If we find it difficult to directly differentiate always factorize .  Your question can be written as follows :

 $ y \equal{} (x^{2}\plus{}x\plus{}1)(\frac{x^{2}\minus{}x\plus{}6}{x^{2}\plus{}1}) \equal{} ( x^2 \minus{} x \plus{} 6 ) \plus{} x \plus{} \frac{ 5x\minus{}x^2 }{x^2 \plus{} 1}$

Now differentiate it term by term.

$ \equal{} 2x \minus{} 1 \plus{} 1 \plus{} \frac{ \minus{}5x^2 \minus{} 2x \plus{} 5 }{ (x^2 \plus{} 1)^2 }$

Please check my arithmetic and do  correct me if its wrong .


[b]Έστω Α και Β πεπερασμένα υποσύνολα των φυσικών αριθμών.
Αν           $\sum_{k \in A}\frac{1}{2^{k}}=\sum_{m \in B}\frac{1}{2^{m}}$  τότε δείξτε ότι Α=Β.[/b]

Συγνώμη Νίκο ΔΗΛΑΔΗ ΘΕΣ ΝΑ ΜΑΣ ΠΕΙΣ ΟΤΙ ΔΕΝ ΕΛΥΣΕΣ ΕΣΥ ΑΣΚΗΣΗ ΠΟΥ ΣΟΥ ΕΒΑΛΑΝ ΣΤΟ ΣΧΟΛΕΙΟ????? :rotfl:  :rotfl:  :rotfl:  Ε ΡΕ ΚΑΖΟΥΡΑ ΠΟΥ ΘΑ ΣΟΥ ΚΑΝΩ ΕΤΣΙ ΚΙ ΤΗΝ ΛΥΣΩ (ΠΟΤΕ ΔΗΛΑΔΗ  :D )
Τα λέμε

Ego den eipa oti den tin elysa!Fysika kai tin elysa.....   :D  :lol:

Αν δεν την έλυνες θα έπρεπε να ανυσηχούμε τώρα  και ο δάσκαλος να πάει για trainer. :lol:
Αλλά όπως ήταν φυσιολογικό τίποτα από αυτά δε συνέβη .
Λοιπόν πολύ ωραία άσκηση (σαν εκφώνηση τουλάχιστον είναι πολύ ωραία )
Να η λύση μου
Έστω $A=\left\{x_{1},...,x_{n}\right\}$  και $B=\left\{y_{1},...,y_{n}\right\}$  .

Δίχως βλάβη της γενικότητας υποθέτουμε ότι $x_{1}<x_{2}<...<x_{n}$ και $y_{1}<y_{2}<....<y_{n}$  και άρα
$x_{n}=x_{1}+p_{1}$  , $x_{n}=x_{2}+p_{2}$ ,....$x_{n}=x_{n-1}+p_{n-1}$
$y_{n}=y_{1}+q_{1}$  , $y_{n}=y_{2}+q_{2}$ ,....$y_{n}=y_{n-1}+q_{n-1}$

Άρα $\sum_{x_{i}\in A}\frac{1}{2^{x_{i}}}=\frac{2^{p_{1}}+...+2^{p_{n-1}}+1}{2^{x_{n}}}$ (1)

$\sum_{y_{i}\in A}\frac{1}{2^{y_{i}}}=\frac{2^{q_{1}}+...+2^{q_{n-1}}+1}{2^{y_{n}}}$ (2)

Από την υπόθεση (1)=(2)
Πολλαπλασιάζοντας χιαστά έχουμε ότι
$2^{p_{1}+y_{n}}+....+2^{y_{n}}=2^{q_{1}+x_{n}}+....+2^{x_{n}}$

Λοιπόν τώρα είναι όλο το <<μανίκι>> . Ελπίζω να εξηγήσω σωστά τη σκέψη μου .
Αν υπάρχουν ίσοι αριθμοί στο πρώτο και δεύτερο μέλος τους διαγράφουμε προφανώς .Άρα εστω ότι δεν υπάρχουν . Τότε κανενας εκθέτης του πρώτου μέλους δεν ίσος με κανέναν του δευτέρου προφανώς .Άρα υπάρχει ο ελάχιστος εκθέτης .
Έστω χωρίς βλάβη της γενικότητας ότι ανήκει στο πρώτο μέλος ο ελάχιστος και έστω ότι αυτός είναι ο $p_{i}+y_{n}$ .Τότε διαιρώντας με $2^{p_{i}+y_{n}}$  το πρώτο μέλος γίνεται
$1+2^{p_{1}-p_{i}}+....=2^{q_{1}+x_{n}-p_{i}-y_{n}}+....+2^{x_{n}-q_{i}-y_{n}}$

Τώρα όμως το πρώτο μέλος είναι περιττός και το δεύτερο μέλος άρτιος άτοπο .
Άρα δύο εκθέτες είναι ίσοι ....Τους διώχνουμε και εργαζόμαστε ομοιώς ώσπου στο τέλος θα καταλήξουμε ότι κάθε εκθέτης του πρώτου μέλους έχει τον αντίστοιχό του στο δεύτερο άρα Α=Β

Ελπίζω να το εξήγησα καλά  :)

Είναι απαραίτητη μια τόσο αναλυτική λύση silouan? :?:

Έστω $z = \max_{r \in A \cup B}r$. Πολλαπλασιάζοντας τη σχέση με $2^{z}$ προκύπτει η νέα σχέση (ή αλλιώς: θεωρούμε $z$ έναν αυθαίρετα μεγάλο αριθμό):

$\sum 2^{a_{i}}= \sum 2^{b_{i}}$

με $a_{i}, b_{i}\geq 0$ και τα $a_{i}$ διαφορετικά μεταξύ τους όπως και τα $b_{i}$.

Το κάθε μέλος αποτελεί αναπαράσταση ενός αριθμού στο δυαδικό σύστημα, και άρα τα μέλη είναι ίσα αν και μόνο αν οι όροι είναι ίσοι, δηλαδη $(a) = (b)$ δηλαδή $A = B$.

Απλά έβαλα αναλυτική λύση για να κατανοηθεί η ιδέα .
Η δική σου λύση είναι πιο ωραία βέβαια και χρησιμοποιεί και το ωραίο αυτό με το δυαδικό σύστημα . :wink:

Η πρόταση δεν ισχύει ως έχει: Πάρε Α={2,3,...} και Β= {1}. Αν Α και Β είναι πεπερασμένα υποσύνολα, τότε η πρόταση συνάγεται από τη μοναδικότητα ανάτυξης πραγματικού αριθμού σε βάση 2 (δυαδικά ψηφία).

Nai k.Du Lac exete dikio!Einai dikia mou i paraleipsi,dioti kai i ekfonisi pou mou edose o kathigitis elege oti ta synola einai peperasmena.
Diko mou to lathos-tha to diorthoso amesos!

Ρε φιλαράκο
ούτε σας προλαβαίνω μια δικιά μου πιθανή λύση είναι
έστω Α' υποσύνολο Α κι Β' Υποσύνολο Β τότε αν ισχύει η σχέση (θ α ήθελα να ελ΄γξτετε ) τότε αν πάμε με μονοσύνολα θα αποδείξουμε πανευκολα επαγωγικά ότι Α=Β  θα ήθελα να δείτε αυτό γιατί αν ισχύει θα ολοκρηρώσω την λύση μου
Χαιρετώ

Για το λόγο που ανέφερα πιο πάνω, η πρόταση ισχύει και όταν και τα δύο υποσύνολα Α και Β είναι άπειρα.


Does anyone have a link to the solutions or answers of AMC12B 2003?

I greatly appreicate it.  Thanks.

I don't think there are any solutions on the web. You could try searching in an old math jam that discusses some of those problems.

Here's a link to the answers:
http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2003-12a/03answers12b.html

But I don't know of anywhere to get the solutions.


calculus
derivative
function
analytic geometry
graphing lines
slope
calculus computations

1. A company has cost function $ C(x) \equal{} 2000 \plus{} 5x \plus{} x^2$ and demand function $ p(x)\equal{}100$. How many units should it make to maximize its profit?

2. The cost of operating a bus from Azusa to Yreka is $ 200 \plus{} 8x$, where $ x$ is the number of passengers. If a ticket is $ \$40$, there will be $ 10$ passengers, while if a ticket is $ \$35$, there will be $ 15$ passengers. Assuming the demand function is linear, what should the price of a ticket be to maximize its profit?

[color=violet][TeXed by mod][/color]

1. I understand that the demand function here gives the price in terms of sales volume, not the other way around. The profit of selling $ x$ units is $ xp(x)\minus{}C(x)\equal{}95x\minus{}2000\minus{}x^2$. Take the derivative and equate it to zero.

2. If the price is $ p$, then from the given data we get $ x\equal{}50\minus{}p$. Since the cost is $ 200\plus{}8(50\minus{}p)$, the profit is $ px\minus{}(200\plus{}8(50\minus{}p))\equal{}p(50\minus{}p)\minus{}200\minus{}8(50\minus{}p)$. Again: derivative, equate, zero.

Are we allowed to assume that x = 50-p just from the two given data like that?

That's the only line that fits the two points. Remember, slope = rise/run = (-5/5) = -1. The constant term 50 is made so that the line passes through the first point.


The daily high temperatures in degrees Celsius for Monday
through Friday were 3, -1, -6, -5 and 2, respectively. What is
the arithmetic mean of these five temperatures? Express your
answer as a decimal to the nearest tenth.

The arithmetic mean can be solved by sum of temps/# of days.
Thus, there are 5 days and a sum of 3-1-6-5+2=-7.
-7/5=[b]1.4[/b]


we have a 10*10 puzzle that have 9 black&81 white 1*1 square we can make a wite house black if ther are 2 black square near it.
prove that we cant have 10*10 square.


geometry
perimeter
geometric transformation
reflection
rotation
Pythagorean Theorem
AMC

[quote="Syntax Error"]Two squares side 1 are placed so that their centers coincide. The area inside both squares is an octagon. One side of the octagon is 43/99. Find its area.

help me quickly![/quote]

Visualize the squares.  The intersection is an equilateral octagon, as described by the problem.  The union, exclusive of the intersection consists of 8 congruent little right triangles.

Let x and y be the two legs of the little right triangles.  The hypotenuse is given by the problem as 43/99.

We can write two equations.  Following one side of a square, we have (1) x+43/99+y=1.  Using the Pythagorean Theorem, we have (2) x^2+y^2=(43/99)^2.

Rewriting and squaring equation (1):
  x+43/99+y=1
  x+y=56/99
  x^2+2xy+y^2=(56/99)^2

Subtract equation (2):
  2xy = (56/99)^2 - (43/99)^2
  2xy = 13/99

No need to solve for x or y, because the area of each little right triangles is just xy/2.

The area of four little right triangles is 2xy=13/99.  So the area of the octagon is the area of a square minus the area of four little right triangles.

If I'm not mistaken, the answer is 1-13/99 = 86/99.

[quote="Syntax Error"]Two squares side 1 are placed so that their centers coincide. The area inside both squares is an octagon. One side of the octagon is 43/99. Find its area.

help me quickly![/quote]

I presume the octagon is regular?

No!  Not a regular octagon!  It's an equilateral octagon, but it's not regular!

My previous assertion that the octagon is both equilateral and equiangular was incorrect.

Thank you for your correction, Joel.  [I should have known not to disagree with you.  :-)]

rcv, you're wrong -- the resulting octagon will not be equiangular.  It would only be equiangular if the side of the octagon were ( :rt2: - 1) times the side of the square.  As it is, half of the angles of the octagon have measure roughly 158 degrees, while the other have measure approximately 112 degrees.

The problem never stated that the octagon is equilateral or regular. That means you have to prove it.

  Draw a circle around the two squares. Take out the octagon first, we have a non-convex 16-gon. Note that for two consecutive sides x and y of 16-gon, x(99-x)=y(99-y). 99(x-y)=x :^2:-y :^2: . Two cases, x+y=99 or x=y.
Suppose that x+y=99, the perimeter of the 16-gon=99(8)=perimeter of the two squares-perimeter of octagon.
so 99(8)=99(8)-P(octagon), so P(octagon)=0 and that's not very good.
 so x=y, means all the right triangles outside of the octagon are  :cong:, so all the sides of the octagon are  :cong: .

Okay.  Alternatively, a reflection about a diameter passing through one of the intersections of the two squares, along with a rotation of 90 degrees, will also establish the equilateral nature of the octagon.  It is definitively not equiangular, which you can prove by looking and the relationships between the angles of the right triangles and the octagon.

Wow.  I'm really surprised I've never seen that solution before.  Go figure.


Let $ P$ be any  point in the interior of a $ \triangle ABC$.Prove That
$ \frac{PA}{a}\plus{}\frac{PB}{b}\plus{}\frac{PC}{c}\ge \sqrt{3}$.

Darij Grinberg's proof : http://www.mathlinks.ro/viewtopic.php?p=136745#136745

I think you have used too strong a condition,even stronger than this problem(Its an IMOTC 2010 problem).Its like proving that $x^3=2y^3$ has no integer solutions using FLT....

Bye..

Sayantan....

