modular arithmetic
number theory unsolved
number theory

You mean, when $ k$ is any [i]nonnegative [/i]integer. Otherwise you can make it as small as you like.

$ k\equal{}0$: impossible since $ 5^n$ is divisible by $ 5$ an odd number of times when $ n$ is odd.

$ k\equal{}1$: impossible since $ 5^n\plus{}1 \equiv 2 \pmod{4}$. Squares are $ 0$ or $ 1$ modulo $ 4$. $ n$ can be anything.

$ k\equal{}2$: impossible since $ 5^n\plus{}2 \equiv 3 \pmod{4}$. Ditto. $ n$ can be anything.

$ k\equal{}3$: impossible since $ 5^n\plus{}3 \equiv 3 \pmod{5}$ when $ n>0$. Squares are $ 0,1$ or $ 4$ modulo $ 5$.

So far, we never used the fact that the exponent is $ n\equal{}4m\plus{}3$, we've only used the facts that $ n$ is odd and greater than $ 0$. What's weird is that for $ n\equal{}4m\plus{}3$ I'm not at all sure that $ k\equal{}4$ is possible at all; it certainly is for $ n\equal{}1$. Are you sure about the wording of the problem?


Triangle ABC is isosceles with AB=AC. Point P is chosen on AC such that AP=BC. Also, BPC=40. What are the angle measure of the initial triangle?

Do you mean the area of BPC is 40 or the length of BPC is 40?

I'm pretty sure that he meant $\angle BPC=40^\circ$.

Hmm... There's a really nice and short solution if $\angle BPC=30^\circ$, but that doesn't work in this problem...

$10^{o}$ , $70^{o}$ , and $100^{o}$
Maybe I did something wrong, because I came to this answer very quickly.

that must be wrong, because ABC is isoceles -> two angles are the same.
I was thing about:
[hide]
rotate ABC and concat together on one of the legs to make a parallogram, then you get bunch of symetries.
[/hide]
Although I can't finish it.

Yeah that's it. Sorry for the typo. :blush:  :oops:

[quote="kimby_102"][quote="john lai"][hide]

Is angle ABC=ACB+80, angle BAC=20?[/hide][/quote]

Do u mean $\angle ABC = \angle ACB = 80^\circ$?[/quote]
That gives $\angle BPC=30^\circ$.

Now I have found like 3 different solutions if $\angle BPC=30^\circ$, not $40^\circ$... :|

[quote="10000th User"]Is the condition AP=BC correct? Because if you use AP=PC, then the problem becomes easier, considering that he got the problem from a geometry class...[/quote]

Yeah...what kind of geometry class is this...??? (I spent like 1 hour)


MATHCOUNTS
modular arithmetic
number theory
Diophantine equation

There have been a couple questions like this in various Mathcounts competitions, mostly nationals.

17x+13y=262
Find the only solution set (x,y) where x and y are positive integers.
Is there a quick way to do this without using modulo, or going as far as solving it as a diophantine?

it's really easy with a graphing calculator.
is this one a target round? you'd get to use a calculator.
it's [hide]x=7, y=11[/hide]

[hide="maybe..."]$4x\equiv 2 \pmod{13}$. The first few multiples of 4 are
4
8
12
16
20
24
28, for x=7, which is $\equiv 2 \pmod{13}$. Plugging it back in to the equation,it works, and we see that y=11.[/hide]

I'm pretty sure they wouldn't give you something like this on a sprint.

Then again, it might be like number 20 on a national one  :D

I'm pretty sure it comes up, just maybe slightly simpler, or disguised.  Always make use of the graphing calculator, it has solved many problems for me without doing brute force.

I would use guess and check on sprint, taking the largest possible multiple of 17, then listing the multiples of 13, then seeing which ones can get to the right units digit, and then check if the rest works.

This was one of the old school "individual," or target rounds, back in the day when calculators weren't allowed in the 10 question target rounds.
Now a days, these end up mostly on state team rounds, around 8 or 9.

[hide]I get that you should try modulo 13.

Then you get that 4x is congruent to 2(mod 13).

So this is pretty simple to see that the first value would be x=7 and y=11

(7,11)

I can't think of anything without mods, sorry.[/hide]

[quote="jclarkemathL314159"]This was one of the old school "individual," or target rounds, [b]back in the day when calculators weren't allowed in the 10 question target rounds.[/b]
Now a days, these end up mostly on state team rounds, around 8 or 9.[/quote]

[quote="SplashD"]Therefore, you can use a calculator and solve it easily :coool:[/quote]

Oops, guess not.

I think it is best to try to find a way to solve a problem without a calculator, and then use your calculator if needed to help you with calculations.

Well, you learn arithmetic in school.  I bet the testwriters made it with brute force + listing in mind.  Otherwise, I would

[hide]subtract 13y over, and see of what modulo 17 class 262 is in, then try to match that with a multiple of 13.[/hide]

[hide]
If y and x's average was z, then 30z=262.
So, z is almost 9, which means that y (or x) is farthur away from their average than x.
If they both equal 8, their sum is 240.
Using guess and check, you get the answer: (7,11)[/hide]

[quote="SonicaSaraf"]x=7
y=11

        you have to change the equation to 13x+13y= 234

use trial and error to get 234[/quote]

Why couldn't y=18 and x=0 then?

[quote="Ignite168"][quote="SplashD"][quote="jclarkemathL314159"]Find the only solution set (x,y) where x and y are positive integers[/quote]

Does that answer your question, Ignite?[/quote]

No, it could be (17,1) (16,2) (15,3) etc. according to the equation SonicaSaraf made.[/quote]

Sry, I wasn't looking closely. I didn't know you were talking of SonicaSaraf's equation

A basic Diophantine equation can be understood by advanced middle schoolers.  I will post a solution using simple Diophantine equations:

$17x+13y=262$
Now we put it in mod 13, so we can eliminate the y term, and get:
$4x=2 (mod 13)$
Therefore, $x=7$.  Plug in and see that $y=11$.  This is very similar to other solutions, but I thought I would show it with mods...

This was an old nationals team round question.

okay, so I guess it is a good thing that I did this without a calculator. :lol:

[quote="mathgeniuse^ln(x)"][quote="pgpatel"]A really old one by the way. I think it was before calcs were allowed on the team round.[/quote]

uM, TAHT WAS ALSO THE SAME TIME THAT CALCS WERE ALLOWED ON THE TARGET ROUND AS WELL, FYI.[/quote]

This is random, but you should turn your Caps Lock off.


x,y,z are positive real numbers
$x+y+x y=8$
$y+z+y z=15$
$z+x+z x=35$

d e tr mi n e
$x+y+z+x y z$

[hide="Solution using SFFT"]
Using Simon's Favorite Factoring Trick:
$A=(1+x)(1+y)=3^{2}$
$B=(1+y)(1+z)=4^{2}$
$C=(1+z)(1+x)=6^{2}$

Since $A,B,C>0$ and $(1+x),(1+y),(1+z)>1$, we can do this:
$\sqrt{\frac{AB}{C}}=1+y=2$
$\sqrt{\frac{BC}{A}}=1+z=8$
$\sqrt{\frac{CA}{B}}=1+x=\frac{9}{2}$

Now, if I computed correctly, $x+y+z+xyz=36$[/hide]

[hide="solution"]We can use Simons Favorite Factoring Trick to get

(x+1)(y+1)=9
(y+1)(z+1)=16
(x+1)(z+1)=36

We multiply all together and take the square root:

(x+1)(y+1)(z+1)=72

so z+1=8, z=7
y+1=2, y=1
9/2=x+1, x=3.5

We compute x+y+z+xyz:

11.5+24.5=36[/hide]


Take $8\times 8$ chessboard. Remove one black and one white square of it. Prove that it is possible to cover the board by 31 $2\times 1$-dominoes.

[quote="Hokkage"]Since each domino covers 1 black and 1 white, if there are 31 black and 31 white then 31 dominos can cover it. (Not sure if this is rigorous enough to constitute a proof.)[/quote]

It isn't i don't think...That fact is [b]necessary[/b] to cover the board, but not [i] sufficient [/i].  It's important to make that distinction.  By your logic, it's fine to simply cover the board as you wish.  But then you may end up with a white and black on opposite sides of the board.  I'm thinking a proof along the following lines.  Suppose you covered an 8x8 chessboard with dominoes so that each one covers two horizontally adjacent squares (so like a 4x8 square of 2x1 dominoes). Then remove the two squares, and the two dominoes (or one in the trivial case) which are on those squares.  Prove that you can manipulate the remaining 30 dominoes to get those two uncovered spaces on the board next to each other.  I think I've developed a strategy to move an uncovered space horizontally, as well as one to move it vertically, but I don't know exactly how to prove it...


Any hints are appreciated. thank you^^

1. There are 15 rabbits, 12 mice, and 10 deer. and they are caught in random order, calculate the probability that mice go extinct first.

2. Roll a fair, six sided die 6 times,
  (a) what is the probability that each face comes up at least once?
  (b) each roll is >= the previous roll?


function
algebra
polynomial
calculus
integration
group theory
abstract algebra

which functions on $ [0,1]$ can be approximated via the polynoms  with integer coefficients ?

It's known that the polynomials with real coefficients are dense in the continuous functions on an interval.  I believe that the polynomials with rational coefficients have the same property; unfortunately, the polynomials $ P(x)$ with integer coefficients are constrained at least by the obvious requirement that $ P(0)$ and $ P(1)$ both be integers, so I'm not sure what the proper answer to this question is.

A function $ f$ in the closure of $ \mathbb Z[X]$ preserves the algebraic integers in $ [0,1]$. Moreover, if $ \alpha\in[0,1]$ is an algebraic integer, then $ f(\alpha)\in\mathbb Z[\alpha]$. It seems very likely to me that this is the defining property for $ \overline{\mathbb Z[X]}$ not having a rigorous proof in stock at the moment.

That doesn't seem right to me; the polynomials with rational coefficients are dense in the polynomials with real coefficients, hence in the continuous functions, but on the other hand the rational polynomials preserve the algebraic numbers.  I don't think algebraic statements hold up in closures (with respect, I assume, to uniform convergence).

Well, the only restriction (for uniform approximability) is $ f(0),f(1)\in\mathbb Z$. Indeed, subtracting an appropriate linear polynomial, we can assume that $ f(0) \equal{} f(1) \equal{} 0$. Now, let us use the Bernstein polynomials $ \sum_k{n\choose k}f(k/n)x^k(1 \minus{} x)^{n \minus{} k}$. Note that the coefficients in front of $ x^k(1 \minus{} x)^{n \minus{} k}$ can be changed by $ 1$ or less and the polynomial will change by just $ 1/n$ or less (here it is important that $ k \equal{} 0$ and $ k \equal{} n$ do not really appear in the sum). Thus, we can safely replace the coefficients by the nearest integers. Since the Bernstein polynomials converge to $ f$, we are done.

Thank you fedja , I was thinking about Bernstein Polynoms , But I did not understand as to achieve the integer coefficients

[quote="t0rajir0u"][quote="Third Edition"]Perhaps there is a kind of confusion here: I am talking about algebraic [b]integers[/b], i.e. the integral closure of $ \mathbb Z$ in the field $ \mathbb Q^*$ of algebraic [b]numbers[/b].[/quote]

I was talking about a different set of polynomials (the ones with rational rather than integer coefficients).  The point still stands - a limit of functions that preserve the algebraic integers need not preserve the algebraic integers.[/quote]@ t0rajir0u: Thanks for insisting here. I was led by the idea that a number ring is algebraically isomorphic to $ \mathbb Z^n$, hence discrete, such that every convergent sequence gets eventually constant. But this is not true. That's why Minkowski embedding come into play in analytic number theory. And the prominent counterexample $ \mathbb Z\plus{}\mathbb Z\sqrt{2}$ is not discrete, but a dense subgroup of $ \mathbb R$. Thanks again - I should better have "embedded" myself instead of posting in the early morning. ;)

This is what fedja meant  :) .
it's obviusly that $ f(0), f(1) \in \mathbb{Z}$ and $ f$ is bounded on $ [0,1]$
there exist $ a,b \in \mathbb{Z}$ such that : $ f(0) \plus{} a*0 \plus{} b \equal{} f(1) \plus{} a*1 \plus{} b \equal{} 0$
consider $ g(x) \equal{} f(x) \plus{} ax \plus{} b$
there exist $ m \in \mathbb{N}$ such that $ \max_{0\leq x\leq 1}\frac {g(x)}{m} \leq 1$
lets denote $ F(x) \equal{} \frac {g(x)}{m}$
OK.
I use theorem (about berstein Polynoms):
for all $ \epsilon > 0$ there exist $ N \in \mathbb{N}$ such as for all $ n > N$ and  for all $ x \in [0,1]$ we have
\[ |F(x) \minus{} \sum_{0\leq k \leq n}\binom{n}{k}F \left( \frac {k}{n} \right)x^{k}(1 \minus{} x)^{n \minus{} k} | < \epsilon
\]
Now  fix $ n$ and consider $ a_{k}$ such as $ a_{k}\binom{n}{k} \in \mathbb{Z}$ where $ k \in \{1,2,,n \minus{} 1\}$ and $ a_{0} \equal{} a_{n} \equal{} 0$
Now consider
\[ \max_{k}|F\left( \frac {k}{n} \right) \minus{} a_{k}|
\]
we can choose rational $ a_{k}$  , it's easy to note that only $ a_{1},a_{n \minus{} 1}$ has minimal denominator since  $ ( \equal{} n)$
since from the set of numbers $ \{\binom{n}{1},\binom{n}{2},...,\binom{n}{n \minus{} 1} \}$ minimal integer is $ \binom{n}{1} \equal{} \binom{n}{n \minus{} 1} \equal{} n$
lets divide $ [0,1]$ $ n$ parts . Hence for $ F\left( \frac {k}{n} \right)$ there exist $ j$ such that $ F\left( \frac {k}{n} \right) \in [\frac {j}{n},\frac {j \plus{} 1}{n}]$
hence by choosing $ a_{i} \equal{} \frac {j}{n}$ we can see that
\[ \max_{k}|F\left( \frac {k}{n} \right) \minus{} a_{k}| \leq \frac {1}{n}
\]
now consider :
\[ | F(x) \minus{} \sum_{0\leq k \leq n}\binom{n}{k}a_{k}x^{k}(1 \minus{} x)^{n \minus{} k}| \leq
\]

\[ |F(x) \minus{} \sum_{0\leq k \leq n}\binom{n}{k}F\left( \frac {k}{n}\right)x^{k}(1 \minus{} x)^{n \minus{} k}| \plus{} \sum_{0\leq k \leq n}\binom{n}{k}|a_{k} \minus{} F\left( \frac {k}{n} \right)|x^{k}(1 \minus{} x)^{n \minus{} k}\leq
\]

\[ \epsilon \plus{} \max_{k}|F\left( \frac {k}{n}\right) \minus{} a_{k}| \sum_{1\leq k \leq n \minus{} 1}\binom{n}{k}x^{k}(1 \minus{} x)^{n \minus{} k} \leq \epsilon \plus{} \frac {1}{n}
\]
since
\[ \sum_{1\leq k \leq n \minus{} 1}\binom{n}{k}x^{k}(1 \minus{} x)^{n \minus{} k} \equal{} 1 \minus{} x^{n} \minus{} (1 \minus{} x)^{n} \leq 1
\]
for $ x\in[0,1]$
thus $ f(x)$ can be approximated by this way
\[ f(x) \approx m\sum_{0\leq k \leq n}\binom{n}{k}a_{k}x^{k}(1 \minus{} x)^{n \minus{} k} \minus{} ax \minus{} b
\]
 where error will be $ \approx  m \epsilon \plus{}m/n$  last tends to $ 0$ for small $ \epsilon$ and $ n \to \infty$


