function
number theory
relatively prime
totient function

[quote="xxxyyyy"][u]$Question 1$[/u]
$x$, a perfect square, is also the product of four consecutive odd integers. Find $x$ and prove it's unique.

[u]$Question 2$[/u]
$x$ is the product of three primes, whose squares add up to $2331$. There are $7560$ numbers less than $x$ and relatively prime to $x$, and the sum of all positive  divisors of $x$ is $10560$. Find $x$.

[u]$Question 3$[/u]
Anna and Brenda are sisters whose ages are such that if Brenda's age is written after  Anna's, the number so formed is a perfect square. If the phenomenon repeats after $13$ years  could one determine their current ages?

[u]$Question 4$[/u]
Prove that the following equality is true for every positive integer $n$:
$\frac{1^{3}+3^{3}+5^{3}+......+(2n-1)^{3}}{1+3+5+...+(2n-1)}= 2{n^{2}}-1$[/quote]
[hide="1"]$\sqrt{(x-2)x(x+2)(x+4)}=y$
$\sqrt{(x)(x+2)(x+4)(x+6)}=y$
$\sqrt{(x^{2}+6x)(x^{2}+6x+8)}=y$
$(x^{2}+6x+4-4)(x^{2}+6x+4+4)=y^{2}$
$(x^{2}+6x+4)^{2}-4^{2}=y^{2}$
$(x^{2}+6x+4)^{2}-y^{2}=16$
This resembles the 3-4-5 triangle so $x^{2}+6x+4=5$ is one solution and gives us radicals so that doesn't work. Hmm.
I'll try to prove that there is only one answer.
$n^{2}-(n-1)^{2}=2n-1$
This is an odd number so there are no consecutive perfect squares with a difference of 16.
$n^{2}-(n-2)^{2}=4n-4$
This value is 16 when n=5, but we already tested that.
$n^{2}-(n-4)^{2}=8n-16$
which works when n=4, where the roots are x=0, -6 and those are even.
$n^{2}-(n-6)^{2}=12n-36$
Well, I'm stuck...
Perhaps someone could help me with this...[/hide]
[hide="4"]The bottom is equal to $n^{2}$ and the top is equal to $(\frac{(2n-1)(2n)}{2})^{2}-8(\frac{n(n+1)}{2})^{2}$
Manipulating this we get
$(2n-1)^{2}n^{2}-2n^{2}(n+1)^{2}$
Dividing our $n^{2}$, we get
$(2n-1)^{2}-2(n+1)^{2}$
Expanding and simplifying, we get
$4n^{2}-4n+1-2n^{2}-4n-2$
But that puts me 8n to low when simplified :([/hide]

[hide="1"]
Let $n$ be some even integer:
$x=(n-3)(n-1)(n+1)(n+3)$
$=(n^{2}-9)(n^{2}-1)=((n^{2}-5)-4)((n^{2}-5)+4)=(n^{2}-5)^{2}-16$

So we need two perfect squares that differ by 16.
The only such are $0,16$ and $9,25$.
So, the first expression ($x$) must be either $0$ or $9$.

However, for the expression to equal $0$, we have
$n=\pm 1\qquad\text{or}\qquad n=\pm 3$,
contradicting the assumption that $n$ was even. So, $x$ is not $0$.

What about $9$? This can be done easily: $(-3)(-1)(1)(3)=9$.

So, $\boxed{x=9}$[/hide]

[hide="#2, the cheap way"]
Well, we have that:

$\sigma(x)=10560$
$\phi(x)=7560$
$a^{2}+b^{2}+c^{2}=2331$

I did it cheaply because I used only sum of the divisors and then used the other two to check my answer.

Because there are only primes in x and no powers of primes, we can derive from the sum of the divisors function that $(a+1)(b+1)(c+1)=2^{6}*3*5*11$

From here, I realized that no $11*2^{n}-1$ was prime except for $11*2^{2}$.  This means that $2^{4}$ is distributed to the 3 and 5.  Because the only compositions of 5s that are prime are $2^{2}*5-1$ and $2^{4}*5-1$, we know it has to be the former, because a 3 is still necessary in the final product and a $2^{4}$ would remove our 3.  So the only sequence that works is $5*2^{2}$, $3*2^{2}$, and $11*2^{2}$, giving us 19, 11, and 43 respectively, which multiply to 8987.

A similar procedure could have been done with the totient function, as $(a-1)(b-1)(c-1)=7560$, although the prime composition of this would require more combinations to analyze.
[/hide]


Diophantine equation
number theory unsolved
number theory

Let $m,n$ be positive integers. Prove that $(m^{2}+2549)(n^{2}+2549)-(m+n)$ can be perfect square.

What program did you use to get that (or you could claim to be a modern Ramanujan and say that you did that in your head... in a dream...)?

I fixed one of the variables to obtain a generalized Pell equation over the other variable. To solve the latter equation I used [url=http://www.alpertron.com.ar/QUAD.HTM]Dario Alpern's java applet[/url].

Nice, what made you think of doing that? (I know I would have spent a week trying to prove it...)


Prove that

$ a^2\plus{}b^2\plus{}c^2\leq4r^2\plus{}8R^2$

[quote="Ligouras"]Prove that

$ a^2 \plus{} b^2 \plus{} c^2\leq4r^2 \plus{} 8R^2$[/quote]

$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2\left( {s^2 \minus{} 4Rr \minus{} r^2 } \right),$

by Gerretsen Inequality:

$ s^{2}\le 4R^{2} \plus{} 4Rr \plus{} 3r^{2},$

we have

$ LHS \le 2\left( {4R^{2} \plus{} 4Rr \plus{} 3r^{2} \minus{} 4Rr \minus{} r^2 } \right) \equal{} 4(2R^2 \plus{} r^2) \equal{} RHS,$

done.


When three numbers are added two at a time the sums are 29, 46, and 53. What is the sum of all three numbers?

$ x\plus{}y\equal{}29$
$ y\plus{}z\equal{}46$
$ x\plus{}z\equal{}53$
adding these gives:
$ 2x\plus{}2y\plus{}2z\equal{}128$
dividing by 2 gives:
$ x\plus{}y\plus{}z\equal{}64$


Two people are playing a game. One of them says,
"either I will win or you will. The sample space contains 2 outcomes, so each has a probabilty of one-half." What is the error?

Because the winning is not random (in general). It depends on the rules of the game and on abilities of the players.

The error is in assuming each outcome is equally likely.

For example, were we to roll a six-sided die, and I would win if 1-4 were the result, then you would only have a 1/3 chance of winning (ie. if a 5 or a 6 were rolled).

well that would only have one outcome, right?

No, see in oceans 13 the die only lands on 3 when the guy flips his lighter. :lol:

if there are 10 numbers (1 to 10)  we can play a game. the rule is that if we get 1 to 8 , i will win. and 9 to 10, you will win.the chance of my win is 8/10. but it depend on times we play.  if we play one time ,
we get 10 and the chance of my winning is 0  but when the time grows up .  the chance will  approach 8/10.  like we can play 1 b times  :wink:


Let $ a,b,c$ are positives and $ \sum\frac{1}{a\plus{}1}\equal{} 2$.

Prove that :

$ \sum\frac{1}{a}\ge 4(a\plus{}b\plus{}c)$

 :)

[code]edited[/code]

For positive reals a,b,c
we have $ \sum\frac{1}{a\plus{}2}\leq\frac{3}{2}$ so your problem is wrong .

[quote="quangpbc"][quote="silouan"]For positive integers a,b,c
we have $ \sum\frac{1}{a\plus{}2}\leq\frac{3}{2}$ so your problem is wrong .[/quote]

@silouan, i don't understand your answer, my problem said that :$ a,b,c$ are positives, not integer positives  :([/quote]
For non-negative reals $ a,$ $ b$ and $ c$ we obtain: $ \sum_{cyc}\frac{1}{a\plus{}2}\leq\frac{3}{2}$ :wink:

oh sorry, i have edited it . Please solve it  :D

This is a well known one  :)
Let $ a \equal{}\frac{x}{y\plus{}z}$ etc
then we have to prove that
$ \frac{y\plus{}z}{x}\plus{}\frac{z\plus{}x}{y}\plus{}\frac{x\plus{}y}{z}\geq 4(\frac{x}{y\plus{}z}\plus{}\frac{y}{z\plus{}x}\plus{}\frac{z}{x\plus{}y})$
This is easy
$ \frac{1}{x}\plus{}\frac{1}{y}\geq\frac{4}{x\plus{}y}$ so  $ \frac{z}{x}\plus{}\frac{z}{y}\geq\frac{4z}{x\plus{}y}$

Adding the similar ones get the desired  :wink:


Need help please. Please show work, would really appreicated if you showed work and gave explanations. I would like to know how to solve other problems similar to these. Usually when someone solves an example equation and explains what he does and the general methods to solving similar equations and such, I can catch on very quickly. Staring into a book, that doesn't give any more examples than 3 can hardly teach you about solving equations that may slightly differ from the lesson itself. Any input will help a lot. Thank you very much.

X^5 + X - 3 = 0 Solve for "X" Please show your method, factoring techniques.  :)

Another problem is:

X^3 - X^2 + 2X - 4 = 0 Solve for "X" Pleasae show your method, factoring techniques.

All help is welcome. I been using the "Rational-Zero test" for most of the book's equations, but I have no clue as to what to do when that test "fails". Am an amateur when it comes to math, but am really want to understand this, instead of skipping it and moving on without knowing how to solve those general problems, I want to master this subject. I am practicing everyday and studying math often, even when it takes me a long time to understand it, I still love it. Thank you for reading my post, and for your input and help.

Heh, if you can simplify your explanations/theorems/methods/techniques into simple terms, I will love you. I been spending so much time trying to understand this material.


inequalities
function
algebra
domain
linear algebra
matrix

Need help with this one.

If [tex]x_i > 0[/tex] and [tex]x_i y_i > z_i^2[/tex], then show that

[tex]
\displaystyle \frac{n^3}{(\Sigma x_i)(\Sigma y_i) - (\Sigma z_i)^2} \leq \Sigma \frac{1}{x_i y_i - z_i^2}
[/tex]

where the sums are taken from [tex]i=1...n[/tex].

I'm pretty sure we can use Jenson's inequality to do this. The function [tex]f(x,y,z)=\frac{1}{xy-z^2}[/tex] is convex on the given domain in each of its three variables, and so by jenson's inequality,

[tex]\displaystyle f\left(\tfrac{1}{n}\Sigma x_i,\tfrac{1}{n}\Sigma y_i,\tfrac{1}{n}\Sigma z_i\right)\le\tfrac{1}{n}\sum f(x_i,y_i,z_i)[/tex].

I figured it out just using AM>GM>HM and Chebycheff's. Proof going up.

Edited: Oh crap, used Chebycheff's in the wrong direction. Should have known it wouldn't be that easy, being an IMO problem!

Both sides are positive so we can take the reciprocal of both sides and switch the direction of the inequality. Now the right side = [tex]HM(x_i y_i - z_i^2)/n[/tex]

[tex]
\displaystyle \frac{(\Sigma x_i)(\Sigma y_i) - (\Sigma z_i)^2}{n^3} \geq HM(x_i y_i - z_i^2)/n}
[/tex]

is true if

[tex]
\displaystyle \frac{(\Sigma x_i)(\Sigma y_i) - (\Sigma z_i)^2}{n^2} \geq AM(x_i y_i - z_i^2)}
[/tex]

Now if we rearrange the terms so they are both in increasing order, the right side (to be shown to be smaller) gets bigger and the left side stays the same therefore if our inequality is true in this case, it will be true in general.

With this order, we can apply Chebycheff's: [tex](\Sigma x_i)(\Sigma y_i) \geq  n(\Sigma x_i y_i)[/tex] on the left side,
and apply AM>GM: [tex](\Sigma z_i)^2 \geq n(\Sigma z_i)^2[/tex] on the right side, so LHS  :ge: RHS

[b]Question[/b]: where do you learn about convexity? I learnt all my inequality stuff from USSR Olympiad Problem Book which doesn't cover it.

Zach,

What form of Jensen's Inequality are you using there? You use a function of multiple variables, while the typical form of Jensen's inequality employs a function of a single variable. Do you know where I can find more information on the form of Jensen you are using?

Dont take my word entirely on this, but the way I see it, is that he treated the jensen uniformly as a function in x, y, and z respectively, just ahving to check if each variable is convex over the domain.

Since f(x,y,z) is convex in each of its three variables, it is a convex function. Jenson's inequality applies to [i]any[/i] convex function, even ones in multiple variables. Think graphically what convex means and you'll understand.

I'm not sure you're right -- couldn't a two-variable function be concave in the direction of both the x- and y-axes but convex in the direction of the line x = y?

(I'd like to see a proof is what I mean, rather than a "look at a picture.")

Hm, you might be right. I haven't taken any multivar calc, so i don't know. I was just kinda guessing it was true from intuition.

