Two distinct numbers $ a$ and $ b$ are chosen randomly from the set $ \{ 2, 2^2, 2^3, \ldots, 2^{25} \}$. What is the probability that $ \log_{a} b$ is an integer?

$ \textbf{(A)}\  \frac {2}{25} \qquad \textbf{(B)}\  \frac {31}{300} \qquad \textbf{(C)}\  \frac {13}{100} \qquad \textbf{(D)}\  \frac {7}{50} \qquad \textbf{(E)}\  \frac {1}{2}$

[hide="I just did this"]

I really like this combination of NT + Probability. There are 25 options for $a$ and 24 options for $b$. Let $a = 2^m$ and $b = 2^n$. Since $\log_a b = \frac{m \log_{10} 2}{n \log_{10} 2} = \frac{m}{n}$, the whole thing is rational if and only if $n$ divides $m$.

If we set $n = 1$, there are 24 values of $n$. If we set $n = 2$, there are 11 values. Repeat this and we'll get total of 62 favorable outcomes.

The total number of outcomes is 600 so our probability is $\frac{62}{600} = \frac{31}{300}$, B. [/hide]

[hide="another_way_to_approach_the_problem "]
as mentioned before, it is obvious that if $a=2^m$ then $b=2^{km}(k>1)$

now the number of multiples of a number less than 25 would be:

$\sum_{n=1}^{25}\lfloor \frac{25}{n}\rfloor-1$ which equals 62.
so the answer is $\frac{62}{2\binom{25}{2}}=\frac{31}{300}$[/hide]

Since, when you do $\binom{25}{2}$, the sets $1,2$ and $2,1$ are equal, but they correspond to different a and b values in the problem.  So you multiply by 2.  An easier way is to just note there are 25 ways to choose a and subsequently 24 ways to choose b, so $25 \cdot 24 = 600$ ways to choose a and b.  When I did this, I got $\frac{31}{150}$ since I just did $\binom{25}{2}$ for the total possible... although I noted the similarity to choice B and figured out my problem.

Here's my solution:
Alternately, we have $ x|y$ for $ x\ne y$ and $ x,y \in S$ where $ S$ is the given set, we have the probability of $ 1/25 * 24/24$ for $ x \equal{} 1$, $ 1/25 * 11/24$ for $ x \equal{} 2$ and so on... therefore the answer is
$ \frac {24 \plus{} 11 \plus{} 5 \plus{} 4 \plus{} 3 \plus{} 2 \plus{} 2 \plus{} 1 \plus{} 1 \plus{} 1}{25*24} \equal{} 62/(25*24) \equal{} 31/300$

We go case by case.
If $a=2^1$, then $b$ can be anything in the form $2^k$, where $k$ is an integer. There are $24$ of such numbers.
If $a=2^2$, then $b$ can be anything in the form $2^{2k}$, where $k$ is an integer. There are $11$ of such numbers.
If $a=2^3$, then $b$ can be anything in the form $2^{3k}$, where $k$ is an integer. There are $7$ of such numbers.
If $a=2^4$, then $b$ can be anything in the form $2^{4k}$, where $k$ is an integer. There are $5$ of such numbers.
If $a=2^5$, then $b$ can be anything in the form $2^{5k}$, where $k$ is an integer. There are $4$ of such numbers.
If $a=2^6$, then $b$ can be anything in the form $2^{6k}$, where $k$ is an integer. There are $3$ of such numbers.
If $a=2^7$, then $b$ can be anything in the form $2^{7k}$, where $k$ is an integer. There are $2$ of such numbers.
If $a=2^8$, then $b$ can be anything in the form $2^{8k}$, where $k$ is an integer. There are $2$ of such numbers.
If $a=2^9$, then $b$ can be anything in the form $2^{9k}$, where $k$ is an integer. There are $1$ of such numbers.
If $a=2^{10}$, then $b$ can be anything in the form $2^{10k}$, where $k$ is an integer. There are $1$ of such numbers.
If $a=2^{11}$, then $b$ can be anything in the form $2^{11k}$, where $k$ is an integer. There are $1$ of such numbers.
If $a=2^{12}$, then $b$ can be anything in the form $2^{12k}$, where $k$ is an integer. There are $1$ of such numbers.
Summing this up, we have $62$ of such valid pairs, but $600$ pairs total, so our answer is $\frac{62}{600} = \boxed{\frac{31}{300}}$ probability.


I'm obsessed with these type of problems!
This is probably boring if you're not, but here we go...

I started toying with these in hypergraph Ramsey theory, but I stepped
back and thought about boring old graph Ramsey questions.

R2(4,4) =18, i.e. if you color every edge of a complete 18 point graph
red or blue, however you do it, you will always end up with a 4 point subset
of the 18 points all of whose 6 edges are either red or blue.

Consider the sub-problem:
How many points do you need to guarantee that at least one 4 point subset
has an EVEN number of red and blue edges, i.e. in this case:
all red, no blue OR
2 red, 4 blue OR
4 red, 2 blue OR
all blue, no red.

This is obviously bounded by R2(4,4) and the answer is...
well its a lot smaller, I think.
I think I can prove it without any pictures, using simple combinatorics.
I would be interested to see how other people would prove it.

What if any, is the connection between this number (any naming conventions?)
and R2(4,4).
Please don't just say "look in a book", but be specific if possible...
(I'll post what I think is a solution later so it can be ripped apart).

Richard.

The answer would seem to be 6 points.

To see 5 points succeeds, consider the convex pentagon with the border 5 edges one color (red) and the interior 5 edges the other (blue).
Take away any point and you are left with 3 red border edges and 3 blue edges from the interior of the pentagon.

Why does 6 points fail?
A simple proof based on ‘double counting’, i.e. counting the same sum in two different ways.

Without loss of generality pick a color, say, red.
Formulate the ways in which we must have an odd number of red edges in a 4-set.
This would require, in any 4-set, 1 red edge, 3 red edges or 5 red edges.
Within a 6-set, create a sum for the number of 4-sets of each type:

1a + 3b + 5c = X.

a is the number of 4-sets with one red edge, b the number with 3, c the number with 5.

There are 6!/4!2! = 15 4-sets within a 6-set, so:

 a+b+c = 15.

We require every 4-set to be one of either a,b or c type to prevent any even numbers of  colors of edges in a 4-set.

However, double counting, we can see that:
X = 6Y.

This is because X is just a total of the number of red edges of a 6-set as counted within 4-sets.
When we count edges in 4-sets for a 6-set, each edge must occur 6 times:
Any edge must appear in 4!/2!2! = 6 4-sets.

The key consideration is that X is therefore even.
To get an even number from 1a + 3b + 5c, either:

a, b and c are all even, or
two of them are odd and one is even.

If they are all even, they cannot sum to 15, so this cannot happen.
If two of them are odd and one is even, again, their sum is even, and cannot be 15.
Therefore there is no combination to fit the bill.

Q.E.D.


Solve the system of equations
\[\begin{cases}2^x+3^y=4\\5^x+6^y=7\end{cases}\]

It's my idea to solve it:just in $N$.

let change the system equation in this form
$5^x-2^x+6^y-3^y=3 \Longrightarrow (5-2)(5^{x-1}+...+2^{x-1})+(6-3)(6^{x-1}+...+3^{x-1})=3
 \Longrightarrow 5^{x-1}+...+2^{x-1}+6^{x-1}+...+3^{x-1}=1$
I think this equation without have the roots in $N$ ....... :(  :?

Hey,Eshan2004 you can't do that.How did you deduce that $x\in\mathbb{N}$ :huh:

I am not sure this is the appropriate place to post this problem. I mean: is it a problem "you couldn't solve and to which you know that there is a solution (i.e. a problem from a contest, etc.) but you don't know it", or a problem "that you have already solved and you are interested in second opinions or solutions", or, finally, "an open question, that is, a question that has no known solution up to this moment, and it is not known wheter the problem has one or not".
I write this because the system, apart from the obvious solution $(0,1)$ has another solution, which I doubt that can be expressed in a simple form.

In short, what's the source of this problem?

what's another solution??

I have problem from my friend  :lol: . I don't solve it but I think exist a beauty solution. :|


Distixws logw tis fisis twn eksetasewn pou exoume stin 3h Likeiou, exoume ksexasei teleiws ta oloklirwmata, ena pedio to opoio exei arketes omorfes askiseis. Sti Roumania akomi vazoun se dikous tous diagwnismous themata apo auti tin perioxi parolo pou gnwrizoun oti stous megalous mathimatikous diagwnismou (BMO, IMO ktl) den mpainoun...

Oriste mia askisi gia na ksekinisoume kai auto to pedio. Thermi paraklisi na min ginei aformi gia na vazoume sxolikes askiseis (ennowntas tis askiseis gia proetoimasia gia Panellinies) alla omorfes askiseis pou den einai anagi na einai ta SOS themata twn panelliniwn. (Profanws gia na mi pareksigithw, den simainei oti epeidi mia askisi einai sxoliki, den exei omorfia. Elpizw na katalavate ti ennoousa).

Ek twn proterwn zitw siggnwmi apo ekeina ta meli tou forum pou den exoun didaxthei akomi ta oloklirwmata... Sta epomena xronia omws euxomai na to exoume emploutisei gia ta kala to forum me tetoies askiseis opote tha mporoun na tis silleksoun sinolika.

[color=red]ASKHSH 1[/color] (Andreescu 1982)

Na vrethei sinartisi $f: [0,1]\to \mathbb{R}$, sinexis sto diastima $[0,1]$ kai tetoia wste na einai

$\int_{0}^{1}f(x) dx=\frac{1}{3}+\int_{0}^{1}f(x^{2})dx.$

Alexandros

Ολες οι συναρτησεις της μορφης f(x)=α*(χ^ν)+c, ν-->φυσικο αριθμο διαφορο του μηδεν, α=(2*ν + 1/ν + 3)/3
π.χ f(x)=2*x, f(x)=5/2*(x^2)

εχω φτασει να πιστευω οτι f(x)=x^3+f(x^2)... ειμαι στο σωστο δρομο? γιατι πεδευομαι ωρα τωρα... :(
βγαινει οτι f(x)=x^3+f(x^2)=x^3+x^6+f(x^4)=...
μοιαζει απειρου βαθμου :( a little help could be useful

H sinartisi pou vgainei den einai poliwnimo kai einai poli apli (den simainei oti vriskontas tin sinartisi omws, exei lithei i askisi, apla meta eisai se kalo dromo gia na tin ftiakseis).

Tip: Polles fores i antistrofi poreia mas voithaei gia na lisoume ena provlima. Gia paradeigma (den exei sxesi me tin askisi) ean deis kapou $\sin^{2}x+\cos^{2}x$ tha to antikatastasiseis amesws me $1$. An deis omws to $1$ tha skefteis na to antikatastiseis me $\sin^{2}x+\cos^{2}x$ ??

Gia na katalaveis kalitera auto pou thelw na sou pw prospathise na vreis to

$\int \frac{1}{\sin^{2}x\cdot\cos^{2}x}dx$

Pws sindeontai ola auta me tin askisi?

Alexandros

[quote]Distixws logw tis fisis twn eksetasewn pou exoume stin 3h Likeiou, exoume ksexasei teleiws ta oloklirwmata, ena pedio to opoio exei arketes omorfes askiseis. Sti Roumania akomi vazoun se dikous tous diagwnismous themata apo auti tin perioxi parolo pou gnwrizoun oti stous megalous mathimatikous diagwnismou (BMO, IMO ktl) den mpainoun[/quote]

To idio mporwna pw oti isxyei kai gia allous ogkolithoys twn mathimatikwn
p.x Pinakes. Exei ginei ena ksekinima me tous migadikous pou didaskontai stin kateythinsi.
Dystyxws apo to gegonos oti ta oloklhrwmata einai ektos ylhs, kanenas
ellhnas den pairnei miroudia kai apo diaforikes eksiswseis pou perigrafoun ta fysika fainomena (ektos apo 4 eksamino kai sto polytexneio).

Auto exei san apotelesma ena xaos anamesa stis yles twn foititikwn
kai twn mathitikwn mathimatikwn. Akoma kai anthrwpoi thetikwn Episthmwn me Phd dyskoleuontai se Askhseis Thali B Gymnasiou!

συμφωνω απολυτα με τον mimmis...μαλιστα εχω να πω οτι απο τη σκοπια της 1ης λυκειου εχοντασ καποιες πολυ στοιχειωδεις γνωσεις περι διαφορικων και ολοκληρωματων κατανοω κυριως το μαθημα της φυσικης πολυ καλυτερα απο ενα μαθητη που δεν τα εχει καν ακουστα(κατ αρχην δεν παπαγαλιζω οοοολες αυτες τις ισοτιτς με τις δυναμεις..ολες βγαινουν απο του νευτoνα την F=ma...)

[quote="cretanman"]Mano exw tin entipwsi oti oi sinartiseis pou egrapses den einai liseis, ektos ean exasa kapou. An theleis koitakse tin kai pali grafontas+ton tropo skepsis sou.

Alexandros[/quote]

Οι συναρτήσεις που έχει γράψει είναι λύσεις.

Μήπως έχεις γράψει κάτι λάθος στην εκφώνηση ?  :)

Asfalws kai einai liseis! Einai diko mou to lathos stis prakseis!! Evgala alli sinartisi i opoia fisika den einai swsti.
Sorry Mano. Euxaristw Ramanujan

Alexandros

[quote="silouan"]ΑΣΚΗΣΗ 2
Βρείτε το $\alpha$  έτσι ώστε to
  $\int_{0}^{2}\frac{ x^\alpha dx}{ \sqrt[3]{(x-1)(x-2)^{2}}}$

na συγκλίνει.[/quote]
μηπως ειναι λιγο δυσκολο ? δεν την ειδα απλα ειναι αλυτη στο φορουμ της αναλυσης ? :P

[quote="kostas"]   απλα ειναι αλυτη στο φορουμ της αναλυσης ? :P[/quote]

Και ?Οι Έλληνες είναι πιο μάγκες από τους ξένους  :D
Let's try... :wink:

Σιλουανέ  δεν  έχω  βρεί  ακόμα  λύση  κάποιες  ιδέες  έχω  μονάχα  ......  Το  ολοκλήρωμα  αυτό  παρατηρούμε  οτι  καθώς  η  μεταβλητή  ολοκλήρωσης  παίρνει  τιμές  από  0  έως  2  δεν  ορίζεται    στα  σημεία  1  και  2  .... Συνεπώς  το  ολοκλήρωμα  αυτό  είναι  γενικευμένο  και  μάλιστα  β  είδους  ....... Αν  το  γράψουμε  δηλαδή   σε  δυο  ολοκληρώματα  ως  εξής

 $I(a) = \int_{0}^{1}{ \frac{x^{a}dx}^{((x-1)(x-2)^{2})^{\frac{1}{3}}}$ +  $\int_{1}^{2}\frac{x^{a}dx}^{((x-1)(x-2)^{2})^{\frac{1}{3}}$  τότε

   εχουμε  δηλαδή   ένα  $I_{1}(a) = \int_{0}^{1}\frac{x^{a}dx }^{((x-1)(x-2)^{2})^{\frac{1}{3}$    και  το  $I_{2}(a) = \int_{1}^{2}\frac{x^{a}dx}^{((x-1)(x-2)^{2})^{\frac{1}{3}$

έτσι  θα  τα  αντιμετωπίσουμε  ευκολότερα  πιστεύω  ........  Ακόμα  δεν  τo  έχω  δει  πώς  δουλέυει   όταν  βρώ  κατι  θα  συνεχίσω...

