Given a full rank n x m matrix, X, decompose the matrix into a sum of three matrices which have ranks 1, 1 and min(n,m).  [X = P + Q + R, Rank(P) = Rank(Q) = 1, Rank(X) = Rank(R) = min(n,m)].  What conditions are sufficient to make the decomposition unique?  Do these conditions change if the norm of R is minimized?

I dont understand the problem. There are always infinitely many ways of doing that:

P=-Q=some n x m matrix of rank one, R=X.  Are there further conditions in the problem?


abstract algebra
function
probability
integration
probability and stats

Consider the $ \sigma$-algebra $ B: \equal{}B(I)$ of Borel of $ I: \equal{} [0,1]$. $ B(\mathbb{R})$ the Borel of $ \mathbb{R}$ We say that a function $ Q: I \times B \longrightarrow{I}$ is a transition kernel if:

For each $ x\in I$ $ Q(x,.)$ is a probability measure on B.
For each $ A\in B$ $ Q(.,B)$ is a function measurable B-B.

a) Prove that if $ f: I \times I\longrightarrow{\mathbb{R_\plus{}}}$ is $ B\otimes B\minus{}B(\mathbb{R})$ measurable, then the function $ x\rightarrow{\int\limits_I {f(x,y)Q(x,dy)}}$ defined for each $ x\in{I}$ is $ B\minus{}B(\mathbb{R})$ measurable.

Suppose further that $ \int\limits_I {f(x,y)dx \equal{} 1}$ for each $ y\in{I}$:

b) Prove that $ (y,A)\rightarrow{\int\limits_I {f(x,y)Q(x,B)dx \equal{} 1}}$ defined on $ I \times B$ is a transition kernel.

Help!

Hello

Well there may be some simpler approach but here is one which I think this might work

Show the statement for functions $ f$ of the form $ g(x)h(y)$ then use Monotone class Theorem to conclude

For the second question if I understood the question properly, you have for $ y \in I$ (by the first part that of the exercise) :

$ \int_{I}{f(x,y)Q(x,A)dx}$ is measurable in $ y$ for every $ A \in B$

To prove the second claim you hve to prove that for fixed $ y \in I$ , $ m_y(A) = \int_{I}{f(x,y)Q(x,A)dx}$ is a probability measure.
$ m_y(B) = 1$ by hypothesis and it is easily seen from the fact that $ Q$ is a kernel, that $ m_y(.)$ it is a measure.

Hope this helps

bon courage


Fie numerele reale pozitive $\ a,b,c$ cu suma $\ 3$ si a caror patrate sunt mai mici decat $\ 2$.Demonstrati inegalitatea:
$\sum_{cyc}\frac{1}{\sqrt{\frac{a}{2}}+4\sqrt{\sqrt{\frac{a^2}{6}}}+4\sqrt{\sqrt{\frac{a}{4}}}+2\sqrt{\sqrt{\frac{a}{2}}}+4\sqrt{\frac{a}{3}}}>\frac{3}{41}$
 @kosmin:stiu ca nu-ti plac inegalitati de tipul asta, dar te rog frumos s-o incerci. :D

eu unu  cred ca nu o sa postez nimeni la inegalitatea asta :)))) promit ca maine ma uit pe ea :) :) :)

Poate ca e prea lunga, d-aia nu prea cred nici eu ca o sa posteze cineva. :(

Atentie, s-ar putea sa fie spoiler (daca e corect... :D )
Am rezolvat-o...aoleu ce lunga e!

Aplicam la numitor inegalitatea mediilor in fiecare radical:
$\sqrt{\frac{a}{2}}\leq\frac{a}{4}+\frac{1}{2}$
$\ 4\sqrt{\sqrt{\frac{a^{2}}{6}}}\leq\frac{a^{2}}{6}+3$
$\ 4\sqrt{\sqrt{\frac{a}{4}}}\leq\frac{a}{4}+3$
$\ 2\sqrt{\sqrt{\frac{a}{2}}}\leq\frac{a}{4}+\frac{3}{2}$
$\ 4\sqrt{\frac{a}{3}}\leq\frac{2a}{3}+2$
Atunci, am avea de demonstrat:
$\sum_{cyc}\frac{1}{\frac{a^{2}}{6}+\frac{17a}{12}+10}\leq\frac{3}{41}$
Impartim cu un 12 si obtinem ca avem de demonstrat ca:
$\sum_{cyc}\frac{1}{{2a^{2}}+17a+120}>\frac{1}{164}$
Aplicam in membrul stang inegalitatea mediilor (MA>MH)
Atunci avem ca membrul stang este mai mare decat:
$\frac{9}{2(a^{2}+b^{2}+c^{2})+17(a+b+c)+360}$
$\ a+b+c=3$ si $\ 2(a^{2}+b^{2}+c^{2})\leq\ 12$, de unde, inlocuind, obtinem ca membrul stang al inegalitatii este mai mare decat:
$\frac{9}{2(a^{2}+b^{2}+c^{2})+17(a+b+c)+360}$
care, la randul sau, este mai mare decat $\frac{9}{423}$, care este, evident, mai mare decat $\frac{1}{164}$.

cum ziceam :) mie imi e si frica sa citesc rezolvarea :)

Slava cerului ca mi-a iesit in sfarsit scrisul in $\ Latex$!

Nu stiu cati isi dadeau seama de asta, dar...
  [i]Solutie[/i]
    Se observa ca $\ LHS$ se rescrie ca
$\sum_{cyc}\frac{1}{(\sqrt{\sqrt{\frac{a}{2}}}+2\sqrt{\sqrt{\frac{a}{3}}}+1})^{2}-1$.Dupa ce aplicam $\ x^{2}-1=(x-1)(x+1)$, inegalitatea se poate solutiona usor, cu MA-MG si Cauchy.


There are general inequality

\sum_{k=1}^{n}(\frac{\sum_{i=1}^{k}a_i^p}{k})^(1/p) <
(\frac{1}{1-p})^{1/p}\sum_{k=1}^{n}a_k

for all real p<1, a_k>=0.

Your inequality arises if p=1/2


modular arithmetic
combinatorics unsolved
combinatorics

We wish to construct a circular sequence $ a_{0},...,a_{7}$ (indices $ mod8$) in such a way that a sliding window $ a_{i},a_{i\plus{}1},a_{i\plus{}3}$ $ i\equal{}1,2,...,7$ will contain every possible three-tuple once. Show(not just by trial or error) that this is impossible.

could you elaborate where does the sliding window come from ? and what does this mean "will contain every possible three-tuple once"?

Wish, i knew it too. That's why i posted this problem here. Because i cant think anything to solve it and of course because i don't understand what does the problem ask :(  :(

I guess either u have got d question wrong or maybe I am missing something (dont take it otherwise..i normally do)

For infact, the thing that u ask is possible(regarding debrujin sequences) if i think i understand the problem. Anyway for clarity sake I restate the problem.

(The ideas below are taken from d wonderful book [b]Mathematics: The man-made universe by Sherman K stein[/b])

The problem:

You are supposed to make a 2^n-letterd wheel using numbers 0's and 1's only. Can u make this wheel in such a fashion that if i start off rreading every n consecutive symbols in the wheel (mod-2^n) i end up reading evey possible 3-bit combination of 0's and 1's. Let us call them magic wheels.

Ur case sets n=3. Here we will see 8 strings (and in general 2^n-strings) like a0.a1.a2...a1.a2.a3...a2.a3.a4...a3.a4.a5...a4.a5.a6....a5.a6.a7....a6.a7.a0....a7.a0.a1...

U get d picture...

Now u may want to know that the answer to the above question is [b]YES.[/b].

[b]Here's the solution[/b]..(It can be genaralised readily)

let us represent the every 3-digit combination of 0's and 1's as cities on map. U get 8 cities. Now there are roads (directed edges) between 2 cities if u find the last two bits of a city common with first 2 bits of another. clearly there are 2 roads that go out of a city and that go into a city. (There are roads leading back to the city itself or self loops)

And now the fun begins......(Please try to think for Stein's power-packed punch is about to come.)

This graph has an Eulerian circuit as every edge has indegree and outdegree both even and infact same. Thus, if I start at a particular city, I can come back to it following a certain Eulerian path visting every edge exactly once...and if I start out with city1 writing its first 3-letters down and attaching to it the third letter of the city i vist, i will come up with an 8-leterd magic wheel.

I hope that the above outline suffices (if i recollect it properl), please point out if i made any mistakes..

Thanks and have a nice day

Hi all,

I thought this could be a nice treat for u all. In spirit with the above problems (well, in continuation of his paper on de-bruijn sequences by de-bruijn) the following problem can be tried.

You need to prove that the number of de-bruijn sequences (or magic wheels) when u desire the magic wheel to read through all n-bit combinations of 1's and 0's is 2^n-1 - n.

I do not know the solution myself.

Hello...anybody...any ideas...(Please go thru d previous few posts)

I think the problem statement is inaccurate. Let me restate the problem:

We wish to construct a circular sequence $ a_0,a_1,a_2,...,a_7$(indices $ \pmod 8$) in such a way that a sliding window $ a_ia_{i\plus{}1}a_{i\plus{}3}$ for $ i\equal{}0,1,2,...,7$ will contain every possible three-tuple once. Show that this is impossible.

The edit I made was the index $ i$ must run from 0 to 7 and not 1 to 7 as previously stated. This is because there are 8 possible 3-tuples. Let the explain what the term "sliding window" means. The problem asks if there are binary variables $ a_0,a_1,a_2,...,a_7$ such that $ a_0a_1a_3$,$ a_1a_2a_4$,$ a_2a_3a_5$,$ a_3a_4a_6$,$ a_4a_5a_7$,$ a_5a_6a_0$,$ a_6a_7a_1$ and $ a_7a_0a_1$ are 000,001,010,011,100,101,110 and 111 in some order. We wish to show that this is not possible. Suppose it were there must be some $ a_ia_{i\plus{}1}a_{i\plus{}3}$ which is 000. Without loss of generality let $ a_0a_1a_3$ be 000.(because if i were not 0 we could have cyclically shifted the sequence to make i=0).
$ a_4a_5a_7$ is the only term that doesn't contain any of $ a_0$,$ a_1$ or $ a_3$. In other words $ a_4a_5a_7$ is the only term that doesn't contain a 0. This implies $ a_4a_5a_7$=111. $ a_1a_2a_4\neq a_2a_3a_5$ implies $ 0a_21\neq a_201$ which implies $ a_2\neq 0$. Also $ a_2a_3a_5\neq a_7a_0a_2$ implies $ a_201\neq 10a_2$ which implies $ a_2\neq 1$. Since $ a_2$ can neither have values 0 and 1 we get the desired contradiction. Hence its impossible to have such $ a_i$'s.


limit
real analysis
real analysis unsolved

Let $ (a_n)_{n\ge{1}}$ , $ a_n\equal{}\sum_{k\equal{}1}^{n}\frac{k}{2^k}$.
Prove that $ \lim_{n\to\infty} a_n\equal{}2$ and $ \lim_{n\to\infty}\frac {2^n}{\sqrt[n]{n!}}\cdot{(2\minus{}a_n)}\equal{}e$

$ a_n \equal{} \sum_{k \equal{} 1}^n \frac {k}{2^k} \equal{} \sum_{k \equal{} 1}^n \left(\frac {k \plus{} 1}{2^{k \minus{} 1}} \minus{} \frac {k \plus{} 2}{2^k}\right) \equal{} 2 \minus{} \frac {n \plus{} 2}{2^n}$

$ \therefore \lim_{n\to\infty} a_n \equal{} 2.$ Then we have $ \frac {2^n}{\sqrt [n]{n!}}(2 \minus{} a_n) \equal{} \frac {n \plus{} 2}{\sqrt [n]{n!}}$

I wonder we would use $ A.M. \minus{} G.M.$ and squeezing theorem. :|

And how about using the Stirling approximation to $ n!$? Then we have:

$ \frac {n \plus{} 2}{\sqrt [n]{n!}} \approx e \frac {n \plus{} 2}{n \sqrt [2n]{2 n \pi} }$

And so, when $ n \rightarrow \infty$ , we know that $ \sqrt [n]{n}$ tends to 1 as well as the fraction part which leaves us with $ e$.

Stirling's formula is a little more than we need: the limit $ \frac{n}{(n!)^{1/n}}\to e$ is easy to prove directly, using the fact that the sequence $ a_{n}\equal{}\frac{n^{n}}{n!}$ satisfies $ \frac{a_{n\plus{}1}}{a_n}\equal{}\left(1\plus{}\frac{1}{n}\right)^n$.


Let $n \in N$;$q>1;q \in R$. Put $\displaystyle P(x)=\sum_{k=1}^n (-1)^{k+1}\binom {n}{k}\frac {x^k-1}{q^k-1}$.
Prove that $\forall x \leq q;1 \leq k \leq n$ then: $\displaystyle (-1)^{k+1}{P(x)}^{(k)} >0$.

hxtung, could you specify what P(x)<sup>(k)</sup> is?

It can't be powers, because then take n=1, we have P(x)=(x-1)/(q-1). Then LHS=P(x)<=(x-1)/(q-1) if x<=1. But it's supposed to hold for all x<=q, so something has gone wrong...

I think it could me the $k$'th derivative. That's the usual notation for it, isn't it?


geometry
geometric transformation
rotation
trigonometry
similar triangles
geometry proposed

Let $ABC$ be a triangle with $AC\neq BC$, and let $A^{\prime }B^{\prime }C$ be a triangle obtained from $ABC$ after some rotation centered at $C$. Let $M,E,F$ be the midpoints of the segments $BA^{\prime },AC$ and $CB^{\prime }$  respectively. If $EM=FM$, find $\widehat{EMF}$.

