
\[ \{1,1,2\},
\{2,2,998\},
\{32,32,1247\},
\{2^{55},2^{55},1322\},
\{2^{221},2^{221},1328\}.
\][/quote][/quote]
nice solution


Let $ W_1$ and $ W_2$ be subspaces of a vector space $ V$. How does one prove that $ W_1\cup W_2$ is a subspace of $ V$ if and only if $ W_1\subseteq W_2$ or $ W_2\subseteq W_1$?

Lemma: if $ W$ is a subspace, $ x\in W,$ and $ y\notin W,$ then $ x \plus{} y\notin W.$ Why? Because if $ x \plus{} y\in W,$ then $ y \equal{} (x \plus{} y) \minus{} x$ would also be in $ W.$

Now, suppose that there exist $ x_1\in W_1\setminus W_2$ and $ x_2\in W_2\setminus W_1.$ Then $ x_1$ and $ x_2$ are both in $ W_1\cup W_2.$ However, by the lemma above, $ x_1 \plus{} x_2\notin W_2$ and $ x_1 \plus{} x_2\notin W_1.$ Hence, $ x_1 \plus{} x_2\notin W_1\cup W_2,$ so $ W_1\cup W_2$ is not closed under addition and thus not a subspace.

If $ W_1\cup W_2$ is a subspace, then either we can't find $ x_1\in W_1\setminus W_2,$ in which case $ W_1\subseteq W_2,$ or we can't find $ x_2\in W_2\setminus W_1,$ in which case $ W_2\subseteq W_1.$

[quote="Kent Merryfield"]If $ W_1\cup W_2$ is not a subspace, then either we can't find $ x_1\in W_1\setminus W_2,$ in which case $ W_1\subseteq W_2,$ or we can't find $ x_2\in W_2\setminus W_1,$ in which case $ W_2\subseteq W_1.$[/quote]
Is this proving the reverse implication? The contrapositive would be: If $ W_1\cup W_2$ is not a subspace, then $ W_1\not\subseteq W_2$ and $ W_2\not\subseteq W_1$, so could you explain what you mean?

Can you just say something like: Suppose that $ W_1\subseteq W_2$ or $ W_2\subseteq W_1$. Then $ W_1\cup W_2\equal{}W_2$ in the first case, and $ W_1\cup W_2\equal{}W_1$ in the second case. Either way, the union is a subspace since $ W_1$ and $ W_2$ are subspaces.

Remove a "not" from the quoted sentence and it works. The proof (using the contrapositive of the original statement) was in the previous paragraph.

Thanks I'm new to the subject so I can't catch typos  :blush: .

So basically Kent only proves the forward direction? Does my proof of the reverse direction work?


Use the figure to fill in the blanks in the following statements about the function $ g$ at point $ B$.

g(___) = ___

g'(___) = ___


I am just starting a physics class, so I was wondering if anyone could post anything they want to that would be something that I could learn with an explanation that is simple enough for a 6th grade boy to understand (you can use complicated formulas and everything, but please explain how you got to that step), thanks    :)

[quote="bus"]I don't think that learning random physics stuff is a very good way to learn physics. However, if you do anything interesting / something you don't understand / something you would like to know more about in your class, feel free to post here and I'm sure there will be somebody to help you.[/quote]
Yes, but I'm interested in laws of motions, gravity, etc., sorry for not specifying. :oops:

Well, there's not too much to say there. You have the three Newton's laws of motion:

1. and 2.:
$F=ma$

where $m$ is mass of the object, $F$ is force applied to it and $a$ is acceleration of the object. This means that if you apply force $F$ to object with mass $m$, it will accelerate in the direction of the applied force with acceleration $a$. If there are no foces being applied to an object, this object can't accelerate. If an object accelerates, there must be some force $F=ma$ being applied to it.

3rd law says that for every action there is an equal opposite reaction. This means that if you apply force $F$ on object A, object A must also apply the same force (but of opposite direction) $F$ on you. When a baseball player hits a ball with his bat, not only the ball accelerates in the direction of the swing, but also the player accelerates in opposite direction - you just don't see it because he's standing on the ground and compensates for this acceleration with his muscles.

Gravity is something that makes things which have nonzero masses attract one another. You don't really see this with small things like pens, pencils or bottles of Coca-Cola - usually you only experience Earth's gravitational force. For everyday purposes, Earth's gravity makes objects near the surface of Earth accelerate towards the center of Earth with acceleration $g=9.81m/s^2$.

The most important thing, however, is to know what the acceleration [i]is[/i]. Acceleration is how much an object changes it's velocity with time. If an object accelerates with acceleration of $a=1m/s^2$, it gains $1m/s$ of velocity every second. And what is velocity hopefully doesn't need to be explained - velocity of $10m/s$ means that you move $10m$ every second. Finally - acceleration, force, velocity and position are all vectors. It is a good idea to learn something more about vectors before studying physics.

Anyway, I still think that you'd be better off learning all of this in your class. Or could just try to google some of this stuff, or get yourself some nice book about physics such as the Feynman Lectures on Physics.

How is gravity 'formed', is it a natural property of mass?

Just learn general theory right now, dont worry about problems.
(Learn Newton's laws, learn the concept behind them)
(Learn basics of other physics chapters like magnetism)

V=IR.
where:
V: potential difference between two points of a material
I: current across the material
R: resistance between the two points

The most fundamental expression in electrical field.

Welcome to the amazing world of physics.  :)


linear algebra
matrix
abstract algebra
linear algebra unsolved

$ 0$.

Whatever the wedge product is, it's alternating.

Now, what space does it live in? That depends on what definition you're using. I would probably default to reading $ n\times n$ matrices as a $ n^2$-dimensional space, in which case the wedge lives in a space of dimension $ \frac{n^2(n^2\minus{}1)}{2}$.

to build on what was said before, the wedge product whatever space its in is probably a quotient of the tensor product space, so that is if you are in some space $ T$, the space is $ T\otimes T/(A\otimes A)$ so the subspace of elements of the form $ A\otimes A$ is modded out.


how to take a photograph of a camera by itself?  :?:
Except using a mirror. :?:
could you offer me some new idears?

use an other camera, (use glass, water, etc instead of a mirror:)

but those (glass, water) are kinda still like the "mirrors" that Liu Ce is talking bout...

ok then do the following:
find a gas (transparent) where you can move your camera faster than light,
place the camera at A, then move it (faster than light) at B and take a photo.
But its a little difficult to find such a gas :)
Note: there is a liquid where scientists managed to accelerate some protons so that finally their velocity was biggers than light's
Why do not you use a mirror?


Re:  Topic #13351, in the Advanced Problems forum:  http://www.artofproblemsolving.com/Forum/viewforum.php?f=151

The home page of each forum has a table of topics that shows "Topics", "Replies", "Author", "Views", and "Last Post".

The entry for this particular topic (#13351) is out of date.  It shows the "Last Post" on Wednesday, July 7.  There have been two posts since that date.