[quote="Marian Dinca"]$
\mbox{Following well-known inequality: }\\
 PB.PC.a+PA.PC.b+PA.PB.c\ge abc\  \mbox{(Hayashy inequality)}\\
\mbox{have: }\\
\frac{PA}{a}+\frac{PB}{b}+\frac{PC}{c}\ge\sqrt{3}\Leftrightarrow \sum_{cyclic}{\frac{PA}{a}.\frac{PB}{b}}\ge 1;\\
\mbox{if pose: } \frac{PA}{a}=x; \frac{PB}{b}=y, \frac{PC}{c}=z\mbox{ and using the inequality: }\\
xy+xz+yz\le \frac{1}{3}.(x+y+z)^2\mbox{, obtain:}\\
1\le xy+xz+yz\le\frac{1}{3}.(x+y+z)^2\Rightarrow\\
\Rightarrow 1\le\frac{1}{3}.(x+y+z)^2\Rightarrow x+y+z\ge \sqrt{3}\mbox{, and proved the inequality.}
$[/quote]
See also here
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=49&t=24803

[quote=ith_power]Let $ P$ be any  point in the interior of a $ \triangle ABC$.Prove That
$$ \frac{PA}{a}\plus{}\frac{PB}{b}\plus{}\frac{PC}{c}\ge \sqrt{3}$$.[/quote]

\[ \boxed{ \frac {PA}a + \frac{PB}b + \frac{PC}c \ge \frac {a + b + c}{PA + PB + PC}  }\]
https://artofproblemsolving.com/community/c6h576972p8751142
[url=https://artofproblemsolving.com/community/c6h1491070p8751807]here[/url]

Isn't the minimum achieved when P is the incenter

[quote=ith_power]Let $ P$ be any  point in the interior of a $ \triangle ABC$.Prove That
$$ \frac{PA}{a}\plus{}\frac{PB}{b}\plus{}\frac{PC}{c}\ge \sqrt{3}$$.[/quote]
Let $P$ be a point in a triangle $ABC$ with circumradius $R$. [url=https://artofproblemsolving.com/community/c6h122569p695339]Prove that[/url]$$\frac{PA}{bc}+\frac{PB}{ca}+\frac{PC}{ab}\geq\frac{1}{R}$$
Given a point $P$ inside triangle $ABC$, [url=https://artofproblemsolving.com/community/c6h352768p1907065]Prove the inequality[/url]
$$\frac{PA}{\sqrt{a}} +\frac{PB}{\sqrt{b}} +\frac{PC}{\sqrt{c}} \ge \sqrt{a+b+c}.$$
Point $P$ is inside an acute triangle $\triangle ABC$,[url=https://artofproblemsolving.com/community/c6h133798p757214]show that[/url] $$(PA+PB+PC)^{2}\geq\sqrt{3}(BC\cdot PA+CA\cdot PB+AB\cdot PC)$$
$x$,$y$, $z$ are real numbers,$ P$ is any point of the plane with a  $ \triangle ABC$ . [url=https://artofproblemsolving.com/community/c6h1331723p7184295]Prove that [/url]
$$\left(\frac{x}{a}PA+\frac{y}{b}PB+\frac{z}{c}PC\right)^2\ge 2(xy+yz+zx)-(x^2+y^2+z^2)$$
$$PA+PB+PC\ge \sqrt{ 2(ab+bc+ca)-(a^2+b^2+c^2)}$$
If $S$ is the area of triangle ABC, then $$PA+PB+PC\geq \sqrt{\frac12\cdot\left(a^2+b^2+c^2+4\sqrt3S\right)}$$ Equality holds if and only if the point P is the Fermat point of triangle ABC, and all angles of triangle ABC are $\leq 120^{\circ}$
Let $P$ be a point inside the triangle $ABC.$ Prove that$$3R\geq PA+PB+PC\ge 6r$$
[url]https://artofproblemsolving.com/community/c6h19316p129377[/url]
[url]https://artofproblemsolving.com/community/c6h155428p874119[/url]
[url]https://artofproblemsolving.com/community/c6h2540603p21650961[/url]
[url]https://artofproblemsolving.com/community/c6h131011p742251[/url]
[url]https://artofproblemsolving.com/community/c6h2550p348780[/url]
[url]https://artofproblemsolving.com/community/c6h73465p424436[/url]
[url]https://artofproblemsolving.com/community/c4h180219p991479[/url]
[url]https://artofproblemsolving.com/community/c4h526706p21965090[/url]

Let $ P$ is any point of the plane with  a regular  $ \triangle ABC$. Prove that$$\frac{7}{5}\leq\frac{x^2}{x^2+y^2}+\frac{y^2}{y^2+z^2}+\frac{z^2}{z^2+x^2}\leq\frac{8}{5}.$$
Where $x=PA,y=PB,z=PC.$


Tom drove to Easton from home at 50 mph. His return trip took 30 minutes longer because he ran into traffic and could only drive 45 mph. How many miles is it from Tom's house to Easton? Express your answer as a decimal.

why is it (t+30)??? The question doesn't say it took 30 minutes longer for each hour...
???

In dragon96's solution, $t$ is being measured in minutes, not hours.


Find integer solution of $(x+y)^x$-x^(x+y) =2

Now it is better
[quote="mathquark"]Find integer solution of $(x+y)^x-x^{x+y} =2$[/quote]

$(1,2)$ -- by inspection, no proof it is the only one.

[quote="Altheman"]if you look at it mod x, then it reduces to $y=\sqrt[x]{2}$, and clearly (1,2) is the only solution because the LHS is an integer so the RHS is an integer, and that happens only when x=1, if x=0,2,3 it is either undefined, or irrational[/quote]
 :?:  :?:  :?:  Taking roots after modulus $x$? I don't think this is valid...

That means it is equivalent to this: $(x+y)^x-x^{x+y}=y^x$


Find all functions $f: \mathbb R \to \mathbb R$ such that \[ f(x\plus{}y\plus{}f(y))\equal{}4x\minus{}f(x)\plus{}f(3y)\] for all $x, y \in \mathbb R$.

[quote="danilorj"]Find all functions f: $ \Re\to\Re$ such: $ f(x \plus{} y \plus{} f(y)) \equal{} 4x \minus{} f(x) \plus{} f(3y)$.[/quote]

Let $ P(x,y)$ be the assertion $ f(x\plus{}y\plus{}f(y))\equal{}4x\minus{}f(x)\plus{}f(3y)$
Let $ f(0)\equal{}a$

$ P(0,0)$ $ \implies$ $ f(a)\equal{}0$
$ P(a,0)$ $ \implies$ $ f(2a)\equal{}5a$
$ P(2a,0)$ $ \implies$ $ f(3a)\equal{}4a$
$ P(0,a)$ $ \implies$ $ f(3a)\equal{}a$

So $ 4a\equal{}a$ and $ a\equal{}0$.

Then $ P(x,0)$ $ \implies$ $ f(x)\equal{}4x\minus{}f(x)$ and so $ \boxed{f(x)\equal{}2x}$
And putting back this value in the original equation shows that this indeed is a solution.


In $ \mathbb{R}^n$,balls of radii $ R_1,\cdots,R_m$ centered at $ A_1,\cdots,A_m$ have a common point.Prove that if  $ B_1,\cdots,B_m$ are points such that $ |B_iB_j| \le |A_iA_j|$,then the balls of radii $ R_1,\cdots,R_m$ centered at $ B_1,\cdots,B_m$ have a common point.

It seems to be obvious. (although it's not)

You solve it for the case $ m\equal{}3$ first. Then, the problem is just one-step generalization.

[quote="mathangel"]It seems to be obvious. (although it's not)

You solve it for the case $ m \equal{} 3$ first. Then, the problem is just one-step generalization.[/quote]
I think it is not easy to generalization...It is a problem from pravalov's book 《geometry》,but I can't understand the proof.


$A,\ B,\ C$ and $D$ are points on a given straight line, in that order. Show how to construct a square $PQRS$, with all of $P,\ Q,\ R$ and $S$ on the same side of $AD$, such that $A,\ B,\ C$ and $D$ lie on $PQ,\ SR,\ QR$ and $PS$ produced respectively.

Let the angle formed by QRC and ABC be X.

It is not hard to see one side of the triangle is CD sin X and the other is AB cos X.  It implies tan X  = AB/CD, which is known.

So from C construct A'C so that |AB| = |A'C| and A'C is perpendicular to ABCD.  Then DA' is a required line, and the rest of the lines can be found from this one  (translate, take perpendicular and translate some more)


A data set with ten numbers has an arithmetic mean of 8.5. When one of the numbers is removed from the data set, the arithmetic mean of the remaining nine numbers is 8. What number was removed?

The sum of the original ten numbers was $ 8.5(10)\equal{}85$. The sum of the remainig nine numbers is $ 9(8)\equal{}72$. Hence, the number that was removed is $ 85\minus{}72\equal{}\boxed{13}$.


If I am login I will see blog1, but if i am a guest I will see blog2. Why? Help me! I want gues so see all.

2nd file , I can not add it to 1st post! Why?

Maybe only 4 of your posts are open to guests.

But I want guest so see all. Please help me  :(


algebra
polynomial
real analysis
real analysis unsolved

Locate the turning points of the curve [i]y=x(x^2 - 12)[/i] and determine their nature.

How I define "turning point" might be different than what you have in mind.  Do you want where the graph changes from increasing to decreasing (or vice-versa) or where it changes concavity?

For changing direction,
$ f'(x)\equal{}3x^{2}\minus{}12\equal{}0\Rightarrow x^{2}\equal{}4\Rightarrow x\equal{}\pm 2$
These are your critical points.  You still need to check whether or not they are extrema and I'll leave that to you.

For changing concavity,
$ f''(x)\equal{}6x\equal{}0\Rightarrow x\equal{}0$


How many factors of $ 2^3\cdot 3^4$ are perfect squares?

Just out of curiosity, do you do the same thing for cubes except divide by 3?

Yes, think about it! You have to have either 0 2's, 2 2's, 4,6, etc.

Similarly with cubes you need 0 2's, 3 2's, 6, 9, etc.

is there a shortcut or formula for this?

[quote=AIME15]We want $ (\lfloor\tfrac{3}{2}\rfloor \plus{} 1)(\lfloor\tfrac{4}{2}\rfloor \plus{} 1) \equal{} 2 \cdot 3 \equal{} \boxed{6}$.[/quote]

How did you get that?

$2^3$ so $\frac{3}{2}$ because they want a power that is even and same for the other floor

So the formula would be $(\lfloor{\frac{n+1}{2}}\rfloor+1)(\lfloor\frac{m+1}{2}\rfloor+1)$??

A formula isn't necessary. You can just use multiplication.

[hide=Solution]
We list out all the total possibilities for each prime factor:
\[ 2 \rightarrow 2^0, 2^2 \qquad (2)\]
\[ 3 \rightarrow 3^0, 3^2, 3^4 \qquad (3)\]
There are 2 possible perfect squares for 2 and 3 for 3. Therefore, there is a total of $2 \times 3=\boxed{6}$ perfect squares in the expression.
[/hide]


Let $ u_n$ be a sequence of real numbers. If $ \frac{1}{u_n}$ diverges what can we say about $ u_n$?

if $ u_n \to c \ne 0$ then we get $ \frac{1}{u_n} \to \frac{1}{c}$ automatically.. so

we know $ u_n$ either converges to 0 or diverges itself.

if $ |u_n| \to \infty$, then $ \frac{1}{u_n} \to 0$ hence this case is excluded.


Because of $i\cdot(1-i)=1+i$, this simplifies to $\frac{i^n (1-i)^n}{2(1-i)^{n-3}} = \frac{i^n (1-i)^3}{2} = \frac{i^n (1-i)^3}{i\cdot(1-i)^2} = i^{n-1} (1-i) = i^{n-1} - i^n$.
But what are you asking for¿

Alternately: $1+i=\sqrt{2}\cdot e^{i\pi/4}$ and $1-i=\sqrt{2}\cdot e^{-i\pi/4}$

So $\frac{(1+i)^n}{2(1-i)^{n-3}}=$ (see next post for correct finish)


Twelve sections of fence, each 1 yard long, are used to enclose a rectangular area. How many different areas can be obtained?

Perimeter = 12 yards, so L + B = 6 both being integers.
1x5, 2x4, 3x3 are possible dimensions giving areas 5, 8 and 9.

There are twelve sections, and a minimum of 1 must be on opposite ends so there are from 1x5 to 5x1. Of these, 3x3 is repeated once, but the other four are repeats, so the total possible areas is 4/2+1=3.

There are $12$ pieces
 $\dfrac{12}{2}$=$6$

$6=5+1$

$6=4+2$

$6=3+3$

There are $3$ possiblities. I agree.

$\boxed{3}$

have a great day!!! :)


geometry
circumcircle
incenter
geometry unsolved

A line through the incenter $I$ of triangle $ABC$, perpendicular to $AI$, intersects $AB$ at $P$ and $AC$ at $Q$. Prove that the circle tangent to $AB$ at $P$ and to $AC$ at $Q$ is also tangent to the circumcircle of triangle $ABC$.

It is just very easy converse of the problem from [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=120938]here[/url].

I think something error in this problem,(Draw these two circles as large as possible).

Dear Mathlinkers,
this circle is the A-mixtilinear incircle of ABC. (when perpendicular at I)
You can see
http://jl.ayme.pagesperso-orange.fr/   vol. 4 a new mixtilinear incircle adventure I, p. 10
Sicerely
Jean-Louis

Oh sorry,I think two different circles tangent at $P$ and $Q$,I make mistake to read the question.

let $\omega$ be a circle which is tangent to the sides $AB,AC$ at $X,Y$ and circumcircle of $ABC$ at $T$.Draw $TX$ and $TY$ and they meet the circumcircle at $P$ and $Q$,note that $P$ and $Q$ are the midpoint of the arcs $AB$ and $AC$.Apply pascal to $BACPTQ$ and we see that $X,I,Y$ are collinear.Hence we get the result.


$53 = (8 \cdot 5)+(8+5)$
Find the lowest number after $53$ that cannot be displayed this way.

my question is....what if one of the numbers is 0?  you can get any number then.

They tell you they msut be positive integers.

This question was on yesterday's NEML. Since everyone was supposed to have already taken it by now...

And yes, it says that they are [i]positive[/i].
[hide]Let $n=xy+x+y$. Note that $(x+1)(y+1)=xy+x+y+1$, so $n+1=(x+1)(y+1)$. If (and only if) $n+1$ is prime, then it cannot be expressed in the form, so the next unexpressible number is 59-1=[b]58[/b].[/hide]


algebra
polynomial
linear algebra
matrix
vector
LaTeX
algebra unsolved

Sorry all,

I reposted this problem here after noticing it on the Algebra Open Question. I was mistakenly posted it there, meant to be here (Algebra Unsolved Problems). Thanks for understanding.

[quote]
I've just learn Jordan Normal Form(JFN) and found out too confusing, hence, I need some help. Sorry I don't know LATEX.

Say, given a matrix A =[(8 8 4), (-1 2 1), (-2 -4 -2)], note 8, 2, -2 are in the diagonal entries. Find the Jordan Normal form of A. A is a square (nxn) matrix and has linearly independent vectors of rank less than n.

Please do show your work, espcially when you find the Eingenvalues and Eingenvectors, how do you proceed from this point to obtain the JNF of A.

Thanks.[/quote]

I've incompletely solved this problem.: with the eingenvalues found are 0 and 4 and the eingenvectors to these eingenvlues are (0 -1 2) and (2 -1 0) respectively.

I stuck on to find the JFN.

PS.: Please moderator, remove this post from the "Algebra Open Problem" section.

[color=red]Edit by Megus: I've delated your post in the "Algebra Open Problem" section - thanks for notifying - also I'm moving this topic to more appropiate place [/color]

Eigenvalues have both a geometric multiplicity and an algebraic multiplicity. The geometric multiplicity is the dimension of the eigenspace, and the algebraic multiplicity is the number of times the factor $(x-\lambda)$ occurs in the characteristic polynomial.

In this case, both eigenvalues have geometric multiplicity 1. (You said as much when you said "the eigenvectors to these eigenvalues are ... ") However, the characteristic polynomial can be written as $x(x-4)^2$ so while $0$ has algebraic multiplicity one, $4$ has algebraic multiplicity two.

(Side note: I didn't actually find the characteristic polynomial for myself. It's enough to know that the sum - counting multiplicities - of the eigenvalues must be the trace of the matrix. The trace is $8$, so the eigenvalues must be $0,4,4.$)

This is already enough information to determine the Jordan form. Each "simple" eigenvalue (algebraic multiplicity one) must be associated to a simple element on the main diagonal, while an eigenvalue of algebraic multiplicity two and geometric multiplicity one must be associated with a $2\times2$ Jordan block. The Jordan form will be:

\[C=\begin{bmatrix} 4&1&0\\0&4&0\\0&0&0 \end{bmatrix}\]

For some more details: Let's return to the eigenvalue $4.$ You have found a vector $v=[2,-1,0]^T$ such that $(A-4I)v=0.$ What we now need to do is find a vector $w$ such that $(A-4I)w=v.$ This can be done. The answer is not unique, but we need only one such vector, so $w=\left[\frac34,0,-\frac14\right]^T$ will do.

