[quote="Valentin Vornicu"][quote="KBabe"]I'm 11, about to be 12 in two days.

and i have a question, are homeschooled ppl usually smarter than public school ppl?[/quote]That mean you're gonna receive two big presents in the same day, one for your birthday and another one from Santa :)

Homeschooled vs School should be discussed (if it hasn't already been) in another topic ... so go ahead and make one.  Sorry for being off-topic  :blush:[/quote]

Isn't Christmas in 4, not 2 days?

[quote="h_s_potter2002"]Isn't Christmas in 4, not 2 days?[/quote]Since it's 22nd (well at least in Romania :D) and usually Christmas presents are offered on the late 24th night (again at least in Romania :) ) my post was (Romanian-ly speaking) correct :)

[quote="KBabe"]it is fun to get double presents. what's your favorite holiday? (anyone)

anyone want to talk about..maybe their fantasies, or what they wish their life was like?[/quote]

my birthday is 11 days after Christmas, so it's like this one big present fest and then you have to wait another whole year. i would prefer to have my b-day and Christmas spread out, but i usually get the day off cuz of Christmas break, but not this year!  :(  lol. my favorite holidy is, in fact, Christmas because it is a very important holiday in regards to my faith, and the presents are simply an added bonus.  :lol:

What I really want for christmas is a new calculator but thats not happening since I got a new one last year for christmas that a TI 83+.  I'd love anything thats horse related and I know that I am getting my show clothes for Class A horse shows since I had to come with my trainer and my mother to try them on and pick them out.  I would say I wanted a horse like I always have but I got my horse a weeks before my b-day so I suppose I should be content lol, but with all the things horses need I'll alwaysbe getting supplies.


integration
calculus
analytic geometry
function
inequalities
linear algebra
matrix

Calculate :
  $ V_n(R) \equal{} \int\int\int_{x_1^2 \plus{} x_2^2 \plus{} ... \plus{} x_n^2 \leq R^2}...\int{dx_1dx_2...dx_n} \equal{} ?$ Here ($ R > 0$)
[hide="Answer"]
  $ V_{n}(R) \equal{} \frac {2(2\pi)^{\frac {n \minus{} 1}{2}}}{1.3.5....n}R^n$ If $ n$ is odd
   $ V_n{(R)} \equal{} \frac {\pi^{\frac{n}{2}}}{(\frac {n}{2})!}R^n$ If $ n$ is even
(True ? or false ?)[/hide]

Key words: n-volume of n-ball.

Here are my two favorite methods. One thing to note: by similarity and scaling, $ V_n(R)\equal{}V_n(1)R^n.$ (I've probably posted both of these before on this site but I'm too lazy to look.)

1. Assume $ n\ge 2$ and slice off two dimensions:

$ V_n(R)\equal{}\iint_{x_1^2\plus{}x_2^2\le R^2}\left(\underset{x_3^2\plus{}\cdots\plus{}x_n^2\le R^2\minus{}x_1^2\minus{}x_2^2}{\int\cdots\int}dx_3\dots dx_n\right)dx_1\,dx_2$

$ \equal{}\iint_{x_1^2\plus{}x_2^2\le R^2}V_{n\minus{}2}\left(\sqrt{R^2\minus{}x_1^2\minus{}x_2^2}\right)dx_1\,dx_2$

Now do that integral in polar coordinates:

$ \equal{}\int_0^R\int_0^{2\pi}V_{n\minus{}2}\left(\sqrt{R^2\minus{}r^2}\right)r\,d\theta\,dr$

$ \equal{}2\pi V_{n\minus{}2}(1)\int_0^R\left(R^2\minus{}r^2\right)^{\frac n2\minus{}1}r\,dr$

You can do that integral by the standard first-year calculus substitution:

$ V_n(R)\equal{}2\pi V_{n\minus{}2}(1)\left.\left(\minus{}\frac12\cdot\frac2n\left(R^2\minus{}r^2\right)^{\frac n2}\right)\right|_{r\equal{}0}^R$

$ \equal{}\frac{2\pi}{n}V_{n\minus{}2}(1)R^n$

In particular, $ V_n(1)\equal{}\frac{2\pi}{n}V_{n\minus{}2}(1).$

From $ V_1(R)\equal{}2R,$ we get $ V_3(R)\equal{} \frac{4\pi R^3}3,\ V_5(R)\equal{}\frac{8\pi^2R^5}{3\cdot 5},$ and so on.

Assuming $ V_0(R)\equal{}1$ is consistent with the correct $ V_2(R)\equal{}\pi R^2.$ Continuing, $ V_4(R)\equal{}\frac{\pi^2R^4}{2}$ and in general, $ V_{2k}(R)\equal{}\frac{\pi^kR^{2k}}{k!}.$

2. Use polar coordinates (the $ n$-dimensional analogue of spherical coordinate) to derive that $ V_n(R)\equal{}\frac1nA_nR^2,$ where $ A_n$ is the $ (n\minus{}1)$-dimensional measure of the unit sphere. Integrate a Gaussian in both rectangular and polar coordinates. Find a gamma function in the resulting radial integral. (Supply the details yourself.) The result should be that $ V_n(R)\equal{}\frac{\pi^{n/2}R^n}{\Gamma\left(\frac n2\plus{}1\right)}.$

I learned method 2 a long time ago (call it "folklore"). I don't think I ever saw method 1 anywhere - and I like it because it's more elementary. Yes, the answers in the form QuyBac wrote them above are correct.

[quote="Kent Merryfield"]Key words: n-volume of n-ball.

Here are my two favorite methods. One thing to note: by similarity and scaling, $ V_n(R) \equal{} V_n(1)R^n.$ (I've probably posted both of these before on this site but I'm too lazy to look.)

1. Assume $ n\ge 2$ and slice off two dimensions:

$ V_n(R) \equal{} \iint_{x_1^2 \plus{} x_2^2\le R^2}\left(\underset{x_3^2 \plus{} \cdots \plus{} x_n^2\le R^2 \minus{} x_1^2 \minus{} x_2^2}{\int\cdots\int}dx_3\dots dx_n\right)dx_1\,dx_2$

$ \equal{} \iint_{x_1^2 \plus{} x_2^2\le R^2}V_{n \minus{} 2}\left(\sqrt {R^2 \minus{} x_1^2 \minus{} x_2^2}\right)dx_1\,dx_2$

Now do that integral in polar coordinates:

$ \equal{} \int_0^R\int_0^{2\pi}V_{n \minus{} 2}\left(\sqrt {R^2 \minus{} r^2}\right)r\,d\theta\,dr$

$ \equal{} 2\pi V_{n \minus{} 2}(1)\int_0^R\left(R^2 \minus{} r^2\right)^{\frac n2 \minus{} 1}r\,dr$

You can do that integral by the standard first-year calculus substitution:

$ V_n(R) \equal{} 2\pi V_{n \minus{} 2}(1)\left.\left( \minus{} \frac12\cdot\frac2n\left(R^2 \minus{} r^2\right)^{\frac n2}\right)\right|_{r \equal{} 0}^R$

$ \equal{} \frac {2\pi}{n}V_{n \minus{} 2}(1)R^n$

In particular, $ V_n(1) \equal{} \frac {2\pi}{n}V_{n \minus{} 2}(1).$

From $ V_1(R) \equal{} 2R,$ we get $ V_3(R) \equal{} \frac {4\pi R^3}3,\ V_5(R) \equal{} \frac {8\pi^2R^5}{3\cdot 5},$ and so on.

Assuming $ V_0(R) \equal{} 1$ is consistent with the correct $ V_2(R) \equal{} \pi R^2.$ Continuing, $ V_4(R) \equal{} \frac {\pi^2R^4}{2}$ and in general, $ V_{2k}(R) \equal{} \frac {\pi^kR^{2k}}{k!}.$

2. Use polar coordinates (the $ n$-dimensional analogue of spherical coordinate) to derive that $ V_n(R) \equal{} \frac1nA_nR^2,$ where $ A_n$ is the $ (n \minus{} 1)$-dimensional measure of the unit sphere. Integrate a Gaussian in both rectangular and polar coordinates. Find a gamma function in the resulting radial integral. (Supply the details yourself.) The result should be that $ V_n(R) \equal{} \frac {\pi^{n/2}R^n}{\Gamma\left(\frac n2 \plus{} 1\right)}.$

I learned method 2 a long time ago (call it "folklore"). I don't think I ever saw method 1 anywhere - and I like it because it's more elementary. Yes, the answers in the form QuyBac wrote them above are correct.[/quote]
 Thanks you very much [b]Pro. Kent Merryfield[/b] !
 Now If can please you for a the [b]answer [/b]
 Let $ A \in M_n{\mathbb{(R)}}$ , $ A \equal{} A^T$ and $ 0 < x^TAx \leq 1,\forall x \in M_{n1}{\mathbb{(R)}} \minus{} {O_{n1}}$ (or $ 0 < A \leq I_n$)
 Here $ x \equal{} \begin{bmatrix}x_1 \\
x_2 \\
.. \\
x_n\end{bmatrix},x_k \in \mathbb{R},k \equal{} 1,2,..,n$
Calculate : $ I \equal{} \int\int_{0 < x^TAx \leq 1}...\int{e^{x^TAx}dx_1dx_2..dx_n}$
(thanks ! I hope can learned method 1 of Pro to have answer in $ S_n^{\plus{}\plus{}}(\mathbb{R})$  :)   )

Hint: find positive definite $ S$ such that $ S^2\equal{}A.$ Make the substitution $ u\equal{}Sx,$ also written as $ x\equal{}S^{\minus{}1}u.$ What happens if you try that?

[quote="Kent Merryfield"] What happens if you try that?[/quote]
 Oke I want a result for calculate ?
 $ A \in S_n^{ \plus{} \plus{} }{\mathbb{(R)}}$ the exist $ \omega \in O_n(\mathbb{R})$ such that : $ A \equal{} \omega^Tdiag(\lambda_1,\lambda_2,..,\lambda_n)\omega$ , $ \lambda_1,\lambda_2,..,\lambda_n$ are positive eigenvalues of $ A$
Let $ y \equal{} \omega x$ then $ Jacobi (J)$ : $ |J| \equal{} |\det(\omega)| \equal{} 1$ and :
 $ x^TAx \equal{} y^Tdiag(\lambda_1,...,\lambda_n)y \equal{} \lambda_1y_1^2 \plus{} ... \plus{} \lambda_ny_n^2$ and :
 $ I \equal{} \int\int_{0 < \lambda_1y_1^2 \plus{} ... \plus{} \lambda_ny_n^2 \leq 1}...\int{e^{\lambda_1y_1^2 \plus{} ... \plus{} \lambda_ny_n^2}dy_1dy_2...dy_n}$
 Put $ z_k \equal{} \sqrt {\lambda_k}y_k,1 \leq k \leq n$ ...... then  use method 1 of Pro to calculate .....
 Are you Agree ? :)

1)Edit : $ |J| \equal{} |\det(\omega)| \equal{} 1$ are $ |J| \equal{} |\det(\omega^T)| \equal{} 1$
2) $ A\equal{}S^2$ are clear but If $ A \geq O$
3) A  nice result  for ($ \mathbb{R}^n$ , $ n \geq 2$) eliptic : $ I \equal{}\int\int_{0 < x^{T}Ax\leq 1}...\int{dx_{1}dx_{2}..dx_{n}}$
4) Thanks you very much  If you are  for a  result  calculate in $ \mathbb{R}^n$,  :roll:

[quote="1234567a"]May be a harder question :

On $ \forall x\in\mathbb{R}^{n}$ define $ \parallel{}x\parallel{}_{p} = (|x_{1}|^{p} + \ldots + |x_{n}|^{p})^{\frac {1}{p}}$
then we know minkowski inequality : $ \parallel{}x + y\parallel{}_{p}\leq \parallel{}x\parallel{}_{p} + \parallel{}y\parallel{}_{p}$ then we have $ (\mathbb{R}^{n},\parallel{}.\parallel{}_{p})$ is a normed space. Find the Volume of the "unit ball" $ B_{n} = \{x : \parallel{}x\parallel{}_{p} < 1\}$?[/quote]

We use the following formula: If the function f is homogeneous of degree k, i.e, $f(tx)=t^kf(x)$ with $x\in \R^n$ and $t>0$ then
\[\int_{R^n}f(x)e^{-||x||_p^p}dx=C_{n,p}\int_{B_p}f(x)dx\]
where $B_p$ is the unit ball of norm $||.||_p$, and
\[C_{n,p}=\Gamma(1+\frac{n+k}{p})\]
Proof of this formula: We have
\[e^{-||x||_p^p}=\int_{||x||_p^p}^{\infty}e^{-t}dt\]
apply the Fubini theorem, we get
\[\int_{R^n}f(x)e^{-||x||_p^p}dx=\int_{R^n}f(x)\int_{||x||_p^p}^{\infty}e^{-t}dt=\int_{R^n}\int_0^{\infty}f(x)e^{-t}1_{||x||_p^p\leq t}dtdx.\]
Change of variable $x=t^{1/p}y$, because of the homogeneous of f we have
\[\int_{R^n}f(x)e^{-||x||_p^p}dx=\int_{B_p}f(y)\int_{0}^{\infty}t^{\frac{n+k}{p}}e^{-t}dtdy=\Gamma(1+\frac{n+k}{p})\int_{B_p}f(y)dy\]
Apply this result for the function $f=1$ with the note that
\[\int_Re^{-|t|^p}dt=2\int_0^{\infty}e^{-t^p}dt=\frac{2}{p}\int_0^{\infty}e^{-u}u^{1/p-1}du=\frac{2}{p}\Gamma(\frac{1}{p})=2\Gamma(1+\frac{1}{p})\]
Hence
\[V(B_p)=2^n\frac{(\Gamma(1+\frac{1}{p}))^n}{\Gamma(1+\frac{n}{p})}\]


Τα προβλήματα είναι:
      1)Σε ένα ορθογώνι τρίγωνο ΑΒΓ με μήκη κάθετων πλευρών α=2 και β=1 φέρουμε την εωτερική και εξωτερική διχοτόμο της γωνίας Α. Οι διχοτόμοι τέμνουν την ευθεία γραμμή ΒΓ εσωτερικά στο σημείο Π και εσωτερκά στο σημείο Σ. Ποια είναι τα μήκη των ΓΠ και ΓΣ?
      2)Σε ένα τετράγωνο με πλευρά μήκουσ 2 εγγράφουμε ένα ισοσκελές τρίγωνο. Ποιο είναι το μήκος της ακτίνας του εγγεγραμμένου στο τρίγωνο κύκλου? (Μπούμε αν θέλουμε να χρησιμοποιήσουμε το αποτέλεσμα του προηγούμενου προβλήματος).
-Για την επίλυση των προβλημάτων να μη χρησιμοποιηθούν τριγωνομετρικές συναρτήσεις
Όποιος δύναται να βοηθήσει ας μου στείλει.
Ευχαριστώ.