For acute triangle ABC, D is a perpendicular foot of A to BC, and E is the intersection of AD and triangle ABC's circumcircle(E is not A).
Also, F is a perpendicular foot of E to AC.
Prove that $ [\text{Circumcircle\ of\ ADF}]\plus{}[\text{Circumcircle\ of\ CEF}]\equal{}[\text{Circumcircle\ of\ ABC}]$.

It is given that $ \angle B\neq \angle C$.

The circumradius of triangle FEC is $ R cosC$, while the one of ADF is $ R sinC$. By adding the squares of these formulas, we get the desired relation.

Best regards,
sunken rock


algebra
polynomial
Vieta
geometry
3D geometry
calculus
trigonometry

Complete the cube and then use the method of vieta to find all roots of x^3 + 6x^2 +9x + 3 = 0 in the form a cos(w) + b for suitable a,b and w

I moved this out of the calculus-analysis forum because it's really more of an algebra/trigonometry problem then a calculus problem.

I'm teaching a course this semester on "Early History of Mathematics." When we get to the 16th century, shortly after we talk about the whole Tartaglia - Cardano - Ferrari soap opera (and Bombelli and complex numbers) we will then talk about Vieta and I will teach this specific method.

$ x^3 \plus{} 6x^2 \plus{} 9x \plus{} 3 \equal{} 0.$

The first step is to get rid of the $ 6x^2$ term. Let $ y \equal{} x \plus{} 2.$ After multiplying everything out, we get

$ y^3 \minus{} 3y \plus{} 1 \equal{} 0.$

The Vieta method comes from the identity $ \cos 3\theta \equal{} 4\cos^3\theta \minus{} 3\cos\theta.$

In general, if we have $ y^3 \minus{} py \plus{} q \equal{} 0,$ we're going to let $ y \equal{} 2\sqrt {\frac {p}{3}}\,\cos\theta.$ In this case, that's $ y \equal{} 2\cos\theta.$

$ 8\cos^3\theta \minus{} 6\cos\theta \plus{} 1 \equal{} 0$

$ 4\cos^3\theta \minus{} 3\cos\theta \plus{} \frac12 \equal{} 0$

$ \cos3\theta \equal{} \minus{} \frac12.$

Hence either $ 3\theta \equal{} 120^{\circ}$ or $ 3\theta \equal{} 240^{\circ}.$ It turns out that either one of these alone will give us all of the answers, so let $ 3\theta \equal{} 120^{\circ}$. Divide the angle by $ 3$ to get that $ \theta \equal{} 40^{\circ}$ or $ 160^{\circ}$ or $ 280^{\circ}.$

Put that back into $ y \equal{} 2\cos\theta,$ then subtract $ 2$ to get $ x.$ The three roots are:

$ x \equal{} \minus{} 2 \plus{} 2\cos 40^{\circ}\approx \minus{} 0.4679$

$ x \equal{} \minus{} 2 \plus{} 2\cos 160^{\circ}\approx \minus{} 3.8794$

$ x \equal{} \minus{} 2 \plus{} 2\cos 280^{\circ}\approx \minus{} 1.6527$

One assumes that a 16th century equation solver would have consulted a previously prepared table of trigonometric functions.

------

Assuming we are talking about real cubic equations, this method can only be used to solve equations with three real roots. These are precisely the equations that give the Cardano-Tartaglia method the most trouble. When there are three real roots, the Cardano-Tartaglia method has a negative under a square root in the interior of the formula, and only gives the correct answers if you interpret this as a complex number and are able to take the cube root of a complex number. When there is only one real root, the Cardano-Tartaglia method produces that root through real arithmetic.

Think about what this says about constructing (with geometry tools) the roots of cubic equations. Cardano-Tartaglia says that for cubic equations with one real root, we could construct that root if only we could construct cube roots. (If we could do that, that would include the classical problem of duplicating the cube.) Vieta says that for cubic equations with three real roots, we could construct those roots if only we could trisect angles. (Another classical problem.)

It turns out that we can do both with the compass, straightedge, and the Archimedes marked straightedge.

thank you very much ....excellent answer


Friends,

Over the past few days i was trying to solve the following problem which demands me to devise an algortihm for the following problem---

Given a connected graph which may or may not be biconnected determine the minimal edge set inclusion which makes the graph biconnected (if it was not) else leaves it as such. Please tell me if u know of some efficent algorithm.  This problem has already haunted me enough. So please help me out. Thanx in advance.

HAVE A NICE DAY...

1) Find the block tree of the graph (in linear time; use DFS).
2) Connect the leaf blocks by a path, i.e. the nuumber of edges we need to add equals the number of leaves minus one.


[quote="mets501"][/quote]
$b^{4}=a^{-\frac{16}{5}}$
[/quote]

How does that happen?

He took both sides to the power of $\frac{4}{5}$.  Remember that $(x^{y})^{z}= x^{yz}$.

[quote="ckck"]1. If $a^{4}b^{5}=1$, evaluate $log_{a}(a^{5}b^{4})$.

2. $2^{x}=3^{y}=\sqrt{10^{z}}$, find $\frac{(x+y)z}{xy}$.[/quote]

I think the second one is just [b]log 36[/b]..I just set everything equal to a c, and found them in logs.  You can then use base change and then mutiply everything by 1/log c, wehre c is what everything equals, and you get to 2(log2+log3) and then 2(log6) so log 36..but Im too tired(late) to type full solution..

[quote="ckck"]2. $2^{x}=3^{y}=\sqrt{10^{z}}$, find $\frac{(x+y)z}{xy}$.[/quote]

Let $S = \frac{(x+y)z}{xy}$

Solve for y:
$2^{x}= 3^{y}$
$x \cdot \text{ln}{2}= y \cdot \text{ln}{3}$
$y = \frac{x \cdot \text{ln}{2}}{\text{ln}3}$

Solve for z:
$2^{x}= \sqrt{10^{z}}$
$2^{2x}= 10^{z}$
$2x \cdot \text{ln}2 = z \cdot \text{ln}10$
$z = \frac{2x \cdot \text{ln}2}{\text{ln}10}$

Substitute:
$S = \frac{(x+\frac{x \cdot \text{ln}2}{\text{ln}3}) \cdot \frac{2x \cdot \text{ln}2}{\text{ln}10}}{x \cdot \frac{x \cdot \text{ln}2}{\text{ln}3}}$
$S = \frac{(x+\frac{x \cdot \text{ln}2}{\text{ln}3}) \cdot 2 \cdot \text{ln}3}{x \cdot \text{ln}10}$
$S = \frac{(1+\frac{\text{ln}2}{\text{ln}3}) \cdot 2 \cdot \text{ln}3}{\text{ln}10}$
$S = \frac{(\text{ln}3+\text{ln}2) \cdot 2}{\text{ln}10}$
$S = \frac{2 \cdot \text{ln}6}{\text{ln}10}$
$S = \frac{\text{ln}36}{\text{ln}10}$
$S = \text{log}_{10}36$

$S = \boxed{\text{log }36}$


[quote="hien"]$\frac{(x+y+z)^{2}}{2(x+y+z)}=\frac{1}{2(x+y+z)}$

[/quote]

This is not true.

And when $x=y=z=1$, expression is equal to $\frac{3}{2}$

This is not a strict problem. The better one is that (also familiar)

If $a^{4}+b^{4}+c^{4}=3$ then
\[\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\ge \frac{3}{2}.\]

[quote="GacKiem"]If $a,\,b,\,c$ be positive numbers such that $a^{4}+b^{4}+c^{4}=3$ then
\[\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\ge \frac{3}{2}. \]
[/quote]
And the better one:
[quote][color=indigo]If $a,\,b,\,c$ be positive numbers such that $a^{4}+b^{4}+c^{4}=3$ then \[\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\ge 3\][/color][/quote]

[quote="N.T.TUAN"]Find the least value of $\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}$, where $x,y,z$ are positive reals satisfy $\sqrt{x^{2}+y^{2}}+\sqrt{y^{2}+z^{2}}+\sqrt{z^{2}+x^{2}}=3\sqrt{2}$.[/quote]
[b] Solution.[/b]
Let $a=\sqrt{x^{2}+y^{2}},\,b=\sqrt{y^{2}+z^{2}},\,c=\sqrt{z^{2}+x^{2}}$
Then: $\sqrt 2a\ge x+y,\,\sqrt 2b\ge y+z,\,\sqrt 2c\ge z+x$
Hence we have: \[\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}\ge\frac{1}{2\sqrt 2}\left(\sum\frac{b^{2}+c^{2}-a^{2}}{a}\right)=\frac{1}{2\sqrt 2}\left(\sum a^{2}\sum\frac{1}{a}-2\sum a\right)\\ \ge\frac{1}{2\sqrt 2}\left(\frac{\left(\sum a\right)^{2}}{3}\cdot\frac{9}{\sum a}-2\sum a\right)=\frac{a+b+c}{2\sqrt 2}=\frac{3}{2}\]

That's right!  :wink: This also is solution from M&Y.

[quote="GacKiem"]This is not a strict problem. The better one is that (also familiar)

If $ a^{4}+b^{4}+c^{4}=3$ then
\[ \frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\ge \frac{3}{2}. \]
[/quote]
I agree with you, but $ a^{2}+b^{2}+c^{2}=3$ enough, because
$ \sqrt{6(a^{2}+b^{2}+c^{2})}\geq\sum_{cyc}\sqrt{a^{2}+b^{2}}$ already. :wink:


real analysis
inequalities
triangle inequality
real analysis theorems

Maybe I am missing something simple, but I am having trouble understanding the proof of the Heine-Cantor Theorem written here:

http://planetmath.org/encyclopedia/ProofOfHeineCantorTheorem.html

I do not understand why the two subsequences $\{x_{k_j}\}$ and $\{y_{k_j}\}$ must converge to the same limit if $d(x_k, y_k) \rightarrow 0$. If someone could put in a few more words to explain this part, that would be very helpful. Thanks.

My guess is that if you let $a_k=d(x_k,y_k)$ and $a_k\to 0$, and if $a$ is a limit point of $a_k$, $a=0$. Since $x,y$ are the limit points of respective $x_k,y_k$, $a=|x-y|=0$ or $x=y$ by definition of a metric... I hope this is correct...

Maybe it has something to do with $|a_k-a|=|(x_k-y_k)-(x-y)|\le|x_k-x|+|y_k-y|<\frac12\epsilon+\frac12\epsilon=\epsilon$

This is Thm 4.19 in Rudin, and is handled in a different way.

If $\overline{x}\not=\overline{y}$, put $t=d(\overline{x},\overline{y})$.  By the triangle inequality, $t\leq d(y,y_n)+d(y_n,x_n)+d(x_n,x)$ for all $n$.  Given $\epsilon>0$, let $n$ be an index of the aforementioned subsequence such that $d(y,y_n)<\epsilon$ and $d(x_n,x)<\epsilon$ and also such that $d(y_n,x_n)<\epsilon$ (why do such $n$ exist?).  It follows that $0\leq t\leq 3\epsilon$ for every $\epsilon>0$, whence $t=0$, a contradiction.


In ciuda, "neplacerii" domnului harazi de a vedea inegalitate pe forumul romanesc, eu o mai pun pe aceasta, caci mi-a placut cum mi-a iesit:

Sa se arate ca pentru $a,b,c$ reale pozitive respectand relatia: $a^{2}+b^{2}+c^{2}+2abc=1$ are loc [b]inegalitatea[/b] :D :

$2 \cdot (\frac{a^{3}}{a+bc}+\frac{b^{3}}{b+ca}+\frac{c^{3}}{c+ab})\geq (\frac{a^{2}}{a+bc}+\frac{b^{2}}{b+ca}+\frac{c^{2}}{c+ab})^{2}$

Cosmine, exista o mare diferenta intre "numai inegalitate" si "inegalitate". Exista unele care apar doar de dragul de a mai scrie o chestie enorma si fara sens. Altele, chiar mai au un sens si chiar pot fi folositoare. In fine, cele mai multe sunt doar combinatii, de multe ori nereusite de chestii foarte simple, dar aspectul lor monstruos creeaza senzatia unor teribile rezultate matematicii. Sa luam acest exemplu, pe care din pacate chiar am pierdut 2 minute: evident, nu poate sa se ascunda decat Cauchy-Schwarz in spatele unei astfel de chestii. Ok, si dupa ce aplicam Cauchy-Schwarz, ce observam? Ca se "reduce" la o chestie numita $\sum{\frac{bc}{a+bc}}\geq 1$. Care oricum este cu mult mult mai draguta decat cea initiala (atunci, why not giving it directly???). Culmea, ca asta nu iese direct. Dar... un mic calcul arata ca e de fapt o egalitate! De asta nu a fost data direct!

