logarithms
trigonometry
analytic geometry
graphing lines
slope
quadratics

If $ \log_{4}{(x\plus{}2y)}\plus{}\log_{4}{(x\minus{}2y)}\equal{}1$ then find the minimum value of $ |x|\minus{}|y|$

[quote="RAGHAV GROVER"]If $ \log_{4}{(x \plus{} 2y)} \plus{} \log_{4}{(x \minus{} 2y)} \equal{} 1$ then find the minimum value of $ |x| \minus{} |y|$[/quote]

$ \log_{4}{(x \plus{} 2y)} \plus{} \log_{4}{(x \minus{} 2y)} \equal{} \log_4 {(x\plus{}2y)(x\minus{}2y)}\equal{} 1 \Rightarrow x^2\minus{}4y^2\equal{}4$

WLOG, $ y$ is non-negative, so $ |x|\minus{}|y|\equal{}2 \sqrt{y^2\plus{}1}\minus{}y\equal{}f(y)$

$ f'(y)\equal{}\sqrt{\frac{4y^2}{y^2\plus{}1}}\minus{}1$

So for $ y \le \frac{1}{\sqrt{3}}$, $ f'(y)<0 \Rightarrow f(y)$ is decreasing.

for $ y \ge \frac{1}{\sqrt{3}}$,  $ f'(y)>0 \Rightarrow f(y)$ is increasing.

$ \Rightarrow$ min. value of $ f(y)$ occurs when $ y\equal{}\frac{1}{\sqrt{3}}$, $ x\equal{}\frac{4}{\sqrt{3}}$, which satisfies the original condition and gives that min. value of $ |x|\minus{}|y|\equal{}\sqrt{3}$

hello, the minimum value is $ \sqrt{3}$ and will be obtained at the point $ x\equal{}\minus{}\frac{1}{\sqrt{3}}\minus{}\sqrt{3}$ and $ y\equal{}\frac{1}{\sqrt{3}}$.
Sonnhard.

$ \log_4 (x + 2y) + \log_4 (x - 2y) = 1\Longleftrightarrow x^2 - 4y^2 = 4$ and $ x + 2y > 0,\ x - 2y > 0$.

Thus we can set $ x = \frac {2}{\cos \theta},\ y = \tan \theta$, we have $ |x| - |y| = - \frac {|\sin \theta| - 2}{|\cos \theta| - 0}$.

$ \frac {\sin \theta - 2}{\cos \theta - 0}$ is the slope of the secant line passing through the points $ (0,\ 2)$ and $ (\cos \theta ,\ \sin \theta)\ (0\leq \theta < 2\pi)$ on the unit circle $ x^2 + y^2 = 1$.

Therefore we have $ |\frac {|\sin \theta| - 2}{|\cos \theta| - 0}|\geq \sqrt {3}|$.

It is sufficient for $ 0\leq \theta \leq \pi, \theta \neq \frac {\pi}{2}$. The answer is $ \boxed{\sqrt {3}}$ when $ \theta = \frac {5}{6}\pi \Longrightarrow x = - \frac {4}{\sqrt {3}},\ y = - \frac {1}{\sqrt {3}}$.

hello kunny, we have $ |x|\minus{}|y|\equal{}\frac{4}{\sqrt{3}}\minus{}\frac{1}{\sqrt{3}}\equal{}\sqrt{3}$.
Sonnhard.

hello kunny, errare humanum est.
Sonnhard.

what the heck did you just say????? :huh:

That's Latain's saying which means making mistakes is humanistic.

From Seneca the Younger. The full quote is [b]errare humanum est, sed perseverare diabolicum[/b]: 'to err is human, but to persist is diabolical.'

Mine!

$ x\plus{}2y>0$,$ x\minus{}2y>0$ so
$ x>2|y|\geq0$
and we have $ x^{2}\minus{}4y^{2}\equal{}4$
Let $ x\minus{}y\equal{}a$,or $ x\equal{}y\plus{}a$
now the equation becomes $ (a\plus{}y)^{2}\minus{}4y^{2}\equal{}0$
this is a quadratic in $ y$ so $ D\geq0$ so we have $ a\geq\sqrt{3}$

Tha just satisfies the necessary condition.


geometry
3D geometry
sphere
tetrahedron
pyramid
inequalities
geometry solved

Find the sphere of maximal radius which can be placed into every tetrahedron which has all altitudes $\geq 1$.

Hi grobber,

Let $A_i$, $h_i$ be the areas of the faces and their corresponding altitudes. Let r denote the radius of the insphere. Then we have $3V=\left(A_1+A_2+A_3+A_4\right)\cdot r=A\cdot r$ (by dissecting into four pyramids all with height r). Also we find $3V=A_1\cdot h_1=A_2\cdot h_2=A_3\cdot h_3=A_4\cdot h_4$. Thus $3V\geq A_1$, ... which sums to $12V\geq A$. With $12V=4A\cdot r$ this becomes $r\geq\frac14$. This inequality becomes an equality for $h_1=h_2=h_3=h_4=1$, what is the case for a regular tetrahedron with all altitudes equal to 1.

Thus, $\frac14$ the maximal radius of a sphere fitting into every tethrahredron all of whose altitudes are $\geq 1$.

-Eckard

[color=red][Moderator edit: I have edited this and the previous post because of a mistake in the formulation of the problem. Now, the problem and the solution both are corrected. Thanks to Myth for his help. -- Darij][/color]


Given $x,y$ real numbers and $x\neq 0$.
Find the minimum of $x^{2}+y^{2}$ if
$xy(x^{2}-y^{2})=x^{2}+y^{2}$


This is more of a logic puzzle than a problem, but I don't know where to put it.
Suppose N prisoners are lined in a row.  Then, they are each given a hat, either black or white, and they can't move or exchange hats.  They can see the colors of all hats in front of them but can't see the colors of their own hats, however.  The guard moves to the back of the line and asks the prisoner to say the color of his own hat, and if he is right, he is set free, but if the prisoner answers incorrectly, he is shot.  The prisoners in front of him can all hear his answer.  If all of the prisoners are able to meet beforehand and discuss the strategy, what should it be such that all prisoners in front of the prisoner in the way back can be saved?
I solved the above part.  The solution is that he notes the parity of the hats, and if white hats are even, he says "white," and otherwise, he says black.  That way, all of the prisoners in front of him can save themselves.
I cannot solve the problem extended to k colors of hats, in which k>2.  Help would be appreciated.'

Oh, and the guard moves up from the last prisoner, asking each consecutive prisoner the color of his hat.

The solution I have come up with generalizes the case where $k=2$.  Essentially there is a strategy for $k$ different colored hats as long as there is a maximum of $k$ acceptable casualties.  Basically the first prisoner picks one of the $k$ colors, counts the number of hats that color and takes the residue of that number $\mod k$.  Each residue would correspond to a different color guess.  Realize that regardless of this prisoners fate, he has just saved all prisoners with a hat of the color guessed.  So for any $k$ we can reduce the problem to the $k-1$ case which ultimately boils down to the case $k=2$ which you have already given a strategy for.  I hope this is complete enough.  I can be more thorough if I am unclear anywhere.

Oooh ooh ooh we did this problem at the math camp I just returned from.
[hide]You mentioned the parity of the hats.  Basically this is mod two.  For n hats, think about in mod n.  Assign each color a different member of mod n (0,1,2,3,4,...n-1). This may involve a lot of mod arithmetic though.  [/hide]

Forgive me, but I don't quite follow.  Suppose we have colors 1, 2, and 3 and the last prisoner sees in front of him, 223332111111112233.  Then, he picks the color 2 and says "2" since it there are 2mod3 2's.  How the heck does he save all of the 2's?  The guy in front of him doesn't know if there are 2mod3 number of 3's or 2's or 1's.

that's what they arrange during their strategy session.  That way when the first prisoner to guess says "2" everyone else will know it's because there are 2 mod 3 2's.

Doh!  I forgot about the strategy session... :blush:

[quote="amd2357"]The solution I have come up with generalizes the case where $k=2$.  Essentially there is a strategy for $k$ different colored hats as long as there is a maximum of $k$ acceptable casualties.[/quote]
Why would anyone bother with $k$ acceptable casualties when there is a strategy which guarentees at most $1$ casualty, no matter what number of colour is :)

Let the colours be numbered $0,1,2,\dots,k-1$. The first prisoner (the last one in the row) sums [u]all the colours he sees[/u] and says the result mod $k$. Knowing that, the second one could easily find out his colour. The third one can now determine his colour too, because now he knows the colour of the second one, and so on.

Of course yours is a much better solution.  Which goes to show why impatience and math don't mix when lives are on the line.  I was trapped into thinking I had to generalize more than just the number of colors.  When I found my solution, I didn't stop to think of any alternatives.  Nice work Bojan Basic!

My favorite statement of this is the "Judgment of the Council of Sages": the king wants to test his sages, so he sets this challenge before them, ginving them a chance to choose the order they line up and the order they answer. All of the answers will be collected, and those who answer wrong will be killed. The night before the challenge, the sages meet and formulate this strategy.

Postscript:
On the morning of the challenge, the junior sage chosen to stand last and answer first decides he really doesn't like all that hazing, and chooses to reverse his answer, dooming all of the other sages.


limit
induction
logarithms
function
calculus
calculus computations

[u][i]   Problem : [/i][/u]

  [i]Let the sequence $ (x_{n})$  defined :

 $ x_{1} >  0$  and    $ x_{n\plus{}1} \equal{}  ln(1\plus{} x_{n})$ with all $ n \in N, n \geq 1$

  Find      $ \boxed{\lim_{n\rightarrow\infty} x_{n} }$[/i]
     [i]  I am waiting for Argady , Nayel , Virgil Nicula , Scorpius119 for their nice solution[/i] :lol:

 [i]  Love An Forever :blush:
 Huỳnh Võ Phương An[/i]

Exactly $ lim_{n\to \infty }nx_n\equal{}2$.

$ x_{n} > 0$ for every $ n$ by induction and decreasing for every $ n$ again by induction. The sequence converges since it is bounded below and is decreasing.

Let $ x \equal{} \lim_{n\rightarrow\infty}x_{n}$

$ \lim_{n\rightarrow\infty}x_{n \plus{} 1} \equal{} \lim_{n\rightarrow\infty}ln(1 \plus{} x_{n})$

$ x \equal{} ln(1 \plus{} x)$

$ \lim_{n\rightarrow\infty}x_{n} \equal{} x \equal{} 0$

For $ f(x) \equal{} \ln (1 \plus{} x)\ (x > 0)$, Let $ \alpha$ be positive real numbers such that $ \alpha \equal{} f(\alpha)$.
$ x_{n \plus{} 1} \equal{} f(x_n)$
$ \alpha \equal{} f(\alpha)$

Using Mean value theorem,

$ |x_{n \plus{} 1} \minus{} \alpha| \equal{} |f(x_n) \minus{} f(\alpha)| \equal{} \left|\frac{f(x_n)\minus{}f(\alpha )}{x_n\minus{}\alpha}\right\parallel{}x_n \minus{} \alpha|$

$ \equal{} |f'(\alpha)\parallel{}x_n \minus{} \alpha|\Longrightarrow |x_{n \plus{} 1} \minus{} \alpha| \equal{} \frac {1}{1 \plus{} \alpha}|x_n \minus{} \alpha|$.

$ \therefore 0\leq |x_{n \plus{} 1} \minus{} \alpha| < \frac {1}{1 \plus{} \alpha}\left|x_n \minus{} \alpha \right |$, yielding $ 0\leq |x_n \minus{} \alpha | < \left(\frac {1}{1 \plus{} \alpha}\right)^{n \minus{} 1}|x_1 \minus{} \alpha |$.