στο 1ο προβλημα, οταν λες φερουμε τη διχοτομο της γωνιας Α, ενοεις τη διχοτομο της 90αρας γωνιας;

1)εστω εσωτερικο το P,εξωτερικο το S.θα εφαρμοσουμε θεωρηματα διχοτομων.
Πιο συγκεκριμενα,
$ \frac{BP}{PC}\equal{}\frac{AB}{b} < \equal{} > \frac{a}{PC}\equal{}\frac{AB \plus{} b}{b}$
$ < \equal{} >$
$ \frac{2}{PC}\equal{}\sqrt{5} \plus{}1< \equal{} >PC\equal{}\frac{2}{\sqrt{5} \plus{}1}$,αφου
$ AB^2\equal{}1 \plus{} 4\equal{}5$
αλλα και
$ \frac{SC}{SB}\equal{}\frac{b}{AB} <\equal{} >\frac{SC}{a}\equal{}\frac{1}{\sqrt{5}\minus{}1}<\equal{}>SC\equal{}\frac{2}{\sqrt{5}\minus{}1}$
ελπιζω να μην εκανα καποιο λαθος στις πραξεις. :)

[quote="theo_1"]Όχι αν δεις οι κά8ετες πλευρές ενια οι α και β οπότε καταλαβαίνουμε ότι ειναι η Γ ορθή γωνία[/quote]

α οκ, οπως και να εχει παντος το 1ο ειναι απλη εφαρμογη του Θεωρηματος διχοτομων, το 2ο ειναι λιγο ζορικο, θα το προσπαθησω αυριο γιατι τωρα πρεπει να διαβασω ΑΟΔΕ για της Πανελληνιες, εχω 20 μερες να μαθω απ εξω 27 σελιδες...

Λοιπον, τελικα βρηκα λυση για το 2ο προβλημα η οποια στηριζεται στη λυση του πρωτου (δηλαδη εγαρμογη Θεωρηματος Διχοτομων), ομως το αποτελεσμα της λυσης εξαρταται απο τα στοιχεια του τριγωνου που εγγραφουμε, και επηδη υπαρχουν παρα πολλα διαφορετικα μεταξυ τους ισοσκελη τριγωνα εγραψιμα σε ενα συγκεκριμενο τετραγωνο, δεν μπορω να δωσω μια συγκεκριμενη λυση στο προβλημα αν δεν δωθει πρωτα καποια πληροφορια για το εγγεγραμενο τριγωνο. Ειναι γραμμενη σε κανενα βιβλιο αυτη η ασκηση;

[quote="theo_1"]Όχι αυτή η άσκηση είναι μια εργασία η οποία δεν είναι από το βιβλίο.
Η εκφώνηση δεν λέει τίποτα περισσότερο και τιποτα λιγότερο από αυτό που ήδη σας είπα.[/quote]

Τοτε μαλον υπαρχει ασαφεια στη διατυπωση, διοτι υπαρχουν αμετριτα διαφορετικα εγραψιμα τριγωνα στο συγκεκριμενο τετραγωνο με διαφορετικα υψη, βασεις και πλευρες, και δεν νομιζω ολα να εχουν σταθερη ακτινα εγγεργαμενου κυκλου

[quote="NickNafplio"]....
Τότε μάλλον υπάρχει ασάφεια στη διατύπωση, διότι υπάρχουν αμέτρητα διαφορετικά εγγράψιμα τρίγωνα στο συγκεκριμένο τετράγωνο με διαφορετικά ύψη, βάσεις και πλευρές, και δεν νομίζω όλα να έχουν σταθερή ακτίνα εγγεγραμμένου κύκλου[/quote]
Θα μπορούσε ίσως το ζητούμενο να είναι η μέγιστη τιμή της ακτίνας του εγγεγραμμένου κύκλου ενός μεταβαλλόμενου ισοσκελούς τριγώνου,  με την κορυφή του κάποιο σταθερό σημείο στην περίμετρο του δοσμένου τετραγώνου.

$ \bullet$ Έστω $ ABCD$ το δοσμένο τετράγωνο και $ \bigtriangleup KLM,$ ένα ισοσκελές τρίγωνο με $ KL \equal{} KM$ και την κορυφή του $ K,$ τυχόν σημείο επί της πλευράς $ AB.$

Με δεδομένο ότι $ a \equal{} 2,$ όπου $ a$ είναι η πλευρά του $ ABCD,$ υπάρχει κάποια μέγιστη τιμή για την ακτίνα $ r$ του εγγεγραμμένου κύκλου στο $ \bigtriangleup KLM,$ με αφετηρία την τιμή $ r \equal{} 0$ $ ($ όταν  $ K\equiv C\equiv L$ $ ),$ που αφορά στο συγκεκριμένο σημείο $ K.$

Μπορεί τελικά το πρόβλημα να μας ζητάει να προσδιορίσουμε κάποιο σημείο $ K$ στην περίμετρο του δοσμένου τετραγώνου, ή αν υπάρχουν περισσότερα του ενός να βρούμε ποια είναι, για τα οποία προκύπτει το μέγιστο δυνατό εύρος τιμών, μιας και κάθε φορά έχουμε αφετηρία την τιμή $ r \equal{} 0.$

Δεν ξέρω άλλο ανάλογο πρόβλημα και προφανώς δεν έχω κάποια λύση υπόψη μου. Προσπαθώ μόνο να καταλάβω το ζητούμενο.

Κώστας Βήττας.


function
calculus
integration
limit
real analysis
Support
calculus computations

Please help!!! Thanks.

$f: R\rightarrow$ $R$ be a continuous function and suppose that
$\int_{-\infty}^{\infty}$ $|f|$ exists.
Show that $\lim_{h\to0} \int_{-\infty}^{\infty}$ $|f(x+h)-f(x)|dx= 0$

$f$ is Lebesgue integrable on $\mathbb{R}$. Now use the density of the continuous functions with compact support in $L_{1}(\mathbb{R})$ and the fact that a fonction continuous on a compact set is in particular uniformly continuous.

Could anyone explain in a simple way or others since the "Lebesgue integrable" by "julien_santini" is sort of dehind of my knowledge? Sorry!!

Thank.

Since $\lim_{K\to\infty}\int_{-K}^K|f(x)|\,dx=\int_{-\infty}^{\infty}|f(x)|\,dx,$

given $\epsilon>0$ we can find $K$ such that $\int_{|x|>K}|f(x)|\,dx<\frac{\epsilon}4.$

Create a function $g$ as follows:

For $-K\le x\le K,\,g(x)=f(x).$

On the intervals $[-K-1,-K]$ and $[K,K+1],$ define $g(x)=f(x)(K+1-|x|).$ Then let $g(x)=0$ for $|x|<K+1.$

$g$ is continuous and compactly supported. $f(x)=g(x)+h(x),$ where

$\int_{\mathbb{R}}|h(x)|\,dx\le\int_{|x|>K}|f(x)|\,dx<\frac{\epsilon}4.$

$\int_{\mathbb{R}}|f(x+h)-f(x)|\,dx\le\int_{\mathbb{R}}|g(x+h)-g(x)|\,dx +\int_{\mathbb{R}}|h(x+h)-h(x)|\,dx$

$\int_{\mathbb{R}}|h(x+h)-h(x)|\,dx\le2\int_{\mathbb{R}}|h(x)|\,dx<\frac{\epsilon}2.$

We have that $g$ is uniformly continuous. Find $\delta<1$ such that if $|h|<\delta$ then $|g(x+h)-g(x)|<\frac{\epsilon}{4(K+1)}$

Then $\int_{\mathbb{R}}|g(x+h)-g(x)|\,dx=\int_{-K-1}^{K+1}|f(x+h)-f(x)|\,dx$

$<\int_{-K-1}^{K+1}\frac{\epsilon}{4(K+1)}\,dx=\frac{\epsilon}2.$

So, for $|h|<\delta,\,\int_{\mathbb{R}}|f(x+h)-f(x)|\,dx<\epsilon.$

This is actually the same proof that julien_santini proposed, with two modifications:

1. I don't mention Lebesgue integration because I don't have to. This should give Disco Very fewer reasons to panic.

2. julien_santini used the abstract assertion of the density of continuous compactly supported functions in $L^1.$ I replaced that by explicitly constructing the approximating function.

A more elementary(?) proof, with lower standpoint. We don't need "compact support" here. Recall the following statement, which is a standard exercise of Riemann integral:
Lemma. $f$ is integrable on $[a-d,b+d] (d>0)$. Show that $\int_{h\to 0} \int_a^b |f(x+h) - f(x)| dx = 0$.

Proof.
[hide]
We are going to prove that, for $\epsilon > 0$, there exists $\delta>0$ such that $\int_a^b |f(x+h) - f(x)| dx < \epsilon$ for all $0 < h < \delta$. In order to use the condition of integrability, we construct a partition of $[a,b]$ by lettting $x_i = a + i\frac{b-a}{n}$.  Thus
$\int_a^b |f(x+h) - f(x)| dx = \sum \int_{x_i}^{x_{i+1}} |f(x+h)-f(x)|dx$
Let $0 < h <\frac{b-a}{n}$. Suppose $x$ is in the $i$-th interval of the partition, then $|f(x+h)-f(x)|$ is in $i$-th or $i+1$-th interval. If $x+h$ is in the $i$-th one, then
$|f(x+h)-f(x)| < \omega_i$
If it is in the $i+1$-th one,
$|f(x+h)-f(x)| < |f(x+h)-f(x_i)| + |f(x_i)-f(x)| < \omega_{i+1} + \omega_i$
So we always have
$|f(x+h)-f(x)| < |f(x+h)-f(x_i)| + |f(x_i)-f(x)| < \omega_{i+1} + \omega_i$

Since $f$ is integrable on $[a-d,b+d]$, it is bounded, suppose $|f|\leq M$. Then we take $\omega_{n+1}=2M$.
Hence
$\sum \int_{x_i}^{x_{i+1}} |f(x+h)-f(x)|dx < (\omega_0 + 2(\omega_1 + \cdots + \omega_n)\frac{b-a}{n} + \omega_{n+1}\frac{b-a}{n}$

$f$ is integrable, we can choose $n$ large enough such that
$\sum \omega_i \frac{b-a}{n} < \epsilon / 4$
and
$2M\frac{b-a}{n} < \epsilon / 2$

Then
$\int_a^b |f(x+h) - f(x)| dx = \sum \int_{x_i}^{x_{i+1}} |f(x+h)-f(x)|dx < \epsilon/2 + \epsilon/2 < \epsilon$.
[/hide]

Now we can apply the lemma to $[-K-1,K+1]$ in the Kent Merryfield's proof.

Thank you all that halps me to understand alot.


Ring Theory
superior algebra
superior algebra unsolved

show that if R is a PID and S is a multiplication subset of R, then the localization S^(-1)R is also a PID.

I am thinking of defining some Dedekind-Hasse norm. But I can only define it as non-integer norm. Is it not allowed????

Perhaps try to look at lifting of ideals directly through the localization? I don't have a lot of experience with localization but that just seems like a logical place to start.


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Thank you.

all of u befor posting please go thru these rules by Rushil...specifically that u shud show ur work :)

I've began to post problems not as homework problems, but as problems I've created and wanted to share. What should I do?

[quote]Recreational and 'fun' problems should also be posted here.[/quote]

Okay. Thank you. But what if they are high-level and original?

[quote="Rushil"]For more advanced theoretical, Calculus-based as well as Olympiad problems, kindly use the [url=http://www.artofproblemsolving.com/Forum/index.php?f=332]Advanced Physics Forum[/url]. [/quote]

Did you read the first post?

It seems as if there are contradicting directions. The part you just quoted indicates that those problems should go in the Advanced forum. However, your other quote suggests that they should be posted in this forum. Therefore, I feel that there is a contradiction.

As far as I can tell, you haven't posted anything that uses calculus, has advanced theoretical material, or is Olympiad level.

So far, you've been posting them in the right place.  I see no contradiction.

High-level ("Olympiad", or "advanced theoretical") problems should be posted in the Advanced Physics forum.

[quote="Rushil"]Kindly note the following rules for posting in this forum.

1) This Forum is for [color=red]High School (non-AP) , Non- Calculus problems and simple College level[/color] discussion. Homework problems at College level can be discussed here. For more advanced theoretical, Calculus-based as well as Olympiad problems, kindly use the [url=http://www.artofproblemsolving.com/Forum/index.php?f=332]Advanced Physics Forum[/url].

2) Problems and exercises of simple nature ( 'cookbook' problems) mainly  requiring application of formulae and [color=orange]homework problems [/color] can be posted here. Recreational and 'fun' problems should also be posted here. [b]However note that no problem shall be fully solved for you and you should try to explain what you didnt understand or were unable to do while solving the problem.[/b]

3)[color=darkblue] Clarification of doubts [/color]on specific High school or Freshman College level  Physics Theory are welcome here. You can also seek help on course material. Advanced theoretical discussions should be done at the Advanced Physics Forum.

For more information, kindly contact the Moderator. Suggestions are welcome.

Thank you.[/quote]
By non-AP, that means no discussion of the AP practice problems or real exam questions, right?

Given that the person you quoted has not touched AoPS for almost 2.5 years, I don't think you'll be getting a response from him.

Just do as the rule says.  Any AP stuff goes in Advanced Physics.


geometry
analytic geometry
geometry proposed

Let $ ABC$ be an isosceles triangle with $ AB \equal{} AC$, and let $ D$ be the midpoint of $ BC$. Let $ E$ be the foot of the perpendicular from $ D$ to $ AB$ and $ F$ the midpoint of $ DE$. Prove that $ AF$ is perpendicular to $ CE$.

I tried coordinate bashing this and it got really ugly. Hints are appreciated.

[hide]Let M be the midpoint of EB. As $ MF \parallel{} BC \Rightarrow MF\perp AD$. So F is orthocentre in $ \Delta DMA$ (as $ DE\perp AM$). Then $ AF \perp DM$. But $ DM\parallel{}EC \Rightarrow AF\perp EC$.[/hide]

This used to be my favorite geometry problem, I have seen the solution delta posted and I think it is the best. Here is another solution:
Let $ G$ be the midpoint of $ DC$. Note that $ \triangle AED \sim \triangle ADC$, and thus $ \triangle AEF \sim \triangle ADG$. Thus, since $ \angle DAG\equal{}\angle EAF$, $ \angle FAG\equal{}\angle DAE$. Also, we have $ \frac {AG}{AF} \equal{} \frac {AD}{AE}$ from the similarity, so $ \triangle AFG \sim \triangle AED$ since they have a same angle and the corresponding sides are in the same ratio. Thus, $ \angle AFG \equal{} 90$, but $ FG \parallel EC$, so $ AF \perp CE$.

For a generalisation you may see [url]https://artofproblemsolving.com/community/q1h202893p1119923[/url] , post #6.