I'm not sure, but this is my intento of solution:
Let A'' and B'' /     A''CA'A and CB''B'B are paralelogramos (A'' and A' in distintos semiplanes respecto AC, and B and B'' in distintos semiplanes respecto B'C).
Then A''B // ME, A'B'' // MF, and A''B=2.ME, A'B''=2.MF... then A''B=A'B''
CA=x=CA', CB=y=CB', then A''C=x.k and B''C= yk (because ACA' and BCB' are similar isosceles). Además <A''CB=A'CB''. See A''CB and A'CB triangles, we deduce that k=1 or x=y (if k no is 1, rotando the triangles obtenemos un trapecio with diagonals =, then is trapecio isosceles, and x=y). Then k=1, the triangles A''CB, A'CB are similars, and ACA' & BAB' are equilaterals... well, the angle is 60 because A''CA'=120 and A''CA'P (with P=A''B_A'B'', reference:  _=intersection) is ciclic.
Sorry for the redaction and word, I don't know english.

Well, I am not sure to understand all what you said above, Maybe because of the language  :?
 Can anyone translate, or post a solution in English ??  :lol:

Well, sorry for the difficult, It's very very difficult to me, too. However I try again, this time is with moooore details, I hope that you understand.
references: GH // RT = GH paralel with RT,      <ABC= angle ABC = <(AB,BC)

Solution:
Let R in the line A'E that RE=EA' (E bettwen  R and A').
Let S in the line BF that BF=FS.
Then, RAA'C and CBB'S are 'paralelogramos' (RC//AA', etc).
And AC=CA', BC=BC' (because A'CB' is a rotation of ACB), then CA=RA and CB'=B'S.
Let L and N the midpoints of CB and CA'.
2.EN=AA'=RC and EN//RC, 2MN=CB and MN//CB, then RCB and ENM are homotetic, then RB=2.EM and RB//EM.
Analogal, 2.LF=BB'=CS and CS//LF, 2.LM=CA' and CA'//LM, then FLM and CSB are homotetic, then A'S=2.MF and MF//A'S
Then RB=A'S and the angle EMF is <(RB,SA')

Let x=AC and y=CB. Let 1/k=AC/AA'=AC/RC, then k=CB/BB' (because ACA' and BCB' are similar isosceles). Then RC=kx and CS=ky.

If k not is 1, I will prove that AC=AB (contradiction).
The rotation of A' and S, with center C and angle A'CR--> A'' and S', with A'' in the line RC and S' in the line CB. Then RS'//A''B, because RC=kx, CS'=ky, and CA''=x, CB=y.
But RB=A''S', then RS'BA'' is a 'trapecio' isosceles, then x=y and contradiction.

Then k=1. Finally, RCA, ACA', BCB' and CB'S are equilateral triangles. Well, not is difficult to see that <(RB,SA')=EMF=60º (because RCB and A'CS are similar triangles, with RCB=A'CS=2.60=120).

Well, I hope that the problem of the language not imposibility the comunication, and I hope that the solution will be good, jeje.
Yes, sorry, my redaction and words are terrible, but I never studied english.

Let $AC=2a$, $BC=2b$, $\angle ACB= \gamma$ and $\alpha = \angle ACA' = \angle BCB'$.
Since $AC=A'C$ and $BC=B'C$: triangles $ACA'$ and $BCB'$ are similars, we put $AA'=2ak$ and $BB'=2bk$.
Let $E'$ and $F'$ the midpoints of $A'C$ and $BC$, respect.

Lemma: $\underline{ \angle EE'M= \angle FF'M}$
Since $EE'//AA'$, $E'M//CB$ ; $FF'// B'B$, $MF'//A'C$ and $\angle A'AC= \angle BB'C$, $\angle ACB=\angle B'CA'$ the lemma  follows. $\blacksquare$

Clearly: $EE'=ak$, $E'M=b$ ; $FF'=bk$, $F'M=a$ ; using the lemma and the fact that $EM=MF$ we get (from the cosine law) : \[ \cos \angle EE'M= \frac{b^2+(ak)^2-EM^2}{abk}=\frac{a^2+(bk)^2-MF^2}{abk}=\cos \angle FF'M \] \[ \Rightarrow b^2+a^2k^2=a^2+b^2k^2 \] \[ \Rightarrow (k^2-1)(a^2-b^2)=0 \] since $a \neq b$ we get: $k=1$ $\Rightarrow$  $AA'=2a$ , and triangles $ACA'$ and $BCB'$ are equilaterals. Since $\angle EE'A=120^ \circ$ and $\angle A'E'M= \angle A'CB= \gamma-60^ \circ$ we get: $\angle EE'M=60^ \circ + \gamma$ , with this, clearly the triangles $EE'M$ and $ECF$ are congruents, so $EF=EM$ and since $ME=MF$, the triangle $EMF$ is equilateral, so $\angle EMF=60^ \circ$.

$Tipe$  :lol:

Well, sorry for the difficult, It's very very difficult to me, too. However I try again, this time is with moooore details, I hope that you understand.
references: $GH // RT = GH$ paralel with $RT$,      $<ABC= angle ABC = <(AB,BC)$

Solution:
Let $R$ in the line $A'E$ that $RE=EA'$ ($E$ bettwen $R$ and $A'$).
Let $S$ in the line $BF$ that $BF=FS$.
Then, $RAA'C$ and $CBB'S$ are 'paralelogramos' ($RC//AA'$, etc).
And $AC=CA'$, $BC=BC'$ (because $A'CB'$ is a rotation of $ACB$), then $CA=RA$ and $CB'=B'S$.
Let $L$ and $N$ the midpoints of $CB$ and $CA'$.
2.$EN=AA'=RC$ and $EN//RC$, $2MN=CB$ and $MN//CB$, then $RCB$ and $ENM$ are homotetic, then $RB=2.EM$ and $RB//EM$.
Analogal, $2.LF=BB'=CS$ and $CS//LF$, $2.LM=CA'$ and $CA'//LM$, then $FLM$ and $CSB$ are homotetic, then $A'S=2.MF$ and $MF//A'S$
Then $RB=A'S$ and the angle $EMF$ is $<(RB,SA')$

Let $x=AC$ and $y=CB$. Let $1/k=AC/AA'=AC/RC$, then $k=CB/BB'$ (because $ACA'$ and $BCB'$ are similar isosceles). Then $RC=kx$ and $CS=ky$.

If k not is 1, I will prove that $AC=AB$ (contradiction).
The rotation of $A'$ and $S$, with center $C$ and angle $A'CR-\to  A''$ and $S'$, with $A''$ in the line $RC$ and $S'$ in the line $CB$. Then $RS'//A''B$, because $RC=kx$, $CS'=ky$, and $CA''=x$, $CB=y$.
But $RB=A''S'$, then $RS'BA''$ is a 'trapecio' isosceles, then $x=y$ and contradiction.

Then $k=1$. Finally, $RCA$, $ACA'$, $BCB'$ and $CB'S$ are equilateral triangles. Well, not is difficult to see that $<(RB,SA')=EMF=60º$ (because $RCB$ and $A'CS$ are similar triangles, with $RCB=A'CS=2.60=120$).

Well, I hope that the problem of the language not imposibility the comunication, and I hope that the solution will be good, jeje.
Yes, sorry, my redaction and words are terrible, but I never studied english.

[size=75][color=red]edited by Davron Latipov[/color][/size]

Nice problem :lol:
WLOG we can assume that the the angle of rotation $\theta$ is between $0$ and $\pi$.Other cases can be proven analogously.
Let $c$ be the origin.
Then,$a'=ae^{i\theta}$,$b'=be^{i\theta}$,$f=\frac{be^{i\theta}}{2}$,$e=\frac{a}{2}$ and $m=\frac{b+ae^{i\theta}}{2}$,$|b|\neq |a|$.
The given condition is $|f-m|=|e-m|\Rightarrow (f-m)(\overline{f-m})=(e-m)(\overline{e-m})$.This simplifies to(after not-so-long calculations :D ) to $(|b|^2-|a|^2)(e^{i\theta}+e^{-i\theta}-1)=0$.Since $|b|\neq |a|$,this implies $e^{i\theta}+e^{-i\theta}=1\Rightarrow\cos\theta=\frac{1}{2}\Rightarrow\theta=\frac{\pi}{3}$ since $\theta\in (0,\pi]$.
Now,$\frac{f-m}{e-m}=\frac{be^{i\frac{\pi}{3}}-b-ae^{i\frac{\pi}{3}}}{a-b-ae^{i\frac{\pi}{3}}}=\frac{-b-b\omega+a}{a\omega-b\omega+a}$(multiplying the nominator and the denominator by $\omega=e^{i\frac{2\pi}{3}}$).
Further,$\frac{-b-b\omega+a}{a\omega-b\omega+a}=\frac{b\omega^2+a}{-a\omega^2-b\omega}=\frac{b\omega^2+a}{-a\omega^2-b\omega^4}=\frac{-1}{\omega^2}\left(\frac{b\omega^2+a}{b\omega^2+a}\right)=\frac{e^{i\pi}}{e^{i\frac{4\pi}{3}}}=e^{-i\frac{\pi}{3}}$.So,$\angle EMF=\frac{\pi}{3}$.


linear algebra
matrix
linear algebra unsolved

Let A is a triangle matrix (A in M_n(R)), and [A]ii=0 , for 1<=i<=n.
Prove that : exist k such that A^k=0

$Tr(A^k)=0$ for all $k$, so it is nilpotent  :D  or you can just compute and see ..


integration
logarithms
calculus
trigonometry
calculus computations

[quote="RaduSorici"]Sum the series $1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\frac{1}{13}-....$[/quote]

it's >1 and <4/3.. more exactly around 1.110720734

Have you tried using fouriersums on this? I think the closed form is $\dfrac{\pi}{2\sqrt{2}}$, and it can be found with fourier.

$S = \sum_{k=0}^{\infty}\frac{(-1)^k}{4k + 1} + \sum_{k=0}^{\infty}\frac{(-1)^k}{4k + 3} = S_1 + S_2$

Start with 2 geometric sequences $\{a_n\}, \{b_n\}$

$\frac{1}{1 + x^4} = 1 - x^4 + x^8 - x^{12} + ...$

$\frac{x^2}{1 + x^4} = x^2 - x^6 + x^{10} - x^{14} + ...$

Both have radius of convergence 1. Inside of the radius, we can integrate term by term. The radius of convergence does not change in the process. On the radius of convergence, we have 2 alternating sign series. For convergence, it is sufficient $a_n \rightarrow 0, b_n \rightarrow 0$ and $\frac{a_{n+1}}{a_n} \rightarrow A$, $\frac{b_{n+1}}{b_n} \rightarrow B$, where $A, B$ are finite limits.

$\int \frac{\text dx}{1 + x^4} = \sum_{k=0}^{\infty} \int (-1)^k x^{4k}\ \text dx = \sum_{k=0}^{\infty} \frac{(-1)^k x^{4k + 1}}{4k + 1}$

$\int \frac{x^2\ \text dx}{1 + x^4} = \sum_{k=0}^{\infty} \int (-1)^k x^{4k + 2}\ \text dx = \sum_{k=0}^{\infty} \frac{(-1)^k x^{4k + 3}}{4k + 3}$

$x^4 + 1 = 0$ has 4 complex roots $\frac{\sqrt 2}{2}(1 + i), \frac{\sqrt 2}{2}(1 - i), \frac{\sqrt 2}{2}(-1 + i), \frac{\sqrt 2}{2}(-1 - i)$, hence, it can be factored as

$x^4 + 1 = (x^2 + x \sqrt 2 + 1)(x^2 - x \sqrt 2 + 1)$

$\int \frac{\text dx}{1 + x^4} = \frac{\sqrt 2}{4} \int \frac{x + \sqrt 2}{1 + x \sqrt 2 + x^2}\ \text dx - \frac{\sqrt 2}{4} \int \frac{x - \sqrt 2}{1 - x \sqrt 2 + x^2}\ \text dx = $

$\frac{\sqrt 2}{8} \int \frac{2x + \sqrt 2}{1 + x \sqrt 2 + x^2}\ \text dx - \frac{\sqrt 2}{8} \int \frac{2x - \sqrt 2}{1 - x \sqrt 2 + x^2}\ \text dx\ +$

$+ \frac 1 4 \int \frac{\text dx}{1 + x \sqrt 2 + x^2} + \frac 1 4 \int \frac{\text dx}{1 - x \sqrt 2 + x^2} = $

$= \frac{\sqrt 2}{8} \ln{\frac{1 + x \sqrt 2 + x^2}{1 - x \sqrt 2 + x^2}} + \frac{\sqrt 2}{4}\left[ \arctan{\left(x \sqrt 2 + 1\right)} + \arctan{\left(x \sqrt 2 - 1\right)}\right] =$

$= \frac{\sqrt 2}{8} \ln{\frac{1 + x \sqrt 2 + x^2}{1 - x \sqrt 2 + x^2}} + \frac{\sqrt 2}{4}\arctan{\frac{x \sqrt 2}{1 - x^2}} + C$

The integration constant is determined for $x = 0$:

$\frac{\sqrt 2}{8} \ln 1 + \frac{\sqrt 2}{4} \arctan 0 + C = 0,\ \ C = 0$

The sum $S_1$ of the 1st series is determined for $x = 1$:

$S_1 = \frac{\sqrt 2}{8} \ln{\frac{2 + \sqrt 2}{2 - \sqrt 2}} + \frac{\pi \sqrt 2}{8}$

$\int \frac{x^2\ \text dx}{1 + x^4} = \frac{\sqrt 2}{4} \int \frac{x}{1 - x \sqrt 2 + x^2}\ \text dx - \frac{\sqrt 2}{4} \int \frac{x}{1 + x \sqrt 2 + x^2}\ \text dx = $

$= \frac{\sqrt 2}{8} \int \frac{2x - \sqrt 2}{1 - x \sqrt 2 + x^2}\ \text dx - \frac{\sqrt 2}{8} \int \frac{2x + \sqrt 2}{1 + x \sqrt 2 + x^2}\ \text dx +$

$+ \frac 1 4 \int \frac{\text dx}{1 - x \sqrt 2 + x^2} + \frac 1 4 \int \frac{\text dx}{1 + x \sqrt 2 + x^2} = $

$= \frac{\sqrt 2}{8} \ln{\frac{1 - x \sqrt 2 + x^2}{1 + x \sqrt 2 + x^2}} + \frac{\sqrt 2}{4}\left[ \arctan{\left(x \sqrt 2 + 1\right)} + \arctan{\left(x \sqrt 2 - 1\right)}\right] =$

$= \frac{\sqrt 2}{8} \ln{\frac{1 - x \sqrt 2 + x^2}{1 + x \sqrt 2 + x^2}} + \frac{\sqrt 2}{4}\arctan{\frac{x \sqrt 2}{1 - x^2}} + C$

The integration constant is again determined for $x = 0$:

$\frac{\sqrt 2}{8} \ln 1 + \frac{\sqrt 2}{4} \arctan 0 + C = 0,\ \ C = 0$

The sum $S_2$ of the 2nd series is determined for $x = 1$:

$S_2 = \frac{\sqrt 2}{8} \ln{\frac{2 - \sqrt 2}{2 + \sqrt 2}} + \frac{\pi \sqrt 2}{8}$

and the sum of the entire sequence is

$S = S_1 + S_2 = \frac{\pi \sqrt 2}{4}$

Could anyone tell me what is "fouriersums" please?
Very nice solution!!!

