logarithms
LaTeX
inequalities
real analysis
real analysis solved

Let $x_0=1$ and let $x_{n+1}= \displaystyle 1+x_n+\frac 1{x_n}$. Prove that we can find a real $a$ such that
\[ \frac n{\ln n}( a+n+\ln n-x_n)=1. \] where $\ln n$ is the natural logarithm.

[Admin Edit: Made Latex compliant]

Sorry, the task is that the sequence tends to 1, not equal to 1.

No one dares to prove this one? It's not that hard.

Sorry but it seemes too easy for me.
It's sufficient to prove x[n]-n-ln(n)-a-1/n ln(n) tends to 0 we can ignore 1/n ln n and the rest follows by induction. Pleade notify me if I have made a mistake

Oh noy, oh boy, oh boy! I almost made a heart-attack while reading your post. But I have a little question: why does it follow from $ x_n-n-lnn-a-\frac{1}{n\cdot lnn} $ tends to 0 that the sequence $\frac{n}{lnn}\cdot( a+n+lnn-x_n) $ tends to 1. I really do not understand.

I think iura has already realized his mistake.

I can prove harazi's statement step-by-step. I hope it works.

(1) First of all, $x_{n}-x_{n-1}=1+1/x_{n-1}$. So $x_n=n+1+1/x_0+1/x_1+...+1/x_{n-1}$. It follows that $x_n\geq n+2$. Now we able to prove $x_n-n-\ln n \to a$ as $n\to \infty$. Actually, $(x_{n+1}-(n+1)-\ln(n+1))-(x_n-n-\ln n)$ = $1/x_n-\ln\frac{n+1}{n}\leq 1/(n+2)-\ln\frac{n+1}{n}<0$. It implies $x_n-n-\ln n$ is decreasing sequence. On the other hand, $x_n=n+1+1/x_0+...+1/x_{n-1}\leq n+1+1/1+1/2+...+1/n \leq n+2+\ln n$, so $x_n-n-\ln n \geq 1+1/x_0+...+1/x_{n-1}-(1+1/2+...+1/n)$ = $(1/x_1-1/2)+...+(1/x_{n-1}-1/n) \geq c\cdot\sum \ln n/n^2\geq C$ for some c<0. Finally, $x_n-n-\ln n$ is decreasing and bounded. More accurate estimation gives us $x_n-n-\ln n>0$ (exercise!).

(2) Second step. We will prove $x_n-n-\ln n - \ln n/n$ is decreasing sequence. Actually, we need to prove $1/x_n-\ln\frac{n+1}{n}-\ln(n+1)/(n+1)+\ln n/n<0$. Using $x_n-n-\ln n>0$ we obtain $1/(n+\ln n)-\ln\frac{n+1}{n}-\ln(n+1)/(n+1)+\ln n/n<0$. The last inequality can be proved with asymptotic technique. In the other words, we can show LHS is $-c\cdot n^{-2}+o(n^{-2})<0$ (where c>0) for sufficiently large $n$.

(3) Third step. We will prove $x_n-n-\ln n - \ln n/n-3/n$ is increasing sequence. in that case we need to prove $1/x_n-\ln\frac{n+1}{n}-\ln(n+1)/(n+1)+\ln n/n+3/n(n+1)<0$. Now we use $x_n<n+2+\ln n$ and obtaing LHS is $n^{-2}+o(n^{-2})>0$ for sufficiently large $n$.

(4) From (2) and (3) follows that $3/n\geq x_n-(n+\ln n + \ln n/n + a)\geq 0$. It implies harazi's statement.


For all $ x\geq 1$ and $ n\in\mathbb{N}$ , the inequality $ x^n > n(x \minus{} 1)$ , we have consequentially that for $ 0 < x < 1$ and $ n\in\mathbb{N}$, the inequality:

$ x^{n \minus{} 1} < [n(1 \minus{} x)]^{ \minus{} 1}$

Prove this fact.

Is induction the right way to go?

it would help if you called the variable in your second proposition something else, like $ y$  :wink:
then you could say $ y\equal{}\frac{1}{x}$ for some $ x\geq 1$
and then its just simple algebraic manipulation ;)

So what exactly do we call $ y$, $ n(1\minus{}x)$? Sorry, I didn't follow.


inequalities
trigonometry
geometry
circumcircle
geometry proposed

[quote="Virgil Nicula"]In an [u]acute[/u] triangle $ABC$ exists the relation $\boxed{\ \sum a\cos^{2}A\ge \frac{S}{R}\ }\ ,$ where $S$- the area and $R$- the circumradius for $ABC\ .$[/quote]

As we have that $S=\frac{1}{2}R^{2}(\sin{2\alpha}+\sin{2\beta}+\sin{2\gamma})$,then we need to prove that
\[\ \sum a\cos^{2}\alpha\ge\frac{1}{2}R(\sin{2\alpha}+\sin{2\beta}+\sin{2\gamma})\]
<=>
\[\sum{\sin{\alpha}\cos^{2}{\alpha}}\geq \frac{1}{4}(\sin{2\alpha}+\sin{2\beta}+\sin{2\gamma})\]
we know that $ABC$ is acute,so we can let that $\alpha=\frac{\pi-x}{2},\beta=\frac{\pi-y}{2}$ and $\gamma=\frac{\pi-z}{2}$,then we need to prove the following inequality
\[\sum{\cos{\frac{x}{2}}\sin^{2}{\frac{x}{2}}}\geq \frac{1}{4}\sum{sin{x}}\]
<=>
\[\sin{x}\sin{\frac{x}{2}}+\sin{y}\sin{\frac{y}{2}}+\sin{z}\sin{\frac{z}{2}}\geq \frac{1}{2}(\sin{x}+\sin{y}+\sin{z})\]
this inequality is equivalent to the following one
\[a\sin{\frac{A}{2}}+b\sin{\frac{B}{2}}+c\sin{\frac{C}{2}}\geq p\]
For the last one see your nice proof here[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=462855#p462855[/url]
 :D

O.K., Tiks ! You "pledged" quickly. So that I composed it, but conversely, from the that "old" inequality to the posted inequality.


As of now there are 299800 posts on this site, almost to 300,000.  :D

Just FYI.

[quote="hello"]Where do you get the number of posts? Btw, it is a better measure of the number of use of the site then members since a lot of people never posted.[/quote]

scroll down to the bottom of the index page.

At first I thought u were reffering to your postcount, 236factorial. :lol:

  Then again, it wouldnt be that hard to believe, considering your a post-aholic. :rotfl:

assuming he maintains his average-day-post-count :lol:

Of course, not counting the deleted ones :D

[quote="MysticTerminator"]ayup, 300000

we should all wait around for 500000 now[/quote]Hehe, whoose post was? Mine or yours? The 300000th :P (I need to know so I know if there's one to delete to claim the prize :D :D)

[quote="MysticTerminator"]I think it was 300000 when I posted so that would make mine 300001 (which is an EVEN COOLER number, of course)[/quote]Aren't you trying to trick your way out of me deleting your post (I have a perfectly valid reason for that too :P spamming, double posting - you name it :D) and make me the holder of the record?  :D

:?: are you denying that 300001 is a cool number? :|:huh::? I mean, sheesh, it's divisible by 13 - that alone makes it enormously cooler than 300000

[quote="noneoftheabove"]administrators have all the power  :D[/quote]

nope, moderators have some too

:rotfl:
this topic is full of spam
but its still fun to read
 :rotfl:

Now it's at 300339  :P

That's about 319.5 per day (counting the inactive times too), there's a lot online  :P

I think AoPS is the biggest math community.  :)

[quote="236factorial"]I think AoPS is the biggest math community.  :)[/quote]You're right about that.

Topic locked. See you in 50,000 posts :)


Mn in MnO4 has an oxidation number of +4

[quote="mr. math"]I was learning about oxidation numbers and read that oxygen almost always comes with the oxidation number of -2. I also read that there is an exception to this rule when considering peroxide. When exactly does this exception come into play? Is it only considered for hydrogen peroxide?

For example, take MnO2

Would you consider O2 as a whole to have an oxidation number of -2, leaving O with an oxidation number of -1?

Or would O still have an oxidation number of -2?[/quote]
for MnO2, the oxidation number of O is -2 and that of Mn is +4.
one common exceotion of oxygen is hydrogen peroxide H2O2, where the oxidation number of H is +1 and that of O is -1

Would there also be an exception for BaO2?

This substance would be Barium peroxide. Because peroxide has an oxidation state of -2, would the oxidation state of O in this substance be -1?

Oxygen has a greater electronegativity than barium. There is a difference of 2.6.

This is what I have reasoned:

Because the electronegativity difference is so high, we can deduce that the bond is ionic. Also since we know that Barium is an alkaline earth metal and always has an oxidation number of +2. Therefore, peroxide as a whole would have an oxidation number of -2. Oxygen by itself will have an oxidation number of [b]-1.[/b]

Also, after reading information from extra books and the internet, I have come to the conclusion that oxygen will always have oxidation state of -2 except for when we see the peroxide ion. An example of this is BaO2.

But with, for example, CO2, even though we have O2, its not the peroxide [b]ion[/b] and therfore oxygen would still have an oxidation state of -2 while carbon would have a state of +4.

Is all of this correct?

Actually there are two exceptions: in peroxides, where it is -1, and when combined with fluorine ($ \ce{F2O}$) where it is +2.

The compound $ \ce{KO3}$ also exists.

[quote="jdlkem"]Let's make things even more complicated!

KO2 exists (potassium superoxide). What is the oxidation number of potassium, and what is the oxidation number of the oxygens in superoxide?[/quote]
for KO2, the oxidation number of potassium is +1 and that of O is -1/2.
is that right?


game
algebra
game strategy
combinatorics
IMO Shortlist

Two students $ A$ and $ B$ are playing the following game: Each of them writes down on a sheet of paper a positive integer and gives the sheet to the referee. The referee writes down on a blackboard two integers, one of which is the sum of the integers written by the players. After that, the referee asks student $ A:$ “Can you tell the integer written by the other student?” If A answers “no,” the referee puts the same question to student $ B.$ If $ B$ answers “no,” the referee puts the question back to $ A,$ and so on. Assume that both students are intelligent and truthful. Prove that after a finite number of questions, one of the students will answer “yes.”

I'm confused...  :| If no one knows the other number, how can this kind of question show the answer?

Example: Suppose $ A$ picked $ 4$, $ B$ picked $ 2$, and the referee wrote down $ 6$ and $ 8$. $ A$ says he doesn't know, because $ B$ could have picked $ 2$ or $ 4$. $ B$ says he knows, because he knows $ A$ had to pick $ 2$ or $ 6$, and $ A$ would have known if he picked $ 6$.

Let the two numbers on blackboard be $ X<Y$. Also use $ A$ and $ B$ to represent the number from students A and B, respectively.
(1) Suppose no "yes" in round 1. A knows $ B<X$, otherwise B would have said "yes" and solve $ A\equal{}Y\minus{}B$. Similarly B knows $ A<X$.
(2) Suppose no "yes" in round 2. If B saw $ Y\minus{}B>\equal{}X$, he would have known $ A$ could not be $ Y\minus{}B$ since he knew $ A<X$ and then he would have said "yes' by solving $ A\equal{}X\minus{}B$. Hence, A knows $ Y\minus{}B<X$, or $ B>Y\minus{}X$. Similarly B knows $ A>Y\minus{}X$.
(3) Suppose no "yes" in round 3. A knows $ B<2X\minus{}Y$. B knows $ A<2X\minus{}Y$.
...
Each round without "yes" will tighten A's knowledge on B, also B's knowledge on A. Here knowledge means both upper bound and lower bound. Let us call the series of upper bounds $ x_n$ and lower bounds $ y_n$. We see that $ x_{n\plus{}1}\equal{}X\minus{}y_n$ and $ y_{n\plus{}1}\equal{}Y\minus{}x_n$. Obviously $ x_n$ are strictly decreasing and $ y_n$ are strictly increasing. So in a finite number of rounds, A or B have to answer yes. The stopping rule is one of the following four:
(i) $ X\minus{}A<x_n\leq Y\minus{}A$, A say yes and solve $ B\equal{}X\minus{}A$.
(ii) $ X\minus{}A\leq y_n<Y\minus{}A$, A say yes and solve $ B\equal{}Y\minus{}A$.
(iii) $ X\minus{}B<x_n\leq Y\minus{}B$, B say yes and solve $ A\equal{}X\minus{}B$.
(ii) $ X\minus{}B\leq y_n<Y\minus{}B$, B say yes and solve $ A\equal{}Y\minus{}B$.

Suppose $A$'s number is $a$, $B$'s number is $b$, and the two numbers on the board are $r_1 < r_2$. Suppose $A$ and $B$ both keep saying $``$no.$"$

Each person always has two guesses as to what the other person's number is. $A$ knows that $b$ must either be $r_1-a$ or $r_2-a$. Likewise, $B$ knows that $a$ must either be $r_1-b$ or $r_2-b$.

Meanwhile, at every stage of the game, there are certain restrictions on what the value of $a$ and $b$ could be. For example, the only thing known initially is that $a,b > 0$. A person will answer $``$no$"$ if and only if both of his guesses satisfy the known restrictions at the given time.

Suppose $\alpha_n$ denotes all the restrictions on $a$ after $A$ has said $``$no$"$ for the $n$th time, and likewise for $\beta_n$. Define $\alpha_0 = [a>0]$ and $\beta_0 = [b>0]$.

When $A$ says $``$no$"$ for the $n$th time, he is saying $``$ I don't know which of my guesses, $r_1-a$ or $r_2-a$, is correct, since both of them satisfy $\beta_{n-1}$", and in saying this, he is conveying useful information about the value of $a$, which contributes to the restriction $\alpha_n$. Try this out for the first couple of times, and you'll see the following pattern for $n>0$:

$\alpha_n = [(n-1)(r_2-r_1) < a < (n)(r_1-r_2) + r_2]$
$\beta_n = [(n)(r_2-r_1) < b < (n)(r_1-r_2) + r_2]$

It is now just an exercise to prove this using induction. Though the details are not terribly important, I give them [hide = "here."]