I think that's because the post was moved out of that forum - that was the last post made while the post lived there (as to why this is the behavior of phpbb for this, I'm not so sure)


calculus
integration
LaTeX
geometry
3D geometry
sphere
calculus computations

Can anyone shed some light on this? I hope so, because I couldn't get that Latex thing to install so I had to do this in paint (as if you couldn't tell):

Show that [img]http://www.mathlinks.ro/Forum/album.php?mode=attach&id=7118[/img] where [b]r[/b] = (t, t^2, t^3) and S is the sphere [i]r = a[/i] with outward unit normal [b]n[/b].

P.S. apparantly I can't get an image to display either.  No wonder I can't solve math problems.  Just copy and paste the URL if you're so inclined.

You don't need to install $\LaTeX$ to use it on the forum- just type the code and put it between dollar signs. The forum software interprets it for you.

[quote="DogToggler"]$\frac16\oint_{S}\mathbf{n}\cdot\nabla\left[r^{2}+\left(1+\frac{a^{3}}{2r^{3}}x\right)\right]\,dS =\frac43\pi a^{3}$[/quote]
Mouse over or click on that to see the code; if that's not exactly what you meant, please explain the differences.

Yes, that's exactly what I was trying to say, thanks.   So can anyone show that
$\frac16\oint_{S}\mathbf{n}\cdot\nabla\left[r^{2}+\left(1+\frac{a^{3}}{2r^{3}}x\right)\right]\,dS =\frac43\pi a^{3}$
where r = (t, t^2, t^3) and S is the sphere r = a with outward unit normal n.


number theory
prime factorization
number theory unsolved

Let $n$ natural number and $n<2007$ . Find all $n$ that all divisor's sum is odd.

[hide]
If we let $n=\prod_{p|n}p^{a_{p}}$ (where the product runs over primes) be the prime factorization of $n$, then the sum of its divisors equals
\[\prod_{p|n}\left(\sum_{k=0}^{a_{p}}p^{k}\right)\]
This number is odd when all factors are odd. This occurs when $p=2$ or $a_{p}$ is even, so the $n$ that work are the ones that can be expressed as the product of a power of 2 and a perfect square, which is equivalent to $n$ being a square or twice a square.

I really don't feel like listing all 75 of those less than 2007...
[/hide]


Express $ k$ as a common fration in terms of $ n,$ where $ n$ is an integer greater than 1.

$ \frac{k(n\minus{}2)!}{(n\plus{}1)!}\equal{}\frac{(n\minus{}1)!}{(n\plus{}2)!}$
[hide="Answer"]$ \frac{n\minus{}1}{n\plus{}2}$[/hide]

DonkeyKong, it's [b]fraction[/b] :lol:
[hide]
Simplifying, we have

$ \dfrac{k}{(n\plus{}1)(n)(n\minus{}1)} \equal{} \dfrac{1}{(n\plus{}2)(n\plus{}1)(n)}$

Thus $ k\equal{}\dfrac{n\minus{}1}{n\plus{}2}$.[/hide]

[quote="dingzhou"]ugh..are you askign this many questions for fun or do you have a ton of questions you need help on?[/quote]
Sorry, if it seems like it. I am posting questions that I missed and that I didn't understand from several years of Nationals.
[quote="distracted523"]DonkeyKong, it's [b]fraction[/b] :lol:
[hide]
Simplifying, we have

$ \dfrac{k}{(n \plus{} 1)(n)(n \minus{} 1)} \equal{} \dfrac{1}{(n \plus{} 2)(n \plus{} 1)(n)}$

Thus $ k \equal{} \dfrac{n \minus{} 1}{n \plus{} 2}$.[/hide][/quote]
EDIT: Now I get it..


Λύστε την εξίσωση στο $R$
$(x+1)^{x-1}=(x-1)^{x+1}$

http://www.mathlinks.ro/Forum/viewtopic.php?t=46280

 :P

:blush:

Να μια που δεν την έχω βάλλει, αφιερωμένη σε σένα Θανάση.

Έστω μια συνάρτηση δύο φορές παραγωγίσιμη από το R -> R
me δεύτερη παράγωγο μη αρνητική.

Αποδείξτε ότι  $f(x+f'(x))\geq f(x)$   , x ανήκει R.    :)

Τι σας λέει ?

Αφού έχει καιρό άλυτη και μου ζητήθηκε να βάλλω λύση την παρουσιάζω.
Αν ο $x$ είναι ρίζα της f' τότε ισχύει η ισότητα .
Έστω $x$ ένας πραγματικός ώστε $f'(x)<0$ . Από το θεώρημα μέσης τιμής στο διάστημα [x+f'(x),x] παίρνουμε $f(x)-f(x+f'(x))=-f'(x)f'(c)$ για κάποιο c sto [x+f'(x),x)]
Τώρα επειδή η δεύτερη παράγωγος είναι μη αρνητική η f' δεν είναι φθίνουσα άρα
f'(c)<f'(x)<0 .Kai άρα $f(x)-f(x+f'(x))<0$  άρα το ζητούμενο.

Όμοια και αν υπάρχει x ώστε $f'(x)>0$.

 :)  :lol:

Ωραία άσκηση Σιλουανέ!

Και ώραία λύση!  :)

Αντί για $f'(c)<f'(x)<0$ δεν έπρεπε να πάρουμε $f'(c)\leq f'(x)<0$ ?

(δεν αλλάζει έτσι κι αλλιώς το αποτέλεσμα)


abstract algebra
calculus
integration
algebra
function
domain
Ring Theory

Background: Let $ R$ be a ring, $ S \subseteq R$ a multiplicatively closed subset, and let $ M$, $ N$ be $ R\minus{}$modules. Let $ \phi : M \rightarrow N$ be an $ R\minus{}$module homomorphism, and let $ \phi_s : S^{\minus{}1}M \rightarrow S^{\minus{}1}N$ denote the $ S^{\minus{}1}R\minus{}$module homomorphism, and let $ \phi_s(\frac{m}{s})\equal{}\frac{\phi(m)}{s}.$ You may assume without proof that $ \phi_s$ is a well-defined $ S^{\minus{}1}R\minus{}$module homomorphism.

Prove:
(a) Prove that if $ \phi$ is injective, then $ \phi_s$ is injective.
(b) Prove that if $ \phi$ is surjective, then $ \phi_s$ is surjective.
(c) Prove that if $ M$ is a finitely generated free $ R\minus{}$module with basis $ \{ b_1,\ldots, b_m \}$ then $ S^{\minus{}1}M$ is a finitely generated free $ S^{\minus{}1}R\minus{}$module with basis $ \{ \frac{b_1}{1},\ldots, \frac{b_m}{1} \}.$
(d) Now suppose $ R$ is an integral domain and that $ m$, $ n$ are positive integers. Prove that if $ R^m \cong R^n$, then $ m\equal{}n$. (Hint: consider $ S\equal{}R\minus{}\{0\}$.)

(a) and (b) follow directly from the definition of a quotient ring. For (c) apply (a) and (b) to an isomorphism $ M\simeq R^n$. For (d) note that $ S^{\minus{}1}R$ is a field and you are in the world of linear algebra over fields.


Hello.