:) aveti dreptate, oricum, am cam supraapreciat-o , mi-am dat seama, imediat dupa ce scrisesem postul de solutia aceasta, oricum ideea de baza era rescrierea conditiei din ipoteza in : $\sum{\frac{a}{a+bc}}=2$


(i) Prove that the product of $ 3$ consecutive positive integers cannot be a perfect square.

(ii) Prove that the product of $ 4$ consecutive positive integers cannot be a perfect square.

[hide="Solution for ii"]

Let $ (x)(x\plus{}1)(x\plus{}2)(x\plus{}3) \equal{} k^2$
$ \Rightarrow (x)(x\plus{}3)(x\plus{}2)(x\plus{}1) \equal{} k^2$
$ \Rightarrow (x^2 \plus{} 3x \plus{}2)(x^2 \plus{} 3x) \equal{} k^2$
$ \Rightarrow (x^2 \plus{} 3x \plus{}1 \plus{}1)(x^2 \plus{} 3x \plus{}1 \minus{}1) \equal{} (x^2 \plus{}3x \plus{}1)^2 \minus{}1 \equal{} y^2$

No such perfect square exists that is one less than another perfect square.

[/hide]

Thanks.

I had $ (n\minus{}1)n(n\plus{}1)(n\plus{}2) \equal{} (n^{2} \plus{} n \plus{}1)^{2} \minus{} 1$, which is the same as yours.

For (ii)

[hide="Solution"]First we prove that $ \sqrt {(n \minus{} 1)n(n \plus{} 1)(n \plus{} 2) \plus{} 1} \in \mathbb{N}, \forall n \in \mathbb{N}$ which is rather easy:

$ \sqrt {(n^2 \plus{} n \minus{} 2)(n^2 \plus{} n) \plus{} 1} \equal{} \sqrt {(n^2 \plus{} n \minus{} 1)^2} \equal{} n^2 \plus{} n \minus{} 1 \in \mathbb{N}$.

But if $ \sqrt {k \plus{} 1} \in N$, then $ \sqrt {k}$ is not an integer ($ k \neq 0$).

(Proof: Let $ k > 4$ and $ k \plus{} 1 \equal{} n^2, n \in \mathbb{N}$, then $ n^2 > k \equal{} n^2 \minus{} 1 > n^2 \minus{} 2n \plus{} 1 \equal{} (n \minus{} 1)^2$ . It's obvious that the product of any 4 consecutive positive integers is larger than 4)[/hide]

Prove that the product of 5 consecutive positive integers cannot be a perfect square?

Here's a solution for i.

Let $ (x\minus{}1)(x)(x\plus{}1) \equal{} k^2$
Then $ (x)(x^2 \minus{} 1) \equal{} k^2$
Now $ x$ and $ (x^2 \minus{}1)$ are coprime; which means each has to be a perfect square. Now $ x^2 \minus{}1$ can't be a perfect square and thus we are done.

[quote="harekrishna"]Prove that the product of 4 consecutive naturals can't be a perfect cube.[/quote]

[hide="Hint"]
$ (n \minus{} 1)(n)(n \plus{} 1)(n \plus{} 2) \equal{} m^{3}$  to get

$ (n^{2} \plus{} n \minus{} 1)^{2} \equal{} m^{3} \plus{} 1$

[/hide]

For (i):

$ n^{2}$ and $ n^{2}\minus{}1 \equal{} (n\minus{}1)(n\plus{}1)$ are co-prime since $ n^{2}$ and $ n^{2}\minus{}1$ are consecutive integers.

This means that $ (n\minus{}1)n(n\plus{}1)$ can  only be a square if $ n$ and $ n^{2}\minus{}1$ are squares. But $ n^{2}\minus{}1$ is not a square.

QED

[quote="harekrishna"]No such perfect square exists that is one less than another perfect square.[/quote]0 and 1 are consecutive perfect squares ;) This will only work if it is nonzero perfect square.

[quote="Maths Mechanic"]$ (n^2\plus{}n\minus{}1)^2\equal{}m^3\plus{}1$[/quote]$ 3^2\equal{}2^3\plus{}1$ ;) Can $ n^2\plus{}n\minus{}1\equal{}3$?

[quote="AndrewTom"]For (i):

$ n^{2}$ and $ n^{2} \minus{} 1 \equal{} (n \minus{} 1)(n \plus{} 1)$ are co-prime since $ n^{2}$ and $ n^{2} \minus{} 1$ are consecutive integers.

This means that $ (n \minus{} 1)n(n \plus{} 1)$ can  only be a square if $ n$ and $ n^{2} \minus{} 1$ are squares. But $ n^{2} \minus{} 1$ is not a square.

QED[/quote]

Isn't this very similar to mine?

@cnyd, how do you prove that the product of 5 consecutive positive integers cannot be a perfect square?


show that 1+2+3+4+..+n never equals x^n, with x > 1, n > 1, naturals

[hide="Quick"]$ 1 \plus{} 2 \plus{} \ldots \plus{} n < 1 \plus{} (1 \plus{} 2 \plus{} 4 \plus{} \ldots \plus{} 2^{n\minus{}1}) \equal{} 2^n \leq x^n$ when $ x < 1$[/hide]


function
vector
superior algebra
superior algebra unsolved

Let $K$ be a field and let $I$ be an infinite set. Consider the set of functions $K^I=\{f\colon I\to K\}$. $K^I$ is a vector space over $K$ (w.r.t. common addition of functions and multiplication on elements from $K$). Evaluate the dimension of $K^I$ over $K$.

It is easy to find an infinite set of independant functions : $f_i=1_i$ where $i$ is an element of $I$. Thus this is an infinite demensional vector space. I think you can also guess the answer when $I$ is finite  :)

The dimension is here the cardinality of the vector space ... you can find the rather complicated proof in

Gottfried Koethe, Topologische lineare Räume, Springer-Verlag 1960

Yes, that's right $dim_KK^I=|K|^{|I|}$. I have not read that book. May be my own proof is even more complicated.

Eh, -oo- , you have spoiled my experiment. I have proposed this problem many times and the common (incorrect) answer was $2^{|I|}$. Even some good specialists have made such a mistake. So my test failed because of Your remarkable erudition.  :clap2:

The proof is based on two (easy?) lemmas.

[b]Lemma 1.[/b] Let $V$ be a vector space over a field $K$ with a basis $B$, then
\[ |V|= \left\{\begin{array}{ll} |K|^{|B|}&\mbox{\em if } B\mbox{ \em is finite}\\ |K|\cdot|B|&\mbox{\em if } B\mbox{ \em is infinite} \end{array}\right. \]
[b]Lemma 2.[/b] The dimension of $K^{\mathbb N}$ ($\mathbb N=\{1,2,3,\ldots\}$) over $|K|$ is greater or equal $|K|$.

Let $I$ be an infinite set and let $B$ be a basis of $K^{I}$ over $K$. $B$ is infinite hence in view of Lemma 1 $|K^I|=|K|\cdot|B|$. There exists a monomorphism $K^{\mathbb N}\to K^I$ so by Lemma 2 $|B|=dim_K K^I\geqslant dim_{K}K^{\mathbb N}\geqslant |K|$. Finally $|B|=|K|\cdot|B|=|K^I|$.


A rectangle has diagonal length 13 units and perimeter 38 units.  Find the rectangle's area without first finding either the rectangle's length or width.

Thanks.

If the side lengths are $ a$ and $ b$, that means we are looking for $ ab$. We can write the equation \[ \sqrt{a^2 \plus{} b^2} \equal{} 13\] Now, we can factor the expression under the radical as $ (a \plus{} b)^2 \minus{} 2ab$, so we have \[ \sqrt{(a \plus{} b)^2 \minus{} 2ab} \equal{} 13\] From what we're given about the perimeter, we can deduce that $ a \plus{} b \equal{} 38 \div 2 \equal{} 19$. That means we have \[ 361 \minus{} 2ab \equal{} 169\] Rearranging gives \[ ab \equal{} \colorbox{red}{96}\]


If ( n - 6 ) ! = 72 , then n is equal the ?

How to do ?

Hi Luis. Is there a typo here: $ 72 \equal{} (3)(4!)$?

hello, i also think there is a typo in your equation.
Sonnhard.

$ 72$ isn't a number that is a factorial of anything so there must be something wrong here [b]luiseduardo[/b]

By 72 does luis mean 720, because $ 720\equal{}6!$


If $ x@y\equal{}xy\minus{}2x$, what is the value of $ (5@3)\minus{}(3@5)$?


If each number in a set of 15 numbers is doubled, how is the average changed?

If each number in a set of 15 numbers is multipied by 10, how is the average changed?

If each number in a set of 15 numbers is multiplied by $n$, how is the average changed?

If a set of 100 numbers has the members ${1, 2, 3, 4, \cdots, 99, 100}$ how is their average changed when the number $1,000,000$ is added to the set?

If a set of 5 numbers has the members ${1, 5, 8, x, 2}$ and their average is 4, what is $x$?

Triangular number $n$ is made of $1+2+3+\cdots+m$. If each of the numbers whose sum equal $n$ is multiplied by $m$, what will their new sum equal, in terms of $n$?

Enjoy!  :)

This problem has been here a while now. I suppose I'll move it to the Mathcounts forum.  ;)

Hope you don't mind.


Concerning the AP Physics test B and C, coming up in May 9th, I will be talking to the admins about locking the forum, because of the time zone differences and the possible alternate testing dates. So I urge you AP test takers not to talk about the tests until those are clear.

By the way, good luck to all the AoPS AP Physics Test Takers!!! Only few days away, to study and prepare.

That would easily extend to most of the forums in this category, and we're not even about Calculus forums. Do you really think it's necessary?

oh, no......how long will it last?? do we have to lock the forum for the test??

not that i have any say, but i think as long as its noted to not talk about the tests there will be no problems.  AoPS is a very good community and I think we can count on everyone to not mention the tests.  all you have to do is put up a sticky saying "PLEASE DO NOT TALK ABOUT THE AP PHYSICS TESTS" or something of the sort.

well I understand what you guys are talking about, and i havent even considered actually locking it yet.... noticed that i said i will be talking to the admin about it....

but for the last test(I think it was like Collaborative Problem Solving Contest), there were some issues about somebody copying down the problem and posting it... and I remember the director/moderator did not do a good job...(I personally know him).... if not posting about the test is not an issue, i guess it is not necessary. But all this AFTER i hear what the admin says.

[quote="shobber"]well, i don't know why you cannot talk about the tests......
if you have finished the exams, then it doesn't matter if you discuss the problems. You have hand in the papers so you cannot change the results.[/quote]

there are people who, under the circumstances, will be taking the test in an alternate dates.... that is May 19th...(although i doubt any of those people will be on AoPS)

[quote="1234567890"]Let me clear this up:

-There is a different test on the Alternate Day then the First Day
-No one is allowed to discuss multiple choice for any AP Test EVER. (They are reused)
-You can discuss the free response 2 days after.[/quote]
Why?? I don't think it will cause trouble if we discuss the problems. So why do they have this kind of law??

[quote="xxreddevilzxx"]well I understand what you guys are talking about, and i havent even considered actually locking it yet.... noticed that i said i will be talking to the admin about it....
 But all this AFTER i hear what the admin says.[/quote]Since this is not a Physics dedicated forum, I doubt people visiting [b]this[/b] particular corner of AoPS - MathLinks would do so to cheat in AP Physics test :) And as I see the activity around here, my last thought would be to lock this (already) low posting forum :)

If we get something official from the organizers, then we'll lock it of course. Other than that it's up to you to delete any inappropriate posting of the questions ...

Yes, just re-emphasizing what 123456890 said:
The CollegeBoard AP website releases publicly the questions 48 hours after the test. When the test takers themselves release the problems for open discussion, I do not think there is a reason for us not to discuss the free response. The multiple choice are NEVER released so we can never talk about a specific multiple choice question EVER anyway.

For the people taking it on the second day, it is NOT the same test. It is a second version, a form B.

I see. Thank you anyways. I will note that for the future.


With exactly 8 coins, each a penny, a nickel, or a dime, their total CANNOT be:

A.25¢	B.26¢	C.27¢	D.28¢	E.29¢

[hide=" click to reveal hidden content"]

$ 25$ cents as all the other amounts can be made.

[/hide]

[hide="@Andrew:"]1 dime, 2 nickels, 5 pennies.[/hide]

[hide=" click to reveal hidden content"]

Thanks $ 1\equal{}2$.

$ 25: 5p\plus{}2n\plus{}1d$;

$ 26: 6p\plus{}2d$;

$ 28: 3p\plus{}5n$;

$ 29: 4p\plus{}3n\plus{}1d$.

Therefore the amount which can't be made must be $ 27$ cents.

[/hide]


geometry
rectangle
perimeter
Pythagorean Theorem

An 8 cm by $ N$ cm rectangle has a diagonal of length 10 cm. What is the number of centimeters in the perimeter of this rectangle?

First, you have to find the length of the other side.
We have to use the pythagorean theorem backwards.

