calculus
integration
logarithms
LaTeX
calculus computations

Can someone walk me through these two problems ? I'm self studying Calculus and I've run into trouble (again).

1. Find $ f'(x)$, if
 $ f(x)\equal{}\left|\frac {x^2\left((3x \plus{} 2)^{\frac {1}{3}}\right)}{(2x \minus{} 3)^3}\right|$

2. Perform the integration:
(a) $ \int(x \minus{} 1)e^{ \minus{} x^2 \plus{} 2x}dx$
(b) $ \int\frac {1}{x}e^{ \minus{} 2\log_3(x)}dx$

Edit: with Sonnhard's help, I have changed the contents of this message so that they are in latex

hello, do you mean
1) Find $ f^'(x)$, if $ f(x)=\left|\frac{x\left((3x+2)^{\frac{1}{3}}\right)^2}{(2x-3)^3}\right|$?
Sonnhard.

hello
2) Perform the integration
    (a)  $ \int(x\minus{}1)e^{\minus{}x^2\plus{}2x}dx$
    (b)  $ \int\frac{1}{x}e^{\minus{}2\log_3(x)}dx$.
Sonnhard.

i meant
f(x)=$ \left|\frac {x^2\left((3x \plus{} 2)^{\frac {1}{3}}\right)}{(2x \minus{} 3)^3}\right|$

The other two are indeed what i intended,
thanks sonnhard.

If anyone could help, that would be great.

hello, to 2 (a) .
Put $ t=-x^2+2x$ then $ dt=-2(x-1)dx$ or ${ (x-1)}dx=-\frac{1}{2}dt$ and
your integral is $ -\frac{1}{2}\int e^t dt=-\frac{1}{2}e^t+C$
Sonnhard.

hello,in the case of 2b) we have:
$ \log_3(x) \equal{} \frac {\ln(x)}{\ln(3)}$. Put $ t \equal{} \minus{} \frac {2\ln(x)}{\ln(3)}$ and from here we have $ \minus{} \frac {\ln(3)}{2}dt \equal{} \frac {1}{x}dx$ and your integral will be
$ \minus{} \frac {\ln(3)}{2}\int e^tdt \equal{} \minus{} \frac {\ln(3)}{2}e^t \plus{} C$. Note that $ \ln(x)$ is
the natural logarithm to base $ e$ the Eulerian number.
Sonnhard.

hello, the answer to 2.a. is
$ \int(x-1)\cdot e^{-x^2+2x}dx=-\frac{1}{2}e^{-x^2+2x}+C$.
b. $ \int\frac{1}{x}e^{-2\log_3x}dx=-\frac{\ln(3)}{2}e^{-\frac{2\ln(x)}{\ln(3)}}+C=-\frac{\ln(3)}{2}x^{-\frac{\ln(3)}{2}}+C$.
Your first exercise i'm just checking.
Sonnhard.

hello, should your first problem be
$ \frac{\minus{}x(12x^2\minus{}26x\minus{}27)}{(2x\minus{}3)^{\minus{}4}(3x\plus{}2)^{\minus{}\frac{2}{3}}}\geq0$?
Sonnhard.


analytic geometry
geometry
geometric transformation
vector
geometry open

Hello,

I have a parametrized curve in R^3.
Now I want to construct two surfaces whose intersection give me that curve.
I need the curves in coordinate form.
Does anybody know something about that problem?
Would be very nice, greets, Dennis.

I might be wrong, but:
take the parametrized curve c(t)=(x(t),y(t),z(t));
then define two directions (let's say (1,0,0) and (0,1,0)) and simply translate the curve. first using the first direction, and then, returning to where the curve was..  translate the scene using the other direction.
I am defining translations like this
c(t)+u*V, where V is the direction (vector)
t sweeps in R or wherever you like and u sweeps in R.. or .. wherever you like.
Am i wrong :D?


α)Υπάρχουν ακέραιοι αριθμοί a,b,c τέτοιοι ώστε: (a-b)(b-c)(c-a)=3370370...370 (με συνολικά 2005 ψηφία) ?
β)Υπάρχουν ακέραιοι αριθμοί a.b.c τέτοιοι ώστε: (a-b)(b-c)(c-a)=370370...370 (με συνολικά 2004 ψηφία) ?

                                                                                [i]Ν.Ράπανος[/i]

Όχι ρε Αντωνακούδη.Για σένα έχω κάιτ καλύτερο:την εικασία του Γκολντμαχ!!!!!

Υποθέτω ότι  απαγορεύεται και για μένα :lol:
Πάντως αν κάποιος λύσει το πρώτο , για το δεύτερο τα πράγματα γίνονται πιο εύκολα

ΩΩΩΩΩΩΩ ΤΟΝ ΑΓΑΠΗΤΟ ΜΟΥ ΝΙΚΟ ΡΑΠΑΝΟ ΕΚ ΠΑΤΡΑΣ  :rotfl:  :rotfl:  ΧΑΙΡΕΤΩ ΕΣΕΝΑ ΚΑΙ ΤΟΝ ΣΙΛΟΥΑΝΟ  :)
ΚΑΙ ΟΣΟΝ ΑΦΟΡΑ ΤΗΝ ΑΣΚΗΣΗ ΔΕΝ ΛΕΩ ΤΙΠΟΤΑ ΑΦΟΥ ΜΟΥ ΤΗΝ ΕΧΕΙΣ ΔΩΣΕΙ

Σιλουανέ πολύ θα ήθελα να δω τη λύση σου στο β ( σε mail φυσικά!!!)

