function inequalities AMC USA(J)MO USAMO calculus derivative

Call a real-valued function $ f$  very convex if

$ \frac {f(x) \plus{} f(y)}{2} \ge f\left(\frac {x \plus{} y}{2}\right) \plus{} |x \minus{} y|$

holds for all real numbers $ x$ and $ y$. Prove that no very convex function exists.

[hide="random comment"] Why would they give it a name if none of them exist?? o.0[/hide]

[hide="hint"] Think of concave up and concave down functions. [/hide]

[hide="my solution please critique!"]
We divide the problem into two cases:

Case 1: $ f$ is concave down.

If the function is concave down, then it is defined that

$ \frac {f(x) \plus{} f(y)}{2} \ge f\left(\frac {x \plus{} y}{2}\right)$.

By adding a positive value, $ |x \minus{} y|$ to the right side of the above inequality, we are only strengthening it. The theorem thus holds true for concave down functions.

Case 2: $ f$ is concave up.

We have Jensen's Inequality applied to a convex function $ f$ with two variables, $ x$ and $ y$.

$ \frac {f(x) \plus{} f(y)}{2} \ge f\left(\frac {x \plus{} y}{2}\right)$

Now, assume that the given inequality,

$ \frac {f(x) \plus{} f(y)}{2} \ge f\left(\frac {x \plus{} y}{2}\right) \plus{} |x \minus{} y|$

 is true for all $ x$ and $ y$. We proceed by contradiction.

Subtract Jensen's Inequality from the given inequality. We are left with

$ 0\ge|x \minus{} y|$

which is not true for all $ x$ and $ y$. Contradiction!

Therefore, there are no very convex functions. Poor guy who made the definition of a very convex function. :(

QED

[/hide]

Please critique my solution! Any tips would be GREATLY appreciated. Thanks! :D

That solution isn't true. You can't subtract an inequality like that. For example, if you have $ 3\geq 2$, then $ 5\geq 5$, you couldn't subtract the first from the second to conclude that $ 2 \geq 3$.

[quote="Arvind_sn"]Why would they give it a name if none of them exist?? o.0[/quote]
Mathematicians do this all the time!  Odd perfect numbers, for example.  Also, it makes it easier to say "suppose $ f$ is very convex" as part of a proof by contradiction.

[quote="Arvind_sn"]We divide the problem into two cases:[/quote]
The problem statement already implies that $ f$ is convex (that is, concave up).

Anyway, as Brut3Forc3's comment should make clear, this problem is on the USAMO for a reason :)

[quote="t0rajir0u"]Odd perfect numbers, for example.[/quote]  Getting a little ahead of yourself there, aren't you?  ;)

[quote="Arvind_sn"]
Case 2: $ f$ is concave up.

We have Jensen's Inequality applied to a convex function $ f$ with two variables, $ x$ and $ y$.

$ \frac {f(x) \plus{} f(y)}{2} \ge f\left(\frac {x \plus{} y}{2}\right)$

Now, assume that the given inequality,

$ \frac {f(x) \plus{} f(y)}{2} \ge f\left(\frac {x \plus{} y}{2}\right) \plus{} |x \minus{} y|$

 is true for all $ x$ and $ y$. We proceed by contradiction.

Subtract Jensen's Inequality from the given inequality. We are left with

$ 0\ge|x \minus{} y|$

[/hide]
[/quote]

Using this step, couldn't you just subtract the equations in reverse order and and up with $ 0\ge \minus{} |x \minus{} y|$
 $ 0\le|x \minus{} y|$?

USAMO problems are generally not trivialized by one observation.

With such a condition, you probably want to approach it by creating a constant term on one side and making an increasing term on the other.

This is kind of moot at this point, but you forgot a case.
Case 3: $ f$ is neither concave up nor concave down.

These cases were a little silly though, because $ \frac {f(x) \plus{} f(y)}{2} \geq f\left( \frac {x \plus{} y}{2} \right) \plus{} |x \minus{} y| \geq f\left( \frac {x \plus{} y}{2} \right)$ means a very convex function has to be concave up. Rather than using cases you only needed to show this and then proceed with case 2. Although this ends up being absolutely no help in finding a solution.

Related question: why does the given condition imply convexity? When I looked this up on wikipedia, it categorizes this as midpoint convex which is weaker than convexity.

I would rewrite this considering a "center" and a "radius", a and d respectively, such that the equation becomes:
$ \frac {f(a \minus{} d) \plus{} f(a \plus{} d)}{2}\ge f(a) \plus{} 2d$

Now call f(a+d)-f(a), m, and f(a)-f(a-d), n.  We'll assume that the function happens to be increasing on (a-d,a+d).
So $ m \minus{} n\ge 4d$.

Unfortunately I have no clue how this might help besides Limiting as d->0 and doing some work with derivatives because my brain just died.

If we do choose to follow my dead brain method, however, we end up with $ f'(x)\ge 4 \plus{} f'(x)$ which doesn't seem right.  EDIT: this equation was wrong

Also, can anyone just make sure for me that $ f(x) \equal{} \frac {e^{3 x}}{3} \minus{} \frac {x}{3 a}$ is in fact NOT very convex?

Well, here are some [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47401]solutions[/url].

[quote="CircleSquared"]I would rewrite this considering a "center" and a "radius", a and d respectively, such that the equation becomes:
$ \frac {f(a \minus{} d) \plus{} f(a \plus{} d)}{2}\ge f(a) \plus{} 2d$

Now call f(a+d)-f(a), m, and f(a)-f(a-d), n.  We'll assume that the function happens to be increasing on (a-d,a+d).
So $ m \minus{} n\ge 4d$.
[/quote]

First, I apologize for not knowing latex.

If d = 0, the inequality becomes 2f(x)/2 >= f(x) + 0, which is obvious.
Then, divide by d on both sides and take the limit as d ==> 0.  (I think) this results in:
f' = lim(as d->0) (f(x)/d) + 2.
This means that f' is +- infinity (depending on f(x)).  Thus we have a vertical line(?), which is not a function.

However, I'm unsure if this works, and my brain, as was circle^2's, is quite dead.

You can probably make that work, but as stated it's not complete.  Functions don't have to have derivatives.


inequalities
geometry
parallelogram
Euler
geometric transformation
reflection
triangle inequality

Let $ ABCDEF$ be a convex hexagon such that $ AD \equal{} BC \plus{} EF$, $ BE \equal{} AF \plus{} CD$, $ CF \equal{} DE \plus{} AB$. Prove that:
\[ \frac {AB}{DE} \equal{} \frac {CD}{AF} \equal{} \frac {EF}{BC}.
\]

It can be proved by using vectors. Actually the same idea was used in IMO 2003 #3.

I had a similar idea, but I was very DUMB not to see what happened if I went on this path.

I got the INEQUALITIES, but I thought that adding them would never give something useful:

$\left| \overrightarrow{AD} \right| = \left| \overrightarrow{BC} \right| + \left| \overrightarrow{EF} \right| \geq \left| \overrightarrow{FC} + \overrightarrow{BE} \right|$. (I don't know why, but this inequality seemed too natural for me to yield anything; really don't know why)

Squaring and adding: $\left( \overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{CF} \right)^2 \leq 0$. Thus, $BC \| EF$ etc. Also, $BC \| DA$. The conclusion is now obvious.

Thanks, Sung-yoon Kim.

Yeah I think you're on the right way..

[quote]$\left| \overrightarrow{AD} \right| = \left| \overrightarrow{BC} \right| + \left| \overrightarrow{EF} \right| \geq \left| \overrightarrow{FC} + \overrightarrow{BE} \right|$[/quote]Consider a parellelogram BCFX. Then $BC+EF=XF+FE \geq XE=|\overrightarrow{XB}+\overrightarrow{BE}|=|\overrightarrow{CF}+\overrightarrow{BE}|$. The rest is same as yours.

Guys I think we shoulld find sinthetical solution for this one,I havn't do it still :( ,but I will try again ;).

Yes it is a good idea to find a synthetic proof for this problem.

Davron

That's really hard to find a such solution.

Yes,it is realy hard job,if that can help you I know that ther are sintetic solution,I know a man who has found it,but I want find it from misalf,so I havn't ask him about that solution. :)

[quote="perfect_radio"]The conclusion is now obvious.

[/quote]

Well, after proving all these parallels, I still can't see why it's obvious...

Let $AD \cap CF = \left\{ T \right\}$. The following statements hold:
- $ABCT$ is a parallelogram;
- $DEFT$ is a parallelogram;
- $\triangle ATF \sim \triangle DTC$;
- the conclusion.

[quote="perfect_radio"] Also, $BC \| DA$. [/quote]

And why is this?

$\left( \overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{CF}\right)^{2}\leq 0$ implies $\overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{CF}= \overrightarrow 0$. Also note that $\overrightarrow{EB}+\overrightarrow{CF}= \overrightarrow{CB}+\overrightarrow{EF}$.
Hence,
\[\overrightarrow{AD}= \overrightarrow{BC}+\overrightarrow{FE}.\]

A simple construction of the configuration, as the problem states, is as follows:

A triangle $\bigtriangleup KLM$ is given and let $A$ be, a fixed point on the extension of the sideline $KL$ $($ $K,$ between $A,$ $L$ $).$

The line through the $A$ and parallel to $LM,$ intersects the sideline $KM$ at a point, so be it $B.$

The line through the $B$ and parallel to $KL,$ intersects the sideline $LM$ at a point, so be it $C.$

The line through the $C$ and parallel to $KM,$ intersects the sideline $KL$ at a point, so be it $D.$

The line through the $D$ and parallel to $LM,$ intersects the sideline $KM$ at a point, so be it $E.$

Through $A,$ $E$ now, we draw to lines parallel to $KM\equiv BE$ and $KL\equiv AD$ respectively, which intersect the sideline $LM,$ at points $F,$ $F'$ and we will prove that $F'\equiv F.$

It is easy to show that $CF = CM+MF = DE+AB$ $,(1)$ and $CF' = CL+LF' = AB+DE$ $,(2)$

From $(1),$ $(2)$ $\Longrightarrow$ $CF' = CF$ $\Longrightarrow$ $F'\equiv F.$

So, it has already been constructed the configuration as the problem states.

$\bullet$ From $AB\parallel DE$ $\Longrightarrow$ $\frac{AB}{DE}= \frac{KA}{KD}= \frac{KB}{KE}$ $,(3)$

But, $\frac{KA}{KD}= \frac{EF}{BC}$ $,(4)$ and $\frac{KB}{KE}= \frac{CD}{AF}$ $,(5)$

From $(3),)$ $(4),$ $(5)$ $\Longrightarrow$ $\frac{AB}{DE}= \frac{CD}{AF}= \frac{EF}{BC}$ and the proof is completed.

Kostas Vittas.

Lemma: Let $ABCD$ be arbitrary quadrilateral. Denote
$a,b,c,d$ it's sides and $e,f$ diagonals, then $a^{2}+c^{2}+2bd\geq e^{2}+f^{2}$.

Proof: Denote $M, N, P$ midpoints of $BC, AD, BD$, from
triangle inequality $MN\leq MP+NP$ or $2MN\leq b+d$. We will
remember Euler theorem for quadrilateral and midpoint of opposite
sides or diagonals: $b^{2}+d^{2}+e^{2}+f^{2}=a^{2}+c^{2}+4MN^{2}\leq a^{2}+c^{2}+(b+d)^{2}$. Equality is hold when $BC \parallel AD$ and our lemma is
proved.

Now we will solve the problem.

 Denote $AB=a,BC=b,CD=c, DE=d, EF=e, AF=f, AC=x_{1}, BD=x_{2}, CE=x_{3}, DF=x_{4}, EA=x_{5}, FB=x_{6}, AD=l_{1}, BE=l_{2}, CF=l_{3}$.

We have
$l_{1}=b+e, l_{2}= c+f, l_{3}=a+f$.  Apply our theorem for $ABCD$ and
$DEFA$ we have $a^{2}+c^{2}+2bl_{1}\geq x_{1}^{2}+x_{2}^{2}$ and
$d^{2}+f^{2}+2el_{1}\geq x_{4}^{2}+x_{5}^{2}$. Adding all of them:
\[a^{2}+c^{2}+d^{2}+f^{2}+2l_{1}^{2}\geq x_{1}^{2}+x_{2}^{2}+x_{4}^{2}+x_{5}^{2}\]
Summing all such quadrilaterals we get
\[2(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2})+2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2})\geq 2(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}+x_{6}^{2}) \]
Next step is to apply our lemma for quadrilateral $BCEF$:
$x_{3}^{2}+x_{6}^{2}+2be\geq l_{2}^{2}+l_{3}^{2}$. Summing with inequalities for
$CDAF$, $DEAB$:
\[(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}+x_{6}^{2})+2ad+2be+2cf\geq 2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2}) \]
Multiplying second inequality by $2$ and adding with first one
we get
\[2(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2})+4(ad+be+cf)\geq 2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2}) \]
or just $2(a+d)^{2}+2(b+e)^{2}+2(c+f)^{2}\geq 2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2})$

 As we see the equality is hold and thus from lemma's case
of equality we get $BC \parallel AD \parallel EF$ and similarly other lines are
parallel. Let a line through $C$ parallel to $AB$ intersects $AD$ at
$P$, then $AP=b$ and thus $PD=e$ and $PD\parallel EF$. Hence $PDEF$ is
parallelogram and $C,P,F$ are collinear as are $A,P,D$. So triangles
$APF$ and $DPC$ are similar and from here
\[\frac{AP}{DP}=\frac{PF}{PC}=\frac{AF}{CD}\]
Rewriting in hexagon's sides:
\[\frac{BC}{EF}=\frac{DE}{AB}=\frac{AF}{CD}\]

This was problem was proposed for Mathematical Reflections 2 in 2006,
by Nairi Sedrakyan (Armenia).

Mathematical Reflections:  http://www.awesomemath.org

in a convex hexagon with AD=BC+EF, CF=DE+AB,BE=AF+CD.THERE WE CAN FIND EASILY.A simple construction of the configuration, as the problem states, is as follows:

A triangle \bigtriangleup KLM is given and let A be, a fixed point on the extension of the sideline KL ( K, between A, L ).

The line through the A and parallel to LM, intersects the sideline KM at a point, so be it B.

The line through the B and parallel to KL, intersects the sideline LM at a point, so be it C.

The line through the C and parallel to KM, intersects the sideline KL at a point, so be it D.

The line through the D and parallel to LM, intersects the sideline KM at a point, so be it E.

Through A, E now, we draw to lines parallel to KM\equiv BE and KL\equiv AD respectively, which intersect the sideline LM, at points F, F' and we will prove that F'\equiv F.

It is easy to show that CF = CM+MF = DE+AB ,(1) and CF' = CL+LF' = AB+DE ,(2)

From (1), (2) \Longrightarrow CF' = CF \Longrightarrow F'\equiv F.

So, it has already been constructed the configuration as the problem states.