Γιώργη  είσαι σωστός .Για να είμαι ακριβής δεν έχω πλήρη λύση.
Απλά έχω βρει ότι

$I_{2}(a) = \int_{1}^{2}\frac{x^{a}dx}{((x-1)(x-2)^{2})^{\frac{1}{3}}}=\frac{\pi\cdot 2^{\frac{4}{3}+a}Hypergemetric2F[\frac{1}{3},\ 1+a,\ 1,\-1]}{\sqrt{3}}$

Ε  Σιλουανέ    για  δες  τι  έγραψες  ........  κατι  παίζει  με  το  Latex

Δεν παίζει κάτι με το Latex  :P  .See here
http://mathworld.wolfram.com/HypergeometricFunction.html

Είναι προφανές ότι κοντά στο μηδέν η υπό ολοκλήρωση συνάρτηση συμπεριφέρεται όπως η -x^α/2^(2/3) και έτσι το ολοκλήρωμα συγκλίνει τότε και μόνο τότε αν α > -1. (Στο -1 και στο -2 δεν υπάρχει πρόβλημα γιατί οι εκθέτες είναι -1/3 και -2/3 αντίστοιχα, ήτοι > -1.)

Lancelot DuLac

[color=red][b]ΑΣΚΗΣΗ 3[/b][/color]
Να υπολογιστεί το ολοκλήρωμα:

$\int_{0}^{1}\sqrt[3]{2x^{3}-3x^{2}-x+1}dx$

=0. Οι λεπτομέρειες επαφίονται στον αναγνώστη!

Λανς

[quote="mostel"][b][color=red]ΑΣΚΗΣΗ 4[/color][/b]

Έστω $a$ και $b$ δύο θετικοί πραγματικοί. Να υπολογιστεί:

$\int^{b}_{a}\frac{e^{\frac{x}{a}}-e^{\frac{b}{x}}}{x}dx$[/quote]

[b] Καλό ολοκλήρωμα ![/b]

Πολύ βιαστικά και σε μια άκρη του χαρτιού έθεσα :

  $x= ab \frac{1}{t}$

    και φαίνεται ότι πάει για μηδέν ! Θα το ξανακοιτάξω αύριο όμως με προσοχή, γιατί μπορεί να έκανα λάθος σε καμιά πράξη. Βγαίνουν μερικά πλην και θέλουν έλεγχο.Τώρα είναι αργά και το πρωί φεύγω για Αθήνα .Μπορεί να συναντήσω και κανένα γνωστό που δίνει στον Αρχιμήδη !
[u] Μπάμπης[/u]

Όντως είναι 0, πολύ ωραία  :lol:

[quote="cretanman"][b][color=red]ΑΣΚΗΣΗ 4 - LYSH[/color][/b]

Thetoume $\frac{x}{a}=\frac{b}{y}$ . Tote $\frac{1}{a}dx=-\frac{b}{y^{2}}dy$ ap'opou $dx=-\frac{ab}{y^{2}}dy$. Telos otan $x=a$ exoume $y=b$ enw otan $x=b$ tote $y=a$ Ara gia to dosmeno oloklirwma exoume:

\begin{eqnarray}I &=& \int^{b}_{a}\frac{e^{x / a}-e^{b / x}}{x}dx \nonumber \\ &=& \int^{a}_{b}\frac{e^{b / y}-e^{y / a}}{\frac{ab}{y}}\left(-\frac{ab}{y^{2}}dy\right) \nonumber \\ &=&\int^{b}_{a}\frac{e^{b / y}-e^{y / a}}{y}dy \nonumber \\ &=&\int^{b}_{a}\frac{e^{b / x}-e^{x / a}}{x}dx = I \nonumber \end{eqnarray}

Ara xrisimopoiwntas diadoxika tin prwti sxesi gia to $I$ kai tin teleutaia, kai prosthetwntas, exoume:

$I+I= \int^{b}_{a}\frac{e^{x / a}-e^{b / x}}{x}dx+\int^{b}_{a}\frac{e^{b / x}-e^{x / a}}{x}dx =0$

Etsi $I=0$

Alexandros[/quote]

 [b]Αλέξνδρε !!![/b]

   Με πρόλαβες στη λύση κατά τρία ακριβώς λεπτά !
  Αλλά χαλάλι σου , γιατί εσύ είσαι το μέλλον και εγώ ... το ηλιοβασίλεμα !!!

 Μπάμπης

Kirie Stergiou,

Pragmatika! Den to paratirisa (dioti tin wra pou egrafa to post den eixe apantisei kaneis).

Alexandros

[color=red]ASKHSH 5[/color]

Estw $g: \mathbb{R}\to \mathbb{R}$ sinartisi artia kai sinexis sto $\mathbb{R}$ kai $f: \mathbb{R}\to \mathbb{R}$ sinartisi sinexis sto $\mathbb{R}$, peritti kai periodiki me periodo $T>0$ kai paragwgisimi me sinexi paragwgo sto $\mathbb{R}$. Na deiksete oti einai \[\int_{T-a}^{T+a}xf'(x)g\left(f(x)\right)dx=2T\int_{0}^{f(a)}g(x)dx\]

Alexandros

Ισχυει:
\[2{\rm T}\int_{0}^{f(a)}{g(x)}dx ={\rm T}\int_{-f(a)}^{f(a)}{g(x)dx}\]
με αλλαγη μεταβλητης εχουμε:
\[x = f(u) \Rightarrow \frac{{dx}}{{du}}= f^' \left( u \right) \Rightarrow dx = f'\left( u \right)du \]
για:
\[x = f(a) \Rightarrow u = a+T \]
\[x =-f(a) \Rightarrow u = T-a \]
αρα:\[2{\rm T}\int_{0}^{f(a)}{g(x)}dx ={\rm T}\int_{-f(a)}^{f(a)}{g(x)dx}\[ =T\int_{T-a}^{T+a}{g(f(x))f^' }(x)dx \]
αρκει να δειξω οτι:
\[\int_{T-a}^{T+a}{xf^' }(x)g(f(x))dx = T\int_{T-a}^{T+a}{f^' }(x)g(f(x))dx \]
Για το:
\[\int_{T-a}^{T+a}{xf^' }(x)g(f(x))dx \]
με αλλαγη μεταβλητης:
\[x = T+\omega\]\[x = T+a \Rightarrow \omega = a\]\[x = T-a \Rightarrow \omega =-a\]\[dx = d\omega \]
αρα:
\[\int_{T-a}^{T+a}{xf^' }(x)g(f(x))dx = \int_{-a}^{a}{(\omega+T)f^' }(\omega+T)g(f(\omega ))d\omega = \int_{-a}^{a}{\omega f^' }(\omega+T)g(f(\omega ))d\omega+T\int_{-a}^{a}{f^' }(\omega+T)g(f(\omega ))d\omega \]
ομως εχουμε οτι η συναρτηση:

\[\ \[ W(\omega ) =\omega f^' }(\omega+T)g(f(\omega )) \]

ειναι περιττη, και αυτο γιατι αφου $f$ ειναι περιττη η πρωτη της παραγωγος ειναι αρτια ακομα η $g(f(x))$ ειναι αρτια.
Aρα:
\[\int_{-a}^{a}{\omega f^' }(\omega+T)g(f(\omega ))d\omega=0 \]
ΤΕΛΟΣ


function
calculus
integration
logarithms
trigonometry
limit
real analysis

Let $ a\in [0,1]$. Find the sums:

1) $ \sum\limits_{n\equal{}1}^{\infty}\frac{1}{n}\left(\frac{1\cdot 3\cdots (2n\minus{}1)}{2\cdot 4\cdots (2n)}\right)a^{n}\equal{}$

2) $ \sum\limits_{n\equal{}1}^{\infty}\frac{(\minus{}1)^{n}}{n}\left(\frac{1\cdot 3\cdots (2n\minus{}1)}{2\cdot 4\cdots (2n)}\right)a^{n}\equal{}$

3) $ \sum\limits_{n\equal{}1}^{\infty}\left(\frac{1\cdot 3\cdots (2n\minus{}1)}{2\cdot 4\cdots 2n}\right)a^{n}\equal{}$

4) $ \sum\limits_{n\equal{}1}^{\infty}(\minus{}1)^{n}\left(\frac{1\cdot 3\cdots (2n\minus{}1)}{2\cdot 4\cdots 2n}\right)a^{n}\equal{}$

5) $ \sum\limits_{n\equal{}1}^{\infty}(\minus{}1)^{n}\left(\frac{2\cdot 4\cdots 2n}{1\cdot 3\cdots (2n\plus{}1)}\right)\frac{a^{n}}{n}\equal{}$

6) $ \sum\limits_{n\equal{}1}^{\infty}\left(\frac{2\cdot 4\cdots 2n}{1\cdot 3\cdots (2n\plus{}1)}\right)\frac{a^{n}}{n}\equal{}$.

I would assume that functions such as $ \sqrt{1\minus{}x^2},$ $ \sqrt{1\plus{}x^2},$ or the reciprocals of those, or the integrals of those (which brings in arcsines among other related functions) will play a large role here.

many o this serie looks like this $ f(x)\equal{}\sum \frac{(2n\minus{}1)!}{(n!)^{2}}x^{n}$ where $ |x|\leq 1$
I have already post solution for this serie in this forum :
hint was
you must consider $ f(x)$
$ f'(x)$
$ f''(x)$
and after few calculation you take some differential equation of this form I(for this case)
$ (A_{11}f(x)\plus{}(A_{21}\plus{}A_{22}x)f'(x)\plus{}(A_{31}\plus{}A_{32}x\plus{}A_{33}x^{2})f''(x)\plus{}C(x)\equal{}0$
you only must find this coeficient. I think you can do it yourself  :wink:

didi, how you doing ? :D i will show 3 and 4 first, then 1 and 2, and then finally, 6 and 5.

we know,
 $ \frac{1}{\sqrt{\;1\minus{}x\;}}\minus{}1\;\;\;\equal{}\;\;\;\sum_{n\equal{}1}^\infty\;\frac{1\cdot 3\cdot 5\;\;\ldots\;\;\cdot (2n\minus{}1)}{2\cdot 4\cdot 6\;\;\ldots\;\;\cdot (2n)}\;x^n\;\;\;\quad(|x|<1)$

following that, we have,
3. $ S\equal{}\boxed{\frac{1}{\sqrt{\;1\minus{}a\;}}\minus{}1}$

4. $ S\equal{}\boxed{\frac{1}{\sqrt{\;1\plus{}a\;}}\minus{}1}$

also,
1. $ S\equal{}\int_0^a\left(\frac{1}{\sqrt{\;1\minus{}x\;}}\minus{}1\right)\;\frac{dx}{x}\;\;\;\equal{}\;\;\;\boxed{\ln\left(\frac{2}{a}\right)\;\minus{}\;\ln\left(\frac{1\plus{}\sqrt{\;1\minus{}a\;}}{1\minus{}\sqrt{\;1\minus{}a\;}}\right)}$
[put $ 1\minus{}x\equal{}t^2$ to solve the first integral. second integral is simply $ \ln\,x$. but the integrals need to be considered [i]together [/i]as both are [b]improper[/b] integrals of the second kind.]

2. $ S\equal{}\int_0^a\left(\frac{1}{\sqrt{\;1\plus{}x\;}}\minus{}1\right)\;\frac{dx}{x}\;\;\;\equal{}\;\;\;\boxed{\ln\left(\frac{2}{a}\right)\;\minus{}\;\ln\left(\frac{\sqrt{\;1\plus{}a\;}\plus{}1}{\sqrt{\;1\plus{}a\;}\minus{}1}\right)}$
[put $ 1\plus{}x\equal{}t^2$ to solve the first integral. method for solving the integrals is the same as 1.]

we also know,
$ \left(\frac{\sin^{\minus{}1}\,x}{x}\right)\minus{}1\;\;\;\equal{}\;\;\;\sum_{n\equal{}1}^\infty\;\frac{1\cdot 3\cdot 5\;\;\ldots\;\;\cdot (2n\minus{}1)}{2\cdot 4\cdot 6\;\;\ldots\;\;\cdot (2n)}\;\frac{x^{2n}}{2n\plus{}1}\;\;\;\quad(|x|<1)$

now note that
$ \left(\frac{\sin^{\minus{}1}\sqrt x}{\sqrt x}\cdot\frac{1}{\sqrt{\;1\minus{}x\;}}\right)\minus{}1\;\;\;\equal{}\;\;\;\sum_{n\equal{}1}^\infty\;\frac{2\cdot 4\cdot 6\;\;\ldots\;\;\cdot (2n)}{3\cdot 5\cdot 7\;\;\ldots\;\;\cdot (2n\plus{}1)}\;x^n$

following that, we have,
6. $ S\equal{}\int_0^a\left[\left(\frac{\sin^{\minus{}1}\sqrt x}{\sqrt x}\cdot\frac{1}{\sqrt{\;1\minus{}x\;}}\right)\minus{}1\right]\;\frac{dx}{x}\;\;\;\equal{}\;\;\;\boxed{2\left(1\;\minus{}\;\left(\frac{\sqrt{\;1\minus{}a\;}}{\sqrt a}\cdot\sin^{\minus{}1}\sqrt a\right)\right)}$
[put $ x\equal{}\sin^2\,u$ and then do integration by parts for $ \int\;u\cdot\text{cosec}^2\,u\;\;du$ to solve the first integral. second integral is simply $ \ln\,x$ as before. but the integrals again need to be considered [i]together [/i]as both are [b]improper[/b] integrals of the second kind. also, we need to use the fact $ \lim_{u\to 0}\;u\cdot\cot\,u\;\equal{}\;1$]

and in a similar fashion,
5. $ S\equal{}\int_0^a\left[\left(\frac{\sinh^{\minus{}1}\sqrt x}{\sqrt x}\cdot\frac{1}{\sqrt{\;1\plus{}x\;}}\right)\minus{}1\right]\;\frac{dx}{x}\;\;\;\equal{}\;\;\;\boxed{2\left(1\;\minus{}\;\left(\frac{\sqrt{\;1\plus{}a\;}}{\sqrt a}\cdot\sinh^{\minus{}1}\sqrt a\right)\right)}$
[put $ x\equal{}\sinh^2\,u$ and then do integration by parts for $ \int\;u\cdot\text{cosech}^2\,u\;\;du$ to solve the first integral. second integral is again simply $ \ln\,x$. and likewise, the integrals need to be considered [i]together [/i]as both are [b]improper[/b] integrals of the second kind. also, we need to use the fact $ \lim_{u\to 0}\;u\cdot\coth\,u\;\equal{}\;1$]


Factor 2ax²-8ax-12a. Is this possible to factor?