Αν δεν κάνω λάθος , πρέπει να την προσπαθήσουμε όπως ο Στέργιος είχε λύσει  τη πρώτη. Αλλά θα το κοιτάξω. Έχεις χρόνο να κάνουμε τσατ?

Θα δείτε τη λύση στο Attachment , μόνο που για το δείτε νομίζω πως πρέπει να κάνετε login .
Νίκο , τι λές για τη λύση ?

Νίκο καλησπέρα!Η λύση σου είναι άψογη!Βασιζόμαστε και οι δύο περίπου στο ίδιο πράγμα...
Πάντως, αν δεν κάνω κάποιο λάθος (γιατί είμαι και λίγο κουρασμένος αυτή τη στιγμή) δεν έχεις αποδείξει το β ερώτημα!!
Και το κακό για σένα είναι ότι μάλλον δεν κάνω λάθος!
 :D
 Χαιρετώ

Είσαι πολύ παρατηρητικός Νίκο !!!
Όντως δεν ασχολήθηκα ,επειδή είπε και ο Σιλουανός οτι το β ερώτημα είναι εύκολο .
Διαβάζω αυτή την περίοδο όπως και εσύ για τα selection tests .

Δυστυχώς ο Σιλουανός κάνει λάθος.....Θα το διαπιστώσεις αν προσπαθήσεις ν α λύσεις το β!!!!!!! :P
Ρε ποιός Νίκος είσαι γιατί μάλλον υπάρχουν πολλοί επιτυχόντες με το ίδιο όνομα...??

Πράγματι είχα κάνει λάθος........Δεν είναι και τόσο απλό.......
Το Νίκο στο επίθετο τον λένε Ρωμανίδη αν δεν κάνω λάθος..........

Υ.Γ. Νίκο (Ρωμανίδη) ήσουνα στο πρώτο μάθημα στην ΕΜΕ ? Γιατί αν ήσουν πάλι δεν γνωριστήκαμε...........

Σιλουανέ , δεν πήγα σε κανένα μάθημα της ΕΜΕ .Είναι μακριά . Είπαν ότι θα μας ταχυδρομήσουν τις σημειώσεις των
 μαθημάτων, αλλά τίποτα μέχρι τώρα .
 Εσείς που έχετε πάει , πείτε μας τις εντυπωσεις σας .
 Νίκο Ράπανε , εκείνο το Σάββατο του Αρχιμήδη συναντηθήκαμε στο δρόμο και είχαμε μιλήσει .
 Θυμάσαι ?
 Όσον αφορά το β ερώτημα  , θα το προσπαθήσω .
 Με την άσκηση Γεωμετρίας  που έχω βάλει  ασχολείται κανείς ?
 Την έχει βάλει και ο Virgil στο αγγλικό φόρουμ .

Για Ρωμανίδη!
Ναι το θυμάμαι!Με τη γεωμετρία σου θα ασχοληθώ μετά το TST............ :huh:
Δεν την έχω δει αλλά θα το έχω υπόψη μου.
Ρε αν έχεις χρόνο κοίτα το β!
Bye


number theory
real analysis
real analysis unsolved

For any $f: N\longrightarrow\mathbb{C}$ we can define the series
\[L(f,s)=\sum_{n=1}^\infty\frac{f(n)}{n^{s}}\]
those series have well known properties for example:
1. If $L(f,s)$ converges for $s_{0}$ it converges for $s>s_{0}$. (the same for absolute convergence).
2. If $f*g=\sum_{d|n}f(d)g(n/d)$ and $L(f,s)$, $L(g,s)$ converge then $L(f*g,s)$ converges and $L(f*g,s)=L(f,s)L(g,s)$.

My question is, if $f*g(1)=1$ and $f*g(n)=0$ for $n>1$, does it implies that $L(g,s)$ converges where $L(f,s)$ converges? certainly in this case $L(g,s)=L(f,s)^{-1}$.

I'm asking this because I think is a gap in Apostol's analytic number theory book, and I couldn't prove it by myself  :P .

Certainly, not: just take $1^{-s}-2^{-s}$ for the first series and $1^{-s}+2^{-s}+4^{-s}+8^{-s}+\dots$ for the second one. The first converges for all $s\in\mathbb C$ while the second converges only for $Re\,s>0$.


The following is from part of an undergrad complex analysis [practice] midterm. What do people think?
--------

Consider the functions defined by

g(z) = (e^(i*pi*(z/2))-1)/(e^(i*pi*(z/2))+1) and
h(z) = (e^(pi*(z/2))-1)/(e^(pi*(z/2))+1).

Let C represent the complex field. Show that g(z) maps the set G:={z in C|-1< Re(z)< 1} to the unit disk D(0,1) while
h(z) maps the set H:={z in C|-1<Im(z)<1} to the unit disk D(0,1).

Hence or otherwise, prove the following:

(a) Let f: D(0,1)-->C be an analytic function that satisfies f(0)=0. Suppose

|Re(f(z))|<1 for all z in the unit disk D(0,1). Prove that
|f'(0)|[u]<[/u]4/pi

(b) Let S be the set of functions defined by

S={f: H-->C| f analytic, |f|<1 on H, and f(0)=0}.

Prove that for f in the set S,

sup|f(1)|= (e^(pi/2)-1)/(e^(pi/2)+1).


calculus
limit
logarithms
complex analysis

Determine the radii of convergence of the following series:

a) $ \sum_{n\equal{}0}^{\infty}(\frac{n\plus{}1}{n\plus{}2})^{n^2}z^2$

b) $ \sum_{n\equal{}0}^{\infty}( n\plus{}3 ) ^{n^4}z^{2^n}$

c) Analyse what happens on the circle of convergence of $ \sum_{n\equal{}0}^{\infty}(\frac{n}{2n^2\plus{}3})z^{4n}$

in the first case should the exponent of z be 2n ?? :maybe:
(instead of z^2)
in that case it seems that the radius of convergence is e^-1/2....

