Hmm.. I haven't actually seen this written down officially anywhere, but I suppose someone has defined it.
For what real a is (-2)^a defined?

This question needs concepts that are beyond mathcouts level, I believe.  Like converging sequences.

Okay, I was thinking reals.  But either way, it's still not for mathcounts.

Actually it's not that difficult and the definition of exponents are very important in MATHCOUNTS.

Answer: [hide]When a is the multiplicative inverse of an even number[/hide]

Er, I don't know why it's wrong...Let me restate it: [hide]For all a element of real number (-2)^a is an element of real number if and only if a is the multiplicative inverse of an even number, excluding 0.[/hide]

It's wrong because it fails in every case.  (-2)^(1/2), the simplest case of those you suggested, is not a real number.

Hmm.. I was thinking of non complex-stuff. What I thought was that a had to be rational with its denominator odd (in lowest terms) if a is positive or numerator odd if a is negative.. maybe I was wrong..

But how do you prove, say, if (-2)^:rt2: is real or not?  On the rational numbers, you're correct, but it's not clear what to do on irrational numbers.

Oh whoops, I meant it's [b]not[/b] defined for the cases I mentioned, so the answer would be "For all a element of real number (-2)^a is not an element of real number if and only if a is the multiplicative inverse of an even number."

Teehee :oops:

I'm still incorrect anyways since irrational power of a negative number is undefined no matter what.

Yes.  But proving that irrational powers are undefined for negative numbers requires knowledge of sequences and such that is definitely not mathcouts level.

That is true...oh, so that's why you said this is beyond MATHCOUNTS level...

...this is harder than I thought :?

[quote="Tare"]
I'm still incorrect anyways since irrational power of a negative number is undefined no matter what.[/quote]

Like 0 of what you said here is correct. Assume w is a complex constant and z != 0, then z^w is defined for all complex z. It's singlevalued when w is an integer, takes finitely many values when w is rational, takes infinitely many numbers in all other cases.

Mmm, I agree. After all this is just MATHCOUNTS level, or close to it, so I doubt complex numbers come into play.

I finally got it right! Whoopy! :)

[quote="JBL"]The problem only wants real numbers, not comlex numbers.  I know it isn't explicitly stated, but in general when you ask a question of this sort, the answers are taken over the field of the real numbers.  So for this question, what Tare said was correct: (-2)^[i]a[/i] will only be defined in real numbers if [i]a[/i] is a rational number with odd denominator.[/quote]

Don't think thats quite right if a is negative :)

It just came up when, like, you are asked to solve something like x^(2/3) = 4, then you go x = 4^(3/2) but then you may miss the negative solution.. or is there a negative solution? Well thats what I was thinking about that particular moment of the day anyway..

You're going to have to explain to me why having a negative [i]a [/i]changes anything here.

[quote="ComplexZeta"]It's defined for all real (and complex) a. Uniquely defined and real definitions are another story.[/quote]

what do ya mean by defined ?


[hide="does this work?"]
let $ f(n)=z_{2n}$
let x=f(n-1),ax=f(n) and abx=f(n+1)
then $ b=a+\frac{2007}{ax^{2}}$
the part on the right rapidly approaches zero, so f(x) should be approximatly geometric
since we are finding the greatest integer less than or equal to
$ \frac{f(1004)}{f(1003)}+\frac{f(1003)}{f(1004)}-\frac{1}{3}$ we just have to find the approximate ratio

for n=2, we have x=1, and a=3, so b=672
then for n=3,  x=3, a=672, so b=672.331...
the ratio should be a bit over 672.3, so
$ \frac{f(1004)}{f(1003)}+\frac{f(1003)}{f(1004)}-\frac{1}{3}\approx672.3+\frac{1}{672.3}-\frac{1}{3}$
the greatest integer less than or equal to that is 671

Addmitably, this is the same reasoning that made me miss this problem on the real AIME
[/hide]

Well done. This is essentially SAME problem as AIME 2007 #14. However, there was one slight variation. As previous poster noted, the answer was actually 1/3 less than what AIME asked (in the format that is).

Now just a thought that came to my mind... Is it possible to answer same question for $ \frac{z^{2}_{2008}+z^{2}_{2006}}{z_{2008}z_{2006}}$?

All we know for that is that it's less than 672.33333333333... I haven't been able to figure this "additional" problem out so any great idea will be appreciated.

wouldn't that be even easier? It would just be 672

Hmm. That's what I think but how do you know how small the difference is? In original AIME problem, we knew that the answer was slightly less than 225 so we choose 224 but like in this case, is there anyway to confirm that the answer is at least 672?

Like... By any chance, can it be something like 671.99999999999999999997?

Then the answer is 671.

I DO think that 672 is right (because I'm assuming the difference is very small from 672+1/3) but this is just a thought... which is not 100%.

what exactly are you asking?

beucase if the question is $ 672.3+\frac{1}{672.3}$, then the answer is obivously 672

I realized something now that I looked into your solution more carefully. It's a lot different from way I solved it.

I solved method that probability used in this thread http://www.artofproblemsolving.com/Forum/viewtopic.php?p=784670&search_id=259307746#784670

Going with what he did, in this problem, my work was:

$ G_{2008}+1/G_{2008}< G_{2008}+1/G_{2006}= 672+1/3$

Thus, that's where my idea came. Honestly, I'm not sure if your idea is 100% accurate.. I don't know, I don't see any definite fault on it but it just feels like it's just "idea" rather than "proved."

well my solution isn't exactly rigorous
For example, I never proved that $ \frac{z_{n+2}}{z_{n}}$ converged


Could anyone tell me what Lambek-moser Theorem is, and how to prove it?
I would be very very thankful if anyone could answer my question.

Look at the following link
[url]http://www.mathsoft.com/mathresources/constants/discretestructures/article/0,,2267,00.html[/url]

Actually, quite a nice and deep result; see [url]http://en.wikipedia.org/wiki/Lambek%E2%80%93Moser_theorem[/url].


vector
linear algebra
matrix
linear algebra unsolved

Let $X$ be a vector space of dimension $n$, $B(X)$ be the space of all endomorphisms of $X$, $T \in B(X)$. Prove that the following statements are equivalent:

(1) all eigenvalues of $T$ are simple;
(2) $(T^{k}x)_{0 \le k \le n-1}$ are linearly independent in $X$ for some $x \in X$;
(3) $(T^{k})_{0 \le k \le n-1}$ are linearly independent in $B(X)$.

\[Let \;T=\begin{pmatrix}0&1\\0&0\end{pmatrix}\]
$T$ satisfies (2),(3) but doesn't satisfy (1).

Indeed.  :blush:  Here is the right one: prove that (1) implies (2) and (3).

[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=minimal&t=84742]Here[/url]'s something a little stronger.


calculus
derivative
Pythagorean Theorem
geometry
calculus computations

I started to learn calculus following a book but I don’t understand the explanation of a derivative example. Here is the excerpt of the book:

[color=darkblue]Then the Pythagorean theorem tells us that

h(t)^2 + b(t)^2 = 13^2

We may differentiate both sides of this equation with respect to the variable t
(which is time in minutes) to obtain

2 • h(t) • h’(t) + 2 • b(t) • b’(t) = 0[/color]

It seems that the author applied the Chain Rule but I am not sure about the intermediary steps of this differentiation. Could someone show me intermediary steps and explain them  ?

Yes, the author applied the chain rule to the expressions $ h(t)^{2}$ and $ b(t)^{2}$.

Let $ g(x) = x^{2}$.  Then

$ h(t)^{2}= g(h(t))$

So that

$ \frac{d}{dt}h(t)^{2}= g'(h) h'(t)$

Where

$ g'(x) = 2x$

So that

$ g'(h) = 2h$

And

$ g'(h) h'(t) = 2h(t) h'(t)$


function
number theory
greatest common divisor
modular arithmetic
prime numbers
number theory proposed

Hi,

I have some difficulty to prove this statement (see the picture)
Is it true?
Thank you for any help or comment.

I checked the statement for some hundreds values it holds.
Even if it is not true for all values p,q there is a way to check numbers by a set of primes.
I tried with 561 (Carmichael number) checking with 89 it shows that 561 is composite (2^560 mod 561=1)
I tried with 341 (Carmichael number) checking it with 89 it shows that 341 is composite. (2^340 mod 341=1)
My goal now it to find the set of prime numbers to use as witness.

If I haven't misunderstood the question , try $ p\equal{}23,q\equal{}11$, the statement will be wrong.

[quote="stephencheng"]If I haven't misunderstood the question , try $ p \equal{} 23,q \equal{} 11$, the statement will be wrong.[/quote]

Thank you for your comment.
I checked 11 and 23.
The statement is wrong. Because 23=2*11 +1 and p=11.
So I have to introduce a condition of co-primality between s and p.
[u]My idea is to check the primality using other known certified prime.[/u] I have found that [b]the twin primes [/b] as numbers for checking out the primality of any dubious number are efficient.  Why? I do not know.
It works often.
I think that there is a way to solve the primality if we exploit this path.
What is sure is that if we find a function f(p,q) easily computable giving an output t [u]if and if only if[/u] p and q are both prime then the primality problem is done.
I have found another function f(p,q) I'm testing it now.

Is this one working under the conditions that (p,s)=(q,t)=1 ?
Thank you for any comment.

Here is the rectified statement

p=2t+1 >3
q=2s+1 >3

gcd(s,p)=1
gcd(t,q)=1

a=(t*s)^(t*s) mod (p*q)

If p and q are both prime then

a mod p = 1 or p-1
a mod q = 1 or q-1

But you have to understand that the statement is the basis of the test because you have to check Carmichael numbers.
That means that if you find the 1 or p-1 and 1 or q-1 you have to check any dubious number by using a set of known certified primes.
That is the new test.
341 is going to pass but if you test with 89 which is a certified prime it is not going to pass.

Sorry the simplified statement I erased now is wrong.
I made a mistake on my calculations.

There's such a thing as an edit button...

[quote="Matharith"]p=2t+1 >3
q=2s+1 >3

gcd(s,p)=1
gcd(t,q)=1

a=(t*s)^(t*s) mod (p*q)

If p and q are both prime then

a mod p = 1 or p-1
a mod q = 1 or q-1[/quote]
The statement is correct.

In the first place, we are going to prove that
\[ a^2 \equal{} (ts)^{\frac {(p \minus{} 1)(q \minus{} 1)}2}\equiv 1\pmod{pq}.
\]In order to do this, it is enough to prove that $ (ts)^{\frac {(p \minus{} 1)(q \minus{} 1)}2}\equiv 1\pmod p$ and $ (ts)^{\frac {(p \minus{} 1)(q \minus{} 1)}2}\equiv 1\pmod q$. Of course, these are analogous to each other, and therefore it suffices to prove the first one. Since $ p$ does not divide neither $ t$ nor $ s$, it follows $ (ts)^{p \minus{} 1}\equiv 1\pmod p$ (by Fermat's little theorem), and therefore $ \left((ts)^{p \minus{} 1}\right)^{\frac {q \minus{} 1}2}\equiv 1\pmod p$, what was to be proven.

Therefore, $ a$ is a square root of $ 1$ modulo $ pq$. Since $ pq$ is the product of two different prime numbers, there are four different square roots of $ 1$ modulo $ pq$, all of which could be found by solving a system of the form
\[ \left.\begin{array}{c}a\equiv\pm 1\pmod p \\
a\equiv\pm 1\pmod q\end{array}\right\},
\]what is precisely your conclusion.


We have non-negative numbers such as: $ w\plus{}x\plus{}y\plus{}z\equal{}2$

Find the maximum value of: $ A\equal{}\frac{w^2}{(w^2\plus{}1)^2}\plus{}\frac{x^2}{(x^2\plus{}1)^2}\plus{}\frac{y^2}{(y^2\plus{}1)^2}\plus{}\frac{z^2}{(z^2\plus{}1)^2}$

I have only:
* $ f(x)\equal{}\frac{x^2}{(x^2\plus{}1)^2}\equal{}\frac{1}{(x\plus{}\frac{1}{x})^2}$ rises iff  $ x \in \langle 0, 1 \rangle$ and comes down iff $ x \in \langle 1,\plus{}\infty \rangle$

** it's easy to get that A is maximum iff $ w,x,y,z \in \langle 0,1 \rangle$ (because $ f(x)\equal{}f(\frac{1}{x})$)

*** so: f is convex iff $ x \in \langle 0, \sqrt{\frac{4\minus{}\sqrt{13}}{3}} \rangle$ and concave iff $ x \in \langle \sqrt{\frac{4\minus{}\sqrt{13}}{3}}, 1 \rangle$

[quote="Sylwek"]We have non-negative numbers such as: $ w \plus{} x \plus{} y \plus{} z \equal{} 2$

Find the maximum value of: $ A \equal{} \frac {w^2}{(w^2 \plus{} 1)^2} \plus{} \frac {x^2}{(x^2 \plus{} 1)^2} \plus{} \frac {y^2}{(y^2 \plus{} 1)^2} \plus{} \frac {z^2}{(z^2 \plus{} 1)^2}$

I have only:
* $ f(x) \equal{} \frac {x^2}{(x^2 \plus{} 1)^2} \equal{} \frac {1}{(x \plus{} \frac {1}{x})^2}$ rises iff  $ x \in \langle 0, 1 \rangle$ and comes down iff $ x \in \langle 1, \plus{} \infty \rangle$

** it's easy to get that A is maximum iff $ w,x,y,z \in \langle 0,1 \rangle$ (because $ f(x) \equal{} f(\frac {1}{x})$)

*** so: f is convex iff $ x \in \langle 0, \sqrt {\frac {4 \minus{} \sqrt {13}}{3}} \rangle$ and concave iff $ x \in \langle \sqrt {\frac {4 \minus{} \sqrt {13}}{3}}, 1 \rangle$[/quote]
We have :
$ \frac {t^2}{(t^2 \plus{} 1)^2} \le\ \frac{48}{125}t\minus{}\frac{4}{125}$

$ \rightarrow A \le\ \frac{16}{25}$

Thanks :) , I solved it, but I had to check four conditions, and your solution is simpler (we have only to check: $ t < 0,12$, which is easier than my sollution (I checked: if three, two, one and none of numbers w,x,y,z are smaller than $ \sqrt{\frac{4\minus{}\sqrt{13}}{3}}$), because your inequality is true f.e. for $ t \in \approx \langle 0.119455591306, 1 \rangle$ ).

P.S. $ 4f(\frac{1}{2})\minus{}3f(\frac{2}{3}) \approx 0,001$, so we can't pass over this condition.

