geometry
circumcircle
incenter
angle bisector
IMO
IMO 1962

On the circle $K$ there are given three distinct points $A,B,C$. Construct (using only a straightedge and a compass) a fourth point $D$ on $K$ such that a circle can be inscribed in the quadrilateral thus obtained.

There are always 3 such points D, D', D'' on the circumcircle $K$ f the triangle $\triangle ABC$, depending on which triangle side is taken as one of the diagonals of the bicentric quadrilateral with the vertices A, B, C, D (or D' or D''). WLOG, assume that CA is selected to be a diagonal, so that the point D lies on the circumcircle arc CA opposite to the vertex B . Let I be the incenter of the bicentric quadrilateral ABCD, which lies on the known bisector BI of the angle $\angle ABC$ and on the unknown bisectors of the angles $\angle BCD, \angle DAB$. The angle

$\angle CIA = 360^\circ - (\angle ICD + \angle CDA + \angle IAD) = 360^\circ - \left(\frac{\angle BCD}{2} + \angle CDA + \frac{\angle DAB}{2}\right)$

Since the quadrilateral ABCD is also cyclic, $\angle BCD + \angle DAB = \angle ABC + \angle CDA = 180^\circ$. Substituting this to the above equation,

$\angle CIA = 360^\circ - (90^\circ + 180^\circ - \angle ABC) = \angle ABC + 90^\circ$

First, assume that the angle $\angle ABC < 90^\circ$ is acute. Then $\angle CIA = \angle ABC + 90^\circ < 180^\circ$ and we have to construct a circle (E) passing through the points C, A, such that the angles spanning its minor arc CA on the same side of the line CA as the vertex B are equal to the known angle $\angle CIA$. The incenter of the quadrilateral ABCD lies on this arc. To do this, we draw tangents to the circumcircle $K$ at the vertices C, A. Since we assume the angle $\angle ABC$ to be acute, these tangents intersect at a point E on the opposite side of the line CA than the vertex B. The angles $\angle CAE = \angle ACE = \angle ABC$, hence, the angle $\angle CEA = 180^\circ - 2 \angle ABC$ and the angles spanning the minor arc CA if the circle (E) centered at E and passing through C, A are equal to

$180^\circ - \frac{180^\circ - 2\angle ABC}{2} = \angle ABC + 90^\circ$

The incenter I of the bicentric quadrilateral ABCD is the intersection of the bisector of the angle $\angle ABC$ with the minor arc CA of the circle (E). Once we have the incenter I, we can construct the circle (I) tangent to the lines BA, BC and then draw the other tangents to (I) from the vertices C, A intersecting at the desiren point $D \in K$.

If the angle $\angle ABC = 90^\circ$ is right, the circle (E) simply degenerates into the line CA (which is now a diameter of the circumcircle K) and the incenter I of the bicentric quadrilateral ABCD is the foot of the bisector of the right angle $\angle ABC$.

Second, assume that the angle $\angle ABC > 90^\circ$ is obtuse. Then $\angle CIA = \angle ABC + 90^\circ > 180^\circ$ and we have to construct a circle (E) passing through the points C, A, such that the angles spanning its minor arc CA on the opposite side of the line CA thans the vertex B are equal to the known angle $180^\circ - \angle CIA < 180^\circ$. The construction of the circle (E) is exactly the same as before.

Here is another construction:

We will use the theorem:
[b][color=blue]A convex quadrilateral ABCD is tangential iff  AB+CD = BC+DA[/color][/b]

(see http://www.mathlinks.ro/Forum/viewtopic.php?t=58984, see also the nice proof from nttu)

So, it's enough to find a point $D$ on the arc $AC$ (not containing $B$), such that $AB+CD = BC+DA$
(similarly for the points $D', D''$ specified by yetti)

If $AB=BC$, then the problem is easy.
We bring the bisector of $\angle{ABC}$, which intersects the circle at a point $D$.
Then $DA=DC \Rightarrow AB+CD=BC+DA \Rightarrow ABCD$ is tangential.
The incenter $I$ of $ABCD$ can be found on the intersection of $BD$ with the angle bisecor of $\angle{DAB}$.

Suppose that $AB>BC$ (I use nttu's proof)
$AB+CD = BC+DA \Rightarrow DA>CD$

Let $M \in BA$, such that $BM = BC$
and $N \in DA$, such that $DN = DC$

$AB+CD = BC+DA \Rightarrow$

$AM+MB+CD = BC+DN+AN \Rightarrow$

$AM+BC+CD = BC+CD+AN \Rightarrow AM=AN$

Let $P$ be the midpoint of the arc $AC$ (the arc whics contains the point $B$)

The triangles $PDN,PDC$ are equal
Indeed:
$PD=PD$
$DN=DC$
$\widehat{PDN} = \widehat{PDC}$, because $DP$ is the angle bisector of $\widehat{ADC}$

As a result, $PN=PC$

[size=150][color=blue]Construction[/color][/size]

We construct the points:
the midpoint $P$ of the arc $AC$ which contains the point $B$
$M \in BA$, such that $BM = BC$
the intersection $N$ of the circles $(A,AM)$ and $(P,PA)$ ; from the two points of intersection we choose the one which is in the interior of the circle $K$.

The ray $AN$ intersects again the circle $K$ at a point $D$

The triangles $PND,PCD$ are equal, because
$\widehat{NDP} = \widehat{CDP}$
$PD=PD$
$\widehat{PND} = 180^0- \widehat{ANP} = 180^0- \widehat{PAN} = \widehat{PCD}$

Hence $DN=DC$

$ \begin{array}{cc} AB+CD = BM+AM+CD \\ BC+AD = BC+AN+DN \end{array}\}\Rightarrow AB+CD=BC+AD \Rightarrow ABCD$ is tangential

We bring the angle bisector of the angle $\widehat{BAD}$, and where it intersects the segment $PD$, there is the incenter $I$.

[hide="Another Solution"]Note that a circle can be inscribed within a quadrilateral $ABCD$ if and only if $AB+CD = AC+BD.$ Also, note that given a triangle $ABC$ and a point $D$ such that $AB+CD = AC+BD,$ the locus of $D$ is a line. So, given points $A,B,C$ on the quadrilateral, let $AB = b, BC = a.$ Now, draw a circle with centre $A$ and radius $2b+a$ and another circle with centre $C$ and radius $2a+b.$ Choose the intersection points of these circles that is farther away from $B$ and call this point $D_{1}.$ Then, draw a circle with centre $A$ with radius $3b/2+a/2$ and another circle with centre $C$ with radius $3a/2+b/2.$ Take the intersection points of these circles that is farther away from $B$ and call this point $D_{2}.$ Draw $D_{1}D_{2}$ and extend this segment infintitely. Take the intersection point between $D_{1}D_{2}$ and the circle circumscribing $ABC$ which makes $ABCD$ convex and call this point $X.$ $X$ is the desired point.[/hide]


Consider the mechanical system  consisting of a child in a pram on a rough horizontal floor.

Is it possible for the child,by moving within the pram,to move the centre of mass of the pram and child horizontally relative to the floor?
Assume that frictional forces act on all moving parts of the pram.

Yeah! It's possible. The system child+pram isn't isolated. There is external forces acting on it.

And to be more examplefic, here's one way to do it:

Suppose the pram is little longer than you'd expect - long enough for the child to comfortably move from one of pram's ends to the other. The child starts at one end and very slowly and very carefully moves to the opposite end so that the pram stays at the same place due to static friction. Then when it reaches the other end, it jumps back to the spot from where it started and thus moves the pram in the opposite direction (assuming it overcame the static friction this time; just imagine Clark Kent when he was a baby).

no it isn't, there are no external forces. net force on the combined child and pram is zero (as you consider them to be attached, any force the child exerts on the pram will cause a force of equal magnitude to be exerted by the pram on the child), so the centre of mass of the child and pram will maintain a constant velocity.


induction
factorial
inequalities
Pascal\u0027s Triangle

Scrambled wrote:is it possible to prove that C(n,r) will always be an integer when n is greater than r?(without saying that combinations will always be an integer)

This is an informational question, right? (If it's a challenge problem, I'd rate its level at Intermediate, because of the proof technique involved, but it is certainly an interesting Getting Started level general question.) The short answer is (in spoiler text):

[hide]yes, there is a proof by mathematical induction that C(n,r) is always an integer when n is greater than r. This can also be read off Pascal's triangle--which you may prefer to call Yang Hui's triangle. [/hide]

Hope this helps. With some time I can edit this to flesh out the proof.

Well, I'm not sure whether it's right or not but if n is greater than r, it means certain factorial divided by certain factorial, and since n is greater than r, the factorial of n will be greater than r unless in one condition:  r is the negative. But we know it's impossible to have value for r is negative because you can't pick negative amount from the certain amount.  So, my point is that since n is greater than r, n's factorial will be greater than r's factorial...

yes, but this is nCr, not nPr. Since it's nCr, instead of being n!/r!, it's n!/(r!x(n-r)!)

Silverfalcon, the combinations formula is nCr = n! / r!(n-r)! and its not just "one factorial divided by another factorial"

[hide]

You can show it will always come out to a positive integer because since n>r, let r=n-x, where x is a positive integer.

nC(n-x)= n! / [(n-x)!(n-(n-x))!]

So its:

nC(n-x) = [(n)(n-1)(n-2)...(n-x)!] / [(n-x)!x!]

nC(n-x) = n(n-1)(n-2)...(n-x+1)! / x!

n > r , n> n-x

So if the last factorial is (smaller+1)! / (even smaller)! the answer is a positive integer....

[/hide]

um ok this was not that great...

yep, Just forgot to say it.  But whether it's permuation or combination, I think the answer is integer, right?

But, that's what you're trying to prove!

Right, exactly.  My "proof" is very bad, so I figured I would answer a rhetorical question...

[quote="JS1527"]
nC(n-x) = [(n)(n-1)(n-2)...(n-x)!] / [(n-x)!x!]
nC(n-x) = n(n-1)(n-2)...(n-x+1)! / x!
[/quote]

Not quite right.. if you remove the first ! sign from the second row, then that would be right :)

By the way, one way you could prove its an integer is just that it equates to the number of ways of choosing r objects from n objects! OK, so that doesn't seem nice to you, but in fact its even possible to prove Fermats Little Theorem in the same way - by constructing a scenario of counting something.

[hide]

So you've got:

n!/(r!*(n-r)!), and without a loss of generalization, we'll say r! :ge: (n-r)!

So, canceling, we're left with:

(n*(n-1)*...*(r+2)*(r+1))/((n-r)*(n-r-1)*...*(2)*(1))

Clearly, there are (n-r) terms on the bottom, and (n-r) terms on the top.  For any set of k consecutive integers, k must divide at least one of those integers, because they will all be congruant to each of the possible remainders 0 to k-1 mod k, in some order.  So 1 will divide at least one of those terms, as long as there is one term, 2 will divide at least one, as long as there are two terms, so on and so forth, until you get to n-r will divide at least one of those terms, as long as there are at least n-r terms.[/hide]

That doesn't necessarily cut it ... I mean, think about 16/(8 * 4 * 2).  Every term on the bottom divides some term on the top, but that doesn't give us an integer in the end.

Hint 1: [hide]I have one particular property from Pascal's triangle in mind.[/hide]

Hint 2: [hide]It's aCb + aC(b+1) = (a+1)C(b+1).[/hide]

Wait... don't you mean Pascal's Triangle? Or is there that property of Fermat's Triangle which I have never heard of?

Also - what is Fermat's triangle - never heard of it. I'm pretty sure you mean Pascal's triangle - yup.

Can anyone go to this site and explain what Fermat's triangle is?

http://www.earthmatrix.com/serie56/maya56.htm

http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Fermat%27s_last_theorem.html

Ack, I can't believe I wrote something as stupid as that. Fixed. ;)

sorry to reply an old thread but does this imply chebyshevs theorem (prime between n and 2n) or am i just deluding myself?

Um.  What it does say is that if p is prime, then p|C(p,r) for 0<r<p


4 cubes of volumes 1, 8, 27, and 125 are glued together at their faces. What is the numba of squ. cm of the  smallest  possible surface area of da solid in da resulting figure.

[hide]194[/hide]

[hide="if your new at mathcounts my solution might be confusing"]ok so we know that the cubes have a side length of 1,2,3, and 5 respectively so whe can place the 5 on the bottom and then try to cover up as much of the top face as possible.  (if im right this will result in the minimal surface area) so we can use the formul to find the surface area of the 5 side length square which is 150 and we can subtract the area of each of the faces of the other cubes so - 9-4-1 = 136 and then we have to add on the surface areas of the other three cubes minus the sides where they are glued.  Ok lets start with the  3 cube.  If we glue the other cubes next to it we have 5*9-1-4 which is 40 and the 2 cube we have 4*4-1 which is 15 and the 1 cube we have 3*1 whcih is 3 adding these togather we get 136+40+15+3=$ \boxed{194}$[/hide]

[hide="hint"]Diagrams are really helpfull!!!!!!!![/hide]

umm yea this is a pretty famous problem, considering when I was doing practice AHSMEs (now AMC 12s), a problem with volume 1,8,27, and 64 (not 125) came  up and it was lke a number 15ish...

its a pretty fun problem and that was my first time seing it but i was afraid that my solution was alittle wordy!!!

remember when you put 2 faces together, you have to subtract twice the area of the intersecting area.