The essence of these problems are more calculus than complex analysis. I'm also going to assume that (a) is $ \sum_{n\equal{}0}^{\infty}\left(\frac{n\plus{}1}{n\plus{}2}\right)^{n^2}z^n.$

(a) is a straightforward root test:

$ \lim_{n\to\infty}\left[\left(\frac{n\plus{}1}{n\plus{}2}\right)^{n^2}\right]^{\frac1n}\equal{}\lim_{n\to\infty}\left(1\minus{}\frac{1}{n\plus{}2}\right)^n\equal{}e^{\minus{}1}.$

So the radius of convergence is $ e.$

(b) The appropriate version of the root test will involve:

$ \lim_{n\to\infty}\left[(n\plus{}3)^{n^4}\right]^{\frac1{2^n}}\equal{}\lim_{n\to\infty}(n\plus{}4)^{n^42^{\minus{}n}}.$

The logarithm of that is $ n^42^{\minus{}n}\ln(n\plus{}4),$ which tends to zero. So the original limit tends to $ 1$ and the radius of convergence is $ 1.$

(c) The radius of convergence is $ 1.$ We have that $ \frac{n}{2n^2\plus{}3}$ is positive, decreasing to 0, and not absolutely summable by comparison to $ \sum\frac1n.$ Then the power series will diverge whenever $ z^4\equal{}1$ bud converge (by a summation by parts argument) whenever $ z^4\ne1.$ So the power series:

-- diverges when $ z\equal{}1,i,\minus{}1,\minus{}i$
-- converges conditionally when $ |z|\equal{}1, z\ne 1,i,\minus{}1,\minus{}i.$

sorry forgot to invert to get the radius of convergence.... :oops:

hi thanks so much for your help!

i have two questions: when you say the appropriate version of the root test, are you just taking the reciprocal of the power of z, so in this case $ \frac{1}{2^n}$?

also, how do you show that the other points around the unit circle converge with this (summation by parts idea)??

Theorem:

If the sequence $ (b_n)$ is eventually real, positive, and monotone nonincreasing and if $ b_n\to0$ as $ n\to \infty,$ then $ \sum_{n \equal{} 0}^{\infty}b_nz^n$ converges for all $ |z| \equal{} 1,z\ne 1.$

Preliminary comments:

We might as well assume the hypotheses of real, positive and decreasing apply for all $ n.$ Changing from that to "eventually" is a matter of ignoring or altering finitely many terms, which does not affect convergence. This result is uninteresting if $ \sum b_n < \infty;$ to get something out of it, apply it in cases where $ \sum b_n \equal{} \infty.$ This theorem contains the usual alternating series test as a special case - just let $ z \equal{} \minus{} 1.$

Summation by parts:

$ \sum_{n \equal{} 0}^Na_nb_n \equal{} A_Nb_N \plus{} \sum_{n \equal{} 0}^{N \minus{} 1}A_n(b_n \minus{} b_{n \plus{} 1})$ where $ A_n \equal{} \sum_{k \equal{} 0}^na_k.$

Check this formula for $ N \equal{} 0$ and $ N \equal{} 1$ to be sure it's right, then prove it by induction.

Proof of theorem:

Assume $ |z| \equal{} 1,z\ne 1.$ Note that $ \sum_{k \equal{} 0}^nz^k \equal{} \frac {1 \minus{} z^{n \plus{} 1}}{1 \minus{} z}.$

$ \sum_{n \equal{} 0}^Nb_nz^n \equal{} \frac {1 \minus{} z^{N \plus{} 1}}{1 \minus{} z}\cdot b_N \plus{} \sum_{n \equal{} 0}^{N \minus{} 1}\frac {1 \minus{} z^{n \plus{} 1}}{1 \minus{} z}\cdot (b_n \minus{} b_{n \plus{} 1})$

Note that $ \left|\frac {1 \minus{} z^{n \plus{} 1}}{1 \minus{} z}\right|\le \frac2{|1 \minus{} z|}$ for all $ n.$ Take the limit in the identity above as $ N\to\infty,$ using the fact that $ b_n\to0.$

$ \sum_{n \equal{} 0}^{\infty}b_nz^n \equal{} \sum_{n \equal{} 0}^{\infty}\frac {1 \minus{} z^{n \plus{} 1}}{1 \minus{} z}\cdot (b_n \minus{} b_{n \plus{} 1})$

Check the sum on the right for absolute convergence:

$ \left|\sum_{n \equal{} 0}^{\infty}\frac {1 \minus{} z^{n \plus{} 1}}{1 \minus{} z}\cdot (b_n \minus{} b_{n \plus{} 1})\right|\le\frac {2}{|1 \minus{} z|}\sum_{n \equal{} 0}^{\infty}|b_n \minus{} b_{n \plus{} 1}|$

$ \equal{} \frac {2}{|1 \minus{} z|}\sum_{n \equal{} 0}^{\infty}(b_n \minus{} b_{n \plus{} 1})$ (where we used the monotonicity of $ b_n$)

$ \equal{} \frac {2b_0}{|1 \minus{} z|}$ as a telescoping series.

By a slight alteration of the same argument, we also get the error estimate

$ \left|\sum_{n \equal{} N \plus{} 1}^{\infty}b_nz^n\right|\le\frac {2b_{N \plus{} 1}}{|1 \minus{} z|},$ which contains the usual alternating series error estimate as the $ z \equal{} \minus{} 1$ case.