First consider $n=1$.

When $A$ says $``$no$"$ for the first time, he is saying $``$both of my guesses satisfy $\beta_0 = [b>0]"$, or in other words, $r_1-a>0 \text{ and } r_2-a > 0$. Hence the restrictions on $a$ are now $[0<a<\text{min}(r_1,r_2)]$. Thus, it is verified that $\alpha_1 = [0<a<r_1]$ as required.

When $B$ says $``$no$"$ for the first time, he is saying $``$both of my guesses satisfy $\alpha_1"$, which conveys that $0<r_1-b<r_1$ and $0<r_2-b<r_1$. As before, this implies $0<b<r_1$, but the second inequality adds the restriction that $b>r_2-r_1$. Hence it is verified that $\beta_1 = [r_2-r_1 < b < r_1]$ as required.

Now for the induction.

When $A$ says $``$no$"$ for the $nth$ time, he is saying $``$both of my guesses satisfy $\beta_{n-1}"$. In other words, $[(n-1)(r_2-r_1) < r_1-a < (n-1)(r_1-r_2) + r_2]$ and $[(n-1)(r_2-r_1) < r_2-a < (n-1)(r_1-r_2) + r_2]$. The first halves of these inequalities taken together mean $[a<(n-1)(r_1-r_2) + \text{min}(r_1,r_2)]$ which gives $[a<(n)(r_1-r_2) + r_2]$. Meanwhile, the second halves taken together give $[a>(n-1)(r_2-r_1)]$. This gives the required value for $\alpha_{n}$.

When $B$ says $``$no$"$ for the $nth$ time, he is saying $``$both of my guesses satisfy $\alpha_n"$. In other words, $[(n-1)(r_2-r_1) < r_1-b < (n)(r_1-r_2) + r_2]$ and $[(n-1)(r_2-r_1) < r_2-b < (n)(r_1-r_2) + r_2]$. The first halves of these inequalities taken together mean $[b<(n-1)(r_1-r_2) + \text{min}(r_1,r_2)]$ which gives $[b<(n)(r_1-r_2) + r_2]$. Meanwhile, the second halves taken together give $[b>(n)(r_2-r_1)]$. This gives the required value for $\beta_n$.

[/hide]

Since $r_1<r_2$, it is clear that the bounds are shrinking in tighter and tighter, which cannot go on indefinitely. Hence, at some point, one must answer $``$yes$"$.


Solve for real numbers $x$ and $y$,
\begin{eqnarray*} \\ xy^2 &=& 15x^2 + 17xy +15y^2 ; \\ \\ x^2y &=& 20x^2 + 3y^2. \end{eqnarray*}

If x or y is zero, then the other is.  So suppose they aren't.

Then let m = y/x; divide first eqn by y^2, second by x^2, put everything in terms of m.

Then put the second equation over the first.  You get $m = \frac{20 + 3m^2}{15m^{-2} + 17m^{-1} + 15}$.

It implies $m^3 - 5m^2 + m - 5 = 0$ after expanding.  Factoring, $(m-5)(m^2 + 1) = 0$, so that m = 5.

Finishing, $y = 20 + 3m^2$ from the 2nd eqn where $y = 95$, and thus $x = 19$.

very simple problem indeed
multiplying first eq. by $ x$ and second by $ y$
$ x^{2}y^{2}\equal{} 15x^{3}\plus{}17x^{2}y\plus{}15xy^{2}$
$ x^{2}y^2 \equal{} 20x^{2}y\plus{}3y^{3}$

subtracting second from first,
$ 0 \equal{} 5x^3 \plus{} 5y^x \minus{} y^3 \minus{} x^2y$

thus, $ y \equal{} 5x$

by mere substittion you get the result

Sorry to revive an old thread. I tried a different approach which leads to no solutions. I think there might be something wrong, so if someone could check it, I'd be really grateful.

So we have the two equations $15x^2 + 17xy - xy^2 + 15y^2 = 0$ and $20x^2 - x^2y + 3y^2 = 0$. Solving using the quadratic formula, we have the discriminants $(17y-y^2)^2 - 30^2y^2$ and $-4(20 - y)(3y^2)$.

The first one implies that $(17y - y)^2 \geq 30^2y^2 \implies 17y - y^2 \geq 30y \implies -13 \leq y \leq 0$

The second one implies that $20 - y \leq 0 \implies y \geq 20$. So we have no coinciding ranges which implies that there are no solutions for $y$. Is there something wrong with this?

EDIT: Made an error: forgot about the $y^2$ in $30^2y^2$ in the first discriminant

[quote=smo]Sorry to revive an old thread. I tried a different approach which leads to no solutions. I think there might be something wrong, so if someone could check it, I'd be really grateful.

So we have the two equations $15x^2 + 17xy - xy^2 + 15y^2 = 0$ and $20x^2 - x^2y + 3y^2 = 0$. Solving using the quadratic formula, we have the discriminants $(17y-y^2)^2 - 30^2$ and $-4(20 - y)(3y^2)$.

The first one implies that $(17y - y)^2 \geq 30^2 \implies 17y - y^2 \geq 30 \implies 2 \leq y \leq 15$

The second one implies that $20 - y \leq 0 \implies y \geq 20$. So we have no coinciding ranges which implies that there are no solutions for $y$. Is there something wrong with this?[/quote]

You have square rooted incorrectly.
if b^2<k then -sqrtk<b<sqrtk

[quote=Potla][b]Singular[/b], you have not really completed your proof. You also need to consider $ m \equal{} 0\implies x \equal{} 0\implies y \equal{} 0$. Thus all possible solutions;
$ (x, y) \equal{} (0,0),(19,85)$
[hide]
My first solution was just phymax's solution (about 9 months ago) :roll: [/hide][/quote]

it should (19,95) in place of (19,85)


[quote="ComplexZeta"]Write n=2^a*3^b*5^c*..., where a, b, c, ... are nonnegative integers. Then the sum of the factors of n is (1+2+4+...+2^a)(1+3+9+...+3^b)(1+5+25+125+...+5^c)...

I'm not sure if there's an easy way to sum the first n factorials (I'll work on this later), but it's possible to sum n*n! from 0 to n: write n*n! = (n+1)! - n!, and everything telescopes nicely, so we get  :Sigma: n*n! = :Sigma: (n+1)! - n! = (n+1)! - 1! = (n+1)! - 1.[/quote]

Just on a technical note, you should call your bound of summation by a different variable than the one you use in your summand. Like say, sum k*k! from k= 0 to n equals (n+1)! - 1.

Another way of calculating the sum of the divisors of n.

Let n=(p_1)^(e_1)*(p_2)^(e_2)*...*(p_n)^(e_n), where p_1,2,...,n are the prime factors and e_1,2,...,n are the powers of each respective prime.

The sum of the positive divisors of n is the product of all numbers of the form [((p_i)^(e_i+1))-1]/(p_i-1), where i ranges from [1,n].


integration
calculus
calculus computations

1.)    int \frac{7x^6dx}{7 + x^7}^3

 2.)    int \frac{dx}{xln(x)^4}

Thank you

Try u substitution. Example $ \int \frac{3x^2dx}{(x^3\plus{}4)^2}$ Let $ u\equal{}x^3\plus{}4$ then $ du\equal{}3x^2dx$ and the integral becomes $ \int\frac{du}{u^2} \equal{}\minus{}\frac{1}{u}\plus{}C\equal{}\minus{}\frac{1}{x^3\plus{}4}\plus{}C$.

How did u get this please explain then i will be able to do them by myself  --->
 -\frac{1}{u}+C

$ \int\frac {du}{u^2} \equal{} \int u^{ \minus{} 2}du \equal{} \minus{} u^{ \minus{} 1}\plus{}C\equal{} \minus{} \frac {1}{u} \plus{} C$


can u pls explain....is there any ortho effect in salicyclic acid......( ortho hydroxy -benzoic acid)

plz.....

arnav could explain in which context are u asking for the ortho effect

i wanted to know that does OH group make the COOH group out of plane.....as in the case of ortho methyl benzoic acid...

yes..of course..it will be able to easily get $ COOH$ out of the plane and thereby increasing the stability of the conjugate base....thereby making it a really strong acid...
(Since $ COO^\minus{}$ is highly stabilised because of having identical resonating structures)

[quote="Rijul saini"]yes..of course..it will be able to easily get $ COOH$ out of the plane and thereby increasing the stability of the conjugate base....thereby making it a really strong acid...
(Since $ COO^ \minus{}$ is highly stabilised because of having identical resonating structures)[/quote]

dude i also just saw some excpetions to it.....like the possibility of hydrogen bonding at o- position...

so will this happen that for forming a hydrogen bond the COOH group will not go out of plane.......
or the H bond can be formed even though the COOH GROUP IS OUT OF PLANE.....

pls give some more cases in which H bonding takes place....!!

dude i reaaly cant remember much examples over here........but hydrogen bond always forms between very electo-ve elements like F, O, and sometimes N.....
also buddy i think the same case with $ \minus{}COCl$ will also be like wise.....(although i cant confirm it)


inequalities
triangle inequality
complex analysis

Prove that |z1 + z2|² <= |z1|² + |z2|² + 2|z1||z2|.

Note: |z| is the modulus of a complex number z.

EDIT: I would be very interested to see the actual proof but not only the formulation you mentioned.

However, try to prove it using the fact that |z1 + z2|² = (z1 + z2) conjugate(z1 + z2) = (z1 + z2)(conjugate(z1) + conjugate(z2)) and z + conjugate(z) = 2 Re z, and Re z <= |z|.

Call me crazy, but this seems just like Triangle Inequality, just squared both sides (possible, because both sides are positive and real).

But if you wish to prove it (without knowing the triangle inequality) you can do it just like any other metrical space, going from the definition, and using Cauchy-Schwarz in one of the steps.

huh ?  :(

$ |z_1 \plus{} z_2|^2 \equal{} (z_1 \plus{} z_2) \overline {(z_1 \plus{} z_2)} \equal{} (z_1 \plus{} z_2)(\bar z_1 \plus{} \bar z_2)$

$ \equal{} |z_1|^2 \plus{} z_1 \bar z_2 \plus{} \overline { z_1 \bar z_2 } \plus{} |z_2|^2 \equal{} |z_1|^2 \plus{} 2\;\mathbb Re (z_1 \bar z_2) \plus{} |z_2|^2$

$ ( \because z_n \plus{} \overline {z_n} \equal{} 2\;\mathbb Re (z_n) )$

$ \le |z_1|^2 \plus{} 2|z_1\parallel{}z_2| \plus{} |z_2|^2 \equal{} (|z_1| \plus{} |z_2|)^2$

$ ( \because \mathbb Re (z_n) \le | z_n| \qquad \text{and}\qquad|z_n| \equal{} |\bar z_n| )$

$ \therefore |z_1 \plus{} z_2| \le |z_1| \plus{} |z_2|$

[quote]huh ?  :( [/quote]

I thought it was pretty understandable what I've said, misan. The described procedure is a way of proving the triangle inequality in any normed space.

Here's how they've written it on wikipedia (pretty same proof as we all learn in our courses: although the use of Cauchy's inequality can be substituted with notion of Schwarz' inequality in the given context)

$ \|x \plus{} y\|^2 \equal{} \|x\|^2 \plus{} 2|\langle x, y \rangle| \plus{} \|y\|^2 \equal{} \|x\|^2 \plus{} \langle x, y \rangle \plus{} \langle y, x \rangle \plus{} \|y\|^2 \equal{} \|x\|^2 \plus{} 2|\langle x, y \rangle| \plus{} \|y\|^2\le \|x\|^2 \plus{} 2\|x\|\|y\| \plus{} \|y\|^2\le \left(\|x\| \plus{} \|y\|\right)^2$

Of course, this idea applied in $ \mathbb {C}$ looks like your proof. But it is important to remember that it can be extended to any space with defined inner product.

no, i didn't mean about what you said, hsiljak.  :)  i was referring to the original poster's question, his/her questions sometimes make no sense to me.

e.g.

EDIT: I would be very interested to see the actual proof but not only the formulation you mentioned.

 :?:


here is a very simple one:

prove that $ \forall x,y,z \geq 0$ we have:
$ \sum x^3 \geq \sum  x^{2} \sqrt{yz} $.

cheers! :D :D

This is almost exactly the same as the other problem posed, but slightly easier.

sum(x^3)=sumcyc(2/3x^3+1/6y^3+1/6z^3)>=sum(x^2(yz)^.5) by the AM-GM inequality.

another  solution:
Σ[(x 2  \sqrt (yz)]  \leq Σx 2 * [(y+z)/2] =
= 1/2 Σxy(x+y)  \leq 1/2( Σχ^3 + 3xyz) \leq Σχ^3

This is also a immediate consquence of Muirhead's inequality since (3,0,0) > (2, 1/2, 1/2).

Pierre.


Let $ a,b,c$ be nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}3$. Prove that
\[ \sum \sqrt{9a\plus{}7(b\minus{}1)^2} \ge 9\]
Equality holds for $ a\equal{}b\equal{}c\equal{}1$ or $ a\equal{}0,b\equal{}1,c\equal{}2$.
:)

I have no elementary solution for this inequality. :(

[quote="Domi"]Where is c ?
Domi[/quote]
$ \sum\sqrt{9a\plus{}7(b\minus{}1)^2}\equal{}\sqrt{9a\plus{}7(b\minus{}1)^2}\plus{}\sqrt{9b\plus{}(c\minus{}1)^2}\plus{}\sqrt{9c\plus{}(a\minus{}1)^2}$

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
\[ \sum \sqrt {9a \plus{} 7(b \minus{} 1)^2} \ge 9
\]
Equality holds for $ a \equal{} b \equal{} c \equal{} 1$ or $ a \equal{} 0,b \equal{} 1,c \equal{} 2$.
:)

I have no elementary solution for this inequality. :([/quote]
Any solution for it?  :(


We say that two or more numbers, with the same amount of digits, are members of the same family who all have at least one digit in common. For example, the numbers 32,25 and 22 are in the same family, but 123,245 and 568 aren't in the same family, because 123 and 568 not belong to the same family. What is the largest number of members of one family, whose elements have three digits ?