[quote="onlylove_math"][quote="Sylwek"]We have non-negative numbers such as: $ w \plus{} x \plus{} y \plus{} z \equal{} 2$

Find the maximum value of: $ A \equal{} \frac {w^2}{(w^2 \plus{} 1)^2} \plus{} \frac {x^2}{(x^2 \plus{} 1)^2} \plus{} \frac {y^2}{(y^2 \plus{} 1)^2} \plus{} \frac {z^2}{(z^2 \plus{} 1)^2}$

I have only:
* $ f(x) \equal{} \frac {x^2}{(x^2 \plus{} 1)^2} \equal{} \frac {1}{(x \plus{} \frac {1}{x})^2}$ rises iff  $ x \in \langle 0, 1 \rangle$ and comes down iff $ x \in \langle 1, \plus{} \infty \rangle$

** it's easy to get that A is maximum iff $ w,x,y,z \in \langle 0,1 \rangle$ (because $ f(x) \equal{} f(\frac {1}{x})$)

*** so: f is convex iff $ x \in \langle 0, \sqrt {\frac {4 \minus{} \sqrt {13}}{3}} \rangle$ and concave iff $ x \in \langle \sqrt {\frac {4 \minus{} \sqrt {13}}{3}}, 1 \rangle$[/quote]
We have :
$ \frac {t^2}{(t^2 \plus{} 1)^2} \le\ \frac {48}{125}t \minus{} \frac {4}{125}$

$ \rightarrow A \le\ \frac {16}{25}$[/quote]
Are you sure this inequality always true?
By the way, the max is $ \frac{16}{25}$.
And the exists a strange proof for it.


geometry
geometric transformation
reflection
inequalities unsolved
inequalities

In which country is the magazine [i]Mathematical Reflections [/i]published?

This journal is published in USA (Texas university)
But it is free. You can look any issues you want with the following website:

[url]http://reflections.awesomemath.org/[/url]


Let $ a,b,c$ be positive reals. Prove that

$ \frac{3a^2\minus{}2c^2\plus{}bc}{b\plus{}c}\plus{}\frac{3b^2\minus{}2a^2\plus{}ca}{c\plus{}a}\plus{}\frac{3c^2\minus{}2b^2\plus{}ab}{a\plus{}b}\geq a\plus{}b\plus{}c$

taking a,b,c  from RHS o LHS ,  we get

$ 3\sum{\frac{c^{2}\minus{}b^{2}}{a\plus{}b}}$ which  is  greater than or equal to 0.

hence proved


[quote="xxreddevilzxx"][quote="paladin8"][hide]
Well let $f(x) = x(x+1)\cdots(x+n)$. We have $f(0) = 0$ and $f\left(\frac{1}{n!}\right) = \frac{\left(1+\frac{1}{n!}\right)\left(2+\frac{1}{n!}\right)\cdots\left(n+\frac{1}{n!}\right)}{n!} > 1$ so there must be some solution between them (since $f$ is continuous).[/hide][/quote]

is this the mean value theorem or the intermediate value theorem?[/quote]

Didn't know it had a name, but it's the Intermediate Value Theorem. The Mean Value Theorem is for calculus.

[quote="m509"]prove that the equation:

$x(x+1)(x+2)...(x+n)=1$ has a postive solution that is less then $\frac{1}{n!}$[/quote]

Let x=1: we get on the LHS $(n+1)!$; so divide that by n! and we get $1<n+1=\prod_{i=0}^n \frac{1}{n!}+\frac{i}{n!}<\prod_{i=0}^n \frac{1}{n!}+i$, so the conclusion follows

[quote="m509"][quote="paladin8"]
 $f\left(\frac{1}{n!}\right) = \frac{\left(1+\frac{1}{n!}\right)\left(2+\frac{1}{n!}\right)\cdots\left(n+\frac{1}{n!}\right)}{n!} > 1$ [/quote]

prove this ;)[/quote]

The $LHS =\frac{\left(n!+1\right) \left(2\cdot n!+1\right) \left(3 \cdot n!+1\right) ... \left(n \cdot n!+1\right)}{\left(n!\right)^2}$

Then $LHS > \frac{n!(n!)^n}{\left(n!\right)^2} = (n!)^{n-1} \ge 1$

So $LHS > 1$

[hide]
Let $f(x) = x(x + 1)(x + 2) \dots (x + n) - 1$.  Clearly, $f(x)$ is an increasing function for $x \ge 0$, $f(0) = -1 < 0$, and $f(1) = (n + 1)! > 0$, so $f(x)$ has exactly one positive root $r$.

Furthermore,
\[ r = \frac{1}{(r + 1)(r + 2) \dots (r + n)} < \frac{1}{1 \cdot 2 \dots n} = \frac{1}{n!}. \]
[/hide]


trigonometry
inequalities
geometry open
geometry

Prove to all the triangle ABC, we have: $ \tan^3\frac{A}{2}\plus{}\tan^3\frac{B}{2}\plus{}\tan^3\frac{C}{2}\ge\frac{1}{\sqrt3}$

$ \tan^3\frac{A}{2}\plus{}\tan^3\frac{B}{2}\plus{}\tan^3\frac{C}{2}\ge\frac{(tan\frac{A}{2}\plus{}\tan\frac{B}{2}\plus{}\tan\frac{C}{2})}{9}^3$
And by using Jensen inequality we have
$ tan\frac{A}{2}\plus{}\tan\frac{B}{2}\plus{}\tan\frac{C}{2}\ge\sqrt{3}$ :lol:

[quote="TRAN THAI HUNG"]$ \tan^3\frac {A}{2} \plus{} \tan^3\frac {B}{2} \plus{} \tan^3\frac {C}{2}\ge\frac {(tan\frac {A}{2} \plus{} \tan\frac {B}{2} \plus{} \tan\frac {C}{2})}{9}^3$
And by using Jensen inequality we have
$ tan\frac {A}{2} \plus{} \tan\frac {B}{2} \plus{} \tan\frac {C}{2}\ge\sqrt {3}$ :lol:[/quote]
Differentiate $ tan(x)$ twice and get: $ tan''(x) \equal{} \frac{2 \cdot sin(x)}{(cos(x))^3}$, which is not always positive!

The proof should be more like: $ (tan^3(x))'' \equal{} \frac{6 \cdot sin(x) \cdot (sin^2(x)\plus{}1)}{cos^5(x)}$. So we have $ \tan^3\frac {A}{2} \plus{} \tan^3\frac {B}{2} \plus{} \tan^3\frac {C}{2} \ge \frac{1}{\sqrt{3}}$ when $ A,B,C < \pi$.
Now assume $ A \ge \pi \ge B \ge C$ wlog because of symmetry.
Now $ \tan^3\frac {B}{2} \plus{} \tan^3\frac {C}{2} \ge 2\tan^3\frac{B\plus{}C}{4} \equal{} 2\tan^3\frac{180\minus{}A}{4}$.
So it is necessary to prove:
$ \tan^3\frac{A}{2} \plus{} 2\tan^3\frac{180\minus{}A}{4} \ge \frac{1}{\sqrt{3}}$.
But that's obvious since $ \tan^3\frac{A}{2} \ge \tan^3\frac{90}{2} \equal{} 1$.

[quote="Mathias_DK"][quote="TRAN THAI HUNG"]$ \tan^3\frac {A}{2} \plus{} \tan^3\frac {B}{2} \plus{} \tan^3\frac {C}{2}\ge\frac {(tan\frac {A}{2} \plus{} \tan\frac {B}{2} \plus{} \tan\frac {C}{2})}{9}^3$
And by using Jensen inequality we have
$ tan\frac {A}{2} \plus{} \tan\frac {B}{2} \plus{} \tan\frac {C}{2}\ge\sqrt {3}$ :lol:[/quote]
Differentiate $ tan(x)$ twice and get: $ tan''(x) \equal{} \frac {2 \cdot sin(x)}{(cos(x))^3}$, which is not always positive!

The proof should be more like: $ (tan^3(x))'' \equal{} \frac {6 \cdot sin(x) \cdot (sin^2(x) \plus{} 1)}{cos^5(x)}$. So we have $ \tan^3\frac {A}{2} \plus{} \tan^3\frac {B}{2} \plus{} \tan^3\frac {C}{2} \ge \frac {1}{\sqrt {3}}$ when $ A,B,C < \pi$.
Now assume $ A \ge \pi \ge B \ge C$ wlog because of symmetry.
Now $ \tan^3\frac {B}{2} \plus{} \tan^3\frac {C}{2} \ge 2\tan^3\frac {B \plus{} C}{4} \equal{} 2\tan^3\frac {180 \minus{} A}{4}$.
So it is necessary to prove:
$ \tan^3\frac {A}{2} \plus{} 2\tan^3\frac {180 \minus{} A}{4} \ge \frac {1}{\sqrt {3}}$.
But that's obvious since $ \tan^3\frac {A}{2} \ge \tan^3\frac {90}{2} \equal{} 1$.[/quote]

  at first for 0<x<90 $ tan''(x) \equal{} \frac {2 \cdot sin(x)}{(cos(x))^3}$ - will always positive!!!!!! :wink: (so TRAN THAI HUNG's solution is right)
secondary, why you assume  that $ A \ge \pi$  :huh: (we already have that $ \frac{\pi}{2} > \frac{A}{2},\frac{B}{2},\frac{C}{2} > 0$  - becouse A,B,C - are angles of triangle)

[quote="laisac"]Prove to all the triangle ABC, we have: $ \tan^3\frac {A}{2} \plus{} \tan^3\frac {B}{2} \plus{} \tan^3\frac {C}{2}\ge\frac {1}{\sqrt3}$[/quote]
Do not use a powerful tool :lol:
We have:$ \tan^3\frac{A}{2}\plus{}\tan^3\frac{B}{2}\ge 2\sqrt{\tan^3\frac{A}{2}.\tan^3\frac{B}{2}}$ (*)
Because: $ \tan\frac{A}{2}.\tan\frac{B}{2}\equal{}\frac{\sin\frac{A}{2}.\sin\frac{B}{2}}{\cos\frac{A}{2}.\cos\frac{B}{2}}$$ \equal{}\frac{\cos\frac{A\minus{}B}{2}\minus{}\cos\frac{A\plus{}B}{2}}{\cos\frac{A\minus{}B}{2}\plus{}\cos\frac{A\plus{}B}{2}}$
$ \equal{}1\minus{}\frac{2\cos\frac{A\plus{}B}{2}}{\cos\frac{A\minus{}B}{2}\plus{}\cos\frac{A\plus{}B}{2}}$
$ \ge 1\minus{}\frac{2\cos\frac{A\plus{}B}{2}}{1\plus{}\cos\frac{A\plus{}B}{2}}\equal{}\frac{1\minus{}\cos\frac{A\plus{}B}{2}}{1\plus{}\cos\frac{A\plus{}B}{2}}\equal{}\tan^2\frac{A\plus{}B}{4}$
 Inequalities (*) Write another:$ \tan^3\frac{A}{2}\plus{}\tan^3\frac{B}{2}\ge 2\tan^3\frac{A\plus{}B}{4} (1)$
 Marked "=" when: A=B.
Same $ \tan^3\frac{C}{2}\plus{}\tan^3\frac{60^0}{2}\ge 2\tan^3\frac{C\plus{}60^0}{4} (2)$
 Marked "=" when:$ C\equal{}60^0$
(1)+(2):
$ \tan^3\frac{A}{2}\plus{}\tan^3\frac{B}{2}\plus{}\tan^3\frac{C}{2}\plus{}\tan^3\frac{60^0}{2}\ge 2(\tan^3\frac{A\plus{}B}{4} \plus{}\tan^3\frac{C\plus{}60^0}{4} )\ge 4\tan^3\frac{A\plus{}B\plus{}C\plus{}60^0}{8}\equal{}4\tan^330^0$
Or $ \tan^3\frac{A}{2}\plus{}\tan^3\frac{B}{2}\plus{}\tan^3\frac{C}{2}\ge 3\tan^330^0\equal{}\frac{1}{\sqrt3}$
 Marked "=" when: $ A\equal{}B\equal{}C\equal{}60^0$.
Help!
http://www.mathlinks.ro/Forum/viewtopic.php?t=240992

[quote="laisac"]Do not use a powerful tool :lol:
We have:$ \tan^3\frac {A}{2} \plus{} \tan^3\frac {B}{2}\ge 2\sqrt {\tan^3\frac {A}{2}.\tan^3\frac {B}{2}}$ (*)
Because: $ \tan\frac {A}{2}.\tan\frac {B}{2} \equal{} \frac {\sin\frac {A}{2}.\sin\frac {B}{2}}{\cos\frac {A}{2}.\cos\frac {B}{2}}$$ \equal{} \frac {\cos\frac {A \minus{} B}{2} \minus{} \cos\frac {A \plus{} B}{2}}{\cos\frac {A \minus{} B}{2} \plus{} \cos\frac {A \plus{} B}{2}}$
$ \equal{} 1 \minus{} \frac {2\cos\frac {A \plus{} B}{2}}{\cos\frac {A \minus{} B}{2} \plus{} \cos\frac {A \plus{} B}{2}}$
$ \ge 1 \minus{} \frac {2\cos\frac {A \plus{} B}{2}}{1 \plus{} \cos\frac {A \plus{} B}{2}} \equal{} \frac {1 \minus{} \cos\frac {A \plus{} B}{2}}{1 \plus{} \cos\frac {A \plus{} B}{2}} \equal{} \tan^2\frac {A \plus{} B}{4}$
 [/quote]
i'm sorry you're wrong here but it should be:
$ 1 \minus{} \frac {2\cos\frac {A \plus{} B}{2}}{1 \plus{} \cos\frac {A \plus{} B}{2}}$
$ \ge1 \minus{} \frac {2\cos\frac {A \plus{} B}{2}}{\cos\frac {A \minus{} B}{2} \plus{} \cos\frac {A \plus{} B}{2}}$

[quote="lizhi"]

i'm sorry you're wrong here but it should be:
$ 1 \minus{} \frac {2\cos\frac {A \plus{} B}{2}}{1 \plus{} \cos\frac {A \plus{} B}{2}}$
$ \ge1 \minus{} \frac {2\cos\frac {A \plus{} B}{2}}{\cos\frac {A \minus{} B}{2} \plus{} \cos\frac {A \plus{} B}{2}}$[/quote]
Sorry, I was mistaken. Thank you.
In other  :lol:
We have   $ \frac{{tg^3 \frac{A}{2} \plus{} tg^3 \frac{B}{2}}}{2} \ge \left( {\frac{{tg\frac{A}{2} \plus{} tg\frac{B}{2}}}{2}} \right)^3$(1)
We have $ tg\frac{A}{2} \plus{} tg\frac{B}{2}\equal{}\frac{{\sin \frac{{A \plus{} B}}{2}}}{{\cos \frac{A}{2}\cos \frac{B}{2}}} \equal{} \frac{{2\sin \frac{{A \plus{} B}}{2}}}{{\cos \frac{{A \minus{} B}}{2} \plus{} \cos \frac{{A \plus{} B}}{2}}}$
$ \ge \frac{{4\sin \frac{{A \plus{} B}}{4}\cos \frac{{A \plus{} B}}{4}}}{{1 \plus{} \cos \frac{{A \plus{} B}}{2}}} \equal{} 2tg\frac{{A \plus{} B}}{4}$
So: $ tg^3 \frac{A}{2} \plus{} tg^3 \frac{B}{2} \ge 2tg^3 \frac{{A \plus{} B}}{4}$	(2)
Similarly, we have:$ tg^3 \frac{C}{2} \plus{} tg^3 \frac{{60^0 }}{2} \ge 2tg^3 \frac{{C \plus{} 60^0}}{4}$ 	(3)
The Community (2) and (3) we have
$ tg^3 \frac{A}{2} \plus{} tg^3 \frac{B}{2} \plus{} tg^3 \frac{C}{2} \plus{} tg^3 \frac{{60^0 }}{2} \equal{} 2(tg^3 \frac{{A \plus{} B}}{4} \plus{} tg^3 \frac{{C \plus{} 60^0 }}{4})$
$ \ge 4tg^3 \frac{{A \plus{} B \plus{} C \plus{} 60^0 }}{8} \equal{} 4tg^3 \frac{{60^0 }}{2}$
Or : $ tg^3 \frac{A}{2} \plus{} tg^3 \frac{B}{2} \plus{} tg^3 \frac{C}{2} \ge 3tg^3 30^0  \equal{}\frac{1}{{\sqrt 3 }}$(*)
Equality, when $ A \equal{} B \equal{} C \equal{} 60^0$

[quote="laisac"] [color=darkred]Prove that for any triangle ABC, we have $ \tan^3\frac {A}{2} \plus{} \tan^3\frac {B}{2} \plus{} \tan^3\frac {C}{2}\ge\frac {1}{\sqrt3}$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] $ \sum\tan^3\frac A2 \equal{} \sum\left(\frac {r_a}{p}\right)^3 \equal{} \frac {\sum r_a^3}{p^3}\ge\frac {\left(\sum r_a\right)^3}{9p^3} \equal{} \frac {(4R \plus{} r)^3}{9p^3}\ge \frac {\left(p\sqrt 3\right)^3}{9p^3} \equal{} \frac {1}{\sqrt 3}$ .