How many pairs of vertical angles are formed by five distinct lines that have a common point of intersection?

[hide=" :nhl: "]
Each pair makes 2 pairs of vertical angles.
$\binom 52=10$ possible pairs.
2*10=20[/hide]

:juggle: [hide]Five with two lines, five with three lines, five with four lines, and five with five lines. 5+5+5+5=20[/hide]


Για κάθε x,y,z πραγματικούς νδο

$ \frac{x(y\plus{}z)^3}{(y\plus{}z)^2\plus{}1}\plus{}\frac{y(x\plus{}z)^3}{(x\plus{}z)^2\plus{}1}\plus{}\frac{z(x\plus{}y)^3}{(x\plus{}y)^2\plus{}1}\leq\frac{(x\plus{}y\plus{}z)^4}{(x\plus{}y\plus{}z)^2\plus{}1}$

Just a hint

[hide]$ \sum\frac{z(x\plus{}y)^{3}}{(x\plus{}y)^{2}\plus{}1}\leq \sum\frac{z(x\plus{}y)^{2}}{2}$[/hide]

Για χ,ψ,ζ  αρνητικούς νομίζω δεν ισχυει η ανίσωση σου

[quote="xp"]Για κάθε x,y,z πραγματικούς νδο

$ \frac {x(y \plus{} z)^3}{(y \plus{} z)^2 \plus{} 1} \plus{} \frac {y(x \plus{} z)^3}{(x \plus{} z)^2 \plus{} 1} \plus{} \frac {z(x \plus{} y)^3}{(x \plus{} y)^2 \plus{} 1}\leq\frac {(x \plus{} y \plus{} z)^4}{(x \plus{} y \plus{} z)^2 \plus{} 1}$[/quote]

H anisotita einai arketa ews polu xalarh. Mia kapws pio isxurh (alla kai pali asthenhs) anisotita pou tha mas boithisei einai :

Gia olous tous pragmatikous a,b isxuei

$ \frac{ab^3}{b^2\plus{}1}\leq\frac{1}{3}\frac{(a\plus{}b)^4}{(a\plus{}b)^2\plus{}1}$      (*)

H (*) isxuei an $ a<0$ i $ b<0$. Episis isxuei an $ a\equal{}0$ i $ b\equal{}0$ kai exoume isotita otan $ a\equal{}b\equal{}0$. An $ a<0$ kai $ b<0$ tote i (*) einai isodinami me to an eixame $ a>0$ kai $ b>0$. Opote estw $ a>0$ kai $ b>0$. Tote i (*) grafetai:

$ (a\plus{}b)^4(b^2\plus{}1)\geq3ab^3((a\plus{}b)^2\plus{}1)  \Leftrightarrow$
$ (a^4\plus{}4a^3b\plus{}6a^2b^2\plus{}4ab^3\plus{}b^4)(b^2\plus{}1)\geq3ab^3(a^2\plus{}b^2\plus{}2ab\plus{}1)  \Leftrightarrow$
$ a^4b^2\plus{}a^4\plus{}4a^3b^3\plus{}4a^3b\plus{}6a^2b^4\plus{}6a^2b^2\plus{}4ab^5\plus{}4ab^3\plus{}b^6\plus{}b^4\geq3a^3b^3$$ \plus{}3ab^5\plus{}6a^2b^4\plus{}3ab^3\Leftrightarrow$
$ a^4b^2\plus{}a^4\plus{}a^3b^3\plus{}4a^3b\plus{}6a^2b^2\plus{}ab^5\plus{}ab^3\plus{}b^6\plus{}b^4\geq0$

which is always true for $ a,b>0$. Efarmozoume tin parapanw treis fores, gia kathe oro tis arxikis anisotitas, prosthetoume kai pernoume to zitoume. Isotita isxuei otan exoume pantou isotita opote $ x\equal{}y\equal{}z\equal{}0$.

[quote="xp"]H λυση (και κατασκευη) μου στηριζεται στην ανισότητα αf(β+γ)<=αf(α+β+γ) για οποιαδ. γνησίως αύξουσα συνάρτηση f  και α,β,γ πραγμ. Το χαλαρό ειναι προφανές απ την κατασκευη της αλλά ειναι μια καλή ασκηση με ερωτήματα για την Γ λυκείου (καταγήγει σε ολοκλήρωμα δεν έχω χρόνο να την γραψω) Δεν ξέρω πως θα λυνόταν με τον τρόπο σου (ωραίος)αν οι δυνάμεις ηταν αντι 2 n kai anti 3 n+1 opoy n αρτιος[/quote]

Geia sou...ontws o tropos sou einai pio elegant!..An eixe perissoterous orous tote den ksero an tha itan problima auto afou kai pali tha ekana mia anisotita kai meta tha tin prostheta n fores. Tora an itan pio poliploki den ksero...elpizw kati na mporousa na kanw. :(  Pantos einai wraia anisotita dioti einai gia olo to $ R$ kai exeis kai isotita. :) Ostoso o logos pou ebala auti ti lisi einai apla kai gia tin anisotita (*).


An equilateral triangle is tiled with $n^2$ smaller congruent equilateral triangles such that there are $n$ smaller triangles along each of the sides of the original triangle.  For each of the small equilateral triangles, we randomly choose a vertex $V$ of the triangle and draw an arc with that vertex as center connecting the midpoints of the two sides of the small triangle with $V$ as an endpoint. Find, with proof, the expected value of the number of full circles formed, in terms of $n.$

[img]http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/497b4e1ef5043a84b433a5c52c4be3ae.png[/img]

[quote="Silverfalcon"]Joe, you should go ahead to the transcript and add the drawing to this problem!  :)[/quote]

Great idea!

[quote="calc rulz"]I'm guessing you want an answer posted? Anyway here goes...

Notice that the expected value does not depend on which particular possibility a particular circle is found in. Ie, if one more circle was formed in one possibility and one less in another, the expected value would be the same. Anyway we have to find the number of circles in total and divide it by the total number of possibilities (which is trivially $3^{n^2}$) to get the expected value.
Now, the total number of circles (including those of different possibilities) results from summing the total number of circles that take place for a given vertex that can have circles around it. Well, any of these vertices has six small triangles around it. The six small triangles must have a particular arrangement for there to be a circle, while the other $n^2-6$ could have any arrangement. So the number of circles a vertex forms is $3^{n^2-6}$.
Now for the number of such vertices. First notice that starting at n=3, there's one such vertex. Now notice that when there are n triangles, there are n-1 vertices at the bottom that will become circle-forming vertices for n triangles. So we'll basically add n-1 to the number of such vertices for n triangles to get the amount for n+1 triangles. Basically, we see that this is the $(n-2)$th triangle number for n triangles, which $= \frac{(n-2)(n-1)}{2}$.
So combining everything,  we end up with \[ \frac{3^{n^2-6}}{3^{n^2}} \times \frac{(n-2)(n-1)}{2} = \frac{(n-2)(n-1)}{2 \times 3^6}  \]

Hehe that was a pretty quick and sweet write-up of a solution. The one I sent in, although being correct and having the same basic ideas, is probably too long and complicated to get commended  :( ...[/quote]

Wow, that is essentially what I did!
Yay to see I didn't miss that problem!

Expected values some for independant events

Sorry, I mean they sum for dependant events

[quote="Scrambled"][quote="calc rulz"]I'm guessing you want an answer posted? Anyway here goes...

Notice that the expected value does not depend on which particular possibility a particular circle is found in. Ie, if one more circle was formed in one possibility and one less in another, the expected value would be the same. Anyway we have to find the number of circles in total and divide it by the total number of possibilities (which is trivially $3^{n^2}$) to get the expected value.
Now, the total number of circles (including those of different possibilities) results from summing the total number of circles that take place for a given vertex that can have circles around it. Well, any of these vertices has six small triangles around it. The six small triangles must have a particular arrangement for there to be a circle, while the other $n^2-6$ could have any arrangement. So the number of circles a vertex forms is $3^{n^2-6}$.
Now for the number of such vertices. First notice that starting at n=3, there's one such vertex. Now notice that when there are n triangles, there are n-1 vertices at the bottom that will become circle-forming vertices for n triangles. So we'll basically add n-1 to the number of such vertices for n triangles to get the amount for n+1 triangles. Basically, we see that this is the $(n-2)$th triangle number for n triangles, which $= \frac{(n-2)(n-1)}{2}$.
So combining everything,  we end up with \[ \frac{3^{n^2-6}}{3^{n^2}} \times \frac{(n-2)(n-1)}{2} = \frac{(n-2)(n-1)}{2 \times 3^6}  \]

Hehe that was a pretty quick and sweet write-up of a solution. The one I sent in, although being correct and having the same basic ideas, is probably too long and complicated to get commended  :( ...[/quote]
pardon me for stupidity (as this was pretty much what the AoPS transcript has) but how did you know that the total amount of circles * probability =expected value? Because this argument doesnt work for some shapes (like 3 disjoint hexagons)[/quote]

Consider it this way: It doesn't matter which possibility a circle is formed in, so let's just assume all circles form in the same possibility. If there were one circle-forming hexagon, the expected value would simply be the probability of forming a circle. If there were another circle-forming hexagon (with the same probability), we'd just multiply that number by two. Get it? That's a different way of explaining it than in my post above. Also note that in my post above "total number of circles" means total number in all possibilities, not just total number of circle-forming hexagons.


When 39,500 is divided by an integer $ n$, the quotient is 123 and the remainder is 17. Find $ n$.


A point $E$ is chosen at random from within square $ABCD$.  Express as a decimal to the nearest hundredth the probability that $\triangle ABE$ is obtuse.

Simple.

[hide]

Draw a circle centered at the midpoint of AB and passing through points A and B. Erase the part outside of ABCD.

If E is on that semicircle, then angle AEB is right. If E is outside the semicircle, then angle AEB is acute. If E is inside the semicircle, then angle AEB is obtuse.

The probability is pi/8.

Note that if E is inside square ABCD, then neither angle ABE or EBA can be obtuse.

[/hide]

[quote=RadiantCheddar]@above since the problem asks for decimal to the nearest hundredth, the answer would be [hide=this]0.39[/hide][/quote]

nice 16 year bump!


Solve in $\mathbb C $ the following equation:
\[
3x^3-3x+1=0
\]

$x=\frac{2}{\sqrt{3}}\cos 50^\circ$ or $x=\frac{2}{\sqrt{3}}\cos 170^\circ=-\frac{2}{\sqrt{3}}\cos 10^\circ$ or $x=\frac{2}{\sqrt{3}}\cos 290^\circ=\frac{2}{\sqrt{3}}\cos 70^\circ$.

It's an irreducible cubic- this is as nice as the answers will get. The method here was to transform into an angle trisection equation $4y^3-3y=\cos\theta$.

Or to give one solution in algebraic terms :

$ x=\frac{y}{6}+\frac{2}{y}$
where
$y= (-36+12i \cdot \sqrt 3)^{ \frac{1}{3} }$


What's special about MIT's Course 18 Option 3 (Bachelor of Science in Mathematics and computer science)?  Anyone care to share their experience?  also where can I find more information about it besides http://web.mit.edu/catalogue/degre.scien.ch18.shtml which makes the course look like any other course.

thank you.

There's nothing particularly special about it except that I believe your diploma will say 'Mathematics with Computer Science'.  It's basically just a way of mixing course 6-3 (computer science) with 18 (math) and getting a degree out of it.  If you want to know more about it you can look at the course description for each individual class required.

I'm personally biased toward just doing Course 6 (because, well, I am...but of course I did have reasons for choosing so).  I've been able to take the math classes that interested me while being able to undertake a thorough study of EECS.  That said, I'm still debating doubling in math since it wouldn't be too hard for me...


ok, one of my colleagues walked into my office a few minutes ago and posed this question to me... he claims he found a rather nifty way to show the following infinite product for $e$ (thought i would post it here, 'cause i am drawing a complete blank!  :huh: ) :

$\frac{e}{2}\;=\;\left( \frac{2}{1}\right)^{\frac{1}{2}}\cdot \left( \frac{2\cdot 4}{3\cdot 3}\right)^{\frac{1}{4}}\cdot \left( \frac{4\cdot 6\cdot 6\cdot 8}{5\cdot 5\cdot 7\cdot 7}\right)^{\frac{1}{8}}\cdots$

notice that it bears a resemblance to wallis' product (except for the [b]exponents[/b]):

$\frac{\pi}{2}\;=\;\left( \frac{2}{1}\right)\cdot \left( \frac{2\cdot 4}{3\cdot 3}\right)\cdot \left( \frac{4\cdot 6\cdot 6\cdot 8}{5\cdot 5\cdot 7\cdot 7}\right)\cdots$


Let $a,b,c$ be positive real numbers. Prove that :
                  $\frac{a}{b+2c} +\frac{b}{c+2a} +\frac{c}{a+2b} \geq 1$

[quote="matematikator"]Take x>=y>=z(inequality is symetric)
a/[b+2c] + b/[a+2c] + c/[a+2b]=A
Using the rearragment inequality:
2A>=[b+c]/[b+2c] + [a+c]/[a+2c] + [b+c]/[b+2c] and clearly, each term on the left side >2/3 since x>=y>=z[/quote]

It's not symmetric.