Best regards,
sunken rock


calculus integration geometry function limit algebra domain

\[ \int_{\minus{}\frac{\pi}{2}}^{\frac{\pi}{2}} csc(x) dx\]

Does it converge?  I fell it is symmetry. it should be zero , if integral is the area of the function including.

thanks

Ah. Every time I teach improper integrals in Calculus II, someone in the class wants to argue this point with me.

With every improper integral, I ask the question, "for how many different reasons is this improper?" Or perhaps the question should be asked, at how many different places is this improper. The integral you give is improper for two reasons: the right hand side of $ 0$ and the left hand side of zero.

If an integral is improper for more than one reason, we [i]must[/i] cut it into pieces, where each piece is only improper for one reason. Then, the only way we can accept the integral as convergent is if each piece converges independently and separately.

In the particular case, $ \int_{\minus{}\frac{\pi}2}^0\csc(x)\,dx\equal{}\minus{}\infty$ and $ \int_0^{\frac{\pi}2}\csc(x)\,dx\equal{}\infty.$

Since they both diverge, we must say that this improper integral diverges. We don't get to combine them. You use the (incomplete) phrase, "the integral is the area of the function including ..." Well, in this case, that notion of area is undefined.

But you want to combine them. Students always do that - that's why it's an argument.

It is true that $ \lim_{a\to 0^\plus{}}\left(\int_{\minus{}\frac{\pi}2}^{\minus{}a}\csc(x)\,dx\plus{}\int_a^{\frac{\pi}2}\csc(x)\,dx\right)\equal{}0.$

There's your symmetry. But what I just did isn't the improper integral. It's something called a "principal value". Any attempt to balance the bad parts of an integral improper for two reasons by making the domain converge "at the same rate" is called a principal value. And to show how dependent this is on the balance, let me give you some other possibilities:

$ \lim_{a\to 0^\plus{}}\left(\int_{\minus{}\frac{\pi}2}^{\minus{}a}\csc(x)\,dx\plus{}\int_{a^2}^{\frac{\pi}2}\csc(x)\,dx\right)\equal{}\plus{}\infty.$

$ \lim_{a\to 0^\plus{}}\left(\int_{\minus{}\frac{\pi}2}^{\minus{}a^2}\csc(x)\,dx\plus{}\int_a^{\frac{\pi}2}\csc(x)\,dx\right)\equal{}\minus{}\infty.$

$ \lim_{a\to 0^\plus{}}\left(\int_{\minus{}\frac{\pi}2}^{\minus{}a}\csc(x)\,dx\plus{}\int_{2a}^{\frac{\pi}2}\csc(x)\,dx\right)\equal{}\minus{}\ln 2.$

And so on.

Let me ask you this: do you think that $ \int_{\minus{}\infty}^{\infty}x^3\,dx$ ought to be zero?

On the other hand, $ \int_{\minus{}\frac{\pi}2}^{\frac{\pi}2}\sqrt[3]{\csc x}\,dx$ is a convergent integral, and does equal zero.

[quote="Kent Merryfield"]Ah. Every time I teach improper integrals in Calculus II, someone in the class wants to argue this point with me.

With every improper integral, I ask the question, "for how many different reasons is this improper?" Or perhaps the question should be asked, at how many different places is this improper. The integral you give is improper for two reasons: the right hand side of $ 0$ and the left hand side of zero.

If an integral is improper for more than one reason, we [i]must[/i] cut it into pieces, where each piece is only improper for one reason. Then, the only way we can accept the integral as convergent is if each piece converges independently and separately.

In the particular case, $ \int_{ \minus{} \frac {\pi}2}^0\csc(x)\,dx \equal{} \minus{} \infty$ and $ \int_0^{\frac {\pi}2}\csc(x)\,dx \equal{} \infty.$

Since they both diverge, we must say that this improper integral diverges. We don't get to combine them. You use the (incomplete) phrase, "the integral is the area of the function including ..." Well, in this case, that notion of area is undefined.

But you want to combine them. Students always do that - that's why it's an argument.

It is true that $ \lim_{a\to 0^ \plus{} }\left(\int_{ \minus{} \frac {\pi}2}^{ \minus{} a}\csc(x)\,dx \plus{} \int_a^{\frac {\pi}2}\csc(x)\,dx\right) \equal{} 0.$

There's your symmetry. But what I just did isn't the improper integral. It's something called a "principal value". Any attempt to balance the bad parts of an integral improper for two reasons by making the domain converge "at the same rate" is called a principal value. And to show how dependent this is on the balance, let me give you some other possibilities:

$ \lim_{a\to 0^ \plus{} }\left(\int_{ \minus{} \frac {\pi}2}^{ \minus{} a}\csc(x)\,dx \plus{} \int_{a^2}^{\frac {\pi}2}\csc(x)\,dx\right) \equal{} \plus{} \infty.$

$ \lim_{a\to 0^ \plus{} }\left(\int_{ \minus{} \frac {\pi}2}^{ \minus{} a^2}\csc(x)\,dx \plus{} \int_a^{\frac {\pi}2}\csc(x)\,dx\right) \equal{} \minus{} \infty.$

$ \lim_{a\to 0^ \plus{} }\left(\int_{ \minus{} \frac {\pi}2}^{ \minus{} a}\csc(x)\,dx \plus{} \int_{2a}^{\frac {\pi}2}\csc(x)\,dx\right) \equal{} \minus{} \ln 2.$

And so on.

Let me ask you this: do you think that $ \int_{ \minus{} \infty}^{\infty}x^3\,dx$ ought to be zero?

On the other hand, $ \int_{ \minus{} \frac {\pi}2}^{\frac {\pi}2}\sqrt [3]{\csc x}\,dx$ is a convergent integral, and does equal zero.[/quote]

Thanks, it is very detailed.


Voi calcula doar $J=\int x\sqrt{x^{2}+4}dx$
$J=\int x\frac{x^{2}+4}{\sqrt{x^{2}+4}}dx$
$J=\int\frac{x^{3}}{\sqrt{x^{2}+4}}dx+4\int\frac{x}{x^{2}+4}dx$
$J=\int x^{2}(\sqrt{x^{2}+4})'dx+4\sqrt{x^{2}+4}$
$J=x^{2}\sqrt{x^{2}+4}+4\sqrt{x^{2}+4}-2\int x\sqrt{x^{2}+4}dx$
$3J=x^{2}\sqrt{x^{2}+4}+4\sqrt{x^{2}+4}$
De aici concluzia rezulta rapid. Pt. varianta ta, cu $x^{3}$, faci analog, si vei ajunge la J


calculus
derivative
calculus computations

use implicit differentiation to find an equation of the tangent line

(x^2 + y^2)^3=8(r^2)(y^2) at (1,-1)


Prove that if x,y are real numbers therefore:

x^4 + y^4 + (x^2 +1)(y^2 +1) >= (y+1)*x^3 + (x+1)*y^3 + x +y

 :ninja:

first let's prove a lemma:

[b]lemma:for real numbers $ x,y$ we have:

$ (|y|\plus{}1)|x|^3\plus{}(|x|\plus{}1)|y|^3\plus{}|x|\plus{}|y|\geq (y\plus{}1)x^3\plus{}(x\plus{}1)y^3\plus{}x\plus{}y$[/b]

[hide="proof of the lemma"]
if $ x\equal{}0$ or $ y\equal{}0$ then it's easy to see,now we have two particular cases:

$ \boxed{\textrm{Case 1}: x<0\textrm{ and }y>0}$

in this case we have $ |x|\equal{}\minus{}x,|y|\equal{}y$ so our lemma becomes:

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

$ \iff 0\geq 2(y\plus{}1)x^3\plus{}2xy^3\plus{}2x$

which is true because $ x<0,y>0$

--------------------

$ \boxed{\textrm{Case 2}: x<0\textrm{ and }y<0}$

in this case we have $ |x|\equal{}\minus{}x,|y|\equal{}\minus{}y$ and our lemma is equivalent to:

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

$ \iff 0\geq 2x^3\plus{}2y^3\plus{}2x\plus{}2y$

which is true because $ x<0,y<0$
----------------------------------------------------------[/hide]

now according to this lemma its sufficinet to show that:

$ x^4\plus{}y^4\plus{}(x^2\plus{}1)(y^2\plus{}1)\geq(|y|\plus{}1)|x|^3\plus{}(|x|\plus{}1)|y|^3\plus{}|x|\plus{}|y|$

now we know that $ x^4\equal{}(\minus{}x)^4,y^4\equal{}(\minus{}y)^4,x^2\equal{}(\minus{}x)^2,y^2\equal{}(\minus{}y)^2$ so its sufficient to prove this problem for $ x>0,y>0$...

now by AM-GM we have:

$ x^4\plus{}x^2y^2\geq 2x^3y$

$ y^4\plus{}x^2y^2\geq 2y^3x$

$ x^4\plus{}x^2\geq 2x^3$

$ y^4\plus{}y^2\geq 2y^3$

$ x^2\plus{}1\geq 2x$

$ y^2\plus{}1\geq 2y$

now sum up all these inequalities and devide bothe sides by $ 2$ to derive at your problem...

or you can use the fact that:

$ x^4 \plus{} y^4 \plus{} (x^2 \plus{} 1)(y^2 \plus{} 1)\geq (y \plus{} 1)x^3 \plus{} (x \plus{} 1)y^3 \plus{} x \plus{} y$

$ \iff (x^2 \minus{} x)^2 \plus{} (y^2 \minus{} y)^2 \plus{} (x^2 \minus{} xy)^2 \plus{} (y^2 \minus{} xy)^2 \plus{} (x^2 \minus{} 1)^2 \plus{} (y^2 \minus{} 1)^2 \geq 0$

which is obviously true...


trigonometry
number theory proposed
number theory

Prove that the equation $4x^3-3x+1=2y^2$ has at least $31$ solutions in positive integers $x$ and $y$ with $x\leq 2005$.

suppose that $x = 2k+1$. ($k \leq 1002$)

$(x+1)(2x-1)^2 = 2y^2 \Longrightarrow (k+1)(4k+1)^2 = y^2$.

so, if $(k+1)$ is a square, our equation has a solution. now, we easily see that we
have 31 perfect squares $\leq 1002$

This was given at an Australian training in 2004.  As well as the straightforward solution posted by e.lopes, there's a weird one which uses the fact that substituting $x=\cos \alpha$ and $y=\cos\beta$ into the equation yields $\cos 3\alpha=\cos 2\beta$.  I've just tried to reconstruct it but I can't remember exactly how it goes.

I hope this is the solution Agrippina had in mind.

Notice that $\cos (ix)=\frac{e^x+e^{-x}}{2}$.

Thus from the previous post, it is sufficient to prove that there are at least $31$ values of $x$ such that $\frac{e^{2x}+e^{-2x}}{2}\le 2005$ and $b=\frac{e^{3x}+e^{-3x}}{2}$ and $a=\frac{e^{2x}+e^{-2x}}{2}$ are both integers.
Let $c=\frac{e^x+e^{-x}}{2}\in \mathbb Z$.  Thus, $a=2c^2-1\in \mathbb Z$ and $b=4c^3-3c\in \mathbb Z$.
Notice that $1\le c\le 31 \le \sqrt{1003}$.  Thus, there are at least $31$ such values of $x$, as desired.

oh yes!the key step is to notice $LHS=(x+1)(2x-1)^2$.
besides,what's 'Donova Olympiad'?


Let $ a,b,c,d > 0$ satisfy $ abc \plus{} bcd \plus{} cda \plus{} dab \equal{} a \plus{} b \plus{} c \plus{} d$. Prove that :
$ (\sqrt {a^2 \plus{} 1} \plus{} \sqrt {b^2\plus{} 1})^2 \plus{} (\sqrt {c^2 \plus{} 1} \plus{} \sqrt {d^2 \plus{} 1})^2\leq (a \plus{} b \plus{} c \plus{} d)^2$

[quote="conan_naruto236"]Let $ a,b,c,d > 0$ satisfy $ abc \plus{} bcd \plus{} cda \plus{} dab \equal{} a \plus{} b \plus{} c \plus{} d$. Prove that :
$ (\sqrt {a^2 \plus{} 1} \plus{} \sqrt {b^2 \plus{} 1})^2 \plus{} (\sqrt {c^2 \plus{} 1} \plus{} \sqrt {d^2 \plus{} 1})^2\leq (a \plus{} b \plus{} c \plus{} d)^2$[/quote]
Nobody... :(


[quote="Svejk"]hi. can you help me ? solve this equation:

 (3+sqrt(3))^x+(3-sqrt(3))^x=(3*sqrt(2)+2*sqrt(3))^x+(3*sqrt(2)-2*sqrt(3))^x[/quote]
I think the only solution is x=0:
let $a=3+\sqrt{3},b=3\sqrt{2}+2\sqrt{3}$ it give:
$3-\sqrt{3}=\frac{6}{a},3\sqrt{2}-2\sqrt{3}=\frac{6}{b}$
therefore:
$a^{x}+\frac{6^{x}}{a^{x}}=b^{x}+\frac{6^{x}}{b^{x}}\Rightarrow (a^{x}-b^{x})(a^{x}b^{x}-6^{x})=0$but ab not equal 6 and a not equal b therefore x=0

Thanks  :laugh:

PS:Could you tell me how did you write the symbols ?


induction
number theory unsolved
number theory

a real number call interesting if it can be written to $ \sqrt{n}$ $ \plus{}$ $ \sqrt{n\plus{}1}$ $ (n \in N)$.prove that if X be an interesting number, then $ X^r$ is an interesting number for all natural r.

[quote="behdad.math.math"]a real number call interesting if it can be written to $ \sqrt {n}$ $ \plus{}$ $ \sqrt {n \plus{} 1}$ $ (n \in N)$.prove that if X be an interesting number, then $ X^r$ is an interesting number for all natural r.[/quote]

If $ x\equal{}\sqrt n\plus{}\sqrt{n\plus{}1}$, then $ n\equal{}\left(\frac{x^2\minus{}1}{2x}\right)^2$. So the question is to show that $ M\equal{}\left(\frac{x^{2p}\minus{}1}{2x^p}\right)^2$ is an integer too.

$ M\equal{}\left(\frac{x^{2p}\minus{}1}{2x^p}\right)^2$ $ \equal{}\left(\frac 12(x^p\minus{}\frac 1{x^p})\right)^2$ $ \equal{}\left(\frac 12((\sqrt{n\plus{}1}\plus{}\sqrt n)^p\minus{}(\sqrt{n\plus{}1}\minus{}\sqrt n)^p)\right)^2$ and so :

$ M\equal{}\left(\sum_{\text{odd } k\in[0,p]}{p \choose k} (\sqrt{n\plus{}1})^{p\minus{}k}(\sqrt n)^k\right)^2$

The quantity inside the parenthesis is either $ A\sqrt n$ if $ p$ is odd, either $ A\sqrt n\sqrt{n\plus{}1}$ if $ p$ even, for some integer $ A$.