Being new to the Olympiad this year, I have enjoyed using the tests to challenge my students. My question is if I have a group of 22 students for a team, what would be considered a good team score after 5 tests for kids who never did this before? They are 7th graders, mostly ELL students, and highly motivated to do well. Would 100 correct questions for the year be an acceptable goal to achieve?

Thanks for any input here.

Good luck with test 4 next week.

Jeff

Frankly, it depends on the team. However, 100 would certainly be an excellent score for a team of 22 new kids, in my opinion.


quadratics
number theory unsolved
number theory

Let $ p$ be a prime of the form $ 4k \plus{} 1$ such that:$ p^2|2^p \minus{} 2$.Prove that:the greatest prime divisor $ q$ of $ 2^p \minus{} 1$ such that: $ 2^q > (6p)^p$
can anyone help me?(not using QUADRATIC RECIPROCITY)

Let me guess, you saw it in Problems from the Book¿ At least it's in there (and thus in the corresponding subforum here).
And I think I told you often enough about meaningfull titles.

(By the way, restrictions like "not using ..." are often stupid and have no reason to exist)


Next year, I will be taking Pre-Calculus in my middle school along with 4 other kids.
So what are some of the best Pre-Calculus textbooks for my school to use?
Please post your replies.

Thanks :ninja:

Uh...this is classroom math if you haven't noticed.  Classroom math is math from middle school.  If you are accelerated, I still suggest you to put this in the High School forum--it belongs there!

AoPS is coming out with a new trig/complex numbers book, which might help. I hear that the pre-calc course spends half the year reviewing algebra 2. So a firm grasp on algebra 2 before pre-calc should help  :)

The book my school (and myself) uses for Precalculus is called [url=http://www.amazon.com/Advanced-Mathematics-Precalculus-Discrete-Analysis/dp/0618250379/ref=pd_bbs_sr_1?ie=UTF8&s=books&qid=1205531202&sr=1-1]Advanced Mathematics[/url].
I found it to be very useful, it covers a lot of material. If you want to look more at what it has, that Amazon link has a table of contents.

my class uses [u]precalculus with limits: a graphing approach[/u], it is a good book that integrates in the homework problems some calculus problems; in addition it provides proofs of the formulas in the back of the book.

Advanced Mathematics is really good - we used that when I took pre-calc. I recommend it for sure; it proves most of the stuff before you see it, which is [b]really[/b] helpful.


[quote="rOadToAIME"]Hey guys,
I know that a^(1/n)= nth root of a if a>0 and m,n(n>=2) are integers, but how can I derive this if a is negative?
And what do I do if I don't know if the base is positive or not?[/quote]

If $ a < 0$, and $ n$ is even then $ \sqrt[n]{a}$ is complex.

Else if $ a < 0$ and $ n$ is odd then $ \sqrt[n]{a} \equal{} \minus{}\sqrt[n]{|a|}$

It works when $ a < 0$ too...
$ ( \minus{} 27)^{1/3} \equal{} \minus{} 3$.

Edit: Beaten. Yes, if the exponent is even the number will be imaginary.

It's as the previous posters said, it's exactly as when $ a$ is positive except that you'll have special cases. You can't take the $ 2n$th root of a number when $ n$ is an integer and $ a$ is a negative number. An even number of factors always ends up with a positive number. This is not true for odd numbers because an odd number of negative numbers multiplied together yield a negative.

First of all, thanks for the reply!
Book says I could only change fractional exponent into a radical form if the base is positive. So, what about (-27)^(1/3)? How can you change this into a radical form; the base is NEGATIVE?
The book says I can change x^(1/n) to nth root of x if x>0.
What I don't get is changing x^(1/n) to nth root of x when x<0 because as I said many times book says x^(1/n) to nth root of x when x>0.
How can you change negative base to the fractional power into a radical form?

$ \sqrt[3]{\minus{}27}\equal{}\minus{}3$

It's not so hard; think about it, what is $ (\minus{}3)^3$?

You would do it like this:

$ ( \minus{} 27)^{\frac {1}{3}} \equal{} \sqrt [3]{ \minus{} 27} \equal{} \minus{} 3$

Another example:

$ ( \minus{} 64)^{\frac {2}{3}} \equal{} (\sqrt [3]{ \minus{} 64})^2 \equal{} ( \minus{} 4)^2 \equal{} 16$

Finally, here's an example of a case where you get imaginary numbers involved.

$ ( \minus{} 16)^{\frac {3}{2}} \equal{} (\sqrt { \minus{} 16})^3 \equal{} (4i)^3 \equal{} 64i^3 \equal{} \minus{} 64i$.

[quote="rOadToAIME"]But how can you change a^(1/n) to a radical form if a is negative?
My book says you can change a^(1/n) to a radical form when a is POSITIVE.
So, how can you change a to the fractional power to a radical form if a is negative?
Help me out here please.[/quote]

The definition of $ a^{\frac1n}$ [b]is[/b] $ \sqrt [n]{a}$.

The definition holds regardless of whether $ a\ge 0$ or not.

For a negative number $ a$, $ a \equal{} \minus{} 1\cdot |a|$.

Further, if $ n$ is odd then $ \minus{} 1 \equal{} ( \minus{} 1)^n$, so that
\[ a^{\frac1n} \equal{} ( \minus{} 1)^{n\cdot\frac1n}|a|^{\frac1n} \equal{} \minus{} 1\cdot\sqrt [n]{|a|}
\]
But if $ n$ is even then $ a^{\frac1n} \equal{} ( \minus{} 1)^{\frac1n}|a|^{\frac1n} \equal{} ( \minus{} 1)^{\frac1n}\sqrt [n]{|a|}$.

An even root of $ \minus{} 1$ is a complex number: it has both real and imaginary parts.

For $ n$ even you can't pull the same trick as you can for $ n$ odd, because the product of an even number of factors of $ \minus{} 1$ equals $ \plus{} 1$.

The reason your book restricts the definition of $ a^{\frac1n}$ to positive values of $ a$ is specifically to avoid the use of complex numbers. Typically in an Algebra II course you'll get a brief introduction to complex numbers, but this usually doesn't include a lot of their applications. In as much as that is the case, you just have to accept that are some operations that you cannot perform, e.g. calculating the n-th root of a negative number when n is even.

Thanks you all!
Could someone prove a^(m/n) = nth root of a^m, where m and n are integers and a is all real numbers?

$ a^{\frac {m}{n}} \equal{}  {\sqrt [n] {a^{m}}}$
I don't know what have here to prove  :|

[b]ee"[/b]

[quote="enndb0x"]$ a^{\frac {m}{n}} \equal{} {\sqrt [n] {a^{m}}}$
I don't know what have here to prove  :|[/quote]

Perhaps just $ a^{\frac{m}{n}} \equal{} \left(a^m\right)^\frac{1}{n} \equal{} \sqrt[n]{a^{m}}$

I would say that $ a^{\frac {m}{n}} \equal{} {\sqrt [n] {a^{m}}}$ is a definition, usually with the restriction $ a>0$.

We could prove that
$ (\sqrt [n]{a})^m \equal{} \sqrt [n]{a^{m}}$ and that $ a^{\frac {m}{n}} \equal{} \left(a^m\right)^\frac {1}{n} \equal{} (a^{\frac{1}{n}})^m$.

