modular arithmetic
number theory unsolved
number theory

For all primes $p \geq 3,$ define $F(p) = \sum^{\frac{p-1}{2}}_{k=1}k^{120}$ and $f(p) = \frac{1}{2} - \left\{ \frac{F(p)}{p} \right\}$, where $\{x\} = x - [x],$ find the value of $f(p).$

USe the lemma $1^m+2^m+..(p-1)^m \equiv 0 (mod p)$ for all $m$ satisfies $0 \leq m \leq (p-2)$ :)  :)

A permutation is a bijective mapping of a set to itself. Like $(4,2,1,3)$ is a permutation of $(1,2,3,4)$.

can explain this one please??
[b]\[
\frac{1}
{2}\sum\limits_{k = 1}^{p - 1} {k^{120} }  \equiv \frac{1}
{2}\sum\limits_{i = 1}^{p - 1} {r^{120i} } \left( {\bmod p} \right)
\][/b]

Notice that It is enough to calculate $\sum\limits_{k = 1}^{p - 1} {k^{120} }  (\text{mod} \ p)$.   If there  is some $1\leq i\leq p-1$  so that $i^{120}\not\equiv 1(\text{mod}\ p)$, then  the mentioned sum is 0 modulo p, since after multiplying by $i^{120}$, it remains the same. If  $i^{120}\equiv 1(\text{mod}\ p)$ holds for each value of $i$,  then   $ \sum\limits_{k = 1}^{p - 1} {k^{120} } =-1 (\text{mod} \ p)$. This precisely means that $p-1|120$, due to the primitive root theorem.  A simple check shows that the primes satisfying the latter congruence are in the set $\{2,3,5,7,11,13,31,41,61\}.$


Hey guys,

can someone tell me what happens when cis 1-bromo-2-methyl-cyclohexane reacts in KOH? THe question hints that a hydrocarbon is formed, but how does this happen? I thought the OH displaces the bromide via SN2?

OH- is a base as well...
elimination can occur instead of substitution. plus, it is a 2' alkyl halide with a methyl group near it, adding steric interference to the factors that affect the reaction.

Plus remember that the methyl group can migrate to form a 3' carbocation, thus favoring elimination almost completely.

We have a secondary halide and a strong base: so we will have an E2 reaction and an alkene as the product. Since the leaving group is neutral, Saytzeff rule applies and the major product will have the most substituted double bond, if possible. Drawing a chair conformation for this molecule, with the bromine axial and the methyl equatorial, we see that both carbons 2 and 6 have hydrogens suitably placed for E2 to occur via the preferred antiperiplanar transition state. So, the major product is probably 1-methylcyclohexene, together with some 3-methylcyclohexene (actually, this one is stereochemically favoured).


can someone plzz mail all the inmo problems 2007
c`mon don`t be lazy they are only six

Hello,
    I have already posted them on the respective forums in the Olympiad Section.
Sathej

Just out of curiosity, how many people take part of INMO?

In Brazil, there are two Olympiads: one had around 300000 participants and the other, for public schools, had around 10000000.

Hi cyshine the INMO is not open for everyone but to only those who have qualified in the RMO. So around 350 students participate in the INMO of which around 35 are ultimately selected.


Ξέρει κανείς που μπορώ να προμηθευτώ ή να παραγγείλω τα 4 τεύχη του περιοδικού Απολλώνιος?Ευχαριστώ πολύ...

Στο τεύχος 3 (Απρίλιος 2004) που έχω στα χέρια μου  έχει την παρακάτω διεύθυνση επικοινωνίας :
 ΙΩΣΗΦΙΔΗΣ ΝΙΚΟΣ , ΤΡΕΜΠΕΣΙΝΑΣ 6 , 59100 ΒΕΡΟΙΑ
  ΤΗΛΕΦΩΝΑ: 23310-20143 , 23310-29504
  FAX : 23310 - 20408   E-MAIL : iossifid@otenet.gr


From the set {1, 2,..., 2n} is chosen a subset S of n+1 elements. Prove that exist x,y∈S such that y | x.

[quote="s1ck_psych0"]From the subset {1, 2,..., 2n} is chosen a subset S of n+1 elements. Prove that exist x,y∈S such that y | x.[/quote]

Every number in the set can be written as $2^pk$ with $k \leq n$. If you choose $n+1$ numbers then there will be two with equal $k$. One of these divides the other.

Daniel


[quote="mathwizarddude"]Steam at $ 100^{\circ} C$ is bubbled into 0.250 kg of water at $ 20^{\circ} C$ in a calorimeter cup.  How much steam will have been added whe the water in the cup reaches $ 60^{\circ}C$? (Ignore the effect of the cup.)

It seems to me that the ans is just $ \frac {c_w*0.25kg*(60C \minus{} 20C)}{c_s*(100C \minus{} 60C)}$(c is the specific heat), but do we have to consider latent heat since the temperature will not change as the steam is condensing into water?[/quote]The steam looses a lot of heat as it condenses into water (latent heat), and then looses a bit more heat as the condensed water cools a bit. I bet the latent heat part is a larger quantity than the part for the cooling of the condensate.


Omul asta (Marius Ghergu) propune numai probleme grele.Spre exemplu, urmatoarea:
   Punctele spatiului se coloreaza cu $\ n\geq 2$ culori date.Aratati ca putem gasi intotdeauna un triunghi nedegenerat avand ortocentrul diferit colorat fata de fiecare din varfurile triunghiului.

sunt tare curios sa vad si eu o rezolvare eleganta si scurta pentru cazul $n=2$, ca pentru $n>2$ e imediat. eu am o solutie destul de lunga si complicata, nici macar nu stiu sigur daca este corecta.

Scuze, Lucian, dar nu pot sa dau o solutie la aceasta problema.Pur si simplu rezolvarea e foarte mare.Sunt cam
doua pagini de caiet studentesc! :ninja: Imens!Nu pot sa o postez. :( Poate pentru cazul $\ n\geq 3$.

incearca macar sa schitezi ideile principale..
eu am facut asa
cazul $n\geq3$
[hide="solutie"]exista punctele $A,B,C$ astfel incat au doua cate doua culori diferite. (deoarece pentru orice culoare exista cel putin un punct colorat cu acea culoare). fie $H$ orocentrul triunghiului $ABC$. aunci, exista un varf al triunghiului sa fie colorat diferit fata de $H$. pesupunem ca acel varf este $A$. Atunci, deoarece $A$ este colorat diferit fata de $B,C,H$ si este ortocentrul triunghiului $BCH$, rezulta concluzia. Observatie, daca triunghiul $ABC$ este dreptunghic, atunci alegem alt punct $C$. Multimea punctelor $C$ colorate diferit de $A$ si $B$ astfel incat triunghiul $ABC$ este dreptunghic sau degenerat nu afecteaza problema, deoarece este compusa din maxim drepte, deci ramane "destul" din plan.[/hide]
cazul $n=2$
[hide="solutie"]vom face intai urmatoarele observatii
pentru inceput, vom defini colorarea ca fiind $f: \mathcal{P}\rightarrow\left\{0,1\right\}$, unde $\mathcal{P}$ este planul.
fie acum un cvadruplet de puncte $(A,B,C,D)$ astfel incat $D$ este ortocentrul triunghiului $ABC$. atunci oricare punct din cvadruplet este ortocentrul celorlalte trei. de aici va rezulta ca fie $f(A)=f(B)=f(C)=f(D)$, fie doua din aceste puncte sunt colorate cu o culoare si celelalte doua cu cealalta culoare. (adica nu putem avea exact trei puncte din cele patru colorate la fel). Pe scurt, aceasta conditie poate fi scrisa ca $f(A)+f(B)+f(C)+f(D)=0$ (adunarea este facuta modulo $2$, adica $1+1=0$)
fie acum un paralelogram $ABCD$. vom alege un punct $O$ pe mediatoarea lui $AD$, astfel incat $OA>AB$. fie punctul $Q$ din plan astfel incat $\overrightarrow{OQ}=\overrightarrow{AB}$. fie acum un punct $X$ pe mediatoarea lui $OQ$, si $Y$ simetricul sau fata de $OQ$. atunci $\overrightarrow{OX}+\overrightarrow{OY}=\overrightarrow{OQ}$. cum aceasta relatie are loc pentru orice $X$, alegem acel $X$ pentru care $OX=OA$. (acest punct exista din conditia $OA>AB$, adica $OA>OQ$). fie $Y$ simetricul lui $X$ fata de $OQ$. atunci avem $OA=OD=OX=OY$, si $\overrightarrow{OX}+\overrightarrow{OY}=\overrightarrow{OQ}=\overrightarrow{AB}$. fie $H$ si respectiv $K$ ortocentrele triunghiurilor $XYA$ si $XYD$. atunci, deoarece $O$ este centrul cercului determinat de punctele $A,D,X,Y$, avem relatiile $\overrightarrow{OH}=\overrightarrow{OX}+\overrightarrow{OY}+\overrightarrow{OA}$ si respectiv $\overrightarrow{OK}=\overrightarrow{OX}+\overrightarrow{OY}+\overrightarrow{OD}$ . dar cum $\overrightarrow{OX}+\overrightarrow{OY}=\overrightarrow{AB}=\overrightarrow{DC}$, avem ca $\overrightarrow{OH}=\overrightarrow{OA}+\overrightarrow{AB}=\overrightarrow{OB}$ si analog $\overrightarrow{OK}=\overrightarrow{OC}$. deci $H=B$ si $K=C$. fie acum cvadrupletul $(A,B,X,Y)$. avem ca $f(A)+f(B)+f(X)+f(Y)=0$. analog pentru cvadrupletul $(C,D,X,Y)$, avem $f(C)+f(D)+f(X)+f(Y)=0$. adunand aceste doua relatii vom obtine $f(A)+f(B)+f(C)+f(D)+2(f(X)+f(Y))=0$, dar cum $2(f(X)+f(Y))=0$, rezulta ca $f(A)+f(B)+f(C)+f(D)=0$. deoarece paralelogramul $ABCD$ a fost ales arbitrar, rezulta ca aceasta proprietate este adevarata pentru orice paralelogram.
fie acum punctele $M,N,P$ in plan, astfel incat $f(M)=0$ si $f(N)=f(P)=1$. (evident acestea exista). fie $Q$ astfel incat $MNQP$ este paralelogram ($NP$ este diagonala). atunci, rezulta ca $f(Q)=0$. construim si $S,T$ mijloacele lui $[MN]$ si $[PQ]$. atunci este adevarat ca $MTQS$ este paralelogram, dar si $STQN$ este parelelogram. deci $f(M)+f(T)+f(S)+f(Q)=f(S)+f(T)+f(Q)+f(N)=0$. cum $f(M)=f(Q)=0$, $f(N)=1$, rezulta ca $f(T)+f(S)=f(T)+f(S)+1$, adica $0=1$, contradictie. aceasta ultima contradictie incheie problema.
am atasat si niste desene mai jos[/hide]

...imi pare rau, dar la cazul $\ n=2$ nu am inteles asa bine.In primul rand eu nu stiu inca geometria vectoriala.
Cat despre primul caz, este OK.Multumesc!

pai am zis ca solutia mea e lunga si complicata...
zi si tu ideile de baza ale solutiei pe care o stii


i dont think there is any exception for any seniors about the top12 rule... if there is, then any seniors would want it.... btw, USAMO looks good on the college app... i forgot... that you have already applied.

[quote="wow346"]I got into the USAMO for the first time this year and I'm a senior. Does this mean there's practically no way for me to go to MOSP unless I make the top 12? If so, it doesn't seem like there's much point in even taking the test...[/quote]

Well, you're right about MOP...but I think you should still take it for the heck of taking it. ^_^

Probably the easiest year in recent history to try to get in the top 12 though.... A big zero returning IMO team members and like one or two returning winners.

Ameya will return; I think that's the only winner...

Brian Lawrence (2nd IMO Alternate last year) would return too...

for me, getting into MOSP is not the only reason to take the USAMO.

I dont know why you said there is not much point in even taking the exam.

On a related topic, if as a senior I qualify for the USAMO, is there any way this can help me in getting into a college, since I will have already applied?  Maybe a transfer after the spring semester?

[quote="wow346"]I got into the USAMO for the first time this year and I'm a senior. Does this mean there's practically no way for me to go to MOSP unless I make the top 12? If so, it doesn't seem like there's much point in even taking the test...[/quote]

According to the letter I got, it's the top 30 who go to MOSP.  It's just the top 12 that have a chance for IMO.

[quote="dts"][quote="rrusczyk"]That top 12 has been adjusted in the past for things like Canadians in the top 12 (i.e. students who are not eligible for the US IMO team).  They may also bring in students who are tied for the last slot, but I'm uncertain of that.[/quote]

Err... yes.  It really says approximately 30 students, plus about 30 9th-and-below graders.[/quote]

I thought it is only 9th graders(and possibly 10th graders)

[quote="jackb1115"]In MOP selection, do lower classmen have a better chance of getting in.  I'm a spoh, so if I score, say 50th overall, but in the top 5 sophs, what do they look at ... ?

JB[/quote]

I'm pretty sure they don't. They look at the MOP index = (USAMO index)+9*(USAMO score), from which they take the top 12 and then the top about 18 non-seniors. But top 50th overall (including seniors that are not winners) I would think has a good chance of making MOP anyway. So in essence, underclassmen have no advantage over juniors but do over seniors.

[quote="rrusczyk"] They may also bring in students who are tied for the last slot, but I'm uncertain of that.[/quote]

That was indeed the case last summer. I think it was a pretty good thing too.

For 11th graders or below, you can pretty much count on making MOP if you're within the top 30 - 35 scores (due to high-scoring, non-winner seniors, qualifiers without permanent residency in the U.S., Canadians and other international writers, etc.). Below that, it's a tough call.

[quote="Geehoon"]in order for seniors to make into MOSP,

top 12 among seniors or just top 12 among all USAMO takers?[/quote]

Top 12 among all USAMO takers.