And so $ M$ is either $ A^2n$, either $ A^2n(n\plus{}1)$ and so is an integer.

Q.E.D.

Thanks dear Pco.
Is it possible to solve it using induction,too?

if $ A \equal{} \sqrt {n \plus{} 1} \plus{} \sqrt {n}$ and $ B \equal{} \sqrt {n \plus{} 1} \minus{} \sqrt {n}$

and if a sequence $ a_r$ is given by $ 4a_r \plus{} 2 \equal{} A^{2r} \plus{} B^{2r}$

use the fact that $ AB \equal{} 1$ so $ A^{2r} B^{2r} \equal{} 1$

so we can solve to get $ A^r \equal{} \sqrt {a_r \plus{} 1} \plus{} \sqrt {a_r}$

[ $ X \equal{} A^{2r}$ and $ Y \equal{} B^{2r}$ then $ X \plus{} Y \equal{} 4a_r \plus{} 2$ and $ XY \equal{} 1$
then  solve to get $ X \equal{} 2a_r \plus{} 1 \plus{} \sqrt { (2a_r \plus{} 1)^2 \minus{} 1}$ and  $ Y \equal{} 2a_r \plus{} 1 \minus{} \sqrt { (2a_r \plus{} 1)^2 \minus{} 1}$
so $ \sqrt {X} \equal{} \sqrt {a_r \plus{} 1} \plus{} \sqrt {a_r}$ ]

also $ A^2$  and $ B^2$ are roots of $ x^2 \minus{} 2(2n \plus{} 1)x \plus{} 1 \equal{} 0$ [ because AB = 1 and $ A^2 \plus{} B^2 \equal{} 2(2n \plus{} 1)$]

so a_r satisfies a recursion relation which we can check by induction
$ 4a_r \plus{} 2 \minus{} 2(2n\plus{}1)(4a_{r\minus{}1} \plus{} 2) \plus{} (4a_{r\minus{}2} \plus{} 2) \equal{} 0$
which simplifies to $ a_r \equal{} 2(2n \plus{} 1)a_{r \minus{} 1} \minus{} a_{r \minus{} 2} \plus{} 2n$ with $ a_0 \equal{} 0$ and $ a_1 \equal{} n$
so a_r is always an integer


[color=green]
prove that

if a,b,c are positive reals with sum 1

((a^3)/(a^2+b^2)) + ((b^3)/(b^2+c^2)) + ((c^3)/(c^2+a^2))

is >= 1/2

[/color]

[hide="hint"]substitute a with 1-b-c[/hide]

[hide="(Ugly) solution"]
Let us write it in a more promising manner,

$\sum_{cyc} \frac{2a^3}{a^2 + b^2} \ge \sum_{cyc} a = \sum_{cyc} \frac{a^3 + ab^2}{a^2 + b^2}$
$\sum_{cyc} a \frac{a^2 - b^2}{a^2 + b^2} \ge 0$
$\sum_{cyc} a \frac{ ( \frac{a}{b} )^2 - 1}{ ( \frac{a}{b} )^2 + 1} \ge 0$

Jensen's on $f(x) = \frac{x^2 - 1}{x^2 + 1}$ yields

$\sum_{cyc} a \frac{ ( \frac{a}{b} )^2 - 1}{ ( \frac{a}{b} )^2 + 1} \ge \frac{ u^2 - 1}{u^2 + 1}$

Where

$u = \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a}$

By Rearrangement,

$u \ge a + b + c = 1$,

Completing the proof.  QED.
[/hide]

Nice Solution:

     Write   a^3/(a^2 + b^2)  = a - (a.b^2)/(a^2 + b^2) and similar expressions for the other two terms on the left side .

The inequality to be proved now takes the form

a+b+c - [ (a.b^2)/(a^2+b^2) + (b.c^2)/(b^2 +c^2) + (c.a^2)/(c^2 +a^2) ] >= 1/2, or equivalently

[ (a.b^2)/(a^2 +b^2)  +  (b.c^2)/(b^2 +c^2)  + (c.a^2)/(c^2 +a^2) ] <= 1/2 which is true  because a^2 +b^2 >=2ab,..... ...and a+b+c =1

Sorry this isn't in Latex.This was not typed on my PC.

[quote="Centy"]

If $a,b,c \in \mathbb{R^+}$ such that $a + b + c = 1$.

$\frac{a^3}{a^2+b^2} + \frac{b^3}{b^2+c^2} + \frac{c^3}{c^2+a^2} \geq 1/2$.[/quote]

Use:
$\frac{a^3}{a^2+b^2} \geq a-\frac{b}{2}$ (It is equivalent to $\frac12 b (a-b)^2 \geq 0$
Sum up all term and use $a+b+c=1$


What exactly are they and where can they applied in mathematics?

Thanks in advance

[url]http://en.wikipedia.org/wiki/Logistic_function[/url]

However, I'm too lazy and intimidated to actually read the article and summarize it, so someone else would have to do that.

A logistic function models situations where a quantity (y) grows (or decreases) at a rate jointly proportional to its size (y), and proportional to the difference of its size from some fixed value (L-y).

In other words, the rate of change of y equals $ k \cdot y(L\minus{}y)$ for some constant of proportionality k.

For example, some simple diseases grow logistically in closed populations. They grow proportional to the number of people who have the disease because the number of carriers (y) directly affects the rate at which it spreads. But, if there are only L people in the population, then it also spreads proportional to (L-y), the number of people in the population who are not yet infected.

Some other examples of things that grow logistically are species in closed ecosystems (L is called the carrying capacity of the ecosystem), and rumors in closed populations.

A neat way to test logistic growth is to take a room full of people and give everyone in the room a card. Put red X's on a few of the people's cards and have them randomly mingle around, talking to different people for 2 minutes. Every time someone with a red X talks to someone without a red X, they put a red X on the other person's card. If the person they talk to already has a red X, they make no change. Every 6 minutes or so have everyone close their eyes and take a poll of how many people have red X's. The number of people with red X's as a function of time will be roughly logistic.

Another such example is Newton's rate of cooling, where the rate that an object cools is proportional to the difference in its temperature and the temperature of its surroudings.
That is
$ \dfrac{dT}{dt} \equal{} k(T_0 \minus{} T)$    where $ T_0$ is the temp of the room

That is not logistic, because the rate at which the temperature changes is only proportional to the difference between the temperature and the surrounding atmosphere - not jointly proportional as is required of a logistic function.

Temperature cooling/heating is a stricly exponential (decay) function with respect to the ambient temperature.


This is going to be a very vague question because I do not know many specifics.

Where do I learn programming (CS I think)? I need to learn it for free; preferably online. Furthermore, I don't know where to ask this.

Um... any answers will be appreciated.

This might better belong in the computers forum.

When you say CS, I think you mean CSS. A good site for learning HTML, CSS, etc. is http://www.w3schools.com . You can also get various books from the library. The SAMS Teach Yourself books are pretty good, as are the "for the absolute beginner".

[quote="isabella2296"]When you say CS, I think you mean CSS.[quote]

Probably Computer Science, not Cascading Style Sheets.

[quote="isabella2296"]The SAMS Teach Yourself books are pretty good, as are the "for the absolute beginner".[/quote][/quote][/quote]

I highly doubt it. Anyway, you haven't really given us any information to work on. What purpose do you expect your programming to serve (i.e. what do you expect to get out of it)?

Well, I don't know much about it except that you can use it to solve math problems. Basically, I would prefer my entire focus on learning programming to be related to math because I already much about computers and I don't to spend more time learning about them.

You can use almost any computer language to perform mathematical calculations. A general answer to a general question.  :lol:
Then there is the question of efficiency, and you run into easy and inefficient languages like basic compared to C and C++ which might be a bit harder to learn (but much more efficient and robust).
(by efficiency, I mean how they make use of computer resources).

It also depends on what level of mathematics you will be computing (ie, how complex it is). The more complex it is, the better it will be for you to use a more efficient language.
I suggest you search online for computer languages such as c, c++, c#, java, basic, etc. Then you weigh them yourself wrt what you want to accomplish, the amount of time you have, the complexity of the language (the learning curve), etc.
On a side note,You could also use math cad to perform you're mathematical calculations. But it is geared more towards engineering, which is largely the reason why I mentioned it.  :D
Hope that answers atleast part of your question.  :)

[quote="Differ"]I suggest you learn MATLAB. You can pick up a free copy by winning a local competition.

[quote="Inspired By Nature"]Then there is the question of efficiency, and you run into easy and inefficient languages like basic compared to C and C++ which might be a bit harder to learn (but much more efficient and robust).
(by efficiency, I mean how they make use of computer resources).[/quote]

I would disregard this as the advice of someone who has not touched either BASIC, C, or C++.[/quote]

I agree, matlab is indeed an excellent suggestion. It completely slipped my mind.  :)

As for the efficiency statement, I stand by it. Higher level programming languages such as BASIC are generally less efficient and consume more resources (more mem consumption, slower exec speed, etc) than low level programming languages such as C and C++.
As for my level of competence in programming, I do believe I have 'touched' all three of the aforementioned languages amongst others.

[quote="Differ"][quote="pytheagle"]So... since I need something free, will C++ be useful?[/quote]

If you can't win MATLAB, I would recommend Java or C#.[/quote]

I'm getting tired of Differing from you Differ, But I believe C and C++ have a much wider base (resources, industry use, support, IDE's, etc) than Java or C#. Interestingly enough, they are all quite similar.
Also, with all due respect, 'winning MATLAB' is not an immediate or relavent solution to the issue provided by pytheagle.

[b]@pytheagle[/b]: Yes, there are many different sources you can get a IDE (integrated development environment) from. C++ is an excellent language to start of with.
IDE's are basically a complete package (all-in-one) including editor (for writing code), compiler (converts your code to something the computer can understand), and debugger (for catching your errors).
My current personal favourite is codeblocks,
http://www.codeblocks.org/

Bloodshed is a close second,
http://www.bloodshed.net/download.html
But I do believe they do not update it anymore.

There is also Visual C++ express,
http://www.microsoft.com/Express/VC/

Also, if you just want to view code of different languages, an excellent little program is notepad++,
http://notepad-plus.sourceforge.net/uk/site.htm
(pick the binary files if you are interested)

Note, that the initial learning curve may be quite steep so be patient. There are many tutorials online to help you start off. Here's an excellent site to start you off,
http://www.cplusplus.com/doc/tutorial/
Best of luck.  :)


[quote="mihail911"]Show that given any 17 integers, it is possible to choose 5 whose sum is divisible by 5.[/quote]

If there are $ 5$ of the given integers congruent $ \pmod{5}$ then choose the five integers and we are done.

If not, then by pigeonholeprinciple we can find an integer $ k \in \{0,1,2,3,4 \}$ such that there are $ 4$ of the given integers congruent to $ k \pmod{5}$. And we can definitely find another two integers which are congruent to $ k\minus{}1$ and $ k\plus{}1$ respectively (because $ \frac{17\minus{}4}{4}>3$). Now choose $ 3$ integers $ \equiv k \pmod{5}$ and the two integers $ \equiv k\minus{}1,k\plus{}1 \pmod{5}$ and we are done.


Can anyone plz verfity if these two lines have a common spot (or whatever you call it in english) in (1+(3/16);(31/32))

l: x=5+4y ^ m: 8y=(1/2)x-4

Do you mean that they intersect there?

If so, I believe that the solution is $(4,-\displaystyle\frac{1}{4})$.  If we substitute the value for x into the second equation, we can simplify to find $y=-\displaystyle\frac{1}{4}$.  Then this value gives x=4 in the first equation.

When I solve the equation, I get the same thing as E^(pi*i)=-1 got. DId you do an incorrect operation?

Hum. Could you (just to make stupid thing on a wednesday) show me how you would find x if you didn't know y, plz...

Multiply the first equation by 2, move the 8y to the left side and the 2x to the right side. Now does it look obvious?

Sorry i've been away for some time. Sometimes i just have other things to attend to (a lot of work lately moved my math study to a 2. priority).

Could you show it using latex, I just don't get it.


geometry
circumcircle
incenter
geometry proposed

Given are acute triangle $ABC$ with circumcenter $O,BO\cap CA={B'},CO\ca AB={C'}$ prove that $ABC$ is isoceles equivalent $BB'=CC'$  :)

Gemath, verify the case $A=90^{\circ}$ and $B\ne C\ !$

ok Virgil It needs an acute triangle! I have fixed it ! Sorry :)

[color=darkblue][b][u]An easy extension.[/u][/b] Given are a triangle $ABC$ and a point $P\not\in BC$ for which $\widehat{PBC}\equiv\widehat{PCB}\ .$
Denote the intersections $B'\in BP\cap AC$ and $C'\in CP\cap AB\ .$
Prove that $BB'=CC'$ $\Longleftrightarrow$  the triangle $ABC$ is isosceles .
[b]Proof.[/b] Prove easily that $B<90^{\circ}$ and $C<90^{\circ}\ .$ Therefore,
$BB'=CC'$ $\Longrightarrow$ $BCC'\equiv CBB'$ $\Longrightarrow$ $B=C\ .$ Reciprocally is obviously.[/color]

ok Virgil Nicula that is a good solution and extension! Really I solve it by very long proof - a algebraic menthod, The solution by synthetic geometry is too nice and short but algebraic menthod can answer the question: which do points inside triangle have a property $AA'=BB'\Rightarrow \triangle ABC$ is isosceles some known point: $H$(orthorcenter),$I$(incenter Steiner-Lehmus),$N$(Nagel point -a paper in forum geometricorum),$O$(circumcenter),$G$(centroid)..et cetera
I think $L$(Lemoine point) is too, hope you are interested in! :)


geometry
geometric transformation
reflection
power of a point
radical axis

Here's another relatively easy problem:

Suppose that circles $S_1$ and $S_2$, with centers $O_1$ and $O_2$ respectively,  meet at points $A$ and $B$.  Extend $O_2B$ to meet $S_1$ at $C$ and $S_2$ at $F$.  Extend $O_1B$ to meet $S_2$ at $D$ and $S_1$ at $E$.  Prove that $AB, CE, DF$ are concurrent.

The circles can be different sizes.  If the circles are congruent, then you can just show that they're all reflections over the line AB.  But if they aren't...you first have to assume that both of the lines OB touch both circles.  This is an interesting problem...i'll look into it a little bit...

But (or so I thought when I last posted in this thread) they are reflections across $O_1O_2$...