BTW, when I said it was convex as a function of x, i meant it was convex in x for ANY choice of y and z, not just on the x axis.

Right -- but if you fix y and z, and then move x, you are restricting yourself to movement in one direction, namely parallel to the x-axis.  The problem with multivariable problems is that you have many more degrees of freedom -- you can move parallel to an axis or in any combination of other, complicated ways. The shape that I'm imagining would curve downwards (be concave) in any plane perpendicular to an x- or y-axis, but would simultaneously curve upwards along the line y = x.  Something vaguely saddle-shaped.  I don't know if it could really have the property that I'm talking about, but it might.  I don't know enough about concavity to prove it one way or the other, unfortunately.  Definitely I wouldn't claim such a fact on an olympiad, though, unless I saw it somewhere I could reference to (or knew how to prove it).  I'm waiting for Guass or Ravi or fuzzylogic or someone to come tell me whether us works or not :)

Ok, after a bit of searching around I found that it is NOT sufficient. The easiest (only) way I found to show a function in many variables is concave is with a hessian test. Basically, to show that the function [tex]f(x_1,\ldots,x_n)[/tex] is concave, we have to prove that the matrix

[tex]\displaystyle H_{ij}=\frac{\partial^2}{\partial x_i \partial x_j}[/tex]

is positive definite. Sounds fun! (Notice that the condition I was stating is just saying that the main diagonal has positive entries :))

[quote="ComplexZeta"]What if you do it in one variable at a time if it's convex in each variable?[/quote]

I already toyed with this idea for a few minutes, and it introduces too many values we dont want.

[tex]
\displaystyle \left((\Sigma x_i)(\Sigma y_i) - (\Sigma z_i)^2\right)\Sigma \frac{1}{x_i y_i - z_i^2}
[/tex]
[tex]
\displaystyle \geq \left(\left(\Sigma \sqrt{x_iy_i}\right)^2 - (\Sigma z_i)^2\right)\Sigma \frac{1}{x_i y_i - z_i^2}
[/tex] by Cauchy-Schwarz
[tex]
\displaystyle = \Sigma(\sqrt{x_iy_i}-z_i)\Sigma (\sqrt{x_iy_i}+z_i)\Sigma \frac{1}{x_i y_i - z_i^2}
[/tex]
[tex]
\displaystyle \geq \left( \Sigma(\sqrt{x_iy_i}-z_i)^{1/3}(\sqrt{x_iy_i}+z_i)^{1/3}\left( \frac{1}{x_i y_i - z_i^2} \right)^{1/3} \right)^3 = \left(\Sigma 1\right)^3 = n^3
[/tex] by Holder.

Nice solution.

By the way, there is a way to generalize the original IMO problem in a different direction.  You can think of the original problem as having to do with the determinants of 2-by-2 matrices.  We can generalize to larger matrices.

For example, let A and B be two positive definite k-by-k matrices.  Show that 2k+1/det(A + B) :le: 1/det(A) + 1/det(B).

Hint: [hide]There is a way to use Jensen here.[/hide]

[Edited to fix the constant in the inequality.]


Verify that $\forall$ $n\in N$  $19\times 8^{n}+17$ is composite

[hide="Possibly not the shortest one"]$N=19\cdot 8^{n}+17\equiv 8^{n}-1\equiv (-1)^{n}-1\pmod{3}$, hence if $n=2k$, we have $3|N$

If $n=2k+1$, then $N=152\cdot 64^{k}+17$. Since $64=5\cdot 13-1$, we get $N\equiv 152(-1)^{k}+17\pmod{13}$ and $\pmod{5}$. For $k=2l$ we have $N\equiv 169\equiv 0\pmod{13}$ and for $k=2l+1$ we have $N\equiv-135\equiv 0\pmod{5}$.[/hide]


If $ a_1,a_2,\cdots,a_n$ are positive numbers and $ k$ non-negative integer, prove that:

\[ \frac{a_1^k\plus{}a_2^k\plus{}\cdots\plus{}a_n^k}{n} \leq \frac{a_1^{k\plus{}1}\plus{}a_2^{k\plus{}1}\plus{}\cdots\plus{}a_n^{k\plus{}1}}{a_1\plus{}a_2\plus{}\cdots\plus{}a_n}\]

[quote="Obel1x"]If $ a_1,a_2,\cdots,a_n$ are positive numbers and $ k$ non-negative integer, prove that:
\[ \frac {a_1^k \plus{} a_2^k \plus{} \cdots \plus{} a_n^k}{n} \leq \frac {a_1^{k \plus{} 1} \plus{} a_2^{k \plus{} 1} \plus{} \cdots \plus{} a_n^{k \plus{} 1}}{a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n}\]
[/quote]

Clear denominators and you have this:

$ \sum_{i \equal{} 1}^{n}{a_i}\sum_{i \equal{} 1}^{n}{a_i^k} \leq n\sum_{i \equal{} 1}^{n}{a_i^{k \plus{} 1}}$ which is true by $ Chebyshev$ inequality.

If you don't know that inequality, $ Chebyshev$ $ inequality$ is:

If $ a_1\geq ... \geq a_n$ and $ b_1\geq ... \geq b_n$ or
$ a_1\leq...\leq a_n$ and $ b_1\leq...\leq b_n$ than holds:

$ n(a_1b_n \plus{} a_2b_{n \minus{} 1} \plus{} ... \plus{} a_nb_1) \leq (a_1 \plus{} ... \plus{} b_n)(a_n \plus{} ... \plus{} b_n) \leq n(a_1b_1 \plus{} ... \plus{} a_nb_n)$


(a)
This one is trivial.

(b)
I don't get this. What is this? I don't see the pattern... :|

[quote="Arne"]Why did you need two separate posts for that?[/quote]
Edit: well, now with the resource page it's clear why I need it. ;)


trigonometry
number theory
relatively prime

If $\tan{x}=\frac{2ab}{a^{2}-b^{2}}$, find $\sin{x}$ in terms of $a$ and $b$

 [hide="hint"]draw a picture[/hide]

[hide="Am I right?"]$\sin(x) = \frac{2ab}{a^{2}+b^{2}}$[/hide]Am I right? If so, I will post my solution.

Well, I looked at your hint.. Anyways, Let the triangle be $ABC$, with right angle at $B$. Also, let $\tan(C) = \frac{2ab}{a^{2}-b^{2}}$. Now it's easy to determine the hypotenuse. The hypotenuse is $a^{2}+b^{2}$. So $\sin(C) = \frac{2ab}{a^{2}+b^{2}}$. It's interesting to note that if $a$ and $b$ are relatively prime integers, the triangle has side lengths of a primitive Pythagorean Triple.

[hide]Using your hint we get that the side lengths are

a^2-b^2, 2ab, a^2+b^2

So the sine of the angle is just 2ab/(a^2+b^2)[/hide]

[quote="poit"]hey can any one help me im stuck on a pretty basic trig problem, mental blank[/quote]
Hi. Welcome to the forums. Generally begin a new topic for new questions. Also your post might be considered spam because it doesn't add to the discussion. Simply posting your questions instead of asking if you can makes much more sense.
[quote="daermon"]whats your question?

Why was this moved to intermediate I could do it and i'm in elementary school  :maybe:[/quote]
There has been some concern about the difficulty level of some of the proposed problems in the lower level forums (fora?). I guess not everyone can do what you can.


Here's a cool problem I found in a school math team lecture packet.

$a_1,a_2,a_3,...,a_{100}$ are real numbers satisfying \[ a_1 + ... + a_n = n(1 + a_{n+1}+ ... + a_{100})  \]
for all integers $1 \le n \le 100$. What is the value of $a_{13}$?

[hide]I'm going to assume that you meant for those to be non-strict inequalities, because when I look at the post, it shows that, but when I reply, it shows strict inequalities.
Look at n=100.
$a_1+a_2+...+a_{100}=100$
Look at n=99.
$a_1+a_2+...+a_{99}=99+99a_{100}$
Substituting, $1=100a_{100}$ or $a_{100}=1/100$.
Look at n=1
$a_1=1+a_2+a_3+...+a_{100}$
Look at n=2
$a_1+a_2=2(1+a_3+a_4+...+a_{100})$
Using substitution, $2a_2=1+a_3+a_4+...+a_{100}$.
Therefore, $3a_2=a_1$
Trying to find $a_3$ in terms of $a_1$, using a similar process results in $6a_3=3a_2=a_1$.
Trying to find $a_4$ in terms of $a_1$, using a similar process results in $10a_4=6a_3=3a_2=a_1$.
It seems that $\frac{n(n+1)}{2}a_n=a_1$ is true.
The base case n=1 is true.
Assume it true for some given integer n as well as every integer before it that's strictly between 1 and 100.
To prove it true for n+1, first set up the equations for n and n+1.
$a_1+a_2+...+a_n=n(1+a_{n+1}+...+a_{100})$ and $a_1+a_2+...+a_{n+1}=(n+1)(1+a_{n+2}+...+a_{100})$.  Substituting equation one into two and rearranging, $(n+1)a_{n+1}=1+a_{n+2}+...+a_{100}$.  Then, add up all the terms on both sides so the right side equals $a_1$.  $(n+2)a_{n+1}+a_n+...+a_2=a_1$.  Since $a_n=\frac{2(a_1)}{n(n+1)}$, $(n+2)a_{n+1}=\frac{2}{n+1}a_1$.  This rearranged is $\frac{(n+1)(n+2)}{2}a_{n+1}=a_1$
Since $a_{100}=1/100$ and $5050a_{100}=a_1$.  Therefore, $a_1=50.5$.  Since 13(14)/2=91, $91a_{13}=a_1=50.5$.  Therefore, $a_{13}=101/182.$[/hide]


function
logarithms
inequalities
inequalities unsolved

As 1/sqrt(1+x) is convex, in descends straightfowardly by jensen's  :lol:

[quote="Harry Potter"]let a,b,c are 3 positive real numbers  such that : abc=1 . Show that :
 $ \frac{1}{\sqrt{1\plus{}a}}\plus{}\frac{1}{\sqrt{1\plus{}b}}\plus{}\frac{1}{\sqrt{1\plus{}c}}\leq\frac{3}{\sqrt{2}}$[/quote]
Let $ a\equal{}\frac{x}{y},$ $ b\equal{}\frac{y}{z},$ $ c\equal{}\frac{z}{x}$ and see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17549

[quote="quangpbc"][quote="Harry Potter"]Did you solve it ? Can you post your show[/quote]

By Jensen'inequality

$ \frac{1}{\sqrt{1\plus{}a}}\plus{}\frac{1}{\sqrt{1\plus{}b}}\plus{}\frac{1}{\sqrt{1\plus{}c}}\le\frac{3}{\sqrt{\frac{a\plus{}b\plus{}c}{3}}}$

Then we will prove $ \frac{3}{\sqrt{\frac{a\plus{}b\plus{}c}{3}}}\le\frac{3}{\sqrt{2}}$(*)

But $ \frac{a\plus{}b\plus{}c}{3}\ge\sqrt [3]{abc}\equal{} 1$ so (*) is  right . :)[/quote]

Actually, $ \frac{1}{\sqrt{1\plus{}x}}$ is concave up. We have
\[ \frac{1}{\sqrt{1\plus{}a}}\plus{}\frac{1}{\sqrt{1\plus{}b}}\plus{}\frac{1}{\sqrt{1\plus{}c}}\geq\frac{3}{\sqrt{1\plus{}\frac{a\plus{}b\plus{}c}{3}}}\]
which doesn't help.

A more natural function to look at would be $ \frac{1}{\sqrt{1\plus{}e^{x}}}$, which is concave down for $ x\geq\ln 2$ but concave up for $ x\leq\ln 2$. (Actually, you can use Majorization with this to equalize two of $ a,b,c$, but there are more elementary ways to do this one.)

[quote="scorpius119"]

Actually, $ \frac{1}{\sqrt{1\plus{}x}}$ is concave up. We have
\[ \frac{1}{\sqrt{1\plus{}a}}\plus{}\frac{1}{\sqrt{1\plus{}b}}\plus{}\frac{1}{\sqrt{1\plus{}c}}\geq\frac{3}{\sqrt{1\plus{}\frac{a\plus{}b\plus{}c}{3}}}\]
which doesn't help.

A more natural function to look at would be $ \frac{1}{\sqrt{1\plus{}e^{x}}}$, which is concave down for $ x\geq\ln 2$ but concave up for $ x\leq\ln 2$. (Actually, you can use Majorization with this to equalize two of $ a,b,c$, but there are more elementary ways to do this one.)[/quote]

The truth is that the function $ f(x)\equal{}\frac{1}{\sqrt{1\plus{}x}}$ is CONVEX not concave.But [b]quangpbc[/b]'s solution is still false ,because the inequality inverse holds as scorpius observed.