We can take out an $ a$ to make it $ a(2x^2 \minus{} 8x \minus{} 12) \equal{} a(2x \plus{} 4)(x \minus{} 3)$ Yes, that is correct..... So our answer is $ \boxed{a(2x \plus{} 4)(x \minus{} 3)}$. I think.......  :maybe:

Edit: Oops.... GJ Awesome Toad.

I don't think you factored correctly... it doesn't multiply out
You can also factor out a $ 2$.
$ 2a(x^2\minus{}4x\minus{}6)$
I don't think this is factorable any further with rational numbers.

[quote="pascal12"]I don't think you factored correctly... it doesn't multiply out
You can also factor out a $ 2$.
$ 2a(x^2 \minus{} 4x \minus{} 6)$
I don't think this is factorable any further with rational numbers.[/quote]

It's not.
You can check by finding the discriminant: $ \sqrt{(\minus{}4)^2\minus{}4(1)(\minus{}6)}\equal{}\sqrt{16\plus{}24}\equal{}2\sqrt{5}$
If it is factorable, it has rational roots.
But $ 2\sqrt{5}$ is not rational

I agree, except that $ \sqrt{16\plus{}24} \equal{} \sqrt{40} \equal{} \sqrt{4 \times 10} \equal{} 2 \sqrt{10}$.

This is not factorable with rational numbers. :starwars:  :spam:


How many three-digit numbers fit the property that the second
digit is greater than or equal to the first digit and the third digit is
greater than or equal to the second digit?

If the 1st number was 1, there would be 9 + 8 + 7 + 6 +... +1, or 45 choices for the 3rd number, depending on the 2nd number (if the 2nd number was 1, the 3rd number would have 9 choices, if the 2nd number was 2, the 3rd number would have 8 choices, and so on.) If the 1st digit was 2, there would be 8 + 7 + 6 +...+1, or 36 choices. And so on. So, there would be (9+8+7+...+1) + (8+7+6+...+1) + (7+6+5+...1) + (6+5+4+...+1) +...+(2+1) + (1), or 45 + 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1, or 165 number? Is that correct?


[quote="stephencheng"]$ xF(x) \minus{} 1$ is divisible by $ (x \minus{} 1)(x \minus{} 2)....(x \minus{} 9)$ as $ xF(x) \minus{} 1 \equal{} 0$ has roots $ 1,2,3,...,9$

So $ xF(x) \minus{} 1 \equal{} (x \minus{} 1)(x \minus{} 2)...(x \minus{} 9)(x \plus{} \frac {1}{9!})$
$ 10F(10) \minus{} 1 \equal{} (9!)(10 \plus{} \frac {1}{9!}) \Leftrightarrow F(10) \equal{} 9! \plus{} \frac {1}{5}$[/quote]

I have a question: How did you know the last term in the second line was $ (x \plus{} \frac{1}{9!})$?

The Idea is simple.          product(roots)=-1.
                                      hence the x independent part of the polynomial should be -1....So 1/9factorail is a root of the equation

[quote="sagar1331"]F(x)= 1/x for all integers 1- 9. Nine degree polynomial. Find f(10)[/quote]

Let $ g(x)\equal{}xf(x)\minus{}1$ ,hence $ g(x)$ is a $ 10$ degree polynomial which can be written as
$ g(x)\equal{}p(x)(x\minus{}1)(x\minus{}2)\cdots(x\minus{}9)$ where $ p(x)$ is a linear polynomial of x [b]not necessarily monic.[/b]

So
$ 10f(10)\minus{}1\equal{}p(10)\times 9!$
or $ f(10)\equal{}(p(10)9!\plus{}1)/10$.
Check that, $ p(10)$ is of the form $ k(x\minus{}1/9!)$ where k is a constant,in fact the leading coefficient of $ f$.


conics
parabola
calculus
derivative
quadratics
logarithms
analytic geometry

Hi all!

I graphed the following set of values:

d (x-axis)                 T in years (y-axis)
0.387                                 0.241
0.723                                 0.615
1                                       1
1.523                                 1.881
5.203                                 11.861
9.541                                 29.457
19.19                                 84.008
30.086                               164.784

ANd it turned out to be a parabola. I am trying to find a, h and k. But when I fit the other numbers in the equation, it doesn't work out!!!

What am I supposed to do??? PLEASE HELP! ASAP!

It's not a parabola- not even close. If we assume enough smoothness, the values of the second derivative range from more than $ 0.9$ early in the interval to less than $ 0.2$ late in the interval.

Growth is too slow for quadratic, too fast for linear ... my next guess was a perfect match.  The fact that you're looking for something passing through $ (0, 0)$ and $ (1, 1)$ narrows down the range of reasonable guesses anyhow.

Ok I think it's this:

y = x ^ 3/2

by trial and error.

But how do I EXPLAIN to the prof that this is the only closest equation I could get. I don't wanna show "trial and error" method...is there any logical solution to this problem??

[quote="Narjis"]How?  :maybe:[/quote]

If $ y\equal{}x^\alpha$, then $ \log y \equal{} \alpha \log x$, so plotting $ \log y$ against $ \log x$ will give you a straight line with a slope of $ \alpha$.

but it's not a straight line! I swear!

:(

ok bear with me folks.

In my paper, it says, "Make a scatter plot of the data and fit a [b]power[/b] model to the data"

blahblahblah, you're telling me to graph a logx vs log y graph?

but what if she deducts marks off my question, if I do it according to your method? I mean we weren't asked to take logarithms to graph the values..

ok so technically, the equation is
[img]http://alt2.mathlinks.ro/Forum/latexrender/pictures/b/4/8/b4894665dfafff79df91e091158e49ccbc759960.gif[/img] ?

and alpha is around 0.66 ??

ahh..this is getting me dizzy   :noo:

Note that blahblahblah was giving you $ \frac{\log x_i}{\log y_i},$ which is upside down from the way you probably want to think if you want to write $ y\equal{}Cx^{\alpha}.$ (In general, you would need such a constant; in this particular case, since we have the point $ (1,1),$ we will get $ C\equal{}1.$)

So turn blahblahblah's quotients over, and get $ \alpha$ approximately equal to to the reciprocal of $ .66,$ namely $ 1.5.$

---

The general explanation: if $ y\equal{}Cx^{\alpha},$ then $ (\log y)\equal{}(\log C)\plus{}\alpha(\log x).$ Letting $ u\equal{}\log x$ and $ v\equal{}\log y,$ that's $ v\equal{} \alpha u\plus{}\beta.$ That's a straight line. Two points suffice to determine a line, but to an expermentalist, five (or eight) points determine a better line than two do. Plot $ v$ versus $ u$ and find the best-fitting line. There's a whole theory out there about what "best-fitting" means, but we don't need this here. (The main reason we don't need it: this is blatantly "fake" data, and the fit is [i]too[/i] good.) The slope of the line is then $ \alpha,$ and the intercept is $ \beta\equal{}\log C.$

Do I mean natural logarithm or base-10 logarithm. Doesn't matter. Just pick one and stay consistent.

When I was teaching BC (business calculus), I used to my students that there are 3 types of dependencies that most commonly occur in physical processes: linear, power, and exponential. The only one that our eye can easily see on the plot and that a simple computation can reveal from not equally spaced data is the linear one. So, if we suspect that the dependence is power or  exponential and wish to see it, the best thing we can do is to reduce the problem to the linear case. Fortunately, in both cases, it is not hard by taking logarithms.

Why am I saying all that? Just to emphasize that looking at $ \log y$ versus $ \log x$ is [b]the best[/b] way to discern a power dependence and to find its parameters and I usually told my students that it should be done automatically once the power dependence is mentioned or suspected, meaning that I would never explicitly state "take logarithms to ..." in any of my problems. I don't know if your teacher has the same opinion but, once you have done the problem by a correct method and got a right answer, I do not see why you should be afraid of your grade being reduced. (The only thing I would deduct the points for is your doing your homework using external help, but it is a different story :P)

If I were to rummage around in the undisturbed depths of some drawer or file cabinet I might find some semi-log or log-log graph paper. Maybe they still even manufacture that stuff, although I would think that with calculators, spreadsheets, and specialized data-graphing programs there's not much call for it.  (If I actually intended to do anything with the original poster's data, my very first step would have been to enter that data into a spreadsheet. I suspect that's the first thing jmerry did before he wrote his first reply.)

There are (or were) three basic types of graph paper with straight lines ("with straight lines" was intended to rule out polar coordinate paper): evenly spaced ("ordinary"), semilog, and log-log.

[quote="fedja"]... there are 3 types of dependencies that most commonly occur in physical processes: linear, power, and exponential.[/quote]
So, graph the linear dependency on the ordinary graph paper, the power dependency (including this problem) on the log-log paper, and the exponential dependency on the semilog paper. Or, without the specialized paper but with your calculator or spreadsheet handy:

If you think you have $ y\equal{}ax\plus{}b,$ graph $ y$ versus $ x.$

If you think you have $ y\equal{}Cx^{\alpha},$ graph $ \ln y$ versus $ \ln x.$

If you think you have $ y\equal{}Ce^{kx},$ graph $ \ln y$ versus $ x.$

As fedja said, your eyes can really only make good sense of straight lines. Now assume the points you're graphing are actual experimental data. If you have more than two points - and you [i]want[/i] to have more than two points - then the points don't lie on a line, but you're going to use the line that does the best job of compromising among the points.

There are other kinds of dependencies, of course. My wife is a biochemist with an interest in enzymes and enzyme kinetics. There are many equations of the general form $ y\equal{}\frac{ax}{b\plus{}x}.$ In that case, one possibility is to graph $ \frac1y$ versus $ \frac1x.$


[quote="chien than"]Prove that:$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+48(ab+bc+ca) \geq 25$
where $a;b;c$ are positive real numbers satisfying $a+b+c=1$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=119301


In this series of problems, I think it might be a good idea to focus on acid-base, complexation, redox, and precipitation titrations. Also, simpler problems regarding acid-base, precipitation, redox, and complexation reactions can also be posted. Anyone is invited to contribute.
___________________________________________

- Consider the titration of 25 mL of an 0.1 M acid solution of Sn2+ with an 0.05 M acid solution of Fe3+.

(a) Show that this titration is possible.
(b) Compute the concentration of non-oxidised Sn2+ in the equivalence point.
(c) What will be the titration error if one use an indicator that turns at 0.6 V (the indicator redox reaction involves one electron)?

Eº(Sn4+/Sn2+) = 0.14V; Eº(Fe3+/Fe2+) = 0.70V.

[quote="Carcul"](a) Show that this titration is possible.[/quote]

What does this mean?  Is it asking for one to write a reaction equation and to use the given cell potentials to show that it is spontaneous?

[quote="Zakary"]What does this mean?[/quote]

For a given titration to be analytically useful it must met the following two criteria (among others):

1) the titration reaction must be complete ($K \geq 10^{8}$).
2) there must be a sudden jump in some property of the system at the equivalence point - frequently, the concentration of some species.

1. The reation is $ 2\ce {Fe^{3+}} + \ce {Sn^{2+}}\rightarrow 2\ce {Fe^{2+}} + \ce {Sn^{4+}}$
$ E_0=0.56V=\frac {RT}{2F} \ln K$ (Nernst)

This gives $ K\approx 10^{19}$

2. $ K=\frac {[\ce {Fe^{2+}}]^2[\ce {Sn^{4+}}]}{[\ce {Fe^{3+}}]^2[\ce {Sn^{2+}}]}$
Initial stannous amount is 2.5 millimoles or 5 meqs, while at equivalence, ferric amount is 5 millimoles or 5 meqs.
This gives $ [\ce {Sn^{2+}}]_{eq} \approx 1.16 \times 10^{-6} M$

3. I don't understand this part.


function
algebra
rational function
complex analysis
complex analysis unsolved

Let $ f$ be a rational function of degree 2 and assume $ f$ has only simple poles. Prove that there exists a Mobius transformation $ T$ and affine linear map $ S$ with $ S\circ f \circ T(z)\equal{}z\plus{}\frac{1}{z}$. What about the case $ f$ has a double pole?

[quote="math14"]Let $ f$ be a rational function of degree 2 and assume $ f$ has only simple poles. Prove that there exists a Mobius transformation $ T$ and affine linear map $ S$ with $ S\circ f \circ T(z) \equal{} z \plus{} \frac {1}{z}$. What about the case $ f$ has a double pole?[/quote]

What is the degree of a rational function?

[quote="math14"]The degree of rational function $ f$ is $ \max\{\deg P, \deg Q\}$, where $ f \equal{} \frac {P}{Q}$ with $ P$ and $ Q$ are coprime.[/quote]I would rather expect $ \deg f  \minus{} \deg g$ regardless if $ f,g$ are coprime or not. This is inspired by the degree of a divisor associated to a meromorph function on the Riemann sphere.


Let $ x, y, z$ be positive reals satisfying:

$ x\plus{}y\plus{}z\plus{}2xyz\equal{}xy\plus{}yz\plus{}zx\plus{}1$

Determine the maximum value of the expression:

$ S\equal{}\sqrt{x}\plus{}\sqrt{y}\plus{}\sqrt{z}$

Επειδή είναι αρκετά βράδυ θα δώσω απλά ένα hint που απλοποιεί πολύ την κατάσταση (πιστεύω μετά γίνεται "εύκολη", αλλά δεν το εγγυώμαι κιόλας 1 η ώρα τη νύχτα :P )
[hide]Η δοσμένη ισότητα γίνεται:
$ (1 \minus{} x)(1 \minus{} y)(1 \minus{} z) \equal{} xyz$ κι έτσι θέτοντας $ a \equal{} \frac {1 \minus{} x}{x}$ και λοιπά έχουμε $ abc \equal{} 1$ και θέλουμε να βρούμε το max του:
$ \sum{\sqrt {\frac {1}{a \plus{} 1}}}$[/hide]

Υ.Γ. Μετά από παρατήρηση του Σιλ, καταλαβαίνω ότι μάλλον από αυτό το σημείο δεν είναι καθόλου "εύκολη", οπότε μη παρασυρθείτε από αυτά που έγραψα πριν... απλά δεν είναι πλέον "ασυνήθιστη"!