Let $P$ be the intersection of $CE, DF$. Since $BE, BF$ are diameters of the circles $(O_1), (O_2)$, the angles $\angle BCE = \angle BDF = 90^o$ are right. Hence, the angles $\angle PDE = \angle PCF = 90^o$ are also right. Consequently, the 4 right angle triangles

$\triangle BCE \sim \triangle PDE \sim  \triangle PCF \sim  \triangle BDF$

are all similar, each pair in this sequence having one acute angle in common. In particular, from similarity of the triangles $ \triangle PDE \sim \triangle PCF$, we get

$\frac{PD}{PE} = \frac{PC}{PF},\ \ \ PD \cdot PF = PC \cdot PE$

Thus the power of the point $P$ to the circles $(O_1), (O_2)$ is the same, which means that this point is on the radical axis $AB$ of these 2 circles.


calculus
integration
search
vector
function
real analysis
real analysis unsolved

Four flies sit at the corners of a square card table, side a, facing inward. They start simultaneously walking at the same rate, each directing its motion steadily toward the fly on its right. Find

(a) the equation of the path traced by one of the flies.
(b) the total distance traced by each fly (without calculus).

im pretty sure they all walk the logarithmic spiral towards the middle,

(mostly because i remember seeing a picture of it)

They indeed spiral in towards the centre. kenn4000 is right.

Someone should answer part (b), which is an old-time classic. I've been holding off because I don't want to spoil the fun.

[hide="Note:"]"Without calculus" might be a slight overstatement as arguments about what happens instantaneously or in infinitesimal increments may play a role. But no explicit integral needs to be computed.[/hide]

The last time it was posted was here: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=48756274&t=214248

I've always felt a bit uneasy about the solution to this classical problem.  The way its worded in this case especially, "each directing its motion steadily toward the fly on its right", seems most naturally to ascribe to each fly an acceleration vector pointing at the next fly, not a velocity vector as is assumed in the classical solutions.  If each fly were actually moving (rather than merely 'directing its motion') toward the next fly, it seems the resulting implied acceleration vector would be some function requiring knowledge on part of the fly of not only the position, but also the dynamics, of the other flies.  Does anyone else feel the usual solution/formulation of this problem is unnatural, or am I just being pedantic?

Would you prefer the following wording?

Four flies walk on a table.  Each travels at a constant speed along a certain smooth path.  A passing observer notes that at each instant of their walk, every fly is facing directly towards the next fly in clockwise order around the center of the table.


ratio
geometry
number theory
relatively prime

A chord of a circle is perpendicular to a radius at the midpoint of the radius. The ratio of the area of the larger of the two regions into which the chord divides the circle to the smaller can be expressed in the form $\frac{a\pi+b\sqrt{c}}{d\pi-e\sqrt{f}}.$, where $a, b, c, d, e,$ and $f$ are positive integers, $a$ and $e$ are relatively prime, and neither $c$ nor $f$ is divisible by the square of any prime. Find the remainder when the product $abcdef$ is divided by $1000$.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=403[/img]

[hide="Solution"]
Let the radius of the circle be $r$.  We know that $M$ is the midpoint of the radius $\overline{AB}$.  So $AM=\frac r2.$ Additionally, $AD=r$.  Thus, right triangle $AMD$ has its hypotenuse equal to twice one of its legs and is thus a 30-60-90 right triangle.  This gives us $m\angle MAD=m\angle MAC=60^\circ.$  Therefore the smaller sector bounded by arc $CD$ is one third the area of the circle.

The area of the smaller section formed by the cord is one third the area of the circle minus $[ACD].$  This gives $\frac 13\pi r^2-\frac{r^2\sqrt{3}}4.$  The area of the large section formed by the chord is $\pi r^2$ minus the area of the smaller section.  This comes out to $\frac 23 \pi 4^2+\frac{r^2\sqrt{3}}4.$

The ratio of these two is

\[ \frac{\frac 23\pi r^2+\frac{r^2\sqrt{3}}4}{\frac 13\pi r^2-\frac{r^2\sqrt{3}}4}=\frac{8\pi+3\sqrt{3}}{4\pi-3\sqrt{3}}. \]

Thus, $abcdef=2592$ which gives $\boxed{592}$ for our answer.[/hide]

[hide]Let's call the radius $2$, now if you know that $M$ cuts the radius in half the then it cuts chord $DC$ in half too since $DC$ is perpendicular to $AE$, so now we know that $AM=1$ and $DA=2$ and so we now have $60,30,90$ triangles, and we know that the circular section $ACED$ is $\frac{1}{3}$ of the circle, so $DCE=\frac{1}{3}circleA-DAC \Longrightarrow DCE=\frac{4}{3}\pi-\sqrt{3}$ so then the rest of the circle is $\frac{8}{3}\pi+\sqrt{3}$. Thus the ratio is $\boxed{\frac{8\pi+3\sqrt{3}}{4\pi-3\sqrt{3}}}$[/hide]


algebra
polynomial
geometry
3D geometry
quadratics
search
calculus

How do you factor cubic polynomials such as this:

 $ x^3\minus{}21x\plus{}20\equal{}0$, where grouping doesn't work ?

Unless I've mistaken your interpretation by factoring a cubic polynomial:
One way includes applying the [url=http://www.artofproblemsolving.com/Wiki/index.php/Rational_Root_Theorem]rational[/url] [url=http://en.wikipedia.org/wiki/Rational_root_theorem]roots[/url] theorem and testing the possible answers, which quickly deduces this cubic polynomial into
[hide]
$ x^3 \minus{} 21x \plus{} 20 \equal{} (x \minus{} 1)(x^2 \plus{} x \minus{} 20) \equal{} \boxed{(x \minus{} 1)(x \plus{} 5)(x \minus{} 4)} \equal{} 0$[/hide]
It sometimes may be tedious and time consuming (depending on the polynomial) but it's definitely worth to try out depending on the polynomial (it becomes more easier to use when working with lesser degree polynomials like this cube, but it doesn't always work).

I have one question, what did you mean by grouping?

$ x^3\minus{}21x\plus{}20\equal{}0$
[hide="rational root theorem"]
Using [url=http://mathworld.wolfram.com/RationalZeroTheorem.html]rational root theorem[/url] we know to test $ \pm$ versions of the divisors of $ 20\equal{}2^25^1$, of which there are $ 12$. After we find one, we can use synthetic division to finish the problem in a more familiar format. Luckily, it didn't take long.

$ f(1)\equal{}(1)^3\minus{}21\plus{}20\equal{}0$

then...
$ \frac{x^3\minus{}21x\plus{}20\equal{}0}{x\minus{}1}\equal{}x^2\plus{}x\minus{}20$

$ x^2\plus{}x\minus{}20\equal{}0$
$ (x\minus{}4)(x\plus{}5)\equal{}0\Longrightarrow x\equal{}\minus{}5,4$

So now we have the three solutions: $ \boxed{x\equal{}\minus{}5,1,4}$
[/hide]

Well there is another complex method called Cardons method for solving a cubic eqn
Refer http://en.wikipedia.org/wiki/Cubic_equation
 :)

[quote="igiul"][quote="xpmath"][quote="erdogankerem123"]How do you factor cubic polynomials such as this:

 $ x^3 \minus{} 21x \plus{} 20 \equal{} 0$, where grouping doesn't work ?[/quote]
A really useful thing is that, if the coefficients add to 0, x=1 is a solution, and then it becaomes a quadratic.[/quote]

Does this hold for all polynomials?[/quote]

The bit about it become quadratic isn't, but $ x\equal{}1$ being a solution for polynomials who's coefficients sum to 0 is.

Yeah, I meant that in this problem, after factoring out x-1, it becomes a quadratic.

Usually you should try the rational root theorem.. but to narrow the search a little, you could use Descartes' rule of signs.

But Descartes rule of signs can only give you the interval of the roots and if there demand a situation to use it then the roots cant be found easily .
 :wink:

Well, most of the time you won't [i]need[/i] to use it, but when you have a lot of possible rational roots it might help a little.

Thanks everyone for the replies! They help a lot.

[quote="Patterns_34"]
I have one question, what did you mean by grouping?[/quote]

According to my Pre-Cal and Alg II teachers, grouping is when you take a cubic polynomial such as $ x^3\plus{}x^2\minus{}4x\minus{}4$ and you "group" the terms that are easily seen to have a common factor, in this case $ x^3$ and $ x^2$, and $ \minus{}4x$ and $ \minus{}4$, then factor out the common terms so that it becomes $ x^2(x\plus{}1)\minus{}4(x\plus{}1)$, which can obviously be factored further.

[quote="t0rajir0u"]Who ever gave you the idea that grouping doesn't work here?

$ (x^3 \minus{} x) \minus{} (20x \minus{} 20) \equal{} 0 \Leftrightarrow$
$ (x^2 \plus{} x)(x \minus{} 1) \minus{} 20(x \minus{} 1) \equal{} 0 \Leftrightarrow$
$ (x \minus{} 1)(x^2 \plus{} x \minus{} 20) \equal{} 0 \Leftrightarrow$
$ (x \minus{} 1)(x \plus{} 5)(x \minus{} 4) \equal{} 0$

Just because it's harder to see doesn't mean it's not there.  :)[/quote]
ahh... didn't see that, thanks!

so, can all polynomials be factored by grouping if rearranged?

[quote="xpmath"][quote="erdogankerem123"]How do you factor cubic polynomials such as this:

 $ x^3 \minus{} 21x \plus{} 20 \equal{} 0$, where grouping doesn't work ?[/quote]
A really useful thing is that, if the coefficients add to 0, x=1 is a solution, and then it becaomes a quadratic.[/quote]

what exactly is the proof of that?

@erdogankerem i think you can factor all polynomials by grouping only if it has rational (maybe even integral) roots. im not exactly sure how you would factor a polynomial if it only has irrational and/or complex roots. although if, in the course of your factoring, you get $ x^2\plus{}1$ as a factor, you know that $ \pm i$ are two roots.

[quote="mihail911"][quote="xpmath"][quote="erdogankerem123"]How do you factor cubic polynomials such as this:

 $ x^3 \minus{} 21x \plus{} 20 \equal{} 0$, where grouping doesn't work ?[/quote]
A really useful thing is that, if the coefficients add to 0, x=1 is a solution, and then it becaomes a quadratic.[/quote]

what exactly is the proof of that?[/quote]

The sum of the coefficients of a polynomial $ P(x)$ is $ P(1)$.  If $ P(1) \equal{} 0$, then $ 1$ is a root by definition.

[quote="mihail911"]@erdogankerem i think you can factor all polynomials by grouping only if it has rational (maybe even integral) roots. im not exactly sure how you would factor a polynomial if it only has irrational and/or complex roots. although if, in the course of your factoring, you get $ x^2 \plus{} 1$ as a factor, you know that $ \pm i$ are two roots.[/quote]

My previous post was somewhat misleading.  Here is the issue:

Grouping is a [i]strategy[/i] for making the factorization of certain kinds of polynomials more obvious.  [b]Any[/b] polynomial can be "factored by grouping" [i]if you already know the roots in advance[/i], but if the roots aren't "nice" (generally, integer), then the groups themselves will be almost impossible to find without knowing the roots in advance.

erdogan -- is this a problem from your calculus textbook by any chance?
I seem to remember it, and if youre trying to integrate, you dont need to factor.

If its not a calc problem, then assume i am insane and ignore me  :)

[quote="erdogankerem123"]How do you factor cubic polynomials such as this:

 $ x^3 \minus{} 21x \plus{} 20 \equal{} 0$, where grouping doesn't work ?[/quote]

[hide="Inspection"](NYSML has spoiled me in that aspect. :P) gives that the sum of the coefficients is 0, therefore 1 is a root.

The product of the other two roots is $ \frac{\minus{}20}{1}\equal{}\minus{}20$, and the sum is $ 0\minus{}1\equal{}\minus{}1$. We now have the quadratic $ x^2\plus{}x\minus{}20$. Then you can bash with the [[Quadratic Formula]], or inspect some more and factor:

$ x^3\minus{}21x\plus{}20\equal{}(x\minus{}1)(x\plus{}5)(x\minus{}4)$[/hide]


Given $n$ a natural number greater than $1$ and $p$ a prime, where $n|p-1$ and $p|n^{3}-1$, show that $4p-3$ is a square number.

This is not true for $n=1, p=2,$ or $n=1,p=5,$ and so on. I guess $n > 1$ is a necessary condition.

[quote="FieryHydra"]This is not true for $n=1, p=2.$[/quote]

Surely it is NOT true!! :P  Now we assume that $n>1$.

[quote="probability1.01"]$0 \equiv kn^{2}+kn+k \equiv-n+k-1 \mod p \implies k = n+1$.[/quote]

Wouldn't it have the case that $k=n+1+np$?


inequalities
geometry
circumcircle
inequalities unsolved

[quote="Jaswinder"]Given x, y, z are positive reals such that x + y + z = 1, show that:

32.xyz.(1 - xy - yz - zx) <= [ 3.(1 - x).(1 - y).(1 - z) ]^2

 :wink:[/quote]
Id est,we need to prove that $ 32xyz(x\plus{}y\plus{}z)(x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}xz\plus{}yz)\leq9(x\plus{}y)^2(x\plus{}z)^2(y\plus{}z)^2,$ which is equivalent to
$ \sum_{sym}(9x^4y^2\minus{}7x^4yz\plus{}9x^3y^3\minus{}10x^3y^2z\minus{}x^2y^2z^2)\geq0,$ which is obviously true..

arquady could u plz tell how the last expression is true i am not able to understand :P

plz give the complete soln that shows that the symmetric expression that u have written is greater than or equal to zero......

here is my soln..........

Homonogise and write x+y=a etc and it'll come down to a^2 + b^2 + c^2
<= 9R^2 as a,b,c are sides of a triangle and R the circumradius.  I
proved this result in a couple of ways. First setting a=2R sinA which
then becums a trigo inequality and also is true by jensens but can be
proved other wise easily as well. Another proof is that OH^2 = 9R^2 -
a^2 - b^2 - c^2 >= 0 where OH is distance btween circumcenter n
orthocenter. QED.

[quote="arqady"]Id est,we need to prove that $ 32xyz(x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2 \plus{} xy \plus{} xz \plus{} yz)\leq9(x \plus{} y)^2(x \plus{} z)^2(y \plus{} z)^2,$ [/quote]
This inequality holds for any real numbers $ x,y,z.$ ;) :)


probability
expected value
probability and stats

You have two decks of 26 cards. Each card in each of the two decks has a different letter of the alphabet on it. You pick at random one card from each of the two decks. A vowel is worth 3 points and a consonant is worth 0 points. Let X be the sum of the values of the two cards picked. Find E(X), V(X), and the standard deviation of X.