[quote="mitchan88"]As 1/sqrt(1+x) is [b]convex[/b], in descends straightfowardly by jensen's  :lol:[/quote]
Stupid mistake  :blush:

Thank all ' I use Mixing variable for sove this problem . But I have some matter when show that :
                   $ f(a,b,c)\leq f(a,t,t)$ (  $ t\equal{}\sqrt{bc}$)
It is ineq : $ \frac{1}{\sqrt{1\plus{}b}}\plus{}\frac{1}{\sqrt{1\plus{}c}}\leq\frac{2}{\sqrt{1\plus{}t}}$
Who can help me

[quote="Svejk"]
The truth is that the function $ f(x) \equal{}\frac{1}{\sqrt{1\plus{}x}}$ is CONVEX not concave.But [b]quangpbc[/b]'s solution is still false ,because the inequality inverse holds as scorpius observed.[/quote]

Oh i see  :blush: .I can't edit my post above . But we can replace $ \frac{3}{\sqrt{\frac{a\plus{}b\plus{}c}{3}}}$ by  $ \frac{3}{\sqrt{1\plus{}\frac{a\plus{}b\plus{}c}{3}}}$  :) .

[quote="Harry Potter"]Thank all ' I use Mixing variable for sove this problem . But I have some matter when show that :
                   $ f(a,b,c)\leq f(a,t,t)$ (  $ t \equal{}\sqrt{bc}$)
It is ineq : $ \frac{1}{\sqrt{1\plus{}b}}\plus{}\frac{1}{\sqrt{1\plus{}c}}\leq\frac{2}{\sqrt{1\plus{}t}}$
Who can help me[/quote]

$ \frac{1}{\sqrt{1\plus{}b}}\plus{}\frac{1}{\sqrt{1\plus{}c}}\minus{}\frac{2}{\sqrt{1\plus{}t}}\equal{}\frac{(x\minus{}y)^{2}[xy\minus{}1\minus{}\sqrt{1\plus{}xy}]}{\sqrt{1\plus{}x^{2}}\sqrt{1\plus{}y^{2}}(\sqrt{1\plus{}x^{2}}\plus{}\sqrt{1\plus{}xy})(\sqrt{1\plus{}y^{2}}\plus{}\sqrt{1\plus{}xy})\sqrt{1\plus{}xy}}$

with $ x \equal{}\sqrt{b},y \equal{}\sqrt{c}$

Try $ xy\le 1$


geometry
circumcircle
Euler
geometry unsolved

In ABC triangle points A1,B1,C1 are on BC,CA,AB sides respectively.consider M is a point in the plane.
line BM, intersect the circumcircle of A1BC1 for the second time at B2.CM intersect the circumcircle of
A1B1C at C2 and AM intersect the circumcircle of AB1C1 at A2.prove that A2,B2,C2,M are on a circle.

Dear Mathlinkers,
the circle in question is knowed like a Mannheim's circle.
Look my website : http://perso.orange.fr/jl.ayme   vol. 2  Les cercles de Morley, Euler, Mannheim et Miquel.Sincerely
Jean-Louis

[hide="Solution"]Let $ P$ be the intersection of the circumcircles of $ AB_1C_1$ and $ CA_1B_1$ other than $ B_1$

We shall show that $ P$ is on the circumcircle of $ BA_1C_1$ as well.

$ \angle PC_1B \equal{} \angle PB_1A \equal{} \angle PA_1C$. Hence $ PA_1BC_1$ concyclic.

Thus $ PB_2BA_1$ concyclic also.

$ \angle PB_2M \equiv \angle PB_2B\equal{} \angle PA_1C \equal{} \angle PC_2M$. Hence $ PC_2B_2M$ concyclic.

It can be similarly proved that $ PA_2B_2M$ concyclic.

Hence $ MA_2B_2C_2P$ all lie on a circle.

QED[/hide]


[hide="Hint"] Let $x = 123...235$.  Then we want to calculate $\sqrt{(x-1)(x+1)+1}$. [/hide]

[quote="perfect628"][hide="simple"]The big numbers I'm not in the mood to type differ by 2, so that product can be written as ${12309182375120394817235}^{2}-1$ Adding the 1, and taking the square root yields $\boxed{12309182375120394817235}$ as the answer.[/hide][/quote]

hmm wait are you saying that,

For example two numbers- x and y differ by 2...so to solve we use the ones digit of z, which is less than y by 1 and greater than x by 1...leaving you to solve $z^{2}-1$?

[quote="kstan013"]Not necessarily, what if the numbers are xxxx9 and xxxx1[/quote]

true but what I was referring to if they had a difference of 2

[quote="perfect628"]What I meant was that, because they differed by 2(so they could be expressed as $x+1$ and $x-1$, so their product is $(x+1)(x-1)=x^{2}-1$[/quote]

So the only reason why you did not have to calculate $x^{2}$ is because of the square root sign...hehe

$12309182375120394817234 \times 12309182375120394817236$


I can no longer open attachments of emails after installing Latex. I am not totally sure that this is because of Latex, but I believe (and am not 100% sure that) I started having this difficulty right after installing Latex. I have uninstalled it, but still cannot open attachments of emails.

Is it possible for Latex to cause something like this? If it is, what can do?

I am clueless on computers and would appreciate any help.

If it helps: I have a windows computer about a year old

Thanks

I can't see how LaTeX can have any effect on opening attachments so it looks like a coincidence.
Use System Restore to get back to when attachments would open.
If that doesn't work you need to tell us exactly what is happening (or not happening), what version of Windows you are using, what firewall you have, what email program you are using and the exact sequence of what you do when opening attachments, any messages you get and the complete file name of the attachments you've tried.

Sorry for the slow reply.

Windows Restore didn't work.

We have Windows Vista Home Premium.

We use hotmail.

The error say something like "Internet Explorer cannot download. No such interface supported"s when we try to right click on the link and then save (the pdf file).

We use Norton Security System.

If we try to save it regularly by left clicking on the link and then hitting save, it shows it downloading, but the progress bar that appears on my other computer (my other computer can opoen files just fine) never appears - it doesn't load at all I think.

We have to save it first to open it.

Once again, I would appreciate any help.

Thanks

[quote="hurdler"]If we try to save it regularly by left clicking on the link and then hitting save, it shows it downloading, but the progress bar that appears on my other computer (my other computer can opoen files just fine) never appears - it doesn't load at all I think.

We have to save it first to open it.[/quote]
These statements don't seem to match up as you say you can't save the file.
Is it just hotmail or is it any linked file anywhere? If it happens everywhere then it sound like an IE problem, so see what happens in another browser such as Firefox. If you have to use IE (you are using the latest version 7 aren't you?) then go to Uninstall/Change programs (see [url=http://www.zolved.com/synapse/view_content/24893/Why_cant_I_find_Add_Remove_programs_in_Windows_Vista]here[/url]), choose IE then Repair.


Let a,b,c>=0 such that ab+bc+ca=3. Then a^2-a+b^2-b+c^2-c>=1-abc. I have a difficlut solution to this one. Is it easier than I think?

It is not only right, but also extremely beautiful. Congratulations!


Let $ a,$ $ b,$ $ c,$ $ d,$ and $ e$ be five consecutive terms in an arithmetic sequence, and suppose that $ a \plus{} b \plus{} c \plus{} d \plus{} e \equal{} 30.$ Which of the following can be found?

$ \textbf{(A)}\ a \qquad \textbf{(B)}\ b \qquad \textbf{(C)}\ c \qquad \textbf{(D)}\ d \qquad \textbf{(E)}\ e$

[quote="DiscreetFourierTransform"][hide="Solution"]
In an arithmetic series, each term varies from the previous and next by a common difference.

Let's call that difference K.

We can represent the sum like this:
$a+(a+K)+(a+2K)+(a+3K)+(a+4K)$
However, that fails to eliminate K.

Try writing it like this:

$(c-2K)+(c-K)+c+(c+K)+(c+2K) = 30$
We have equal amounts of adding and subtracting K's, so we are left with $5c = 30$, and thus we can find that $\boxed{c=6}$

Choice C is correct.
[/hide][/hide][/quote]

[hide]$a+(a+d)+(a+2d)+(a+3d)+(a+4d) = 5a+10d = 30$

$a+2d = 6$.

$a+2d$ is the value of $\boxed{c}$[/hide]

[hide]
The sum of an arithmetic series is the first term plus the last term times the number of terms divided by 2.

So $\frac{5}{2}(a+e)=30$.  However $e=a+4k$ so $(a+a+4k)=12$ or $(a+2k)=c=6$.

$\boxed{C}$
[/hide]

[hide]Guessing since c is the median and the mean and stuff, C[/hide]

Solution:
Let $x$ be the common difference. Then $a = c - 2x$, $b = c - x$, $d = c + x$, and $e = c + 2x$, so \[a + b + c + d + e = (c - 2x) + (c - x) + c + (c + x) + (c + 2x) = 5c.\]But this sum is also 30, so $5c = 30$, which means $c = 6$. Therefore, the answer is $\boxed{\text{(C)}}$.

To see that the values of the other terms cannot be found, note that the sequences 4, 5, 6, 7, 8 and 10,8,6,4,2 both satisfy the given conditions.

Just by looking at it, if you can find a, you can find e, and same with b and d--the answer must be [b]c[/b]

Why revive a 8.75 year old post especially if 4 people came up with a solution already?


calculus
integration
geometry
calculus computations

The line $ y\equal{}mx$ has 3 intersection points with the curve $ y\equal{}|x(x\minus{}1)|$. Find the value of $ m$ such that the area of the 2 regions bounded by the line and the curve are equal.

[hide="one"]
$ mx \equal{} x(1 \minus{} x) \to x_1 \equal{} 1 \minus{} m$

$ mx \equal{} x(x \minus{} 1) \to x_2 \equal{} m \plus{} 1$

$ A_1 \equal{} \int_0^{1 \minus{} m} x(1 \minus{} x) \minus{} mx dx \equal{} \int_0^{1 \minus{} m} (1 \minus{} m)x \minus{} x^2 dx \equal{}$

$ \frac {1 \minus{} m}{2} (1 \minus{} m)^2 \minus{} \frac {1}{3}(1 \minus{} m)^3 \equal{} \frac {(1 \minus{} m)^3}{6}$

$ A_2 \equal{} \int_{1 \minus{} m}^{1} mx \minus{} x(1 \minus{} x) dx \plus{} \int_{1}^{1 \plus{} m} mx \minus{} x(x \minus{} 1) dx \equal{}$

$ \int_{1 \minus{} m}^{1} (m \minus{} 1)x \plus{} x^2 dx \plus{} \int_{1}^{1 \plus{} m} (m \plus{} 1)x \minus{} x^2 dx \equal{} \frac {(m \plus{} 1)^3 \plus{} (1 \minus{} m)^3 \minus{} 2}{6}$

$ A_2 \minus{} A_1 \equal{} \frac {(m \plus{} 1)^3}{6} \minus{} \frac {1}{3} \equal{} 0 \to m \equal{}  2^\frac {1}{3} \minus{} 1$  fixed [/hide]

did i fix it? i think my signs on 1/3 cancelled somewhere they shouldnt?

i dont see it, maybe im blind. ill let someone else figure it out =P

The equation of the first integral of $ A_2$ in your solution seems to be incorrect, kenn4000.

The same one with result $ m\equal{}2^\frac{1}{3}\minus{}1$.

If i did not make any computational errors, then the answer is $ m\equal{}\frac 15$

The answer is obtained by equating the following 2:

$ I_1\equal{}\int_0^{1\minus{}m} [x\minus{}x^2\minus{}mx]dx$

$ I_2\equal{}\int_{1\minus{}m}^1 [mx\minus{}(x\minus{}x^2)]dx\plus{}\int_1^{1\plus{}m} [mx\minus{}(x^2\minus{}x)]dx$

 :P

[quote="kunny"][quote="ppt"]The same one with result $ m \equal{} 2^\frac {1}{3} \minus{} 1$.[/quote]

That's right, but what's the range of $ m$?[/quote]
Dear kunny, you can look at kenn4000's solution again. He (She) fixed the result yesterday. My solution is like that.

Sorry, I haven't noticed the correction by kenn4000.