Το τελειωμα σε hint  :P . [hide]Κάνουμε πρώτα το μετασχηματισμό $ x \equal{} \frac {b}{a}$ στην ανισότητα του mim . Μετά
Από  Cauchy-Schwarz
\[ \left( a(b \plus{} c) \plus{} b(c \plus{} a) \plus{} c(a \plus{} b)\right)\cdot\left(\frac {1}{(a \plus{} b)(b \plus{} c)} \plus{} \frac {1}{(b \plus{} c)(c \plus{} a)} \plus{} \frac {1}{(c \plus{} a)(a \plus{} b)}\right)
\]

\[ \geq\left(\sqrt {\frac {a}{a \plus{} b}} \plus{} \sqrt {\frac {b}{b \plus{} c}} \plus{} \sqrt {\frac {c}{c \plus{} a}}\right)^{2}
\]
Άρα αρκεί να δείξουμε ότι
\[ 2(ab \plus{} bc \plus{} ca)\cdot\left(\frac {1}{(a \plus{} b)(b \plus{} c)} \plus{} \frac {1}{(b \plus{} c)(c \plus{} a)} \plus{} \frac {1}{(c \plus{} a)(a \plus{} b)}\right)\leq\frac {9}{2},
\]
που ισοδυναμεί με
\[ (a \plus{} b)(b \plus{} c)(c \plus{} a)\geq\frac {8}{9}\cdot(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca),
\]
που είναι γνωστή . QED  :) [/hide]

Εαν κάποιος δεν θυμάτε αυτόν το μετασχιματισμό του silouan μπορεί να την αποδείξει εύκολα ως εξείς:
$ \sum\frac{ab\plus{}bc\plus{}ca}{ab\plus{}bc\plus{}ca\plus{}a^2} \le \frac{9}{4} \Longleftrightarrow \sum\frac{ab\plus{}bc\plus{}ca}{ab\plus{}bc\plus{}ca\plus{}a^2}\minus{}1 \le \minus{}\frac{3}{4} \Longleftrightarrow \sum\frac{a^2}{ab\plus{}bc\plus{}ca\plus{}a^2} \ge \frac{3}{4}$ που ισχύει από Andreescu.

Πολύ ωραίες οι λύσεις και πολύ όμορφη άσκηση Κώστα :wink:

Πρέπει να τονιστεί όμως (και να προσεχθεί) ότι σε αυτή, τη μορφή η άσκηση δεν έιναι σωστή! Μπορεί π.χ. το x να τείνει στο άπειρο κι έτσι η παράσταση να μην έχει μέγιστο!
Η άσκηση γίνεται σωστή αν πρστεθεί ο επιπλέον περιορισμός $ x,y,z<1$ ώστε τα $ a,b,c>0$!

Νice και πάλι :wink:

Πολύ σωστά! Ο μετασχηματισμός εμπερικλείει και έναν περιορισμό! Δεν το είχα σκεφτεί. Νόμιζα ότι έφτανε να είναι θετικοί. Ωραίες όλες οι ιδέες. Ήταν αρκετά δύσκολη άσκηση πάντως. :)

[quote="Dimitris X"]Εαν κάποιος δεν θυμάτε αυτόν το μετασχιματισμό του silouan μπορεί να την αποδείξει εύκολα ως εξείς:
$ \sum\frac {ab \plus{} bc \plus{} ca}{ab \plus{} bc \plus{} ca \plus{} a^2} \le \frac {9}{4} \Longleftrightarrow \sum\frac {ab \plus{} bc \plus{} ca}{ab \plus{} bc \plus{} ca \plus{} a^2} \minus{} 1 \le \minus{} \frac {3}{4} \Longleftrightarrow \sum\frac {a^2}{ab \plus{} bc \plus{} ca \plus{} a^2} \ge \frac {3}{4}$ που ισχύει από Andreescu.[/quote]

Γινεται και πιο απλα, αν ανοιξεις τα αθροισματα στην ανισοτητα του sil προκειπτει μια μορφη AM-GM για 6 αριθμους...

btw sil το 9/8 καντο 8/9 :P


i believe i^6 is -1, and i^16 is 1, but i dont know about the third term.

The third term is $ \minus{}1$ too, because of the cycles of $ i$.

We add them up to get...$ \boxed{\minus{}1}$.

PS: If you don't know what the cycles of $ i$ are, please ask. I have a feeling that you might not know what they are, as you asked this on the forum, but I don't want to be "annoying", do I? *cough* *cough* mz94!!! *cough* *cough*

*cough* cough* ernie *cough* *cough*

 :rotfl:  :rotfl:  :rotfl:


How many triangles are in this figure?

[img]7748[/img]

[hide="answer is right herererererererer"]around 35 you get it by literally counting every trianle u see!![/hide][/hide]

for this i honestly dont see one :P  though in other kinds of diagrams (such as a bunch of squares and u have to find rectangles, there might be more "slick ways"

[hide]
1 triangle = 10
2 triangles = 10
3 triangles = 5
triangles formed by two small triangles and the pentagon in the middle = 5
triangles formed by a side of the larger pentagon and the far vertice = 5

add together to get 35.
[/hide]

I got 35 but the answer book says it's 34 :read:

[hide="Solution with two cases"]The answer is the same as if this were a regular pentagon.

In the regular pentagon, count the number of triangles pointing straight up (4), plus the number of triangles pointing straight down (3).  Then multiply by 5 to allow for various rotations.

Finally, ask yourself if there are any additional triangles pointing obliquely.  I don't see any.  I get $(4+3)\times 5=35.$[/hide]

I think that if the general shape is [b]convex[/b], you can count the number of interior angles.(I remember one on 2004 national sprint where the shape was concave and this didn't work)

the all time greatest mathcounts problems clearly says that its [hide]35, because of the 5-way rotational symmetry[/hide]


pigeonhole principle
number theory proposed
number theory

Let the number $ x\equal{}n,a_1a_2a_3....$  and it is known that for every $ k$ there is a block of $ k$ numbers at the decimal part of $ x$ which can be found at least two times at the decimal part of $ x$ . Prove that the same property holds also for the number $ x^2$

May it true for $ x\equal{} \sqrt{0.5}$? I think it's not when $ x$ is irrational and by Pigeonhole principle easy to see $ x$ satisfies the conditon of problem, but $ x^{2}\equal{}0.5$ is not


probability geometry ARML AMC AIME MATHCOUNTS

Here is a set of problems that you can solve.  PM me the answers and i will post the names of the people that got it right.  You have to use the answer to the first problem in the second problem, you have to use the answer to the second problem in the third problem, and so on.  There a 4 total problems.  Let the race begin!

RELAY QUESTION 1
$n=2^3 \cdot 3^2$
Let $k$ be the number of positive integers which are factors of $n$ including 1 and $n$.  Use $k$ in problem 2

RELAY QUESTION 2
Let $k$ be the answer from question 1.
Let $p=k(3-k)(2+k)+k^2(k-1)-k$

use $\frac{p}{2}$ in problem 3.

RELAY QUESTION 3
Let $k$ be the answer from question 2.

A jar contains $k$ black marbles, 12 blue marbles, 6 red marbles, and 2 yellow marbles.  Three marbles are drawn from the jar without replacement.

The probability of all three marbles being black can be written as $\frac{m}{n}$, a fraction in lowest terms.  Find $\sqrt{m+n}$

RELAY PROBLEM 4
Let $k$ be the answer to question 3.

In a circle, there is a right triangle inside.   The right triangle is triangle ABC where angle c is a right angle.  AB is the diameter, BC=8k, CA=6k.
Let F= the area of the circle.
Submit $\frac{f}{\pi}$ as your final answer(pm me $\frac{f}{\pi}$
if you need a clearer explanation just tell me.

Does anyone want to do it?  nat mc already got it.

good job nat mc[hide="the answer"]just kidding you have to do it yourself[/hide]
i'm going to make another round so if you get this correct you move onto the next round.  The next round will posteed next saturday.

It's a mainly high school math competition.
15 people per team different states and cities compete.

[quote="jli"][quote="chess64"][quote="nat mc"]cool but I'll be at ARML[/quote]

how can you be in arml in 7th grade[/quote]

im in  arml but im in 6th grade[/quote]

how?

they are probably taking high school math.

In Texas you have to get invited to be in ARML which is dependant on how well you do on varoius competitions throughout the year (AMC, AIME, MATHCOUNTS, etc.).

It's different from state to state...

[quote="nat mc"]Yeah, I was invited, I did it last year too.  Last year I was on the Indiana blue team (like the C team) and this year I was on the Indiana Gold  team (like the A team)[/quote]

I did it this year... I was put on the SFBA B Team, although I did better than half the people on the A team...

what happened to the 2nd round? And how come you didn't write my name?  :(

[quote="nat mc"]lol,  What was your individual score[/quote]

5

[quote="bob123"]1,2,3,5,7[/quote]
This year seemed really easy, and I made 3 stupid mistakes.
I got a 4, 1,2,3,4, but I easily got theanswer 2 seconds afterward for 5,6,7, but I couldn't factor number 8


Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ (a\plus{}b\plus{}c)^8\geq128(a^5b^3\plus{}a^5c^3\plus{}b^5a^3\plus{}b^5c^3\plus{}c^5a^3\plus{}c^5b^3)\]

[quote="2424"]
Hi materzzi,I think you have bad solution! :P[/quote]

[hide] tớ chỉ muốn thử làm kiểu này xem nó thế nào thôi :)[/hide]

[quote="2424"][quote="Materazzi"][quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ (a \plus{} b \plus{} c)^8\geq128(a^5b^3 \plus{} a^5c^3 \plus{} b^5a^3 \plus{} b^5c^3 \plus{} c^5a^3 \plus{} c^5b^3)(*)\]
[/quote]

Assume : $ c \equal{} min\{a;b;c\}$

With :  $ a \equal{} x \plus{} c;b \equal{} y \plus{} c$ ( $ x,y \geq 0$ )

We have :

$ (*) < \equal{} > 5793c^8 \plus{} 15488c^7(x \plus{} y) \plus{} c^6(17084(x^2 \plus{} y^2) \plus{} 36984xy)$

$ \plus{} c^5(10792(x^3 \plus{} y^3) \plus{} 35064xy(x \plus{} y))$

$ \plus{} c^4(4390(x^4 \plus{} y^4) \plus{} 18200xy(x^2 \plus{} y^2) \plus{} 26340x^2y^2)$

$ \plus{} c^3(1256(x^5 \plus{} y^5) \plus{} 5640xy(x^3 \plus{} y^3) \plus{} 10000x^2y^2(x \plus{} y))$

$ \plus{} c^2(252(x^6 \plus{} y^6) \plus{} 1128xy(x^4 \plus{} y^4) \plus{} 1860x^2y^2(x^2 \plus{} y^2) \plus{} 2480x^3y^3)$

$ \plus{} c(24(x^7 \plus{} y^7) \plus{} 168xy(x^5 \plus{} y^5) \plus{} 200x^3y^3(x \plus{} y) \plus{} 120x^2y^2(x^3 \plus{} y^3))$

$ \plus{} (x \minus{} y)^2(x^6 \plus{} y^6 \plus{} 10xy(x^4 \plus{} y^4) \plus{} 47x^2y^2(x^2 \plus{} y^2) \plus{} 12x^3y^3) \geq 0$

Done !

:)[/quote]
Hi materzzi,I think you have bad solution! :P[/quote]
my solution:
Let $ f(a,b,c) \equal{} \sum_{cyc}a^5(b^3 \plus{} c^3)$
Assume $ a\ge b\ge c$  $ (a,b,c\geq0)$
$ \equal{} > f(a,b,c) \minus{} f(a,b \plus{} c,0) \equal{} \sum_{cyc}a^5(b^3 \plus{} c^3) \minus{} a^5(b \plus{} c)^3 \minus{} (b \plus{} c)^5a^3\leq0$
And $ 128(a^3(b \plus{} c)^3(a^2 \plus{} (b \plus{} c)^2))\leq (a \plus{} b \plus{} c)^8$ (by Am-Gm)
So,(*) is true! :lol:
@materazzi:[hide]minh nghi ban chi nen su dung cach cm dat $ a \equal{} x \plus{} c,b \equal{} y \plus{} c$ cho nhung bdt chat va bac k cao.[/hide]

[quote="2424"]
@materazzi:[hide]minh nghi ban chi nen su dung cach cm dat $ a \equal{} x \plus{} c,b \equal{} y \plus{} c$ cho nhung bdt chat va bac k cao.[/hide][/quote]

[hide] ấy ấy ..... tớ nói rồi ..... ý tưởng về việc giải nó thì tớ ko thiếu đâu cậu ;) ...... chỉ là kiểu làm này trước giờ tớ chỉ thấy ngta xài chứ chưa bao giờ áp dụng cả ........ nên chỉ muốn dùng " thử " 1 lần cho thỏa mãn " khao khát " .... thế thôi :)    [/hide]

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ (a \plus{} b \plus{} c)^8\geq128(a^5b^3 \plus{} a^5c^3 \plus{} b^5a^3 \plus{} b^5c^3 \plus{} c^5a^3 \plus{} c^5b^3)\]
[/quote]
We can see that
$ a^5b^3 \plus{} a^5c^3 \plus{} b^5a^3 \plus{} b^5c^3 \plus{} c^5a^3 \plus{} c^5b^3 \equal{}\sum a^3b^3(a^2\plus{}b^2) \le \sum a^3b^3(a^2\plus{}b^2\plus{}c^2) \equal{}(a^3b^3\plus{}b^3c^3\plus{}c^3a^3)(a^2\plus{}b^2\plus{}c^2).$
On the other hand,
$ a^3b^3\plus{}b^3c^3\plus{}c^3a^3 \le (a^2b^2\plus{}b^2c^2\plus{}c^2a^2)(ab\plus{}bc\plus{}ca),$
and
$ (ab\plus{}bc\plus{}ca)(a^2\plus{}b^2\plus{}c^2) \le \frac{1}{8}[2(ab\plus{}bc\plus{}ca)\plus{}a^2\plus{}b^2\plus{}c^2]^2\equal{}\frac{1}{8}(a\plus{}b\plus{}c)^4,$
so
$ a^5b^3 \plus{} a^5c^3 \plus{} b^5a^3 \plus{} b^5c^3 \plus{} c^5a^3 \plus{} c^5b^3 \le \frac{1}{8}(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)(a\plus{}b\plus{}c)^4,$
and the inequality is reduced to
$ (a\plus{}b\plus{}c)^4 \ge 16(a^2b^2\plus{}b^2c^2\plus{}c^2a^2).$
The last one can be proved easily using expanding or mixing variables or something similar.