[b][u]Remark[/u].[/b] Il used the following well-known relations :

$ \odot\ \boxed {\ r_a \equal{} p\tan\frac A2\ }$ a.s.o. ; $ \begin{array}{c}
\boxed {\ S\equal{}pr\equal{}(p\minus{}a)r_a\equal{}(p\minus{}b)r_b\equal{}(p\minus{}c)r_c\ }\\\\
 \boxed{\ \begin{array}{ccc}
r_a \plus{} r_b \plus{} r_c & \equal{} & 4R \plus{} r\\\
r_ar_b \plus{} r_br_c \plus{} r_cr_a & \equal{} & p^2 \\
\ r_ar_br_c & \equal{} & p^2r\end{array}\ }\end{array}$ .

$ \odot\ \boxed {\ \{x,y,z\}\subset (0,\infty )\ \implies\ (x \plus{} y \plus{} z)^3\le 9\left(x^3 \plus{} y^3 \plus{} z^3\right)\ }\ (*)$ .

Indeed, $ \left\|\ \begin{array}{ccc} \left(x \plus{} y \plus{} z\right)\left(x \plus{} y \plus{} z\right) & \le & 3\left(x^2 \plus{} y^2 \plus{} z^2\right) \\
 \\
\left(x \plus{} y \plus{} z\right)\left(x^2 \plus{} y^2 \plus{} z^2\right) & \le & 3\left(x^3 \plus{} y^3 \plus{} z^3\right)\end{array}\ \right\|\ \bigodot\ \implies\ (*)$ .

$ \odot\ \boxed {\ p\sqrt 3\ \le\ 4R \plus{} r\ }\ (**)$ .

Indeed, $ 3\left(r_ar_b \plus{} r_br_c \plus{} r_cr_a\right)\le\left(r_a \plus{} r_b \plus{} r_c\right)^2\ \implies\ 3p^2\le (4R \plus{} r)^2\ \implies\ (**)$ .[/color]


Prove that

In any integer sided right triangle, the product of the altitude and base is a multiple of 12.

let the sides be $ a,b,c$
$ a^{2} \plus{} b^{2} \equal{} c^{2}$ by Pythagoras theorem
when a perfect square is divided by 12 the possible remainders are $ 0,1,9,4$
out of these 4 remainders
only when $ a^{2}$ leaves remainder 4 and $ b^{2}$ leaves remainder 9 ,when $ a^{2}$ leaves 0 and $ b^{2}$ leaves 0,when $ a^{2}$ leaves 0 and $ b^{2}$ leaves 1 ,only out of these three cases $ c^{2}$ has the required form when a perfect square is divided by 12
and for the remaining combinations of remainders of a and b , $ c^{2}$ has a remainder which a perfect sqauare cant have when it is divided by12.
and out of the three valid cases in all $ a*b \equiv 0\mod12$.it is obvious that a can be the base or altitude ,b can be base or altitude, and c is the hypotenuse
hence proved

also you can use [url=http://en.wikipedia.org/wiki/Pythagorean_triple]primitive pythagorean triplets[/url]

one of $ 2mn$ or $ m^2\minus{}n^2$ is divisible by $ 3$ and one is divisible by $ 4$ for all $ m,n$


geometry
circumcircle
inradius
trigonometry
incenter
Pythagorean Theorem
geometry proposed

Given triangle $ABC$ circumcircle $(O),A',B',C'$ are circumceter of triangle $OBC,OCA,OAB$ and $A'',B'',C''$ are symetric point of $A',B',C'$ with lines $BC,CA,AB$ resp prove that in four circles (three excircles and incircle of triangle $A''B''C''$) always exist a circle pass through $E$ - ninepoint center of $ABC$ and in four number (three exradius and inradius of triangle $A''B''C''$) always exist a number equal to $\frac{1}{2}OH$ with $H$ is orthocenter of $ABC$.

Compare with the 1st solution of the Furhmann triangle problem at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=101687[/url] (no hint was given about Fuhrmann at the time).

Assume $\triangle ABC$ is acute. B'C', C'A', A'B' are perpendicular bisectors of OA, OB, OC, $\triangle A'B'C'$ is similar to the tangential triangle with similarity center O and coefficient 1/2. With usual notation,

$A'B' = R\ \frac{\tan A+\tan B}{2}= R\ \frac{\sin C}{2 \cos A \cos B}$

$OA'' = \frac{R}{2 \cos A}-2R \cos A = R\ \frac{1-4 \cos^{2}A}{2\cos A}$

$OB'' = R\ \frac{1-4 \cos^{2}B}{2\cos B}$

By the cosine law for the $\triangle OA''B'',$

$\frac{A''B''^{2}}{A'B'^{2}}= \frac{OA''^{2}+OB''^{2}+2\ OA'' \cdot OB''\ \cos C}{A'B'^{2}}=$

$\frac{(1-4 \cos^{2}A) \cos^{2}B+(1-4 \cos^{2}B) \cos^{2}A+2 (1-4 \cos^{2}A)(1-4 \cos^{2}B) \cos A \cos B \cos C}{\sin^{2}C}=$

(substituting $2 \cos A \cos B \cos C = 1-(\cos^{2}A+\cos^{2}B+\cos^{2}C)$)

$= \frac{(1-\cos^{2}C)-4 \cos^{2}A (1-\cos^{2}C)-4 \cos^{2}B (1-\cos^{2}C)+16 \cos^{2}A \cos^{2}B (1-\cos^{2}C)+4 (\cos^{2}A-\cos^{2}B)^{2}}{\sin^{2}C}=$

$= 1-4 \cos^{2}A-4 \cos^{2}B+16 \cos^{2}A \cos^{2}B+4 \sin^{2}(A-B) =$

$= 1+8 \cos A \cos B (\cos A \cos B-\sin A \sin B) = 1-8 \cos A \cos B \cos C$

If D is foot of the A-altitude, power of H to (O) is

$R^{2}-HO^{2}= 2 HD \cdot HA = 8R^{2}\cos A \cos B \cos C$

Therefore $\frac{A''B''}{A'B'}= \frac{HO}{R}$ and similarly, $\frac{B''C''}{B'C'}= \frac{C''A''}{C'A'}= \frac{HO}{R},$ $\triangle A''B''C'' \sim \triangle A'B'C'$ with coefficient $\frac{HO}{R}.$ If parallel to BC through H meet A'A'' at X,

$XH = \frac{|b^{2}-c^{2}|}{2a}= R\ \frac{|\cos^{2}B-\cos^{2}C|}{\sin A}= R |\sin (B-C)|$

$XA'' = \frac{R}{2 \cos A}-R \cos A-2R \cos B \cos C = R\ \frac{1-2 \cos^{2}A-4 \cos A \cos B \cos C}{2\cos A}$

$OA' = \frac{R}{2 \cos A}$

By Pythagorean theorem for the $\triangle HXA'',$

$\frac{HA''^{2}}{OA'^{2}}= \frac{XH^{2}+XA''^{2}}{OA'^{2}}=$

${= (1-2 \cos^{2}A-4 \cos A \cos B \cos C)^{2}+4 \sin^{2}(B-C)}\cos^{2}A=$

$= 1-8 \cos A \cos B \cos C-4 \cos^{2}A (1-\cos^{2}A)+16 \cos^{2}A \cos^{2}B \cos^{2}C+$
$+4 \cos^{2}A (2 \cos A \cos B \cos C)+4 \cos A (1-\cos^{2}A-\cos^{2}B-\cos^{2}C)+$
$+4\cos^{2}A \cos^{2}B+4 \cos^{2}A \cos^{2}C-8 \cos^{2}A \cos B^{2}\cos C^{2}-$
$-8 \cos^{2}A \sin B \sin C \cos B \cos C=$

$= 1-8 \cos A \cos B \cos C+4 \cos^{2}A (2 \cos A \cos B \cos C)+$
$+8 \cos^{2}A \cos B \cos C (\cos B \cos C-\sin B \sin C) =$

$= 1-8 \cos A \cos B \cos C = \frac{HO^{2}}{R^{2}}$

Therefore $\frac{HA''}{OA'}= \frac{HO}{R}$ and similarly, $\frac{HB''}{OB'}= \frac{HC''}{OC'}= \frac{HO}{R}.$ Since O is the incenter of $\triangle A'B'C',$ H is the incenter of the $\triangle A''B''C''.$ R is inradius of the tangential triangle, R/2 inradius of the $\triangle A'B'C'$ and OH/2 inradius of the $\triangle A'' B''C''.$ Incircle of the last triangle cuts OH in half, passing through the 9-point center of the original $\triangle ABC.$ If the $\triangle ABC$ is obtuse, O is the appropriate excenter of the tangential triangle and of the $\triangle A'B'C'$ and H is the corresponding extenter of the $\triangle A''B''C''.$


Let the roots of the cubic equation $ x^3 \plus{} ax^2 \plus{} bx \plus{} c \equal{} 0$ be all real.
Show that the difference between the greatest and last of them is not less than $ \sqrt {a^2 \minus{} 3b}$ and not greater than $ 2\sqrt {a^2 \minus{} 3b}$.

Let $ x_1, x_2, x_3$ be roots of this polynomial and $ x_1 \leq x_2 \leq x_3$. Applying Viete'a we have
 \[ \sqrt{a^2\minus{}3b}\equal{}\sqrt{x_1^2\plus{}x_2^2\plus{}x_3^2\minus{}x_1x_2 \minus{} x_2x_3 \minus{} x_3 x_1} \le x_3 \minus{} x_1\]
after squaring we get $ (x_2 \minus{} x_1 )(x_2 \minus{} x_3 ) \le 0$

\[ x_3 \minus{} x_1 \le 2\sqrt{a^2\minus{}3b}\equal{}2\sqrt{x_1^2\plus{}x_2^2\plus{}x_3^2\minus{}x_1x_2 \minus{} x_2x_3 \minus{} x_3 x_1}\]

after aquaring we get $ 0\leq 2(x_1\minus{}x_2)^2 \plus{} \left(2x_2 \minus{}x_1 \minus{} x_3 \right)^2$


calculus
integration
number theory
prime factorization

What is the smallest number that contains precisely 18 integral factors other than one or the number itself.

[hide] 18 factors not including 1 and itself gives 20 factors including everything.  So, if the number was $p_{1}^{e_{1}}+p_{2}^{e_{2}}...p_{n}^{e_{n}}$, the number of factors would be $(e_{1}+1)(e_{2}+1)(e_{3}+1)...(e_{n}+1)=20$, since there are 20 factors.  $20=2^{2}*5$, so the lowest number would be $2^{5-1}*3^{2-1}*5^{2-1}=16*3*5=\boxed{240}$ [/hide]

[hide]
i think the lowest would be something to the fourth times two other primes

16*3*5=240

is that it?

[/hide]
jorian

My answer and complete explanation:

[hide]
I got $240$.

The number of factors a number has including 1 and itself is found by taking the prime factorization to a number, adding one to each of the exponents, and multiplying the exponents together.

However, here we are not including 1 and itself, so add 2 to 18, making up for 1 and itself.

Now we need to find numbers that multiply to get 18.
I got: 2 x 2 x 5

Now subtract 1 from each.

We have 1, 1, and 4.

Now what is the smallest prime? 2.
So lets give 2 the power of 4, since we are aiming for the smallest possible number.

Then what are the next to primes? 3 and 5. They both get a power of 1.

$2^{4}* 3^{1}* 5^{1}= 240$

[/hide]

I hope this is right :)

[hide]
$x^{x_{1}}\cdot y^{y_{1}}\cdot z^{z_{1}}=n$ where x, y, and z are prime factors. The number of prime factors is $(x_{1}+1) \cdot (y_{1}+1) \cdot (z_{1}+1)=20$, since we have to add two to 18 for one and the number. The possible values are 1, 1, 20; 1, 2, 10; 2, 2, 5; 1, 4, 5.

The one that has the closest terms is the smallest one.
$2^{4}\cdot 3^{1}\cdot 5^{1}= \boxed{240}$[/hide]

[quote="bpms"][quote="happyme"]English please?
 :blush:[/quote]
Okay, here is a specific case.
Consider how many factors 100 has.
100 is prime factorized as $2^{2}\cdot 5^{2}$.
Any factor of 100 can be constructed with 2s and 5s.
You can have either 0, 1, or 2 2s. You can have either 0, 1, or 2 5s.
3*3=9
So 100 has 9 factors.[/quote]

I see. Thanks.


inequalities
trigonometry
algebra unsolved
algebra

$0 \leqslant x \leqslant \frac{\pi}{2}$
$\frac{\pi}{6} \leqslant \alpha \leqslant \frac{\pi}{3}$
Prove that:
$\tan({\frac{\pi\sin{x}}{4\sin{\alpha}}) + \tan({\frac{\pi\cos{x}}{4\cos{\alpha}}) > 1}}$


algebra polynomial calculus integration search AMC USA(J)MO

Let $ a$, $ b$, and $ c$ denote three distinct integers and let $ P$ denote a polynomial having all integral coefficients.  Show that it is impossible that $ P(a)\equal{}b$, $ P(b)\equal{}c$, and $ P(c)\equal{}a$.

Suppose $ a$, $ b$, and $ c$ are distinct integers such that $ P(a)\equal{}b$, $ P(b)\equal{}c$, and $ P(c)\equal{}a$.  Then each of $ a$, $ b$, and $ c$ are (distinct) roots of the polynomial $ P(P(P(x))) \minus{} x$, and are not roots of the polynomial $ P(x)\minus{}x$.  Since each value of $ y$ with $ P(y)\equal{}y$ also clearly has $ P(P(P(y)))\equal{}y$, the polynomial $ P(P(P((x)))\minus{}x$ is divisible by the polynomial $ P(x)\minus{}x$.  They have the same constant coefficient by inspection, and hence the quotient $ Q(x) \equal{} \frac{P(P(P((x)))\minus{}x}{P(x)\minus{}x}$ will have a constant coefficient of 1.

By assumption, each of $ a$, $ b$, and $ c$ are integers and are roots of the polynomial $ Q(x)$ which has integral coefficients by the division algorithm.  [It is also possible to verify directly that $ Q(x)$ has integral coefficients by writing it out in terms of $ P(x)$.] The rational root theorem implies that each of $ a$, $ b$, and $ c$ divides the constant coefficient of $ R(x)$, which is 1.  But since there are only two distinct integers which divide 1, two of $ a$, $ b$, and $ c$ must be equal.  This is a contradiction, and therefore no such $ a$, $ b$, and $ c$ can exist.