Portuguese question (if necessary) :
Dizemos que dois ou mais números, com a mesma quantidade de algarismos, são membros da mesma família, quando todos possuem pelo menos um algarismo em comum. Por exemplo, os números 32, 25 e 22 pertencem à mesma família, enquanto que 123, 245 e 568 não pertencem à mesma família, pois 123 e 568 não pertencem à mesma família. Qual é a maior quantidade de membros de uma família, cujos elementos têm três algarismos?

How to do ?

WLOG assume that the common digit is $ 1$, let count all $ 3$ digits numbers with at least one $ 1$ digit
Type $ 1bc$ there are $ 10$ possibilities for $ b$ from $ 0$ to $ 9$ and the same for $ c$ thus $ 10 \times 10\equal{}100$
Type $ a1c$ not yet included, there are $ 8$ possibilities for $ a$ from $ 2$ to $ 9$ and $ 10$ for $ c$ thus $ 8 \times 10\equal{}80$
Type $ ab1$ not yet included, there are $ 8$ possibilities for $ a$ from $ 2$ to $ 9$ and $ 9$ for $ c$ from $ 0,2,3,\cdots,9$ thus $ 8 \times 9\equal{}90$

TOTAL $ \boxed{270}$


geometry
circumcircle
Ross Mathematics Program
3D geometry
prism
Euler
incenter

Give $ \bigtriangleup$ABC has incircle (I) and cicumcircle (O).( (I) and (O) are given before)
Prove that there are infinity triangles ABC in this condition and their intersection of 3 median (I don't known call it in E) are on a circle.
I'm had a bad solution for it. Hope you will help me. :)

For the first part, there is a nice solution published in the book of Ross Honsberger "Mathematical Gems II" (MAA 1976), pp.50-53
By the way, the name in english for the intersection of the medians is centroid.

The following link maybe will be useful to you: Emelianov & Emelianova: Euler's formula and Poncelet's porism, in Forum Geometricorum vol.1 (2001) pp.137-140. It is accesible from the net.

I have located the following nice proof for the second part of your question, at the book of W. Gallatly, The Modern Geometry of the Triangle, London 1913:

Consider the line OI (which is fixed) and the Euler line OH. On OH we have the points G(centroid), and O' (incenter of the medial triangle A'B'C', being A',B',C' the mid points of the sides BC,CA,AB, respectively). It is well know that OO'=O'H and GH=2OG.
Moreover, let F be the point of Feuerbach of the incircle (that is, the point of tangency of the incircle with the circle O'(R/2). Note R/2 is the radius of this circle, also called Euler circle, Nine points circle or Feuerbach circle.

Draw parallel to O'I through H, G and I', which meet the line OI at H_1,G_1 and K, respectively.
Then G moves in a circle, because
3OG=2OO', whence OG_1 = (2/3)OI, therefore G_1 is fixed, and GG_1=(2/3)O'I= (1/3)(R-2r), which is constant.
In the same way you can find the locus of the points F, H, O' and I'.


For a survey, 850 homeowners gave one answer to the question, What would you most like to improve about your home? The pie chart shows the percentage of the homeowners with particular responses, expressed to the nearest whole number. To the nearest 10 people, how many responded that they would improve their bedroom?

[asy]import graph;
draw(Circle((0,0),50));
draw((0,0)--(0,50));
draw((0,0)--(0,-50));
draw((0,0)--(43.3,25));
draw((0,0)--(48,-14));
draw((0,0)--(-15,-47.7));
label("Bathroom",(4,48),E);
label("$20\%$",(-2,34),E);
label("Other",(50,0),E);
label("$9\%$",(25,0),E);
label("Living Room",(56,-26),S);
label("$20\%$",(40,-16),S);
label("Bedroom",(10,-50),W);
label("$13\%$",(13,-37),W);
label("Kitchen",(-35,35),S);
label("$38\%$",(-26,20),S);[/asy]

13% of people wanted bedroom.
Since 850 were surveyed, 850*.13=110.5
With nearest ten, its 110.


All the possible three-digit numbers with distinct digits are formed using the digits 2, 3, 4 and 5. What is the probability, expressed as a common fraction, of choosing a number at random that is divisible by 4?

We ignore all but the last two digits.

Out of the following:
23 24 25
32 34 35
42 43 45
52 53 54
Out of the 12 choices, there are 3 that are divisible by 4. [24,32,52]. So, our answer is 3/12=1/4


function
integration
trigonometry
LaTeX
inequalities
advanced fields
advanced fields unsolved

I have a solution of this question but I am looking for a more
elegent proof. Can somebody please help me with this?

Show that, for any continuously differentiable function $ f$ on $ [\minus{}\pi,\pi]$,
\[ \left|\int_{\minus{}\pi}^{\pi}f(t)\cos t\minus{}f'(t)\sin t\ dt\right|\le\sqrt{2\pi}\left\{\int_{\minus{}\pi}^{\pi}
|f(t)|^2\plus{}|f'(t)|^2\,dt\right\}^\frac12.\]\\

I have shown this by saying proving that

 Let $ a<b$ in $ \mathbb R$. Show that the space $ W[a,b]$ of
continuously differentiable functions on $ [a,b]$, with values in
$ \mathbb C$,
is an inner product space with respect to pointwise addition and
scalar multiplication, and inner product
\[ (f,g)_W \equal{} \int_a^b f(t)\overline{g(t)}\plus{} f'(t)\overline{g'(t)}\,dt.\]

But please help me with a better proof of this.

Wouldn't applying the chain rule on the modulus content yield 0 and since the right side is always positive, the assumption would be true?

Can you please explain in more detail, because I dont undersatnd you point. Thanks

Well, first of all, I'm not experienced in Latex, so apologies in advanced!

This is what I ment:

π
∫f'(t) sin(t) dt =
-π

 π
-∫f(t) cos(t) dt + f(π) sin(π) - f(-π) sin(-π) =
 -π

 π
-∫f(t) cos(t) dt
 -π

Now, plugging that into:

π
∫f(t) cos(t) + f'(t) sin(t) dt =
-π

π
∫f(t) cos(t) - f(t) cos(t) dt = 0
-π

And since the right side of the inequality is always positive, it is true!

That doesn't work. The absolute values matter, and the left-hand side is not zero.

Pointwise, $ |f(t)\cos t\minus{}f'(t)\sin t|\le \sqrt{f(t)^2\plus{}f'(t)^2}$ by Cauchy-Schwarz. Next, $ \left|\int_I g(x)\cdot 1\,dx\right|\le\left(\int_I (g(x))^2\,dx\right)^{\frac12}\cdot |I|^{\frac12}$, also by C-S. Equality is only possible if $ \frac{f(t)}{f'(t)}\equal{}\frac{\cos t}{\minus{}\sin t}$ everywhere, and $ f(t)^2\plus{}f'(t)^2$ is constant, which occurs when $ f(t)\equal{}c\cos t$ for some constant $ c$.


Here's another nice and hard problem that I can't solve it. Please help me.
Prove that for all positive real numbers $ a_{1},a_{2},...a_{n}$ :
$ \frac{1}{a_{1}}\plus{}\frac{2}{a_{1}\plus{}a_{2}}\plus{}\frac{3}{a_{1}\plus{}a_{2}\plus{}a_{3}} \plus{}...\plus{}\frac{n}{a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n}} \leq 2(\frac{1}{a_{1}}\plus{}\frac{1}{a_{2}}\plus{}...\plus{}\frac{1}{a_{n}})$

I remember an easier problem from Russian 1986
Prove that
$ \frac {1}{a_{1}} \plus{} \frac {2}{a_{1} \plus{} a_{2}} \plus{} \frac {3}{a_{1} \plus{} a_{2} \plus{} a_{3}} \plus{} ... \plus{} \frac {n}{a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n}} < 4(\frac {1}{a_{1}} \plus{} \frac {1}{a_{2}} \plus{} ... \plus{} \frac {1}{a_{n}})$
where $ a_1;a_2;...;a_n$ be positive real numbers


MUAHAHA! The Parting Curse of the aops has been laid upon you. Many of you wise old sages must be about to leave behind ignorant young foals like us in the wilderness and embark upon your journey into the world of IITs, BITS, NIITs and other holy places of enlightenment.
But BEWARE! You dare not forget where you roots lie.... Or else.... the Parting Curse shall be invoked and you shall fail every exam of your life!  :diablo:

[hide]
C'mon stop abusing me like that! I told you not to open this thread :nhl:
Some people just don't understand...  :ewpu: [/hide]

Nothing to wory about  :) . I alredy know i will fail of my exam in 12 class(As per according to my prep)  :(

I just thought, weren't you the first one to read it. :rotfl:

[quote="mittal"]I just thought, weren't you the first one to read it. :rotfl:[/quote]
Lol, but it applies only to you guys, not to our new batch of 12-graders :P

Ahh I have been cursed.. :ninja:

but the topic of the thread is very tempting. I guess curiosity kills the...erm... is it cat or rat..

I'm not parting. After Chem screwed every* exam I appeared for, I am seriously considering repeating JEE and then work up in Chem.

*yes, every single exam. Even the boards. Just got 60 off 100 in in Chem to miss 80% threshold for BITS.


function
quadratics
conics
parabola
inequalities
inequalities unsolved

[b][size=150]Let f(x)=ax^2 - 4ax + b, (a>0) be defined in 1<=x<=5.
Suppose the average of the maximum value and the minimum value of the function is 14, and the difference between the maximum value and minimum value is 18. Find the values of a (a>0) and b. [/size][/b]:?:

Mine was a little different:

$f(x)=ax^2-4ax+b$
If the avg of the max and min values is 14, their sum is 28.  The problem also gives the difference so
$f(M)+f(m)=28 , f(M)-f(m)=5$ $\rightarrow$ $f(M)=23, f(m)=5$
Trying to complete the square, we get
$f(x)=ax^2-4ax+b=ax^2-4ax+4a+b-4a=a(x-2)^2+(b-4a)$
b and a are constants, so the minimum occurs when the quadratic is zero, when $x=2$, so $f(m)=f(2)=5 \rightarrow b-4a=5 \to b=4a+5$
Thus, $f(x)=a(x-2)^2+4a+5-4a=a(x-2)^2+5$
$f$ is a parabola, so it's symmetric about $x=2$.  Since 5 is farther from 2 than 1 and $a>0$, $f(5)$ is the maximum.  Plugging in and solving the equation for a, then b, $a=2 , b=13$
Just cuz this is the inequalities section doesn't mean we can't be different :oops:


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose $ H$ is a subgroup of $ G$ such that whenever $ Ha \neq Hb$
then $ aH \neq bH$. Prove that for all $ g \in G$ $ gHg^{\minus{}1} \subseteq H$.

(This one is from Herstein)

Again, I can't type set. After I return from vacation I will try learn some.

The idea is to assume the contrary. We want to show gH is contained in Hg for all g in G. Suppose some gh1(=g*) in gH is not in Hg then since cosets are either equal or disjoint we have Hg* doesn't equal Hg so by hypothesis
g*H doesn't equal gH, but I leave it to you show that indeed g*h=gH ( this is easy )....QED

Note $ aH \equal{} bH \iff b \in aH$ and similarly $ Ha \equal{} Hb \iff b \in Ha$.
Hence for every $ g,x \in G$ $ x \in gH$ imply $ x \in Hg$, so $ gH \subseteq Hg$.
But this imply the thesis cause then for every $ h \in H$ $ ghg^{ \minus{} 1} \equal{} h_2$ for some $ h_2 \in H$.

(I've edited my note on your typeset)

Before you shoot off your mouth, at least get it right. What you posted is garbage and I'll leave you to think about why ( pensare ). Hint: In your proof every subgroup would be normal and you know that is not correct.

Uhm, I still don't understand your point.
Maybe you mean the error is here:
[quote="vl4d"]
Hence for every $ g,x \in G$ $ x \in gH$ imply $ x \in Hg$, so $ gH \subseteq Hg$.
[/quote]

But I'm not saying this for every subgroup $ S\le G$... I'm using the fact:
"if $ aH\equal{}bH$ then $ Ha\equal{}Hb$"
which is the contrapositive on our initial hypothesis...

?!?

Exactly. The point is your conclusion would say every subgroup is normal. You never used the hypothesis Herstein gave you.

I'm using it. The following is a little expansion of the implication:

$ x \in gH$ imply $ xH \equal{} gH$ but, by Herstein's contrapositive hypotesis
then $ Hx \equal{} Hg$, so $ x \in Hg$. I'm saying this is true just for the subgroup $ H$,
not for every subgroup.