It's also my proof  :lol:
By this way, we can prove the stronger:
Given $ a, b, c \geq\ 0$ and $ a\plus{}b\plus{}c\equal{}2$. Prove that:
$ \sum\ a^5(b^3\plus{}c^3)\plus{}21a^2b^2c^2 \leq\ 2$  :)

[quote="Guest"][quote="nguoivn"]It's also my proof  :lol:
By this way, we can prove the stronger:
Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 2$. Prove that:
$ \sum\ a^5(b^3 \plus{} c^3) \plus{} 21a^2b^2c^2 \leq\ 2$  :)[/quote]
This is not true, for a counter-example, $ a \equal{} \frac{28}{39},\ b \equal{} \frac{2}{3},\ c \equal{} \frac{8}{13}.$[/quote]
However, it holds the following inequality:
Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 2$. Prove that

$ \sum\ a^5(b^3 \plus{} c^3) \plus{} \frac{1931}{96}a^2b^2c^2 \leq\ 2,$
and the constant $ \frac{1931}{96}$ is the best possible.

[quote="Guest"]
However, it holds the following inequality:
Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 2$. Prove that

$ \sum\ a^5(b^3 \plus{} c^3) \plus{} \frac {1931}{96}a^2b^2c^2 \leq\ 2,$
and the constant $ \frac {1931}{96}$ is the best possible.[/quote]
Of course!
An equality holds also when $ a \equal{} b \equal{} c.$
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3.$
Hence, $ \sum\ a^5(b^3 \plus{} c^3) \plus{} \frac {1931}{96}a^2b^2c^2 \leq\ 2\Leftrightarrow Aw^6 \plus{} Bw^3 \plus{} C\leq0,$
where $ A\equal{}A(u,v^2) \geq 0,$ $ B\equal{}B(u,v^2)$ and $ C\equal{}C(u,v^2).$
Thus, we need to check only $ b \equal{} c$ and $ w^3 \equal{} 0.$
In  both cases your inequality is true.


conics
parabola
analytic geometry
graphing lines
slope
calculus
calculus computations

How many nomal lines can be drawn to a palabola $ y^2 \equal{} 4px\ (p > 0)$ from the point $ P(a,\ b)$?

i answered it before. in general, it is going to be $ 3$.

parametric equations of the given parabola are $ (pt^2,2pt)$. so, at any point on the parabola, slope of tangent is $ \frac{dy}{dx}\equal{}\frac{2p}{2pt}\equal{}\frac{1}{t}$,
and therefore, slope of the normal at that point will be $ \minus{}t$. hence, equation of normal is $ y\minus{}2pt\equal{}\minus{}t(x\minus{}pt^2)$

$ \Rightarrow \;\;pt^3\plus{}t(2p\minus{}x)\minus{}y\equal{}0$

this is clearly a cubic in $ t$, and so in general there are going to be $ 3$ points $ t_1\;, t_2\;, t_3$
where a normal from a point $ (a,b)$ can be drawn. sure, a cubic could also have
one real and two complex roots, in that case, there will be points $ (a,b)$ from where
only $ 1$ [i]real[/i] normal can be drawn.


1. Prove that 11^(n+2) + 12^(2n+1) is divisible by 133
2. Prove that 1.2 + 2.5 + .... + n(3n-1) = n^2(n+1)
3. Prove that 1^2 + 3^2 + ..... + (2n-1)^2 = n(4n^2 - 1)/3
4. Prove that for any positive natural number n , 1/(n+1) + 1/(n+2) + ... + 1/(3n+1) > 1
5. Prove that for all natural number n > 1, 4^n/(n+1) < (2n)!/((n!)^2).

Can give me some clue how to do these question

[quote="dreamer.ande"]
2. Prove that 1.2 + 2.5 + .... + n(3n-1) = n^2(n+1)

Can give me some clue how to do these question[/quote]

dreamer.ande,

do not use periods for multiplication.

Instead, suggestions are:

"Prove that 1(2) + 2(5) + ... + n(3n - 1) = n^2(n + 1)" or

"Prove that 1*2 + 2*5 + ... + n(3n - 1) = n^2(n + 1)."

If you know and/or (mimic from other posts) at least a relatively
smaller percent of Latex phrasing/code as I do, then there are also,
but not limited to, these:

"Prove that $ 1\cdot2 \plus{} 2\cdot5 \plus{} ... \plus{} n(3n \minus{} 1) \equal{} n^2(n \plus{} 1)$" or

"Prove that $ 1(2) \plus{} 2(5) \plus{} ... \plus{} n(3n \minus{} 1) \equal{} n^2(n \plus{} 1)$."

[quote="dreamer.ande"]1. Prove that 11^(n+2) + 12^(2n+1) is divisible by 133
[/quote]
Prove the base case (n=1):

$ 11^{1 \plus{} 2} \plus{} 12^{2 \plus{} 1} \equal{} 3059$
$ 3059/133 \equal{} 23$

Then assume that it holds true for natural number k ($ 11^{k \plus{} 2} \plus{} 12^{2k \plus{} 1} \equal{} 133h$, where h is an integer) and prove for k+1.  If you want to see it, [hide="click here"]

$ 11^{k \plus{} 1 \plus{} 2} \plus{} 12^{2(k \plus{} 1) \plus{} 1}$

$ 11 \cdot 11^{k \plus{} 2} \plus{} 12^{2} \cdot 12^{2k \plus{} 1}$

$ 11 \cdot 11^{k \plus{} 2} \plus{} (133 \plus{} 11) \cdot 12^{2k \plus{} 1}$

$ 11 \cdot (11^{k \plus{} 2} \plus{} 12^{2k \plus{} 1}) \plus{} 133 \cdot (12^{2k \plus{} 1})$

$ 11 \cdot (133h) \plus{} 133 \cdot (12^{2k \plus{} 1})$

$ 133 \cdot (11h \plus{} 12^{2k \plus{} 1})$

..which is a multiple of 133.

Q.E.D.[/hide]

[quote="dreamer.ande"]3. Prove that 1^2 + 3^2 + ..... + (2n-1)^2 = n(4n^2 - 1)/3[/quote]

We prove this by induction. Base case $ n\equal{}1$ we have $ 1\equal{}\frac{1(4\minus{}1)}{3}\equal{}1$ so base case is done.

Assume it is proven for $ k$ and prove the result for $ k\plus{}1$

We have $ \frac{n(4n^2\minus{}1)}{3}\equal{}\frac{n(2n\plus{}1)(2n\minus{}1)}{3}$ so for $ k\plus{}1$ we want to show that $ LHS\equal{}\frac{(k\plus{}1)(2k\plus{}1)(2k\plus{}3)}{3}$

$ LHS\equal{}1^2\plus{}3^2\plus{}5^2...\plus{}(2k\minus{}1)^2\plus{}(2k\plus{}1)^2\equal{}\frac{k(2k\minus{}1)(2k\plus{}1)}{3}\plus{}(2k\plus{}1)^2$

Factoring gives: $ (2k\plus{}1)(\frac{k(2k\minus{}1)}{3}\plus{}(2k\plus{}1))$

$ \frac{1}{3}(2k\plus{}1)(k(2k\minus{}1)\plus{}3(2k\plus{}1))$

Multiplying out:

$ \frac{1}{3}(2k\plus{}1)(2k^2\minus{}k\plus{}6k\plus{}3)\equal{}\frac{1}{3}(2k\plus{}1)(2k^2\plus{}5k\plus{}3)\equal{}\frac{(2k\plus{}1)(2k\plus{}3)(k\plus{}1)}{3}$

This is what we were trying to prove.

[quote="dreamer.ande"]2. Prove that 1.2 + 2.5 + .... + n(3n-1) = n^2(n+1)[/quote]

Again, by induction.

Base case: $ 2\equal{}1^2(1\plus{}1)\equal{}2$ base case done.

Assume true for $ n\equal{}k$ and prove for $ n\equal{}k\plus{}1$

$ 1 (2) \plus{} 2(5) \plus{} 3(8)... \plus{} k(3k\minus{}1) \plus{} (k\plus{}1)(3k\plus{}2)\equal{}k^2(k\plus{}1)\plus{}(k\plus{}1)(3k\plus{}2)$

Factoring: $ (k\plus{}1)(k^2\plus{}3k\plus{}2)\equal{}(k\plus{}1)(k\plus{}1)(k\plus{}2)\equal{}(k\plus{}1)^2(k\plus{}2) \equal{} n^2(n\plus{}1)$

This is what we were trying to prove.

EDIT: This was originally Q4 but my solution was incorrect. Ignore this post.

[quote="KrazyFK"]EDIT: Ignore, was incorrect.[/quote]

both your solutions looked correct to me.

NOTE: the other problems can all be done by inducting. I will type up solutions if you want me to.

If you can, please do so so that I can check with the solution that I have done. I will like to view how to present a proof more than a solution.

[quote="dreamer.ande"]4. Prove that for any positive natural number n , 1/(n+1) + 1/(n+2) + ... + 1/(3n+1) > 1[/quote]

Let us call this $ f(n)$

$ f(1) \equal{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} \frac {1}{4} \equal{} \frac {13}{12} > 1$

$ f(k) \equal{} \frac {1}{k \plus{} 1} \plus{} \frac {1}{k \plus{} 2}... \plus{} \frac {1}{3k \plus{} 1}$.

$ f(k \plus{} 1) \equal{} \frac {1}{k \plus{} 2} \plus{} \frac {1}{k \plus{} 3}... \plus{} \frac {1}{3k \plus{} 4} \equal{} f(k) \plus{} \frac {1}{3k \plus{} 2} \plus{} \frac {1}{3k \plus{} 3} \plus{} \frac {1}{3k \plus{} 4} \minus{} \frac {1}{k \plus{} 1}$

After putting fraction over common denominators and multiplying out etc. we obtain:

$ \frac {1}{3k \plus{} 2} \plus{} \frac {1}{3k \plus{} 3} \plus{} \frac {1}{3k \plus{} 4} \minus{} \frac {1}{k \plus{} 1} \equal{} \frac {2}{3(k \plus{} 1)(3k \plus{} 2)(3k \plus{} 4)}$   which is positive.

Thus we have shown that $ f(k \plus{} 1) \equal{} f(k) \plus{} \frac {2}{3(k \plus{} 1)(3k \plus{} 2)(3k \plus{} 4)}$

Thus $ f(k \plus{} 1) > f(k)$ for all $ k$. Since $ f(1) > 1$, this means that $ f(n) > 1$ for all $ n$.

[quote="KrazyFK"]@hurdler.

Sorry for ambiguity, the post where I said incorrect I had actually posted a solution to question 4. Shortly after posting I realised it was incorrect so I deleted it and left that message in itis place.[/quote]

Ah, I probably should have realized that myself :blush:

[hide="number 5"]
I don't know if my solution presentation skills are any good but here goes.

We proceed by induction: In the base case $ n \equal{} 2$, $ \dfrac{16}{3} < \dfrac{24}{4} \equal{} 6$. Assuming

\[ \dfrac{4^n}{n \plus{} 1}<\dfrac{(2n)!}{(n!)^2} \Longleftrightarrow 4^n(n!)^2 < (2n)!(n \plus{} 1)
\qquad \qquad (1)\]

is true for $ n \equal{} k$, we wish to show the statement for $ n \equal{} k \plus{} 1$, that is

\[ \dfrac{4^{n \plus{} 1}}{n \plus{} 2} < \dfrac{(2n \plus{} 2)!}{((n \plus{} 1)!)^2} \Longleftrightarrow 4^{n \plus{} 1}(n \plus{} 1)!^2 < (2n \plus{} 2)!(n \plus{} 2)
\qquad \qquad (2)\]

Dividing $ (2)$ by $ (1)$, we see that it suffices to show that

\[ 4(n \plus{} 1)^2 <\dfrac{ (2n \plus{} 2)(2n \plus{} 1)(n \plus{} 2)}{n \plus{} 1} \equal{} 2(2n \plus{} 1)(n \plus{} 2)
\qquad \qquad (3)\]

However, $ (3)$ is true for $ n > 1$, as can be shown by expanding.
Thus $ (3)$ implies $ (2)$ implies $ (1)$.

Hmm, maybe there is a combinatorial solution.
[/hide]


geometry
calculus
analytic geometry
vector
geometric transformation
reflection
conics

Just branching out from the other topic. What are the subjects that you are passionate about? In those subjects what are the areas you like? Why do you like them?

I also want to feed a general idea. Usually one likes a subject/area one is good at, and since one likes that he becomes better at that. What is your opinion about this?

And please no bashing of subjects. No saying, bio is waste.
[hide]$ We all know that is the universal truth$[/hide]

shreyas-- very nice of u to start this discussion...

me interested particularly in calculus in maths... and also rot mech and optics in physics...

IN chem i don like nving lessons...i like only biomolecules( this might bve a contradiction for some beings).... and alll togetger i like watching the sky and learning abt the stars(like astronomy)... and i am particulary interested in heliocentric orbits around the earth ... and my ambition is to build a specially designed rocket for india so that we like the US can study diff planets...

a question to all of u

HOW DID THE CASSINI MAKE IT TO SATURN?????

THIS WAS A TOP QUESTION...i was asked by a NASA scientist at the VSSC(VIkram sarabhai space centre) when i went there for an interview...

[quote="valeriummaximum"]a question to all of u

HOW DID THE CASSINI MAKE IT TO SATURN?????

THIS WAS A TOP QUESTION...i was asked by a NASA scientist at the VSSC(VIkram sarabhai space centre) when i went there for an interview...[/quote]

[hide="Answer"] Obviously by space :P :D :rotfl:
[/hide]

using N2H4 and liquid hydrogen and oxygen as fuel is this rite!!! :rotfl:  :rotfl:

bio is waste
Yes i agree today--its practical is just a big timewaster
i like Physics:vectors,waves&electricity

my favourite area of interest is chemistry (all) physics except electricity and maths (except COJA)! :D ofcourse PE typewriting and electrical gadgets and electronics and comsci to add!!!! :ninja:

Better rite everything!!!!!!!!! :rotfl:  :D  :blush:

[quote="Albert Einstien"]indirectly i mean that i am a socialist ie i dont differentiate between any subject and always integrate them !!!! i am the bee the wasp!!!!! i hope u understand the pun shreyas!!! :rotfl:[/quote]

[hide="actually"] That was adhu A-M-M-A-N  TRY :P :D
[/hide]

[quote="madness"]My major areas of interest are
Math: Everything except coordinate geometry specifically and geometry in general :D
Phy: Nothing except optics + electricity :D
Chem: Everything except bits of inorganic :)[/quote]
we hate the same subjects da. :D the only thing i like in phy is mechanics. and i totally hate co-geo, and inorganic.
anda as for interest- organic chemistry, specially organic synthesis.