Let me say that we should reffer that $ 0<m<1$. :lol:

Yes, I understand you. It's a condition of the problem.


function
limit
induction
real analysis
real analysis unsolved

Let $ f: N* \minus{} \minus{} > R$ be a function such that :
$ f(1) \equal{} \frac {2007}{6}$ and $ \frac {f(1)}{1} \plus{} \frac {f(2)}{2} \plus{} ... \plus{} \frac {f(n)}{n} \equal{} \frac {n \plus{} 1}{2}.f(n)$ for all n
Determine the limit : $ lim_{n \minus{} \minus{} \minus{} > \infty } (2008 \plus{} n)f(n)$

[hide="hint"]
Let
$ a_n \equal{} \frac {f(n)}{n}$
then
$ a_1 \plus{} .. \plus{} a_n \equal{} \frac {n(n \plus{} 1)}{2}a_n$
then
$ a_1 \plus{} ... \plus{} a_{n \minus{} 1} \equal{} \frac {n(n \plus{} 1) \minus{} 2}{2}a_n$
so
$ \frac {(n \plus{} 1)(n \plus{} 2) \minus{} 2}{2}a_{n \plus{} 1} \equal{} a_1 \plus{} .. \plus{} a_n \equal{} a_n \plus{} \frac {n(n \plus{} 1) \minus{} 2}{2}a_n \equal{} \frac {n(n \plus{} 1)}{2}a_n$
So $ n(n \plus{} 3)a_{n \plus{} 1} \equal{} a_{n} n(n \plus{} 2)$
So $ \frac {a_{n \plus{} 1}}{a_n} \equal{} \frac {n \plus{} 2}{n \plus{} 3}$
But
$ \lim_{n\to\infty} \frac {n \plus{} 2008}{n} \equal{} 1$ then
$ \lim_{n\to\infty}(n \plus{} 2008)f(n) \equal{} \lim_{n\to\infty}a_n$
[/hide]

Thank , But i don't understand : How to find the limit of $ a_{n}$ ?


probability
integration
probability and stats

Hey, I have a rather difficult (and interesting) probability problem. Perhaps someone could help me out here.

Let's say you have the interval [0,1]. You generate 3 uniformly random numbers between 0 and 1 and place them on that interval. You now have 4 line segments. Sort the 4 segments in increasing lengths. What is the probability that the sum of the lengths of the two middle segments is greater than 1/2?

e.g.
If your random numbers were: 0.5, 0.3, and 0.1
The lengths of the line segments would be: 0.1, 0.2, 0.2, and 0.5
The two medium-length segments would be of lengths 0.2 and 0.2, whose sum is 0.4

I know the answer to this problem via computer simulation, but an elegant proof would be really interesting to see.
Answer: [hide]1/4[/hide]

keyword: [url=http://en.wikipedia.org/wiki/Order_statistic]order statistics[/url]

suppose that you have $ n$ iid random variables: $ X_1, X_2, \ldots, X_n$ with common density $ f(x) dx$. Rearrange them in increasing order $ X_{(1)}, \ldots, X_{(n)}$. Then the joint distribution of $ (X_{(1)}, \ldots, X_{(n)})$ is
\[ F(x_1, x_2, \ldots, x_n) = n! f(x_1) f(x_2) \ldots f(x_n) 1_{x_1 < x_2 < \ldots < x_n}\]
(can you see why?)
hence the probability you are looking for is
\[ P = \int_{x=0}^1 \int_{y=x}^1 \int_{z=y}^1 1_{z > x+\frac{1}{2}} \, dx \, dy \, dz\]
which is routine computation. You should find $ P=\frac{1}{4}$.

ps: this is the obvious approach. There might (must) be a more clever approach  :)

Alekk, the question was about two [b]medium length[/b] segments, not about two [b]middle[/b] segments!


I just installed Ubuntu 7.04 on my laptop.  I dual boot with Win XP Pro, but I can't get my Microsoft Intellimouse to work on Ubuntu.  It's a 5 button mouse - left click, right click, back, forward, and scroller - but only the left and right mouse buttons work.  I've tried to use the solutions on Ubuntoforums but haven't found anything effective or easy to understand.  Can anyone offer a solution?

The problem is most likely that you don't have the right driver for it. You might have some kind of general driver that recognizes the mouse but not all of its functions. You need to find the right driver. If there isn't one, then you just have to find some people who write drivers and suggest it and hope that they do it. If you're a programmer, you could try to do it yourself.


How many natural number $ n$ with $ n\leq2009$ which satisfies $ lcm(n,2009)\plus{}gcd(n,2009)\equal{}n\plus{}2009$? ?

Write $ g\equal{}\gcd(2009,n)$. Then, $ \text{lcm}(2009,n)\equal{}2009n/g$ so we have:
$ 2009n/g\plus{}g\equal{}n\plus{}2009$ or $ g^2\plus{}2009n\equal{}gn\plus{}2009g$, or $ (g\minus{}2009)(g\minus{}n)\equal{}0$

So $ g\equal{}2009\iff2009\mid n$, or $ g\equal{}n\iff n\mid2009$
Thus there are 6 solutions: 1,7,41,49,287,2009

[quote="GTK X Hunter"]How many natural number $ n$ with $ n\leq2009$ which satisfies $ lcm(n,2009) \plus{} gcd(n,2009) \equal{} n \plus{} 2009$? ?[/quote]

-factor 2009
-notice that for n big enough and n not divisor of 2009,it won´t work.
-notice that n|2009.


Does anyone use this?  It's basically the Google version of Myspace.

If you do, then you should give me an invite.  My e-mail is shravas@gmail.com.

Orkut? That sounded ike you were trying to pronounce Urquhart..no offense to people with that name..Google has everything. Couldn't you just go there without an invite? JW

Hmm..that's strange..well sorry I am not a member of Orkut.

I'll invite you. Anyone else interested?

me too

a=haron_ix
b=@
c=yahoo.com

abc

thanks

okrut is some service from google...but i don't know much more

By the way, to clear up some confusions, the name orkut comes from an individual with the frightening name of Orkut Buyyukkoten. He is probably dutch or belgian, and the creator of orkut.

Among the vast number of peope in orkut,  64.30% are Brazillians, 13.63% from the USA, 9.00% ar Indians, 1.83% are Pakistanis, 1.27% are Iranians, 0.70% British, 0.55% Japanese, 0.47% Portugese, 0.44% Canadians and 0.40% Italians.

By the way, deimos is yet to join.

[quote="bubka"]
By the way, deimos is yet to join.[/quote]

not any more, I managed to found password for the mail.  :lol:


Let X be a set with n elements, and 0  k  n. We denote by an;k (bn;k) the maximum
number of permutations of the set X such that every two of them has at least (at most) k
common components. (An element x 2 X is called a common component of permutations f
and g if f(x) = g(x).)
(a) Show that an;k  bn;k  n!.
(b) Find the exact value of ap;2 where p is a prime number.

Bomb

1.: dont' post it twice  ( http://www.mathlinks.ro/Forum/viewtopic.php?t=42258 )
2. use $\LaTeX$ !
3. what shall we prove at a) ¿

ZetaX is right, try to follow his hints ;)


Freeeeeeeeeeeeeeee Music Engelish Irani arabic turky ...

 :oops:

http://www.iranmusic.eu/
http://www.iranmusic.eu/
http://www.iranmusic.eu/

:P its not half bad(some of it is horrible though!)

Threeeeeeeeeeeeee times is enough times to post the same link...


geometry rectangle perimeter ratio circumcircle trigonometry

A rectangle and an equilateral triangle with equal area are inscribed in a circle with radius $2$. Find the dimentions of the rectangle.

[b]EDIT[/b]- I changed the problem a bit (from "perimeter" to "dimentions")

This problem looks awfully familiar to me, so here's
[hide="a little hint"]The altitude of the equilateral triangle is 3 (do you know why?). [/hide]

[quote="236factorial"]
The altitude of the equilateral triangle is 3 (do you know why?). [/quote]

Well obviously because the radii can break it down to $6$ $30,60,90$ triangles, and by solving for them, you see the height is $3$, or by knowing the ratio of the segments that make up the circumcenter is $2: 1$.

[quote="Bictor717"][quote="riddler"][hide="another hint"]
$\text{area of a triangle}=\dfrac{1}{2}r^2\sin{\theta}$
[/hide][/quote]

Not for this triangle.[/quote]
You didnt listen to me, I never said for THIS triangle. This applies for isoscelese triangles, So taking circles and joining the centre with the vertices....

You're on the right track, Iversonfan2005. If you use substitution, it shouldn't be very hard to find $l$ and $w$ (although algebra gets kinda messy):)

No need to know trig for this problem ;)

The reason I changed the problem (asking dimentions rather than perimeter) is because perimeter gets ugly. You'll see what I mean if you get $l$ and $w$ :roll:

[quote="frt"]You're on the right track, Iversonfan2005. If you use substitution, it shouldn't be hard to find $l$ and $w$ :)

No need to know trig for this problem ;)

The reason I changed the problem (asking dimentions rather than perimeter) is because perimeter gets ugly. You'll see what I mean if you get $l$ and $w$ :roll:[/quote]
Yes the perimeter is a square root under square root $2\sqrt{16+6\sqrt{3}}$

[quote="Iversonfan2005"]This problem is kinda tough :( do you need to use trig?

Here's what I've done so far:

I got the area of the triangle of $3\sqrt{3}$, so I assigned the sides $l$ and $w$, so I have:

$lw=3\sqrt{3}$
$l^2+w^2=4^2$

I can get the perimeter this way but not the individual values of $l$ and $w$ :([/quote]
You can get the individual values quicker.
$l=\dfrac{3\sqrt3}{w}$
Substituting in the other equation,
$27+w^4=16w^2$
$w^4-16w^2+27=0$
Solve for $w$
 :D

[quote="Altheman"][quote="riddler"]
$27+w^4=16w^2$
$w^4-16w^2-27=0$
Solve for $w$
 :D[/quote]

you mean $w^4-16w^2+27=0$ and using the quadformula,
we get $\sqrt{8\pm\sqrt{37}}$[/quote]
Yeah, that is the width (and length) of the rectangle :yup:


$ P(x),Q(x),R(x),S(x),G(x) and T(x)$ polinom $ P(x^6)\plus{}x.Q(x^6)\plus{}x^2.R(x^6)\plus{}x^3.S(x^6)\plus{}x^4.G(x^6)\equal{}(x^5\plus{}x^4\plus{}x^3\plus{}x^2\plus{}x^1\plus{}1).T(x)$ so $ P(x),Q(x),R(x),S(x),G(x) and T(x)$ polinom each other divided by $ (x\minus{}1)$

[quote="yildobekir"]$ P(x),Q(x),R(x),S(x),G(x) and T(x)$ polinom $ P(x^6) \plus{} x.Q(x^6) \plus{} x^2.R(x^6) \plus{} x^3.S(x^6) \plus{} x^4.G(x^6) \equal{} (x^5 \plus{} x^4 \plus{} x^3 \plus{} x^2 \plus{} x^1 \plus{} 1).T(x)$ so $ P(x),Q(x),R(x),S(x),G(x) and T(x)$ polinom each other divided by $ (x \minus{} 1)$[/quote]

$ x^6\equal{}(x\minus{}1)(x^5 \plus{} x^4 \plus{} x^3 \plus{} x^2 \plus{} x^1 \plus{} 1)\plus{}1$

So $ x^5 \plus{} x^4 \plus{} x^3 \plus{} x^2 \plus{} x^1 \plus{} 1| P(1)\plus{}xQ(1)\plus{}x^2R(1)\plus{}x^3S(1)\plus{}x^4G(1)$
$ \Rightarrow P(1)\equal{}Q(1)\equal{}R(1)\equal{}S(1)\equal{}G(1)\equal{}0$
$ \Rightarrow x\minus{}1|P(1),Q(1),R(1),S(1),G(1) \Rightarrow x\minus{}1|T(x)$ also.


How many four-digit numbers are there that consist of different digits, end in 5 and are divisible by each of their digits?

[hide]Six: $1395, 1935, 3195, 3915, 9135, 9315$
We already can limit our search to 900 numbers, since that is how many four digit numbers end with 5. Because the numbers end in 5, we know that it cannot be divisible by 2, therefore, no even number can be used, which greatly limits our search. Also remembering that each digit can only be used once makes it very easy just to test out the numbers that we have not already eliminated. Didn't take that long at all, could be done without a calculator pretty easily.[/hide]

I did the same, but I was too lazy to check all the combinations (24). I got to 1395 and stopped a few numbers later  :blush:


inequalities
ratio
geometry
geometry unsolved

A point M is given in the interior of the triangle ABC; consider the intersection points of AM, BM, CM with the corresponding sides A', B', C'. Show that in the set

            [size=150]{AM/MA', BM/MB', CM/MC'}[/size]

there is at least one ratio greater than or equal to 2 and at least one ratio less than or equal to 2.

It's simple to prove :
$ \frac{MA'}{AA'} \plus{} \frac{MB'}{BB'} \plus{} \frac{MC'}{CC'} \equal{} 1$
(Use area is a simple way to prove it)
Therefore, there is at least one ratio greater then one third and one ratio smaller than one third.


calculus
integration
function
trigonometry
calculus computations

Find the functions $ f(x),\ g(x)$ such that $ f(x)\plus{}\int_0^x g(t)\ dt\equal{}\sin x(\cos x\minus{}\sin x),\ \{f'(x)\}^2\plus{}\{g(x)\}^2\equal{}1$.


Could all you guys tell me how can you type mathematical expression in such a good one? How can I learn to do that, because I have many maths problems to post, but I can not manage to type it in the computer. thank you in advance.

LaTeX is the mathematics typing system that most people use.  Use the forum or the links on the left of AoPS for more information.  But don't ask me if you have any questions about it, I'm just learning it myself.  You can download the enitre thing for free.