This is a special case of a problem I posted a few months ago [url=http://www.artofproblemsolving.com/Forum/topic-21316.html]here[/url]. For any nonzero $k$ other than the reciprocal of an integer, $\sum_{n=-\infty}^\infty \frac{(-1)^n}{1+nk}=\frac{\pi/k}{\sin \pi/k}.$

This problem was the $k=4$ case.


solve this equation:
$tgx=tg2x .tg3x .tg4x$ :)

With the condition $\cos x\cos 2x\cos 3x\cos 4x\neq 0$, the equation can be written as

$(\sin x\cos 4x)(\cos 2x\cos 3x)=(\cos x\sin 4x)(\sin 2x\sin 3x)$

Transforming the products in the parentheses into the sums, we get

\begin{eqnarray*}\frac{1}{2}(\sin(-3x)+\sin 5x)\frac{1}{2}(\cos(-x)+\cos 5x)=\frac{1}{2}(\sin 5x-\sin(-3x))\frac{1}{2}(\cos(-x)-\cos 5x)\end{eqnarray*}

Expanding and simplifying gives

$\sin 5x\cos 5x-\sin 3x\cos x=0$

Transforming the second term into the sum we get

$\frac{1}{2}\sin 10x-\frac{1}{2}(\sin 2x+\sin 4x)=0$, or

$\sin 10x-\sin 2x-\sin 4x=0$

Transforming the first two terms into a product we get

$2\cos 6x\sin 4x-\sin 4x=0$

or

$\sin 4x(2\cos 6x-1)=0$

This gives solutions $x=\frac{k\pi}{4}\lor x=\pm\frac{\pi}{18}+\frac{k\pi}{3}$, but the initial condition about the product of the cosines narrows this down to $x=k\pi\lor x=\pm\frac{\pi}{18}+\frac{k\pi}{3},k\in\mathbb{Z}$


I go from point A to point B at 50mph, and from B to A at 30 mph. If the distance between A and B is 15 miles. Find the average rate for the whole trip.

Now try this one. =o

A car travels downhill at 72mph on the level at 63 mph, and uphill at
only 56mph. The car takes 4 hours to cross from town A to town . The
return trip is 4 hours 40 minutes. Find the distance between the two
towns.

[hide="Solution to #1"]We travel for $ \displaystyle\frac{15}{50}$ hours on the way there, then $ \displaystyle\frac{15}{30}$ on the way back. We travel a total of $ 30$ miles. The desired answer is $ \displaystyle\frac{30}{\displaystyle\frac{15}{30}\plus{}\displaystyle\frac{15}{50}} \equal{} \boxed{37.5}$ miles per hour.[/hide]

[hide="Solution to #2"]We travel for $ 40$ more minutes going back than up. That implies we travel uphill on the way back. If we travel $ x$ miles uphill, we want $ \displaystyle\frac{x}{56} \minus{} \displaystyle\frac{x}{72} \equal{} 2/3$, so $ x \equal{} 168$. On the trip there, the car spends $ \displaystyle\frac{7}{3}$ hours traveling to the other town, so it must spend $ \displaystyle\frac{5}{3}$ hours traveling on level ground. This covers a distance of $ 105$ miles. The total distance is $ 168 \plus{} 105 \equal{} \boxed{273}$ miles.[/hide]


I was doing an aime geometric probability problem and i brute-forced it and came across a final answer of $ \frac {5\cdot34\cdot156}{137\cdot163 \plus{} 169}$; then i realized the common factor of $ 5$ (since 7*3=21;1+9=10) and i got

$ \frac {34\cdot156}{27\cdot163 \plus{} 65 \plus{} 35 \plus{} 1}$

but now what? Are there any obvious divibility rules that apply here?

(BTW I cheated using a calculator and it showed my answer agrees with the official one, but with one more step of simplifying it to $ p/q,(p,q) \equal{} 1$ and finding $ p \plus{} q$; for me, ironically and frustratingly, this is the HARDEST part of the problem + the dumb and messy computations :mad:  :( the general approach is obvious but in my opinion this problem is testing your computational skills + error making skills instead of mathematical ingenuity)
)

Also generally what are the fast and but with less tendency of making dumb error ways of simplifying large fractions as such (of course WITHOUT calculator)? Thanks for any suggestions... :D

A nice start would be actually writing out the denominator, which is 4502.  Pull out a 2 from the top and bottom to get $ \frac{2652}{2251}$.  2251 is prime so you can't do any more reducing.  Are you sure you have everything correct?

Hmm I seem to have typed out the reciprocal of the correct answer, which is 1 over $ \frac{5\cdot34\cdot156}{137\cdot163\plus{}169}$ The offical answer is $ 375/442$... and I verified using my calculator that my answer is the same...
Are you sure $ \frac{2652}{2251}$ is correct? (In that case you should have gotten $ 442/375$ due to my incompetence of determining whether a number is a valid probability value...  :D

In your new expression, the denominator (really numerator according to you) is 22500 and the numerator (really denominator) is 26520.  Pull out the 5 to get $ \frac{4500}{5304}\equal{}\frac{2250}{2652}\equal{}\frac{1125}{1326}\equal{}\frac{375}{442}$, which is what you got.  All I did was multiply everything out (a little tedious on paper, but not difficult), and then divide by 2, 2, and 3, which is easy.

hello, we get ${ \frac{5\cdot34\cdot156}{137\cdot163+\cdot169}}=\frac{26520}{22500}$ calculating $ \gcd(26520,22500)=60$ we get the answer $ \frac{442}{375}$ as above.
Sonnhard.

I see... however, making computational error's always haunting me when I'm doing calculations as such (making me want to do back and check every now and then and wasting time used for doing other problems)? Any remedies please?

Also if I commit a little computational error but get the whole method right does it even worth my time to go through my work and just find that small error (sometimes even hours if I'm frustrated)? Or should I just ignore (since my method is right) and move on? :maybe:

(BTW I would have notice the common factor of 3 if I didn't misread my writing of $ 33$ instead of $ 35$... seriously I MUST stop making those dumb mistakes... :( )

Work more slowly and double-check your work each step of the way.


Let $n$ be a natural number divisible by $4$. Determine the number of bijections $f$ on the set $\{1,2,...,n\}$ such that $f (j )+f^{-1}(j ) = n+1$ for $j = 1,..., n.$

$ j \equal{} f(x)$ $ \Rightarrow$ $ f(f(x)) \equal{} n \plus{} 1 \minus{} x$ (notice here that $ f(x)\neq x$ since n is even).

So, if I call $ y \equal{} f(x)$, we have, applying $ f()$ many times,  4-numbers cycles $ x,y,n \plus{} 1 \minus{} x, n \plus{} 1 \minus{} y, x, y, n \plus{} 1 \minus{} x, n \plus{} 1 \minus{} y, ...)$

In each cycle, the four numbers are different and exactly two of them are in $ \{1,2,...,\frac {n}{2}\}$.
There are $ (\frac {n}{2} \minus{} 1)(\frac {n}{2} \minus{} 3)...3*1$ ways for grouping $ (x,y)$s and each of the $ \frac {n}{4}$ couples may lead to 2 cycles $ (x,y,n \plus{} 1 \minus{} x,n \plus{} 1 \minus{} y)$ or $ (x,n \plus{} 1 \minus{} y,n \plus{} 1 \minus{} x,y)$

So the requested number is $ ((\frac {n}{2} \minus{} 1)(\frac {n}{2} \minus{} 3)...3*1)2^{\frac {n}{4}}$
And, since
$ (\frac {n}{2} \minus{} 1)(\frac {n}{2} \minus{} 3)...3*1 \equal{} \frac {(\frac {n}{2})!}{2^{\frac {n}{4}}(\frac {n}{4})!}$, the result is $ \frac {(\frac {n}{2})!}{(\frac {n}{4})!}$
--
Patrick


function
inequalities
limit
calculus
calculus computations

Find if this function is increasing or decreasing

$ f(x)\equal{}ln(1\plus{}x^2)\minus{}e^{\minus{}x}\plus{}1$

hello, we have to prove that $ f^'(x) = \frac {2x}{x^2 + 1} + e^{ - x} > 0$ for all
real $ x$. This condition ist equivalent with the following inequality:
$ \frac {2xe^x}{x^2 + 1} > - 1$. Lets define $ h(x) = \frac {2xe^x}{x^2 + 1}$, then we have
$ h(0) = 0$,$ \lim_{x \to - \infty}h(x) = 0$ and $ \lim_{x\to + \infty}h(x) = + \infty$. From
$ h^'(x) = \frac {2e^x(x^3 - x^2 + x + 1)}{(x^2 + 1)^2} = 0$ we have
$ x_{Min} = - \frac {1}{3}\sqrt [3]{17 + 3\sqrt {33}} + \frac {2}{3\sqrt [3]{17 + 3\sqrt {33}}} + \frac {1}{3}$, easily we conclude $ h(x_{Min})\geq - 0,48 > - 1$.
Sonnhard.

$ f'(x) \equal{}\frac {2x}{x^2\plus{}1} \plus{} e^{\minus{}x} > 0 \  \Longleftrightarrow \  e^{\minus{}x}  > \minus{}\frac {2x}{x^2\plus{}1}$
The last inequality is easy  because of :
$ x \geq 0 \Longrightarrow e^{\minus{}x} > 0 \geq \minus{}\frac {2x}{x^2\plus{}1}$ , and  , $ x < 0 \Longrightarrow e^{\minus{}x} > 1 \geq \minus{}\frac {2x}{x^2\plus{}1}$
 :cool:


trigonometry
inequalities
inequalities unsolved

Hi , Let $ a,b,c >0$ and $ ab\plus{}ac\plus{}bc\equal{}1$. show that :
$ \frac{a}{\sqrt[]{a^2 \plus{}1}} \plus{} \frac{b}{\sqrt[]{b^2 \plus{}1}} \plus{} \frac{c}{\sqrt[]{c^2 \plus{}1}} \leq \frac{3}{2}$

[quote="mahanmath"]Hi , Let $ a,b,c > 0$ and $ ab \plus{} ac \plus{} bc \equal{} 1$. show that :
$ \frac {a}{\sqrt []{a^2 \plus{} 1}} \plus{} \frac {b}{\sqrt []{b^2 \plus{} 1}} \plus{} \frac {c}{\sqrt []{c^2 \plus{} 1}} \leq \frac {3}{2}$[/quote]
$ \sum \frac {a}{\sqrt {a^2 \plus{} 1}} \equal{} \sum \frac {a}{\sqrt {(a\plus{}b)(a\plus{}c)}} \leq \sum \frac{1}{2} \left( \frac{a}{a\plus{}b} \plus{} \frac{a}{a\plus{}c} \right) \equal{}\frac{3}{2}$

Nice solution  :wink:  What about this :
$ a,b,c>0$ , $ a\plus{}b\plus{}c\equal{}1$ Prove that $ \frac{ab}{\sqrt[]{ab\plus{}c}} \plus{} \frac{ac}{\sqrt[]{ac\plus{}b}} \plus{} \frac{bc}{\sqrt[]{bc\plus{}a}}\leq \frac{1}{2}$

[quote="mahanmath"]Nice solution  :wink:  What about this :
$ a,b,c > 0$ , $ a \plus{} b \plus{} c \equal{} 1$ Prove that $ \frac {ab}{\sqrt []{ab \plus{} c}} \plus{} \frac {ac}{\sqrt []{ac \plus{} b}} \plus{} \frac {bc}{\sqrt []{bc \plus{} a}}\leq \frac {1}{2}$[/quote]

$ LHS\equal{}\sum\sqrt{\frac{ab}{a\plus{}c}}\sqrt{\frac{ab}{b\plus{}c}}\leq\frac{1}{2}\sum(\frac{ab}{a\plus{}c}\plus{}\frac{ab}{b\plus{}c})\equal{}\frac{1}{2}$ :)

[quote="mahanmath"]Hi , Let $ a,b,c > 0$ and $ ab \plus{} ac \plus{} bc \equal{} 1$. show that :
$ \frac {a}{\sqrt []{a^2 \plus{} 1}} \plus{} \frac {b}{\sqrt []{b^2 \plus{} 1}} \plus{} \frac {c}{\sqrt []{c^2 \plus{} 1}} \leq \frac {3}{2}$[/quote]
There exists $ x,y,z \in [0;\frac{\pi}{2}], x\plus{}y\plus{}z \equal{} \pi$ such that $ a \equal{} \frac{1}{\tan x}, b \equal{} \frac{1}{\tan y}, c \equal{} \frac{1}{\tan z}$. Inserting gives $ \cos x \plus{} \cos y \plus{} \cos z \le \frac{3}{2}$ which is wellknown. (Jensen solves it)


13. If f(x) = (4x-1)/3, what is the value of (f^-1(1))^-1

I got 2...the answers my coach gave me last year says 5/3...any1 bother to help?