So 8^2 + N^2  = 10^2

We know that: 8^2=64  and  10^2=100 so 100-64 = 36. The square root of 36 is 6

Now the second part: Now that we know both of the sides, we know all of the sides.
8X2 + 6X2 = 28
Therefore, the perimeter is 28

hello, we have $ 8^2\plus{}N^2\equal{}100$, from here we get $ N\equal{}6$ and the perimeter of the rectangle is $ p\equal{}(6\plus{}8)\cdot2cm\equal{}28cm$.
Sonnhard.


topology
function
real analysis
real analysis unsolved

Prove that every continuous function $ f: \mathbb{R}^{n}\rightarrow X$ ($ X$ is a set with the discrete topology) is constant (ie. $ f\left(\mathbb{R}^{n}\right)\equal{}\left\{x\right\}$ for some $ x\in X$).

Let $ x$ be an arbitrary member of $ X.$ The set $ \{x\}$ is both open and closed in $ X.$ Therefore $ f^{\minus{}1}(\{x\})$ is both open and closed in $ \mathbb{R}^n.$ Since $ \mathbb{R}^n$ is connected, the only sets that are both open and closed are the empty set and the whole space. Hence $ f^{\minus{}1}(\{x\})\equal{}\emptyset$ or $ f^{\minus{}1}(\{x\})\equal{}\mathbb{R}^n.$

The only property of $ \mathbb{R}^n$ used here was its connectedness. The same result would be true for any connected domain.


Restrictions:

I.  Exactly 8 ones and exactly 2 exclamation points (for factorials)

II.  The ones must be kept apart from each other (i.e. there can be no
      11, 111, 1111, etc. formations).

III.  You may use as many pairs of parentheses as possible as long
        as they don't violate I & II.

IV.  You may use any number of plus signs as long as there are
       enough of them and their number does not violate I, II, & III.

V.  No other characters/symbols and/or operations are allowed; only
     the ones, exclamation points, plus signs, and parentheses are to
     be used.

VI.  The expression should give valid values on graphics calculators
       TI-82, TI-83, TI-83 Plus and similar operating models.

Note:  Your solution is one of many that is out of bounds of the calculator.  It produces
an "overflow" when entered.  However, I did not define in rule VI if I considered an
overflow a valid "value."  Take it as such if the order of the symbols is not incorrect.

At this time, I am not commenting on whether you have the correct answer.

The number $(18!)!$ is truly massive.  Using Stirling's approximation (and taking logarithms, of course), we can see that it has nearly $10^{17}$ digits!.  In comparison, a TI-83 will overflow at around $100$ digits.  That's why you need to calculate with logarithms, in which case you should be able to get an answer on your TI-83 (the TI-83 cannot do large integer computations anyways, unlike, say, the TI-89).

Saying a number is "truly massive" is nonsensical because, what, I can throw ANY number into the mix, e.g. a centillion, a googol, a googolplex, etc.

If you HAVE a better (larger) value OR a tie expression for (18!)!, then fine.  Otherwise, your opinions aren't valid solutions.

I gave you an accurate estimate on the size of $(18!)!$, which is actually of reasonable interest here and hardly qualifies as an opinion.  I suspect, although I have not given it much thought, that $(18!)!$ is the best you can do.

It is particularly relevant to speak of a number as being 'massive' in the context of a computer science thread, where computations on the order of $10^{30}$ operations are computationally infeasible.  Of course, $10^{30}$ is much, much smaller than $10^{10^{17}}$

But then minus one times(fill in some positive value greater than one)^(fill in some positive value greater than one) is "much, much smaller than" 10^(10^17), or countless or examples, so you've said nothing!

the number truely is massive!  this being an information section of the forum is a good point.  i tried putting the number into my ti-89 and it gave me $\infty$, i e too large.  the ti-83 plus cant do a factorial higher than around 70 i think and the 89 cant go higher than 90 or so.  both of these cant do more than 2 digits, we have 6,402,373,705,728,000!, also i tired it on maple, and it was too large, too large for maple!!!

(picture a plot of n! from n=0 to n=10...we have n=6,402,373,705,728,000)


The surface area and volume of a prisim r given find the surfase area and volume of a prisim with (a) double ,(b) triple,(c) half the dimensions of the given prisim?
S.A= 23 ft sq
Volume= 12 ft cubed
1a)
1b)
1c)

S.A = 61.7 metrs sq

volume=28.2 meters cubed
2a)

2b)
2c)

Let's say the prisms had edge lengths of $ a, b,$ and $ c$. The surface area of this prism is then $ 2(ab \plus{} ac \plus{} bc) \equal{} 2ab \plus{} 2ac \plus{} 2bc$ and the volume is $ abc$. Then if we scale each dimension by a constant $ n$, the new dimensions would be $ na, nb,$ $ nc$, and the surface area would be $ 2((na)(nb) \plus{} 2(na)(nc) \plus{} 2(nb)(nc)) \equal{} n^2(2ab \plus{} 2ac \plus{} 2bc)$. This is $ n^2$ times the original surface area. In addition, the volume would be $ (na)(nb)(nc) \equal{} n^3(abc)$, which is $ n^3$ times the original volume.

One way to tell whether you multiply by $ n^2$ or $ n^3$ is to tell whether you're dealing with plane (area, surface area, etc.) or space (volume).

Therefore, if you are given the surface area and if each dimension of a 3-D solid is scaled by a certain amount $ n$, the new surface area will be $ n^2$ times the original surface area.

Similarly, the new volume would be $ n^3$ times the original volume.

1a) SA = $ 23 \cdot 2^2 \equal{} 92$ sq ft, Volume = $ 12 \cdot 2^3 \equal{} 96$ cu ft.
1b) SA = $ 23 \cdot 3^2 \equal{} 207$ sq ft, Volume = $ 12 \cdot 3^3 \equal{} 324$ cu ft.
1c) SA = $ 23 \cdot \left(\frac 12\right)^2 \equal{} 5.75$ sq ft, Volume = $ 12 \cdot \left(\frac 12\right)^3 \equal{} 1.5$ cu ft.

2a) SA = 246.8 sq m, Volume = 225.6 cu m.
2b) SA = 555.3 sq m, Volume = 761.4 cu m.
2c) SA = 15.425 sq m, Volume = 3.525 cu m.


Prove that the four vertices of a square cannot lie on four concentric
circles, the radii of which form an arithmetic progression.

[hide="Solution"] Let the common center of the circles be $O$.  Then we have $OA = a, OB = a+d, OC = a+2d, OD = a+3d$ for some reals $a, d$ (the vertices are not in order).  But by the British Flag Theorem we have

$a+(a+3d)^{2}= (a+d)^{2}+(a+2d)^{2}\implies$
$2a^{2}+6ad+9d^{2}= 2a^{2}+6ad+5d^{2}\implies$
$d = 0$

Which is the degenerate case. [/hide]


function
integration
calculus
real analysis
real analysis unsolved

Let $f:[0,1]\to\mathbb R$ be a derivable function s.t. $f'$ is continuous on $[0,1]$.
 Prove that if there exists $a\in[0,1]$ s.t. $f(a)\le 0$ then
$\int_0^1f(x)dx\le\int_0^1\mid f'(x)\mid dx$

$f(x)=f(a)+\int_{a}^{x}f'(t)dt$

$A=\int_{0}^{1}f(x)dx = f(a) + \int_{0}^{1} ( \int_{a}^{x}f'(t)dt )dx$

we apply Fubini in the double integral we get

$A=f(a) + \int_{a}^{1}( \int_{t}^{1} dx ) f'(t)dt$

$A=f(a) +\int_{a}^{1}(1-t)f'(t)dt\leq f(a) + \int_{a}^{1}|1-t||f'(t)|dt$

$|1-t|$ is decreasing on [a,1]

$A\leq f(a) + |1-a|\int_{a}^{1}|f'(t)|dt\leq \int_{0}^{1}|f'(t)|dt$

What was the official solution ?

I don't know the official solution... I tried to solve it by myself (all day) and managed to do some things but I don't think that I have an entire proof... actually I don't know yet if it is entirely correct...

Moubi.. could you please tell me what is Fubini?
Thank you.

Fubini's Theorem is the name for a family of theorems that state that given certain sufficient conditions, one may interchange the order of integration in a double (iterated) integral. In this case, we have a very stong condition justifying the interchange: on this bounded set, the integrand is continuous.

The procedural "meta-theorem" is that if you write down an iterated integral (or a double sum, or an integral of a sum), then it is extremely likely that you intend to invoke Fubini or one of its relatives at some point in your proof.

Kent, would you please write the complete statement for Fubini's Theorem... thx

As I said, it's really a family of theorems rather than a single theorem. I'll give you a fairly general version, but there are certainly other versions. Let $f$ be a function defined on the cartesian product $A\times B$. Here, you can think or $A$ and $B$ as being measureable subsets of $\mathbb{R}^n$ and $\mathbb{R}^m$, but we can actually use any $\sigma$-finite measure spaces. Assume that $f$ is measuable with respect to product measure. In what follows, use $dx$ for the measure on $A$, $dy$ for the measure on $B$, and $dm$ for the product measure on $A\times B.$

[b]Tonelli's Theorem:[/b] If $f\ge 0$ on $A\times B$, then

$\int_{A\times B} f(x,y)\, dm=\int_A\int_Bf(x,y)\,dy\,dx= \int_B\int_Af(x,y) \,dx\,dy.$

(The conclusion includes that if any of these three integrals is equal to $\infty$, then all three are equal to $\infty.$)

[b]Fubini's Theorem:[/b] If $\int_{A\times B}|f(x,y)|\, dm<\infty$ then

$\int_{A\times B} f(x,y)\, dm=\int_A\int_Bf(x,y)\,dy\,dx= \int_B\int_Af(x,y) \,dx\,dy.$

Note that Tonelli's Theorem can be useful in determining whether $f$ is absolutely integrable over $A\times B$

Note also that we can think of something like $\int_0^1\int_x^1f(x,y)\,dy\,dx$ as the integral over $[0,1]\times [0,1]$ of $f(x,y)\chi_E(x,y)$ where $\chi_E$ is the characteristic or indicator function of the set $E=\{(x,y) : 0\le x\le y \le 1\}.$

[quote="Moubinool"]
$A=\int_{0}^{1}f(x)dx = f(a) + \int_{0}^{1} ( \int_{a}^{x}f'(t)dt )dx$

we apply Fubini in the double integral we get

$A=f(a) + \int_{a}^{1}( \int_{t}^{1} dx ) f'(t)dt$
[/quote]
We don't need Fubini's theorem here, I think, because they are proper integrals.

here is what I did.. I don't know if it correct
It is enough to prove the relation for those functions $f$ with $\inf_{x\in[0,1]}f(x)=0$.
Let's say that $f(a)=0$ where $a\in [0,1]$. Now I wanted to prove that $\int_0^1\mid f'(x)\mid dx\ge \sup_{x\in[0,a]}f(x)+\sup_{x\in[a,1]}f(x)$.
Let $x_0\in[0,a]$ be so that $f(x_0)=\sup_{x\in[0,a]}f(x)$ and $x_1\in[a,1]$ so that $f(x_1)=\sup_{x\in[a,1]}f(x)$.
Now $\int_0^1\mid f'(x)\mid dx\ge\int_{x_0}^{x_1}\mid f'(x)\mid dx$.
$A=\{x\in(x_0,a)\mid f(x) \textrm{ is a maximum (local)}\}$
$A'=\{x\in(x_0,a)\mid f(x) \textrm{ is a minimum (local)}\}$
$B=\{x\in(a,x_1)\mid f(x) \textrm{ is a maximum (local)}\}$
$B'=\{x\in(a,x_1)\mid f(x) \textrm{ is a minimum (local)}\}$
Now
$\int_{x_0}^{x_1}\mid f'(x)\mid dx=\int_{x_0}^a\mid f'(x)\mid dx+\int_a^{x_1}\mid f'(x)\mid dx=
f(x_0)+f(x_1)+2\sum_{x\in A}f(x)-2\sum_{x\in A'}f(x)+2\sum_{x\in B}f(x)-2\sum_{x\in B'}f(x)>f(x_0)+f(x_1)$.
This is from the fact that if $x'$ is a point of minimum then it is followed by a point $x''$ of maximum (local) and $f(x'')-f(x')\ge0$

Now $\int_0^1f(x)dx\le af(x_0)+(1-a)f(x_1)\le f(x_0)+f(x_1)\le\int_0^1\mid f'(x)\mid dx$

Am I doing something wrong?


calculus
integration
logarithms
trigonometry
LaTeX
calculus computations

[hide="Integral (a)"]Changing variables according to $ x \equal{} \frac{\pi}{2} \minus{} t$, $ dx \equal{} \minus{} dt$, we have

$ I \equal{} \int_0^{\pi/2} \ln(\sin x)\,dx \equal{} \minus{} \int_{\pi/2}^0 \ln\left[\sin \left(\frac{\pi}{2} \minus{} t\right)\right]\,dt \equal{} \int_0^{\pi/2} \ln(\cos t)\,dt$.

Therefore,

$ \int_0^{\pi/2} \ln(\sin x)\,dx \plus{} \int_0^{\pi/2} \ln(\cos x)\,dx \equal{} 2I \equal{} \int_0^{\pi/2} \ln(\sin x \cos x)\,dx$

$ \equal{} \int_0^{\pi/2} \ln\left(\frac{\sin(2x)}{2}\right)\,dx \equal{} \underbrace{\int_0^{\pi/2} \ln(\sin (2x))\,dx}_{\equal{} J} \minus{} \ln 2 \int_0^{\pi/2}dx \equal{} J \minus{} \frac{\pi}{2} \ln 2$

To evaluate J let's make the substitution $ 2x \equal{} \frac{\pi}{2} \minus{} t$, $ dx \equal{} \minus{} \frac12$:

$ J \equal{} \minus{}\frac12 \int_{\pi/2}^{\minus{} \pi/2} \ln\left[\sin\left(\frac{\pi}{2} \minus{} t\right)\right]\,dt \equal{} \frac12 \int_{\minus{} \pi/2}^{\pi/2} \ln(\cos t)\,dt$

$ \equal{} \int_0^{\pi/2} \ln(\cos t)\,dt \equal{} I$.