[quote="rOadToAIME"]
Could someone prove a^(m/n) = nth root of a^m, where m and n are integers and a is all real numbers?[/quote]

\[ \textrm{Let } r \equal{} a^{\frac{m}{n}}\textrm{ likewise let }s \equal{} \sqrt[n]{a^m}.\]
\[ \textrm{Then }r^n \equal{} (a^{\frac{m}{n}})^n \equal{} a^m.\]
\[ \textrm{Also }s^n \equal{} (\sqrt[n]{a^m})^n \equal{} a^m.\]
\[ \therefore r^n \equal{} s^n \implies r \equal{} s \quad\mbox{Q.E.D.}\]

[quote="rOadToAIME"]Dr. no,
I have a counterexsmple to your proof: 2^2=(-2)^2, 2 does not equal -2.[/quote]

It looks like a counterexample, but it's not. In this case, you're saying that $ \sqrt{a^m} \equal{} \minus{}2$; that's a number that doesn't exist.

\[ \forall a\in \mathbb{R}, \sqrt[n]{a^m} > 0 \mbox{ or }\sqrt[n]{a^m}\in \mathbb{C}\mbox{ for n even}\]

Dr.N0 - imaginary numbers are still alive  :lol:

[b]ee"[/b]

[quote="enndb0x"]Dr.N0 - imaginary numbers are still alive  :lol:[/quote]

To say that a number, real or complex (imaginary numbers are a subset of the complex numbers), is "alive" is ambiguous.

The point of my last post was that starting with a real base, no n-th root of any real power of said base is negative if n is even. Furthermore, if n is odd such that the n-th root of some real power of a real base [b]is[/b] negative, then that root is unique. (You're not going to get positive and negative solutions simultaneously.)

I think that comment was regarding you saying "that's a number that doesn't exist."
Imaginary/complex numbers are numbers in every way (complex numbers form a field, in which all the operations defined on the reals are defined; in fact, $ \mathbb{R}\subseteq\mathbb{C}$).

So it's not true to say that $ \sqrt{\minus{}2}$ doesn't exist; it's simply not defined in $ \mathbb{R}$.

[quote="grn_trtle"]I think that comment was regarding you saying "that's a number that doesn't exist."
Imaginary/complex numbers are numbers in every way (complex numbers form a field, in which all the operations defined on the reals are defined; in fact, $ \mathbb{R}\subseteq\mathbb{C}$).

So it's not true to say that $ \sqrt { \minus{} 2}$ doesn't exist; it's simply not defined in $ \mathbb{R}$.[/quote]

$ \sqrt{\minus{}2}$ is not an example of what I was talking about. $ \sqrt{\minus{}2} \equal{} i\sqrt{2}$ is [b]not[/b] negative, it's imaginary. This is because $ \mathbb{C}$ is not well ordered. Not well ordered means for $ u,v\in \mathbb{C}$ one cannot say that $ u > v$ nor can one say $ u < v$.

[b]Negative[/b] denotes that a particular number is less than zero, but the complex numbers are not well ordered so "negative" is a meaningless concept as applied to them.

More succinctly,

[quote="I once"]$ ...\sqrt{a^m} \equal{} \minus{}2;$ that's a number that doesn't exist.[/quote]

[quote="Dr. No"][quote="rOadToAIME"]Dr. no,
I have a counterexsmple to your proof: 2^2=(-2)^2, 2 does not equal -2.[/quote]

It looks like a counterexample, but it's not. In this case, you're saying that $ \sqrt {a^m} \equal{} \minus{} 2$; that's a number that doesn't exist.
\[ \forall a\in \mathbb{R}, \sqrt [n]{a^m} > 0 \mbox{ or }\sqrt [n]{a^m}\in \mathbb{C}\mbox{ for n even}\]
[/quote]
what if a^(m/n) was -2?
Does it not exist either? if so, why not?

If by existing you mean it's not real, then a solution $ a$ to the equation $ a^{\frac {m}{n}} \equal{} \minus{} 2$ only exists when $ n$ is even.


[quote="perfect_radio"]Prove that for all $\displaystyle x,y,z,t > 0$ we have that \[ \displaystyle \frac{x}{x+y+z+t} + \frac{x+y}{x+y+z} > 2 \cdot \frac{2x+y}{2x+2y+2z+t} \, .  \][/quote]

[hide]
Let $a=x$, $b=x+y+z+t$, $c=x+y$, $d=x+y+z$ for some $a,b,c,d \in \mathbb{R}^+$. Then $b>d>c>a$.
We can rewrite the inequality as \[ \displaystyle \frac{a}{b}+\frac{c}{d} > \frac{2(a+c)}{b+d}  \]
\[ \Rightarrow abd+ad^2+b^2c+bcd > 2abd+2bcd  \]
\[ \Rightarrow ad^2+b^2c > abd+bcd  \]
\[ \Rightarrow ad(d-b) > bc(d-b)  \]
and since $d-b<0$,
\[ \Rightarrow ad<bc  \]

By rearrangement, the last inequality is true.
[/hide]

General point about inequalities that obviously needs to be made more often: what you've posted there is only a proof if the arrows go the opposite direction that you have them pointing.  Starting from something and arriving at something true tells you nothing about the truth of the things you started with.

From:
\[ \frac{1}{a}+\frac{1}{b} \ge \frac{4}{a+b},\ a,b \ge 0,  \]
the ineq. easily shown.

This is very easy.
Lovasz, I don't know if my solution is what you aimed with your hint, I also used AM-HM in the last part of the solution. I'm sorry if I'm just posting the same solution ...

Use Chebyshev:
\[ \displaystyle \frac{x}{x+y+z+t} + \frac{x+y}{x+y+z} \geq \frac{2x+y}{2}\cdot \left( \displaystyle \frac{1}{x+y+z+t} + \frac{1}{x+y+z}\right) > 2 \cdot \frac{2x+y}{2x+2y+2z+t} \]

The last step is AM-HM.  :)

I don't understand the part of the AM-HM...I think you have an extra 2 there

It's just plain AM-HM  :huh:

\[ \left( \displaystyle \frac{1}{x+y+z+t} + \frac{1}{x+y+z}\right) > \frac{4}{2x+2y+2z+t} \]

:blush:  :oops: Yes its ok....i stupidly miscounted....


Quadrilateral ABCD is cyclic. E and F are points on sides AD and AB such that DE = DC and BF = BC. G is the mid-point of EF.
Prove that <DGB is a right angle.

PROOF：
Let EH∥BF and EH=BF So quadrilateral HEBF is parallelogram
G is the midpoint of EF   =>  HG=BG
∠AEH=180-∠DEH   ∠DAB=180-∠DCB   EH=BF=BC   DE=DC
=>  △DEH and △DCB are  congruence => DH=DB  △DGH and △DGB are  congruence => DG⊥BG

Thank you guys it is better later than never.


geometry
geometric transformation
reflection

The official results for the 2004 Ciphering Time Trials were published earlier this month by e-mail to the coaches of all participating schools, as well as on our website: http://www.natassessment.com. These results reflect the scores of participants after double-checking. Many scores changed by a few points during this process, usually downwards but occasionally upwards.