Let $P=[v\,w\,u]$ where $u$ is the eigenvector such that $Au=0.$

We can write $P$ as $P=\begin{bmatrix}2&\frac34&0\\-1&0&-1\\0&-\frac14&2\end{bmatrix}$

We can now show that $AP=[4v,v+4w,0]$ and $P^{-1}AP=C,$ given above.

Thanks very much Kent,

Your explanation has given me some lights. Ohnestly, I'm new to all these stuffs(through self-learning) but  grdually harvesting on them as confusions of abstract presentation by lituratures slowly faded with ease. Probably, thats the price for self-learning!

I will look into your solution espcially where you obtained the JNF and if the "why" query comes to mind, I'll be back to post this "why" here. Otherwise, am satisfied with your solution (much better than my books). I got almost to your conlusion but failed.

I'll make sure to learn LATEX too along with my next reading on QR-Decomposition.


There is a fortune-teller at the annual village fair. Nine fortune-tellers out of ten are always right; the other ones are always wrong. Nobody at the fair, of course, ever knows which sort is there this year. Another annual feature of the fair is the lottery. Each of the twenty boys in the village has an equal chance to be chosen as the Harvest King; and each of the twenty girls has an equal change to be chosen as the Harvest Queen.

(a) Henry went into the fortune-teller's tent. He asked the fortune-teller whether he would be chosen, and was told that he would. What are his chances now?
(b) As he came out, Henry met his friend Anne. He told her about the prediction. She immediately went into the tent and asked whether she would be chosen as Harvest Queen. The fortune-teller told her that she would be chosen. What are her chances now?

[hide]
a) Henry will become Harvest King iff the fortune teller is always right. This has a 90% chance of happening.
b) Anne's case is independent of Henry's, so her's is also 90%.
[/hide]

Heh, these problems are always so tricky.

[hide="a"]There is a $ .045$ chance that, simultaneously, Henry is to be the king and the fortune teller was right. There is a $ .095$ chance that Henry is not to be the king and the fortune teller was wrong. So $ \frac{.045}{.045 \plus{} .095} \equal{} \frac{9}{28}$.[/hide]

[hide="b"]Now there is a $ .00225$ chance that the fortune teller is right (that Henry and Anne are both the King and Queen is very improbable). However, there is a $ .09025$ chance that the fortune teller is wrong. So we get $ \frac{.00225}{.00225\plus{}.09025} \equal{} \frac{9}{370}$. Note that Henry's chances also changed when Anne got this answer.[/hide]

Just in case you're wondering how come we have two radically different but apparently reasonable answers...  :)

[hide="More details"]
The thing that's missing from zephyredx solution is this: before the fortune teller made any prediction, it is indeed true that there's a 90% chance they are the real deal. But afterwards, the chances necessarily change. For instance, suppose another kid went into the tent and heard that he, too, is the harvest king - what are the chances that the fortune teller is truthful? Zero, of course.

Now, since a-priori the chances that Henry really is the king are relatively small (5%), as soon as he hears the fortune teller telling him he's the king his first thought should be - this is more likely to be a bad fortune teller than a good one. The probability is now modified, and is no longer 90%.

Another way to assist our intuition is to take more extreme values. Suppose that the chances of winning the lottery and becoming the harvest king were 1 in a billion instead of 1:20. You walk into the tent and hear you are the winner. Do you really think there's a 90% chance it's actually you? No, it's [b]way[/b] more likely that this is a bad fortune teller and she's telling the good news to pretty much everyone who walks in - including you.
[/hide]

See [url=http://en.wikipedia.org/wiki/Bayes_theorem#Further_examples]some material here[/url] for more information on seemingly nonintuitive results like this. In fact my solution is basically Bayes theorem in disguise.

I'd read about this once before...but I completely forgot it. Thanks! :lol:


Equation is
\[ 3cos2x \plus{} cos6x \equal{} 1/2
\]
find the (0 and pi)interstice ,totality origin?
answer is pi  :o
i didnt solve i need to see solution way  :(

I dont understand what you want nevertheless im giving a soln to the eqn.

We know that $ \cos {3x} \equal{} 4 \cos^3 x \minus{} 3 \cos x$
Therefore the given equation is equivalent to $ 3 \cos {2x} \plus{} 4 \cos^ {2x} \minus{} 3 \cos {2x} \equal{} \frac {1}{2}$
$ \implies \cos^3 {2x} \equal{} \frac {1}{8} \implies \cos {2x} \equal{} \frac {1}{2}$

Now i think you can handle it  :)

Thank you very much.i solved:)
\[ cos2x\equal{}1/2 \minus{}\minus{}\minus{}>2cos^2x\minus{}1\equal{}1/2\]
\[ 2cos^2x\equal{}3/2\minus{}\minus{}>cos^2x\equal{}3/4\]
\[ cosx_1\equal{}\plus{}sgrt3/2 \minus{}\minus{}\minus{}\minus{}>cosx_2\equal{}\minus{}sgrt3/2\]
\[ x_1\equal{}30\minus{}>pi/6\]    \[ x_2\equal{}150\minus{}>5pi/6\]
\[ x_1\plus{}x_2\equal{}pi\]


Ένα ενδιαφέρον υπολογίστικό πρόβλημα από τον Καναδικό φοιτητικό διαγωνισμό "The Bernoulli Trials" του 1998.

Ποιος από τους αριθμούς    $2^{3^{2^{3^{2^{3}}}}}$    και    $3^{2^{3^{2^{3^{2}}}}}$    είναι μεγαλύτερος;

[hide]$3^{2}>2^{3}$.....ontos... :oops:
ligo mathematica 8eli tlk...[/hide]

 :D

Μια μικρή υπόδειξη
[hide] Χρησιμοποιήστε απαγωγή σε άτοπο[/hide]

Λύση (με λίγη σκέψη & καθόλου mathematica)

[quote="bouzoukman"]Με αφορμή την άσκηση θα ήταν καλό να γράψουμε σε ποιες περιπτώσεις ισχύει $\alpha^{\beta}>\gamma^{\delta}$ με $\alpha,\beta,\gamma,\delta>0$...

Πολλές φορές μου έτυχε να κάνω λάθος σε μια άσκηση επειδή τα έχω μπερδέψει... :wink:[/quote]

$f(x)=a^{x}$ ειναι αυξουσα για $0<x<\infty$

κατι τετοιο δεν εννοεις ?
μηπως εννοεις γενικα για τους πραγματικους αριθμους ,

@ kostas

Σε ό,τι αφορά τη συνάρτηση $f(x)=a^{x}$ είναι γνησίως αύξουσα  μόνο για $a>1$. Για $a<1$ είναι γνησίως φθίνουσα (σκέψου για παράδειγμα την $h(x)=(1/2)^{x}$). Αυτό προκύπτει και από την παραγώγιση, αφού $f'(x)=a^{x}ln(a)$. Για $a>1$ είναι $f'(x)>0$ ενώ για $a<1$ είναι $f'(x)<0$

@ bouzoukman

Νομίζω ότι αυτό που ρωτάς είναι πολύ γενικό. Για παράδειγμα, μπορεί να βρεθεί μια γενική ανισότητα σε αυτό που ρωτάς για $0<a<c<d<b$; Ή ακόμα και σε πιο απλές περιπτώσεις, τι ανισοτική σχέση ισχύει για τα $a^{b}, b^{a}$ (πχ $e^\pi, \pi^{e}$);

ουπς δικιο εχεις το ειχα γραψει αλλα το εβησα . :blush:

@alexandros_plus

Βασικά ήθελα να ρωτήσω αν για $a>b$ και $c>d$ ισχύει $a^{c}>b^{d}$.Και με αφορμή αυτό είπα να το γενικεύσω.. :wink:
Ισχύει το παραπάνω;;;
Η αλλίως για εκθετικές παραστάσεις με διαφορετική βάση και διαφορετικό εκθέτη πότε είναι μεγαλύτερη η μία από την άλλη??

Το παραπάνω ισχύει πάντα μόνο στην περίπτωση που ικανοποιείται και η επιπλέον συνθήκη $b>1$ και ακέραια $c, d$, γιατί αλλιώς πάρε την περίπτωση $a=0,5, b=0,25, c=8, d=2$ και θα δεις ότι δεν ισχύει.

Πραγματικά θα ήταν χρήσιμο να βρεθούν γενίκοτεροι τύποι που να ισχύουν για τις δυνάμεις. Υποπτεύομαι ότι (ενδεχομένως να) γίνεται με τη βοήθεια της ανάλυσης.

Αλήθεια, μπορεί κάποιος να αποδείξει ποιο από τα $e^\pi$ και  $\pi^{e}$ είναι μεγαλύτερο (χωρίς βεβαίως να κάνει τους υπολογισμούς);

[quote="alexandros_plus"]
Αλήθεια, μπορεί κάποιος να αποδείξει ποιο από τα $e^\pi$ και  $\pi^{e}$ είναι μεγαλύτερο (χωρίς βεβαίως να κάνει τους υπολογισμούς);[/quote]

Theleis na siggrineis to $e^{\pi}$ kai to $\pi^{e}$.

Ara  $e^{\pi}\ \ ? \ \ \pi^{e}$  (opou $?$ einai to simvolo $<$ 'h $>$)
 diladi logarithmizontas thelw   $\pi \ \ ? \ \ eln{\pi}$

Pairnw loipon ti sinartisi $f(x)=elnx-x, x>0$

kai ti meletw. $f'(x)=\frac{e}{x}-1$ i opoia gia $x<e$ einai gnisiws auksousa

ara epeidi $\pi<e \Rightarrow f(\pi)<f(e) \Rightarrow \pi^{e}>e^{\pi}$

Alexandros

Y.G. Enas allos tropos tha itan apo tin arxi na thewriseis ti sinartisi f(x)=e^x-x^e alla den tha itan toso eukoli i meleti opws prin.

[quote="alexandros_plus"]
Αλήθεια, μπορεί κάποιος να αποδείξει ποιο από τα $e^\pi$ και  $\pi^{e}$ είναι μεγαλύτερο (χωρίς βεβαίως να κάνει τους υπολογισμούς);[/quote]
προσφατα ειδα μια πολτυ ομορφη λυση με αναπτυξη του $e^{x}$ κατα Taylor.αλλα ειναι μεγαλυτερη σε εκταση απο του Αλεξανδρου

Μία λύση που είχα παραθέσει παλαιότερα στην ιστοσελίδα του Πανεπιστημίου της Πάτρας βρίσκεται εδώ:

http://in.math.upatras.gr/viewtopic.php?t=98

Θα ήθελα να ρωτήσω αν ξέρει κανείς αν τέτοιου είδους υπολογιστικά προβλήματα μπορούν να πέσουν σε διαγωνισμούς της ΕΜΕ όπως στην Μεσογειάδα... :?:

Kostas, μήπως μπορείς να γίνεις λίγο πιο συγκεκριμένος για την ανάπτυξη κατά Taylor;

Βασικα δεν εχει πολυ Taylor.

εχουμε οτι :$e^{x}=1+x+\frac{x^{2}}{2!}+....>1+x$.

$\pi>e\Rightarrow \frac{\pi}{e}>1$

Εστω $x=\frac{\pi}{e}-1,x>0$

Αρα : $e^{(\pi/e)-1}>1+\frac{\pi}{e}-1$

$\frac{e^{\pi/e}}{e}>\frac{\pi}{e}$

$e^{\frac{\pi}{e}}>\pi$

$e^{\pi}>\pi^{e}$

Ουσιαστικα αυτο τυο τεχνασμα ειναι του Polya,με το οποιο εδωσε μια φοβερη αποδειξη του  αριθμητικου-γεωμετρικου μεσου

Στην παραπάνω ο Taylor δεν χρειάζεται . Απλά θεωρείστε τη συνάρτηση $f(x)=e^{x}-x-1$   :wink:

@Αλέξανδρο : Έχεις ένα τυπογραφικό στο τέλος της λύσης σου
γράφεις $\pi <e$

[quote="silouan"]Έχεις ένα τυπογραφικό στο τέλος της λύσης σου
γράφεις $\pi <e$[/quote]

Swsta, gi'auto to logo einai lathos kai i teliki anisotita!
To swsto loipon einai:

$f'(x)=\frac{e}{x}-1$ i opoia gia $x>e$ einai gnisiws fthinousa

ara epeidi $\pi>e \Rightarrow f(\pi)<f(e) \Rightarrow \pi^{e}<e^{\pi}$

Euxaristw silouan! (Distixws den mporw na to allaksw)

Alexandros


[quote="orl"]Let $P(x)$ be given a polynomial of degree 4, having 4 positive roots. Prove that the equation
\[(1-4 \cdot x) \cdot \frac{P(x)}{x^2} + (x^2 + 4 \cdot x - 1) \cdot \frac{P'(x)}{x^2} - P''(x) = 0\]
has also 4 positive roots.[/quote]

My solution:

$\Leftrightarrow \frac{1-4x}{x^2}[P(x)-P'(x)]+[P'(x)-P''(x)]=0   (*)$

Put $Q(x)=P(x)-P'(x)$ then $deg[Q(x)]=4$. We will prove that $Q(x)$ has 4 positive roots.

Assuming that: $P(x)=(x-a)(x-b)(x-c)(x-d), (0<a<b<c<d)$.

$\Rightarrow P'(x)=(x-a)(x-b)(x-c)+(x-a)(x-b)(x-d)+\\+(x-a)(x-c)(x-d)+(x-b)(x-c)(x-d)$

So:

$Q(0).Q(a)=[abcd+abc+abd+acd+bcd][(a-b)(a-c)(a-d)]<0 \Rightarrow Q(x)$ has s root in $(0,a)$

Similar: $Q(x)$ has roots in $(a,b), (b,c), (c,d)$. So, $Q(x)$ has 4 positive roots.

Assuming that: $ Q(x)=(x-A)(x-B)(x-C)(x-D), (0<A<B<C<D)$

$(*)\Leftrightarrow \frac{1-4x}{x^2}Q(x)+Q'(x)=0$.

Put: $f(x)=(1-4x)Q(x)+x^2Q'(x)/R$.

Then: $f(0).f(A)=ABCD.[A^2(A-B)(A-C)(A-D)]<0\Rightarrow f(x)$ has s root in $(0,A)$.

Similar: $f(x)$ has roots in $(A,B), (B,C), (C,D)$. So, $f(x)$ has 4 positive roots.

Because $f(0)\neq 0\Rightarrow \frac{f(x)}{x^2}=0$ has 4 positive roots,too.

$\Leftrightarrow \frac{1-4x}{x^2}[P(x)-P'(x)]+[P'(x)-P''(x)]=0$ has 4 positive roots.

$\Rightarrow Q.E.D$

[quote="BaronShadeNight"][quote="xxxxtt"]What if the roots are not distinct?[/quote]

$P(x)=0$ has 4 roots, if the roots are not distinct then $P(x)$ has <4 roots![/quote]

Yes, but many books count a double root as two roots, a triple root as 3 roots, etc.
Even the fundamental theorem of algebra says: an equation of degree 4 has 4 roots.(counted with multiplicity, by default).
If we speak about  "4 roots" and not "4 distinct roots", it's ambiguous. There may be even a multiple root of order 4 counted as 4 roots.
And, does the problem remain true if some roots are multiple?


trigonometry
function
number theory
least common multiple
logarithms
calculus
integration

Find the period of $ 2\sin{(4\pi x + \pi/2) + 3\cos{(5\pi x)}}$.