[quote="limkimsrun"]Where can I download Latex for free!!!!!![/quote
look at the top of the mathlinks skin.
by the way, this topic shouldn't be here.

you don't need to download it...
just put dollor sign , behind and after math expression you want to write, for example if istead of writing sqrtx, if you put dollor sign with a "\"behid it you get: $\sqrt{x}$ :)

If you want to use it OUTSIDE of Mathlinks/AoPS (yes, there does exist math outside of here) you want to download a compiler as well as the actual language itself. Just go to the LaTeX section on AoPS for information on how to install and where to download it, like bubala said.


modular arithmetic
graph theory
combinatorics unsolved
combinatorics

Can you colour the edges of a graph with 2004 vertices with 2003 colours each one used 1002 times such that for every 2 colours used there exists a cycle through the graph which passes through all the points and has it's edges exclusively of these 2 colours?

Consider a complete graph on $2003$ vertices, labeled with $0,1,\ldots,2002$. Color the edge $ij$ with color $i+j\pmod{2003}$. We can prove that, for each $i\ne j$, there is a hamiltonian path beginning in vertex $i$ and ending in vertex $j$ which only goes along the edges $2i,2j\pmod {2003}$. This is equivalent to saying that $i,2j-i,3i-2j,4j-3i,\ldots,2001i-2000j,2002j-2001i$ are distinct $\pmod{2003}$, which is easy, considering that $2003$ is prime.

Now, in order to complete the construction, add another vertex $x$ (we must have $2004$ vertices in the graph, but we only have $2003$ up until now), and, for all $i\in\overline{0,2002}$, color the edge $xi$ with color $2i\pmod{2003}$. For each $i,j$, the hamiltonian path beginning at $i$ with the edge $2j$ and ending at $j$ with the edge $2i$, and going only along the edges $2i,2j$, can be completed with the two edges $xi,xj$ to give a hamiltonian cycle, going only along the edges $2i,2j$ (all the indices are modulo $2003$).

I hope it's clear.


For $ a,b,c$ positive reals such that $ a\plus{}b\plus{}c\equal{}1$ prove that

$ \sum a\sqrt{1\minus{}bc} \geq \frac{2\sqrt{2}}{3}$

[quote="manlio"]For $ a,b,c$ positive reals such that $ a \plus{} b \plus{} c \equal{} 1$ prove that

$ \sum a\sqrt {1 \minus{} bc} \geq \frac {2\sqrt {2}}{3}$[/quote]
$ \left(\sum_{cyc} a\sqrt {1 \minus{} bc} \right)^2 \equal{} \sum_{cyc}a^2(a^2 \plus{} b^2 \plus{} c^2 \plus{} 2ab \plus{} 2ac \plus{} bc) \plus{}$
$ \plus{} 2\sum_{cyc}ab\sqrt {(a^2 \plus{} b^2 \plus{} c^2 \plus{} 2ab \plus{} 2ac \plus{} bc)(a^2 \plus{} b^2 \plus{} c^2 \plus{} 2ab \plus{} ac \plus{} 2bc)}\geq$
$ \geq\sum_{cyc}\left(a^4 \plus{} 2a^3b \plus{} 2a^3c \plus{} 2a^2b^2 \plus{} a^2bc \plus{} 2ab(a^2 \plus{} b^2 \plus{} c^2 \plus{} 2ab \plus{} ac \plus{} bc \plus{} c\sqrt {ab})\right) \equal{}$
$ \equal{} \sum_{cyc}(a^4 \plus{} 4a^3b \plus{} 4a^3c \plus{} 6a^2b^2 \plus{} 7a^2bc \plus{} 2abc\sqrt {ab}).$
Id est, it remains to prove that $ 9\sum_{cyc}(a^4 \plus{} 4a^3b \plus{} 4a^3c \plus{} 6a^2b^2 \plus{} 7a^2bc \plus{} 2abc\sqrt {ab})\geq8(a \plus{} b \plus{} c)^4.$ But
$ 9\sum_{cyc}(a^4 \plus{} 4a^3b \plus{} 4a^3c \plus{} 6a^2b^2 \plus{} 7a^2bc \plus{} 2abc\sqrt {ab})\geq8(a \plus{} b \plus{} c)^4\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^4 \plus{} 4a^3b \plus{} 4a^3c \plus{} 6a^2b^2 \plus{} 18abc\sqrt {ab} \minus{} 33a^2bc)\geq0.$
By AM-GM we obtain: $ \sum_{cyc}(a^4 \plus{} 4a^3b \plus{} 4a^3c \plus{} 6a^2b^2 \plus{} 18abc\sqrt {ab}) \equal{}$
$ \equal{} \sum_{cyc}(a^4 \plus{} 4a^3b \plus{} 4a^3c \plus{} 3a^2b^2 \plus{} 3a^2c^2 \plus{} 9abc\sqrt {ab} \plus{} 9abc\sqrt {ac})\geq$
$ \geq33\sum_{cyc}\sqrt [33]{a^{67}b^{32.5}c^{32.5}}\geq33\sum_{cyc}a^2bc,$ where the last inequality is true by Muirhead.
By the way, for all $ 0\leq k\leq4$ and $ a,$ $ b$ and $ c$ are non-negative such that $ a \plus{} b \plus{} c \equal{} 1$ the following inequality is true too:
\[ a\sqrt {1 \minus{} k bc} \plus{} b\sqrt {1 \minus{} k ac} \plus{} c\sqrt {1 \minus{} k ab}\geq\frac {\sqrt {9 \minus{} k}}{3}
\]

How is your solution, Pvthuan?

Thank you very much.

After working a lot on this inequality I cannot think a solution different from Arqady, but perhaps it exists a more simple solution  :wink:

Is your solution, Pvthuan similar to Arqady's solution?

Thank you very much.

Is your solution Pvthuan similar to Arqady's solution? :maybe:

Can you, please, answer by YES or NOT. :wink:

Thank you very much.  :)

Another way is to apply Chebyshev Inequality and it suffices to show

$ \sqrt {1 \minus{} ab} \plus{} \sqrt {1 \minus{} bc} \plus{} \sqrt {1 \minus{} ca} \ge 2\sqrt {2}$

See

http://www.mathlinks.ro/viewtopic.php?t=198498

In this way we can prove the arqady's generalization for $ k \le \frac {32}{9}$


Let $ ABC$ be a triangle. Let points $ D$, $ E$, $ F$ in $ BC$, $ CA$, $ AB$, respectively such that $ AD$, $ BE$, $ CF$ are concurrent at $ P$. Prove that $ BD=DC\implies EF//BC$.

Solution with projective geometry:
[hide]
We will consider a projective transformation that sends $ A$ to infinity. Suppose that we have this diagram in a plane $ P$. Then take another plane $ Q$ that passes through $ BC$. Project the diagram onto $ Q$. $ B$, $ M$, and $ C$ are fixed. $ A$ is off at infinity, so $ AB//AM//AC$. $ P$ is a point on $ AD$ and $ E$ and $ F$ are the intersections of $ AB$, $ CP$, etc. It trivially follows that $ BCFE$ is a parallelogram. Our projection preserves parallelism to $ BC$, so we are done.[/hide]

Well of course I know that. I thought that that solution was interesting.

Here is another quick way to solve it:

If $ EF$ and $ BC$ intersect at $ X$, then $ (BC, DX)=-1$. Since $ D$ is a midpoint, then $ X$ is at infinity; i.e. $ EF//BC$.

Of course these solutions are relatively useless when the trivial ceva solution is out there. But I wanted to start applying my newly learned projective geometry knowledge to previously known problems.

Ah...okay.

When you say $ (BC, DX)=-1$, do you mean that the circles with diameters $ DX$ and $ BC$ are orthogonal? (ie we have harmonic conjugates?)

The notation for cross ratio is always a source of confusion...I basically mean that $ B,C$ and $ D,X$ are harmonic conjugates wrt. the other pair. In otherwords,

$ (B,C; D,X)=-1\iff \frac{\frac{BD}{DC}}{\frac{BX}{XC}}=-1$.

[I got lazy with my punctuation, but it is clear what I mean.]

Oh...so that's what cross-ratio is...
I should keep that notation in mind.

I never really understood why Zabel used directed lengths when he taught us cross ratio... Ivan's was so much clearer

Well...with out directed lengths, the theory is inconsistent. Like if you have a point such that $ AB: BC=2$ where $ A$ and $ C$ are fixed, then there are two possibilities for point $ B$ if we use undirected lengths.


Let k be a positive real number such that

[code]\frac{1}{k+a} + \frac{1}{k+b} + \frac{1}{k+c}\le{1}

for any positive real numbers a, b, c with abc= 1. Find the minimum value of k.[/code]

Assume $ 0\le k<2$
Then $ \frac{1}{k\plus{}1}\plus{}\frac{1}{k\plus{}1}\plus{}\frac{1}{k\plus{}1}>1$
It is easy to show, that inequality holds for $ k\equal{}2$.


Let be a $\triangle {ABC}$ with $AB=100$ and $AC=156$. Let be $M$ the midpoint of $AB$, and $K$ a point in $AC$ such that $MK \perp AC$. If $AK=14$, find $BC$.

In $\triangle{AMK}$, we can find $\cos{A}=\frac{7}{25}$. Thus use the law of cosine in $\triangle{ABC}$.

[hide]Yes, and $\boxed {BC=160}$. These problem was from an Argentinian Competition... beautiful!!!!.[/hide]


A composite (positive integer) is a product ab with a and b not necessarily distinct integers. Show that every composite is expressible as xy+xz +yz+1, with x, y, and z positive integers.

This was problem 1988 Putnam B1 for anyone who searches it.


What is the sum of the numerators in the partial fraction decomposition of:

$ \dfrac{11x^2\minus{}25x\plus{}1}{x^3\minus{}x^2\minus{}8x\plus{}12}$

This is to be done without a calculator. Thanks in advance!

Alternatively, from $ \frac{11x^2\minus{}25x\plus{}1}{x^3\minus{}x^2\minus{}8x\plus{}12}\equal{}\frac A{(x\minus{}2)^2}\plus{}\frac B{x\minus{}2}\plus{}\frac C{x\plus{}3}$,

we could have multiplied both sides by $ (x\minus{}2)^2(x\plus{}3)\equal{}x^3\minus{}x^2\minus{}8x\plus{}12$ to get

$ 11x^2\minus{}25x\plus{}1\equal{}A(x\plus{}3)\plus{}B(x\minus{}2)(x\plus{}3)\plus{}C(x\minus{}2)^2$,

so plugging in the values of $ x\equal{}2,\minus{}3$ gives us

$ \minus{}5\equal{}5A\implies A\equal{}\minus{}1$ and $ 175\equal{}25C\implies C\equal{}7$, respectively, and then we have

$ B(x\minus{}2)(x\plus{}3)\equal{}11x^2\minus{}25x\plus{}1\minus{}A(x\plus{}3)\minus{}C(x\minus{}2)^2$

$ \equal{}11x^2\minus{}25x\plus{}1\plus{}x\plus{}3\minus{}7x^2\plus{}28x\minus{}28\equal{}4x^2\plus{}4x\minus{}24\equal{}4(x\minus{}2)(x\plus{}3)$

$ \implies B\equal{}4$, giving us

$ (A,B,C)\equal{}\boxed{(\minus{}1,4,7)}$.


Let $ p$ be a prime and suppose that $ (a_1, b_1)\in\mathbb { N }\times\mathbb { N }$ is a solution of $ 5p\equal{}2a^2\plus{}3b^2$. Prove or disprove that there is exactly one other solution $ (a_2, b_2)\in\mathbb { N }\times\mathbb { N }$. :)

[hide]If this is true, then prove or disprove that if $ a_1>a_2$, then $ a_1\plus{}b_1\ge a_2\plus{}b_2$. :D

Smallest primes:
$ p\equal{}7, (4,1)$ and $ (2,3)$
$ p\equal{}31, (8,3)$ and $ (2,7)$
$ p\equal{}73, (13,3)$ and $ (1,11)$
[/hide]

The equation is equivalent to $ 10p \equal{} (2a)^2 \plus{} 6b^2 \equal{} (2a \plus{} \sqrt { \minus{} 6}b)(2a \minus{} \sqrt { \minus{} 6}b)$. So we are interested in solving $ 10p \equal{} x^2 \plus{} 6y^2 \equal{} N(x \plus{} \sqrt { \minus{} 6} y)$ (note that $ \mod 4$ tells us that $ x$ is necessary even if we have that equation). Well, this could be done elementary, but just see http://www.mathlinks.ro/viewtopic.php?t=232507 and make the adaptions (it's perfectly the same).

You end with
$ S \equal{} \{ p \in \mathbb P | p \equiv 1,7 \mod 24\}$
$ T \equal{} \{ p \in \mathbb P | p \equiv 2,3,5,11 \mod 24\}$
$ U \equal{} \{ p \in \mathbb P | p \equiv 13,17,19,23 \mod 24\}$
and the same result:

Let $ n \in \mathbb N$. Then $ n$ is of type $ x^2 \plus{} 6y^2$ iff the following conditions hold all together:
a) $ 2|v_p(n)$ for all $ p \in U$
b) $ 2|\sum_{p\in T} v_p(n)$.