Anyways, the usual Jensen's method gives us, since $f(x)=\frac {1}{x}\rightarrow f''(x)\geq 0$, with $a+b+c=1$,

$\sum \frac {a}{b+2c}\geq \frac {1}{\sum a(b+2c)}=\frac {1}{3\sum ab} \geq \frac {1}{(\sum a)^2}=1$


What is the sum of the first 15 triagular numbers?

Hint (formula): [hide]in order to figure out the nth triangular number, use this formula: n(n+1)/2[/hide]

Thats the 15th triangular number, you want the sum of the first 15.

Tare wants the SUM of the first fifteen numbers, not the fifteenth number.  With just 15 terms, it would be easy enough to brute-force this, but in general, if you can get a polynomial formula for what you're trying to sum (in our case .5n^2 + .5n), you can always break it up into the sum of a constant (0 in our case), the sum of the first n numbers (times .5 in this case), etc.  We have formulae for many of the smaller powers that are fairly easy to memorize.

Ok ok ok I'll do it.
*All the sums displayed by the sigma notations are of the n from 1 to 15.*

15
:Sigma:  (n:^2: + n) / 2 =  1/2 * :Sigma:  (n:^2: + n) =
n = 1

1/2 *(:Sigma: n:^2: + :Sigma: n) =

1/2 * (15*16*31*1/6 + 15*16*1/2) = 680

Therefore 680 is our answer.

Don't you just love neat work???


function
calculus
derivative
calculus computations

In proving Darboux theorem we consider the function $ h(x) \equal{} f(x)\minus{}rx$. Then we take the derivative $ h'(x) \equal{} f'(x)\minus{}r$. Is it faulty then to assume that $ h'$ has a maximum [b]inside[/b] an interval $ [a,b]$ as opposed to the endpoints? Because assuming that $ f(a) < r < f(b)$ then $ h'(a) <0$ and $ h'(b) >0$. Since neither of these are $ 0$, they cannot be maxima. But consider $ e^x$. In every compact interval their is a local maximum. But nowhere is the derivative $ 0$. So this is a bad proof right?

You're talking about the mean value theorem; "Darboux" tends to refer to the intermediate value property/theorem around here.

Divide into cases. Is there a maximum in the interior? A minimum in the interior? Neither?


Give $ \alpha$ is a irrational. Prove that $ \{n!\alpha\}$ is dense in $ (0,1)$ with $ n \in N*$

Seems to be wrong for $ \alpha\equal{}e$...

why it's wrong for $ e$ ? actually i   think it's wrong for $ \frac {1}{e}$
we have : $ \frac {1}{e} \equal{} \sum_{p \equal{} 0}^{\infty}( \minus{} 1)^p \frac {1}{p!}$
so for some $ n$ we will get : {$ \frac {n!}{e}$}$ \equal{} n! \sum_{p \equal{} N \plus{} 1}^{\infty} ( \minus{} 1)^p \frac {1}{p!} \le \frac {1}{n\plus{}1}$
so for $ n \ge 1$ we can't get numbers between 1 and 1/2 for exemple we can't get 3/4

It's wrong for $ e$ for the same reason. $ \{e n!\} \leq \frac{1}{n\minus{}1}$.


just curious if any teachers out there have a good guess as to which topics the 6 free response questions will cover on this year's free response

well we're not allowed to talk about the test, but I think i did well, i knew almost every part of the free response and I only had trouble with about 5 or 6 m/c u?

We can talk about it now, it's been 48 hours.

I think the mechanics section was much easier than it could have been.  The first two E&M problems weren't a surprise at all. #3 was about as difficult I've seen the AP exam be, although it was still doable and much more interesting it was harder.  Still, I got $(2L+2vt)$ in the denominator instead of $(L+2vt)$ because I didn't catch the part about the bar having negligible resistance.


analytic geometry
geometry
quadratics
trigonometry
conics

[quote="Rushil"]Does anybody know any good websites where I can find [b]interesting and difficult [/b]problems in coordinate geometry from [b]straight lines to conics(all) and general quadratics[/b]. Please [b]don't suggest overly tedious [/b]calculative problems where basically no insight is neede and you just have to solve very very long equations. I want questions which[b] require considerable thought [/b]before progressing with the solution.

Any suggestions :coolspeak:  :spider:[/quote]

Try to get an exhaustive set of IIT-JEE problems over the period 1990-2005...u'll get an immensly interesting set of analytical geometry problems..Also try to procure the book on co-ord. geometry by S.L. Loney who's authored a brilliant book on elementary trigonometry as well

I have S.L Looney's book and also Arihant(which by the way merely copies all of looney;s problems) However, most of them require either more knowledge(diameters etc. which are not in IIT) or are very very tedious(believe me!!)

I am also looking for internet resources. Looney is probably the ONLY book available in India. It is good but simple put, too archaic. Doesn't involve much thinking. Smply solving of eqns. I had seen the MAA book on COnics on net. Wonder if it is available online and if t meets my requirement???

[quote="Michael Niland"]Hello Rushil!

If you are looking for a nice coordinate geometry problem that requires thought but minimal calculations you might like to try a problem submitted by me on the Olympiad Geometry Forum(Propsed and Own Problems) under the heading "Tangents and Normals of a Parabola".It was posted March 27th  this year and nobody has tried to answer it.
I have several dozen good problems in this area of maths but I have not posted them because of the lack of interest shown in this problem.
Good Luck!

Michael Niland[/quote]

I have answered your problem.


geometry
trapezoid
FTW
Pythagorean Theorem

I always see people on FTW solve the Isosceles Trapezoid areas really quickly, when they're just given the 4 sides. Is there a way to do that?

[asy]pair A,B,C,D;
A=(0,0); B=(5,12); C=(10,12); D=(15,0);
draw(A--B--C--D--cycle);
draw(B--(5,0)); draw(C--(10,0));
label("$A$",A,SW); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,SE);
label("$X$",(5,0),S); label("$Y$",(10,0),S);[/asy]
When $ ABCD$ is isosceles, $ AX \equal{} DY$ (assume that $ \overline{BX}$ and $ \overline{CY}$ are perpendiculars to $ \overline{AD}$).  This makes it easy to find them and then use the Pythagorean Theorem to get the height.

An alternate method which may or may not be faster is to use the fact that since an isosceles trapezoid is cyclic, the area is $ \sqrt {(s \minus{} a)(s \minus{} b)(s \minus{} c)(s \minus{} d)}$ (someone confirm this is correct or otherwise correct it).  This is probably not very appropriate for this forum though.

Yes, the formula works, although I'm not sure if it's faster than dropping the altitude, seeing as FTW problems usually involve nice Pythagorean Triples.

Does AX=XY=YD in an isosceles trapezoid?

AX=YD, we don't know about XY though.  It happens to be the case in that specific diagram though.


Putnam
inequalities
algebra
polynomial
LaTeX
arithmetic sequence

hello, i am looking for some advise regarding studying math. i have finished one year of undergraduate math and i plan to continue studying math. One thing that is starting to bother me is that till recently i have never seen and attempted olympiad level problems. i was wondering if this will hinder my progression in learning math or, by doing undergraduate level courses i will learn much of the mathematical thinking that i have missing from not practicing olympiad problem solving skills. i have a feeling that without doing olympiad math im at an eternal disadvantage to those who do/did.. so i would like to know what you people think?

i actually really feel it does disadvantage me...  beacuse olympiad math builds a sort of core of mathematical thinking and knowledge. coming to college with this more developed mind will enable a student to effectively absorb more content then a regular student will. i would compare this to having so called 'good fundamentals' which enable you to achieve more then someone who is already building on flawed fundamental techniques.  you know... the tower is only as good as its base kind of analogy..
do you all agree?

p.s to gauss202: im not talking about participating in an actual olympiad im talking about learning olympiad problem solving skills and material.. if your friends might have not done well (whatever your standard of well is...) in olympiads but if they participated i would imagine they would have studied some olympiad problem solving in preparation/selection.

[quote="gauss202"]You are not at a significant disadvantage. I've known many great mathematicians who did not participate in (or did not do exceptionally well in) High School Olympiads.[/quote]

Conversely, I seem to find out fairly frequently that various graduate students/faculty at my university or top researchers in my subfield were IMO medalists or Putnam high scorers/fellows.  And it's a bit disconcerting (if completely unsurprising) that they are the ones who come off as being really, really smart.

I mean, I scored in the top 200 on the Putnam as an undergraduate and graduated at the age of 20, and now that I've completed almost two years of graduate school, I feel like I knew nothing as an undergrad.  In comparison to most of my graduate school peers I feel like my problem solving skills are quite good.  But all of the olympiad superstars and whatever that I know are just scary smart, and it's hard not to feel like they are in another league from you, and that you just don't have the intellectual capacity to accomplish the things that they will or that they have.

On the other hand, unless your happiness or sense of self-worth is predicated on being one of the top x mathematicians in the world, all of this doesn't matter that much.  You can be a good mathematician without being terry tao, after all.  So if you are disadvantaged by not having an IQ of 180 like some olympiad superstars or because you picked up math later than others, well, accept it and go on with your life.

[quote="blahblahblah"]

Conversely, I seem to find out fairly frequently that various graduate students/faculty at my university or top researchers in my subfield were IMO medalists or Putnam high scorers/fellows.  And it's a bit disconcerting (if completely unsurprising) that they are the ones who come off as being really, really smart.

I mean, I scored in the top 200 on the Putnam as an undergraduate and graduated at the age of 20, and now that I've completed almost two years of graduate school, I feel like I knew nothing as an undergrad.  In comparison to most of my graduate school peers I feel like my problem solving skills are quite good.  But all of the olympiad superstars and whatever that I know are just scary smart, and it's hard not to feel like they are in another league from you, and that you just don't have the intellectual capacity to accomplish the things that they will or that they have.

On the other hand, unless your happiness or sense of self-worth is predicated on being one of the top x mathematicians in the world, all of this doesn't matter that much.  You can be a good mathematician without being terry tao, after all.  So if you are disadvantaged by not having an IQ of 180 like some olympiad superstars or because you picked up math later than others, well, accept it and go on with your life.[/quote]
Does John Conway get IMO anytime in his life?No(?). Is he one of the best mathematicians compared to Terry Tao? Absolutely! But we don't need to always mention Terry Tao though. He's a genius and everyone knows that. However, I believe we already knew many such "genius" people in this forum(Vasc, Harazi, or Iurie Boreico(5 times IMO medalist). We should note that each mathematician has his/her own strength, and h/she only works on the fields that he/she is an expert on it.  I agree that Terry Tao discovers the significant proof for Green-Tao theorem, but Ngo Bao Chau-a Vietnamese mathematician who improved the Fermat's last theorem further and got a prize(don't remember whether it's a Field Medal or not)- is not well-known as Terry is. It's okay because that's life and sometimes people don't know your work until you die! If you consider the world of inequality, Vasc and Harazi should be the "Mozart" in that world. Will Terry Tao can beat them in this field? I suspect that possibility despite the fact that he's the greatest mathematician in our time. Like Bill Gate versus Steve Jobs in the entertainment, who wins?Steve Jobs is leading even though Bill Gate is a genius and beat Steve Jobs in computer field easily

I'm not sure why you decided to fly off the handle just because I mentioned Terry Tao.  My point was that your decision to study math, or your happiness, or whatever, should not be dependent on your ability to be one of the top x mathematicians in the world.  I mentioned Terry Tao because as long as x>1, he is probably among those top x mathematicians.

[quote="ghjk"]
Does John Conway get IMO anytime in his life?No(?). Is he one of the best mathematicians compared to Terry Tao? Absolutely! But we don't need to always mention Terry Tao though. He's a genius and everyone knows that. However, I believe we already knew many such "genius" people in this forum(Vasc, Harazi, or Iurie Boreico(5 times IMO medalist).
[/quote]

There are lots of great mathematicians (especially older ones!) who never participated in the IMO or on the Putnam, and some who have managed to be great without having a skill-set suited to success on olympiads and the like.  I'm just observing that most of the people who have been very successful in competition math often seem almost 'untouchably' smart in person.  So are you at a disadvantage because you didn't do olympiad math per se?  No.  But you probably are at a significant comparative disadvantage because they have probably put much more time into math than you, have received specialized training than you haven't, and are inherently more capable than you.  But math isn't a competitive sport, and life isn't fair, so this isn't something you should waste much time thinking about.

[quote="ghjk"]
We should note that each mathematician has his/her own strength, and h/she only works on the fields that he/she is an expert on it.  I agree that Terry Tao discovers the significant proof for Green-Tao theorem, but Ngo Bao Chau-a Vietnamese mathematician who improved the Fermat's last theorem further and got a prize(don't remember whether it's a Field Medal or not)- is not well-known as Terry is.  It's okay because that's life and sometimes people don't know your work until you die!
[/quote]

So what are you trying to argue here?  Ngo Bao Chau is certainly a very very good mathematician.  He's also a multiple IMO medalist.  By any reasonable measure he's not as good of a mathematician as Terry Tao is.  That, too, is life.