\bullet From AB\parallel DE \Longrightarrow \frac{AB}{DE}= \frac{KA}{KD}= \frac{KB}{KE} ,(3)

But, \frac{KA}{KD}= \frac{EF}{BC} ,(4) and \frac{KB}{KE}= \frac{CD}{AF} ,(5)

From (3),) (4), (5) \Longrightarrow \frac{AB}{DE}= \frac{CD}{AF}= \frac{EF}{BC} and the proof is completed.

[quote="prowler"]Lemma: Let $ABCD$ be arbitrary quadrilateral. Denote
$a,b,c,d$ it's sides and $e,f$ diagonals, then $a^{2}+c^{2}+2bd\geq e^{2}+f^{2}$.

Proof: Denote $M, N, P$ midpoints of $BC, AD, BD$, from
triangle inequality $MN\leq MP+NP$ or $2MN\leq b+d$. We will
remember Euler theorem for quadrilateral and midpoint of opposite
sides or diagonals: $b^{2}+d^{2}+e^{2}+f^{2}=a^{2}+c^{2}+4MN^{2}\leq a^{2}+c^{2}+(b+d)^{2}$. Equality is hold when $BC \parallel AD$ and our lemma is
proved.[/quote]

Sorry for reviving the old topic, but your lemma is not true, and therefore the whole solution as well. From Euler theorem we have not as you wrote $b^{2}+d^{2}+e^{2}+f^{2}=a^{2}+c^{2}+4MN^{2}$ but $ a^{2}+b^{2}+c^{2}+d^{2}=e^{2}+f^{2}+4MN^{2} $. So your lemma rewrites as $ e^2+f^2+2bd\geq a^2+c^2 $.  :wink:


Har n zelei mohaddab ra mitavan ba tarsime (n-3) ta az ghotrhayash ke hich 2taei az anha joz dar rasha moteghate nistand be n-2 mosalas efraz kard.sabet konid hameye zelha va ghotrhaye chenin efrazi ra mitavan rooye masire chandzeli baste va peyvasteri bi anke az ghesmati az masir bish az 1 bar oboor konim tey kard,agar va faghat agar n mazrabi az 3 bashad. :wink:

dg soal nadashti ke ino dadi?ye chand ta soale joondare combinatorics ham age mitooni bede.

chashm saman joon shoma herfei hasti man ke herfei nistam soale khafan bezaram.. :D
motenahi noghte siah va motenahi noghte sefid dar safhe darim .har 4 noghte ke dar nazar migirim khati vojood darad ke noghate sefid dar yek samt va noghate siah dar samte digar bashand.sabet konid khati vojood darad ke tamame noghate siah dar yek samt va tamame noghate sefid dar samte digar vaghe shavad. :P

in soale riaziat tahadobe 3 sal pish too dore 40 nafar boode. :wink:
kafiye sabet konim pooshe mohadabe do majmoe ra mitavan ba khate rasti az ham joda kard.agar pooshe mohadabe do majmooe ba ham tadakhol dashte bashad ke be rahati be tanaghoz miresad agar ham yeki daroone digari bashad baz ham badihi mishavad va agar pooshe mohadab ha in do halat ra nadashte bashand az ghaziyeye helli estefade mikonim. :roll:

:wacko:
nemidoonam doroste ya na vali man ke hichi nafahmidam/. :P


Assume $ a,b,c$ be posutive numbers such that $ a\plus{}b\plus{}c\equal{}1$.
Prove that  $ 4(ab\plus{}bc\plus{}ca)\minus{}9abc\leq1$

after expanding
It's equivalent to prove that $ \sum_{cyc}a^3\plus{}3abc \geq \sum_{sym}a^2b$
which is Schur's inequality.


projective geometry
geometry proposed
geometry

[quote="dreammath"]Let $ABCD$ be a quadrilateral. Let $M$, $N$, $P$, $Q$ be points on the lines $AB$, $BC$, $CD$, $DA$, respectively.
Prove that: The lines $MQ$, $NP$, $BD$ are concurrent if and only if the lines $MN$, $PQ$, $AC$ are concurrent.[/quote]

Easy indeed:

We will use directed segments. Let the line MQ meet the line BD at a point U. Then, the lines MQ, NP and BD are concurrent if and only if the point U lies on the line NP.

Now, after the Menelaos theorem for the triangle ABD and the collinear points M, Q, U, we have

$ \frac{DU}{UB}\cdot \frac{BM}{MA}\cdot \frac{AQ}{QD}=-1$.

Thus,

$ \frac{DU}{UB}=-\frac{QD}{AQ}\cdot \frac{MA}{BM}=-\frac{DQ}{QA}\cdot \frac{AM}{MB}$.

Now, we have the following equivalence:

{The lines MQ, NP and BD are concurrent}
$\Longleftrightarrow$ {The point U lies on the line NP}
$\Longleftrightarrow$ {The points N, P, U lie on one line}
$\Longleftrightarrow$ $ \frac{DU}{UB}\cdot \frac{BN}{NC}\cdot \frac{CP}{PD}=-1$

(the last equivalence follows from the Menelaos theorem for the triangle BCD). And since $ \frac{DU}{UB}=-\frac{DQ}{QA}\cdot \frac{AM}{MB}$, we can rewrite this last equation as

$-\frac{DQ}{QA}\cdot \frac{AM}{MB}\cdot \frac{BN}{NC}\cdot \frac{CP}{PD}=-1$,

i. e. as

$ \frac{DQ}{QA}\cdot \frac{AM}{MB}\cdot \frac{BN}{NC}\cdot \frac{CP}{PD}=1$.

Thus, we have showed that the lines MQ, NP and BD are concurrent if and only if $ \frac{DQ}{QA}\cdot \frac{AM}{MB}\cdot \frac{BN}{NC}\cdot \frac{CP}{PD}=1$. Similarly, we can prove that the lines MN, PQ and AC are concurrent if and only if $ \frac{DQ}{QA}\cdot \frac{AM}{MB}\cdot \frac{BN}{NC}\cdot \frac{CP}{PD}=1$. Hence, it follows that the lines MQ, NP and BD are concurrent if and only if the lines MN, PQ and AC are concurrent. Proof complete.

Another problem that can be solved by the same trick:

[i]Let A, B, C, D be four points in space, not lying on a common plane. Let M, N, P, Q be four points on the lines AB, BC, CD, DA, respectively. Then prove that the points M, N, P, Q lie on one common plane if and only if $ \frac{DQ}{QA}\cdot \frac{AM}{MB}\cdot \frac{BN}{NC}\cdot \frac{CP}{PD}=1$.[/i]

  Darij

this seems to me be just Desargue's theorem

Oh yes you are right! If the lines MQ, NP and BD are concurrent, then we can apply Desargues to the triangles MNB and QPD and see that the points of intersection MN /\ QP, NB /\ PD and BM /\ DQ are collinear; but NB /\ PD = C and BM /\ DQ = A, so that it follows that the points MN /\ QP, C and A are collinear, i. e. the lines MN, QP and CA concur; in other words, the lines MN, PQ and AC are concurrent. The converse is analogous...

I really don't understand why I didn't see this simple solution when I posted my previous message...

  Darij


How do you do this problem?

If $ x\plus{}y\equal{}5$ and $ xy\equal{}1$, then what is $ x^3\plus{}y^3$?

[quote="cf249"]Thank you!  Where did you get this equation from?[/quote]
That's what you get by expanding out $ (x\plus{}y)^3\equal{}x^3\plus{}3x^2y\plus{}3y^2x\plus{}y^3$ and factoring.

We could derive this directly from the binomial theorem,
\begin{align*}
(x + y)^3 &= x^3 + 3x^2y + 3xy^2 + y^3 = x^3 + y^3 + 3xy(x + y) \\
x^3 + y^2 &= (x + y)^3 - 3xy(x + y)
\end{align*}
In fact, if you let $ e_1 = x + y$, $ e_2 = xy$, and $ s_n = x^n + y^n$ (the elementary symmetric polynomials and the power sum, respectively), the recursion
\[ s_0 = 2, s_1 = e_1, s_n = e_1 \cdot s_{n - 1} - e_2 \cdot s_{n - 2}
\]
holds true. From this recursion we can find $ s_3$.


is it $9^{98} \binom{100}{98}$ by the binomial theorem?

[quote="cats..."]$P(x)=(x+9)^{100} \cdot Q(x)$.   $P$ and $(x+9)^{100}$ are of degree 100, so $Q(x)$ is of degree 0, i.e $Q(x)=c$.  since we're given $P$ is a monic polynomial, $c$=1, and $P(x)=(x+9)^{100}$, so expand and match coefficients to get answer.[/quote]

Can you show how? thanks

what do you want to be shown?  how to expand and match the coefficients?  to do that just use binomial theorem.


[quote="ghjk"]Give Triangle ABC(A IS RIGHT ANGLE) inscribed Circle (o).Then take any M on thecircle.Draw opposite to BA AND CA some lines such as BE=BC=CF.PROVE:MA+MB+MC>=EF[/quote]

Can you phrase it more explicitly. I don't understand the problem

Daniel


calculus
integration
function
limit
real analysis
real analysis unsolved

Prove that there are NO continous functions $f: \Re\to\Re$ such that if $F\in\int f(x)dx$ then $e^{F}\in\int e^{f(x)}dx$.

it's wrong...the solution in the book is flawed :huh:

The conditions are $F'=f$ and $fe^{F}=e^{f}$. Necessarily $\forall{x},f(x)>0$ and $F$ is an increasing function.
$1^{st}$ case: $\lim_{x\to-\infty}F(x)=-\infty$; $\frac{e^{f(x)}}{f(x)}\to{0}$. $\frac{e^{f(x)}}{f(x)}\ge\frac{1}{f(x)}$ thus $f(x)\to \infty$, a contradiction.
$2^{nd}$ case: $\lim_{x\to-\infty}F(x)=l$. Then $\frac{e^{f(x)}}{f(x)}\sim{e^{l}}$. Necessarily $l\ge1$ and $\exists{\alpha,\beta}$ s.t. $0<\alpha\leq{1}\leq\beta$ and $e^{l}=\frac{e^\alpha}{\alpha}=\frac{e^\beta}{\beta}$. $f$ is a continuous function, therefore, for example, $\lim_{x\to-\infty}f(x)=\alpha$. $\exists{m}>0$ s.t. $\forall x\in]-\infty,0],f(x)\ge{m}$ and $\lim_{x\to-\infty}F(x)=-\infty$, a contradiction.


I have just realized I have been a member of AoPS for 3 years and 4 days. Wow. So like, the point of this thread: go through your first posts and pick out any funny, interesting, self-deprecating first posts and post them.
[quote="The Neverending Story"]The lady gets run over by a random bus. Her family thanks the bus driver, who pulls out a golf club...  :evilgrin:[/quote]


calculus
integration
calculus computations

If $ \sinh x\cdot \sinh y\equal{}1$, find

$ \int\limits_{0}^{\infty}ydx\equal{}?$

Using the substitution $ x = \arcsin h(%Error. "csch" is a bad command.
y)$, we have $ dx = -\frac{dy}{\sinh y}$, hence $ \int_{0}^{\infty} y \, dx = \int_{0}^{\infty} \frac{y}{\sinh y} \, dy$. Then

\begin{eqnarray*}
& & \int_{0}^{\infty} \frac{y}{\sinh y} \, dy\\
& = & \int_{0}^{\infty} \frac{2ye^{-y}}{1 - e^{-2y}} \, dy\\
& = & \int_{0}^{\infty} 2y \sum_{n=0}^{\infty} e^{-(2n+1)y} \, dy\\
& = & 2 \sum_{n=0}^{\infty} \int_{0}^{\infty} y e^{-(2n+1)y} \, dy\\
& = & 2 \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}\\
& = & \frac{\pi^2}{4}
\end{eqnarray*}


A circle having center (0,k), with k>6, is tangent to the lines y=x, y=-x, and y=6. what is the radius of the circle?

A.6 x square root of 2 -6
B. 6
C. 6 x square root of 2
D12
E.6 + 6 x square root of 2

I keep on getting the radius to be 12 can someone please tell me if its right and if not pleasea demonstrate how you do it?[/list][/code]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=372639[/url]

Please use the search function or go through the contests section. Post in the old thread if you have questions.


Source: Alabama State Math Contest, 2001 Comprehensive Division III, #24

Three circles, each with radius 6, intersect so that each circle passes through the centers of the other two circles.  Find the area of the region that is the intersection of the interiors of the three circles.

(A) 9:rt3:
(B) 6:pi:
(C) 9:pi:
(D) 18(:pi:-:rt3:)
(E) 48:pi: - :rt3:

For this problem, I got 18(:pi:-:sqrt:3) which is D.  Now, someone please do the backbreaking labor of typing a solution for this problem, preferrably one with a diagram of some sort.  Sorry, I don't have enough time to type up a solution.

[img]http://confuted.0catch.com/diagram.GIF[/img]
And if that doesn't start working, http://confuted.0catch.com/diagram.html has the diagram, with a bonus banner ad for y'all.

Uhh...confuted...the diagram isn't showing up on your website.
:-:theta:

yeah...
you forgot to end the quotation in <img src="diagram.GIF">

nice.... i finally got wat da problem meant!

I'm not sure why confuted's diagram didn't post in this forum, so I'm going to retry it...

Quotes or no quotes, no banana.  I think the site you're using doesn't allow remote users to view pictures hosted by them.  (But I thought you were using webhop to host directly off your own computer!)

[i]Edited[/i]: I know I'm not using the BBCode tags here - img src is HTML....  But evidently neither one works.

1) You're using the image tags incorrectly
2) The site it's hosted on probably doesn't allow remote linking, which is why I made a page
3) I am, but this isn't on my computer because I figured people might want to look at it when my computer isn't on ;) http://confuted.webhop.org is still hosted on my computer.


Let $ a$, $ b$, $ c$ be nonnegative reals such that $ a \plus{} b \plus{} c \equal{} 1$. Show that
  $ \frac {1}{a \plus{} b^{2}} \plus{} \frac {1}{b \plus{} c^{2}} \plus{} \frac {1}{c \plus{} a^{2}}\geq\frac {27}{4}$.