Ba farz \[ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\equal{}2\]  \[ x,y,z >1\]  sabet konid :

\[ \sqrt{x\plus{}y\plus{}z} \geq (\sqrt{x\minus{}1} \plus{}\sqrt{y\minus{}1} \plus{}\sqrt{z\minus{}1})\]

 :)

bana bar $ cauchy-shwarz$ ineq
$ (\sqrt{x+y+z})(\sqt{\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}) >= \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$

midunim ke $ \sum \frac{x-1}{x} =1$

chun ke $ \sum \frac1{x}=2$

bana bar $ cauchy - shwarz$ ineq
$ (\sqrt {x + y + z})(\sqrt {\frac {x - 1}{x} +\frac {y - 1}{y} +\frac {z - 1}{z}}> = \sqrt {x - 1} + \sqrt {y - 1} + \sqrt {z - 1}$

midunim ke $ \sum \frac {x - 1}{x} = 1$

chun ke $ \sum \frac1{x} = 2$ :wink:

kamyar kheili khoob hal kardi . hala ye soal dige daram ke khodam halesh nakardam :
\[ a,b,c>0\] \[ abc\equal{}8\]

Prove that :
\[ \sum\frac{1}{\sqrt{a^3\plus{}1}}\geq 1\]

nice one:
eqaul to:$ \sum \sqrt \frac {bc}{8a^2 \plus{} bc}\geq 1$ equal to $ \sum \sqrt \frac {y^2z^2}{8x^4 \plus{} y^2z^2}\geq 1$
$ m \equal{} \frac {yz}{x}$,$ n \equal{} \frac {xz}{y}$,$ p \equal{} \frac {xy}{z}$
pas darim
$ \sum \sqrt \frac {y^2z^2}{8x^4 \plus{} y^2z^2} \equal{} \sum \frac {m}{\sqrt {m^2 \plus{} 8np}}$
vali darim $ \frac{m}{\sqrt{m^{2}\plus{}8np}}\geq \frac{m^{\frac{4}{3}}}{m^{\frac{4}{3}}\plus{}n^{\frac{4}{3}}\plus{}p^{\frac{4}{3}}}$age jam bezanim dar miad.
ghesmate akhar soale IMO2001e.
mitunin inja ro bebinin http://www.mathlinks.ro/viewtopic.php?p=119168#119168

kheili soale ghashangi bud vali rahesh ghashangtar khube say konim ye seri soal entekhab konim va 1 hafte vaght bedim bad az 1 hafte biaim rush harf bezanim. chetore? mitunin once get online per week beshin? :lol:

ba in lemme ke
$ xyz \equal{} t^3$  va $ t > \equal{} 8$

$ \sum\frac{1}{\sqrt{x \plus{} 1}} > \equal{} \frac {3}{\sqrt {1 \plus{} t}}$

age $ x\longrightarrow x^3$ badihi mishe :lol:

excellent !!  man baraye soale apmo 2005 in az in idea estefade kardam.

$ abc\equal{}8$ and $ a,b,c>0$
$ \sum \frac{a^2}{\sqrt{(1\plus{}a^3)(1\plus{}b^3)}}\geq \frac{4}{3}$


Let $\triangle ABC$ be a triangle and denote $a=BC, b=CA, c=AB$. The internal angle bisectors intersect the sides $BC,CA,AB$ respectively at $L,M,N$. If $L', M', N'$ are on the sides $CA,AB,BC$ such that $LL'\parallel AB, MM'\parallel BC, NN'\parallel CA$ show that

\[
\frac{1}{{LL'}} + \frac{1}{{MM'}} + \frac{1}{{NN'}} = 2\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)
\]

\[LL_{1} = c\frac{LC}{a} = c\frac{\frac{ab}{b+c}}{a} = \frac{bc}{b+c} \Rightarrow \frac{1}{LL_{1}}=\frac{1}{b} + \frac{1}{c}.\]
Similarly we have $\frac{1}{MM_{1}} = \frac{1}{c}+\frac{1}{a}$ and $\frac{1}{NN_{1}} = \frac{1}{a} + \frac{1}{b}$, after adding up both side we get the desired result.


function
number theory
greatest common divisor
number theory proposed

Let $m,n$ are positive integers.
a)Prove that $(m,n)=2\sum_{k=0}^{m-1}[\frac{kn}{m}]+m+n-mn$.
b)If $m,n\geq 2$, prove that $\sum_{k=0}^{m-1}[\frac{kn}{m}]=\sum_{k=0}^{n-1}[\frac{km}{n}]$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=59747
The second is a trivial corollary.


inequalities
calculus
derivative
calculus computations

how do we prove the AM-GM inequality using lagrange multipliers?

I don't think it works particularly well because you don't have a constraint (assuming you're minimizing the difference between the arithmetic and geometric mean, for example). It's probably easier just to set the partial derivatives equal to zero, with the assumption that $ a_{i}\ge 0$ for all $ i$:

$ f(a_{1}, a_{2},\ldots, a_{n}) \equal{}\frac{a_{1}\plus{}a_{2}\plus{}\cdots\plus{}a_{n}}{n}\minus{}\sqrt[n]{a_{1}a_{2}\cdots a_{n}}$

$ f_{a_{i}}(a_{1}, a_{2},\ldots, a_{n}) \equal{}\frac{1}{n}\minus{}\frac{1}{na_{i}}\sqrt[n]{a_{1}a_{2}\cdots a_{n}}\equal{} 0$.

Solving the equation immediately gives $ 1\minus{}\frac{G}{a_{i}}\equal{} 0\Rightarrow a_{i}\equal{} G$ where $ G$ is the geometric mean for all $ i$. Thus $ f$ has a critical point whenever $ a_{1}\equal{} a_{2}\equal{}\cdots \equal{} a_{n}$ which evaluates to $ f \equal{} 0$. We can also check the boundary, i.e. where some of the $ a_{i}$'s are zero, but clearly $ f\ge 0$ in this case. I think this is sufficient to show that $ f\ge 0$ for all points in our domain.

[quote="paladin8"]I don't think it works particularly well because you don't have a constraint[/quote]

An easy constraint: set the arithmetic mean to a constant.  :)


MATHCOUNTS
geometry
trigonometry
analytic geometry

Sometimes I look at some of the stuff that you guys post in this forum, and it makes me want to go back to the MATHCOUNTS Forum and post there.  Is this the kind of work that I'll really see in High School, or are you guys doing advanced work on this forum.  I'm really starting to get scared.  A little Algebra 1 KId accidently stumbled into this forum, and now he's scarred for life.

Well, I know for me here in Canada, most of the topics that are in the basics forum are really grade 11+ material.. the intermediate Id say 12+ and beyond.. and the other forums are beyond beyond pretty much.

Most of what gets posted though is stuff you wont really learn in a classroom (atleast I have not yet), you may learn some fundamentals but the rest is using your noodle to put two and two together most of the time.

GlenthemanN is right, there is no need to be intimidated; some of the stuff here is a little advanced, but manageable.
If you do not understad one of the threads then wait until someone posts a solution, and if you still are unsure then ask them how they got what they did.

Most of the stuff in HSB is Geometry-Basic trig (this includes a lot of Alg 2). So don't be discouraged when you go to HSB.

thanks, i feel better about this now.  how do i get a topic locked?

That is good  :) .

About locking the thread I think that a moderator must do it.

Oooooh, thanks, i think.
hey, try to become a member of the DUH! Club everybody.  Look at my sig.

[quote="diophantient"][quote="rd5493"]so, national mathcounts is coming up in two weeks.  anybody have any sort of advice/help/comments/questions/etc. (advice really helps)

(To any moderator reading this- never mind about locking this topic.  i want it open)

Can anyone tell me what type of geometry i should give myself crash course in to be able to compete better at mathcounts?[/quote]
Analytic geometry[/quote]

Well on Mathcounts? Analytic geometry is pretty cheap though I use it when necessary, but learning the pure geometry is more important and more fulfilling. I myself am not good at geometry but I find it pretty intriguing, and I appreciate nice geometric solutions to stuff.

I can do most of the stuff in HSB and some in intermediate, but when I go to a mathcounts practice drill page I suck hard at most of the topics..

[quote]Is math class too easy for you? Looking for a greater challenge?

You've come to the right place.[/quote]

[quote="rd5493"]thanks, i feel better about this now.[/quote]

I know how you feel, at first I thought HSB was something I learn in High school, and I did not know it included Algebra II (which what I was learning at that point)..I kept posting in Classroom Math lol cuz it said "Middle school"


Game:

Put your iPod or iTunes to "shuffle" (or any other music playing device with this functionality), play the next song, and post what song that was here.

I'll go first:

Clarinet Quintet (III Adagio) -- Johannes Brahms

Wake me up When September Ends- Green Day

Variation: X Fugetta
Daniel Barenboim, Piano
Goldberg Variations, BWV 988

.......

Ready Steady Go! -- (Full Metal Alchemist) -- L'Arc~en~Ciel

Back The F*** Up - Fear Factory

what do you know

Idomeno, Re Di Creta, K.  366:"Quando Avran Fine Omai"  - "Padre, Germani, Addio!"
Anna Netrebko, Gianandrea Noseda & Weiner Philharmoniker
Mozart 250(iTunes Exclusive)

Piano Sonata No. 12 in F, K.332:111.Allegro Assai
Mitsuko Uchida
Mozart 250(iTunes Exclusive)

Le Nozze Di Figaro, K. 492:"Giunse Alfin II Monumento" - "Deh Vieni Non Tardar"
Dawn Upsaw, James Levine & Metropolitan opera orchestra

Anon: Improperium
The Benedictine Monks of Santo Domingo De Silos
Chant

Finally, something Not in the 250 mozart file.

The Uruk-Hai, LOTR: Two Towers, Howard Shore

[quote="ckck"]It Ends Tonight-The All-American Rejects[/quote]

haha that song is awesome!

Riot by Three Days Grace

Song: I'll Sue Ya
Artist: Weird Al Yankovic
Album: Straight Outta Lynwood

Song: Problem Child
Artist: AC/DC
Album: Let There Be Rock

[quote="mwpl11"]Mr. Brightside - The Killers[/quote]

we need a :murder: emoticon specifically for occasions such as these

Get On Top - Red Hot Chili Peppers [Californication]
Unrepentant - Trivium [The Crusade]
Lips of Deceit - Avenged Sevenfold [Sounding The Seventh Trumpet]
Heaven - Kamelot [Dominion]
I Like Dirt - Red Hot Chili Peppers [Californication]
Pure Hatred - Chimaira [The Impossibility of Reason]

Estranged - Guns N' Roses [Use Your Illusion II]

Nickelback-Someone That You're With
Daughtry-Home

Tijaniyya - Youssou N'Dour(Artist)

Look What You've Done - Ragtime: The Musical - Act II

Dancing for a Vision - David & Steve Gordon (artists)

Samba da Bençao - Bebel Gilberto(artist)

Hamana Nale (Lotus Mix) - One At Last

Shallow Bay - Breaking Benjamin [Saturate]
Elizabeth: III. Fall From Grace - Kamelot [Karma]
Sp lit - Chimaira [Pass Out of Existence]
Scumgrief - Fear Factory [Soul of a New Machine]


Let $ ABC$ be a triangle inscribed circle $ (O,R)$ and its incircle is $ (I,r)$, the lines $ IA,IB,IC$ meet $ (O)$ at the second points $ A',B',C'$. Let $ k \equal{} \sqrt {\frac {R}{2r}} \minus{} 1$. Contruct the points $ A_1,A_2$ such that $ \vec{C'B} \equal{} k\vec{BA_1}$ and $ \vec{B'C} \equal{} k\vec{CA_2}$ the same we get $ B_1,B_2,C_1,C_2$ prove that the lines $ A_1A_2,B_1B_2,C_1C_2$ are concurrent.

I only have the following "result": Let $ M_1$, $ M_2$, $ M_3$ be the midpoints of $ A_1 A_2$, $ B_1 B_2$, $ C_1 C_2$, respectively.  Then the point of concurrency seems to lie on the circumcircle of triangle $ M_1 M_2 M_3$.

I am very puzzled by the presence of the square root in $ k$.  I have no idea what to do with it.


ratio
function
calculus
derivative
integration
trigonometry
calculus computations

Just looking at (i) for now; I suspect related arguments apply to all four. One thing to note is that the ratio test shows each of the four series as having radius of convergence 1. We should be looking for formulas with singularities at points with $ |x|=1.$

Let $ f(x)$ be the function in (i). Playing around with derivatives, I get that

$ \left[\frac1xf(x)\right]'=xf'(x).$

This can be rearranged to $ \frac{f'(x)}{f(x)}=\frac1{x-x^{3}}.$

Integration of that differential equation yields $ f(x)=\frac{x}{\sqrt{1-x^{2}}}.$

This does satisfy the basic predictions of being an odd function with singularities one unit from the origin.

Checking: the binomial series (Newton's series) shows that

$ (1-x^{2})^{-1/2}=1+\frac12x^{2}+\frac12\left(-\frac12\right)\left(-\frac32\right)x^{4}+\frac1{3!}\left(-\frac12\right)\left(-\frac32\right)\left(-\frac52\right)\left(-x^{6}\right)+\cdots$

Do the appropriate rearrangements and then multiply by $ x,$ and you do have the same series for $ f(x).$

With Kent's hard work, it is now easy to determine that the answer to (iii) is:

$ f(x)=1-\sqrt{1-x^{2}}$  :)   (just integrate and solve for the constant)

[quote="misan"]
$ \textbf{(ii)}\;\;\;x+\frac{2}{3}\cdot\frac{x^{3}}{3}+\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{x^{5}}{5}+\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\frac{x^{7}}{7}+\cdots$

[/quote]

for ${ \text(ii)}$, i get $ \int_{0}^{\;x}\;\frac{\sin^{-1}x}{x\cdot \sqrt{1-x^{2}}}\;\;\textbf dx\;=\;\int_{0}^{\;\sin^{-1}x}\;\frac{u}{\sin u}\;\;\textbf du$

but what next ?  :maybe:

all i know is for $ x=1$, i get $ 2\textbf K$, where $ \textbf K$ is catalan's constant :|


trigonometry
inequalities
inequalities proposed

[quote="ehsan2004"]I hope it's true

By using the couchy-schwartz we'll have \[ \sum(\sqrt{(1+a^2)(1+b^2)} \sum\sqrt{\frac{(1+ab)^2}{(1+a^2)(1+b^2)}}\geq (\sum (1+ab))^2  \]

We know that $\sum\sqrt{(1+a^2)(1+b^2)} \leq 1+\sum a^2$ so that we must prove that \[ \frac{(\sum (1+ab))^2}{1+\sum a^2}\geq 1  \] and I think it's true  :huh:[/quote]
sorry,I think you are wrong.....
first,you use the caughy in wrong way.