Thanks!

[quote="facis"]The period is just the LCM of the periods of the parts, so the period here is $ 20 \pi$[/quote]

The solution says that the answer is [hide]2[/hide], but I don't understand the solution.

good point, t0rajir0u.  I definitely confused frequencies and periods.  Anyway, on the topic of combining periods, here I was saying the overall period is the lcm of the individual functions' periods because we have a sum - I was not claiming this held under multiplication and division. (isn't it exactly the LCM with addition?)

So anyway I guess I should do my arithmetic on $ 1/2$ and $ 2/5$.  First, lets multiply through be $ 10$, so we are working with integers: $ 5$ and $ 4$.  The LCM of theses is $ 20$, and then, when divided by $ 10$ to undo our change to integers, gives us the period of $ 2$, as desired.

[quote="facis"](isn't it exactly the LCM with addition?)[/quote]
There isn't any particular reason to suspect that addition is special.  For example, $ \sin^2 x, \cos^2 x$ both have period $ \pi$ but their sum, $ 1$, is constant.  Less trivial counterexamples are just as easy to construct.  As I mention [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=207609]here[/url], there are natural ways to translate addition to multiplication and vice versa, so the intuition you should keep in mind is that $ \ln \tan x \equal{} \ln \sin x \minus{} \ln \cos x$ (even though these functions are not defined half of the time).

If it were true in general that when $ f(x)$ has (positive integral) period $ a$ and $ g(x)$ has period $ b$ that $ f(x) + g(x)$ had integral period $ \text{lcm}(a, b)$ then it would follow that whenever $ u(x)$ has period $ a$ and $ v(x)$ has period $ b$ that $ u(x)v(x)$ has period $ \text{lcm}(a, b)$ -- $ \log u(x)$ and $ \log v(x)$ have the same periods, respectively, as $ u(x)$ and $ v(x)$, and $ \log u(x)v(x)$ has the same period as $ u(x)v(x)$.  (Okay, so there are some issues with negative values here, but the point is fairly clear.)

[Edit: t0r posted while I was writing, so this comment is now largely redundant.]

Addition is actually the easiest case to think about.  To think about the sum of two functions with commensurable periods $ a, b$ (that is, periods such that $ \frac{a}{b}$ is rational), we consider their [url=http://en.wikipedia.org/wiki/Fourier_transform]Fourier transform[/url] with respect to a period of $ [a, b]$ (the unique positive real such that $ \frac{[a,b]}{a}$ and $ \frac{[a,b]}{b}$ are coprime integers).  The Fourier coefficients of the sum are just the sum of the Fourier coefficients, so in the generic case that lots of Fourier coefficients are left (which tends to be the case if we just add simple trig functions) we will have a generic function of period $ [a,b]$, but it may well be that most of the low-frequency Fourier coefficients add to zero and we get a function of period, say, $ \frac{[a,b]}{7}$.

[quote="t0rajir0u"][quote="facis"]The period is just the LCM of the periods of the parts, so the period here is $ 20 \pi$[/quote]
First, you've found the LCM of the frequencies rather than the periods.  The periods are $ \frac {1}{2}$ and $ \frac {2}{5}$, and the notion of LCM must be modified slightly for rational periods.

Second, the period of a function of two periodic functions is at best known to [i]divide[/i] the LCM of the individual periods.  A well-known example is the fact that $ \tan x$ has period $ \pi$ even though its constituents, the cosine and sine, have period $ 2 \pi$.  A slightly different argument needs to be used to verify that the period (in any particular case) is [i]exactly[/i] the LCM of the individual periods.

[b]314.math:[/b]  What don't you understand about the solution?[/quote]

The solution says that the period of the $ 2\sin{(4\pi x + \pi/2)}$ is $ 1/2$ and the period of $ 3\cos{(5\pi x)}$ is 2/5.  How would I find this?

Also, how would find the LCM of fractions?

Thanks!

[quote="314.math"]The solution says that the period of the $ 2\sin{(4\pi x + \pi/2)}$ is $ 1/2$ and the period of $ 3\cos{(5\pi x)}$ is 2/5.  How would I find this?
....[/quote]
You might see it more clearly if you consider that:
the values of $ (4\pi x + \pi/2)$ for x=0 and x=1/2 differ by $ 2\pi$ (as x increased by 1/2 you went a whole turn around the unit circle, and sine has the same value again)
the values of $ 5\pi x$ for x=0 and x=2/5 differ by $ 2\pi$ (same idea)
In general if you have a confusing expression f(x) inside a trigonometric function, like sin[f(x)], look at f(x).  In the case of sin[f(x)], you can find the period (p) by solving f(p)-f(0)=$ 2\pi$. If it were tan[f(x)], you would solve f(p)-f(0)=$ \pi$.

metafor, I agree with your definition of LCM but I think it is a bit confusing.  I would suggest to find the LCM of fractions that you multiply both fractions by some integer (the LCM of the denominators works great) to get two integers, find the LCM of those, and then divide it by the number you multiplied by.  Thus, $ lcm(1/2, 2/5) \equal{} 1/10 lcm(5,4) \equal{} 20/10 \equal{} 2$.

The following two conditions are equivalent:

- $ \frac{a}{b}$ is rational (in other words, $ a$ and $ b$ are [i]commensurable[/i]), or
- There exists a real number $ r$ such that $ ra$ and $ rb$ are coprime integers.

Intuitively, because the period of a real function is dimensionless, we are simply measuring the period by the appropriate "unit" (in this case, $ \frac{1}{10}$) to get everything to work out in nice multiples.  The second condition is more useful; we can define the "LCM" of two commensurable reals $ a, b$ as $ rab$.  (Note that this particular definition restricts exactly to the ordinary definition of LCM on two integers.  In this case, $ r \equal{} \frac{1}{(a,b)}$.)

[b]Follow-up:[/b]  A related result is that if a function $ f(x)$ satisfies $ f(x) \equal{} f(x \plus{} a) \equal{} f(x \plus{} b)$ for two commensurable periods $ a, b$, then $ f(x) \equal{} f(x \plus{} (a, b))$ (again for an appropriate definition of GCD).  What happens when $ a, b$ aren't commensurable?

[b]facis[/b] — Yes, the method you suggested is simpler. :P

t0rajir0u, are you saying that if the integers $ a$ and $ b$ are commensurable, then $ gcd(a,b) \neq min(a,b)$? (that would be equivalent to neither $ a/b$ or $ b/a$ being integers).

If I understand this correctly: the answer to your follow up is that if $ a$ and $ b$ are not commensurable, the result still holds.  But if we notice then that $ gcd(a,b) \equal{} min(a,b)$ then we end up with either $ f(x) \equal{} f(x \plus{} a)$ or $ f(x) \equal{} f(x \plus{} b)$, both of which are given.

EDIT: I definitely think I just misinterpreted that.  (integers are also considered rationals here, right?) so then the extension to your follow-up would be that $ f(x)$ is constant if it has some non-commensurable periods.  Not exactly sure how to word a proof, but it relies on the idea that we can choose integers $ A$ and $ B$ such that we can make $ Aa \plus{} Bb > 0$ arbitrarily small, and every possible choice of $ A$ and $ B$ generates a new period for the function. (any ideas for a more rigorous way to argue this?)

[quote="facis"]t0rajir0u, are you saying that if the integers $ a$ and $ b$ are commensurable, then $ gcd(a,b) \neq min(a,b)$? (that would be equivalent to neither $ a/b$ or $ b/a$ being integers).[/quote]
The appropriate definition of GCD for commensurable reals $ a, b$ is the unique positive real $ g$ such that $ a \equal{} ga', b \equal{} gb'$ where $ a', b'$ are relatively prime integers.  Again, this definition restricts to the usual definition of GCD when $ a, b$ are integers.  I don't understand what $ \text{min}(a,b)$ has to do with anything.

[quote="facis"]so then the extension to your follow-up would be that $ f(x)$ is constant if it has some non-commensurable periods.[/quote]
This is actually [i]false[/i] and something of a common misconception (whenever this problem comes up, anyway):  the rational vector space spanned by any finite set of non-commensurable periods has finite dimension over $ \mathbb{Q}$, but $ \mathbb{R}$ has infinite dimension over $ \mathbb{Q}$, so it is not in fact the case that $ f(x)$ must be constant.  Equivalently, one is countable and the other is not.

Take, for example, the case that $ a \equal{} 1$ and $ b \equal{} \pi$.  If we claimed that the set of periods $ \{ ax \plus{} by | x, y \in \mathbb{Z} \}$ included the reals (which is necessary and sufficient for $ f(x)$ to be constant) then there would exist integers $ x, y$ such that $ x \plus{} \pi y$ was equal to, say, half an integer.  But this implies $ \pi$ is rational, which is absurd.  In other words, the set $ \{ x \plus{} \pi y : x, y \in \mathbb{Z} \}$ is dense in the reals but is far from being the entire set.

It does give that if $ f(x)$ is continuous, it must be constant.

As far as I can tell, no one has actually posted a correct solution of the original problem!

JBL I think you missed metafor's solution:

[quote="metafor"]Are you familiar with transformations?

The graphs of $ \sin \left(4\pi x\right)$ and $ \cos (5\pi x)$ are basically transformations of the graphs of $ \sin (x)$ and $ \cos (x)$. Now, the graph of $ f(Bx)$ scales $ f(x)$, horizontally, by a factor of $ \frac {1}{|B|}$ — this implies that the period of a function, if any, is also scaled by $ \frac {1}{|B|}$. Applying this to our problem, we can conclude that

$ period\; \{\sin(4\pi x)\} \equal{} \frac {period\; \{\sin(x)\} }{4\pi} \equal{} \frac {2\pi}{4\pi} \equal{} \frac {1}{2}$ and $ period\; \{\cos (5\pi x)\} \equal{} \frac {period\; \{\cos(x)\} }{5\pi} \equal{} \frac {2\pi}{5\pi} \equal{} \frac {2}{5}$.

Now, we know that $ period\; \{\sin(4\pi x)\} \equal{} period\; \left\{2\sin \left(4\pi x \plus{} \frac {\pi}{2} \right)\right\}$ and $ period\; \{\cos (5\pi x)\} \equal{} period\; \{3\cos (5\pi x)\}$, because the graph of $ f(x \pm h)$ is a horizontal translation of the graph of $ f(x)$, and the graph of $ A f(x)$ is a vertical scaling of the graph of $ f(x)$, neither of which alters the period of a function.

I believe that the least common multiple can be defined as $ lcm (a, b) \equal{} \{\ \min \{ x \} \in \mathbb{Z^{ \plus{} }} : (x \equal{} ak \equal{} bq)\ \cap \ (k, q \in \mathbb{Z^{ \plus{} }})\ \}$ for $ a, b \in \mathbb{Q^{ \plus{} }}$ (if that makes any sense); if so,  $ lcm \left(\frac {1}{2}, \frac {2}{5} \right) \equal{} 2$.[/quote]

The only confusing part is how to find the LCM.  Anyway, I have another suggestion: we can prime factorize rationals just as we prime-factorize integers with the difference that we can now get negative powers.  The LCM will still be the product of the prime factors to the greatest power used in either prime factorization (the power is 0 if the prime does not show up in the prime factorization).  Hence: $ 1/2 \equal{} 2^{\minus{}1}5^0$ and $ 2/5 \equal{} 2^1 5^{\minus{}1}$.  The greatest exponent on the $ 2$'s is $ 1$ and the greatest exponent on the $ 5$'s is a $ 0$, so the lcm is $ 2^1 5^0 \equal{} 2$.

@t0rajir0u/JBL: I think I did assume continuity.  t0rajir0u, I understand what you are trying to say but what do you mean when you talk about the dimension over rationals?

metafor says a lot of true facts about the functions involved, then computes the LCM of their periods.  Nowhere does metafor deal with the fact that [quote="t0rajir0u"]the period of a function of two periodic functions is at best known to [i]divide[/i] the LCM of the individual periods.[/quote]

An example of what t0r is talking about: suppose that $ f$ is known to have periods of 1 and $ \sqrt{2}$.  It follows that $ f$ is constant on the set of values $ \{a \plus{} b\sqrt{2} \mid a, b \in {\bf Z}\}$.  This set is dense in the reals (or in nontechnical terminology, if you tried to draw its picture you'd just have to draw the whole real line), so if $ f$ is continuous then $ f$ is constant everywhere -- however, if $ f$ is [i]not[/i] given as continuous, these values are not enough to determine the value of $ f$ at, say, $ \pi$ or $ \sqrt{3}$ or a lot of other values.

[b]facis[/b] — I think you can turn what you said into a definition of the LCM that can be extended to the rationals (which, in my opinion, is as far a stretch as you can make with the concept).

[b]JBL[/b] — Actually, I wasn't making a serious attempt at a solution; I was just answering [b]314.math[/b]'s questions. I currently have no idea of how to solve the problem.

[quote="t0rajir0u"]the period of a function of two periodic functions is at best known to [i]divide[/i] the LCM of the individual periods.[/quote]

Since we have been talking about rationals, I suppose we need to define what you mean by divide, since that is usually defined in terms of integers.  Here's my best guess: If we have a period $ p$, then for some $ a \in \mathbb{Z\plus{}}$ we have $ ap \equal{} 2$.  That would then mean $ p\equal{}2/a$.  So the possible values of $ p$ are $ {2, 1, 2/3, 1/2, 2/5, 1/3, 2/7, 1/4 ...}$.  Basically, we need to show that the period is exactly $ 2$ and cannot be any smaller.  First notice that if $ f(x) \equal{} f(x\plus{}1/n)$ (for some $ n \in \mathbb{Z\plus{}}$), then we also have $ f(x)\equal{}f(x\plus{}1)$ so to disprove half of the possible periods we merely need to show that $ 1$ is not a period.  Easy enough to do, but we are stuck for the other half of the cases.

... Better idea.  Lets take a different approach.  If the function has period $ p$, then $ f(0)\equal{}f(p)$.  Substituting,  we have $ 2 \sin(\pi/2) \plus{} 3\cos(0) \equal{} 5 \equal{} 2\sin(4\pi p \plus{} \pi/2) \plus{} 3\cos(5\pi p)$.  Since this value is the maximum possible value of the function, we know that both $ \sin(4\pi p \plus{} \pi/2) \equal{} 1$ and $ \cos(5\pi p) \equal{} 1$ must be true.

For the first equation, we have $ 4 \pi p \plus{} \pi/2 \equal{} \pi/2 \plus{} 2\pi m$ for $ m \in \mathbb{Z}$, so $ p \equal{} m/2$.

For the second equation, we have $ 5 \pi p \equal{} 2 \pi n$ for integer $ n$ and hence $ p\equal{}2n/5$.