16. What is the sum of all positive two-digit integers with exactly 12 positive factors

I got 318 (60,72,90,96) but the answers say 312...any1 bother clarifying?

Thanks,

EP

For the function problem I'm pretty sure that 2 is the right answer.

For #16 I found that 60,72,84,90, and 96 work. The sum of these is 402.

13. The original problem has the function is $ f(x) \equal{} \frac {4x \plus{} 1}3$. We want to find $ \frac1{f^{ \minus{} 1}(1)}$. Remember that $ f(x)^{ \minus{} 1}\neq f^{ \minus{} 1}(x)$! $ f^{ \minus{} 1}(x)$ is the inverse function of $ f(x)$ that "undoes" it, so that if we let $ f^{ \minus{} 1}(x) \equal{} y$, then we can take the $ f$ of both sides to get $ x \equal{} f(y) \equal{} \frac {4y \plus{} 1}3$, and solve for $ y$. This means that we can let $ f^{ \minus{} 1}(1) \equal{} y$, and then take the $ f$ of both sides to get $ 1 \equal{} f(y) \equal{} \frac {4y \plus{} 1}3\implies3 \equal{} 4y \plus{} 1\implies y \equal{} \frac12$. Thus, $ \frac1{f^{ \minus{} 1}(1)} \equal{} \frac1y \equal{} \boxed2$.
*If you assumed that $ f^{ \minus{} 1}(1) \equal{} f(1)^{ \minus{} 1}$, then you would get $ \frac53$.

16. $ 12$ factors implies that the numbers we want are of the form
a. $ a^{11}$--no possibilities.
b. $ a^5b$--$ (a,b) \equal{} (2,3)\implies96$.
c. $ a^3b^2$--$ (a,b) \equal{} (2,3)\implies72$.
d. $ a^2bc$--$ (a,b,c) \equal{} \{(2,3,5),(2,3,7),(3,2,5)\}\implies\{60,84,90\}$
Adding up, we have $ 96 \plus{} 72 \plus{} 60 \plus{} 84 \plus{} 90 \equal{} \boxed{402}$.

[quote="EcstaticPotter"]13. If f(x) = (4x-1)/3, what is the value of (f^-1(1))^-1

I got 2...the answers my coach gave me last year says 5/3...any1 bother to help?[/quote]

First, find $ f^{-1}(x)$. First, we know that $ f(f^{-1}(x))=x$, so: \begin{align*}x &= \frac {4f^{-1}(x)-1}3 \\
3x &= 4f^{-1}(x) - 1 \\
3x + 1 &= 4f^{-1}(x) \\
\frac {3x + 1}4 &= f^{-1}(x)\end{align*} So $ f^{-1}(1)=\frac {3+1}4=1$, so $ (f^{-1}(1))^{-1}=1$.

[color=darkred]Edit: Ah-ha! There was a problem with your original problem! It's $ \frac {4x+1}3$, not $ \frac {4x-1}3$.[/color]


[b][u]Lemma.[/u][/b] $x,y,z\in R\Longrightarrow xy+yz+zx\le x^{2}+y^{2}+z^{2}\ .$

[b][u]Proof.[/u][/b] $a,b,c\in R^{*}$ and $x: =bc,\ y: =ca,\ z: =ab$ $\Longrightarrow$ $bc\cdot ca+ca\cdot ab+ab\cdot bc\le (bc)^{2}+(ca)^{2}+(ab)^{2}$ $\Longrightarrow$
$abc(a+b+c)\le b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}$ $\Longrightarrow$ $\frac{a+b+c}{abc}\le \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\ .$
Therefore, $\boxed{a,b,c\in R^{*}\Longrightarrow \frac{a+b+c}{abc}\le \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}\ .$ [b][u]Briefly:[/u][/b] $x: =\frac{1}{a},\ y: =\frac{1}{b},\ z: =\frac{1}{c}\Longrightarrow$
$\frac{1}{a}\cdot \frac{1}{b}+\frac{1}{b}\cdot \frac{1}{c}+\frac{1}{c}\cdot\frac{1}{a}\le \left(\frac{1}{a}\right)^{2}+\left(\frac{1}{b}\right)^{2}+\left(\frac{1}{c}\right)^{2}\Longrightarrow\frac{a+b+c}{abc}\le\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\ .$

[quote="Virgil Nicula"][b][u]Lemma.[/u][/b] $x,y,z\in R\Longrightarrow xy+yz+zx\le x^{2}+y^{2}+z^{2}\ .$

[b][u]Proof.[/u][/b] $a,b,c\in R^{*}$ and $x: =bc,\ y: =ca,\ z: =ab$ $\Longrightarrow$ $bc\cdot ca+ca\cdot ab+ab\cdot bc\le (bc)^{2}+(ca)^{2}+(ab)^{2}$ $\Longrightarrow$
$abc(a+b+c)\le b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}$ $\Longrightarrow$ $\frac{a+b+c}{abc}\le \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\ .$
Therefore, $\boxed{a,b,c\in R^{*}\Longrightarrow \frac{a+b+c}{abc}\le \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}\ .$[/quote]
nice solution.
you could've also used rearrangement inequality to solve this quickly.

with rearrangement is very easy, because WLOG $a \ge b \ge c$ and we have to show that
$\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\le \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}$, that is the definition of rearrangement

[quote="kimby_102"]Let $a,b,c>0$. Show that
$\frac{a+b+c}{abc}\leq \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}$.[/quote]
How about this short proof? :wink:

$\frac{a+b+c}{abc}\leq \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}$

$\Leftrightarrow \frac2{ab}+\frac2{bc}+\frac2{ca}\leq \frac{2}{a^{2}}+\frac{2}{b^{2}}+\frac{2}{c^{2}}$

$\Leftrightarrow 0 \leq{(\frac1a-\frac1b)}^{2}+{(\frac1b-\frac1c)}^{2}+{(\frac1c-\frac1a)}^{2}$

[quote="hydro"][quote="Kurt Gödel"]Yeah, it's also homogenous, so assume WLOG that $abc=1$. We have to prove that $a+b+c \geq \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}$. Multiply the left hand side with $abc=1$ to get the equivalent inequality $a+b+c \geq \frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}$. But as $\left(1, 0, 0\right) \succ \left(1,1,-1\right)$, we're done thanks to Muirhead.[/quote]

Doesn't Muirhead hold only for [i]symmetric[/i] sums?[/quote]
Multiply both isdes by $2$ and you will get symmetric sums. :wink:


I had this problem on a test.  It was the 'bonus' problem.  No one in the class got the answer and the teacher won't supply one.  I think she wrote the problem wrong and won't admit it.

The problem:

(4x+5)^1/2 = (2x-2)^1/2 - 3

I tried to expand the first expression by subtracting and adding 4:
(4x-4+4+5)^1/2 = (2x-2)^1/2 - 3    then
(4(x-1) + 9)^1/2 = (2(x-1))^1/2 -3

but I couldn't get any further.   Help!

Maybe your problem is solving equation.
$ (4x\plus{}5)^{\frac{1}{2}}\equal{}(2x\minus{}2)^{\frac{1}{2}}\minus{}3$
<=>$ \sqrt{4x\plus{}5}\equal{}\sqrt{2x\minus{}2}\minus{}3$ (1).
Condition $ x\geq1$, we have:
(1)<=>$ \left(\sqrt{4x\plus{}5}\plus{}3\right)^2\equal{}2x\minus{}2$
<=>$ 4x\plus{}5\plus{}6\sqrt{4x\plus{}5}\plus{}9\equal{}2x\minus{}2$
<=>$ 6\sqrt{4x\plus{}5}\equal{}\minus{}2x\minus{}16$
<=>$ 3\sqrt{4x\plus{}5}\equal{}\minus{}x\minus{}8$
With condition $ x\geq1\equal{}>\minus{}x\minus{}8<0\equal{}>$ this equation doesn't have root.

It would be somewhat quicker to note that $ 4x \plus{} 5 > 2x \minus{} 2$ whenever both expressions are positive, so $ \sqrt {4x \minus{} 5} > \sqrt {2x \minus{} 2} > \sqrt {2x \minus{} 2} \minus{} 3$.


Rules:
1.When guessing a number you may ask a question
eg:12345, Is the number lower or higher or 54321, about how far away am I
I will answer with either it is higher/lower or with it is about (i will round to the nearest 1000) 14000 away

2.You may NOT guess more than 2 number at one time or ask more than 1 question

3.The winner gets something, interesting

4.do not ask me what the number is

5.please bold the guess and question or I might miss it

START (I hope this doesn't fail miserabley)

I made this before  :dry:
Is the number prime?

[b]Is it a two or one digit or three digit or four digit or 5 digit or 6 digit or 7 digit or 8 digit number?
MEH[/b]

[b]Is the number an integer?[/b]

Dude, keep up with this.

0?
is it a positive number or a negative number?

ok in order

no
no
none of the choices
9
no and higher
none of the choices
yes and i dont check 10 times every day
it is even
it is posotive

REMINDER:try to guess a number first, it makes it less confusing for people to read

er, no and only guess 1 number each time
you also should ask a question

[quote="Sabien5"]Rules:
2.You may NOT guess [b]more than 2 number [/b]at one time or ask more than 1 question
[/quote]
.....

Facts: Positive even 9-digit integer, not a palindrome

[b]What is the digital sum (sum of digits) of this number?[/b]

Is it greater than or less than 133713371?

product of digits?
sum of digits when in binary?

So far we have:
- positive integer
- 9 digits
- even
- not a prime
- not a palindrome

[b]Is it greater than or equal ($ \ge$) to 500,000,000?[/b]

Ok then...  gotta stick to the rules

1.The digital sum is 49
2.Greater
3.Hmm, im sure i said only one question so ill answer the first one.  The product is 1143072
4.No
5.Ill answer first one, don't think you can get away with 2 QUESTIONS.   No

Maybe i should share the answer so someone else may answer while i cant?

That was one question buddy.  I was looking for only one answer.

- positive integer
- 9 digits
- even
- not a prime
- not a palindrome
- sum of digits is 49
- product of digits is 1143072
- it is between 300,000,000 and 500,000,000
- it does not have a 0

the last thing must be true because if it did have a 0, the product of its digits would be 0.

The prime factorization of 1143072 is $ 2^5 \cdot 3^6 \cdot 7^2$. Thus, it must have exactly two $ 7$s and no $ 1$s or $ 5$s.

So now we have
[hide="Restrictions"]
- positive integer
- 9 digits
- even
- not a prime
- not a palindrome
- sum of digits is 49
- product of digits is 1143072
- it is between 300,000,000 and 500,000,000
- exactly two 7s
- no 0s, 1s or 5s
[/hide]

Since factors of 2 can combine with factors of 3, it is possible for the number to have a 6 as a digit.

The sum of the digits of the number must also be 49. Subtracting the two 7s we already have, we get the sum of the other 7 digits to be 35.

Experimenting, I got a combination of digits that work: 2, 2, 3, 4, 6, 7, 7, 9, 9

They add up to 49 and have a product of 1143072. So the number is just an arrangement of these numbers. Since the number is between 300,000,000 and 500,000,000, the first digit must be a 3 or a 4.

My two guess are: [b]322467799[/b] and [b]422367799[/b]

My question is: [b]Is it greater than 400,000,000?[/b]

Ooops. Well, then my guesses are useless.

oh no..there are too many combinations
ok
[b]What number does it start with?[/b]

1.no, no and no
2.no, no and , no but ill tell you that 6 of the digits you named are in there
3.it starts with a 3

k00lperson may be answering some questions now

yep..i can start answering questions, bcuz sabien told me the number....:D
So i guess I won't be guessing ne more lol :P

He did the same with me, k00lperson. :lol:

yeah, ill verify that musicinmath will also be helping, i wont be needing more helpers thank you

kk...:D
Vera nice.....no more helpers....3 people is more zan enouf. :P

[quote="ksun48"]322467979 is it greater than 350000000[/quote]
[b]Your guess isn't right :(
Sorry

It is not greater than 350000000[/b]

no, there are not consecutive digits

and your guess is wrong.

some digits are repeated, but none are consecutive.

Are the digits 2, 2, 3, 4, 6, 7, 7, 9, and 9 in some order?

Guess: 346927972


geometry
circumcircle
symmetry
combinatorics unsolved
combinatorics

Let $P$ be a regular 2006-gon. A diagonal is called good if its endpoints divide the boundary of $P$ into two parts, each composed of an odd number of sides of $P$. The sides of $P$ are [i]not[/i] good.
Suppose P has been dissected into triangles by 2003 diagonals, no two of which have a common point in the interior of P. Find the maximum number of isosceles triangles having two good sides that could appear in such a configuration.