The first-place school was the Academy for the Advancement of Science and Technology in Hackensack, New Jersey.

The first-place ninth-grader was Edward Hong of the Academy for the Advancement of Science and Technology in Hackensack, New Jersey.

The first-place tenth-grader was Adam Hesterberg of Garfield High School in Seattle, Washington.

The first-place eleventh-grader (and the highest score in the entire contest) was Chris Kennedy of the Academy for the Advancement of Science and Technology in Hackensack, New Jersey.  Zach Abel of Greenhill School in Addison, Texas tied with Chris, but placed second due to tie-breakers (highest score in the latest round wins ties).

The first-place twelfth-grader was Roger Li of Mark Keppel High School in Alhambra, California.

Please take the time to check out the results and congratulate your peers on their accomplishments.

Thank you,

[quote]The first-place twelfth-grader was Roger Li of Mark Keppel High School in Alhambra, California.[/quote]
Hooray!


linear algebra
matrix
linear algebra unsolved

we have a matrix ($n\times n$) with the zero sum of each row or column  and main backdiagonal.prove it's determinant is $0$.

dashmiz,
The zero sum of each row seems enough (no need for the other two asumptions I think).
$^{T}(1, \ldots , 1)$ is an eigenvector ($0$ being the eigenvalue), hence determinant is $0$

yes,of course the first problem is so easy,as a pie.
but i t's a part of this (i think open)problem:
suppose we have $n*n$ magic square i.e the square with the same sum of each row or column and the main diagonal and back diagonal($=s$)
prove that  $s$ is the greatest eigenvalue of the matrix.


linear algebra
matrix
function
induction
algebra
polynomial
topology

If $A,B,C \in\mathcal l M_2(\Re)$ such that they commute and $\det(AB+BC+AC)<0$ then $\det(A^2+B^2+C^2)\geq 0$.This can be proven....

I don`t understand. SorryI made a mistake in what I said the order should be $2$.

let's consider the function $f$ such that for $A$and $B$ in $M_n(R)$ we have $f(A,B)=AB-BA$ $f$ is a continuous function Thus $f^{-1}(0)$ is a closed set of $M_n(R)^2$ then let $A_m$ and $B_m$ be the sequences of diagonalisable elements of that set (density,in fact it's sufficient to consider only the $A_m$ matrices diagonalisable) such that $A_m\to A$ and $B_m\to B$ .we know that $A_n$ and $B_n$ have a triangularisation  base in commun( a diagonalisation one if the $A_m$ have got exactly $n$ different eigenvalues)  so the conclusion follows easily.

if $A$ and $B$ be matrices $\in M_n(R)$ and $I_n$ be the matrix unit $\in M_n(R)$ such that $AB=BA$ then
$det(A^2+B^2+ I_n) \geq 1$
It's better.NO?

MoHaMed, I thnik what you write is not correct. Maybe it was not what you meant ?

i think mohamed want say if $AB=BA$ so $A$ and $B$  have a commun trigonalisation base

in fact if $A$ is diagonisable and $AB=BA$ then $A$ and $B$ have a commun triangularisation base in $M_n(C)$

I think that alekk hinted to the density argument right ?? i guess u're right alekk

[quote="K09"]Prove that if $A$ and $B$ be matrices $\in M_n(R)$ and $I$ be the matrix unit $\in$ $M_n(R)$ such that $AB=BA$  then
$det(A^2+B^2+ I) \geq 0$[/quote]

A,B commmute then
$A^2+B^2=(A+iB)(A-iB)=CC*$   where $C*$ is the complex conjugate of C
so
$det(C.C*+I_n)\geq 0$

Let me try to present my solution, although it's long and awkward.

Lemma 1:
Let $a_1,\ldots,a_n$ be positive real numbers. Let $f \in \mathbb{R}[\sqrt{x+a_1}, \sqrt{x+a_2},\ldots, \sqrt{x+a_n}]$. If there is a non-degenerate interval $I\subset \mathbb{R+}$ with $f(x)=0, x\in I$ then $f$ is identically zero.

Proof: We do induction on $n$. For $n=1$, we write the polynomial as $\sqrt{x+a_1}p(x)+q(x)$ where $p(x),q(x) \in \mathbb{R}[X]$. Multyplying by the conjugate we get $(x+a_1)p^2(x)=q^2(x)$, which holds for all $c$ because it is a polynomial identity, hence $p=q=0$ because otherwise the power of $(x+a_1)$ in the factorisation of $(x+a_1)p^2(x)=q^2(x)$ would be even from one hand, and odd from the other hand. The base is proven. Now let's prove the induction step, from $n-1$ to $n$. WLOG suppose $a=a_n$ is the smallest of $a_1,a_2,\ldots,a_n$. We can write $f$ as $\sqrt{x+a}p(x)+q(x)$, where $p,q \in \mathbb{R}[\sqrt{x+a_1},\ldots, \sqrt{x+a_{n-1}}]$. Multiplying by the conjugate we get $(x+a)p^2(x)-q^2(x)=0$. Clearly $(x+a)p^2(x)-q^2(x)$ belongs to $\mathbb{R}[\sqrt{x+a_1},\ldots, \sqrt{x+a_{n-1}}]$ thus it is identically zero from the induction assumption. Particularly, the terms that contain no radical also equal zero. Now for a subset $J \in \{1,2,\ldots,n-1\}$ and $P \in \mathbb{R}[\sqrt{x+a_1},\ldots, \sqrt{x+a_{n-1}}]$ denote $[J]P$ be the sum of the $u(x)$ which belong to terms of form $\sqrt{\prod_{j \in J}(x+a_j)}u(x), u(x) \in \mathbb{R}[X]$. We see that $P=\sum_J \sqrt{\prod_{j \in J}(x+a_j)} [J]P$.

However, we deduce $[\{\} ](x+a)p^2=(x+a)\sum_J \prod_{j \in J}(x+a_j) ([J]p)^2, [\{\}] q^2(x)=\sum_J \prod_{j \in J}(x+a_j) ([J]q)^2$. As $J[\{\} ] ((x+a)p^2(x)-q^2(x))=0$ we then deduce $\sum_J \prod_{j \in J}(x+a_j) ([J]q)^2$ is divisible by $x+a$ hence it vanishes if setting $x=-a$. However in it all terms are noonegative, hence all terms must be zero which implies that $q(x)=0$ and then $p(x)=0$, implying the induction step.

The Lemma is proven.

Lemma 2: If $M$ is a $n \times n$ matrix with all elements above the main diagonal zero and all elements on the main diagonal positive, then there is a matrix $R$ with $R^2=M$. Moreover $R$ also has all elements above the main diagonal zero and all elements on the main diagonal positive.