My first USAMO proof!!! :D

All the terms of a given polynomial are divisible by x except for the constant term, so we can write the polynomial as $ P(x)\equal{}nx\plus{}r$, where n is an expression in terms of x and r is the constant term. By the problem, n and r are both integers. So, we are trying to prove that $ P(P(P(a)))$ cannot equal a. $ P(x)$ is $ nx\plus{}r$, so $ P(P(P(a)))\equal{}n(n(na\plus{}r)\plus{}r)\plus{}r\equal{}n^3a\plus{}[n(n\plus{}1)\plus{}1]r\equal{}n^3a\plus{}(n^2\plus{}n\plus{}1)r$.

Setting that equal for a and looking for a contradiction, if $ n^3a\plus{}(n^2\plus{}n\plus{}1)r\equal{}a, then (n^3\minus{}1)a\plus{}(n^2\plus{}n\plus{}1)r\equal{}0$. This factors to $ (n\minus{}1)(n^2\plus{}n\plus{}1)a\plus{}(n^2\plus{}n\plus{}1)r\equal{}0$, or $ [(n\minus{}1)a\plus{}r](n^2\plus{}n\plus{}1)\equal{}0$. This implies that either $ n^2\plus{}n\plus{}1\equal{}0$ or $ (n\minus{}1)a\plus{}r\equal{}0$. Checking the roots of $ n^2\plus{}n\plus{}1\equal{}0$, we find that neither is an integer, so that is eliminated. Working with $ (n\minus{}1)a\plus{}r\equal{}0$, this expands to $ na\minus{}a\plus{}r\equal{}0$, or $ na\plus{}r\equal{}a$. That means, $ P(a)\equal{}a$. But wait, $ P(a)\equal{}b$, and that would mean $ b\equal{}a$. But the problem states that a, b, and c are distinct, which is a contradiction.

[quote="Textangle"]$ P(P(P(a))) \equal{} n(n(na \plus{} r) \plus{} r) \plus{} r \equal{} n^3a \plus{} [n(n \plus{} 1) \plus{} 1]r \equal{} n^3a \plus{} (n^2 \plus{} n \plus{} 1)r$.[/quote]
This is valid if and only if $ n$ is a constant; otherwise, because $ n$ is a polynomial, substituting $ P(a)$ into $ P(x)$ changes $ n$ itself.  In other words, you've only proven the result if $ P$ is a linear polynomial.

@chaucerchu: Your claim that $ P(P(P(x))) \minus{} x$ and $ P(x) \minus{} x$ have the same constant term is invalid -- take $ P(x) \equal{} x \plus{} 1$.

yes it only true if $ P(P(P(0))) \equal{} P(0)$


inequalities
algorithm
inequalities proposed

[quote="yao"]Suppose x1,x2,x3,x4,x5,x6 and x7 are non-negative integers and S denotes the sum of these seven integers. Then we have the following inequality:
              (x4+x7)/(S-x3)+(x5+x7)/(S-x1)-(x6+x7)/(S-x2)<=1.
  My question is: can we prove this inequality mechanically? i.e., an algorithm can be given to test whether this inequality holds.[/quote]

$\frac{x_{4}+x_{7}}{S-x_{3}}+\frac{x_{5}+x_{7}}{S-x_{1}}-\frac{x_{6}+x_{7}}{S-x_{2}}\leq$
$\leq\frac{x_{4}+x_{3}+x_{7}}{S}+\frac{x_{1}+x_{5}+x_{7}}{S}+\frac{x_{2}+x_{6}-x_{7}}{S}=1.$ :)

Thank arqady. But I know this proof, my question is if I dot not know how to prove, whether can I verify this inequality only using computer?


Prove or disprove:
For positive reals $a_1,\ldots,a_n$ and $x_1,\ldots,x_n$ the inequality

\[ \frac{a_1}{x_1}+\ldots+\frac{a_n}{x_n}\geq \frac{n(a_1+ \ldots + a_n)}{x_1 + \ldots +x_n} \]

holds.

No, this is not true. Say $a_i=x_iy_i$ and think about the condition for Chebyshev Inequality.


calculus
integration
trigonometry
function
complex analysis
real analysis
real analysis unsolved

1.) The integral of sin(tx)/(pi*(1+x^2)) dx from negative infinity to positive infinity. (should be 0 but don't know how to get it.)

2.) The integral of cos(tx)/(pi*(1+x^2)) dx from negative infinity to positive infinity. (should be e^-abs.(t), but again don't know how to get there.)

(1) is $ \int_{ \minus{} \infty}^{\infty}\frac {\sin tx}{\pi(1 \plus{} x^2)}\,dx$ and (2) is $ \int_{ \minus{} \infty}^{\infty}\frac {\cos tx}{\pi(1 \plus{} x^2)}\,dx$

Both integrals converge absolutely by comparison to $ \int_{ \minus{} \infty}^{\infty}\frac {1}{\pi(1 \plus{} x^2)}\,dx,$ so we may take that for granted.

We can do (1) very easily: the integrand is an odd function, and the interval of integration is symmetric. In such a case, as long as the integral converges, the integral is zero.

We need to do more work for (2). This is, of course, a Fourier transform. One pathway is to assume that we know the appropriate theorems and formulas for the inversion of Fourier transforms and simply compute the inverse Fourier transform of $ e^{ \minus{} |t|}.$ I regard that as a perfectly legitimate way to do the problem - but it does rely on two things: knowing the formula for Fourier inversion, and knowing the answer to this particular problem ahead of time.

My second suggestion is to use the residue theorem from complex analysis to compute the integral along the real axis of $ \frac {e^{itz}}{\pi(1 \plus{} z^2)}.$ Why do we get one answer for $ t > 0$ and a different answer for $ t < 0?$ Because we must take the return path through whichever of the upper or lower half planes that exponential is small in.


1)Let $ A,B$ be complex matrices of size $ 3$x$ 3$ .Prove that :
          $ \det(AB\minus{}BA)\equal{}Tr(AB(AB\minus{}BA)BA)$
 2) Let $ A$ and B be 2 by 2 matrices with integer entries such that $ AB\equal{}BA$ and $ \det(B)\equal{}1$
Prove that if $ \det(A^3\plus{}B^3)\equal{}1$ then $ A^2\equal{}I$
 3) Let $ A$ be a $ 2$x$ 2$ matrix with complex entries and $ C(A)\equal{}\{B \in M_2{(\mathbb{C})}\}$
 Show that $ |\det(A\plus{}B)| \geq |\det(B)| , \forall B \in C(A)$
 If and only if $ A^2\equal{}O$

[quote="Qruni"]For problem number $ 2$, note that, if $ A \equal{} 0$, $ B \equal{} I$, then $ A$ and $ B$ are matrix with integer entries, $ AB \equal{} BA$, $ det(B) \equal{} 1$ and $ det(A^3 \plus{} B^3) \equal{} 1$, but $ A^2$ is not an identity matrix, so maybe we should add some statement, such as "... $ A$ is an invertible matrix", or other statement.
I am sorry if I am wrong...[/quote]
 Sorry Prove that $ A^2 \equal{} O_2$  :oops:
 [color=red]I Edited [/color]: 2) Let $ A$ and $ B$ be $ 2$ x $ 2$ matrices with integer entries such that   $ AB \equal{} BA$ and $ det(B) \equal{} 1$
Prove that if $ det(A^3 \plus{} B^3) \equal{} 1$  then  $ A^2\equal{}O_2$

[quote="QuyBac"] 2) Let $ A$ and $ B$ be $ 2$ x $ 2$ matrices with integer entries such that   $ AB \equal{} BA$ and $ det(B) \equal{} 1$
Prove that if $ det(A^3 \plus{} B^3) \equal{} 1$  then  $ A^2 \equal{} O_2$[/quote]

The "edited" second:

First we notice that $ det(B^{ \minus{} 1}) \equal{} 1$

Then we have:

$ det(A^3 \plus{} B^3) \equal{} det(B^{ \minus{} 1})^3 det((AB^{ \minus{} 1})^3 \plus{} I) \equal{} 1 \rightarrow det((AB^{ \minus{} 1})^3 \plus{} I) \equal{} 1$

Setting $ X \equal{} AB^{ \minus{} 1}$, we have $ det(X^3 \plus{} I) \equal{} det(X \plus{} I)det(X \plus{} wI)det(X \plus{} w^2I) \equal{} 1$ where w is a third root of unity.

We know that:
$ det(X \minus{} sI) \equal{} s^2 \minus{} tr(X)s \plus{} det(X)$

Setting det(X)=D, tr(X)=T we see that T is an integer since $ B^{ \minus{} 1} \equal{} \frac {1} {detB} Adj(B)$ and therefore has integer entries... (obviously D is an integer as well)

$ det(X \plus{} wI) \equal{} w^2 \plus{} Tw \plus{} D \equal{} \overline{w} \plus{} T \overline {w^2} \plus{} D \equal{} \overline{det(X \plus{} w^2I)} \\
\rightarrow det(X \plus{} wI)det(X \plus{} w^2I) \in R_ \plus{}$

Then $ det(X \plus{} I) \equal{} 1 \equal{} det(X \plus{} wI)det(X \plus{} w^2I)$

and we have:

$ 1 \plus{} T \plus{} D \equal{} 1 \rightarrow D \equal{} \minus{} T \\
(w^2 \plus{} Tw \plus{} D)(w \plus{} Tw^2 \plus{} D) \equal{} 1 \rightarrow D^2(2 \minus{} w \plus{} w^2) \equal{} 0 \rightarrow D \equal{} T \equal{} 0$

Then from Caley-Hamilton:
$ X^2 \minus{} X \cdot trX \plus{} det(X) \equal{} 0 \rightarrow X^2 \equal{} 0 \rightarrow (AB^{ \minus{} 1})^2 \equal{} 0$

But $ AB \equal{} BA \rightarrow B^{ \minus{} 1}A \equal{} AB^{ \minus{} 1}$ yielding A^2B^{-2}=0

Multiplying on the right by B we get $ A^2 \equal{} 0_2$

For 2) new version.
$ A^3\plus{}B^3\equal{}(A\plus{}B)(A^2\plus{}B^2\minus{}AB)$, then $ det(A\plus{}B)\equal{}\epsilon,det((B\plus{}jA)(B\plus{}j^2A))\equal{}\epsilon$ where $ j^2\plus{}j\plus{}1\equal{}0$ and $ \epsilon\equal{}\pm{1}$. $ \epsilon\equal{}1$ because the last det is $ >0$.
Let $ f: \lambda\rightarrow{d}et(B\plus{}\lambda{A})\equal{}a\lambda^2\plus{}b\lambda\plus{}1$ where $ a\equal{}det(A),b$ are real numbers. $ f(1)\equal{}1,f(j)f(j^2)\equal{}1$. Therefore $ b\equal{}\minus{}a,3a^2\plus{}1\equal{}1$ and $ f(\lambda)\equal{}1$. $ A,B$ are simultaneously triangularizable over $ \mathbb{C}$; then $ spectrum(A)\equal{}\{0,0\}$ and $ A^2\equal{}0$.
More precisely: if $ A\not\equal{}0$ then we may assume that (over $ \mathbb{C}$):
$ A\equal{}\begin{pmatrix}0&1\\0&0\end{pmatrix},B\equal{}\begin{pmatrix}u&v\\0&u\end{pmatrix}$ with $ u^2\equal{}1$, that is $ u\equal{}\epsilon$. Finally $ B\equal{}\epsilon{I}_2\plus{}vA$, where $ A\not\equal{}{0},A^2\equal{}0$.

About the first...

I think there is a mistake in this as well (unless i am wrong somewhere)

Anyways:

Caley-Hamilton for C=AB-BA gives:

$ C^3\minus{}tr(C) C^2 \plus{}aC\minus{}det(C)I\equal{}0$

Tracing this with tr(C)=0 we get:

$ tr((AB\minus{}BA)^3)\minus{}3det(AB\minus{}BA)\equal{}0$

$ (AB\minus{}BA)^3\equal{}(AB)^3 \minus{} ABABBA\minus{}ABBAAB\plus{}ABBABA\minus{}BAABAB\plus{}BAABBA\plus{}BABAAB\minus{}(BA)^3$

Now if i have done the multiplications correctly, tracing this and using the commutativity inside the trace we get $ tr((AB\minus{}BA)^3)\equal{}3tr(AB(BA\minus{}AB)BA)$

Therefore i think the result needed (and proved) is:

$ det(AB\minus{}BA)\equal{}tr(AB(BA\minus{}AB)BA)$

For 1),  lemonidas is right.
For 3),  I assume that $ C(A)\equal{}\{B;BA\equal{}AB\}$.
-Quybac, your 3 exercises were incorrectly stated, that is remarkable-
The result is true in dimension $ n$:
Assume that $ A$ is not a nilpotent matrix. Then $ A$ admits a non zero eigenvalue $ \lambda$. We take $ B\equal{}\minus{}\lambda{I_n}$. Therefore $ 0\geq|\lambda|^n$, a contradiction.
Conversely, if $ A$ is a nilpotent matrix anf if $ AB\equal{}BA$, then we may assume that $ A$ is a strictly upper triangular matrix and $ B$ is an upper triangular matrix. Therefore $ det(A\plus{}B)\equal{}det(B)$.

[quote="loup blanc"]
For 3),  I assume that $ C(A) \equal{} \{B;BA \equal{} AB\}$.
-[/quote]
 Oke ! :blush:


What is the sum of the solutions of the equation $ x(2x\minus{}9) \equal{} 0$?
Express your answer as a common fraction.

hello, this equation has exactly two solutions, $ x_1\equal{}0$ or $ x_2\equal{}\frac{9}{2}$, so the sum is given by
$ x_1\plus{}x_2\equal{}\frac{9}{2}$.
Sonnhard.

Or you can quickly expand in your head, and use Vieta's to get $ \frac{9}{2}$, but seeing as its already factored for you, yours is a bit faster.

We expand and get $2x^2 - 9x = 0$.  Hence, by Vieta's, the sum of the roots of $ax^2 + bx + c$ is $-\frac{b}{a}$, so in this case the sum would be $-\frac{-9}{2} = \boxed{\frac{9}{2}}$.


integration
logarithms
calculus
calculus computations

$ \int_{a}^{3a} (e^{3x} \plus{} e^{x})\, dx \equal{} 100$.

[b](i)[/b] Show that $ a \equal{} \frac {1}{9}\ln (300 \plus{} 3e^{a} \minus{} 2e^{3a})$.

[b](ii)[/b] Use an iterative process, based on the equation in part [b](i)[/b], to find the value of $ a$ correct to 4 decimal places. Use a starting value of $ 0.6$ and show the result of each step of the process.

[I make it $ a\equal{}0.6309 \, (4dp)$.]

hello, we have $ \int e^{3x}\plus{}e^x\,dx\equal{}\frac{e^{3a}}{3}\plus{}e^x$ and the definite integral is
$ \frac{e^{9a}}{3}\plus{}e^{3a}\minus{}\frac{e^{3a}}{3}\minus{}e^a\equal{}100$ from here we get by multiplying with 3
$ e^{9a}\plus{}2e^{3a}\minus{}3e^a\equal{}300$ or $ e^{3a}\equal{}300\plus{}3e^a\minus{}2e^{3a}$ or $ a\equal{}\frac{1}{9}\ln(300\plus{}3e^a\minus{}2e^{3a})$
Sonnhard.