Now, what if we have $ \sum_{n \equal{} 0}^{\infty}b_nz^{4n}?$ Just apply this theorem to $ w \equal{} z^4$ to get convergence when $ |z| \equal{} 1,z^4\ne 1.$


hi, my school is doing mandelbrot for the first time this year, and i was wondering how many individual tests are sent to the school?  Does it depend on how many registered teams you have?  I was hoping that even if everyone cant do the team test, everyone can do the individual, but does it cost more money to order extra individuals, or does the package come with extras so that those not doing the team can still participate?  Thanks.

uhh i just realized that this is in the wrong section, i meant to put it under other contests.  sorry.

Anyone who wants to can take the Mandelbrot. They send a few, but then the math team teacher can make photocopies, and then the top few scores are sent in.

oh alright thanks.  I wasnt sure if they would accept photocopies.

You don't send them back. You enter in the scores online, and they just trust you. I doubt anyone would lie about it though.

[color=cyan]All information about the contest should be somewhere on their website, http://www.mandelbrot.org.  Check out the problem corner -- it's a good place to be.[/color]

thanks for answering my questions.  The problem corner looks neat, i see you and simon did a lot of them.

I don't remember sending in a solution to more than one of them actually.

my bad.  Joel did pretty much every one though.

[color=cyan]You can also see a few earlier years of them, I believe.[/color]


geometry
3D geometry
sphere
analytic geometry

A variable plane passes through a fixed point $ (a,b,c)$ and cuts the co-ordinate axes at $ P,Q,R$.Then find the locus of the center$ (x,y,z)$ of the sphere passing through $ P,Q,R$ and the origin.

[hide]The plane $ {x\over p} \plus{} {y\over q} \plus{} {z\over r} \equal{} 1$ must pass through $ M(a,b,c)$, hence $ r \equal{} \frac {c}{1 \minus{} {a\over p} \minus{} {b\over q}}$.

Then $ P(p,0,0), Q(0,q,0), R\left(0,0,\frac {c}{1 \minus{} {a\over p} \minus{} {b\over q}}\right),O(0,0,0)$, hence for the coordinates of the desired center we get

$ x \equal{} {p\over 2},y \equal{} {q\over 2},z \equal{} \frac {c}{2\left(1 \minus{} {a\over p} \minus{} {b\over q}\right)}$.

Eliminating $ p,q$, we get $ z \equal{} {cxy\over 2xy \minus{} ay \minus{} bx}\iff ayz \plus{} bxz \plus{} cxy \equal{} 2xyz\iff{a\over x} \plus{} {b\over y} \plus{} {c\over z} \equal{} 2$[/hide]


trigonometry
geometry
angle bisector
trig identities
Law of Cosines

In $ \triangle{ADC}$, angle bisector $ DB$ is drawn.  If $ AB \equal{} 3$, $ AD \equal{} 6$, and $ CD \equal{} 8$, find $ BD$.

--

Find $ \tan{C}$, where $ C$ is the angle opposite side $ c$ of a triangle whose side lengths $ a$, $ b$, and $ c$ satisfy
\[ \frac{a^{3}\plus{}b^{3}\plus{}c^{3}}{a\plus{}b\plus{}c}\equal{}c ^{2}
\]

Thanks in advance!

Solution to 1

[hide]By Angle Bisector Theorem, $ \dfrac{CD}{AD}\equal{}\dfrac{BC}{AB}$, suggesting that $ BC\equal{}4$.  Now we use Stewart's Theorem to find the length of the angle bisector, which is $ 6$.[/hide]

I believe both are from Aops 2
[hide="Solution to 2"]
The given is equivalent to
$ c^2 \equal{} a^2 \plus{} b^2 \minus{} ab$
So by the Law of Cosines, $ cos C \equal{} 1/2$, $ \angle C \equal{} 60$
So $ tan C \equal{} \sqrt {3}$[/hide]


Dunno what forum specifically to post this in so I guess I'll just post here.

Does anyone here know how to get rid of headers/footers in MS Word 2007 from the 2nd and 3rd pages of a 3-page paper? Because I only want the header to appear on the first page, and not on the 2nd and 3rd pages, but whenever I tried to delete the header from the 2nd and 3rd pages, the header on all the pages disappears.

I also tried googling stuff like this for help, but none of those work. Help please? :(

This probably isn't exactly what you want but I would go about it in a more "brute force" way.  By that, I mean manually create headers and footers where you need them, through text boxes or expanding the margins.  I'm not sure if you can selectively choose the pages where the headers and footers appear.

I think you're confused about the notion of [url=http://en.wikipedia.org/wiki/Page_header]headers[/url] and footers.  Generally, they are things that go at the top of [i]every[/i] page, except possibly "important" pages (e.g., first pages).


Can anyone help me solve the following?

y*csc(xy)cot(xy) + x*csc(xy)cot(xy)(dy/dx) = 0

I know the equation is exact, but it's the procedure after that that I am tripped up with.

Thanks

Are you referring to

$ y\csc(xy)\cot(xy)\plus{}x\csc(xy)\cot(xy)\frac{dy}{dx}\equal{} 0$

or to

$ y^{\csc(xy)\cot(xy)}\plus{}x^{\csc(xy)\cot(xy)}\frac{dy}{dx}\equal{} 0$?

Then, as long as $ \csc(xy)\cot(xy)\neq 0$, we have $ \frac{dy}{dx}\equal{}\minus{}\frac{y}{x}$.


How can you go from the number 11 to 25 by only multiplying with 2 or decreasing with 3 in a minimum number of steps?