So we get that such $ a,b$ exists iff $ p \equiv 1,7 \mod 24$ and that then there are two of them (the later by a carefull study how $ 10p$ factors).

Thank you very much. Now I see it more clearly...

By the way, more generally for $ p\in S$ and $ k\in \mathbb { N }$ odd, the equation $ 5^kp \equal{} 2a^2 \plus{} 3b^2$ has exactly $ k \plus{} 1$ solutions $ \in\mathbb { N }\times\mathbb { N }$. :)
[hide]And in the case $ k \equal{} 1$, my conjecture

[quote="spanferkel"]If this is true, then prove or disprove that if $ a_1 > a_2$, then $ a_1 \plus{} b_1\ge a_2 \plus{} b_2$. :D

[/quote]
is indeed true. :)[/hide]

Likewise, for $ p\in\mathbb{P},p\equiv 5,11 \mod 24$ (except $ p \equal{} 5$) and $ k\in \mathbb { N }$ even, the equation $ 5^kp \equal{} 2a^2 \plus{} 3b^2$ has exactly $ k \plus{} 1$ solutions $ \in\mathbb { N }\times\mathbb { N }$. :)


Prove that for every odd integer $ n$ greater than 1, there is a primitive pythagorean triple with shortest leg with length $ n$.

[hide]
It is well known, that all the primitive pythagorean triples $ (x,y,z)$ so that $ x^2 \plus{} y^2 \equal{} z^2$ are generated so, that

$ x \equal{} m^2 \minus{} n^2$
$ y \equal{} 2mn$
$ z \equal{} m^2 \plus{} n^2$

for coprime integers $ m,n$ one of which is even.

(also well known that one of $ (x,y)$ *must* be even and the other and $ z$ odd)

Now let $ m\equal{}n\plus{}1$ they are coprime and one of them is even, so the resulting triple is primitive.

$ x \equal{} (n\plus{}1)^2 \minus{} n^2 \equal{} 2n\plus{}1$ which is an odd integer, as $ n$ goes up every odd integer is included.
$ y \equal{} 2(n\plus{}1)n$
$ z \equal{} (n\plus{}1)^2 \plus{} n^2$

Now clearly $ 2(n\plus{}1)n > 2n\plus{}1$ if $ n \ge 1$ (for odd integers $ \ge 1$)
[/hide]

[hide="In a more elementary fashion"] Recall the identity

$ (k \plus{} 1)^2 \equal{} k^2 \plus{} (2k \plus{} 1)$

where $ 2k \plus{} 1$ can be an arbitrary odd number.  Let $ 2k \plus{} 1 \equal{} n^2$.  The conclusion follows. [/hide]


If a cevian AQ of an equilateral triangle ABC is extended to meet the circumcircle at P, then prove 1/PB+1/PC=1/PQ.

$\left\{\begin{array}{ccc}\triangle ABP\sim\triangle CQP & \Longrightarrow & PB\cdot PC=PA\cdot PQ\\\\ PB+PC=PA & \Longrightarrow & (PB+PC)\cdot PQ=PA\cdot PQ\end{array}\right\|$ $\Longrightarrow$ $(PB+PC)\cdot PQ=PB\cdot PC$ $\Longrightarrow \boxed{\ \frac{1}{PQ}=\frac{1}{PB}+\frac{1}{PC}\ }\ \ (*)\ .$

[b]Remark (question).[/b] Given is a triangle $ABC$ for which  $b<a<c\ .$ Ascertain the position of the point $Q\in BC$ which verifies  verifies the relation $(*)\ .$
[hide="Answer."] $\frac{QB}{QC}=\frac{c(b-a)}{b(a-c)}\ .$[/hide]


Um why are there two different values for white paint? Is there a typo here? One of the paint amounts is supposed to be blue, right? :maybe:

Yeah, it is 130 ounces of blue paint and 188 ounces of white paint.

so it is
Blue: 130
Red: 130+34
White:130+58

so regions of paint are each 92
$ 130\minus{}92\equal{}38$
$ 38 \cdot 3\equal{}114$

$ \boxed{114}$

I'm retarded. I forgot that the ounces of paint for each of the paint colors would be the same. I put down the answer of how many ounces of each of the paint colors was left before the pink stripe was even made. :furious:

[quote="worthawholebean"]Before starting to paint, Bill had $ 130$ ounces of blue paint, $ 164$ ounces of red paint, and $ 188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stirpe. Pink is a mixture of red and white, not necessarily in equal amounts. When Bill finished, he had equal amounts of blue, red, and white paint left. Find the total number of ounces of paint Bill had left.[/quote]

[quote]
Let the blue stripe = x, the red stripe = x, the white stripe = x and the pink stripe = 1/n x red + (n-1)/n x white.
The blue paint left = 130 - x
The red paint left = 164 - (x + 1/n x)
The white paint left = 188 - ( x + (n-1)/n x)

Bill had equal amounts of blue, red, and white paint left
130 - x = 164 - x - 1/n x ,, 34 - 1/n x = 0 ... (1)
130 - x = 188 - x - (n-1)/n x ,, 58 - x + 1/n x = 0 ... (2)

From (1) and (2)
92 - x = 0 ,, x = 92
34 - 1/n (92) = 0 ,, 1/n = 34/92 = 17/46

Then
The blue paint left = 130 - 92 = 38
And Bill had equal amounts of blue, red, and white paint left

Thus ...
the total number of ounces of paint Bill had left = [u]114 ounces [/u]

 :lol:
[/quote]

Since all the stripes are of equal size, the pink stripe must have brought both the red and white paint amounts to 130 ounces, making the pink stripe using (164-130)+(188-130)=34+58=92 ounces. This amount must have been the amount of the other stripes, meaning the amount of each color paint is (130-92)=38 ounces. Multiplying by 3 (there are three colors of paint) gives 114, the total number of ounces of paint that Bill has left.

[b]114 ounces[/b]

[quote="worthawholebean"]Before starting to paint, Bill had $ 130$ ounces of blue paint, $ 164$ ounces of red paint, and $ 188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stirpe. Pink is a mixture of red and white, not necessarily in equal amounts. When Bill finished, he had equal amounts of blue, red, and white paint left. Find the total number of ounces of paint Bill had left.[/quote]

[hide="Solution"]
Lets say one strip takes $x$ ounces of paint. After painting the four strips, Bill has $130 - x$ ounces of blue paint left, so he must have $130-x$ of red and white paint left as well. The red and white paints, after their respective strips, have $164 - x$ and $188 - x$ ounces left, so they must have used $34$ and $58$ ounces to make the pink strip, meaning the pink strip is made up of $x = 92$ ounces. Therefore, Bill has $130 + 164 + 188 - 4(92) = \boxed{114}$ ounces of paint left.
[/hide]

[hide=Solution]After drawing the pink stripe, there's an equal amount of all three colors, so the amount of red and white paint would be both $130$. Let $a$ be the fraction of red in the pink stripe and let $x$ be the size. Then we have $ax=164-130$ and $(1-a)x=188-130$, so $x=92$. The answer is hence $3(130-92)=\boxed{114}$.[/hide]

[color=#38761D][b]BJUMPS AIME GRINDS #10[/b][/color]
Since we want the final amounts to be equal and $164 = 130+24$ and $188 = 130 +58$ we want $38$ more ounces of red paint and $58$ more ounces of white paint to be used than blue paint but since we have equal stripes that extra $24$ and $58$ will need to go there. Thus each color including pink will use $34+58=92$ ounces subtracting this from each and adding gives us $\boxed{114}$ $\blacksquare$


geometry
circumcircle
incenter
geometry proposed

[quote="stergiu"]Let  $ABC$ an isosceles triangle  ($AB = AC$) , its circumcenter  $O$  and its  incenter  $I$ . A line which passes through  $O$ and is perpendicular to line  $CI$ meets side   $AC$ at point  $D$ . Prove that   $DA = DI$ .

 [u]Babis[/u][/quote]
Assume that the line meet $CI$ at point $E$
and the mid-point of BC be $F$

$D,E,F,C$are concylic
Therefore
$\angle EOI=\angle ECF=\angle ICD$

Then we find that $O,I,D,C$are also concylic...

Let $M$ be the intersection point of $AI$ with $BC$.
$\angle AIC = 180^o - (90^o - \angle MCI) = 90^o + \angle MCI = \angle ODA$.
So $ODIC$ is cyclic and $\angle OCD = \angle OID$.
Since $OA = OC$, we have that $\angle OCD = \angle OAD = \angle OID$.
Hence $DA = DI$.


Consider two triangles $\triangle ABC$ with sides $a,b,c$, and equilateral $\triangle PQR$ with side $x$, such that $\triangle PQR > \triangle ABC$.

Three cevians of $\triangle ABC$ meet at $D$. $AD = u$, $CD = w$, $BD =  v$.

Likewise, Three cevians of $\triangle PQR$ meet at $M$. $PM = a$, $RM = c$, $QM = c$.

If for $\triangle ABC$, $\angle ADB = \angle BDC = \angle CDA = 120$, Prove that $x=v+u+w$

I don't quite understand the question. Can't you make $\triangle ABC$ arbitrarily large? Then $x$ can't ALWAYS equal $v+u+w$... is there a missing condition?

sorry, forgot some conditions. I should have added the diagrams, but i dont know how to draw one very well.

the key to this problem is to rotate $\triangle BDC$ (or the other two triangles) by $60^\circ$.

Would you mind telling us which year this is?  Maybe we could look at the wording of the original problem.

[quote="liyi"]USAMO 1974[/quote]
how do you know that??
Have you done this before, or you have a collection of the USAMO problems?

I have a book written by Luo Zengru.
This problem is listed in the book.


using Minscowki inequality

some people in Vn want to solve it by Torixelly point

[quote="cuong"]using Minscowki inequality[/quote]

How??
I tried minkowski, however unsuccessfully.........

[quote="cuong"]some people in Vn want to solve it by Torixelly point[/quote]

I solved it using this point  $(T)$ and the incenter $I$ of the that triangle

$AT+BT+CT \leq AI+BI+CI$ etc....

[quote="Peter VDD"][quote="shyong"]If $x,y,z\in \mathbb{R^+}$ , prove

$\sqrt{x^2+y^2+xy}+\sqrt{x^2+z^2+xz}+\sqrt{y^2+z^2+yz}\geq \sqrt{3}(x+y+z)$[/quote]

I think I have a very nice solution to this one:
\[\sqrt{x^2+y^2+xy}+\sqrt{x^2+z^2+xz}+\sqrt{y^2+z^2+yz}\\
\geq \sqrt{\frac{3}{4}}\left(\sqrt{x^2+y^2+2xy}+\sqrt{x^2+z^2+2xz}+\sqrt{y^2+z^2+2yz}\right)\\
= \sqrt{3}(x+y+z)\]

:)[/quote]

Very nice Peter!
Really clever solution, well done!! :lol:

I don't understand the first step of Peter's proof. Can someone explain it?


geometry
3D geometry
IMO Shortlist
number theory
number theory proposed

Long time ago I have seen the following at some older Romanian TST. I suppose this is quite standard:
"Prove that exist some integer (I got $128$) such that for all integers greater than it we are allowed to represent each of them as a sum of distinct perfect squares."
What numbers cannot be represented in this manner?
There is a similar statement similar for $n-th$ powers?
Any link is welcomed...

Oh, I am really surprised to hear that the most difficult problem from that year's TST's and also from the IMO shortlist is so standard. Congratulations, you must be really damn good in number theory to say that. ;)

The only Idea I have got is that iff all numbers in the interval $(k,4(k+4))$ can be written in that form, then all numbers bigger than $k$ can.

[quote="harazi"]Oh, I am really surprised to hear that the most difficult problem from that year's TST's and also from the IMO shortlist is so standard. Congratulations, you must be really - good in number theory to say that. ;)[/quote]

IMO shortlist 2000 - looks "classical", maybe not standard - I did not claim the proof being easy but it seems to me be "forged"... Obviously am I not really good at all...  ;)
A similar statement works for perfect cubes or greater powers? Does anyone know these?


Am nevoie de ajutor pentru urmatoarea problema. Am intalnit-o in gazeta si nu mi-a iesit.
Sper sa ma puteti ajuta.
Fie G un grup cu proprietatea ca exista doua numere naturale nenule a,b prime intre ele cu proprietatea ca
x^{a}y^{b}=y^{b}x^{a}
pentru orice elemente x,y din G.
Atunci G este abelian.

Asa cum apare aici problema nu poate fi corecta - vezi de exemplu grupul cuaternionilor, in care proprietatea are loc pentru $ a\equal{}2$ si $ b$ impar oarecare, grupul nefiind insa abelian.


calculus
integration
calculus computations

I need to find a way to simplify an equation. I have the original equation and the answer, I just don't know how to get there.