I agree, Sonnhard. But is my answer to part (ii) correct?


calculus
integration
function
trigonometry
LaTeX
calculus computations

Hi :) !!

$ \int \sqrt {1 + cos^2(x)}dx = \int \sqrt {2 - sin^2(x)} = \sqrt {2} \int \sqrt {1 - \frac {1}{2}sin^2(x)}dx = \sqrt {2}E (x | \frac {1}{2})$

with $ E(x| h) = \int \sqrt {1 - hsin^2(x)}dx$ is a function $ Elliptic \ integral$

and thx :)

[quote="yeahnigga"]That doesnt make sense what u wrote mathema  :wink:
Also check the definition of the elliptic integral of the second kind again...  :wink:[/quote]

Hi

you can check that the function $ x \rightarrow 1 \plus{} cos^2(x)$ is a derivate of$ x \rightarrow \sqrt {2} E(x;1/2)$

BUT i don't understand what do you want to say !!!

thx

hello, your integral looks so better $ \int\sqrt{1\plus{}\cos^2(x)}\,dx$.
Sonnhard.

Lieber Sonnhard, Ihr Integral sieht gut aus aber Sie wollten wohl sagen "your integral would look better as ".

[quote="AndrewTom"]Thanks mathema*: the guy who asked me the question said it was really easy which sent me on some wild goose chases.[/quote]

YOU ARE WELCOME :) ;)

hello, i wanted to say the integral would look so in $ \text{\LaTeX}$  better as yours.
Thank your for your hint.
Sonnhard.

Lieber Sonnhard,

danke für den Ratschlag; er war sehr nützlich und ich habe ihn gern angenommen. (Übrigens ist "as" in Deutsch "wie" und "than" das deutsche "als" im Komparativ.)


Let $ f : \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable, $ 2 \pi$-periodic even function. Prove that if \[ f''(x)\plus{}f(x)\equal{}\frac{1}{f(x\plus{} 3 \pi /2 )}\] holds for every $ x$, then $ f$ is $ \pi /2$-periodic.

[i]Z. Szabo, J. Terjeki[/i]

Since $f$ is even, we get that \[\frac{1}{f(x+3\pi/2)}=f''(x)+f(x) = f''(-x)+f(-x)=\frac{1}{f(-x+3\pi/2)}=\frac{1}{f(x-3\pi/2)}\]

Therefore $f$ is $3\pi$-periodic and $2\pi$-periodic, therefore $f$ must be $\pi$-periodic
That means
\begin{align*}
&f''(x)+f(x)=\frac{1}{f(x+\pi/2)}\\
&f''(x+\pi/2)+f(x+\pi/2)=\frac{1}{f(x+\pi)}=\frac{1}{f(x)}\
\end{align*}
That means that \[f''(x)f(x+\pi/2)-f''(x+\pi/2)f(x)=0 \Rightarrow (f'(x)f(x+\pi/2)-f(x)f'(x+\pi/2))'=0\]
Therefore \[f'(x)f(x+\pi/2)-f(x)f'(x+\pi/2) = c\]
Since $f(x+\pi/2)=f(x-\pi/2)$ ($f$ is even and has period $\pi$), by setting $x\rightarrow -x$ we get that \[-f'(x)f(x+\pi/2)+f(x)f'(x+\pi/2) = c\]

So, we got that $c=0$.
\begin{align*}
&f'(x)f(x+\pi/2)-f(x)f'(x+\pi/2) = 0 \\\\
\Rightarrow &(f(x)/f(x+\pi/2))'=0 \\\\
\Rightarrow &f(x)=kf(x+\pi/2)=k^2f(x+\pi)=k^2f(x) \\\\
\Rightarrow &k^2=1 \Rightarrow k\in\{1,-1\}\
\end{align*}
Since $f$ doesn't vanish and is continuous, $f$ must have the same sign on $\mathbb{R}$. That forces $k$ to be $1$, which means $f$ has period $\pi/2$.

[quote=ehsan2004]Let $ f : \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable, $ 2 \pi$-periodic even function. Prove that if \[ f''(x)\plus{}f(x)\equal{}\frac{1}{f(x\plus{} 3 \pi /2 )}\] holds for every $ x$, [/quote]

Can one give an example $f$ function wich satisfy above

Since $f$ is $\pi$ periodic
$ f : \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable, $ \pi$-periodic even function $\[ f''(x)\plus{}f(x)\equal{}\frac{1}{f(x)}\]$ holds for every $ x$

Can one give an example $f$ function wich satisfy this condition

Probably not what you are looking for but $f(x) = 1$ works...

A function satisfying $\frac{df}{dx}^2=C+2ln(f)-f^2, where C is a constant

[quote=SassTramm]A function satisfying $\frac{df}{dx}^2=C+2ln(f)-f^2$where C is a constant[/quote]

Do you an explicit form, I mean with usual functions ? Does it involve trigonometry functions since f is pi- periodic?


Hello,
I am having a problem with the following problem:

"Find an equation of the tangent line to the circle $x^{2}+y^{2}=25$ at the point $(4,-3)$."

Please give some pointers - thank you.

edit - couldn't find latex button

This is way too hard for Classroom. This should be moved to Getting Started.

Oh? I am sorry. It was in my Equations of Lines packet.
Apologies.


What was the highest score in your chapter?

3 perfects in Santa Clara Valley, California. I got ONE sprint problem wrong and I ended up tied for 6th.  :?  Stupid tiebreakers.

2nd in Countdown due to sudden panic after the guy I was against in the final beat me in the first question.

In the Central Florida Chapter, there were 5 perfect scores, mine being one of them, hehe! :) :) I came up 2nd because this guy went from 6th to first.

I got full score, as the only one.  So it was :winner_first::winner_first::winner_first::winner_first::winner_first::winner_first::winner_first::winner_first:!!!

One of 3 perfect scores in Lake Washington Chapter (in Washington, of course).  I won the first ever tiebreaker in our chapter! YAY!

sorry boys... In our chapter.. I got first with a 35 =)

The first place in our chapter got a 35 as well.

wow in my chapter some guys got 40s and 42s, but the 40s were 5th to 6th

Aren't you not allowd to post your score?

EDIT: whoops, new announcement sry.

I got 1st with 44 in my chapter.

I got first with a 43.  Next was like 20 points down, :rotfl:  :rotfl:

Sounds like Delaware state will be this year

[hide]
[hide]
[hide]

just kidding

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

I was 4th w/ 45. The lowest score at our Seattle chapter was 41. But I think at the Lake Washington chapter, the lowest score was 42.

Why? What happened at your chapter? I was in a 3 way tie, but a girl beat everyone and went from 10th to 1st in the countdown round. At the Lake Washington Chapter, there were 3 perfect scores.

our 10th got 40 (can't remember what the others got).

there are people out there who are...mathmatically challenged.

[quote="robinhe"]Our 83rd place got 3  :rotfl:  :rotfl:[/quote]
A 3!!!
If someone is that bad my coach wouldn't have even taken them. She'd rather not take the maximum if it means taking someone who isn't good at these problems.

keep in miond it mioght have been an inexperienced coach who didn't know what the competition was like...or the kid was having a really bad day!

I'm sure there were multiple 3- at my chapter. Its amazing how low some scores are.

Our 10th was either 45 or 44.
83rd would've been somewhere around 32.

A 3 would've gotten you a solid 190th place.

There were only like 50 people at my chapter.  190 is a lot for just one chapter.

Yeah, our chapter has like 60 people.  I got a  :first:  with a 43 though, but last year i won it with only 37.  I can't believe that people even got perfect scores.  That's crazy that more than one person would from one chapter.  Our 10th place about 30.  I'm scared of state, but really want to go to nationals.  Good luck to everyone else just as screwed as i am  :lol:

I do Mathcounts! It Rules! NEW YORK, BABY! :lol:

Embarassed I only came in third individually in my chapter cause I made four retarded mistakes
I knew the answers, just wrote down in wrong unit Sad

My coach got the results then took them while he's on a bussiness trip and he still hasn't come back. His son told me that I got top ten but I want to know which place. Anyone who went to Georgia Tech's chapter with the results, please email me at jingesu9@gmail.com
thanks.
Oh yeah, my team got first in chapter.

Which chapter were you in? There were 5 different chapters at the competition at GA Tech. :o I think GA's the only place with multiple competitions at the same place at the same time.....

actually no. my uncle proctors a chapter, (actually two) on the border of illinois and iowa. there's one chapter from each state there, so they have to do countdown rounds seperately.

Colorado Northern Chapter, 45 by me!
I know it's a bit late to be saying this (State was today), but it's true!


Please put $ 6$ coins on $ 5 \times 5$ grid points so that all each distance is different.

[asy]for(int n=0; n <= 5; ++n)
draw((n,0)--(n,5)^^(0,n)--(5,n));

filldraw(circle((1,2),0.25));
filldraw(circle((1,4),0.25));
filldraw(circle((3,1),0.25));
filldraw(circle((3,3),0.25));
filldraw(circle((4,2),0.25));
filldraw(circle((5,4),0.25));[/asy]

This problem is questions from "Takeshi Kitano presents Comaneci University Mathematics".
It is a Japanese TV program.

[quote="bobma99"]oh nvm 5x5
6 points means you need 15 distances i believe and on a 5x5 thats possible....
hm.....[/quote]

Yes, that's right :)
$ \binom{6}{2}\equal{}15$.

And, ｔhe existing distances is $ \sqrt{1}$, $ \sqrt{2}$, $ \sqrt{4}$, $ \sqrt{5}$, $ \sqrt{8}$, $ \sqrt{9}$, $ \sqrt{10}$, $ \sqrt{13}$, $ \sqrt{16}$, $ \sqrt{17}$, $ \sqrt{18}$, $ \sqrt{20}$, $ \sqrt{25}$, $ \sqrt{26}$, $ \sqrt{29}$, $ \sqrt{32}$, $ \sqrt{34}$, $ \sqrt{41}$, $ \sqrt{50}$.
($ 19$ distances.)


linear algebra
matrix
algebra
polynomial
LaTeX
function
absolute value

Solve for $ x$:

$ \begin{pmatrix}2 & 1 & x \\
0 & 1 & 1 & \\ 1 & x & 0\end{pmatrix} \minus{} \begin{pmatrix}5 & 1 \\ x & 2\end{pmatrix} \equal{} 8$.

As stated, that is nonsense. One cannot add, or equate, matrices of different sizes.

Do you perhaps mean to take the determinant of each of those matrices? That would make this a polynomial equation.

I was looking for a latex to put "vertical lines" (like absolute value) instead of the "parentheses"... I am not familiar with the difference between matrices and determinants.

[quote="jeez123"]$ \begin{vmatrix}2 & 1 & x \\ 0 & 1 & 1 & \\ 1 & x & 0\end{vmatrix} \minus{} \begin{vmatrix}5 & 1 \\ x & 2\end{vmatrix} \equal{} 8.$[/quote]

It's vmatrix.
I usually use the notation $ \det(\cdot)$; it makes it clear that the determinant is a function from matrices to numbers in the base field.

At this point, it's really just a grind-through-it calculation. The equation turns out to be linear.

$ (1\minus{}3x)\minus{}(10\minus{}x)\equal{}8$ yields $ x\equal{}\minus{}\frac{17}{2}.$

For some reason, I kept making computational errors, so it took a while for me to get this to check.


Let $ A_1,A_2,...,A_n$ be finite sets. If $ |A_i \cap A_j| \equal{} i \plus{} j$, $ (\forall) i\neq j$, find $ min|A_1 \cup A_2 \cup ... \cup A_n|$.

Where $ |A|$ is the number of elements of A.

Sorin Dumitrica

[hide]One needs to craft a careful inductive argument with the following ingredients:
1. Lets call the set $ \{A_1,A_2,\dots,A_n\}$ optimal if $ |A_1\cup A_2\cup \dots \cup A_n|$ is minimum. For an optimal set $ A_{n\minus{}1}\subset A_n$.

2. For a set $ \{A_1, \dots, A_{n\plus{}1}\}$, the number $ |A_1\cup A_2\cup \dots \cup A_{n\plus{}1}|$ is at least $ n\plus{}1$ more than $ min\{|A_1\cup A_2\cup \dots \cup A_n|\}$[/hide]


inequalities
Cauchy Inequality
inequalities proposed

Let $ a, b, c$ positive real numbers. Prove or disprove that

$ \sum[\frac{ab \plus{} 3c}{a \plus{} b \plus{} 2c}]\le \frac{a \plus{} b \plus{} c \plus{} 9}{4}$.

My answer for your inequality is absolutely true. Well, firstly, the inequality can be divided into 2 parts.
The first part is to prove
$ \sum_{cyc}\frac{ab}{a\plus{}b\plus{}2c} \le \frac{a\plus{}b\plus{}c}{4}$
and the second part need proving like
$ \sum_{cyc}\frac{3c}{a\plus{}b\plus{}2c} \le \frac{9}{4}$
They are both trivial after using Cauchy inequality according to the standard form
$ \frac{1}{a} \plus{}\frac{1}{b} \ge \frac{4}{a\plus{}b}$


limit
trigonometry
calculus
calculus computations

Evaluate the limit :
\[ \lim_{n \to \infty}{\sqrt[3]{n\plus{}1} \cdot \cos{\sqrt{n\plus{}1}}}\minus{} \sqrt[3]{n} \cdot \cos{\sqrt{n}}.\]

[quote="§outh§tar"]Stolz-Cesaro does the trick.[/quote]
Stolz-Cesaro doesn't do the trick.
In this case it only shows that the limit is 0, IF the limit exists.


modular arithmetic
number theory
least common multiple
greatest common divisor
function

hello, we have $ 3^{12} \equiv 381 \mod 2004$ and $ 381^{167}\equiv 381 \mod 2004$ so your result is
$ 3^{2004}\equiv 381\ mod 2004$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, we have $ 3^{12} \equiv 381 \mod 2004$ and $ 381^{167}\equiv 381 \mod 2004$[/quote]

Why is the last one true?

$ 381^{167}\equiv 3x\pmod{2004}\implies 381^{166}\cdot 127\equiv x\pmod{668}$.

We conveniently have $ 668\equal{}2^2\cdot 167\implies\lambda(668)\equal{}\text{lcm}(\lambda(2^2),\lambda(167))\equal{}166$. Therefore, because $ \text{gcd}(381,668)\equal{}1$, we have $ 381^{166}\equiv 1\pmod{668}$, so $ x\equiv 127\pmod{668}\implies 3x\equiv 381\pmod{2004}$.

I suppose Dr Sonnhard Graubner simply left that computation to the reader  :D.

(If anyone doesn't know, $ \lambda(n)$ denotes the [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael Function[/url]).


calculus
integration
geometry
trigonometry
calculus computations

Find the values of $ a,\ b$ such that the area bounded by the curve $ y \equal{} \cos x\ \left(0\leq x\leq \frac {\pi}{2}\right)$, the $ x$-axis and $ y$-axis are devided into three equal parts by the curves $ y \equal{} a\sin x,\ y \equal{} b\sin x\ (a > b)$.