I posted this same one [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=113874]here[/url]

[quote="2006"]$\ a,b,c\geq0$ and $\ a+b+c=1$ show that
  $\frac{1}{a+b^{2}}+\frac{1}{b+c^{2}}+\frac{1}{c+a^{2}}\geq\frac{27}{4}$
 :)  :D[/quote]
I think you should work on your topic naming skills! :P

[quote="me@home"][quote="arqady"][quote="2006"]$\ a,b,c\geq0$ and $\ a+b+c=1$ show that
  $\frac{1}{a+b^{2}}+\frac{1}{b+c^{2}}+\frac{1}{c+a^{2}}\geq\frac{27}{4}$
 :)  :D[/quote]
I think, it's wrong. Try $a=0.6,$ $b=0.4$ and $c=0.$ :wink:[/quote]
Are you sure?
I'm getting:
$f(.4,.6,0)=9.232456...>\frac{27}{4}$
maybe I typed it in wrong... but it seems to hold![/quote]

I get $\frac{2255}{342}<\frac{27}{4}$

Must have typed it in wrong, now I'm getting $6.593567...=\frac{2255}{342}$
I can prove that $\sum_{CYC}\frac{1}{a+b^{2}}\geq \sum_{CYC}\frac{1}{a+b}\geq \frac92$
actually, I think that the original question with the $13/2$ is the correct lower bound, because playing around with the numbers gives a minimum around $a=.356,b=.644,c=0$ and the quantity equals $6.5176...$
[quote="vietnam2007"]I think that it is easy to prove it by using twice AM-GM .[/quote]
obviously this isn't right, but I am curious what your idea was? I don't see how AM-GM works here at all

[quote="me@home"]I think that the original question with the $13/2$ is the correct lower bound, because playing around with the numbers gives a minimum around $a=.356,b=.644,c=0$ and the quantity equals $6.5176...$
[/quote]
Maybe $a=0.644,$ $b=0.356$ and $c=0$ $?$ :wink:

Yes thats what I meant...
I am wondering if the inequality:\[\sum_{cyc}\frac{1}{a+b^{2}}-\frac23\geq \sum_{cyc}\frac{1}{a+b}\]
holds,
what do you think arqady?

That "double Am-Gm" is just called Cauchy, Engel form:\[\sum\left(\frac{X}{Y}\right)\geq \frac{(\sum\sqrt{X})^{2}}{\sum Y}\]
and that is a very weak bound - just as weak as $\sum\frac{1}{a+b^{2}}\geq \sum\frac{1}{a+b}$
because:\[\sum\frac{1}{a+b}\geq \frac{9}{\sum(a+b)}\text{ from your \" double AM-GM\" }\\ = \frac{9}{2\sum a}=\frac92\]

interesting to note: $\frac{1}{a+b^{2}}=\frac{1}{(a+b)^{2}+(c-b)a}$ so the quantities $\frac{1}{a+b^{2}}$ and $\frac{1}{(a+b)^{2}}$ approximately equal each other. I don't have a solution for the proposed problem, it is more difficult than it appears.

[quote]$\frac{9}{a^{2}+b^{2}+c^{2}+a+b+c}\geq \frac{9}{2}$[/quote]
are you sure vietnam 2007?

[/quote]

It' s equivalent to $a^{2}+b^{2}+c^{2}+a+b+c \leq 2$ wich is true because $a+b+c=1$ and $a^{2}+b^{2}+c^{2}\leq (a+b+c)^{2}=1$

I don't see what's wrong with it, equality even at (1,0,0)... :maybe:

Any solutions to the actual inequality? I am really curious :!:

[quote="me@home"]I think that the original question with the $ \frac {13}{2}$ is the correct lower bound. Any solutions to the actual inequality?[/quote]Because the original question occurred equality when $ a = b = c$, the following recension is nice :

[color=blue]If $ a, b, c$ be nonnegative reals such that $ a + b + c = 1$, then

$ \frac {1}{a + b^2} + \frac {1}{b + c^2} + \frac {1}{c + a^2}\geq\frac {13}{2 - 2abc},$

with equality if and only if $ a = b = c = \frac {1}{3}.$[/color]
[hide="Proof."]$ \sum_{cyc}{\frac {1}{a(a + b + c) + b^2} - \frac {13(a + b + c)}{2(a + b + c)^3 - 2abc}}$

$ \equiv\frac {F(x,y,z)}{2[a(a + b + c) + b^2][b(a + b + c) + c^2][c(a + b + c) + a^2][(a + b + c)^3 - abc]}.$

$ F(a,b,c) = F(a,a + s,a + t)$

$ = 136a^5(s^2 - st + t^2) + a^4(260s^3 - 323s^2t + 223st^2 + 260t^3)$

$ + a^3[3(s - t)^2(75s^2 + 52st + 14t^2) + 362st^3 + 183t^4]$

$ + a^2[5(s - 2t)^2(13s^3 + 25s^2t + 19st^2 + 5t^3) + 37s^5 + 17s^4t + 2s^3t^2 + 2s^2t^3 + 3st^4 + 2t^5]$

$ + \frac {a}{4}[(2s - 3t)^2(23s^4 + 49s^3t + 38s^2t^2 + 3st^3 + 2t^4) + s^4t^2 + 3s^3t^3 + 2s^2t^4 + 341st^5 + 74t^6]$

$ + \frac {(s + t)^2}{3375}[(5s - 9t)^2(270s^3 + 297s^2t + 329st^2 + 83t^3)$

$ + 10s^3t^2 + 103s^2t^3 + 1071st^4 + 27t^5] > 0,$

which is clearly true for $ a = \min\{a,b,c\}.$[/hide]
With the same conditions, the stronger inequality

$ \frac {1}{a + b^2} + \frac {1}{b + c^2} + \frac {1}{c + a^2}\geq\frac {13 + 918abc}{2 + 134abc}$

holds too.

See also here : http://www.mathlinks.ro/viewtopic.php?t=89008

Use Titu's Lemma.
The problem gets vastly simplified... :D  :)


find all pairs a,b of positive integers satisfying
$2a^{2}= 3b^{3}$


geometry
trigonometry
trig identities
Law of Sines
geometry proposed

Let $ ABC$ be a right triangle with $ \angle C\equal{}90^o$ and center of incircle is $ O$.$ E$ is midpoint of $ BC$.$ OE$ cuts $ AC$ at $ M$.Incircle $ (I)$ touch $ AB;AC$ at $ P;Q$,respectively.
$ CH \perp AB(H \in AB)$
$ CH$ cuts $ PQ$ at $ N$
Prove that $ CM\equal{}CN$

PROOF：
1.It is known that AM=r=OG
2. △NHP～△AQO   NH/AQ=HP/OQ  (1)
3.In △NHP and △NCQ , use law of sines , then  NH/HP=NC/QC (2)
4.According to (1) and (2) , AQ=CN, we already have AQ=CM
So CM=CN

[quote="chien than"] [color=darkred] Let $ ABC$ be a $ C$ - right triangle with incircle $ w = C(I,r)$ . Circle $ w$ touches $ AB$ , $ AC$ in  $ P$ , $ Q$ respectively. Denote :

midpoint $ E$ of $ BC$ ; $ M\in IE\cap AC$ ; projection $ H$ of $ C$ to $ AB$ ; $ N\in CH\cap PQ$ .  Prove that $ CM = CN$ .[/color] [/quote]

[quote="Virgil Nicula"][color=darkred][u][b]Lemma (memorize !).[/b][/u] Consider $ \triangle\ ABC$ and $ \|\begin{array}{ccc} M\in (AB) & , & N\in (AC) \\
 \\
P\in (MN) & , & S\in BC\cap AP\end{array}\|$ . Then $ \boxed {\ SC\cdot\frac {MB}{MA} + SB\cdot\frac {NC}{NA} = BC\cdot \frac {PS}{PA}\ }$ .

[b]Particular cases[/b] : $ \{\begin{array}{cccc} 1\ \blacktriangleright & P: = G & \implies & \frac {MB}{MA} + \frac {NC}{NA} = 1 \\
 \\
2\ \blacktriangleright & P: = I & \implies & b\cdot \frac {MB}{MA} + c\cdot\frac {NC}{NA} = a\end{array}$ .[/color][/quote]

[color=darkblue][b][u]Proof (metric).[/u][/b] Apply [b]second particular case  of upper lemma[/b] to $ E$ , $ M$ and $ I\in EM$ :

$ b\cdot\frac {EB}{EC} + a\cdot \frac {MA}{MC} = c$ $ \implies$ $ \frac {MA}{c - b} = \frac {MC}{a} = \frac {b}{a + c - b}$ $ \implies$ $ \boxed {\ CM = \frac {ab}{2(p - b)}\ }$ .

Ascertain easily that $ PH = \frac {(a - b)(p - c)}{c}$ . Apply [b]Menelaus' theorem[/b] to transversal $ \overline {PNQ}$ for $ \triangle\ ACN$ :

$ \frac {PH}{PA}\cdot \frac {QA}{QC}\cdot\frac {NC}{NH} = 1$ $ \implies$ $ \frac {NH}{NC} = \frac {(a - b)(p - c)}{c(p - a)}\cdot\frac {p - a}{p - c}$ $ \implies$ $ \frac {NC}{c} = \frac {NH}{a - b} = \frac {h_c}{2(p - b)}$ $ \implies$ $ \boxed {\ CN = \frac {ab}{2(p - b)}\ }$ .

[b][u]Remark.[/u][/b] Observe that $ CM = CN = r_b=p-c$ . [b]Plane geometry[/b] used [b]nicely[/b] an well-known property $ AM = r$ . Here is its proof.

$ MA = r$ $ \Longleftrightarrow$ $ b - \frac {ab}{2(p - b)} = p - c$ $ \Longleftrightarrow$ $ \frac {ab}{2(p - b)} = p - a$ $ \Longleftrightarrow$ $ 2ab = c^2 - (a - b)^2$ $ \Longleftrightarrow$ $ c^2 = a^2 + b^2$ what is truly.

[b]Here is an interesting and very nice (in my opinion !) extension of [u]Chien Than[/u]'s proposed problem :[/b][/color]

[quote="Virgil Nicula"] [color=darkred][u][b]Generalization.[/b][/u] Let $ ABC$ be a nonobtuse triangle with $ c\ge a\ge b$  and incircle $ w = C(I,r)$ . Circle $ w$ touches $ AB$ , $ AC$ in  $ P$ , $ Q$ respectively. Denote :

midpoint $ E$ of $ BC$ ; $ M\in EI\cap AC$ ; projection $ H$ of $ C$ to $ AB$ ; projection $ R$ of $ M$ to $ BC$ ; $ N\in CH\cap PQ$ .  Prove that $ RM = CN = r_b$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote $ L\in AI\cap BC$ . Ascertain easily that $ EL = \frac {a(c - b)}{2(b + c)}$ and $ PH = \frac {(a - b)(p - c)}{c}$ .

Apply [b]Menelaus' theorem[/b] to transversal $ \overline {EIM}$ in $ \triangle\ ALC\ : \ \frac {EL}{EC}\cdot\frac {MC}{MA}\cdot\frac {IA}{IL} = 1$ $ \implies$ $ \frac {c - b}{b + c}\cdot \frac {MC}{MA}\cdot\frac {b + c}{a} = 1$ $ \implies$

$ \frac {MC}{a} = \frac {MA}{c - b} = \frac {b}{2(p - b)}$  $ \implies$ $ MC = \frac {ab}{2(p - b)}$ . Since $ MR = MC\sin C =$ $ \frac {ab\sin C}{2(p - b)} =$ $ \frac {S}{p - b}$ $ \implies$ $ \boxed {\ MR = r_b\ }$ .

Apply [b]Menelaus' theorem[/b] to transversal $ \overline {PQN}$ in $ \triangle\ AHC\ : \ \frac {PH}{PA}\cdot\frac {QA}{QC}\cdot\frac {NC}{NH} = 1$ $ \implies$ $ \frac {(a - b)(p - c)}{c(p - a)}\cdot \frac {p - a}{p - c}\cdot\frac {NC}{NH} = 1$ .

Therefore,  $ \frac {NC}{c} = \frac {NH}{a - b} = \frac {h_c}{2(p - b)}$ $ \implies$ $ NC = \frac {ch_c}{2(p - b)} = \frac {S}{p - b}$ $ \implies$ $ \boxed {\ NC = r_b\ }$ . In conclusion, $ RM = CN = r_b$ . [/color]

[color=red][b]I wait a synthetical proof for $ \boxed {\ RM = CN = r_b\ }$ .[/b][/color]


Sorry if this has been asked before.  I am going to be sending my solutions for year 18 in $\LaTeX$ format, and I have a question on the solutions template.  Do we just fill in the information like name and id and the solutions and rename the file, or fill in the information, rename the file, but take out the comments?  In other words, do the comments have to be there?

It doesn't matter.  The comments are ignored anyway -- they're only there to help you fill out the form.

If you're using last year's template, make sure you change to year to 18.  We'll also be provided an updated template when we release the new Round 1 problems.

As long as we're on the subject, let me remind everyone: submitting your solutions in LaTeX is [b]not[/b] required.  The only "advantage" you have by submitting your solutions in LaTeX is that we might selection one of your solutions for publication on the USAMTS website.  But it will not affect your scoring in any way if you don't send in LaTeX solutions.

Thanks for your help, and I'm only using $\LaTeX$ because I'm really accustomed to it.


calculus
real analysis
real analysis unsolved

I need help with this question. I'm not in college but i got no other response from the other post i made in the olympiads section. can someone help me?

[b][u]The question is:[/u][/b]

A martial arts Instructor devised a way to train the martial arts academy students in the art of mass combat. These battles are divided into 2 rounds. Rounds A and B.

[b][u]Round A :[/u][/b] The students are allowed to fight against anyone and as many opponents they choose. (i.e. 1 Student can go against 5, or 3 can go against 7) However, the students cannot change opponents.

[b][u]Round B :[/u][/b] The students must not spar with the opponents encounted in the previous round and must spar with all those they didn't fight in the first round.

This is to ensure that the weakest student will have a chance to fight the strongest of the lot and that the rest will be trained in fighting as a part of a group or against many. Each student is tagged numerically and their sensei draws up a table that lists the opponents of each student.

[b][u]Question:[/u] In one particular battle, thier sensei discovered that the table he gets in the 2nd round could be made identical to that in the 1st round simply by re-labelling the students. If all the students are now divided into groups of 4, what is the possible numbers of students that will be left over? [/b]

I need to know how many and why?   :?:
And I also need a good and reasonable explination so that I may study from it.
I would be very grateful to anyone who could help. Thank you.  :D

[quote="Kent Merryfield"]Please don't double post. This is not calculus or analysis and does not belong in this forum.[/quote]

Would you please tell me kin which forum it belongs? Sorry for double posting. :|


[quote="anirudh"]Figure out the pattern, and then write the next three terms in the series.

$98, 72, 77, 49, 54, 20\dots$[/quote]
Are you sure the last one isn't 24?

if it was 24 the next 3 would be 29, -3, 2, -32
-jorian

Its not 24; that's why I posted it here :roll:


probability
function
integration
rate problems
probability and stats

Calls for dial-in connections to a computer center arrive at an average rate of four per minute. The calls follow a Poisson distribution. If a call arrives at the beginning of a one-minute interval, what is the probability that a second call will not arrive in the next 20 seconds?

My answer:

I know that if events are occurring in time according to a Poisson distribution with mean $ \lambda t$, then the interarrival times between events have an exponential distribution with mean $ 1/\lambda$ and density function of $ f(y)\equal{} \int^\infty_0 \frac{1}{\beta}e^{(\minus{}y/\beta)} dy$. So

P(call not arriving in next 20 seconds)= 1 - P(call will arrive in next 20 seconds) = $ 1\minus{} \int^{1/3}_0 4e^{\minus{}4y} dy\equal{} e^{\minus{}4/3}$

Is this correct?


conics
parabola
calculus
integration
function
calculus computations

Is there a way to write the value of $ c$ without using $ x$ when $ 8x\minus{}27x^3\equal{}c$?
Thanks for any input.