[quote="paladin8"][quote="MithsApprentice"]For 11th graders or below, you can pretty much count on making MOP if you're within the top 30 - 35 scores (due to high-scoring, non-winner seniors, qualifiers without permanent residency in the U.S., Canadians and other international writers, etc.). Below that, it's a tough call.[/quote]

Only 30-35? Don't they take 30 people total? I would think it would be a little lower than that...[/quote]

60 counting the freshmen. What Miths was pointing to is that around 5 of these top 35 USAMO scorers will not be able to go to MOP.

[quote="paladin8"]Yes I understand that, but I would suspect more than 5 to be in the cateogories he mentioned...  I could be wrong, though.[/quote]

We can't always assume a full 30 will be invited. My 30 - 35 number is based on the case of 24 students being invited. If it is indeed 30, you can probably expect to make MOP if you're within 45 or so.

[quote="Geehoon"]I got two answers for my question.

Im assuming picounts corrected what xxreddevilzxx had answered?[/quote]

Yes. A senior must be a USAMO winner (top 12 among all test takers) to go to MOP.

[quote="Geehoon"]I got two answers for my question.

Im assuming picounts corrected what xxreddevilzxx had answered?[/quote]

indeed


irrational number
continued fraction
calculus
calculus computations

let $ \alpha$ be a irrational number, try to prove:
$ \exists m,n \in \mathbb{N}$, make $ m\alpha \minus{}n<\epsilon$ $ (\epsilon>0)$

[quote="Extremal"][quote="tobeno_1"]let $ \alpha$ be a irrational number, try to prove:
$ \exists m,n \in \mathbb{N}$, make $ m\alpha \minus{} n < \epsilon$ $ (\epsilon > 0)$[/quote]
It's follows from Dirichlet theorem . also continued fraction expansion.[/quote]
can you tell me something more about it?


What do you think about taking both AIME classes in the fall. Do you think this would be to much, it would be too confusing ect... like are there any cons besides time if you take both classes

I do not think that it's a bad idea.  Both classes cover completely different topics.

However, the things covered are geared more toward one wishing to score above 8 on the AIME, so you might want to check the difficulty.

[quote="AIME15"]However, the things covered are geared more toward one wishing to score above 8 on the AIME, so you might want to check the difficulty.[/quote]

I am not sure what you mean by that. Do you mean one who can get to AIME pretty easily and wants to get an 8+ or one that wants to get an 8+ on AIME?

If you can get ~1-3 problems on AIME, you'll have to be a really, really, really, [b]REALLY[/b] faster learner to keep up with the class.

Doesn't the class description state that the class is intended for students who score ~1-2 on the AIME? I'm probably wrong so don't quote me on this one  :lol:

[quote="AIME15"]The class is intended for students who score [b]at least[/b] 1-2 on the AIME, but if a student routinely scores 11 or higher, they'd probably not learn too much from this class.[/quote]

Indeed, if you're notching 11+ consistently, you shouldn't be doing AIME problems much anymore.  Check out WOOT instead.


geometry
3D geometry
tetrahedron
inequalities
rotation
geometric transformation
sphere

Given a tetrahedron $ABCD.$ Let $x=AB\cdot CD,$ $ y=AC\cdot BD$ and $z=AD\cdot BC.$ Prove that there exists a triangle with the side lengths $x,$ $y$ and $z.$

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[quote="grobber"]This is simply Ptolemy's inequality, which holds for a tetrahedron as well (a tetrahedron is just another kind of "quadrilateral").[/quote]

I bet the usual proof of the planar Ptolemy inequality won't hold for this case. :D

  Darij

I wasn't thinking of a direct, Ptolemy-like proof.

I'll try to justify it, but I doubt I'll make any sense :D:

Assume $AC\cdot BD\ge AB\cdot CD+AD\cdot BC$. The triangle $BDA$ can be rotated around $BD$ in two positions s.t. $A$ reaches positions $A',A''$, coplanar with $B,C,D$. I claim that one of $CA',CA''$ is $>CA$. After showing this, in the plane $BCD$ we will have $A^*C\cdot BD>A^*B\cdot CD+A^*D\cdot BC$, where $*$ is either $'$ or $''$ (because $A^*B=AB,A^*D=AD$).

Consider the circle described by $A$ during a full rotation around $BD$, and let $C_{\perp}$ be the projection of $C$ on the plane of this circle. $A'A''$ is the line passing through $C_{\perp}$ and the center of the circle, and it's well-known that the maximal distance between a fixed point $C_{\perp}$ and a point on a fixed circle is reached for a point on the line connecting $C_{\perp}$ and the center of the circle. Assume $C_{\perp}A'$ is this maximal distance. This means that $CX=\sqrt{CC_{\perp}^2+C_{\perp}X^2}$ reaches its maximum when $X=A'$. In particular, $CA'>CA$.

Grobber, I really admire your spatial imagination which makes you able to imagine such things which I can't even properly draw on a piece of paper! Yes, your solution does make sense and it is very beautiful.

Actually, when I first saw the problem, I thought what a pity it is that in $ \mathbb{R}^3$, we don't have something like complex numbers to make such a problem trivial. Anyway, now I see that the problem is not too difficult to solve algebraically:

[b]Lemma 1.[/b] If u, v and t are real numbers, then $ \left(u^2 \plus{} t^2\right)\left(v^2 \plus{} t^2\right)\geq \left(uv \plus{} t^2\right)^2$.

[i]Proof.[/i] We have $ \left(u^2 \plus{} t^2\right)\left(v^2 \plus{} t^2\right) \minus{} \left(uv \plus{} t^2\right)^2 \equal{} \left(t\left(u \minus{} v\right)\right)^2$, what is clearly $ \geq 0$. This proves Lemma 1.

[EDIT: As an alternative, just notice that Lemma 1 is a particular case of Cauchy-Schwarz.]

Now, in our problem, we have to prove that there exists a triangle with sidelengths x, y, z. This is equivalent to showing the three inequalities y + z > x, z + x > y and x + y > z. Actually, we will only show the inequality x + y > z; the other two inequalities will follow analogously.

Since $ x \equal{} AB\cdot CD$, $ y \equal{} AC\cdot BD$ and $ z \equal{} AD\cdot BC$, proving x + y > z reduces to showing $ AB\cdot CD \plus{} AC\cdot BD > AD\cdot BC$.

Let P be the orthogonal projection of the point D on the plane ABC. Then, by the Pythagoras theorem for the right-angled triangles APD, BPD and CPD, we have $ AD \equal{} \sqrt {AP^2 \plus{} PD^2}$, $ BD \equal{} \sqrt {BP^2 \plus{} PD^2}$ and $ CD \equal{} \sqrt {CP^2 \plus{} PD^2}$. Thus, the inequality in question, $ AB\cdot CD \plus{} AC\cdot BD > AD\cdot BC$, rewrites as

$ AB\cdot\sqrt {CP^2 \plus{} PD^2} \plus{} AC\cdot\sqrt {BP^2 \plus{} PD^2} > \sqrt {AP^2 \plus{} PD^2}\cdot BC$.

In order to prove this inequality, we square it (yes, both sides are nonnegative... ;) ):

$ \left(AB\cdot\sqrt {CP^2 \plus{} PD^2}\right)^2 \plus{} \left(AC\cdot\sqrt {BP^2 \plus{} PD^2}\right)^2$
$ \plus{} 2\cdot AB\cdot\sqrt {CP^2 \plus{} PD^2} \cdot AC\cdot\sqrt {BP^2 \plus{} PD^2} > \left(\sqrt {AP^2 \plus{} PD^2}\cdot BC\right)^2$.

In other words, we have to show:

$ AB^2\cdot\left(CP^2 \plus{} PD^2\right) \plus{} AC^2\cdot\left(BP^2 \plus{} PD^2\right)$
$ \plus{} 2\cdot AB\cdot AC\cdot\sqrt {\left(CP^2 \plus{} PD^2\right)\left(BP^2 \plus{} PD^2\right)} > \left(AP^2 \plus{} PD^2\right)\cdot BC^2$.

But, applying Lemma 1, we see that $ \left(CP^2 \plus{} PD^2\right)\left(BP^2 \plus{} PD^2\right) \geq \left(CP\cdot BP \plus{} PD^2\right)^2$, so that

$ \sqrt {\left(CP^2 \plus{} PD^2\right)\left(BP^2 \plus{} PD^2\right)} \geq \sqrt {\left(CP\cdot BP \plus{} PD^2\right)^2} \equal{} CP\cdot BP \plus{} PD^2$,

and it remains to show

$ AB^2\cdot\left(CP^2 \plus{} PD^2\right) \plus{} AC^2\cdot\left(BP^2 \plus{} PD^2\right)$
$ \plus{} 2\cdot AB\cdot AC\cdot\left(CP\cdot BP \plus{} PD^2\right) > \left(AP^2 \plus{} PD^2\right)\cdot BC^2$.

But this is just the sum of the two inequalities

$ AB^2\cdot CP^2 \plus{} AC^2\cdot BP^2 \plus{} 2\cdot AB\cdot AC\cdot CP\cdot BP \geq AP^2\cdot BC^2$

and

$ AB^2\cdot PD^2 \plus{} AC^2\cdot PD^2 \plus{} 2\cdot AB\cdot AC\cdot PD^2 > PD^2\cdot BC^2$;

the first of these two inequalities is just the square of the inequality $ AB\cdot CP \plus{} AC\cdot BP \geq AP\cdot BC$, which is the Ptolemy inequality applied to the quadrilateral ABPC, and the second of these inequalities, divided by $ PD^2$, takes the form

$ AB^2 \plus{} AC^2 \plus{} 2\cdot AB\cdot AC > BC^2$,

what is clearly the square of the triangle inequality AB + AC > BC. This completes the proof of our inequality.

  Darij

[quote="darij grinberg"]Grobber, I really admire your spatial imagination which makes you able to imagine such things which I can't even properly draw on a piece of paper! Yes, your solution does make sense and it is very beautiful.
[/quote]

Thanks! I just wanted to say that I couldn't make a proper drawing for this one either :D.

Here's another approach:

Let $S$ be a sphere passing through the points $B,C,D$ and intersecting the line segments $AB,AC,$ and $AD$ at the points $B^{\prime},C^{\prime},$ and $D^{\prime},$ respectively. Clearly, the quadrilateral $BCC^{\prime}B^{\prime }$ is cyclic, so triangles $ABC$ and $AC^{\prime}B^{\prime}$ are similar. This yields
\[
\frac{B^{\prime}C^{\prime}}{BC}=\frac{AC^{\prime}}{AB}.
\]
The power of a point theorem gives
\[
AC^{\prime}\cdot AC=AB^{\prime}\cdot AB=AD^{\prime}\cdot AD=r,
\]
therefore
\[
B^{\prime}C^{\prime}=\frac{rBC}{AB\cdot AC}.
\]
Similarly,
\[
B^{\prime}D^{\prime}  =\frac{rBD}{AB\cdot AD},
\]
\[C^{\prime}D^{\prime}  =\frac{rCD}{AC\cdot AD}.
\]
The triangle inequality
\[
B^{\prime}C^{\prime}>B^{\prime}D^{\prime}+C^{\prime}D^{\prime}
\]
becomes
\[
\frac{rBC}{AB\cdot AC}>\frac{rBD}{AB\cdot AD}+\frac{rCD}{AC\cdot AD},
\]
which is equivalent to
\[
BC\cdot AD>BD\cdot AC+CD\cdot AB,
\]
as desired.

A nice problem related to this one is:

In the tetrahedron $ABCD$ the angles between the opposite sides are equal. Prove that they are right angles.

Hint: Use the well known identity
\[
\overrightarrow{AB}\cdot\overrightarrow{CD}+\overrightarrow{AC}\cdot
\overrightarrow{DB}+\overrightarrow{AD}\cdot\overrightarrow{BC}=0,
\]
and express the dot product using the common angle.

[quote="enescu"]Here's another approach:

Let $S$ be a sphere passing through the points $B,C,D$ and intersecting the line segments $AB,AC,$ and $AD$ at the points $B^{\prime},C^{\prime},$ and $D^{\prime},$ respectively. Clearly, the quadrilateral $BCC^{\prime}B^{\prime }$ is cyclic, so triangles $ABC$ and $AC^{\prime}B^{\prime}$ are similar. This yields
\[
\frac{B^{\prime}C^{\prime}}{BC}=\frac{AC^{\prime}}{AB}.
\]
The power of a point theorem gives
\[
AC^{\prime}\cdot AC=AB^{\prime}\cdot AB=AD^{\prime}\cdot AD=r,
\]
therefore
\[
B^{\prime}C^{\prime}=\frac{rBC}{AB\cdot AC}.
\]
Similarly,
\[
B^{\prime}D^{\prime}  =\frac{rBD}{AB\cdot AD},
\]
\[C^{\prime}D^{\prime}  =\frac{rCD}{AC\cdot AD}.
\]
The triangle inequality
\[
B^{\prime}C^{\prime}>B^{\prime}D^{\prime}+C^{\prime}D^{\prime}
\]
becomes
\[
\frac{rBC}{AB\cdot AC}>\frac{rBD}{AB\cdot AD}+\frac{rCD}{AC\cdot AD},
\]
which is equivalent to
\[
BC\cdot AD>BD\cdot AC+CD\cdot AB,
\]
as desired.[/quote]

Thanks, this is a beautiful solution, and moreover, it is similar to the famous inversion proof of Ptolemy's inequality in $\mathbb{R}^2$ !

[quote="enescu"]A nice problem related to this one is:

In the tetrahedron $ABCD$ the angles between the opposite sides are equal. Prove that they are right angles.

Hint: Use the well known identity
\[
\overrightarrow{AB}\cdot\overrightarrow{CD}+\overrightarrow{AC}\cdot
\overrightarrow{DB}+\overrightarrow{AD}\cdot\overrightarrow{BC}=0,
\]
and express the dot product using the common angle.[/quote]

Indeed, this is already almost the whole solution. In fact, since we are given that the angles between the opposite edges of the tetrahedron ABCD are all equal, we can denote them by $\phi$, and then the scalar products of these opposite edges are

$\overrightarrow{AB}\cdot\overrightarrow{CD} = \pm AB\cdot CD\cdot\cos\phi = \pm x\cdot\cos\phi$;
$\overrightarrow{AC}\cdot\overrightarrow{DB} = \pm AC\cdot BD\cdot\cos\phi = \pm y\cdot\cos\phi$;
$\overrightarrow{AD}\cdot\overrightarrow{BC} = \pm AD\cdot BC\cdot\cos\phi = \pm z\cdot\cos\phi$

(the $\pm$ signs are due to the fact that we don't know the exact orientation of the angles).