$a, b, c$ are real numbers. Given two functions: $f(x)=ax^2+bx+c$ and $g(x)=ax+b$. When $|x| \leq 1$ we have $|f(x)|\leq 1$.
(1) Prove: $|c| \leq 1$
(2) Prove: When $|x| \leq 1$, prove $|g(x)| \leq 2$
(3) Let $a>0$. When $|x| \leq 1$, the maximum value of $g(x)$ is $2$. Find $f(x)$.

For 1:
Assume the contrary that $|c|>1$ then $f(0)>1$ contradiction.
For 2:
\[1\geq |f(x)|\geq |ax^2+bx|+c\geq |g(x)|+c\]
\[1-c\geq |g(x)|\]
\[2\geq |g(x)|\]
from 1

For 3:$f(x)=2x^2-1$. This problem's background is Tchevichev's polynomial.

[quote="amd2357"]$1\geq |f(x)|\geq |ax^2+bx|+c\geq |g(x)|+c$[/quote]
However, the inequality $|ax^2+bx+c|\geq |ax^2+bx|+c$ is not always true. It is true only when $c$ is negative.
[hide="solution to part 2"]Since $\begin{cases}f(0)=c\\f(-1)=a-b+c\\f(1)=a+b+c\end{cases}$, we have $a=\frac{1}{2}[f(1)+f(-1)-2f(0)]$, $b=\frac{1}{2}[f(1)-f(-1)]$.
So when $-1\leq x\leq 1$,
$|g(x)|=|ax+b|$
$=|\frac{1}{2}[f(1)+f(-1)-2f(0)]x+\frac{1}{2}[f(1)-f(-1)]|$
$=\frac{1}{2}|(x+1)f(1)+(x-1)f(-1)-2xf(0)|$
$\leq\frac{1}{2}(|x+1||f(1)|+|x-1||f(-1)|+|2x||f(0)|)$
$\leq\frac{1}{2}(|x+1|+|x-1|+2)$
$=\frac{1}{2}((x+1)+(1-x)+2)$
$=2$[/hide]
[hide="solution to part 3"]Since $a>0$, $g(x)=ax+b$ is increasing on the interval $[-1,1]$. When $x=1$, $g(x)$ attains its maximum. From the conditions,
$1=g(1)=a+b=f(1)-f(0)$
Therefore $-1\leq f(0)=f(1)-2\leq 1-2\leq -1$
So $c=f(0)=-1$
When $|x|\leq 1$, $|f(x)|\leq 1$, so
$f(x)\geq -1=c=f(0)$
This shows that on the interval $[-1,1]$, $f(0)$ is the minimum of $f(x)$. $x=0$ must be the axis of symmetry of $f(x)$. Therefore, $b=0$, $a=2$ and $f(x)=2x^2-1$.[/hide]

Actually, $g_{max}(x)=2 \Longleftrightarrow |f(1)|=|f(-1)|=|f(0)|=1$. From here we can easily get $f(x)$.


geometry
perpendicular bisector
geometry unsolved

let triangle $ ABC$ be a triangle and construct squares $ ABB_1A_2$, $ BCC_1B_2$, $ CAA_1C_2$ externally on its sides.prove that perpendicular bisector of $ A_1A_2,B_1B_2,C_1C_2$ are  concurrent


calculus
integration
number theory
prime factorization
geometric sequence

Prove that if p is a prime number, then $p^n$ is never a perfect number.

It comes directly if you write :
p^n = p*p....*p (n times)

:)

[quote="Altheman"][hide]since $p$ is prime, then we have that the prime factorization is $p^n$

so the definition of a perfect number is that the sum of the divisors is double the number,

clearly the divisors are 1,p,p^2,...p^n, and that is a geometric sequence so

$\sum_{k|p^n}k=\frac{p^{n+1}-1}{p-1}$

then using the definition of a perfect number, we have that

$\frac{p^{n+1}-1}{p-1}=2p^n$
$p^{n+1}-1=2p^{n+1}-2p^n$
$p^{n+1}-2p^n+1=0$
and the only integral answer we get for that is 1, and that is not prime, so there are no perfect numbers in that form[/hide][/quote]

I'm not entirely convinced by your proof. Why can't primes be roots of your equation? Your final equation can be obtained by multiplying $p^n=p^{n-1}+p^{n-2}+\cdots+1$ (which is what we want to prove) by $p-1$. Basically, it's restating the given.

That's rational root theorem..  Basically you can write, his last line as $p-2+ \frac 1{p^n}=0$  now $p-2$i integer and last part can not be integer.


polynomial
algebra
function
domain
Ring Theory
superior algebra
superior algebra theorems

Through some research, I found that for any N, there can only be up to one Gaussian prime with a norm of N, up to associates and their conjugates. Is this true for Eisenstein primes as well? How can this be proven?

More specifically, any rational prime congruent to 3 modulo 4 is a Gaussian prime, as are its associates. Any Gaussian integer with a rational prime norm congruent to 1 modulo 4 is a Gaussian prime, as are its associates and their conjugates - and for any rational prime congruent to 1 modulo 4, there is always [i]gauranteed[/i] to be a Gaussian integer with a norm equal to it. With Eisenstein integers, all is in terms of congruence to 1 modulo 3 versus to 2 modulo 3. Does the same existence and uniqueness apply to Eisenstein primes that does to Gaussian ones?

Lastly, can any of this be extended to arbitrary Unique Factorization Domains?[/i]

Yes; this is a consequence of unique factorization, which holds in both the Gaussian and Eisenstein integers.

Your second question doesn't quite make sense as stated.  For one thing, not every UFD has a natural notion of norm.

Generally a number ring is a UFD iff PID iff all irreducibles have prime-power norms (i.e. norms divisible by only one rational prime).


group theory
abstract algebra
superior algebra
superior algebra unsolved

Find "all" the Sylow subgroups of S5 and S6

I don't know how to find 1.Sylow 2-subgroups of S5
                                 2.Sylow 2-subgroups of S6
                                 3.Sylow 3-subgroups of S6
                                 4.Sylow 5-subgroups of S6

I use Sylow thms to get the possible numbers of Sylow subgroups but don't know how to find the right one.
Even I know the right number.
How to find them all?

You may want to work out the Sylow 5-subgroups of S6 first.  Make sure you know how to find the Sylow 3-subgroups of S3, S4, and S5 too.  You should be able to find the Sylow 5-subgroups of S5, S6, S7, S8, and S9.

Let n_5 = #Sylow 5-subgroups of S5 ;
then n_5|5!/5 = 24 and is congruent to 1 mod 5, so n_5 = 1 or 6.
But a (sub)group of order 5 is cyclic, so generated by some 5-cycle;
and the 5-cycles are of the form (abcde), of which there are 5!/5 = 24.
Since the Sylow 5-subgroups intersect trivially,
there are n_5 (5 - 1) = 24 elements of order 5 in S_5,
so we must have n_5 = 6.

Similarly for S6.
Let n_5 = #Sylow 5-subgroups of S6 ;
then n_5|6!/5 = 144 and is congruent to 1 mod 5, so n_5 = 1 or 6 or 36.
Again,a (sub)group of order 5 is cyclic, so generated by some 5-cycle;
and there are 24*6=144 5-cycles.
Since the Sylow 5-subgroups intersect trivially,
there are n_5 (5 - 1) = 144 elements of order 5 in S_5,
so we must have n_5 = 36.

Simiarly for Sylow 3-subgroups of S5.

How to proceed to find
1.Sylow 2-subgroups of S5
2.Sylow 2-subgroups of S6
3.Sylow 3-subgroups of S6 ??

Yup.

The next easiest is probably the Sylow 3-subgroups of S6.

< (1,2,3), (4,5,6) > has order 9, and 9 is the highest power of 3 dividing 6!.

Any conjugate < (a,b,c), (d,e,f) > of < (1,2,3), (4,5,6) > is also a Sylow 3-subgroup.  How many conjugates are there?  Well choosing the conjugate amounts to choosing the set of orbits {{a,b,c},{d,e,f}}.  There are Binomial(6,3)/2=10 ways to do this (choosing a,b,c is Binomial(6,3) ways, then {d,e,f} is determined, but instead of choosing {a,b,c} it is the same to choose {d,e,f}, so one has double counted).

One can also check that the normalizer of < (1,2,3), (4,5,6) > is < (1,2,3), (4,5,6), (2,3), (5,6), (1,4)(2,5)(3,6) > of order 72 and index 10.

This should also describe how to find the Sylow 3-subgroups of S7 and S8.

Describing the Sylow 2-subgroups of S5 is the same as describing the Sylow 2 subgroups of S4.

< (1,2), (3,4), (1,3)(2,4) > has order 8, so it is a Sylow 2-subgroup of S4 and S5.

So is any conjugate < (a,b), (c,d), (a,c)(b,d) >.  Again choosing such a conjugate comes down to choosing {{a,b},{c,d}}, and so there are Binomial(4,2)/2=3 ways to do this in S4 and Binomial(5,2)/2*Binomial(3,2)=15 ways to do this in S5.  In S4 {a,b} determines {c,d}, but in S5 you still have three points left, so there is the factor of Binomial(3,2).

Alternatively one can calculate that the Sylow 2-subgroup is its own normalizer, and so has index 3 in S4 and and index 15 in S5.

The Sylow 2-subgrous of S6 and S7 are somewhat similar.  They are all < (a,b), (c,d), (a,c)(b,d), (e,f) >.  There are Binomial(6,2)*Binomial(4,2)/2=45 ways of choosing {{a,b},{c,d}} for S6 and Binomial(7,2)*Binomial(5,2)/2*Binomial(3,2)=315 ways of choosing ({{a,b},{c,d}},{e,d}) for S7.  Notice that in S6 {e,f} is determined by {{a,b},{c,d}}.  Again you can also calculate that the Sylow 2-subgroups are self-normalizing, and so have index 45 and 315.

I know the groups of the form < (a,b,c),(d,e,f)> are all Sylow 3-subgroups of S6.
But why all Sylow 3-subgroups of S6 are of the form < (a,b,c),(d,e,f)> ?
I'm confused.

Similar problems in the case Sylow 2-subgroups of  S5 and S6.

[quote="laipou"]I know the groups of the form < (a,b,c),(d,e,f)> are all Sylow 3-subgroups of S6.
But why all Sylow 3-subgroups of S6 are of the form < (a,b,c),(d,e,f)> ?[/quote]

This follows from the "conjugacy" part of Sylow's theorem and a description of conjugacy in Sn.

All Sylow 3-subgroups of S6 are conjugate to < (1,2,3),(4,5,6) > by Sylow's theorem.  In other words, if P is a Sylow 3-subgroup of S6, then there is a g in S6 such that P = $ \langle (1,2,3),(4,5,6) \rangle^g$.

In S6, (1,2,3)^g = ( 1^g, 2^g, 3^g ) and (4,5,6)^g = ( 4^g, 5^g, 6^g ) where g is the permutation that takes 1 to 1^g, 2 to 2^g, 3 to 3^g, 4 to 4^g, 5 to 5^g, and 6 to 6^g.  In other words, P = < (a,b,c), (d,e,f) > where a,b,c,d,e,f are the images under the permutation g of 1,2,3,4,5,6.


Hi! Mathlinkers.

How many people are there  those who can understand or speak Japanese,in this site?

kunny

Hi kunny! :)

I've lived in Japan for 15 yrs since I was born, and came to the U.S. 2 yrs ago. My English skill has imporved a lot since I came here, but I still have to work on it. Anyway, I can definitely understand Japanese because it's my native language ;)

I saw your posts everywhere, and it seems like you're really fluent in both English and Japanese... When did you come to the U.S.?

I'm sure there are a couple of other Japanese AoPSers, but maybe they haven't found this topic yet.

BTW, I'm new to the AoPS, and I really feel like I should've found this web site earlier. This is definately one of the most excellent math web sites that I've ever seen!!! :)

Hello,frt.

Thank you for your reply.I'm very glad to receive that.

[quote="frt"]
I've lived in Japan for 15 yrs since I was born, and came to the U.S. 2 yrs ago. My English skill has imporved a lot since I came here, but I still have to work on it. Anyway, I can definitely understand Japanese because it's my native language ;) [/quote]

Me,too. :)  ;)   :lol:

[quote="frt"]
I saw your posts everywhere, and it seems like you're really fluent in both English and Japanese... When did you
come to the U.S.?[/quote]

No, my English is not good. I can speak English a little bit. I live in Tokyo now.If there is something wrong about my English, please correct for me without restraint, because I want to learn plain English. :)

[quote="frt"]
I'm sure there are a couple of other Japanese AoPSers, but maybe they haven't found this topic yet.
[/quote]

 :(  Come on, every one! :)

[quote="frt"]
BTW, I'm new to the AoPS, and I really feel like I should've found this web site earlier. This is definately one of the most excellent math web sites that I've ever seen!!! :)[/quote]

Yes, I think this site is very excellent. It's a pity that there are few site such as this in Japan.

Best regards.

kunny

[quote]No, my English is not good. I can speak English a little bit. I live in Tokyo now.[/quote]

Your English is much better than me ;)  Just curious, are you a high school student? (or...)

[quote]It's a pity that there are few site such as this in Japan.
[/quote]

Yeh, I agree. We need more math-related web sites in Japan.

Also, why is it that there's no Japanese forum in AoPS while there are Korean and Chinese ones? Is it just because there are very few Japanese people here?

Yuichi (my real first name) :)

I think Kunny's purpose of making this post is to have a Japanese community built in ML.

[quote="frt"]Also, why is it that there's no Japanese forum in AoPS while there are Korean and Chinese ones? Is it just because there are very few Japanese people here?

Yuichi (my real first name) :)[/quote]Yup, that's the reason. As soon as you guys gather (3-4 persons at least) we can easily make one for you :)

I can speak.. lets see... a few words/phrases and i can count in Japanese! I visited Japan for a month when i was 5 and thats all I remember...

O'renkideskaw = How are you?

Konnichiwa, ToBe.  :)

[color=blue]Ogenkidesuka?[/color] Message [color=red]arigatou[/color]. Where did you visit in Japan?

I visited Canada several years ago.

kunny

Sugoi...

I used to live in Japan for four years, but I'm nowhere near fluent.  I haven't spoken Japanese with someone else in a REALLY LONG time so my skill with it kind of faded away :-(.  And my accent sucks.

I know how to count to five in Japanese, sort of  :blush: , and know how to say hello and goodbye. Are japanese symbols related in anyway to chinese characters?

[quote="236factorial"]Are japanese symbols related in anyway to chinese characters?[/quote]
Yeah, Japanese letters originally came from Chinese letters. A long time ago, Japanese adopted Chinese letters and started using them. That's the origin of one of the three Japanese characters ([i]kanji[/i]). Then, for convenience, Chinese letters were simplified to create the second type of Japanese characters ([i]hiragana[/i]). The third type of Japanese characters, [i]katakana[/i], was also created from Chinese letters.

Katakana is soooo easy to learn though, then hiragana, then kanji...well...I won't even go there :-o.  2000+ characters...