[quote="ghjk"]
If you consider the world of inequality, Vasc and Harazi should be the "Mozart" in that world. Will Terry Tao can beat them in this field? I suspect that possibility despite the fact that he's the greatest mathematician in our time. Like Bill Gate versus Steve Jobs in the entertainment, who wins?Steve Jobs is leading even though Bill Gate is a genius and beat Steve Jobs in computer field easily[/quote]

This is pretty nonsensical and irrelevant.  The fact is that there are very few mathematicians who have any sort of interest in proving 3-variable polynomial inequalities.  It's stupid to compare work done in a subfield that virtually no one studies seriously to the kind of things Terry Tao does (harmonic analysis and arithmetic combinatorics are among the most active subfields of mathematics today).  Basically you have no clue what you're talking about.

I was perusing through this section of AoPS, and I accidentally managed to stumble upon this thread, it seems an interesting discussion is going on. I apologize for reviving this thread again.

[quote] You are not at a significant disadvantage. I've known many great mathematicians who did not participate in (or did not do exceptionally well in) High School Olympiads. [/quote]

I wholly second this post, an instance for the above 'quote' is - Ramanujan. (Though, I would like to concede that IMO and Putnam didn't exist when he was there, but he would have easily won many IMO medals for sure).

[quote] i actually really feel it does disadvantage me... because Olympiad math builds a sort of core of mathematical thinking and knowledge. coming to college with this more developed mind will enable a student to effectively absorb more content then a regular student will. i would compare this to having so called 'good fundamentals' which enable you to achieve more then someone who is already building on flawed fundamental techniques. you know... the tower is only as good as its base kind of analogy..
do you all agree?
[/quote]

I agree with this post, though I hadn't competed at the Olympiads, I have interacted with people who had competed; If you manage to solve Olympiad level problems it deeply strengthens your understanding of Mathematics, because 'some of the research problems in Math' (I mean some) seem pretty much like Olympiad problems (but they require more patience and perseverance). If you managed to compete at the IMO and also managed to get a medal, you're at a significant advantage than your peers at your undergraduate institution both in terms of mathematical ability, and proficiency.

[quote]  You can be a good mathematician without being terry tao, after all. [/quote]

[quote] So what are you trying to argue here? Ngo Bao Chau is certainly a very very good mathematician. He's also a multiple IMO medalist. By any reasonable measure he's not as good of a mathematician as Terry Tao is. That, too, is life.
[/quote]

So, you seem to greatly concur with your delusional notion that 'Terry Tao' is the greatest mathematician of our time (or whatever). I bet Ramanujan, and as well as Perelman, Grothendieck, Charles Fefferman are as good as Tao if not better. It is unfortunate that some people with as much talent and ability don't get fame like Tao does. (But I acknowledge that Tao is a great mathematician).

[quote="Gen8"]So, you seem to greatly concur with your delusional notion that 'Terry Tao' (I respect him greatly) is the greatest mathematician of our time (or whatever). I bet Ramanujan, and as well as Perelman, Grothendieck, Charles Fefferman are as good as Tao if not better.[/quote]

no, that's not what I said.  i said that terry tao was probably a better mathematician than ngo bao chau.  tao is younger, has proved more important results (slightly subjective), and been awarded more prestigious prizes by his fellow mathematicians.  both of them are easily in the top 1% of research mathematicians.  just for my own personal amusement, i'd like to hear why calling terry tao the best mathematician of our time (which I never did) would be 'delusional', though.

[quote="Gen8"]I bet Ramanujan, and as well as Perelman, Grothendieck, Charles Fefferman are as good as Tao if not better.[/quote]

you know [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=276166]so[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=276175]little[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=275175]mathematics[/url] that this opinion is completely worthless.  your understanding of mathematics is so rudimentary that it is unlikely that you could follow a single important paper of tao, grothendieck, or fefferman, so why do you think you could correctly judge which of them is the better mathematician (if, indeed, there is even a well-ordering on mathematicians of this type)?

[quote] you know so little mathematics that this opinion is completely worthless. your understanding of mathematics is so rudimentary that it is unlikely that you could follow a single important paper of tao, grothendieck, or fefferman, so why do you think you could correctly judge which of them is the better mathematician (if, indeed, there is even a well-ordering on mathematicians of this type)?
 [/quote]

I am a major in computer science (and not Mathematics!); I had taken just 3 courses in Math in my under-grad school (I'm not from the U.S.A.) so it might be that I have a rudimentary level of proficiency in Math. I also don't know LATEX (I'm learning it though).

Those links that you had cited were- in specific, some doubts I had , (I committed a mistake in one of the arithmetic progression problems, it is quite natural to commit mistakes and have doubts).

 [quote] you know so little mathematics that this opinion is completely worthless. your understanding of mathematics is so rudimentary that it is unlikely that you could follow a single important paper of tao, grothendieck, or fefferman, so why do you think you could correctly judge which of them is the better mathematician (if, indeed, there is even a well-ordering on mathematicians of this type)? [/quote]

Well, As I have reiterated my major is in CS and not Math! So, you can't expect me to know math as you do, similarly I can't expect you to know CS as I do. I bet your understanding or proficiency in CS is similar to what I have in Math.

I never said that I correctly judged any of them (or my opinion was correct), it was an opinion of mine (just as you had regarding Tao), given the fact that fefferman won a fields medal at 28, and Perelman solved an 'very difficult open problem in Math' and Tao didn't.

I hope you get it into your head that these are highly subjective opinions based upon achievements (like solving open problems).

[b] P.S.: [/b] With due respect, your posts almost always (I mean non-technical posts) exhibit characteristics of a 'troll'. :lol: I think you delight in 'flame wars' in your pastime.

[quote="Gen8"]Well, As I have reiterated my major is in CS and not Math! So, you can't expect me to know math as you do, similarly I can't expect you to know CS as I do. I bet your understanding or proficiency in CS is similar to what I have in Math.[/quote]

i think being able to turn on a computer would indicate a proficiency in CS roughly equal to your proficiency in mathematics.

[quote="Gen8"]and Perelman solved an 'very difficult open problem in Math' and Tao didn't.[/quote]

the question resolved by the [url=http://en.wikipedia.org/wiki/Green-Tao_theorem]green-tao theorem[/url] was asked over a hundred years before the [url=http://en.wikipedia.org/wiki/Poincaré_conjecture]poincare conjecture[/url] was ever posed.  you're just embarassing yourself (that is, more than usual) now.

Back to the [b]original topic:[/b]  a few observations.

1.  You don't have to be good at solving Olympiad problems to be good at research mathematics.  They occur on very different time scales - being good at solving Olympiad problems requires that you can solve difficult problems on the scale of hours, whereas being good at research mathematics takes place on a scale of days, weeks, months, or years, and with very different resources available to you.  They're not necessarily comparable activities; it is possible to be good at one and not the other.

2.  Nevertheless, there is a strong correlation between the two.  Why?  As Kiran Kedlaya puts it, if you like Olympiad math that probably means you prefer to solve problems that take hours instead of minutes (as in high school), so it's likely that you prefer to solve problems that take weeks instead of hours.  How [b]interested[/b] (as opposed to how good) you are in Olympiad mathematics is a good indicator of how interested you will be in a professional mathematics career, although it is by no means a foolproof indicator.

3.  I also think that a second advantage in having done a lot of Olympiad mathematics is that you become acquainted with a wide variety of elementary (and not so elementary) problem-solving tools; having many tools in your toolbox will of course make it easier to do research math.  It is possible, however, that the rise of the [url=http://www.tricki.org/]Tricki[/url] will make this less of an issue.

So, short answer:  it is not [b]necessary[/b], but it can be [b]useful.[/b]

[quote] i think being able to turn on a computer would indicate a proficiency in CS roughly equal to your proficiency in mathematics. [/quote]

So, you mean you can only turn on a computer?  :rotfl:  Since my proficiency in Math is similar to your proficiency in CS.

Since I'm a graduate in CS, I would be much better off in comprehending abstruse concepts of CS than learning elementary/advanced Math (or whatever); But I like to dabble in Math ( and also comment on Math-related topics).


vector
factorial
algebra unsolved
algebra

hey man , free family on the Q-vector space : IR . i don't know , isn't 'free' the term in use?

ya correct thanx ill soon reply
gotta go know
bye

A more appropriate terminology would be "show that $ \{1,e,e^2\}$ is an independent set over $ \mathbb{Q}$".

[hide="solution"]

Just like in the standard proof that $ e$ is irrational, we proceed to arrive at a contradiction by multiplying by a large factorial and estimating. If $ 1, e, e^2$ are dependent over $ \mathbb{Q}$ then there are rational numbers $ p, q, r$ satisfying

$ p \plus{} q e \plus{} r e^2 \equal{} 0$

by clearing denominators we can find integers $ a, b, c$ so that

$ a \plus{} b e \plus{} c e^2 \equal{} 0$

or

(*) $ a e^{\minus{}1} \plus{} b \plus{} c e \equal{} 0$.

Now fix some integer $ n$ and write

$ e \equal{} 1 \plus{} \frac{1}{1!} \plus{} \frac{1}{2!} \plus{} \frac{1}{3!} \plus{} \cdots \plus{} \frac{1}{n!} \plus{} R_n$

where $ R_n$ is the remainder

$ R_n \equal{} \frac{1}{(n\plus{}1)!} \plus{} \frac{1}{(n\plus{}2)!} \plus{} \cdots$

similarly,

$ e^{\minus{}1} \equal{} 1 \minus{} \frac{1}{1!} \plus{} \frac{1}{2!} \plus{} \cdots \plus{} (\minus{}1)^n \frac{1}{n!} \plus{} S_n$

where the remainder $ S_n$ is

$ S_n \equal{} (\minus{}1)^{n\plus{}1} \frac{1}{(n\plus{}1)!} \plus{} (\minus{}1)^{n\plus{}2} \frac{1}{(n\plus{}2)!} \plus{} \cdots$

Now multiply (*) by $ n!$ and note that everything other than the remainders is now an integer, so for some $ M \in \mathbb{Z}$:

(**) $ M \plus{} c n! R_n \plus{}a  n! S_n \equal{} 0$

it remains to estimate $ c n! R_n \plus{} a n! S_n$.

$ | c n! R_n \plus{} a n! S_n | \leq |c| |n! R_n| \plus{} |a| |n! S_n | \leq$

$ |c| \left( \frac{1}{n\plus{}1} \plus{} \frac{1}{(n\plus{}1)(n\plus{}2)} \plus{} \frac{1}{(n\plus{}1)(n\plus{}2)(n\plus{}3)} \plus{} \cdots\right) \plus{}$
$ \plus{} |a| \left( \frac{1}{n\plus{}1} \plus{} \frac{1}{(n\plus{}1)(n\plus{}2)} \plus{} \cdots \right) \leq$
$ |c| \left( \frac{1}{n\plus{}1} \plus{} \frac{1}{(n\plus{}1)^2} \plus{} \frac{1}{(n\plus{}1)^3} \plus{} \cdots \right) \plus{}$
$ \plus{} |a| \left( \frac{1}{n\plus{}1} \plus{} \frac{1}{(n\plus{}1)^2} \plus{} \cdots \right) \leq$
$ \frac{|a|\plus{}|c|}{n}$

Finally, if $ n$ is sufficiently large, that last quantity is smaller than 1 which contradicts (**). QED

[/hide]

Of course, if you know the proof that $ e$ is transcendental, it is obviously a much stronger statement: it is equivalent to saying that $ \{1, e, e^2, \ldots, e^n\}$ is independent over $ \mathbb{Q}$ for any $ n$. You can find the proof in any number of books (e.g. "Proofs from the Book") and web sources, for example:

http://www.math.cmu.edu/~bobpego/21236/eistrans.pdf

Thanks so much [b]Randomgraph[/b] for your comprehensive reply  :roll: .


Prove that if $a$ and $b$ are positive integers, then\[(a+\frac{1}{2})^n+(b+\frac{1}{2})^n\]is an integer for only finitely many positive integers $n$

[quote="blahblahblah"]I never understood why this question made it to Naoki Sato's number theory notes.  If we write it as

$\frac {(2a+1)^n+(2b+1)^n}{2^n}$

It's obvious that $4\not | \ (2a+1)^n+(2b+1)^n$ and we're done.[/quote]

what if a,b are of different parity and n is odd?

In that case, it's obvious that I didn't think my solution over very well the first time.  Give me a minute (or a day . :blush: )

Okay, lets try this

WLOG let $n$ be odd (otherwise write $(2a+1)^{2q}=[(4a^2+4a)+1]^q$

We may write

$(2a+1)^n+(2b+1)^n=(2a+2b+2)((2a+1)^{n-1}\pm\ldots\pm(2b+1)^{n-1})$

There are $n$ summands of odd numbers in that last factor; hence we are left with an odd number.  This gives the result immediately (e.g, pick $n$ much larger than $a$ or $b$ and we see that we are done immediately)

I find it hilarious that I can't even remember the factorization for $a^n+b^n$, $n$ odd, anymore off hand (yeah, I remembered roughly how it telescoped, but I still had to look it up for comfort's sake).  In conclusion, the older I get, the dumber I get, and I'm not looking forward to this semester of school very much  :?