Therefore, we have

$ 2I \equal{} I \minus{} \frac{\pi}{2} \ln 2 \equal{}> I \equal{} \minus{} \frac{\pi}{2} \ln 2$.[/hide]

[hide="Integral (b)"]Making $ x \equal{} \tan t$, $ dx \equal{} \sec^2t\,dt$, we have

$ I \equal{} \int_0^{\plus{} \infty} \frac{\arctan x}{(1\plus{}x^2)^{3/2}}\,dx \equal{} \int_0^{\pi/2}t \cos t\,dt$,

and I think you can proceed from here.[/hide]

[quote="kabi"]a part can solved simply by using the
integral(a to b){ f(x) }= integral(a to b) {f(a+ b -x)}
for part c
take substitution ::
1-x=t^2
dx=-2tdt
thus I = integral (1 to 0)  {  2 dt/(t^2 +1)  }
         = pi / 2

I know its really difficult to understand the way in which I m posting my answer.
Sorry for that.[/quote]

You may learn how to use Latex to write the Math. Formulae and symbols. :thumbup:

[quote="vinskman"]

You may learn how to use Latex to write the Math. Formulae and symbols. :thumbup:[/quote]
I know it is too hard to understand when someone post in such manner.
I m reallyyyyyyyyyyyy sorry if someone finds difficulty.
 vinskman Thanks for  suggestion

[quote]
I know it is too hard to understand when someone post in such manner.
I m reallyyyyyyyyyyyy sorry if someone finds difficulty.
 vinskman Thanks for  suggestion[/quote]

No worries!! :jump:


modular arithmetic
function
number theory

I don't get these kinds of questions. An example would be\
99^36/ 24.

What is the remainder?
Can someone solve this and give me the trick to these kinds of questions.

Well, the remainder would be:

$ 29017217418732239040597418558388566293399113957161837107334498473595683$

I doubt any competition, unless you are allowed a calculator, would ask that...(i used my Ti-84 to fins this)

And this probably doesn't belong in Classroom Math also...

Since 24=3*8, we can consider 99^36(mod 3), and 99^36(mod 8). Obviously, 99^36=0(mod 3). Euler's formula states that a^phi(n)=1(mod n), if gcd(a,n)=1. Since phi(8) is 4, 99^4=1(mod 8), which means that 99^36=1(mod 8). Now, suppose that 99^36=m(mod 24). Then m=1(mod 8) and m=0(mod 3), and the smallest such m is 9. Hence (99^36)/24 has a remainder of 9.
The above solution can be extended for a general case.

To find $ a^b\pmod c$ in general, we can proceed in several ways. Two standard ways are [url=http://www.artofproblemsolving.com/Wiki/index.php/Euler%27s_totient_theorem]Euler's Totient Theorem[/url] and the [url=http://www.artofproblemsolving.com/Wiki/index.php/Carmichael_function]Carmichael Function[/url] (the latter of which is more powerful; a more specific case is [url=http://www.artofproblemsolving.com/Wiki/index.php/Fermat%27s_Little_Theorem]Fermat's Little Theorem[/url]). One different method (that can use the aforementioned methods in it) is to utilize the [url=http://www.artofproblemsolving.com/Wiki/index.php/Chinese_Remainder_Theorem]Chinese Remainder Theorem[/url].

However, this is Classroom Math, so here's an elementary solution:

$ 99^{36}\equiv3^{36}\equiv(3^3)^{12}\equiv27^{12}\equiv3^{12}\equiv(3^3)^4\equiv27^4\equiv3^4\equiv3^3\cdot3\equiv27\cdot3\equiv3\cdot3$

$ \equiv\boxed9\pmod{24}$, where the key fact used to reduce calculation is that $ 3^3\equiv3\pmod{24}$.

There. None of those fancy tools used. :P

Edit: Beaten, but my solution is simpler. Note that Twin Prime Conjecture used the Chinese Remainder Theorem and Euler's Totient Theorem.

[quote="AIME15"]THe remainder can't be over 24...

99==3 mod 24, so this is 3^36 mod 24.  From here it's easier.[/quote]

Oh I misread the question, I thought it asked for teh real value... :blush:


vector
number theory unsolved
number theory

Clearly, the $a_{0}\neq 0$ case is trivial.

In the $a_{0}=0$ case, we simply take a generator $g$, and in this way replace the sequence $\{1, 2, 3, ..., p-1\}$ of residues with a permutation of $\{0, 1, 2, ..., \}$, namely the powers of the generator which produce the residues.  From here, it suffices to show that we cannot permute the vector $(0, 1, 2, ..., p-1)$ two different ways, add them, reduce mod $p-1$, and then get a third permutation of the vector.  But the sum of the constituents of the sum of the two permutations will have value $p(p-1) \equiv 0\mod{p-1}$.  But the sum of the components of any permutation of the vector has sum $\frac{p(p-1)}{2}$ which is not divisible by $p-1$ because $p$ is odd, thus the two vectors cannot be the same, hence we cannot add two permutations of this vectors and reduce mod $p-1$ to get a third permutation, and so the result follows that two of these elements must be congruent or one is $0$, which is the value of $a_{0}$ anyways.


integration
calculus
derivative
calculus computations

I am trying to prove $E = mc^2$ to see how the famous physics formula is derived. I started by using the notion that $E$ is Force, $F$ in the direction of motion times distance, and I denoted motion times distance as $x$, where $x = \int^x_0 dx$, so I have:

$E = \int^x_{0} F dx$

$F = ma$ and we know that $a = \frac{dv}{dt}$ and $v = \frac{dx}{dt}$ so simplifying the integrand I got:

$\int^{mv}_{0} vd(mv)$

I then used the definition of relativistic mass in which $m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$ and I ended up with:

$\int^v_{0} vd\left(\frac{m_{0}v}{\sqrt{1-\frac{v^2}{c^2}}}\right)$

From here do I just do the derivative of the relatavistic mass portion and simplify the rest or do I do something else? Any help appreciated!

This site explains the proof, you seem to be on the right track.

http://webphysics.iupui.edu/webscience/courses/math163/honors/HP13.pdf

[quote="Jimmy"]Einstein said that everything moves at the speed of light, but only a small portion of that speed takes place in the 3 spacial dimensions. So since kinetic energy is defined as $\frac{1}{2}mv^2$, then if $v=c$ the kinetic energy would be $\frac{1}{2}mc^2$. If the potential energy somehow is $\frac{1}{2}mv^2$, then $E=K+U=mc^2$.  I hope this helps.  Also, what's the proof for the definition of relativistic mass?[/quote]

Your claim above is wrong. Simply stated, there are no potential energies at relativistic frames, there are only relativistic energies and relativistic momenta.

To derive the relativistic mass of a particle in relativistic motion from its rest mass, start from the conservation of momentum

To derive $E=mc^2$, start by relating the kinetic energy of a particle and its momentum: $K_b-K_a=\int_a^b \frac{d\mathbf{p}}{dt}\cdot d\mathbf{r}$

Thanks for the link, Jimmy. I see now that I need to simplify the differential by doing the derivative and then just play with the integrad until I get $E = mc^2 - m_{0}c^2$ where $m$ is still equivalent to ${\frac{m_0}{\sqrt{1- \frac{v^2}{c^2}}}}$ I see now how I can get it to that, too. Also, the proof for the definition of relatavistic mass in on that site that you gave to me. Thanks!


The average value of all the pennies, nickels, dimes, and quarters in Bobert's pocket is 20 cents. If he had one more quarter, the average value would be 21 cents. How many dimes does he have?

[hide]Suppose Robert has $ p$ pennies, $ n$ nickels, $ d$ dimes, and $ q$ quarters and that there are $ N$ coins altogether.

So, $ p\plus{}n\plus{}d\plus{}q \equal{} N$ and $ p\plus{}5n\plus{}10d\plus{}25q \equal{} 20N$. We are told that with one more quarter, the average value would be 21 cents. So $ p\plus{}5n\plus{}10d\plus{}25(q\plus{}1) \equal{} 21(N\plus{}1)$. Substituting the second equation into the third, we have $ 20N\plus{}25 \equal{} 21N\plus{}21$, so $ N \equal{} 4$. So Robert has $ 20\cdot 4 \equal{} 80$ cents in his pocket and this is made up of four coins. We easily see that the only way to make this up is with 3 quarters and a nickel, so he has $ \boxed{0}$ dimes. Notice that we didn't need the first equation![/hide]


HAPPY NEW YEAR!!!

DON't SPEND IT ON AOPS!!!

TO INFINITY AND BEYOND!! THE FUTURE!!!! :lol:


Let a_1, a_2, ....a_n be positive real numbers. prove that

$ \sum_{i \equal{} 1}^{n}\left({\frac {1}{a_i^n} \plus{} a_i}\right) \ge 2 \sum_{i\equal{}1}^{n} \frac{1}{a_i^{n\minus{}1}}$

[quote="realnumber"]$ a_1 \equal{} a_2 \equal{} 0.8 \equal{} a_3$,n=3[/quote]

yes, wrong, thx...


Show that if 14 distinct integers are chosen from the sequence 100,101,102,103,.........,123,124 there must be two of them whose difference is 4.

Could someone please give me an answer and how you arrived at the answer by using the pigeon hole principal(could the answer be correct as well??)

Thanks

yea i think basically what it comes down to is if you only choose $13$ of the $24$ integers, it is possible to just choose one of the numbers in each of the $13$ sets , but since you are forced to choose $14$, you will have to pick the 2nd number from one of those sets


Let $n$ be a positive integer. Find all $n$ such that $\phi(n)=p^{4}$, and p is a prime number.

Since p=2, we must have a number whose only prime factors are 2 or a Fermat prime as phi(n)=n(1-1/p)... and n is to be a power of 2. Say n=(2^m)(2^(2^n)+1). Then phi(n)=(2^(m-1))(2^(2^n))=16. So m-1+2^n=4. Giving m+2^n=5. We thus have the solutions (4,0), (3,1), (1,2). These correspond to 48, 40, and 34 respectively. Finally we have the two cases where there is either no power of two or no Fermat prime which gives us 17 and 32. Thus 17, 32, 34, 40, and 48 are all solutions.


I know some of the official solutions for A>=9/4, but I do not like them, so I won't give a solution for that. But the other inequalities in the chain are simple:
   For D>=C, observe that (sum a/(b+c))^2<=(sum a^2)(sum 1/(b+c)^2)<=(sum a^2)*(sum 1/4bc)=1/(4abc)*sum a*sum a^2<=3 sum a^3/4abc, from Cebasev.
   For C>=B. It is a trivial consequence of Cebasev inequality.
   For B>=A. Observe that sumab*sum 1/(a+b)^2=sum ab/(a+b)^2+sum a/(b+c)<=3/4+sum a/(b+c). So, it sufices to prove that (3+sum a/(b+c))^2>=9(3/4+sum a/(b+c)) which is (sum a/(b+c)-3/2)^2>=0.


geometry
circumcircle
geometric transformation
homothety
ratio
incenter
geometry proposed

Dear Mathlinkers,
Let ABC be a triangle, (0) the circumcircle of ABC, (1) the incircle of ABC, I the center of (1), D the point of contact of (1) with BC,
(2) the circle with diameter AI, A* the second point of intersection of (0) and (1),
Prouve that the circle tangent to (0) at A* and passing through D is tangent to (1).
Sincerely
Jean-Louis

Use this lemma: " Given (O) and AB is a chord. If circle (W) is tangent to (O) at X and AB at Y then XY is the bisector of angle XAB".
So the main idea in this problem is to show that $ A'D$ and $ AI$ meet at $ (O)$.
It's easy to have $ \Delta A'KB\sim \Delta A'JC$
Then $ \frac {A'B}{A'C} \equal{} \frac {BK}{CJ} \equal{} \frac {BD}{CD}$ which follow $ A'D$ is the bisector of angle $ BA'C$
And we are done!

[b]LEMMA-[/b] [b][color=darkblue] Let $ ABC$ be a triangle with $ H$ be its orthocenter. Let $ A_0,B_0,C_0$ repsetively be the projections of $ A,B,C$ onto $ BC,CA,AB$. Denote $ (\mathcal {E})$ be the $ 9$- point circle wrt $ \triangle ABC$. Let $ (O_a)$ be the circle passing through $ A,A_0$ such that $ (O_a)$ internally tangents to $ (\mathcal {E})$ at $ A_0$. Prove that $ (O_a)$ also internally tangents to the circumcircle $ (O)$ of $ \triangle ABC$ at $ A$ $ (*)$[/b][/color]

Indeed, consider the homothety through center $ A_0$, ratio $ k \equal{} \frac {R}{r}$ where $ R$ is the radii of $ (O_a)$ and $ r$ is the radii of $ (\mathcal {E})$. We have $ \mathcal {H}(A_0,k): O_a\mapsto \mathcal {E}$, $ A\mapsto N_a$. Therefore $ N_a\in (\mathcal {E})$. Hence, it is followed that $ N_a$ is the midpoint of $ AH$, which implies that $ N_a$ will be center of $ (AC_0HB_0)$ $ \Longrightarrow$ $ N_aC_0 \equal{} N_aB_0$. Thus, $ N_a$ will be the midpoint of arc $ B_0C_0$, which implies $ \mathcal {E}N_a\perp B_0C_0$ $ \Longrightarrow AO_a\perp B_0C_0$. But it is well- known that $ AO\perp B_0C_0$, hence $ A,O,O_a$ are collinear. Therefore $ (O_a)$ internally tangents to $ (O)$.