Well, SO many of u here seem to hate coordinate from the bottom of ur hearts.
But i think its pretty lovely, especially if u study it frm loney s book. I think all u guys do it like some statistics or sm s**t like that, so u find it computationally heavy or sth like dat. But if u do things geometrically, its seriously wonderful n luvable. Dats da speciality of all of my coordinate problems i ve posted on da jee forum. :D

[quote="valeriummaximum"]i did not ask that u stupid madness....

the casiini was to got to saturn... but it was launched towards venus????

can u tell me why it was launched towards VENUS ????[/quote]

[hide="Answer"]
By mistake :P . But god knew man's intentions n turned it towards saturn :P :rotfl:
[/hide]

i think maybe it was sani desai for that satellite and thus even when it tried to go to venus it went to saturn!!! :D  :D  :D

a deflection from asteroid belt :D  :rotfl:

you are all use less people...

the answer is the cassini used the venus's gravitational field and used the planet ,Venus and Jupiter as a flyby... to reduce the fuel of course....

that is it gets into the orbit of venus and is finaalyy thrown of along the tangential direction of the orbit...

[quote="valeriummaximum"]you are all [u][b]use less[/b][/u] people...

the answer is the cassini used the venus's gravitational field and used the planet ,Venus and Jupiter as a flyby... to reduce the fuel of course....

that is it gets into the orbit of venus and is [u]finaalyy[/u] thrown of along the tangential direction of the orbit...[/quote]

1st line tragic error da :rotfl: yes we all use ourselves less and dont exert ourselves too much...

and last line, brilliant spelling da :D

Well nice info thanks . Meet and talk with soumya if you want :D

we appreciate your comments :rotfl:

now rohit then albert einstein!!!! :D

On a related note see my signature :D :rotfl:

yeah rohit we are two canonical forms

A RESONANCE HYBRID DA!!!!!

NOW ROHIT THEN ALBERT EINSTEIN!!! :D

Well, I prefer Maths and Physics (and ofc Programming) to Chemistry anyday. Nothing against chemistry but just that I can't really the grasp of higher Organic. ;)

Maths - Combinatorics, Calculus
Physics - Most of it 'cept waves.

And I'm right now doing co-geo. Rags is right. Coordinate Geometry has been stereotyped as a bad villain from time immemorial that people are scared of it without even trying it. The fun in co-geo is to identify the right shortcut to use to solve a sum. Ofcourse proficiency in geometry does help. You can always do co-geo sums the hard way but some tips and tricks always help. :D

Maybe I like it because I haven't come to ellipse/hyperbola yet.  :rotfl:

ya even me ... i am now in parabolas... will complete ellipse by tomorrow..

Yay!!! ive convinced sm1 of my belief... which i generally am bad at. :!:

ok now talking about cassini tangential can u tell me howlong did it took for cassini and how many asteroid deflections did it took ?
and which part of cassini was broken?
ans ..................dudes :D

perhaps 14 deflections

and perhaps rear part was broken

 :rotfl:  :D  there is no harm in guessing:P

i am sorry but all those answers r wrong
deflection :987
asteroid hits :43
part broken :front axial!

that's good but i wanted 2 kno which book u read it in

astronomy encyclopaedia lectures given at regional science city by Prof.jayant.v.narlikar

perhaps i've heard the name of that prof.

correct! i usually read his books on astronomy and astrophysics

he is quite quite famous

perhaps u will read story in eng by him in class 11!! :D

ok but i thought he was a professor in astronomy and astrophysics

most of such personalities r good writers

and the eng story is somewhat related 2 ur area of interest :D

o yes
thats why i wondered i've heard his name

i haven't read his any book on astronomy :D

dude he is very famous author!
i hope u might hav heard Mr. narlikar's name before

i've read about him in some magazine i suppose

though i am not much interested in astronomy,i am interested in readind science related magazine :D

but rituraj only interested in science !
oh sorry no science onl;y bbbbbbbbbbbiiiiiiiiiiiiiiiiiiiiiiioooooooooooooooooooooolllllllllllooooooooooooooooggggggggggggyyyyyyyyyyyyyyy!!!!!!!!!

when did  IIIIII say dat???? :?:

and astronomy is also "science"

hey, science include everything

physics,chem,bio,maths,astronomy,astrology,blah blah blah

and social science as well :rotfl:


Simply AM-GM \[\frac{2a^{4}+b^{4}+c^{}}{4}\geq a^{2}bc\]

PS:and I think Muirhead only works with symetric sums ,so in this case we can't apply it

[quote="Svejk"]Simply AM-GM
\[\frac{2a^{4}+b^{4}+c^{}}{4}\geq a^{2}bc \]
PS:and I think Muirhead only works with symetric sums ,so in this case we can't apply it[/quote]
$f(a,b,c)=f(b,c,a)=f(c,b,a)=f(c,a,b)=f(a,c,b)=f(b,a,c)$
where $f(x)=a^{4}+b^{4}+c^{4}-abc(a+b+c)$
I thiink it's symetric or.. I don't remember  :wink:

[quote="Extremal"][quote="Teixeira"]please, someone could help with this?

prove that a^4+b^4+c^4[u]>[/u]abc(a+b+c)

Thanks for attencion.

Teixeira.[/quote]
By muirhead ... it's simpl
(4,0,0)+(0,4,0)+(0,0,4)>=(2,1,1)+(1,2,1)+(1,1,2)
  :|[/quote]
What [b]Svejk[/b] said about cyclic and symmetric sums is true, but writing $\left(4,0,0\right)\geqslant \left(2,1,1\right)$ is enough, becuase $\left(4,0,0\right)= 2\left(a^{4}+b^{4}+c^{4}\right)$ and $\left(2,1,1\right)= 2abc\left(a+b+c\right)$.
If someone finds the Muirhead solution brutal, the same result follows instantly by the rearrangement inequality.

Yes you're right Extremal.It didn't seem to me like a symetric one ,but from what Rzeszut wrote I noticed that they're equivalent.Thank you for your correction.

well. thanks for help but i need something more basical because it will be presented to high school students.
If someone could help me this way willl be wonderfull.

regards,
Teixeira.

But using AM-GM it's basical.Maybe if I post a more datailed solution you will understand.
\[\frac{2a^{4}+b^{4}+c^{4}}{4}\geq\sqrt[4]{a^{8}b^{4}c^{4}}=a^{2}bc,(1)\]
Analogously we have
\[\frac{a^{4}+2b^{4}+c^{4}}{4}\geq ab^{2}c,(2)\]
and
\[\frac{a^{4}+b^{4}+2c^{4}}{4}\geq abc^{2},(3)\]
And summing up the relations $(1),(2),(3)$ we obtain \[\frac{4(a^{4}+b^{4}+c^{4})}{4}\geq abc(a+b+c)\]
$QED$

[quote="Svejk"]But using AM-GM it's basical.[/quote]
Any inequality which can be done by Muirhead can be done also by $\mathrm{AM-GM}$, because the proof of Muirhead is based on an application of $\mathrm{AM-GM}$.


Just recently got a bank account and going to get a paypal account soon. Just wondering what other students feel about these books. Did it help alot in the AMC etc? Also I live overseas (Australia), so how can I organise the shipping?

Also when setting up paypal account do u have to have a credit card?

I can tell you how paypal works in US and Romania: you have to have a bank account (normal one) - it can also work with a Credit Card. They will deposit a small (random) amount of money 2 times in that account, and then you will have to confirm those amounts to prove the account is actually yours (it takes a couple of minutes if you have an online banking account).

We ship books overseas, but it takes a couple of weeks to get there. Just put in your address (the correct one, international format) and we will ship it.

As for the books, I think the testimonials page (and this site) proves the point :) If you're training for the AMC (or similar competitions) you can also check out our online classes.


calculus integration algebra function domain polynomial factorial

[b]Problem [/b]
Determine whether $ \sqrt {\pi}$ and $ \pi^2$ are algebraic or transcendental over the field $ \cal{Q}(\pi)$. If one is algebraic, find its degree.

[b]Attempt at the Solution[/b]
So, we know that $ \cal{Q}(\pi)$ will just be the field of quotients of the integral domain $ \cal{Q}$[x] evaluated at $ \pi$. I just cannot think of a technique for determining whether or not something is algebraic over that rather large, complicated field!

[quote="darij grinberg"]The technique is called "reading the definition". What does "algebraic" mean?

  darij[/quote]

An element $ \alpha$ of an extension field E of a field F is [b] algebraic over[/b] F if $ f(\alpha)$=0 for some nonzero $ f(x) \in F[x]$.

The degree of the $ \pi^2$ will be 1 for the following reason. f(x)=x^2 is a polynomial in $ \mathbb{Q}[x]$  and $ f(\pi) \equal{} \pi^2$. $ p(x) \equal{} x \minus{} \pi^2$ is a polynomial with coefficients in $ Q(\pi)$ such that $ p(\pi^2) \equal{} 0$.

[quote="FibonacciFan"]The degree of the $ \pi^2$ will be 1 because if f(x)=x^2, then $ f(\pi) \equal{} \pi^2$. That is obvious. What about the other one?[/quote]

The degree of $ \pi^2$ is $ 1$ indeed, but your argument is wrong. You need a polynomial $ f$ with coefficients in $ \mathbb{Q}\left(\pi\right)$ such that $ f\left(\pi\right) \equal{} 0$ (not $ \equal{}\pi^2$).

Figure this out and you will see how to get $ \sqrt {\pi}$ as well.

  darij

[quote="darij grinberg"]
The degree of $ \pi^2$ is $ 1$ indeed, but your argument is wrong. You need a polynomial $ f$ with coefficients in $ \mathbb{Q}\left(\pi\right)$ such that $ f\left(\pi\right) \equal{} 0$ (not $ \equal{} \pi^2$).
  darij[/quote]

See my edited post above. Does that work better?

$ \sqrt {\pi}$ must be algebraic over $ \mathbb{Q}(\pi)$ for the following reason. $ p(x) \equal{} x$ is a polynomial in Q[x] and $ p(\pi) \equal{} \pi$. $ f(x) \equal{} x^2 \minus{} \pi$ is then a polynomial with coefficients in $ \mathbb{Q}(\pi)$ with the property that $ f(\sqrt {\pi}) \equal{} 0$. The degree of $ \pi$ is then less than or equal to 2. How do you know it is not one? Because that would require $ \sqrt {\pi}$ to be an element of $ \mathbb{Q}(\pi)$. Why can't $ \sqrt {\pi}$ be an element of $ \mathbb{Q}(\pi)$? That seems intuitive, but how do you prove it?

Suppose $ \sqrt{\pi}\in\mathbb{Q}(\pi)$. Then what does an arbitrary element look like in there? Suppose $ \sqrt{\pi}$ has this form and then derive a contradiction from facts about $ \mathbb{Q}$.

Note: I didn't try this. This is the usual technique for these types of problems, though.

If ${ \sqrt {\pi}}\in \mathbb{Q}(\pi)$, then ${ \sqrt {\pi}} = \frac {\sum_{i = 0}^{\infty}q_i \pi^i}{\sum_{j = 0}^{\infty}q_j \pi^j}$

Then we can square both sides. Multiply both sides by $ \sum_{j = 0}^{\infty}q_j \pi^j$, square both sides, convolve, subtract the LHS from both sides, and get a polynomial in $ \mathbb{Q}[x]$ that has $ \pi$ as a zero. Thanks. That completes the problem, right?

[quote="FibonacciFan"]See my edited post above. Does that work better?[/quote]

Yes.

[quote="FibonacciFan"]$ f(x) = x^2 - \pi$ is then a polynomial with coefficients in $ \mathbb{Q}(\pi)$ with the property that $ f(\sqrt {\pi}) = 0$. The degree of $ \pi$ is then less than or equal to 2. How do you know it is not one? Because that would require $ \sqrt {\pi}$ to be an element of $ \mathbb{Q}(\pi)$. Why can't $ \sqrt {\pi}$ be an element of $ \mathbb{Q}(\pi)$? That seems intuitive, but how do you prove it?[/quote]

Right up to here. Proving that $ \sqrt {\pi}\not\in\mathbb{Q}\left(\pi\right)$ can be done as you said:

[quote="FibonacciFan"]If ${ \sqrt {\pi}}\in \mathbb{Q}(\pi)$, then ${ \sqrt {\pi}} = \frac {\sum_{i = 0}^{\infty}q_i \pi^i}{\sum_{j = 0}^{\infty}q_j \pi^j}$

Then we can square both sides. Multiply both sides by $ \sum_{j = 0}^{\infty}q_j \pi^j$, square both sides, convolve, subtract the LHS from both sides, and get [b]a polynomial in $ \mathbb{Q}[x]$[/b] that has $ \pi$ as a zero. Thanks. That completes the problem, right?[/quote]

but you need to show that this polynomial is not identically zero. Maybe you left this out because you found it clear (this has a rather simple reason), but you should write it out on your homework sheet.

  darij

[quote="darij grinberg"]
but you need to show that this polynomial is not identically zero. Maybe you left this out because you found it clear (this has a rather simple reason), but you should write it out on your homework sheet.[/quote]

No. The reason I left it out was because I completely forgot about it.

If the this polynomial is identically zero, then the following must be true:

 ${ \pi} \left(\sum_{j = 0}^{\infty}q_j x^j\right)^2 = \left(\sum_{i = 0}^{\infty}q_i x^i\right)^2$

Then all of their coefficients have to be equal but this is impossible because all the coefficients on the LHS have a factor of pi and all the coefficients on the RHS are rational, so in fact none of the coefficients can be equal except the ones that are zero.

Is that good?

Not right.
You get
 $ x \left(\sum_{j \equal{} 0}^{\infty}q_j x^j\right)^2 \equal{} \left(\sum_{i \equal{} 0}^{\infty}q_i x^i\right)^2$,
not
 $ \pi \left(\sum_{j \equal{} 0}^{\infty}q_j x^j\right)^2 \equal{} \left(\sum_{i \equal{} 0}^{\infty}q_i x^i\right)^2$.
Note that the $ q_j$'s on the left hand side and the $ q_i$'s on the right hand side are different - please use better notations next time.

  darij

Sorry about the notation.

Consider the smallest power of x on the LHS and smallest power of x on the RHS. It will be an integral multiple of 2 on RHS but it will be odd on the LHS.

Is that good?