Since $ 0 < \frac {1}{1 \plus{} \alpha} < 1$, we have $ \lim_{n\rightarrow \infty} \left(\frac {1}{1 \plus{} \alpha}\right)^{n \minus{} 1}|x_1 \minus{} \alpha | \equal{} 0$. $ f(x)$ is an increasing function and $ y \equal{} x$ is the tangent line at $ x \equal{} 0$ for $ y \equal{} f(x)$.

$ \therefore \lim_{n\to\infty}|x_n\minus{}\alpha|\equal{}0\Longleftrightarrow \lim_{n\to\infty}x_n\equal{}\alpha \equal{} 0$.

if  you said that :
$ \frac {f(x) \minus{} f(\alpha)}{x \minus{} \alpha} \equal{} f'(\alpha)$ then you recognize that, $ \lim_{n \to infty } x_n \equal{} \alpha$
 hence i think that solution's not complete  exactly, but  ideal's very  interesting
 also :  if $ x_1 < 0.$ then  $ \lim nx_n \equal{}2$,  this problem in journal Titu's

You are right.

Sorry, I made a typo. Here is an edited version.

For $ f(x) \equal{} \ln (1 \plus{} x)\ (x > 0)$, Let $ \alpha$ be positive real numbers such that $ \alpha \equal{} f(\alpha)$.
$ x_{n \plus{} 1} \equal{} f(x_n)$
$ \alpha \equal{} f(\alpha)$

Using Mean value theorem,

$ |x_{n \plus{} 1} \minus{} \alpha| \equal{} |f(x_n) \minus{} f(\alpha)| \equal{} \left|\frac {f(x_n) \minus{} f(\alpha )}{x_n \minus{} \alpha}\right\parallel{}x_n \minus{} \alpha|$

$ \equal{} |f'(c)\parallel{}x_n \minus{} \alpha|\ (there\ is\ c\ between\ x_n\ and\ \alpha)\Longrightarrow |x_{n \plus{} 1} \minus{} \alpha| \equal{} \frac {1}{1 \plus{} c}|x_n \minus{} \alpha|$.

$ \therefore 0\leq |x_{n \plus{} 1} \minus{} \alpha| < \frac {1}{1 \plus{} c}\left|x_n \minus{} \alpha \right |$, yielding $ 0\leq |x_n \minus{} \alpha | < \left(\frac {1}{1 \plus{} c}\right)^{n \minus{} 1}|x_1 \minus{} \alpha |$.

Since $ 0 < \frac {1}{1 \plus{} c} < 1$, we have $ \lim_{n\rightarrow \infty} \left(\frac {1}{1 \plus{} c}\right)^{n \minus{} 1}|x_1 \minus{} \alpha | \equal{} 0$. $ f(x)$ is an increasing function and $ y \equal{} x$ is the tangent line at $ x \equal{} 0$ for $ y \equal{} f(x)$.

$ \therefore \lim_{n\to\infty}|x_n \minus{} \alpha| \equal{} 0\Longleftrightarrow \lim_{n\to\infty}x_n \equal{} \alpha \equal{} 0$.

[color=green]Unnecessary quote removed by moderator. It never makes sense to quote the previous post in full[/color]
    now, ur solution's good.
thanks ! development of method . you  have some problems


I don't know how you call this in English: it is called' processus de renouvellement' in French, and it  is just a generalisation of Poisson process.

Let $ {N\{t\}, t >=0 }$ be a ' renouvellement' process with: $ X_1, X_2,...,X_n$ having a Poisson-distribution with parameter
  $ \lambda$.  That means: $ P\{ X_n =k\}= \frac{{ \lambda ^k}e^_{-\lambda} }  {k!} ,k=0,1,...$

Find the distribution of $ S_n$ (the sum of$ X_i$).
I found: $ S_n$ has a Poisson distribution with parameter $ n\lambda$
determine $ P\{N(t)=n\}$
I don't know how to proceed.  Can anyone solve this for me?
Thanks!

$ \mathbb{P}(N(t)=n)=\mathbb{P}(T_{n}\leq t<T_{n+1})$, where $ (T_n)_{n\in\mathbb{N})}$, are arriving times of (let's say) payouts to an insurance company. Of course $ (T_n)_{n\in\mathbb{N})}$ it's a nondecreasing process...  , I'm sure you know what I'm talking about, if not please say so  :lol: .

And it's called "renewal process".


LaTeX
least common multiple
greatest common divisor
number theory
prime numbers
number theory solved

Let's define the infinite sequence [tex]{a_n}[/tex] for each n from N as follows:
1) [tex]a_p = 1/p[/tex] if [tex]p = 1[/tex] or [tex]p[/tex] is a prime number.
2) [tex]a_i_j = a_L_C_M_(_i_,_j_) a_G_C_D_(_i_,_j_)[/tex] for each natural i,j
Determine the value of [tex]a_n [/tex] commensurate with natural number [tex]n[/tex].
NOTE:
[tex]LCM(i,j)[/tex] denotes the least common multiple of numbers [tex]i,j[/tex]
[tex]GCD(i,j)[/tex] denotes the greatest common divisor of numbers [tex]i,j[/tex]

We can easily see that if [tex]n = p^k[/tex], where [tex]p[/tex] is a prime number and [tex]k[/tex] natural, then [tex]a_p_^_k[/tex] = [tex](1/p)^k[/tex], but if n has different form I think that
we can't express [tex]a_n[/tex] - my question is how to prove it?

Thank you very much for any help...

Well, If I'm not wrong and if you assume that $a_{p_1p_2...p_n} \neq 0$ for each integer $n > 1$ and distinct primes $p_1,p_2,...,p_n$ then it is not difficult to prove that $a_n = \frac {1} {n}$ for all $n$.
Otherwise, you may have something like $a_n = \frac {1} {n}$ when $n = p^k$ for some prime and $a_n = 0$ for the other $n$ (and probably many others, according to the choices of primes $p_i$ for which $a_{p_1p_2...p_n} = 0$

Pierre.

a quick latex tip :P $\neq$ is \neq (from not equal, as \leq is less than or equal, geq is greater than or equal :D).

I think that we can prove the following statement:
if [tex]n = p_1^r^1 p_2^r^2 ... p_k^r^_k[/tex] and
[tex]m = p_1^r^1^-^1 p_2^r^2^-^1 ... p_k^r^k^-^1[/tex]
then [tex]a_n = a_(_n_/_m_) / m[/tex]
where [tex]p_1, p_2,..., p_k[/tex] are distinct prime numbers and [tex]r_1,r_2,...,r_k[/tex] natural numbers.

But then the value of every [tex]a_n_/_m[/tex] can be arbitrary, I think that to distinguish two cases [tex]a_n_/_m = 0[/tex] and "otherwise" is too little, or may I say that value of each [tex]a_n_/_m[/tex] can be arbitrary?

Well, the sequence is entirely defined from the values of the numbers $a_{p_1p_2...p_n}$ for any $n > 1$ and any distinct primes $p_i$ (that is the sequence is defined from its values on the squarefree integers), and for such an integer $k$ we have only two possibilities : $a_k = 0$ or $a_k = \frac {1} {k}$.

Pierre.

I'm sorry, but I still don't understand... or I can't see it. In your first reply you say that if [tex]a_k \neq 0[/tex] where [tex]k=p_1 p_2 ... p_i[/tex] for each integer [tex]i>1[/tex] then it isn't difficult to prove that [tex]a_n=1/n[/tex] for all n. But I can't prove it and I'm not really sure that it's possible to prove it. For example if we take [tex]a_6=1[/tex], it seems that this value doesn't mind to the requests of task (at least according to my opinion). Thanks.

Suppose that $a_6 \neq 0$.
Then $a_{36} = a_{6 \times 6} = a_6^2$ and in another hand $a_{36} = a_{2 \times 18} = a_2 \times a_{18} = a_2 \times a_{3 \times 6} = a_2a_3a_6$.
It follows that $a_6 = a_2a_3 = \frac {1} {6}$.
Thus $a_6 = 1$ is impossible. the general case is similar to that one above, using induction.

Pierre.


How many whole numbers between 99 and 999 contain exactly one 0?

We have two cases to consider, where x represents a non-zero digit.

Case 1: The number is in the form xx0.

There are $ 9 \cdot 9 \cdot 1 \equal{} 81$ possible ways.

Case 2: The number is in the form x0x.

There are $ 9 \cdot 1 \cdot 9 \equal{} 81$ possible ways.

Thus our answer is $ 81 \plus{} 81 \equal{} \boxed{162}$.

I get the solution (it's the one I came up with myself), but I feel like it's not explained in a way that someone who didn't already get it could understand easily. Anyone agree?

Can you elaborate? What parts seem confusing to you?

Hi.  I got this problem correct, but I wanted to tell you another way to do it.  I listed all the numbers from 99-199 with one zero, then counted them.  There were 18.  There's 9 groups of hundreds between 99 and 999, so I multiplied 18x9 and I got the right answer, 162.

[hide=Amc8 Solution]This list includes all the three digit whole numbers except 999. Because the hundreds digit cannot be 0, there are $2$ ways to choose whether the tens digit or the ones digit is equal to 0. Then for the two remaining places, there are $9$ ways to choose each digit. This gives a total of $(2)(9)(9)=\boxed{\text{(D)}\ 162}$.[/hide]

Numbers with exactly one zero have the form $\_ 0 \_$ or $\_ \_ 0$, where the blanks are not zeros. There are $(9\cdot1\cdot9)+(9\cdot9\cdot1) = 81+81 = \boxed{162}$ such numbers.


algebra
polynomial
geometry
algebra solved

let f(z) be a polynomial with complex roots a_1,a_2,a_3.....a_n.prove that htere xists a complex z_0 such that lz_0l=1 and
(3^n) lf(z_0)l >=(product j=1 to n ) (1+la_jl) .this solution will be in the 1998-99 book so can ne1 post it?
i'd be grateful
best regards

c'mon friends,this problem has been lying around long enuf??

The problem is like this: prove that 2 n-1|f(z 0)| \geq (prod j=1,...,n)(1+|aj|). Hint: first suppose that all aj are on the unit circle and prove that there exists z0, |z0|=1 so that |f(z 0)| \geq 2. If not then put bj=aj/|aj|  and prove that |z-aj|/(1+|aj|) \geq |z-bj|/(1+|bj|), for all z, and then apply the first case.

hey calin
ur hint actually answers two questions
a geometry question on the forum by bugzpodder and this one.but i've still to finish the proof

to which geometry question are you reffering to galois ?

sum1  posted a question which went .M is a point on the unit circle and ABC  is a triangle if MA.MB.MC  \leq  2  for all M,prove that ABC  is an equilateral triangle.but pcalin gave the hint that there exists a z such that
lf(z)l \geq  2 for sum z on the unit circle..and as i stated earlier the property is true for any polygon inscribed in a circle


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Suppose $A\in\mathcal M_{n\times m}(\mathbb F)$ and $B\in\mathcal M_{m\times n}(\mathbb F)$. Prove that $x^{m}ch_{AB}(x)=x^{n}ch_{BA}(x)$, whcih $ch_{T}(x)$ is charcteristic polynomial of matrix $T$.

I have posted a Hint but I'd like to see different solutions

[hide="Hint"]
Suppose $C=\left( \begin{array}{cc}xI_{n}&I_{m}\\B&I_{m}\end{array}\right)$ and $D=\left( \begin{array}{cc}I_{n}&0\\-B&xI_{m}\end{array}\right)$. we know that $\det (CD)=\det(DC)$
[/hide]

Add $m$ rows and $n$ columns to $A$ consisting only of zeros, and to the same for $B$ (with $m$ instead of $n$ and vice-versa). Call the two matrices $\widetilde A$ and $\widetilde B$ respectively. $x^m$ ($x^n$) times the haracteristic polynomial of $AB$ ($BA$) is precisely the characteristic polynomial of $\widetilde A\widetilde B$ ($\widetilde B\widetilde A$). Now the conclusion follows from the well-known fact that given square matrices $C,D$ over a field, $CD$ and $DC$ have the same characteristic polynomial.