Back to our problem. First let me denote $ I$ by the incenter of $ \triangle ABC$, $ r$ be its radii. Let $ K,J$ respetively be the tangency points of $ (I)$ with $ AB,AC$. Denote $ M_a,M_b,M_c$ respetively be the midoints of $ JK,KD,DJ$. Let $ O$ be the circumcenter $ \triangle ABC$, $ (O_a)$ be the circle passing through $ A',D$ such that $ (O_a)$ internally tangents to $ (I)$ at $ D$, where $ A'$ be the second intersection of $ (AKIJ)$ and $ (O)$. Now, consider the inversion through pole $ I$, power $ k \equal{} r^2$, we have $ \mathcal {I}(I,k): A\mapsto M_a$, $ B\mapsto M_b$, $ C\mapsto M_c$, $ K\mapsto K$, $ J\mapsto J$, $ D\mapsto D$. Therefore $ \mathcal {I}(I,k): (ABC)\mapsto (A_0B_0C_0)$, $ (AKIJ)\mapsto KJ$, $ (I)\mapsto (I)$. Therefore $ A'\mapsto A_0$, such that $ A_0\equiv (M_aM_bM_c)\cap KJ$, $ (A_0\neq M_a)$ and $ (O_a)\mapsto (O'_a)$ such that $ (O'_a)$ passes through $ A_0$ and $ D$ and $ (O'_a)$ internally tangents to $ (M_aM_bM_c)$ at $ A'$. We claim that $ (O'_a)$ also internall tangents $ (I)$ at $ D$. Indeed, note that $ (M_aM_bM_c)$ is the $ 9$- point circle wrt $ \triangle KJD$ hence $ A_0$ will be the projection of $ D$ onto $ KJ$. Then apply the [b]lemma[/b], we shall get $ (O'_a)$ internally tangents to $ (I)$ at $ D$, which leads to the fact that $ (O_a)$ also internally tangents to $ (I)$ at $ D$. $ \square$


A frog is placed on each cell of a $n \times n$ square inside an infinite chessboard (so initially there are a total of $n \times n$ frogs). Each move consists of a frog $A$ jumping over a frog $B$ adjacent to it with $A$ landing in the next cell and $B$ disappearing (adjacent means two cells sharing a side). Prove that at least $ \left[\frac{n^2}{3}\right]$ moves are needed to reach a configuration where no more moves are possible.

I somehow showed the solution for all n at least 48 because I claim that no frog can jump out of the (n+8)*(n+8) square centered at the center of the original square (a famous result). But the small cases still remain elusive.

I guess that the solution to this problem shouldn't to too long and complex at all. (as many other Russian problems)

Here, processes number 4 : http://www.mit.edu/~alexrem/MiscellaneousProblemsSolutions.pdf

The minimum value of n to have this process possible is 3 and in an 3×3 square with at least 3 moves we can reach a configuration that no more moves are possible by having our three moves in three different lines so to count the total number of needed moves in an n×n square we can count the number of 3×3 squares and multiply it by 3 to get the number of moves so it will be [(n^2)/(3^2)×3]=[(n^2)/3]

[hide = Solution]

Call a move [b] efficient [/b] if the frog lands outside of the initial $n \times n$ square and [b] inefficient [/b] otherwise. Let there be $x, y$ efficient, inefficient jumps respectively. Clearly, there must be at most $\left \lfloor \frac{n^2}{2} \right \rfloor$ frogs in the $n \times n$ square in the end as otherwise there are two adjacent ones. This gives us that
$$x+2y \ge \left \lceil \frac{n^2}{2} \right \rceil. \qquad (1)$$
Now, let's consider two cases on the parity of $n$.

[b] Case 1. [/b] $n$ is even

Partition the $n \times n$ into $2 \times 2$'s so that there are $\frac{n^2}{4}$ of them. Now, say that a move [b] operates [/b] on a $2 \times 2$ if the number of frogs in the $2 \times 2$ strictly decreases as a result of the move. Note that the Observe that any efficient move operates on at most $1$ square, and any inefficient move at most $2$. Furthermore, note that every $2 \times 2$ must be operated on at most twice. This is enough to imply that
$$2x + y \ge \frac{n^2}{2},$$
which added with $(1)$ gives the result.

[b] Case 2. [/b] $n$ is odd

When $n=3$ it's easy to verify, suppose now that $n > 3.$ Assign a labelling $\{1, 2, \cdots, n\} \times \{1, 2, \cdots, n\}$ of the squares of the grid in the obvious manner. We will partition the $n \times n$ into an assortment of shapes as follows. First partition $(3, 3)$ into it's own $1 \times 1.$ Next partition the remaining squares of the third rows and columns into pairs, e.g. $(3, 1)$ and $(3, 2)$, $(3, 3)$ and $(3, 4)$, etc., and similarly for the column. Finally, partition the remaining squares into $2 \times 2$'s. Define [b] operates [/b] similarly as before, and observe that the $2 \times 2$'s must be operated on at least twice and the $1 \times 2$'s at least once. Notice that each inefficient move operates on at most two shapes, and each efficient move at most one. This implies that
$$x + 2y \ge \frac{n^2-1}{2}.$$
Adding this with $(1)$ again gives the result.

$\square$

[/hide]


function
algebra
domain
complex analysis

Suppose that $ f$ is analytic on a domain containing $ \{z: |z|\le 1\}$ and that $ |f(e^{i\theta})|<1, 0\le \theta\le 2\pi$.  Show that $ f$ has exactly one fixed point (ie exactly one solution to $ f(z)\equal{}z$) in the disk $ |z|<1$.

If $ f$ has two fixed points, then you can move one of them to $ 0$ and obtain a contradiction with the equality case in the Schwarz lemma.

alternatively its one line with Rouche's thm :D


Find the remainder of 2^87 divided by 3.

For the problem given by Arvind_sn, I get
[hide]
2
2^2=4 4-3=1
2^3=8 8-6=2
2^4=16 16-15=1
2^5=32 32-30=2
2^6=64 64-63=1
odd powers have remainder of 2, so answer is 2
[/hide]

[quote="zserf"]odd, I could have sworn I already posted on this topic...

anyway...

[hide]
proof of pattern

2(X+2/3)=(2x+1)+1/3
2(X+1/3)=2x+2/3[/hide][/quote]

[hide="There were somehow two identical threads  :D"]
Another one: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=138206[/url]

[hide="briefly testing LaTeX"]$3^{87}\equiv (-1)^{87}\equiv-1\equiv 2 \bmod{3}$[/hide][/hide]

[quote="funcia"][hide]this isn't the right question the right one is 2^87+3/7' s remainder [/hide][/quote]
Right. I got that one. 5, right?

[quote="vishalarul"][hide="FAST WAY (Arvind)"]

$2^{2}\equiv 1\pmod{3}$
$2^{86}\equiv 1\pmod{3}$
$2^{87}\equiv 2\pmod{3}$

$2$

If you didn't know, and had to guess, you have a $\frac{1}{3}$ chance of getting it right!
[/hide]

[hide="FAST WAY (funcia)"]
$2^{3}\equiv 1\pmod{7}$
$2^{87}\equiv 1\pmod{7}$
$2^{87}+3\equiv 4\pmod{7}$

$4$
[/hide][/quote]
why are the two answers different? isn't it the same problem?

@vishalarul: did u ever go to happy hollow? cuz i remember someone named Vishal Arul from like 5th grade and we were both on the math bowl team?

[quote="mwpl11"][quote="vishalarul"][hide="FAST WAY (Arvind)"]

$2^{2}\equiv 1\pmod{3}$
$2^{86}\equiv 1\pmod{3}$
$2^{87}\equiv 2\pmod{3}$

$2$

If you didn't know, and had to guess, you have a $\frac{1}{3}$ chance of getting it right!
[/hide]

[hide="FAST WAY (funcia)"]
$2^{3}\equiv 1\pmod{7}$
$2^{87}\equiv 1\pmod{7}$
$2^{87}+3\equiv 4\pmod{7}$

$4$
[/hide][/quote]
why are the two answers different? isn't it the same problem?

@vishalarul: did u ever go to happy hollow? cuz i remember someone named Vishal  from like 5th grade and we were both on the math bowl team?[/quote]

FUNCIA the super math/spell/grammar police checker person thing had to correct the problem in the thread-read the former posts for more info

[hide]4 is correct[/hide]


whoops..
[quote="nutz_for2.718281828"]
[Unparseable or potentially dangerous latex formula. Error 6 ]
[/quote]

how is $\tan{x}=\frac{\sin{x}}{\cos{x}}$ not a simple identity? If you read the problem, you would have found that it says you can use the quotient identities, which I'm assuming means the above.

ya i know..well before i read his whole solution i posted it.. :blush:

EDIT: Nice SOlution (@nutz)..sorry about before..i clearly shouldve read ur solution more carefully..its very simple..

[quote="7h3.D3m0n.117"]Using only the Quotient Identities, the Reciprocal Identities, and the Pythagorean Identities, simplify:

$\frac{\text{tan}^{2}\theta-\text{sin}^{2}\theta}{\text{tan}^{2}\theta\text{sin}^{2}\theta}$

YAY My 300th post! :ddr:

Please show work.[/quote]

[hide]Divide the numerator and denominator by $\tan^{2}\theta \sin^{2}\theta$

We then get

$\csc^{2}\theta-\cot^{2}\theta$.

Multiplying both sides by $\frac{\sin^{2}\theta}{\sin^{2}\theta}$

We get $\frac{1-\cos^{2}\theta}{\sin^{2}\theta}$

And proceed with the way rjt's solution to get $1$

$1$[/hide]

hmm the solution i like the best was mathgeniuse^ln(x)'s because it actualyl explained some parts  :) ...thanks everyone for helping  :P


LaTeX
number theory proposed
number theory

How many positive integers $ n$ are there, such that
\[ \sqrt[41]{\sqrt{n}\plus{}\sqrt{n\plus{}2009^{2009}}}\in\mathbb{N}?\]

Hi Rust! :)

The result is correct: the total numbero of $ n$ such that respect the condition is 4950/2 = 2475.
I solved it in a slight different way...

Could you please try to correct the latex warning (I get "[Unparseable or potentially dangerous latex formula.Error: 7]" and can't se the formula) on the third line?

Thank you! :D

[quote="Rust"]
\[ \sqrt n \plus{} \sqrt {n \plus{} N} \equal{} x^{41},N \equal{} 2009^{2009} \equal{} 7^{4018}41^{2009}.\]

\[ 2n \plus{} N \plus{} 2\sqrt {n(n \plus{} N)} \equal{} x^{82},\]

\[ n \equal{} \frac {x^{82} \minus{} 2N \plus{} \frac {N^2}{x^{82}}}{4}.\]
Therefore $ x \equal{} 7^a41^b$, were $ 0\le a\le 98, 0\le b\le 49$. There are $ 99*50 \equal{} 4950$ solutions for x.
But $ x$ and $ N/x$ give same solution for $ n$.[/quote]

I solved it too without looking at answers !
Be Successful


calculus
integration
trigonometry
Galois Theory
calculus computations

[hide="hint"]What Pythagorean identity involving $ \sin{x}$ relates $ 2\theta$ to a square?[/hide] [hide="the answer"]is found by doing $ \sin^2{x} \equal{} \frac{1\minus{}\cos{2x}}{2}$, which can easily be finished via u-substitution. The result I obtained (been awake for far too long) is \[ \int \sin^2{x}\text{ }dx\equal{}\frac{2x\minus{}\sin{2x}}{4}\plus{}C\][/hide]

igiul, that's not the problem that nayel stated. $ \sin^2x$ is not the same thing as $ \sin x^2.$

If you think about it for a moment, you'll realize that $ \sin x^2$ must be related to $ e^{x^2},$ and $ \int e^{x^2}\,dx$ is perhaps the most famous example of an antiderivative that cannot be expressed as an elementary function.

[quote="Kent Merryfield"]If you think about it for a moment, you'll realize that $ \sin x^2$ must be related to $ e^{x^2},$ and $ \int e^{x^2}\,dx$ is perhaps the most famous example of an antiderivative that cannot be expressed as an elementary function.[/quote]

How do we prove that $ e^{x^2}$ does not have an elementary antiderivative?  And how do we use that fact to prove that $ \sin x^2$ does not have an elementary antiderivative, either?

The question of how to even define an "elementary" antiderivative is one that belongs to the field of [url=http://en.wikipedia.org/wiki/Differential_Galois_theory]differential Galois theory[/url].  As for the second question, $ \sin x^2 \equal{} \frac {e^{ix^2} \minus{} e^{ \minus{} ix^2}}{2i}$, so an elementary antiderivative for $ \sin x^2$ can be extracted from an elementary antiderivative for $ e^{x^2}$.

You can't calculate that integral, but [url=http://books.google.com/books?id=MtU8uP7XMvoC&printsec=frontcover&dq=handbook+mathematical+functions&ei=bJfGSMXTGIGSyATxsPC0Aw&sig=ACfU3U13gKsFw0azkOtd_FGQnR3i5uH0Sg#PPA78,M1]Abramowitz & Stegun[/url](page 78) has this one:

$ \int_0^\infty\sin(t^2)\,dt\equal{}\int_0^\infty\cos(t^2)\,dt\equal{}\frac{1}{2}\sqrt{\frac{\pi}{2}}$


geometry
parallelogram
circumcircle
trigonometry
perpendicular bisector

As shown in the figure, quadrilateral $ ABCD$ is inscribed in a circle with $ AC$ as its diameter, $ BD \perp AC,$ and $ E$ the intersection of $ AC$ and $ BD.$ Extend line segment $ DA$ and $ BA$ through $ A$ to $ F$ and $ G$ respectively, such that $ DG \parallel{} BF.$ Extend $ GF$ to $ H$ such that $ CH \perp GH.$ Prove that points $ B, E, F$ and $ H$ lie on one circle.