[quote="dEUS"]I had the same solution :-), have you already solved question 4 of the juniors?[/quote]

Post that one and you'll find out ;)

[quote=Peter]In fact you could just write out the pattern, but I'll try to write down a "clean" solution: (please note: this question is meant for younger pupils :D)

Consider the reverse progress: from 25 to 11; move A is dividing by 2, move B is adding 3.
And we need to prove this is impossible in 6 moves. (7 is possible: BAABABB)

Looking at the problem mod 3 shows us that the number of A moves will be odd - so 1,3 or 5.
1 is impossible, since that cannot make the number drop to or below 11.
5 is impossible, since then we only have 1 B step and we must start with B for sure so 28 is not a multiple of 32.
so there must be 3 A's and 3 B's.

The pattern begins with a B (28). if a number is even we mustn't do B cuz that will take all twi B's to get it back even, and we ain't getting a multiple of 11.4^k
==> if the i'th term is even, we must do A
(and of course if it's odd, B is the only option)

So if there were a 6-move-pattern ending in 11 it would be: 25-28-14-7-10-5-8 and 8 is not 11, contradiction.[/quote]

can you explain how does mod 3 show that the number of A moves will be odd please?


Find the sum of:

$\displaystyle S_{n}=\frac{2^2+1}{2^2-1}+\frac{3^2+1}{3^2-1}+\cdots\cdots+\frac{(n+1)^2+1}{(n+1)^2-1}$

It's very easy (my homework) but don't get the answer.

I tried different values of n on my calculator, but it still seems to diverge

[quote="Jimmy"]If the sum is from 1 to $\infty$, I'm pretty sure it diverges.[/quote]

You are right if the sum is from 1 to $\infty$ but here,  the sum is from 1 to n. ,Hint:  it is obviously a little more than n - see my first  post here...

[hide]If you do the math it may look something  like $S_n = \frac {n(n+3)(2n+3)}{2(n+1)(n+2)} $

(When you simplify  something like $n+1+1/2 - 1/(n+1)-1/(n+2)$ .. where middle terms cancle out .. [/hide]

Thanks, the problem makes much more sense now.

Take the limit of your function a(n) as n goes to infinity. You will realize that the limit is 1. So you are always adding numbers very close to 1 forever near the end of your series. Therefore, the series diverges - it's sum is infinity.[/u]

[b]
Thanks!
Always have a little trouble with sequence.[/b]


I don't know if this has been asked before but anyways.
Suppose I don't want to read a particular topic ever again. How do I remove that topic from my list of post???

... Just delete it by pressing the right button (the "x" in the right corner under the post).

You cannot delete any topics. Only a moderator or admin can delete topics. You can make your own personal list of topics by adding them to your favourite topics and then clicking on the favorites link.

I'm not talking about deleting topics. I'm just asking how to remove the topic from the 'View Your posts' list????

Maybe you didn't read my above reply. You cannot delete those topics and they will always appear in that list. If you want a personal list of your own with the topics you choose click on the Favorites link. You can add and delete there any topic by easily clicking on "Add this topic to Favorites" link.

[quote="Rushil"]I'm just asking how to remove the topic from the 'View Your posts' list????[/quote]

The reason it's called "View Your Posts" is because it is a complete and exclusive list of your posts. This means that a topic appears there if and only if you make a post in that topic, and that furthermore there are no other ways of adding or removing items from that list.

If there has not been a reply to a post you've made, then you can delete it by clicking on the X button located in the lower right-hand corner of the post. Otherwise, notify a moderator of that forum and if one is not available, try an administrator.

For future reference, "delete" as far as posts and topics are considered means to erase them. What you wish to do is to modify a list of shortcuts. Providing as much information as possible (including page names, URLs, etc.) helps us answer your question the first time around.


function
trigonometry
algebra unsolved
algebra

prove that the function $f$ wich satisfy
$f(x+1)+f(x-1)=\sqrt{2}f(x)$ is periodic .
x is reel number. :)

More general , if  $r$ is rational but [u]non integer[/u], and $\forall_{x\in R},f(x+1)+f(x-1)=2\cos (r\pi)f(x)$, then $f$ is periodic.
$r=\frac{p}{q}$;; $p=even \Longrightarrow f(x+q)=f(x)$ ;; $p=odd \Longrightarrow f(x+2q)=f(x)$
See you .. :cool:


I'm looking to learn data structure. Does anyone have any recommendations for online courses/books, etc.? Thank you for your feedback.

MIT OCW has complete video course on data structures.

Google is a good source of *code* in many languages and just about any data structure know to man.


maybe today this ineq can be posted:
if it 's banned ,i 'll be very sorry to all vietnamese mathlinkers :blush:
for a,b,c>0 then prove :

$\sum\frac{b+c}{a^2+bc}\leq\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ :ninja:  :rotfl:  :huh:  :idea:

I had a short solution , but I couldn't remember it  :huh:
  Maybe this solution is not appropriate for lower secondary schools :
 the ineq is equivalent to $ \sum \frac{(a-b)(a-c)}{c(ab+c^2)} \geq 0 $
  After that , use Vornicu-Schur ineq .