Let $ a,b,c$ be sides of trangle. Prove that $ \frac{\sqrt{b\plus{}c\minus{}a}}{\sqrt{b}\plus{}\sqrt{c}\minus{}\sqrt{a}}\plus{}\frac{\sqrt{c\plus{}a\minus{}b}}{\sqrt{c}\plus{}\sqrt{a}\minus{}\sqrt{b}}\plus{}\frac{\sqrt{a\plus{}b\minus{}c}}{\sqrt{a}\plus{}\sqrt{b}\minus{}\sqrt{c}}\leq3$

It's Imo shortlist 2006.
http://www.mathlinks.ro/viewtopic.php?p=741368#741368


$ a_{n} \equal{} \sqrt {\frac {1 \plus{} a_{n \minus{} 1}}{2}}$
$ \minus{} 1 < a_{0} < 1$
$ A_{n} \equal{} 4^{n}(1 \minus{} a_{n})$
 find
 $ lim A_{n}$
$ n\rightharpoonup\infty$

Let $ \alpha\in(0,\pi)$ such that $ a_0\equal{}\cos\alpha$

$ a_1\equal{}\sqrt{\frac{1\plus{}\cos\alpha}{2}}\equal{}\cos\frac{\alpha}{2}$

By induction $ a_n\equal{}\cos\frac{\alpha}{2^n}$

Then $ A_n\equal{}4^n\left(1\minus{}\cos\frac{\alpha}{2^n}\right)\equal{}2\cdot 4^n\sin^2\frac{\alpha}{2^{n\plus{}1}}\equal{}\frac{1}{2}\left(\frac{\sin\frac{\alpha}{2^n}}{\frac{1}{2^n}}\right)^2$

$ \lim_{n\to\infty}A_n\equal{}\frac{\alpha^2}{2}\equal{}\frac{\arccos^2 a_0}{2}$

From our recursive identities, 0<$ a_1$<1, $ \sqrt {1/2} < a_2 < 1$, etc.  We see that the upper bound remains $ 1$, and the lower bound is consistently increasing.  As $ n$ grows, the numerator of the fraction of the lower bound is approaching 2, and so the lower bound is approaching 1.  As $ n$ tends to infinity, therefore, $ a_n$ gets in a bound which is increasingly strapped between two very close numbers- $ 1$ and $ (1 \minus{} \text{small number})$.  1-$ a_n$ then approaches $ 1$ from both sides, and $ A_n$ approaches 0.

[b]@ J.Mayers:[/b] Everything you said was correct except for your last sentence. Note that, while $ 1\minus{}a_n$ is decreasing, the term it's multiplied with, $ 4^n$, is increasing.

For example, what if you were to find out that $ a_n \equal{} 1\minus{} \frac{1}{4^n}$? (it isn't in this case).

Then, $ a_n$ would be approaching $ 1$, just like in this case, but $ A_n$ would just be a constant $ 1$. In other words, $ \lim (ab) \equal{} (\lim a) (\lim b)$ only holds true when $ \lim a$ and $ \lim b$ exist.

Here's another problem you can try --

$ b_n \equal{} \frac{1}{2\minus{}b_{n\minus{}1}}$
$ 0 \leq b_0 \leq 1$
$ B_n \equal{} n (1\minus{}b_n)$.

Find $ \lim_{n \to \infty} B_n$

From the values I determined, $ a_4$ or $ a_5$ (one of them) got to a value of .99...So I assumed that, over time, this would over-do the $ 4^{n}$ term, but I wasn't able to confirm it...

Thanks  to all ! :lol:
I suppose the lim is related to a(0).and I have read the answer provided by red_dog,I think his answer is correct
TZF is right for "only holds true when  and  exist. "


search
combinatorics unsolved
combinatorics

You and your friend are each given a piece of cord that are of different integer lengths. Both of your task is to cut the cord into any number of pieces of any length, in such a way that when you multiply the length of your pieces, the result shall be the maximum.

After doing a few calculations, both of you cut the cords successfully satisfying the condition.

The length of your cord is 6 times the length of the cord that your friend has. However, the number of pieces that you obtain is 5 times the number that your friend has.

Find the length of the cords that have been given to you and to your friend.

Notes:

*The number of the pieces is greater than or equal to 2.
*The lengths of the pieces need not to be integers.
*Example: If you have a 5 m. cord, you can divide it into two pieces of 2 and 3 meters, and when you multiply the length of these pieces you will get 6 as the result. If you divide it into 3 pieces of 1.5, 1.5 and 2 meters, then you will get 4.5 as the result. But, both of these results are not the maximum.

(Enter the length of your cord. Unit is in meters.)

I'm so happy to have found solution !!! I hope it's correct
I have found that the lengths are 7 et 42
Solution is long and hard (for me !), I start to write it ...

We know that $\prod_1^n x_i\leq \Big(\frac{\sum x_i}{n}\Big)^n$ and equality holds iff $x_1=x_2=...=x_n$
So we cut the cord in equal parts
Let n be the number of parts of my cord and m the number of parts of the second cord
L is the length of my cord and 6L the length of the second cord

So I've chosen n for having $\Big( \frac{L}{n} \Big)^n$ be maximal
We set $f$ such that $f(x)=\Big( \frac{L}{x} \Big)^x$ and solve $f'(x)=0$ we have $x=\frac{L}{e}$
thus $n=\Big[ \frac{L}{e} \Big]$ or $n=\Big[ \frac{L}{e} \Big]+1$
using same reasonment we get $m=\Big[ \frac{6L}{e} \Big]$ or $m=\Big[ \frac{6L}{e}+1 \Big]$

Now we know that $m=5n$ and $m,n$ have been chosen maximal
We set $n=\Big[ \frac{L}{e} \Big]+\delta_1$ and $m=\Big[ \frac{6L}{e} \Big]+\delta_2$ with $\delta_1,\delta_2$ are $0$ or $1$

$m=5n$
$\Leftrightarrow \Big[ \frac{6L}{e} \Big]+\delta_2 = 5 \times (\Big[ \frac{L}{e} \Big]+\delta_1)$
$\Leftrightarrow  5 \times \Big[ \frac{L}{e} \Big] - \Big[ \frac{6L}{e} \Big]= \delta_2 - 5 \delta_1$
$\delta_2 - 5 \delta_1$ can take values $1,0,-4,-5$

We know that $L\geq 2$, we will show that $L\leq [11e]=29$
$5 \times \Big[ \frac{L}{e} \Big] - \Big[ \frac{6L}{e} \Big]<5 (\frac{L}{e}+1) - \frac{6L}{e} + 1=-\frac{L}{e}+6$
$-\frac{L}{e}+6<-5 \Leftrightarrow L > 11e \Rightarrow L\geq 30$
If $L$ is greater than 30 we can't have $\delta_2 - 5 \delta_1$ equals $1,0,-4,-5$ so $2\leq L\leq 29$

Now by computations we search $L$ such that $5 \times \Big[ \frac{L}{e} \Big] - \Big[ \frac{6L}{e} \Big]$ equals $1,0,-4$ or $-5$
We have :
- $L=2, \delta_1=1, \delta_2=1$
- $L=7, \delta_1=1, \delta_2=0$
- $L=9, \delta_1=1, \delta_2=0$
- $L=11, \delta_1=1, \delta_2=1$
- $L=14, \delta_1=1, \delta_2=0$

$f(x)=\Big( \frac{L}{x} \Big)^x$ and $g(x)=\Big( \frac{6L}{x} \Big)^x$
If $\delta_1=1$ we must have $f(\Big[ \frac{L}{e} \Big]+1)>f(\Big[ \frac{L}{e} \Big])$
If $\delta_1=0$ we must have $f(\Big[ \frac{L}{e} \Big]+1)<f(\Big[ \frac{L}{e} \Big])$
If $\delta_2=1$ we must have $g(\Big[ \frac{6L}{e} \Big]+1)>f(\Big[ \frac{6L}{e} \Big])$
If $\delta_2=0$ we must have $f(\Big[ \frac{6L}{e} \Big]+1)<f(\Big[ \frac{6L}{e} \Big])$
There is only $L=7$ which works

Yes. The answer is "7 and 42".Your solution is great.Thanks.


If $ P(x) \equal{} 4x^4 \plus{} 2x^2 \plus{} x$ has roots $ r_1,r_2,r_3,r_4$, find $ \sum_{cyc} \dfrac{r_1 \plus{} r_2}{r_3 \plus{} r_4}$ (VHHS)

I am wondering if there is a way to do this avoiding a bunch of algebra [hide="for example"] after using vieta's, putting everything under one denominator, etc. Perhaps the fact that one of $ r_i$ is 0 is useful[/hide]. Thanks.

This has already been posted before but I'll tell you the gist of this problem. You have to note that the sum of the roots is $ 0$. Therefore each cyclic sum as two of the roots in the numerator and the other two in the denominator. I'll give you one example to show why this is important. Take a look at the first sum $ \frac{r_1\plus{}r_2}{r_3\plus{}r_4}$, and noting that $ r_1\plus{}r_2\plus{}r_3\plus{}r_4\equal{}0$ or $ r_1\plus{}r_2\equal{}\minus{}(r_3\plus{}r_4)$ therefore this term is $ \minus{}1$. This works for every term and since this is a cyclic sum there are $ 4$ of these so the answer is $ \minus{}4$.

oh thanks, i feel silly for not seeing that now :)

The problem is easy!

W.L.O.G. $ r_4 \equal{} 0$, then by Vieta, we have $ r_1 \plus{} r_2 \plus{} r_3 \equal{} 0$.

[hide="Answer"]$ \minus{} 4$[/hide]

Sorry, ckck has already posted it.

I can't delete my post now.

Already posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=318325

Check out the second post!  :wink:

Assume the roots of the polynomial $\boxed{P(x)=4x^4+2x^2+x\implies x(4x^3+2x+1)}$ are $\alpha  ,\beta  ,\gamma  ,\delta.$ By Vieta's Relation we get$$(\alpha+\beta)+(\gamma+\delta)=0$$$$(\alpha+\beta)(\gamma+\delta)+\alpha\beta+\gamma\delta=2$$$$(\alpha+\beta)\gamma\delta+(\gamma+\delta)\alpha\beta=-1$$$$(\alpha\beta)(\gamma\delta)=0$$ we have to find the value of $$\sum_{cyc}{}\frac{\alpha+\beta}{\gamma+\delta}=\frac{\alpha+\beta}{\gamma+\delta}+\frac{\beta+\gamma}{\alpha+\delta}+\frac{\gamma+\alpha}{\beta+\delta}+\frac{\alpha+\delta}{\beta+\gamma}=-4$$
$$\boxed{\sum_{cyc}{}\frac{\alpha+\beta}{\gamma+\delta}=-4.}$$ Is the correct answer!!!!
Vieta's Relation!!!!!!! :oops:

[quote=djmathman]Two things @above:

[list]
[*]The cyclic sum only has four terms, not six as you wrote.  (No $\alpha + \gamma$ term in the numerator, for instance.)
[*]A bit of a minor point, but you don't need to write out all four Vieta relations if you're only going to use one of them!  You can just say "by Vieta, $\alpha + \beta + \gamma + \delta = 0$, so $\tfrac{\alpha + \beta}{\gamma + \delta} = -1$, etc."
[/list][/quote]

Ok thanks!!


let A $ \in M_{m.n}$ ,Let B $ \in M_{n.p}$.Prove that
 $ rank(AB) \leq$ $ min$ {$ rankA,rankB)$}.