I've got it!

In the following, we call an isosceles triangle with two odd sides a good triangle.

The answer is 498, achieved by the following construction:

A 511-511-1022 triangle.
Under each odd side, a 255-255-510 triangle.
Under each odd side of each 255-255-510 triangle, a 127-127-254 triangle.
And so on, down to 128 3-3-6 triangles at the bottom.
That's a total of 255 good triangles.
Repeat this construction in other parts of the polygon with:
A 255-255-510 triangle "on top"
A 127-127-254 triangle "on top"
A 63-63-126 triangle "on top"
A 31-31-62 triangle "on top"
A 15-15-30 triangle "on top"
A lonely 1-1-2 triangle which isn't good.

That covers 1022+510+254+126+62+30+2=2006 edges with 255+127+63+31+15+7+0=498 good triangles. We triangulate the central 7-sided polygon any old way- all edges are even anyway.

A short proof that we can't have more than 500 good triangles:
Each of the 2004 triangles has at least one even side, so there are at most 4008 odd sides of the triangles. 2006 of these are the sides of the polygon, and each odd diagonal contributes 2 to this count, so there are at most 1001 odd diagonals. Each odd diagonal can be part of only one good triangle, so there are at most $\frac{1001}2$ good triangles.

That's the easy part. The hard part is getting from 500 to 498.

The proof:
Generalize the situation to an $n$-gon, where $n$ is even.
We use the same notion of "below" as in the construction; a good triangle $ABC$ ($AB,BC$ odd sides) is below another good triangle $DEF$ ($DE,EF$ odd sides) if either $DABCE$, $DCBAE$, $EABCF$, or $ECBAF$ are in cyclic order in a $180^\circ$ arc of the circumcircle of $P$. In the first two cases, $ABC$ is under $DE$; in the last two, $ABC$ is under $EF$.
This is a partial order on the good triangles; each odd side has a unique choice of "length" less than $\frac{n}{2}$, and the odd sides of $ABC$ must be less than half the "length" of the odd sides of $DEF$ if $ABC$ is below $DEF$. In addition, every good triangle in the region cut off by an odd side $DE$ is below $DEF$ (if $DEF$ is good).

Lemma 1: If $ABC$ is a good triangle, there is a triangulation with at least as many good triangles in which the good triangles under $AB$ are congruent to those under $BC$.
Proof: If there are more good triangles under $AB$ than under $BC$, copy the triangulation of the region cut off by $AB$ to the region cut off by $BC$. Otherwise, copy the triangulation of the region cut off by $BC$ to the region cut off by $AB$.

Lemma 2: Let $ABC$ be a good triangle satisfying the property of Lemma 1, and suppose the triangulation has a maximal number of good triangles. If $k$ good triangles under each of $AB$ and $BC$ are maximal $(k>1)$, there is a triangulation of the region cut off by $AC$ with $2k-1$ maximal triangles and as many total good triangles.
Proof: Let $2m+1$ be the length of $AB$. Since the outer edges of the maximal triangles under $AB$ are all even, we can move those triangles (and everything under them) around so that the entire figure is symmetric around $B$, and the vertices $D,G$ adjacent to $P$ on each side are each vertices of one of these maximal triangles. Replace $ABC$ with $FBI$, where $DEF,GHI$ are these previously maximal triangles. Draw any sides necessary to complete the triangulation; this cannot introduce any new good triangles by maximality.

Lemma 3: There exists a triangulation with a maximal number of good triangles with the following property: If $ABC$ is good, there are either 0 or exactly 2 good triangles maximal among those below $ABC$.
Proof: Start with a maximal triangulation, and give it the symmetry properties of Lemma 1. By that lemma, each good triangle has an even number of good triangles dirctly under it. If the property does not hold, there is some minimal good triangle $ABC$ for which it fails and there are at least 4 good triangles directly below $ABC$; apply Lemma 2 to it and all other such minimal triangles. This can be done in a way that preserves symmetries.
Now there may be triangles for which the property fails, all of which were above the triangles in the previous step. Apply Lemma 2 to these. Repeat the process- it must terminate, since all chains of good triangles have finite length.

Corollary: There exists a maximal triangulation with maximal good triangles $T_{i}$ such that each $T_{i}$ has $2^{a_{i}}-1$ triangles under it, including itself.
Proof: Apply Lemma 3. Moving down from a maximal triangle $T_{i}$, note that if any triangle in a level has triangles under it, all must by symmetry. There are $1+2+\dots+2^{a_{i}-1}=2^{a_{i}}-1$ triangles in this tree.

Lemma 4: If there are $m$ maximal good triangles in a triangulation, there are at most $\frac{n}{4}-\frac{m}{2}$ good triangles.
Proof: This is very similar to the "500" proof. If we remove the good triangles and everything under them, the remainder is a $k$-gon, where $k\ge m$. This remainder is dissected into $k-2$ triangles, so the removed portion contains $n-k$ triangles, for a total of at most $2n-2k$ odd sides. $n-k+m$ of these are sides of $P$, so at most $n-k-m$ of these are odd diagonals. These are double-counted, so there are at most $\frac{n-k-m}2\le \frac{n-2m}2$ odd diagonals in the relevant portion of the triangulation. Each good triangle has two of these as sides, and each odd diagonal is a side of at most one good triangle, so we have at most $\frac{n}{4}-\frac{m}{2}$ good triangles.

Now we prove that there are at most 498 good triangles when $n=2006$. If there were 499 or 500, there would be at most 5 maximal good triangles by Lemma 4, and therefore either $499=2^{a}+2^{b}-2^{c}-2^{d}-2^{e}-5$ or $500=2^{a}+2^{b}-2^{c}-2^{d}-2^{e}-5$. Since 504= 1 111 110 000 and 505= 1 111 110 001 in base 2, this is impossible; we would need at least 6 terms.

It isn't to hard to do any particular case with this method, but a general statement is very messy.


Hello,

During the last ten years I have written several small articles of "Math Olympiad Related" topics, most of the time to give some lecture. Most of them are still just in handwritten notes, but I am interested in the creation of an extensive sillabus about relevant results for Math Olympiads, probably divided into levels. Is someone interested in trying to do it with me? Probably it would take a while. And I am not talking about writting every resutl with the proof, but having at least a nice struture!

Best regards,

I think many people here would be interested in your project.

I am the father of RC-7th.  I am also interested in your project.

Hello,

Well, if someone is interested, please send me a PM. What I am going to do, probably on weekend or next week ('cause I am behind in my thesis research, I need to cach up this week) is start one by myself, and we could discuss, add new topics, reorganize, etc.

Best regards,


Rezolvati in $\mathbb R$ ecuatiile:
$a)$ $\sqrt[4]{97-x}+\sqrt{9+x}=8$
$b)$ $\{\begin{array}{c}\sqrt{3x+y}+\sqrt{x-y}=6 \\\ 3x+\sqrt{x-y}=17 \end{array}$

Pentru sistemul de ecuatii:
Notam $ \sqrt{3x+y}=a$ si $ \sqrt{x-y}=b, a,b>0$ si $ a+b=6$;
Avem $ 3x=a^{2}-y$ si $ x=b^{2}+y$, de unde $ 4x=a^{2}+b^{2}\Longleftrightarrow 3x=\frac{3a^{2}+3b^{2}}4$.
A doua ecuatie a sistemului devine $ \frac{3a^{2}+3b^{2}}4+b=17$. Cum $ a=6-b$, ecuatie devine, dupa prelucrare, $ 3b^{2}-16b+20=0$, cu solutiile $ b_{1}=2$ si $ b_{2}=\frac{10}3$.
Pentru $ b=2$ avem $ a=4$ si se obtine solutia solutia in $ x$ si $ y$ $ (5;1)$, iar pentru $ b=\frac{10}3$, avem $ a=\frac{8}{3}$ si solutia $ (\frac{41}9;\frac{-59}9)$.


Given two lines $ l$ and $ m$ intersecting at an acute angle, and a point $ P$ in the region contained by this acute angle, construct a circle passing through $ P$ and tangent to lines $ l$ and $ m$ using a ruler and compass.

You draw one random circle tangent to both lines lying in the same region bounded by the lines as the region containing $ P$. Let $ l$,$ m$ intersect at $ O$. Draw $ OP$. Then $ OP$ will intersect the random circle we drew in two points, say $ A$ and $ B$. Then let the circle be tangent to the lines $ l$,$ m$at pts $ X$ and $ Y$ resepectively.
Then draw the angle bisector of the acute angle, let it be line $ t$.
Then we construct 2 new circles that satisfy the requirements of the problem:
1. Find pt $ R$ on line $ l$ in the region containing the pt $ P$ so that $ \frac{OR}{OX}\equal{}\frac{OP}{OA}$ -- just do this by constructing a triangle $ OPR$ similar to $ OAX$. Let perpendicular to $ l$ at $ R$ intersect $ t$ at $ O_{1}$. Then the circle with centre $ O_{1}$ and radius $ O_{2}P$ satisfies the condition because it will be tangent to the two lines because it is a dialation of the random circle we drew in the beginning.
2. Find pt $ T$ on on line $ l$ in the region containing the pt $ P$ so that $ \frac{OT}{OX}\equal{}\frac{OP}{OB}$ -- just do this by constructing a triangle $ OPT$ similar to $ OBX$. Let perpendicular to $ l$ at $ T$ intersect $ t$ at $ O_{2}$. Then the circle with centre $ O_{2}$ and radius $ O_{2}P$ satisfies the condition because it will be tangent to the two lines because it is a dialation of the random circle we drew in the beginning.


Factor:
$ P(x,a,b,c) \equal{} \sum_{cyc} (x \minus{} a)(x \minus{} b)(a^2 \minus{} b^2)$

$ P(a) \equal{} a^{10} \plus{} a^5 \plus{} 1$

Simplify:
$ E \equal{} \frac {x^{35} \plus{} x^{25} \minus{} x^{10} \minus{} 1}{x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1}$

[hide="2, assuming in reals"]Let $ a$ be a third root of unity that is not 1. Then, $ a^{10}\plus{}a^5\plus{}1\equal{}a\plus{}a^2\plus{}1\equal{}0$.

Hence, by Remainder Theorem, we know that $ a\minus{}e^{\frac{2\pi i}{3}}$ and $ a\minus{}e^{\frac{4\pi i}{3}}$ are factors of $ a^{10}\plus{}a^5\plus{}1$.

We want to factor $ a^{10}\plus{}a^5\plus{}1$ in reals, so multiply two factors to get $ a^2\plus{}a\plus{}1$.

Dividing $ a^{10}\plus{}a^5\plus{}1$ by $ a^2\plus{}a\plus{}1$ via long division, we find that $ a^{10}\plus{}a^5\plus{}1\equal{}(a^2\plus{}a\plus{}1) (a^8\minus{}a^7\plus{}a^5\minus{}a^4\plus{}a^3\minus{}a\plus{}1).$[/hide]

[quote="hasan4444"]Factor:
$ P(x,a,b,c) \equal{} \sum_{cyc} (x \minus{} a)(x \minus{} b)(a^2 \minus{} b^2)$[/quote]

Any idea  :huh:

Hmm, I'm not sure but $ P(0,a,b,c)\equal{}(a\plus{}b\plus{}c)(a\minus{}b)(b\minus{}c)(c\minus{}a)$ from here [url]http://www.mathlinks.ro/viewtopic.php?p=571945#571945[/url], maybe that helps?


function
real analysis
real analysis unsolved

assume that $f,g$ are continuous function from unit disk to itself .if $f$ and $g$ commute ,does it follows that they have common fixed point?

By commute, you mean $(g\circ f)(z)=(f\circ g)(z)$ for all $z$?

you want us to prove that there exists a point $z$ such that $f(z)=g(z)$ or that there exists a point $z$ such that $f(z)=g(z)=z$ ?

anyway, in dimension $1$, this is of course true that there exists $x$ such that $f(x)=g(x)$.
But I've read somewhere it was false that there is a common fixed point (and hard to find a counter example).
But suppose we've found a counter example in dimension $1$. Then there exist $2$ continuous functions $F,G: [0;1]\to [0;1]$ such that $F(G(x))=G(F(x))$ but with no common fixed point. Hence, if one define $f(x+iy)=F(x)$ and $g(x+iy)=G(x)$, this is a counter example in dimension $2$.

I was absolutely convinced that the problem had a positive answer in dimension $1$, and I thought I had posted it, but apparently, the problem I was thinking about was [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=fixed+point&t=1434]this one[/url], which has the extra condition that $f$ be monotonic.

It would be interesting to see a counterexample.


LaTeX
inequalities unsolved
inequalities

prove that   :D  following assertion :If  $ a_{1},a_{2},...a_{n}$ are real numbers such that
   $ (n-1)(c_{1}^{2}+c_{^}2+...+c_{n}^{2})=(c_{1}+c_{2}+...+c_{n})^{2}$
the either all these numbers are nonnegative or all these are non positive

   ( learning TEX)

[quote="NDMinh"]prove that   :D  following assertion :If  $ a_{1},a_{2},...a_{n}$ are real numbers such that
   $ (n-1)(c_{1}^{2}+c_{^{}}2+...+c_{n}^{2}) = (c_{1}+c_{2}+...+c_{n})^{2}$
the either all these numbers are nonnegative or all these are non positive

   ( learning TEX)[/quote]

  :(  .NDMinh, You let $ a_{1},a_{2},...,a_{n}$ but in the question, you have typed $ c_{1},c_{2},..,c_{n}$ .

and, your $ \text{\LaTeX}$ is wrong :lol:


New leaderboard after 3 rounds:
theone-9
MysticTerminator-7
ChinaBoy-6
MathFiend-1

1. What is the product of
(1 + 1/5)(1 + 1/6)(1 + 1/7)(1 + 1/8)(1 + 1/9)?

2. If a^2 + a = b^2 + b and a~=b, then
a.)a=-b
b.)a + b = 1
c.) a + b = -1
d.) a- b = 1
e.) a - b = -1

3. Each step of an escalator is 8 inches above the rpevious step, and the total height of the escalator is 20 feet.  The escalator moves up one step in 2 seconds.  If a person steps on the lowest steip of the escalator when it is level with the lower floor and walks up the escalator at the rate of 1 step a second, then what is the number of steps taken by the rider on the way up to the second floor?