What do you mean?  $ c$ is given in terms of $ x$, so if you don't know anything else about $ c$ then no.

I'll try to show the assigned problem:

[url=http://img210.imageshack.us/my.php?image=mathxh2.jpg][img]http://img210.imageshack.us/img210/9881/mathxh2.th.jpg[/img][/url][url=http://g.imageshack.us/thpix.php][img]http://img210.imageshack.us/images/thpix.gif[/img][/url]

We were to find a $ c$ that would make A1 = A2
parabola: $ 8x\minus{}27x^3$

So what you're actually asking us is how to solve for $ x$.  That's not the question you asked initially.  Unfortunately, the expression for $ x$ in general is very, very ugly, so don't use it.  But since you're asked to compute $ c$, not to find $ x$ in terms of $ c$, you don't need to talk about $ c$ at all until the last step.

well, I figured you'd needa know where y=c in order to have bounds for your integral, and I figured you couldn't figure that out numerically so I would therefore needa write c in terms of x, which I saw as impossible, but then I remembered the Lambert W function, and I thought you might have some jacked-up way of writing the value of c in terms of x. But I guess what I really wanna know is the proper way to solve for c in this problem..

${ A_2(c) = ca_c - 4a_c^2 +\frac{27}{4}a_c^4 }=\frac{3ca}{4}a_c-2a_c^2$
That last step doesn't make any sense to me :(

just  using $ 27a_c^3\minus{}8a_c\plus{}c\equal{}0$

I may find the manipulation in a moment, but please explain the steps in your process. Thanks. Sorry for the algebraic slowness.

Oh, I should have seen this earlier; it's simply plugging in what $ c$ equals. Once again, my apologies for being so too lazy to input true effort.


function
real analysis
real analysis unsolved

(a) Build a function $f : \mathbb R \to \mathbb R_+$ with the property $\left( \mathcal P \right)$, i.e. all $x \in \mathbb Q$ are local, strict minimum points.

(b) Build a function $f : \mathbb Q \to \mathbb R_+$ such that every point is a local, strict minimum point and such that $f$ is unbounded on $I \cap \mathbb Q$, where $I$ is a non-degenerate interval.

(c) Let $f: \mathbb R \to \mathbb R_+$ be a function unbounded on every $I \cap \mathbb Q$, where $I$ is a non-degenerate interval. Prove that $f$ doesn't have the property $\left( \mathcal P \right)$.

for 1/ I think the following works. Take an increasing function $g: \mathbb{R}_+\to \mathbb{R}^*_+$ such that $g(x)<2$ for all $x\geq 0$. ($arctan(x)+1$ for example).
Now one can define $f$ by:
*if $x=\frac{p}{q}$, $(p,q)=1$, $p \geq 0$, then $f(x)=g(q)$
*if $x$ is irrational, $f(x)=2$

I was thinking of almost the same thing: $f(x) = \mathcal C - \frac{1}{q}$ for $\displaystyle x = \frac{p}{q}$, $p,q \in \mathbb Z$, $(p,q) = 1$, $q > 0$, and $f(x) = \mathcal C$ for $x \in \mathbb R \setminus \mathbb Q$.

The challenge (for me) is of course (c) ...

Assume that there exists such functions. Construct the sequence of intervals as follows. There exist a rational $x$ number s.t. $f(x)\ge 1$ and define $[a_1,b_1]$ an interval s.t. $x$ is local minimum on $[a_1,b_1]$, Hence $f(x)\ge 1, \forall x\in [a_1,b_1]$.
Since $f$ is unbounded on $[a_1,b_1]$ we can find $x_1\in[a_1,b_1]$ s.t. $f(x)\ge 2$ $[a_2,b_2]$ an interval s.t. $x$ is local minimum on $[a_2,b_2]$, Hence $f(x)\ge 2, \forall x\in [a_2,b_2]$. Following the same method define inductively $[a_n,b_n]\subset [a_{n-1},b_{n-1}]$ such that $f(x)\ge n$ on $[a_n,b_n]$.

From Cantor th we can find $a\in \cap_{k=1}^\infty [a_k,b_k]$. Since $a$ is  in every $[a_k,b_k]$ we get that $f(a)\ge k,\forall k\in \mathbb{N}$. Contradiction.
Hence there not exists such functions.


find out all $f$ such that
$1)$$f:N \longrightarrow N$
$2)$ $f(n+1)>f(f(n))$

I think It's a very very famous problem.
Only one root $f(n)=n$.

Yes, it is IMO 1977 #6.
A solution can be found almost everywhere...for instance :
http://www.kalva.demon.co.uk/imo/isoln/isoln776.html

Pierre.

i agree with [b]hungkhtn[/b]
for example, see this pro:
[b]PRO[/b]: find all $f:\mathbb {N^*}\to\mathbb {N^*}$ such that: $f(1)=1$ and $f(f(n)).f(n+2)+1=f(f(n+1))\cdot{f(n+1)}$


geometry
number theory
number theory proposed

[i]Author of the problem is Adrian Satja Kurdija (although, of course, it may be possible that there already exists a very similar or same problem author didn't know about). This was Problem 4 (Number Theory) for 10th grade on 1st Balkan Student Mathematical Competition 2008 - a competition 11 schools from Balkan area took part of during November 2008. All information about the competition can be found on http://www.mathlinks.ro/viewtopic.php?t=243614.[/i]

$ a$, $ b$ and $ c$ are natural numbers. It is known that $ a^2 \plus{} b^2 \plus{} abc$ has no more than $ 2008$ natural divisors and that it is divisible by $ (c \plus{} 2)^{1004}$. Prove that $ a$ and $ b$ are not relatively prime.

Note that $ (c\plus{}2)^{1004}$ divide $ S\equal{}a^{2}\plus{}b^{2}\plus{}abc$ so it has less than $ 2008$ divisors
From here easy to see $ c\plus{}2$ must be prime $ p$ And $ a^{2}\plus{}b^{2}\plus{}ab(p\minus{}2)\equal{}p^{k},2008 \geq k \geq 1004$
Hence $ (a\minus{}b)^{2}\equal{}p(p^{k\minus{}1}\minus{}ab)$
$ \rightarrow a\minus{}b \vdots p \rightarrow (a\minus{}b)^{2} \vdots p^{2} \rightarrow ab \vdots p$
Hence $ a \vdots p$ and $ b \vdots p$
The conclusion is followed

Sorry  by the ignorence but y don't undersand how do you no that $ a^2\plus{}b^2\plus{}abc$ it's a power of a prime. :P

[quote="Raúl"]Sorry  by the ignorence but y don't undersand how do you no that $ a^2 \plus{} b^2 \plus{} abc$ it's a power of a prime. :P[/quote]

It 's easy to show that $ c \plus{} 2$ is prime . Because if not, $ a^2 \plus{} b^2 \plus{} abc$ will have more than 2008 natural divisors, contradiction. :roll:
Assume that $ a^2 \plus{} b^2 \plus{} abc \equal{} (c \plus{} 2)^{1004}.p$
If $ p$ is not divisible by $ c \plus{} 2$, we also get $ a^2 \plus{} b^2 \plus{} abc$ have more than 2008 natural divisors.
$ \Rightarrow p \equal{} (c \plus{} 2)^m$
That 's explanation why we got the result $ a^2 \plus{} b^2 \plus{} abc$ is a power of prime. :lol:


inequalities
induction
logarithms
inequalities proposed

Prove that for every positive integer $ n$,

$ n^n \le (n!)^2 \le \left( \frac{(n\plus{}1)(n\plus{}2)}{6} \right) ^n.$

[quote="M@ri@n"][quote="moldovan"]Prove that for every positive integer $ n$,

$ n^n \le (n!)^2 \le \left( \frac {(n \plus{} 1)(n \plus{} 2)}{6} \right) ^n.$[/quote]

If I'm not wrong then both inequalities can be proved by induction on $ n$ and first year calculus.[/quote]
you had prove how?When I  using induction prove:$ n^n\leq (n!)^2$,I need prove:$ n^n\geq (n\plus{}1)^{n\minus{}1}$.
To prove this problem ,I must using :$ f'(x)$,where $ f(x)\equal{}x^x\minus{}(x\plus{}1)^{x\minus{}1}$
It is not nice.

[quote="WiZ@Rd"]Can you post the proof, please.[/quote]
(1) Obviously, for all positive integers $ n > 1$ we have the inequality
$ n\cdot\ln \left (1 \plus{} \frac {1}{n}\right )\le n\cdot\frac {1}{n} \equal{} 1 < \ln (n \plus{} 1).$
This implies that $ \left (1 \plus{} \frac {1}{n}\right )^n\le n \plus{} 1$ where $ n\in\mathbb{N}$. This can be transformed to $ (n \plus{} 1)^{n \plus{} 1}\le (n \plus{} 1)^2\cdot n^n$.

(2) Induction.
For $ n \equal{} 1$ the situation is simple and the statement holds. Now suppose that $ n^n\le (n!)^2$ holds. Multiplying both sides by $ (n \plus{} 1)^2$ gives the inequality
$ (n \plus{} 1)^2\cdot n^n\le ((n \plus{} 1)!)^2$.
The result form (1) finishes the proof.


ratio
email
number theory
superior algebra
superior algebra unsolved

$Aut(G)$ refers to the group of automorphisms of a group $G$.

Define $''Aut^{n}(G)'' = Aut(Aut^{n-1}(G))$ for $n \ge 1$ and $Aut^{0}(G) = G$.

Is it true* that

$\limsup_{n \to \infty}\frac{|Aut^{n+1}(G)|}{|Aut^{n}(G)|!}= 0$

Intuitively speaking, the question is about how often we have a large number of automorphisms.

*Note that if $|G|$ is 1 or 2, then every automorphism group is of the same order as $G$ originally was, so the ratio is always 1. The question is asking for any other $G$.

For groups with $Z(G)=\{1\}$ this is false, because a theorem of Wielandt says that there is a n s.t. $Aut^{n}(G)=Aut^{n+1}(G)$, so for all $m>n$ we have $\frac{|Aut^{m+1}(G)|}{|Aut^{m}(G)|!}=\frac{1}{(|Aut^{m}(G)|-1)!}=\frac{1}{(|Aut^{n}(G)|-1)!}$ so the sequence becomes constant.

In fact it seems that there's a fair amount of literature about this. For example this [url=http://www.jstor.org/view/00029939/di970911/97p0214g/0]short paper[/url]. (If anyone can't access this PM me and I'll email the file.)

But what about when the center of $G$ is not trivial? Can we say anything? What about for $G = \mathbb{Z}/n\mathbb{Z}$? (The latter is more of a number theory question I guess.)


Union of infinite number of closed sets is not closed in general.
Can the following be an example??

[0,1] U [1,2] U ... U [n,n+1] ... = [0,$ \infty$)

Yes.  You've taken advantage of the fact that $ \mathbb{R}$ is not compact; does the statement still hold in a compact space, say $ [0, 1]$?

Did you mean this example:
$ \bigcup_{n \equal{} 1}^{\infty}{[0,1 \minus{} 1/n]} \equal{} [0,1)$

Perfect. Just keep in mind that $ [0,\plus{}\infty)$ [b]is[/b] a closed subset of $ \mathbb R$, so your original example doesn't work ;).

But, t0rajir0u says yes, you say no. Which one is true  :lol:

Think yourself a bit :P. (just recall the definitions and proceed directly from them; if you get stuck, post your questions :)).

Definition 1: A set S is open if every point in S is interior.
Definition 2: A set S $ \subset$ X is said to be closed if the complement of set S is open.
Definition 3: A set is closed if and only if it contains all of its limit points.

So, You say that $ [0,\infty)$ is closed. Then complement of it in R is $ ( \minus{} \infty,0)$ should be open, right?(Def 2)
Then all the points in this set must be interior. But, if we accept the points near $ \minus{} \infty$ to be interior(Def 1), why do we accept the points near $ \infty$ as boundary(limit) points(Def 3)? They're interior, too. So, $ [0,\infty)$ must be neither open nor closed.

Also, I'm asking why t0rajir0u accepted my example, by the way?

Because I was mistaken.  :P  My apologies; $ [0, \infty)$ is in fact closed since the set of its limit points is none other than $ [0, \infty)$ (review the definition of a limit point), and $ (\minus{}\infty, 0)$ is in fact open since the set of its interior points is $ (\minus{}\infty, 0)$ (review the definition of an interior point).

OMG, I had thought that near $ \infty$, there should be limit points. That's the reason of my paradox. Thanks a lot :)

Well, there should be, but they're not there - that's what [url=http://en.wikipedia.org/wiki/One-point_compactification]one-point compactification[/url] is for.


I'd recommend reading up on the basic rules for rounding to significant figures if that's a big part of the assignment.  Here is a link: [url]http://en.wikipedia.org/wiki/Significant_figures#Rounding_conventions[/url].  Also read up on your properties of logs and exponents that I have used below.

$ 1)$ $ \ln x \equal{} 6$

[hide]$ \ln x \equal{} 6$
$ \iff e^{\ln x} \equal{} e^6$
$ \iff \boxed{x \equal{} e^6}$[/hide]

$ 2)$ $ 5 \ln x \equal{} 35$

[hide]$ 5 \ln x \equal{} 35$
$ \iff \ln x \equal{} 7$
$ \iff e^{\ln x} \equal{} e^7$
$ \iff \boxed{x\equal{}e^7}$[/hide]

$ 3)$ $ \ln \left( \frac{1}{x^2} \right) \equal{} 5$

[hide]$ \ln \left( \frac{1}{x^2} \right) \equal{} 5$
$ \iff \ln x^{\minus{}2} \equal{} 5$
$ \iff \minus{}2 \ln x \equal{} 5$
$ \iff \ln x \equal{} \frac{\minus{}5}{2}$
$ \iff e^{\ln x} \equal{} e^{\minus{}\frac{5}{2}}$
$ \iff \boxed{x\equal{} e^{\minus{}\frac{5}{2}}}$[/hide]

$ 4)$ I'm going to assume you mean: $ \frac{e^{1\minus{} 2x}}{4} \equal{} 12$

[hide]$ \frac{e^{1\minus{} 2x}}{4} \equal{} 12$
$ \iff e^{1\minus{}2x}\equal{}48$
$ \iff \ln{e^{1\minus{}2x}}\equal{}\ln 48$
$ \iff 1\minus{}2x \equal{} \ln 48$
$ \iff 2x \equal{} 1\minus{} \ln 48$
$ \iff \boxed{x\equal{}\frac{1\minus{} \ln 48}{2}}$[/hide]

$ 5)$ Again I'm going to assume you mean: $ 4.5e^{x\plus{}1} \minus{}145 \equal{} 0$

[hide]$ 4.5e^{x\plus{}1} \minus{}145 \equal{} 0$
$ \iff 9e^{x\plus{}1} \minus{} 290 \equal{} 0$
$ \iff e^{x\plus{}1} \equal{} \frac{290}{9}$
$ \iff \ln{e^{x\plus{}1}} \equal{} \ln \left( \frac{290}{9} \right)$
$ \iff x\plus{}1 \equal{} \ln \left( \frac{290}{9} \right)$
$ \iff \boxed{x \equal{} \ln \left( \frac{290}{9} \right) \minus{}1}$[/hide]

[quote="Magic Man"]Solve the following. Give your answer to three significant figures.