Note:
E(X) stands for expected value
V(X) stands for variance

The variate X has a discrete probability distribution: (note: I take the vowels to be A,E,I,O,U, and Y.)
There are then 20 consonants and 6 vowels in each deck of cards. The probabilities are

$ P(X\equal{}0)\equal{}P(CC)\equal{}\frac{20}{26}\cdot\frac{20}{26}\equal{}\frac{100}{169}$

$ P(X\equal{}3)\equal{}P(CV\mbox{ or }VC)\equal{}\frac{20}{26}\cdot\frac{6}{26}\plus{}\frac{6}{26}\cdot\frac{20}{26}\equal{}\frac{60}{169}$

$ P(X\equal{}6)\equal{}P(VV)\equal{}\frac{6}{26}\cdot\frac{6}{26}\equal{}\frac{9}{169}$

giving

\[ \begin{array}{c|c|c|c}X&P(X)&X\cdot P(X)&X^2\cdot P(X)\\\hline0&{\scriptstyle \frac{100}{169}}&0&0\\\hline 3&{\scriptstyle \frac{60}{169}} & {\scriptstyle \frac{120}{169}}&{\scriptstyle \frac{360}{169}}\\\hline 6&{\scriptstyle \frac{9}{169}}&{\scriptstyle \frac{54}{169}}&{\scriptstyle \frac{324}{169}}\end{array}\]

so that $ \boxed{E(X)\equal{}\sum_X X\cdot P(X) \equal{} \frac{174}{169}}$ and $ \boxed{V(X)\equal{}\sum_X X^2\cdot P(X)\minus{}E(X)^2 \equal{} \frac{85320}{28561}}$.


Find all integers h such that there exist infintely many positive integers n for which $[\sqrt{h^2+1}n]$ is a perfect square.

Well i dont know if the following is truth, but anyway...

I think that the set ${m^2/\sqrt{h^2+1}}$ has fractional part wich is dense in $(0,1)$. So there are infinitely many $m$, such that the interval $(m^2/\sqrt{h^2+1},(m^2+1)/\sqrt{h^2+1})$ contains  an integer, also this integer can be expressed as $\sqrt{n}/\sqrt{h^2+1}$ with $n$ between $m^4$ and $m^4+2m^2+1$ so we get that $n=t^2(h^2+1)$, so this $t$ gives a value for which $t\sqrt{h^2+1}$ has it integer part equal to $m^2$.

Sow e only need to prove that the set is dense, or at least it aproaches as we would like to integers, I think it is a known fact, at least I saw it in a problem from Iran or something like that!


We can think of it as "one diver is chosen to not be in each event."

We want louie's name to be drawn at most once.  It is drawn 0 times with probability $(2/3)^3$.  It is drawn 1 times with probability $3(1/3)(2/3)^2$.  So total probability 20/27.

[quote="Tharinor"]Huey, Dewey, and Louie are competing in the diving competitions. There are three diving events, but only two of the three divers are allowed to compete in each of the events and the diver's names are drawn randomly. What is the probability that Louie's name is drawn to compete in at least two of the three diving events?

I think I'm missing something obvious but I just can't seem to get the answer from an algebraic approach using combinations/permutations etc...[/quote]The probability of n successes in N trials is a binomial distribution: $P(n|N) = \binom{N}{n} p^n(1-p)^{N-n}$

The probability that Louie's name is chosen for at least two event is the probability that his name will be chosen for either two or three of the events.

$P(n \ge 2|N =3) = P(n =2|N =3) + P(n =3 |N =3) \\
\\
= \binom{3}{2}(\frac{1}{3})^2(1-\frac{1}{3})^1 + \binom{3}{3}(\frac{1}{3})^3(1-\frac{1}{3})^0 \\
\\
= 3(\frac{1}{9})(\frac{2}{3}) + 1(\frac{1}{27}) \\
\\
= \frac{7}{27}$

well think of it this way, if we want the times that he is chosen 2 times or more, then he can be chosen 2 or 3 times
the probability that he will be chosen 2 times is $\frac{2}{3}*\frac{2}{3}*\frac{1}{3}*3$ because (win for chosen, loose for not chosen) win, win, lose, and then the order of those has 3 ways (wwl wlw lww)
then the chances that he wins is $\frac{2}{3}*\frac{2}{3}*\frac{2}{3}$
and then adding those, you get $\boxed{\frac{20}{27}}$


if  $a,b\in\mathbb{Z}$ then $(a,b)=(0,0)$
a and b are even. infinite descent.


What comes next in this sequence:
fawn bear lion mole ==?==

Select from: seal duck stag vole wolf.


its just like the top admin election BUT with a slight differnece. this time you can take away one point OR! add one point.I only know 3 mods on FTW though.Oh and all the mods start at 0.Remeber you can only choose one choice. take away or add a point. I'll take the first vote. I will vote for izzy

isabella:1

vallon22:0

iin77:0

i vote izzy. yes, vallons a mod, dragon96.

I would vote all three if I could.

However, I cannot, so I will remain neutral and not vote.

I vote iin77

isabella2296: 4
vallon22: 0
iin77: 1

Umm.. why don't you type the scores...
that way everyone can see it.

isabella2296: 4
vallon22: 1
iin77: 1

isabella2296: 5
vallon22: 1
iin77: 1

The only one I THINK has never muted me in FTW. (Before the update I can't talk now  :( )

I will take a point away for once. Sorry Izzy.

Izzy:4.5

iin77:3

vallon22:1.5

gah
i'm losing...

i vote myself  :P

Izzy:4.5

iin77:3

vallon22:2.5

gah
i'm still losing  :D

izzy- 4.5
iin77- 4
vallon22-2.5

yay! go iin! sorry izzy =[

okay gonna subtract your votes again. if you keep doing this i'm gonna ask a mod to lock this and the competition is over.

Izzy:9

iin:6

vallon:4

Well you might as well ask a mod to lock it because earlier you said in 24 hours its over. The scores are 0,0, and -12. No winner.

[quote="jjx1"]oka w/e mod please lock this. THE WINNER IS IZZY. this is the final score

Izzy:9

iin:6

vallon:4[/quote]
yeah, this topic is locked down i can't even post  :rotfl:

ERNIE THE GAME IS OEVER. IZZY WON. I DID NOT COUNT YOUR VOTES OR JAMES OR DRAGONS. ALSO JOIN FFC!!!!!!!

guys
it's just a game
stop yelling at each other about it
plus this is mostly a popularity vote
and for now, i really don't care about my popularity on AoPS

also, there is no mod for FF, so no one can lock it

vallon you're right. guys you can do whatever you want to this forum. But i thought there were two mods for ff.

Oh well... IDC.

izzy: 0
vallon22: 0
iin77: -16


function
real analysis
real analysis unsolved

Given that  $ A$ and $ \mathbb R\setminus A$ are both dense subsets of $ \mathbb R$ does it follow that there exists either a function that is  continuous only on $ A$ or a function that is discontinuous only on $ A$.
P.S. I have no idea whether or not this is true - I just came across it thinking about a problem that mlok posed in response to an (easy) problem I posted. (The answer is yes for countable sets.)

The set of continuity of a function must be a $ G_{\delta}$ set (countable intersection of open sets). This is a more or less direct consequence of the $ \epsilon$-$ \delta$ definition. If I remember correctly, any $ G_{\delta}$ set is the set of continuity for some function.

So this amounts to showing that there are no dense sets (with dense compliment) that are neither $ F_{\sigma}$ or $ G_{\delta}$? This seems unlikely to be true and even if it is it may be quite hard.
Anyway, the reason I did this was to see if such a set $ A$ could satisfy the hypothesis of that previous problem by invoking a theorem of Volterra that there do not exist two pointwise discontinuous (continuous on a dense set) functions with one continuous only where the other is discontinuous.
(Supposing the existence of an $ F$ as in the previous problem we could take $ f$ a function continuous only on $ A$ and then achieve a contradiction by considering $ g=f(F)$.)

Of course density has little to do with being $ G_{\delta}$. Consider the set formed by all positive rationals and negative irrationals: it is dense, has dense complement, and is neither $ G_{\delta}$ nor $ F_{\sigma}$. Thus, the answer to your question is no.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=162163]Link to the earlier thread that motivated this one[/url].


Given $ a, b, c \geq\ 0$ satisfy $ ab \plus{} bc \plus{} ca \equal{} 1.$
Prove that: $ \sqrt []{a^2b \plus{} b^2c \plus{} c^2a} \plus{} \sqrt []{ab^2 \plus{} bc^2 \plus{} ca^2} \plus{} 3 \sqrt []{abc} \geq\ 2$
 :)

@mitdac123: : Good luck in love.

[quote="nguoivn"]I haven't solved the general problem:
Find the best constant k satisfy:
$ \sqrt []{a^2b \plus{} b^2c \plus{} c^2a} \plus{} \sqrt []{ab^2 \plus{} bc^2 \plus{} ca^2} \plus{} k \sqrt []{abc} \geq\ 2$
(with the same condition)
I only have proof for $ k \geq\ 3$
 :wink:[/quote]
Here is the hint: $ (a^2b\plus{}b^2c\plus{}c^2a)(ab^2\plus{}bc^2\plus{}ca^2) \ge (a^2b^2\plus{}b^2c^2\plus{}c^2a^2)(ab\plus{}bc\plus{}ca)$. ;) :)

[quote="nguoivn"]Nice hint, can_hang. But after using this hint, we'' ll have to prove:
$ 2 \sqrt [4]{3(9 \minus{} 2pr)} \plus{} k \sqrt []{r} \geq\ 2$
I think we must to use calculator  :wink:
Can you show me the best constant $ k \equal{} ?$
Besides, with $ k \geq\ 3,$ I have a nice and short proof by Am-Gm  :)[/quote]
No, I didn't mean to prove it by that way. ;)
My hint used for another way. ;) Try it!

I want to see the solution by AM-GM.Thanks you.


inequalities
algebra
polynomial
inequalities proposed

This is inequality was proposed by Dinu Serbanescu at jbmo test selection in 2004.


inequalities
number theory unsolved
number theory

I just calculated this . It says that the error between the actual and the computed value of pi(N) using sieve is atleast in the order of ( √N/log(N) ) i.e O(√N/logN).
This is the proof for the above the statement .
I have taken take a specific case and proved it . However it can be generalised .
Consider 25<N<49 then pi(√N) = 3. The primes are 2,3 and 5.
The actual value for pi(N) = N - [N/2] - [N/3] - [N/5] + [N/6] + [N/10] + [N/15] - [N/30] + ( pi(√N) -1).
The computed value using sieve theory for which i use the formula N*(p1-1)*(p2-1)*(p3-1)/( p1)*(p2)*(p3).
I expand it to get N - N/2 - N/3 - N/5 + N/6 + N/10 + N/15 - N/30 = computed value ( Note : I have substituted p1,p2,p3 as 2,3,5 ).
Then computed - actual = - (N/2 + N/3 + N/5 + N/30) + N/6 + N/10 + N/15 + [N/2] + [N/3] + [N/5] +[N/30] - [N/6] - [N/10] - [N/15] - ( pi(√N) -1).
( i use this inequality that [a] + [b] < [a+b] thus [N/2] + [N/3] + [N/5] +[N/30] < [N/2 + N/3 + N/5 + N/30] , [N/2] + [N/3] + [N/5] +[N/30] -(N/2 + N/3 + N/5 + N/30) < [N/2 + N/3 + N/5 + N/30] - (N/2 + N/3 + N/5 + N/30) < -1 ).
Thus computed - actual < - 1 - ( pi(√N) -1) + N/6 + N/10 + N/15 - [N/6] - [N/10] - [N/15] . - eqn1
In the same way actual - computed < ( pi(√N) -1) -1 + N/2 + N/3 + N/5 + N/30 - ( [N/2] + [N/3] + [N/5] +[N/30] ). - eqn2.
eqn1 - eqn2 implies computed - actual < 1 - pi(√N) .
Note : I have not done every step because i thought most of it was simple arithmetic in case you have any doubts let me know .
Thus it is trivial to conclude that the error is at least in the order of √N/logN . In other words pi(N) cannot be computed with a error less than √N/logN . However R.H states that the maximum error is N^( .5 +ξ) . So if R.H is true we have both the upper and lower limits of the error.
I am not sure if this result has already been found out .

I believe it's spelled Einst[b]ei[/b]n, Einstein!  :P

I believe it's spelled Einstein, Einstein!.
I do know i intentionally did this to say that i am not einstein :).
If u are gonna ask why that name with an error . The answer is that i wish to to be close to him.
PS: Closeness is not defined as N-> infinity . for example one knows how close 9.99 is to 10.
But how close is 10^(10^10^10^10^10) close to 10^(10^10^10^10^9.99) ???
I am sorry for replying to the above post . Though i am myself not very interested in discussing these kinds of things.


I was viewing the online people and I noticed that several guests were listed as "posting a message."  Screenshot attached for clarification if need be.

That looks like another side effect of the server issues last night; see also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135229]here[/url]. The posts from 2013 are a dead giveaway.


function
quadratics
algebra
quadratic formula
floor function

Greatest Integer Function Problem...

Solve for all positive integer $ x$ : $ [x]^2 \equal{} xh$ where $ [x]$ represents GIF, $ x$ is the number and $ h$ is the fractional part.

I've reached this part of the solution: since $ x \equal{} h\plus{}[x]$, $ [x]^2 \minus{} [x]h \minus{} h^2\equal{}0$. using quadratic formula, $ [x] \equal{} \frac{h\plus{}h\sqrt{5}}{2}$. what next?

[hide="Partial solution?"]
We know $ [x] \equal{} x\minus{}h$ so $ (x\minus{}h)^2 \equal{} xh$.
It follows that $ x^2 \minus{} 3hx \plus{} h^2 \equal{} 0$.
Then $ D \equal{} 5h^2$.

to be continued...
[/hide]

[hide](You meant Solve for all positive reals, not integers (it doesn't make sense for ints)).

"I've reached this part of the solution: since $ x \equal{} h \plus{} [x]$, $ [x]^2 \minus{} [x]h \minus{} h^2 \equal{} 0$. using quadratic formula, $ [x] \equal{} \frac {h \plus{} h\sqrt {5}}{2}$. what next?[/quote]"

Well, as $ h<1$, $ [x]\leq 2$.

Case 1. $ [x]\equal{}0$. Simply putting it into the equation, we have $ x\equal{}0$ (not quite positive...).
Case 2. $ [x]\equal{}1$. Put it in the last equation you've got and voila, you have one solution.
Case 3. $ [x]\equal{}2$. Same as above.
[/hide]

Ok, I was going to post a really long and ugly solution (was planning to finish it now because I was interrupted last time). I guess yours is like a hundred times better.

[quote="hsiljak"][hide](You meant Solve for all positive reals, not integers (it doesn't make sense for ints)).

"I've reached this part of the solution: since $ x \equal{} h \plus{} [x]$, $ [x]^2 \minus{} [x]h \minus{} h^2 \equal{} 0$. using quadratic formula, $ [x] \equal{} \frac {h \plus{} h\sqrt {5}}{2}$. what next?[/quote]"

Well, as $ h < 1$, $ [x]\leq 2$.