4. If 3/(2-x/2) = 2, then what does x equal?

5. If 3x + 7 = x^2 + k = 7x + 15, then what does k equal?

[quote="topper"]Mystic is in Fun & Games, and theone was logged in when I started, but doesn't seem to be on line anymore.  Such is the luck![/quote]

lol... the fun and games is very amusing

Incorrect for #2.  Can't give another answer since it was multiple choice.  I'll hold off on the last round unless one other user says they feel like doing it tonight...

Mystic-no problem.

I'll hold off until tomorrow night for the final round.

theone gets #2.

After 4 rounds:
ChinaBoy & Theone-10
Mystic Terminator-7
MathFiend-1

I wonder topper, what's with the tilde you put on question 2, where a~=b?

Its meant to be not equal to. ~ is not in logic. Could also use != (programming style), or =/= or anything. But yeah.

Normally, if using the logic notation, one would write ~(a = b), since you can only not-ify a statement with a truth value.

Oh! :) Thank you.  I already assumed a is not equal to b.


conics parabola ellipse hyperbola trigonometry rotation invariant

This is a problem from an old Introductory Analysis book. It at the end of the section on oblique conics.

In the general equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F$, when replacing $x$ with $xcos\phi + ysin\phi$ and $y$ with $-xsin\phi + ycos\phi$, you are rotating the graph of the equation by $-\phi$. Expanding this equation, it is evident that:

$A' = A cos^2 \phi - Bcos\phi sin\phi + Csin^2\phi$
$B' = 2(A-C)sin%Error. "hicos" is a bad command.
\phi + B(cos^2\phi - sin^2\phi)$
$C' = Asin^2\phi+Bcos\phi sin\phi + B(cos^2\phi)$

Prove that $B'^2 - 4A'C' = B^2 - 4AC$

Furthermore, with the given information, prove that:

If $B^2 - 4AC = 0$, then the equation is of a parabola
If $B^2 - 4AC < 0$, then the equation is of a circle or ellipse
If $B^2 - 4AC > 0$, then the equation is of a hyperbola

[quote="breez"]This is a problem from an old Introductory Analysis book. It at the end of the section on oblique conics.

In the general equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F$, when replacing $x$ with $xcos\phi + ysin\phi$ and $y$ with $-xsin\phi + ycos\phi$, you are rotating the graph of the equation by $-\phi$. Expanding this equation, it is evident that:

$A' = A cos^2 \phi - Bcos\phi sin\phi + Csin^2\phi$
$B' = 2(A-C)sin%Error. "hicos" is a bad command.
\phi + B(cos^2\phi - sin^2\phi)$
$C' = Asin^2\phi+Bcos\phi sin\phi + B(cos^2\phi)$

Prove that $B'^2 - A'C' = B^2 - AC$[/quote] 4 typoes found

Correct information is:

In the general equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$, when replacing $x$ with $x\cos\phi+y\sin\phi$ and $y$ with $-x\sin\phi + y\cos\phi$, you are rotating the graph of the equation by $-\phi$. Expanding this equation, it is evident that:

$A'=A\cos^2\phi-B\cos\phi\sin\phi+C\sin^2\phi$
$B'=2(A-C)\sin\phi\cos\phi+B(\cos^2\phi-\sin^2\phi)$
$C'=A\sin^2\phi+B\cos\phi\sin\phi+C\cos^2\phi$

Prove that $B'^2-4A'C' = B^2-4AC$

to prove it, all you have to do is expand them...not that hard, and for the discriminant being a certain way, consider when you rotate it so B=0, then if you have the product of a*c ?0, then you can see which form you can put it into by completing the square, in a hyperbola, they have opposite signs, ellipse, the same, and parabolas have 0 for either A or C (all of these i am talking about when b=0)

Okay, so after expanding, how would you prove that $B^2 - 4AC$ is invariant under rotation?


Let $r$ and $s$ be two rational numbers. Find all functions $f: \mathbb Q \to \mathbb Q$ such that for all $x,y\in\mathbb Q$ we have \[ f(x+f(y)) = f(x+r)+y+s. \]

To solve this beautifull problem you have to solve the pseudo causchy's equation :
$g(x+y+s)=g(x)+g(y)$ where $g(x)=f(x)-r$  :D

the only solutions are f(x)=x+r+s and f(x)=-x+r-s.f(x+f(y))=f(x+r)+y+s(*)
int (*) let x=0.then f(f(x))=f(r)+x+s.now in (*)y=f(y)=>f(x+f(f(y)))=f(x+y+f(r)+s)=f(x+r)+f(y)+s.(1)in (1) let x=x-r.<=>f(x+y+f(r)-r+s)=f(x)+f(y)+s.denote w=f(r)-r+s.so we have f(x+y+w)=f(x)+f(y)+s.let g(x)=f(x)+s.then (1) is equivalent with g(x+y+w)=g(x)+g(y).(2).in (2)let x=y=0.then g(w)=2g(0).let g(0)=a,where a is a rational number(we will find a later in the solution).then g(w)=2a.in(2) let y=0.then g(x+w)=g(x)+a(3).but using (3) (2)becomes g(x+y)+a=g(x)+g(y).putting h(x)=g(x)-a (2)becomes h(x+y)=h(x)+h(y) with x,y rational numbers.but this is just cauchy so h(x)=ex,where e is a rational number(we will find e later).so g(x)=h(x)+a=ea+a.if we plug this in relation (2)then we find that e=a/w.so g(x)=(a/w)x+a.so f(x)=g(x)-s=(a/w)x+a-s.we plug this into (*) and we get after some easy computations that(a^2/w^2)y+a^2/w-s(a/w)-s=(a/w)r+y.so a^2/w^2 must be 1.so w=a or w=-a.
case 1:w=a.we get a=2s+r.we easily get f(x)=x+s+r for every rational x.
case 2:w=-a.=>f(x)=-x+r-s.that's all :showoff:  :blush:
ps:could someone put this into latex pls :oops:

Yes that is right :first:
Who is the designer of this problem?

[quote="bodom"]the only solutions are $f(x)=x+r+s$ and $f(x)=-x+r-s$

$f(x+f(y))=f(x+r)+y+s$  (*)
int (*) let $x=0$ Then  $f(f(x))=f(r)+x+s$
now in (*)   $y=f(y)$ and we have $f(x+f(f(y)))=f(x+y+f(r)+s)=f(x+r)+f(y)+s$  (1)
in (1) let $x=x-r$ <=> $f(x+y+f(r)-r+s)=f(x)+f(y)+s$
Now denote $w=f(r)-r+s$
 So we have $f(x+y+w)=f(x)+f(y)+s$  Let $g(x)=f(x)+s$  .Then (1) is equivalent with    $g(x+y+w)=g(x)+g(y)$ (2)
in (2)let $x=y=0$ and then  $g(w)=2g(0)$
Let $g(0)=a$,where a is a rational number(we will find a later in the solution).then
     $g(w)=2a$  In(2) let $y=0$ Then  $g(x+w)=g(x)+a$   (3).but using (3) (2)becomes $g(x+y)+a=g(x)+g(y)$ .Putting $h(x)=g(x)-a$  (2)becomes $h(x+y)=h(x)+h(y)$ with x,y rational numbers.But this is just cauchy so $h(x)=ex$,where e is a rational number(we will find e later).so  $g(x)=h(x)+a=ea+a$
If we plug this in relation (2) then we find that $e=\frac{a}{w}$
so $g(x)=(\frac{a}{w})x+a$.
so $f(x)=g(x)-s=(\frac{a}{w})x+a-s$.
 we plug this into (*) and we get after some easy computations that             $\frac{a^2}{w^2}y+\frac{a^2}{w}-s(\frac{a}{w})-s=\frac{a}{w}r+y$
.so $\frac{a^2}{w^2}$ must be 1 .so $w=a$ or $w=-a$.
case 1:w=a.we get a=2s+r.we easily get $f(x)=x+s+r$ for every rational x.
case 2:w=-a.=> $f(x)=-x+r-s$
.that's all :showoff:  :blush:
ps:could someone put this into latex pls :oops:[/quote]

Adding z-r to both sides of the definning equation and taking f, we get
f(z-r+f(x+f(y)))=f(z+y+s-r+f(x+r)).
Applying the definning equation to each side, we have
f(z)+x+f(y)+s=f(z+y+s)+x+r+s, which simplifies to
f(y)+f(z)=f(y+z+s)+r.
Let g(x)=f(x-s)-r, then
g(u)+g(v)=g(u+v), where u=y+s and v=z+s.
Hence g(x)=cx where c is a constant.
Substituting f(x)=c(x+s)+r into the defining equation, we find c=1 or -1.

This solution may reveal how the problem was created.

Let $g(x)=f(x)-r-s$

So now we have $g(x+g(y))=g(x-s)+y+s$ and $g^2(x+g(y))=x+g(y)$

Now as we can express all rational in terms of $x+g(a)$ with $a$ fixed. So $g$ is bijective.

So now replacing $y$ by $g(y)$ we have by previous result $g(x+y)=g(x-s)+g(y)+s=g(y-s)+g(x)+s$

Hence $g(x)-g(x-s)=c$ , where $c$ is constant.

Hence finally we get $g(x+y)=g(x)+g(y)-c$ so $g(x)=ax+b$ for some constant $a,b$.

Now putting this into original equation we get $(a,b)=(1,0),(-1,-2s)$

So clearly all solutions are given by $f(x)=x+r+s$ or $f(x)=-x+r-s$.


hi,
calculate
\[ \lim_{n \rightarrow \infty}{\sum_{k\equal{}0}^{n}{\frac{k(n\minus{}k)! \plus{} (k\plus{}1)}{(k\plus{}1)!(n\minus{}k)!}}}.\]


function
number theory unsolved
number theory

Find all surjective function f:N->N satisfying the condition mln <=>f(m) l f(n)

[hide="Two hints for a path to the solution"]
$f$ is in fact bijective;
$f$ is multiplicative.[/hide]

1: $f$ is bijective
in fact if $f\left(n\right)=f\left(m\right)$:then$m=n$
 2:$f$ is multiplicative
we have $\forall{n}\in\mathbb{N}$ we have $f\left(1\right)|f\left(n\right)$   then$f\left(1\right)=1$.
let $(m,n)=1$  we have the only divisor of $f\left(nm\right)$ are $f\left(n\right)$.and $f\left(m\right)$ ,then $f\left(mn\right)=f\left(m\right)f\left(n\right)$.
3:$f\left(p\right)$ is a prime .and $f\left(p^{n}\right)=(f\left(p\right))^{n}$.

in bref we have infinitly many solution detireminite by$f\left(p\right)$
 for example $f\left(p_{i}\right)=p_{i+1}$ :maybe:

$f$ isn't necessarily bijevtive
the solution is:
$f(\prod_{k=1}^{m}{p_{i}}^{a_{i}})=c\prod_{k=1}^{m}{h(p_{i})}^{b_{i}}$
with $f(1)=c\in N^{*}$ and $h: \{primes\}\to \{primes\}$ is an application injective

[quote="fermat3"]no ! $f$ is injective.
sopose that $f\left(n\right)=f\left(m\right)$.then $f\left(m\right)|f\left(n\right)$.=>$m|n$. and vise versa we get $n|m$. so $m=n$.
and $f$
 is also surjective => $f$ is bijective. and we have $f\left(1\right)|f\left(n\right) .\forall{n}\in\mathbb{N}$ => $f\left(1\right)=1$. :lol:[/quote]
I did not pay attention that F must be surjective ,therefore I gave a general solution which is surjective for $c=1$ ,thank you to have me to announce that $f$ is surjective according to the statement of the problem because the property (m|n <=> f(m)|f(n)) gives only the injection

so,
$f(\prod_{k=1}^{m}{p_{i}}^{a_{i}})=\prod_{k=1}^{m}h(p_{i})^{a_{i}}$
with $h: \mathbb{P}\to \mathbb{P}$ an application bijectif
$f$ is bijective
$\forall (m,n)\in N$ we can write $m$ and $n$ like $n=\prod_{k=1}^{c}{p_{k}}^{a_{k}}$ and $m=\prod_{k=1}^{c}{p_{k}}^{b_{k}}$ with $a_{k}$ or $b_{k}$ can be null

$n|m$ [u][b]<=>[/b][/u] $\forall 0\le k\le c$; ${p_{k}}^{a_{k}}|{p_{k}}^{b_{k}}$ [u][b]<=>[/b][/u] $\forall 0\le k\le c$; ${h(p_{k})}^{a_{k}}|{h(p_{k})}^{b_{k}}$ [u][b]<=>[/b][/u] $f(n)=\prod_{k=1}^{c}{h(p_{k})}^{a_{k}}|\prod_{k=1}^{c}{h(p_{k})}^{b_{k}}=f(n)$


Este multimea $A={ \cos x, \cos 2x, \ldots , \cos nx }$ libera? ;)

Fie $a_1,\ldots,a_n$ astfel ca $f(x)=\sum_{k=1}^na_k\cos{kx}=0,\forall x\in\mathbb{R}.$
Avem insa
\[ \int_0^{\pi}f(x)\cos{kx}dx=a_k\frac{\pi}{2}, \]
deci $a_k=0,\forall k=1,\ldots n.$


If $n$ is an integer, then find all values of $n$ for which $\sqrt{n}+\sqrt{n+2005}$ is an integer as well.