So we are left to find the smallest $ p$ such that there exists $ m,n \in \mathbb{Z}$ that $ m/2 \equal{} 2n/5$.  For $ m\equal{}1,2,3,4...$ we have $ p\equal{}1/2, 1, 3/2, 2...$ and for $ n\equal{}1,2,3,4,5...$ we have $ p\equal{}2/5, 4/5, 6/5, 8/5, 2...$.  Since $ 2$ is the smallest number in both of these lists, it is our answer.

[quote="metafor"][b]facis[/b] — I think you can turn what you said into a definition of the LCM that can be extended to the rationals (which, in my opinion, is as far a stretch as you can make with the concept).[/quote]
From a number-theoretic perspective, perhaps, but this particular definition of LCM is tailored to making statements about periods of periodic functions.  Given a set of commensurable periods, we can scale by the appropriate constant to make them all simultaneously rational (or, for that matter, integer) and apply the "rational" (or integer) concept of LCM to solve a problem, or we can define LCM in a scale-invariant way and skip that step.


MATHCOUNTS
AMC
AIME
AMC 10
AIME II
AIME I
AMC 8

how many math trophies/plaques/medals do you have? please don't count things like certificates and pins and stuff.

I probably don't have more than 15. idk they're all in a box in lafayette :| though my recent amc 10 medal is on my desk :)

I have 26+, but only because my state has so many tournaments.

yeah i have 16

3 plaques (2 ICTM, 1 IML)
6 trophies (All mathcounts)
7 medals (2 ICTM, 2 Mathcounts, 2 CML, 1 OML)

i just counted those triangular mathcounts awards as trophies.

Well, I didn't know anything about AMC or MathCounts until 7th grade, so it makes sense that you, cognos, were far superior to me in 7th grade in those competitions.
But anyways, with some good catching up (math competition-wise) in 8th grade, I think I have about 5-10 trophies (although I can't exactly count, seeing as I'm in a foreign country right now).

But I WILL try to get a perfect score on AMC 10 next year, and then we'll see who's laughing (directed to logic812 and MustItAlwaysBeMe, because cognos will most likely get perfect as well if I do).

And they made the ICTM competition's Comprehensive Test the hardest on the one and only official year that I can take it (compare a score of 31/32 on a past exam I took to 11/32...).

Whenever I hear the words 'math competition,' I think of humiliating defeat from logic812 and MustItAlwaysBeMe (AMC 10 and IML, respectively), not-so humiliating defeat from cognos (a [i]lot[/i] of competitions, but he's smart so it's fine), and just pure bad luck (17th place at State MathCounts; 1 place away from an individual trophy and 1 place away from participating in the Countdown Round).  Oh well, I have four more years left.

good job cognos you have 16 pieces of cheapo plastic indicating your failures

did you get one?

MathAndKnowledge and i are gonna own next year.

[quote="MathAndKnowledge"]Ok, I have to say that cognos and I can at least pull off a 144, and if we can do that, why not a 150?
I got an 8 on the AIME II for this year (I took it unofficially about three weeks ago), and apparently everyone says that was much harder than the AIME I, which I got a 5 on.
If my AIME score improves by 3 in a matter of 3 months, then how fast do you think my AMC score improves (and I've been getting 138's and 144's on those, recently)?  I've still got 7-8 months; I think a perfect is achievable.[/quote]

i got a 19 on AMC 8 this year and i think making IMO next year is perfectly achievable.  The only problem is that i actually have to do it.

i can at least pull off a 100, so why not 106. Maybe i'll guess a couple right and get 118. Heck, i could just happen to have seen all those problems before and get a 150.

8 on aime doesn't mean 150. I got more than 8 and I didn't get 150. AIME II isn't much harder (if harder) than AIME I. The average is the same.

Irrelevant comments removed. Let's keep all future comments on-topic ;)

yeah, just a question, how hard is the Rose Hulman competition?

I don't remember too much, but it was probably AMC 10-12 level.

Is it like something that I'll completely crumble under?
(AMC 10A: 120 AIME I: 1  :P  :P )

EDIT: please check your PM box if you already haven't

[color=green]Comment removed by moderator. [i]Please[/i] keep on-topic. Locked for two days.[/color]

MathAndKnowledge: How can he not take hard feelings from that post? You insulted his ability to think logically multiple times within your post... and you're assuming he was being sarcastic (which he probably was but  :P).


induction function logarithms inequalities

Let $n>2$ a positive number.
Prove that :

$n^{n+1}>(n+1)^n.$

------------------
Hint: Induction Principle!!

Faby      ;)
Lima -Peru

$n^{n+1}>(n+1)^n \Rightarrow n^n \times n > (n+1)^n$ dividing over, $n> (\frac{n+1}{n})^n$ and (assuming we know that the right goes to $e$, which can be proven easily enough) we know this is true $\forall n>3$ since $3>e \approx 2.718$ since the right is always less than $e$, and the left is always greater than 3.

Hello,fabycv :)

I don't find solution using Induction Principle yet,but another solution  :blush:
[hide]
use function $f(x)=\frac{\log x}{x}$ .this declines $x>e$,so we can say
　　　　$\frac{\log (n+1)}{n+1}<\frac{\log n}{n}$
　　　　$\longleftrightarrow n^{(n+1)}>(n+1)^n$
[/hide]

[hide="A long proof"]
As Hokkage wrote, we need to prove $n^{n+1}>(n+1)^n\Leftrightarrow n>\left(\frac{n+1}{n}\right)^n \Leftrightarrow \left(1+\frac{1}{n}\right) ^n <n$ for all $n>2$, or $\left(1+\frac{1}{n}\right) ^n <3$.

The key here is to use binary expansion.
$\left(1+\frac{1}{n}\right)^n =\sum_{k=0}^{n} \frac{{}_n C_k}{n^k} =1+\sum_{k=1}^{n} \frac{{}_n C_k}{n^k}$

Then we use $k!\ge 2^{k-1}$ to prove that $\frac{{}_n C_k}{n^k}\le \frac{1}{2^{k-1}}$.
$k!\ge 2^{k-1}\iff \frac{2^{k-1}}{k!}\le 1$
So $2^{k-1}{}_n C_k =\frac{2^{k-1}}{k!}\cdot \frac{n!}{(n-k)!}\le \frac{n!}{(n-k)!} =n(n-1)(n-2)...(n-k+1) \le n^k$
Hence $\frac{{}_n C_k}{n^k}\le \frac{1}{2^{k-1}}$.

Therefore,
$1+\sum_{k=1}^{n} \frac{{}_n C_k}{n^k} \le 1+\sum_{k=1}^{n} \frac{1}{2^{k-1}} =1+2\left(1-\left(\frac{1}{2}\right)^n\right)<3$

$\therefore \left(1+\frac{1}{n}\right)^n <3$
[/hide]
Maybe a proof with induction partially uses this idea...

to prove that $\left(1+\frac{1}{n}\right)^n<3$ , we can use Bernoulli Inequality

$3^{\frac{1}{n}}=(1+2)^{\frac{1}{n}}>1+\frac{2}{n}>1+\frac{1}{n}$

 :)

[quote="fabycv"]Let $n>2$ a [b]positive number[/b].
Prove that :

$n^{n+1}>(n+1)^n.$

Faby      ;)
Lima -Peru[/quote]
NOT Always true. It is true if $n$ is  a positive [b]integer[/b]
(Try $n=2.1$ and you would see this is not true) :)  (for a mere [i]number[/i] n has to be > e)

[quote="scorpius119"][quote="shyong"]to prove that $\left(1+\frac{1}{n}\right)^n<3$ , we can use Bernoulli Inequality

$3^{\frac{1}{n}}=(1+2)^{\frac{1}{n}}>1+\frac{2}{n}$

 :)[/quote]

wait, wouldn't that 'prove' $e^2\leq 3$ (which is clearly false)?![/quote]

opps ! I forget that the sign is reverse in Bernoulli inequality when the exponent is between $0$ and $1$ .. :oops:

[hide="Solution using induction"]
Assumption $n^{n+1}>(n+1)^n$
Thesis: $(n+1)^{n+2}>(n+2)^{n+1}$
Let's transform assumption and thesis equivalently:

We multiply the assumption by $(n+1)^{n+2}$ and we get:
A: $(n+1)^{n+2}n^{n+1}>(n+1)^{2n+2}$
Now multiply thesis by $n^{n+1}$
T:$(n+1)^{n+2}n^{n+1}>(n+2)^{n+1}n^{n+1}$
Now it suffices to prove, that:
$(n+1)^{2n+2}>(n+2)^{n+1}n^{n+1}$, which is equivalent, to: $(n+1)^2>n(n+2)$, which is obviously true.

  [/hide]

Like Gyan said , the result is only true for $n\geq e$ since the function $f(x)=x^{\frac{1}{x}}$ is increasing in $(0,e]$ but decreasing in $[e,\infty)$


1,Find all $ n$ is belong to N satisfy $ sin n$degree isbelong to Q
2,Find all $ n$ is belong to N satisfy $ cos n$degree is belong to Q
3,Find all $ n$ is belong to N satisfy $ tan n$degree is belong to Q

 :D  :D   My English is weak    :D  :D

[hide="Hints"]$ \sin x \equal{} \frac{e^{ix} \minus{} e^{\minus{}ix}}{2i}$

$ \cos x \equal{} \frac{e^{ix} \plus{} e^{\minus{}ix}}{2}$[/hide]


It really reminds of the type proposed by Vasc to be solved by his LCF/RCF theorem :)


probability
advanced fields
advanced fields unsolved

A message it's send from one of the n chanels of comunication, grouped in 4 groups, on their  grad of qualityes based on the power of the transmition. Those  groups countain n1, n2, n3, n4 chanels, and n1+n2+n3+n4=n. The probability that one message is correctly transmited for a chanel from i group it is Pi, where i=1,2,3,4. For a good transmition, the chanel is the same for each one of them, and this selection is indipendent from the quality of the transmition. Find the probability that the message to be transmited correctly on at last  one chanel. :?:

Unless I missed something that makes this problem more complicated, the probability of at least one successful transmission is 1 minus the probability of all transmissions failing. That is, your probability should be $1 - P_1P_2P_3P_4$.

Oh..Yes....That's right!
Thank uuuuuuuuuuuuuuuu :D  i didn't realise at first  :P


A  ball 1.001 kg is in 30 m distance from the earth. What is the cynetic energy spended when the ball is in 2m distance from the earth.

As the gravitational field is conservative, then Em = Ec + Ep or $ \Delta E_c \equal{} \minus{} \Delta E_p \equal{} mg(h_1 \minus{} h_2) \equal{} 1.001 \cdot 9.8 \cdot (30 \minus{} 2) \equal{} 275\,J$

what is the real force who the earth pull back the things in to it $ 10N/kg$ or $ 9.8N/kg$ ?

The acceleration of gravity g on the Earth's surface, which is given by $ g \equal{} \frac{G \cdot M_T}{r_T^2}$, equals 9.80665 m/s^2, but this value is commonly approximated to 10 m/s^2.


Prove that if roots of the quadratic equation $ ax^2 +\ (\ a +\ 2 )\ x +\ 1 = 0 $ are of the form $ \alpha +\ k \ and\ -\ \beta -\ k $, then the roots of $ ax^2 +\ (\ 2a +\ 3 )\ +\ (\ 1 -\ a )\ = 0 $ can not be of the form $ \alpha -\ s \ and \ \beta -\ s $ and vice-versa for any real value of a.

What is $k$, for example? And $s$? And $\alpha$?

$\alpha \not= a $. Alos $k$ and $s$ are arbitrary and just indicate the form of the roots of the eqn

So basically we just need to show that the sum of the roots of the first equation is never equal to the difference of the roots of the second equation, when $a$ takes all real values. The sum of the roots of the first equation is equal to $-\frac{a + 2}{a}$. The difference of the roots of the second equation is equal to $\pm\frac{\sqrt{\Delta}}{a}.$ So we need to show (after squaring) that $\Delta = (a + 2)^2$ does not have any real solutions. Since $\Delta = (2a + 3)^2 + 4a(a - 1) = 8a^2 + 8a + 9$ and $(a + 2)^2 = a^2 + 4a + 4$ this reduces to \[7a^2 + 4a + 5 = 0\] which indeed doesn't have any real solutions. Tell me, is this the correct interpretation of this question? I am still not sure about its meaning.

I do believe you are interpreting it correctly . i don't know the correct solution but your does seem right!!!


geometry
angle bisector
geometry proposed

Let $ \triangle ABC$ be an isosceles triangle with $ AB\equal{}AC$. Suppose that the angle bisector of $ \angle B$ meets $ AC$ at $ D$ and that $ BC\equal{}BD\plus{}AD$. Determine $ \angle A$.

1996 Canada, #4:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77673[/url]


The population of Laughland increased from 711,117 to 732,666. What was the percent increase in the population? Round your answer to the nearest percent.

The percentage increase is given by $ \frac{732666\minus{}711117}{711117}\equal{}.\overline{03}\approx .03$.  So about $ 3\%$.


Prove that for any two positive integers then $ (36a\plus{}b)(a\plus{}36b)$ can't be a power of $ 2$.

[hide="Solution"]
For $ (36a \plus{} b)(a \plus{} 36b)$ to be a power of 2, both factors must be also power of two, since $ 36a \plus{} b, 36b \plus{} a \geq 37$. If $ 36a \plus{} b$ is a power of two, $ 4(9a \plus{} b')$ where $ 4b' \equal{} b$ and $ b' \in \mathbb{N}$ must be also true. Since $ 9a \plus{} b' > 1$, $ 9a \plus{} b'$ must also be a power of two. From this, we find that $ a$ and $ b'$ must be of the same parity.

$ a \plus{} 36b \equal{} a \plus{} 144b'$ must also be a power of 2. If $ a$ and $ b'$ are both odd, we immediately find a contradiction, since $ a \plus{} 144b'$ is odd.

If $ a$ and $ b'$ are both even, it is possible to precipitate more powers of two. The same logic applies to the other case, as well. Similar logic continues indefinitely. Since both $ a$ and $ b$ cannot "equal" infinity. By infinite descent, the statement cannot be true.
[/hide]

[quote="scai1"]I don't understand aqua's solution. Why does it have to be greater than 37?

Sorry I am not so good at math ;)[/quote]
It should be greater than 37 not greater than or equal to since and b are positive integers. Since these two numbers are integers 36a+b is greater than 37. The minimum of this equation occurs when a=b=1 or at 38.

Let me rewrite the second part of the proof, since last time I pretty much handwaved.

[hide="Solution+"]
Proved: a, b' must have same parity

$ a\plus{}36b\equal{}a\plus{}144b'$ must be also power of 2. Therefore, $ a$ and $ b'$ cannot be both odd.

Thus, we assume that they are even. Let $ a\equal{}2^m\alpha$ and $ b'\equal{}2^n\beta$, where $ \alpha$ and $ \beta$ are both odd numbers. Plugging these into $ a\plus{}144b'$ gives $ 2^m\alpha\plus{}9\cdot 2^{n\plus{}r4}\beta$. If $ m>n\plus{}4$, this factor will always be odd ($ 2^{m\minus{}n\minus{}4}\alpha \plus{} 9\beta$). The similar applies to $ m<n\plus{}4$. We conclude that $ m\equal{}n\plus{}4$.