[quote="seamusoboyle"]Is it not obvious? Look at the decimal expansion of some number ending in a half. After n=1, the deciaml expansion is less than a half, so only for n=1 is it a integer.[/quote]

No, I don't think it is that obvious (if I understand what you are saying correctly).  Are you saying that the decimal part of a number with a decimal part of .5 raised to a power is always less than .5?  That is how I interpreted what you said, but that is not correct.  For example, consider $1.5^5=7.59375$.

Whoops! My bad!

But even still, look at the number of digits after the decimal point. If you can prove that a number ending in .5 to the power of n is always the same, no matter the integer part, then the sum must look like

0.500000...+0.5000000, contradiction. n=1

[quote="seamusoboyle"]Is it not obvious? Look at the decimal expansion of some number ending in a half. After n=1, the deciaml expansion is less than a half, so only for n=1 is it a integer.[/quote]

No.  By looking at my post, you may easily construct counterexamples for which $a^n+b^n$ is divisible by a power of $2^k$, $k$ arbitrary.

So Seamus, say we want to make $a^k+b^k$ divisible by $2^4$.  Pick $a,b$ such that $a+b+1=2^3$, say, $a=5,b=2$ and $k$ arbitrary but odd ($k=3$ works)

In this case we get $11^3+5^3=1456=2^4\cdot 7\cdot 13$


trigonometry inequalities algebra polynomial

Let $ m \in \mathbb{N}$
prove that
$ 1.cos{\frac{\pi}{2m\plus{}1}}cos{\frac{2\pi}{2m\plus{}1}}...cos{\frac{m\pi}{2m\plus{}1}}\equal{}\frac{1}{2^m}$
$ 2.cos{\frac{\pi}{2m}}cos{\frac{2\pi}{2m}}...cos{\frac{(m\minus{}1)\pi}{2m}}\equal{}\frac{\sqrt{m}}{2^{m\minus{}1}}$

[hide="hint"]The problem becomes fairly trivial if you put in form $ \cos{\theta}\equal{}\frac{e^{i\theta}\plus{}e^{\minus{}i\theta}}{2}$...[/hide]

Can you explain how to use hint?  :blush:

[quote="Anaxerzia"][hide="hint"]The problem becomes fairly trivial if you put in form $ \cos{\theta} \equal{} \frac {e^{i\theta} \plus{} e^{ \minus{} i\theta}}{2}$...[/hide][/quote]
Would you explain,please?  :omighty: Thanks

you should to selection your text and clik on 'HIDE'  :D

Consider the coefficients of the [url=http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html]Chebyshev polynomials of the first kind[/url].  :)

[hide="1"]Let $ \omega = e^{i \frac {2\pi}{2m + 1}}$

$ z^{2m + 1} - 1 = (z - \omega)(z - \omega^2)\cdots (z - \omega^{2m + 1})$

Dividing by $ z - 1$

$ 1 + z + \cdots + z^{2m} = (z - \omega)(z - \omega^2)\cdots (z - \omega^{2m})$

Subbing in $ z = - 1$, then taking modulus of both sides,

$ 1 = |z + \omega| |z + \omega^2| \cdots |z + \omega^{2m}|$

It can be shown that, $ |z + \omega^i| = 2 | \cos\left(\frac {i\pi}{2m + 1}\right)|$

$ 1 = \prod_{k = 1}^{2m} 2 | \cos\left(\frac {k\pi}{2m + 1}\right)| = \left( \prod_{k = 1}^{m} 2 \cos\left(\frac {k\pi}{2m + 1}\right) \right)^2$

$ \therefore \prod_{k = 1}^{m} \cos\left(\frac {k\pi}{2m + 1}\right) = \frac {1}{2^{m}}$
[/hide]

[hide="2"]
Let $ \omega = e^{i\frac {2\pi}{2m}}$

$ z^{2m} - 1 = (1 - \omega)(1 - \omega^2)\cdots (1 - \omega^{2m})$

Dividing both sides by $ z^2 - 1$

$ 1 + z^2 + (z^2)^2 + \cdots + (z^2)^{m - 1} = (1 - \omega)\cdots (1 - \omega^{m - 1}) (1 - \omega^{m + 1}) \cdots (1 + \omega^{2m - 1})$

Setting $ z = - 1$ then taking modulus of both sides,

$ m = \left( \prod_{k = 1}^{m - 1} 2 \cos \left( \frac {k\pi}{2m} \right) \right)^2$

$ \implies \prod_{k = 1}^{m - 1} 2 \cos \left( \frac {k\pi}{2m} \right) = \frac {\sqrt {m}}{2^{m - 1}}$
[/hide]


probability
geometry
number theory
complementary counting
conditional probability
AMC 10

Anyone from China on this forum taking the AMC 10 in 2009?

The first time take this test, and i am lost in those questions, can anyone give me some topics to review?

Well, as you're from China, I think you might want to focus on some basics of counting and probability as well as number theory (geometry and algebra will probably not be a problem); for example, complementary counting, balls-and-urns, conditional probability, the formula for the number/sum of divisors of a number, properties of mods (Fermat's little theorem/Euler's theorem are especially helpful for those "find last 3 digits"-type questions), etc.

Lol there's no point on teaching chinese people who like math how to study for AMC 10
most people could get 120+ on AMC 12 when they are grade 9


AMC
AIME
geometry
geometric transformation
reflection
symmetry
search

On the alternate AIME math jam, it was mentioned that a solution to #14 was possible using Menelaus and the Angle Bisector Theorem.   I can see how these theorems might be helpful, but am having difficulty relating the given congruent angles to the problem.  Please advise.

HINT: Reflect part of the triangle over the angle bisector. Then use menelaus.

I see it now! -- very clever.  What made you think of this?   I'm guessing it was the symmetry of the situation  -- probably the same kind of thinking leads to the solution of problem 12 using Pythagoras.

In the search for a "synthetic" technique, you generally need to 'stick' the angles together in one way or another (make an isosceles, make them coincident, etc.)


geometry
circumcircle
angle bisector
geometry solved

Let ABC be a triangle inscribed in cirle with center O and point
M inside it .AM,BM,CM cut (O) respectively at A1,B1,C1.A2,B2,C2 are
on the segments of BC,CA,AB and satisfy the condition
        angle : A2A1B=A2A1C;B2B1C=B2B1A;C2C1A=C2C1B
Prove that AA2,BB2 ,CC2 are concurrent

Well, A1A2 is an angle bisector of triangle BA1C, hence BA2 / A2C = BA1 / A1C. Now, if R is the radius of the circumcircle of triangle ABC, then BA1 = 2R sin BAA1 and A1C = 2R sin A1AC, hence BA2 / A2C = BA1 / A1C = sin BAA1 / sin A1AC. Similarly, CB2 / B2A = sin CBB1 / sin B1BA and AC2 / C2B = sin ACC1 / sin C1CB.

However, the lines AA1, BB1, CC1 are concurrent, and thus the trigonometric form of the Ceva theorem yields

  sin BAA1 / sin A1AC * sin CBB1 / sin B1BA * sin ACC1 / sin C1CB = 1.

Thus,

  BA2 / A2C * CB2 / B2A * AC2 / C2B = 1,

and the Ceva theorem proves the lines AA2, BB2, CC2 concurrent.

  Friendly,
  Darij Grinberg

yes,That's right!
Simple problem and solution


fairyofwind wrote:If 0<P(A)<1 and 0<P(B)<1

1. Must P(A and B) >= P(A)P(B)?
2. Must P(A and B) < P(A or B)?

1. [hide]No.  Counterexample: suppose that event A occurs if you roll a 1 or a 2 using a 6-sided die, and event B occurs if you roll a 3 or a 4.  Define P(x) as the probability that event x occurs after rolling 1 die.  P(A and B) = 0, since you cannot roll both a 1 or a 2 and a 3 or a 4.  But P(A)*P(B) = 1/3*1/3 = 1/9.  Since 0 is not :ge: 1/9, P(A and B) need not be :ge: P(A)P(B).[/hide]

2. [hide]No.  Suppose that A and B represent the same event.  Then P(A and B) = P(A or B), which contradicts P(A and B) < P(A or B).[/hide]

fairyofwind wrote:3. If P(A and B) >= P(A)P(B), show that P(A and B) < P(A or B).
4. If P(A and B) < P(A)P(B), show that P(A or B) < P(A and B) + 2(P(A) + P(B) - 2P(A)P(B)).

#3's pretty easy... haven't really looked at #4 yet. It isn't really easy to word the sol'n to 3 though, unless we're allowed to assume that P(A) and P(B) are independent and mutually exclusive. Spoiler: [hide]in fact, unless the events are mutually exclusive, there is a case in which that inequality does not hold: A=B. For example, Event A = rolling a 1 on a 6 sided dice = Event B. P(A and B) = 1/6 = P(A or B)[/hide]


Does anyone know of a webpage that defines all uses involving the :Sigma: sign? The only method I used to know was the one that was taught to me in algebra 2, the one with a max number on top and an iterator with an initial value on the bottom. HOwever, I have seen uses with no value on the top, no valuie on the bottom, more than one expression on the top/bottom, among others. I would just like a definitive guide, if such a thing is available on some web page. Thanks.

Fierytycoon

I don't know of a website, but here are some of the things I have seen.

Sometimes there are no bounds. This means that it starts from 1 (or possibly 0) and goes to infinity usually. It has to do with context. You probably won't see a sigma without bounds unless it's fairly clear what it means.

You might see something like :Sigma: _{p prime} 1/p. That means you only count the ones that are prime. (In this case it's 1/2+1/3+1/5+1/7+..., which happens to diverge.)

Sometimes instead of putting an upper and a lower bound, you'll see everything on the bottom. Such an expression would look like :Sigma: _{1 :le: n :le: 10} n, which would be 1+2+3+...+8+9+10=55.

Sometimes people get lazy and don't write the variable name if it is clear. So you might see something like :Sigma: _1 ^10 n, which would also mean 1+2+3+...+8+9+10=55.

Hope this helps.

Aside from the notation that you already know, where the numbers below and above the sigma mean the bounds of the summation; you can also just reference the set you are summing over underneath the sigma. For example, if I wanted to sum over all of the integers x between 0 and y, I could simply write 0 :le: x :le: y underneath the sigma. It is implicitly understood that the sum is taken over integers in this range.

It also makes for a compact way of expressing double sums (or more) using just one sigma. So if I wanted to sum over all pairs of integers (x,y) with say, 0 :le: x < y :le: 10; I could just write that under the sigma and again it's implicitly understood that this means integers in that range. You can also do a similar thing using multiple sigmas, where the order of operations is done from the rightmost sigma to the left. So for example the same summation I did above could be written equivalently as sigma sub 1 < y :le: 10 sigma sub 0 :le: x < y.

Another notation attributed to Einstein, and commonly used in physics and differential geometry, is to drop the sigma altogether and just sum over any repeated indices. For example x_i*y_i would mean to sum x_1*y_1 + x_2*y_2 + ... + x_n*y_n for whatever bounds your indices are understood to be over.

Here's a neat problem using this notation. Evaluate  :Sigma:_{m :ge: 0 n :ge: 0} 1/(2^m*3^n).

I do not know if these are "formal" uses of the sigma notation, but in AoPS 2 (and in other places) you can sometimes write sum only over d|n, or the numbers that are divisors of n (the same way Zeta defined it, I was just pointing out another way you could use sigma over specific numbers)

Also, cyclic sums (there was another topic somewhere about it, you can check that) use sigma notation, I believe.

Hmm, I just saw a sum resembling

n
 :Sigma: ak, where a is a constant
k=m

What would this mean?

This is equal to am+a(m+1)+a(m+2)+...+a(n-1)+an.

Oh, whoops! Of course that's right: m is simply a constant. Silly me. Sorry about that, don't know what I was thinking.

Fierytycoon


MATHCOUNTS
probability
geometry
3D geometry

Take approximately 25 minutes with a calculator.

Target

1. -1<x<4 -4<y<2 What is the probability that the product xy is positive?

2. f(x)=f(x-1)*f(x-2)-f(x-1) for all x>2. If f(1)=3 and f(2)=5, what is the value of f(6)?

3. A cylinder has height 7 and base radius 4. A cone has height 12 and base radius 5. What is the volume of the solid with a greater volume:surface area proportion? Express your answer in terms of pi.

4. A runner is practicing for a marathon. He runs 40 miles each day. Today, he ran 40 miles in 6 hours. If his pace for the first 5 miles was 12 mph, his pace for the next 10 miles was 7 mph, and his pace for the next 15 miles was 6 mph, then what was his speed for the final 10 miles?

5. A baseball pitcher can throw his fastball at 100 miles per hour. If he throws it from 60 feet away from home plate, how long does the hitter have before it goes by him? Express your answer to the nearest 100th of a second.

6. Quadrilateral ABCD is inscribed in circle P. If AB=17, BC=13, CD=13 and DA=7, then what is the area of circle P? Express your answer in terms of pi.

7. How many palindromic numbers less than 1000 are divisible by 6?

8. Jamie is drawing a path on a cartesian plane. He begins at 0,0 and only moves one unit right or up. If there is an equal probability of him drawing a segment up or right at each intersection, what is the probability his path passes 4,4?