[quote="FibonacciFan"]Consider the smallest power of x on the LHS and smallest power of x on the RHS. It will be an integral multiple of 2 on RHS but it will be odd on the LHS.

Is that good?[/quote]

Yes, this was what I had in mind (though you probably mean "greatest" rather than "smallest").

  darij

[quote="FibonacciFan"]I think smallest or greatest would work.[/quote]

Oh, you are right.

  darij

for a prime $ p$ in a factorial ring $ R$, $ x^2\minus{}p$ is irreducible over $ Q(R)$ by eisenstein. in particular, $ \sqrt{p} \notin Q(R)$. apply this to $ p \equal{} \pi$ and $ R \equal{} \mathbb{Q}[\pi] \cong \mathbb{Q}[x]$, and you get the claim.


floor function
superior algebra
superior algebra unsolved

Prove that an unitary ring that has  a finite and greater than $ 1$ number of zero's divizors(on left or right) is finite.Also, is $ |R|\equal{}n$,prove that $ |U(R)| \leq n\minus{}[\sqrt{n}]$(where $ U(R)$ is the set of invertible elements of $ R$).

[quote="gordon_c"]
Prove that an unitary ring that has  a finite and greater than $ 1$ number of zero's divizors (on left or right) is finite. Also, is $ |R| \equal{} n$. prove that $ |U(R)| \leq n \minus{} [\sqrt {n}]$ (where $ U(R)$ is the set
of invertible elements of $ R$).[/quote]

The first part of your problem is an old result from N.Ganesan. The proof is based on this simple observation that $ R$ must have at least one two-sided zero-divisor (just consider all products

in the form $ uv$, where $ u$ is a right and $ v$ is a left zero-divisor). So let $ a$ be a two sided zero-divisor and consider $ I \equal{} \{r \in R: \ ra \equal{} 0 \}.$ Let $ D$ be the set of all left zero-divisors of $ R$ union

with {0}. Looking at $ R$ as an additive group, define the map $ f: R/I \longrightarrow D$ by $ f(s \plus{} I) \equal{} sa.$ See that $ f$ is a well-defined one-to-one function. So $ I$ and $ R/I$ are both finite and thus $ R$

is finite too. It's clear now that $ n \equal{} |R| \leq |D|^2.$ Thus $ |D| \geq \lfloor \sqrt {n} \rfloor.$ The second part of your problem follows quickly from the fact that $ U(R) \cup D \subseteq R$ and $ U(R) \cap D \equal{} \emptyset.$


calculus
integration
trigonometry
calculus computations

hello, use the formula
$ \cos(t)^4\equal{}\frac{1}{8}(3\plus{}4\cos(2t)\plus{}\cos(4t))$
Sonnhard.

Let $ I_n \equal{} \int_0^{\frac {\pi}{4}} \tan ^ n u\ du\ (n\geq 0)$, we have $ \boxed{\boxed{I_n \equal{} \frac {1}{n \minus{} 1} \minus{} I_{n \minus{} 2}\ (n\geq 2)}}$,

so we get $ I_2 \equal{} \int_0^{\frac {\pi}{4}} \tan ^ 2 u\ du \equal{} \int_0^{\frac {\pi}{4}} \left(\frac {1}{\cos ^ 2 u} \minus{} 1\right)\ du \equal{} \left[\tan u \minus{} u\right]_0^{\frac {\pi}{4}} \equal{} 1 \minus{} \frac {\pi}{4}$.

Set $ \frac {\pi}{2} \minus{} t \equal{} u$, $ 8\int_{\frac {\pi}{4}}^{\frac {\pi}{2}}\cot ^ 4 t\ dt \equal{} 8\int_0^{\frac {\pi}{4}} \tan ^ 4 u\ du \equal{} 8I_4 \equal{} 8\left(\frac 13 \minus{} I_2\right) \equal{} \boxed{2\pi \minus{} \frac {16}{3}}$


Everybody knows the solution of the 2nd order ODE $ aD^2(x)_t\plus{}bD(x)_t\plus{}cx_t\equal{}f$ when $ a,b,c,f\not\equal{}0$

However, what is the solution of the above when $ a,b,f\not\equal{}0$ but $ c\equal{}0$?
i.e. $ aD^2(x)_t\plus{}bD(x)_t\equal{}f$?   :oops:

There's no real difference- it just means that zero is one of the roots of the characteristic polynomial. This leads to constant solutions of the homogeneous equation; the inhomogeneous part can be handled by the method of undetermined coefficients. In this case, the duplication leads to a degree increase in the $ e^{0t}$ part and a linear term in the solutions.


From $(x^2-y^2)(x-y) \geq0$, we have $x^3+y^3 \geq xy(x+y)$.

Applying this and AM-GM, we get

$LHS \leq \frac{1}{a+b}+\cdots +\frac{1}{d+e} \leq \frac{1}{2\sqrt{ab}}+\cdots +\frac{1}{2\sqrt{de}} = \frac{1}{2}$


Prove that in a classroom of n rows and n columns of students, it is impossible to move seats so that
               -everyone has to move to a new seat
               -a student may only move one seat left, right, front, or back
   if n is odd.

[i]BAMO Q#1 part 2[/i]

Color the seats in a checkerboard pattern; everyone moves to a seat of a different color, but if $n$ is odd then there is a different number of white seats than black seats, which establishes a contradiction.

nice use of invariants

was it this year's bamo?

[quote="krustyteklown"]nice use of invariants[/quote]

The problem screams for it.  I am horrible with coloring problems, but I can sometimes recognize when to use the checkerboard :)

yep, that's how I did it too! :)
What was really cool was I just learned about that trick days before taking the BAMO!


real analysis
function
calculus
integration
trigonometry
real analysis unsolved

Let "*" be an operator as described in http://en.wikipedia.org/wiki/Convolution.
Supposed that $ f: R\minus{}>R$ is a Riemann integrable function and $ f*f\equal{}f$, for all $ x \in R$, show that $ f(x)\equal{}0$ for all $ x$ in $ R$.

That creates some interesting difficulties. For instance, what do you mean by Riemann integrable? Any integral over all of $ \mathbb{R}$ is not a Riemann integral, since that integral is only defined over bounded intervals. Are you willing to accept improper integrals, even if they don't converge absolutely?

Let $ f(x) \equal{} \frac {\sin\pi x}{\pi x}$ (with the removable discontinuity removed, so that it's a continuous function at $ x \equal{} 0.$)

$ f*f(x) \equal{} \int_{ \minus{} \infty}^{\infty}\frac {\sin\pi(x \minus{} y)}{\pi(x \minus{} y)}\cdot\frac {\sin \pi y}{\pi y}\,dy$

The integral of $ f$ over the whole line fails to converge absolutely - but it does converges as an improper integral. And then the integral for $ f*f$ even converges absolutely (note that the integrand is $ O(y^{\minus{}2})$ for large $ y$) - and if we accept that, we do get that $ f*f(x) \equal{} f(x).$

Now let me back up to where I got that.

Look at the Fourier transform (in some sense - don't get too picky yet). My favorite convention is $ \widehat{f}(\xi) \equal{} \int_{ \minus{} \infty}^{\infty}f(x)e^{ \minus{} 2\pi i\xi x}\,dx.$

The key identity is then that $ \widehat{(f*g)}(\xi) \equal{} \widehat{f}(\xi)\widehat{g}(\xi).$

So to get an idempotent $ f$ under convolution, we need an idempotent $ \widehat{f}$ under pointwise multiplication. We need $ \widehat{f}(\xi)$ to take on only values of $ 1$ and $ 0$ in some mixture, with no other values.

I got the example I gave of $ f(x)$ (the "sinc" function) by taking the inverse Fourier transform of the "box function": $ \widehat{f}(\xi)$ is $ 1$ on the interval from $ \minus{} \frac12$ to $ \frac12$ and zero otherwise.

What happens if we restrict $ f$ to be absolutely integrable? That is, $ f\in L^1(\mathbb{R})?$ Then $ \widehat{f}$ would be continuous, and a continuous function cannot take on both $ 0$ and $ 1$ as values without taking on any other values. And we can rule out being identically $ 1$ on the whole line by the Riemann-Lebesgue lemma: if $ f\in L^1,$ then $ \widehat{f}(\xi)\to 0$ as $ |\xi|\to\infty.$ That leaves identically zero as the only possibility.

So the correct statement of the theorem is that if $ f$ is [b]Lebesgue[/b] integrable and $ f*f \equal{} f,$ then $ f$ is the zero function. Either that, or interpret the words "Riemann integrable" in such a way as to require absolute convergence.

Followup: Let $ \mathbb{T}$ be the unit circle in the complex plane, considered as a multiplicative group. Or (and this is the same thing), let $ \mathbb{T}\equal{}\mathbb{R}/(2\pi\mathbb{Z})$ as a quotient group, with the appropriate topology. Or just say that a function on $ \mathbb{T}$ is a function on $ \mathbb{R}$ that is periodic of period $ 2\pi.$

Define $ L^1(\mathbb{T})$ by the finiteness of the norm $ \|f\|_1\equal{}\frac1{2\pi}\int_0^{2\pi}|f(x)|\,dx.$

Now define $ f*g(x)\equal{}\frac1{2\pi}\int_0^{2\pi}f(x\minus{}y)g(y)\,dy.$

In that, if an arithmetic operation (such as $ x\minus{}y$) carries you outside of $ [0,2\pi],$ just perform that arithmetic $ \mod{2\pi}.$

(a) Can you find a nontrivial $ f\in L^1(\mathbb{T})$ such that $ f*f\equal{}f\ ?$

(b) Can you find any general characterization of such functions $ f\ ?$


Hey,

In Disney's beauty and the beast, isn't there a huge evil talking piano (or organ or whatever) that tries to kill everyone? Its not on wikipedia but I .. am SO sure of it...

The image is so firm in my mind... The beast or someone eventually killing a giant talking piano with a face like that guy from power rangers except eviler... also possibly made with the aid of computer graphics..

Maybe some Disney spinoff?

Putting the words "serious discussion" into the title does not, in fact, make your post about serious discussion.

ummm i really don't think there is an evil piano/organ in beauty and the beast. just some crazy wolves who try to attack belle and then the beast comes and fights them off. don't ask how i know this :D

hm...nope. i dont remember any evil, talking piano. just some evil, talking trees. oh wait, was that even FROM beauty and the beast? maybe ur piano is from a different movie, too.

Sorry for the confusion. http://www.ilhawaii.net/~beast/pages/bb_xmas.htm

I suppose we can't have a discussion if no one's parents actually paid the money for their kids to watch an evil talking piano...

nope. i agree its so confusing between movies/books/stories, but my sister keeps on watching the movie and reading the book, so i cant exactly say im not sure...

EDIT: guineapiggies u are right, and dreambold, no, i think the trees were from the lion, the witch and the wardrobe, but there are so many stories, maybe something else.

[quote=Nerd_of_the_Ages]Believe it or not, I know the evil talking organ, since I think my sister got the enchanted xmas thing on Netflix or something.[/quote]

I am awesome and please ban me


algebra
polynomial
calculus
derivative
number theory open
number theory

I wanna know where this question is from...
\[\sum_{p=2}^{n}[n^{(n-p)}(p-1)\binom{n+1}{p}]=n^{n}-1\]

Consider the polynomial identity
\[\sum_{p=1}^{n}{n+1\choose p}x^{p-1}= \frac{(1+x)^{n+1}-x^{n+1}-1}{x}.\]
Take derivative of both sides and multiply by $x^{2}$:
\[\sum_{p=2}^{n}{n+1\choose p}(p-1) x^{p}= (xn-1) (1+x)^{n}-n x^{n+1}+1.\]
Now put $x=\frac{1}{n}$ and multiply both sides by $n^{n}$:
\[\sum_{p=2}^{n}{n+1\choose p}(p-1) n^{n-p}= n^{n}-1.\]
QED.


Let $ r,s,t,u$ be roots of the fourth degree polynomial $ x^{4}+Ax^{3}+Bx^{2}+Cx+D=0$. Prove that if $ rs=tu$, then $ A^{2}D=C^{2}$

[hide]
We have
$  \left\{\begin{array}{llll}r+s+t+u=-A\\(r+s)(t+u)+rs+tu=B\\rs(t+u)+tu(r+s)=-C\\rstu=D\end{array}\right.$
From the first relation we have $  rs(r+s+t+u)=-C\Rightarrow rs=tu=\frac{C}{A}$ and replacing in the fourth $  \frac{C^{2}}{A^{2}}=D\Rightarrow A^{2}D=C^{2}$[/hide]


function
inequalities
algebra unsolved
algebra

We have $ \frac1{a^3}\plus{}\frac1{b^3}\ge\frac{a\plus{}b}{a^2b^2}$.
If we denote $ x_i$ as the remaining numbers on board, we see $ \sum\frac1{x_i^3}$ is decreasing with the operation. Then it is trivial to see the final number is no less than $ \frac1{\sqrt[3]{n}}$.

[quote="Inequalities Master"]There are $ n$  $ 1$'s written on a board. At each step, we can select two of the numbers on the board and replace them by $ \sqrt [3]{\frac {a^2b^2}{a \plus{} b}}$. We keep applying this operation until there is only one number left. Prove that this number is greater or equal than $ \frac {1}{\sqrt [3]{n}}$.[/quote]
Let $ X_n$ is the set of numbers we obtain after step $ n$ . Consider the function :
$ f(X_n)\equal{}\sum_{x\in X_n} \frac{1}{x^3}$
It is easy to prove that :
$ \frac{1}{x^3}\plus{}\frac{1}{y^3}\geq \frac{x\plus{}y}{x^2.y^2}$
Thus $ f(X_{n\plus{}1})\leq f(X_n)$ . At the last step we obtain $ f(X_j)\equal{}\frac{1}{c^3}$ . From the inequality we have $ \frac{1}{c^3}\leq f(X_0)\equal{}n$ , so $ c\geq \frac{1}{\sqrt[3]{n}}$.
Problem has been proved .


At a slumber party, five 9-year-old boys tried to get to sleep. The least amount of time it took for any of them to fall asleep was one minute. The mode was four minutes. The range of times was nine minutes. One boy took two minutes more than the median. What was the average (mean) number of minutes it took the boys to fall asleep?

The set has to be $ \{1,4,4,6,10\}$.  Thus, the answer is $ \boxed{5}$.


1. Using digits 0 through 9 only once in each letters, complete the multiplication below.
\begin{align*}
7ab\\
\times 4c\\
\end{align*}
_______
$ defgh$

Do you guys know an easier way than guess and check?
[hide="Answer"]$ 715\times 46$[/hide]

2. Find the following sum:
$ \frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}+\cdots+\frac{1}{18\cdot20}$
[hide="Answer"]$ \frac{9}{40}$ or $ 0.225$[/hide]

3. For what value(s) of $ k$ will the graphs of the equations $ y=x^2+2$ and $ y=kx^2+4$ intersect in two distinct points?