The original equation is [(1/5)((x^2)+9)^(5/2)]-[3((x^2)+9)^(3/2)]

The answer is [((x^2)-6)((x^2)+9)^(3/2)]/5

if you find a way, please show how you did it  :lol:

thanks

You're trying to turn

$\frac15\left(x^2+9\right)^{\frac52}-3\left(x^2+9\right)^{\frac32}$

into

$\frac{\left(x^2-6\right)\left(x^2+9\right)^{\frac32}}5.$

Note that you have a common factor of $\left(x^2+9\right)^{\frac32}.$ Factor that out:

$\left(x^2+9\right)^{\frac32}\left[\frac15\left(x^2+9\right)-3\right].$

Can you see how to proceed from here?

[Of course, this problem is in no sense calculus or analysis. It's pure manipulative algebra. I'll let it stay in this section on the thin premise that it probably arose in an integration problem.]

thanks. actually it arose from taking the antiderivative of a integrate :)

edit: yea, I got it. I must've been braindead today  :blush:


Add up all the inside numbers (e.g. for a 4x4 square add up 1-16) and divide by the number of rows/columns. So for a 4x4 square it would be (1+2+3+...+16)/4 which is 34.

A useful formula:
The sum of the elements in the rows/columns is $\frac{n(n^2+1)}{2}$
This follows directly from ubemaya's reasoning.
You add up $1+2+\cdots+n^2-1+n^2$
The sum of these numbers is $\frac{n^2(n^2+1)}{2}$
Then divide by $n$ for each row/column
$\frac{n^4+n^2}{2n}=\frac{n^3+n}{2}=\frac{n(n^2+1)}{2}$

You can apply it for 4 to get
$\frac{4(17)}{2}=2(17)=34$

[quote="mathwiz101"]does anyone know how to solve for the sum of any row or column of a magic square without filling it out?[/quote]

add the numbers together (for example 1+2+3...9). Then divide the total numbers 45/9=5). That number is in the middle. The number(s) preceding it and suceeding it are on the diagonal.


Although this may not be very important, because most people know what the answers are anyway, but there are 2 mistakes on the website. It says for number 4, that the Answer is A) 0.5, while it should say (A) 4, and also for number 21, where it says C) 3, instead of (C) 8. I'm really obsessive about these things, so I was checking my answers again. :blush: Anyways...

[quote="davidyko"]Must be the 10A. It makes not an ounce of sense if he's talking about the 12A.[/quote]

She's*.

And yeah, it's 10A.


calculus Harvard college vector Yale MIT parameterization

I'm applying to a transition school, where i would take high school in a year, then go to college, what is your opinion on this?  Do you think i should do this, or go to high school?

[quote="tokenadult"] My opinion on this is that high school is nearly always a waste of time for a high-ability student, unless the high school is very unusual. If you think you are academically ready for college the year after next, you should go for it, I think.[/quote]

There are various tradeoffs, and there is an argument for optimizing the time/cost/value of highschool-plus-college rather than either one taken in isolation.   For example, taking a large number of AP exams in high school can shorten the time in college, but at lower cost, or one might want to jump (or attempt to jump) certain hurdles that are bottlenecks in the college program, such as organic chemistry or multivariable calculus, at high school age.

well, you might not have enough experience in 1 year of high school... :huh:

[quote="tokenadult"][quote="justinian"]I'm applying to a transition school, where i would take high school in a year, then go to college, what is your opinion on this?[/quote]

Is this a long-established program, or a new one? My opinion on this is that high school is nearly always a waste of time for a high-ability student, unless the high school is very unusual. If you think you are academically ready for college the year after next, you should go for it, I think.[/quote]
I disagree. With my high school, and it's flexible period system, many more and interesting classes are availible instead of the old core classes. I would use this time to explore other things that interest me and get an introduction to higher-level education.

Grade school to college isn't good for social interaction.

[quote="tokenadult"][quote="justinian"]I'm applying to a transition school, where i would take high school in a year, then go to college, what is your opinion on this?[/quote]

Is this a long-established program, or a new one? My opinion on this is that high school is nearly always a waste of time for a high-ability student, unless the high school is very unusual. If you think you are academically ready for college the year after next, you should go for it, I think.[/quote]

Unfortunately, I don't think it's this simple.  Perhaps I'll write more extensively on this issue soon, but I can think of scads of reasons not to skip high school for college.  Some quick ones:

1) Socially it can be a disaster.  If you find an early-entry program in which everyone is younger than usual, this problem can be mitigated.  But I'd argue that it's way better to go to a top-tier school than do this.

2) If your high school doesn't have a high workload, the self-study you do during all that free time in high school will be way better than anything you'll do in college (on this, I speak from personal experience).  Learning how to find your own way is very important.

3) Much of your education comes from your peers.  If you jump into college 3 years early, odds are it will be at a college in which the other students are neither your social or intellectual peers.  That's a double whammy.  Instead, find ways in high school to be around other interesting and interested people.  Then go to college with social and intellectual peers.

I could go on and on on this topic - when I have more time, I will. . .   Quick acceleration may work for some, but in most cases, I would bet heavily against it.  Life is not a race.

What worked for me, and is probably a very good alternative, was to find a program that allowed me to take a few college courses cheaply, while still a high school student. Courses taken this way may also be transferable when you go off to college full time, and the early depth allows a great deal of flexibility later.

I don't think I would have handled the social aspects of early college well- I had enough problems at the normal time.

[quote="tokenadult"]Hi, Richard,

Your point number 3 is actually one of the best reasons to get out of high school, if the correct peer group can't be found there. Peer groups do matter. I agree that the ideal is to part of an age-similar (but MORE age diverse than age-graded school) group most of the time growing up, as long as that group provides intellectual challenge too, but there are a LOT of school districts that evidently are worse than the one you attended for finding that. I think my first reply admitted the possibility of the "unusual" suitable high school, but that is a rare possibility.[/quote]

I go to a public middle school with 500 kids. Even though I am the only one there who has a stong passion for math, there are a bunch of other interesting, smart kids that are fun to talk to for nothing other than philisophical discussion.

What's not important is being the best mathmatician ever, even though you think that's what you want to do, it's being an intellingent individual, well rounded and able to have an intelligent conversation about almost anything.

Hm, I don't completely agree. Why would I [i]want[/i] to have an intelligent conversation in something like History, in which I see no point in studying as the subject itself is uninteresting to me? I would much rather spend my time pursuing math, something that I enjoy. Although society, or maybe you are just generalising from yourself, might say that you need to be a well-rounded individual, you should only listen to that to a certain extent. If, in your attempt to become a 'well-rounded individual' you find yourself not enjoying the subject itself or feeling not challenged or motivated to go on, and you see an opportunity that will help you go on, I say take it.

[quote="tokenadult"]Hi, Richard,

Your point number 3 is actually one of the best reasons to get out of high school, if the correct peer group can't be found there. Peer groups do matter. I agree that the ideal is to be part of an age-similar (but MORE age diverse than age-graded school) group most of the time growing up, as long as that group provides intellectual challenge too, but there are a LOT of school districts that evidently are worse than the one you attended for finding that. I think my first reply admitted the possibility of the "unusual" suitable high school, but that is a rare possibility.[/quote]

I don't think going to college necessarily solves this problem, though.  It may, in fact, make it worse by placing the student with people at a very different place in life socially.  A better solution would be to find extracurricular programs to find interesting/challenging peers.

I'll note that my high school had a <65% graduation rate, with no one going to a top tier school in the 25 years before I went there.  Yet I had a few very interesting and intellectually challenging friends.  Sure, they weren't winning math contests and didn't go to Harvard, but they were perfectly good intellectual peers to me.  So, basically, I agree with Penguin here.  Of course, for those stuck with a small school, this doesn't help.

But then, there's always AoPS.

What I'd suggest instead of rapid acceleration is to pursue a hybrid education involving both part of the day in school, and part of the day in independent study (i.e. essentially self-homeschooling).  This, I think, would be far more educational, and do much more to prepare the student for future life challenges, then just following the beaten path a little more quickly.  A great many things in life worth doing don't have a beaten path.  Rapid acceleration doesn't help with those at all (and from what I've seen from many rapidly accelerated students, it hinders them more than helps in this regard - not working on their own prevents them from developing their own insight and creativity, I think).

[quote="WindSlicer"]Hm, I don't completely agree. Why would I [i]want[/i] to have an intelligent conversation in something like History, in which I see no point in studying as the subject itself is uninteresting to me? [/quote]

You're not studying the right history. I had a great discussion on communism today.

Also, if you take this attitude towards many other subjects, how should other people react to you liking math? Should they go "Who cares about math, I'm a civil war buff..."
or should they converse with you on their ideas, opinions, and questions about it?

Having friends is essiantial for social well being. I really feel sad for the kids who homeschool and do nothing but math, with cusory attention to the other subjects.
Yes, I like math, but part of being an intelligent individual is being able to pick up things quickly.

Diverse Talents = Good.

No, I've had some great history teachers, I could just care less about what happened in the past. And that's just the way I am.  There are people who don't have their fortes in math and who just generally dislike it, and that's going to happen. I think wasting time pursuing something that you will never use again as compared to using that time productively towards something you like and enjoy and are good at is a bad idea. And don't misinterpret what I'm saying.. I have an A in my IB History HL class and have always gotten A's in History, I just don't like the subject and spending time on that instead of math, when not needed is pretty pointless.

I really get offended by people who say "I feel sad for ____" when talking about someone's lifestyle. Those students have most likely chosen the things that make [i]them[/i] the happiest. So often, people want to judge others based on their own views of enjoyment, but that's not the way everyone operates. I say to each his own and if you are happiest doing this (and it doesn't violate any laws) then go for it - pursue the happiness. Now, myself, I am generally diverse with a forte in math - I run cross country/play tennis, I play the alto saxophone in my concert band and jazz band, I love to help out in church and do drama (and perhaps sing occasionally, but I'm not that good so :P). I also like many subjects including bio,chem,math,psychology,philosophy, spanish and physics (History is pretty much the only subject that I specifically don't care for, and English is tolerable), but I also understand that this is not for everyone so I think that if someone chooses to focus solely on math, it is fine and good for them.

Now, aside from hijacking this thread, I'll give my point of view on the going to college early. I generally believe the same as mr. rusczyk that it is better for a student ready for college to spend that time further developing math problem solving skills that one might not necessarily be exposed to when getting into the higher level math, and just either assumed you know it or just learning more complex formulae and ideas. I also think that AOPS has a great program that would allow one to do this. But, of course, if one ends up being at the Olympiad level problem solving and is doing well overall and is content with skipping their senior year, I have no problem with that.

How about this potential sad scenario: A bright kid enters a middle-school-to-college program, does not enjoy it, and is now stuck in a virtual nowhere land. The kid probably feels too embarrassed (or too beneath him/her) to go back to high school, and no college would accept such a transfer.

I personally don't really like those quick - acceleration to college.  Your teen years are your only in high school, there are a lot of skills (social skills, maturity develpment, becoming more responsible) that can be developed with high school, best around a similar age group.  Sure, there are a lot of jerks in a normal public high school that will make fun of you, but its important to learn how to stand up for yourself and react camly with those people, because everyone will deal with people like that in real life.

There was one kid a few years ago in my middle school who skipped 6th grade.  He acted very immature and was basically socially isolated.  So I feel skipping is usually not a good idea.  If the intellectaul challenge is not great enough in High school, like Mr. Rusczyk (sp.?) said, there's always AOPS and self-study, maybe take classes ata community college or something.

[quote="thisisme"]"Most things doable at college can be done at high school, but with more flexibility and at lower cost.  "

This is certainly NOT true.  There is no way to access in high school most curricular offerings found in college, whether they be in math or other areas.[/quote]

Books and internet reach the post-baccalaureate level and beyond in almost any subject.
If you specifically need a lab, PhD-level advising, or rare manuscripts, the venue is grad school and these resources are not necessarily found at most colleges, either.

Young accelerated kids I know turn out to be somewhat arrogant...

[quote="fleeting_guest"]Parents taking 11-year-olds to class at Yale is an academically worthless ego trip.  Too bad for the kid.[/quote]
I could not agree more. Parents should really have better sense. Tokenadult has emphasized many times on READING in this forum. Those students who are bored in their high schools can certainly put their extra time doing more classical literature reading.

Speaking as a college freshman, I'm much more prepared as a person now then I would have been 3 years ago.  One fewer year probably would have been good.  But three would not.

[quote="fleeting_guest"]The social opportunities of college are easier to exploit at the later ages, and if admission at later age means access to more selective universities (sending Junior to MIT at 18 rather than Local State College at 15) that will further enhance the social differential. e, but to conquer, university in their teenage years, it may be a different story.[/quote]

This would be my number 1 reason for waiting to enter university (at least until one is able to be accepted at the top schools).  The experiences I had in university were fantastic mainly because of the other very high caliber students there.  It would have been a shame to go to a lesser college earlier, even though high school was not a particularly great place for me.  Those undergrad years are a once-in-a-lifetime opportunity, and they would be squandered going to a "lesser" institution.  A student who attends a mediocre college at 15 might be with "peers" in terms of people ready to study the same level of classes, but would likely be with fewer true intellectual peers (most of whom attend the top colleges).