The proof is again by induction. The base is obvious. Now the induction step. By the induction assumption there is a $(n-1) \times (n-1)$ matrix $R'$ with $R'^2=M'$ where $M'$ is obtained from $M$ by deleting the last row and the last column. denote by $y_i$ the element $R_{i,n}$ and let $a_i=R'_{i,i}, a_n=\sqrt{M_{n,n}}$. Set $y_n=a_n$. Equalizing the enty $(n-1,n)$ in $R^2$ and $M$ we get $y_{n-1}(a_{n-1}+a_n)=M_{n-1,n}$ so $y_{n-1,n}=\frac{M_{n-1,n}}{a_{n-1}+a_n}$. We can continue computing analogously $y_{n-2},\ldots,y_1$ the condition for $y_k$ being $y_k(a_k+a_n)+C=M_{k,n}$ where $C$ is already known to us. By this method, we find out all entries of $R$ and its obvious that $R^2=M$.

The Lemma is proven.

Lemma 3: If $B$ is any matrix with real entries and $\lambda\geq 0$ then there is a matrix $C$ with $C^2=B^2+\lambda I_n$.

It directly follows from the previous lemma. If we write $B$ is Jordan from, we obtain $S^{-1}BS+\lambda I_n$ satisfying the conditions in Lemma 2, thus there is a matrix $R$ with $R^2=S^{-1}BS+\lambda I_n$. Then $C=SRS^{-1}$ is the desired matrix. Moreover, it's clear that all entries of $R$ can be written as $f(\lambda)$ where $f \in \mathbb{R}[\sqrt{x+a_1}, \ldots,\sqrt{x+a_n}]$, $(a_i)$ being the diagonal elements of $S^{-1}BS+\lambda I_n$.

The Lemma is proven.

Lemma 4: If $A,B$ commute then $det(A^2+B^2)\geq 0$.

Proof: Indeed, $det(A^2+B^2)=det(A+Bi)det(A-Bi)$, which is the product of two conjugated complex numbers so is nonnegative.

The Lemma is proven.

Now, let's pass to the solution of the problem. By Lemma 4 we find $C_{\lambda}$ with $C_{\lambda}^2=B^2+\lambda I_n$, then $det(A^2+B^2+\lambda I_n)=det(A^2+C_{\lambda}^2)$. Call a number $\lambda$ good if $A$ commutes with $C_{\lambda}$. If we find a sequence of good numbers converging to $1$, then by continuity($C_{\lambda}$ is continuous in $\lambda$ from the last sentence of the proof of Lemma 4) $det(A^2+C_1^2)\geq 0$ and we are done.

Otherwise there is a neighborhood $V$ of $1$ containing only bad numbers. Let $P_{\lambda}(x)$ be the characteristic polynomial of $C_\lambda$, clearly having coefficients in $\mathbb{R}[X][\sqrt{\lambda+a_1}, \ldots,\sqrt{\lambda+a_n}]$(see the end of Lemma 4).

Then $A$ commutes with $P_{\lambda}(C_{\lambda})$ and with $C_{\lambda}^2$. As $P_{\lambda}=C_{\lambda} f_{\lambda}(B)+g_{\lambda}(B)$, if $det (f_{\lambda}(B)\neq 0)$ we could deduce that $A$ and $C_{\lambda}$ commute. Since it's not the case for $\lambda \in I$, we would have $det(f_{\lambda}(B))=0$ for all $\lambda \in I$. As it is a polynomial from $\mathbb{R}[\sqrt{x+a_1}, \ldots,\sqrt{x+a_n}]$, by Lemma 1 we deduce that $det(f_{\lambda}(B))$ is identically zero. By setting $\lambda=0$ we get $det (f_0(B))=0$ where $f_0\in R[x]$. Actually one can compute $f_0(X)$ to be exactly $\frac{h(X)-h(-X)}{2X}$ where $h$ is the characteristic polynomial of $B$.  Now let $B_y=B+y I_n$ and $h_y(x)$ be the characteristic polynomial of $B_y$. If $h_y(x)-h_y(-x)$ is not identically zero, then $det (f_0(B_y))$ is a polynomial in $y$ not identically zero, so for all $y$ except a finite number of them we would have $det (f_0(B_y))\neq 0$ so $det(A^2+B_y^2+I_n)\geq 0$, and taking $y \rightarrow 0$ and passing to limit we would have $det(A^2+B^2+I_n)\geq 0$, as desired. This is indeed the case, because $h_y(x)=h_y(-x)$ can hold only for finitely many values of $y$, as letting $x=t\neq 0$ we get $det(B+It+I_ny)-det(B-It+I_ny)=0$ but this is a polynomial in $y$ with leading coefficient of $y^{n-1}$ non-zero (it is $Tr(B+tI_n)-Tr(B-tI_n)=2nt\neq 0$) so is satisfied by at most $n-1$ values of $y$.

Concluding, there are at most $n+(n-1)$ values for $y$ that yield $det(f_0(B_y))=0$, so we can take a sequence of $y$ coverging to 0 with $det(f_0(B_y))\neq 0$ which implies, as we saw,  $det(A^2+B_y^2+I_n)\geq 0$, so passing to the limit we obtain the desired conclusion.

I think that $det(CC*+I)\geq 0$ right.But proof?


$a_1, a_2, ...$ and $b_1, b_2, ...$ are integers such that $a_1=b_1=0$ and $a_n = nb_n+a_1b_{n-1}+a_2b_{n-2}+...+a_{n-1}b_1$ for $n\geq 2$. Prove that $a_p$ is divisible by $p$ when $p$ is a prime number.

It's been posted a long time ago by sam-n, I believe. I remember posting something (not a proof, but a general term for $a_n$, from which the conclusion followed, but without proof), and then Myth solved it.


vector
algebra
polynomial
Functional Analysis
advanced fields
advanced fields unsolved

Here's a "quickie" :):

Let $\mathcal M$ be a dense linear subspace of a Hilbert space $\mathcal H$. Is it always true that $\mathcal M$ contains an orthogonal basis of $\mathcal H$?

if it is separable, the Gram-Schmidt orthonormalization process gives a basis isn't it ? If it is not separable, take a maximal orthogonal family $\mathcal F$ included in $\mathcal M$. Such a family exists according to Zorn lemma. And if the closure of $Vect (\mathcal F)$ does not contain $\mathcal M$, there would exists a vector $v \in \mathcal M \setminus \overline{Vect \mathcal F}$. This is impossible, because if $w$ is the orthogonal projection of $v$ on $\overline{Vect (F)}$, then $(v-w) \cup \mathcal F$ is a larger ortogonal family included in $\mathcal M$.

$\mathcal M$ is not complete, so you can't always take orthogonal projections onto subspaces of $\mathcal M$. Try again :).

I think not. Take $H=L^{2}([0,1])$ and $M=$ linear subspace of all polynomials $p\colon[0,1]\to\mathbf R$. Then $M$ is dense in $H$ but doesn't satisfy the hypothesis.

That's not a counterexample. Start with $\{1,x,x^{2}\dots\}$ and apply Gram-Schmidt, arriving at a suitably translated and scaled version of the Legendre polynomials. That's an orthonormal basis of $L^{2}.$ (This is just the first sentence of alekk's post.)