Let, $ L_A : \mathbb{R}^n \rightarrow \mathbb{R}^m$, $ L_B : \mathbb{R}^p \rightarrow \mathbb{R}^n$
($ L_{AB} : \mathbb{R}^p \rightarrow \mathbb{R}^m$)

$ \text{rank} (AB) \equal{} \dim (\text{Im} \ L_{AB}) \equal{} \dim (\text{Im} \ (L_A \circ L_B)) \equal{} \dim \ L_A(L_B \mathbb{R}^p)$

$ \dim \ L_A(L_B \mathbb{R}^p) \leq \dim (\text{Im} \ L_A)$
and
$ \dim \ L_A(L_B \mathbb{R}^p) \leq \dim \ L_B(\mathbb{R}^p)$

Thus, $ \text{rank} (AB) \leq \min \{ \text{rank}(A), \text{rank}(B) \}$


What is the remainder when $ 2^{87} \plus{}3$ is divided by $ 7$?

How do we turn the 8^29 into 1^29? Thanks!

hello, since $8\equiv 1 \mod 7$.
Sonnhard.

Even if you guess this, you have a 1/7 chance of getting this correct.


Using latex :)

See [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]here[/url] for example ;)

Or if you want to use it only on the forum, just check the links in my signature below.


algebra
polynomial
superior algebra
superior algebra unsolved

Let E be an extension field of F. Then the [b]algebraic closure of F in E[/b] is {$ \alpha \in E | \alpha$ is algebraic over F}.

Let E be an algebraically closed extension field of a field F. Show that the algebraic closure $ \bar{F}_E$ of F in E is algebraically closed.

If $ \sum_{i \equal{} 0}^n x^i \bar{f}_i$ is a polynomial in  $ \bar{F}_E$[x], we need to find a zero of it that is contained in $ \bar{F}_E$[x]. Now, $ \bar{f}_i$ must be a zero of some polynomial of in F[x], but I don't see how you use that fact.

use: $ E$ is algebraically closed $ \Leftrightarrow$ every finite extension of $ E$ is $ E$.


logarithms
vector
probability
expected value
combinatorics proposed
combinatorics

[u][b]The author of this posting is : manuel[/b][/u]
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Let R(r) be the minimum number of vertices such that a graph in R(r) vertices contains a chromatic Kr, for a 2-edge coloring.
Prove that: 2^(r/2) < R(r).

I would like to see different approaches.

[u][b]The author of this posting is : grobber[/b][/u]
____________________________________________________________________

Here's one solution (and I read it in an article which wasn't even concerned with this subject: the proof was just mentioned in a hurry, but I never forgot it  :) ):

Color the edges of the graph randomly (assume the number of vertices is $n$). The expected number of sets of $r$ vertices, all joined by edges of th same color, is $2\cdot 2^{-\binom r2}\cdot \binom nr$. It's easy to see that this is $<1$ if $n=2^{\frac{r}{2}}$. However, if this expected number is $<1$, then there must be a coloring with no monochromatic $K_r$'s. This means that we must have $n>2^{\frac{r}{2}}$.

[u][b]The author of this posting is : pbornsztein[/b][/u]
____________________________________________________________________

Yes, it is the first time the so-called probabilistic method has been used.  :P

Pierre.

I would like to see example of graph with $\approx 2^{r/2}$ vertices, which doesn't contain monochromatic $K_r$. I think, it is not so difficult to construct it.  :?

Good luck.  I believe the best result for explicit constructions of Ramsey graphs is due to Frankl and Wilson (1981).  They constructed a graph with $r^{c \ln r / \ln\ln r}$ vertices without a monochromatic $K_r$.  That's a lot weaker than what you are hoping for.

[quote="grobber"]A related issue: there is a construction showing that $r(t+1,t)\ge \binom{t-1}3$.[/quote]
What is it?

Identify the points of the graph with the three element subsets of any set with $t-1$ elements. Now color the edge $AB$ blue if $|A\cap B|=1$ and red otherwise. It should be fairly easy to show that there's no red $K_t$ and no blue $K_{t+1}$ based on parity arguments and/or linear algebra.

Actually, this works for $R(t,t)$ as well.

We have to show that there are no red or blue $K_t$'s, i.e. there are at most $t-1$ sets with $3$ element s.t. the intersection of each two has a)$1$ element, b)$0$ or $2$ elements (a) and b) are two different problems).

For the second one, just look at the characteristic vectors as vectors with components in $\mathbb Z_2$ over $\mathbb Z_2$, and consider the billinear form ${\bf x}\cdot {\bf y}=\sum x_i\cdot y_i$. We easily find the vectors to be independent, so there can be at most $t-1$.

For the first one, do exactly the same thing, only that we do it for real vectors over $\mathbb R$ (and consider the usual scalar product). We also find the vectors to be independent, so, again, there are at most $t-1$.

Is there anything wrong here?


Say that a positive integer [i]n[/i] is a sum of consecutive integers if there exist positive integers [i]m[/i] and [i]k[/i] such that $n=m+(m+1)+...+(m+k)$.Prove that [i]n[/i] is so expressible if it is not a power of $2$.

I think this proof is valid, I wrote it like two days ago.

Edit: That proof was bad. I uploaded a new version.


ABC be triangles.
Prove that:
[tex]sinA+sinB+sinC+cosA+cosB+cosC+cotgA+cotgB+cotgC\ge 3+\sqrt{3}[/tex].
[tex]\sqrt{3}(sinA+sinB+sinC+cosA+cosB+cosC)+cotgA/2+cotgB/2+cotgC/2\ge 6\sqrt{3}[/tex].

well, for the first one, here's the first step:

[hide]cot(A)+cot(B)+cot(C) >= Sqrt(3)[/hide]

for the second one, just note that:

[hide]cot(A/2)+cot(B/2)+cot(C/2) >= 3*Sqrt(3). :) [/hide]


a,b,c are positive numbers
prove that:
a^2/(2b+c) + b^2(2c+a) + c^2(2a+b) >= (a^2+b^2+c^2)/(a+b+c)

Apply cauchy so we have $\sum_{cyc}\frac{a^{2}}{2b+c}\geq\frac{(\sum_{cyc}a^{2})^{2}}{ 2\sum_{cyc}a^{2}b+\sum_{cyc}b^{2}a}\geq\frac{\sum a^{2}}{\sum a}$ so we need to prove $ (\sum a^{2})(\sum a)\geq 2\sum_{cyc}a^{2}b+\sum_{cyc}b^{2}a$ so we have $ \sum_{cyc}a^{3}\geq \sum_{cyc}a^{2}b$ which is rearangements ineq. :)


We can realize that $f(x)$ is surjective .
  So there exists $x_0$ such that : $f(x_0)=0$
 Put $x =x_0$ , we have : $f(f(y)) = y$ (1)
 Put $x=0$ , we have : $f(f(y)) =y+f^2(0)$ (2)
 From (1) and (2) , we infer $f(0)=0$
  So $f(f(x))=x$ (3)
 Put $y =0$ , we have : $f(xf(x)) =f^2(x)$ (4)
 In (4) replaceing $x$ by $f(x)$ , combining with (3) , we have : $f(xf(x))=f^2(x) =x^2$
 After some normal computations , we get : $f(x) =x \forall x$ or $f(x)=-x \forall x$

That's funny :) see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=56415]www.mathlinks.ro/Forum/viewtopic.php?t=56415[/url]

This is also the problem from 2012 Raffles Invitational Math Olympiad in Singapore. :lol:

[quote="applepi2000"]Let $P(x, y)$ be this assertion. Then $P(0, y)\implies f(x)$ is surjective. Let $f(a)=0$, so $P(a, y)\implies f(f(y))=y$. Then $P(x, y)$ and $P(f(x), y)$ together give $f(x)^2=x^2$, and thus $\boxed{f(x)=\pm x}$ which easily works.[/quote]
I think you left the proof of showing either $f(x)=x$ for all $x$ or $f(x)=-x$ for all x.

My proof:
[hide]
Let $P(x,y)$ be the assertion. $P(0,x)\implies f(x)$ bijective and $f(f(x))=f(0)^2+x$. Let $f(c)=0$. Now, $P(c,x)\implies f(f(x))=x\implies f(x)=f^{-1}(x)$. It implies $f(f(x)^2+y)=xf(x)+f(y)......(i)$ Pluging $f^{-1}(x)$ instead of $x$ finally implies $f(x)^2=x^2$.Now, Let $x_0, y_0 \neq 0$ but $f(x_0)=x_0$ and $f(y_0)=-y_0$ then plugging $P(x_0,y_0)$ we'll get either $x_{0}^2-y_0=x_{0}^2+y_0$ or $-(x_{0}^2-y_0)=x_{0}^2+y_0$ but both are contradiction. So, either $f(x)=x$ for all $x$ or $f(x)=-x$ for all x[/hide]

Let $P(x,y)$ be $f(xf(x)+f(y))=(f(x))^2+y$.Clearly $f$ is surjective.Let $f(c)=0$,then $P(c,y)\rightarrow f(f(y))=y$.$P(f(x),y)\rightarrow f(xf(x)+f(y))=x^2+y$.Thus $(f(x))^2=x^2\Leftrightarrow f(x)=\pm x$.We suppose that ∃$a\neq 0$, ∃$b\neq 0$ s.t. $f(a)=a, f(b)=-b$.$P(a,b)\rightarrow f(a^2-b)=a^2+b$.This implies $a^2-b=a^2+b\Leftrightarrow b=0$ or $-a^2+b=a^2+b\Leftrightarrow a=0$ which are both absurd.Therefore $\boxed{f(x)=x(\forall x\in \mathbb R) or f(x)=-x(\forall x\in \mathbb R)}$ which both satisfy the condition$\blacksquare$ :D

I apologize if my solution is the same as others above me.
Let $P(x,y)$ be the assertion
$f(xf(x)+f(y)) =(f(x))^2+y$
$P(0,x)$ $\Longrightarrow$ $f(f(y))=(f(0))^2+y$ so $f$ is surjective. So there exists real number $a$ such that $f(a)=0$. $P(a,x)$ $\Longrightarrow$ $f(f(x))=x$ so.$f(0)=0$ here $P(x,0)$ $\Longrightarrow$ $f(xf(x))=(f(x))^2$ here x---->>f(x) we ll get $f(xf(x))=x^2=(f(x))^2$ so there are two cases.
___________________________________
Case 1. $f(x)=x$ is solution 1.
___________________________________
$f(x)=-x$ is solution 2.

Hello, [b]Murad.Aghazade[/b]. :lol:

From $(f(x))^2=x^2$,you can only say that $\forall x\in \mathbb R:f(x)=x$ or $f(x)=-x$.You cannot say that $\forall x\in \mathbb R:f(x)=x$ or $\forall x\in \mathbb R:f(x)=-x$.For example,there is a possibility that $f(1)=f(-1)=1$. :w00t:

My solution:
Let $P(x,y)$ be the assertion $f(xf(x)+f(y))=f(x)^2+y.$
$\exists a$ such that $f(0)=a.$
$P(0,a)\implies f(f(a))=a^2+a (1).$
$P(x,-f(x)^2)\implies 0=f(xf(x)+f(-f(x)^2))\to \exists c$ such that $f(c)=0.$
$P(c,x)\implies f(f(x))=x\to f$ is bijective $(2).$
$$P(c,a)\implies a=f(f(a))=^{(1)} a^2+a\to a=0.$$
$P(x,0)\implies f(xf(x))=f(x)^2,$ let $x\to f(x),$ we find
$$x^2=f(f(x))^2=f(f(x)f(f(x)))=^{(2)} f(f(x)x)=f(x)^2.$$
Then $f(x)\in \{-x,x\}.$

Suppose that
$\exists a$ such that $f(a)=a\not= 0,$ and
$\exists b$ such that $f(b)=-b\not= 0.$
$P(a,b)\implies f(a^2-b)=a^2+b.$
Case 1: $a^2-b=a^2+b\to b=0,$ Impossible.
Case 2:  $a^2-b=-a^2-b\to a=0,$ Impossible.

Solution:

Let $P(x,y)$ be the assertion of the problem statement.