Maybe people get some of that "intellectual peer" interaction right here, these days, but I still say that it is a fantastic opportunity to go to college with other young adults "like" yourself, and you should try to make sure that that happens.

We did have some younger students in my "top notch" program.  Probably half were a year younger than "normal", and that was fine.   Of those 3 years younger than "normal" I can think of only one who succeeded, and several who dropped out.

Good luck to all students/parents grappling with this issue!

Actually it is very difficult to obtain access to good resources in high school. The internet is terrific for many things, but learning advanced mathematics is absolutely not one of them. There are lots of lecture notes online, but can you really learn mathematics properly by reading sketchy lecture notes? I usually cannot. And as you get even more advanced, it even becomes difficult to search for math notes. (For example, if I want to learn about the field $\mathbb{C}_p$, what am I going to type on Google to get the results I want?) But at a university it's easy. Just spend an hour at the library, and you can find books on what you're looking for (unless it's something extremely obscure). Of course, I can get them from Amazon as well, but that is expensive if you like books as much as I do.

I think I would have been better off going to college instead of high school, especially here. High school killed something in me that I haven't been able to revive even after more than two years.

[quote="ComplexZeta"] There are lots of lecture notes online, but can you really learn mathematics properly by reading sketchy lecture notes? I usually cannot.[/quote]

No, you can't -- you'd need to be willing to wait.  Or...

[quote]But at a university it's easy. Just spend an hour at the library, and you can find books on what you're looking for (unless it's something extremely obscure).[/quote]

If you live anywhere near a good university, you may be able to get access to their library while in high school.

[quote="ComplexZeta"]A library card at Stanford costs several hundred dollars a quarter for non-students. So one would have to read books there instead. Who wants to do that?[/quote]

A library is as good a place as any to read/study books. :)

I wonder if your local public library can get academic books students want to borrow on inter-library loan?  This will depend on the quality of the ILL system, and it is a pain, and it does take time....

It's a trade-off, and if someone can't stand to wait to learn more advanced material in a specific area, then college may be the best bet for them.  But if going early means you are going to get only, say, 80% of the quality of learning and peer experiences as going later, is it worth that sacrifice?

ComplexZeta: High school took something out of you that you couldn't revive? I disagree... So I didn't take college courses to challenge me further during my final 2 years of high school... instead I had a ton of fun, and I made friends that will last me for the rest of my life most probably, friends that I can rely on. And then I got to college, and I made MORE friends. If I got to college at the age of 16, I'm fairly certain that I couldn't have done as well socially. So big deal, I would've been progressing further mathematically, while I spent my Fridays and Saturdays in my room doing problem sets instead of going out and partying because I was immature and everyone hated me. Do you know why you go to Ivy League Schools? It's because of the enviroment. There was an article written I believe, in the New Yorker, that explains what makes Ivy League Schools so great... Actually, its funny, I don't really agree with the author 100%, but I think that the environment is definetly a huge factor, and so you want to make the best out of your college years. Life isn't a race. It's not about who took analysis first. It won't fly too well if you say "oh I took analysis at 12 and I graduated from college at 16." That doesn't make you a better person then say someone who graduated at 22. Why? Chances are that the 22 year old will have better people's skills than the 16 year old. He will also be able to explain himself better. I don't see 16 year olds working at i-banking firms in NYC. So unless you plan on pursuing some sort of academic career, just think that you might actually want to have a future, with friends.

What right do you have to disagree with my assessment of my own past given that you don't even know me? Okay I'll explain what I meant.

When I was young (i.e. before high school), I went to excellent private schools. Most people I know say they had a lousy time in middle school; I didn't. My classmates were all very intelligent people, and there weren't any idiots who bullied me for being intelligent. (If anything, it was the moderately less intelligent people who were picked on, in general.) I learned a lot, and I was very happy to be learning a lot, to the point that I was disappointed when we had days off because I wouldn't learn as much.

Then I went to a public high school since my parents didn't want to pay the astronomical sums that private high schools charge. It was a huge shock for me: there were plenty of ``academically challenged'' students in my classes, and the teachers frequently didn't know what they were talking about. I was insulted to be asked to do many of the assignments, so frequently I wouldn't do them, and I would return home depressed by the stupidity of my courses. Okay, so I had some friends, and there was even one student at my school who was better at math than I was. His presence was the main reason I got anything done at all in high school, I think -- I worked hard at math until I was as good as he was. But in everything other than math, I learned little. My study habits atrophied, and I would come home after school and play chess on the internet for hours and spend little if any time on my assignments.

Then I went to college. Not an Ivy League school (since my grades obviously weren't good enough for that), but a lousy state school that happens have a special program for talented students -- students motivated to work hard and learn a lot. It's wonderful being around students who actually want to learn again. But after four years of prison (er, I mean high school), it's hard to revive my former study habits, and while I am doing very well in my courses (which are much more advanced than math majors here usually do), I still feel that something is missing.

I know life isn't a race. But I'm an impatient person. There are lots of things I want to learn, and I don't want to wait to learn them. I'm not competing with anyone else when it comes to taking a course first. There are just a lot of things I want to learn. Not being able to learn things makes me very depressed.

And with respect to social life, it's not as though I have much of a social life now. I take twice as many classes as an average student because there are always so many classes that look fascinating. And I don't go to parties on Friday nights -- I don't like going to parties. And since I usually take graduate classes, my classmates tend to be at least a few years older than I am. So socially I'm in about the same situation as I would have been had I skipped high school and gone directly to college. But then I wouldn't have had to go through high school.

Maybe it wouldn't have worked for you, but I'm pretty sure it would have for me.

[quote="Hashbrownie"]So I didn't take college courses to challenge me further during my final 2 years of high school... instead I had a ton of fun, and I made friends that will last me for the rest of my life most probably, friends that I can rely on. And then I got to college, and I made MORE friends. If I got to college at the age of 16, I'm fairly certain that I couldn't have done as well socially.[/quote]

What works for one person (socially, academically, or in any area) does not work for everyone.  Some people end up doing better socially among older people, and for them, going to college early might work better.  Similarly for those whose high school classmates simply don't have much in common with them.

I guess I fell somewhere between you and Zeta.  I didn't have a lot of friends in High School, and it certainly wasn't because I was working too hard.  I just didn't click/fit with many people there.  I did have a great time in college, and I think that I, personally, would have done worse if I'd been much younger, so I was glad to be there when I was.  I was also glad that I was in as competitve a program as I was, as that contributed to my finding so many compatible friends (and to better learning opportunities), and since I don't think I would have a) qualified for that program, or b) chosen to live that far from home if I'd been much younger, I am glad, for myself, that I waited.

I do think it's a shame when people see math/schooling as a race, and end up taking inferior quality courses as soon as they possibly can, rather than waiting until they can take top-quality courses, but I do understand why for some people high school can be a toxic environment best avoided.

Good luck to anyone making this decision.  As you can see, there's not a clearcut right answer, and the best solution for one person is not necessarily the best for another.

[quote="tokenadult"]Rejoining the fray here, I will simply point out that if someone finds their current high school environment dissatisfactory, and there is a chance to get into a more satisfactory environment, that is a good idea, even if the new environment is college. Which is the gist of what I said way back when at the time this thread first opened. These decisions are exquisitely sensitive to local circumstances and to family financial resources.[/quote]

Good point.  There's no one-size-fits-all.

The first thing to consider is what your high school option is like, I think this makes a big difference, if you have the option to go to a high school with smart students and good teachers that will probably be a better option than colllege but if the high school is going to crush your soul college is probably better.  (A quick test is to look at the aps offered, a school that doesn't offer ap calc or at least one ap science is probably not a good choice while a school where at least half the senior class takes ap calc is probably very good, this is just something i've noticed from talking to people)

In general i do not recommend completely skipping high school because there are a lot of options and activities only available to high school students.  However skipping senior (or possibly junior and senior although i would not recommend missing three yeas of high school) year may be a very good idea (I know many people who have done this).  In general being one year younger than the rest of your class won't be a big deal, several kids wind up a year younger anyway.

In my case I have decided not to skip my senior year for several reason even though socially high this year and next year are probably worse for me socially than going to college would be (most of my friends are freshman in college or seniors).
1.  Math contests-  most competitions are only available to high school students and senior year will be a chance for me to hopefully do well on some of the various contest I take (for example qualifying for usamo and trying to get an 8 at arml)
2.  Light schedule-  this year(junior) I'm taking a fairly heavy workload but as a result next year my schedule will be very relaxed and full of classes that I want.  This will give me time to do #1.
3.  Choice of school-  hopefully next year I get into my top choice schools while if I applied this year I probably couldn't
4.  teaching younger kids- i've learned from other students in my school and i feel obligated to be there to teach younger kids in the school and help our math team
I'm sure there are more reasons but im too tired to keep typing so....

last remarks if you want to skip grades skip middle school


I assume that $ \log_2 2^0$ is also a factor, hence the product sum is zero.

No, archimedes you aren't right, it's $ 2^1024\equal{}\frac{1}{2^{\minus{}1024}}$ not yours.
$ 2^{\frac{1}{1024}}\equal{}\sqrt[1024]{2}$

[quote="Kyoshiro"][quote="vinskman"][quote="Kyoshiro"]$ log_2\frac {1}{2^{512}}\cdot log_2\frac {1}{2^{256}}\cdot log_2\frac {1}{2^{128}}\cdot...\cdot log_22^{1024} \equal{} ?$[/quote]
I don't agree with Jarle10's work .
[hide="This may be true"]$ log_2\frac {1}{2^{512}}\cdot log_2\frac {1}{2^{256}}\cdot log_2\frac {1}{2^{128}}\cdot...\cdot log_22^{1024}$

$ \equal{} ( \minus{} 512)( \minus{} 256)( \minus{} 128).....( \minus{} 4)( \minus{} 2)( \minus{} 1)(1)(2)(4)......(256)(512)(1024)$

$ \equal{} ( \minus{} 1)^{10}.2^{9 \plus{} 8 \plus{} \cdots \plus{} 1} . 2^{1 \plus{} 2 \plus{} 3 \plus{} \cdots \plus{} 10} \equal{} 2^{45 \plus{} 55} \equal{} 2^{100}$[/hide][/quote]
No,you're wrong! his work is right!  :wink:
Note that $ \log_2 1 \equal{} 0$[/quote]

Yes,I agree my work is not correct !
But logically, we cannot get the last term $ 2^{1024}$and the interior term $ 0$

the orginal question$ \equal{} ( \minus{} 512)( \minus{} 256)( \minus{} 128).....( \minus{} 4)( \minus{} 2)( \minus{} 1)( \minus{} 1/2)( \minus{} 1/4)( \minus{} 1/16)...$
:D  :blush:


LaTeX
function
calculus
integration
calculus computations

[quote="colin_java"]The question is (D'Alemberts Solution):

Solve u_xx = u_tt
With intial conditions
u(x,0) = |x|^3 + x^6
u_t(x,0) = e^(-x^2)

So I just plugged it into the formula:

u(x,t)=(1/2)[f(x+ct)+f(x-ct)]+(1/2c)INTEGRAL[x-ct...x+ct]{g(s)ds}

using c=1, to get:

u(x,t)=(1/2)[|x+t|^3+|x-t|^3]+x^6+15x^4t^2+15x^2t^4+t^6+(1/2)INTEGRAL[x-t...x+t]{e^(-s^2)ds}

Is that all I should do?

Thanks for any help.  :)[/quote]

 :huuh:  :huuh:  :huuh:

Could you write your problem by Latex ?

Hi,

Well I don't remember the language, and I think it would take a while, and might not be worth it, just for this problem.

All I have done is plugged the functions f and g into the formula given by D'alembert u(x,t)=...

I just wanted to know if thats all thats required, I guess a formula is to be used instead of reinventing the wheel, and doing it from scratch.

You can see I expanded out the (x+t)^6 and (x-t)^6, and collected like powers, but with the |x+t|^3 + |x-t|^3, I don't see any way forward with that.

And with the last integral, I don't see any way to simplify that either.

So is my solution ok?

Seems fine to me (at least, you made all the right steps but I was too lazy to check the binomial coefficients for degree 6).

Thanks, that helps

Just out of interest, what would one do with the last integral generally?

Isn't it impossible to integrate with normal methods?

I know the integral over R of e^(-x^2) is sqrt(pi), but its not quite the same as what I've got.

Would it just be done numerically, giving an approximation to whatever desired accuracy is required?

Thanks

Yes, the antiderivative is not elementary, so numerical methods are the only way to compute it "exactly". You may figure out what happens asymptotically though.