Yiffie, did you edit my previous post?  I didn't know the test was being taken by pm's.

Oops. I thought you were still giving the test by PMs! I'm really sorry about the misunderstanding, so you can repost it now.  :blush:

I never got the pm.  Can someone tell me why?

and of course you bumped another 16 year old thread, and wasting your 200th post in doing so

Bruh! Stop bumping threads if you have nothing to contribute! It's not even funny! :mad:

[hide="idk might as well contribute"]
[hide="S7"]
Notice that there are $4$ options for the first digit, and $3$ for the middle, and one is fixed for the last, but we have to add one for $606$. That gives $13$ for 3-digit. For two-digit, we can examine, and the only one is $66$. Obviously one-digit is $6$. So the answer is $13+1+1=\boxed{15}$
[/hide]
can OP confirm? lol jk  :-D
[/hide]


[quote="akhil0422"]Please give me a solution for this problem; I know it's very simple, but I don't undertand the solutions manual solution. Sometimes the easy problems seem hard to me?

Find $ x$ if $ 2^{{16}^x} \equal{} 16^{2^x}$.[/quote]

$ 2^{{16}^x} \equal{} 16^{2^x}$ -- $ 2^{{16}^x} \equal{} {2^4}^{2^x}$
Now since the bases are equal..solve for x

$ 2^{16x}\equal{}16^{4x}$.

Don't believe it? Try it for yourself. $ 2^8\equal{}4^4$

So now we have $ 16^{4x}\equal{}16^{2x}$

We can take out of this $ 4x\equal{}2x$ which means that x=0.

I don't know; I keep getting infinite solutions? :?:

Can someone post a step-by-step solution?

Or you could do it this way:
$ 2^{16^x}\equal{}16^{2^x}$
$ 2^{16^x}\equal{}2^{4\cdot{2^x}}$
$ 2^{16^x}\equal{}2^{2^{x\plus{}2}}$
$ 16^x\equal{}2^{x\plus{}2}$
$ 2^{4x}\equal{}2^{x\plus{}2}$
$ 4x\equal{}x\plus{}2$
$ 3x\equal{}2$
$ x\equal{}\frac{2}{3}$

So the answer would be $ \frac{2}{3}$. I don't know if I did something wrong, but I guess 0 works too. So there might be two answers.

PowerofPi is correct.
akhil0422 - i kept getting infinite solutions too until i realized $ 2^{{16}^{x}}\equal{} 16^{2^{x}}$ reduces to $ 2^{16^{x}}\equal{}2^{4\cdot{2^{x}}}$ , not $ 2^{{16}^{x}}\equal{}{2^{4}}^{2^{x}}$.
Also, i don't think 0 works because then you get 2=16

Using what AOPS23 did, 2^16^x=2^[4(2^x)], so 16^x=4(2^x)16^x => (2^4)^x => 2^(4x)=2^2*2^x => 2^(x+2). Therefore, 4x=x+2. Solve this to get that [b]x = 2/3[/b] is the answer. Whew! Sorry, but I don't have Latex.

to rts2007 - i don't have latex either, but you can still use commands to make your solutions show up nicely: http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Commands


Can anyone help me with these.... sure it will be quite easy 4 u all... I already know the answer but can't understand why it is right... any help...

a) -2 -3 + 4x = 5 - 6x

x=?

b) 2x -1 + 4x -8 = 5x -9

x=?

In any single variable solving attempt, always isolate the variable. Just remember that whatever operation you perform on one side of the equation must be done on the other side of the equal sign.
A
$ \minus{}2\minus{}3\plus{}4x\equal{}5\minus{}6x$
Combine like terms
$ \minus{}5\plus{}4x\equal{}5\minus{}6x$
Add to both sides
$ 4x\equal{}10\minus{}6x$
Add again
$ 10x\equal{}10$
Divide both sides by 10
$ x\equal{}1$
See if you can do the other one now

[hide="If Not"]
$ 2x\minus{}1\plus{}4x\minus{}8\equal{}5x\minus{}9$

$ 6x\minus{}9\equal{}5x\minus{}9$ (Combining like terms)

$ 6x\equal{}5x$ (Adding 9 to both sides)

We see that the only case in which this is true is $ x\equal{}0$
[/hide]

where you say 'add both sides starting from here

-5 + 4x = 5 -6x

-5+5= 0
5+5=10
so i end up like this (if this makes sense)

-5 + 4x = 5 -6x
0 + 4x = 10 - 6x
4x = 10 -6x

or

-5 + 4x = 5 -6x
-10 + 4x = 6x
-10 = 2x
-5 = x

I don't get why you can add -5 + 5 and get 10?

Can you explain further or is it just me that is thick...?
p.s. havent looked at the other one yet... hoping I can get it on my own:-)

[quote="electerr"]
-5 + 4x = 5 -6x
-10 + 4x = 6x
-10 = 2x
-5 = x
I don't get why you can add -5 + 5 and get 10?

Can you explain further or is it just me that is thick...?
p.s. havent looked at the other one yet... hoping I can get it on my own:-)[/quote]

There is a mistake on your second line the negative sign needs to be kept in front of the $ 6$.

You add $ 5$ to both sides not $ \minus{} 5$, so step by step it would look like this
$ \minus{} 5 \plus{} 4x \equal{} 5 \minus{} 6x$
$ \plus{} 5 \qquad \plus{} 5$
$ 4x \equal{} 10 \minus{} 6x$
       $ \plus{} 6x \qquad \plus{} 6x$
$ 10x \equal{} 10$
$ \frac {10x}{10} \equal{} \frac {10}{10}$
$ x \equal{} 1$

(Sorry if there are any spacing problems)

Some exercise problems for you to do:

1. $ 9 \minus{} x \equal{} 6 \plus{} 2x$
2. $ 12 \plus{} 7x \equal{} 6x \plus{} 14$
3. $ 5 \minus{} 3x \equal{} 3 \plus{} x$

Ok, I think I am starting to get this but I am still a bit unsure...
I think that most of my problems are comming from not knowing if I am suppose to add or subtract and which number to add or subtract from which... Are there rules that say how this is suppose to work?
I did the 2nd prob I posted and it turned out like this...

2x - 1 + 4x - 8 = 5x - 9
6x - 9 = 5x - 9
6x = 5x
-5x  -5x
x = 0

I think that that worked out right...
and I did the other problems that you posted...

9 - x = 6 + 2x
-6       -6
3 - x = 2x
   +x   +x
3 = 3x
x = 1

12 + 7x = 6x + 14
-12                 -12
7x = 6x + 2
-6x  -6x
x = 2

5 - 3x = 3 + x
-3         -3
2 - 3x = x
  +3x   +3x
2 = 4x
x = 2

I hope these are right... Ill be holding my breath to find out...
Do you have any more exercises or know where I might find a good site on the net that has ?'s and the answers so I can check if I am right?

Thanks a lot for all your help!
If your curious I am starting school soon and havent studied math in 9yrs! So I am trying to refresh my memory and all I got is a book which dosent explain everything so well... So your help is greatly appreciated!! Thanks again!![/b]

[quote="electerr"]
5 - 3x = 3 + x
-3         -3
2 - 3x = x
  +3x   +3x
[b]2 = 4x
x = 2[/b][/quote]
You did all of that problem right until the very end. If $ 2\equal{}4x$, then $ \frac{2}{4}\equal{}\frac{4x}{4} \iff \frac{1}{2}\equal{}x$. All the other ones are right though, so I'm guessing that was just a mistake?

Uh... I guess I'll just make some problems up. (Each line is a new problem.)

$ 4x\minus{}7\equal{}3x\\
10x\plus{}13\equal{}14x\plus{}17\\
3x\minus{}4\equal{}8\minus{}x\\
5\minus{}5x\equal{}4\minus{}4x$

[hide="To Check Your Answers"]

1. $ 4x \minus{} 7 \equal{} 3x$

First we get the $ x$ terms all on one side.

$ \minus{}7 \equal{} \minus{}x$

$ 7 \equal{} x$

2. $ 10x \plus{} 13 \equal{} 14x \plus{} 17$

We get the $ x$ terms on one side and the constants on the other.

$ 10x \minus{} 14x \equal{} 17 \minus{} 13$

$ \minus{}4x \equal{} 4$

$ x \equal{} \minus{}1$

3. $ 3x \minus{} 4 \equal{} 8 \minus{} x$

$ 3x \plus{} x \equal{} 8 \plus{} 4$

$ 4x \equal{} 12$

$ x \equal{} 3$

4. $ 5 \minus{} 5x \equal{} 4 \minus{} 4x$

$ 5 \minus{} 4 \equal{} \minus{}4x \plus{} 5x$

$ 1 \equal{} x$[/hide]

Sorry, I don't quite understand your question. Can you please rephrase it? Thanks.

In the second one, you are doing

[quote]10x + 13 = 14x + 17
14x - 10x = 17 - 13 [/quote]

If you subtract 10x AND 13 in the same line, you should get

$ 0\equal{}14x\plus{}17\minus{}10x\minus{}13$

However, you are moving the 17-13 to the other side. To fix this problem, you should just subtract to isolate the constant and the variable. So, instead of subtracting one whole side from the equation, you should subtract 10x (left hand side) and 17 (right hand side) to get

$ 13\minus{}17\equal{}14x\minus{}10x\equal{}\minus{}4\equal{}4x$

From here, you divide by $ 4$ to get $ x\equal{}\minus{}1$

Hope that made sense

I think the problem here is with signs.

For example, 10x+4 is the same as 4+10x, but 10x-4 is different than 4-10x.

Generally, if you combine like terms before proceeding to solve, it will eliminate a lot of this problem, but the other rule you should keep in mind is that

Signs automatically "stick to" the following term.

[quote="electerr"]One pattern that I think I am noticing that is causing me to make mistakes sometimes (I think) is that when moving x terms and constants from one side to the other that the constant or x term that is not being moved always holds its place in the equation.

i.e.
10x + 13 = 14x + 17
10x - 14x = 17 - 13

right, because when moving 14x then 10x keeps its place (in this case 1st) in the equation...

instead of...

10x + 13 = 14x + 17
14x - 10x = 17 - 13

wrong, because when moving 14x then 10x changes places...

is this a rule for all problems or....

Thanks[/quote]
1) Try to think of the minus sign as belonging to the number. That way you are always adding terms and you can change the order.
17 - 13 is not the same as 13 - 17, but if you think of it as 17 + (-13), you can see that
17 + (-13) = (-13) + 17, which you would just write as
17 - 13 = -13 + 17 to save the trouble of writing so many parentheses.

2) Talking about "moving" terms around, just confuses the issue. Look at it a different way. What you are doing is adding the same number to the expressions on both sides of the equal sign, and then simplifying, doing several steps at once. so from
10x + 13 = 14x + 17
you add (-13) and (-14x) to both sides to get
10x + 13 - 14x - 13 = 14x + 17 - 14x - 13 and then you simplify it to
10x - 14x = 17 - 13

Finally got it!! Thanks for everyones help!! The last post was very helpful!!


Suppose you want to place $N$ distinguishable books on two bookshelves. Suppose that each of the two must contain at least $M$ books, where $1 \le M < \frac{N}{2}$. How many ways can you place the $N$ books?

How about the same problem if there are $p$ bookshelves and each must contain $M$ books? Under what conditions can this problem be solved?

[hide="failedattempt"]Let the number  of books on the shelves be $x$ and $y$.  Then $x+y=N$, and $x>M-1$, and $y>M-1$.  Then $x=M-1+a$ and $y=M-1+b$ for some integers $a,b,M\geq1$.  Substituting we have $M-1+a+M-1+b=N$, so $2M+a+b=N+2$  Hmmm...I know how to get the number of solutions to this but still need to factor in the fact that the books are distinguishable... :? [/hide]

For two bookshelves, first consider how many you'll place on each bookshelf.  Then consider every position different.  So  you have $N!$ permutations.  the number of ways to choose the books is just $\binom{N}{M}$.  So that makes it $\frac{(N!)^2}{M!(N-M)!}$.

Perhaps I need to reword what I am asking.

Find the number of ways to distribute the books on the bookshelves - not arrangement. Just distribution.

Shouldn't it be $M < \frac{N}{2}$?

But anyway, I think you first put $M$ on each so you have $N-2M$ left. Then each book has $2$ places to go: the first bookshelf or the second. So it's $2^{N-2M}$?

And for $p$ bookshelves, it's $p^{N-pM}$ I guess.

but the books are distinguishable.let the ]  be bookshelves' seperator.we have 3 books and 2 bookshelves.for n=3 m=1
A]BC,A]CB,AC[B,CA[B then 4 different cases we have but  $2^{n-2m}$ gives us 2 diffenent cases.

Sorry you are right it is $ \frac{N}{2}$. But, the problem wouldn't make sense otherwise...  ;)

i think you need to multiply 2m choose m and n choose 2m to paladin's solution, then it should be good

edit: hm that still doesn't meet gunes' example

Here was my thinking tho:

You first choose the 2m out of n to put on the bookshelfs first. Then you choose the n that will go on the first bookshelf. The rest go on the other bookshelf. THen you use paladin's number for the rest of the n-2m

actually i think gunes is wrong in a way

A]BC and A]CB should be the same.

However, there are still 6 ways in that case.