[asy]
defaultpen(linewidth(0.8)+fontsize(10));size(150);
real a=4, b=6.5, c=9, d=a*c/b, g=14, f=sqrt(a^2+b^2)*sqrt(a^2+d^2)/g;
pair E=origin, A=(0,a), B=(-b,0), C=(0,-c), D=(d,0), G=A+g*dir(B--A), F=A+f*dir(D--A), M=midpoint(G--C);
path c1=circumcircle(A,B,C), c2=Circle(M, abs(M-G));
pair Hf=F+10*dir(G--F), H=intersectionpoint(F--Hf, c2);
dot(A^^B^^C^^D^^E^^F^^G^^H);
draw(c1^^c2^^G--D--C--A--G--F--D--B--A^^F--H--C--B--F);
draw(H--B^^F--E^^G--C, linetype("2 2"));
pair point= E;
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$F$", F, dir(point--F));
label("$G$", G, dir(point--G));
label("$H$", H, dir(point--H));
label("$E$", E, NE);[/asy]

Circle $ \mathcal C$ with center $ C$ and radius $ CG$ cuts the rays $ AB, AD, AF$ at $ B', D', F'.$ $ B'D'GF'$ is cyclic isosceles trapezoid. Since $ CB \perp B'G$ and $ CD \perp F'D',$ $ B, D$ are midpoints of its diagonals $ B'G, F'D',$ $ BD$ is its midline. It cuts the segment $ D'G$ at its midpoint $ M,$ so that $ DM$ is the D-median of $ \triangle DD'G.$ $ CM$ is perpendicular bisector of the chord $ D'G$ $ \Longrightarrow$ $ M$ is on a circle $ \mathcal K$ with diameter $ CG.$ The angles $ \angle CBG, \angle CHG$ are right $ \Longrightarrow$ $ B, H \in \mathcal K.$

Parallel to $ DA$ through $ B$ cuts $ CG$ at $ P.$ $ BPDF$ is a parallelogram, its diagonals $ BD, PF$ cutting each other in half at $ E,$ so that $ BE$ is the B-median of $ \triangle BPF.$ Triangles $ \triangle BPF \sim \triangle DD'G$ are centrally similar, having parallel 2 pairs of corresponding sides $ BF \parallel DG, BP \parallel DD'$ and corresponding medians $ BE \parallel DM$ $ \Longrightarrow$ $ EF \parallel MG.$ Since quadrilateral $ BMGH$ is cyclic with circumcircle $ \mathcal K,$ quadrilateral $ BEFH$ with equal internal angles is also cyclic.

Let $CH$ meet $BD$ at $I$, and let $CE$ meet $FG$ at $J$. $FHEB$ cyclic is equivalent to $JI || FB$, since $\angle HJI = \angle HEI$, and $\angle BFH = 180^{\circ} - \angle HEI$ if and only if $BFHE$ is cyclic and if and only if $JI || BF$.

Since $BF || GD$, to show $IJ || BF$ it suffices to show $\frac{FJ}{JG} = \frac{BI}{ID}$. Since $\angle FHC = \angle FDC = 90^{\circ}$, $FHDC$ is cyclic, and since $\angle GHC = \angle GBC = 90^{\circ}$, $GHBC$ is cyclic. Therefore, $\angle AFG = \angle ICD$ and $\angle AGF = \angle BCI$. In addition, since $CA$ bisects $\angle FAG$, we have $\frac{FA}{AG} = \frac{FJ}{JG}$. Thus,
\[ \frac{FJ}{JG} = \frac{FA}{AG} = \frac{\sin \angle AGF}{\sin \angle AFG} = \frac{\sin \angle BCI}{\sin \angle ICD} = \frac{CB \cdot CI \sin \angle BCI}{CD \cdot CI \sin \angle ICD} = \frac{2[BCI]}{2[ICD]} = \frac{BI}{ID}, \] as desired.

Good problem , i have same solution as mitdac123

$\frac{AF}{AD}=\frac{AB}{AG} \Rightarrow AB^2=AF \cdot AG=AE \cdot AC \Rightarrow \frac{AE}{AF}=\frac{AG}{AC} \Rightarrow \triangle AFE \sim \triangle ACG$, and quad $BHGC$ is cyclic then problem solved, I don't know why in the given figure $BHGC$ is not cyclic...

WRONG SOLUTION:
We know $BHGC$$(1)$ is cyclic,then $$\angle BHG=90+\angle BHC=^{(1)} 90+\angle BGC=90+\angle AGD=90+(90-\angle DAG)=180-\angle DAG=180-\angle BAF\to \angle BHG+\angle BAF=180,$$ then $BHFA$ is cyclic.Also we know $\angle BFA=\angle BEA=90\to BFAE$ is cyclic.Then $BHFAE$ is cyclic.As desired.

[quote=Ferid.---.][quote=RC.][quote=Ferid.---.]My easy solution:
We know $BHGC$$(1)$ is cyclic,then $$\angle BHE=90+\angle BHC=^{(1)} 90+\angle BGC=90+\angle AGD=90+(90-\angle DAG)=180-\angle DAG=180-\angle BAE\to \angle BHE+\angle BAE=180,$$ then $BHEA$ is cyclic.[/quote]
How can you tell that $BHEA$ is cyclic? $B , E , F$ and $H$ are to be proved cyclic not $BHEA$.[/quote]

I am sorry,it is my mistake,and I write full and true solution in the last post.[/quote]

$\angle BGC \not =\angle AGD$ I'm afraid  :maybe:

[quote=CeuAzul][quote=Ferid.---.][quote=Ferid.---.]My easy solution:
We know $BHGC$$(1)$ is cyclic,then $$\angle BHE=90+\angle BHC=^{(1)} 90+\angle BGC=90+\angle AGD=90+(90-\angle DAG)=180-\angle DAG=180-\angle BAE\to \angle BHE+\angle BAE=180,$$ then $BHEA$ is cyclic.[/quote]
How can you tell that $BHEA$ is cyclic? $B , E , F$ and $H$ are to be proved cyclic not $BHEA$.[/quote]

I am sorry,it is my mistake,and I write full and true solution in the last post.[/quote]

$\angle BGC \not =\angle AGD$ I'm afraid :maybe:

That's exactly what I was trying to say. I have a simple proof using similarity which I will add tomorrow as it is IMO today.  :P  :jump:

[quote=RC.][quote=CeuAzul][quote=Ferid.---.]My easy solution:
We know $BHGC$$(1)$ is cyclic,then $$\angle BHE=90+\angle BHC=^{(1)} 90+\angle BGC=90+\angle AGD=90+(90-\angle DAG)=180-\angle DAG=180-\angle BAE\to \angle BHE+\angle BAE=180,$$ then $BHEA$ is cyclic.[/quote]
How can you tell that $BHEA$ is cyclic? $B , E , F$ and $H$ are to be proved cyclic not $BHEA$.

I am sorry,it is my mistake,and I write full and true solution in the last post.[/quote]

$\angle BGC \not =\angle AGD$ I'm afraid :maybe:

That's exactly what I was trying to say. I have a simple proof using similarity which I will add tomorrow as it is IMO today.  :P  :jump:[/quote]

I am sorry ,I don't understand the condition of question.I think $G$ lies on $CD.$ Then my solution is wrong.I am sorry. :(

Since, $AC$ is a diameter of \(\odot ABCD\) and \(AC\perp BD\)  we know obviously that \(ABCD\) is a kite with \(AB = AD\) and also \(\angle BAC = \angle CAD => \angle EAF = \angle GAC - (1)\)

Since, \(DG || BF => \dfrac{AF}{AD}=\dfrac{AB}{AG} => AB\times AD = AF\times AG.\)
In right \(\Delta ABC\) \(AB^{2} = AE * AC => AB \times AD = AE \times AC => AF \times AG = AE\times AC =>\dfrac{AF}{AE}=\dfrac{AG}{AD}\) and from [1] \(\angle EAF = \angle GAC => \Delta EAF \sim \Delta GAC => \angle FEA = \angle AGC = \angle BHC.\) \(\angle GHB + \angle FEB = \angle GHC +\angle BEA = 180^{\circ}\)

[quote=mitdac123]$ BF\parallel{}DG\rightarrow \frac{AF}{AD}\equal{}\frac{BA}{AG}\leftrightarrow AF.AG\equal{}AB^2\equal{}AE.AC$
$ \rightarrow \frac{AF}{AE}\equal{}\frac{AC}{AG}\rightarrow \triangle AFE \sim \triangle ACG$
$ \rightarrow \angle{AFE}\equal{}\angle{ACG}$
$ \angle{CHF}\equal{}\angle{FDC}\equal{}\angle{CBG}\equal{}90$
$ CHFD,CBHG$ are cyclic
$ \rightarrow \angle{GFE}\equal{}\angle{GFD}\plus{}\angle{AFE}\equal{}\angle{HCD}\plus{}\angle{ACG}$
$ \equal{}\angle{HCG}\plus{}\angle{ACD}\equal{}\angle{HBG}\plus{}\angle{ABD}\equal{}\angle{ABE}$
=>$ HFEB$ is cyclic[/quote]

How did you get the first line? Where did you get that $AF\cdot AG=AB^2=AE\cdot AC$

Since $\angle{FHD}+\angle{FCD}=180$
So $GF.GH=GC.GD=GA.GB$
and $\angle{AFB}+\angle{AEB}=180$
then HFAEB is cyclic


calculus
integration
real analysis
real analysis solved

Let E(a,b) = { f:R--->R, C^1, f(0) =a, f(1)=b}

K(f) = int_{0^1} (f'(t))^2dt

Find

inf _{ f in E(a,b)} K(f)

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

f is C^1 ( f' exist and is continous)

Oral examination classes prparatoires MPSI

this problem seems easy

b-a = f(1)-f(0) = \int _0^1 f'(t) dt.

by Cauchy-Schwartz ineq,

(\int _0^1 f'(t)2 dt) (\int _0^1 12 dt) \geq (\int _0^1 f'(t) dt)2  = (b-a)2

hence

\int _0^1 f'(t))2 dt \geq (b-a)2
and the equality is obtained when f is a line.


Let $ a\le b\le c$ be the sides of a right triangle, and let $ 2p$ be its perimeter. Show that
\[ p(p \minus{} c) \equal{} (p \minus{} a)(p \minus{} b) \equal{} S,
\]
where $ S$ is the area of the triangle.

Let $ r$ be the radius of the incircle of the right triangle. Then: $ S \equal{} rp$. Hence:

$ r \equal{} \frac {S}{p} \equal{} \frac {ab}{a \plus{} b \plus{} c} \equal{} \frac {ab(a \plus{} b \minus{} c)}{(a \plus{} b \plus{} c)(a \plus{} b \minus{} c)} \equal{} \frac {ab(a \plus{} b \minus{} c)}{(a \plus{} b)^2 \minus{} c^2}$

$ \equal{} \frac {ab(a \plus{} b \minus{} c)}{a^2 \plus{} b^2 \minus{} c^2 \plus{} 2ab} \equal{} \frac {a \plus{} b \minus{} c}{2} \equal{} p \minus{} c$.

So $ p(p \minus{} c) \equal{} rp \equal{} S$. Since $ S^2 \equal{} p(p \minus{} a)(p \minus{} b)(p \minus{} c)$, so $ (p \minus{} a)(p \minus{} c) \equal{} S \equal{} p(p \minus{} c)$.

If ABC is the triangle, with C=90 degs, I - incenter, M and N the contact point of the incircle with BC, CA, then CNIM is a square, hence r=CN=p-c; further, follow shobber's.

Best regards,
sunken rock


A train travels its own length is 3 seconds. it approaches a station which is 336 feet long. From the time the front of the train reaches the station until the end of the train is even with the other end of the station, 10 seconds elapse. What's the length in feet of the train?

[hide] 144 feet?
Well, the distance the train traveled is 336 + it's length, which takes 10 seconds. Since it's length takes 3 seconds, the train takes 7 seconds to go 336 feet. That means the train goes 48 feet per second, multiply that by 3, that means 144 feet per 3 seconds. And since the train takes 3 seconds to travel it's length, it's length is 144 feet.
[/hide]


I sent a PM to another member of the forum, but in my sentbox it does not show up.  Does this mean that my message didn't go to him?

It goes to the outbox first.  Once the other user opens the message it will go from the outbox to the sentbox.


The sequence of positive integers $ \{a_n, n\geq 1\}$ is such that $ a_n\leq a_{n\plus{}1}\leq a_n\plus{}5$ and $ a_n$ is divisible by $ n$ for each $ n\geq 1$. Determine all possible values of $ a_1.$

[quote="rogue"]The sequence of positive integers $ \{a_n, n\geq 1\}$ is such that $ a_n\leq a_{n \plus{} 1}\leq a_n \plus{} 5$ and $ a_n$ is divisible by $ n$ for each $ n\geq 1$. Determine all possible values of $ a_1.$[/quote]
Very nice problem !!
[hide="My solution"]
1) $ \forall n\geq 5$ knowledge of $ a_n$ implies knowledge of $ a_{n \plus{} 1}$
$ a_{n \plus{} 1}\in[a_n,a_n \plus{} 5]$ and the result is an immediate consequence of the fact that there is at most one multiple of $ n \plus{} 1$ in $ [a_n,a_n \plus{} 5]$ if $ n\geq 5$,

2) $ \forall n > 5$ knowledge of $ a_{n \plus{} 1}$ implies knowledge of $ a_{n}$
$ a_{n}\in[a_{n \plus{} 1} \minus{} 5,a_{n \plus{} 1}]$ and the result is an immediate consequence of the fact that there is at most one multiple of $ n$ in $ [a_{n \plus{} 1} \minus{} 5,a_n]$ if $ n > 5$,

3) For $ n$ great enough, either $ a_n \equal{} n$, either $ a_n \equal{} 2n$, either $ a_n \equal{} 3n$, either $ a_n \equal{} 4n$, either $ a_n \equal{} 5n$
We obviously have $ a_n\in[\max(a_1,n),a_1 \plus{} 5(n \minus{} 1)]$ and so $ \frac {a_n}{n}\in[max(\frac {a_1}n,1),5 \plus{} \frac {a_1 \minus{} 5}{n}]$
And so, for $ n$ great enough, $ \frac {a_n}{n}\in[1,5]$

4) $ a_6\in\{6,12,18,24,30\}$
We know that the sequences $ n$, $ 2n$, $ 3n$, $ 4n$, $ 5n$ are acceptable sequences.
We know that knowledge of $ a_n$ for some $ n > 6$ implies unique knowledge of $ a_6$ (from point 2 above)
So, if $ a_n \equal{} kn$ (with $ k\in[1,5]$) for some great $ n$, there is a unique possible $ a_6$. But $ a_6 \equal{} 6k$ is a possibility, hence this is the unique one.
And since (point 3 above) we know that for $ n$ great enough, either $ a_n \equal{} n$, either $ a_n \equal{} 2n$, either $ a_n \equal{} 3n$, either $ a_n \equal{} 4n$, either $ a_n \equal{} 5n$, we get the result.