You could also prove it using an identity: For any positive reals a, b, c, we have

$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) - \left(\frac{b+c}{a^2+bc}+\frac{c+a}{b^2+ca}+\frac{a+b}{c^2+ab}\right)$
$=\frac{(c^2a^2-a^2b^2)^2+(a^2b^2-b^2c^2)^2+(b^2c^2-c^2a^2)^2}{2abc(a^2+bc)(b^2+ca)(c^2+ab)}\geq 0$,

so that

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\frac{b+c}{a^2+bc}+\frac{c+a}{b^2+ca}+\frac{a+b}{c^2+ab}$.

  darij

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63262]The folowing inequality is also true[/url]
$ \sum\frac {a^2 \plus{} bc}{b \plus{} c}\geq\ a \plus{} b \plus{} c$
Try it
Darji,I really like the way you solve some inequalties, it is amazing :!:

[quote="cuong"]
for a,b,c>0 then prove :

$\sum\frac{b+c}{a^2+bc}\leq\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$[/quote]
I found another solution
assume that $a\ge b\ge c$,then
$RHS-LHS=(\frac{1}{a^3+abc}+\frac{1}{c^3+abc}-\frac{1}{b^3+abc})(a-b)(b-c)+\frac{(a-b)^2}{a^3+abc}+\frac{(b-c)^2}{c^3+abc}\ge 0$


What is $ \frac{1}{1\plus{} \frac{1}{2\plus{} \frac{1}{3\plus{} \frac{1}{4\plus{}...}}}}$
What is $ \frac{1}{\frac{2}{\frac{3}{\frac{4}{\frac{5}{...}}}}}$

I just learned about such functions. Is there a book or field in math that covers this more in enough detail for me to learn alot about them?

[hide="2"]$ \frac{1}{n!}$, where $ n$ is the number of terms in the denominator.[/hide]

i actually think i like pie, by your solution, if n is the number of terms in the denominator, the solution is $ \dfrac{1}{(n\plus{}1)!}$

Eh... You're right. Can anybody solve 1? I've seen it before, but don't remember/can't reproduce the solution.

I know the answer is this, I had to show it once, but I'm still trying to remember the method

[url]http://www.research.att.com/~njas/sequences/A052119[/url]


geometry
geometric transformation
reflection

Does anybody has pdf document of the book <Mathematics Reflection>?
Thanks!!


Can some one explain to me how to add and multiply letter exponents?

letter exponents?? :huh: do you mean irrational exponents that are written with letters. could you please give an example

Sure.
For example:
something like 10 to the A power.

ok. well,
$(10^a)(10^{a+1})=10^{2a+1}$
you just add the exponents if it's the same base.
is that what you're looking for? i doubt it.

Pretty close. I will try and find a example from another section on the forum.

Another example:
$A^b+B^c+C^d$

This is rediculious. I can do trig, physics ,and some of Einsteins work but I dont remember how to do this  :oops:


geometry
linear algebra
matrix
inequalities
trigonometry
complex numbers
absolute value

Let $||A||=max_{i,j=1,n}{|a_{ij}|}$ and $A_1,...,A_k$ in $M_n(C)$.
  a) If $|A_1^p+...A_k^p||\leq\frac{1}{p!}$ for all natural $p$ then the $k$ matrices are nilpotent.
  b) If $||A_1^p+...+A_k^p||\leq k$ for all $p$ then $|detA_i|\leq 1$.

Easy? :( Not for me.

I can show that in a) the equality holds $A_1^p+...A_k^p=0$ for all large $p$.

My supervisor have an approach via Jordan form, but he didn't check any details.

No, I do not think it is easy, it was just an irony at the address of one of my friends, who said the problem is trivial, even if his solution was wrong. I would be glad to see your supervisor's approach.

[quote="fedja"]A hint: replace the norm by the trace and, if everything has not become clear, click
[hide="here"] Use the fact that for any complex numbers $z_1,\dots,z_N$ there exist arbitrarily large $p$ such that $|\arg (z_j^p)|<\pi/4$ for all $j$.[/hide][/quote]It is absolutely unclear for me. I don't know any relations with traces.

If Fedor want to make a hint then I don't know for whom it was suppossed. For harazi? or someone who already know a solution?

I never studied algebra at all. I try to reveal this area for myself, but I see only too "experienced" men, who play with each other in game "-I know it? and you?  -Me too!  -Ok. We are so clever!"

[quote="Myth"]
If Fedor want to make a hint then I don't know for whom it was suppossed. For harazi? or someone who already know a solution?

I never studied algebra at all. I try to reveal this area for myself, but I see only too "experienced" men, who play with each other in game "-I know it? and you?  -Me too!  -Ok. We are so clever!"[/quote]
Don't get frustrated or offended, Myth! My hints are usually meant just to give a slight push in the direction of the solution, not to reveal the entire thing. And, you are right, the unhidden part of the hint was for those who know basic linear algebra and diophantine approximations well (in this particular case, those, who can almost automatically switch from the trace to the sum of the eigenvalues and so on). I should be more careful about wording though, you caught me on that once and, seems, are catching again. As to the "I know it, and you?" or, rather, "I can do it, and you?" game, it is, in a sense, played all the time here. Every time you post a problem whose solution you know, you challenge all others to solve it and some people may choose to answer with "I have a solution based on such and such ideas, let's see if someone else can recover it from what I said or come up with an alternative approach". Of course, many, including yours truly, will choose a more "playful" way to express this and I see nothing wrong with this game as long as one is always ready to go into every detail when being asked to clarify things, which most of the time is the case. What seems really impolite is to refuse to reveal a solution when directly asked and not to give any hints for a problem that nobody but you can solve. I hope I'm not guilty of that. As to abstract algebra, it is not my strong point either... Not that I don't know basic things, but I seem just to be unable to think that way. Again, my sincere apologies if my reckless wording offended you or someone else.

As to the solution itself, there is not much left after the hint I gave. One needs to know, of course, that the trace of a matrix equals the sum of its eigenvalues and that the eigenvalues of $A^p$ are the $p$-th powers of the eigenvalues of $A$. So, if $z_1,\dots,z_N$ are all eigenvalues of all matrices $A_j$, then (for a))we have the inequality $\left|\sum z_j^p\right|\leqslant \frac 1{p!}$. Since we can choose arbitrarily large $p$ such that all $|\arg(z_j^p)|<\frac \pi 4$, we have $\sum |z_j|^p\leqslant \frac 2{p!}$ for such $p$, from where one can conclude that all $|z_j|=0$. The last step is to recall that nilpotent matrices are exactly those whose all eigenvalues are $0$. b) is similar only requires another "well-known" fact that the determinant is the product of the eigenvalues.