Now go back to $ 36a\plus{}b\equal{}36a\plus{}4b'\equal{}4(9a\plus{}b')$. Our new equations tell us that $ 9a\plus{}b\equal{}9\cdot 2^{n\plus{}4}\alpha\plus{}2^{n}\beta$, which would also have to be a power of two. Interestingly, this factors to $ 2^n(16\alpha\plus{}\beta)$. This cannot be a power of two, since $ \beta$ was defined earlier to be odd.

Thus, $ (36a\plus{}b)(a\plus{}36b)$ cannot a power of two.
[/hide]


analytic geometry
graphing lines
slope
vector
LaTeX
trigonometry

A truck weighing 30,000 pounds is parked on a slope. the slope is d degrees from the horizontal. Assuming that the only force to overcome is gravity, find the force required to keep the truck from rolling down the hill in terms of d.

Are we assuming no friction on the slope?

If we eliminate friction from the problem:

[hide]
The force of gravity is mg. But since we're on a slope, we must consider the downward vector of the force. This is equal to mg sind. This is the force of gravity acting on car and you must counteract with the same force (in the opposite direction). I will leave you to plug in the numbers.

Note: When working with this problem, make sure to set a correct positive direction. This will determine the signs involved with the forces.
[/hide]

If you want to include friction in the problem, you will have to use the frictional constant, mew, and make some changes to this approach.

Ok, sorry I'm a beginner on this forum so I don't know how to use Latex or the "View hidden content" thingy whatever it is, but the answer is (mgsin(D)) because assuming there is no friction or outside forces, there are three forces on the truck: the force of gravity mg, the normal force, and the force downwards mgsinD (it is sine and not cosine since it is opposite to the angle D). Now, your answer is 18000*sinD whatever the angle D is. [/code][/hide]


Clearing the denominators of the given condition and desired inequality, we are to prove $abc\sum a^2b\ge\sum ab$ if $abc\sum ab\ge\sum a^2b^2$. It suffices to consider only the following\[abc\sum a^2b(\sum a^2b^2)^4\ge\sum ab(abc\sum ab)^4\]This follows trivially from\[\sum a^2b^2\ge\frac{(\sum ab)^2}{3}\]

[quote="silouan"]Let $a,b,c>0$ such that $ab+bc+ca\geq\frac{ab}{c} +\frac{bc}{a} +\frac{ca}{b}$.Prove that:

 $a^2b+b^2c+c^2a\geq\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$[/quote]

 the given condition is equivalent to
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$
We have
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq \frac{1}{3}\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)^2$
hence $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq 3 \Rightarrow 3abc\geq ab+bc+ca\geq 3\sqrt[3]{(abc)^2}\Rightarrow abc\geq1$
 $a^2b+b^2c+c^2a\geq 3abc \geq 3\ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
.

[quote="mecrazywong"][quote="silouan"][quote="mecrazywong"] It suffices to consider only the following\[abc\sum a^2b(\sum a^2b^2)^4\ge\sum ab(abc\sum ab)^4\]This follows trivially from\[\sum a^2b^2\ge\frac{(\sum ab)^2}{3}\][/quote]
Could you use more details? I din't understand it[/quote]
Use AM-GM for $\sum a^2b$. I think it is [i]really[/i] rather easy now.[/quote]
After this I found that we should prove that $3(a^2b^2+c^2a^2+b^2c^2)^4\geq (ab+bc+ca)^5(abc)^2$  :?:

[quote="silouan"][/quote][quote="mecrazywong"][quote="silouan"][quote="mecrazywong"] It suffices to consider only the following\[abc\sum a^2b(\sum a^2b^2)^4\ge\sum ab(abc\sum ab)^4\]This follows trivially from\[\sum a^2b^2\ge\frac{(\sum ab)^2}{3}\][/quote]
Could you use more details? I din't understand it[/quote]
Use AM-GM for $\sum a^2b$. I think it is [i]really[/i] rather easy now.[/quote][quote="silouan"]
After this I found that we should prove that $3(a^2b^2+c^2a^2+b^2c^2)^4\geq (ab+bc+ca)^5(abc)^2$  :?:[/quote]
Man, at least give a try. This is absolutely trivial.

This is the first time and also the last time I'm explaining something like this in mathlinks.
$3(\sum a^2b^2)^4\ge3(\sum ab)^8/3^4=(\sum ab)^5(\frac{\sum ab}{3})^3\ge(\sum ab)^5(abc)^2$.


Solve equation:

$ sinx\plus{}sin2x\plus{}sin3x\equal{}1$

I've tried but i couldn't get the solution...however here it is

$ \sin x \plus{} \sin 2x \plus{} \sin 3x \equal{} 1$
$ \sin x \plus{} 2\sin x \cos x \plus{} \sin x(3\minus{}4\sin^2 x) \equal{} 1$
$ \sin x (1 \plus{} 2\cos x \plus{} 3 \minus{} 4 \sin^2 x) \equal{} 1$
$ \sin x (4 \plus{} 2\cos x\minus{} 4 (1\minus{}\cos^2 x)) \equal{} 1$
$ \sin x (4 \plus{} 2\cos x\minus{} 4 \plus{} 4\cos^2 x)) \equal{} 1$
$ \sin x (2\cos x \plus{} 4\cos^2 x)) \equal{} 1$
$ 2\sin x\cos x (1 \plus{} 2\cos x) \equal{} 1$
$ \sin 2x (1 \plus{} 2\cos x) \equal{} 1$
and then?

or

$ \sin x (4 \plus{} 2\cos x\minus{} 4 \sin^2 x) \equal{} 1$
$ 2\cos x \equal{} \frac{1}{\sin x} \minus{} 4 \plus{} 4 \sin^2 x$
$ (2\cos x)^2 \equal{} (\frac{1}{\sin x} \minus{} 4 \plus{} 4 \sin^2)^2$
$ 4(1 \minus{} \sin^2 x) \equal{} \frac{1}{\sin^2 x} \minus{}16 \plus{} 16\sin^4 x \minus{} \frac{8}{\sin x} \plus{} 8\sin x \minus{} 32 \sin^2 x$
$ 1 \minus{} 16\sin^2 x \plus{} 16 \sin^6 x \minus{} 8 \sin x \plus{} 8 \sin^3 x \minus{}32 \sin^4 x \minus{} 4 \sin^2 x \plus{} 4 sin^4 x\equal{}0$
$ 16\sin^6 x \minus{} 28 \sin^4 x \plus{} 8 \sin^3 x \minus{} 20\sin^2 x \minus{} 8\sin x \plus{} 1 \equal{} 0$

But i can't solve this 6th degree equation...

hello, applying the addition formulas we get
$ 2\sin(x)\cos(x)\plus{}4\sin(x)\cos^2(x)\minus{}1\equal{}0$.
Substituting $ \tan\left(\frac{x}{2}\right)\equal{}t$ we have to solve
$ t^6\minus{}4t^5\plus{}3t^4\plus{}16t^3\plus{}3t^2\minus{}12t\plus{}1\equal{}0$.
Sonnhard.

well, the only possible rational roots of that are 1 and -1, neither of which work.


Let a_1...a_n be n complex numbers.Prove that if for any subset A of{1;2...n} we have
$|\prod_{i\in S}(1+a_i)-1|\leq\frac{1}{2}$ then $\sum_{i=1}^{n}|a_i|\leq\ 3$


$ 6\left(x \minus{} \frac {y}{3}\right)\left(x \plus{} \frac {y \minus{} 1}{2}\right)$.


What do people think the index cutoff for USAMO will be?  I would say this AIME is about the same difficulty as the 2006 AIMEI.  That year the index was 217.5, but this year has a harder AMC and a larger USAMO, so I would predict maybe around 210?  Of course, I have a 216, so I'm biased . . .  :P

I say the floor will be 8 and the index cutoff will be 210. The AIME was pretty easy this year.

I would say like 212

ill add 20 points to last years AIME because it was about 2 questions easier, but ill subtract like 5 points because the AMC 12 was harder.

And i know someone that got a 144/7, which means that for 10th graders, the floor is likely 7, but if there are too many people, they might bump the index up to like 200ish..

Nuts....my index sucks but my AIME was ok...

mmmm...last year the index was 186, and that was the first year the field expanded, so I'm going with 195, because the indices can't really be compared to years past besides last year. (probably because i got a 197)

Last year, I think it was a 197, so I'm pretty sure that a 216 will make it.  I would be very confident.

In terms of the AIME, I think it was of comparable difficulty to previous years, although a bit easier.  And the AMC was harder.  So I'm even thinking around 205 or so, but I could be wrong.

i think the floor will be 7 or 8 and the index will be like 220.

[quote="E^(pi*i)=-1"]Um, if the index is 220, it's impossible for the floor to be 7 . . .[/quote]It is possible...$ 150\plus{}7\cdot 10\equal{}150\plus{}70\equal{}220$.

Hah, I hope it's like 200, that way, I'll qualify via index, and that'd be so cool. But it won't be. Oh well, I can still qualify via floor.
Maybe around 210, yeah, that's what I would think. It's hard to tell now though.

It's probably lower than people think, but I would guess 200-210.

[quote="123456789"][quote="E^(pi*i)=-1"]Um, if the index is 220, it's impossible for the floor to be 7 . . .[/quote]It is possible...$ 150 \plus{} 7\cdot 10 \equal{} 150 \plus{} 70 \equal{} 220$.[/quote]

Yeah, you're right.  I'm not sure what I was thinking.

But anyway, I really don't think the index will be 220.  People don't seem to be scoring much higher overall than in 2006, and other factors are dragging it down.

I shouldn't qualify through the 197 index, but probably through the 7-8 floor

I'm not gonna qualify with an index of 189.5 and an AIME of 5.
...

If this AIME is the same difficulty as 2006 AIME's, with a harder AMC (-2)+ lower scoring system(-3.5)+ 100 extra ppl(-5), index cutoff will likely be 205.

You don't need to meet the regular index, but lately they have been setting a secondary index attached to the floor, so that not everyone who gets the floor makes it.

[quote="Zmastr"]And for sophomores and below, only the floor needs to be met to make the USAMO, not the index, right?[/quote]

Yes, unless a restriction is added like last year to keep the number of people that qualify to around 500.
(last year it was 6 AIME + 171 index or something)

I completely failed AIME, could've gotten 11 but I was being dumb so 7. I'm hoping someone with a high AMC score fails like me...
Cutoff will probably be 8...a lot of random people I know who aren't that good at math got 9s, but hopefully someone will have made tons of dumb mistakes, then I get into usamo woot.

Index 210, Floor 8 would be my guess. This was probably one of the easier AIMEs ever but the floor has only been 9 once and that was before the expansion.


inequalities
induction
inequalities proposed

a1,a2,...,an  distinct  positive  integers. Prove  that
Σai^7 + Σai^5  \geq 2[Σai^3] 2

i think this inequality can be proved by induction
for $n=1$
it becomes $a^7+a^5 \geq 2a^6$
assume that it's true for $n$ we will prove it for $n+1$
$\displaystyle \sum_{i=1}^n a_i^7 + \sum_{i=1}^n a_i^5 +a_{n+1}^7+a_{n+1}^5 \geq 2(\sum_{i=1}^n a_i^3)^2 + 2a_{n+1}^6+4(\sum_{i=1}^n a_i^3)a_{n+1}^3$
but we have $\displaystyle \sum_{i=1}^n a_i^7 + \sum_{i=1}^n a_i^5 \geq 2(\sum_{i=1}^n a_i^3)^2$
so all we have to do is to show that $a_{n+1}^7+a_{n+1}^5 \geq 2a_{n+1}^6+4 (\displaystyle \sum_{i=1}^n a_i^3)a_{n+1}^3$
which is equivalent to
$a_{n+1}^4+a_{n+1}^2 - 2a_{n+1}^3-4\sum_{i=1}^n a_i^3 \geq 0$
we may assume that $a_1\leq a_2\leq ... \leq a_n = k \leq a_{n+1}$
so the left hand side is bigger than
$(k+1)^4+(k+1)^2-2(k+1)^3-4\sum_{i=1}^n i^3 = 0$
so it is done.
 :D


linear algebra
matrix
vector
algebra
polynomial
modular arithmetic
trigonometry

i have two new matrix which i have to find eigenvalues without using det(sI-A)=0 but in formal way.
Matrix are here
1.
0 1 1 1 1
1 0 0 0 0
1 0 0 0 0
1 0 0 0 0
1 0 0 0 0

2.
0 1 0 0 1
1 0 1 0 0
0 1 0 1 0
0 0 1 0 1
1 0 0 1 0

Let's write the first one in a block-structured way. Let $ j$ be the $ 4\times 1$ column vector of all $ 1$'s. We can write

$ A=\begin{bmatrix}0&j^T\\j&0\end{bmatrix}.$

Then $ A^2=\begin{bmatrix}{j^Tj}&0\\0&jj^T\end{bmatrix}=
\begin{bmatrix}4&0\\0&J\end{bmatrix}$ where $ J$ is the $ 4\times 4$ matrix lf all $ 1$'s.

Then $ A^3=\begin{bmatrix}4&0\\0&J\end{bmatrix}\begin{bmatrix}0&j^T\\j&0\end{bmatrix}
=\begin{bmatrix}0&4j^T\\4j&0\end{bmatrix}=4A.$

Then $ A$ satisfies the polynomial equation $ A^3-4A=0.$ Its eigenvalues must be among the roots of that polynomial, namely $ 0,2,-2.$ But $ A$ is evidently of rank $ 2.$ That means the eigenspace associated to $ 0$ is three dimensional. And the trace of $ A$ is zero, so the eigenvalues add up to zero.

The final list of eigenvalues of $ A:$ $ 0,0,0,-2,2.$

I'll leave the second for a little while, but I will note that it is a "circulant" - a matrix constant on "wrap-around" diagonals. That is, $ a_{ij}=a_{kl}$ iff $ j-i\equiv l-k\pmod{n}.$ Such matrices have numerous special properties.

Kent Merryfield thank you for solving.
I am waiting for the second.

Let $ B$ be the matrix in your problem.

Let $ C \equal{} \begin{bmatrix} & 1 & & & \\
& & 1 & & \\
& & & 1 & \\
& & & & 1 \\
1 & & & & \end{bmatrix},$ where all of the missing entries are zero.

This is essentially the simplest possible nontrivial circulant matrix; all other circulants are polynomials in it.