5) solution
He know that $ a_6\in\{6,12,18,24,30\}$
Hence $ a_5\in\{5,10,15,20,25,30\}$
Hence $ a_4\in\{4,8,12,16,20,24,28\}$
Hence $ a_3\in\{3,6,9,12,15,18,21,24,27\}$
Hence $ a_2\in\{2,4,6,8,10,12,14,16,18,20,22,24,26\}$
Hence $ a_1$ any integer in $ [1,26]$

[/hide]


Solve the following equation:
$ 2x^{2}y{'}\equal{}y^{3}\plus{}\sqrt{x}y^{2}$

Use substitute $ y(x)\equal{}u(x)\sqrt{x}$.

[quote="Carcul"]V.V., how did you found that substitution?[/quote]

Let $ y(x)\equal{}u(x)x^\alpha$. Then $ y^3\equal{}u^3x^{3\alpha}$, $ \sqrt{x}y^2\equal{}u^2x^{1/2\plus{}2\alpha}$.

I want $ 3\alpha\equal{}2\alpha\plus{}1/2$.

[quote="Carcul"]Ok, thanks, but why did you think in making the substitution $ y \equal{} ux^{\alpha}$?[/quote]

Because this is method of solutions ODE.

It seems to me that the purpose of that substitution was to transform the original dif. eq. into a separable variables dif. eq., and in this case that only worked because we also have $ \alpha \plus{} 1 \equal{} 3 \alpha$. However, what about if the dif. eq. were

$ 2xy' \equal{} y^3 \plus{} \sqrt{x}y^2$?

[quote="Carcul"]
$ 2xy' \equal{} y^3 \plus{} \sqrt {x}y^2$?[/quote]

I think this equation is not integrable.


[quote="marko avila"]jejejeje si mi ortografía es mala pero mas bien porque me da flojera esccribir bien. pero aparte de la redaccion de la solucion de los problemas en que tiene que ver mi pobre ortografia con la olimpiada? mi ortografia en el peor de los casos son cosas que mi delegado apuntaría durant las coordinaciones. pero si tienes razón en lo anterior estoy muy de acuerdo contigo los hechos siempre hablan mejor¡¡¡ pero honestamente solo estaba jugando¡¡¡¡¡ hay que tomar las cosas un poco mas tranquilas no crees?? y si pascual lo que dices pasa mucho a mi me ha pasado mucho especislmente el ultimo año¡¡ yo siempre era de los mejores y ultimamente no he estado muy inspirado en los examenes. bueno espero que en el futuro los hechos hablen mejor¡¡ suerte a todos¡¡¡[/quote]

Creo por tu bien, que deberías hacer un gran esfuerzo por aprender a leer y a escribir correctamente, quizás te parezca tonto, pero una buena comprensión de lo leído te ayudará mucho y sobre todo en matemáticas, para entender lo  que te preguntan y saber contestarlo. No está en mi animo ser grosero o prepotente, pero estás a tiempo para aprender. Por otro lado, la carencia de buena ortografía es, en general y en la mayor parte de los idiomas, considerada como algo tosco y que demuestra bajo nivel de educación. Te lo digo honestamente, yo he sido una de las personas que más ha sufrido por su ortografía. Por suerte, dada la persistencia de algunos de mis profesores, fue algo que corregí durante la secundaria (Bueno, a veces me canso del Alt+XXX para las acentos, puesto que viviendo en gringolandia, la mayor parte de los teclados no tienen tildes ni eñes, y por el uso de algunos lenguajes, incluyendo $\LaTeX$, es mejor tener el teclado configurado en inglés), sin embargo conozco casos de gente a quien se les ha negado un buen trabajo por tener faltas de ortografía en su resumé o que han sido degradadas y hasta despedidas por tener faltas de ortografía o malos usos gramaticales en algun informe que presentan!

Honestamente, a mi me daría vergüenza como delegado llegar a defender una solución plagada de horrores de ortografía y con una pobre redacción que hace casi intelegible el texto. Normalmente algunos de los problemas en Olimpiadas son lo suficientemente confusos como para agregarle otras confuciones.

bueno creo honestamente que a pesar de algunos posts anteriores mi ortografía ha mejorado, en el sentido en que me preocupo por escribir bien para que se entienda lo que quiero decir. y tienes razón en lo que dices pero ya te expliqué porqué escribía mal antes¡¡¡¡ porqué insistes???¡¡¡ mira mejor nos preocupamos por el propósito principal que son las matemáticas?? y tu actitud si parece ser grosera cuando insistes tanto en cosas que no tiene caso insistir no crees?? mira yo creo q en otros posts que he escrito ultimamente que mi ortografía no está ni mejor ni peor que la de los demás. asi que vamos a dejar este tema así no???

ajaj muy bn marko, es cierto hablando de ortografia en un post de matematicas! qes eso! en los posts en ingles en otros lados de mathlinks la gente no sabe hablar ingles bien jaja y la matematica se entiende.. aqui estamos para hacer matematicas el colegio se ocupa de lo otro.. si alguien no entiende algo que pregunte y ya! asi  q trankilos to el mundo!

[quote="marko avila"]bueno creo honestamente que a pesar de algunos posts anteriores mi ortografía ha mejorado, en el sentido en que me preocupo por escribir bien para que se entienda lo que quiero decir. y tienes razón en lo que dices pero ya te expliqué porqué escribía mal antes¡¡¡¡ [b]porqué insistes???¡¡¡[/b][/quote]

Bueno, no era insistencia en si, pues yo tome como una pregunta a la que pretendias que respondiera cuando dijiste

[quote="marko avila"]en que tiene que ver mi pobre ortografia con la olimpiada?[/quote]

Pues simplemente estaba dando mi opinion al respecto (perdon por los acentos, pero tengo mucha pereza de ponerme a escribir con Ctl+XXX a cada rato).

[quote="marko avila"]mira mejor nos preocupamos por el propósito principal que son las matemáticas?? y tu actitud si parece ser grosera cuando insistes tanto en cosas que no tiene caso insistir no crees??[/quote]

Bueno, por supuesto que si, preocupemonos por el objetivo principal, que en este foro en especifico es el de Discuciones generales, y en el foro en general, pues si es matematicas. Ahora, si te parece grosera la actitud, pues lo lamento. La verdad dicen en mi pais que [b]la verdad no peca pero incomoda[/b], y lo que me parece es que estas recibiendo una critica y como actitud infantil estas simplemente sintiendote atacado, en lugar de tomar una actitud madura y tomar consejo de quien venga. Ahora, yo estoy consciente que yo no soy nada sutil para hacer criticas. Si has mejorado tu ortografia, pues muy bien por vos, eso a nadie va a beneficiar mas que a vos mismo.

[quote="marko avila"]mira yo creo q en otros posts que he escrito ultimamente que mi ortografía no está ni mejor ni peor que la de los demás.[/quote]

Pues al menos estas acentuando, despues de ello, con toda honestidad, pues yo no veo tanta mejoria, ademas, la escritura no es solo ortografia, intervienen gramatica, concordancia, coherencia, y muchas otras cosas...

[quote="marko avila"]asi que vamos a dejar este tema así no???[/quote]

Pues si queres podemos hacerlo. La verdad a mi ni me va ni me viene si vos queres o no recibir criticas. De todo corazon, cada critica que hago es con intencion de critica constructiva, y de la misma manera, si alguien desea criticarme, mientras no se pierda la compostura (lo que principalmente me preocupa es vocabulario soez, o insultos personales, en realidad mas por la imagen del foro que porque a mi me valga algo que me insulten o no, pero de todas maneras tengo que moderar el foro) pues adelante.

Espero no ofender a nadie.

actitud infantil?? porqué??? en algun momento te insulté?? recibo tus criticas y las tendré en mente. no digo que  no quizá no tengas razón. pero yo creo críticas de mi ortgrfaía en la escuela. como dice manuel. críticas de matemáticas son muy bienvenidas¡¡¡ yo quiero recibir críticas pero que realmente parezcan constructivas. y si realmente te interesas de la apariencia de el foro que administras pues realmente el hecho que hayas creado una gran discusión y empezado ciertas ofensas ante mi por mi ortografía no se ce muy bien que digamos no crees?? y cuando q dejemos este asunto así, no es porque no recibo críticas sino que es muy fastidioso recibir lecciones de mi vida y de mi futuro como profesional de alguien que ni me conoce no crees?? ahora recibir criticas de ti en asuntos de matemáticas es muy diferente. ahora te fijas en la ortografía de los demás?? es igual que la mia o bueno casi. tu exiges ortografía casi perfecta lo cual es absurdo y casi imposible. espero que realmente dejes de ser tan insistente en temas y asuntos no relacionados con las matemáticas. ahora si te fijas la opinion de otras personas sobre este asunto no están muy de acuerdo contigo. espero que estos comentarios sirvan solo para mejorar este foro y tu manera de administrarlo. y honestamente espero que podamos trabajar juntos porqué creo que como matemático eres muy bueno y respeto tu opinión.

[quote="marko avila"]actitud infantil?? [b]porqué???[/b] en algun momento te insulté??[/quote] Bueno, si es pregunta, deberia ser por que, separado. Y la respuesta a tu ultima pregunta alli es no, no me has insultado en ningun momento.
[quote="marko avila"]y si realmente te interesas de la apariencia de el foro que administras pues realmente el hecho que hayas creado una gran discusión y [b]empezado ciertas ofensas ante mi[/b] por mi ortografía no se ce muy bien que digamos no crees??[/quote]Pues no, no creo. La verdad me parece que todo tema esta sobre el tapete de discusion. Ahora, me gustaria que me dijeras en que momento he [i]empezado ciertas ofensas ante vos[/i] (me imagino que querras decir que he [i][b]cometido[/b] ciertas ofensas [b]contra[/b] vos[/i]), puesto que considero no haberlo hecho. Soy medio criticon, cierto, de la misma manera me gusta ser criticado (y me gusta discutir a quien me critica :P , el debate es uno de mis entretenimientos favoritos), pero en general procuro por todos los medios no ser ofensivo, lo cual no quiere decir que me ande por las ramas y siempre trate de ser sutil e indirecto, son cosas totalmente distintas.
[quote="marko avila"]y cuando q dejemos este asunto así, no es porque no recibo críticas sino que es muy fastidioso recibir lecciones de mi vida y de mi futuro como profesional de alguien que ni me conoce no crees??[/quote]Pues tampoco creo. Estas en lo correcto cuando decis que soy alguien que basicamente no te conoce, pero no estoy criticando ni tu modo de vestir, ni tus habitos alimenticios, ni nada para lo que en realidad necesite conocerte, simplemente lo que es evidente al leer tus posts, que es algo que si conozco. Con respecto a tu futuro profesional, pues en realidad simplemente estoy haciendo una afirmacion, que corregir ciertos malos habitos de escritura es profundamente beneficioso en el ambito [u]profesional[/u], en cualquier trabajo en el que verdaderamente apliques cualquier titulo universitario, y simplemente he dado ejemplos de cosas que se que han pasado. No es que te tengan que pasar a vos tambien, pero pues, si te los podes evitar mejor.
[quote="marko avila"]ahora recibir criticas de ti en asuntos de matemáticas es muy diferente. ahora te fijas en la ortografía de los demás?? es igual que la mia o bueno casi. tu exiges ortografía casi perfecta lo cual es absurdo y casi imposible.[/quote]Bueno, entonces pongamonos a resolver problemas (eso si, en el foro de olimpiadas, no en el de discusiones generales). Ahora, no exijo ortografia perfecta, pues excepto por moderacion en lenguaje y contenido de los posts, no estoy en posicion de exigirle nada mas a nadie en este foro. Absurdo, en realidad no creo que lo sea, casi imposible, pues tampoco, aunque he de confesar que entre mas tiempo pase antes de que una persona conscientemente trate de corregirlo, pues mas dificil le sera.
[quote="marko avila"]ahora si te fijas la opinion de otras personas sobre este asunto no están muy de acuerdo contigo.[/quote]Pues si, solo que mientras me sigan haciendo preguntas, pues yo seguire respondiendolas, y me parece que cada vez que haces un post, pues enuncias preguntas :D .
[quote="marko avila"]espero que estos comentarios sirvan solo para mejorar este foro y tu manera de administrarlo. [/quote]Yo tambien lo espero.
[quote="marko avila"]y honestamente espero que podamos trabajar juntos porqué creo que como matemático eres muy bueno y respeto tu opinión.[/quote]Bueno, voy a ver si propongo algunos problemillas por alli, y empezamos a discutir. Y pues, tambien, no solo de matematica vive el hombre, por eso tenemos un foro de Discusiones Generales.

Voy a ver si puedo partir el topic en dos, para que la discusion de Olimpiadas 2005 sigue su curso inicial.

bueno espero que todo esto por fin haya terminado¡¡¡


How many different prime factors does $ 20!$ have?

One way to answer this is to count the prime numbers less than 20!. We know that all of primes divides 20.

So we have.. $ 2,3,5,7,11,13,17$ and $ 19$.

Hence we have $ \boxed{8}$

[hide="Short solution"]$20!=20\times19\times18 ... 3\times 2\times 1$ is divislble by primes less than 20. There are $\boxed{8}$ primes!
         2, 3, 5, 7, 11, 13, 17, 19.[/hide]