[quote="grobber"]Here's a "quickie" :):
[/quote]
Every time I see grobber's :) on a proposed problem, I am tempted to abuse my moderator's priviliges and change it to ;). The question seems, indeed, not so hard once you know the solution but it took me several days to figure it out.  :lol:  :blush:

The answer is "No". Take a Hilbert space $H$ with an orthonormal basis of cardinality continuum. Choose a countable subset $\{e_{k}\}_{k\in\mathbb N}$ of this basis. Now, consider the set of all pairs $(e,\delta)$ where $e$ is an element of our orthonormal basis of cardinality continuum and $\delta>0$. For each such pair choose an infinite set $A=A(e,\delta)\subset \mathbb N$ in such a way that every two subsets corresponding to different pairs have finite intersection (the fact that it is possible is a classical problem that, probably, has been discussed on the forum before). Now choose a vector $v=v(e,\delta)$ such that $\|v-e\|<\delta$ and $(v,e_{k})\ne 0$ exactly for $k\in B(e,\delta)$ where $B(e,\delta)$ is $A(e,\delta)$ to which, maybe, one extra $k$ for which $e=e_{k}$ is added. Now take $M$ to be the linear span of all such $v$. Clearly, it is dense in $H$. On the other hand, each non-zero vector in $M$ has a non-zero scalar product with at least one of $e_{k}$ (this is where the finite intersection property is crucial). But in any orthonormal system there may be only countably many vectors having a non-zero scalar product with a given vector. Thus $M$ cannot contain more than countable orthonormal sets.


topology
geometry
advanced fields
advanced fields theorems

Can anyone recommend some french books on differential topology?
Thanks!
Actually some introduction on Riemann Geometry or A-S index theorem or anything at this level will do, the only constraint is I want it in French.

Hi!

Gallot, Hulin and Lafontaine: Géométrie Riemanienne, Springer. This is an excellent book (in french) about Riemannian geometry.

Thanks a lot~~
is there any free electronic version?


geometry
geometric transformation
homothety

[quote="krishkoushik"]i want to learn homothety. can anyone suggest some books or resources for that?

thanks in advance[/quote]
You could try Geometry Unbound by Kiran Kedlaya

See [url=http://www.cut-the-knot.org/pythagoras/Transforms/ProblemsByHomothety.shtml]this[/url] and I've also attached the Mathematical Excalibur article.


Just write y=mx+b forms for both and then subtract the mx+b parts of both to find difference.

adding on to the other problem:

The Hawking has a gas tank capacity of 15 gallons. If it is driven for 50% of the time at 30 mph and at 60 mph for the remaining time, how far, in miles, will it traveled when it runs out of gas?

[quote="mr. math"]The Hawking has a fuel efficiency of 20 miles per gallon (mpg) while traveling at 30mph and a fuel efficiency of 40 mpg while traveling at 60mph. The Gabe has a fuel efficiency of 25 mpg while traveling at 30 mph and a fuel efficiency of 35 mpg while traveling at 60 mph.

now here is the problem:

Both the hawking and the Gabe follow a linear model for fuel effiency with respect to speed. After you find equations for fuel efficiency, E, in terms of speed, S, for both the Hawking and the Gabe, find an expression for the difference between the Hawkings fuel efficiency and the Gabe's fuel efficiency in terms of speed.

according the the information given, how would you do this problem?[/quote][hide] The Hawking has points $(30,20)$ and $(60,40)$ in its graph. $m=\frac{40-20}{60-30}=\frac{2}{3}$.
Plugging in the point $(30,20)$, we have $20=\frac{2}{3}\cdot30+b\Longrightarrow b=0$.
So, $E_{h}=\frac{2}{3}S_{h}$.

The Gabe has points $(30,25)$ and $(60,35)$ in its graph. $m=\frac{35-25}{60-30}=\frac{1}{3}$.
Plugging in the point $(30,25)$, we have $25=\frac{1}{3}\cdot30+b\Longrightarrow b=15$.
So, $E_{g}=\frac{1}{3}S_{g}+15$.

The difference is $|\frac{2}{3}S-(\frac{1}{3}S+15)|=|\frac{1}{3}S-15|$. [/hide]

The Hawking drives at 20 mpg at 30 mph, and 40 mpg at 60 mph.
If it drives at 30 mph for half the time and 60 mph for the other half, then it travels at 20 mpg for half the time and 40 mpg for the other.

Thus, because the time driven at each fuel efficiency is equal, we can say that the car has travelled at an average fuel efficiency of $\frac{20+40}{2}$ mpg, or
30 mpg.

$\frac{Gallons}{Mile}\cdot Miles = Gallons Used$
$Miles = \frac{Gallons Used}{(\frac{Gallons}{Mile})}$
Miles driven = 15 / 30
The driver can go $\boxed{0.5}$ miles with 15 gallons of gas.


I need some TI-89 Titanium Math Programs for math class (Pre-Calculus, Algebra 2, and etc.)

I know ticalc.org has a huge collection of math programs
but
I don't know which ones are worth downloading

My Preferences:
The program shows all work (well as much as possible)
The program is accurate

Please give advice on certain programs I should download!

Try downloading "Competition Suite"
[url=http://www.ticalc.org/pub/89/basic/math/]In here[/url]

About 20 kB

My poor TI-89 didn't have enough memory to start the program, so I broke it up into four separate programs [and I added some extra stuff so that each of the programs has about 10kB  :D ].

[quote="vishalarul"][quote="Aznwithabrain"][quote="Danbert"]I would suggest writing the programs yourself.  It helps a lot with understanding if you have to explain to your calculator what to do.

If you don't have time, on the other hand, please give more specific examples of what you want the program(s) to do.[/quote]

Can you give advice on writing a program (yeah I'm a total noob at that on the TI-89)

I just want any work-showing program

ANY![/quote]

Download the competition suite that I showed before. It's written in basic, so you can look at the source code. Look at the formulas the author uses and try to derive them on your own.[/quote]

your program didn't seem to work on my calculator...
though
it worked before (i deleted it one day on accident)

[quote="Aznwithabrain"][/quote][quote="vishalarul"][quote="Aznwithabrain"][quote="Danbert"]I would suggest writing the programs yourself.  It helps a lot with understanding if you have to explain to your calculator what to do.

If you don't have time, on the other hand, please give more specific examples of what you want the program(s) to do.[/quote]

Can you give advice on writing a program (yeah I'm a total noob at that on the TI-89)

I just want any work-showing program

ANY![/quote]

Download the competition suite that I showed before. It's written in basic, so you can look at the source code. Look at the formulas the author uses and try to derive them on your own.[/quote][quote="Aznwithabrain"]

your program didn't seem to work on my calculator...
though
it worked before (i deleted it one day on accident)[/quote]

Ok, then.

[url=http://education.ti.com/guidebooks/graphing/8992p/8992bookeng.pdf]Try this website.[/url]

Click Chapter 17.

Also, [url=http://www.ticalc.org/pub/text/misc/]try this[/url]

Download the zip file "8992advg.zip"; it should contain two word documents.


calculus
trigonometry
calculus computations

Is there an easy proof using only calculus for:

"the number of real numbers in the interval $ [0,1]$ is equal to the number of real numbers in the interval $ (\minus{}\infty,\plus{}\infty)$"???