My friend speak Japanese, and i know a little bit as well.


remember that tan(x)=sin(x) divided by cos(x)

[quote="Farcus"]cos(x)+4sin(x)=19/5

determine the value of sec(x)

and a similiar
2cos(x)+sin(x)=2

determine value of tan(x)

give me some hints please i want to actually figure this out[/quote]
What well-known ratios could be the values for $\cos x$ and $\sin x$ (Hint: they're very well-known :D)
[hide="Answers (kinda)"] $\cos x=\frac{3}{5}$
$\cos x =1$
:D [/hide]

[quote="mysmartmouth"]How can we do this with identities/algebraically?[/quote]
Use $\sin x =\pm\sqrt{1-\cos^{2}x}$. Isolate the square root, then square both sides. Solve the quadratic for $\cos x$. :D


This problem is from AIME 1983.
w and z are complex numbers such that w^2 + z^2=7, and w^3 + z^3=10, what is the LARGEST REAL VALUE OF w + z.

The solution was posted, along with lots of other problems, at http://www.kalva.demon.co.uk but I dont see how to come up with it.
Post EXPLANATIONS.

A Try :

X = w + z

X^3 = w^3 + z^3 + wz * X = 10 + wz * X

X^2 = w^2 + z^2 + 2 wz = 7 + 2 wz

Replacing wz in the first equation by (X^2 - 7)/2 gives a real cubic equation with (presumably, too lazy to compute it) obvious real roots.

Let x = w + z, and y = wz.

(w + z)^2 = w^2 + 2wz + z^2 = 7 + 2wz
--> x^2 = 7 + 2y, y = (x^2 - 7)/2

(w + z)^3 = w^3 + 3w^2z + 3wz^2 + z^3 = 10 + 3wz(w + z)
--> x^3 = 10 + 3xy

So 2x^3 = 20 + 3x(x^2 - 7)
x^3 - 21x + 20 = 0
(x-1)(x^2 + x -20) = 0
(x-1)((x+5)(x-4) = 0
--> x = 1, 4, or -5

So the largest possible real value for x = w + z is 4.

In general whenever you see a problem involving symmetric terms and sums of powers, think of looking at the expansions of their powers. The sums usually appear somewhere in there.

This is similar to i/3:

Let X=w+z and Y=xz. Then

X^2-2Y=7
X^3-3XY=X(X^2-2Y)-XY=7X-XY=10.

Y=(X^2-7)/2

7X-X^3/2+7X/2=10, or

X^3-21X+20=0.

1 is a root, so dividing gives X^2+X-20, which has roots at 4 and -5, so the maximum for w+z is 4.


Let $ U \equal{} 2\times2004^{2005}$, $ V \equal{} 2004^{2005}$, $ W \equal{} 2003\times2004^{2004}$, $ X \equal{} 2\times2004^{2004}$, $ Y \equal{} 2004^{2004}$ and $ Z \equal{} 2004^{2003}$. Which of the following is the largest?

$ \textbf{(A)}\  U \minus{} V \qquad \textbf{(B)}\  V \minus{} W \qquad \textbf{(C)}\  W \minus{} X \qquad \textbf{(D)}\ X \minus{} Y \qquad \textbf{(E)}\ Y \minus{} Z$

[hide="Solution"]A.

Some quick factoring shows us that A is largest.  :) [/hide]

[hide]W-X is my quick guess looking at the numbers...
edit: should stop making quick guesses, hehehe[/hide]

[hide]A. U-V is the largest.[/hide]

Did you guys use a calculator or is there an easier way to do it?

[quote="XCgirl"]Did you guys use a calculator or is there an easier way to do it?[/quote]
You can factor them. These numbers would be too big for most calculators anyway. ;)

[quote="catcurio"]These numbers would be too big for most calculators anyway. ;)[/quote]Make that "for all calculators."  :D


Let be $ a,b,c\in \mathbb{R}_\plus{}$ such that $ \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}\equal{}1$ . Prove that :

$ a(\sqrt{b}\plus{}\sqrt{c})\plus{}b(\sqrt{c}\plus{}\sqrt{a})\plus{}c(\sqrt{a}\plus{}\sqrt{b})\le \frac{1}{4}$

[quote="alex2008"]Let be $ a,b,c\in \mathbb{R}_ \plus{}$ such that $ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \equal{} 1$ . Prove that :

$ a(\sqrt {b} \plus{} \sqrt {c}) \plus{} b(\sqrt {c} \plus{} \sqrt {a}) \plus{} c(\sqrt {a} \plus{} \sqrt {b})\le \frac {1}{4}$[/quote]
Put $ \sqrt{a}\equal{}x,,\sqrt{b}\equal{}y,,\sqrt{c}\equal{}z$.This inequality becom:
$ x^3\plus{}y^3\plus{}z^3\plus{}6xyz \ge x^2(y\plus{}z)\plus{}y^2(z\plus{}x)\plus{}z^2(x\plus{}y)$
Which is obvious true :)


Either the book I am using right now (NOT aops) has made a very big typo, or something is wrong with me... :(

"However, since there are n distinct nth roots of each complex number, the answer is not unique. Thus {complex numbers} is not closed under the operation [nth root of - I don't know how to LaTeX that, sorry]"

I thought complex numbers were closed...

It's more that it is not a function than that the complex numbers aren't closed under it.

Cheers,

Rofler

Without more context it's not clear what is being said.  The map "find all complex $ n$th roots of $ z$" is a map from complex numbers to [i]sets[/i] of complex numbers -- it looks to me that this may be the (not very interesting) point.  It is true that every algebraic equation with complex coefficients has the "right" number of complex solutions (i.e., the Fundamental Theorem of Algebra).


Let $\{a_{n}\}$ be a strictly increasing positive integers sequence such that $\gcd(a_{i}, a_{j})=1$ and $a_{i+2}-a_{i+1}>a_{i+1}-a_{i}$. Show that the infinite series \[\sum^{\infty}_{i=1}\frac{1}{a_{i}}\] converges.

maybe the problem is not exatly same writen here...because the condition $ \gcd(a_i,a_j) \equal{} 1$ is extra...
$ a_n \minus{} a_{n \minus{} 1} > a_{n \minus{} 1} \minus{} a_{n \minus{} 2} > .... > a_2 \minus{} a_1\geq1$ therefore $ a_n \minus{} a_{n \minus{} 1}\geq{n \minus{} 1}$...and also:

$ a_k \equal{} (a_k \minus{} a_{k \minus{} 1}) \plus{} (a_{k \minus{} 1} \minus{} a_{k \minus{} 2}) \plus{} .... \plus{} (a_2 \minus{} a_1) \plus{} a_1\geq{(k \minus{} 1) \plus{} (k \minus{} 2) \plus{} .... \plus{} 1 \plus{} 1}\Rightarrow{a_k > \frac {k(k \minus{} 1)}{2}} \\
\Longrightarrow\sum^{\infty}_{i \equal{} 1}\frac {1}{a_{i}} < \frac {1}{a_1} \plus{} \sum^{\infty}_{i \equal{} 2}\frac {2}{i(i \minus{} 1)} \equal{} a_1 \plus{} 2\sum^{\infty}_{i \equal{} 2}{\frac {1}{i \minus{} 1} \minus{} \frac {1}{i}} \equal{} a_1 \plus{} 2$

Very good point, and nice solution. I have no idea what $ \gcd(a_i,a_j)\equal{}1$ is doing there.  :maybe:

Maybe the $ >$ should be a $ \geq$ (and the $ \gcd(a_i,a_j)\equal{}1$ still be there). Otherwise it isn't number theory at all.

[quote="ZetaX"]Maybe the $ >$ should be a $ \geq$ (and the $ \gcd(a_i,a_j) \equal{} 1$ still be there). Otherwise it isn't number theory at all.[/quote]That looks more like a NT problem indeed. Do you have a solution for that problem?

A solution for when we only assume $a_{i+2} - a_{i+1} \ge a_{i+1} - a_i$ for all $i \ge 1$, as suggested by ZetaX almost 15 years ago:

For a constant $c \in \mathbb{N}$ let $S_c = \{a_k, a_{k+1}, \ldots, a_{k+l-1}\}$ be the set of elements such that the difference $a_{k+i} - a_{k+i-1}$ is exactly equal to $c$ for all $i$ with $1 \le i \le l$. I claim that $|S_c| = l < 2c-3$ for all $c \ge 3$. Otherwise, note that $a_k, a_{k+1}, \ldots, a_{k+c-2}$ are all distinct modulo $c-1$, and so are $a_{k+c-1}, a_{k+c}, \ldots, a_{k+2c-3}$. That means that in each of these two subsequences there exists an element divisible by $c-1$, which is not allowed by the assumption $\gcd(a_i, a_j) = 1$. Likewise, $|S_1| < 3$ and $|S_2| < 5$, as otherwise there must be two integers divisible by two or three respectively. \\

The inequalities $|S_1| < 3$ and $|S_2| < 5$ imply $a_5 - a_4 \ge 2$ and $a_{10} - a_9 \ge 3$. The inequality $|S_c| < 2c-3$ for $c \ge 3$ can now be used to show via induction, that $a_{i+1} - a_i \ge n$ holds for all $i \ge n^2$ in general. Or, stated differently, $a_{i+1} - a_i \ge \left \lfloor \sqrt{i} \right \rfloor$. This, in turn, implies that $\displaystyle a_{i+1} = a_1 + \sum_{j=1}^{i} (a_{j+1} - a_j) \ge 1 + \sum_{j=1}^i \left \lfloor \sqrt{j} \right \rfloor > \sum_{j=1}^i \frac{1}{2} \sqrt{j} > \int_0^i \frac{1}{2}\sqrt{j} \, dj = \frac{i^{3/2}}{3}$.
Now the convergence of $\displaystyle \sum_{i=1}^{\infty} \frac{1}{a_i}$ is clear.


Let $ a,b,c>0$. Prove that

$ \frac{a^3\plus{}4a\plus{}2}{b\plus{}c}\plus{}\frac{b^3\plus{}4b\plus{}2}{c\plus{}a}\plus{}\frac{c^3\plus{}4c\plus{}2}{a\plus{}b}\geq\frac{21}{2}$

${ \{{a}^{3} + 1 + 1 > = 3a }$

and it becomes well-known equality.\


floor function
inequalities
algebra
polynomial
function
number theory unsolved
number theory

Find all real numbers $x$ such that $\lfloor x^3 \rfloor = 4x + 3$.

This is rather boring "go through all cases" type solution for this
problem, but it is correct and totaly doable in about 10 minutes, so
never mind. First what we can see that $4x$ is an integer, in other
words $x\in \{\frac{k}4|k\in Z\}$.

Suppose that $[x^3]=0$ then $x=-\frac34$. If $x>0$ then $x>1$ (or
else $[x^3]=0$) so we have $4x+3=[x^3]>x^3-1$ so $4x+4>x^3$. From
this we easily get $x<3$ so $1<x<3$. $x=\frac{a}4$ where $a$ is
integer so $a\in \{5, 6, 7, ..., 11\}$, and just try them all.

If $x<0$ then $x^3+1>[x^3]=4x+3$ so $x^3>4x+2$. Again $-1<x<-3$ so if $x=\frac{a}4$ now $a\in \{-5, -6, ..., -11\}$, again, try them
all, and that's it.

In order to determine all solutions of the equation

$(1) \; \lfloor x^3 \rfloor = 4x + 3$

we write

$(2) \; x^3 = n + \theta$

there n = $\lfloor x^3 \rfloor$ is an integer and $\theta = x^3 - \lfloor x^3 \rfloor$ is a real number in the interval [0,1). Thus by (1)

$(3) \; x = \textstyle{\frac{n - 3}{4}},$

which inserted in (2) gives

$(4) \; \theta = (\textstyle{\frac{n - 3}{4}})^3 - n.$

Since $\theta$ can be any real number in [0,1), the solutions of (1) is the integers n which satisfies the two inequalities

$0 \leq (\textstyle{\frac{n - 3}{4}})^3 - n < 1$

which is equivalent to

$(5) \; 0 \leq (n + 1)(n^2 - 10n - 27) = f(n) < 64$.

The polynomial function $f(n) \geq 0$ if and only if $n \in [5 - 2\sqrt{13},-1] \cup [5 + 2\sqrt{13},\rightarrow)$, i.e. $n \in \{-2,-1\} \cup \{13, 14,15, \ldots\}$. Since f(n) is increasing when $n \geq 13$ and f(13) = 168, (5) has no integer solution for $n \geq 13$. Futhermore f(-2)=3 and f(-1)=0, so n=-2 and n=-1 are solution of (5). In other words, the only integer solution of (5) is n=-2 and n=-1. According to (3), the equation (1) has only two solutions, namely

$x = \textstyle{\frac{-2 - 3}{4}} = -\textstyle{\frac{5}{4}} \;\; \& \;\; x = \textstyle{\frac{-1 - 3}{4}} = \textstyle{\frac{-4}{4}} = -1.$


You've probably posted this before, but I can't make my way through the site very smoothly for now,so could you pleas explain to me what to do to post the source in that nice way? I mean, what do I if I want "Source:blabla" to appear in the upper part of my post, like a title? I've seen some posts which are like that.

you do not posts that, it is automatically posted.
now if you have opened a new topic in the advanced section a new field will apear: Source. there you put the source of the problem.

Don't you guys ever read the news forum ? :D

Also posting source rules can be found at the beginning of each forum Algebra, etc.

For now source field required only in Advanced Section and some of the forums in College Playground.

A complete guide to how to post the source can be found here
http://www.mathlinks.ro/viewtopic.php?t=3468

Also please edit your older posts (only one source / topic is available) to put the source string.

You can also search only for source (look carefully in the search page menu).


search
number theory proposed
number theory

let an increasing secuence of positive integers satisfy:
1: if a_n is a prime, then n is a prime
2: a_2n=a_n+n

find a_2004!

I think I've seen this problem before, but I kind of remember the sequence being strictly increasing. Is it just increasing? (without the "strictly"?). If the sequence is strictly increasing the problem really is trivial, so maybe it's as you've posted it.

yes I distinctly remember that pbornsztein solved it with dirichlet theorem do the search  ;)

here it is: http://www.mathlinks.ro/viewtopic.php?t=4093

without the assumption that it is strictly increasing i believe it is really hard...maybe one should concentrate to determine a_2004 and not more...

Well, I don't think it's determined without this assumption..

well i forgot to say it, in the original problem i solved and that was very easy it was strictly increasing, but if you want it we can think about it without that assumption :cool:

I have a solution wich only uses that there are arbitrary large groups of consecutive numbers such that none of them are primes. Maybe ill post it later if you want, but still, the problem is to easy!


Billingsly and Josephus each arrange to arrive at the bus stop at a random time between 10 and 11 o'clock and to stay there fore twelve minutes. What is the probability that they will meet each other there?

[hide="hint"]  A similar problem to this one is as follows:

Two people arrive in a restaurant independently. Each arrives a random time between 5pm and 6pm, distributed uniformaly (no moment in this range is any more likely for arrival than another). What is the probability they arrived within 10 minutes of each other?

The solution:

Draw a 6 by 6 square. Plot the arrival time of one person vertically. Plot the arrival time of the other person horizontally. The times they arrive within 10 minutes of each other can be represented by a diagonal stripe across the square. The area of the entire square is 36. The area of the portions outside the stripe is 25. Thus the area of the stripe is 11. Considering the uniformity of the arrival times the probability of arriving within ten minutes is the area of the stripe divided by the area of the square = 11/36.

  Mabye that will help.