Claim 1: $f$ is bijective.
Proof: If $f(a)=f(b)$, $P(x,a)$ and $P(x,b)$ gives $f(x)^2+a=f(x)^2+b$ which gives $a=b$. So, $f$ is injective. $P(0,y)$ gives $f(f(y))=f(0)^2+y$. So, $f$ is surjective.

Claim 2: $f(0)=0$.
Proof: Let $f(0)=c$ and let $f(a)=0$. $P(a,a)$ gives $c=f(0)=a$. $P(0,0)$ gives $f(c)=f(f(0))=c^2$. So, we have \[c^2=f(c)=f(a)=0.\] So, $f(0)=c=0$.

Main problem: $P(x,0)$ gives $f(xf(x))=f(x)^2$ and $P(0,y)$ gives $f(f(y))=y$. Putting $x=f(x)$ we have, \[x^2=f(f(x))^2=f(f(x)f(f(x)))=f(f(x)x)=f(x)^2.\] So, $f(x)=\pm x$.
Suppose $f(a)=a$ and $f(b)=-b$ for $a,b\neq 0$. $P(a,b)$ gives $a^2+b=f(a^2-b)=a^2-b$ or $b-a^2$. So, $a=0$ or $b=0$ which gives us a contradiction.

Solutions: $f(x)=x$ for all $x$ and $f(x)=-x$ for all $x$.

[quote=Kunihiko_Chikaya]Find all functions $f : \mathbb{R} \mapsto \mathbb{R}$ such that $f(xf(x)+f(y)) =(f(x))^2+y$ for all $\ x,y \in \mathbb{R}.$[/quote]
[rule]
[b]Answer.[/b] The only functions which satisfy the given assertion are $f(x) = \pm x$.
[rule]
[b]Proof.[/b] It is clear that $f(x) = \pm x$ works as both sides equal $(x^2 + y)$, in each case. It remains to show that these are the only such functions. First, let us denote the given assertion by $P(x,y)$

[b]Claim.[/b] $f$ is surjective.

(Proof). From, $P(0,y)$ we get $f(f(y))=f(0)^2+y \implies f$ is surjective. $\square$ (By varying $y$ with $x$ fixed we get that $f$ is surjective, and thus we can pick $x_0$ so that $f(x_0) = 0.$)

Next, $$\begin{cases}&P(0,0): f(f(0))=f(0)^2\\ &\text{ (Claim)}:f(f(0))=0\end{cases} \implies f(0)^2=0 \implies \boxed{f(0)=0}.$$

From $P(x,0)$ we get: $$f(xf(x))=f(x)^2.$$And from $P(f(x),0)$: $$f(f(x)f(f(x)))=f(f(x))^2 \implies f(f(x)x)=x^2.$$

Hence, $$f(x)^2 = x^2 \implies f(x) = \pm x. \quad\square$$
[rule]
[b](Check).[/b] Now, $$A \overset{\text{def}}{=} \{x \in \mathbb{R}: f(x)=x\}$$ and $$B \overset{\text{def}}{=} \{y \in \mathbb{R}: f(y)=-y\}$$
From $P(x,y)$ we have: $$f(xf(x)+f(y))=f(x)^2+y \implies f(x^2-y)=x^2+y.$$ Now, $f(x)=x$ or $f(x)=-x$. Hence $f(x^2-y)=(x^2-y)$ or $-(x^2-y)$.

Now, for the first one we get: $x^2-y=x^2+y \implies y=0$, and in the second case we get: $y-x^2=x^2+y \implies x=0$. Hence, the only possibility is that when one of them is $0$ i.e. either $A=\{0\}$ or $B=\{0\}$. $\quad \blacksquare$
[rule]

$x = 0, y = y$ in the problem gives $f(f(y)) = f(0)^2 + y$ so $f$ is bijective. Hence there exists $c$ such that $f(c) = 0$. Now letting $x = c, y = 0$ gives $f(f(y)) = y$ but also previously $f(f(y)) = f(0)^2 + y$ so $f(0) = 0$. Now setting $x = x, y = 0$ in the original question gives $$f(xf(x)) = f(x)^2 \qquad (1)$$ but since $f(f(x)) = x$, we replace $x$ with $f(x)$ in $(1)$ to get $f(xf(x)) = x^2$. So by $(1)$, $f(x)^2 = x^2$ for all $x \in \mathbb{R}$. Now to solve the pointwise trap, suppose $f(a) = a$ and $f(b) = -b$ for distinct non-zero $a, b$. Setting $x = a, y = b$ in the question gives $f(a^2 - b) = a^2 + b \implies b = 0$, contradiction. Hence, either $\boxed{f(x) = x \forall x \in \mathbb{R}}$ or $\boxed{f(x) = -x \forall x \in \mathbb{R}}$ and we see that both solutions work

$P(0,x) : f(f(x)) = f(0)^2 + x$.
$P(0,x-f(0)^2) : f(f(f(0)^2 - x)) = x \Longrightarrow f$ is surjective.
Hence there exists a real number $r$ such that $f(r) = 0$.

$P(r,x) : f(f(x)) = x = f(0)^2 + x \Longrightarrow f(0) = 0$ and $f$ is bijective.

$P(x,0) : f(xf(x)) = f(x)^2$.

Substituting $f(x)$ in place of $x$ in the above equation : $f(xf(x)) = x^2 \Longrightarrow \forall x \in \mathbb{R}$ either $f(x) = x$ or $f(x) = -x$

Assume that $\exists a,b \in \mathbb{R}$ such that $f(a) = a$, $f(b) = -b$ and $a,b \neq 0$\\
$P(a,b) : f(a^2 - b) = a^2 + b$\\
If $f(a^2 - b) = a^2 - b$ then $b=0$. Contradiction .\\
If $f(a^2 - b) = b -a^2$ then $a=0$. Contradiction . \\
Hence the functions that work are $f(x) = x \forall x \in \mathbb{R}$ and $f(x) = -x \forall x \in \mathbb{R}$.\\
These functions satisfy the original equation. \\

Please someone check this :maybe:
[b]Solution :[/b] Let $P(x,y)$ denotes the assertion.

Now $P(0,y):f(f(y))=f(0)^2+y~(\star)$ so $f$ is surjective. Now assume $f(a)=0$ for some $a \in \mathbb{R}$.
Then $P(a,a):f(0)=0 \implies a=0$. So $f(f(y))=y$  from $~(\star)~$.
Now $P(f(x),y):f(f(x) \cdot x + f(y))=x^2+y$. Comparing with $P(x,y)$ we get that $f(x) = \pm x$. It's easy to check that these infact, works. $\qquad \blacksquare$

@above your proof is correct. However I think you should consider the pointwise trap, which is very easy to deal with here.

I am quite new at FEs and I still don`t have that "intuitions" that experienced FE solvers have
So i don`t totally get this step

[quote=nttu]f(xf(x)) =f^2(x) (4)
 In (4) replaceing x by f(x) , combining with (3) , we have : f(xf(x))=f^2(x) =x^2$
 After some normal computations , we get : f(x) =x \forall x or f(x)=-x \forall x[/quote]
I thought we can`t but f(x) instead of x, because f(x) may NOT be equal to x for any x in R, or can we? I`ve always been putting some constants in assertion, like 1, 0, f(0), (f(2)-2)/3 or whatever
Please help me to build up my intuition

@above
$x$ is defined as any real number(!) and since $f(x)\in \mathbb R$ (note that it might not take all the values of $x$, but important is that all the values it takes, those all are in $\mathbb R$ (in domain)), thus we can plug $f(x)$ in place of $x$.
I hope this helps.

[quote=Kunihiko_Chikaya]Find all functions $f : \mathbb{R} \mapsto \mathbb{R}$ such that $f(xf(x)+f(y)) =(f(x))^2+y$ for all $\ x,y \in \mathbb{R}.$[/quote]

We claim that the only solutions are $\boxed{f(x)=\pm x}$ which clearly works. We will show that these are the only solutions. Let $P(x:y)$ denote the assertion in the above equation. Suppose there $\exists a,b$  such that $f(a)=f(b)$. Then
$\begin{cases} P(x:a)\implies f(xf(x)+f(a))=f(x)^2+a \\ P(x:b)\implies f(xf(x)+f(b))=f(x)^2+b\end{cases} \implies a=b\implies f\text{ is injective. }$

Now fix $x$ and vary $y$. We get $f$ is surjective hence bijective too. Let $f(k)=0$ for a unique real number $k$. Now $P(k:y)\implies f(f(y))=y\implies f$ is Involutive. So, $P(0:0)\implies f(f(0))=0=f(0)^2\implies f(0)=0$. So, $P(x:0)\implies f(xf(x))=f(x)^2\implies f(xf(x))=x^2\implies \boxed{f(x)=\pm x}$ by plugging $x=f(x)$.

Now suppose there $\exists p,q\in\mathbb R$ such that $f(p)=p$ and $f(q)=-q$. So, $P(p:q)=f(p^2-q)=p^2+q$ or $q-p^2$. So, $p=q=0$. Hence, we get that the only solutions are $\boxed{f(x)=\pm x}$. $\qquad\blacksquare$

Let $p(x,y)$ be the assertion for  $f(xf(x)+f(y))=f(x)^2 + y$
$p(0,0)$,we get that $f(f(0))=f(0)^2$
Also from $p(x,0)$ we get that $f(xf(x)+f(0))=f(x)^2$
$p(x,-f(x)^2)$,
we get that $f(xf(x)+f(-f(x)^2))=0$
This tell us that $\exists a\in\mathbb R$ ,such that $f(a)=0$.
$p(a,a)$,means that $f(0)=a$,which gives us $f(f(0))=0$,and from $p(0,f(0))$,we get that $f(0)=0$
Now from $p(0,x)$ we get that $f(f(x))=x$--involutive, so $f$ is injective and surjective.
Now let us put $p(f(x),y)$ and we get that
$f(xf(x)+f(y))=x^2+y$,substracking $p(x,y)$ from it we get that
$f(x)^2=x^2$ , and we get that $f(x)= x$ and $f(x)=-x$

Quick exercise :D
$\color{black}\rule{25cm}{1pt}$
The only solutions are $f(x)=x$ and $f(x)=-x$ for all reals.

Let $P(x,y)$ be the given assertion from $P(0,y)$ we get that $f(f(y))=y+f(0)^2$.
Thus this implies that $f$ is a bijection.
Let $c=f(0)$.
From $f(f(y))$, we get that $f(f(-c^2))=0$.
But from $P(-c^2,y)$, we get that $f(f(y))=y$, which implies that $y=y+c^2$ from which we get that $c=0$.
So now we get that $P(x,y)$ and $P(f(x),y)$ are the same thing.
Thus we have that $f(x)^2=x^2$
This devolves into a classic pointwise trap and we easily get that we have that the solutions are $f(x)=x$ and $f(x)=-x$ for all real numbers $x$.

[quote=EulersTurban]Quick exercise :D
$\color{black}\rule{25cm}{1pt}$
The only solutions are $f(x)=x$ and $f(x)=-x$ for all reals.

Let $P(x,y)$ be the given assertion from $P(0,y)$ we get that $f(f(y))=y+f(0)^2$.
Thus this implies that $f$ is a bijection.
[/quote]

How did you know f(x) is injective?

See page 3: [url]https://web.evanchen.cc/handouts/FuncEq-Intro/FuncEq-Intro.pdf[/url] :)

Let $P(x,y)$ be the assertion of $f(xf(x)+f(y))=f(x)^2+y$
$P(0,x)\implies f(f(x))=f(0)^2+x$ hence f is bijective
So there exist $a\in\mathbb{R}$ such that $f(a)=0$
$P(a,x)\implies f(f(x))=x\quad\forall x\in\mathbb{R}$
$P(x,0)\implies f(xf(x))=f(x)^2$ and let $x\to f(x)$ we have $f(x)^2=x^2\quad\forall x$.
Assume $f(a)=a$ and $f(b)=-b$ for some $a,b\neq 0$ then $P(a,b)$ we get that
$$f(a^2-b)=a^2+b\implies b=0\quad\text{or}\ a=0$$ which is the contracdition and so
$$\boxed{f(x)=x\ \text{or}\ -x \quad\forall x\in\mathbb{R}}$$

The answer is $f(x) = x$ and $f(x) = -x$ only. It's easy to check that these work.