A]BC
AB]C
B]CA
BC]A
C]AB
CA]B

there are $\binom{N}{M}$ ways to choose M books for the first shelf
there are $\binom{N-M}{M}$ ways to choose M books for the second shelf out of the ones that remain
there are $2^{N-2M}$ ways to distribute the remaining books between the two shelves.
together they give:
$2^{N-2M}\binom{N}{M}\binom{N-M}{M}$ arrangements

with $p$ shelves, solutions are only possible when $M\le \frac{N}{p}$ in which case we get:
$2^{N-pM}\binom{N}{M}\binom{N-M}{M}\binom{N-2M}{M}...\binom{N-(p-1)M}{M}$ arrangements

well, that gives 12 for the case n=3 and m=1, and i only found 6 when listing them out

maybe that way makes ordering unique?

cuz lets say its ABC

You choose A to go on shelf one, B to go on shelf two, C to go on shelf one.

That's the same as C on 1 first, B on 2, and A on shelf 1 last.

L_Li is correct, according to what I was asking.


[quote="ehsan2004"]find all [b]m,n,l[/b] in natural number such that :
[b]m+n=gcd(m,n) 2 , n+l=gcd(n,l) 2 , m+l=gcd(m,l) 2 .[/b][/quote]

Well, nice proble, Is simple to show that $m+n=gcd(m,n)^2$ iff exists integers $\alpha$ and $\beta$  such that
$(\alpha m+\beta n)^2=m+n$, so maybe this problem is a good start to solve the problem.

Best Regards.

RX the Crazy Math

sorry but i dont agree, i think a good way to construct such numbers is taking integers $m',n'$ such that $(m',n')=1$. and then $m=m'(m'+n')$ and $n=n'(m'+n')$ satisfy $m+n=(m,n)^2$ in fact, it gives all those couples.

I have solved this problem before, so i will try to remember

write $m=m'pqr$, $n=n'psr$, $l=l'qsr$, such that: $r=(m,n,l)$, $pr=(m,n)$, $qr=(m,l)$ and $sr=(n,l)$ then we must have:

$m'q+n's=pr$, $m'p+l's=qr$ , $n'p+l'q=sr$ and the truth i dont know how to finish this

[quote="Pascual2005"]sorry but i dont agree, i think a good way to construct ....   it gives all those couples.
[/quote]
Sorry, but I can't see WHY they form all couples?

RX

[quote="Pascual2005"]because let $d=(m,n)$ then $m=m'd$, $n=n'd$ with $(m',n')=1$. and we must have $m'd+n'd=d^2$ from where my claim follows.[/quote]
Well   :? mmmmm  :blush:   you are right!!!! :blush:


function
real analysis
real analysis solved

if $f_1$'s smallest positive period is $1$,and $f_2$'s smallest positive period is $a$,and $a$ is a  irrational number.
prove $f_1+f_2$ is not a period function

This looks very strange. Are you positive the functions are not continuous or something? Because I think it's not true in general.

Are you sure?  :lol:

Consider the Hamel basis $H$ on $\mathbb{R}$ over $\mathbb{Q}$ containing elements 1 and $a$.
Now define $f_1(x)=-\{y_1\}+y_2a+\sum x_i\alpha_i$ and $f_2(x)=y_1-a\{y_2\}-\sum x_i\alpha_i$, where $x=y_1+ay_2+\sum x_i\alpha_i$, $\alpha_i\in H$, $y_1,y_2,x_i\in \mathbb{Q}$.
Then $f_1(x)+f_2(x)=[y_1]+a[y_2]$. So the arbitrary element $\alpha\in H$ distinct from 1 and $a$ is a period of $f_1+f_2$.

I knew it :). I had a proof that we could construct a counterexample, but it was harder to explain than this, and it was not a really explicit construction :).

I think theres some condition.This type of problem came in one of the tests in our mathematical olympiad training camp.I think a problem like this also came in the olympiad of some country like China,Romania or Iran.I am not very sure..
There the version was stronger .1 being a period of f1 is not important.Any rational no will do.

What does that mean? :? Do you mean to say they're continuous?

[quote="zhaobin"]mistake again.
$f_1,f_2$ are both sequence.[/quote]

What do you mean $f_1,$f_2 are sequences...

[quote="Myth"]How could irrational number $a$ be a period of a sequence?[/quote]
I dont think they are sequences.They must be [b]functions[/b]..
The problem didnot explicitly say what $f_i are.
In mathematical logic we call these type of problems as "not well-posed problems"..Sorry I didnot mean to upset zaobin,but its just a standard terminology in mathematical logic.

I've see zhaobin use the term "sequence" to describe a function in another thread as well (in the Algebra section, I think). I really think that by "sequence" function (:)) he means continuous function.

[quote="grobber"]If the function is continuous, then the problem is very easy, so it would be more than helpful :).[/quote]
yes.you are right.sorry for my poor english. ;)  ;)
but can you kindly post the solution? :)  :)  :)
thank you

Assume $f_1+f_2$ has some period $T>0$. We then have $g(x)=f_1(x+T)-f_1(x)=f_2(x)-f_2(x+T)=h(x),\ \forall x$. However, $g(x)$ has $m$ as its period for all $m\in\mathbb Z$, while $h(x)$ has $na$ as its period for all $n\in\mathbb Z$. This means that the values $na,\ n\in\mathbb Z$ are periods for $g$, which means that $g$ is a continuous function having all the values $m+na,\ m,n\in\mathbb Z$ as its periods.

Now try to show that a continuous function having a dense set of periods is, in fact, constant (it's not hard, and I won't prove it here :)). We have thus found that $g(x)=f_1(x+T)-f_1(x)=c,\ \forall x$, for some constant $c$. Since $f_1$ is continuous and periodic, it must be bounded on $\mathbb R$, so $c=0$, and $T$ is a period of $f_1$. This means that $T\in\mathbb Z$, because $1$ was a principal period of $f_1$. In exactly the same way we show that $T$ is a period of $f_2$, so it must be of the form $an,\ \n\in\mathbb Z$, and from these two facts we get a contradiction, since the only common element of the sets $\mathbb Z,a\cdot\mathbb Z$ is $0$.


y = sin(f(x,i)) such that the maximum value for y only occurs at x=i*n for every n
such n is a natural number and x a real number.

we could have
Maxima of sin(x) = Pi/2 + 2x*Pi
Maxima of sin(ax) = (Pi/2 + 2x*Pi)/a
If I want the nth maximum to be the nth multiple of i, that is to say,
(pi/2 + 2n*Pi)/a = n*i

If I solve this for a, then
a=(pi/2 + 2n*Pi)/(n*i)
However, if I graph this function
y=sin(((Pi/2 + 2x*Pi)/(x*i))x)
I do not get the expected values at all??


calculus
integration
trigonometry
algebra
partial fractions
calculus computations

Hi,

Who know how to integrate explicitly
$\int\sqrt{\tan x}\,dx$

It's a monster, I'll try and walk you through it a bit:

1.  Substitute $u=\tan x$
2.  Substitute $u=x^2$
3.  Partial fractions

i am not getting by the method blahblah blah said.

can anyone post the complete solution.   how  can we substitute for  x two times  ?

hello, setting $ t\equal{}\sqrt{\tan(x)}$ we get $ x\equal{}\arctan(t^2)$ and $ dx\equal{}\frac{2t}{1\plus{}t^4}\,dt$, so our integral is $ 2\int\frac{t^2}{1\plus{}t^4}\,dt$.
Sonnhard.

$ \tan x \equal{}t^2 \\ \\
\sec^xdx \equal{} 2tdt \\ \\
(1\plus{}\tan^2x)dx\equal{}2tdt \\ \\
dx\equal{}\frac{2tdt}{1\plus{}t^4} \\ \\
\int\sqrt{\tan x} dx \equal{} \int\frac{2t^2dt}{1\plus{}t^4}$

[hide="And"]$ \int \frac{2t^2}{1\plus{}t^4}\,dt \equal{} \int \frac{2}{t^2 \plus{} \frac{1}{t^2}}\,dt \equal{} \int \frac{1 \minus{} \frac{1}{t^2}}{\left(t \plus{} \frac1t\right)^2 \minus{} 2}\,dt \plus{} \int \frac{1 \plus{} \frac{1}{t^2}}{\left(t \minus{} \frac1t\right)^2 \plus{} 2}\,dt$

$ \equal{} \int \frac{1}{u^2\minus{}2}\,du \plus{} \int \frac{1}{v^2 \plus{} 2}\,dv$[/hide]


Does no one have a solution to my problem?

[quote="SplashD"]For any positive integer $ N$, prove that
\[ \sum_{i \equal{} 1}^{N} (2 \plus{} i)^{N} \le (3 \plus{} N)^{N}
\]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=158897


calculus
integration
derivative
function
real analysis
real analysis theorems

I don't really understand how the integration power rule is derived from the differentiation power rule. Can someone post a proof?

If $ \frac {d}{dx}[x^n] = nx^{n - 1}$ for any $ n$, then $ \int x^n \; dx = \frac {x^{n + 1}}{n + 1} + C$, for $ n \neq - 1$

Proof: $ \int x^n \; dx$ stands for the class of functions whose derivative with respect to $ x$ is $ x^n$. By the sum rule, constant multiple rule, and power rule for differentiation,
$ \frac {d}{dx}\left[\frac {x^{n + 1}}{n + 1} + C\right] = \frac {1}{n + 1}\frac {d}{dx}[x^{n + 1}] + 0 = \frac {1}{n + 1}[(n + 1)x^n] = x^n$, provided $ n \neq - 1$.

Hence, $ \frac {x^{n + 1}}{n + 1} + C$ is contained in the class of anti-derivatives of $ x^n$.

To show that these are the only anti-derivatives of $ x^n$, let $ g$ be some anti-derivative of $ x^$ and let $ h(x) = \frac {x^{n + 1}}{n + 1} - g(x)$.

Then $ \frac {dh}{dx} = x^n - x^n = 0$ and so, by the Mean-Value Theorem, $ h(x) =$ constant. So all anti-derivatives of $ x^n$ differ by a constant, and so $ \frac {x^{n + 1}}{n + 1} + C$ are the only anti-derivatives of $ x^n$.

I think I didn't present my case properly. I don't want a proof per say, I want a deriviation :D


LaTeX
inequalities unsolved
inequalities

Let A, B diferent positives integers. Find the minimun value of a-b (a>b) iff
(A+B)/2, (AB)^1/2 and 2AB/(A+B) are integers too.

Wait your solutions...

let's latex: [quote="conejita"]Let $a>b$ different positives integers. Find the minimun value of $a-b$ iff $\frac{A+B}{2}, \sqrt{AB}, \frac{2AB}{A+B}$ are integers too. [/quote]


Triangle $ ABC$ has $ D$ on $ AB$ such that $ AD: DB\equal{}1: 3$, $ E$ on $ BC$ such that $ EC: EB\equal{}1: 4$, and $ F$ on $ DE$ such that $ DF: FE\equal{}1: 5$.  $ G$ is the intersection point of $ BF$ and $ AC$.  Find the values of $ BF: FG$ and $ AG: GC$.

A solution using mass points would be nice.

[quote="cosinator"]Triangle $ ABC$ has $ D$ on $ AB$ such that $ AD: DB \equal{} 1: 3$, $ E$ on $ BC$ such that $ EC: EB \equal{} 1: 4$, and $ F$ on $ DE$ such that $ DF: FE \equal{} 1: 5$.  $ G$ is the intersection point of $ BF$ and $ AC$.  Find the values of $ BF: FG$ and $ AG: GC$.

A solution using mass points would be nice.[/quote]

[hide]Let's use [url=http://mathworld.wolfram.com/BarycentricCoordinates.html]barycentric coordinates[/url](equivalent to mass points)

$ A \equal{} (1,0,0)$
$ B \equal{} (0,1,0)$
$ C \equal{} (0,0,1)$
So $ D \equal{} (3/4,1/4,0)$. (See the equivalence? Cause $ D \equal{} 3/4A \plus{} 1/4B$)
$ E \equal{} (0,4/5,1/5)$
This means $ F \equal{} 5/6D \plus{} 1/6E \equal{} (5/8,41/120,1/30)$
Alternatively, we can write $ F \equal{} tB \plus{} (1\minus{}t)G$(because F is on the line $ BG$). Now, see that $ F$ has 2nd coordinate $ 41/120$, See that $ B$ has second coordinate $ 1$. This means that $ F$ is distanced $ 79/120$ from $ B$, so $ BF/FG \equal{} 79/41$. So $ F \equal{} 41/120B \plus{} 79/120G$. $ F \minus{} 41/120B \equal{} (5/8,0,1/30)$. So $ (5/8,0,1/30) \equal{} 79/120G \implies G \equal{} (75/79,0,4/79)$ and so $ AG: GC \equal{} 4/75$.
[/hide]

Does anyone have a solution that explicitly uses Mass Points?

That's the method that I'm trying but I'm not making progress.