[quote="ehsan2004"] and I think it's true  :huh:[/quote]
I don't think :|

[quote="zhaobin"][quote="Tiks"]Let$a=tgx_1,b=tgx_2,c=tgx_3,d=tgx_4,e=tgx_5$wher$x_i\in(0;\pi/2),i=(1,2,3....)$

We have that$\sqrt{\frac{(1+ab)^2}{(1+a^2)(1+b^2)}}+\sqrt{\frac{(1+bc)^2} {(1+b^2)(1+c^2)}}+\sqrt{\frac{(1+cd)^2}{(1+c^2)(1+d^2)}}+\sqrt{\frac{(1+de) ^2}{(1+d^2)(1+e^2)}}+\sqrt{\frac{(1+ea)^2}{(1+e^2)(1+a^2)}}\geq1$<=>

$cos(x_1-x_2)+cos(x_2-x_3)+cos(x_3-x_4)+cos(x_4-x_5)+cos(x_5-x_1)\geq 1$ :(  :)[/quote]
I will do follow you approach.
we can know  one of the following must be true:
$Lemma$,if   $-\frac{\pi}{2} \leq A,B,C,A+B,B+C,C+A,A+B+C \leq \frac{\pi}{2}$
we have $\cos A+\cos B+\cos C \geq \cos(A+B+C)$
proof:wlog,$A \geq 0,B \geq 0$ we should only prove:$\cos A +\cos B+ \cos C \geq \cos(A+B+C)$
and because $\cos A +\cos B \geq 1+\cos (A+B)$(easy to prove)
and because $\cos (A+B) \geq 0,\cos C \geq 0, \cos(A+B+C) \leq 0$
so the lemma is proved.
so we have $cos(x_1-x_2)+cos(x_2-x_3)+cos(x_3-x_4)+cos(x_4-x_5)+cos(x_5-x_1)\geq cos(x_1-x_4)+cos(x_4-x_5)+cos(x_5-x_1) \geq \cos 0 =1$

 and this is a so nice inequality :)[/quote]
Nice soloution ZHAOBIN :thumbup:  ;)

$Tiks$, and it is based on your nice substitution. :)
 and I think the inequality is true for $n$ is odd.

You are rigth zhaobin,it is true for $n$ is odd :)
and I think  don't true for $n$ is even.

[quote="zhaobin"][quote="Tiks"]Let$a=tgx_1,b=tgx_2,c=tgx_3,d=tgx_4,e=tgx_5$wher$x_i\in(0;\pi/2),i=(1,2,3....)$

We have that$\sqrt{\frac{(1+ab)^2}{(1+a^2)(1+b^2)}}+\sqrt{\frac{(1+bc)^2} {(1+b^2)(1+c^2)}}+\sqrt{\frac{(1+cd)^2}{(1+c^2)(1+d^2)}}+\sqrt{\frac{(1+de) ^2}{(1+d^2)(1+e^2)}}+\sqrt{\frac{(1+ea)^2}{(1+e^2)(1+a^2)}}\geq1$<=>

$cos(x_1-x_2)+cos(x_2-x_3)+cos(x_3-x_4)+cos(x_4-x_5)+cos(x_5-x_1)\geq 1$ :(  :)[/quote]
I will do follow you approach.
we can know  one of the following must be true:
$Lemma$,if   $-\frac{\pi}{2} \leq A,B,C,A+B,B+C,C+A,A+B+C \leq \frac{\pi}{2}$
we have $\cos A+\cos B+\cos C \geq \cos(A+B+C)$
proof:wlog,$A \geq 0,B \geq 0$ we should only prove:$\cos A +\cos B+ \cos C \geq \cos(A+B+C)$
and because $\cos A +\cos B \geq 1+\cos (A+B)$(easy to prove)
and because $\cos (A+B) \geq 0,\cos C \geq 0, \cos(A+B+C) \leq 0$
so the lemma is proved.
so we have $cos(x_1-x_2)+cos(x_2-x_3)+cos(x_3-x_4)+cos(x_4-x_5)+cos(x_5-x_1)\geq cos(x_1-x_4)+cos(x_4-x_5)+cos(x_5-x_1) \geq \cos 0 =1$

 and this is a so nice inequality :)[/quote]
Very nice; It was better than my ugly solution ;)

[quote="ehsan2004"]I hope it's true

By using the couchy-schwartz we'll have \[ \sum(\sqrt{(1+a^2)(1+b^2)} \sum\sqrt{\frac{(1+ab)^2}{(1+a^2)(1+b^2)}}\geq (\sum \sqrt{(1+ab)})^2  \]

We know that $\sum\sqrt{(1+a^2)(1+b^2)} \leq 1+\sum a^2$ so that we must prove that \[ \frac{(\sum\sqrt{ (1+ab)})^2}{1+\sum a^2}\geq 1  \] and I think it's true  :huh:[/quote]

Ok ,I edit it  :ninja:


k = 3wx     m = (w - 1)k    If k and m are defined by the equations above, what is the value of m when w = 4 and  x = 1?

a)  0   b)  3  c)  12   d)  24   e)  36

$ k\equal{}3(4)(1)\equal{}12$

$ m\equal{}(4\minus{}1)(12)\equal{}3(12)\equal{}\boxed{36}$

So the answer is $ \boxed{E}$.


In $ \triangle ABC, \tan A \equal{} 3.$ The altitude from A divides BC into segments of length 4 and 16. Find the area of $ \triangle ABC.$

123456789, how can you assume it is right, I think there is only one such triangle fitting the conditions.
[hide] Call the height x, the foot of the altitude H. We have 3=tanBAC=tan(BAH+HAC)=tanBAH+tanHAC/(1-tanBAHtanHAC)=$ (4/x \plus{} 16/x)/(1 \minus{} 64/x^2)$. Solving for x gives x=12, so the area is 120. [/hide]

Oh yea, sorry. Stupid moment there. I'll delete my "solution".


real analysis
integration
function
real analysis unsolved

Let $ \mu : \Upsilon\to [0,\infty ]$ be a measure on X. Let $ f\ge 0$ be $ \Upsilon$-measurable. Define

$ \nu (E) \equal{} \int^{}_E f\,d\mu$ for all $ E\in\Upsilon$.

Show that:

$ \int g\,d\nu \equal{} \int fg\,d\mu$         $ \forall$ measurable g $ \ge$0.

I can show that $ \nu$ is a measure, but I cannot get the final result.

It's clearly true for simple functions. Then do it for positive functions by monotone convergence theorem, and then you're essentially done.


function
calculus
derivative
calculus computations

Why is there no way to write down an explicit solution for $ ye^{x} \equal{} \minus{}xe^{y}$ around the neighborhood of the origin?

The set of functions with explicit, elementary names is but a tiny sliver of the set of all functions (or all smooth functions, or whatever). If the question is, "Can the solution to __ be given explicitly?" my attitude would be that the answer "No" doesn't normally need an explanation, since that is the ordinary order of things - it's the "Yes" answer that requires justification or details.

My book says to "observe" that there is no way to write an explicit solution y = y(x) of the above equation. I didn't understand what was the book's reasoning. Is there something with the behavior the graph that I don't know about ( it's kind of hard to graph....)

I'm guessing that the intent isn't too deep - just to note that you know of no algebraic techniques to solve the equation for $ y$ in terms of $ x.$ That isn't a proof - it's a statement of lack of knowledge - but there's some chance that's all the authors meant.

Got it. However, then the book goes on to say that there still exists a $ C^{\infty}$ solution $ y\equal{} y(x)$ near (0,0) for the above equation. Is that just to say that the equation is differentiable? Could we take partial derivatives of the equation at the point (0,0)? ( I remember that partial derivatives can still exist even though a function may not be differentiable at a certain point (0,0), i.e.

$ f(x,y) \equal{} \begin{cases}\frac{xy}{x^{2}\plus{}y^{2}}&if (x,y) \ne (0,0)\\0&if (x,y) \equal{} (0,0)\end{cases}$

Does this particular chapter of the book contain the statement (and perhaps also proof) of the Implicit Function Theorem?

That example - that the existence of partial derivatives may not imply smoothness - is basically irrelevant to this situation. If $ F(x,y)\equal{}ye^x\plus{}xe^y,$ then $ F$ is smooth on all of $ \mathbb{R}^2,$ and in particular is $ C^{\infty}$ in a neighborhood of the origin.

What does the Implicit Function Theorem say?

The Implicit Function Theorem seems to say ( at least to me) that evven though you don't have an explicit formula $ f(x,y)$, you can still find its derivative.....( i think)


function
trigonometry
calculus
calculus computations

My math teacher revealed to me one day in a class that hyperbolic cotangent function is the basis of the calculator, which means that hyperbolic cotangent solely is employed in a calculator to calculate all values you put in. However, I'm interested in how that function calculates all values of functions, and if anyone knows how the function works to produce values, please tell me the mechanism precisely. That will be greatly helpful to me. My math teacher went down to Texas to investigate about information on calculators, and the fact I just wrote down came from his investigation. I think it's pretty interesting information to know.

http://en.wikipedia.org/wiki/Tangent_half-angle_formula

Cheers,

Rofler

Thanks for the information, but would you know how hyperbolic cotangent function calculates any values you want to figure out? The gist of my posting was that hyperbolic cotangent function is programmed in a calculator to calculate everything. But why is that so?

Because all the trig functions can be written in terms of cotangent function as shown in the article. (It's actually shown with the tangent function but you can apply a transformation in variables to make it in terms of cotangent.)

Or just scroll down 1 page on the link that I provided... :P

Cheers,

Rofler


Each pirate lied in making the following statements: \\
\textbf{Long}: I stole the map. \\
\textbf{John}: Silver did not steal the map. \\
\textbf{Silver}: John stole the map. \\
Who stole the map? \\

Well if john lied, then silver did steal the map.


[quote="Leon"]For positive reals $x$, $y$, $z$ show that
$x^3 +y^3 + z^3 +2(x^2z+y^2x+z^2y) \geq 3(x^2y+y^2z+z^2x)$
Leon[/quote]
My ugly proof:
Let $x=\min\{x,y,z\}.$ Then $y=x+u,$ $z=x+v,$ where $u\geq0$ and $v\geq0.$
Hence, our inequality equivalent to $2(u^2-uv+v^2)x+u^3-3u^2v+2uv^2+v^3\geq0,$
which obviously true.
What  your proof?

It is similarly possible to find maximal k, for which
$x^3 +y^3 + z^3 +k(x^2z+y^2x+z^2y) \geq (k+1)(x^2y+y^2z+z^2x)$ for all non-negative $x,$ $y$ and $z$
since, $x^3 +y^3 + z^3 +k(x^2z+y^2x+z^2y) \geq (k+1)(x^2y+y^2z+z^2x)\Leftrightarrow$
$\Leftrightarrow2(u^2-uv+v^2)x+u^3-(k+1)u^2v+kuv^2+v^3\geq0.$

[quote="arqady"]Hence, our inequality equivalent to $2(u^2-uv+v^2)x+u^3-3u^2v+2uv^2+v^3\geq0,$
which obviously true. [/quote]

Can you explain this ?

I have not yet a proof.

Leon

[quote="Leon"][quote="arqady"]Hence, our inequality equivalent to $2(u^2-uv+v^2)x+u^3-3u^2v+2uv^2+v^3\geq0,$
which obviously true. [/quote]
Can you explain this ?
Leon[/quote]
$2(u^2-uv+v^2)x\geq0.$
And $u^3-3u^2v+2uv^2+v^3\geq0$ since, if $v=0$ then $u^3\geq0.$
If $v\neq0$ then let $\frac{u}{v}=t.$
Hence, remain to prove that $t^3-3t^2+2t+1\geq0,$ where $t\geq0.$
Let $f(t)=t^3-3t^2+2t+1.$ Then $f'(t)=3t^2-6t+2.$
Hence, $t_{min}=\frac{3+\sqrt{3}}{3}.$
Id est, $f(t)\geq f\left(\frac{3+\sqrt3}{3}\right)=0.61509982...>0.$ :)


Hi everyone,

I am new to the forum and i joined because it looks really friendly - and i am sure i will be coming back here regularly for answers to things, so i have bookmarked this site  :D

I have a question about an assignment that i have. I generally understand it, except one question has me in a muddle.

The question is (and i know there has been a topic on this, but i have a different question related to the same topic.):

1 g of hydrogen will react with 19 g of fluorine to form HF. By Contrast, 1 g of hydrogen will react with 80 r of bromine to form HBr.

a. i understand the question

b. Why are different amounts of each element required to react with the same amount to hydrogen? - i am stuck here

c. i understand the question.

So, basically, i am trying - and unsucessfully - trying to work out why different amounts are needed. Studing the computer for to long i bet - make my brain go a bit unconcentrated  :lol:

Any help you guys can give me would be greatly appreciated.

Kind Regards,
Ash

hey u seem like ur in big trouble. so i shall help i think u need to use molar masses. and for b u need to use like different mass of the element. haha ash this is just a bunch of jiberish! anyone reading this please help???

hey i think that you can solve your trouble with the mass the bromine is bigger than the flourine then in 19 g wont the necesary ammount for the reaction with H.


Can someone explain these two concepts for me please?
1) Radian.
2) Reflexive angle.