By fixing $x$ and varying $y$ over the reals, we get that $f$ is surjective. Since $f$ is surjective, there exists some $\ell$ such that $f(\ell) = 0$. Plugging in $x=\ell$ in our original equation, we have $$f(\ell f(\ell) + f(y)) = f(\ell)^2 + y \Rightarrow f(f(y))=y.$$
So, $f$ is an involution from which it follows that $f$ is bijective. Substituting $x=f(v)$, we have $$f(f(v)f(f(v))+f(y))=f(f(v))^2 + y$$
$$\Rightarrow f(vf(v)+f(y))=v^2+y$$

However, by the original equation, we have $$f(vf(v)+f(y))=f(v)^2+y$$
Therefore, $$f(v)^2+y=v^2+y \Rightarrow f(v) = \pm v$$

Now, assume that $f(a) = +a$ and $f(b) = -b$ for some nonzero $a,b$. Substituting $x=a$ and $y=b$ into the original equation, we have $$f(af(a)+f(b)) = f(a)^2+b \Rightarrow f(a^2-b) = a^2+b$$
This does not hold for nonzero $a,b$ so we cannot have $f(x)=+x$ for some values of $x$ and $f(x)=-x$ for others.

$\spadesuit \color{red}{\textit{\textbf{ANS:}}}$ $f(x)=x \quad \textrm{and} \quad f(x)=-x \quad \forall x\in \mathbb{R}$.
$\blacklozenge \color{blue}{\textit{\textbf{Proof:}}}$ It's easy to see that these are indeed solutions to the given FE. Let $P(x,y)$ denote the given assertion, we have \[P(0,x): f(f(x))=f(0)^2+x\] which clearly implies that $f$ is bijective. There exists $t\in \mathbb{R}$ such that $f(t)=0$, we use this to get \[P(t,x): f(f(x))=x\] comparing to $P(0,x)$, yields $f(0)=0$ and $t=0$. Then, \[P(x,0): f(xf(x))=f(x)^2\]\[P(f(x),0): f(f(x)x)=x^2\] and so $f(x)=\pm x \quad \forall x\in \mathbb{R}$. Let $a,b \in \mathbb{R} \setminus \{0\}$ such that $f(a)=a, f(b)=-b$, we have \[P(a,b): f(a^2-b)=a^2+b\] if $f(a^2-b)=a^2-b,$ then $b=0$ a contradiction. If $f(a^2-b)=b-a^2$, then $a=0$ which is also a contradiction. Thus, $f(x)=x \quad \textrm{and} \quad f(x)=-x \quad \forall x\in \mathbb{R}$ are the only solutions. $\quad \blacksquare$

[hide = Incomplete]
[b][color=#f00]Answer: [/color][/b]$f(x) = x \quad \text{or} \quad f(x) = -x$

Proof:
Let $P(x,y)$ denote the given assertion.
$P(0,y) \to f(f(y)) = y + f(0)^2 = y + \text{constant}$
$\implies$ [b][color=#f00]f is bijective[/color][/b]
$\implies \exists \ a$ such that $f(a) = 0$
$P(a,y) \to \textbf{\color{red}{f(f(y)) = y}}$
In the other hand $P(0,y) \to f(f(y)) = f(0)^2 + y$
$\implies f(0) = 0$
$P(x,0) \to f(x \cdot f(x)) = f(x)^2$
$x \to f(x)$
$\implies f(f(x) \cdot f(f(x))) = f(f(x))^2$
$\implies f(f(x) \cdot x) = x^2$ (Since $f(f(x)) = x$)
$\implies f(x)^2 = x^2$
$\implies f(x) = \pm x$
Plugging it back in our original equation we see that both the solutions work.
[/hide]
:oops: thanks for pointing out the mistake.

The proof is incomplete. $f(x)$ could be $x$ for some $x$ and $-x$ for other $x$.

by gussing we can see here that $f(x)=\pm x$ is the only solution.
[b][color=#f00]Claim1:[/color][/b] $f(x)$ is bijective

so suppose if $f(a)=f(b)$ we have
$P(x,a)$
\[f(xf(x)+f(a))=(f(x))^2+a\]
$P(y,a)$
\[f(xf(x)+f(b))=f(x)^2+b\]
since $f(a)=f(b)$ we can say that
\[\implies f(x)^2+a=f(x)^2+b\]
which gives $a=b$ this proves that $f$ is injective\\
$P(0,y)$
\[f(f(y))=f(0)^2+y\] so we can see the $f$ is surjective too so since out assumption is proved that $f$ is bijective.$\square$
so since $f$ is bijective so $f(f(y))=y$ so we have $f(0)=0$\\
So in the main problem we have \\
$P(x,0)$\[\implies f(xf(x))=f(x)^2\]
$P(f(x),0)$\[\implies f(f(x)f(f(x)))=f(f(x))^2=x^2\]
\[\implies f(f(x)x)=x^2\implies f(x)^2=x^2\]
so we can see that we get $f(x)=\pm x$
so here we can see we have a point wise trap so suppose $f(a)=a$ and $f(b)=-b$ so by putting\\
$P(a,b)$ we have
\[f(af(a)+f(b))=f(a)^2+b\implies f(a^2-b)=a^2+b\]
and since we know that answer should be $\pm x$ so $f(a^2-b)=a^2-b\text{ or }b-a^2$ so by this $a=0,b=0$ which is contradiction which give the solution is $f(x)=x$ $\forall x\in\mathbb{R}$
and $f(x)=-x$ $\forall x\in\mathbb{R}$
and we are done $\blacksquare$

[quote=Kunihiko_Chikaya]Find all functions $f : \mathbb{R} \mapsto \mathbb{R}$ such that $f(xf(x)+f(y)) =(f(x))^2+y$ for all $\ x,y \in \mathbb{R}.$[/quote]

Let $P(x,y)$ the assertion [color=#f00][color=#f00][/color][/color]to the given F.E.
[b][color=#f00]Claim 1:[/color][/b] $f$ is bijecitive
[b][color=#f00]Proof:[/color][/b] Note that it follows from $P(0,x)$
$$f(f(x))=f(0)^2+x$$
[b][color=#f00]Claim 2:[/color][/b] $f(0)=0$
[b][color=#f00]Proof:[/color][/b] Since $f$ is surjective there exists $c$ such that $f(c)=0$.
$P(c,x)$
$$f(f(x))=x \implies f(0)=0$$
Becuase on the last proof we used that $f(f(x))=f(0)^2+x$
[b][color=#f00]Claim 3:[/color][/b] $f(x)=\pm x$ is the only solution to this F.E.
[color=#f00][b]Proof:[/b][/color] The following replaces and some stuff show us the desired claim.
$P(f(x),y)-P(x,y)$
$$f(x)^2=x^2 \implies f(x)=\pm x$$
Assume that there exists $a,b \ne 0$ such that $f(a)=a$ and $f(b)=-b$.
$P(a,b)$
$$\pm(a^2-b)=a^2+b \implies \text{contradiction!!!}$$
Thus by [b]Claim 3[/b] the only solution is:

$\boxed{f(x)=\pm x \; \forall x \in \mathbb R}$

Thus we are done  :blush:

A nice problem :)

The answers are $f\equiv x$ and $f\equiv -x$. It is easy to see that these work.

Now let $P(x,y)$ denote the assertion. $P(0,y)$ gives that $f$ is surjective, and then choosing $x$ such that $f(x)=0$ gives that $f$ is an involution, so $f$ is injective as well. Then $P(x,y)$ and $P(f(x), y)$ give $f(xf(x)+f(y))=f(x)^2+y=x^2+y,$ so we know that $f(x)=\pm x$.

To avoid pointwise trap, suppose $f(a)=a$ and $f(b)=-b,$ for $a,b\neq 0$. Then $P(a,b)$ gives $f(a^2-b)=a^2+b,$ and testing both possibilities gives at least one of $a,b$ is $0,$ so we are done.

We claim that the only two solutions are $f(x) \equiv x$ and $f(x) \equiv -x$. Plugging these in they evidently work, so we will show that no other solutions exist.

    By substituting $(x, -f(x)^2)$, we discover that there exists a value $\kappa$ such that $f(\kappa) = 0$. (Why?) Thus, by substituting $(\kappa, x)$, we discover that \[f(f(x)) = x \ \forall \ x \in \mathbb{R}.\] Thus, $f$ is an involution! We use the trick of substituting $f(x)$ for $x$ to reduce the number of total $f$'s to get \[f(f(x)x + f(y)) = x^2 + y.\] Comparing this with the given equation, we get that \[f(x)^2 = x^2.\] Thus, $f(x) \in \{-x, x\}$ for $x \in \mathbb{R}$. Suppose that there existed nonzero real numbers $a$ and $b$ such that $f(a) = a$ and $f(b) = -b$. Then, substituting $(a, b)$, we get that \[\pm (a^2 - b) = a^2 + b,\] which is a contradiction. Thus, the only solutions are $f(x) \equiv x$ and $f(x) \equiv -x$, as desired.


greatest common divisor
number theory proposed
number theory

a,b are elements of N numbers. a+1/b + b+1/a is N and GCF(a,b)=d.
Prove that d2<=a+b

$ x^3\minus{}y^3 <x^3 \plus{}y^3$
$ x^3\minus{}y^3<x\minus{}y$ we know that x>y because x,y are both positive
$ x^2\plus{}xy\plus{}y^2<1$  so $ x^2\plus{}y^2<1$

And I would like you to explain the title "PLEASE SOLVE THIS PROBLEM UNDERSTANDABLE AND QUICKLY"!
a) No shouting/full capital letter titles!
b) We are not your personal army and not there to solve your problems "as quickly as possible".
c) Your title says nothing about the problem itself.
Next time, use a title that says something about the problem itself and don't shout.

i think he is in an exam now  :lol:  so excited :D


[quote="SplashD"][quote="wobster109"]i was wondering if anyone has a little green check mark next to their round 1 on the USAMTS site?  the little check means that the answers have been received, and i don't have one yet.  should i try to resend my answer pages?[/quote]

I have it. If you don't have it yet, it doesn't necessarily mean they didn't receive. Besides, I don't think they would accept it now either since it's past the deadline and the math jam[/quote]
Didn't you send in your sols like 10 days ago...

I don't have mine yet, but I did send it in by snail mail.

I emailed my solutions and don't have a check mark yet, either. I doubt it's anything to worry about.

Nor I. Don't bother with it. Remember, the USAMTS probably had a thousand people submit, each of which has to get logged, so it takes a while to get all of them logged.

ok, thank you!  i did send it a few days ago by fax, or rather, my mother sent it for me because i don't know how to work the fax machine.  i was afraid the fax machine may have malfunctioned or something.  thanks!

I faxed; no green check-mark.
EDIT:
Oh, now I have it!

Don't worry about the lack of check mark.  Not all papers are logged.  We should have them all logged by the end of the week.

I just got mine logged, and I sent it about 6 hrs before the deadline (I emailed them).

I got mine today, sent 1 hour before the deadline (email)

I have the checkmark, and I emailed my solutions a few hours before the deadline.

I have a checkmark in, and I sent mine in on snail mail on the morning of the 10th. I think that they are probably all done now.

[quote="lotrgreengrapes7926"]I sent mine a few days before the deadline, and still haven't gotten a check mark. :(   :lol:[/quote]
Did you send yours  by email or mail or fax? It seems that everyone who sent theirs by email got theirs logged first, (not saying that they favor tex solutions).