Dont know if it helps, but if you set all that equal to $m$, you get

$n=(\frac{m^2-2005}{2m})^2$

Not too hard from there...

Yep, that's the solution. I found this the easiest question of the set, but apparently it is hard to see under stressful circumstances :)

[ from what you said it follows that m|2005 so we only research m=1, m=5, m=401, m=2005 -- and that yields m=198^2 or m=1002^2 ]

The thing i came up with after i handed in:

You know $n$ should be one of the following numbers: $1,4,9,16,25,36,49,64,etc.$
You know that $n+2005$ should also be one of those numbers.
As you can see, $1+3=4, 4+5=9, 9+7=16, 16+9=25, etc.$ You have to add an extra $2$ to obtain the next number. After a while, you should have that one number is the following $+2005$. That is the $1002th$ term in line, so $n=1002^2$
You can also obtain two perfect squares with difference $2005$ if you add $5*401$ to the smallest. So both $n$ and $n+397+399+401+403+405$ are square if $n=198^2$
Since $2005=5*401$ and $5$ and $401$ are prime and $n>0$, there is no other way to obtain two perfect squares with difference $2005$.

[quote="Peter VDD"][quote="Psycho"]You know $n$ should be one of the following numbers: $1,4,9,16,25,36,49,64,etc.$
You know that $n+2005$ should also be one of those numbers.[/quote]

You would have to prove this thing.

Who says for example, that you won't have sqrt(n) ending in .62 adn sqrt(n+2005) ending in .38 ?
These are idiot examples of course since it is correct, but it is not trivial so you would have to prove it.[/quote]
seriously?  :?
meh, i would've been glad if this stuff was on my answering sheet  :P

It is a detail (because it is true) but it is not trivial ;) so you would have to prove it.

Just know that $n$ must be some perfect square and $2005 + n$ must be some perfect square so $2005 = p^2 - m^2$ or $2005 = (p+m)(p-m)$ and choose the factors of 2005 in any combination and solve for $m^2$ to get values of $n$.

[quote="Peter VDD"][quote="tarquin"]Just know that $n$ must be some perfect square and $2005 + n$ must be some perfect square so..[/quote]Your solution is very short and clean, but what we're saying is exactly that you would have to prove that part.

It is not true in general that from a+b in N follows that a in N and b in N. You would have to prove it is true for sqrts of integers.[/quote]

$\sqrt{a}+\sqrt{b}$ in N an a,b in N => a,b are perfect squares.
This is not hard i suppose...
a-b is in N
$\sqrt{a}+\sqrt{b}$ in N => $\sqrt{a}-\sqrt{b}$ in Q
=> $\sqrt{a}$ and $\sqrt{b}$ are in Q.
a,b are in N => their root must be in N or R
so we hv the conclusion..

[quote="siuhochung"]$\sqrt{a}-\sqrt{b}$ in Q
=> $\sqrt{a}$ and $\sqrt{b}$ are in Q.[/quote]
How do you do this step? It looks like the same again :)

But you're quite close to the solution arne gave me. :P

No, he's right, he just doesn't go into detail.

$\sqrt{a} + \sqrt{b} \in \mathbb{N}$
$a - b \in \mathbb{N}$
implies
$\sqrt{a} - \sqrt{b} = \frac{a-b}{\sqrt{a} + \sqrt{b}} \in \mathbb{Q}$
so $\sqrt{a} + \sqrt{b}, \sqrt{a} - \sqrt{b} \in \mathbb{Q}$
so their sum, $2\sqrt{a}$, is rational as well and so a is the square of a rational and hence a perfect square.

yes, that is exactly the solution arne gave me (with that part added) :)

[ as soon as he spoke of difference and rationals I knew he got it but still on a contest not stating it would cost you points -- I don't think anyone will have a 20/20 in this way. The other way 20's will be frequent.]

Since the final result must be an integer,then both terms must be integers as well,which implies that n and n+2005 are also perfect squares.
Let n=a^2 and n+2005=b^2. Then,(b-a)(b+a)=2005.
Since 2005=5x401 and 401 is prime,we have (a,b)=(1002,1003),(198,203),(1002,-1003),(198,-203).
Solving,we get n=1004004 and n=39204.   :)

This is indeed a public forum, and this is a mathematics problem, but this mathematics problem has been resolved for more than five years. Please don't post on topics that have been long since resolved unless you have a beautiful result, extension, etc. You are free to work on these problems, but please don't post on these old resolved ones.

[quote="BarbieRocks"]for the most part, volume 1 can help with 1-15 on the AMC10...maybe up to #20.[/quote]

umm....what aree you referring to...

Basically $ n(n+2005)=a^2 $, from which we can plug in that discriminant is a perfect square, to get the reqd values...

If $\sqrt n + \sqrt{n+2005}$ is rational (its given to be an integer), then so is $\frac{1}{\sqrt n + \sqrt{n+2005}} = \frac{\sqrt n - \sqrt{n+2005}}{2005}$ and hence $\sqrt n - \sqrt{n+2005}$ is rational. Hence $\sqrt n$ and $\sqrt{n+2005} $ are rational, which means $n$ and $n+2005$ are both perfect squares.

So if $n=x^2, n+2005 = y^2$ then $(y-x)(x+y) = 2005 = 5 \ \times \ 401 = 1 \ \times \ 2005$ and thus $n=198^2$ or $1002^2$

edit: didnt see that JBL had already explained this in detail. sorry


find the minimum of f(a,b,c) for arbitrary real numbers a,b,c
 \sqrt [a 2 +(1-b) 2 ] + \sqrt [b 2 +(1-c) 2 ] + \sqrt [c 2 +(1-a) 2 ]
if any of the moderators (of anyone for that matter) considers this too easy move it to the beginners section,k!


inequalities
Cauchy Inequality
inequalities solved

Let a, b, c be positive numbers with abc = 1, and let n >= 1 be an integer. Prove the inequalities

[b](a)[/b] [tex]\frac{1}{1+2a}+\frac{1}{1+2b}+\frac{1}{1+2c}\geq 1[/tex];
[b](b)[/b] [tex]\frac{a^{n}}{b+c}+\frac{b^{n}}{c+a}+\frac{c^{n}}{a+b}\geq \frac{3}{2}[/tex].

  Darij

These are really well known.
(a)Let $ a=x/y;b=y/z;c=z/x$ and we must prove that:
   [tex]  A= \frac {y}{y+2x} \ + \frac {z}{z+2y} \ + \frac {x}{x+2z} \ \geq 1 [/tex]
By Cauchy inequality we have:
       [tex] [y(y+2x)+z(z+2y)+x(x+2z)]A \geq (x+y+z)^2 [/tex]
           [tex]  <=> A \geq 1 [/tex]
(b)Let $ a \geq b \geq c $ $ and n=m+1 $.
By Chebysev inequality,we have:
   [tex]  \sum a^m.\frac{a}{b+c}\ \geq \sum \frac{a^m}{3}\  \sum \frac{a}{b+c}\ \geq 3/2 [/tex]
 :)

This one remind me of a nice inequality that my teacher told me before:
Let $ a,b,c>0$
i)Prove that:
       $ \sum a^k/(b\plus{}c)^k \geq 3.(1/2)^k$
 where $ k \geq \frac 32$.
ii)Find the least value of k that satisfied:
       $ \sum a^k/(b\plus{}c)^k  \geq 3.(1/2)^k$
I haven't  solution for ii) yet.Could everyone help me? :)


[hide]Only x=2. For possible $1\le x \le 3$ we have encreasing LHS and decreasing RHS [/hide]


jut, zur information für alle hier mal die liste:

tobias elbinger
darij grinberg
paul jonas hamacher
daniel harrer
annika heckel
martin herdegen
christian hirsch
michael holl
lars kühne
martin montag
michael müller
matthias ohst
georg schönherr
peter scholze ( yeah :D )
kolja szillat
christopher wulff

her(t)zlichen glückwunsch an alle! insbesondere an daniel harrer(zetaX), annika heckel(annika) und darij grinberg von ml(und an imo-seminar-teilnehmer geschnitten ml-ler minus eben aufgezählte, die ich leider vergessen habe).

insgesamt gibt es also mindestens eine echte überraschung: christian sattler ist nicht dabei, obwohl er letztes jahr einen festen zweiten platz in der gesamtrangliste hatte! weiterhin hat es friedrich feuerstein auch nicht geschafft, das war aber nach den auswahlklausuren relativ klar... der rest der letztjährigen imo-mannschaft hats geschafft(also alle außer christian sattler und michail shkolnikov, der sowieso zu alt ist), und christopher wulff ebenfalls, der eigentlich auch zur mannschaft gehört hätte nach punkten, jedoch keine zeit für die imo hatte.

[quote="Valentin Vornicu"]hertzlichen glückwunsch an alle!! :D[/quote]

Dein Deutsch ist perfekt, Valentin !!!  :)

Wenn ich Gronau waere, wuerde ich auf jeden Fall die besten Leute vom letzten Jahr einladen (d.h. IMO) , insbesondere wenn sie wie Christian Sattler noch sehr jung sind und sehr erfolgreich waren. Speziell haben junge Teilnehmer wie in den USA einen Bonus, um am MOP teilzunehmen.

[quote="orl"]Kann jemand die Liste nach Klassenstufen gruppiert, veroeffentlichen, bitte.  :)[/quote]

Klasse 11:
matthias ohst
georg schönherr
peter scholze

Klasse 12:
darij grinberg
paul jonas hamacher
daniel harrer
lars kühne
michael müller
kolja szillat

Klasse 13:
tobias elbinger
annika heckel
martin herdegen
christian hirsch
michael holl
martin montag
christopher wulff

ich habe übrigens 36 von 60 punkten.

christopher

danke peter daß du die namen gepostet hast... gratuliere zur teilnahme. daß christian nicht reingekommen ist finde ich auch sehr schade, insbesondere weil er wirklich eine gute chance hat bei der IMO mindestens silber zu holen wenn nicht gold... die vorauswahlklausuren waren verdammt unräpresentativ, die aufgaben sind eher mit der DeMO als mit der IMO zu vergleichen und vieles hängt sehr vom glück ab...

BTW, bei den auswahlklausuren hab ich 38 von 60 punkten. keine ahnung warum es so viele sind (ich hätte weniger erwartet, so etwa 32-35, aber irgendjemand muß wohl mitleid mit mir gehabt haben ;) )...

falls es noch jemand braucht, hier eine liste der teilnehmer samt herkunftsorten:

[b]Tobias Elbinger[/b], Fürth
[b]Darij Grinberg[/b], Karlsruhe
[b]Paul Jonas Hamacher[/b], Mönchengladbach
[b]Daniel Harrer[/b], Wenzenbach
[b]Annika Heckel[/b], Immenstadt
[b]Martin Herdegen[/b], Bischberg
[b]Christian Hirsch[/b], München
[b]Michael Holl[/b], Rheine
[b]Lars Kühne[/b], Weil am Rhein
[b]Martin Montag[/b], Wittlich
[b]Michael Müller[/b], Freiburg i.Br.
[b]Matthias Ohst[/b], Burg
[b]Georg Schönherr[/b], Dresden
[b]Peter Scholze[/b], Berlin
[b]Kolja Szillat[/b], Langenfeld
[b]Christopher Wulff[/b], Bad Aibling

und hier die 5 seminare:

[b]10.-17. Februar 2005[/b] in Rostock
[b]18.-21. März 2005[/b] in Bad Homburg
[b]22.-25. April 2005[/b] in Bad Homburg
[b]5.-8. Mai 2005[/b] in Bad Homburg
[b]28. Mai - 5. Juni 1005[/b] in Oberwolfach

beim dritten seminar wird übrigens christian reiher einer der betreuer sein. ;)

  grüße,
  darij

[quote="orl"][quote="Valentin Vornicu"]herzlichen glückwunsch an alle!! :D[/quote]
Dein Deutsch ist perfekt, Valentin !!!  :)
[/quote]Danke Orl :)

[quote="darij grinberg"]danke peter daß du die namen gepostet hast... gratuliere zur teilnahme. daß christian nicht reingekommen ist finde ich auch sehr schade, insbesondere weil er wirklich eine gute chance hat bei der IMO mindestens silber zu holen wenn nicht gold... die vorauswahlklausuren waren verdammt unräpresentativ, die aufgaben sind eher mit der DeMO als mit der IMO zu vergleichen und vieles hängt sehr vom glück ab...
[/quote]

Insbesondere weil die Aufgaben nicht notwendigeweise fuer die IMO representativ sind und Christian Sattler bereits sein aussergewoehnliches Talent bei der IMO bewiesen hat, insbesondere durch Loesen der herausfordernden Kombinatorikaufgabe, sollte man Feingefuehl zeigen. Und nicht nur strikt nach Ergebnissen in dem pre-TST gehen, d.h. man sollte nicht unbedingt eine feste Anzahl an Teilnehmern fuer das TrainingCamp festlegen, sondern eine ungefaehre Anzahl.  :)

es war so klar...  :(
schade für Friedrich

Ich schließe mich übrigens deiner Meinung an, ich weiß nicht was Christian Sattlers Problem war, aber Friedrich war eindeutig beim 2ten Test krank...

christian sattler meint, er hätte viele fehler gemacht bei den aufgaben... wie schon gesagt, der human factor ist bei den klausuren immens, wenn man einmal in schlechter form ist hat man wenig chancen...