 :)

To be more precise, the cardinality of $ [0,1]$ is the same as cardinality of $ \mathbb{R}$. That can be shown if you can build a bijection between them.

There are few such bijections I know by heart:

$ y \equal{} \tan\left(\pi\left(x \minus{} \frac {1}{2}\right)\right)$

$ y \equal{} \frac {x}{|x| \plus{} 1}$

First maps $ ( \minus{} 1,1)$ in $ \mathbb{R}$ and second one $ \mathbb{R}$ in $ ( \minus{} 1,1)$, now you just need to map $ ( \minus{} 1,1)$ to $ [ \minus{} 1,1]$ or vice versa (think of mapping $ \frac {1}{x} \to \frac {1}{x \plus{} 1}$ as a part of the map to remove the points 1 and -1, and leave the other points intact-it's a known procedure-if you simply map the rationals of the form $ \frac {1}{n}$, you'll lose the 1, (other numbers in interval should be fixed points of the mapping) and applying a similar mapping will exclude 0 too.)

If I made some lapsus writing this, just correct it  :lol:

[hide]P. S. I didn't know that plural of lapsus is lapsus.[/hide]


integration
limit
real analysis
real analysis unsolved

Let $ f: (0,\infty) \to [0,1[$ be measurable such that $ \int_0^\infty \frac {f(t)}{t \plus{} 1} dt < \infty$.

Show that for all $ \epsilon > 0$ $ \lim_{T \to \infty} T^{ \minus{} 1} | \{ 0 < t < T \big| f(t) < \epsilon \}| \equal{} 1$, where $ |A|$ is the Lebesgue measure of $ A \subset (0,\infty)$.

Corrected (thanks, Leonard and mlok)

are you sure  everything is ok in this limit ?

The limit should be $ \lim_{T \to \infty} T^{\minus{}1} | \{0<t<T \big| f(t) < \epsilon \}| \equal{} 1$.


superior algebra
superior algebra unsolved

prove that the number of elements in conjugacy class of finite simple group can never be a prime powerlarger than $1$.

A proof can be found in most books on representation theory (I have one such proof right in front of me, but I don't like copying proofs from books to the forum; it defies the purpose of browsing the forum in order to solve problems by myself :)).

Burnside used this as a lemma for the theorem which is actually known as [url=http://en.wikipedia.org/wiki/Burnside_theorem]Burnside's Theorem[/url], stating that any group having an order with at most $2$ different prime factors is not simple. I read that Thompson subsequently found an "elementary" proof for the theorem, but that it's very tedious.

Anyway, for the record :), Burnside's Theorem follows immediately from the statement posted by sam-n:

Assume that $a,b\ge 1$ in $(*)$, consider the action of $G$ on itself by conjugation, and write the corresponding class equation. There are two cases:

Either every non-trivial orbit (i.e. conjugacy class of a non-central element) has a number of elements divisible by $p$, in which case the group must have a non-trivial center (because the center must have $kp$ elements for some $k\ge 1$), meaning that the group is not simple, or there is some non-trivial orbit with a number of elements not divisible by $p$, which must hence have a number of elements of the form $q^t$ for some $t\ge 1$, which leads, again, to the conclusion that the group is not simple (by using the statement posted here).


geometry
circumcircle
trigonometry
trig identities
Law of Sines
geometry unsolved

In triangle ABC, angle bisectors AA[size=59]1[/size] and CC[size=59]1[/size] intersect at point J. Find the circumradius of triangle AJC if angle ABC=Z and circumradius of triangle ABC=R.

by the extended law of sines, the circumradius of $ \triangle AJC$ is $ \frac {b}{\sin \angle AJC} \equal{} \frac {2R\sin Z}{\sin(90 \plus{} Z/2)} \equal{} \frac {2R \sin Z}{\cos(Z/2)}$


Let $ a,b,c\geq0$. Prove that

$ \sum(\sqrt{a\plus{}1}\plus{}\sqrt[3]{b\plus{}1}\plus{}\sqrt[6]{c\plus{}1})\leq a\plus{}b\plus{}c\plus{}9$

[quote="Ligouras"]Let $ a,b,c\geq0$. Prove that

$ \sum(\sqrt {a \plus{} 1} \plus{} \sqrt [3]{b \plus{} 1} \plus{} \sqrt [6]{c \plus{} 1})\leq a \plus{} b \plus{} c \plus{} 9$[/quote]

$ \sqrt{a\plus{}1}\plus{}\sqrt[3]{a\plus{}1}\plus{}\sqrt[6]{a\plus{}1}\le \frac{1\plus{}a\plus{}1}{2}\plus{}\frac{1\plus{}1\plus{}a\plus{}1}{3}\plus{}\frac{1\plus{}1\plus{}1\plus{}1\plus{}1\plus{}a\plus{}1}{6}\equal{}a\plus{}3$

=> Q.E.D

[quote="gigaman"][quote="Ligouras"]Let $ a,b,c\geq0$. Prove that

$ \sum(\sqrt {a \plus{} 1} \plus{} \sqrt [3]{b \plus{} 1} \plus{} \sqrt [6]{c \plus{} 1})\leq a \plus{} b \plus{} c \plus{} 9$[/quote]

$ \sqrt {a \plus{} 1} \plus{} \sqrt [3]{a \plus{} 1} \plus{} \sqrt [6]{a \plus{} 1}\le \frac {1 \plus{} a \plus{} 1}{2} \plus{} \frac {1 \plus{} 1 \plus{} a \plus{} 1}{3} \plus{} \frac {1 \plus{} 1 \plus{} 1 \plus{} 1 \plus{} 1 \plus{} a \plus{} 1}{6} \equal{} a \plus{} 3$

=> Q.E.D[/quote]

simple and beautiful your proof :wink:


Joe's bowling scores were 112, 182 and 142. What does he
need to score in his fourth game to get a mean score of 132?

112 is -20 from mean
182 is 50 from mean
142 is 10 from mean
to balance we need -40 from mean, or 92.


[quote="Calculus"]How do I join? What are the qualifications/requirements? Who should I contact?[/quote]

Hi, Calculus.  If you're in regular school, you might try contacting Eric Walstein at Montgomery Blair High School.  He's the coach for the Montgomery County teams.  If you're homeschooled, I can probably answer your questions.

Good luck!  :-)

[quote="Calculus"]How do I join? What are the qualifications/requirements? Who should I contact?[/quote]

The top scorers in the Montgomery County Math League (middle-school and high-school level) are invited to join the ARML team.  Generally, anyone is welcome to join the team, as they usually field 4-6 teams (generally, the A, B, and Jr. teams are competitive, while the other teams do it for fun).

They used to meet at Montgomery Blair High School on Wednesdays from 7-9pm starting around Feb./Mar.  (Check with Walstein)

I'm pretty sure Howard County also fields a team or two, but I don't know who the contact is.

[quote="mdk"]I live in D.C, do you think I can join ?
To my knowledge we don't have any arml teams.[/quote]

not sure.  You should contact the people as stated above.