Thanks in advance

Radians exist to make the calculus of trigonometric functions nice. If we try to differentiate $ \sin$, we find that $ \frac{d}{dx}\sin x\equal{}c\cdot\cos x$, where $ c$ is some constant depending on how we scale the angle measurement. Radians make that constant 1, and other constants in calculus similarly nice.
For any non-calculus application, the choice of angle measurement systems is basically arbitrary. We wouldn't choose radians in that context, because multiples of $ \pi$ fell weird. They're brought up in school so they won't feel too strange when you get to calculus. The definition makes sense without calculus; the radian measure of an angle is just the arclength of its piece of a unit circle.

Is Radian also a measuring tool for angles?

Yes, if I remember correctly $ \pi$ radians is $ 180$ degrees.

Radians are one of three units for measuring angle size: degrees, radians, and grads.

Yes, AIME15, $ \pi$ radians is indeed $ 180$ degrees. I don't know what grads are, though.  :oops:

Radians are usually used instead of degrees in functions.

And I think that a reflexive angle is one that is >180 degrees (or PI radians)

Sorry I didn't understand what you mean by the reflexive angle?
Can you show me a picture or something?
Thanks


inequalities
integration
calculus
real analysis
real analysis unsolved

[size=150]Which is greater, √(∫[1,2] x^x dx) or ∫[1,2] √(x^x) dx?[/size]

Using Chebyshev inequality:

http://mathworld.wolfram.com/ChebyshevIntegralInequality.html

and setting $ n \equal{} 2$, $ b \equal{} 2$, $ a \equal{} 1$, $ f_1 (x) \equal{} f_2(x) \equal{} x^\frac {x}{2}$

we get

$ \sqrt {\int^2_1x^x\,dx} \geq \int^2_1\sqrt {x^x}\,dx$

It's easy to prove that $ x^\frac{x}{2}$ is nonnegative and monotonic increasing.

This is a classic Jensen; $ \sqrt{x}$ is concave, so the average of the square roots is less than the square root of the average.

Where can I find a proof of Chebyshev Integral Inequality, and what is a classic Jension?  Thanks.

[url=http://www.artofproblemsolving.com/Wiki/index.php/Jensen%27s_inequality]Jensen's inequality[/url].

It generalizes to integrals (with total weight 1) as well as discrete averages, with the same proof.


Sorry if this has been asked before, I have looked through the topics but couldn't find it.

I am doing a project that involves writing a lot of matrices and tableaux at the same time.  However, I can't get it them to go across the pages instead of just listing.

To create the tableaux I am using the package young and using the following code,
\[ \begin{Young}
1&2&3&4\cr
5&6\cr
\end{Young}\]

To create matrices I have used the code
\[ \sigma=
\begin{pmatrix}
1 & 2 & 3 & 4 \\
2 & 4 & 1 & 3
\end{pmatrix}\]

Any help would be gratefully received as find it very difficult to wade through the list of over 25 tableaux.

Thanks

Sorry, thought the code would be ok,

Think this should be better:

The tableaux code is [code]$$\begin{Young}
1&4\cr
2\cr
3\cr
5\cr
\end{Young}$$[/code]

and the matrices code is [code]$$\sigma=
\begin{pmatrix}
1 & 2 & 3 & 4 \\
2 & 4 & 1 & 3
\end{pmatrix}$$[/code]

fingers crossed it works!

I'm not sure what the problem is the matrices are side by side by default so that you can write
\[ \begin{pmatrix} 2 & 1 \\
\minus{} 3 & 0 \end{pmatrix} \begin{pmatrix} 3 & \minus{} 2 & 0 \\
1 & 1 & 4 \end{pmatrix} \equal{} \begin{pmatrix} 7 & \minus{} 3 & 4 \\
\minus{} 9 & 6 & 0 \end{pmatrix}
\]
You don't say what packages you are using that may alter this behaviour.


ARML geometry calculus integration trigonometry

1.
In right triangle $ABC$, $\angle C = 90^\circ$, $c=3$, and $a+b=\sqrt{17}$.  Compute the area of the triangle.

2.
If $(a+b):(b+c):(c+a) = 6:7:8$, and $a+b+c=14$, compute the value of $c$.

3.
Let $z$ be a root of $z^5-1=0$, with $z \not= 1$.  Compute the value of

\[z^{15} + z^{16} + z^{17} + \cdots + z^{50}. \]

4.
A regular polygon of $2n$ sides is inscribed in a circle of radius 5.  The sum of the squares of the distances from any point on the circle to the vertices of the polygon is 600.  Compute the value of the integer $n$.

5.
The circles, each of radius 5, lie in perpendicular planes.  They intersect in two points, $X$ and $Y$, thus determining a common chord $\overline{XY}$.  If $XY = 8$, compute the distance between the centers of the circles.

6.
How many positive integers less than 54 are equal to the product of their proper divisors?

7.
Find all positive integers $n$, less than 17, for which $n! + (n+1)! + (n+2)!$ is an integral multiple of 49.

8.
In triangle $ABC$, $a \ge b \ge c$.  If $\displaystyle \frac{a^3+b^3+c^3}{\sin^3{A} + \sin^3{B} + \sin^3{C}} = 7$, compute the maximum possible value of $a$.


Nay moi nguoi, phan "bai toan trong ngay" (Problem of the day) chung ta co the xem loi giai o dau vay?  :?:
Cam on

Click vo phan Discuss problem of the day sau bai toan


[quote="ZyXel"]Given triangle ABC. Let Y i Z be the feet of the altitudes from B and C. Suppose that the bisectors of angles BYC and BZC meet at X. Prove that triangle BXC is isosceles.[/quote]
BZXY inscriptible, XYZC inscriptible, because <BZX=<BYX, <CZX=<CYX .
Then B,Z,X,Y,C conclics , <BZX = <XYC -> XB=XC, then BXC issoceles

[hide]( It is only to [b][size=100]ZyXel[/size][/b] and only because of he posted this problem in unsolved area ).

For every circle $ (C)$ taken as cord the segment $ BC,$ which intersects the sidelines $ AC,\ AB$ at points $ Y,\ Z$ respectively, we can also say  that the angle bisectors of $ \angle BYC,\ \angle BZC$ meet at the midpoint $ X$ of the arc $ BC$ of $ (C),$ not containing the points $ Y,\ Z.$[/hide]
Kostas Vittas.

Thx for your comment. I had put it in "unsolved area" by mistake. My solution is same as aev5peru`s and I can say that it`s easy problem. :roll:


geometry
incenter
circumcircle
trigonometry
geometry proposed

Hi,

R. Honsberger in [i]From Erdos to Kiev[/i], pp 199-201 treats a problem from [i]Crux Mathematicorum[/i], 1987, 160:

"As usual, let I and O be the incenter and circumcenter, respectively, of triangle ABC. Suppose angle C is 30 degrees, and that the side AB is laid off along each of the other two sides to give points D and E so that

EA = AB = BD.

Prove that the segment DE is both equal and perpendicular to IO."

Now, I think that the problem is a mixture of two:

1. DE is perpendicular to IO,
2. DE = IO.

The first one is independent of the magnitude of angle C: DE is perpendicular to IO anyway. The second one is the consequence of the angle requirement.

As a preliminary, the construction of points D and E can be repeated for the other two sides AC and BC, giving three lines. Perhaps, curiously, the three are parallel, see

http://www.cut-the-knot.org/Curriculum/Geometry/ThreeParallelsInTriangle.shtml

Best,
Alexander Bogomolny

Let D be on CB, E on CA. Let S, T be the tangency points of $(I)$ with CB, CA, let M, N the midpoints of CB, CA and let d = IO.

$\angle SIO+\angle TIO = \angle SIT = 180^\circ-\angle C = \angle CDE+\angle CED$

$\sin \widehat{SIO}= \frac{MS}{IO}= \frac{|b-c|}{2d}\,\ \ \sin \widehat{TIO}= \frac{NT}{IO}= \frac{|a-c|}{2d}$

$\frac{\sin \widehat{SIO}}{\sin \widehat{TIO}}= \frac{|b-c|}{|a-c|}= \frac{CE}{CD}= \frac{\sin \widehat{CDE}}{\sin \widehat{CED}}$

It follows that $\angle SIO = \angle CDE,\ \angle TIO = \angle CED.$ Since $IS \perp BC$ and $\angle SIO = \angle CDE,$ $DE \perp IO.$ Moreover,

$\frac{DE}{CE}= \frac{\sin C}{\sin{CDE}},\ \ DE = CE \frac{\sin C}{\sin \widehat{SIO}}= |b-c| \cdot \frac{2d}{|b-c|}\sin C = 2d \sin C$

Thanks for that too. It quite nicely avoids the trouble of having to consider signed lengths and angles, except perhaps in the final step of the proof of orthogonality.


For a given N, consider the problem of choosing a sequence of positive integers with sum N so as to maximise their product. Call this maximum product P(N).

[i]For example,
P(5) = 2 x 3 = 6
P(6) = 3 x 3 = 9
P(7) = 3 x 4 = 12[/i]

[b]Calculate P(1000).[/b]

[hide="Hint"]
Consider the inequality $k\leq 2(k-2)$
[/hide]

threes are good. The answer should be $3^{332} \cdot 2^2$ btw

BTW, just in case u want an elaborate proof or whatever, look at IMO 1976, Day 2, problem 1

Ideally, you would want to include as many $e$s as possible, but since you need integers, you want to use the next best thing, which is 3.  Once you can't use any more 3s, you use 2s.  And that's how you get the answer Iversonfan2005 gave you.

Firstly note that there cannot be any ones in the maximal sequence, since $1\cdot k < (k+1)$, so there would be another sequence with 1,k replaced by (K+1) which would then be greater than the maximal sequence.

For $k \geq 4$, we have $k \leq 2(k-2)$. So any number greater than or equal to 4 need not appear in the maximal sequence (since 2(k-2) has an equal or greater product and the same sum).

So we can only have twos and threes in our sequence. Since $3^2 > 2^3$, we cannot have three or more twos in the maximal sequence (they could be replaced with two threes, having a greater product and the same sum).

We conclude that the maximum sequence consists of as many threes as possible and then just enough twos to make up the remainder mod 3.

$1000 = 332\cdot 3 + 2\cdot 2$
$P(1000) = 3^{332} \cdot 2^2$

First consider a sequence of [i]n[/i] numbers adding up to S: $a_1 + a_2 + ... + a_n = S$.
We want to maximise their product $p = a_1 a_2 ... a_n$.

The AM-GM inequality states that
$\frac{a_1 + a_2 + ... + a_n}{n} \geq \sqrt[n]{a_1 a_2 ... a_n}$ with equality when $a_1 = a_2 ... = a_n$.

Therefore
$\left(\frac{S}{n}\right)^n \geq p$. Since the left hand side is a constant for given n, S, we have the maximum value of p when equality occurs (when $a_1 = a_2 ... = a_n = \frac{S}{n}$). So if we split our list of sum S into n numbers, the maximum possible product is $p = \left(\frac{S}{n}\right)^n$ achieved when every number is equal to $\frac{S}{n}$.

Now we just to decide how many numbers to split our sum into. We maximise p as a function of n:
$p = \left(\frac{S}{n}\right)^n$
$\frac{dp}{dn} = (\ln s - \ln n - 1)\left(\frac{S}{n}\right)^n$
$\frac{dp}{dn} = 0$ when $\ln n = \ln s - 1 \rightarrow n = \frac{S}{e}$ (in this case all the numbers in our list would be equal to [i]e[/i]).

But since we require n to be an integer, we have to round $\frac{S}{e}$ up or down and see which way gives the best result for each individual case (It's not necessarily the [i]nearest[/i] integer because the graph p,n is not symmetrical across the turning point). We then split S into n equal numbers.

For example, if S = 10: $\frac{S}{e} = 3.678...$
$n = 4: P = \left(\frac{10}{4}\right)^4 = 39.0625$
$n = 3: P = \left(\frac{10}{3}\right)^3 = 37.0359$

So the best way to split 10 is $2.5^4$


im not sure if this goes here:

i was reading Music of the Primes, and I read this passage about a guy that figured out that the sum of all the numbers from $ 1\minus{}\infty\equal{}\minus{}\frac{1}{12}$
i know it seems outrageous, but apparently it can be proven when you do the following:
$ 1\plus{}2\plus{}3...\infty\Longrightarrow 1\plus{}\frac{1}{2^{\minus{}1}}\plus{}\frac{1}{3^{\minus{}1}}\plus{}...\infty$
i still wasn't sure how that helped prove the sum tended to $ \minus{}\frac{1}{12}$. so i talked to a friend, and he said it might have something to do with the fact that when you write the sum $ 1\plus{}2\plus{}3..$ the other way, you get a harmonic series with $ k\equal{}\minus{}1$.

what do u guys think? does anybody have any ideas on this?

$ S\equal{}1\plus{}2\plus{}3\plus{}4...\equal{}\zeta (\minus{}1)$
Normally this wouldn't converge (outside the domain for which the Dirichlet series for zeta function converges), but if we use the method of Ramanujan summation (Euler-Maclaurin summation formula), we'll get $ \minus{}\frac{1}{12}$.

Read more on Wikipedia.


I think there are only two pairs of $ (x,\ y) \equal{} (3,1),\ (5,3)$.

Rough sketch of the solution:

For $ x\geq 6\Longrightarrow y \equal{} 16n \plus{} 1$ for some $ n\in\mathbb{Z}$.

Consider mod 17 would give a solution, I think.