$ C^2 \equal{} \begin{bmatrix} & & 1 & & \\
& & & 1 & \\
& & & & 1 \\
1 & & & & \\
&1 & & & \end{bmatrix},$ $ C^3 \equal{} \begin{bmatrix} & & & 1 & \\
& & & & 1 \\
1 & & & & \\
&1 & & & \\
& & 1 & & \end{bmatrix},$ and $ C^4 \equal{} \begin{bmatrix} & & & & 1 \\
1 & & & & \\
&1 & & & \\
& & 1 & & \\
& & & 1 & \end{bmatrix}.$

Then $ C^5 \equal{} I.$ We also have that $ C^T \equal{} C^4 \equal{} C^{ \minus{} 1}.$

$ B \equal{} C \plus{} C^T \equal{} C \plus{} C^{ \minus{} 1}.$ Square that to get $ B^2 \equal{} C^2 \plus{} 2I \plus{} C^{ \minus{} 2}.$

You can determine everything about a circulant just from its top row. The top row of $ B$ is $ \begin{matrix}0 & 1 & 0 & 0 & 1\end{matrix},$ while the top row of $ B^2$ is $ \begin{matrix}2 & 0 & 1 & 1 & 0\end{matrix}.$ Add those together to get that the top row of $ B^2 \plus{} B$ is $ \begin{matrix}2 & 1 & 1 & 1 & 1\end{matrix}$

That is, $ B^2 \plus{} B \equal{} I \plus{} J.$ Multiply that by $ B$ to get $ B^3 \plus{} B^2 \equal{} B \plus{} 2J.$

Subtract $ 2(B^2 \plus{} B \minus{} I) \equal{} 2J$ from $ B^2 \plus{} B^2 \minus{} B \equal{} 2J$ to get $ B^3 \minus{} B^2 \minus{} 3B \plus{} 2I \equal{} 0.$

So we now have a polynomial that this matrix satisfies: $ x^3 \minus{} x^2 \minus{} 3x \plus{} 2.$ This does appear to be the minimal polynomial. This factors as $ (x \minus{} 2)(x^2 \plus{} x \minus{} 1).$

That means the eigenvalues are $ 2$ and $ \frac { \minus{} 1\pm\sqrt {5}}2.$

The only question is how many of each. We do have that the trace of $ B$ equals zero, so the eigenvalues have to add up to $ 0.$ There's only one way to do this with positive integer coefficients adding up to $ 5,$ which is that the eigenvalues of $ B$ are $ 2,\frac { \minus{} 1 \plus{} \sqrt {5}}2,\frac { \minus{} 1 \plus{} \sqrt {5}}2,\frac { \minus{} 1 \minus{} \sqrt {5}}2,\frac { \minus{} 1 \minus{} \sqrt {5}}2.$

And the characteristic polynomial is $ (x\minus{}2)(x^2\plus{}x\minus{}1)^2\equal{}x^5\minus{}5x^3\plus{}5x\minus{}2.$

If you consider $ B$ to be the incidence matrix of a graph, then the graph in question would be a pentagon. Somehow, that makes the appearance of golden means among the eigenvalues all the more plausible.

True this is pentagon and first is star-node graph.
Let B =n-node circle connection matrix. How can we generalize the eigenvalues?
By the way thank you for the solution.

[quote="ozlem81"]Let B =n-node circle connection matrix. How can we generalize the eigenvalues?[/quote]
This time, I'll go after the eigenvectors rather than the eigenvalues, and I'll put on my Fourier analysts hat. All circulants are diagonalized by the discrete Fourier transform.

Number the rows and columns of your $ n\times n$ matrix as $ 0,1,\dots,n\minus{}1.$ This convention works better with arithmetic mod $ n.$ Your "$ n$-node circle connection matrix" I take to be $ A$ whose entries $ a_{jk}$ satisfy $ a_{jk}\equal{}1$ if either $ j\equiv k\plus{}1\pmod{n}$ or $ j\equiv k\minus{}1\pmod{n}.$

For fixed $ m\in\{0,1,\dots,n\minus{}1\}$ let vector $ v^{[m]}$ have elements $ v^{[m]}_j\equal{}e^{2\pi ijm/n}\equal{}\omega^{jm}$ where $ \omega\equal{}e^{2\pi i/n}$ is a primitive $ n$th root of unity.

Then $ (Av^{[m]})_j\equal{}e^{2\pi i (j\plus{}1)m/n}\plus{}e^{2\pi i(j\minus{}1)m/n}$

$ \equal{}e^{2\pi ijm/n}\left(e^{2\pi im/n}\plus{}e^{\minus{}2\pi i m/n}\right)\equal{}
2e^{2\pi ijm/n}\cos\left(\frac{2\pi m}n\right).$

So $ Av^{[m]}\equal{}2\cos\left(\frac{2\pi m}n\right)v^{[m]}.$

Hence the eigenvalues of $ A$ are $ 2\cos\left(\frac{2\pi m}n\right)$ for $ m\equal{}0,1,\dots,n\minus{}1.$ However, since cosine is an even function, we get the same thing for $ n\minus{}m$ as for $ m.$

We always have a single eigenvalue of $ 2.$ If $ n$ is even, we have a single eigenvalue (corresponding to $ m\equal{}n/2$) of $ \minus{}2.$ Then for each $ m,\ 0<m<\frac{n}2,$ we get $ 2\cos\left(\frac{2\pi m}n\right)$ as an eigenvalue of multiplicity 2.

In for the case worked above, with $ n\equal{}5,$ we should have a single eigenvalue of $ 2$ and doubled eigenvalues of $ 2\cos\left(\frac{2\pi}5\right)$ and $ 2\cos\left(\frac{4\pi}5\right).$ Matching this to what we got, we evidently have $ \cos\left(\frac{2\pi}5\right)\equal{}\cos(72^{\circ})\equal{}\frac{\minus{}1\plus{}\sqrt{5}}4$ and $ \cos\left(\frac{4\pi}5\right)\equal{}\cos(144^{\circ})\equal{}\frac{\minus{}1\minus{}\sqrt{5}}4$


Hello!

we define the sequence $(x_{n})_{n\in{\Bbb N}}$ by: $x_{0}=\alpha, 0<\alpha<1$, $x_{1}=\alpha^{2}$, and $x_{n+2}=\alpha(x_{n+1}+\alpha x_{n})$ for all $n\geq 2$. Can you compute the following sum
\[\sum_{n=0}^{+\infty}{{x_{n}}\over{n+1}}.\]

It is a sequence defined by a recurrence of degree 2, simply solve the equation and by using the initial conditions you should get
$x_{n}=A\alpha^{n}+B\beta^{n}$, where $\alpha$ and $\beta$ are the solutions of the characteristic equation. Then find A and B by solving the initial condition.

Now your sum is of the form $\sum_{n=0}^{\infty}\frac{\alpha^{n}}{n+1}$ which can be calculated by integrating the classical sum equality

$\sum_{n=0}^{\infty}x^{n}=\frac{1}{1-x}$ for $|x|<1$.
It seems to me it is a computational problem.


Pentru ca e week-end, fac o exceptie de la regula (a se citi super-oferta) si mai pun doua probleme algebra-analiza, pentru cei interesati (???):
  1) Sa se arate ca daca $n\geq 3$, orice matrice de determinant $1$ peste un corp se poate scrie ca produs de matrici de ordin $4$.
  2) Fie $f_{n}$ un sir de functii crescatoare din $R$ in $[0,1]$. Sa se arate ca putem extrage un subsir $g_{n}$ astfel incat sirul $g_{n}(x)$ sa convearga pentru orice $x$.

La 1) matricile sunt din $M_{n}(K)$? ($n\geq 3$). Ordinul este 4 inseamna $X^{4}=I$?

Da si nu. Da pentru M_n(K), nu pentru ordin: trebuie ca $X^{4}=I$ si $X^{2}$ sa nu fie $I$.

Problema 2 a mai fost postată [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=sequence+increasing&t=21893]aici[/url].

M-am gândit cam la acelaşi lucru, dar m-am blocat la partea asta:
[quote="Myth"]It easily follows that f^k converges in all real points. [/quote]


trigonometry
trig identities
Law of Cosines

A boat travels $ x$ distance to the west, then he turns $ 30^\circ$ respective to the west and goes $ 42m$ and now he is $ 62m$ away from the point where he started. What is $ x$?

[hide="Solution"][geogebra]0035e4948d80fe71492f9994a86bc0b154942c3b[/geogebra]
Using law of cosines, $ 62^2 \equal{} 42^2 \plus{} x^2 \minus{} 2\cdot42\cdot x\cdot\cos150^{\circ}.$

Solving for $ x$, we obtain $ x\approx 46.25$

Answer: $ 46.25$[/hide]

EDIT: Oops, typo.

Thanks for pointing out, tonypr. :)

[quote="FantasyLover"]
Using law of cosines, $ 62^2 = 42^2 + x^2 - 2\cdot42\cdot x\cdot\cos150^{\circ}.$
[/quote]

Small typo: $ 62^2 = 42^2 + x^2 - 2\cdot42\cdot x\cdot\cos150^{\circ}.$, it didn't read the $ x$ since there was no space between x and \cdot.


integration
real analysis
real analysis unsolved

Prove that for fixed $ \alpha\in \mathbb{R}$ when $ n\to\infty$
\[ \sum_{k \equal{} 0}^{n}{k^{n \plus{} \alpha}e^{ \minus{} k}} \equal{} n^{n \plus{} \alpha}e^{ \minus{} n}\left(\sqrt {\frac {\pi n}{2}} \minus{} \frac {1}{6} \minus{} \alpha \plus{} \left(\frac {1}{12} \plus{} \frac {1}{2}\alpha \plus{} \frac {1}{2}\alpha^2\right)\sqrt {\frac {\pi}{2n}} \plus{} O(n^{ \minus{} 1})\right).
\]

the first term is just the first order of:
$ \psi(n) \equal{} \int_0^1 \exp(n [ \ln(x) \minus{} x] ) dx$
after one should work out the difference between $ \int_0^1 \exp(n [ \ln(x) \minus{} x] ) dx$ and its Rieman sum, as well as the second order of $ \psi$


real analysis
advanced fields
advanced fields unsolved

Let $n\in \mathbb{N}$, $U=\{X\subset \mathbb{R}^n|X\text { is Lebesgue measurable}\}$ and  $V=\{X\subset \mathbb{R}^n|X\text { is not Lebesgue measurable}\}$. Can we say now that $card(U)<card(V), card(U)>card(V), card(U)=card(V),$ or does it depends for example from the continuum hypothesis?

They're equal; both sets have cardinality $2^c$, where $c$ is the cardinality of the reals.

Let $S$ be an uncountable set of measure zero contained in the unit ball, and let $Q$ be a nonmeasurable set disjoint from the unit ball. Let $f$ map $\mathbb{R}^n$ injectively to $S$.

Given a set $X$, let $G(X)=f(X)\cup Q$. $G(X)$ is nonmeasurable, and $G$ is injective on the power set of $\mathbb{R}^n$.

Given a set $Y$, let $H(Y)=f(Y)$. $H(Y)$ is measurable with measure zero, and $H$ is injective on the power set of $\mathbb{R}^n$.


inequalities
Cauchy Inequality
inequalities proposed

If  $ a,b,c,d$  are  positive numbers, then
$ \frac{c}{a\plus{}3b}\plus{}\frac{d}{b\plus{}3c}\plus{}\frac{a}{c\plus{}3d}\plus{}\frac{b}{d\plus{}3a} \ge 1$.

After you multiply the first fraction (numerator and denominator) by c, the second by d the third by a and the last one by b you can use the Cauchy inequality to prove that
$ LHS >\equal{} (a \plus{} b \plus{} c \plus{} d)^2/(ac \plus{} 3cb \plus{} bd \plus{} 3cd \plus{} ac \plus{} 3ad \plus{} bd \plus{} 3ab)$
So we need to prove that $ (a \plus{} b \plus{} c \plus{} d)^2/(ac \plus{} 3cb \plus{} bd \plus{} 3cd \plus{} ac \plus{} 3ad \plus{} bd \plus{} 3ab) >\equal{} 1 \equal{}>$
$ (a \plus{} b \plus{} c \plus{} d)^2 >\equal{} ac \plus{} 3cb \plus{} bd \plus{} 3cd \plus{} ac \plus{} 3ad \plus{} bd \plus{} 3ab \equal{}>$
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} 2ab \plus{} 2ac \plus{} 2ad \plus{} 2bc \plus{} 2bd \plus{} 2cd >\equal{} ac \plus{} 3cb \plus{} bd \plus{} 3cd \plus{} ac \plus{} 3ad \plus{} bd \plus{} 3ab \equal{}>$
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 >\equal{} ab \plus{} bc \plus{} cd \plus{} ad$ which is true
(you can verify this by multiplying by two and writing it as
$ (a \minus{} b)^2 \plus{} (a \minus{} d)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} d)^2 >\equal{} 0$ which is clearly true)
Do you have another solution in mind?

[quote="soruz"]If $ x_1,x_2,...x_n > 0, n \ge 2$  then prove

$ \frac {x_{1}}{x_{n \minus{} 1} \plus{} (n \minus{} 1)x_{n}} \plus{} \frac {x_{2}}{x_{n} \plus{} (n \minus{} 1)x_{1}} \plus{} ... \plus{} \frac {x_{n}}{x_{n \minus{} 2} \plus{} (n \minus{} 1)x_{n \minus{} 1}}\ge1$.[/quote]

Cauchy-Schwarz works here  :wink:

[quote="salonikios"]
$ (a \plus{} b \plus{} c \plus{} d)^2 > \equal{} ac \plus{} 3cb \plus{} bd \plus{} 3cd \plus{} ac \plus{} 3ad \plus{} bd \plus{} 3ab \equal{} >$
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} 2ab \plus{} 2ac \plus{} 2ad \plus{} 2bc \plus{} 2bd \plus{} 2cd > \equal{} ac \plus{} 3cb \plus{} bd \plus{} 3cd \plus{} ac \plus{} 3ad \plus{} bd \plus{} 3ab \equal{} >$
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 > \equal{} ab \plus{} bc \plus{} cd \plus{} ad$
$ (a \minus{} b)^2 \plus{} (a \minus{} d)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} d)^2 > \equal{} 0$
[/quote]

This ineq is called Maclaurens Inequality
 :)


Honeybees can lift 300 times their own weight.  If humans could lift weight in this same ratio, how many pounds would a person need to weigh to lift one ton?  One ton equals 2000 pounds.  Express your answer as a decimal to the nearest tenth.

The answer is simply $ \frac{2000}{300}$, which simplifies to $ 6.666...$, so the answer is $ \boxed{6.7}$.


My inequality

Let $ a,b,c$ are positive number. Prove that
$ \frac {\sum_{cyc} ab }{2}\plus{}\frac{(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2}{3} \leq \frac {a^5}{a^3 \plus{} b^3} \plus{} \frac {b^5}{b^3 \plus{} c^3} \plus{} \frac {c^5}{c^3 \plus{} a^3} \leq \frac {\sum_{cyc} ab }{2} \plus{} (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2$

2) $ \sum_{cyc} \frac {a^5}{a^3 \plus{} b^3} \leq \sum_{cyc} \frac {a^4}{a^2 \plus{} b^2} \plus{} \frac {(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2}{3}$

1) first part:
it is moreover true that:
$ \sum_{cyc}\frac {ab}{2} \plus{} \frac {3}{8}\sum_{cyc}(a \minus{} b)^2\le \sum_{cyc}\frac {a^5}{a^3 \plus{} b^3}$
it is equivalent to $ \sum_{cyc}\frac {(a \minus{} b)^2(a^2b \plus{} 3ab^2 \plus{} b^3)}{a^3 \plus{} b^3}\geq 0$


modular arithmetic
number theory unsolved
number theory

If $ a$ and $ b$ are two positive rational numbers such that none of them entirely but $ a \plus{} b$ is integer, determine if it is possible that $ a ^ {20} \plus{} b ^ {20}$ and $ a ^ {21} \plus{} b ^ {21}$ are both integers?