1) lnx = 6
2) 5lnx = 35
3) ln(1/x^2) = 5
4) e^1-2x /4 = 12
5) 4.5e^x+1 -145 = 0

if someone could help me with the answer or explain it to me it would be helpful.
thank you
Magic Man[/quote]
[hide]1. e^6
2. e^7
3. sqrt(1/e^5)
4. 2e-24
5. ln (e^(144/4.5))

These are just the exact values without simplification/approximation.[/hide]

[quote="gauss1181"]Oh well. Isn't my fault.

Magic Man should have inserted parentheses properly, where they're supposed to be. That way, we can avoid ambiguity of orders of operations.[/quote]
you should have showed your work. answers mean nothing without solutions

[quote="skimnc"]I'd recommend reading up on the basic rules for rounding to significant figures if that's a big part of the assignment.  Here is a link: [url]http://en.wikipedia.org/wiki/Significant_figures#Rounding_conventions[/url].  Also read up on your properties of logs and exponents that I have used below.

$ 1)$ $ \ln x \equal{} 6$

[hide]$ \ln x \equal{} 6$
$ \iff e^{\ln x} \equal{} e^6$
$ \iff \boxed{x \equal{} e^6}$[/hide]

$ 2)$ $ 5 \ln x \equal{} 35$

[hide]$ 5 \ln x \equal{} 35$
$ \iff \ln x \equal{} 7$
$ \iff e^{\ln x} \equal{} e^7$
$ \iff \boxed{x \equal{} e^7}$[/hide]

$ 3)$ $ \ln \left( \frac {1}{x^2} \right) \equal{} 5$

[hide]$ \ln \left( \frac {1}{x^2} \right) \equal{} 5$
$ \iff \ln x^{ \minus{} 2} \equal{} 5$
$ \iff \minus{} 2 \ln x \equal{} 5$
$ \iff \ln x \equal{} \frac { \minus{} 5}{2}$
$ \iff e^{\ln x} \equal{} e^{ \minus{} \frac {5}{2}}$
$ \iff \boxed{x \equal{} e^{ \minus{} \frac {5}{2}}}$[/hide]

$ 4)$ I'm going to assume you mean: $ \frac {e^{1 \minus{} 2x}}{4} \equal{} 12$

[hide]$ \frac {e^{1 \minus{} 2x}}{4} \equal{} 12$
$ \iff e^{1 \minus{} 2x} \equal{} 48$
$ \iff \ln{e^{1 \minus{} 2x}} \equal{} \ln 48$
$ \iff 1 \minus{} 2x \equal{} \ln 48$
$ \iff 2x \equal{} 1 \minus{} \ln 48$
$ \iff \boxed{x \equal{} \frac {1 \minus{} \ln 48}{2}}$[/hide]

$ 5)$ Again I'm going to assume you mean: $ 4.5e^{x \plus{} 1} \minus{} 145 \equal{} 0$

[hide]$ 4.5e^{x \plus{} 1} \minus{} 145 \equal{} 0$
$ \iff 9e^{x \plus{} 1} \minus{} 290 \equal{} 0$
$ \iff e^{x \plus{} 1} \equal{} \frac {290}{9}$
$ \iff \ln{e^{x \plus{} 1}} \equal{} \ln \left( \frac {290}{9} \right)$
$ \iff x \plus{} 1 \equal{} \ln \left( \frac {290}{9} \right)$
$ \iff \boxed{x \equal{} \ln \left( \frac {290}{9} \right) \minus{} 1}$[/hide][/quote]

taking his solutions, I will round them to the nearest thousandth, who cares about sig figs?
1. 403.429
2. 1096.633
3. 0.0821
4. -1.436
5. 2.473

Even though there's a link, I want to address Significant Figures   because it's important.
[hide="A Rough Statement on Principles of Significant Figures"]
Significant Figures are very important in indicating the accuracy of a result from an experiment, mostly from labs.
In science labs, when we get some expressions and use them, our  statement [b]can only be as confident as our least accurate measurement.[/b]
Lets say we're measuring an old, smoothed, damaged block with a cm ruler. When we measure, we measure the values we know for sure, and then estimate for the last digit. For example: $ 2.56 cm$ means we know that the $ 2$ and $ 5$ are known for certain and the $ 6$ is a estimate. [b]Always[/b] include $ 1$ estimate.
Example, we add $ 2.01 \plus{} 25.5789 \equal{} 27.5889$, in this we have to change it to $ 27.6$ instead since we can only have $ 3$ significant figures.
When we have a number like $ 0.000027$,or $ 75000000$, we only have $ 2$ significant figures, the other zeros are just place holders. With a number like $ 209$, it has $ 3$ significant figures, since the zero is part of the accurate measurement. A way to tell though that 760 actually has $ 3$ significant figures is to rewrite it as $ 760.0$ or in scientific notation, $ 7.60 \times 10^2$ (include the zero!)
[/hide]
[hide="Edited given solutions for $ 3$ significant figures"]
1. $ 403$ or $ 4.03 \times 10^2$
2. $ 110.0$ or $ 1.10 \times 10^1$
3. Correct as is, or $ 8.21 \times 10^{ \minus{} 2}$
4. $ \minus{} 1.44$ or $ \minus{} 1.44 \times 10^1$
5. $ 2.47$ or $ 2.47 \times 10^1$
[/hide]

OMG @2above You posted when I was just born!


1)$ (CH_3)_3CCH(OH)CH_3$ + HBr ->

[hide="My View"]I believe reaction should be elimination because $ Br^ \minus{}$ is  a weaker nucleophile than $ OH^ \minus{}$ so it cannot substitute it.[/hide]

2)$ CH_3CH_2CHBrCH_3$   + $ EtO^ \minus{}$->

[hide]Substitution?[/hide]

3) Compare the nucleophilicity of $ NH_3$ and $ PH_3$

[hide]$ NH_3$ has more $ e^ \minus{}$ density so $ NH_3 > PH_3$[/hide]

i think ans for third question should be $ PH_3>NH_3$
The outermost electrons on the phosphorous are farther away from the nucleus and more loosely held.  :maybe:  :maybe:  :maybe:

I don't think anything happens in (1) and (2) looks like Williamson ether synthesis

Obviously the 2nd is elimination. $ \ce{EtO^{ - }}$ is a strong base and the halide is secondary, implies it will favour elimination.

For the first I think it's neither. HBr is not enough to substitute the alcohol.

In the first one, HBr is a strong acid, so -OH will be protonated and a molecule of water will depart, leaving behind a secondary carbocation which can undergo an easy methyl shift in order to yield a tertiary carbocation. Now since bromide is a very weak base, its combination with the positive carbon is preferred and so the final product is very likely 2-bromo-2,3-dimethylbutane.

In the second case, we have a secondary halide, and a strong, small base, so elimination would be the preferred path, with but-2-ene being the major product.

Phosphorus is a bigger atom than nitrogen, so its valence electrons feel less atraction from the nucleous, and so they are more readily available to combine with another atom.

first one is easy.
see oxygen have the lone pair which it give to H+ of HBr and form O+H2 now since  H2O is a good leaving group it takes the bond pair electrons.now rearrangement takes place and then methyl shift (simultaneously).
and at last Br will attach to tert carbon.


Give a,b,c,d,e are non negative and a+b+c+d+e=1.Prove that:
$ab+bc+cd+de \leq\frac{1}{4}$ :(

[quote="N.T.TUAN"]We have $ab+bc+cd+de\leq (a+c+e)(b+d)\leq (\frac{a+b+c+d+e}{2})^{2}=\frac{1}{4}$ .

QED.[/quote]
Sorry
can u sayit quite more.
I can't understand

[quote="mysterious"][quote="N.T.TUAN"]Because $a,b,c,d,e$ are nonnegative and $xy\leq (\frac{x+y}{2})^{2}\forall x,y\in\mathbb{R}$   :wink:[/quote]
No say why we have
$ab+bc+cd+de\leq(a+c+e)(b+d)$[/quote]
Because $(a+c+e)(b+d)=ab+bc+cd+de+be+da$


\[\sin(x)=-\sin(360^\circ-x)\] Hence, \[\sin 1^\circ+\sin 359^\circ=0,\]  \[\sin2^\circ+\sin358^\circ=0,\] etc. The sum is equal to $\sin360^\circ$, which is also just zero.

[quote="sinajackson"]$sin1^{\circ}+sin2^{\circ}+\cdots+sin360^{\circ}= ?$[/quote]
Another solution:
Let $z=\mathrm{cis}1^{\circ}$. Then $\sin1^{\circ}+\sin2^{\circ}+\cdots+\sin360^{\circ}= \sin0^{\circ}+\sin1^{\circ}+\cdots+\sin359^{\circ}=$ $\mathrm{Im}(z^{0})+\ldots+\mathrm{Im}(z^{359})= \mathrm{Im}(z^{0}+\ldots+z^{359})= \mathrm{Im}(0)=0$.  :D


probability
LaTeX
trigonometry
real analysis
real analysis unsolved

A field F0 is a collection of subsets of a set Ω that contains the empty set phi , and is closed under complements and finite unions . A finitely additive probability measure P on (Ω,G), where G is a field or sigma -field containing F0, is defined to satisfy P(A) belongs[0,1] for any set A in G, P(Ω)=1,and P( union Ai) = sum of P(Ai) for every disjoint collection A1,A2, …….Ak of sets from G.

(a)	Show that if P is finitely additive and continuous on (Ω,F) where F is a  sigma-field, then P is a probability measure.
(b)	Suppose P1 and P2 are probability measures on the measureable space(Ω,F), where F is the  -field generated by the field F0, show that if P1(A)=P2(A) for every A F, then P1(A)=P2(A) for every A F.

I think it is really hard to prove P(A) less than 1 here. Maybe the question is wrong?

Thanks

No, the question is correct. The key to (b) is the so called monotone class lemma that says that the $ \sigma$-field generated by a field $ F_0$ is the same as the least monotone class containing $ F_0$. (a monotone class is a collection $ M$ of subsets of $ \Omega$ such that together with every increasing sequence of sets $ E_1\subset E_2\subset\dots$, it contains the union $ \bigcup_j E_j$, and together with every decreasing sequence of sets $ E_1\supset E_2\supset\dots$, it contains the intersection $ \bigcap_j E_j$).

By the way, if you spend half an hour learning basic $ \text{\LaTeX}$, your posts will become much more readable so you'll not need to find ascii codes for symbols like $ \Omega$ and we'll not need to make guesses about missing indices, $ \in$ sines, etc. :).

Hi fedja,
Thanks.
I know how to prove (b). But it looks like (a) miss something, how can We prove P is a probability measure?
The question just gave us some definitions. However, it did not show us P(A)<=0.

I will try to learn Latex next time, sorry for the trouble.

(a) may miss the [i]explicit[/i] words "probability measure" after the words "finitely additive" but it is rather clear that they were meant to be there :).

Thanks fedja.
This was my exam questions. And I spent one hour to prove the the question. But I never got the proof P(A)<=1.
With this word"finitely additive measure", I can do it.

This is my first time to post question and reply.
Thanks.


find positive number so that has a positive number k so that
                  2k 2 +2010k+3 =3^n*m :D

Do you mean 3^(n*m) or(3^n)*m??And k,m,n are integer,right?
If you mean (3^n)*m so k=3t is the answer.

You posted an identicall problem yesterday i think......

multiply it by 8, and it cecomes:

(4k+2010) 2 -2010 2 +24=8m*3^n, now 3 divides 4k+2010, so in fact if n>1, 9 divides both sides, but 9 doesnt divides 2010 2 +24, so n<=1 and k=3

I think k=3t and n=1 because we still have m .
The true problem is:
[tex] 2k^2+2004k+5=5^n [/tex] :)


How to prove: If an ordered set A has the least upper bound property, then it has the greatest lower bound property.

Thanks

Let $ B \subset A$ be nonempty and bounded from below. Let $ C$ be the set of all lower bounds of $ B$. Then $ C$ is nonempty and bounded from above (by any element of $ B$). So, by the least upper bound property, $ C$ has a least upper bound, $ \sup C$. We claim that $ \sup C$ is a greatest lower bound of $ B$. We need to show two things 1) that $ \sup C$ is in fact a lower bound of $ B$ (i.e that $ \sup C \in C$) and 2) that there is no lower bound greater than $ \sup C$.

1) Assume on the contrary that there exists $ b \in B$ such that $ b < \sup C$. Then, there must exist an element of $ c \in C$ such that $ b < c$. But this contradicts the fact that everything in $ C$ is a lower bound for $ B$. Therefore, $ \sup C \in C$.

2) By the definition of the set $ C$ and the definition of $ \sup C$, there cannot be a lower bound for $ B$ that is greater than $ \sup C$.

Thanks , How do you think of it ?
This is an exercise of Topology of Munkres , I read that chapter carefully , but I can not give a proof of it .

Am I not suitable to study math ?

[quote="FibonacciFan"]You know you'll have to apply the least upper bound property somewhere in the proof. The only question really is what set to apply it to. Applying it to $ B$ isn't possible and wouldn't help if it were possible. After a little bit of scratching my head, I realized that the set of lower bounds of $ B$ was a natural one to try. And that happened to work.

And don't be discouraged just because you can't solve one problem.[/quote]

Thanks , I will study hard.

I am not quite sure about this question. The Reals (up to some bound, say M) is an ordered set with the least upper bound property but it doesn't have a greatest lower bound (unless infinity is allowed? But that to me seems sketchy, awkward)

What is the question you are unsure of? Greatest lower bound property can only be applied to sets that are bounded from below.

Exactly. THe given set was not specified to be bounded from below. I don't think an ordered set implies its bounded from below, just because the Reals is an ordered set.
So taking $ A$ to be all negative real numbers for instance, it has a lest upper bound, but not a greatest lower bound.

"$ A$ has the greatest lower bound property" does [b]not[/b] mean "$ A$ has a greatest lower bound."  It means "every subset of $ A$ bounded from below has a greatest lower bound," same as in the least upper bound case.


1) $ a_{n \plus{} 1} \equal{} (4/7)a_n \plus{} (3/7)a_{n \minus{} 1}$ and $ a_0 \equal{} 1; a_1 \equal{} 2$: Find $ lim_{n\to \infty}a_n$

2) Given two sequences $ x_n$ and $ y_n$ defined by $ x_0 \equal{} y_0 \equal{} 7$,
$ x_n \equal{} 4x_{n \minus{} 1} \plus{} 3y_{n \minus{} 1}$, and
$ y_n \equal{} 3y_{n \minus{} 1} \plus{} 2x_{n \minus{} 1}$;
Find $ lim_{n\to \infty}x_n/y_n$

[hide="1"]We try to find a closed form for $ a_n$.  The standard solution is in the form $ ac_1^n \plus{} bc_2^n$.  We try to find the two $ c$'s.  If $ c^{n \plus{} 1} \equal{} (4/7)c^n \plus{} (3/7)c^{n \minus{} 1}$, then we divide by $ c^{n \minus{} 1}$ and get $ c^2 \minus{} 4/7c \minus{} 3/7 \equal{} 0$.  We can solve this to get $ c \equal{} 1, \minus{} 3/7$.

We plug this into the solution form to get $ a \plus{} b( \minus{} \frac {3}{7})^n \equal{} a_n$.  We plug in 0 and 1 for $ n$ to find $ a$ and $ b$:

$ a \plus{} b \equal{} 1$ and $ a \minus{} \frac {4}{7}b \equal{} 2$.  Solving this we get $ a \equal{} 1.7$ and $ b \equal{} \minus{} .7$.

Now we get $ a_n \equal{} 1.7 \minus{} .7( \minus{} \frac {4}{7})^n$.  As $ n$ gets higher and higher, the second term gets smaller and smaller, so as $ n\rightarrow \infty$, we ignore that term.  Thus the limit is 1.7.[/hide]

[hide="2"]If you don't understand matrices, you probably will not get this solution at all.

We can write the given equations as $ \begin{array}{|c|c|cc\parallel{}c|} x_n & & 4 & 3 & x_{n \minus{} 1} \\
& \equal{} & & & \\
y_n & & 2 & 3 & y_{n \minus{} 1}\end{array}$.

Using induction, we get $ \begin{array}{|c|c|cc|} x_n & & 4 & 3 \\
& \equal{} & & \\
y_n & & 2 & 3\end{array}^n\cdot \begin{array}{|c|}7 \\
 \\
7\end{array}$.

Now we find the eigenvalues of that matrix.  They are found to be 1 and 6, with corresponding eigenvectors $ \begin{array}{|c|}1 \\
\minus{} 1\end{array}$ and $ \begin{array}{|c|}3 \\
2\end{array}$.  Now we can write the above equation as:

$ \begin{array}{|c|c|cc|}x_n & & 4 & 3 \\
& \equal{} & & \\
y_n & & 2 & 3\end{array}^n\cdot \left(\begin{array}{c|c|c|c|} & 1 & & 3 \\
\minus{} \frac {7}{5} & & \plus{} \frac {14}{5} & \\
& \minus{} 1 & & 2\end{array}\right)$

$ \begin{array}{cc|cc|} & & 4 & 3 \\
& \equal{} \minus{} \frac {7}{5} & & \\
& & 2 & 3\end{array}^n\cdot\begin{array}{|c|c|cc|}1 & & 4 & 3 \\
& \plus{} \frac {14}{5} & & \\
\minus{} 1 & & 2 & 3\end{array}^n\cdot \begin{array}{|c|}3 \\
 \\
2\end{array}$

$ \equal{} \minus{} \frac {7}{5}\begin{array}{|c|}1 \\
\minus{} 1\end{array} \plus{} \frac {14}{5}\cdot6^n\begin{array}{|c|}3 \\
2\end{array}$

$ \equal{} \begin{array}{|c|} \minus{} \frac {7}{5} \plus{} \frac {42}{5}\cdot6^n \\
 \\
\frac {7}{5} \plus{} \frac {28}{5}\cdot6^n\end{array}$

The ratio of these two is influenced less and less by the $ \frac {7}{5}$'s and are dominated by the $ 6^n$'s.  These cancel and give the ratio to be about $ \frac {42}{5}/\frac {28}{5} \equal{} \frac {3}{2}$.[/hide]

The matrix solution is the most meaningful, if you understand it. Actually, if you understand what an "eigenvalue" is, you should immediately see that the sequence of $ \binom{x_n}{y_n}$ approaches the direction of the eigenvector with the highest associated eigenvalue (namely, 6).

I'm not entirely sure why [b]tenniskidperson3[/b] decided to use vertical bars to denote matrices, since that's commonly used for other purposes, such as determinant and absolute value. More commonly, matrices are enclosed in []'s or ()'s.

[b]Method 2:[/b] [i]What are the possible limits, anyway?[/i]

$ \frac {x_n}{y_n} \equal{} \frac {4x_{n \minus{} 1} \plus{} 3 y_{n \minus{} 1}}{2 x_{n \minus{} 1} \plus{} 3 y_{n \minus{} 1}} \equal{} \frac {4\frac {x_{n \minus{} 1}}{y_{n \minus{} 1}} \plus{} 3 }{2 \frac {x_{n \minus{} 1}}{y_{n \minus{} 1}} \plus{} 3}$

So suppose $ \frac {x_n}{y_n}$ actually had a limit. Let this limit be $ L$. We have:

$ L \equal{} \frac {4L \plus{} 3 }{2L \plus{} 3} \iff (L \minus{} 1)(2L \minus{} 3) \equal{} 0 \iff L \equal{} \minus{} 1 \; \; \text{or} \; \; L \equal{} \frac {3}{2}$

But clearly, $ x_n$ and $ y_n$ are always positive for our given inputs, so the limit absolutely cannot be $ L \equal{} \minus{} 1$. Thus, if it exists (as suggested by the problem), it must be equal to $ \boxed{L \equal{} \frac {3}{2} }$

[b]Method 3:[/b] [i]Focus on one variable by decoupling the equations[/i]

Subtract the two equations to get $ x_n \minus{} y_n \equal{} 2 x_{n \minus{} 1} \iff y_n \equal{} x^n \minus{} 2 x_{n \minus{} 1}$

Substitute this into the first equation to get $ x_n \equal{} 7x_{n \minus{} 1} \minus{} 6x_{n \minus{} 2}$.
Using the characteristic equation, we can find that the eigenvalues are $ 1$ and $ 6$, so $ x_n \equal{} A \cdot 6^n \plus{} B$.

Going back to that first thing we found:
$ y_n \equal{} x_n \minus{} 2 x_{n \minus{} 1} \implies \frac {y_n}{x_n} \equal{} 1 \minus{} 2 \frac {x_{n \minus{} 1}}{x_n} \to 1 \minus{} 2 \cdot \frac {1}{6} \equal{} \frac {2}{3}$, since the $ 6^n$ term in $ x_n$ dominates.

Invert to get $ \frac {x_n}{y_n} \to \frac {3}{2}$

Maybe I should have used parentheses, boxed or curved, but I was in a rush to finish the solution before I had to go to tennis, and the bars were the easiest things to use.

By the way, method 3 you used was what my first idea, but for some reason I got the characteristic equation for $ x_n$ to be $ c^2\minus{}5c\minus{}2\equal{}0$ and decided to try it a different way.

I don't understand the matrix solution, so I'll read up on eingen values.

TZF, in your second to last line, why did you let x_{n-1}=1?

I don't think I did. Let's walk through that second to last line:

$ y_{n} \equal{} x_{n} \minus{} 2 x_{n \minus{} 1}$ -- we found this in the first line

$ \implies\frac {y_{n}}{x_{n}} \equal{} 1 \minus{} 2\frac {x_{n \minus{} 1}}{x_{n}}$ -- divide the equation by $ x_n$

But, we know that $ x_n \equal{} A \cdot 6^n \plus{} B$ and $ x_{n \minus{} 1} \equal{} A \cdot 6^{n \minus{} 1} \plus{} B$, so $ x_n \equal{} 6 x_{n \minus{} 1} \minus{} 5B$, and $ \frac {x_n}{x_{n \minus{} 1}} \equal{} 6 \minus{} \frac {5B}{x_{n \minus{} 1}}$, which goes to $ 6$ as the sequence increases without bound. Equivalently, $ \frac {x_{n \minus{} 1}}{x_n}$ goes to $ \frac {1}{6}$.

Therefore, in the limit, we have that
$ \frac {y_{n}}{x_{n}} \equal{} 1 \minus{} 2\left( \frac {x_{n \minus{} 1}}{x_{n}} \right) \to 1 \minus{} 2\cdot \left( \frac {1}{6} \right) \equal{} \frac {2}{3}$


calculus
vector
derivative
calculus computations

Consider the two lines:

$ L_1: x\equal{}4\plus{}5t, y\equal{}5\plus{}5t, z\equal{}1\minus{}4t$

$ L_2: x\equal{}4\plus{}s, y\equal{}\minus{}6\plus{}8s, z\equal{}7\minus{}3s$

Find the two parallel planes in which these lines lie.

The normal to the planes is a vector perpendicular to both $ (5, 5, \minus{}4)$ and $ (1, 8, \minus{}3)$, the direction vectors of the two lines. The cross product is $ (\minus{}3,4,35)$,

So the equation of both planes will be of the form $ \minus{}3x\plus{}4y\plus{}35z\plus{}d\equal{}0$

For equation one, plug in the point $ (4,5,1)$.

$ \minus{}12\plus{}20\plus{}35\plus{}d\equal{}0$
$ d\equal{}\minus{}43$
The equation of the first plane is $ \minus{}3x\plus{}4y\plus{}35z\plus{}\minus{}43\equal{}0$

For equation two, plug in the point $ (4, \minus{}6, 7)$

$ \minus{}12\minus{}24\plus{}245\plus{}d\equal{}0$
$ d\equal{}\minus{}209$
The equation of the second plane is $ \minus{}3x\plus{}4y\plus{}35z\minus{}209\equal{}0$

[quote="b555"]well, does this  involve  any kind  of  calculus ?   :o[/quote]
The problem most likely came from a calculus textbook and would be part of the subject matter for what would be our Calculus III course. But no, it's not itself calculus - if asked to classify it, I'd probably call it linear algebra.

The curriculum for the province of Ontario (containing Toronto) has Calculus and Vectors as one course, half of it dedicated to derivatives and their applications, and the other dedicated to vectors. We still call it 'calc', even though half of it actually deals with vectors, including planes.


trigonometry
function
limit
calculus
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real analysis
real analysis unsolved

This is part of a problem that appears in my book
"
Let $S$ and $C$ be real-valued functions on $\mathbb R$, such that for all $x,y$
$S(x+y) = S(x)C(x)+C(x)S(y)~,$
$C(x+y) = C(x)C(y)-S(x)S(y)~,$
$S^{2}(x)+C^{2}(x) = 1~,$
and $\lim_{x \rightarrow 0}S(x)/x = 1$.

Show that for all $x,y$
$S(x)-S(y) = 2C \left( \frac{x+y}{2}\right) S \left( \frac{x-y}{2}\right)$, and deduce that $S$ is differentiable on $\mathbb m R$ with $S' = C$.
"
Okay, so I can show that functional relation, and see that the derivative at $y$ exists and is equal to
$\lim_{x \rightarrow y}\frac{S(x)-S(y)}{x-y}= \lim_{x \rightarrow y}C \left( \frac{x+y}{2}\right) \frac{S \left( \frac{x-y}{2}\right)}{\left( \frac{x-y}{2}\right)}= \lim_{x \rightarrow y}C \left( \frac{x+y}{2}\right)$,
if this limit exists.

But I am not given continuity for $S$ or $C$.  Should this be an additional condition?

We're supposed to derive continuity.

A [hide="derivation"]First, show that $S(0)=0$ and $C(0)=1$. Then the limit condition implies that $S$ is continuous at zero. Since $\left|C\left(\frac{x+y}2\right)\right|\le 1$, $|S(x)-S(y)|\le 2\left|S\left((\frac{x-y}2\right)\right|$ by the latter functional relation; since $S(t)\to 0$ as $t\to 0$, this implies uniform continuity of $S$.
Continuity of $C$ follows from writing $C(x)=\frac{S(x+t)-S(x)C(t)}{S(t)}$ and fixing $t$ with $S(t)\neq 0$; this gives $S$ as an algebraic combination of continuous functions.[/hide]

evidently $S = \sin$ and $C = \cos$.  I was thinking that one way to prove this would be by developing the series for $S$ and $C$... any trickier ways?

There's the differential equation method- we know that $S$ and $C$ satisfy a second order linear ODE which has $\sin$ and $\cos$ as solutions. When you add in the initial conditions, we can apply the uniqueness theorem for differential equations to conclude that they must actually be $\sin$ and $\cos$.
The power series idea essentially proves a version of this uniqueness.

Another possible reorganization would be to define the complex-valued function of a real variable $E(x)=C(x)+iS(x).$ Then the functional equations given above become simply $E(x+y)=E(x)E(y).$ We're also given that $|E(x)|=1.$

Then $E'(x)=\lim_{h\to0}\frac{E(x+h)-E(x)}h= E(x)\lim_{h\to0}\frac{E(h)-1}h.$

If we can derive that $\lim_{h\to0}\frac{E(h)-1}h=i,$ we'll have what we want; that in turn comes down to proving that $\lim_{h\to0}\frac{C(h)-1}h=0.$


Let the sum of positive numbers x_1, x_2, ... , x_n be 1.
Let s be the greatest of the numbers
x_1
---------,
1 + x_1

  x_2
-----------------,   ...  ,
1 + x_1 + x_2

       x_n
----------------------.
1 + x_1 + ... + x_n

What is the minimal possible s? What x_i correspond it?

It is clear and very nice. We have that
\[ \frac{x_{i}}{1\plus{}x_{1}\plus{}x_{2}\plus{}\cdots\plus{}x_{i}}\equal{}1\minus{}\frac{1\plus{}x_{1}\plus{}x_{2}\plus{}\cdots\plus{}x_{i\minus{}1}}{1\plus{}x_{1}\plus{}x_{2}\plus{}\cdots\plus{}x_{i}}\leq s\]
for all $ i$. Hence
\[ \frac{1\plus{}x_{1}\plus{}x_{2}\plus{}\cdots\plus{}x_{i\minus{}1}}{1\plus{}x_{1}\plus{}x_{2}\plus{}\cdots\plus{}x_{i}}\geq 1\minus{}s.\]
Multiplying the inequalities for $ i \equal{} 1,2,\,\cdots,n$ gives
\[ \frac{1}{2}\equal{}\frac{1}{1\plus{}x_{1}\plus{}x_{2}\plus{}\cdots\plus{}x_{n}}\geq (1\minus{}s)^{n}\]
and hence
\[ s\geq 1\minus{}\frac{1}{\sqrt[n]{2}}.\]
It is easy to see that inequality can occur, in fact, we have equality for \[ x_{1}\equal{}\sqrt[n]{2}\minus{}1,\ x_{2}\equal{}\sqrt[n]{2^{2}}\minus{}\sqrt[n]{2},\ x_{3}\equal{}\sqrt[n]{2^{3}}\minus{}\sqrt[n]{2^{2}},\ \cdots,\ x_{n}\equal{} 2\minus{}\sqrt[n]{2^{n\minus{}1}}.\]

Thank you, Arne, I like your posts. I found many interesting problems solved by you. I think all my problems are "very nice" check them and you'll see.

So who rated my post as spam? Is there ANYTHING wrong with it? ... This is just so annoying!

I think someone is kidding you. I'm not - I still don't have 25 posts :-(. If depends on me it is excellent. And all solutions to my problems are "excellent". I really like mathlinks.ro. I think - mathlinks, imocompendium and kalva's sites are one of the best sites in the Web.


In a triangle $ \triangle{A'B'C'}$ , $ AA' , BB' , CC'$ are bisectors . Prove that $ AA' , BB'$ and $ CC'$ are heights in $ \triangle{ABC}$ .

Obviously your diagram is not fully defined...We can add the condition that BB' is perpendicular to A'C'. Then A,B,C are excenters of triangle A'B'C'. Then it is an easy angle chase.

We denote as $ A'',\ B'',\ C'',$ the intersection points of the circumcircle $ (O'),$ of the triangle $ \bigtriangleup A'B'C',$ from its angle bisectors through vertices $ A',\ B',\ C'$ respectively and let be the point $ H\equiv A'A''\cap B'B''\cap C'C''.$

It is easy to show that the midperpendiculars of the segments $ HA',\ HB',\ HC',$ taken per two of them, intersect each other at points $ A'',\ B'',\ C''.$

We draw the lines through $ A',\ B',\ C'$ and parallel to $ B''C'',\ A''C'',\ A''B''$ respectively and it is easy to show that their intersection points $ A,\ B,\ C,$ taken per two of them, lie on the line segments $ A'A'',\ B'B'',\ C'C'',$ respectively.

Hence, it is formed the triangle $ \bigtriangleup ABC,$ as the [b][size=100]anticevian[/size][/b]  triangle of $ \bigtriangleup A'B'C',$ with respect to the point $ H,$ with $ AA'\perp BC,\ BB'\perp AC,\ CC'\perp AB$ and the proof is completed.

Kostas Vittas.

One line,if you consider a line which is parallel to $ AB$,and passes through $ C$,and its intersection points with $ C'B'$ and $ C'A'$ :wink:


[quote="charlmathnon"]if a b c d belongs to R,and a>=b>=c>=d>0 then what about the minimum of
f(a,b,c,d)=[1+c/(a+b)][1+d/(b+c)][1+a/(c+d)][1+b/(d+a)]
of course it is (3/2)^4,but what about the proof?
thank you[/quote]
If $ a, b, c, d$ belongs to $ \mathbb{R}$, and $ a\geq b\geq c\geq d>0$. Then what is the minimum of:
$ f(a,b,c,d)\equal{}\left(1\plus{}\frac{c}{a\plus{}b}\right)\left(1\plus{}\frac{d}{b\plus{}c}\right)\left(1\plus{}\frac{a}{c\plus{}d}\right)\left(1\plus{}\frac{b}{d\plus{}a}\right)$

or did you mean?:
$ f(a,b,c,d)\equal{}\left(\frac{1\plus{}c}{a\plus{}b}\right)\left(\frac{1\plus{}d}{b\plus{}c}\right)\left(\frac{1\plus{}a}{c\plus{}d}\right)\left(\frac{1\plus{}b}{d\plus{}a}\right)$

If $ a, b, c, d$ belongs to $ \mathbb{R}$, and $ a\geq b\geq c\geq d > 0$. Then what is the minimum of:
$ f(a,b,c,d) \equal{}\left(1\plus{}\frac{c}{a\plus{}b}\right)\left(1\plus{}\frac{d}{b\plus{}c}\right)\left(1\plus{}\frac{a}{c\plus{}d}\right)\left(1\plus{}\frac{b}{d\plus{}a}\right)$

i'm sorry i don't no how to print these, i mean the first one.


find asymptotical expression of the solution of this differential equation when $ k \minus{} > \plus{} oo$
$ x^{2}y'' \plus{} xy' \plus{} k^{2}(x^{2} \minus{} \alpha ^{2})y \equal{} 0$
where $ 0 \leq x \leq \alpha$ :huh:

Well, there is no "[b]the[/b] solution" of a linear second order ODE, to start with. You should specify an initial condition or something else that tells one which of the infinite number of solutions you are interested in for each particular $ k$. We can probably make some "educated guesses" here but it will be better if you formulate your problem as it should be yourself ;)

Look likes Bessel differential equation

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

 Maple 11 will help

yes , you are right fedja and Moubinool .
I forget something :
$ y(0) \equal{} a$
and
$ y'(0) \equal{} bik$
if we put in bessel equation instead $ x$ and $ n$ this $ x \equal{} zk$ and $ n^{2} \equal{} k^{2}\alpha ^{2}$ ( I note that $ n$ is not integer).
then we have this equation
And solution we consider this form :
$ y \equal{} e^{ikt(x)}\sum_{j \equal{} 0}^{ \plus{} oo}\frac{y_{j}(x)}{(ik)^{j}}$ we must find $ t(x)$ and $ y_{j}(x)$ for all $ j$
$ i\in C$


integration
geometry
3D geometry
sphere
geometry unsolved

Why is lenght of segment $AB$ minimum distance betven points $A$ and $B$?

Because in every triangle the lenght of one side is less then the sum of the other one: for n segments apply this fact to arrive to n-1 segments whose sum of lenght is less then the previous one...then go to n-2, n-3, until n-(n-1)=1, which is the least element in $\mathbb{N}$. For the curves let $n\to \infty$

Because for a given curve $\gamma,$ we have \[\int_{a}^{b}|\gamma'(t)|dt\geq\left|\int_{a}^{b}\gamma'(t)dt\right|=|\gamma(b)-\gamma(a)|\] where the $LHS$ is the length of the curve on the interval $[a,b],$ and the $RHS$ is the distance between the points $\gamma(a)$ and $\gamma(b).$ This holds for all metric spaces.

other solution: for each point of the orbit from A to B for what the derivate change let simply draw, for these points, perpendicular lines to the segment: since in an Eucledian space any segment is less or equal his projection on a line or on a plane, the theorem is proved

you can use ''the triangle inequality''. let C be any point the Euclid space. (C is different from A;B) AB<AC+BC so AB is the shortest way from A to B. we can apply this method for all polygonal paths.that is,we can show that AB<AC+CD+DE+......+YZ+ZB. saperately, we can show that length AB is more shorter than any curvilinear path from A to B with similar reasoning.

 but this theorem is not true all spaces.for exaple, on the sphere, the shortset way from A to B is the short arc of the biggest circle that contains points A,B.

[quote="scarface"]but this theorem is not true all spaces.for exaple, on the sphere, the shortset way from A to B is the short arc of the biggest circle that contains points A,B.[/quote]

This theorem [b]is[/b] true for all metric spaces by the very definition of a metric space. In your example a metric is defined on the sphere such that the shortest arc of the biggest circle containing two fixed points is the distance between those two points.


In Ms. Smith's class, each student averages one day absent out of thirty. What is the probability that out of any two students chosen at random, one student will be absent while the other is present? Express your answer as a percent rounded to the nearest tenth.

[hide="Case 1- Student X is absent, Student Y is present"]$ \frac{29}{30} \cdot \frac{1}{30} \equal{} \frac{29}{900}$[/hide]

[hide="Case 2- Student X is present, Student Y is absent"]$ \frac{29}{30} \cdot \frac{1}{30} \equal{} \frac{29}{900}$[/hide]

Summing these gives you the answer of $ \boxed{\frac{29}{450}}$.

The problem asked for a percent, so the answer is $ \boxed{6.4}$.


trigonometry
email
LaTeX
integration
calculus
calculus computations

[quote="the_liong"]Hi All,

Can anybody solve the following equation?

Integrate[Sqrt[w^2 *Cos[w*t] + 1], t]

My email is the_liong@hotmail.com .Please send the derivation to my email if possible. Thank you very much

 :lol:[/quote]

let's put that in LaTeX:

do you mean?

$\int \sqrt{w^{2}\cos(wt)+1},t$?

[quote="davidyko"]And this wouldn't belong in HSB, heck, it wouldn't even belong in Intermediate.[/quote]

It needs to be in a college calculus board  :wink: . Mods please move this.

It doesn't have an elementary antiderivative, except in the special cases $w=0$ and $w=1$. It's an elliptic integral.


inequalities function calculus derivative

$ \text {Suppose } 0<a_1\le a_2\le a_3\le \cdots\le a_{2005}<1 \text{, and for } k\equal{}1,2,3,\cdots ,2005,$

 $ f(x)\equal{}\frac{k}{1\minus{}a_1a_2a_3\cdots  a_k}$

$ \text{Prove that  }f(1) < f(2) <f(3) <\cdots <f(2005).$

f(x) is a constant function, so the statement is actually disproved.

hmm shouldnt it be $ f(k)\equal{}\frac{k}{1\minus{}a_1a_2a_3\cdots a_k}$

[hide="Possibly not the best solution"]
We note that for the statement to be true, all we need is for its forward difference to be positive. Or,
$ \dfrac{k\plus{}1}{1\minus{}a_1a_2 \cdots a_{k\plus{}1}} \minus{} \dfrac{k}{1\minus{}a_1a_2 \cdots a_{k}} > 0$
Multiplying out the denominators,

$ 1\plus{}k(a_1a_2 \cdots a_{k\plus{}1})\minus{}(k\plus{}1)(a_1a_2 \cdots a_{k}) > 0$

$ \Rightarrow a_1a_2 \cdots a_{k} \left(ka_{k\plus{}1} \plus{} \frac{1}{a_1a_2 \cdots a_{k}} \right) \minus{}(k\plus{}1)a_1a_2 \cdots a_{k} > 0$

Dividing out $ a_1a_2 \cdots a_{k}$ and messing around, the statement is equivalent to

$ \dfrac{a_{k\plus{}1} \plus{} \cdots \plus{} a_{k\plus{}1} \plus{} \frac{1}{a_1a_2 \cdots a_{k}}}{k\plus{}1} > 1$

Where we add $ a_{k\plus{}1} \ k$ times. By AM-GM, we must have

$ \dfrac{a_{k\plus{}1} \plus{} \cdots \plus{} a_{k\plus{}1} \plus{} \frac{1}{a_1a_2 \cdots a_{k}}}{k\plus{}1} \geq \sqrt{\dfrac{a_{k\plus{}1}^k}{a_1a_2 \cdots a_{k}}}$

Because $ a_{k\plus{}1}^k > a_1a_2 \cdots a_k$ by the problem statement, we have

$ \sqrt{\dfrac{a_{k\plus{}1}^k}{a_1a_2 \cdots a_{k}}} > 1$
And we are done.

Note that equality is never reached because for this to be true, then $ a_{k\plus{}1} \equal{} \frac{1}{a_1a_2 \cdots a_{k}}$, which is impossible because $ a_{k\plus{}1} < 1$

[/hide]

[hide="Much easier calculus solution"]
Take the derivative of $ f(k)$ with respect to $ k$
$ f'(k) \equal{} \frac{1}{1\minus{}a_1a_2a_3\cdots a_k}$
Which is clearly positive since $ 0<a_1a_2a_3\cdots a_k<1$ by the problem statement, and so the function is increasing on the interval.

The above post was a nice exercise, though, if a bit unnecessary.
[/hide]

$ \text{I don't agree your derivative proof like that,}$

$ \text{because  }1\minus{}  a_1a_2a_3a\cdots a_k \text {is not a constant with respect to f(k) ,??}$

You're right. $ f(k)$ is not actually continuous when you look at it like that. If you can somehow construct a continuous version of $ f(k)$ (such that it equals the discrete case for integers) and then take the derivative and notice it's monotonic and positive, then you might have a solution, but that is probably more involved. Sorry, I wasn't thinking.


Prove that $ \forall n \in N$ we have:
$ (a\plus{}b)^{2n}\plus{}(b\plus{}c)^{2n}\plus{}(c\plus{}a)^{2n} \geq\ \frac{2^{2n}}{3^{2n\minus{}1}\plus{}1}(a^{2n}\plus{}b^{2n}\plus{}c^{2n}\plus{}(a\plus{}b\plus{}c)^{2n})$

[quote="tanpham90"]Prove that $ \forall n \in N$ we have:
$ (a \plus{} b)^{2n} \plus{} (b \plus{} c)^{2n} \plus{} (c \plus{} a)^{2n} \geq\ \frac {2^{2n}}{3^{2n \minus{} 1} \plus{} 1}(a^{2n} \plus{} b^{2n} \plus{} c^{2n} \plus{} (a \plus{} b \plus{} c)^{2n})$[/quote]
Nice, Tan. :)
But I cant find the solution by AM-GM Inequality as your requirement  :blush: , my solution consisted 2 step
Step 1. Assume that $ a(a\plus{}b\plus{}c) \ge 0$, then denote $ f(a,b,c)\equal{}LHS\minus{}RHS$, we can prove that
\[ f(a,b,c) \ge f(a,\frac{b\plus{}c}{2},\frac{b\plus{}c}{2})\]
Step 2. We need to prove
\[ f(a,\frac{b\plus{}c}{2},\frac{b\plus{}c}{2}) \ge 0\]
This inequality is tight since it has equality when $ (a,b,c) \sim (1,1,1)$ and $ (a,b,c) \sim (\minus{}3,1,1)$.
:)

[quote="can_hang2007"][quote="tanpham90"]Prove that $ \forall n \in N$ we have:
$ (a \plus{} b)^{2n} \plus{} (b \plus{} c)^{2n} \plus{} (c \plus{} a)^{2n} \geq\ \frac {2^{2n}}{3^{2n \minus{} 1} \plus{} 1}(a^{2n} \plus{} b^{2n} \plus{} c^{2n} \plus{} (a \plus{} b \plus{} c)^{2n})$[/quote]
Nice, Tan. :)
But I cant find the solution by AM-GM Inequality as your requirement  :blush: , my solution consisted 2 step
Step 1. Assume that $ a(a \plus{} b \plus{} c) \ge 0$, then denote $ f(a,b,c) \equal{} LHS \minus{} RHS$, we can prove that
\[ f(a,b,c) \ge f(a,\frac {b \plus{} c}{2},\frac {b \plus{} c}{2})
\]
Step 2. We need to prove
\[ f(a,\frac {b \plus{} c}{2},\frac {b \plus{} c}{2}) \ge 0
\]
This inequality is tight since it has equality when $ (a,b,c) \sim (1,1,1)$ and $ (a,b,c) \sim ( \minus{} 3,1,1)$.
:)[/quote]
I cant find the solution by AM-GM Inequality as your requirement  too,,with the value of$ n$ that it is small,we have an other proof which be  used some qualities .See them in MYM 351 9/2006,your problem is very nice

[quote="tanpham90"][quote="onlylove_math"][quote="tanpham90"]Well , for example with $ n \equal{} 2$ we must prove that :
$ (a \plus{} b)^{4} \plus{} (b \plus{} c)^{4} \plus{} (c \plus{} a)^{4} \geq\ \frac {4}{7}(a^{4} \plus{} b^{4} \plus{} c^{4} \plus{} (a \plus{} b \plus{} c)^{4})$

Put : $ x \equal{} a \plus{} b$ ; $ y \equal{} b \plus{} c$ and $ z \equal{} c \plus{} a$
The inequality become :
$ 28(x^{4} \plus{} y^{4} \plus{} z^{4}) \geq\ (x \plus{} y \minus{} z)^{4} \plus{} (y \plus{} z \minus{} x)^{4} \plus{} (z \plus{} x \minus{} y)^{4} \plus{} (x \plus{} y \plus{} z)^{4}$
$ \Leftrightarrow (x^{4} \plus{} y^{4} \plus{} z^{4}) \geq\ (x^{2}y^{2} \plus{} y^{2}z^{2} \plus{} z^{2}x^{2})$
Which is obviously true by AM-GM :D[/quote]
Great!
With $ n \equal{} : 2n$ . Put : $ x \equal{} a \plus{} b$ ; $ y \equal{} b \plus{} c$ and $ z \equal{} c \plus{} a$

The inequality become :

$ (3^{2n \minus{} 1} \plus{} 1)(x^{2n} \plus{} y^{2n} \plus{} z^{2n}) \geq\ (x \plus{} y \minus{} z)^{2n} \plus{} (y \plus{} z \minus{} x)^{2n} \plus{} (z \plus{} x \minus{} y)^{2n} \plus{} (x \plus{} y \plus{} z)^{2n}$

( Follow Holder $ 3^{2n \minus{} 1}(x^{2n} \plus{} y^{2n} \plus{} z^{2n}) \ge\ (x \plus{} y \plus{} z)^{2n}$)

$ \Leftrightarrow(x^{2n} \plus{} y^{2n} \plus{} z^{2n} \ge (x \plus{} y \minus{} z)^{2n} \plus{} (y \plus{} z \minus{} x)^{2n} \plus{} (z \plus{} x \minus{} y)^{2n}$

$ \Leftrightarrow x^{2n} \plus{} y^{2n} \plus{} z^{2n} \geq\ x^{n}y^{n} \plus{} y^{n}z^{n} \plus{} z^{n}x^{n}$

This ineq is true by AM-GM[/quote]
Definitely Wrong  :wink:[/quote]
[quote]$ \Leftrightarrow(x^{2n} \plus{} y^{2n} \plus{} z^{2n} \ge (x \plus{} y \minus{} z)^{2n} \plus{} (y \plus{} z \minus{} x)^{2n} \plus{} (z \plus{} x \minus{} y)^{2n}$

$ \Leftrightarrow x^{2n} \plus{} y^{2n} \plus{} z^{2n} \geq\ x^{n}y^{n} \plus{} y^{n}z^{n} \plus{} z^{n}x^{n}$[/quote]
Is this wrong?[/quote]

[quote="tanpham90"]Yes , that's not true . You can check it by expanding . For example try with $ n \equal{} 3$ or $ n \equal{} 6$ .....etc...... :)[/quote]
Thank tanpham90 ; I 'll edit it. But I think this ineq is true
$ x^{2n} \plus{} y^{2n} \plus{} z^{2n} \ge (x \plus{} y \minus{} z)^{2n} \plus{} (y \plus{} z \minus{} x)^{2n} \plus{} (z \plus{} x \minus{} y)^{2n}$

[quote="onlylove_math"]Thank tanpham90 ; I 'll edit it. But I think this ineq is true
$ x^{2n} \plus{} y^{2n} \plus{} z^{2n} \ge (x \plus{} y \minus{} z)^{2n} \plus{} (y \plus{} z \minus{} x)^{2n} \plus{} (z \plus{} x \minus{} y)^{2n}$[/quote]
It's not true , too . Try with $ x\equal{}y\equal{}0$  :wink:

[quote="tanpham90"][quote="onlylove_math"]Thank tanpham90 ; I 'll edit it. But I think this ineq is true
$ x^{2n} \plus{} y^{2n} \plus{} z^{2n} \ge (x \plus{} y \minus{} z)^{2n} \plus{} (y \plus{} z \minus{} x)^{2n} \plus{} (z \plus{} x \minus{} y)^{2n}$[/quote]
It's not true , too . Try with $ x \equal{} y \equal{} 0$  :wink:[/quote]
You are wrong onlylove math,this ineq is true with $ n$ that it is quite small$ 4,6,10,12,8$,can you post your proof tanpham90


integration
trigonometry
function
calculus
calculus computations

$ \int \sqrt{1 \minus{} \sin(2x)}\,dx \equal{} \int \sqrt{\sin^2x \minus{} 2 \sin x \cos x \plus{} \cos^2x}\,dx$

$ \equal{} \int \sqrt{(\sin x \minus{} \cos x)^2}\,dx \equal{} \sqrt{2} \int \left|\sin(x \minus{} \frac{\pi}{4}) \right|\,dx$

$ \int \sqrt {1 \minus{} \sin 2x}dx \equal{} \int \frac {|\cos 2x|}{\sqrt {1 \plus{} \sin 2x}}dx \equal{} \pm\sqrt {1 \plus{} \sin 2x} \plus{} C$,

The answer is sinx + cosx because maybe they take the root of it as cosx-sinx and integrate it , they dont take into notice the -part.

Aye you saying that $ \sqrt{1\minus{}\sin 2x}\equal{}\sqrt{(\cos x\minus{}\sin x)^2}\equal{}\cos x\minus{}\sin x?$

[quote="kunny"]Aye you saying that $ \sqrt {1 \minus{} \sin 2x} \equal{} \sqrt {(\cos x \minus{} \sin x)^2} \equal{} \cos x \minus{} \sin x?$[/quote]
Yea , thats what my teacher told me , we just started on Indefinite integrals.

That's wrong. $ \boxed{\sqrt {A^2} \equal{} |A|}$ for $ A\in\mathbb{R}$

She Said it was something to do with complex values that you attain when you have sinx-cosx , and not cosx-sinx.

[quote="kunny"]That doesn't make sense.

Your teacher would say $ \sqrt {( \minus{} 2009)^2} \equal{} \minus{} 2009?$, wouldn't she?[/quote]
Id love to bash her someday along with you  :lol: But seriously , she denied me an Aplus. Anyway , maybe i can prove it to her , thanks!

Hey Doesnt the function have to have a definite range with all values as real? In That Case cosx-sinx should be the proper root.


I was wondering if AoPS gives recomendation letters or can be written down as one of the activites participated in. This would be in the case of MIT or Stanford.

There's nothing stopping you from putting it down as an activity.  Will it make much of a difference?  No.  But it shouldn't hurt you either.  Being a part of AoPS should allow you the opportunity to achieve high mathematical achievements in high school which will make you a much more competitive applicant to colleges.

For anyone to write a recommendation that will help you get in, they need to know you well enough to write about some aspect of you that doesn't show through in the rest of your application.  Does anyone on AoPS know you that well?

I've invested a lot of time in AoPS, so I put it on my applications. Just make sure you have something worthwhile to say about your time here (rather than, I hang out in G&FF all day or something).

We can write recommendation letters for students who have taken one of our evaluated classes (meaning one of our Intro or Intermediate level classes, or WOOT; not our problems series classes like MC, AMC, or AIME), because then we have some work to judge you by (namely, the challenge sets, and to a lesser extent your class and message board participation in the class).  We can't write letters for "ordinary" members of our site who are not students in our online school, because, as alluded to earlier in this thread, we wouldn't have anything meaningful to say about you.

So if I'm taking Intro Geo, Inter Alg (MC but that doesn't count does it?) and I'm going intio 7th, am i well off?

My general impression is that college recommendation letters from people who knew you in high school are much more useful than those arising from middle school relationships.

While some colleges are encouraging of extra recommendation letters, many (especially those which receive large volumes of applicants) get pissed off if you send more than they ask you to. Be very selective about your letters of recommendation, as they make a large difference.

You want to build as large a case for yourself as you can, within the limits of the application. If you are interested in problem solving, and that shows through your extracurricular activities and your resume, then most likely getting an AoPS letter of rec will not earn you much mileage. Of course, it depends on your individual situation, but I'm sure this holds true for the vast majority of AoPSers.


I've received my apmo score and I've heard others have received theirs as well.  I was just wondering how everyone did and what a good score was.  Mine was 8... I wasn't too disappointed since I only completely solved the first problem.  Also, I would like to see anyone's approach to problem 2, the painful inequality.  Thanks.

You may find APMO discussions here: http://www.mathlinks.ro/Forum/viewtopic.php?t=30956

I got 17... apparently that got me tied with 1 other person for 3rd in Canada...  I got 1, 3, and the algorithm for 4, though I wasn't able to prove it.

Jennifer Park told me that she was able to do #2 with just AM-GM and some re-arranging of things.

AM-GM? That's so simple... wow... I think getting 2 problems is really good... maybe I was just tired that day, but they didn't seem very approachable... I think they required a lot of insight to solve.  A friend from my school took it unofficially and solved problem 5 in maybe an hour... it was just algebraic manipulation from a Cosine Law and Heron's Formula.  The olympiads were really fun to write this year... I hope all of you guys did well! :)

APMO is hard though.  The time limit is so tight, and the problems are much more difficult than CMO problems.

Fun contest though... I wish I could do it again.


Let ABC be a triangle with A< \pi/2 ;B< \pi/2  ;C< \pi/2 ;Prove that:
    (sinA)^(2sinB)+(sinB)^(2sinC)+(sinC)^(2sinA) >2

sorry darkmaster
may be you don't think it carefully
easy to have 0 \leq sinA,sinB,sinC \leq 1
we have o \leq (sinA) 2  \leq 1
and sinB \leq 1
so we have (sinA) 2  \leq (sinA)^(2sinB)
so the last problem is prove that
sinA 2 +sinB 2 +sinC 2 >2
if you can't do that tell me

yes, master, your solution is very intelligent.

Mr.Th will be interested in it. But Dr.Buy now don't. He said he had good solution.

Dr .Buy Now said that he used Benzano Weiertrass to solve it and he didn't accept this proof.

I think Dr.BuyNow is so crazy and stupid.

Now you can try another test from Mr.Th in the topic "Mr th 's problem"


algebra polynomial LaTeX trigonometry function domain calculus

I think I saw this having to do with the roots of polynomials.  Can someone tell me what this means:(also I want to show off my LaTeX skills)

[tex]
\begin{document}
$e^{it} = \cos{t} + i\sin{t}
$\end{document}$
[/tex]

It tells you how to raise a number to a complex value.  It's really important in complex analysis because it allows you to extend the domain of a lot of functions (the obvious being exp and log) to the complex numbers.  The formula is called Euler's Formula, and, as a matter of fact, it's my personal favorite formula.

The proof involves taking the Taylor expansion (it's a calculus method) of cos x and i*sin x and adding them together, which yields the Taylor expansion of e^x.

And for finding roots, just google "Nth root theorem"

Ah, complex numbers - one of my favorite branches of math.  That formula shows the connections between exponential functions and complex numbers.  This can be used to raise any number to a complex power (see the topic on the Other Problem Solving Topics fourum):

[tex]n^{a+bi} = n^a \cdot e^{i \ln n^b} = n^a\cos \ln n^b + n^a i \sin \ln n^b[/tex]

[quote="Osiris"]Robert: you do not have to include the \begin and \end statements. Also, the LaTeX environment on this board assumes you're in math mode, so you don't have to use $ either.[/quote]

How do you know that I did this?

if you put your cursor over the tex you should be able to see it.

Here's another fun one.  By that formula,
[tex]\displaystyle{e^{i\frac{\pi}{2}}=\cos{\frac{\pi}{2}}+i\sin{\frac{\pi}{2}}}[/tex]
[tex]\displaystyle{e^{i\frac{\pi}{2}}=0+i=i}[/tex]
[tex]\displaystyle{(e^{i\frac{\pi}{2}})^i=i^i}[/tex]
[tex]\displaystyle{e^{\frac{-\pi}{2}=i^i}[/tex]

And, in reality, because cos and sin are periodic functions, really [tex]\displaystyle{e^{-\frac{\pi}{2}+2k\pi}=i^i}[/tex] with k as an integer.  Whenever I need a joke answer, for whatever reason, whether it's a math team question I have no idea on, or whatever, I always put [tex]\displaystyle{e^{-\frac{\pi}{2}}[/tex]


function
limit
complex analysis
complex analysis unsolved

If $ f$ is an analytic function that maps unit disk $ D$ onto the unit disk $ D$, then prove that $ |f(z_0)\plus{}f(\minus{}z_0)| \leq 2|z^2|$. Then when the equality occurs ??
NB : I am not sure with my answer.

Could you be more precise and say what is $ z_0$ and $ z$? I can only presume that you mean: "prove that $ |f(z) \plus{} f(\minus{}z)| \leq 2|z|^2, z \in D$. But then the thesis is true only under additional assumption that $ f(0) \equal{} 0$ and follows from the same reasoning as we use in Schwarz lemma.

Sorry, the original problem is :
If $ f$ is an  analytic function that maps unit disk $ D$ onto the unit disk $ D$, then prove that $ |f(z_0)\plus{}f(\minus{}z_0)| \ leq 2|z_0|^2$ for all $ z_0 \in D$, then determine when the equality occurs.

NB : I don't understand how to solve it by using Schwarz Lemma. Can anyone explain to me about it ? Thank you for your explanation before.

We can use Schwarz lemma in the following way. However, we need the assumption $ f(0) \equal{} 0$ because the function $ f(z) \equal{} \frac{2z \minus{} 1}{z\minus{}2}$ maps $ D$ onto $ D$, is analytic, but $ |f(0) \plus{} f(\minus{}0)| \equal{} 1 \geq 2\cdot|0|^2$.

If $ f(0) \equal{} 0$ let us consider
\[ g(z) : \equal{} \frac{f(z) \plus{} f(\minus{}z)}{2z}.\]
$ g$ maps $ D$ onto $ D$ and is analytic (indeed, except for $ z \equal{} 0$ it is obvious; at $ 0$ we have $ \lim_{z \longrightarrow 0} g(z) \equal{} 0$ - it is easily seen expanding $ f(z)$ and $ f(\minus{}z)$ at $ 0$ in Taylor series). Therefore we can put $ g(0) \equal{} 0$ and use Schwarz lemma. Then the problem is solved; moreover, equality holds if $ |g(z_0)| \equal{} |z_0|$ for some $ 0 \neq z_0 \in D$

Please explain the Schwarz Lemma. Thank you.

Here is a more problem.
Supposed $ f$ in analytic, $ f$ maps all elements in the unit disk into unit disk (also), prove that :
$ \frac {f'(z)}{1\minus{}|f(z)|^2} \leq \frac {1}{1\minus{}|z|^2}$.


Let m and n be natural numbers and consider the equation

 $(x+m)(x+n)= x+m+n$

 prove that if it has at least one integer solution than 1/2<m/n<2.

i would solve the 2nd degree equation, and use absurdity. I think it could work.