Thus, the identity

$\overrightarrow{AB}\cdot\overrightarrow{CD}+\overrightarrow{AC}\cdot\overrightarrow{DB}+\overrightarrow{AD}\cdot\overrightarrow{BC}=0$

rewrites as

$\pm x\cdot\cos\phi \pm y\cdot\cos\phi \pm z\cdot\cos\phi = 0$.

In other words,

$\left(\pm x\pm y\pm z\right)\cdot\cos\phi = 0$.

Now, the term $\pm x\pm y\pm z$ can never be equal to 0; in fact, all eight sums x + y + z, - x + y + z, x - y + z, x + y - z, x - y - z, - x + y - z, - x - y + z, - x - y - z are $\neq 0$, as can be easily seen:

[hide="Proof"]At first, x + y + z > 0 is clear since the numbers x, y and z are positive (remember that $x=AB\cdot CD$, $y=AC\cdot BD$, $z=AD\cdot BC$); this immediately yields - x - y - z = - (x + y + z) < 0.

Since the original problem discussed in this thread shows that there exists a triangle with sidelengths x, y, z, we have y + z > x, so that - x + y + z > 0. This yields x - y - z = - (- x + y + z) < 0. Similarly, x - y + z > 0, - x + y - z < 0, x + y - z > 0, - x - y + z < 0.

This proves that all eight sums x + y + z, - x + y + z, x - y + z, x + y - z, x - y - z, - x + y - z, - x - y + z, - x - y - z are $\neq 0$.[/hide]

So, since we have $\left(\pm x\pm y\pm z\right)\cdot\cos\phi = 0$, while the term $\pm x\pm y\pm z$ can never be equal to 0, we must have $\cos\phi = 0$, and thus $\phi = 90^{\circ}$. In other words, the angle between opposite edges of our tetrahedron ABCD is 90°.

Proof complete.

  Darij

I have looked to Imo compendium. So I know that, this problem is in Shortlist 1973 too. One can find there the solution. But I have made a solution with inversion and this is very synthetic I think.


You are right it is not very difficult...

First 1/m  \leq  2f(m) - f(m 2 ) is equivalent to
1/(m+1) + 1/(m+2) + ...+ 1/m 2   \leq  1 + 1/2 + ... + 1/(m-1).
There are m-1 terms in the rhs and m(m-1) in the lhs. Group the terms in the lhs by groups of m consecutive terms :
S_k = 1/((k+1)m+1) + ... + 1/((k+1)m + m) for k = 0,1,...,m-2
Then S_k  \leq  m*1/((k+1)m+1)  \leq  1/k+1.
Sum the inequalities for k = 0,...,m-2 and you get the result.

In the other hand, 2f(m) - f(m 2 )  \leq  1 is equivalent to
1/2 + 1/3 + ... + 1/m  \leq  1/(m+1) + ... + 1/m 2
Now, there are m(m-1) terms in the rhs and m-1 in the lhs. Form groups in the rhs as above
S'_k = 1/((k+1)m+1) + .... + 1/((k+1)m+m)
Then S'_k  \geq  m/((k+1)m+m) = 1/(k+2).
Sum these inequalities, you get the result.

Pierre.


geometry
inradius
trigonometry
geometry proposed

hay
proov that:
in any triangle
tan A/2 tan B/2 + tanA/2 tan C/2 + tanB/2 tan C/2=1

take a look at $7^{th}$ post of the second page of [url]http://mathlinks.ro/Forum/viewtopic.php?t=151066&start=20[/url]...


Can you please post your proof, Ductrung.

Thank you very much. :)

[quote="manlio"]Can you please post your proof, Ductrung.[/quote]
Not yet, Manlio as I want to see a proper solution.

[quote=ductrung]For any positive reals $ a,b$ and $ c$,
\[ \sqrt[4]{\frac{a^2\plus{}bc}{a(b\plus{}c)}}
\plus{}\sqrt[4]{\frac{b^2\plus{}ca}{b(c\plus{}a)}}
\plus{}\sqrt[4]{\frac{c^2\plus{}ab}{c(a\plus{}b)}}
\ge \sqrt{10\minus{}\frac{3\sqrt[3]{abc}}{a\plus{}b\plus{}c}}
\][/quote]

The following inequality is also true.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
\[ \sqrt[4]{\frac{a^2\plus{}bc}{a(b\plus{}c)}}
\plus{}\sqrt[4]{\frac{b^2\plus{}ca}{b(c\plus{}a)}}
\plus{}\sqrt[4]{\frac{c^2\plus{}ab}{c(a\plus{}b)}}
\ge \sqrt{15\minus{}\frac{18\sqrt[3]{abc}}{a\plus{}b\plus{}c}}
\]

I think the Can's solution is true but his way does not help for a proof of my inequality.

[quote=ductrung]For any positive reals $ a,b$ and $ c$,
\[ \sqrt[4]{\frac{a^2\plus{}bc}{a(b\plus{}c)}}
\plus{}\sqrt[4]{\frac{b^2\plus{}ca}{b(c\plus{}a)}}
\plus{}\sqrt[4]{\frac{c^2\plus{}ab}{c(a\plus{}b)}}
\ge \sqrt{10\minus{}\frac{3\sqrt[3]{abc}}{a\plus{}b\plus{}c}}
\][/quote]

[img]http://s13.sinaimg.cn/middle/006ptkjAzy76rcmt6Ik8c&690[/img]


For $a>0$ and $n\in\mathbb{N}$ which of the following numbers is bigger?

$\frac{a^{n}+1}{a^{n+1}+1}\ \ \ \ \ \text{or}\ \ \ \ \ \ \frac{a^{n-1}+1}{a^{n}+1}$


Let $P$ be the product of those roots of $z^{6}+z^{4}+z^{3}+z^{2}+1=0$ that have positive imaginary part, and suppose that $P=r(\cos{\theta}+i\sin{\theta})$, where $0<r$ and $0 \leq \theta < 360$. Find $\theta$.

Please provide full solution if possible.

Basically I figured it had some sort of nice factorization, and I tried pairing up certain terms to see how they'd factor (and trying to get them into a simpler form by doing the finite geometric sum formula)

the first one  Iactually tried was pairing z^6+z^3+1 together, but that didn't work out (different denominators). Trying to pair z^4+z^3+z^2 together yields z-1 in denominator and simple term in top, so I tried factoring z^6+1 into a form with z-1 in denominator, which gave what you saw in the solution.

Then it happened that z^2-z+1 was a pretty obvious factor and you get the factorization as above

Another thing to notice is that it's a pretty easy guess that roots of unity will PROBABLY be a solution, so if you look at the expression you notice that you have all the terms of the fifth roots of unity (z^4+z^3+z^2+z+1) except the z, but since z^5 = 1 for fifth roots of unity, you have z = z^6 and you get the equation given...then it's obvious that the four nonreal fifth roots of unity are all roots and you immediately get a fourth degree factor.

But mainly it's just playing around

yea the fact that $z^{6}=z$ for 5thz is all you really need.  then just divide $z^{4}+z^{3}+z^{2}+z+1$ into the given expression (you know 1 isnt a factor) to get the remaining quadratic factor.

computer_nuke wrote:Elemennop wrote:[hide]

SO!

The roots are the fifth roots of unity, excluding one, plus the third roots of -1, excluding -1. These are  for . The ones with imaginary part positive have . Thus,

[/hide]

Hmmm, I was looking at the official solutions, it had a different route, but your's is good. What was your thought process? why factor it that way?

what was the official solution...?


function algebra polynomial geometry conics parabola domain

Pretty much any line curve-ish thing is an equation.

lol, idk how to explain your question any better without becoming confusing...

We have just finished reviewing equations in school, how cool is that? Pretty much, an equation is a relationship between variable(s). Most graphing equations involve x, y, z, or f(x)--function of x. Now, like iron fist said, an equation can be a "line curvy thingy" (pretty good description, really, I'm not being sarcastic) or any relation, as long as there is a variable. I hope that helps. :)

Umm...I think we need some clarification. There are a lot of equations for a lot of different graphs. Do you mean for linear equations (straight lines) or polynomials (line curvish thing) or something else?

i agree.. in my opinion every graph can generate some sort of equation  :)

In algebra we learned how to graph lines in y=mx+b form and in geometry we learned how to graph parabolas and ellipses.

[quote="t0rajir0u"][quote="needhelp"]this isn't a problem just a mathematical question...
How do you know if any graph appears to represent an equation?
does it appear to represent an equation when it is a
positive or negative corrilation or does it appear to represent
an equation when its in a straight line... or a slope...?
just need to know... thanks<3[/quote]

In theory, any graph can have an equation.  It might be a [b]very complicated[/b] equation.[/quote]

Yes - in math class I was making a parametric equation for my signature XD.

[quote="fireemblem13"]when we were doing equations and stuff(i seldom pay attention), she mentioned something like a graph cannot be modeled by a function if at any $n$ the horizontal line $x=n$ intersects the graph at more than two points. then later i was like, what about parametric equations. maybe i misheard what she said...[/quote]

She's correct.  A graph cannot be modeled by a [b]function[/b] if this is the case.  That's because a function is of the form

$y = f(x)$,

And cannot return two $y$-values for the same $x$.

Now, the graph can still be modeled by an [b]equation[/b].

(It should be "two or more points," though.)

[quote="fireemblem13"]hmmmm... so you're saying parametric equations aren't functions? i didn't know that...[/quote]

When I say "function," I mean very specifically a mapping from one set (the domain) to another (the range).  A pair of parametric equations consists of [b]two[/b] functions.


geometry
perimeter
conics
ellipse
analytic geometry
perpendicular bisector

What would be the easiest way to prove that in triangle $ ABC$, with a fixed length $ AB$ and fixed height from $ C$, that the perimeter would be minimized when $ AC \equal{} BC$, i.e. when $ C$ is lies on the perpendicular bisector of $ AB$?

[hide="hint"]
Maybe you could consider an elipse with focal points at A and B. For a point on an elipse, P, you know that AP+BP is constant. i.e. the perimeter of that triangle constant...  [/hide]

Below, I've written the points in terms of an arbitrary coordinate system to avoid verbal description; however, it is synthetic in spirit. We can assume those particular coordinates because it is always possible to coordinatize into that form.

[hide="Solution"]Consider points $ A : \; (0,0)$, $ B: \; (x,0)$, $ C : \; (c,h)$ and $ B' : \; (x, 2h)$

$ BC \equal{} B'C$, so $ AC \plus{} CB \equal{} AC \plus{} CB' \geq AB'$, with equality iff. $ C \in \overline{AB'} \implies C \equiv \left( \frac {x}{2}, \; h \right) \implies AC \equal{} CB$[/hide]


Find all  integer number such that : $ \frac{x^7\minus{}1}{x\minus{}1}  \equal{}{y^5\minus{}1}$

a) Bad title!
b) http://www.mathlinks.ro/viewtopic.php?t=138053


linear algebra
matrix
analytic geometry
vector
superior algebra
superior algebra solved

let A=(aij)ij be a real valued nxn matrix such that \sumj=1,n |aij| <1.

prove that det (I-A) <> 0.

source: [i]Mihaly Bencze[/i]

Suppose that ker(A-I)=/=0
Take Y \in ker(A-I) where Y=/=0

 \sum _(j=1 to n)a(i,j)y(j)-y(i)=0 where a(i,j) are coefficients of A
y(i) coordinates of Y

||Y||=max|y(i)|=|y(i_0)|
consider the i_0 row
\sum _(j=1 to n)a(i_0,j)y(j) = y(i_0)
|y(i_0)| \leq \sum _(j=1 to n)|a(i_0,j)|.|y(j)|  \leq  \sum _(j=1 to n)|a(i_0,j)|.|y(i_0)| <  |y(i_0)|
we have a contradiction |y(i_0)| <|y(i_0)|
So ker(A-I)={0}
A-I is invertible

For those of you who don't understand what kerM is, for a matrix M, it is the set of vectors X such that MX=0 (the eigenvectors corresponding to the eigenvalue 0).


inequalities
number theory proposed
number theory

prove that if $ n \in N$ such that $ \gcd(n,2) \equal{} 1$.
then there exists a prime $ p$ for which $ p | 2^{\phi(n)} \minus{} 1$ but $ p \nmid n$.
is it true for all $ m > 1$:
there exists a prime $ p$ for which $ p | m^{\phi(n)} \minus{} 1$ but $ p \nmid n$.

It is true. Let
\[ G_k(x) \equal{} \prod_{d|k}(x^{k/d} \minus{} 1)^{\mu(d)}\]
. Consider
\[ G_{\phi(n)}(m).\]

okay,but i want a solution different of RUST's solution(thank u for your reply!) ,
just using lifting the exponent and inequality, or maybe none of them!(the case $ m\equal{}2$ is iranmo 2-nd round)

For the case $ m\equal{}2$ then just consider $ n\equal{}2^k$ then $ \varphi(n) \equal{} 2^{k\minus{}1}$ and we have $ 2^{\varphi(n)} \minus{} 1 \equal{} 2^{2^{k\minus{}1}} \minus{} 1$, which is an odd number so it must have an odd prime divisor $ p$, which is obviously not a divisor of $ n$.


Question:  Take any tetrahedron and draw the are vectors outwards from the faces.  (The length of each vector is equal to area of its face, and its points perpendicular to the face.)  Prove that these four vectors always add up to zero.

[hide="Hint"]Use the properties of the cross product.[/hide]

[hide="And after that"]Generalize for any polyhedron[/hide]


probability
geometry
conditional probability

[quote="Hychu"]The probability that event A occurs is 3/4; the probability that event B occurs is 2/3. Let p be the probability that both A and B occur. The smallest interval necessarily containing p is the interval:

(A) [1/12,1/2] (B) [5/12,1/2] (C) [1/2,2/3] (D) [5/12,2/3] (E) [1/12,2/3]

p.s. I see a lot of past AMC questions in this website, so I thought it was okay to post this problem here. But if there is some technicality concerning the property rights of AMC problems that I did not know, just let me know and I will erase this post asap.[/quote]

Well in math class we never do probability so I suck at it due to lack of practice...anyway the answer is either A or B because the maximum is $\frac{2}{3}\times \frac{3}{4}= \frac{1}{2}$ I don't know how you get $\frac{1}{12}$ or $\frac{5}{12}$

[hide="hint"] Use P(A or B) = P(A) + P(B) - P(A and B).[/hide]

[quote="cincodemayo5590"][hide="Well..."]
Either the two events are conditionally related or not.

If they are independent:
$P(A \cap B)=P(A) \cdot P(B) = \frac{1}{2}$

If they are:
$P(A | B)= \frac{ P(A \cap B) }{ P(B)}= \frac{ 1 / 2}{ 3 / 4}= \frac{2}{3}$

And I figure that if there's some grey area, i.e. the two events are conditionally related [i]sometimes[/i], $p$ will be in this interval. Thus, the answer is $\left[ \frac{1}{2}, \frac{2}{3}\right] \implies \boxed{\text{C}}$.[/hide][/quote]

I meant to say on the second part, "If they are related, (etc)". But now I can't edit my post. Grrr.

[quote="cincodemayo5590"][hide="Well..."]
Either the two events are conditionally related or not.

If they are independent:
$P(A \cap B)=P(A) \cdot P(B) = \frac{1}{2}$

If they are:
$P(A | B)= \frac{ P(A \cap B) }{ P(B)}= \frac{ 1 / 2}{ 3 / 4}= \frac{2}{3}$

And I figure that if there's some grey area, i.e. the two events are conditionally related [i]sometimes[/i], $p$ will be in this interval. Thus, the answer is $\left[ \frac{1}{2}, \frac{2}{3}\right] \implies \boxed{\text{C}}$.[/hide][/quote]

what is conditionally related?

conditional probability?

wtf are you talking about

this is pie

JUST PIE

lol

$\frac{3}{4}\le P(a \, \, \text{or}\, \, b)=P(a)+P(b)-P(a \, \, \text{and}\, \, b)\le 1$

17/12-p<1

p>5/12
p<2/3

D

Lmao.

Pie tastes good.

I meant conditional probability, but w/e.

 :D

(Oh ya, apple pie is really good. If you didn't know.)

It might help to think of it as a Venn Diagram problem if you don't have the formula memorized.   You have 4 sectors in your diagram: P(A and not B), P(B and not A), P(A and B), P(not A and not B), and they all have to add up to a total of 1.  Then figure out what the minimum and maximum can be in the P(A and B) overlap segment and still have all the numbers work out and add up to 1.  This is that same as what Altheman just said, but maybe a different way of thinking about it that is easier to visualize for some people.

There has not been a single right answer it seems so far in this thread:

[hide]Probability they BOTH happen you multiply. 2/3 and 3/4 multiply to 1/2. While all the intervals contain 1/2, the SMALLEST is B.[/hide]

[quote="mathmom"]Ignite, that was my first thought, but I don't think it's correct.  That's only the case if A and B are independent events, but the problem doesn't say that they are.  Altheman's answer looks correct to me.[/quote]

Well usually it's implied unless stated otherwise =/

[quote]Well usually it's implied unless stated otherwise =/[/quote]
Not really, unless it is explicitly  stated, the  two events can not be taken as independent.

A pedestrian way to think about this is  to ask the  question  in the following way: A Math team  of 12 people consists of 9 boys and 3 girls, and out of 12 - 8 are 8th graders (and 4 are 7th graders) ...  what is the probability that a randomly selected person is a boy from  the 8th grade?
[hide]
(Number of boys in 8th grade can be in the range of 5 to 8)
[/hide]

[quote="Altheman"][quote="cincodemayo5590"][hide="Well..."]
Either the two events are conditionally related or not.

If they are independent:
$ P(A \cap B) \equal{} P(A) \cdot P(B) \equal{} \frac {1}{2}$

If they are:
$ P(A | B) \equal{} \frac { P(A \cap B) }{ P(B)} \equal{} \frac { 1 / 2}{ 3 / 4} \equal{} \frac {2}{3}$

And I figure that if there's some grey area, i.e. the two events are conditionally related [i]sometimes[/i], $ p$ will be in this interval. Thus, the answer is $ \left[ \frac {1}{2}, \frac {2}{3}\right] \implies \boxed{\text{C}}$.[/hide][/quote]

what is conditionally related?

conditional probability?

wtf are you talking about

this is pie

JUST PIE

lol

$ \frac {3}{4}\le P(a \, \, \text{or}\, \, b) \equal{} P(a) \plus{} P(b) \minus{} P(a \, \, \text{and}\, \, b)\le 1$

17/12-p<1

p>5/12
p<2/3

D[/quote]

How do we get the two values of $ P(a\,\,\text{and}\,\, b)$?  (I'm assuming we need two values since we have to find an interval.)

Thanks!

this is so easy

all you have to do is multiply3/4x2/3 to get p.

3/4x2/3=1/2

then you have to find the smallest interval that contains 1/2.

the correct answer would be D because1/2 is between 5/12 and 2/3(8/12) is the smallest interval containing 1/2


algebra polynomial function logarithms trigonometry ratio calculus

Nope. Polynomials are very special functions with very specific properties that most other functions don't have. A few of these properties include:

- Is defined for all real inputs.
  Counterexample: $ \ln x$ is not defined at $ x\equal{}0$.

- Is continuous - loosely speaking, you can draw any portion of the graph without picking up your pen.
  Counterexample: $ \frac{1}{x}$ is not continous for $ x$ near $ 0$.

- A finite, positive number of roots, real or complex.
  Counterexample: $ 2^x$ has no roots - there are no real or complex numbers that will make this function 0.
  Counterexample: $ \sin x$ has an infinite number of roots.

- Is differentiable - that is, it has a unique tangent line at every point.
  Counterexample: $ |x|$ has a corner point at $ x\equal{}0$, so we cannot take a tangent line

- The ratio $ \frac{f(x\plus{}n)}{f(x)}$ converges to $ 0$ as $ x$ increases without bound, and as $ x$ decreases without bound, for all real $ n$.
  Counterexample: $ f(x) \equal{} 2^x$ has $ \frac{f(x\plus{}1)}{f(x)} \equal{} 2$ for all $ x$.

In calculus, you will learn about Taylor's Theorem, which let's us approximate certain kinds of non-polynomial functions with a polynomial.

[quote="binomial_4eva"]prove or disprove that any formula can be turned into a polynomial.[/quote]
I wonder what this even means...
What's a formula, and what is "can be turned"...

[quote="The Zuton Force"]Nope. Polynomials are very special functions with very specific properties that most other functions don't have. A few of these properties include:

- Is defined for all real inputs.
  Counterexample: $ \ln x$ is not defined at $ x \equal{} 0$.

- Is continuous - loosely speaking, you can draw any portion of the graph without picking up your pen.
  Counterexample: $ \frac {1}{x}$ is not continous for $ x$ near $ 0$.

- A finite, positive number of roots, real or complex.
  Counterexample: $ 2^x$ has no roots - there are no real or complex numbers that will make this function 0.
  Counterexample: $ \sin x$ has an infinite number of roots.

- Is differentiable - that is, it has a unique tangent line at every point.
  Counterexample: $ |x|$ has a corner point at $ x \equal{} 0$, so we cannot take a tangent line

- The ratio $ \frac {f(x \plus{} n)}{f(x)}$ converges to $ 0$ as $ x$ increases without bound, and as $ x$ decreases without bound, for all real $ n$.
  Counterexample: $ f(x) \equal{} 2^x$ has $ \frac {f(x \plus{} 1)}{f(x)} \equal{} 2$ for all $ x$.

In calculus, you will learn about Taylor's Theorem, which let's us approximate certain kinds of non-polynomial functions with a polynomial.[/quote]

ok if not to a polynomial , how about a series? like

$ e^x \equal{} \sum_{n \equal{} 0}^{\infty} {x^n \over n!} \equal{} 1 \plus{} x \plus{} {x^2 \over 2!} \plus{} {x^3 \over 3!} \plus{} {x^4 \over 4!}
\plus{} \cdots$

anyone plz

Look up the terms [i]Taylor series[/i], [i]Laurent series[/i], and [i]Fourier series[/i].

The short answer is no.  For example, the function $ \ln x$ does not have a Maclaurin series (a Taylor series centered around $ 0$).  A Maclaurin series is essentially an "infinite polynomial."

But if i recall correctly a bounded and continouus function on a bounded interval can be approached arbitrarily well by polynomials... some theorem by Weierstrass :)

That'd be the [url=http://mathworld.wolfram.com/WeierstrassApproximationTheorem.html]Weierstrass Approximation Theorem[/url].  Anyhow, the problem with this question is simply that it doesn't make any sense without a definition of "formula" and "can be turned into," as ZetaX noted.


number theory
prime numbers
number theory open

why does P versus NP appear in my right and why is Riemann Hypothesis so important?

hahaha... The Riemann Hypothesis is intricately linked to the Prime Number Theorem, and since prime numbers are like the foundations of number theory, proving (or disproving) the Riemann Hypothesis would give us immense insight into how the prime numbers really behave. So the Riemann Hypothesis is VERY important!!

What I have just written does not do justice as to why the Riemann Hypothesis is SO important... Perhaps you would want to read some text on the Riemann Hypothesis and its history. There are quite a lot of good books out there covering this.

perhaps this might interest you...
http://www.mathlinks.ro/Forum/viewtopic.php?t=35163


[b]A VERY HAPPY NEW YEAR 2009 TO ALL OF YOU !!

MAY THIS YEAR BRING LOTS OF HAPPINESS TO YOU ALL !!

GOOD LUCK FOR THIS NEW YEAR !![/b]

 :)   :wink:

Thanks! Same to all other members of this awesome community!

HAPPY NEW YEAR!!!!!!!!!!!!!!!!!!!!!!

Hope this year brings u a lot of happiness.

Happy new yea

its gonna b a yr of reckoning...

Thank you to you all. :)
And Happy New Year to you all. :)

[quote="gauravpatil"]thanks and many many happy returns of the year to all[/quote]
lol how many of us have been in 2009 before?
P.S. Gaurav meets Gaurav ;-)

HAppy new year to all
may luck be always with you

[b]a very happy new year to all[/b]

regards,
romil

Couldnt stop myself frm posting it.. :D  :)

[url=http://www.servimg.com/image_preview.php?i=12&u=13090220][img]http://i79.servimg.com/u/f79/13/09/02/20/gif10.gif[/img][/url]

[quote="rituraj007"]Couldnt stop myself frm posting it.. :D  :)

[/quote]
what is it a cake or chocolate (actually i cant think beyond this)

i present frm my teacher.. :rotfl:  :rotfl:

never opened it :rotfl:

great then open and tell us? :wink:  :rotfl:

well as i said...its a book..
and she is a teacher who wud kno which book i lik most :P

"100 jokes that will make u roll over"

this was better than choclatesand cakes :wink:  :rotfl:  :rotfl:

the book wud be too thin indeed.......... :rotfl:  :rotfl:  :P

i wud say it wud be too short and thicker than most :rotfl:  :rotfl:  :P

Correction "101 jokes" :rotfl:
well its not too small!!!

she's our maths teacher,a teacher with whom i hav shared[hide="this"]
Calling all math geniuses and logisticians out there ..

A mother is 21 years older than the child.
6 years from now the mother will be 5 times as old as the child.

Question : Where's the father?

Try first, before you check the answer below.

Solution:

The mother is 21 years older than the child.
M = C + 21

In 6 years from now the mum will be 5 times as old as the child.
M + 6 = (C + 6) x 5

Hence,
C + 21 + 6 = (C + 6) x 5
C + 27 = 5C + 30
-3 = 4C
C = -3/4

The child is -3/4 years old, that is, it'll be born in 9 months.

Thus.. by mathematical deduction....

Right now, the father is on top of the mother...
[/hide]joke :P  :rotfl: [/hide]

sharing this with ur teacher??????? :rotfl:  :P


$\{a,b,c\}\subset R,\ a+b+c=1,\ a^2+b^2+c^2=3\Longrightarrow$
$\{a,b,c\}\subset \left[-1,\frac 53\right],\ abc\in \left[-1,\frac{5}{27}\right]\ .$

Let $a+b+c=1,\ a^2+b^2+c^2=3\Longrightarrow b+c=1-a,bc=a^2-a-1$. Discriminant for b and c:  $D=(1-a)^2-4(a^2-a-1)=-3a^2+2a+5\ge0 \Longrightarrow -1\le a \le \frac 53.$ Find $abc=a(a^2-a-1)$ in interval [-1,5/3]. It give abc in [-1,5/27].


integration
inequalities
function
real analysis
real analysis unsolved

Let $ A: L^1[\minus{}1,1] \rightarrow \mathbb{R}$ be defined: $ Ax(t)\equal{}\int_{\minus{}1}^{1}tx(t)dt$. Prove that $ A$ is linear continuous operator on $ L^1[\minus{}1,1]$ and find $ \parallel{}A\parallel{}$
($ L^1[\minus{}1,1]$ is linear normed space with respect to the norm: $ \parallel{}x\parallel{}\equal{}\int_{\minus{}1}^{1}|x(t)|dt$)

Assuming that my calculations are correct we have the following

Let $ a\geq b>0$, let $ k\geq 1$ be a natural number and let $ F$ be the functional defined by  $ F: L[\minus{}b,a]\rightarrow \mathbb{R},$

$ F(f)\equal{}\int\limits_{\minus{}b}^{a}t^{k}f(t)dt$. Then,

$ \parallel{}F\parallel{}_{L[\minus{}b,a]\rightarrow \mathbb{R}}\equal{}a^{k}$.

The simpler way to say this: if $ A: L^1(E)\to\mathbb{R}$ is given by $ Ax\equal{}\int_Eg(t)x(t)\,dt$ for some $ g,$ then the operator norm of $ A$ is $ \|g\|_{\infty},$  and if $ g$ is not essentially bounded, then the operator is unbounded and not defined on all of $ L^1(E).$

That $ |\int_Egx|\le \|g\|_{\infty}\|x\|_1$ is straightforward and shows that $ \|A\|\le\|g\|_{\infty}.$

What about the opposite inequality?

Given $ \epsilon>0,$ there is a set $ B$ of positive measure such that $ |g(t)|>\|g\|_{\infty}\minus{}\epsilon$ on $ B.$ Define $ x(t)\equal{}\frac1{m(B)}\text{sgn}\,(g(t))$ on $ B$ and zero otherwise. Then $ g(t)x(t)\equal{}\frac1{m(B)}|g(t)|$ on $ B$ and zero otherwise, so that

$ \int_Eg(t)x(t),dt\equal{}\frac1{m(B)}\int_B|g(t)|\,dt\ge\|g\|_{\infty}\minus{}\epsilon.$

Since we built $ x$ so that $ \|x\|_1\equal{}1,$ that gives us the inequality we want.

Which all means that the answer to hien's original question is that the operator norm of $ A$ is the $ L^{\infty}$ norm of the function $ t$ on $ [\minus{}1,1],$ namely $ 1.$

What didilica stated falls within the same framework.


AMC
AIME
USA(J)MO
USAMO
trigonometry
number theory
AoPS Books

For those who made USAMO, when you were preparing for the AMC and AIME, did you find it necessary to use any supplementary books besides AoPS Vol 1 & 2 and your school math class? For example, subject specific books on number theory or more in-depth geometry.

Meh school math classes aren't exactly very helpful unless you happen to be very fortunate. For the AMC, supplementary books weren't really helpful but the titu books - 104NT, 103 Trig, and 102 Combo would help for deeper study. The AoPS volume 2 is around harder AMC/medium AIME level so if you're looking higher such as the USAMO, you would need some other books.

I think the mods should make a sticky thread where people post what helps them... so that these threads don't pop up ALL THE TIME (there's nothing wrong with them; just toooooooooooo many!)

also a sticky thread about facts about AMC/AIME/USAMO qualification/scoring/difficulty/etc

@#H3N41 would v1 + v2 contain enough to make usamo (obviously not to do well on it).? what books have you done? (AcoPs after v2 or the titu books first?)

mathemonster: I would say that if you have done the hardest problems in them (some are from MOP and IMO/Bulgarian MO) you may have a fair shot at making USAMO, presuming that you do not make stupid mistakes. This year, I did not do either of the AoPS books and just did a few practice AIMEs and made 5 stupid mistakes. Without them, I could have made USAMO. :dry:

If you entirely understand most of Aops V1, V2, you should be able to make USAMO ( provided you do not make too many stupid mistakes.)  After that, ACoPs, then Engel's Problem Solving Strategies and Titu 102 Combinatorial Problems, etc. Also, try to do most of the problems in these books.

thanks guys. yea i think after i do v2 and the other general books (PSS etc) ill do some of the titu books. @h3n41- wat are u working from this year and wat have u done in the past? - just out of curiousity...


Given the equation \[ y^4 \plus{} 4y^2x \minus{} 11y^2 \plus{} 4xy \minus{} 8y \plus{} 8x^2 \minus{} 40x \plus{} 52 \equal{} 0,\] find all real solutions.


http://www.artofproblemsolving.com/Wiki/index.php/2003_AMC_10A_Problems

Look at problem 17.
[hide="My incorrect solution"]

In the figure DC is a perpendicular bisector and E is the center of the circumcircle. AE and BE are angle bisectors. Therefore $ \angle EAD = \angle EBD = 30$ and we find that $ \triangle AED$ and $ \triangle EDB$ are 30-60-90. If we let AB=S, then AD = A/2. Let $ ED=a$, then we have that $ AD=a sqrt{3}$. Therefore, $ a sqrt{3}=frac{S}{2}$, therefore $ S=2a sqrt{3}$. From the problem, we have that $ 3S=(2a)^2 \pi$. Substituting, we have, $ 6a sqrt{3} = 4a^2 \pi$. Simplifying yields, $ a=\frac{3 sqrt{3}}{\pi}$, which, according to the solution given is incorrect. Can anybody help? Because I checked it 5 (that's right, read it and weep) times and I keep getting the same answer.
[/hide][/hide]

If I can see well, the answer is B, yours is correct.

oh sorry  :blush: wow. this is really bad.


Let $f(n)$ the nearest integer of $\sqrt[4]{n}$. Find

$\sum_{k=1}^{1996} \frac{1}{f(k)}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66888]try searching next time, this is similar[/url]

Why is this a calculus question?


i guess that it should be very easy, but i just don't get it :maybe:
can some body plz help me with
  1 1
  1 1
and
  1 2
  2 1
thanks!  :)


I have a few questions:

What recreational activities are there at Awesomemath?
Is there a lot of homework every day?
How exactly are the lectures and problem sessions formatted?
Will we be able to access computers there?
The website says that internet access will be availible on campus, but not in the dormitories.  If you bring a laptop computer, where can you take it to access the internet?
Are the classes very far-spaced on the campus?
How will they enforce the "lights-out" rule?
How much money should you bring?
If you get accepted and arrive there, will you be provided with a campus map?

apparently you can get internet access, but dont tell anyone because your not supposed to.

[quote="infinity4ever"]
What recreational activities are there at Awesomemath?
[/quote]
Lots and lots.
Everyday from 7-8ish (I think) there were recreational activities. The main ones were frisbee, tennis (you have to find someone to play with, but there were many people who brought rackets).
[quote="infinity4ever"]Is there a lot of homework every day?
[/quote]
It depends on your teacher, in my 3 weeks, 5 out of my 6 teachers didn't assign any homework (other than to review the material each day if you didn't understand them). The other teacher assigned homework twice, each time it was only one problem, and each time it took about half an hour.
[quote="infinity4ever"]How exactly are the lectures and problem sessions formatted?[/quote]
It might've changed from last year, but the teacher would give an hour and a half of lecturing, going through problems, etc. It varied from teacher to teacher, but last year 5 of the six teachers provided problem sets with lots of practice problems. One of them gave really cool packets that went over the topics in detail. Anyways, you would usually work on a few problems given (the number varying based on the difficulty of the topic) during the time, then go over them with the assistant.
[quote="infinity4ever"]
Will we be able to access computers there?
The website says that internet access will be availible on campus, but not in the dormitories.  If you bring a laptop computer, where can you take it to access the internet?[/quote]
Well, if I remember correctly, they provided internet access in the dormitories to everyone who had a laptop, but this could have changed. I don't remember there being anywhere that you could've gone to in order to get intertubes access.
[quote="infinity4ever"]
Are the classes very far-spaced on the campus?
[/quote]
No, they were all in the same building. Walking from apartments to rooms was about 5 minutes, from rooms to cafeteria was about a 10 minute walk.

[quote="infinity4ever"]
How will they enforce the "lights-out" rule?[/quote]
Once you're back to the apartment, you can stay up as late as you want. It depends on the strictness of your mentor (most didn't really care if you stayed out till late), but technically you were supposed to be back in the apartment by 11.
[quote="infinity4ever"]
How much money should you bring?
[/quote]
I only spent 20 bucks the entire trip, but I was stingy in lots of places where I didn't need to be.
You probably should bring about 150 bucks.
[quote="infinity4ever"]
If you get accepted and arrive there, will you be provided with a campus map?
[/quote]
You might as well print one off of the interblag, but during the mandatory scavenger hunt we had, they provided them.

On the issue of homework, I believe teachers were asked later in the camp to give homework problem(s). However, for problems that need to be turned in: if you understood the material, you should get these pretty easily. They're generally a somewhat straightforward application of one of the theorems proved in class, or it may ask you to write down a proof, in your own words, of such a theorem. This is just to check that you actually are getting something out of the class, and this wasn't a problem (no pun intended) at all, other than for a specific person who was in my track.

You may instead be given hard problem(s) just to think about, in which you aren't expected to solve fully, or even make significant progress. But most of the math you'll do outside of class will be because you want to do math, not because you have to do math. Unless you really want to do math, in which you shouldn't be there in the first place.

[quote="infinity4ever"]Thanks for the great responses guys!  :D I just need one more clarification:

So during the problem sessions, do students work individually in a quiet classroom, or do people work together in groups and discuss the solutions to the problems?[/quote]
You sit in the room and quietly do problems (usually), but you're free to collaborate with your peers, or ask them to check a solution, and the like.

I know last year we had Capture the Flag and Manhunt games at like, 9:30 at night.
Awesome stuff.
Volleyball net marks the halfway point :] and i hid in trees.
(note to self: wear bugspray when hiding in pine trees).

[quote="infinity4ever"]Thanks for the great responses guys!  :D I just need one more clarification:

So during the problem sessions, do students work individually in a quiet classroom, or do people work together in groups and discuss the solutions to the problems?[/quote]
Actually now that I think about it, it varies from teacher to teacher, one time we were divided into two teams of 5 (I had a tiny class) and given 3 Olympiad problems and an hour.


[quote="aidan"]Place 2005 points on the circumference of a circle. Two points $ P, Q$ are said to form a pair of [i]neighbors[/i] if the chord $ PQ$ subtends an angle of at most $ 10^{\circ}$ at the center. Find the smallest number of pairs of neighbors.[/quote]

[hide]
I'm thinking something along the lines of grouping 2005 into 36 groups of 55 and 1 group of 54, and placing those groups 360/37 degrees away from each other, evenly spaced around the circle
[/hide]

ahh im rusty, can't think of anything


Is there an infinite sequence of prime numbers $p_1,p_2,...,p_n,p_{n+1},...$such that $|p_{n+1}-2p_n|=1$ for each $n \in N$ ?

Have a look here :

http://www.mathlinks.ro/Forum/viewtopic.php?t=1661&highlight=beautiful

Pierre.


quadratics
algebra
polynomial
algebra unsolved

Prove that if $n\in\mathbb{N}$ then the equation \[x(x+1)\hdots(x+2n-1)+(x+2n+1)(x+2n+2)\hdots(x+4n)=0\] has no real solution.

Substitute $x=y-2n$ to obtain

$(y-1)(y-2)\dots(y-2n)+(y+1)(y+2)\dots (y+2n)=0$

Both products have even number of factors, and obviously, if we expand them, all odd-powered terms will cancel each other out, leaving us with the polynomial which has only even-powered terms with positive coefficients (including the constant term), and such polynomial is obviously always positive.


Prove that for every polygon , there exist at least 2 sides such that the measures a, b of which satisfy the conditions $ a \leq b \leq 2a  $

Let $a_1\leq a_2\leq ...\leq a_n$ be sides of $n$-gon.
Suppose contary. In this case we have:
$2a_1<a_2$;
$2a_2<a_3$;
...
$2a_{n-1}<a_n$.
Sum up and obtain: $2a_1+a_2+a_3+...+a_{n-1}<a_n$. It implies that $a_1+a_2+a_3+...+a_{n-1}<a_n$, i.e. sides of $n$-gon doesn't satisfy triangle inequality. Contradiction.


Given a natural number $m$ ≥ $2$ and a polynomial $P(x)$of positive degree with integer coefficients. Prove that if $P(x)$ has at least three distinct integer roots, then $P(x)+5^{m}$ has at most one integer root.

I'm sorry about my poor English!!!!

Assume we have $m\geq 2$ and $a,b,c,d,e\in\mathbb{Z}$ with $a<b<c$ and $d<e$,
such that $P(a)=P(b)=P(c)=0$ and $P(d)=P(e)=-5^{m}$,
 then $(x-a)(x-b)(x-c)|P(x)$ and $x=d,e$ gives
$d-a,d-b,d-c,e-a,e-b,e-c\in\{\pm 5^{k}: k\in\mathbb{N}_{0}\}$.
So for $x=d-a,y=d-b,z=d-c,t=e-d$ we have
$x,y,z,x+t,y+t,z+t\in\{\pm 5^{k}: k\in\mathbb{N}_{0}\}$ with $x>y>z$ and $t\in\mathbb{N}$.
And now long and tedious but straight forward case work will give a contradiction. (I have found nothing better :oops: )

For $\alpha,\beta\in \{\pm 5^{k}: k\in\mathbb{N}_{0}\},\alpha\not =\beta$ we have $\min\{|\alpha|,|\beta|\}$ the largest power of $5$ dividing $\alpha-\beta$.

For $y>0$ we then have $x,y,x+t,y+t>0$ with $0\not =x-y=(x+t)-(y+t)$
and $y=\min\{|x|,|y|\}=\min\{|x+t|,|y+t|\}=y+t$. Contradiction!
For $y+t<0$ we have $y,z,y+t,z+t<0$ with $0\not =y-z=(y+t)-(z+t)$
and $-y=\min\{|y|,|z|\}=\min\{|y+t|,|z+t|\}=-(y+t)$. Contradiction!

So $y<0<y+t$ and $t=(y+t)-y=5^{r}+5^{s}$ for some $r,s\in\mathbb{N}_{0}$.
And now from $x,y,z,x+t,y+t,z+t\in\{\pm 5^{k}: k\in\mathbb{N}_{0}\}$ follows $x,y,z\in\{-5^{r},-5^{s}\}$,
as every natural number has a unique base $5$ representation.
Contradiction!


function
inequalities
complex numbers
superior algebra
superior algebra unsolved

$f (C,+) \rightarrow (C,+) $ morphism of group, suppose that there exist $k > 0$
such that for any $z \in C$, $ |f(z)| \leq k|z|$

Prove that there exist $a,b \in C$ such that $f(z)=az+bz^{*}$

where $z^{*}$ is the complex conjugate of $z$

Cristinel Mortici,Târgoviste

Put $a=\frac{f(1)-if(i)}2,\ b=\frac{f(1)+if(i)}2$. The function $\tilde f(z)=f(z)-(az+b\bar z)$ satisfies the condition in the hypothesis, and the extra conditions $\tilde f(1)=0,\tilde f(i)=0$. This means that the restriction of $\tilde f$ to $\mathbb Q[i]$ is null, and, since $\tilde f$ is continuous (we have $|\tilde f(z_1)-\tilde f(z_2)|=|\tilde f(z_1-z_2)|\le k|z_1-z_2|$), it means that $\tilde f(z)=0,\ \forall z\in\mathbb C$, so $f(z)=az+b\bar z,\ \forall z\in\mathbb C$.

Is this wrong? It seems too easy to be Ok. :)

An alternative solution

The inequality imply that $f$ is continous  as grobber noticed

$f(z)=f(x+iy)=a(x,y)+ib(x,y)$ with $a,b$ continous real functions

$f$ is a additive morphism so

(1) $a(x+x',y+y')=a(x,y)+a(x',y')$
(2) $b(x+x',y+y')=b(x,y)+b(x',y')$

Get $a(x,y)=Ax+By$, $b(x,y)=Cx+Dy$

$f(z)=(A+iC)x+(B+iD)y$ to finish $2x=z+\overline{z}$, $2iy=z-\overline{z}$


[b]Problem 1.[/b] Let $ ABC$ be a triangle and incircle touch $ BC,CA,AB$ at $ D,E,F$ the excircles touch $ BC,CA,AB$ at $ D',E',F'$ the excircles of $ DEF$ touch $ EF,FD,DE$ at $ A',B',C'$, incircles of $ D'E'F'$ touch $ E'F',F'D',D'E'$ at $ A'',B'',C''$ prove that $ AA',BB',CC'$ and $ AA'',BB'',CC''$ are concurrent.

[b]Problem 2.[/b] Let $ ABC$ be a triangle and incircle touch $ BC,CA,AB$ at $ D,E,F$ the excircles touch $ BC,CA,AB$ at $ D',E',F'$ incircles of $ DEF$ touch $ EF,FD,DE$ at $ A',B',C'$, the excircles of $ D'E'F'$ touch $ E'F',F'D',D'E'$ at $ A'',B'',C''$ prove that $ AA',BB',CC'$ and $ AA'',BB'',CC''$ are concurrent.

They are all particolar case of the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=6579]Cevian Nests Theorem[/url].

Nice thanks, I haven't known this theorem, yet  :) .


[quote="tranthanhnam"]supose that $a,b,c$be positive real numbers .
 prove that ;
 $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}) \geq \frac{9}{1+abc}$[/quote]

Very nice inequality, tranthanhnam!
The hint see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6044


If 40♦ represents a positive three-digit integer with a ones digit of ♦, and 1♦ is a positive two-digit integer with a ones digit of ♦, what value of ♦ makes the equation 40♦ ÷ 27=1♦ true?

[hide=" click to reveal hidden content"]

$ 5$.

I'm afraid I did it by trial and error.

[/hide]

[b]i guessed and checked and got 5.. Hm thanks aime15usamo for the algebraic way :)[/b]


[quote="GameBot"]What is the second number in the row of Pascal's triangle that has 43 numbers?[/quote]

In general, if a row has $ n$ numbers, its second number, assuming it has one, is $ n\minus{}1$. Thus, our answer is $ 42$. Awesome!  :lol:

[quote=ThriftyPiano]THE LIFE THE UNIVERSE AND EVERYTHING[/quote]

Lol yes, but next time refrain from bumping old threads if you don't have a full/new sol

The answer to life, the universe, and e v e r y t h i ng :rotfl:


Is it true that for every natural $ b>1$, there are 2 consecutive Fibonacci numbers, $ F_{m}$ and $ F_{m\plus{}1}$, that have a zero digit in the same place in a base $ b$? (for example, 101 and 1000 for $ b\equal{}2$)

Yes it's true. infact u can make it any place but last digit and infinitely many pairs of consecutive numbers.
for any natural number $ m$ define sequence $ G_n\equal{}F_n\ \mod{m}$ (the reminder of n-th Fibonacci number upon division by m). It's a known result that this sequence is periodical. i will skip this part of the proof (i can write it down later if u want)
now for any base $ b$ consider $ G_{b^2}$. it is periodical therefore the part $ 0,1,1,2,3$ repeats infinitely many times. but if $ F_k\equal{}1\ \mod{b^2}$ it means that $ F_k$ has a digit $ 0$ on the second plase from behind in a base $ b$. and since we have infinitely many pairs of consecutive $ 1,1$ in the sequence $ G_{b^2}$ the result follows.

if u wanna make it k-th place from behind just consider the sequence $ G_{b^k}$


do you have a preference when it comes to pencil or pen?

As long as it's legible, it doesn't matter.

Having said that, pencil solutions often have a tendency to be insufficiently dark.


AMC
USA(J)MO
USAMO
algebra proposed
algebra

The sequence $(a_n)_{n \in\mathbb{N}}$ is defined by $a_1 = 8, a_2 = 18, a_{n+2} = a_{n+1}a_{n}$. Find all terms which are perfect squares.

they really used to give problems like this on olympiads??

Well, I can say "yes" for this rectoric question :P
But at this time the OBM (Olimpíada Brasileira de Matemática) was less difficult (except in some problems...).
And it is, in certain sense, a way to guarrantee no zero points in the exam :).

yeah, as long as the other ones are challenging, which i see some are, it's good.

on our USAMO, the easiest ones (1 and 4) never get quite this easy, lucky us.

It is obviosly $a_n=2^{b_n}3^{c_n}$, were $b_1=3,b_2=1,c_1=0,c_2=2,b_{n+2}=b_{n+1}+b_n, c_{n+2}=c_{n+1}+c_n.$
Term $a_n$ is perfect square if and only if $b_n,c_n$ are even. Sequence $c_n=2F_{n-1}$ is even for all n. Sequence $b_n=F_n+2F_{n-2}, \ (F_{-1}=1,F_0=0).$
It is easy to show, that $F_n$ is even if and only if 3|n. Therefore $a_n$ is square if and only if 3|n.


integration
real analysis
real analysis unsolved

Let the f(x) be bounded and monoton on [0,1]
Prove that $\int_{0}^{1}f(x)dx-\frac{1}{n}\sum_{k=1}^{n}f(\frac{k}{n})=O(\frac{1}{n})$
Do we really nead the f(x) be monoton and bounded?

Rustam bu misol unch qiyin emas.
F(x) funktsiya bu oraliqda monotonligidan  |S-s|→0, bunda S va s mos holda Darbuning yuqori va  quyi yigindilari.bundan osongina kelibchiqadi

[0,1] ni n ta teng bo’lakka bo’lamiz.U holda S-s=(1/n)│f(1)-f(0)│=O(1/n).
(Bunda S va s Darbuning  yuqori hamda quyi yig’indilari.)


calculus
integration
trigonometry
calculus computations

Prove

\[\left[\int_\pi^\infty \frac{\cos x}{x}\ dx\right]^2< \frac{1}{{\pi}^2}\]

I usually stay away from kunny's problems but I can't resist this one.

[hide="The usual ideas."]
Integrate by parts:

$\int\limits_\pi^\infty\dfrac{\cos x}x\,dx=\dfrac{\sin x}x|_{\pi}^{\infty}+\int\limits_{\pi}^{\infty}\dfrac{\sin x}{x^2}\,dx$

$\dfrac{\sin x}x|_{\pi}^{\infty}=0$ and

$|\int\limits_{\pi}^{\infty}\dfrac{\sin x}{x^2}\,dx|< \int\limits_{\pi}^{\infty} \dfrac{|\sin x|}{x^2}\,dx<\int\limits_{\pi}^{\infty}\dfrac1{x^2}\,dx=\dfrac1{\pi}$[/hide]


Ai có sách về dãy số xin đóng góp vào đây . Em cảm ơn nhiều :P

ở đây anh biết vài cuốn:
1)Phương trình sai phân, tác giả Lê Đình Thịnh.
2)Dãy số và các vẫn đề liên quan, tác giả Lê Sáng.
3)Một số bài toán chọn lọc về dãy số, tác giả N.V. Mậu.

Tạm thời chú cứ đọc các cuốn đó đi đã. :)

Hai quyển đầu giờ em mới nghe tên đó . Chẳng biết thế nào nữa .Ai có eBook thì upload lên nha  :P

Thương chú tuổi trẻ lại ham học , anh cho chú file này, nhưng hai cuốn đầu thì hỏi anh Hưng ấy, cái file dưới đây là của Namdung.

Cảm ơn anh nhiều . :P .Vẫn câu cũ :[hide]Ai còn nữa thì upload lên [/hide] :rotfl:

Hình như còn có quyển : [b]Problems in real analysis [/b]nữa thì phải! Không biết có ebook chưa ạ?

[quote="N.T.TUAN"]Chú này chưa vào mục làm quen và giới thiệu hả em? Giới thiệu đi :D Còn bài tập giải tích thì bên DDTH cũng có một cuốn đấy, bản tiếng Việt.  :)[/quote]
Em giới thiệu rồi thầy ạ (là người đề xướng ai đẹp trai nhất! Thầy không nhớ ạ?). Quyển sách : Bài tập giải tích: Số thực, dãy số và chuỗi số đã được dịch ra tiếng Việt rồi! NXB Đại học SP thì phải, nếu bác nào cần thì có thể bảo em photo cho!

Cuốn đó có cả bản Ebook nữa, bên DĐTH có đưa lên rồi.

thế bác the soul of rock dẫn link cho em luôn cái nhỉ  :D

Tìm một post có tác giả là đoàn chi!  :wink:

Giai tich thuc thi tim sach cua W. Rudin ma doc!

ngày xưa mình cũng có hai cuốn đầu!
sau cho tay đệ tử mất!
giờ không biết kiếm ở đâu!
ai biết xin chỉ giáo!

Anh có hai cuốn đầu đấy :D Nhưng không dám chỉ giáo đâu :D .

[quote="N.T.TUAN"]Anh có hai cuốn đầu đấy Nhưng không dám chỉ giáo đâu .[/quote]

Poto cho em cái nha anh Tuân, để lại địa chỉ em gửi tiền và [hide=":rotfl:"]  vào quả bom :starwars:  :pilot: [/hide]<<== có hiden


Does there exist a sequence of $2005$ consecutive positive integers that
contains exactly $25$ prime numbers?

Yes. It is obviosly, because density of prims ---> 0, and $|t(x+1)-t(x)|\le 1$ were $t(x)=\pi(x+2005)-\pi(x)$.


In isosceles triangle $ ABC$, let $ AB\equal{}AC\equal{}5, BC\equal{}6$. Let $ D$ be a point on $ AC$, and $ E$ be a point on $ BD$ such that $ \angle ABD\equal{}\angle ECB$, and$ \angle AEC$ is a right angle. Find $ AD: DC$.

See the attached diagram for additional notation.

$ \angle CAF\equal{}{\pi\over 2}\minus{}\theta\minus{}\phi\land\angle CAE\equal{}{\pi\over 2}\minus{}\theta\implies\angle EAF\equal{}\phi$

$ \angle BAF\equal{}{\pi\over 2}\minus{}\theta\minus{}\phi\implies\angle BAE\equal{}{\pi\over 2}\minus{}\theta\minus{}2\phi$

$ \triangle BAE: \angle AEB\equal{}{\pi\over 2}\plus{}\phi\plus{}\theta$

Sine Law for $ \triangle ADB$: $ {x\over\sin\phi}\equal{}{b\over\sin\delta}$

Sine Law for $ \triangle BCD$: $ {b\minus{}x\over\sin\theta}\equal{}{a\over\sin\delta}$

From those two we get

$ {x\sin\theta\over(b\minus{}x)\sin\phi}\equal{}{b\over a}\implies x\equal{}\frac{b^2\sin\phi}{a\sin\theta\plus{}b\sin\phi}\equal{}\frac{b^2}{b\plus{}a{\sin\theta\over\sin\phi}}\quad(1)$

$ \triangle AFC: \cos(\phi\plus{}\theta)\equal{}{a\over 2b}$
$ \triangle AEC: AE\equal{}b\sin\theta$

Sine Law for $ \triangle AEB$: $ {AE\over\sin\phi}\equal{}\frac{b}{\sin\left({\pi\over 2}\plus{}\phi\plus{}\theta\right)}\equal{}{b\over\cos(\phi\plus{}\theta)}\implies {b\sin\theta\over\sin\phi}\equal{}{2b^2\over a}$

Hence $ {\sin\theta\over\sin\phi}\equal{}{2b\over a}$, so $ (1)$ gives

$ x\equal{}\frac{b^2}{b\plus{}a\cdot{2b\over a}}\equal{}{b\over 3}$, finally yielding $ AD: DC\equal{}1: 2$

Obviously, $E$ is the intersection of the circle tangent at $B, C$ to $AB, AC$ respectively with the circle having $AC$ as diameter. Call $N$ the midpoint of $BC$ and $P\in AN\cap BE$.
By easy angle chase we get that $AP$ is tangent to the circle $\odot (ABE)\implies AP^2=PE\cdot PB$.
Since $AE$ is the symmedian of $\Delta BEC$, this implying that $\angle AED=\angle NEC=\angle CAN$ and $NE\bot BD$, hence $PN^2=PE\cdot PB$ (from right-angled triangle $BNP$), consequently $P$ is the midpoint of $\overline {AN}$; applying [b]Menelaos[/b] to triangle $\Delta ANC$ with the transversal $\overline {BPD}$ we get the required relation.

Best regards,
sunken rock


[quote="GameBot"]Forty-nine of the first 50 positive integers are in a list whose sum is 1257. Which one of the first 50 positive integers is missing from the list?[/quote]

If we added up all the 50 integers, we would get a sum of $ \frac{50 \times 51}{2} \equal{} 1275$. Thus the number missing is $ 1275 \minus{} 1257 \equal{} \boxed{18}$.


probability
integration
logarithms
calculus
geometry
3D geometry
probability and stats

Q#)

if we choose three numbers in the interval $ [a,b]$ , say $ x,y,and \ z$.

what is the probability that :

a) $ P(x > yz)$

b) $ P(x \plus{} y < z)$

c) $ P(xyz < 1)$

where $ a,b \in \mathbb{R}$.

 :D

all these questions are quite the same. You can even easily find the distribution of these type of random variables. For example, to keep things very simple, suppose that $ X,Y$ are uniformly distributed on $ [0;1]$. Then to find the law of $ XY$ you can do like:
$ E[f(XY)] \equal{} \int_{x \in [0;1]} \int_{y \in [0;1]} f(xy) dx dy \equal{} \int_{x \in [0;1]} \left(\int_{u \in [0;1]} f(u) 1_{u<x}(u,x) \frac{du}{x} \right) dx$
so (Fubini):
$ E[f(XY)] \equal{} \int_{u \in [0;1]} f(u) \log(\frac{1}{u}) du$
which says that $ \log(\frac{1}{u}) 1_{u \in (0;1)}$ is the density of $ XY$.

O.k that 's nice ...

but we could solve these parts using the double integral in 3-Dimension.

Could you know how ?

for example ,if we define such as an e.g $ [a,b] \equal{} [0,1]$

Draw cubic with length 1 .in 3-d

Thus, if we need to calculate $ P(z < yx)$ , we could do that  by calculating  the area under the surface of Cube of the plane $ z \equal{} yx$.

What about details ?

Thank you  :wink:

yes, that is the method. If you want the details, just post it in the "Calculus Computations and Tutorials" section  :)

:D

Thank you ....

I'll post it there in next time.

 :wink:


geometry
3D geometry
function
topology
Functional Analysis
advanced fields
advanced fields unsolved

is there a space, which has

a) exactly 2 different preduals ?

b) infinite number of different preduals ?

Are you referring to Banach spaces? What is the definition of predual that you are working with? The definition I take for granted is the following: For Banach spaces $ X$ and $ Y$, we say that $ X$ is a predual of $ Y^\ast$ if $ X^\ast$ is isometrically isomorphic to $ Y^\ast$. I have seen, however, some people leave the word 'isometrically' out of this definition when defining the notion of a predual. Anyway, assuming that you are working with definition I have given, I do not know off the top of my head what the answer to the first question is; perhaps an 'exotic' Banach space not isomorphic to its square (i.e. $ X$ is not isomorphic to $ X \oplus X$) will work... the reason I suggest this is that if a space $ Y$ has (nonisomorphic) preduals $ X$ and $ Z$ then you want to avoid any renorming of, say, $ X \oplus Z$ being a predual of $ Y^\ast$ non-isomorphic to $ X$ or $ Z$. There are some pretty funky Banach spaces out there - eg, the Gowers-Maurey spaces, but whether there exists a non-reflexive space sharing some of their exotic properties I don't know (you are obviously looking for non-reflexive spaces). Perhaps you could play around with the properties of these recently discovered exotic spaces (or other properties for that matter) to see what properties (if any) are sufficient for a Banach space to have precisely two preduals. For example, there are Banach spaces which are isomorphic to their subspaces of finite even codimension but not to their subspaces of finite odd codimension; can such a Banach space have the property that the even and odd codimension subspaces have the same dual? There are Banach spaces isomorphic to their cube and not their square... does this help? I haven't thought about this at any great length, I am just making some naive (and possibly very wrong!) suggestions.

As far as the second question goes, that's easy to answer - the space $ \ell_1$ has uncountably many non-isomorphic pre-duals: these include the spaces $ C(K)$ of scalar-valued continuous functions on a countable compact Hausdorff space $ K$ (see the paper of [url=http://matwbn.icm.edu.pl/ksiazki/sm/sm19/sm1913.pdf]Bessaga and Pelczynski[/url] on this topic); each such $ K$ is homeomorphic to a countable ordinal equipped with its order topology, and so it suffices to consider spaces of continuous functions on countable ordinals. Each isomorphism class amongst these spaces contains only countably many elements, and so since there are uncountably many countable ordinals there are uncountably many isomorphism classes (again, see the paper of Bessaga and Pelczynski for details). There are also preduals of $ \ell_1$ which are not $ C(K)$ spaces; they're constructed from Choquet simplices and are probably best read about from the paper of Benyamini and Lindenstrauss on the topic.

I hope this helps  :)

Philbert


find all answers of$\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$ in$\mathbb{N}$

$\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$
$xz + yz = xy$
$xy - xz - yz = 0$
$xy - xz - yz + z^2 = z^2$
$(x-z)(y-z)=z^2$

So the $x-z$ and $y-z$ have to be divisors of $z^2$ such that $(x-z)(y-z)=z^2$. But I'm not sure can we go any further? For example: $x=2z$ and $y=2z$ and we have infinitely many solutions.  It would be much easier if the $z$ was a given prime number...

EDIT: Ah... http://www.mathlinks.ro/LaTeX/Examples/DiophantineEqn.pdf


geometry
3D geometry
pyramid
geometry unsolved

We have a pyramid ABCD and we know length of its edges: AB, AC, AD, BC, BD, CD;
What's the ABCD volume ?


Will it be possible that Round 3 Problems to be posted during Thanksgiving?

I don't know why but for Round 2, I had very short amount of time to work on the problems and I'm hoping to get some more if I can :D  on Round 3 and Thanksgiving is a good time to do some USAMTS for me.

Just wondering.  :lol:

Doubtful - usually they take at least 1-2 weeks to get the next round up, and getting them up during Thanksgiving weekend would probably mean a lot of other work during that holiday season, which is not likely to be something the USAMTS people would be happy with  :P


How many distinct prime divisors does 1100 have?

Just prime factorize and see how many different primes come up.

11*100
11*2*50
11*2*2*25
11*2*2*5*5

The answer is $ 3$.


Let x,y,z be positive real numbers. Prove that: x/(x+ (x+y)1/2(x+z)1/2) + y/(y+ (y+z)
1/2(y+x)1/2) +z/(z+ (z+x)1/2(z+y)1/2)<= 1.
I've a solution, And You?

I tried to use the same technique as in problem IMO 2001 2(I think):

Let's try to prove that x/(x+(x+y)^1/2(x+z)^1/2)<=x^a/(x^a+y^a+z^a), and also find a. Well, I actually guessed a:1/2 so I tried to prove that:

x^3/2+x^1/2(x+y)^1/2(x+z)^1/2>=x^3/2+x*y^1/2+x*z^1/2 iff

x*(x+y)*(x+z)>=x^2*y+x^2*z+2*x^2*y^1/2*z^1/2 iff

x^3+xyz>=2x^2*y^1/2*z^1/2 which is true by AM-GM between x^3 and xyz.
I hope it's correct.
I'd like to see your solution.

I found this nice solution in book. :D
\[\sum \frac{x}{x+\sqrt{x+y}\sqrt{x+z}}\leq\sum \frac{x}{x+\sqrt{xy}+\sqrt{xz}}=\sum\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}} =1\]


The number of significant digits in the measurement of the side of a square whose computed area is $ 1.1025$ square inches to the nearest ten-thousandth of a square inch is:
$ \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 1$

[hide="Click for solution"]
If it is to the nearest ten-thousandth of a square inch, there are $ 5$ digits or $ \boxed{\textbf{(D)}}$.
[/hide]

Significant digits are any digits that aren't leading zeroes or trailing zeroes in a decimal are counted as significant. You read more about it here: [url]https://en.wikipedia.org/wiki/Significant_figures#Identifying_significant_figures[/url]

[quote=lucasxia01]Significant digits are any digits that aren't leading zeroes or trailing zeroes in a decimal are counted as significant. You read more about it here: [url https://en.wikipedia.org/wiki/Significant_figures#Identifying_significant_figures] Significant Figures/Digits Wikipedia [/url][/quote]

This link takes me back to the forum??

It would be $1.0500$, which in terms of significant digits is actually different, as there are 5 significant digits instead of 3. There needs to be 5 because if we multiply a number with 5 significant digits by another number with 5 significant digits, then the product should have 5 significant digits as well. Since in this case we are given that the product is $1.1025$, which has 5 significant digits, the side lengths must have 5 significant digits, too.


At around what time will be receive our Round 2 scores?

Also, will we know whether or not we qualified for the AIME via USAMTS before AMC 10/12 A starts?

Based on what DPatrick said in the Math Jam, I would guess mid-December.

[quote="123456789"]At around what time will be receive our Round 2 scores?[/quote]We are aiming for the week of December 17-21.

[quote="123456789"]Also, will we know whether or not we qualified for the AIME via USAMTS before AMC 10/12 A starts?[/quote]Hard to say.  It usually takes us about a month to score a round, so the Round 3 scores might be available during the first week of February.  The AMC A date is Feb 12.


When the width of a rectangular prism is increased by $ 2$ cm and the
length is increased by $ 1$ cm, the volume of the resulting cube exceeds
the original volume by $ 96$ cm$ ^3$. Find the number of centimeters in the sum of the length, width, and height of the original box.

Let the side length of the new cube be $ s$ so that
$ s^3 \equal{} s(s \minus{} 1)(s \minus{} 2) \plus{} 96\implies s^3 \equal{} s^3 \minus{} 3s^2 \plus{} 2s \plus{} 96\implies3s^2 \minus{} 2s \minus{} 96$
$ \equal{} (3s \plus{} 16)(s \minus{} 6) \equal{} 0\implies s \equal{} 6\implies(s \minus{} 2) \plus{} (s \minus{} 1) \plus{} (s) \equal{} 3s \minus{} 3 \equal{} \boxed{15}$.


triangle ABC has sides of length AB = 7, AC = 12, and BC = 10. There is a point D on BC such that the circles inscribed in triangles ABD and ACD are both tangent to line AD at a common point E.
Find the length of BD.

[hide]Let the incircle of $ABD$ be tangent to $AB$ and $AD$ at $F$ and $G$, respectively, and the incircle of $CBD$ be tangent to $DC$ and $BC$ at $H$ and $I$.

Let $DE=DH=DG=a$, $BE=BI=BF=b$, $BH=BI=c$, and $AF=AG=d$. (They're equal because tangents to a common point are equal)

So, $2a+c+d=10$, $b+c=7$, and $b+d=12$. Therefore, $b-a=9/2$. This means that $a+d=7.5$ and $a+c=2.5$. Using Stewart's theorem, $7(7.5)+12(2.5)=10((AB)^{2}-7.5(2.5))$. So, $8.25=(AB)^{2}-18.75 \Rightarrow AB=3\sqrt{3}$.[/hide]


limit
real analysis
real analysis unsolved

How may I compute $ \sum_{n\equal{}1}^\infty \frac{n^2}{n!}$?

Try calculating $ \sum \frac{n^2}{n!}x^n$ instead- is this related to a power series you know?

Let's find constants A, B, and C such that $ \frac {n^2}{n!} \equal{} \frac {A}{n!} \plus{} \frac {B}{(n \minus{} 1)!} \plus{} \frac {C}{(n \minus{} 2)!}$.

[hide="Answer"]We have A = 0, B = C = 1, and so $ \frac{n^2}{n!} \equal{} \frac{1}{(n \minus{} 1)!} \plus{} \frac{1}{(n \minus{} 2)!}$.

Let's consider $ S \equal{} \sum_{n\equal{}1}^{\infty} \frac{n^2}{n!}$, $ S' \equal{} \sum_{n\equal{}2}^{\infty} \frac{n^2}{n!}$, $ a_n \equal{} \frac{1}{(n\minus{}1)!}$, $ A_n \equal{} \sum_{k\equal{}2}^na_k$, and $ A_n' \equal{} \sum_{k\equal{}2}^{n}(a_k \plus{} a_{k\minus{}1})$. Then, note that $ S' \equal{} \sum_{n\equal{}2}^{\infty}(a_n \plus{} a_{n\minus{}1})$. But:

$ A_n \equal{} a_2 \plus{} a_3 \plus{} ... \plus{} a_n$

and

$ A_n' \equal{} a_2 \plus{} a_1 \plus{} a_3 \plus{} a_2 \plus{} ... \plus{} a_n \plus{} a_{n\minus{}1} \equal{} a_1 \plus{} 2(a_2 \plus{} a_3 \plus{} ... \plus{} a_{n\minus{}1}) \plus{} a_n$

$ \equal{}> A_n' \equal{} a_1 \plus{} 2A_n \minus{} a_n$.

Hence: $ S' \equal{} \lim A_n' \equal{} a_1 \plus{} 2 \lim A_n \minus{} \lim a_n \equal{} a_1 \plus{} 2 \sum_{n\equal{}2}^{\infty} \frac{1}{(n\minus{}1)!} \minus{} 0$
$ \equal{} 1 \plus{} 2(e\minus{}1) \equal{} 2e\minus{}1$.

But S = S' + 1: so, $ \boxed{S \equal{} 2e}$.[/hide]

or:
answer is
$ (x(e^{x})')'|_{x \equal{} 1}\equal{}2e$


Here is the first problem posted in this forum.   :)

In year $N$, the $300^{th}$ day of the year is a Tuesday. In year $N+1$, the $200^{th}$ day
is also a Tuesday. On what day of the week did the $100^{th}$ day of year $N-1$
occur?
$\text{(A) Thursday (B) Friday (C) Saturday (D) Sunday (E) Monday}$

Let $x$ be the 300th day of year N. 53 weeks ago, that is 1 year and 2 days ago, the day was a Tuesday. So the 363rd day of year $N-1$ was a Tuesday. Backtracking, the 97th day of year $N-1$ is a Tuesday.
98th is a Wednesday
99th is a Thursday
100th is a Friday

The answer is B.


function
topology
Ring Theory
advanced fields
advanced fields unsolved

Let $ C(X)$ be the space of continuous functions on a compact Hausdorff space $ X$. Let $ \Sigma$ be the set of all nonzero homomorphisms $ C(X)\rightarrow\mathbb{C}$ (i.e., the maximal ideal space).

Let $ \mathcal{E}: X\rightarrow\Sigma$ be the evaluation map that sends $ x\mapsto\epsilon_x$ where $ \epsilon_x: f\mapsto f(x)$.

Prove that $ \mathcal{E}$ is a homeomorphism.

note that $ \Sigma$ corresponds to the set of maximal ideals of $ C(X)$, according to gelfand-mazur. the nontrivial part is that E is surjective, i.e. that every maximal ideal $ I$ of $ C(X)$ has the form $ ker(e_x)$ for some $ x \in X$. if the functions in $ I$ have no common zero, every $ x \in X$ possesses a neighborhood $ U_x$ s.t. there is some $ f \in I$ s.t. $ f \neq 0$ on $ U_x$. since $ X$ is compact, we may choose finitely many $ U_x$ which cover $ X$. if $ f_1,...,f_n$ are the corresponding functions, $ f_1^2\plus{}...\plus{}f_n^2 \in I$ has no zero, thus $ I\equal{}C(X)$ is not maximal. thus, there is some $ x$ s.t. $ I \subseteq ker(e_x)$. maximality yields $ I\equal{}ker(e_x)$.

[quote]$ f_1^2 \plus{} ... \plus{} f_n^2 \in I$ has no zero[/quote]
But $ f_i$s are complex-valued.

... well  $ f_1 \overline{f_1}\plus{}...\plus{}f_n \overline{f_n}$.


For every natural number [i]n[/i], s_n = 1 + 1/(1!) + 1/(2!) + ... + 1/(n!).

(a) Show that {s_n} is a cauchy sequence in Q under the euclidean metric for R.
(b) Show that for every natural number q, we have q!(s_q) is an integer and 0 < q!(e - s_q) < 1, where e = summation^infinity_n=0 (1/n!)
(c) Deduce that e is irrational and that Q is not complete.

[hide="Hint for part (a)"]$ s_{m}\minus{}s_{n}\equal{}\sum_{k\equal{}n\plus{}1}^{m}\frac{1}{k!}<\sum_{k\equal{}n\plus{}1}^{m}\frac{1}{n!(n\plus{}1)^{k\minus{}n}}<\sum_{k\equal{}n\plus{}1}^{\infty}\frac{1}{n!(n\plus{}1)^{k\minus{}n}}$

Use what you know about geometric series.[/hide]


If you downloaded the package at sourcefourge then it should come with one.

They update asymptote every week tho


For fixed integer $n$, given $2n$ non-negative reals $\{a_{i}\}, \{b_{i}\}$ such that

$n \sum_{i=1}^{n}a_{i}^{2}-2 \sum_{i\neq j}a_{i}a_{j}\leq n$
$n \sum_{i=1}^{n}b_{i}^{2}-2 \sum_{i\neq j}b_{i}b_{j}\leq n$

Is it true that

$n \sum_{i=1}^{n}(a_{i}-b_{i})^{2}-2 \sum_{i\neq j}(a_{i}-b_{i}) (a_{j}-b_{j}) \leq 2n+2$?

I am also interested in the equality cases.

I think this should hold for all n. I have a solution for n=2,3 but there are so different that i cannot generalize it to n variables. The equality cases have one 1's and the rest 0 for each set $\{a_{i}\}. \{b_{i}\}$, with $a_{j}=1=b_{k}, j\neq k$. For n=2,3 these are the only equality cases, I'm not sure about larger n.

Hi,

The problem as stated is untrue for large values of n, like n>300++. However, i do not have an explicit counterexample. I am still interested in small cases, especially n=4.


Find a reccurence relation for the no of functions from {1,2,3...n} to {1,2,..k} with $f(n) \neq f(1)$, n>2, k>1.  And hence solve it

I don't know how to set up such a recurrence relation. Instead, I think this problem can be solved by counting.
Let $f(1) = i$ and $f(n) = j$ such that $i \neq j$.
There are totally $k \choose 2$ sets of such $\{i,j\}$. But i and j can be switched.
Therefore, there are $2 \cdot {k \choose 2}$  choices for f(1) and f(n).
Then each of  the remaining values in {2,3,4,...,n-1} has to be assigned to any one of {1,2,...,k}.
This accounts for $k^{n-2}$ choices.
Therefore, the total number of functions as requried should be $2 {k \choose 2} k^{n-2}=(k-1)k^{n-1}$.

Anyway, I also want to know the recurrence relation as demanded.


Been sitting with this problem for more than an hour now and just can't seem to work it out... Anyone care to give me a hand?

[size=200]15 = 2t + 9.82t^2 / 2 [/size]
[size=150]where t > 0[/size]

Thanks

hello, do you want to solve this equation
$ 15\equal{}2t\plus{}\frac{9.82t^2}{2}$ by $ t$?
Sonnhard.

This does not belong in the Intermediate Forums.

The quadratic
$ 15 \equal{} \frac{2t\plus{}9.82t^2}{2}$
has a positive root.

The quadratic
$ 15 \equal{} 2t \plus{} \frac{9.82t^2}{2}$
has no real roots.

Yeah, i want to solve the equation. I know enough to see that the solved equation will have 2 answers and that one of the answers (which is the one I want) will be >0...

Sorry if this 'dosen't belong here'... missed the section that explains the proper location for all diffrent types of problems....

The easiest way to solve the quadratic is to use the quadratic formula. Simply put the equation $ 15 \equal{} \frac{2t\plus{}9.82t^2}{2}$ into the standard form $ at^2 \plus{} bt \plus{} c \equal{} 0$ and plug in $ t \equal{} \frac{\minus{}b \pm \sqrt{b^2\minus{}4ac}}{2a}$.

[quote="leoxnlin"]The quadratic
$ 15 \equal{} 2t \plus{} \frac {9.82t^2}{2}$
has no real roots.[/quote] I think you might want to reconsider this.

electerr, most (if not all) of the forums here have a permanent announcement at the top giving guidelines for posting.  This probably should have been placed in the High School Basics forum originally (but don't go post another copy there now! -- probably a moderator will come along and move it at some point).

hello, i have calculated $ t_1\equal{}\frac{10}{491}(\minus{}10\plus{}\sqrt{7645})$ or
$ t_2\equal{}\frac{10}{491}(\minus{}10\plus{}\sqrt{7465})$ and only $ t_2$ is positive.
Sonnhard.

[quote="JBL"][quote="leoxnlin"]The quadratic
$ 15 \equal{} 2t \plus{} \frac {9.82t^2}{2}$
has no real roots.[/quote] I think you might want to reconsider this.
[/quote]

 :oops_sign:
Oh noes, I've gone stupid again.

I got something similar to Dr. Graubner in my first calculations but my book lists the answer as, t=1.56, which is rounded to the 3rd decimal but I can't figure how they got that....

Someone who can figure it out and has the time to show me how they got the answer...?

hello, writing your equation in the form
$ 0\equal{}\frac{491}{100}t^2\plus{}2t\minus{}15$ and using the formula
$ t_{1,2}\equal{}\minus{}\frac{b}{2a}\pm\frac{1}{2a}\sqrt{b^2\minus{}4ac}$ we get
$ t_{1,2}\equal{}\minus{}\frac{2}{\frac{2\cdot491}{100}}\pm\frac{1}{\frac{2\cdot491}{100}}\sqrt{4\minus{}4\cdot\frac{491}{100}(\minus{}15)}$.
Simplifying this you will get the desired solution.
Sonnhard.


Hi everyone: I will appreciate the solution to these 4 problems:

1. An insulated rigid tank is divided into two equal parts by a partition. Initially one part contains 3kg of an ideal gas at 800 kPa and 50 oC, and the other part is evacuated. The partition is now removed, and the gas expands into the entire tank. Determine the final temperature and pressure in the tank.

2. Air at 2MPa and 30 oC is throttled to the atmospheric pressure 100kPa. Determine the final temperature of the air.

3. Air flows steadily through an adiabatic turbine, entering at 150 psia, 900 oF, and 350 psia and leaving at 20 psia, 300 oF, and 700 ft/s. The inlet area of the turbine is 0.1ft2. Determine:
(A) The mass flow rate of the air.  (B) The power output of the turbine.

4. A Carnot heat engine operates between a source at 1800 R and a sink at 440 R. If the heat engine is supplied with heat at a rate of 1200 Btu min, determine:
(A) The thermal efficiency.  (B) The power output  of this heat engine.

1. Free adiabatic expansion (I don't think it has something with the 2nd law).

  No work, no heat exchange, so internal energy and (for an ideal gas) temperature remains
  constant (1st Law).

  Boyle: pv = constant ==> p = 400 kPa

2.We can take the Joule-Thompson Coefficient as 0.25 K/bar to get the final temperature as 25°C

4. $ \eta\; ={{1800-440}\over{1800}}\; =\;{{68}\over{90}}$

  Assuming a supply of 1200 Btu/min = 20 Btu/s the output would be $ \;{{68}\over{90}}*20*1055\; = 15.94 kW$

Hi Immanuel: you are correct on all the answers. I researched and did them myself before you answered them. Thank you for your time. What do you do for a living that you know about thermodynamics? I'm going on my 2nd bachelor's in mechanical engineering.

I did some Engineering (both Civil and  Nuclear) and also some Physics.

About you, what was your first bachelor?

By the way, could you specify better your 3d question?

I did my first bachelor in: Electronics Engineering Technology. Now I have 1.5-2 years to finish Mechanical Engineering. In regards with the 3erd question;

Find the constant R for air which is
Rair = R(universal)/Molar mass
Rair =  (1545 (ft lbf)/(lbmol R))/(28.97 lbm/lbmol)  = 53.33 (ft lbf)/(lbm R)
The density of air: ρair =  P/RT
ρair =  (150 lbf/〖in〗^2 (〖144 in〗^2/〖ft〗^2 ))/((53.33 (ft lbf)/(lbm R))(1360 R)) = 0.298 lbm/〖ft〗^3
The mass flow rate is:
(A) m = (ρ)(V)(A)  =  (0.298lbm/〖ft〗^3 ) (350 ft/s^2 ) (.1 ft2) = 10.43 lbm/s

With the mass flow rate we can now calculate the power output (work) done:
– W = m[(h2 + (V_2^2)/2) – (h1 + (V_1^2)/2)]

The enthalpy h1 and h2 we can find in the Table A-22E because we have the temperature and pressure. We substitute in the energy equation and thats it.


Let $a_1>a_2>...>a_{2004}$ be 2004 real numbers.They are written on the circle,respectively .The following operations are caried out:
+)Replace $a_1 $by$ a_1'=max\{0,a_2,...,a_{2004}\} - a_1$
+)Replace $a_1 $by$ a_2'=max\{0,a_3,...,a_{2004},a_1'\} - a_2$
To do so continue...
Prove that efter a finite number of the above operations we can obtain $a_1,a_2,...,a_{2004}$,respectively
Sorry if my English is not good :(  :P

Dear Myth,
I think this problem is truth,my teacher who his name is professor Ha Huy Khoai give me

Maybe numbers are integers or something like this?

Myth,I'm sure $a_1,a_2,....,a_n$ are real numbers ! ;)

No,I don't.My teacher suggest that:Replace 2004 by n ;cover all figures on line; $a_1'=a_{n+1},a_2'=a_{n+2},....$.Then prove that sequence$\{a_n\}$ is periodic. However,I haven't completed this problem.Myth,I believe that you'll help me complete this problem! ;)

It is easy if they are integers i think.
Myth did you also found this?

Hi pascual2004,
I'm agree with you, it's easy if they are integers . When they are real,I think I solve it.But my solution is long.I will post it tomorrow,today I take the exam and I'm busy  :( .

Forgive me if i am not seeing something obvious or that easilly follows from what you did, but what if a1>a2>a3....>ak>0>ak+1...>an?

I will read your calculations carefully....But i thought the problem didnt have a solution you could achieve doing just calculations... I dont know if there is solution, but what happen if we remove the ordering of the ai? i think it becomes much more harder

I didn't recieve notification about replies in this topic. So I though everyone forgot it.
I will now read pigfly's solution.

I think pigfly means that case 3 looks like following:

[b]Case3.[/b] If $a_1>a_2>....>a_i\geq 0>a_{i+1}>...>a_n$
Then,
          $a_{n+j}=a_{j+1}-a_j(\forall j=\overline{1,i-1})$,
$\Rightarrow a_{n+i}=-a_i$,
 $a_{n+j}=-(a_{i+1}+...+a_j)(\forall j=\overline{i+1,n})$,
 $\Rightarrow a_{2n+j}=a_{2n}+a_1-a_{j+1}(\forall j=\overline{1,i-1})$
Hence,
$a_{2n+i}=a_{2n}+a_1$,
$a_{2n+j}=-(a_{j+1}+...+a_n)+a_1(\forall j=\overline{i+1,n-1})$,
$a_{3n+j}=a_{j+1}(\forall j=\overline{0,n-1})$
So we conclude that $\{a_n\}$ is periodic with:
      $a_{3n+i}=a_{i+1}(\forall i=\overline{1,n-1})$

if the condition on the monotonicity of the a's is dropped,then in fact there is no guarantee that there is periodicity.i tried to just check it on a computer with 5 numbers but not in order.and i just chose some numbers randomly-the computer gave no repetitions till 10000 iterations!so one gets an idea about how necessary the oordering actually is....

Seshadri! Did you use integer or real numbers?

i used some integers but there was a pi!(i don't remeber the exact numbers!)

dear Myth,
     the process works with integers all right!(Of course i tried with a starting set of 5,6,7 integers).basically i checked to see if there is periodicty within 10000 iterations and if so to return the ith step where periodicity occurs.
it is also possible that that 10000 wasn't large enough when real numbers were used-mostly i input the real numbers as decimals.and if the result is true with integers then it HAS to be true with rationals too(since all the operations are linear one can extract the 'common denominator and make the numbers as integers).

[quote="seshadri"]the process works with integers all right![/quote]
Do you mean you found periodicity?

here are a few examples:
    for the integers 11,9,-5,0,-4,-6 as(a,b,c,d,e,f)(starting values) the periodicity was after 24 steps.
here are some sample results:
                 set                                         iterations
 11,9,-5,0,-4,6                                              270
 11,19,-1,20,-4,1                                         4630(!)
 -11,32,0,-12,-4,1                                        2002(!)

and so on.

the format is that the first 6 entries denote the set and the last number is teh number of iterations..
so for instance for the sequence(11,9,-5,0,-4,6) there were 270 iterations before stopping and so on.
(hope this makes it clearer)

I can't rigorously prove that numbers are bounded in integer case.

i have a conjecture that the least upper bound will be the (largest +ve integer) - (sum of all the negative integers).