[quote="bpms"]Did you send yours  by email or mail or fax? It seems that everyone who sent theirs by email got theirs logged first, (not saying that they favor tex solutions).[/quote]
Email, but I just got the checkmark today. :D

I got my checkmark yesterday and i sent in answers a day early.

[quote="me@home"]I sent in the answers at 11:40 pm pacific time on the 10th, anyone cut the deadline better?[/quote]
I think I sent mine in around 11:50...  :wink:

The correct term is "Procrastinators", not "crazy"
sorry I just finished the psats 20 mins ago and am still on that kind of thinking

I sent them at 1:30 [color=white][size=0]EST.[/size][/color]

I also have a little green checkmark.

...I have a little green check mark, but it also says that I haven't submitted my Entry Form yet. Has USAMTS processed their entry forms yet?

[quote="me@home"]They accepted them an hour and a half after the deadline?[/quote]

Santosh like me is on the East Coast, so we can wait till 3am to send our solutions in :D.

[quote="me@home"]They accepted them an hour and a half after the deadline?[/quote]

Yes, that time is EST. This was indicated by the letters "EST" after the time [size=9]which were displayed in a font size that is smaller than this one.[/size]


Bueno solo para informarles a los participantes de este forum..
que se esta realizando un importante evento en Peru en esta semana...

el [b][color=blue][size=150]CONAMAT 2005 - CONCURSO NACIONAL DE MATEMATICA CESAR VALLEJO[/size][/color][/b]..

organizada por el Colegio Bertolt Brecht - [url=http://www.colegiobertoltbrecht.edu.pe/colegio/index.htm]Web Site[/url]

y que tiene lugar cada año, el cual cuenta con una fase eliminatoria en Setiembre a nivel Nacional
agrupando a los 24 Departamentos del PERU EN 6 SEDES, clasificando a los mejores de acuerdo a
dos modalidades de Procedencia: Colegio Estatal y Colegio no Estatal...

Y eso culmina con la Gran Final, examen que tiene lugar en Lima en uno de los locales de
la Sociedad de Ciencias y Humanidades - ADUNI - Bertolt Brecht.

Mayor Informacion:   [url=http://www.aduni.com.pe/conamat/conamat.html]Aqui[/url]

En este link podra encontrar tambien los examenes completos de las ediciones anteriores 2003-2004
Estos examenes son de alternativas multiples..

Para mejor entendimiento, este es la equivalencia de los niveles del sistema educativo peruano:

Educacion Primaria: 4to Grado, 5to Grado, 6to Grado

Educacion Secundaria (High School):
 - 1er Año = 8th Grade
 - 2do Año = 9th Grade
 - 3er Año = 10th Grade
 - 4to Año = 11st Grade
 - 5to Año = 12nd Grade
--------------------------------------
He elegido algunos problemillas de estos examenes - Edicion 2001

    [size=150] [color=blue]CONAMAT 2001 - PERU[/color] [/size]

1. (10th Grade / Q24) Sea $f: R \to R$, tal que:
$f(x^3 + y^3) = (x+y)[f^2(X) - f(x)f(y) + f^2(y)],$
donde $x,y \in \mathbb{R}$, tal que: $f(x) \neq 0$.
Encontrar el valor de: $\frac{f(2001x)}{2001x}$.

2. (12nd Grade / Q12) Sea la funcion $G$, tal que:
   (1) $G(x+y+z) = \frac{1}{G(x)} + \frac{1}{G(y)} + \frac{1}{G(z)}$, para todol $x,y,z \in R$.\\
   (2) $G(x) > 0$, para todol $x \in R$.
   Encontrar $G(2001)$

3. (10th Grade/Q25) Factorizar el siguiente polinomio:
     $P(a,b,c,x) = (a-b)(x-a)(x-c) + (b-c)(x-b)(x-c) + (c-a)(x-c)(x-a).$

 Good Luck   ;)

Carlos Bravo
Lima-PERU

Cuantos participantes se van a tener en esta oportunidad???


Prove that for n>= 1,
      $ \sum_{k\equal{}1}^n\ (k,n)\equiv\ n\sum_{d|n}\frac{\phi(d)}{d}$

We are told to modify the Gauss Theorem but I still don't understand. Please help! Thanks!!

First, note that $ (k,n)$ is a divisor of $ n$ and second, for any divisor $ d$ of $ n$:

$ \;k\in\{1,\ldots,n\}$ and $ (k,n)\equal{}d\Longleftrightarrow k\equal{}dt$ with $ t\in\left\{1,\ldots,\frac{n}{d}\right\}$ and $ \left(t,\frac{n}{d}\right)\equal{}1$

This allows to rewrite the left hand side as (I carried out all intermediate steps for your benefit):

$ \sum_{k\equal{}1}^n(k,n)\equal{}\sum_{d|n}\left[\sum_{k\equal{}1\atop (k,n)\equal{}d}^nd\right]\equal{}\sum_{d|n}\left[\sum_{t\equal{}1\atop \left(t,\frac{n}{d}\right)\equal{}1}^{\frac{n}{d}}d\right]\equal{}\sum_{d|n}\left[d\cdot\sum_{t\equal{}1\atop \left(t,\frac{n}{d}\right)\equal{}1}^{\frac{n}{d}}1\right]\equal{}$
$ \equal{}\sum_{d|n}d\cdot\phi\left(\frac{n}{d}\right)\equal{}\sum_{d|n}\frac{n}{d}\cdot\phi(d)\equal{}n\cdot \sum_{d|n}\frac{\phi(d)}{d}$

[quote="hjbrasch"]First, note that $ (k,n)$ is a divisor of $ n$ and second, for any divisor $ d$ of $ n$:

$ \;k\in\{1,\ldots,n\}$ and $ (k,n) \equal{} d\Longleftrightarrow k \equal{} dt$ with $ t\in\left\{1,\ldots,\frac {n}{d}\right\}$ and $ \left(t,\frac {n}{d}\right) \equal{} 1$

This allows to rewrite the left hand side as (I carried out all intermediate steps for your benefit):

$ \sum_{k \equal{} 1}^n(k,n) \equal{} \sum_{d|n}\left[\sum_{k \equal{} 1\atop (k,n) \equal{} d}^nd\right] \equal{} \sum_{d|n}\left[\sum_{t \equal{} 1\atop \left(t,\frac {n}{d}\right) \equal{} 1}^{\frac {n}{d}}d\right] \equal{} \sum_{d|n}\left[d\cdot\sum_{t \equal{} 1\atop \left(t,\frac {n}{d}\right) \equal{} 1}^{\frac {n}{d}}1\right] \equal{}$
$ \equal{} \sum_{d|n}d\cdot\phi\left(\frac {n}{d}\right) \equal{} \sum_{d|n}\frac {n}{d}\cdot\phi(d) \equal{} n\cdot \sum_{d|n}\frac {\phi(d)}{d}$[/quote]

THANK YOU!!!!... i understand it now... i feel so stupid that i didn't think of it... thanks a lot!!!!!


linear algebra
matrix
algorithm
induction
superior algebra
superior algebra theorems

an easy one:
let $(a_1,a_2,..,a_n)$ be $n$ integers prime together
(excuse my english, I don't remember the exact expression).
(there exists integers $u_i$ such that $a_1u_1+..+a_nu_n=1$)
Prove that there exists a matrix in $SL_n(Z)$ with first line
$(a_1,..,a_n)$.

Given any matrix with first line $(a_1,\ldots a_n)$, we can perform transformations of this matrix by multiplying it with matrices in $SL_n(\mathbb Z)$ which add a column to a certain column or perform even permutations on columns s.t. we end up with a matrix having the first line $(1,0,\ldots,0)$ (it's some sort of generalized Euclid's algorithm, for $n$ numbers). Now, starting with $I_n$, we can simply perform the inverse transformations, which is done by multiplying $I_n$ with the inverses of the matrices above. What we get must then be a matrix with integer entries (because the described transformations perform only additions and permutations) and determinant $1$ (because the transformations do not alter the determinant). Furthermore, since the transformations are the inverses of those used above, the first line of the new matrix must be $(a_1,\ldots,a_n)$, as required.

Induction.
$n=2$. It is well known and actually comes with alekk's defintion that there exist $b_1,b_2\in\mathbb{Z}$ such that $a_1b_1+a_2b_2=1$.
Consider the matrx $\left(\begin{array}{cc}a_1& a_2\\-b_2& b_1\end{array}\right)$. It has $1$ as det.

$n\rightarrow n+1$. Let $d=gcd(a_2,a_3,\ldots,a_{n+1})$. Let $c_i=\frac{a_i}d\ \forall i=\overline{2,n+1}$.
$(c_i)_{i=\overline{2,n+1}}$ are coprime so by induction there exists $A\in SL_n(\mathbb{Z})$ such that the first row of $A$ is $(c_2,\ldots,c_{n+1})$
Because $(a_1,\ldots,a_{n+1})$ are coprime, $gcd(a_1,d)=1$ so there exist $u,v\in\mathbb{Z}$ such that $ua_1+vd=1$.

Consider the $n+1\times n+1$ matrix:
On the first row place $(a_1,a_2,\ldots, a_{n+1})$.
On the $2^{nd}$ to the $n^{th}$ entry of the first column put $0$.
In the down-right $n\times n$ corner put the matrix obtained from $A$ by interchanging the first row multiplied by $-u$ with the last row.
On the $n+1^{th}$ entry of the first column put $(-1)^{n+1}v$.

Now use Lagrange for the first column to compute the determinant of the obtained matrix. Notice that the relevant minors have dets $u$ and $d$.


[quote="nmtvn"]Prove that in any triangle ABC, we have:
$ \frac {cotA.cotB.cotC}{sinA.sinB.sinC} \leq (\frac {2}{3})^3$[/quote]
YOU SHOULDN'T POST THIS INEQUALITY IN THIS FORUM. :mad:  :mad:  :mad:  :mad:
IT IN MATHEMATICS AND YOUTH MAGAZINE IN VIETNAM

I think this topic should be IMMEDIATELY locked!!!! :mad:
However, I have informed the mod.


1. Let $ p$ be a prime and $ a,b,c\in Z$ Prove that $ \sum_{i \equal{} 0}^{p \minus{} 1} (\frac{ai^2\plus{}bi\plus{}c}{p}) \equal{} \minus{}(\frac{a}{p})$ if $ p \not{|} b^2\minus{}4ac$.

Looks very much like PEN C6: http://www.mathlinks.ro/viewtopic.php?t=150500 .


Okay, the problem asks you to find an equation of the tangent line to the curve at the given point.

8. Y = (2X + 1)^1/2 and the point (4,3)

I then made it into a limit (as x approaches 4) for ((2X + 1)^1/2) - 3)/(X - 4).

The problem is that I don't know where to go from here. I don't know how I can reduce this fraction any more. What do I do?

Ah, genius idea! I'll be sure to keep that trick in mind.

Thanks!  :D

(I got 1/3, correct?)


Find the minimum value of :

$ E \equal{} a^2 \plus{} 4b^2 \plus{} 9c^2 \plus{} 16d^2 \minus{} a \plus{} 2b \minus{} 3c \plus{} 4d\ ,\ a,b,c,d\in \mathbb{R}$

$ E \equal{} (a \minus{} 1/2)^2 \plus{} 4(b \plus{} 1/4)^2 \plus{} 9(c \minus{} 1/6)^2 \plus{} 16(d \plus{} 1/16)^2 \minus{} 3/4 \minus{} 1/16 \ge \minus{} 13/16$

Edit: $ E \equal{} (a \minus{} 1/2)^2 \plus{} 4(b \plus{} 1/4)^2 \plus{} 9(c \minus{} 1/6)^2 \plus{} 16(d \plus{} 1/8)^2 \minus{} 1 \ge \minus{} 1$
Also, another sol.:
$ k^2 \minus{} k \equal{} (k \minus{} 1/2)^2 \minus{} 1/4 \ge \minus{} 1/4$. Letting $ k \equal{} a, \minus{} 2b,3c, \minus{} 4d$ and adding the four inequalities give $ E\ge \minus{} 1$

Sorry , i had a typo . Now the minimum should be more nice  :)


A school organization consists of 5 teachers, 7 parents and 6 students. A subcommittee of this group is formed by choosing
2 teachers, 3 parents and 3 students. How many different
subcommittees can be formed?