  dg

hehe.. jetzt ist einer von euch nicht in der endgültigen auswahl, also müsst ihr alle OM mitspielen, freu mich schon auf OM13.. :D ,

ok, auch von mir gratulation an alle die es geschafft haben, viel spass bei den seminaren!

jo, wir müssen jetzt mitmachen... aua, wird sicher lustig, jedesmal nach einem von den vielen seminaren, treffen und allem möglichen heimzufahren und 30 seiten forumsposts lesen zu müssen um überhaupt nen durchblick zu erhalten... und die seminare werden bestimmt auch witzig, bei sovielen spielern, also wenn da sich einer verplappert, autsch... hoffen wir mal zumindest daß das spiel nach rostock beginnt...

  dg

gute nachrichten!

christian sattlers imo-auswahlklausuren sind nie bei herrn langmann eingetroffen. er soll jetz noch mal in der schule nachfragen. falls sie nicht mehr auffindbar sind, soll er noch mal mit gronau sprechen, welcher ihn mit sehr hoher wahrscheinlichkeit noch dazunehmen wird; in diesem fall wird aber keiner gekickt werden. und selbst wenn die klausuren nicht gefunden werden sollten, stehen die chancen vielleicht auch nicht schlecht, dass er weiterkommt.

die gerechtigkeit siegt doch ;)

[quote="Peter Scholze"]wahrscheinlichkeit noch dazunehmen wird; in diesem fall wird aber keiner gekickt werden. und selbst wenn die klausuren nicht gefunden werden sollten, stehen die chancen vielleicht auch nicht schlecht, dass er weiterkommt.

die gerechtigkeit siegt doch ;)[/quote]

Das finde ich gut. Allgemein denke ich, dass 16 Teilnehmer immer nur eine ungefaehre Zahl sein sollte. Es sei denn, die ersten 16 Teilnehmer sind mit "Abstand" besser. Wie gesagt speziell bei nur 6 Aufgaben, insbesondere wenn der Schwierigkeitsgrad nicht ausgewogen erscheint, sind die Faehigkeiten der Teilnehmer nur hoechst nicht-trivial zu evaluieren. Wenn man ca. 7 Tests schreibt, ist die Entscheidung wesentlich gerechter, da jeder einmal einen unguten Tag erlebt. Und schliesslich muss eine Entscheidung getroffen werden. Es sollte fuer Christian auch keiner hinausgeworfen werden. Jetzt geht es bei 0 los und jeder hat die Moeglichkeit sich fuer das IMO-Team zu qualifizieren.  :)

oh schön :) und siehst du, darij, das war der grund, warum ich die AKTION noch nicht publik machen wollte ;-)

@NoName Nervensägen vergisst man nicht so leicht
@ZetaX Du Dödel wir haben uns doch gestern noch gesehen.[/quote]

hallo friedrich, willkommen auf ML gleichfalls (ich bin hier anscheinend der langsamste)...

  dg

Wenn du Idril meinst: man kann auf der ML-Forumshauptseite nachsehen!!!

Vielleicht versteh ich die Frage total falsch/unkapiere...

sag mal ZetaX was is des für ne Signatur

Keine richtige, ich bin zu unkreativ für sowas, also nehme ich die schönste Formel die mir einfällt

@FFreak: Schon in die Spaßfabrik geschaut?

so jetzt hat Friedrich mir mein Ohr abgekaut  :D *schon fast taub ist*


calculus
integration
logarithms
real analysis
real analysis unsolved

i hope i have not asked this before.:|

show :D

$ \int_{0}^{1}\int_{0}^{1}\;\frac{\textbf dx\;\textbf dy}{\left(\minus{}\ln(x\,y)\right)^{\frac{3}{2}}}\;\;\equal{}\;\;\boxed{\sqrt\pi}$

Substitute $ u\equal{}xy\quad I\equal{}\int_{0}^{1}\int_{0}^{y}\frac{1}{(\minus{}\ln u)}\frac{du}{y}dy$ and reverse integration order

$ I\equal{}\int_{0}^{1}\int_{u}^{1}\frac{dy}{y}\frac{1}{(\minus{}\ln u)^{\frac{3}2}}du\equal{}\int_{0}^{1}\frac{\minus{}\ln u}{(\minus{}\ln u)^{\frac{3}2}}du \equal{}\int_{0}^{1}\left[\ln\frac{1}{u}\right]^{\frac{1}2\minus{}1}du\equal{}\Gamma\left(\frac{1}2\right)$


The perimeter of triangle ABC is 62 feet. The lengths of its sides are AB = 4x + 6, BC = 2x − 3
and AC = 3x + 5 feet. Which angle is the triangle’s largest interior angle?

[hide="Solution"]Obviously, $ x>0$. Therefore, $ 4x\plus{}6 > 3x\plus{}5 > 2x\minus{}3$, so the largest interior angle is angle $ \boxed{C}$.

[/hide]

The largest angle corresponds with the largest side, so say $ AB$ was the largest side, then $ \angle C$ would be the largest angle.  This is the same for the smallest and mediumest angles/sides.  You could solve for $ x$ in this problem (which would be adding up all the equations and setting it equal to 62), or do what the guy above me did.


Find the number of polynomials of degree 5 with [b]distinct[/b] coefficents from the set $\{1,2,3,4,5,6,7,8\}$ that are divisible by $x^{2}-x+1$.

[hide]divide polynomial ax^5+bx^4+cx^3+dx^2+ex+f by x^2-x+1

ax^5+bx^4+cx^3+dx^2+ex+f

ax^5-ax^4+ax^3

(a+b)x^4+(c-a)x^3+dx^2 +ex+f

(a+b)x^4-(a+b)x^3+(a+b)x^2

(b+c)x^3+(d-a-b)x^2+ex+f

(b+c)x^3-(b+c)x^2+(b+c)x

(d-a+c)x^2+(e-b-c)x+f must be divisible by x^2-x+1

so d-a+c=b+c-e=f

d-a=b-e[/hide]


Let there be a sequence where $ a_0 \equal{} 0$, $ a_1 \equal{} 1$ and $ a_n \equal{} \frac{a_{n\minus{}2}\plus{}a_{n\minus{}1}}{2}$. In other words, a term of the sequence is the average of the last two terms.

Define $ f(x)$ such that $ f(n)\equal{}a_n$ of the sequence.

we take $ b_{n}\equal{}a_{n\plus{}1}\minus{}a_n$
$ a_{n\plus{}2}\minus{}a_{n\plus{}1}\equal{}\frac{a_{n}\minus{}a_{n\plus{}1}}{2}$ gives $ b_{n\plus{}1}\equal{}\frac{\minus{}1}{2}b_n$
the rest is easy...

[quote="aviateurpilot"]we take $ b_{n} \equal{} a_{n \plus{} 1} \minus{} a_n$
$ a_{n \plus{} 2} \minus{} a_{n \plus{} 1} \equal{} \frac {a_{n} \minus{} a_{n \plus{} 1}}{2}$ gives $ b_{n \plus{} 1} \equal{} \frac { \minus{} 1}{2}b_n$
the rest is easy...[/quote]
$ b_n\equal{}(\minus{}1/2)^{n}b_0$ and $ b_0\equal{}a_1\minus{}a_0\equal{}1$
so $ \forall n\ge 1: \ a_n\equal{}\sum_{k\equal{}0}^{n\minus{}1}b_k\equal{}\sum_{k\equal{}0}^{n\minus{}1}(\minus{}1/2)^{k}\equal{}...$


[hide]replacing x we have 3(3)+1+3(4)+1=
9+2+12=23 :idea: our answer[/hide]

[quote="236factorial"]Let f(x) = 3x+1
What is f(3)+f(4)?[/quote]

[hide]You could seperate this into two equations:

One where x = 3;

$3x + 1 = 3 (3) + 1$

$9 + 1$

$10$

Then, one where x = 4;

$3x + 1 = 3(4) + 1$

$12 + 1$

$13$

Then you add them up:

$13 + 10 = 23$

$\boxed{\boxed{\text{The Answer is 23}}}$[/hide]


Compute the sum of all real x such that
[2^x-4]^3 + (4^x-2)^3=(4^x+2^x-6)^3

[hide]if we let 2^x-4=a and 4^x-2=b, the problem statement is that a:^3:+b:^3:=(a+b):^3: => a:^3:+b:^3:=a:^3:+b:^3:+3a:^2:b+3ab:^2:=>ab(a+b)=0. This means one or more of the following are true: a=0, b=0, or a=-b. The first two scenarios give x=2 and x=1/2, respectively. The last scenario gives 4^x+2^x-6=0. If we let 2^x=z, this means z^2+z-6=0 => z=2,-3 => 2^x=2,-3 => x=1. All possible solutions: x=.5,1,2. Correct?[/hide]

its half correct

on the meet, u woulda gotten it wrong, cuz the question asks for hte SUM

You know, the answer reminds me of South Park..."three-fifty!" :lol: *rolls on the floor* :lol: *hits the wall* :lol:  :cry:


How many different isosceles triangles have integer side lengths and perimeter 23?

1-11-11
3-10-10
5-9-9
7-8-8
9-7-7
11-6-6
13-5-5
15-4-4
17-3-3
19-2-2
21-1-1

These are all the possibilities, but keep in mind the Triangles conjecture where any two sides of a triangle have to be greater than the third. We eliminate a bunch of these, to remain.

1-11-11
3-10-10
5-9-9
7-8-8
9-7-7
11-6-6

These leave us with the [size=150][b]6[/b][/size], the answer.


A car tyre of volume $1.6*10^-2 m^3$ contains air  at a temperature of $17^0 C$ and a pressure of $2.6*10^5 PA$. A foot pump of volume $3.0*10^-4 m^3$ is used to pump air , at a temperature of $17^0 C$ and an initial pressure of $1.0*10^5 Pa$, ito the tyre:

calculate the number of strokes  of the pump needs to raise the pressure to $3.1*10^5 Pa$
 Assume that: pumping does  not change the temperature of the air in the tyre. :D

good luck! :P

If the temperature doesn't change, shouldn't it be really easy? Well, assuming that air is an ideal gas.[hide="My answer"]$\frac{80}{3}\approx27$ times[/hide]


Is there any way of expressing this in factorial form (or any short form)?
[code]n! * (n-1)! * (n - 2)! * ... * 2! * 1![/code]

It's definitely not very well known, and you'll almost never see it, but it's been called "superfactorial" by Neil Sloane and Simon Plouffe:

http://en.wikipedia.org/wiki/Factorial#Superfactorials

\[ \text{sf}\left(n\right)\equal{}1! \cdot 2! \cdots n!\]

Thanks, I didn't think to check the wikipedia page on factorials. That was exactly what I meant.


MATHCOUNTS USAMTS AMC USA(J)MO USAMO AIME

[quote="mtms5467"][quote="lotrgreengrapes7926"][quote="inom"]What is the number of factors of number 10000?[/quote]
[hide] $2^{4}\cdot5^{4}$ has $(4+1)(4+1)=25$ factors. [/hide]
How many factors are in $10^{n}$?[/quote]
[hide]$(n+1)^{2}$[/hide][/quote]
I have a question. In general when you have an expression like (3x+2)(2x+3) is it better to expand or leave factored?

For $10^{n}$ we have[hide]$10^{n}=2^{n}\times5^{n}$
THe numbers of factors would be $(n+1)(n+1)=(n+1)^{2}$  :lol: [/hide]

[quote="bpms"][/quote][quote="mtms5467"][quote="lotrgreengrapes7926"][quote="inom"]What is the number of factors of number 10000?[/quote]
[hide] $2^{4}\cdot5^{4}$ has $(4+1)(4+1)=25$ factors. [/hide]
How many factors are in $10^{n}$?[/quote]
[hide]$(n+1)^{2}$[/hide][/quote][quote="bpms"]
I have a question. In general when you have an expression like (3x+2)(2x+3) is it better to expand or leave factored?[/quote]
It depends on what you're using the expression for. Usually either way is fine.
But in MathCounts, answers usually aren't in terms of variables though, so you won't have to worry about it too much.  :wink:  :P

[quote="nebula42"][/quote][quote="bpms"][/quote][quote="mtms5467"][quote="lotrgreengrapes7926"][quote="inom"]What is the number of factors of number 10000?[/quote]
[hide] $2^{4}\cdot5^{4}$ has $(4+1)(4+1)=25$ factors. [/hide]
How many factors are in $10^{n}$?[/quote]
[hide]$(n+1)^{2}$[/hide][/quote][quote="bpms"]
I have a question. In general when you have an expression like (3x+2)(2x+3) is it better to expand or leave factored?[/quote][quote="nebula42"]
It depends on what you're using the expression for. Usually either way is fine.
But in MathCounts, answers usually aren't in terms of variables though, so you won't have to worry about it too much.  :wink:  :P[/quote]
What if, lets say, you have derived a formula? Or for example would you leave the formula for $1^{2}+2^{2}+3^{2}+...+x^{2}$ as $\frac{x(2x+1)(x+1)}{6}$ or $\frac{2x^{3}+3x^{2}+x}{6}$?

[quote="bpms"][quote="lotrgreengrapes7926"][quote="bpms"]What if, lets say, you have derived a formula? Or for example would you leave the formula for $1^{2}+2^{2}+3^{2}+...+x^{2}$ as $\frac{x(2x+1)(x+1)}{6}$ or $\frac{2x^{3}+3x^{2}+x}{6}$?[/quote]
I'm pretty sure competitions like USAMTS, USAMO, etc. would accept both forms. Same with work in math class. AMC, MathCounts, and AIME never have solutions in that form, so you won't need to worry there.[/quote]
But which way is simpler and better for solving?[/quote]
I usually see formulas like that in factored form. In my opinion, the factored form is also easier to evaluate.