Tip for #2:

Factor out 4 from each fraction.

I think he means 2 since 6 is not a multiple of 4.

[hide="number 3"]

since y is the same you can combine them
$ x^2\plus{}2\equal{}kx^2\plus{}4$
$ \minus{}2\equal{}kx^2\minus{}x^2$
$ \minus{}2\equal{}x^2(k\minus{}1)$
$ \frac{\minus{}2}{k\minus{}1}\equal{}x^2$
So k can be anything that is 0 or less.
Hmmm....that can't be right, did I make a mistake?[/hide]

No, he means 4, because you can extract 2 from each part of the denominator.

Oh, well I meant 2 from both of them, so yeah 4. Do you know where I went wrong, I can't find it?

Well, you subtract them to make a quadratic, and you want that quadratic to have two roots (those roots represent intersection points). Therefore, the discriminant ($ b^2 \minus{} 4ac$) has to be greater than 0.

$ (k \minus{} 1)x^2 \plus{} 2 \equal{} 0$

Therefore, $ b^2 \minus{} 4ac \equal{} 0 \minus{} 4(k \minus{} 1)(2) \equal{} 8 \minus{} 8k$, so we have the inequality
$ 8 \minus{} 8k > 0$
which resolves to
$ k < 1$ which is the requested answer.

[hide="#2"] Factor out the 4 from the denominator to get $ \frac{1}{4}(\frac{1}{1\cdot{2}}\plus{}\frac{1}{2\cdot{3}}\plus{}\cdots\plus{}\frac{1}{9\cdot{10}})$. Now, to evaluate the part in the parentheses, we use telescoping series. Notice how we can write $ \frac{1}{n({n\plus{}1})}$ as $ \frac{1}{n}\minus{}\frac{1}{n\plus{}1}$, so if we write all the terms like this, we just get $ \frac{1}{4}(1\minus{}\frac{1}{10})$=$ \frac{9}{40}$.

Addtionally, if the last term in the parentheses was $ \frac{1}{n({n\plus{}1})}$, the sum in the parentheses is $ \frac{n}{n\plus{}1}$, which can be proved by either telescoping series or induction.[/hide]


calculus
integration
complex numbers
probability and stats

I met a problem during my reading of  a paper about fractional levy motions.
Let
$ S(y) \equal{} \exp\left\{|y|^{\alpha}\int_{\mathbb{R}\times{\mathbb{R}}}[e^{iuf(s)} \minus{}1 \minus{} iuf(s)\mathbf{1}_{|u|\leq1}]|u|^{\minus{}1\minus{}\alpha}duds\right\}$
Then the integral:
$ \int_{\mathbb{R}}(S(y) \minus{} 1)|y|^{\minus{}1\minus{}\beta}dy$
is convergent for $ 0 < \beta < \alpha$ and divergent for $ \alpha < \beta < 2$.

I don't see why, can someone help me, thank you very much!

This question is original from a paper written for about fractional levy motions (see attachment).
On the Page. 370:
\emph{"`We can now prove the convergence of the $ \beta$-variations for $ 0 < \beta < \alpha$. The integral representation ......
 ......
 We now study the $ \beta$-variations for $ \alpha < \beta < 2$. The integral on the right-hand side of (27) is divergent, so of course the dominated convergence theorem can no longer be applied."'}

The question is then:
Why the dominated convergence theorem can be applied for the case $ 0 < \beta < \alpha$? It must be some bound for the integrand there, how i find it? And maybe the same question: why we can't apply the dominated convergence theoem for the case $ \alpha < \beta < 2$?

Thank you very much!!!

In fact, you have
\[ S(y) \equal{} \exp\{ \minus{} c|y|^\alpha\},\]
where $ c$ is some constant (perhaps, positive).

There are no problem at infinity, only at zero. But at zero we have $ S(y) \minus{} 1\sim c|y|^\alpha$, so this integral is convergent iff the integral $ \int_{0 \plus{} } |y|^{\alpha \minus{} 1 \minus{} \beta}$ is convergent. Is it clear now?
--------------
About the dominated convergence, I cannot see it easily for $ \beta < \alpha$.
--------------
There are no problems with bounding the integrand at infinity (by some constant times $ |y|^{ \minus{} 1 \minus{} \beta}$, but I don't see how their reasoning (sum tending to zero) can lead to bounds at zero. It's tempting to use (5), but we get $ |y|^2$ there, which means that if it works, then it would work in the case $ \alpha < \beta$ as well, but this is nonsense, because it cannot work there (the limit is a divergent integral).
--------------
There is also a similar (but much more noticeable) flaw at the last paragraph of the article (changing $ \mathbb E S_n(y)$ to $ S_n(y)$). It is ridiculous that the authors write tons of calculations but leave some horrible mess like this. And it is also strange, because all three authors are very good mathematicians.

Thanks a lot for your help!

The convergence at zero is now clear.
But i am still not so clear with the convergence at infinity, you said: actually
$ S(y) \equal{} \exp\{\minus{}c|y|^{\alpha}\}$,
you really mean there is a minus before $ c$?

About the dominated convergence also the same question, you said: There are no problems with bounding the integrand at infinity (by some constant times $ |y|^{\minus{}1\minus{}\beta}$),
but how? i.e. by which constant?
Sorry for my  dullness  :)

I did not read the paper you've posted so I can tell you anything about DCT and so on but returning to your question about the problem at infinity: zhoraster of course meant that this constant $ c$ is positive. To observe this one just need to notice that the $ c \equal{} \minus{} (\int \int [cos(uf(s)) \minus{} 1]|u|^{ \minus{} 1 \minus{} \alpha}duds) > 0$ (integral from imaginary part equal to $ 0$ because it is odd function).
P.S I've just understood that I don't need to read the paper(you formulated the question about DCT in your post, sorry :) )

$ |S_n(y)|\le 1$, because it is average of some numbers with absolute $ 1$, so at infinity you can bound the integrand by $ 2|y|^{\minus{}1\minus{}\beta}$.

Now i am totally clear with it!
Thanks again for your friendly help!  :lol:

Thank you as well for posting the paper. Despite the flaws, it's interesting.


If n is a natural number prove that:
$ (1 \plus{} \cfrac{1}{n})^n < (1 \plus{} \cfrac{1}{n \plus{} 1})^{n \plus{} 1}$

For a possible proof - cf. attached file    :arrow:


vector
linear algebra
matrix
algorithm
analytic geometry
geometry
algebra

Hi ;
I have m vectors V1,V2...,Vm  in    Rn space .

I also know that they are linearly dependent . So I have :
A1V1+A2V2+....AmVm=0

The question is :
How can I find those scalars A1,A2...Am ?

regards;
akbar khalifehloo

Create a matrix $M$ whose columns are $V_{1},V_{2},\dots,V_{m}.$ You're now trying to find the nontrivial solutions of the homogeneous equation $Mx=0.$ The entries of the column vector $x$ are the coefficients you're looking for.

You're going to solve that matrix equation by Gaussian elimination: do row operations until you've found a matrix in echelon form that is row-equivalent to $M.$ The set $\{V_{1},\dots,V_{m}\}$ is independent iff the echelon form has a pivot in every column; otherwise it is dependent.

Every elementary linear algebra or matrix theory textbook that I know goes over this, usually in a very early chapter.

Dear Kent ;
 I already know that the vectors are linearly dependent . And I am searching for the values of Ai .
How can I find them ? I have m vectors in Rn space.

regards;

Remember that you're solving a linear system. If $B$ is the row-reduced matrix derived from $A$, $Ax=0$ and $Bx=0$ have the same solutions. The difference is that the latter equation can be solved by choosing some entries of $x$ arbitrarily and then back-substituting.

Dear Jmerry ;
Can I program the method you stated ?
Please tell me more .

Do you have a textbook? This is in there.

When A is reduced to B , some rows would be zeros . That is because the rows of A are not independent .
example :
v1(1,1)    v2(1,2)   v3(2,3)
Easy hand calculations show that A1=1 A2=1 A3=-1 . or v1+v2-v3=0
But in case of larger sets , hand calculations would not work .
I have studied plenty of textbooks , but none of them present an applicable general algorithm to my problem . Would you introduce me one , please ?
regards

[quote="khalif01"]I have studied plenty of textbooks , but none of them present an applicable general algorithm to my problem .[/quote]
I'm wondering what books you have read. For example, in http://www.math.hmc.edu/calculus/tutorials/linearsystems/ there is a general algorithm.

hi

I sure know about solving linear system of equations . The url you introduced took nxn systems , while mine are mxn .
Besides , how can I cope with the parametric approach of that url in programming ?
I have already found an algorithm for the case where m<n . But it would't work for m>n .

thanks

[quote="khalif01"]The url you introduced took nxn systems , while mine are mxn . [/quote]
Only the examples are of the form $n\times n$. The same algorithm works in the case $m\times n$ too.

[quote="khalif01"]But the method of the url would not work for the case m>n ![/quote]
Why not? :maybe:

There are three things that can happen with any system of linear equations: there can be a unique solution; there can be no solution (inconsistent system); or and there can be an infinite number of solutions. When we're trying to solve $n$ equations for $n$ unknowns and $m>n,$ (the coefficient matrix is wider than it is tall), then we cannot have a unique solution. A homogeneous system always has the trivial solution, so it is never inconsistent. Thus, an $n\times m$ homogeneous system with $m>n$ must have infinitely many solutions. In particular, these solutions form a nontrivial vector subspace. A good textbook treatment of Gaussian elimination will deal with the concept of "free variables" and show how to represent the resulting solution in parametric form, or (in what is really the same thing) find a basis for that solution space.

hi ;
To be clear , take this example :
v1(1,2,3)
v2(2,4,6)
running guass elimination shows that v1 and v2 are dependent . But , how can we automatically know that 2v1-v2=0 ?
(i.e. the linear coefficients of a1v1+a2v2+0 are  a1=2 and a2=-1 .
How can guassian elimination show us that a1=2 abd a2=-1  ?

the solution is straight forward for the case we have more vectors that the vector space .
here we have 2 vectors in R3 , 2<3 and I am looking for the solution for a1,a2,...  for this case .

I hope the problem is clear now .
thanks

Hi ;
It is amzing ! No comments yet !
Maybe the problem is too hard or much more than "too easy" !
regards;

In his last post, Kent Merryfield said all there was to say: to find the coefficients in every case is the same as to solve a linear system where the number of unknowns is the number of vectors and the number of equations is the number of elements in each vector (the equations themselves just state that every coordinate of the linear combination is $0$). To find the coefficients, just solve the system by Gaussian elimination. Any decent textbook in computational linear algebra would explain the procedure in detail. I'm personally too lazy to rewrite a few pages from a textbook here and I suspect that everybody else has the same attitude :P. As to your particular example, you get the system $x+2y=0$ (first coordinate), $2x+4y=0$ (second coordinate) and $3x+6y=0$ (third coordinate). After doing the first step in Gaussian elimination, you get $x+2y=0$, $0x+0y=0$ and $0x+0y=0$. Now, you see that $y$ can be taken arbitrary and $x=-2y$. Your choice $x=2$, $y=-1$ is included in the answer, but it is by no means unique: $x=-2468$, $y=1234$, say, is neither worse, nor better. I suspect that what you meant by "Gaussian elimination" was the procedure in which you modified the rows of $2\times 3$ matrix (with rows being the original vectors), while for finding the coefficients, you should rather work with the $3\times 2$ matrix where your vectors are columns, but it is only my guess.

Hi ;
You did great ! Also there is no need to say anything about the guassian elimination .
But how can the procedure you showed be used for programming general cases involving m vectors in n space ?
for example for v1(1,2,3,4) v2(2,4,6,8) and v3(3,6,9,12) we reach x+2y+3z=0 .
then by taking x=y=1 we get z=-1 which satisfies v1+v2-v3=0 and the prblem is solved .
I think there would be some programming difficulties .
any comment . please ?
regards;

1. Implement a matrix class
2. Implement the elementary row operations.
3. Implement the Gaussian algorithm by choosing the right operations in right order.
4. Now pick the vectors from the reduced matrix to find the coefficient.

You can implement exactly the sama algorithm as you have been taught here. Gaussian algorithm is a very straightforward to implement and you can find open source codes in Google.

Hi;
Everything goes very well in hand calculations . But in developing the general algorithm , there seem to be some exceptions which need further refinement .
I believe there must exist a general and much easier method for determination of linear dependent coefficients . I also believe that the gussian elimination  would be the main axis of solution.
I will do my best to work it out !
regards ;

Hi;
I also want to add something else .
The determination of A's in A1V1+A2V2+...+AmVm=0 are not dealt with in a straight forward manner in books , while much simpler subjects are greatly covered by them .
Does this note imply that there is a very limited application for the subject above ? I , as a structural engineer , have a deep need towards determination of those A's . And when I entered the literature , the more I searched , the less I found !

Does anybody know about other applications ?

regards

[quote="khalif01"]
The determination of A's in A1V1+A2V2+...+AmVm=0 are not dealt with in a straight forward manner in books , while much simpler subjects are greatly covered by them .
[/quote]
Let me repeat one more time: [color=red]determination of such $A$'s is [b]exactly[/b] the same as solving the corresponding system of linear equations, period[/color]. I don't know of a subject in elementary linear algebra that is paid more attention to and covered in more details in textbooks than the problem of solving a system of linear equations. Just understand, please, that the phrase in red is [b]all[/b] one needs to say about your problem in any book. :roll:

You are REPEATING my words , but ANGRILY and I don't know why ??!!
I SIMPLY wanted to know it there were other applications for those A's  !
There is no obligation for you and anybody else to answer any question !
I expected this forum a scientific area . But some people here TEACH others and make them UNDERSTAND ! GREAT !
To add the final thing , I developed an absolute algorithm for my problem , which needs ANY of those DETAILED coverage and attensions of the books and suggestions made by some people . THAT IS AS EASY AS 2+2=4 .

Well, there was no anger just desperation in my post :). Perhaps, you [b]meant[/b] the same, but what you [b]wrote[/b] clearly reads as a direct opposite to what I said. Anyway, I apologize for the tone but the point remains: there is no need to specially write about the question you posed in any book, the best way to find a solution to your problem is to look at some textbook in computational linear algebra and to implement one of the algorithms for solving linear systems. As to most of those algorithms being as easy as 2+2=4, I'm inclined to agree with you here :P.

This is quite the fruitful thread.  :rotfl:

You can treat your vectors either as rows or columns in a matrix.