We choose teachers, parents, and students separately.

There are $ \binom{5}{2}$ ways to choose teachers, $ \binom{7}{3}$ ways to choose parents, and $ \binom{6}{3}$ to choose students. Since each choice is independent, we have $ \binom{5}{2}\binom{7}{3}\binom{6}{3}\equal{}10\cdot35\cdot20\equal{}\boxed{7000}$

[quote=bestzack66]23 hour bump (dont get mad at me)[/quote]

Just check your answers, don't submit before  you check.

[quote=eduD_looC][quote=bestzack66]23 hour bump (dont get mad at me)[/quote]

Why the bump though? ...[/quote]

He wants to know how not to silly.

[quote=bestzack66]Imagine adding the combinations instead of multiplying -_-

I am making so much of these silly mistakes, can someone give me some tips for sillies?

(Sorry for the 11 year bump!)[/quote]

Check your work, focus on what you learn NOT RATING


vector
geometry
parallelogram
analytic geometry
rhombus

1) Using vectors, prove that the diagonals of a parallelogram bisect each other.

2) Prove that the vector $w = \parallel u\parallel v+\parallel v\parallel u$ bisects the angle between $u$ and $v$.

[hide="Proof 1"]Take $\mathbf{v}+\mathbf{w}=\mathbf{x}$.  We then have \[\mathbf{v}+\frac{\mathbf{w}-\mathbf{v}}{2}=\frac{\mathbf{w}+\mathbf{v}}{2}=\frac{\mathbf{x}}{2},\] and we're done.[/hide]
[hide="Hint 2"]Since $\textrm{span}\left(\mathbf{v},\mathbf{w}\right)$ is a plane, I believe WLOG it is sufficient to prove this statement for $\mathbf{v},\mathbf{w}\in\mathbb{R}^{2}$.  Therefore you need only set $\mathbf{v}=\left(v_{x},v_{y}\right)$ and $\mathbf{w}=\left(w_{x},w_{y}\right)$ and use the definition of dot product to find the angle between the two vectors.  From here, all you have to do is to show that the angle between the given vector and either of $\mathbf{v}$ or $\mathbf{w}$ is half of the already known angle.[/hide]

[hide="Better hint for 2"]Look for a rhombus.[/hide]

I think JBL's hint is probably far better than what I'm going to suggest, but I'm too lazy to figure out why it works.  You don't need to put down coordinates though, as:

$(w,u) = \|u\|(u,v)+\|v\|\|u\|^{2}$
and
$(w,v) = \|u\|\|v\|^{2}+\|v\|(u,v)$

Or:

${\bold{w}\over uv}={\bold{u}\over u}+{\bold{v}\over v}$, and the summands are unit vectors of $\bold{u},\bold{w}$ which, as [b]JBL[/b] suggested, actually are two adjacent sides of a rhombus with the corresponding diagonal being the sum, and the conclusion follows.


limit
LaTeX
calculus
calculus computations

there we known $ \mbox{lim}(\frac{n^n}{n!})^{\frac {1}{n}}\equal{}e$. I had known a proof which with the help of definition of integration. but could anybody tell me what is the result below, and how to prove it: $ \mbox{lim}\frac{n!e^n}{n^n}$?

hello, do you mean $ \lim_{n \to \infty}\frac{n!e^n}{n^n}$?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, do you mean $ \lim_{n \to \infty}\frac {n!e^n}{n^n}$?
Sonnhard.[/quote]
yes, that's right. thank you very much and i'm sorry that i had forgotten how to post limitation with latex and got you confusion.

[quote="hsiljak"]Stirling's formula will do the trick:
since $ \lim_{n \rightarrow \infty} {\frac {e^n\, n!}{n^n \sqrt {n}}} \equal{} \sqrt {2 \pi}$, your limit is $ \infty$ because of its equivalence to $ \lim \sqrt {2\pi n}$.

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

If you want proof without the use of it, just take the proof of the Stirling's formula itself and adjust it.[/quote]

thank you very much. i'm reading about the website now. the problem comes from a mistake of my calculation, when i got the right way, i found out that this limitation is of somewhat interesting, but i couldn't made it.


Consider points $ M$ and $ N$ on respectively sides $ BA$ and $ BC$ of triangle $ ABC$ such that $ \frac{2CN}{BC}\equal{}\frac{MA}{BA}$. Let $ P$ be a point on line $ AC$. Prove that lines $ MN$ and $ NP$ are perpendicular iff $ PN$ bisects angle $ \angle MPC$.

Let $ MN\cap AC\equal{}{U}$ , by the Menelaus Theorem ; $ |MN|\equal{}|UN|$  $ \Longrightarrow$ $ (NP\perp MN \Longleftrightarrow m(MPN)\equal{}m(NPU))$ :wink:

Could you solve it without using Menelaus Theorem? By the way, I am sure the proof above is true.

Oh it's simple to prove without using Menelaus theorem. Let us draw $ MV\parallel AC$ and $ V\in BC$ then it's easy to see that $ N$ is the midpoint of $ MQ$.


function
algebra
functional equation
algebra unsolved

Find all real valued functions satisfy
$f(x^2+y^2+2f(xy)) = f(x+y)^2$

Put $y = y-x$. The equation then becomes

$f(x^2+(y-x)^2+f(xy-x^2)) = f(y)^2$

For any fixed y, the right side is constant, implying that the left side must remain constant as well; we thus have two possibilities:

1: f is constant:

Only $f(x) = 1$ and $f(x) = 0$ satisfy the equation.

2: f is not constant but $x^2 + (y-x)^2 + 2f(xy-x^2) = k$ , where k is  a constant, for any fixed y.

$2f(x(y-x)) = k - 2x^2+2xy = k +2x(y-x)$

 $f(x) = C+x$ for some constant C. Substituting this function into the original functional equation, we have:

$C+(x^2+y^2+2C+2xy) = 3C + (x+y)^2 = (C+(x+y))^2 = C^2 + 2C(x+y) + (x+y)^2$
 For this equality to hold for all x and y, C must equal zero, so we have f(x) = x.

In summary, we have found the following suitable functions:

$f(x) = 0$
$f(x) = 1$
$f(x) = x$

Did I miss any functions?

mathisfun1 is wrong in step 2 unless he can prove $ f$ is injective, which is not necessarily true.
Also, this is double posted.
http://www.mathlinks.ro/viewtopic.php?p=1361019#1361019


calculus
derivative
algebra
polynomial
function
ceiling function
calculus computations

I'm looking for an explicit formula for the Hermite polynomials (the probabilists' kind, i.e. $ H_n(y) \equal{} ( \minus{} 1)^ne^{ \minus{} \frac {1}{2}y^2}\frac {d^n}{dy^n} \left( e^{ \minus{} \frac {1}{2}y^2}\right)$--opposed to the physicists' Hermite polynomials in which both $ \frac1{2}$'s are dropped.) Any help would be, well, helpfull.

Thanks in advance,
-Ben

Here's my work thus far, I tried 3 different approaches:

[b]i.[/b] It seems clear that
\[ \frac {d^n}{dy^n} \left( e^{ \minus{} \frac {1}{2}y^2}\right) \equal{} e^{ \minus{} \frac {1}{2}y^2}\left[ ( \minus{} y)^n \plus{} \sum_{j \equal{} 0}^{n \minus{} 1}c_{j}y^j \right]
\]
My trouble is determining the coefficients $ c_j$. I tried setting-up a recurrence, but my analysis produces the absurd conclusion that $ c_0 \equal{} 0$. My recurrence was based on this:

Let $ h_n(y) \equal{} \frac {d^n}{dy^n} \left( e^{ \minus{} \frac {1}{2}y^2}\right)$. Then $ h_{n \plus{} 1}(y) \plus{} yh_{n}(y) \plus{} nh_{n \minus{} 1}(y) \equal{} 0$. Now put $ h_n(y) \equal{} e^{ \minus{} \frac {1}{2}y^2}\left[ ( \minus{} y)^n \plus{} \sum_{j \equal{} 0}^{n \minus{} 1}c_{j}y^j \right]$ to get the recurrence.

[b]ii.[/b] I also tried using [url=http://planetmath.org/encyclopedia/FaaDiBrunosFormula.html]Faa di Bruno's formula[/url] for the $ n^{\mbox{th}}$ derivative of a composition of functions, but the integer partitions proved a roadblock for me. I got this down to
\[ \frac {d^n}{dy^n} \left( e^{ \minus{} \frac {1}{2}y^2}\right) \equal{} e^{ \minus{} \frac {1}{2}y^2}\sum_{m_1 \plus{} 2m_2 \plus{} \cdots \plus{} nm_n \equal{} n}\frac {n!}{m_1!m_2!\cdots m_n!}\prod_{j: m_j\neq 0}\left(\frac { \minus{} y\delta_{1j} \minus{} \delta_{2j}}{j!}\right)^{m_{j}}
\]
where $ \delta_{mn} \equal{} \left\{\begin{array}{cc}0, & \mbox{ if } m\neq n \\
1, & \mbox{ if } m \equal{} n\end{array}\right.$ is the discrete delta function.

[b]iii.[/b] Lastly I have tried the brute-force power series approach, namely
\[ \frac {d^n}{dy^n} \left( e^{ \minus{} \frac {1}{2}y^2}\right) \equal{} \frac {d^n}{dy^n}\sum_{k \equal{} 0}^{\infty}\frac {1}{k!}\left( \minus{} \frac {1}{2}y^2\right)^k
\]

\[ \equal{} \sum_{k \equal{} \lceil \frac {n}{2}\rceil}^{\infty}\frac {( \minus{} 1)^k}{2^k \cdot k!} \frac {(2k)!}{(2k \minus{} n)!}y^{2k \minus{} n}
\]

The coefficents depend on two indices $ n$ and $ j$, not just one.

It's not really that hard to find a recurrence: $ H_{n\plus{}1}(y)\equal{}\minus{}e^{y^2/2}\frac{d}{dy} \left(\frac{d^n}{dy^n}\left((\minus{}1)^ne^{\minus{}y^2/2}\right)\right)\equal{} \minus{}e^{y^2/2}\frac{d}{dy}\left(e^{\minus{}y^2/2}H_n(y)\right)$
$ \equal{}\minus{}e^{y^2/2}\left(\minus{}ye^{\minus{}y^2/2}H_n(y)\plus{}e^{\minus{}y^2}H_n'(y)\right)\equal{} yH_n(y)\minus{}H_n'(y)$
With the base case $ H_0\equal{}1$, you can easily determine each polynomial. For a given $ n$, the polynomial is either odd or even.


algebra
polynomial
trigonometry
algebra proposed

For $ n\in\mathbb N$, prove \[ \prod_{k\equal{}1}^{n}(x\minus{}2\cos\frac{k\pi}{n\plus{}1})\equal{}\sum_{k\equal{}0}^{[\frac n2]} (\minus{}1)^k\binom{n\minus{}k}{k}x^{n\minus{}2k}.\]

Have a look [url=http://en.wikipedia.org/wiki/Chebyshev_polynomials]here[/url] first. :)


search
inequalities unsolved
inequalities

For $ a,b,c \geq 0$.Prove that
     $ \sum_{cyc} \frac {a^4}{a^3\plus{}b^3} \geq \sum_{cyc} \frac {a}{ 2}$