I do not know who considers almost any problem he can solve as obvious and plays the game...  :?  :?

[quote="fedja"]
we have the inequality $\left|\sum z_j^p\right|\leqslant \frac 1{p!}$. Since we can choose arbitrarily large $p$ such that all $|\arg(z_j^p)|<\frac \pi 4$, we have $\sum |z_j|^p\leqslant \frac 2{p!}$ for such $p$.[/quote]
Could you please explain how did you get that all $|\arg(z_j^p)|<\frac \pi 4$. $|\arg(z_j^p)|=p|\arg(z_j)|$, ... i couldn't understand how does it follow from $\left|\sum z_j^p\right|\leqslant \frac 1{p!}$

Excuse for (maybe stupid) question

It doesn't follow from that, it's true for any finite collection of complex numbers. Try to show this: given $n$ numbers $a_1,a_2,\ldots,a_n$ and $\varepsilon>0$, there is an $m$ s.t. $ma_i$ is within $\varepsilon$ from an integer for all $i$. After this, you apply this result to the arguments of our $z_i$'s, or, to be precise, if $z_i=r_i e^{i\theta_i}$, we apply it to the numbers $\theta_i$.

I can't be sure, but I believe this is what fedja meant. Anyway, we should probably wait for him :).

Yes, that is exactly what I meant. See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35154]the problem posted by Harazi[/url] for an alternative approach.  ;)

[quote="grobber"]It doesn't follow from that, it's true for any finite collection of complex numbers. Try to show this: given $n$ numbers $a_1,a_2,\ldots,a_n$ and $\varepsilon>0$, show that there is an $m$ s.t. $ma_i$ is within $\varepsilon$ from an integer for all $i$. After this, you apply this result to the arguments of our $z_i$'s, or, to be precise, if $z_i=r_i e^{i\theta_i}$, we apply it to the numbers $\theta_i$.

I can't be sure, but I believe this is what fedja meant. Anyway, we should probably wait for him :).[/quote]

How do you prove that? (the integer approximation thingy) And also, I don't that $\theta_i$ integer helps, I think that $\frac{\theta_i}{2\pi}$ integer helps. Please do correct me if I understood nothing from what grobber said  :ninja:

and, lastly, why is this [quote]we have $\sum |z_j|^p\leqslant \frac 2{p!}$ for such p, [/quote]

Sorry for so many questions  :blush:

[quote="perfect_radio"]
How do you prove that? (the integer approximation thingy) And also, I don't that $\theta_i$ integer helps, I think that $\frac{\theta_i}{2\pi}$ integer helps. Please do correct me if I understood nothing from what grobber said  :ninja:

and, lastly, why is this [quote]we have $\sum |z_j|^p\leqslant \frac 2{p!}$ for such p, [/quote]
[/quote]

I'll let someone else answer your last question, since it's not concerned with what I wrote and I'm too lazy to try to follow the messages in this thread which are not my own :).

Your right about that $\frac{\theta_i}{2\pi}$ thing. Indeed, the result must be applied to the system of reals $\left(\frac{\theta_i}{2\pi}\right)_i$, not $(\theta_i)_i$, like I wrote. I'll edit the post above.

As for the approximation result, you can do something like this:

Assume that at least one of the $a_i$'s is irrational. Otherwise, if $q$ is a common denominator of all the $a_i$'s, then $qa_i$ is an integer for every $i\in\overline{1,n}$, and we're done. Now fix an $m\in\mathbb N^*$ (not the same $m$ as in my previous post). In the "parallelepiped" $\mathcal P=[0,1)^n\subset\mathbb R^n$ consider the $m^n+1$ points $A_0=(0,0,\ldots,0),A_k=(\{ka_1\},\{ka_2\},\ldots,\{ka_n\}),\ k=\overline{1,m^n}$. Because at least one of the $a_i$'s is irrational, all the $A_k$'s are distinct. Divide $\mathcal P$ into $m^n$ smaller parallelepipeds by dividing each $[0,1)$ into $m$ equal segments and taking the product of this division. Two of our points $A_k,\ k=\overline{0,m^n}$ must lie in the same small parallelepiped, and if these two points are $A_u,A_v,\ u<v\in\overline{0,m^n}$, then it's easy to see that $A_{v-u}$ lies in the parallelepiped having $A_0$ as a corner, or, in other words, $\{(v-u)a_i\}<\frac 1m,\ \forall i\in\overline{1,n}$.

Note that this is a stronger, quantitative version of the approximation theorem: basically, it says that for every $m\in\mathbb N$ we can find $q\in\overline{1,m^n}$ such that $\{qa_i\}<\frac 1m,\ \forall i\in\overline{1,n}$.

[quote="perfect_radio"]And, lastly, why is this?[quote]we have $\sum |z_{j}|^{p}\leqslant \frac 2{p!}$ for such $p$ [/quote][/quote]

I think that I can finally answer my own question.