[hide="hint"]Let $ A,D,B,E,C$, have weights of $ 75, 100, (25\plus{}4), 20,$ and $ 16$ so that everything balances nicely. Note the split weight at $ B$.[/hide]
[hide="more"]What is the center of weight of the system?  What do we get when we add $ 75A \plus{} (25\plus{}4)B \plus{} 16C$?[/hide]
[hide="more"]We get $ 75A \plus{} (25\plus{}4)B \plus{} 16C \equal{} (75A \plus{} 25B) \plus{} (4B \plus{} 16C) \equal{} 100D\plus{} 20E \equal{} 120F$.
Thus,
$ 75A \plus{} 16C  \equal{} 120F \minus{} (25\plus{}4)B$.[/hide]

Okay, then what I did originally may in fact be correct, although it yields a different answer than what agolsme got and I don't see any mistake in algolsme's solution.

[hide]I assigned masses of a, c, 4a/3, 5c/4, and a/3 + c/4 to A, C, D, E, and B respectively. In order for F to be the balancing point of DE we must have 5(5c/4) = 4a/3. So c:a = 16/75.

Since F is the balancing point of BG, the mass of G must equal the mass of F minus the mass of B. So the mass of G is (4a/3 + 5c/4) - (a/3 + c/4) = a + c, which implies that G is in deed the balancing point of AC.

Therefore AG:GC = c:a = 16/75.[/hide]

This is the error:[quote="agolsme"]$ E \equal{} (0,4/5,1/5)$[/quote]  $ E$ is supposed to be much closer to $ C$, not to $ B$.

[quote="mysmartmouth"]After assigning all the masses, I got
[hide]
AG:GC = 16/75
and
BF:FG = 91/29
[/hide]

Does that seem right for the second ratio?[/quote]
My solution gives that as well.

Just curious, what would be the best method to solve this type of problem without barycentric coordinates or mass points?

You can do it with similarity by drawing a bunch of parallel lines, which requires clever observation, which can be annoying when you are geometrically impaired. To the contrary, mass points are great: even you suck at geometry, you still can do these problems using them.

[quote="archimedes1"]Argh, my solution was eaten!  I'm not going to type it up again, but look at [url]http://www.sanjosemathcircle.org/handouts/20071114.pdf[/url] Problem 3, Section 5.1, page 5.[/quote]
 :o  The San Jose Math Circle website was just "updated," and this handout is not there anymore!   :(

I believe that area ratio chasing can also be used to solve these types of problems.


[quote="bertram"][quote="razad"](10^3/4 times 4^3/4)^-4[/quote]

[hide="Correct solution"]
$ \left(10^\frac {3}{4}\cdot 4^\frac {3}{4}\right)^{ \minus{} 4} \equal{} \left(10^\frac {3}{4}\right)^{ \minus{} 4}\cdot\left(4^\frac {3}{4}\right)^{ \minus{} 4} \equal{} 10^{ \minus{} 3}\cdot 4^{ \minus{} 3} \equal{} \frac {1}{40^3} \equal{} \boxed{\frac {1}{64000}}$
[/hide][/quote]

Can't think today... :(


Thought I'd share  :D !

[url]http://thrillingwonder.blogspot.com/2006/11/new-breakthroughs-in-mathematics.html[/url]

Yeah it was pretty funny. :rotfl:  :rotfl:  :rotfl:


limit
logarithms
algebra
polynomial
real analysis
calculus
calculus computations

Hello, two infinite products to play with:
\[ \lim_{n\to\infty}\prod_{k \equal{} 1}^{n} (1 \plus{} k/n^2)
\]
and
\[ \lim_{n\to\infty}\prod_{k \equal{} 1}^{n} (1 \plus{} k/n)^{1/k}
\]
cheers

Well I don't  know if there is some sort of formula for infinite products like there is for infinite sums but in the first one, the limit as $ n\rightarrow\infty\equal{}0$.  Of course, that is for each individual term so that may not be correct.

The first product is $ \frac{(n^2\plus{}n)!}{(n^2)!\,n^{2n}}\sim e^{\minus{}n}\to 0$ by [url=http://en.wikipedia.org/wiki/Stirling%27s_approximation]Stirling's formula[/url].

The second limit equals $ e^{\pi^{2}/12}$.

Simply take the logarithm and observe that the sequence in question is the Riemann sum of

$ x\rightarrow \frac{\ln(1\plus{}x)}{x}$ over the interval $ [0,1]$, with the division and the system of intermediary points given by $ k/n$.

[quote="JRav"]Of course, that is for each individual term so that may not be correct.[/quote]

In the case of infinite (rather than finite) products, it is not true that $ \lim \prod f(k) \equal{} \prod \lim f(k)$.  For example, clearly $ \lim_{n \to \infty} 1 \plus{} \frac {1}{n} \equal{} 1$.  On the other hand,

$ \lim_{n \to \infty} \prod_{k \equal{} 1}^{n} \left( 1 \plus{} \frac {1}{n} \right) \equal{} e$

didilica:  can we apply a similar idea to the first limit?  The logarithm is

$ \lim_{n \to \infty} \sum_{k\equal{}1}^{n} \ln \left( \left( 1 \plus{} \frac{k}{n^2} \right)^n \right) \frac{1}{n}$

and of course $ \lim_{n \to \infty} \ln \left( \left( 1 \plus{} \frac{k}{n^2} \right)^n \right) \equal{} \ln e^{ \frac{k}{n} } \equal{} \frac{k}{n}$, so the logarithm of the desired sum is the Riemann sum of $ x$ from $ 0$ to $ 1$, which is $ \frac{1}{2}$.

Hence the original product is $ e^{ \frac{1}{2} }$.

[quote="Timestopper_STG"]I think it obvious that the first product is $ \geq1$  :?:[/quote]
Indeed. $ \frac{(n^2\plus{}n)!}{(n^2)!\,n^{2n}}\sim (1\plus{}1/n)^{n^2\plus{}n}e^{\minus{}n}$. Taking the logarithm and using the Taylor series for $ \log$, find $ (n^2\plus{}n)\log(1\plus{}1/n)\minus{}n\equal{}(n^2\plus{}n)(1/n\minus{}1/(2n^2)\plus{}\dots)\minus{}n\to\frac12$. Thus, the limit is $ \sqrt{e}$. Thanks for the correction.

This is how I would solve the first problem:

Let $ x_{n}\equal{}\sum\limits_{k\equal{}1}^{n}\frac{\ln(1\plus{}k/n^{2})}{k/n^{2}}\frac{k}{n^{2}}$.

Then

$ \min\limits_{k}\frac{\ln(1\plus{}k/n^{2})}{k/n^{2}}\sum\limits_{k\equal{}1}^{n}\frac{k}{n^{2}}\leq x_{n}\leq \max\limits_{k}\frac{\ln(1\plus{}k/n^{2})}{k/n^{2}}\sum\limits_{k\equal{}1}^{n}\frac{k}{n^{2}}$

Take the limit and the result follows as an application of the Squeeze Theorem.

[quote="t0rajir0u"]
and of course $ \lim_{n \to \infty} \ln \left( \left( 1 \plus{} \frac {k}{n^2} \right)^n \right) \equal{} \ln e^{\frac {k}{n} } \equal{} \frac {k}{n}$, so the logarithm of the desired sum is the Riemann sum of $ x$ from $ 0$ to $ 1$, which is $ \frac {1}{2}$.

Hence the original product is $ e^{\frac {1}{2} }$.[/quote]

How did you manage to take a limit as n tends to infinity and obtain k/[b]n[/b] as your result?

Whoops, that's a little problematic, isn't it?  :P

I realize there's some problems with rigor in that step; didilica's method is probably the preferred one in that case, since it's essentially what I would need to do.

It's remarkable that you still managed to be correct though.  :P

In first problem We can easy solve without stirling numbers and e.t.c. ()
Let $ A_{n} \equal{} \prod_{k \equal{} 1}^{n}(1 \plus{} \frac {k}{n^{2}})$
then
$ \ln A_{n} \equal{} \sum_{k \equal{} 1}^{n}\ln(1 \plus{} \frac {k}{n^{2}}) \equal{} \sum_{k \equal{} 1}^{n}\frac {k}{n^{2}} \minus{} \frac {k^{2}}{2n^{3}} \plus{} o()\approx \frac {1}{2}(\frac {1 \plus{} n}{n}) \plus{} \frac {Cn^{3}}{n^{4}} \plus{} o(\frac {1}{n})$
hence $ \lim_{n \minus{} > \infty}A_{n} \equal{} \sqrt {e}$
in seccond problem I agree with didilica (there also is another way ) it's one of the best way by which can be solved problems like this.

How can we show that $ \sum\limits_{k \equal{} 1}^n {k^p }$is a polynomial in n of degree p+1? Those are the sums used to compute the series.

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

See also http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=178381 .


analytic geometry
advanced fields
advanced fields theorems

Problem Statement:
First, make a coordinate transform and write the following PDE in canical form.  Second, solve the resulting equation:

$ u_x \plus{} u_y \equal{} u$

Omitting several steps:

$ (\xi_x \plus{} \xi_y) u_{\xi} \plus{} (\eta_x \plus{} \eta_y) u_{\eta} \minus{} u \equal{}0$

Letting $ A \equal{} (\xi_x \plus{} \xi_y)$ and $ B \equal{}(\eta_x \plus{} \eta_y) \equal{} 0$

I solve the initial curve problem $ \eta_x \plus{} \eta_y \equal{} 0$ and $ \eta(x_0,y) \equal{} y$ and pick $ \xi \equal{} x$

I arrive at the required form...  (I think is right):

Question 1:
$ u_{\xi} \minus{} \frac{1}{A} u \equal{}0$
Is this correct?

Question 2:
Is $ u_{\xi} \equal{} \frac{1}{A} u$ directly integrable w/r/t/ $ \xi$ ?

Yes, I have $ \xi_x\equal{}1$ and $ \xi_y \equal{} 0$

The equation $ u_{\xi} \equal{} u$ has solutions $ u(\xi,\eta) \equal{} g(\eta)e^{\xi}$, where $ g$ is any continuously differentiable function. (You need this assumption on $ g$ to justify the preceding computations with the chain rule.)

Working through the problem I found $ \eta \equal{} g(\eta) \equal{} g(y\minus{}x)$ and so we have

$ u(x,y) \equal{} g(y\minus{}x)e^{x}$

I understand the problem and the solution...  Thank you.


What are the first three digits after the decimal point in (10^2002 + 1)^(10/7)? You may use the extended binomial theorem: (x + y)^r = x^r(1 + 4 (y/x) + r(r-1)/2! (y/x)2 + r(r-1)(r-2)/3! (y/x)3 + ...) for r real and |x/y| < 1

um...unless I'm missing something....isn't this just a little computation?

I suppose that means to the power of 10/7 (and r) instead of times 10/7 (and r)..

Sorry guys, exponents don't carry over.

ANyway, check the revised problem
and forget about the last part.

i don't know the correct revision for it....


Let $ ABC $ be a triangle with $ AB = AC $ . On the perpentdicular to $ AC $ at $ C $ , take a point $ D $ such that : $ B , D $ lie on different sides of the line $ AC $ . Let $ K $ be the point of intersection of the line perpentdicular to $ AB $ at $ B $ and the line passing through the midpoint $ M $ of $ CD $ , perpentdicular to $ AD $ . Compare the measures of $ KB , KD $.

this is not worded very well. perhaps you could tell me the source and ill go look at it.


inequalities
Ring Theory
inequalities unsolved

Suppose that $a,b,c$ are three non-negative real numbers so that $a+b+c=1$. Prove that $a^{2}b+b^{2}c+c^{2}a\leq \frac{4}{27}$.

Wow, thanks Sung-yoon Kim. Your post made me look at the inequality in the correct way. They're so much easier when they are that sharp. :D

Is PID theorem in Hojoo Lee's note good?

[quote="mathematica"][quote="Sung-yoon Kim"]Moreover, there is a sharper one: $a^{2}b+b^{2}c+c^{2}a+abc \leq \frac{4}{27}$.[/quote]
How is it proved?[/quote]
Try it:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100960


abstract algebra
superior algebra
superior algebra unsolved

If $ A$ is a commutative ring with unit, and $ M$ is a noetherian $ A$-module, must it be true that $ \textrm{Ass}_A(M)$ is a finite set?

(Note that the answer is positive if we assume that $ A$ is noetherian and $ M$ a finite $ A$-module.)

Can you even show under your premises that $ Ass_A(M)$ is non-void iff $ M$ is non-trivial? I think this should be fulfilled to get some kind of control on the associated primes of $ M$.


x,y are positive integers. Find all (x,y) satisfying
$(y+1)^{x}-1=y!$


Determine all $(n,k) \in \mathbb{N}^2$ such that $3^{n-1} + 7^{k-1}$ is prime.

(1,1) :?

elseway it's even as you said :?

Is it such a tricky question, or is this one of the things that I'd have in beginner's section? :)

Hi,

I guess the only solutions are $n = k = 1$, as for any other possible combinations of $n$ and $k$, the sum is even and is greater than $2$.

Peter, could you please explain to me your answers? You said that the solutions are
$n = 1, k=0$;  $n = 0, k=1$; $n = 0, k=0$

But if either of $n$ or $k$ is zero, then the sum is not an integer. So what do you mean by prime then?

Thanks,
Sambit

ya, sumtin confused me in the notation. (It's even easier then I thought.)
Still this seems a bit very easy to be possible :?