Case 1. $ [x] \equal{} 0$. Simply putting it into the equation, we have $ x \equal{} 0$ (not quite positive...).
Case 2. $ [x] \equal{} 1$. Put it in the last equation you've got and voila, you have one solution.
Case 3. $ [x] \equal{} 2$. Same as above.
[/hide][/quote]

oh yeah....that's what i lacked... thanks for the help and info:P


[hide="General Solution"]We can prove by induction that $ \frac{1}{x\minus{}1} \minus{} \displaystyle\sum_{i\equal{}0}^{n}{\displaystyle\frac{2^i}{x^{2^i}\plus{}1}}\equal{}\displaystyle\frac{2^{n\plus{}1}}{x^{2^{n\plus{}1}}\minus{}1}$.

Base case: $ n \equal{} 0$. $ \frac{1}{x\minus{}1} \minus{} \frac{1}{x\plus{}1} \equal{} \frac{(x\plus{}1)\minus{}(x\minus{}1)}{x^2\minus{}1} \equal{} \frac{2}{x^2\minus{}1}$

We make the assumption that the claim is true for some $ k$, and seek to prove it for $ k\plus{}1$.

$ \frac{1}{x\minus{}1} \minus{} \displaystyle\sum_{i\equal{}0}^{k\plus{}1}{\displaystyle\frac{2^i}{x^{2^i}\plus{}1}} \equal{} \frac{2^{k\plus{}1}}{x^{2^{k\plus{}1}}\minus{}1} \minus{} \frac{2^{k\plus{}1}}{x^{2^{k\plus{}1}}\plus{}1} \equal{} \frac{2^{k\plus{}2}}{x^{2^{k\plus{}2}}\minus{}1}$.

Therefore, $ \frac{1}{x\minus{}1} \minus{} \displaystyle\sum_{i\equal{}0}^{n}{\displaystyle\frac{2^i}{x^{2^i}\plus{}1}}\equal{}\displaystyle\frac{2^{n\plus{}1}}{x^{2^{n\plus{}1}}\minus{}1}$.

Let $ n\equal{}3$, and the result is $ \boxed{\frac{16}{x^{16}\minus{}1}}$.

[/hide]

[hide="Or"]
We could simply see that if we subtracted the first two equations, then it would become $ \frac {2}{x^{2}\minus{}2}$
And following this we get the same answer.[/hide]


abstract algebra
group theory
permutations
combinatorics
IMO Shortlist

Suppose that $ a_1$, $ a_2$, $ \ldots$, $ a_n$ are integers such that $ n\mid a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n$.
Prove that there exist two permutations $ \left(b_1,b_2,\ldots,b_n\right)$ and $ \left(c_1,c_2,\ldots,c_n\right)$ of $ \left(1,2,\ldots,n\right)$ such that for each integer $ i$ with $ 1\leq i\leq n$, we have
\[ n\mid a_i \minus{} b_i \minus{} c_i
\]

[i]Proposed by Ricky Liu & Zuming Feng, USA[/i]

This problem and the following generalisation appeared 1979 in Ars Combinatoria (thanks to Darij who found it):

Let $ (G, \plus{} )$ be a finite abelian group of order $ n$.
Let also $ a_1,a_2,...,a_{n \minus{} 1} \in G$ be arbitrary.
Then there exist pairwise distinct $ b_1,b_2,...,b_{n \minus{} 1} \in G$ and pairwise distinct $ c_1,c_2,...,c_{n \minus{} 1} \in G$ such that $ a_k \equal{} b_k \plus{} c_k$ for $ k \equal{} 1,2,...,n \minus{} 1$.

[Moderator edit: The Ars Combinatoria paper is:
F. Salzborn, G. Szekeres, [i]A problem in Combinatorial Group Theory[/i], Ars Combinatoria 7 (1979), pp. 3-5.]

so could someone post a proof of either the problem or its generalization?

you can find the proof in file shortlist 2005 which has been posted by orl
The main idea is given two sequence $a_{1}...a_{n}$ and $b_{1}...b_{n}$ s.t $\sum a_{i}\equiv 0(mod n)$ and $\sum b_{i}\equiv 0(mod n)$ and there are exactly two i;j s.t $a_{i}\neq\ b_{i}(modn)$ and $a_{j}\neq\ b_{j}(modn)$.Then if we know two permutation good for the sequence (a_1...a_n) then we can build two permuttionm good for (b_1...b_n)
i will come back with detail if you need

Well, I will post the solution from Ars Combinatoria if a re-find that two sheets of paper...
It's a bit different from the ISL one.

I think you didnt keep promise,Zetax :lol: Please post it here and now!

to spdf: that was also my idea when i first saw the problem but i can't find a good way to contruct those 2 permutations for $ b_j$.you said you can post details.please do so :)

[quote=ZetaX]Sorry for not repsonding (I merely forgot...). But I just saw that problem again: In a slightly different manner (but being equivalent to this one here) it is solved in "The Mathematics of Juggling", called the "Converse of the Average Theorem".

Main ideas:
You show that this property (being a sum of two permutations) is invariant under the operations $ a_{i,j,d}$ that add $ d$ to $ a_i$ and subtract $ d$ from $ a_j$.
For this, you need to do it algorithmically (but describing it is a bit hard without that graphics...).[/quote]

that is what I tried to do, but I can not prove that we can do it for the general case... can someone please help?

Since it's almost twelve years without complete solution, here's the official solution:

Suppose there exists permutations $\sigma$ and $\tau$ of $[n]$ for some sequence $\{ a_i\}_{i\in [n]}$ so that $a_i\equiv_n \sigma (i)+\tau (i)$ for all $i\in [n]$.
Given a sequence $\{ b_i\}_{i\in [n]}$ with sum divisible by $n$ that differ, in modulo $n$, from $\{ a_i\}_{i\in [n]}$ in only two positions, say $i_1$ and $i_2$.
We want to construct permutations $\sigma'$ and $\tau'$ of $[n]$ so that $b_i\equiv_n \sigma' (i) +\tau' (i)$ for all $i\in [n]$.
Recall that $b_i\equiv a_i\pmod{n}$ for all $i\in [n]$ that $i\neq i_1,i_2$.
Construct a sequence $i_1,i_2,i_3,...$ by, for each integer $k\geq 2$, define $i_{k+1}\in [n]$ to be the unique integer satisfy $\sigma (i_{k-1})+\tau (i_{k+1})\equiv_n b_{i_k}$.
Let (clearly exists) $p<q$ are the indices that $i_p=i_q$ with minimal $p$, and then minimal $q$.

If $p>2$. This means $i_j\not\in \{ i_1,i_2\} \implies \sigma (i_j) +\tau (i_j) \equiv_n b_{i_j}$ for all $j\in \{ p,p+1,...,q\}$.
Summing the equation $\sigma (i_{k-1})+\tau (i_{k+1})\equiv_n b_{i_k}$ for $k\in \{ p,p+1,...,q-1\}$ gives us
$$\sum_{j=p-1}^{q-2}{\sigma (i_j) } +\sum_{j=p+1}^{q}{\tau (i_j)} \equiv_n\sum_{j=p}^{q-1}{b_{i_j}} \implies \sigma (i_{p-1}) +\sigma (i_p) +\tau (i_{q-1}) +\tau (i_q) \equiv_nb_{i_p}+b_{i_{q-1}}.$$
Plugging $i_p=i_q$ and use $\sigma (i_p) +\tau (i_p)\equiv_n b_{i_p}$ gives us $\sigma (i_{p-1}) +\tau (i_{q-1})\equiv_n b_{i_{q-1}} \equiv_n \sigma (i_{q-1})+\tau (i_{q-1})$.
Hence, $\sigma (i_{p-1}) \equiv_n \sigma (i_{q-1})\implies i_{p-1}=i_{q-1}$, contradiction to the definition of $p,q$.

So, we've $p\in \{ 1,2\}$. Let $p'=3-p$. Define the desired permutations $\sigma'$ and $\tau'$ as follows:
$$\sigma' (i_l)=\begin{cases}
\sigma (i_{l-1}), & \text{ if } l\in \{ 2,3,...,q-1\} \\
\sigma (i_{q-1}), & \text{ if } l=1
\end{cases} ,\tau' (i_l)= \begin{cases}
\tau (i_{l+1}), & \text{ if } l\in \{ 2,3,...,q-1\} \\
\tau (i_{p'}), & \text{ if } l=1
\end{cases}  $$
and $\sigma' (i) =\sigma (i),\tau' (i)=\tau (i)$ for the rest $i\in [n]$ that $i\not\in \{ i_1,i_2,...,i_{q-1}\}$.
Note that the reason we choose $\tau (i_{p'})$ is just to not use $\tau (i_p)=\tau (i_{(q-1)+1})$ more than one time.
This construction gives us $\sigma' (i)+\tau' (i)\equiv_n b_i$ for all $i\in [n]$ except when $i=i_1$.
But since both $\sigma'$ and $\tau'$ are permutations of $[n]$, we've $\sum_{i\in [n]}{(\sigma' (i)+\tau' (i))} \equiv_n 2\times \frac{n(n-1)}{2}\equiv_n 0\equiv_n \sum_{i\in [n]}{b_i}$.
This guarantee that $\sigma' (i) +\tau' (i)\equiv_n b_i$ when $i=i_1$ too.  This prove the validity of permutations we constructed.

The case with $n$ prime is also resolved in [url=https://services.math.duke.edu/~dasgupta/papers/finallatin.pdf]this paper[/url]


note that a^5+b^5=(a+b)(a^4+b^4-a^2*b^3-a^3*b^2+a^2*b^2)
i'll prove
(a^4+b^4-a^2*b^3-a^3*b^2+a^2*b^2)  \geq  ab(3ab-a^2-b^2)
which is equivalent (a^2-b^2) \geq 0 .thus
(a^5+b^5)/ab(a+b)  \geq (3ab-a^2-b^2)
add two similar inequalitites to get the result
cheers


Prove that this is true :
[img]http://img200.imageshack.us/img200/6218/cotj.jpg[/img]

[hide="Solution"]
$ \frac {\tan x \minus{} \frac {1}{\tan x}}{\tan x \plus{} \frac {1}{\tan x}} \equal{} \frac {\tan^2 x \minus{} 1}{\tan ^2 x \plus{} 1}$.

But $ \sec^2 x \minus{} \tan^2 x \equal{} 1\implies \tan^2 x \plus{} 1 \equal{} \sec^2 x$

$ \frac {\tan^2 x \minus{} 1}{\sec^2 x} \equal{} \sin^2 x \minus{} \cos^2 x \equal{} \minus{} \cos 2x$.

Therefore we have reduced it to

$ 1 \minus{} \cos 2x \equal{} 2\sin^2 x$, which is clearly true.[/hide]

Sorry, grn_trtle, I didn't see your post. :read:


Let $BE$ and $CF$ be the altitudes of an acute triangle $ABC$ with $E$ on $AC$ and $F$ on $AB$. Let $O$ be the point of intersection of $BE$ and $CF$. Take any line $KL$ through $O$ with $K$ on $AB$ and $L$ on $AC$. Suppose $M$ and $N$ are located on $BE$ and $CF$ respectively. such that $KM$ is perpendicular to $BE$ and $LN$ is perpendicular to $CF$. Prove that $FM$ is parallel to $EN$.

[hide]Since $AF || LN$ and $AE || KM$ we have $\angle CAF = \angle CLN = BKM$.

Then $\angle MKF = NLE$.

Since $OFKM$ and $LNOE$ are cyclic we have $\angle FOK = \angle FMK$ and $\angle LON = \angle LEN$.

Since $\angle FOK = \angle LON$ we have $\angle FMK = \angle LEN$.

Now $\angle OMF = 90^o - \angle FMK = 90^o - \angle LEN = \angle NEO$. So $EN$ is parallel to $MK$.[/hide]

[i]I add one more solution  to Andreas nice solution![/i]

We ‘ ll use similar triangles.

 a) Triangles  $OKM$ and  $OEL$  are similar so :
              $\frac{OM }{OE} = \frac{OK}{OL}$

b) Triangles  $OKF$  and  $OLN$  are similar so:
                $\frac{OK}{OL} = \frac{OF}{ON}$

From (1) and (2) we take

                     $\frac{OM}{OE} = \frac{OF}{ON}$

which means that  $FM // EN$

 [u]Babis[/u]

Since, $\angle AFC =\angle LNC=90^{\circ} \implies AF||LN \implies \angle AKL =\angle KLN$, but since, $FKMO$ and $ELNO$ are cyclic quadrilaterals
Hence,$\angle AKL=\angle FMO=\angle KLN=\angle BEN \implies \boxed{FM||EN}$


I'm going to do a test to see how many people view this topic.

Hehe, had to see what it was.  :blush:  :D

Hehe. To make sure a lot of people will view your thread, warn them not to view your thread.

Oh, curiosity killed the cat.  But satisfaction brought it back.   :D

[quote="Bictor717"]"Buuuutttt..... the cat came back, the very next day..."[/quote]

He just couldn't stay away!  :D

[quote="plokoon51"]Basic example of reverse psychology :lol:[/quote]

somehow.. i thought reverse psychology was a canadian thing..  :D
(we always joke around with reverse psychology in math camps)

I guess I couldn't resist... but if someone made a "Please view this topic" thread, I would probably look at that too...

er...i'm sure it says somewhere in the Moderator Bill of Rights that I have the right to look at a possibly inappropriate topic. *nods to self* Thus, I'm sure I didn't fall for your trick. :-)

As always, the best way to make many people to see something, is ask loudly enough not to do it.  :D

This is the first topic I entered out of all I saw.

 :D

the easiest way to get a lot of views? (=

Cats are too smart to be killed by curiosity.  That's why i'm not a cat.....  eh.....

[quote="mcalderbank"]not my cat......... :D[/quote]

BTW, I don't know if it would work with all the cats, but with the cat of a friend, we always play to point a laser to the floor, and move it around, and the cat always tries to catch it. It is very funny. :rotfl:

Best regards,

That's why cats have nine lives. By the way, there's a poem called Curiosity by Alistair Reid:

Curiosity
may have killed the cat; more likely
the cat was just unlucky, or else curious
to see what death was like, having no cause
to go on licking paws, or fathering
litter on litter of kittens, predictably.

Nevertheless, to be curious
is dangerous enough. To distrust
what is always said, what seems,
to ask odd questions, interfere in dreams,
leave home, smell rats, have hunches
do not endear cats to those doggy circles
where well-smelt baskets, suitable wives, good lunches
are the order of things, and where prevails
much wagging of incurious heads and tails.

Face it. Curiosity
will not cause us to die--
only lack of it will.
Never to want to see
the other side of the hill
or that improbable country
where living is an idyll
(although a probable h*ll)
would kill us all

Only the curious have, if they live, a tale
worth telling at all.

Dogs say cats change too much, are irresponsible,
are changeable, marry too many wives,
desert their children, chill all dinner tables
with tales of their nine lives.
Well, they are lucky. Let them be
nine-lived and contradictory,
curious enough to change, prepared to pay
the cat price, which is to die
and die again and again,
each time with no less pain.
A cat minority of one
is all that can be counted on
to tell the truth. And what cats have to tell
on each return from h*ll
is this: that dying is what the living do,
that dying is what the loving do,
and that dead dogs are those who do not know
that dying is what, to live, each has to do.