[/hide]

[quote="behemoth571"]Using G-Unit's technique the solution is
[hide] $\frac{25-16}{25} = \frac{9}{25} $[/hide][/quote]

That is indeed correct. :)


When someone posts a big $ \text{\LaTeX}$ formula, when it gets clicked on, a popup comes up showing the code.  But on small $ \text{\LaTeX}$ like $ 5+2=7$, it doesn't show up, is that the way its supposed to be :?:
:D

Yes. The code needs to have a certain lenght (30 characters) to show up as a popup. For the rest of it, it won't appear (the code is too simple basically).

Can you make it if its < 30 then it shows up as a title thing?

like

[code]
<img src="blah" title="equation here" />[/code]

or something like that?

The alt attribute is supposed to do that. Apparently it doesn't work on Firefox :D Ok, I'll see what I can do.


What is the greatest prime factor of $12!+14!$?

[hide="solution because I just figured it out"]
Factoring is your friend :)

Factor:
$12!(1+14\cdot13)$
Clearly on the left side the biggest prime factor is 11.

$14\cdot13+1=182+1=183$.
It is $3\cdot61$
61 is prime, and thus it is the largest prime factor[/hide]

[hide]You can factor that expression into $12!(1+14*13)$, which equals $12!(183)$. The Greatest prime factor in the $12!$ portion is 11, but in the 183 portion, 183 is 3*61. Sixty-one is prime. Therefore, the greatest prime factor of $12!+14!$ is $61$.[/hide]


Find all $(m,n)$ such that $m,n \in N$ and $(5+3\sqrt{2})^m=(3+5\sqrt{2})^n$

I think it was already posted...
Indeed:
Take conjugates to get the equivalent equation $(5-3\sqrt{2})^m=(3-5\sqrt{2})^n$. But $0<5-3\sqrt{2}<1$ and $-5<3-5\sqrt{2}<-4$, so equality is not possible.

I think the following argument is valid as well:

Working in $\mathbb{Z}_7$ if $\sqrt{2} \equiv 3$ so $5+3\sqrt{2} \equiv 5+9 \equiv 0$ whereas $3+5\sqrt{2} \equiv 4$ so if the equality holds we will have $4^n \equiv 0$ which is not possible.

if $\sqrt{2} \equiv -3$ we take conjugations to both side as ZetaX showed above and arrive at contradiction.

I've already said before that the definitions of $\mathbb{N}$ are different. But the same applies when $0$ is allowed.

I have something that is similar to ZetaX's solution, but a little different.

Work in $\mathbb{Q}[\sqrt{2}]$. Every element can be unique written in the form $a + b \sqrt{2}$ for $a, b \in \mathbb{Q}$ and $f: a + b \sqrt{2} \rightarrow a - b\sqrt{2}$ is an automorphism (the conjugate).

Since $f$ is an automorphism, we know that $f^m(3+5\sqrt{2}) = f^n(5+3\sqrt{2}) \Leftrightarrow (3+5\sqrt{2})^m=(5+3\sqrt{2})^n$.
So we may multiply both sides and get
$f^m(3+5\sqrt{2})(3+5\sqrt{2})^m = f^n(5+3\sqrt{2})(5+3\sqrt{2})^n$
$(3-5\sqrt{2})^m(3+5\sqrt{2})^m = (5-3\sqrt{2})^n(5+3\sqrt{2})^n$
$((3-5\sqrt{2})(3+5\sqrt{2}))^m = ((5-3\sqrt{2})(5+3\sqrt{2}))^n$
$(-41)^m = 7^n$

Since $7$ and $41$ are prime, this equation cannot hold for positive integers $m$ and $n$.


invariant
advanced fields
advanced fields unsolved

Let $ G$ be a finitely generated group that acts on a tree $ T$ with no inversions (no element of $ G$ interchanges the vertices of an edge). We say that such a tree $ T$ is a $ G$-tree. Show that any $ G$-tree contains a minimal (with respect to inclusion) $ G$ sub-tree (subtree with $ G$ action).

Also, the minimal $ G$-subtree is unique unless $ G$ fixes at least two vertices.

Give and example when this fails for an infinitely generated group.

First the uniqueness assertion:

Let $ A,B$ be distinct minimal subtrees of $ T$. They cannot intersect, because then their intersection would be a minimal subtree strictly smaller than one of them. The geodesic connecting the two must be fixed pointwise by $ G$, and it contains at least two vertices, since $ A$ and $ B$ are disjoint.

Existence:

Case 1 - there is an element $ s\in G$ which fixes no vertices of $ T$. In this case there is a unique bi-infinite path $ P\subset T$ on which $ s$ acts as a non-trivial translation. I claim that the smallest tree containing $ \bigcup_{g\in G}gP$, which is obviously a $ G$-tree, is minimal. It suffices to prove that every $ G$-subtree contains $ P$. Indeed, let $ a$ be a vertex of some subtree. Then the path connecting $ a$ and $ sa$ contains a segment of the path $ P$, and this segments iterates by $ s$ (both positive and negative iterates) make up the whole of $ P$; this means that every subtree invariant under the group generated by $ s$ contains $ P$, and we're done.

Case 2 - there is a vertex $ a$ fixed by all the elements of $ G$. Here, there's nothing to prove: $ a$ is a minimal $ G$-subtree.

When $ G$ is finitely-generated cases 1 and 2 cover all the possibilities: if a finitely generated group acts (without inversion) on a tree such that every element fixes some vertex, then the whole group fixes a vertex. I won't prove this. It can be found, for instance, in Serre's book "Trees, amalgams and $ SL_2$".

The counterexample (this was the hardest part, actually :)):

The vertices of $ T$ will be pairs of the form $ (n,\ \omega)$, where $ n$ is a natural number and $ \omega \equal{} (\omega_n,\ \omega_{n \plus{} 1}, \ldots)$ is a binary sequence which is eventually zero. Notice that $ \omega$ is indexed beginning with $ n$; that's the point of the notation: the $ n$ in $ (n,\ \omega)$ is supposed to indicate that we regard the sequence $ \omega$ as starting from position $ n$. For all $ n\ge 0$ and all $ \omega \equal{} (\omega_n,\ \omega_{n \plus{} 1},\ \ldots)$, the corresponding point $ a \equal{} (n,\ \omega)$ is connected to $ (n \plus{} 1,\ \omega)$, where this second $ \omega$ stands for $ (\omega_{n \plus{} 1},\ \omega_{n \plus{} 2},\ \ldots)$. We call the vertices of the form $ (n,\ \omega)$ "vertices at level $ n$". This describes a tree in which the vertices at level $ 0$ are leaves (i.e. are connected only to one other vertex) and those at higher levels have degree $ 3$ (they are connected to two vertices at a lower level and to one vertex at a higher level).

The group $ G$ is the group of automorphisms of $ T$ which preserve the levels of the vertices and, for some $ n\ge 0$, fix the set of vertices at levels $ \ge n$ pointwise. Now, if a $ G$-subtree has a vertex $ (n,\ \omega)$, then it contains all the vertices at the level $ n$, as can easily be seen. It follows that the invariant subtrees are precisely those consisting of vertices at levels $ \ge n$ for $ n\in\mathbb N$, and so that there can be no [i]minimal[/i] $ G$-subtree.


symmetry
geometry
geometric transformation
reflection
perimeter

Prove that the quadrilateral with symetric center is parallelogram.

sorry...what do you mean by symetric center...

You know...... the intersection between the diagonals is the symetric center of the parallelogram.

I hope I understand corectly.Let the quadrilateral be $ ABCD$ and $ O$ the symetric center( If I am right $ AO\equal{}OC$ and $ BO\equal{}BD$).Then
$ \Delta AOB \equiv \Delta  DOC$ and $ \Delta AOD \equiv \Delta BOC$ then it results that $ AB\parallel{}DC$ and $ BC\parallel{}ACD$.

I repeat, I don't think I understand corectly what is the symetric center.It may be also the altitudes from $ O$ to $ AB$ and $ DC$(are equal) and from $ O$ to $ AD$ and $ BC$ are equal(please corect me!)

[color=darkblue][b]Center of symmetry[/b] of a polygon $ P$ is a point, lets call it $ O$, such that for any point $ X \in P$, the reflection of $ X$ wrt $ O$ also lies on the perimeter of $ P$.[/color]

Hi

Are you sure yuor definition is correct . I am trying to translate what is sad in my book : A polygon $ P$ has a center of symetry if there exist a point $ O$  such that every point of this polygon has each own reflaction according to $ O$ in the plane limited by the polygon.

It is also a large possibility that my book may be wrong.

Can somebody insure me about the definition of center of symetry of a polygon


What is the least amount a number can have if it is divided by 3,4, and 5 and yields a remainder of 2 with the exception of 2?

[hide="answer"]Since it leaves a remainder of $2$, it simply is $3 \cdot 4 \cdot 5+2=\boxed{62}$[/hide]

Oh, and by the way, please refrain from using "pretty obvious" or "really easy" etc, because it might make other people uncomfortable if they are learning and trying to get better

i'll keep that in mind for next time. :?


It is given that one root of $2x^2+rx+s=0,$ with $r$ and $s$ real numbers, is $3+2i$ ($i=\sqrt{-1}$). The value of $s$ is:
(A) undetermined
(B) 5
(C) 6
(D) -13
(E) 26

[hide]
For $r$ and $s$ to be real numbers, the second root must be $3 - 2i$
Expanding, we get the quadratic to be $x^2 - 6x + 13$, then to make the lead coefficient 2, $2x^2 - 12x + 26$.  $s = 26$.  $D$.  Tricky there at the end...
[/hide]

I got the same answer as you but a different way,
$\frac{-r+\sqrt{r^2-8s}}{4}=3+2i$
So equating coefficients we get,
$\frac{-r}{4}=3$
$r=-12$
similarily,
$\frac{\sqrt{r^2-8s}}{4}=2\sqrt{-1}$
squaring both sides we get,
$\frac{r^2-8s}{16}=-4$
Solving we get,
$\boxed{s=26}$
So it is $\boxed{E}$
Correct?
My method is not elegant though :?  :(

[hide]the two roots must be 3+2i and 3-2i for coefficients to be real, so the expansion we have to do is (x-3-2i)(x-3+2i), the i's will disappear.
which equals $x^2 - 6x + 13$.

Note that $-(2i)^2 = 4$.

Doubling that, $2x^2 - 12x + 26$.

s = 26. E. [/hide]

Just use the product of two roots = $\frac {s}{2}$ and you are done .


How many 5 digit integers are there consisting of the digits 1, 2 and 3 each appearing at least once?

[hide]We count all numbers with digits limited to 1, 2, 3.  Then we subtract those that don't have a 1, 2, or 3.

The first is equal to $ 3^5\equal{}243$.

The second is equal to $ \{\text{digits are 1 and 2}\}\plus{}\{\text{digits are 1 and 3}\}\plus{}\{\text{digits are 2 and 3}\}\minus{}\{\text{digits all 1}\}\minus{}\{\text{digits all 2}\}\minus{}\{\text{digits all 3}\}\equal{}2^5\plus{}2^5\plus{}2^5\minus{}1\minus{}1\minus{}1\equal{}93$.

The answer is $ 243\minus{}93\equal{}\boxed{150}$.[/hide]

@lingomaniac88, what about a number like $ 99123$? It works...

[hide]A five-digit number has five digits. We need to have $ 1, 2, 3$ each appear at least once, so we need to select $ 3$ of the $ 5$ digits to represent $ 1, 2, 3$. We split it into two cases:
There are $ \binom{5}{3}=10$ ways to do so.
Of these $ 10$ ways, $ 6$ of them fill the first digit.
Then for $ 6$ of the choices, there are $ 10^2$ ways to fill the two remaining digits.
For the $ 10-6=4$ choices, there are $ 9\cdot 10$ ways to fill the two remaining digits.

In total, the answer is $ 6 \cdot 10^2+4 \cdot 9\cdot 10=600+360=\boxed{960}$.[/hide]

[quote="nayel"]No lingomaniac's answer is correct. The number consists of only the digits 1,2,3. Sorry if my statement is troubling.[/quote]

Oh sorry. Ignore my solution then. lol

No it's okay. There's no harm in discussing more problems. :roll:


Is there a formula or pattern to find this?

Obviously, finding all digits of $ 2^1000$ doesn't seem like a good idea

$ 2^{1000} \equal{} 16^{250}$
[hide="Observation"]Observe:
[list]$ 16^0 \equal{} 1 \rightarrow 1$
$ 16^1 \equal{} 16 \rightarrow 7$
$ 16^2 \equal{} 256 \rightarrow 13$
$ 16^3 \equal{} 4096 \rightarrow 19$
$ 16^4 \equal{} 65536 \rightarrow 25$
...
See a pattern?[/list][/hide]
[hide="Possible Solution"]
From the observation, we can conclude that the sum of the digits of $ 16^n$ for $ n >\equal{}0$ is $ 6n \plus{} 1$ (there yet has to be a proof or disproof). The sum of the digits of $ 2^{1000}$, or $ 16^{250}$, is $ 6(250) \plus{} 1 \equal{} \boxed{1501}$.[/hide]

Perhaps taking it modulo 9 would help? Also, this is a project Euler problem!

[quote="Patterns_34"]$ 2^{1000} \equal{} 16^{250}$
[hide="Observation"]Observe:
[list]$ 16^0 \equal{} 1 \rightarrow 1$
$ 16^1 \equal{} 16 \rightarrow 7$
$ 16^2 \equal{} 256 \rightarrow 13$
$ 16^3 \equal{} 4096 \rightarrow 19$
$ 16^4 \equal{} 65536 \rightarrow 25$
...
See a pattern?[/list][/hide]
[hide="Possible Solution"]
From the observation, we can conclude that the sum of the digits of $ 16^n$ for $ n > \equal{} 0$ is $ 6n \plus{} 1$ (there yet has to be a proof or disproof). The sum of the digits of $ 2^{1000}$, or $ 16^{250}$, is $ 6(250) \plus{} 1 \equal{} \boxed{1501}$.[/hide][/quote]

Fails for $ n \equal{} 8$.  You shouldn't expect patterns like this to hold.

If this is a Project Euler problem, it isn't meant to be solved by hand.

Thank you for the insight t0rajir0u.

Just a question, why have a [url=http://projecteuler.net/]Project Euler[/url] [url=http://projecteuler.net/index.php?section=problems]problem[/url]  be posted in High School Basics?

[quote="math92"]Find the sum of the digits of $ 2^{1000}$.[/quote]

1    -    2^0

7    -    2^10

31  -    2^20

37  -    2^30

61   -   2^40

i'm not too experienced with euler...but i think you can find it out that way

i found a cool thing from a wiki page somewhere

[quote="math92"]Find the sum of the digits of $ 2^{1000}$.[/quote]

A quick basic computation with Excel gives a value of $ 1366$

The max value for $ n<1000$ is obtained for $ 2^{996}$ which gives $ 1468$

But I have no idea about how to find these values except just computing $ 2^n$

[quote="Patterns_34"][quote="pco"]A quick basic computation with Excel gives a value of $ 1366$[/quote]
Pco, how did you get Excel to do that kind of computation?

When I imput $ 16^{13}$, or $ 2^{52}$, or greater my Excel program doesn't return an accurate number.[/quote]

You just have to use slices of 6 digits, for example :
Put in A0 value 1 and in the remainder of line A values 0.
Put in B0 the formule "=\mod(2*A0;1000000)"
Put in the remainder of line B the formula "=\mod(2*uppercell+ent(upperleftcell/500000);1000000)"

You just have to extend right and down and you get values of $ 2^n$ by slices of 6 digits.