Let $ a,b,c$ be nonnegative real numbes. Prove that

$ \sqrt {\frac {a}{a^2 \plus{} bc}} \plus{} \sqrt {\frac {b}{b^2 \plus{} ca}} \plus{} \sqrt {\frac {c}{c^2 \plus{} ab}}\ge2\sqrt{\frac2{a \plus{} b \plus{} c}}$

if variables positive we will prove that RS$ \ge 3\sqrt {\frac {3}{2(a + b + c)}}$ its stronger because $ 3\sqrt {\frac {3}{2}}\ge2\sqrt {2}$ if $ A = \frac {a}{a + b + c},B = \frac {b}{a + b + c},C = \frac {c}{a + b + c}$ let $ F(x) = \sqrt {\frac {x^2}{x^3 + abc}}$ then $ A + B + C = 1$ and our ineq $ F(A) + F(B) + F(C)\ge3\sqrt {\frac {3}{2}}$ $ F''(x)\ge0$ then by Jensen ineq $ F(A) + F(B) + F(C)\ge 3F(\frac {1}{3}),$ $ abc\le \frac {1}{27}(a + b + c)^3 = \frac {1}{27}$ then
$ 3F(\frac {1}{3})\ge3\sqrt {\frac {3}{2}}$ if one of variables =0 then $ \sqrt{\frac{1}{a}}+\sqrt{\frac{1}{b}}\ge 2\sqrt{\frac{2}{a+b}}$

[quote="dima ukraine"]we will prove that RS$ \ge 3\sqrt {\frac {3}{2(a \plus{} b \plus{} c)}}$ [/quote]
Try $ a\equal{}b\equal{}1$ and $ c\equal{}0.$

[quote="dduclam"]Let $ a,b,c$ be nonnegative real numbes. Prove that

$ \sqrt {\frac {a}{a^2 \plus{} bc}} \plus{} \sqrt {\frac {b}{b^2 \plus{} ca}} \plus{} \sqrt {\frac {c}{c^2 \plus{} ab}}\ge2\sqrt {\frac2{a \plus{} b \plus{} c}}$[/quote]
It is equivalent to
\[ \sum \frac{1}{\sqrt{x^2\plus{}yz}} \ge \frac{2\sqrt{2}}{\sqrt{xy\plus{}yz\plus{}zx}}\]
my friend.
And I have posted my solution on this forum ;)

[quote="dduclam"][quote="can_hang2007"][quote="dduclam"]Let $ a,b,c$ be nonnegative real numbes. Prove that

$ \sqrt {\frac {a}{a^2 \plus{} bc}} \plus{} \sqrt {\frac {b}{b^2 \plus{} ca}} \plus{} \sqrt {\frac {c}{c^2 \plus{} ab}}\ge2\sqrt {\frac2{a \plus{} b \plus{} c}}$[/quote]
It is equivalent to
\[ \sum \frac {1}{\sqrt {x^2 \plus{} yz}} \ge \frac {2\sqrt {2}}{\sqrt {xy \plus{} yz \plus{} zx}}
\]
my friend.
And I have posted my solution on this forum ;)[/quote]
You're right, Can :)
Exist a nice solution only with AM-GM for it.  :wink:[/quote]
Yeah, I also posted my AM-GM and Cauchy Schwarz proofs for it here. ;)


function
geometry
geometric transformation
reflection
complex numbers
superior algebra
superior algebra unsolved

In  $ R^2$ let
$ (a,b)\plus{}(c,d) \equal{} (a\plus{}c,b\plus{}d)\\
(a,b)*(c,d)\equal{}(ac\plus{}tbd,ad\plus{}bc)$
show that  if it is a filed isomofphic to a complex number field then $ t\equal{}\minus{}1$

Let's think about a function $ f : R^2 \ - > C$ is a bijection such that $ f{(a,b)} = a + bi$. If $ f$ is isomorphic function, $ f{(a,b)*(c,d) = {f(a,b)}*{f(c,d)}}$. The left hand side is $ f{(ac + tbd,ad + bc)} = (ac + tbd) + (ad + bc)i$, by your assumption. The right hand side is $ (a + bi)*(c + di) = (ac - bd) + (ad + bc)i$, by definition of producting complex numbers. By definition of the equality of complex numbers, we get $ t = - 1$.

 This is one way to generate complex number field from real number field.

J.Y.Choi: You have only shown that such an isomorphism isn't $ f(a,b)\equal{}a\plus{}bi$ if $ t \neq \minus{}1$. You may have something like $ f(a,b)\equal{}a\plus{}2bi$ etc., see below.

wiosna: The claimed result is wrong, you have an isomorphism to $ \mathbb C$ iff $ t<0$; it is then given by $ (a,b) \mapsto a\plus{}bi \sqrt{\minus{}t}$. That ring you defined there is $ \mathbb R[x]/(x^2\minus{}t)$ in general.

It's false. You can infer the correct result from my sci.math post
http://groups.google.com/group/sci.math/msg/6a9d3993f60bc4e0
http://google.com/groups?selm=y8z8wuxlbnn.fsf%40nestle.csail.mit.edu

[quote="J.Y.Choi"]Ah, I understand. Then, what is the right proof?[/quote]
Take a look at my previous post for one direction. For the other direction, show that for $ t\geq 0$ the constructed ring isn't even a field (there are too many solutions for $ x^2\equal{}t$).

billwgd: A bit less links to sci.math would sometimes be nice  :wink:
The motivation was probably just the lack of negative square root, the one of $ \minus{}1$ being the canonical one; then looking for the smallest meaningfull thing containing $ \mathbb R$ and $ \sqrt{\minus{}1}$ gives the complex numbers.

[quote="ZetaX"] billwgd: A bit less links to sci.math would sometimes be nice  :wink: .[/quote]

That's a bizarre remark. Why do you dislike links? Helpful links should be encouraged, not discouraged.
Many questions are FAQs and are most efficiently answered via appropriate links. My posts almost
always contain links, and most readers seem to enjoy learning what they find there (judging by the
many replies I've received). Of course YMMV....

[quote="ZetaX"]for $ t\geq 0$ the constructed ring isn't even a field [/quote]

Why it is true?

HINT $ \;(x\minus{}s)(x\plus{}s) \equal{} 0$ in $ \mathbb{R}[x]/(x^{2}\minus{}s^2)$

wiosna: I think you should try to think a little bir for yourself.

billwgd: I agree that helpfull links should be encouraged. It's just that it's sci.math and it's a bit hard to read; I would have prefered to see this posted on the forum in TeX etc.

[quote="ZetaX"]billwgd: I agree that helpfull links should be encouraged. It's just that it's sci.math and it's a bit hard to read; I would have prefered to see this posted on the forum in TeX etc.[/quote]

Occasionally I do reformat important posts, e.g. compare the MathLinks vs. sci.math versions of my post [1] on a little-known ideal-theoretic generalization of Euclid's classic proof that there are infinitely many primes. Notice that I do go to great pains to format even my plain-text posts to reflect innate mathematical structure, e.g. vertically aligning derivations so that syntax reflects semantics. Thus the sci.math version is almost as readable as the MathLinks LaTeXed version. But with many thousands of prior posts on plain-text forums such as sci.math, Ask an Algebraist, etc, this is not feasible generally.

--Bill Dubuque

[1] [b]Infinite fewunit rings have infinitely many prime ideals[/b]
http://www.mathlinks.ro/viewtopic.php?p=1209616
http://google.com/groups?selm=y8zk5f3rn4e.fsf%40nestle.csail.mit.edu


A parking lot for compact cars has 12 adjacent spaces, of which 8 are already occupied. A large SUV arrives, needing two adjacent open spaces. What is the probability that it will be able to park?

So I see that this is a modification of the AMC 12 problem from this year.

[hide]The SUV will not be able to park if none of the 4 empty spaces are adjacent to each other. To find the probability that the SUV will be able to park, we can find the probability that it will NOT be able to park, and subtract this probability from $ 1$.

Imagine $ 8$ occupied spots in a row. Now, there are $ 9$ places where we can insert empty spaces. Thus, there are $ \binom{9}{4}$ ways for the SUV not to be able to park, and there are $ \binom{12}{4}$ ways to pick any four spaces out of 12 spaces.

The requested probability is then $ 1 \minus{} \frac {\binom{9}{4}}{\binom{12}{4}} \equal{} 1 \minus{} \frac {9 \cdot 8\cdot 7 \cdot 6}{12 \cdot 11 \cdot 10 \cdot 9} \equal{} 1 \minus{} \frac {14}{55} \equal{} \boxed{\frac {41}{55}}$[/hide]


I am new to geometry, and I was hoping that someone would be willing to write me a couple of beginners basic type proofs, son that I can improve in that area of math.

                                                         -Thanks

I've posted these two before, but I think they're pretty basic problems so here they are...
~Prove that no right isosceles triangle can have all integer side lengths.
~Prove that a scalene triangle with a side length of 1 cannot have both of its other sides be integers.

Okay, here's an easy one:

Prove that, if a transversal cuts two parallel lines, then the opposite exterior angles are congruent, i.e. prove that <BCA=<HFG:
(Note that, in the figure, AH is not necessarily perpindicular to BD and EG, but BD is parallel to EG.)
[code]
     |
     A
     |
--B--C--D--
     |
     |
--E--F--G--
     |
     H
     |
[/code]

<ACB and <BCF are supplimentary, therefore 180-(<ACB)=<BCF.
<BCF and <CFE are alternate interior angles, therefore 180-(<BCF)=<CFE.
Combining the previous two statements using substitution, we get 180-(180-(<ACB))=<CFE, which simplifies to <ACB=<CFE.
Since <CFE and <HFG are vertical angles, they are congruent, giving us <ACB=<CFE=<HFG.
Since congruence is transitive, we now know that <ACB=<HFG.

Basically, think of a proof as a very detailed showing of your work.

This is pretty easy: Prove that any angle inscribed in a semicircle is right.

 :)

Umm, here's a really easy one:

Prove that a pentagon must have 540 degrees.

Perhaps you might want to try proving that the sum of the interior angles of a triangle? And then, once you've done that, prove that the formula for the sum of the interior angles of an [i]n[/i]-gon is $180(n-2)$. The former is good practice with drawing auxilary lines, and the latter good for practicing applying the former.

Hope you get proofs down! They're fun once you get used to them. ;)

So many people suggesting a lot of things but I wonder if AGeneson checked back his replies yet...
Anyways, give out suggestions with proofs please since those are the core skills in math.

Alrighty. For the proof about the sum of the interior angles of a triangle, here's an example:

[code]X--A--Y
  / \
 /   \
 B---C[/code]

Let $ABC$ be a triangle such that $\theta$ is equal to angle A, $\phi$ equal to angle B, and $\gamma$ equal to angle C. Using the postulate that states that given a line and a point not on the line, there is exactly one line through that point parallel to the original line, construct line j parallel to BC through A, with X and Y being unique points on j such that X is "to the left" of A, and Y to the "right." Then, since alternate interior angles of parallel lines are congruent, <BAX=$\phi$, and <CAY=$\gamma$. Since XAY is a straight line, its degree measure is 180, so $\phi+\theta+\gamma=180$, making the sum of the angles of a triangle 180.

Prove that the angle bisectors of a square are perpendicular bisectors.


Let $a,b,c,x,y,z$ be real numbers so that satifying the following system:
$\left\{\begin{array}{c}aa+b+c=0x+y+z=0\\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\end{array}\right$.
Calculate value of expression $A=xa^{2}+yb^{2}+zc^{2}$

Are you sure there are no typos?  The expression does not appear to be "nice" to me, i.e. I think all one can do is eliminate two of the variables and plug in.


Putnam
function
integration
calculus
college contests

Suppose that $f(x,y)$ is a continuous real-valued function on the unit square $0\le x\le1,0\le y\le1.$ Show that

$\int_0^1\left(\int_0^1f(x,y)dx\right)^2dy + \int_0^1\left(\int_0^1f(x,y)dy\right)^2dx$

$\le\left(\int_0^1\int_0^1f(x,y)dxdy\right)^2 + \int_0^1\int_0^1\left[f(x,y)\right]^2dxdy.$

I'll rebuild the solution I posted here before:

First, we convert the integrals into an alternate form:
$\int_0^1\int_0^1\left[f(x,y)\right]^2dxdy= \int\int\int\int_{[0,1]^4}f(x,y)f(x,y)dx dy du dv$
$\left(\int_0^1\int_0^1f(x,y)dxdy\right)^2= \int\int\int\int_{[0,1]^4}f(x,y)f(u,v)dx dy du dv$
$\int_0^1\left(\int_0^1f(x,y)dx\right)^2dy = \int\int\int\int_{[0,1]^4}f(x,y)f(u,y)dx dy du dv$
$\int_0^1\left(\int_0^1f(x,y)dy\right)^2dx = \int\int\int\int_{[0,1]^4}f(x,y)f(x,v)dx dy du dv$

(Fubini's theorem is applied freely here, since everything is continuous and bounded)

Now we want to show that
$\int_0^1\int_0^1\left[f(x,y)\right]^2dxdy+ \left(\int_0^1\int_0^1f(x,y)dxdy\right)^2$
$-\int_0^1\left(\int_0^1f(x,y)dx\right)^2dy -\int_0^1\left(\int_0^1f(x,y)dy\right)^2dx \ge 0$
In the four-variable form, this is
$\int\int\int\int_{[0,1]^4}f(x,y)[f(x,y)-f(x,v)-f(u,y)+f(u,v)]dx dy du dv=A$
that we wish to prove is positive. Routine substitutions (interchanging $x$ and $u$, interchanging $y$ and $v$) give that
$A=\int\int\int\int_{[0,1]^4}f(u,y)[-f(x,y)+f(x,v)+f(u,y)-f(u,v)]dx dy du dv$,
$A=\int\int\int\int_{[0,1]^4}f(x,v)[-f(x,y)+f(x,v)+f(u,y)-f(u,v)]dx dy du dv$,
$A=\int\int\int\int_{[0,1]^4}f(u,v)[f(x,y)-f(x,v)-f(u,y)+f(u,v)]dx dy du dv$.
Adding these up, $4A=\int\int\int\int_{[0,1]^4}[f(x,y)-f(x,v)-f(u,y)+f(u,v)]^2dx dy du dv$, and this is clearly positive.

We have equality precisely when we can write $f(x,y)=g(x)+h(y)$.

Thanks Kent! I've working on this problem for two hours, later solved jointly with another professor. To me it's really hard.