[quote="jpark"]

somehow.. i thought reverse psychology was a canadian thing..  :D
(we always joke around with reverse psychology in math camps)[/quote]

Hehehe, don't forget Jenn, there's no such thing as reverse psychology.   :D

The point of this post wasn't to trick people. I'm not [i]that[/i] mean(almost). I just wanted to see how people deal with curiosity.

well at least we could use this for consensus purposes
out of 25 posts, 17 are boys 5 are girls, i didnt bother to subtract the overlap (like someone posting twice)


Let $ f: [0 ,1]\minus{}>[0, 1]$ an increasing function.
$ f(0)\equal{}0$
$ f(x) \plus{} f(1\minus{}x) \equal{} 1$
$ 2f(x) \equal{} f(3x)$
Calculate $ f(18/1991)$.

I found that $ f(1\minus{}\frac{1}{3^n}) \equal{} 1\minus{}\frac{1}{2^n}$
I don't know if it helps.

Well if what you found is true, letting $ c \equal{} \frac{18}{1991}$
we get $ f(c) \equal{} 1\minus{}\frac{1}{2^{\frac{\minus{}\log(1\minus{}c)}{3}}}$

You mean: $ f(c) \equal{} 1\minus{}\frac{1}{2^{\frac{\minus{}\log(1\minus{}c)}{log{3}}}}$

I didn't understand this second method you used to find $ f(x)$. Could you explain it in details?

[quote="danilorj"]You mean: $ f(c) \equal{} 1 \minus{} \frac {1}{2^{\frac { \minus{} \log(1 \minus{} c)}{log{3}}}}$[/quote]
But it'll not work. n is integer.

Man, how can you get to the solution if you didn't use the conditions of the problem?
$ f(0)\equal{}0$ Where did you use it?
$ f(x)\plus{}f(1\minus{}x)\equal{}1$ The same
$ 2f(x)\equal{}f(3x)$ and the same.
And returning to your solution, why did you remove the number after the first one and replaced every two number by one ? Then you treat it as a binary expansion ?

[quote="danilorj"]Man, how can you get to the solution if you didn't use the conditions of the problem?
$ f(0) \equal{} 0$ Where did you use it?
$ f(x) \plus{} f(1 \minus{} x) \equal{} 1$ The same
$ 2f(x) \equal{} f(3x)$ and the same.
[/quote]

$ f(0) \equal{} 0$ and $ f(0) \plus{} f(1) \equal{} 1 \implies f(1) \equal{} 1$ -> first condition and second condition
$ 2f(\frac13) \equal{} f(1) \implies f(\frac13) \equal{} \frac12$ -> third condition
$ f(\frac13) \plus{} f(\frac23) \equal{} 1 \implies f(\frac23) \equal{} \frac12$ -> second condition
"f is an increasing function" implies $ f(x) \equal{} \frac12 \forall x \in [\frac13, \frac23]$ -> condition that f is increasing (you forgot about that one).

And that's only in the first line of his solution! ;)

[quote="xxp2000"]$ f(x) \equal{} \frac12$ for $ x\in[\frac13,\frac23]$.
$ \frac {1176}{1991}\in[\frac13,\frac23]$
$ f(\frac {392}{1991}) \equal{} f(\frac {1176}{1991})/2 \equal{} \frac14$
$ f(\frac {1599}{1991}) \equal{} 1 \minus{} f(\frac {392}{1991}) \equal{} \frac34$
$ f(\frac {533}{1991}) \equal{} f(\frac {1599}{1991})/2 \equal{} \frac38$
$ f(\frac {1458}{1991}) \equal{} 1 \minus{} f(\frac {533}{1991}) \equal{} \frac58$
$ f(\frac {18}{1991}) \equal{} f(\frac {1458}{1991})/16 \equal{} \frac5{128}$

Another way to find $ f(x)$.
1) Write $ x$ in base $ 3$.
2) If $ 1$ shows up, then remove all the number after the first $ 1$.
3) replace $ 2$ with $ 1$.
4) treat the new number as $ 2 \minus{}$ base, that will be $ f(x)$.

$ \frac {18}{1991} \equal{} 0.0000201abc...$ in $ 3 \minus{}$ base. So $ f(\frac {18}{1991}) \equal{} 2^{ \minus{} 5} \plus{} 2^{ \minus{} 7}$.[/quote]

(3) Notice that from $ 2f(x) \equal{} f(3x)$
$ f(x)\equal{}\frac{f(3x)}{2}$

(4) $ f(x) \equal{} \frac12 \forall x \in [\frac13, \frac23]$

(5) Notice that for $ 0<a<\frac{1}{3}$,
$ f(\frac{2}{3}\plus{}a)\equal{}1\minus{}f(\frac{1}{3}\minus{}a)\equal{}1\minus{}\frac{f(1\minus{}3a)}{2}\equal{}1\minus{}\frac{1\minus{}f(3a)}{2}\equal{}\frac{1}{2}\plus{}\frac{f(3a)}{2}\equal{}\frac{1}{2}\plus{}f(a)$

Use (3) to remove all the zeros until you get either $ 0.1abc...$ in $ 3 \minus{}$ base then use (4) and forget all the $ abc$ or $ 0.2xyabc...$ then use (5).

PS: I was in that competition unfortunately I could not find either the chain upto $ \frac {1176}{1991}$ or the second method... however manage to solve the other ones to get a bronze...by the way the problem 2 of that competition was very nice [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=831475#831475[/url]


[quote="yaa"]${ \frac {a - 1}{ca} + \frac {b - 1}{ab} + \frac {c - 1}{bc}}\leq \frac {3}{4}$with a,b,c>2[/quote]
Let $ a=2+x,$ $ b=2+y$ and $ c=2+z.$
Hence, your inequality is equivalent to the following:
$ \sum_{cyc}(2xy+xyz)\geq0.$

[quote="yaa"]Sorry my problem is: a,b,c>1[/quote]
If so it's wrong: $ c\rightarrow1^{ \plus{} }$ and $ a \rightarrow \plus{} \infty.$ :wink:


1> I have never chamged my sig.
2> Mmmmmm ... chocolate. was mcuh better.
3> Stop creating so many topics.

1: I only created three!
2: Ok, I'll change my avatar.
3: Really? uh... ok.

I changed my signature three times. However i had two out of four of my signatures for less than three days.

I saved my old signature, check my "new" one out.

My first sig was MATHCOUNTS related.
My second sig was "Penguin!".
My third sig had a link to the MIMC thread.
My fourth sig is "Penguin! / (:>)[O]{".

[quote="Iversonfan2007"]i change my sig every two minutes[/quote]

you dont have a sig right now  :|


function
algebra
polynomial
complex analysis
complex analysis unsolved

Show that any proper entire function is a polynomial.

// A mapping between two topological spaces $X$ and $Y$ is [i]proper[/i] iff the preimage of every compact subset of $Y$ is a compact subset of $X$.

Since the preimage of $\{z\colon |z|\le 1\}$ is compact, the function $1/f(z)$ is bounded near infinity. Therefore, $\infty$ is a removable singularity for $1/f(z)$. From this it follows that $f$ has at most polynomial growth at $\infty$, which by the Cauchy estimates implies that $f$ is a polynomial.


[quote="Allnames"]Find all function continous
$ f (0;\infty ) \minus{} > (0;\infty )$
such that:$ f(2x) \equal{} 2f(x)$
$ f((f(x))^2) \equal{} xf(x)$
$ f(x)\in N^*$ if $ x\in N^*$[/quote]
The two first conditions are enough to show that $ f(x)\equal{}x$ :

Let $ f(x)$ a continuous function $ (0,\plus{}\infty)\to(0,\plus{}\infty)$ such that :
$ f(f(x)^2)\equal{}xf(x)$
$ f(2x)\equal{}2f(x)$

Notice that $ f(f^2(x))\equal{}xf(x)$ implies $ x\equal{}\frac{f(f^2(x))}{f(x)}$ and so $ f(x)$ is a bijection, and so monotonous, and so strictly increasing (since $ f(2x)>f(x)$).

Let $ A\equal{}\{x$ such that $ f(x)\equal{}x\}$

If $ A$ is dense in $ (0,\plus{}\infty)$, continuity implies $ f(x)\equal{}x$ $ \forall x\in(0,\plus{}\infty)$
If $ A$ is not dense in $ (0,\plus{}\infty)$, let $ 0<u<v$ such that $ f(x)\neq x$ $ \forall x\in(u,v)$

If for some $ x_0\in(u,v)$ $ f(x_0)>x_0$, then $ f^2(x_0)>x_0f(x_0)\equal{}f(f^2(x_0))$ and so $ f(y_0)<y_0$ for some $ y_0\equal{}f^2(x_0)$ and so $ \exists z$ such that $ f(z)\equal{}z$ (since $ f(x)$ is continuous).
Same, if for some $ x_0\in(u,v)$ $ f(x_0)<x_0$, then $ f^2(x_0)<x_0f(x_0)\equal{}f(f^2(x_0))$ and so $ f(y_0)>y_0$ for some $ y_0\equal{}f^2(x_0)$ and so $ \exists z$ such that $ f(z1)\equal{}z$ (since $ f(x)$ is continuous).

Then $ f(2x)\equal{}2f(x)$ implies that $ f(2^kz)\equal{}2^kz$ $ \forall k\in\mathbb Z$
So $ A$ is not empty and contains elements as little as we want ($ k\to \minus{}\infty$) and as great as we want ($ k\to \plus{}\infty$).

Let then $ a\equal{}\sup\{x\leq u$ such that $ f(x)\equal{}x\}$. "$ a$" exists since we can have in A elements as little as we want.
Let then $ b\equal{}\inf\{x\geq v$ such that $ f(x)\equal{}x\}$. "$ b$" exists since we can have in A elements as great as we want.
Since $ f(x)$ is continuous, $ f(a)\equal{}a$ and $ f(v)\equal{}v$

So we now have an interval $ [a,b]$ such that $ a\leq u<v\leq b$ and $ f(a)\equal{}a$ and $ f(b)\equal{}b$ and $ f(x)\neq x$ $ \forall x\in(a,b)$

We clearly have $ f(a^2)\equal{}a^2$, $ f(b^2)\equal{}b^2$ and $ f(x)<x$ in $ (a^2,b^2)$ if $ f(x)>x$ in $ (a,b)$ or $ f(x)>x$ in $ (a^2,b^2)$ if $ f(x)<x$ in $ (a,b)$ . Wlog say $ f(x)>x$ in $ (a,b)$ (else consider $ a'\equal{}a^2$ and $ b'\equal{}b^2$).

We then know that $ f(x)<x$ $ \forall x\in(a^2,b^2)$, so $ f(x)>x$ $ \forall x\in(a^4,b^4)$, so $ f(x)<x$ $ \forall x\in(a^8,b^8)$, ...

More generally, $ f(x)<x$ $ \forall x\in(a^{2^{2k\plus{}1}},b^{2^{2k\plus{}1}})$ $ \forall k\geq 0$

Let then $ c\in(a,b)$ We know that $ f(c)>c$ and so that $ f(2^nc)>2^nc$. So, if we find two integers $ n$ and $ k$ such that $ a^{2^{2k\plus{}1}}<2^nc<b^{2^{2k\plus{}1}}$, we'll get a contradiction.

And this is very easy to find :
$ a^{2^{2k\plus{}1}}<2^nc<b^{2^{2k\plus{}1}}$ $ \Leftrightarrow$ $ 2^{2k\plus{}1}\ln(a)<n\ln(2)\plus{}\ln(c)<2^{2k\plus{}1}\ln(b)$

$ \Leftrightarrow$ $ \frac{2^{2k\plus{}1}\ln(a)\minus{}\ln(c)}{\ln(2)}<n<\frac{2^{2k\plus{}1}\ln(b)\minus{}\ln(c)}{\ln(2)}$

And it is easy to find such $ n$ as soon as $ \frac{2^{2k\plus{}1}\ln(b)\minus{}\ln(c)}{\ln(2)}$ $ \minus{}\frac{2^{2k\plus{}1}\ln(a)\minus{}\ln(c)}{\ln(2)}$ $ > 1$

So, as soon as $ \frac{2^{2k\plus{}1}(\ln(b)\minus{}\ln(a))}{\ln(2)}> 1$ and we just have to take $ k$ great enough.

So, $ A$ not dense in $ (0,\plus{}\infty)$ implies a contradiction.

So the only solution is $ f(x)\equal{}x$ $ \forall x\in(0,\plus{}\infty)$

And it's immediate to check that this function matches all the requirements of the initial problem.


Find all positive integers $ n$ such that both $ n$ and $ n^2\plus{}8$ are primes.

$ x^2$ is either congruent to $ 0$ or $ 1$(mod $ 3$). If it is congruent to $ 1$(mod $ 3$), $ x^2\plus{}8$ is divisible by $ 3$, which is impossible. Hence $ x$ is divisible by $ 3$, and since $ x$ is prime, $ x\equal{}3$, $ x^2\plus{}8\equal{}17$. So $ x\equal{}3$ is the only solution.

what about the case $ x^{2}\equiv 2(mod3)$

$ x^2\equiv 0,1\pmod 3$ for all $ x\in\mathbb{Z}$.

Proof:
Working in $ \mathbb{Z}/3\mathbb{Z}$:
$ x\equal{}1,2$.
If $ x\equal{}1$, $ x^2\equal{}1$.
If $ x\equal{}2$, $ x^2\equal{}4$.
$ 4\equal{}1$, so there's no case where $ x^2\equal{}2$.

Hey guys,

Can n=11? If you plug that in, you get 11 and 129. I know 11's prime, but I'm pretty sure 129 is also prime. Therefore, x=3 is NOT the only solution. Can someone write a complete solution for this which finds ALL numbers, because 11 works also.

Thanks,

rts2007

If $ 3 \nmid n$ then $ 3 \mid n^2\plus{}8$ $ \Rightarrow 3 \mid n \Rightarrow n\equal{}3$

OH,

Thanks grn_turtle. Oh wow. I can't believe I made such a stupid mistake like that, overlooking 3.

-rts2007


If the equal sides of an isosceles right triangle are $2^x$ and the hypoteneuse is $2^{2005}$, then $x=?$

[hide]By definition, an isosceles right triangle (45-45-90) has a ratio of side lengths of: $1: 1: \sqrt{2}$.  Since the hypoteneuse has length $2^{2005}$, the equal side lengths must be:

$\frac{2^{2005}}{\sqrt2}=\frac{2^{2005}}{2^{0.5}}=2^{2005-0.5}=2^{2004.5}$

Therefore, if expressed as: $2^x$, $x$ must equal: $\boxed {2004.5}$ or $\frac{4009}{2}$[/hide]