Greetings  :D

$ a \equal{} \frac {p}{q},b \equal{} \frac {c}{d}$ $ (p,q) \equal{} (c,d) \equal{} 1$  $ \implies$
$ q.d|pd \plus{} qc$  $ \implies$  $ q|d$ and $ d|q$ $ \implies$  $ p \equal{} q$.
$ a \equal{} \frac {p}{q},b \equal{} \frac {c}{q}$.
$ q|p \plus{} c$ and $ q^{20}|p^{20} \plus{} c^{20}$  $ \implies$  $ q|2p^{20}$  $ \implies$  $ q|2$.
$ q \equal{} 1$  $ \implies$ $ (p,c)$ is a solution.
$ q \equal{} 2$  $ \implies$  $ a \equal{} \frac {p}{2},b \equal{} \frac {c}{2}$ .
$ \implies$  $ 2^{20}|p^{20} \plus{}c^{20}$  ,but $ p^{20} \plus{}c^{20}\equiv 2\pmod{4}$ conradiction!  :wink:


The vertices of $\triangle ABC$ are $A = (0,0)$, $B = (0,420)$, and $C = (560,0)$.  The six faces of a die are labeled with two $A$'s, two $B$'s, and two $C$'s.  Point $P_1 = (k,m)$ is chosen in the interior of $\triangle ABC$, and points $P_2$, $P_3$, $P_4, \dots$ are generated by rolling the die repeatedly and applying the rule: If the die shows label $L$, where $L \in \{A, B, C\}$, and $P_n$ is the most recently obtained point, then $P_{n + 1}$ is the midpoint of $\overline{P_n L}$.  Given that $P_7 = (14,92)$, what is $k + m$?

Reminds me of the [url=http://math.bu.edu/DYSYS/applets/chaos-game.html]CHAOS GAME[/url]

[hide]
Since the first coordinate was in the triangle, none of the points $P_n(x_n,y_n)$ for $n\ge1$ go outside of the triangle. Using the boundaries for the points, we can eliminate possible outcomes from the last roll of the die. (A, B, or C)

The equation of the hypotenuse is $y=-\frac{3}{4}x+420$
$P_7=(14,92)$ is the midpoint of $P_6$ and either A, B, or C.
If it is A, then by the midpoint formula, $\left(14,92\right)=\left(\displaystyle\frac{0+x_6}{2},\frac{0+y_6}{2}\right)\rightarrow(x_6,y_6)=(28,184)$
$184<-\frac{3}{4}(28)+420$, so it is inside the triangle, satisfying the situation.
A is possible.

If it is B, then by the midpoint formula, $\left(14,92\right)=\left(\displaystyle\frac{0+x_6}{2},\frac{420+y_6}{2}\right)\rightarrow(x_6,y_6)=(28,-236)$
But $y_6$ can't be negative.. so it can't be B.

If it is C, then by the midpoint formula, $\left(14,92\right)=\left(\displaystyle\frac{560+x_6}{2},\frac{0+y_6}{2}\right)\rightarrow(x_6,y_6)=(-512,184)$
But $x_6$ can't be negative.. so it can't be C.
So $P_6=(28,184)$.

Using this method, calculate all points individually:
$P_6=(28,184)$; $P_7$ is the midpoint of $P_6$ and $A$
$P_5=(56,368)$; $P_6$ is the midpoint of $P_5$ and $A$
$P_4=(112,316)$; $P_5$ is the midpoint of $P_4$ and $B$
$P_3=(224,212)$; $P_4$ is the midpoint of $P_3$ and $B$
$P_2=(448,4)$; $P_3$ is the midpoint of $P_2$ and $B$
$P_1=(336,8)$; $P_2$ is the midpoint of $P_1$ and $C$
$P_1=(336,8)$
$\boxed{344}$
[/hide]

[hide="Possibly Quicker?"]
Let $ L_1$ be the $ n^{th}$ roll that directly influences $ P_{n \plus{} 1}$. Note that $ P_7 \equal{} \cfrac{\cfrac{\cfrac{P_1 \plus{} L_1}2 \plus{} L_2}2 \plus{} \cdots}{2\ldots} \equal{} \frac {(k,m)}{64} \plus{} \frac {L_1}{64} \plus{} \frac {L_2}{32} \plus{} \frac {L_3}{16} \plus{} \frac {L_4}8 \plus{} \frac {L_5}4 \plus{} \frac {L_6}2 \equal{} (14,92)$. Then quickly checking each addend from the right to the left, we have the following information (remembering that if a point must be $ (0,0)$, we can just ignore it!): for $ \frac {L_6}2,\frac {L_5}4$, since all addends are nonnegative, a non-$ (0,0)$ value will result in a $ x$ or $ y$ value greater than $ 14$ or $ 92$, respectively, and we can ignore them, for $ \frac {L_4}8,\frac {L_3}{16},\frac {L_2}{32}$ in a similar way, $ (0,0)$ and $ (0,420)$ are the only possibilities, and for $ \frac {L_1}{64}$, all three work. Also, to be in the triangle, $ 0\le k\le560$ and $ 0\le m\le420$. Since $ L_1$ is the only point that can possibly influence the $ x$ coordinate other than $ P_1$, we look at that first. If $ L_2 \equal{} (0,0)$, then $ k \equal{} 2^6\cdot14 \equal{} 64\cdot14 > 40\cdot14 \equal{} 560$, so it can only be that $ L_2 \equal{} (560,0)$, and $ k \plus{} 560 \equal{} 2^6\cdot14\implies k \equal{} 64\cdot14 \minus{} 40\cdot14 \equal{} 24\cdot14 \equal{} 6\cdot56 \equal{} 336$. Now, considering the $ y$ coordinate, note that if any of $ L_3,L_4,L_5$ are $ (0,0)$ ($ L_3$ would influence the least, so we test that), then $ \frac {L_3}{32} \plus{} \frac {L_4}{16} \plus{} \frac {L_5}8 < \frac {420}{16} \plus{} \frac {420}8 \equal{} 79\pm\epsilon < 80$, which would mean that $ P_1 > 2^6\cdot(92 \minus{} 80) \equal{} 64\cdot12 > 42\cdot10 \equal{} 420\ge m$, so $ L_3,L_4,L_5 \equal{} (0,420)$, and now $ \frac {P_1}{64} \plus{} \frac {420}{2^5} \plus{} \frac {420}{2^4} \plus{} \frac {420}{2^3} \equal{} 92\implies P_1 \equal{} 64\cdot92 \minus{} 420(2 \plus{} 4 \plus{} 8) \equal{} 64\cdot92 \minus{} 420\cdot14$
$ \equal{} 64(100 \minus{} 8) \minus{} 14^2\cdot30 \equal{} 6400 \minus{} 512 \minus{} (200 \minus{} 4)\cdot30 \equal{} 6400 \minus{} 512 \minus{} 6000 \plus{} 120$
$ \equal{} \minus{} 112 \plus{} 120 \equal{} 8$, and finally, $ k \plus{} m \equal{} 336 \plus{} 8 \equal{} \boxed{344}$.
[/hide]

Exactly 3 years old! :P


algebra
function
domain
real analysis
integration
topology
real analysis unsolved

Let D be a bounded domain in R^2.
Let f be a real analytic function on D.(i,e f equals to its Taylor series in some nbhd of every pt in D).
Suppose the set {D:f(x)=0} has positive Lebesgue measure.
Must f=0 on D??

Yes.
Every uncountable subset of open set $ U$ (not only bounded) in $ R^{n}$ has a limit point in $ U$. It's common known fact that if an analitic function which vanishes on set with limit point, vanishes in all domain.

How about sin(xy) on R^2?
sin(xy)=0 on {0}XR,but sin(xy) !=0 on R^2.

I think it's true. See the Weierstrass preparation theorem.

$ \{0\} \times \textbf{R}$ is measure 0 in $ \textbf{R}^2$.

I am just saying that "It's common known fact that if an analitic function which vanishes on set with limit point, vanishes in all domain." is not true in general(sin(xy) is an example).

I am not sure under the assumption I made,must f=0 on D?

You're missing something: dimension. That's true for a real analytic function on $ \mathbb{R}$, or a complex analytic function on $ \mathbb{C}$, not a real analytic function on $ \mathbb{R}^2$.

Now, a solution:
First, note that a real analytic function on $ \mathbb{R}$ which is zero on a set of positive measure is identically zero.
Suppose $ f$ is analytic on $ \mathbb{R}^2$, and its zeros are a set $ S$ of positive measure. Define $ S_x \equal{}x\times \mathbb{R}\cap S$ for each real $ x$. This $ S_x$ is measurable for almost every $ a$, and $ \int_0^{\infty}\mu(S_x)\,dx \equal{}\mu(S)$. Therefore, $ A\equal{}\{x: \mu(S_x)>0\}$ has positive measure.
For each $ x\in A$, $ f$ restricted to $ x\times\mathbb{R}$ is real analytic, and must be identically zero. For each $ y$, $ f$ restricted to $ \mathbb{R}\times y$ is real analytic and zero on $ A$. Therefore, $ f$ is identically zero on $ \mathbb{R}\times y$. These sets cover everything, and $ f$ is identically zero.

So the condition "D is bounded" is not necessary?

OK.I see.
Thanks a lot for your help  :) .

Definitely not necessary.

Actually, my argument might get you in trouble if the slices we make aren't connected. For the original version, restrict to some convex subset on which the zero set still has positive measure. Once we know the function is identically zero on that subset, it's easy to extend to the rest of the connected domain.


calculus
derivative
vector
function
limit
real analysis
real analysis unsolved

Let $x_i \in R, i = \overline{1,n}, n \geq 2$ distinct numbers 2 by 2 and $f: R\to R, f(x) = (x-x_1) \cdots (x-x_n)$.
Let $a_{ii} = f''(x_i), a_{ij} = \frac{f'(x_i)-f'(x_j)}{x_i - x_j}, i \neq j, i,j = \overline{1,n}$.
Prove $det(A) = 0$.

Yes sure, I post my solution. Sorry, I wasn't around for a while, a couple of years or more... :)

Let's find a not null vector $ X = (x_i)$ of dimension (n,1) such that $ A \cdot X = O$. From this det A = 0.
We prove that $ x_i = \frac{1}{f'(x_i)}, i = 1,n$
Consider function: $ h(x) = \sum_{i=1}^{n}\frac{f'(x) - f(x_i)}{f'(x_i)\cdot(x - x_i)}, %Error. "foreach" is a bad command.
x \neq x_1,...x_n$
If we can show that h(x) = 0, then: $ 0 = h(x) = \lim[x->{x_k}]{h(x)} = \sum_{i=1}^{n}a_{ki}\cdot\frac{1}{f'(x_i)}, %Error. "foreach" is a bad command.
k = 1,...,n so A \cdot X = O$
Computing: $ h(x) = f'(x)\cdot\sum_{i=1}^{n}\frac{1}{f'(x_i)\cdot(x - x_i)} - \sum_{i=1}^{n}\frac{1}{x - x_i}$
Easily we can see that: $ \frac{f'(x)}{f(x)} = \sum_{i=1}^{n}\frac{1}{x - x_i}$ so:
$ h(x) = f'(x)\cdot(\sum_{i=1}^{n}\frac{1}{f'(x_i)\cdot(x - x_i)} - \frac{1}{f(x)})$
To prove $ \frac{1}{f(x)} = \sum_{i=1}^{n}\frac{1}{f'(x_i)\cdot(x - x_i)}$ first put: $ \frac{1}{f(x)} = \sum_{i=1}^{n}\frac{A_i}{x - x_i}$
multiply by f(x), take the limit $ x -> x_k$ and get $ A_k = \frac{1}{f'(x_k)}$


Simple really: Start at 1000 and countdown to 0!  :coolspeak:
(Replacement for when Count to 1000 reaches 1000: What goes up must come down!  :D )
[size=200][color=darkblue][u][i][b]1000!!![/b][/i][/u][/color][/size]

\[ \color{red} \star \boxed{994} \star\]

How about the person that reaches 0 wins? :P

987.

883 :trampoline:  :trampoline:  :trampoline:

no really! im enjoying this!
This is like the best game in the world!
874

Really don't see the point of this, but...

[color=red][b]871[/b][/color]

 :furious:  :furious:  :furious:

[b]870..[/b] :maybe:

[size=84][color=red]Locked for the same reason as the other thread[/color][/size]


For any prime $ p$, prove that there exist a positive integer $ n$ such that $ p$ divides

$ 1\plus{}(\frac{1}{2^2}\plus{}\frac{1}{3^2}\plus{}...\plus{}\frac{1}{n^2})(2^2\times 3^2\times ...\times n^2)$

Two Lemmas

[b]Wilson's Theorem[/b]
$ (p \minus{} 1)!\ \equiv\ \minus{} 1\ (\mbox{mod}\ p)$

And

[b]Lemma 2[/b]

$ \frac {1}{1^2} \plus{} \frac {1}{2^2} \plus{} \ldots \plus{} \frac {1}{(p \minus{} 1)^2} \equiv\ 0\ (\mbox{mod}\ p)$


Hello!

(sorry if this is a repost)
I have a minor problem, that wont get solved...

Its been posted in an old competition at our university, but I didnt manage to solve it in time. Maybe someone in here could help me?

[url=http://img504.imageshack.us/my.php?image=optimeringts0.jpg][img]http://img504.imageshack.us/img504/4683/optimeringts0.th.jpg[/img][/url]

Where D is a symmetric n x n matrix, and x is an unknown n x 1 vector.

I can only figure out how to test for a maximum later on, but is there something im missing with the charataristic roots?

Regards

Can you solve it in the case that $ D$ is diagonal? Can you convert to that case in general?

I used the fact that D is symmetric, to do the following, but im not sure if its possible to eliminate lamda?

[url=http://img225.imageshack.us/my.php?image=matrixyi9.jpg][img]http://img225.imageshack.us/img225/1716/matrixyi9.th.jpg[/img][/url]

Thanks


real analysis
induction
algebra
polynomial
algebra proposed

We'll first prove by induction that $P_n(x), Q_n(x)$ have degree $2^n-1$ and their first coefficients: $a_0, \ldots, a_{2^n-1}$ are $1$ or $-1$. It is true for $n=0$. Now suppose it is true for $n$. Well... it follows immediately that it's true for $n+1$.

Now we'll prove that any polynomial $a_kx^k+\ldots+a_1x+a_0$ with $a_i= \pm 1$ for $i=0, 1, \ldots, k$ satisfies $|z|<2$ for all its complex roots $z$. Indeed, if $a_kz^k+\ldots+a_0=0$, then $a_kz^k=-a_{k-1}z^{k-1}- \ldots -a_0$. Clearly $|a_kz^k|=|a_k||z^k|=|z^k|$. Then, taking modulus on both sides: $|z^k|=|a_{k-1}z^{k-1}+\ldots+a_0|$. Since $|a+b| \leq |a|+|b|$ and $|ab|=|a||b|$, and also $|a_i|=1$, we have
$|z|^k \leq |z|^{k-1}+\ldots+|z|+1$. But if $|z|>2$ the LHS gobbles up the RHS. To prove $|z| \geq 1/2$, just consider $x^kP(1/x)$ and proceed analogously.