One [url=https://www.zhihu.com/question/396944680/answer/1242792430]Chinese solution[/url] is available (maybe interesting) as well.

$=\int_{x_1=0}^1 \int_{y_1=0}^1 \int_{x_2=0}^1 \int_{y_2=0}^1 (f(x_1,y_1)+f(x_2,y_2)-f(x_1,y_2)-f(x_2,y_1))^2 dx_1 dy_1 dx_2 dy_2$

Motivation: We talk about how to prove the Cauchy-Schwartz inequality, namely

$$(\int_0^1 f(x) dx)^2 \le \int_0^1 f(x)^2 dx$$

Well, the proof is just expansion, symmetry, then factorization:

$$(\int_0^1 f(x) dx)^2 = \int_0^1 \int_0^1 f(x)f(y) dx dy$$ (1)

$$(\int_0^1 f(x)^2 dx) = \int_0^1 \int_0^1 f(x)^2 dx dy =  \int_0^1 \int_0^1 f(y)^2 dx dy =  \int_0^1 \int_0^1 \frac{f(x)^2+f(y)^2}{2} dx dy$$ (2)

Subtracting (2)-(1) yields the problem statement is equivalent to $$\int_0^1 \int_0^1 \frac{(f(x)-f(y))^2}{2} dx dy \ge 0$$

This problem is similar.


integration
real analysis
real analysis unsolved

Let $ z$ is complex number  and $ Re z>0$ prove that
$ \sum_{n \geq N}e^{-z\sqrt{n}}=\frac{2}{z}e^{-z\sqrt{N}}(\sqrt{N-\frac{1}{2}}+\frac{1}{z}+z^{2}O(1+\frac{|z|^{2}}{\sqrt{N}}))$
where  constant from $ O$  not include $ z$ and $ N$

Small point of note, I believe you want a capital $ N$ in the exponent of the exponential on the right side rather than a lower-case $ n$ since that is a dummy index.

Also, are you sure about the leading order term?  Just with quick scribbling I get the following:

$ \sum_{n \geq N}e^{-z \sqrt n}\sim \int_{N}^{\infty}e^{-z \sqrt x}dx$ and since $ \sqrt x = \sqrt N+\frac{1}{2 \sqrt N}(x-N)+\cdots$ we get that the sum is asymptotic (to leading order) to:

$ e^{-z \sqrt N}\int_{N}^{\infty}e^{\frac{-z}{2 \sqrt N}(x-N)}dx = \frac{2}{z}e^{-z \sqrt N}\sqrt N$

I edit $ N$ thanks.//
About term: yes I'm sure.
this step is not so exact:  $ \sum_{n \geq N}e^{-z \sqrt n}\sim \int_{N}^{\infty}e^{-z \sqrt x}dx$
try another


The addition below is incorrect. The display can be made correct by changing one digit $d$, wherever it occurs, to another digit $e$. Find the sum of $d$ and $e$.
\[\begin{tabular}{ccccccc}& 7 & 4 & 2 & 5 & 8 & 6 \\ + & 8 & 2 & 9 & 4 & 3 & 0\\ \hline 1 & 2 & 1 & 2 & 0 & 1 & 6 \end{tabular}\]
$\textbf{(A)}\ 4 \qquad
\textbf{(B)}\  6 \qquad
\textbf{(C)}\ 8 \qquad
\textbf{(D)}\ 10 \qquad
\textbf{(E)}\ \text{More than } 10$

[hide="Answer"]The first three rows from the right are correct, and the fourth looks correct. However, the fifth and sixth are clearly wrong. The only way that the sixth can be fixed while at the same time fixing the fifth is by changing $2$ to $6$, which will still work with the fourth, so we have $d+e=2+6=8\Rightarrow \boxed{C}$.[/hide]

[hide]Obviously, the 3 rows from the left are where the problem occurs.

Changing 2 to 6 yields to desired result so d+e =2+6 =8  $\boxed{C}$[/hide]

[quote="lightning"]Sorry for reviving an old post (is it allowed for ahsme problems?), but I do not understand this problem...

Why is the first two digits of the answer "12"? because would there be two errors instead of one then?[/quote]The problem says the addition is incorrect but that doesn't mean there is only one error. There can be many errors and that's what this problem is about, because the erroneous digit has multiple instances in the sum.

[quote="lightning"]ok, so the purpose of the problem was the digit that causes the most errors?

why couldn't it be changing the "4" in 742586 into "8"?[/quote]The problem didn't specifically say 'most errors', even though it just happened to be most in this problem. Digit $ d$ is making the whole sum incorrect. This $ d$ can appear many times in the addition, not just simply one and only one digit. See what I mean? 4 would not work because if you change the 4 in the second summand to 8, the addition would still be incorrect.

EDIT: The idea is to change every instance of digit $ d$ to digit $ e$ to make the addition correct.

ohhh, ok, i get it now.

so every 2 should be changed to a 6 to be correct....

Thank you. That was really helpful. :)


algebra
polynomial
quadratics
superior algebra
superior algebra unsolved

Let $ p$ be prime, $ a_0,a_1,...,a_k$ integers, and let $ \{x_i\}$ be a sequence such that for all $ i$, we have
\[ x_{i \plus{} k \plus{} 1} \equal{} \sum_{j \equal{} 0}^k a_j x_{i \plus{} j}
\]
Suppose, furthermore, that the polynomial $ x^{k \plus{} 1} \minus{} a_k x^k \minus{} ... \minus{} a_1 x \minus{} a_0$ is irreducible over $ \mathbb{F}_p$.

Prove that for all $ i$, we have
\[ x_i \equiv x_{p^{k \plus{} 1} \minus{} 1 \plus{} i} \mod{p}
\]
[hide="very big hint"]finite fields[/hide]

[hide="concrete example"]What if $ k \equal{} 1$ and $ a_0,a_1 \equal{} 1$?[/hide]

It's a fun result.  The hint gives it away, of course. :)

With the Fibonacci numbers, you can do a little better... I believe the period always divides $ p\minus{}1$ when $ p$ is a square $ (\rm{mod}\ 5)$ and $ 2(p\plus{}1)$ when it isn't.

Ok, I changed it to "very big hint." I'm also thinking of ways to phrase this problem without saying [quote]Suppose, furthermore, that the polynomial $ x^{k \plus{} 1} \minus{} a_k x^k \minus{} ... \minus{} a_1 x \minus{} a_0$ is irreducible over $ \mathbb{F}_p$. [/quote] That is, formulating the problem so that it's not obvious that it involves abstract algebra  :wink:
You could set the $ a_i$ to a specific example that is always irreducible, or you could replace $ p^k \minus{} 1$ with $ \prod_{i \equal{} 1}^{k} (p^i \minus{} 1)$ and work over the splitting field of a possibly reducible polynomial or something like that.

See [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=229442]here[/url] for the result about the Fibonacci numbers.  By the way, the result is repairable when the polynomial isn't irreducible; one just works over its splitting field, which is not necessarily $ \mathbb{F}_{p^k}$ but may be smaller.  It should be possible to strengthen this result at least as much as the Fibonacci result is strengthened.

[quote="t0rajir0u"]It should be possible to strengthen this result at least as much as the Fibonacci result is strengthened.[/quote]

For quadratic sequences, I'd believe it.  But what do you have in mind for higher values of $ k$?


inequalities
monovariant
combinatorics unsolved
combinatorics

Well I think this one is very very hard and I am not sure whether it belongs to combinatorics or to something else (inequalities, algebra?), but I think it doesn't matter:

Let $a_1,\dots,a_{n+1}$ be $n+1$ real numbers from [0,1]. Prove that one can divide numbers $a_1,\dots,a_{n+1}$ into two groups in such a way that the distance between the arithmetic means of the numbers of the first and the numbers of the second group is not greater than $1/n$.

I apologize for my formulation probably isn't so nice, but I hope you can understand problem.

So anyone has idea?

[quote="Wumbate"]Here is a way to do it, but I admit it is not very thorough:
Start by putting the highest number into set A and the lowest into set B.  Then repeat by putting the lowest into set A and the highest into set B.  Repeat the process until you are out of numbers (i.e. you had an even amount) or you only have 1 number left (an odd amount).  Here is the proof for an even amount:
If you had an even amount, the average of set A will be: $(x_1+x_3+...+x_n)/(n/2)$ and the average of set B will be: $(x_2+x_4+...+x_{n+1})/(n/2)$, with $1>=x_1>=x_2>=x_3>=...>=0$.

I got stuck.  I don't really know how to explain it mathematically...

Another idea would be to start out and switch a number from one set to another, and show that the difference between the averages is a monovariant that must get lower if you switch it right...[/quote]

Yes, that's the trivial part of it. Another way is putting the highest & the lowest in A, the second highest & second lowest in B, the third highest and third lowest in C, etc.

The requirement $\in [0,1]$ must be used in somewhere...

By the way, are you sure the problem is correct?

Say n=3

If $a_1=0, a_2=1, a_3=3, a_4=1$

Than the geometric average must be divided 0-1

of which the difference is above 1/3? :?

In case of $a_1=0,a_2=a_3=a_4=1$ group $A$ can be $0,1,1$, and group $B$ is 1. Then we have
$(0+1+1)/3=2/3$ and $1-2/3\leq1/3$, so in that case the statement really holds. I think it
is always true. Anyway it is clear that something like that MUST be true. The question
is only to find out whether $1/n$ is the correct estimate or not. It looks like the case
$a_1=0, a_2=\dots=a_{n+1}=1$ is the worst. In that case $1/n$ obviously cannot be
replaced by a smaller number. We see that the above problem has a natural foundation.

oh Arithmetic mean :D

yes sorry, totally misread :P

ok i'll give it a try as soon as possible, this seems not so hard.


geometry
3D geometry
sphere
circumcircle
geometry solved

Consider a circle, and a fixed point $A$ inside the circle. If $BC$ is a chord of the circle such that $\angle BAC =\frac{\pi}{2}$, and $M$ is the midpoint of this chord, what is the locus of $M$ as the chord $BC$ moves around on the circle?

Here's an analog of this problem in space: if $P$ lies inside a sphere and $PX,PY,PZ$ are variable and mutually perpendicular rays from $P$ ($X,Y,Z$ are on the sphere), what's the locus of $Q$, the vertex opposite to $P$ of the parallelepiped constructed on $PX,PY,PZ$?

[i]Solution to the initial problem:[/i]

I reformulate your problem:

[i]Let A be a fixed point inside a circle k. Consider all chords BC of the circle k such that < BAC = 90°. Find the locus of the midpoints of all of these chords.[/i]

In fact, if O is the center and r is the radius of k, and N is the midpoint of the segment AO, then the required locus is the circle with center N and radius $\frac{1}{2}\sqrt{2r^{2}-AO^{2}}$.

In order to prove this, we will show that for any chord BC satisfying < BAC = 90°, we have $NM=\frac{1}{2}\sqrt{2r^{2}-AO^{2}}$, where M is the midpoint of the chord BC.

The point M is the midpoint of the hypotenuse BC of the right-angled triangle BAC. Therefore, the point M is the circumcenter of this triangle; this implies AM = BM. Now, since N is the midpoint of the segment AO, the segment NM is a median in the triangle AMO; hence, by the well-known formula for the length of a median in a triangle, we have

  $4\cdot NM^{2}=2\cdot AM^{2}+2\cdot OM^{2}-AO^{2}$.

Since AM = BM, this becomes

  $4\cdot NM^{2}=2\cdot BM^{2}+2\cdot OM^{2}-AO^{2}=2\cdot \left( BM^{2}+OM^{2}\right)-AO^{2}$.

Now, if we connect the center of a circle with the midpoint of a chord, the resulting line is perpendicular to the chord. Hence, the line OM is perpendicular to the chord BC, and in the right-angled triangle BMO we can apply the Pythagoras Theorem: $BM^{2}+OM^{2}=BO^{2}=r^{2}$. Thus, $4\cdot NM^{2}=2\cdot r^{2}-AO^{2}$ and $NM=\frac{1}{2}\sqrt{2r^{2}-AO^{2}}$, qed..

  Darij


I'm needing documents  about solve standardize of Inequality
If anyone have it, let's send link for me
thank you very much

[quote="TUYEN_LOVER"]when
$ (abc)^2 \leq{1}$
prove that
$ \frac {a^2}{a^2 \plus{} 2} \plus{} \frac {b^2}{b^2 \plus{} 2} \plus{} \frac {c^2}{c^2 \plus{} 2} \leq{1}$
copy right of meR◙╡♫♫  :)[/quote]
Try $ a\equal{}b\equal{}2$ and $ c\equal{}0.$ :wink:

[quote="TUYEN_LOVER"]when a,b,c >0 and
$ (abc)^2 \leq{1}$
prove that
$ \frac {1}{a^2 \plus{} 2} \plus{} \frac {1}{b^2 \plus{} 2} \plus{} \frac {1}{c^2 \plus{} 2} \leq{1}$
copy right of meR◙╡♫♫  :)
 :D  sorri[/quote]

Try $ a\equal{}b\equal{}c\equal{}\frac{1}{2}$  :wink:

was changed...........
but Why it's not reason with Svac :(

when get 1/2.(3- Left hand side of that equation) $ \leq$ 1/2.(3-1)
If folow Svac , It'll  = or < 1 :(
WHY???

[quote="TUYEN_LOVER"]when a,b,c >0 and
$ (abc)^2 \leq{1}$
prove that
$ \frac {1}{a^2 \plus{} 2} \plus{} \frac {1}{b^2 \plus{} 2} \plus{} \frac {1}{c^2 \plus{} 2} \geq{1}$
copy right of meR◙╡♫♫  :)
 :D  sorri
It's false ,isn't it? :([/quote]

Try $ a \equal{} b \equal{} 2,c \equal{} \frac {1}{4}$  :wink: