floor function
least common multiple
number theory unsolved
number theory

Let A={1,2,3,.....,2n} and S is a subset of A with n elements and for every 2 distint elements of S like a and b [a,b]>2n.([a,b] means the least common multiple

 of a and b).prove that for any element of S like a, a>[2n/3]

[hide]
Consider the $ n$ subsets of $ A$,

$ B_1\equal{}\{1,2,2^2,...\}$
$ B_3\equal{}\{3,3\cdot 2,3\cdot 2^2,...\}$
...
$ B_{2n\minus{}1}\equal{}\{2n\minus{}1\}$

Note how for any two elements $ a$ and $ b$ of a single set, one of them divides the other. Therefore $ lcm(a,b)\equal{}max(a,b)\le 2n$, so $ a$ and $ b$ cannot both be in $ S$. Since there are $ n$ elements total in $ S$, there must be exactly one element from each set in $ S$.

Now suppose one of these elements of $ S$, say $ a\equal{}c\cdot 2^x$, (where $ c$ is odd) is less than or equal to $ \left\lfloor\frac{2n}{3}\right\rfloor$. Then $ 3c\cdot 2^x\le 2n$. Now, we know that there is some element of $ B_{3c}$ that is in $ S$, call it $ 3c\cdot 2^y$. However, $ lcm(c\cdot 2^x,3c\cdot 2^y)\equal{}lcm(3c\cdot 2^{\max(x,y)})$. Now we know that $ 3c\cdot 2^x\le 2n$ by above, and $ 3c\cdot 2^y\le 2n$ since $ 3c\cdot 2^y$ is of $ S$.

Therefore $ lcm(c\cdot 2^x,3c\cdot 2^y)\le 2n$, a contradiction. Hence all elements must be greater than $ \left\lfloor\frac{2n}{3}\right\rfloor$.
[/hide]


[quote="OHmathguy"]How do you get 44%?[/quote]

...read the [b]solution packet[/b] for chapters...not just the answer key...

what was the question. i dont have the test on me right now

consider if he made the 20% error to make it smaller and consider when the error makes the circle larger. pick the one that has the greatest difference from 100% of the circle and you have your answer.

My solution:
[hide]
0.8*0.8 = 0.64
1-0.64 = 0.36 = 36%

1.2*1.2 = 1.44
1.44-1 = 0.44 = 44%, which is the answer.[/hide]

sorry, i can't use LaTeX or however u spell it.

this one could've been tricky if you solved it the other way (reducing instead of enlarging)

Phew... thank got I got it...

Yeah, you had to test both enlarging, and reducing

i didn't know you could do it by reducing...the only way i got 44% is by enlarging it. :huh:

[quote="footballrocks41237"]This was one of those which I made a stupid error on...  I got $144$%, as in it was $144$% of the circle...
[/quote]

that must hurt...

[quote="footballrocks41237"]This was one of those which I made a stupid error on...  I got $144$%, as in it was $144$% of the circle...

[hide]  $120$% of $20$ is $24$, so the area is $24^{2}=576 pi$, instead of $24^{2}=400 pi$  $576/400=1.44$, so it's $144$% of the circle, and thus increased by $44$%[/hide][/quote]

haha i think i got the same answer but then i had 30 seconds left so i double checked my work and quickly erased the 1  :rotfl:


Show that for any positive integer $ n>2$ we can find $ n$ distinct positive integers such that the sum of their reciprocals is $ 1$.

That's easy.
[hide]
Observe $ 1 \equal{} \frac{1}{2} \plus{} \frac{1}{3} \plus{} \frac{1}{6}$,
and that $ \frac{1}{2} \equal{} \frac{1}{3} \plus{} \frac{1}{6}$.

Thus, use mathematical induction, with base case $ n\equal{}3$:
$ n\equal{}3$: 2, 3, 6

Induction hypothesis: we have $ n$ distinct positive integers whose reciprocals sum to 1, and the largest of those numbers is even.
Then, assume the largest number is $ 2k$; replace it with $ 3k$ and $ 6k$. $ 6k$ is obviously the new largest number, and just as obviously even. Then we have $ n\plus{}1$ distinct positive integers whose reciprocals sum to 1.
Q.E.D.
[/hide]


Ok guys - 1st post, so forgive me if this has been posted before - I couldn't find it on the forums anywhere...

BMO 1998 Q. 4

Find a solution of the simultaneous equations
xy + yz + zx = 12
xyz = 2 + x + y + z
in which all of x; y; z are positive, and prove that it is the only
such solution.
Show that a solution exists in which x; y; z are real and
distinct.

Any help would be greatly appreciated thanks!

Oct.

$ xyz\equal{}2\plus{}x\plus{}y\plus{}z \Rightarrow$ there exists $ a,b,c \in \mathbb{R}^\plus{}$ such that $ x\equal{}\frac{a\plus{}b}{c},y\equal{}\frac{b\plus{}c}{a},z\equal{}\frac{c\plus{}a}{b}.$

[b]first part:[/b]

note that by AM-GM, $ xy\plus{}yz\plus{}zx\ge 12.$

Therefore, the only solution is if $ a\equal{}b\equal{}c.$

$ 12\equal{}xy\plus{}yz\plus{}zx\ge3\sqrt[3]{x^2y^2z^2}$, so $ 8\ge xyz$.

$ (x\plus{}y\plus{}z)^2\ge3(xy\plus{}yz\plus{}zx)\equal{}36$, so $ x\plus{}y\plus{}z\ge6$.

Thus $ 8\ge xyz\equal{}2\plus{}x\plus{}y\plus{}z\ge2\plus{}6$. Equality holds everywhere, so $ x\equal{}y\equal{}z\equal{}2$.


On the vertices of a convex $ n$-gon are put $ m$ stones, $ m > n$. In each move we can choose two stones standing at the same vertex and move them to
the two distinct adjacent vertices. After $ N$ moves the number of stones at each vertex was the same as at the beginning. Prove that $ N$ is divisible by $ n$.

Let $ a_1, a_2, \ldots, a_n$ denote number of times we re-distribute $ 2$ stones from a point $ P_n$, with $ a_{n\plus{}i} \equal{} a_i$. Then to restore to initial conditions, we require $ 2a_k \equal{} a_{k\minus{}1}\plus{}a_{k\plus{}1}$.

Summing that with $ 2a_{k\plus{}1}\equal{}a_k\plus{}a_{k\plus{}2}$ yields $ a_{k\plus{}2} \minus{} a_{k\plus{}1} \equal{} a_k \minus{} a_{k\minus{}1} \equal{} m$ for some integer $ m$. For odd $ n$, it follows that $ a_k \minus{} a_{k\minus{}1} \equal{} m$ for all $ k$. Then, $ 0 \equal{} (a_2 \minus{} a_1) \plus{} (a_3 \minus{} a_2) \plus{} \cdots \plus{} (a_1 \minus{} a_n) \equal{} mn \Longrightarrow m \equal{} 0$, so all of the $ a_i$ are equal.

For even $ n$, we have $ a_{2k} \minus{} a_{2k\minus{}1} \equal{} m_1, a_{2k\plus{}1} \minus{} a_{2k} \equal{} m_2$, with $ 0 \equal{} \frac n2 \cdot m_1 \plus{} \frac n2 \cdot m_2 \Longrightarrow m_1 \equal{} \minus{}m_2$. Substituting into $ 0 \equal{} 2a_{2k} \minus{} a_{2k\minus{}1} \plus{} a_{2k\plus{}1} \equal{} m_1 \minus{} m_2 \equal{} 2m_1$, and again all of the $ a_i$ are equal. Clearly $ n | N \equal{} \sum a_i$ then.


The incircle of a triangle $ABC$ touches $AC, BC$ , and $AB$ at $M , N$, and $R$, respectively. Let $S$ be a point on the smaller arc $MN$ and $t$ be the tangent to this arc at $S$ . The line $t$ meets $NC$ at $P$ and $MC$ at $Q$. Prove that the lines $AP, BQ, SR, MN$ have a common point.

Dear TUAN.

It is exactly the well known [b][size=100]Newton's[/size][/b] theorem. So, I think it has been posted in incorrect area.

Kostas Vittas.

[hide=Solution]
[hide=Lemma]
Let $ABCD$ be a tangential quadrilateral with $EFGH$ ($E\in AB$, $F\in BC$, $G\in CD$, $H\in DA$) being the intouch quadrilateral. Then $AC$, $BD$, $EG$, and $FH$ concur.

[hide=Proof]
Brianchon's on $AEBCGD$ gives that $AC$, $EG$, and $BD$ concur. Brianchon's on $ABFCDH$ gives that $AC$, $BD$, and $FH$ concur. Thus, $AC$, $BD$, $EG$, and $FH$ concur.
[/hide]
[/hide]

Apply the lemma on tangential quadrilateral $AQPB$ and we get that $AP$, $BQ$, $SR$, and $MN$ concur.

Q.E.D.
[/hide]


Sa se calculeze:

         $\int\frac{4cosx-6sinx}{5cos2x+12sin2x+4sinx+6cosx+13} dx$ ;)

Nice,kunny.I used the substitution $t=2sinx+3cosx$ and the integral becomes: $\int\frac{dt}{t(t+1)}=\ln|\frac{t}{t+1}| +C$ ;)

Thanks, cezar lupu.:)  I also calculated in your way. :)


Find all integers satisfying the equation $ 2^x\cdot(4\minus{}x)\equal{}2x\plus{}4$.

$ x$ non-negative or $ L.H.S.$ is not integer.
$ x \le 4$ or $ R.H.S.$ is negative.,
$ x\equal{}0,1,2$

[hide="Another approach"]$ \frac{2x\plus{}4}{4\minus{}x}\equal{}2^x>0$. So $ 2x\plus{}4$ and $ 4\minus{}x$ have the same signs and hence $ \minus{}2<x<4$. Checking $ x\equal{}\minus{}1,0,1,2,3$, it is easy to get the solutions $ x\equal{}0,1,2$.[/hide]


search
function
algebra unsolved
algebra
trigonometry
Solve equation

we have the equation $ln\left(1+\sqrt{1+tg^2(\alpha)}\right)=ln\left(ctg(\alpha)+\sqrt{1+ctg^2\alpha}\right)-\frac1{2}ln3$
find the roots of the equation $\in(0,\pi)$

I think it is for the Algebra Section sooner, not Geometry.
And are you sure it is not for Pre-Olympiad? Just an standart equation. Is it your homework? :?

I am preparing for an exam wich is not supose to be very dificult though

First of all lets search for roots on $(0,\frac{\pi}{2})$.
Just note that $LHS$ is a decreasing function and $RHS$ is an increasing. So we have at most 1 root.

Clearly, $x=\frac{\pi}{6}$ is that root. ;)
And second root is $\pi-\frac{\pi}{6}$.


Hello everybody,

Today I continue to have two new paragraphs, help me to repair my mistake, please. Thank you very much.

1.  The authors introduced a new self-contained and rigorous proof of the smoothness of invariant fiber bundles for dynamical equations on measure chains or time Scales, where an invariant fiber bundle is the generalization of an invariant manifold to the non-autonomous case. The main result generalizes in this paper is the "Hadmard-Perron-Theorem" to the time-dependent, infinite-dimensional, non-invertible and parameter-dependent case, in which the linear part is not neccessarily hyperboloic with variable growth rates. Their proof didn't use to be complicated technical tools.

2.  In this paper, the quthors wrote reviews for the recent notion of a motonomous dynamical system (NDS), which has been introduced as an abstraction of both random dynamical systems and continuous skew product flows. They focus on fundamental analogies and discrepancies brought about by these two principal classes of NDS. Additionally, they discussed base dynamics main through almost periodicity and autonomorphy, and they also emphasize the importance of these concepts for NDS which are generated by differenrtial and difference equations. THey presented means of selected examples about nonautonomous dynamics. Finally, they mention several natural yet unresolved questions.

Have a happy day to you.

Hoc.

I'm sorry, but you cannot impose yourself upon a forum in such a way. You should have the willingness and desire which other forum members here have to do this yourself. If there is a smaller portion you need help with, I'm sure somebody will be able to help you with that.

Hi hoctoan.

I seem to remember answering this by private message some time ago. Maybe it was sent to me by some-one else. I cannot answer at the moment as I am in a rush but could get back to you when I return from abroad on Saturday.

The American-English spellings notwithstanding, it appears more or less okay apart from a number of typos. Some I've spotted are : scalene should have a lower-case s, necessarily has only one c, hyperbolic has only one o, "to be" is extraneous, you have quthors instead of authors, main should be (the adverb) mainly, you have one r too many in differential , non-autonomous has a hyphen.

Sorry I haven't got more time.

By the way what is your first language?

1.  The authors introduced a new self-contained and rigorous proof of the smoothness of invariant fiber bundles for [color=#FF0000]dynamic[/color] equations on measure chains or time scales, where an invariant fiber bundle is the generalization of an invariant manifold to the non-autonomous case. The main result [color=#FF0000]generalized[/color] in this paper is the [color=#FF0000][u]Hadmard-Perron Theorem[/u][/color] to the time-dependent, infinite-dimensional, non-invertible and parameter-dependent case, in which the linear part is not neccessarily [color=#FF0000]hyperbolic[/color] with variable growth rates. [color=#FF0000]Their proof did not use complicated technical tools.[/color]

2.  In this paper, the [color=#FF0000]authors[/color] wrote reviews for the recent notion of a motonomous dynamical system (NDS) [color=#FF0000](shouldn't it be MDS?)[/color], which has been introduced as an abstraction of both random dynamical systems and continuous skew product flows. They focus on fundamental analogies and discrepancies brought about by these two principal classes of an NDS. Additionally, they discussed base dynamics main through almost periodicity and autonomorphy, and they also emphasize the importance of these concepts for an NDS which are generated by [color=#FF0000]differential[/color] and difference equations. [color=#FF0000]They[/color] presented means of selected examples about nonautonomous dynamics. Finally, they [color=#FF0000]mentioned[/color] several natural yet unresolved questions.


Sandy typed a positive six digit integer, but the two 9's she typed didn't show. If 2010 appeared, how many different positive six-digit integers could she have typed? I such integer to include is 920910.

[hide="My (wrong) solution"]
a 9 can go in the spots marked by |,
|2|0|1|0|
That's five of em. So, first we put in the first 9 with 5 choices, then the second nine, with 4 choices so the answer should be $ 5 \times 4 \equal{} 20$
The books says its 15.
[/hide]

1) Your way doesnt account for repeats in 9's. ie. it counts 920109 twice. So it shouldnt be 5x4, it should be (5x4)/2, (or just 5C2).

2) Even if it is 5C2, you will still only get 10. Why? You dont take into account if two nines can be next to each other. There are 5 ways that can happen, for a total of 15


vector
analytic geometry
linear algebra
linear algebra unsolved

i know it is not linear algebra but it subject is linear algebra.
$X$ be collection of orthogonal vectors in $R^n$ and suppose the projection  of each of these vectors  on the first k coordinate is of euclidean norm at least $\alpha$ prove $|X|\leq \frac{k}{\alpha^2}$.

You mean orthonormal, right?

Let $x_1,x_2,\dots,x_r$ be the vectors of $x$. We know that $\|x_1+x_2+\dots+x_r\|^2=\|x_1\|^2+\|x_2\|^2+\dots+\|x_r\|^2=r$.

On the other hand, $\|x_1+x_2+\dots+x_r\|^2\ge k\cdot(r\alpha)^2=kr^2\alpha^2$ since the first $k$ coordinates of $x_1+x_2+\dots+x_r$ are all at least $r\alpha$. This gives $r\ge kr^2\alpha^2$, so $r\le\frac1{k\alpha^2}$.
Maybe I'm misunderstanding- what do you mean by "projection on the first $k$ coordinates"?
If that's the sum of the first $k$ entries of $x_i$, I get your result.

no by projection i meant that norm of the vector which is first k element of each of the orthonormal ,is of at least $\alpha$(euclidean norm).


[color=darkred]The incircle of $\triangle ABC$ touches the its sides in the points $D\in BC$, $E\in CA$, $F\in AB$. Let $w$ be the circle which passes through the points $B$, $C$ and which is tangent in the point $T$ to the incircle. Denote the second intersections $X$, $Y$, $Z$ of the lines $TD$, $TE$, $TF$ with the circle $w$ respectively. [b]Construct[/b] the center of the circle $w$ and [b]ascertain[/b] the its radius.

[b]Remark.[/b] Study the circle $w$ and the triangle $XYZ$.[/color]

[quote="Virgil Nicula"] Construct the center of of the circle $w$...[/quote]

easy if $\triangle ABC$ is isosceles. otherwise WLOG let $BD<CD$. denote $x=AE=AF,y=BD=BF,z=CF=CD$. construct the circle $\mathcal C$ with center on $BC$ on $B$'s side, passing trough $D$ and of radius $\frac{yz}{z-y}$. Then $\mathcal C$ will intersect the incircle of $ABC$ at $D$ and $T$. the perpendicular bisectors of $TB$ and $TC$ intersect at $O$. then $O$ is the desired center.


group theory
abstract algebra
superior algebra
superior algebra unsolved

1) Let M and N be normal subgroups of a group G. Prove that the intersection of M and N is normal in G.

 (I managed to show that its a subgroup of G by simply show that for x,y in the intersection xy^(-1) is also in the intersection, but to prove the normality was harder)

2) Let G = Q/Z. Prove that G is an infinite group in which every element has finite order.

(I managed to show that G is infinte with elements 1`/2, 1`/3, 1`/4, ... . But how to prove that every element has finite order? The only thing I can see is that 1`/k has order k.)

1)  Well, the prove of normality is similar:
Let $H= M \cap N$, then for every $x \in H$ we have that $axa^{-1}$ is in $aMa^{-1}=M$ for all $a$ and also in $N$ for all $a$, so $axa^{-1} \in H$. But the immediate reult is that $aHa^{-1} \subset H$, so $H$ is normal.

2) Let be $x \in \mathbb{Q}$, then $x=\frac{a}{b}, a,b \in \mathbb{Z}, b >0$.  But then $b \cdot x \equiv 0 \mod \mathbb{Z}$, so the order of $x$ is finite (and a divisor of $b$).

An alternative proof of (1): consider the kernel of the map $G \rightarrow G/M \times G/N$.


Two lines $ y$ and $ z$ form a 60æ angle at the point $ X$ , and $ Y1$ is a point on y. From $ Y1$ draw a line perpendicular to the line $ y$ meeting the line $ z$ at the point $ Z1$. From $ Z1$ draw a line perpendicular to $ z$ meeting the line $ y$ at $ Y2$. Continue in this manner giving points $ Z2$,$ Y2$,$ Z3$,$ Y3$ and so on. These points are the vertices of right triangles $ X$ $ Y1$ $ Z1$ ,  $ X$ $ Y2$ $ Z2$,•••. If the area of $ X$ $ Y1$ $ Z1$ = 1, determine
area$ X$ $ Y1$ $ Z1$ + area$ X$ $ Y2$ $ Z2$ + ••• + area$ X$ $ Y2007$ $ Z2007$.

Options:

1.) (($ 16^{2007}$) -1)/7
2.) (($ 16^{2006}$) -1)/15
3.) (($ 16^{2007}$) -1)/15
4.) $ 16^{2007}$/7

The ratio between area of $ XY_{n}Z_{n}$ and $ XY_{n\plus{}1}Z_{n\plus{}1}$ is $ 16$.
So the sum is actually equal to:
$ 1\plus{}16\plus{}16^{2}\plus{}...\plus{}16^{2006}\equal{}\frac{16^{2007}\minus{}1}{15}$.
Sol: $ 3).$


Can someone explain to me what the invariant principle is? I looked it up on the AoPS Wiki, but there were no articles about it.

The invariant principle is a very general problem-solving principle.  It can be stated as follows:

[i]Given a problem involving a transformation of some kind, find a quantity or property that remains the same under that transformation.  This is called an invariant.[/i]

The invariant principle can be applied to a wide variety of questions.  One common way it is used is to prove that a certain scenario cannot be obtained using some set of transformations because some invariant needs to change that cannot.

Examples:
- http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1796254710&t=60773
- http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1796254710&t=90046
- http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1796254710&t=168639

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1796254710&t=164506]Engel[/url] has a great chapter on the invariant principle; it's well worth buying.

Do you mean the Invariance Principles that define basic symmetry (and Physics)?

Or is this a math term that I'm not familiar with?


A convex quadrilateral ABCD is inscribed in the circle with diameter AB. Let S be the intersection point of AC and BD and T the foot of the perpendicular from S to AB. Prove that ST bisects angle CTD.

this is a easy problem!
Look that SCBT and DSTA are ciclics, becose SCB=STB=90 AND SDA=STA=90
so DTS=DAS and STC=SBC, using DAS=SBC we must have DTS=STC.

[i]First solution:[/i] Since the points C and D lie on the circle with diameter AB, we have < ACB = 90 and < ADB = 90. Hence, if the lines AD and BC meet at a point X, then the lines AC and BD are two altitudes of the triangle ABX; the third altitude of this triangle is the line ST (since it is perpendicular to its side AB and passes through the point of intersection S of the other two altitudes), and hence the problem reduces to the well-known fact that the altitudes of a triangle bisect the angles of the orthic triangle.

[i]Second solution:[/i] Since the point C lies on the circle with diameter AB, we have < ACB = 90, so < SCB = 90. On the other hand, < STB = 90. Thus, the points C and T lie on the circle with diameter SB. Hence, the points S, B, C and T lie on one circle; thus, < STC = < SBC. In other words, < STC = < DBC. Similarly, < STD = < CAD. But < DBC = < CAD since the points A, B, C, D lie on one circle; therefore, < STC = < STD, and the line ST bisects the angle CTD. This completes the solution.

  Darij


calculus
integration
function
trigonometry
search
real analysis
real analysis unsolved

prove the following 2 integrals using the beta function only! :D

$ \textbf{(i)}\;\;\int_{0}^{\infty}\;\;\frac{x^{p\minus{}1}}{1\plus{}x^{q}}\;\;\textbf dx\;\; \equal{}\;\;\int_{0}^{1}\;\;\frac{y^{p\minus{}1}}{(1\minus{}y^{q})^{\frac{p}{q}}}\;\;\textbf dy\;\; \equal{}\;\;\boxed{\frac{\pi}{q\cdot\sin\left(\frac{p\,\pi}{q}\right)}}$

( where $ 0 < p < q$ )

$ \textbf{(ii)}\;\;\int_{0}^{\infty}\;\;\frac{\textbf dx}{1\plus{}x^{p}}\;\; \equal{}\;\;\int_{0}^{1}\;\;\frac{\textbf dy}{(1\minus{}y^{p})^{\frac{1}{p}}}\;\; \equal{}\;\;\boxed{\frac{\pi}{p\cdot\sin\left(\frac{\pi}{p}\right)}}$

( where $ p > 1$ )

FYI- the beta function states:
$ \textbf B(m,n)\;\equal{}\;\int_{0}^{1}\;\;t^{m\minus{}1}\cdot(1\minus{}t)^{n\minus{}1}\;\;\textbf dt\;\;\equal{}\;\;\frac{\Gamma(m)\cdot\Gamma(n)}{\Gamma(m\plus{}n)}$

Misan, I made a search in mathlinks by word 'integral' and I saw you have biggest average numbers of posted integrals. It make me to think you have experience in integrals. Please if you have time look at the following problem - it is still not solved by anyone. It not seems to be so difficult or it just seems?
http://www.mathlinks.ro/Forum/viewtopic.php?t=167800

(1)
Substitute $ x^q \equal{} u$, then
$ \int_0^{\infty} \frac{x^{p\minus{}1}}{1\plus{}x^q}\ dx \equal{} \int_0^{\infty} \frac{1}{q}\frac{u^{\frac{p}{q}\minus{}1}}{1\plus{}u}\ du$
Now
$ B(x,y) \equal{} \int_0^{\infty}\frac{t^{x\minus{}1}}{(1\plus{}t)^{x\plus{}y}}\ dt$ for $ \mbox{Re}(x)>0$ and $ \mbox{Re}(y)>0$
thus
$ \int_0^{\infty} \frac{x^{p\minus{}1}}{1\plus{}x^q}\ dx \equal{} \frac{1}{q}B(\frac{p}{q},1\minus{}\frac{p}{q})$.
Now use the formula $ B(x,1\minus{}x) \equal{} \frac{\pi}{\sin(\pi x)}$


geometry
incenter
geometric transformation
reflection
geometry proposed

[b]The incircle of the triangle ABC touches the sides BC,CA,AB at D,E,F respectively.
Each pair from the incircle of AEF,DBF,DEC has two common external tangents oneof which is a side of triangle ABC show that the other three tangents are concurrent.[/b]

[u] [color=red][ Looking for a synthetic proof no Transformations and projections please.] [/color][/u]

Jamshed (Jimbo)

Suppose the incircle $(I)$ intersects the segments $AI$, $BI$ and $CI$ at $A', B'$ and $C'$. Then $A'$, $B'$, $C'$ are respectively the incenters of $AEF$, $BDF$, $CFE$.

Let $H$ be the orthocenter of $A'B'C'$. It is easy to see that $DEF$ is the circumcevian triangle of $H$ wrt $A'B'C'$. Hence $H$ is the reflection of $D$ on $B'C'$, $E$ on $C'A'$, $F$ on $A'B'$. Therefore, the other three tangents of the three circles $(A'), (B'), (C')$ are concurrent at $H$.

Excuse me Treegoner, but what do you mean with "circumcevian triangle"? Thanks in advance :)


function
integration
real analysis
real analysis unsolved

Let $f: [0,2]\to \mathbb{R}$ a function such that f is derivable on $[0,2] ,f(0)=f(2)=1$ and $|f'(x)|\leq 1$ for any $x$ in $[0,2]$.
Find the best $K$ such that $\int_{0}^{2}f(x)dx\geq K$ for any function with the above properties.

Find $g$ so that $f\ge g$ pointwise, and $g$ can be closely approximated. It's not hard: what if $|f'|$ is (almost) identically 1?

Thx anik_andrew :) The original problem was with K=1, but I made some errors in my solution and I thought that a better constant exists...Here is my solution to the problem:
$\int_{0}^{2}f(x)dx=2-\int_{0}^{2}f'(x)xdx$. As a result  $|\int_{0}^{2}f(x)dx-2|=$

$|\int_{0}^{2}f'(x)xdx|=$

$|\int_{0}^{2}f'(x)(x-1)dx|\leq $

$\int_{0}^{2}|f'(x)(x-1)|dx\leq $

$\int_{0}^{2}|(x-1)|dx=1$.

From here the result.... :)

Btw...Could anyuone recommend me a good online article about uniform convergence? (not very advanced...just to make a better idea about this topic and maybe to see some problems where the theorems are applied) :)


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

   [i]Let  $ x ,y , z$ be positive real numbers such that

   $ xy \plus{} yz \plus{} zx \equal{}  7xyz$

     Find the least value of the expression :

     $ P \equal{}  \frac{8x^4 \plus{}1 }{x^2}  \plus{} \frac{108y^5 \plus{}1 }{ y^2 } \plus{} \frac{16z^6 \plus{}1 }{z^2}$[/i]

    [i] Nguyễn Vũ Thanh Hà[/i]

[quote="nkht-tk14"]

Let  $ x ,y , z$ be positive real numbers such that

$ xy \plus{} yz \plus{} zx \equal{} 7xyz$

Find the least value of the expression :
$ P \equal{} \frac {8x^4 \plus{} 1 }{x^2} \plus{} \frac {108y^5 \plus{} 1 }{ y^2 } \plus{} \frac {16z^6 \plus{} 1 }{z^2}$
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=177044


probability
trigonometry
integration
probability and stats

P and Q are two points in the plane a distance 1 apart.

Short rods PP' ,QQ' , pivoted at P and Q respectively, are spun and come to rest at random.

Let R be the point where the lines PP', QQ' meet when extended. Let Y be the distance QR.

1.Find the probability that Y is less than x, for x$ \geq$ 0.Leave your answer as an integral.
2.Calculate P(Y$ \leq$ 1).

if $ \alpha \equal{} (PQ,PP')$ and $ \beta \equal{} (QP, QQ')$ are (hypothesis) independant and uniformly distributed on $ [0; 2\pi]$, then the sinus law in the triangle $ PQR$ shows that $ Y$ is distributed as
\[ | \frac {\sin(\alpha)}{\sin(\alpha \plus{} \beta)} |
\]
For example
$ P[Y<x] \equal{} \frac{1}{4\pi^2}\int_{\alpha \in [0; 2\pi[} \int_{\alpha \in [0; 2\pi[} 1_{| \frac {\sin(\alpha)}{\sin(\alpha \plus{} \beta)} |<x} d \alpha d \beta$


Find the coefficient of $x^{2}$ in the expansion of
\[(1+x)(1+2x)(1+4x)\cdots(1+2^{n}x).\]

This is actually very neat. Good problem.

The problem can be expressed as the following summation:

[hide]

$\sum_{k=1}^{n}2^{k}{\sum_{j=0}^{k-1}2^{j}}$

This quickly reduces to $\sum_{k=1}^{n}{2^{k}(2^{k}-1)}= \frac{4^{n+1}-1}{3}-2^{n+1}+1$.[/hide]


[quote="minsoens"][quote="TRAN THAI HUNG"]Prove that $\sqrt[3]{{\sqrt[3]{2}-1}}= \sqrt[3]{{\frac{1}{9}}}-\sqrt[3]{{\frac{2}{9}}}+\sqrt[3]{{\frac{4}{9}}}$[/quote]
[hide]
$\left(\sqrt[3]{{\sqrt[3]{2}-1}}\right)\left(\sqrt[3]{\frac{2}{3}}+\sqrt[3]{\frac{1}{3}}\right)= \left(\sqrt[3]{\frac{2}{3}}+\sqrt[3]{\frac{1}{3}}\right)\left(\sqrt[3]{{\frac{4}{9}}}-\sqrt[3]{{\frac{2}{9}}}+\sqrt[3]{{\frac{1}{9}}}\right)$
$\left(\sqrt[3]{{\sqrt[3]{2}-1}}\right)\left(\sqrt[3]{\frac{2}{3}}+\sqrt[3]{\frac{1}{3}}\right)= 1$
$1=\left(\sqrt[3]{{\sqrt[3]{2}-1}}\right)^{3}\left(\sqrt[3]{\frac{2}{3}}+\sqrt[3]{\frac{1}{3}}\right)^{3}$
$\frac{1}{\sqrt[3]{2}-1}=\frac{2}{3}+\frac{1}{3}+3\sqrt[3]{\frac{4}{27}}+3\sqrt[3]{\frac{2}{27}}$
$1=\left(\sqrt[3]{2}-\sqrt[3]{1}\right)\left(\sqrt[3]{4}+\sqrt[3]{2}+\sqrt[3]{1}\right)$
[/hide][/quote]

Good work, but on the first step, aren't you assuming that what you are supposed to prove is true and then proving the same thing? Can you do this in a more direct way?

Hint: call the right hand side X and simplify and solve for X to be the same as the left hand side.


[color=blue][size=150]PERU TST IMO - 2006[/size]
Saturday, may 20.[/color]

[b]Question 02[/b]
Find all pairs $(a,b)$ real positive numbers $a$ and $b$ such that:

$[a[bn]]= n-1,$

for all $n$ positive integer.
Note: [x] denotes the integer part of $x$.

----------
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88510]Spanish version[/url]

$\text{\LaTeX}{}$ed by carlosbr

[quote="Rust"]$a - irrational 0<a<1,b=\frac 1a.$
Prove: $bn-1<[bn]\le bn$. If a>0 $abn-a-1<[a[bn]]\le abn$. If a<0 $abn-1<[a[bn]]<abn-a$. For all case if $ab\not =1$ exist n: $[a[bn]]\not n-1$. If a<0 exist n: $[a[bn]]=n$. If a>1 exist n: $[a[bn]]<n-1$. If a - rational exist n: $[a[bn]]=n$.[/quote]

[]: denotes the floor function :P


AMC
USA(J)MO
USAMO
AMC 12
AMC 12 A
geometry
AMC 12 B

I know there are still some late papers being graded etc, but I am a little bit concerned...

I took the AMC12A at my highschool and took the 12B at the dallas area math circle. My scores were 124.5 and 126 respectively.
When I took the AMC12B, I put down the math circle's address and they sent an AIME I that I didn't take there (spring break).
I took the AIME II at my highschool. For some reason, we didn't use the actual scantron but the printed off version from the website and our teacher filled in the bubbles later on (idk what actually happened but I thought it was weird... a similar thing happened on the AMC12A because we didn't get the materials in time).

So when I found my name not on the list, I was initially surprised because from the answers online I had a 9 on the AIME. And another guy from my school qualed, so I know that the test papers have been received. If there was confusion as to my identity (because I had two schools listed), that still shouldn't be a problem because I have a score from the 12A at my actual highschool (and it is still enough to qualify)...

I was wondering if there was any way of getting any peace of mind from AMC as to what my actual score was (perhaps stupid bubbling mistakes on teachers part?, because i checked all of bubbling numerous times and transcribed my answers on some scratch paper after the test to compare on AoPS... or some paperwork mix up?) Ironic thing is, last year, I did the same thing (took the B at the math circle, AIME at school... but there was no problem in USAMO qualification last year...)

Any suggestions as to what I might do? Also, anybody else in a similar situation?


Solve the equation $ 2x^2\plus{}4\equal{}5\sqrt{x^3\plus{}1}$.

Thank.

[img]http://alt2.mathlinks.ro/latexrender/pictures/d/4/b/d4bb29f1136b2430a0e40555f0fbd0093cf26b9a.gif[/img]$ \Longleftrightarrow$ $ 4x^{4} \minus{} 25x^{3} \plus{} 16x^{2}\minus{}9 \equal{} 0$

Now,apply Ferrari's method ;)

[quote="mathematikos"][img]http://alt2.mathlinks.ro/latexrender/pictures/d/4/b/d4bb29f1136b2430a0e40555f0fbd0093cf26b9a.gif[/img]$ \Longleftrightarrow$ $ 4x^{4} \minus{} 25x^{3} \plus{} 16x^{2} \minus{} 9 \equal{} 0$

Now,apply Ferrari's method ;)[/quote]

Please explain me about Ferrari's method, notice that you only solve with elementary maths, thank you.

$ (4x^2\minus{}5x\plus{}3)(x^2\minus{}5x\minus{}3)$

Now use the quadratic equation.

hello, your equation is for $ x\geq\minus{}1$ equivalent with $ (4x^2\minus{}5x\plus{}3)(x^2\minus{}5x\minus{}3)\equal{}0$. You can find these two factors by the ansatz
$ 4x^4\minus{}25x^3\plus{}16x^2\minus{}9\equal{}(ax^2\plus{}bx\plus{}c)\cdot(dx^2\plus{}ex\plus{}f)$.
Sonnhard.


I've tried logging out and logging back in multiple times but the problem still won't load. All that appears is how many points I have in the right hand corner and then blank space below that.

Try refreshing the page, and click "Get Problem" again.  Try repeatedly.

maybe your internet connection is slow or something's wrong with it

do other pages load successfully?

What browser and operating system are you using? Include version numbers please.

Seems a [i]tiny[/i] bit similar to this problem.  AIME15's screenshot...

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=241201

Yeah, 007math, that's exactly what the screen looks like.

I tried using AOL today and it didn't work but then I tried internet explorer and it loaded

once I got that problem to load on internet explorer I opened up AOL and was able to get it to work there

The AOL browser has a number of quirks. We highly recommend using a different browser at all times while on the artofproblemsolving.com website.


function
group theory
abstract algebra
algebra
rational function
Galois Theory
superior algebra

Let $ k$ be a field, and let $ K\equal{}k(x)$ be the rational function field in one variable over $ k$. Let $ \sigma$ and $ \tau$ be the automorphisms of $ K$ defined by $ \sigma (f(x)/g(x))\equal{}f(1/x)/g(1/x)$ and $ \tau (f(x)/g(x))\equal{}f(1\minus{}x)/g(1\minus{}x)$, respectively. Dertemine the fixed field $ F$ of $ \{\sigma,\tau\}$ and determine $ \text{Gal}(K/F)$. Find an $ h\in F$ so that $ F\equal{}k(h)$.

a standard result vom algebra: if $ K$ is a field and $ G$ a finite subgroup of $ Aut(K)$, then $ K$ is galois over the fixed field $ F \equal{} \{a \in K : \forall \sigma \in G : \sigma(a) \equal{} a\}$ with galois group $ G$.

in our case, $ \sigma$ and $ \tau$ generate a subgroup $ G$ of $ Aut(K)$, and $ F$ is the fixed field of $ G$. observe that $ \sigma$ and $ \tau$ have order $ 2$, and $ \sigma \tau$ has order $ 4$. thus, $ G$ is isomorphic to the dihedral group $ D_4$, and also $ Gal(K/F)$. it remains to determine $ F$ and a primitive element $ h$ (which must have the form $ f/g$ where $ max(deg(f),deg(g))\equal{}8$).


calculus
integration
function
topology
geometry
rectangle
real analysis

Let $ S$ be a bounded subset of $ \mathbb{R}^n$; let $ f: S \to \mathbb{R}$ be a bounded continuous function; let $ A \equal{}$ Int $ S$. Give an example where $ \int_A f$ exists but $ \int_S f$ does not.

It might help if you described what the definitions are here, and how there's a difference at all. Different values on a nowhere-dense set of measure zero don't change the Riemann integral, and I don't see how the Riemann integral even makes sense on a region with a larger boundary.

The problem is taken directly from Munkres's Analysis on Manifolds; the definition of the Riemann integral is the standard one. It is a theorem that if a function is integrable on a bounded set it is also integrable on that set's interior (and in fact the two integrals agree); this is intended to be a counterexample to the converse. Something weird has to happen with the set $ S$ (probably a boundary with nonzero measure, but I can't make it work).

If it helps, Munkres defines the integral of $ f: S \to \mathbb{R}$ by extending $ f$ to $ f_S: \mathbb{R}^n \to \mathbb{R}$, letting $ f_S$ agree with $ f$ on $ S$ and vanish elsewhere. Then enclose $ S$ in a rectangle $ Q$ and integrate $ f_S$ over $ Q$ in the normal way.

OK. Stick to one dimension, and let $ S$ be a Cantor-type set of positive measure, with $ f$ equal to $ 1$ on that set. Since $ f$ (extended to the full interval) is discontinuous everywhere on $ S$, its Riemann integral doesn't exist. Since the interior of $ S$ is empty, that integral is just zero.

Ah, now I see. If you wanted an example with nonempty interior, how about the constant function 1 on the set $ S \equal{} [0,1] \cup ([1,2] \cap \mathbb{Q})$? Obviously not integrable since enclosed in an interval it includes the indicator function for the rationals on $ [1,2]$, but restricted to Int $ S$ it's just the constant function on the interval $ (0,1)$.

That too. I was thinking more Lebesgue than I had to be.

Here's an even more restrictive question: Find a bounded open set $ A \subset \mathbb{R}^n$ (you can do it in $ \mathbb{R}$, I think) and a bounded continuous function $ f: A \to \mathbb{R}$ such that $ \int_{\bar{A}} f$ (as in the integral over the closure of $ A$) exists but $ \int_A f$ does not.

As you've phrased it, that's completely impossible; the two integrals are by definition exactly the same thing. Expanding the region can't possibly hurt unless we're extending to nonzero values, and it can't possibly help under any circumstances.

The closest you can come is something like my Cantor set example; fine on the open set, but the extended values break it on the closure.

You're right, I misstated it: let $ f: \mathbb{R}^n \to \mathbb{R}$.


inequalities
number theory proposed
number theory

Find all positive integers $ n$ such that there exists some positive integers $ x_1,x_2,...,x_k$ satisfying $ n\equal{}(2x_1\plus{}1)(2x_2\plus{}1)...(2x_k\plus{}1)\equal{}2^{x_1x_2...x_k}\minus{}1$

[hide="hint"]
just use inequality and the fact that $ (2^{x_1}\minus{}1)....(2^{x_k}\minus{}1) \le 2^{x_1x_2...x_k}\minus{}1$(note that we can have the reverse but this occurs in some specific conditions(which can easily be found and) which can easily be checked)
[/hide]


[quote="mim"]There was a topic about preponderance pencils over pens :P
I have got a continue question: do you prefer to write on quad-ruled paper or blank paper when doing your math work? In notebook or loose leaf? What is the size of the page?[/quote]

I really not prefer the quad-ruled paper......

I don't have a preference. the closest peice of paper larger than 1 inch square and the closest pen or semi-sharp pencil will do for me!

Any kind of paper, as long as there's not a lot of writing already on it.
I'll write with anything but a scratchy pencil. (the ones that make that awful scratching noise when you write) I prefer mechanical pencils though.

I prefer to write on a blank paper if I'm working on geometry problems. Other than that, I'm fine with any kind of paper except for a quad-ruled paper.

What's "quad-ruled" paper? Is it graph paper?

For math, I absolutely have to have blank, white, perfectly clean paper. Actually it only has to be clean on one side--I'm fine with scrap paper and use it most of the time. Naturally it's all loose leaf and usually normal-sized, though sometimes I end up with half-pages when my teachers use them for grade sheets.

Any scrap paper at hand, usually an old school newspaper. I usually write over the articles if I have to open the paper to find blank space.
Or Paint, if I'm on the computer.

I use spiral notebooks, they're about 5 by 8 inches, and the spirals are on the top.   I like using these b/c they're easy to carry around and i can work on them anywhere b/c they dont need something underneath them for me to write on them

I use a small little quad-rule book about 5 by 7 inches.

closest piece of paper that I can write on works for general purpose math
low quality computer paper for homework

[quote="karthik_k"]I'll prefer blank paper anyday......I feel you're much more free to write however you want on blank paper... :)[/quote]

Most definitely.  If given a choice, I prefer blank paper.  However, if I'm doing geometry, graph paper can come in handy.  If given the choice, I will almost never use lined paper.

Basically whatever I can get my hands on...  Often the back of my little brother's worksheets from school (elementary) that come home.  Do you have any clue how much paper elementary schools waste?  It is crazy!  After it is filled with scribbles, numbers, etc. it goes into the recycle bin...


calculus
derivative
function
ratio
inequalities
integration
geometric sequence

Yes, of course that is the function, but proving unicity seems extremely hard.My idea was to put $a=inf_{x>0}{\frac{xf'(x)}{f(x)}}$ and $b$ the sup of the same function. Then the functions $ g(x)=\frac{f(x)}{x^a}$ and $ h(x)=\frac{f(x)}{x^b}$ are increasing and respectively decreasing. The problem is that I cannot prove they have finite limits at infinity and in 0. If one can show that, I solved the problem, but something tells me this is the hard part.

This is a nice problem (like most problems by Harazi  :) ). I have a solution but I do not want to spoil the fun so soon. Nevertheless, if you run out of ideas, here is a small [hide="Hint"] Suppose that you can show that $f(x)\geqslant (\leqslant) cx^p$ on $[0,a]$. Then, if you use the identity $f'(x)=f^{-1}(x)$ in a wise way, you can conclude from here that $f(x)\leqslant(\geqslant) c'x^{p'}$ on $[0,a']$ with certain $a',c'$ and $p'$. Then you can do it again and again. Show that $f(x)\geqslant \frac 12 x^2$ on $[0,2]$ and see where it leads. Hope this helps a little bit   :P [/hide]

[quote="grobber"]I know this is the easy part, but it's a start :).

We can prove the existence by taking $f(x)=\alpha x^{\beta}$ with $\beta=\frac{1+\sqrt 5}2$ and $\alpha=\beta^{-\frac \beta{\beta+1}}$.[/quote]

For those of us not so well versed, how do we show that this is true? (Well I presume we can show it by differentiating and finding the inverse although this didn't work for me ;) - how do we find it?)

You just take $ f(x)=Ax^B$, compute it's inverse, which is $ (x/A)^{\frac{1}{B}}$ and impose the condition that $ (x/A)^{\frac{1}{B}}=AB x^{B-1}$. Then you must have $ \frac{1}{B}=B-1$ and then yyou find $A$.

Of course! So obivious  :blush:

Thank you for the prompt reply

Fedja will surely find again a stupid mistake, so I apologise in advance for the following, but this problem really took me mad:
     First, I considered the function $ g(x)=\frac{f(x)}{x}$. It is not difficult to see it is strictly increasing (this reduce to proving that $ f^{-1}(x)x-f(x)>0$ or $ xf(x)-f(f(x))>0$, which follows from the fact that $ h(x)= xf(x)-f(f(x))$ has the derivative $f(x)>0$ and the limit in 0 equal to 0-which is obvious). Moreover, the limit of $g$ in 0 is 0 (l'Hospital roule + the fact that the derivative=inverse of the function which goes towards 0 in zero) and the limit of $g$ in infinity is infinity (the same reason). So, there is $x_0$ a fixed point for $f$. I will prove that $f$ is uniquely determined on the interval $ (0,x_0)$ and the same will hold for $x>x_0$, from where the conclusion. Since $g$ is strictly increasing, for $x<x_0$ we have $ f(x)<x$. Thus $ f^{-1}(x)>x$ for $x$ between $0$ and $x_0$  ( we see that $ f^{-1}(x)<x_0$ for $x<x_0$ since $x_0$ is fixed point).  Thus $ f'(x)>x$ for $x$ in $I=(0,x_0)$. Since the limit of $f$ in 0 is $0$ we find that $ f(x)>\frac{x^2}{2}$. Suppose we have proved that $ f(x)\geq a_n x^{b_n}$. Then it is easy to see that $ f'(x)\leq \frac{x^{\frac{1}{b_n}}}{a_n^{\frac{1}{b_n}}}$ and then $ f(x)\leq\frac{1}{(1+1/b_n) a_n^{\frac{1}{b_n}}} x^{\frac{1+b_n}{b_n}}$. Take again the inverse and then replace it with its derivative and obtain in the end
    $f(x)\geq \frac{1}{1+\frac{b_}{b_n+1}}\cdot a_n^{\frac{1}{1+b_n}} (1+1/b_n)^{\frac{b_n}{b_n+1}} x^{1+\frac{b_n}{b_n+1}}$. Thus, define the sequence $ a_1=1/2, b_1=2$ and $ a_{n+1}=a_n^{\frac{1}{1+b_n}}\cdot\frac{b_n+1}{2b_n+1}\cdot (1+1/b_n)^{\frac{b_n}{1+b_n}}$ and $ b_{n+1}=1+\frac{b_n}{b_n+1}$. Then we have seen that $ f(x)\geq a_n x^{b_n}$ for all $n$ and all $x$ in $I$ and also that $ f(x)\leq\frac{1}{(1+1/b_n) a_n^{\frac{1}{b_n}}} x^{\frac{1+b_n}{b_n}}$.
     The sequence $b_n$ is trivial to show that converge towards the golden number. Then, the fact that $ (1+b_1)...(1+b_k) ln a_{k+1}-(1+b_1)...(1+b_{k-1}) ln a_k= b_k (1+b_1)...(1+b_{k-1})+(1+b_1)...(1+b_k) ln\frac{b_k+1}{2b_k+1}$ showh by summation and application of cesaro-Stolz theorem that $a_n$ also converges. Let $x$ be it's limit. Then if $t$ is the golden ration, we will have $ x=x^{\frac{1}{1+t}}\cdot\frac{1}{t} t^{\frac{1}{t}}$ thus $x=1/t^{\frac{1}{t}}$. Thus $ f(x)\geq g(x)$ on $I$, where $g$ is the function considered by Grobber. if we let $n$ very large in $ f(x)\leq\frac{1}{(1+1/b_n) a_n^{\frac{1}{b_n}}} x^{\frac{1+b_n}{b_n}}$ we see that we also have the opposite inequality. Thus the functions are identical on I.
     Sorry again for the many idiot mistakes I surely have done. At least I know the method must be good...

[quote="harazi"]Suppose we have proved that $f(x)\geq a_n x^{b_n}$. Then it is easy to see that $f'(x)\leq \frac{x^{\frac{1}{b_n}}}{a_n^{\frac{1}{b_n}}}$[/quote]
On what interval? You should be careful here: if you have an inequality for $f$ on $[0,x_0]$, you have an inequality for $f^{-1}$ on a different interval, namely on $[0,a_n x_0^{b_n}]$, which may be smaller if you are unlucky. So, one has to be just a little bit more careful here.   ;)

I supposed that $f(x)\geq a_n x^{b_n}$ (*) on the interval $I=(0,x_0)$.Now it is clear that since $x_0$ is fixed then $f$and $f^{-1}$ define bijective functions on $I$. Then we can replace $x$ by $f^{-1}(x)$  on this interval and the inequality $f^{-1}(x)=f'(x)\leq \frac{x^{\frac{1}{b_n}}}{a_n^{\frac{1}{b_n}}}$ remains true over I. I am sorry for not explaining well what I thought. Is it correct now?

Oh, sorry. This time I was stupid  :blush: . Indeed, it works as you said. I just didn't realize that you essentially use the fact that $f(x_0)=x_0$. One more stupid question then: how do you show uniqueness for $x>x_0$?

No, it's not stupid at all. The idea is the same. We have $ f(x)>x$ on $ J=(x_0,\infty)$, thus since again $x_0$ is fixed we have $ f(x)-x^2/2$ increasing and its value in every point of J is greater than the value in $x_0$, which is greater than 0 (as we have already shown). I think that the same arguments as above (and a little care, but anyway this problem has already taken my breath) can show that the same inequalities as those on $I$ remain true on $J$. Fedja, please post your solution for this problem.

[quote="harazi"] Fedja, please post your solution for this problem.[/quote] Basically it is the same as yours. The main difference is that I start with the inequality $f(x)\geqslant \frac {x^2}2$ on $[0,2]$. To prove it, consider 2 cases: $f(x)\geqslant x$ and $f(x)<x$. In the first case there is nothing to do because $x\geqslant \frac{x^2}2$ on $[0,2]$. In the second case note that $f$ is convex (the derivative is increasing) and, therefore, $f^{-1}$ is concave and $f(x)=\int_0^x f^{-1}(t)\,dt\geqslant \frac 12xf^{-1}(x)>\frac {x^2}2$ since $f(x)<x$ implies $f^{-1}(x)>x$. Now, I use the same technique as you. To make it formal, let us call a triple $(a,b,g)$ a comparison box if $g$ is a power function $cx^p$ with $c>0$, $p>1$, $g(a)=b$ and either $f\leq g$ everywhere on $[0,a]$ or $f\geq g$ everywhere on $[0,a]$. We start with the comparison box $(2,2,\frac12x^2)$. Now, if we can compare $f$ to $g$ on $[0,a]$, we can compare $f^{-1}$ to $g^{-1}$ on $[0,b]$ (with the opposite inequality sign). Integrating, we get a new comparison box $(\widetilde a,\widetilde b, \widetilde g)$ where $\widetilde a=b$, $\widetilde b=\int_0^b g^{-1}(t)\,dt$ and $\widetilde g$ is another power function (the same as in your post). There are two things to take care of: the comparison function $g$ and the size of the box. The first one is done in more or less the same way as in your post. As to the size of the new box, I used convexity of $g$ to conclude that $\widetilde b>ab/2\geqslant a$ if $b\geqslant 2$. So, starting with the box whose both sides are $2$, we get a box whose both sides are strictly greater than $2$ in 2 steps after which the sides grow as a geometric progression, so finally we get $f$ compared to our power functions on the entire $(0,+\infty)$. The only advantage of this solution compared to yours is that it gives the result on the entire $(0,+\infty)$ at once.

I also played for a while with the idea to find a function space in which the mapping $f(x)\mapsto \int_0^x f^{-1}(t)\,dt$ is a contraction, but didn't succeed.

I thougt about this problem and that is my result:
 Because $f:R^+->R^+$ is a biection => $f^{-1}:R^+->R^+$  so $f^{-1}(x)>0$ for all $x$ on $R^+ $

$(*)$ $f^{'}(x)=f^{-1}(x)$
So $f^{'}(x)>0$. We have that $f(x)$ is differentiable and $f^{'}(x)\not=0$, so $(f^{-1}(x))^{'}=\frac{1}{f^{'}(x)}$. So, we can differentiate (*): $f^{''}(x)=\frac{1}{f^{'}(x)}$ or $f^{''}(x)f^{'}(x)=1$ . By integration the last expression we brought $\frac{{f^{'}}^{2}(x)}{2}=x$  or $f(x)=\frac{2\sqrt{2}}{3}x^{\frac{3}{2}}$
But I see that $f^{'}(x)\not=f^{-1}(x)$ . So there is no (!?) function with this properties....

[quote="Genrih"]
$(f^{-1}(x))^{'}=\frac{1}{f^{'}(x)}$. [/quote]
This is a very common mistake. The right formula is $(f^{-1})'(f(x))=\frac 1{f'(x)}$

OOu yes! thank you ,   [b]fedja [/b]. it's a really  big mistake

[quote="harazi"]Show that there exists a unique bijection from the set of positive real numbers with values in the same set that verifies $ f'(x)=f^{-1}(x)$ where $f^{-1}$ is of course its inverse.[/quote]
Oops ! I discovered in my notebook with  some [i] "interesting problems" [/i], an earlier  source of this problem. This is :

           [b][1]   H.L. Nelson , [/b] [i]Problem  E2105[/i], Amer.Math.Monthly 76([b]1969[/b]) .

[It seems that the people of ENS  select many  problems from Amer.Math. Monthly !  [b] Indeed, nice ![/b]]

In the (published)  solution it's used the fact that  $ f^{\prime}\left(f(x)\right)=x ,\; \forall x\in I:=(0,\infty) , $ implies that   $ f\in C^{\infty}(I) $ and moreover that $ f $ is an analytic function [according to a theorem by  S.N. Bernstein].
Then, integrating $ f^{\prime}\left(f(x)\right)f^{\prime}(x)=xf^{\prime}(x) $  one finds
$ \; (1)\; \; \; \; \; \; f\left(f(x)\right)=\int\limits_{0}^{x}tf^{\prime}(t)\; dt .$  Using (1) ,  it's shown that the inequalities $ f(x)<x \; ,\; \forall x \in I ,$ or   $\; \; (2)\; \; \;  f(x)>x ,\; \forall x \in I , $ are not  possible. For instance, (2) implies $ f^{\prime}(t)<t ,\; \forall t \in I, $ and from (1)  $   x<\int\limits_{0}^{x}t^2\; dt= \frac{x^3}{3}\; \; \forall x \in  I,   $ that is a contradiction.    Therefore there exists  $ a \in I $ such that $  f(a)= a . $ Because $ f(x)-x $  is strictly convex  on $ I $ , it follows that there exists at most two fixed points $ a_1,a_2 . $ But one fixed point   is $a_2:=0 \not \in I .$  In conclusion there exists only one fixed point $ a \in I , $  ....  [further a similar augumentation for $ f^{\prime} , $ ....[better , please see  the published solution  of [b][1][/b] .

Indeed, AMM is a frequent source of problems for oral examinations at ENS. And it is normal, when you know how strong teachers and researchers are there. But who thinks to the poor student who has to solve these nice but very difficult problems? You know, someone like me for example?  :(  :(  :D

Hello!
This is my first post. I have decided to join your community when a friend of mine told me that he found this problem on your site :). This is a problem that I enjoyed a lot (thought it was mine:)) ). Here is the solution that I found. Hope it is correct!
[quote]
First of all, let's observe the following: (1) $f(0+)=0$, $f(\infty)=\infty$ and $f\nearrow$.
It is relatively easy to decide to look for a solution for this equation of the form $f(x)=ax^{\alpha}$ after noticing (by l'Hospital) that $\frac{f(x)}{x}\stackrel{x\longrightarrow0}{\longrightarrow}0$; $\frac{f(x)}{x^2}\stackrel{x\longrightarrow0}{\longrightarrow}\infty$; $\frac{f(x)}{x}\stackrel{x\longrightarrow\infty}{\longrightarrow}\infty$; $\frac{f(x)}{x^2}\stackrel{x\longrightarrow\infty}{\longrightarrow}0$.
After solving the equation $f^{'}(x)=f^{-1}(x)$ for $f(x)=ax^{\alpha}$ we find the solution:

(2)$f(x)=\phi^{1-\phi}x^\phi$, where $\phi=\frac{\sqrt{5}+1}{2}$ is the golden number.

Let us now prove that (2) is the unique solution.

Consider the sequence $\alpha_0=1, \alpha_n=1+\frac{1}{\alpha_{n-1}},n\geq1$. It's easy to prove that (3)  $\alpha_{2n}\nearrow$,  $\alpha_{2n+1}\searrow, n\geq0$ and that (4) $\alpha_n\stackrel{n\longrightarrow\infty}{\longrightarrow}\phi$.

In order to prove the uniqueness of the solution we shall prove that $g: (0,\infty) \rightarrow (0,\infty), g(x)=\frac{f(x)}{x^\phi}$ is a constant function.

Let $g_n: (0,\infty) \rightarrow (0,\infty), g_n(x)=\frac{f(x)}{x^{\alpha_n}}, n\geq0$.

Let us notice that $g_0=\frac{f(x)}{x}$ and that (5) $g_0\searrow$.
That is because $g_0(x)\searrow\Leftrightarrow g'(x)<0\Leftrightarrow f(x)-xf'(x)<0$. Denote $h: (0,\infty) \rightarrow (-\infty,\infty), h(x)=f(x)-xf'(x)$. We have $h(0+)=0$ and $h'(x)=f'(x)-f'(x)-xf''(x)=-x(f^{-1}(x))'=-x\frac{1}{(f'\circ f^{-1})(x)}=-x\frac{1}{(f^{-1}\circ f^{-1})(x)}$. This implies $h'<0$, hence $h\searrow$ and $h<0.$

Next we have that $g_n\nearrow\Leftrightarrow g'_n(x)>0\Leftrightarrow \alpha_nf(x)-xf'(x)>0$.
Denote $h_n: (0,\infty) \rightarrow (-\infty,\infty), h_n(x)=\alpha_nf(x)-xf'(x)$. We have $h_n(0+)=0$ and
$h'_n(x)=\alpha_nf'(x)-f'(x)-xf''(x)\Leftrightarrow$ $h'_n(x)=(\alpha_n-1)f'(x)-xf''(x)\Leftrightarrow h'_n(x)=\frac{1}{\alpha_{n-1}}(f'(x)-\alpha_{n-1}xf''(x))$

Let us notice that
$h'_n(f(x))=\frac{1}{\alpha_{n-1}}(f'(f(x))-\alpha_{n-1}f(x)f''(f(x))) \Leftrightarrow h'_n(f(x))=\frac{1}{\alpha_{n-1}}(f^{-1}(f(x))-\alpha_{n-1}f(x)\frac{1}{f^{-1}(f^{-1}(f(x)))})\Leftrightarrow h'_n(f(x))=\frac{1}{\alpha_{n-1}}(x-\alpha_{n-1}\frac{f(x)}{f'(x)}) \Leftrightarrow h'_n(f(x))=\frac{xf'(x)-\alpha_{n-1}f(x)}{\alpha_{n-1}f'(x)} \Leftrightarrow h'_n(f(x))=-\frac{g'_{n-1}(x)}{\alpha_{n-1}f'(x)}$

Hence (6) $h'_n(f(x))>0 \Leftrightarrow g'_{n-1}(x)<0$

But since $f: (0,\infty) \rightarrow (0,\infty)$, it is obviously that $h'_n(f(x))>0 \Leftrightarrow h'_n(x)>0$. From this and (6) we have that $h'_n(x)>0 \Leftrightarrow g'_{n-1}(x)<0$. Following this, $h_n\nearrow \Leftrightarrow g_{n-1}\searrow$. From this, $f(0+)=0$ and the fact that $g_n\nearrow\Leftrightarrow h_n>0$ we get that (7) $g_n\nearrow \Leftrightarrow g_{n-1}\searrow$.

From (5) and (7) we get that (8) $g_{2n}\searrow$,  (9) $g_{2n+1}\nearrow, n\geq0$.

Given $0<x<y$, from (8) we get that $\frac{f(x)}{x^{\alpha_{2n}}}>\frac{f(y)}{x^{\alpha_{2n}}}$. Now, making $n\rightarrow\infty$ we get that (10) $\frac{f(x)}{x^\phi}\geq\frac{f(y)}{x^\phi}$. Also, for the same x and y, we get from (9) that $\frac{f(x)}{x^{\alpha_{2n+1}}}<\frac{f(y)}{x^{\alpha_{2n+1}}}$. Making again $n\rightarrow\infty$ we get that (11) $\frac{f(x)}{x^\phi}\leq\frac{f(y)}{x^\phi}$.

Now, from (10) and (11) we get that $\frac{f(x)}{x^\phi}=\frac{f(y)}{x^\phi}$ for any $0<x<y$ which complets the proof. $\hookleftarrow$
[/quote]

[quote="flip2004"] ...[color=red][b] Wellcome ! [/b] [/color] ...[/quote]
[quote]
[b] Proposed question:    [color=blue]  Denote by ${\mathcal F}$  the set of funtions  $y: (0,\infty)\to (0,\infty) \; ,\; y\in C^{1}(0,\infty) \; ,$  and    let $y_0$ be given in $(0,\infty).$   Investigate if  the (non-linear) differential-equation  $\begin{array}{|c|} \hline y^{\prime}\left(y(x)\right)-x=0 \\ \hline \end{array} \; \; \; ,\; \; y(1)=y_0 \; ,$ has   unique solution $y \in {\mathcal F}$. [/color] [/b]
[/quote]

[quote]
Given the previous problem, I think that this differential-equation has a solution if and only if $y_0$ has the specific value of $\phi^{1-\phi}$ and in that case the solution is unique.

Do you have other examples of solutions of this equation for different values of $y_0$?

[/quote]

[quote="saab"] Do you have other examples of solutions of this equation for different values of $y_0$?
[/quote]
No, I  have'nt . I think that your remark is OK. Perhaps a better formulation is: [b]   to find [/b] $y_0$ such that the diff.equation .... to have a (unique) solution with $y(1) =y_0$ ./Alex

[color=red]One more challenge ![/color]
 After this nice argumentation , I wonder if we can find a real function defined on the set of the real numbers with continous derivative and
 $(f^{-1}(x) )' = (f')^{-1}(x) , \forall x \in IR$

 [u]Babis[/u]

in fact when i was in the third grade of high school,this problem has been come to my mind. :D
see http://www.mathlinks.ro/Forum/viewtopic.php?p=408096#408096
but i coudnt find all those functions,but today see the problem is solved.

[quote="fedja"]This is a nice problem (like most problems by Harazi  :) )[/quote]

Where can those ENS problems be found (is there some webpage for example or a magazine). Or is this only for ENS students? What course was this actually in :S Or are analysis courses at ENS given really at such a high level :p

[quote="harazi"]There is a frech magazine called Revue de Mathematiques Superieures, you can google and find its site. I imagine you can subscribe, because of course it is not free. No, that is not for ENS students only, it is actually the problems they give to select their students, and that is different. As for the course, it was the "analysis course"  :rotfl: . This was not an ENS course of analysis, it was preparatory for entrance. The courses given at ENS are a little bit harder (some of them much harder...), but yes, usually for a first year in school they are pretty challenging. Anyway, the system is different: first year at ENS means third year of university in France, and we still have 3 more years here.[/quote]

I see. So ENS is a school for some kind of a 'master-program' (I don't understand french - so I can't figure out what their site has to say) or are you at the moment 18 years old and are you preparing for your entrance-exam at ENS (so it's a private college in france  :o and only few people are allowed to follow the courses?) . Or have you done already 2 years in France at a regular college and are you finishing your study at ENS?

No, I'm not 18 years old and I'm in the first year at ENS. I had passed the 2 preparatory years and I still have 3 years except this one at ENS. It's pretty strange, I know, but this is the system here: either you go to universities and follow a standard 4 years program, either you first make 2 preparatory years and then 4 years at ENS or Ecole Polytechnique or some other such schools. Of course they are not private, but it is true that only a few people study there: but not because of prices (everything is for free, students are even paid!), because of exams and courses. It's more oriented towards research than engineering. Anyway, if you do have some other questions, please send a private message,  I'm afraid we're just spamming a bit this topic.


number theory
least common multiple
MATHCOUNTS
modular arithmetic

[quote="aayyaa"]can u name the smallest number which
when divided by 10 gives rem 9
when divided by 9 gives rem 8
when by 8 gives rem 7 and so on
and
down to rem of 1 when by 2??????????[/quote]

I saw this problem around a while ago...

[hide="Super Hint"]Add 1 to the number.[/hide]

If the number is $ x$, wouldn't we say $ x\equiv 9 (\mbox{mod } 10)$, $ x\equiv 8 (\mbox{mod } 9)$, $ x\equiv 8 (\mbox{mod } 8)$,..., $ x\equiv 1 (\mbox{mod } 2)$?  Unfortunately, I'm not a particularly skilled number theorist so I'm not really sure where to go from here.

Hmm.
This is very similar to a question for this month's competition at my school.  :maybe:
Though the question itself can be seen other places.

not sure but we may be able to use chinese remainder theorem

I think bertam has hint the nail on the head

[hide="Almost a solution"]
If $ n \equiv 9 (mod 10) \Rightarrow n \equal{} 10k \plus{} 9$ What does that say about n+1?
[/hide]

First of all, we must [hide="Assume..."]The answer we are looking for is not negative and it is not 1.  Otherwise, the answer doesn't exist or the problem is trivial.[/hide]

[hide="Then..."]Let the number be n.  Then n+1 is divisible by 10, 9, 8, ... , 2, 1.  Thus,
$ n \plus{} 1 \equal{}lcm(10,9,8,7,6,5,4,3,2,1) \equal{} 5670$
So $ n \equal{} 5669$[/hide]

Here's how I see it

[hide]

$ n \plus{} 1$ Will be divisible by all numbers  1 - 10, so

$ n \equal{} lcm(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) \minus{} 1$
$ n \equal{} 2519$
[/hide]

I'm quite sure this problem appeared on a 1990-something Mathcounts state competition.

[quote="alanchou"][quote="ZhangPeijin"]You may notice that the number you are looking for is just 1 less than a multiple of 10,9,8,7,6,5,4,3,and 2.
So just subtract one from the LCM of those numbers.[/quote]

That's exactly the same thing as what's above, just worded differently[/quote]
What?

Let's do it the 'dodgy' way:

Let the number be $ N$ and let's find $ N\minus{}1$

Now $ N\minus{}1$ is divisible by $ 2$ so it equals $ 2 \times \text{something}$

Now $ N\minus{}1$ is divisible by $ 3$ so it equals $ 2 \times 3 \times \text{something}$

Now $ N\minus{}1$ is divisible by $ 4$ but it already has a factor of $ 2$ so it equals $ 2 \times 3 \times 2 \times \text{something}$

Now $ N\minus{}1$ is divisible by $ 5$ so it equals $ 2 \times 3 \times 2 \times 5 \times \text{something}$

Now $ N\minus{}1$ is divisble by $ 6$ but it already has $ 2,3$ as its factors

Now $ N\minus{}1$ is divisble by $ 7$ so it equals $ 2 \times 3 \times 2 \times 5 \times 7 \times \text{something}$

Now $ N\minus{}1$ is divisible by $ 8$ but it already has the factor $ 2^{2}$ so it equals $ 2 \times 3 \times 2 \times 5 \times 7 \times 2 \times \text{something}$

Now $ N\minus{}1$ is divisble by $ 9$ but it already has the factor $ 3$ so it equals $ 2^{2}$ so it equals $ 2 \times 3 \times 2 \times 5 \times 7 \times 2 \times 3 \times \text{something}$

Finally $ N\minus{}1$ is divisible by $ 10$ but it already has the factor $ 2 \times 5$. Let $ \text{something}\equal{}1$ to minimise $ N\minus{}1$.

$ N\equal{} 2521$ :D  Sorry for the length.

Close, but we want to find $ N \plus{} 1$ then it will be divisible by all numbers 2, 3, 4, 5, 6, 7, 8, 9, 10.

Think about it

if $ x \equiv 9 \pmod{10}$, then

$ x \plus{} 1 \equiv 0 \pmod{10}$


calculus
integration
real analysis
real analysis solved

For all $a$ in $(\frac{1}{i}:i\in\Bbb N_0)$ calculate $\int_0^{\infty}e^{-x^a}dx$.

I didn't wanted to post it into the Calculus section because it might be well-known to many people with some Calculus experience.

Hope you like it!

Misha

indeed, nice. i got that result by random just one or two months ago...

Change variable $y=x^a$ and related to $\Gamma(z)=\int_{0}^{+\infty}y^{z-1}e^{-y}dz$


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

$ \frac{a^3\plus{}3a\plus{}2}{b\plus{}c}\plus{}\frac{b^3\plus{}3b\plus{}2}{c\plus{}a}\plus{}\frac{c^3\plus{}3c\plus{}2}{a\plus{}b}\geq9$


The  equation $(1+x^3)^4+(1+x^2)^4=2x^4$ has real roots ?

Did you solve it Silouan? I hope so. Please , do not send your solution yet. We must wait the answer of  other friends in the ML forum.

[u]Babis[/u]

[quote="biffanddoc"]It has $12$ roots because it is degree $12$.[/quote]

You are wrong. Soarer's solution is as mine. But $1+x^2\geq 2\left| x\right|$

Solouan , Biffanddoc is right ! You forgot to ask only real roots. Otherwise , in the set of complex numbers , we have 12 roots.

    [u]To Biffanddoc[/u]

 Please , you must forgive Silouan , becouse he  has not been taught Complex numbers untill now in the Greek School.

[u] Babis[/u]

I know about complex numbers. But from the equation we clearly want real solutions.
Sorry for the misunderstanding

I know it asked for real numbers but
[quote="seamusoboyle"]Even for complex numbers?[/quote]
So I replied to that


Μπορεί κάποιος να μου πει τι ώρα είναι ο αρχιμήδης γιατί έχασα το χαρτί που το έλεγε... :blush:

[quote="pontios"][quote="meinstein"]Δεω είμαι σίγουρος αλλα νομίζω στις 9. Το λέει πάντως για να σιγουρευτείς κ στη σελίδα της ΕΜΕ.[/quote]
Δεν το λέει στη σελίδα της ΕΜΕ

[hide="@meinstein"]Όπως βλέπεις, μπορώ να προβλέψω τι θα ειπωθεί σ'αυτό το φόρουμ  :D
Η ώρα είναι 10:27 και ξέρω τι θα γράψεις στις 12:41 [/hide][/quote]xmmm poly parakseno...bas kai eisi i pythia?
tespa pote lete na dimosieytoyn ta themata?


Both x and y are positive real numbers, and the point (x,y) lies above or on the two lines 2x+5y=10 and 3x+4y=12.  What is the least possible value of 8x+12y?

Are you sure it isn't 8x+13y? Because we would have $2x+5y\ge10,3x+4y\ge12$, adding the first to twice the second would give $8x+13y\ge34$, which if I recall correctly was the correct answer.[/hide]

[hide]
I multiplied the second equation by 2: 6x+8y=24.
Then, add the two equations up: 8x+13y=34.
Since they can be on the two lines, 34 is the answer. [/hide]


I asked this problem ,but anyone cared about this problem.
$ a_{1},a_{2},....,a_{2009}> 0$  and $ a_{i}\geq a_{1}\plus{}...\plus{}a_{i\minus{}1}$
$ \implies$  find the maximum value of that
$ \frac{a_{1}}{a_{2}}\plus{}\frac{a_{2}}{a_{3}}\plus{}...\plus{}\frac{a_{2008}}{a_{2009}}$

I have replied in your topic , why don't you post your discussion there ?

 [url]http://www.mathlinks.ro/viewtopic.php?t=297841[/url]


Let $ W$ be a subspace of a vector space $ V$ over a field $ F$. For any $ v\in V$, the set $ v\plus{}W\equal{}\{v\plus{}w: w\in W\}$ is called the coset of $ W$ containing $ V$.

Prove $ v_1\plus{}W\equal{}v_2\plus{}W$ if and only if $ v_1\minus{}v_2\in W$.

Also, for the collection $ S\equal{}\{v\plus{}W: v\in V\}$ of all cosets of $ W$, we define:

$ (v_1\plus{}W)\plus{}(v_2\plus{}W)\equal{}(v_1\plus{}v_2)\plus{}W$ for all $ v_1,v_2\in W$ and $ a(v\plus{}W)\equal{}av\plus{}W$ for all $ v\in V$ and $ a\in F$.

Prove if $ v_1\plus{}W\equal{}v_1'\plus{}W$ and $ v_2\plus{}W\equal{}v_2'\plus{}W$, then $ (v_1\plus{}W)\plus{}(v_2\plus{}W)\equal{}(v_1'\plus{}W)\plus{}(v_2'\plus{}W)$.

This is to be proven from basic principles, since this leads up to the definition of $ V/W$.

Hints for the first part:

[hide]
(->)
Note that if $ v_1 \plus{} W \equal{} v_2 \plus{} W$ then for any $ w_1\in W$ there exists a $ w_2\in W$ such that $ v_1 \plus{} w_1 \equal{} v_2 \plus{} w_2$. Proceed from there (remember that we have closure in $ W$!)

(<-)
Since $ v_1 \minus{} v_2 \in W$, then $ v_2 \plus{} (v_1 \minus{} v_2) \in v_2 \plus{} W$. Also, $ v_2 \minus{} v_1$ better be in $ W$, so $ v_1 \plus{} (v_2 \minus{} v_1) \in v_1 \plus{} W$. Use this to show that any element of $ v_1 \plus{} W$ is found in $ v_2 \plus{} W$ and vice versa.

Hope this helps!
[/hide]

Thanks I actually figured out the first part in class this morning. Could you give a hint for the other problem?

Use part 1 to show $ (v_1 \plus{} v_2) \plus{} W \equal{} (v_1' \plus{} v_2') \plus{} W$


[quote="aidan"]While shopping at the local mall, Ann walked slowly (at a constant speed) down a descending escalator, taking 10 steps to reach the last step. While on the lower level, she heard that free concert tickets were being given away on the next level up. She ran up the same (descending) escalator with a speed five times greater than the speed with which she had walked down, taking 25 steps to reach the top.

Find the number of steps showing on the escalator at any one moment.

Source: Mathematics Challenge for Young Australians, Problem 71(a)[/quote]

[hide]Going up takes half the time so $\frac{25-x}{x-10}=\frac{1}{2}$ Therefore, $x=\boxed{20}$ steps[/hide]

[quote="lotrgreengrapes7926"][quote="aidan"]How do you know that going up takes half the time? Could you please explain in detail? I'd appreciate it and btw, your answer is correct.[/quote]
[hide] She took 25 steps 5 times faster, in the same amount of time as 5 steps at the slower rate. [/hide][/quote]

Yeah, and everybody knows 5 is half of 10

Lol.. please clarify, SplashD. I will post the official solution if desired.


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

Let A,B belong to M_n(R) so that A^mB=A^m+B, where m is natural. Prove that AB=BA

(A m-I)(B-I)=I, so B-I is the inverse of A m-I. But the inverse of every matrix C is a polynomial of C (use Hamilton-Cayley), so B-I is a polynomial of A m-I, so B is a polynomial of A, so A and B commute.


Given the equation x^2 + y^2 + 6x + 2y + 6 = 0 ,
complete the square and write the equation in standard form

A. (x + 3)^2 + ( y +1)^2 = 4

B. (x + 3)^2 + ( y −1)^2 = 4

C. (x −1)^2 + ( y − 3)^2 = 4

D. (x + 6)^2 − ( y + 2)^2 = 4

E. none of these

$ x^2\plus{}6x\plus{}y^2\plus{}2y\plus{}6\equal{}0\implies (x^2\plus{}6x\plus{}9)\plus{}(y^2\plus{}2y\plus{}1)\equal{}4\implies (x\plus{}3)^2\plus{}(y\plus{}1)^2\equal{}4$

Thus, the answer is A


Prove that  for all positive integers $n$, \[\lfloor \sqrt[3]{n}+\sqrt[3]{n+1}\rfloor =\lfloor \sqrt[3]{8n+3}\rfloor.\]

I work on the principle "a few words are worth a paragraph of $ \text{\TeX}$", so let's try at least to give some explanations.
First of all, let's notice that it suffices to prove the inequalities
$ \sqrt[3]{8n + 3} < \sqrt[3]{n} + \sqrt[3]{n+1} < \sqrt[3]{8n + 4}$ $ (\star)$, for all $ n \in \mathbb{N}$.
Hence $ \lfloor \sqrt[3]{8n + 3} \rfloor \le \lfloor \sqrt[3]{n} + \sqrt[3]{n+1} \rfloor \le \lfloor \sqrt[3]{8n + 4} \rfloor$. Now it's a well-known fact that if $ \sqrt[3]{8n + 4}$ is irrational, $ \lfloor \sqrt[3]{8n + 3} \rfloor = \lfloor \sqrt[3]{8n + 4} \rfloor$, which proves our identity. Otherwise, $ \sqrt[3]{n} + \sqrt[3]{n+1} < \sqrt[3]{8n + 4}$ yields, in the view of $ \sqrt[3]{8n + 4} \in \mathbb{N}$, $ \lfloor \sqrt[3]{n} + \sqrt[3]{n+1} \rfloor \le \lfloor \sqrt[3]{8n + 4} \rfloor - 1 = \lfloor \sqrt[3]{8n + 3} \rfloor \le \lfloor \sqrt[3]{n} + \sqrt[3]{n+1} \rfloor$, which was to be proved.
Now $ (\star)$ is equivalent to
$ 2n + \frac{2}{3} < \sqrt[3]{n^2(n+1)} + \sqrt[3]{n(n+1)^2} < 2n + 1$.
At first, we prove the leftmost part. This is implied by
$ n + \frac{1}{5} < \sqrt[3]{n^2(n+1)}$ and
$ n + \frac{7}{15} < \sqrt[3]{n(n+1)^2}$.
The first reduces to $ 15n + 1 < 50n^2$, which is obvious, and the second to $ 343 < 2025n^2 + 1170n$, again clear.
The rightmost is broken even more straightforward into
$ \sqrt[3]{n^2(n+1)} < n + \frac{1}{3}$ and
$ \sqrt[3]{n(n+1)^2} < n + \frac{2}{3}$.
The first reduces clearly to $ 9n + 1 > 0$, and the second to $ 9n + 8 > 0$. The inequalities have been proved.
I have used above a well-known
[b]Lemma.[/b] Let $ n, k$ be positive integers, $ k \ge 2$. Then,
$ \sqrt[k]{n} \in \mathbb{R} \backslash \mathbb{Q} \Rightarrow \lfloor \sqrt[k]{n-1} \rfloor = \lfloor \sqrt[k]{n} \rfloor$.
$ \sqrt[k]{n} \in \mathbb{N} \Rightarrow \lfloor \sqrt[k]{n-1} \rfloor + 1 = \lfloor \sqrt[k]{n} \rfloor$.
The proof is standard as it is left to the reader  :arrow:


Find $ p \in P$ such that equation $ x^3\plus{}y^3\equal{}p^z$ has integer root

It is well-known

Since $ x^3\plus{}y^3\equal{}p^z$ and $ p$ is a prime, we have

$ x\plus{}y\equal{}p^t$ and $ x^2\minus{}xy\plus{}y^2\equal{}p^m$

where $ t\plus{}m\equal{}z$.

Then $ gcd(x\plus{}y,3xy)\equal{}p^{min(t,m)}$

If $ min(t,m)\geq1$ then:
If x, y are both divisible by $ p$, we reduce the problem to $ x'^3\plus{}y'^3\equal{}p^{z\minus{}3}$ and continue the process until they are both not divisible by $ p$. In such case, both $ x,y$ are not divisible by $ p$ and this leads to $ 3xy$ is divisible by $ p$ iff $ 3$ is divisible by $ p$. And hence $ p\equal{}3$.
If $ min(t,m)<1$ then $ x\plus{}y\equal{}1$ and $ x^2\minus{}xy\plus{}y^2\equal{}p^z$ or $ x\plus{}y\equal{}p^z$ and $ x^2\plus{}y^2\minus{}xy\equal{}1$. Such cases are easily handled.


Let $\alpha$ be a real number, $1<\alpha<2$.

(a) Show that $\alpha$ can uniquely be represented as the infinte product \[ \alpha = \left(1+\dfrac1{n_1}\right)\left(1+\dfrac1{n_2}\right)\cdots \] with $n_i$ positive integers satisfying $n_i^2\le n_{i+1}$.

(b) Show that $\alpha\in\mathbb{Q}$ iff from some $k$ onwards we have $n_{k+1}=n_k^2$.

For (a), we can construct $n_k$ inductively: we choose it to be the unique natural for which $\frac{n_k+1}{n_k}<\frac{\alpha}{\left(1+\frac 1{n_1}\right)\cdot\ldots\cdot\left(1+\frac 1{n_{k-1}}\right)}\le\frac{n_k}{n_k-1}$.

One implication of (b) is obvious: for every $n$, the product $\prod_{i\ge 0}\left(1+\frac 1{n^{2^i}}\right)$ is equal to $1+\frac 1n+\frac 1{n^2}+\frac 1{n^3}+\ldots=\frac n{n-1}\in\mathbb Q$, so if $n_i^2=n_{i+1}$ for large enough $i,\ \alpha$ is rational.

Conversely, suppose $\alpha$ is rational. We write it in the form $\alpha=\frac pq,\ p,q\in\mathbb N^*$, and assume $\alpha$ is strictly between $\frac{n+1}n$ and $\frac n{n-1}$ for some $n$. Then we take $n_1=n$, and we get $pn-q(n+1)<p-q$ (easy to check; we simply manipulate $\frac pq<\frac n{n-1}$), i.e. the difference between the numerator and the denominator decreases when we divide $\frac pq$ by $1+\frac 1n$. We repeat this procedure for $\frac{pn}{q(n+1)}$ instead of $\frac pq$, and so on. Since the difference between the numerator and the denominator cannoit decrease indefinitely, at some point we must have $\frac{\alpha}{\left(1+\frac 1{n_1}\right)\cdot\ldots\cdot\left(1+\frac 1{n_{k-1}}\right)}=\frac n{n-1}$ for some $n$, and the conclusion follows from here.


number theory
prime numbers
prime factorization

Find all natural numbers $N$ which satisfy the following properties :
(i)  $N$ has exactly $6$ distinct positive factors $1,d_1,d_2,d_3,d_4,N$
(ii) $1+N=5(d_1+d_2+d_3+d_4)$

so far i have 2299 working out...

im getting two prime numbers, squaring one and then finally multiplying the square of one times the other prime number...
and you also need a units digit of 4 or 9. the only way to get an even units digit is to have one prime equalling 2. It turns out you cant have the other term squared, so you need to use 2 for the squared term. now u need a prime numebr with a 1 units digit. i stopped after 41.

I moved on to the units digit of N being 9. From there, i randomly picked 11 and 19 as the random numbers on my 2nd try (seeing that their product's units digit was gonna be 9). Im not sure how to do the rest...in fact, im not sure if there are a finite amount.

[hide]Well there are two cases when a number has exactly six different factors. It's prime factorization must contain $a^5$ or $a^2 b$. This gives us:

$a^2b + 1 = 5\left(\frac {a^3 -1}{a-1}\right)\left(\frac {b^2-1}{b-1}\right) - 5a^2b - 5$

and

$a^5 + 1 = 5\left(\frac {a^6 - 1}{a - 1}\right) - 5a^5 - 5$

Simplifying

$a^2b +1 = 5(a^2 + ab + a + b + 1)$

becomes

$a^2(b - 5) + 1 = 5(a +1)(b+1)$

Which probably means that $b$ must be greater than 5.
(2299 doesn't work here) We can conclude that $a^2(b-5) - 4$ must be a multiple of 5 and therefore $a^2(b -5)$ must end in either a 9 or a 4. $a^2$ cannot end in a 2, so $b - 5$ cannot end in a 2, $a^2$ cannot end in a 3 which leaves us with $a^2$ and $b - 5$ ending in a 1 and a 9. $b -5$ cannot end in a 1 because then it would be forced to be even and therefore would not be a prime number. $b -5$ cannot end in a 9 either because then it wouldn't be a prime number. 4 and 1 will work, though, because $b$ could then end in a 9 and $a^2$ end in a 1. I'm not sure what numbers will work.

The other equation is

$a^5 + 1 = 5(a^4 + a^3 + a^2 + a)$

Testing the prime numbers 2, 3, 5, and 7, any other prime number would be much too large.
This, as far as I can tell, will not yield any positive integer solutions with $a$ prime.

[/hide]


limit trigonometry calculus derivative ratio function induction

Prove that if $f$ is differentiable and $\lim_{x \to \infty} f(x) = L$, then $\lim_{x \to \infty} f'(x) = 0$

Yes, this problem is very cute, but also very easy. Anyway here is a solution. ;)

 [i] Solution. [/i]

   We have \[ L=\lim_{x\to\infty} f(x)=\lim_{x\to\infty}\frac{e^{x}f(x)}{e^{x}}.  \]

    Now, by applying L'Hospital reules we get

\[ \lim_{x\to\infty}\frac{e^{x}f(x)}{e^{x}}=\lim_{x\to\infty}\frac{e^{x}f(x)+e^{x}f'(x)}{e^{x}}=\lim_{x\to\infty} f(x)+f'(x).  \]

      Clearly, the conclusion follows immediately. ;)

[quote="Cezar Lupu"]\lim_{x\to\infty} f(x)-f'(x).  \]

      Clearly, the conclusion follows immediately. ;)[/quote]
Should be a plus but that doesnt really affect the nice solution!

I wonder if this is correct:

$\lim_{x\to\infty}f'(x)=\lim_{x\to\infty}\lim_{\epsilon\to 0}\frac{f(x+\epsilon)-f(x)}{\epsilon}=$
$=\lim_{\epsilon\to0}\lim_{x\to\infty}\frac{f(x+\epsilon)-f(x)}{\epsilon}=\lim_{\epsilon\to0}\frac{0}{\epsilon}=0$

Should I [hide="spoil the fun?"]Try $f(x)=\frac1x\sin x^3$.[/hide]

All right, then: exactly what is wrong with Cezar Lupu's "proof"? (Of course, Cezar himself knows exactly what's wrong with it, as do jmerry and some of the others who have posted here.)

How do you proceed with Cezaru's proof in the case when $L=0$

The problem with cezar's proof is that he really shows that if $\lim_{x\to\infty} f'(x)$ exists, then it is zero.

The idea I think of is to take some nonzero $f\in C^{\infty}_c(\mathbb{R})$ (think of a 'smooth bump'), and then consider $f_0=f,f_1 = 1/2\cdot f(2x), f_2 = 1/4\cdot f(4x), \ldots, f_n = 1/2^n\cdot f(2^nx)$.  If $\text{supp}(f)\subseteq [-C,C]$, then consider
\[ g(x) = \sum_{j\geq 0} f_j(x-Cj)  \]

ninja edit: just so that doesn't get too complicated, the idea is that we preserve the values of the derivatives by shrinking the height and the width of the bumps at the same time, while ensuring that the heights of the bumps go to zero.

[quote="t0rajir0u"]The problem with cezar's proof is that he uses l'Hopital's rule on a defined limit.   :rotfl:[/quote]
What do you mean?

l'Hopital's rule states that if one has a limit of the form $\lim_{x \to c} \frac{ f(x) }{ g(x) }$ where $\lim_{x \to c} f(x) = 0$ and $\lim_{x \to c} g(x) = 0$ (there are other cases, such as $\infty, \infty$, but this is the basic one), then

$\lim_{x \to c} \frac{ f(x) }{ g(x) } = \lim_{x \to c} \frac{ f'(x) }{ g'(x) }$

The proof is simple - take the Taylor series around $c$.  Then you're just ignoring the constant term of the Taylor series in the case where that term is $0$.  (The proof for the $\infty$ case is similar, I think.)

However, cezar used this rule on a limit that was already defined - to $L$.  This does not work (because the ratio of the constant terms of the Taylor series of numerator and denominator gives the limit, which is then ignored).  For example:

$\lim_{x \to 3} \frac{3 - x}{3 + x} = 0$

But

$\lim_{x \to 3} \frac{ -1 }{ +1} = 1$

t0rajir0u: that is not what is wrong with Cezar's "proof." Your example of $\frac{3-x}{3+x}$ does not qualify for L'Hôpital because it's not a $\frac00$ or $\frac{\infty}{\infty}$ case.

Cezar did make his an $\frac{\infty}{\infty}$ case (at least if $L\ne 0.)$

You must go back to the direction of the logical implication in L'Hôpital:

[b]If[/b] the limit of the quotient of derivatives exists, [b]then[/b] the limit of the original quotient of functions exists and equals the same number.

Cezar was attempting to apply the converse to this, and to use the existence of the limit of the quotient of functions to derive the limit of the quotient of derivatives. But that converse is false. See jmerry's example in this topic (or blahblahblah's, which is essentially the same idea).

False proofs are produced by four different kinds of people:

1. Researchers at the limits of knowledge, trying to do something that is very difficult and not yet known.

2. Confused students.

3. Cranks.

4. Smart-alecks just trying to cause trouble, and/or teachers intentionally falsifying a proof to make a point. People who know what they're doing and are toying with their readers.

By now, I'm sure you realize which category Cezar's post belongs to.

Even if $L=0$, we can repeat the 'reasoning' with $\frac{\exp(-x)f(x)}{\exp(-x)}$

Cezar, I'd say those "smileys" you stuck in the post in question put it in category 4: someone who knows what he's doing who is toying with his readers.

If this is your opinon about me. What can I say?

Ah, I see.   :blush:  It's been awhile since I've done calculus :P

Here's another proof - of something, not necessarily the original problem. I avoid L'Hôpital completely.

Claim: Suppose $f(x)$ is differentiable on $\mathbb{R}$ and that $\lim_{x\to\infty}f(x)=L$ exists. Then there exists a sequence $(x_n)$ such that $lim_{n\to\infty} f'(x_n)=0.$

The corollary to this is that if the limit of $f'(x)$ as $x\to\infty$ exists, then that limit must be zero. That's what Cezar proved.

We won't even need that $f$ approaches a limit: merely being bounded is sufficient. Suppose $|f(x)|\le M$ for all $x\ge 1.$

Choose some arbitrary $x_1>1,$ just to get our induction started. Assume that $x_n$ has been chosen.

Let $a=\max(x_n,n+1).$ Let $b=a+2Mn$ By the Mean Value Theorem, there exists $x_{n+1}\in(a,b)$ such that $f'(x_{n+1})=\frac{f(b)-f(a)}{b-a}.$ But then $|f(b)-f(a)|\le 2M$ and thus $|f'(x_{n+1})|\le \frac{2M}{2Mn}=\frac1n.$

Thus by induction, we have found $x_n$ such that $x_n > n$ and $|f'(x_n)|\le\frac1n.$

The examples by jmerry and blahblahblah can be shown to have this property even though the limit of the derivative does not exist.

[quote="Kent Merryfield"]
Let $a=\max(x_n,n+1).$ Let $b=a+\frac1{2Mn}$ By the Mean Value Theorem, there exists $x_{n+1}\in(a,b)$ such that $f'(x_{n+1})=\frac{f(b)-f(a)}{b-a}.$ But then $|f(b)-f(a)|\le 2M$ and thus $|f'(x_{n+1})|\le \frac{2M}{2Mn}=\frac1n.$
[/quote]
Is it $b=a+2Mn$?

Yes I thinks it is $b=a+2Mn$
I remember this question is the same as i asked before on this forum.
Dr. Kent gave counterexample like Jmerry's example.

And my proof is also using MVT (like Dr. Kent's Proof).
But in my book its appears  on the limit section before MVT section. So are there any other proof?


[size=150]
بچه ها من ایجاد قسمت فارسی را به همه شما تبریک میگویم

[/size]

manam be nobeye khodam eftetahe in ghesmato be hameye hendese doostan tabrik migam :D  ;)


for $ k\ge l \ge 1$, prove : $ \frac{l\minus{}k}{\sqrt{1\plus{}15kl}} \plus{} \frac{k\minus{}1}{\sqrt{l^2\plus{}15k}} \plus{} \frac{1\minus{}l}{\sqrt{k^2\plus{}15l}} \ge 0$

someone told me that Jensen's inequality would help,,,

but I can't see how... could you help please?


$ 1989$ noghte dar safhe mafroozand va hich $ 3$taei az anha rooye yek khat gharar nadarand..chegoone bayad in noghteharaa be $ 30$ goroohe mokhtalef be andazehaye mokhtalef efraz kard ta tedade kolle mosallashayii ke raashayeshan dar goroohhaye mokhtalefand $ max$ shavad..

fek konam javabesh in bashe:
[hide]$ 51, 52, 53, 54, 55, 56, 58, 59,..., 81$
yeni aadade $ 51$ ta $ 81$ bejoz adade $ 57$...[/hide]

babak javabet dorosteeee  :lol:  vali khodet midooni kee ..:d esbaaaaaaaaaat?

khob avval az hame vaseye rahatie fahme rahe hal esme mosalasai ke se raseshoon az se daste motefavetano bezar mosalase [b]khoob[/b],be hamin tartib do noghtaro [b]khoob[/b] minamim age az ye daste nabashan... :D

khob hala farz kon noghteharo be soorate matloob daste bandi kardim,yeni yetori ke tedade mosalasaye khoobemoon maximum shode basheo andaze dasteha ye seri adade motefavet bashe,khob hala farz kon yeki az dastehamoon $ m$ta va yeki dige $ n$ta noghte dashte bashe va $ n - m\geq 1$...

hala tedade mosalasaie khoobi ke hich rabti be in do daste nadarano bezar $ S$ (yeni mosalasai ke se raseshoon khareje $ m,n$e)...

hamchenin farz kon tedad rahaye entekhabe do noghte khoob kharej az $ m,n$ barabare $ k$ bashe...

khob pas tedade mosalasaye khoobemoon barabar mishe ba:

$ \boxed{S + k(m + n) + (1989 - m - n)mn}$

khob hala chon halati ke bala dar nazar gerefte boodimo maximum farz karde boodim pas adadi ke be das oomade maximum tedade mosalasaye khoobe momkene...

khob hala bia az too daste $ n$tai ye noghte vardaro bendazesh too daste $ m$tai ta dastehamoon beshan $ n - 1,m + 1$ age too in halatam andaze hich do dastei barabar nashe oonmoghe tedade mosalasaie khoobe in halat bayad az tedad mosalasaye khoobe halate ghabli kamtar beshe (chon halate ghabli max bood)

hala tedade mosalasaye khoobemoon too in halat mishe:

$ S + k(m + 1 + n - 1) + (1989 - m - 1 - n + 1)(m + 1)(n - 1)$

(deghat konid ke ma ye noghte az $ n$ be $ m$ ezafe kardim pas tedade mosalasaie khoobi ke khareje $ m,n$ hastan taghir nemikoneo hamoon $ S$ mimoone)

khob pas tebghe oon chizi ke goftim bayad dashte bashim:

\begin{eqnarray*}S + k(m + n) + (1989 - m - n)mn\geq S + k(m + 1 + n - 1) + (1989 - m - 1 - n + 1)(m + 1)(n - 1)\end{eqnarray*}

$ \Rightarrow mn\geq (m + 1)(n - 1)$

$ \Rightarrow\boxed{1\geq n - m}$

ke inam ba farze avvaliamoon dar tanaghoze,pas too halate maximum,NEMISHE az ye daste ye noghte vardashto be oonyeki daste ezafe kard (yeni age bekhaim inkaro bokonim do daste ham size bayad peida beshe)...

pas $ 3$ halat darim:

[u]$ \boxed{\textrm{halate 1 : }}$[/u]
halati ke andazeye dastehamoon ye seri adade motevalian (ke dar natijash age az ye daste bekhaim yeki kam konimo be ye daste dge ezafe konim daste hamsize bevojood miad)

yeni dar vaghe size dastehamoon be soorate ziran:

$ a,a + 1,a + 2,...,a + 29$

amma midoonim jame andazeye dasteha bayad beshe $ 1989$ pas:

$ a + (a + 1) + (a + 2) + ... + (a + 29) = 1989$

$ \Rightarrow a = 51.8$

pas in halat hichvakh pish nemiad...

[u]$ \boxed{\textrm{halate 2 : }}$[/u]
halati ke dombale andazeye aazamoon az ye seri aadad tashkil shode ke avval ye tedadishoon motavalian bad yeja ye fasele darim bad dobare ye seri adade motavalian yeni:

$ a,a + 1,...,a + (k - 1),a + (k + 1),...,a + 30$

baz too in halatam darim:

$ a + (a + 1) + ... + (a + 30) - (a + k) = 1989$

$ \rightarrow 30a - k = 1524$

amma chon $ k$ ye adadie beine 0,30 pas tanha halati ke momkene $ k = 6,a = 51$e...
ke dastehamoon too in halat mishan:

$ \boxed{51,52,53,54,55,56,58,59,...,81}$

[u]$ \boxed{\textrm{halate 3 : }}$[/u]
in halat shabihe halate ghable faghat ba in fargh ke too ghabli daghighan yeja fasele dashtim inja bishtar az yeja fasele darim yeni haddeaghal doja in [b]fasele[/b] rokh mide,pas farz konid andazeye dombalehamoon be soorate zir bashan:

$ a,a + 1,...,a + (i - 1),a + (i + 1),...,a + (j - 1),a + (j + 1),...,a + k$

khob dar in soorat daste $ a + j + 1$ ozvi ro dar nazar begirid, az in daste yeki bardarido be daste $ a + i - 1$ ezafe konid oon moghe dastehamoon be soorate zir mishe:

$ a,a + 1,...,a + i,a + (i + 1),...,a + (j - 1),a + j,...,a + k$

pas bazam andaze dasteha motafavete,vali tebghe oon algortihmi ke avvale esbat goftim in kare ma tedade mosalasaye khoobo afzayesh mide ke inam tanaghoze...
pas in halatam hichvakh rokh nemide...

pas tanha halate maximume momken ine:

$ 51,52,53,54,55,56,58,59,...,81$

:wink:  dorosteeee bacheehaa enghad be hendesee gir nadiin faghataam ke soalaye khoshkhaan copy mishe injaa khoskhano hame daran khodeshoon bekhoonan dgggg.. yezare ham soalaye nazarye adad bezarin:-?? :P


Let {b_n}_n>=0 be a monotone decreasing sequence of positive real numbers, such that the series [sum]_n=0 to oo (b_n) is convergent.

Prove that the lim_n->oo (nb_n) = 0.

Give an example of a strictly decreasing positive sequence {c_n}_n>=0 such that lim_n->oo (nc_n) = 0, but [sum]_n=0 to oo (c_n) is divergent.

If possible, please provide explanations along with your proofs . Thank you.

Let $[n/2]$ denote the integer part of $n/2$. Let $s_{n}=b_{0}+\dots b_{n}$. Since the sequence $(s_{n})$ has a limit, it follows that $s_{n}-s_{[n/2]}\to 0$. On the other hand, $s_{n}-s_{[n/2]}\ge nb_{n}/2$.

For the second part, try $c_{n}=\frac{1}{(n+1)\ln(n+e)}$.

The line about explanations got old quickly.


algebra
polynomial
trigonometry
modular arithmetic

I have dificulties with some polynomial problems. :blush:

$1)$ Find $a$ and $b$ such as the polynom $f=(X+1)^n+aX+b$ is divisible by $g=X^2+1$

$2)$ Let $n\in N$. Show that polynom $f=(X+1)^{3n+2}+X+2$ is divisible by $g=X^2+3X+3$

That's all.  :lol:

[b]1)[/b]
$a=-\sqrt 2 ^n \cdot sin(\frac{n \cdot \pi}{4})$
$b=-\sqrt 2 ^n \cdot cos(\frac{n \cdot \pi}{4})$

[quote="Myth"]It follows that $0=(1\pm i)^n\pm ia+b=\sqrt{2^n}\left(\cos \frac{\pm \pi n}{4}+\sin \frac{\pm \pi n}{4}\right)\pm ai+b=0$.
[/quote]

You forgot an $i$  :P .  So the solutions of a and b are those which ZetaX wrote ? :lol:

Opss! The problems are easy  :D ... now that I found the equivalence between 2 polynoms $(a\equiv b\pmod p)$  :P


function
integration
topology
probability
calculus
Functional Analysis
real analysis

Let $F$ be a family of continuous functions on $(0,1)$ such that any linear combination of them has a zero in $(0,1)$. Prove the existence of a nondecreasing function $g$ such that for all $f\in F$ one has $\int_{0}^{1}{f(x)dg(x)}=0$.

Something's wrong here:

First of all, any constant function $g$ will do, but this is a silly remark :). I assume you're looking for non-constant functions $g$ and simply forgot to mention it. If we state the problem in this way, however (asking for [i]non-constant[/i] non-decreasing functions $g$), then it's no longer true. Just take $F$ to be the linear space of all continuous functions on $(0,1)$ which vanish on some interval of the form $(0,a),\ a\in(0,1)$. It should be easy to see that no non-constant $g$ meets the requirements.

If $(0,1)$ were to be replaced by $[0,1]$, things would be different.. :)

and $\int_{[0;1]}f(x) dg(x)$ does not necessarily exist  :maybe: and the case of $[0;1]$ follows easily from Hahn-Banach I think ..

Ooops, sorry: $g$ is indeed nonconstant and we work on $[0,1]$. Sorry again. However, I don't see how it follows easily from Hahn-Banach... :(

Here's the proof I had in mind (and using Hahn-Banach is indeed a big part of it):

The left-continuous non-decreasing functions $g: I=[0,1]\to\mathbb R$ with $g(0)=0$ are in bijection with the positive Borel measures $\mu$ through the correspondence $\mu([0,x))=g(x),\ \forall x\in(0,1]$. This means that in order to find our $g$ it suffices to prove the following:

Let $X$ be a compact Hausdorff topological space, and let $\mathcal F\le\mathcal C(X,\mathbb R)$ be a linear subspace with the property that every $f\in\mathcal F$ has a zero. There is then a regular Borel probability measure $\mu$ on $X$ such that $\int_{X}f\ \mbox d\mu=0,\ \forall f\in\mathcal F$ ($\mathcal C(X,\mathbb R)$ is the Banach space of continuous real-valued functions defined on $X$ with the supremum norm).

Let ${\bf 1}$ be the constant function equal to $1$ everywhere on $X$. Notice that the distance from ${\bf 1}$ to $\mathcal F$ is $1$, so by Hahn-Banach, we can find a functional $\Lambda\in\mathcal C(X,\mathbb R)^{*}$ which is zero on $\mathcal F$ and such that $\|\Lambda\|=\Lambda(1)=1\ (*)$. If we manage to prove that $(*)$ implies that $\Lambda$ is positive (i.e. $\Lambda(f)\ge 0,\ \forall f\ge 0$), then we're done, since every positive functional of norm $1$ on $\mathcal C(X,\mathbb R)$ is given by integration with respect to a regular Borel probability measure.

$\Lambda$ is, in any case, given by integration with respect to a regular Borel measure $\mu$ on $X$. Because the relation $\left|\int f\ \mbox d\mu\right|\le\sup|f|\ (**)$ holds for all continuous functions $f$ on $X$, it also holds for all bounded measurable functions. In particular, it holds for characteristic functions $\chi_{A}$ of measurable sets $A$. If we take $A$ to be the (essentially unique; see the [url=http://planetmath.org/encyclopedia/HahnDecompositionTheorem.html]Hahn Decomposition Theorem[/url]) measurable subset of $X$ for which $\mu$ is positive on $A$ and negative on $X\setminus A$, then $(**)$ translates to $\mu(A)\le 1$. Coupled with $1=\mu(X)=\mu(A)+\mu(X\setminus A)\le\mu(A)$, this gives $\mu(X\setminus A)=0$, i.e. $\mu$ is indeed a positive measure.

Alternatively, just use the fact that your linear space and the open set of everywhere positive functions can be separated by a linear functional. Of course, this approach doesn't differ much from grobber's one.


I have used Asymptote's graph3 package to draw graphs of surfaces, but I need to draw a graph that contains several intersecting surfaces.  Is there a simple way to cause Asymptote to display whichever surface is closer to the front, rather than the one that happens to have been the last one drawn?

The problem is that "it is difficult [b]not[/b] to do what you want".
What's your code currently ?
What is your version of Asymptote ?

I have Asymptote version 1.43.
[code]import graph3;
size(200);
currentprojection=orthographic(8,-4,2);

bbox3 b=limits((-2,0,0),(2,4,4));
xaxis("$x$",b,RightTicks);
yaxis("$y$",b,RightTicks);
zaxis("$z$",b,RightTicks);

triple y0(pair t) {
  return (t.x, t.x^2, t.y);
}

triple y1(pair t) {
  return (t.x, 4, t.y);
}

triple z0(pair t) {
  return (t.x, t.y, 0);
}

triple z1(pair t) {
  return (t.x, t.y, t.y);
}

add(surface(y0,(-2,0),(2,4),outward=true,40,40));
add(surface(y1,(-2,0),(2,4),outward=true,40,40));
add(surface(z0,(-2,0),(2,4),outward=true,40,40));
add(surface(z1,(-2,0),(2,4),outward=true,40,40));
[/code]

With the version 1.51 the code...
[code]settings.outformat="pdf"; // Produce pdf file with PRC embedded
settings.render=8; // Embed bitmap picture for other pdf viewer than acrobat.

import graph3;

size(200);
currentprojection=orthographic(2,-4,-2);

limits((-2,0,0),(2,4,4));
xaxis3("$x$",OutTicks);
yaxis3("$y$",OutTicks);
zaxis3("$z$",OutTicks);

triple y0(pair t) {
  return (t.x, t.x^2, t.y);
}

triple y1(pair t) {
  return (t.x, 4, t.y);
}

triple z0(pair t) {
  return (t.x, t.y, 0);
}

triple z1(pair t) {
  return (t.x, t.y, t.y);
}

draw(surface(y0,(-2,0),(2,4),40,40),blue);
draw(surface(y1,(-2,0),(2,4),40,40),red);
draw(surface(z0,(-2,0),(2,4),40,40),green,nolight);
draw(surface(z1,(-2,0),(2,4),40,40),yellow);[/code]
... produces the attached pdf file.
With acrobat reader you can rotate, zoom etc the picture.


inequalities
function
analytic geometry
probability
algebra
polynomial
inequalities unsolved

[quote="nayel"]
sorry but what is this theory called? :oops:[/quote]
Draw the graf of linear function in the cartesian system of coordinates on  $[0,1].$ The maximum is reached either in $0$ or in $1.$ :wink:

[quote="weiquan"][quote="nayel"]sorry arquady but i dont understand your solution. its my problem that i haven't yet studied these topics. :blush:[/quote]

The expression is a linear function of $x_{i}$ for any $i$, and the $x_{i}$s are independent, so the extrema must occur at the endpoints of the graph, since it is a straight line.[/quote]

oookey, now i think i'm getting this. thanks everyone! :D

Here's a probabilistic approach to solving the inequality.  Let $A_{1}$, $A_{2}$, ..., $A_{n}$ be a sequence of independent events such that the probability of $A_{i}$ is $1-x_{i}$.  It's a basic inequality in probability theory that
\[\Pr(A_{1}\cup A_{2}\cup \cdots \cup A_{n}) \le \Pr(A_{1})+\Pr(A_{2})+\cdots+\Pr(A_{n}) \, . \]
By independence, $\Pr(A_{1}\cup A_{2}\cup \cdots \cup A_{n})$ is clearly $1-x_{1}x_{2}\ldots x_{n}$.  So we get the desired inequality.

[quote="me@home"]For case of two numbers:
\[a+b-ab\leq 1 \iff (a-1)(b-1)\geq 0,\text{ clearly true by condition}\]
Case of three numbers ($\sum$ always taken cyclically):
\begin{eqnarray*}\sum a-abc\leq 2 &\iff& 2+abc-\sum a\geq 0 \\ &\iff& (a-1)(b-1)(c-1)+\sum ab-2\sum a+3\geq 0 \\ &\iff& (a-1)(b-1)(c-1)+\sum (a-1)(b-1) \geq 0 \\ &\iff& \frac{\sum (a-1)(b-1)(c+2)}{3}\geq 0 \end{eqnarray*}
The last is clear.

I'm not sure, but in general I think it is
\begin{eqnarray*}x_{1}+x_{2}+...+x_{n}-x_{1}x_{2}...x_{n}\le n-1 &\iff& \prod_{i=1}^{n}(x_{i}-1)+\sum \prod_{i=1}^{n-1}(x_{i}-1)\geq 0 \\ &\iff& \frac{\sum (x_{1}-1)(x_{2}-1)(x_{3}-1)(x_{4}-1)\cdots (x_{n}+n-1)}{n}\geq 0 \end{eqnarray*}
which is clear.

I still need to verify these guesses...
Of course, this is a really easy problem with $\text{arqady's}$ and $\text{kunny's}$ methods... :roll:

[hide="{Edit:}"] 1111 posts :lol: [/hide][/quote]

I have reworked and solved the problem using the methods I outlined above, the generalization I posted is incorrect. It is a much more complicated but beautiful polynomial... I challenge people to find one of the two polynomial factorizations I found which proves the relation....


I don't think setting numerical goals is necessarily the thing to do. Just keep doing math because you enjoy it, and when these tests roll around, just try to solve as many problems as you can.

Goal setting never really helped me either, but I think it does help some others.

That said, there is absolutely no way that anyone except you can set your goals. Ask yourself what score you would be really happy to get next year, and whether it is really doable.

I set myself goals subconsciously but they're always relatively easy for me so they make me feel good.

Work hard. Get some actual sleep. Take the test. Do well.  :lol:

Personally, math becomes boring for me once I do it just because of a goal.

Being the uber-competitive person that I am, I usually set goals such as "defeat so and so on the AMC, AIME, etc. by next year" or something similar.
Right now, I want to somehow catch up to Eric Larson, but my chances seem somewhat low. :(

[quote="mathemonster"]i can get a 126 on the amc 10 now, and am on chapter 15 of aops v1. im hoping to work through both v1 and v2 by next year and am doing the online amc 10 course. What do u think good goals are aime usamo etc.[/quote]

Well, it seems you've already set your goals. :wink:
I would personally set goals just so you have some objective. Don't obsess over them, however, as they do not help your mathematical understanding of concepts.

My main goal: make blue MOP (~21), since it'll be my last chance next year...
Other goals: AMC 12 - 138, AIME - 12. (Somewhat unrealistic, but you can always try!)
Necessary goal: OMG STOP MAKING STUPID MISTAKES!

That being said, having fun is the most important part! Even if it is hard work!


Prove that all primetive root of prime number p be root $\Phi_{p-1}(x)$

I've edited the first post such that it is correct. And to prove it, just count roots (of $x^{p-1}-1$ and of the $\Phi_{d}(x)$, $d|p-1$); or use the main theorem on cyclotomic polynomials.

[hide="Original problem statement was"]Prove that all primetive root of prime number p be root $\Phi_{p}(x)$[/hide]

Uh...$1+x+x^{2}+...+x^{r-1}\equiv 0$ where $r$ is the order of $x$ or is divisible by the order of $x$...


A projectile of mass m =3kg is fired with an initial speed of 120m/s at an angle of 30 degree with the horizontal . At the top of its trajectory, the projectile explodes into two fragment of masses 1 kg and 2kg. The 2 kg fragment lands on the ground directly below the point of explosion 3.6s after the explosion.
a) Determine the velocity of the 1 kg fragment immediately after the explosion?
b) Find the distance between the point of firing and the point at which the 1kg fragment strikes the ground.
c) Determine the energy released in the explosion.

$u$ -the velocity of 1kg mass
$v_{0}$ -the velocity of 3kg mass
$v$ -the mass of 2kg mass
$\alpha$-angle between the initial direction of 1kg mass and the direction of 3kg mass
$\theta$ -30 degree angle

$3v_{0}cos\theta=1ucos\alpha$

$usin\alpha=2v$

$h_{m}=\frac{v_{0}^{2}sin^{2}\theta}{2g}=180m$

$180=3.6v+5*3.6^{2}$

$v=32\frac{m}{sec}$

$t=\frac{v_{0}sin\theta}{g}=6sec$

$x=tv_{0}cos\theta=623m$

$sin^{2}\alpha+cos^{2}\alpha=1$

$\frac{4v^{2}}{u^{2}}+\frac{9v^{2}_{0}cos^{2}\theta}{u^{2}}=1$

$u=318\frac{m}{sec}$

$Q=E_{1}-E_{2}$

$Q=\frac{3v^{2}_{0}cos^{2}\theta}{2}-\frac{1u^{2}}{2}-\frac{2v^{2}}{2}$

I have no idea what you have done. Can you write explicitly what you are trying to do?

You can start off by parameterizing the path of the particle in terms of time using vectors.

For the first part, the law of Conservation of Momentum says that the sum of the product of each mass with its velocity is constant.

For the second, you can use $d = v_{x}t$, where $t$ is how long the object will stay airborne and $v_{x}$ is the velocity obtained in (a).

Finally, I think the energy released during the explosion would just be the difference in the kinetic energy of the one particle before the explosion and the sum of the kinetic energies of the two particles after the explosion.

Why does it have this name.i,e Killer...? @gibberenergy:chuyen qua hoc ly ha dai ca.

Do you know the solution or you just don't know it?So you want me to explain it?

To Namiqfizik: I try to decode your soution. However, it may be a little hard for me to keep up with your solution. I wish you give me  your argument in each step. That 's all

To Cuong: The problem named "Killer" because my physics professor gave it to me on the final exam. Well, I have done it but it took me 1 hours.


What's the maximum number of triangulars that can be formed by $ 8$ lines in a plane if any $ 2$ lines are concurent?

what is a triangular?
what does it mean for any 2 lines to be concurrent?

Sorry for my bad english :oops:  I mean "triangles" and "any 2 lines intersect"

since every line intersects we know all points are connected together by a line. No points are collinear since then there would exist two lines that are parallel.

Thus my hunch is the optimal figure to look at would be the regular octagon.
Count the triangles obtained from connecting all the points.

is the answer by any chance 16? no wait.....

My guess is simply taking a triangle, which would take up 3 lines, and using the five lines to make up the maximum amount of regions.
which would be 16.
But then, some of those regions are probably not quadrilaterals, but whatever.

I would be inclined to interpret this as allowing overlapping triangles. In that case, any three lines give exactly one triangle, unless two are parallel (ruled out by the statement) or all three intersect at a point. That allows for a simple answer.


What is the largest perfect square that divides 64,800?

Notice that the number given divided by two, $ \fbox{32400}$, is already a perfect square.