[color=blue]For all $z_{1}, \ldots, z_{n}$ whose arguments are $\leq \frac{\pi}4$ in absolute value, we have that $\sum_{j=1}^{n}\left| z_{j}\right| \leq \sqrt 2 \left| \sum_{j=1}^{n}z_{j}\right|$.[/color]

[i]Proof.[/i] Pick some $z_{k}$ and change its argument so that the $LHS$ is as small as possible. Using the formula $|a+b| = \sqrt{|a|^{2}+|b|^{2}+2|ab| \cos \measuredangle(a,b)}$ for $a = z_{k}$ and $b =-z_{k}+\sum_{j=1}^{n}z_{j}$, we see that what we want happens when the cosine is the smallest. But the angle between $a,b$ is in $\left[-\frac{\pi}2, \frac{\pi}2 \right]$, so the minimum is reached when the angle is the biggest in absolute value, i.e. when $\textrm{arg}(z_{k}) = \pm \frac{\pi}4$. Note that the $RHS$ remains the same after changing $z_{k}$'s argument.
Repeating this for all $k$ in $\overline{1,n}$, we see that it suffices to prove the inequality when $\textrm{arg}(z_{k}) = \pm \frac{\pi}4$.
Let $x_{1},\ldots,x_{u}$ be the ones with argument $\frac{\pi}4$ and $y_{1},\ldots,y_{v}$ be the ones with argument $-\frac{\pi}4$.
Then, $LHS = \sum_{j=1}^{u}|x_{j}|+\sum_{j=1}^{v}|y_{j}|$ and $RHS = \sqrt 2 \cdot \sqrt{\left( \sum_{j=1}^{u}|x_{j}| \right)^{2}+\left( \sum_{j=1}^{v}|y_{j}| \right)^{2}}$.
But $LHS \leq RHS$ is true by Cauchy, so the inequality is proven.

I hope that I haven't made any fatal mistake.


probability
LaTeX
calculus
calculus open

This question involves a seemingly-simple coin-flipping game, in which each player (there are only two players) flips a coin. Each player that flips heads gets 1 point. If both players flip tails, the game ends. What I am trying to determine are the odds of player 1 getting between [A_1] and [A_2] points, and player 2 getting between [B_1] and [B_2] points. I am also taking into account the odds of you getting heads/tails (ie, it's not assumed to be 50%), so O_1 (Odds of player 1 getting a point), O_2 (Odds of player 2 getting a point), O_12 (Odds of both players getting a point), and O__ (Odds of neither player getting a point) are also included.

The farthest that I've gotten, so far, are the odds of player 1 getting [A] points and player 2 getting [B] points. (Given the four O's):

[code]       A  (B + i)! * (O_1 ^ i) * (O_2 ^ (-A + B + i)) * (O_12 ^ (A - i))
O__ * Sum --------------------------------------------------------------
      i=0                 (A - i)! * (-A + B + i)! * i![/code]

Letting that be O(A, B, O_1, O_2, O_12, O__), the original problem could be gotten by:
 B_2   A_2
 Sum ( Sum ( O(A, B, O_1, O_2, O_12, O__) ))
B=B_1 A=A_1

Somewhat expanded, and semi-"factored":
[code]      B_2          A_2                     A
O__ * Sum (O_2^b * Sum ( (O_12 / O_2)^a * Sum ( nCr(B + i, A) * nCr(A, i) * (O_1 * O_2 / O_12)^i )))
     B=B_1        A=A_1                   i=0[/code]

I'm very uncomfortable with the three summations, as they make it essentially impossible to work out without a calculator or computer. I also cannot expand A_2 or B_2 to infinity (ie, Odds of player [X] getting "more than [A_1]"). My goal is to consolidate it down to one (or  the best-case none) summation, by I don't know where to start, other than expanding expanded expansions by hand and looking for patterns (which probably isn't too practical). Can you provide any suggestions or help?

(If this isn't in the appropriate forum, I apologize; I wasn't quite sure where to put it)

And for everything else, there's $\LaTeX$


Points $ a$ and $ b$ are in nonadjacent vertices in a 8 by 8 grid square. This square is then sliced in half by cutting through the line $ ab$. Find the number of ways to get from $ a$ to $ b$ in this new triangle.

How do I approach this problem? Where do I start?

It obviously depends on $ a$ and $ b$; is it assumed that the line $ ab$ does divide the square into two equal areas, and that we can only travel vertically and horizontally?


If $x \in R$ then solve this equation
$2x^{2}-3x=2x \sqrt{x^{2}-3x}+1$


What is the arithmetic mean of $ x$, $ y$ and $ z$ if $ 3x\plus{}10\equal{}9$, $ 8y\minus{}4 \equal{} 12$, and $ \frac{1}{2}z \plus{} 3 \equal{} 2$?

From the equations we get $ x\equal{}\minus{}1/3,y\equal{}2,z\equal{}\minus{}2$. Their AM is -1/9

[quote="Bugi"]From the equations we get $ x \equal{} \minus{} \frac13$, $ y \equal{} 2$, and $ z \equal{} \minus{} 2$. Their AM is $ \boxed{ \minus{} \frac19}$.[/quote]

Could anyone see the Tex?  I couldn't...

EDIT: Really?  I couldn't see the Tex...

The problem was different on ftw. IT said 3x+10=19

[hide=sol]we get $24x=-8$ $24y=48$ and $24z=-48$
adding we get $24(x+y+z)=-8$ so $\frac{x+y+z}{3}=\boxed{-\frac{1}{9}}$ note that on ftw the first equation is $3x+10=19$ and the answer is 1
[/hide]


USAMTS
AMC
AIME
USA(J)MO
USAMO
ceiling function
AMC 10

could i know how your usamts score is calculated along with your aime score for qualifying for the usamo?

i couldn't find the usamts discussion board.......

I was told you can't qualify for USAMO with just USAMTS and AIME.

It is possible, but if you qualify for AIME via USAMTS, your AMC score is considered to be 100. Thus for an upperclassman, if the index was x, you would need to get $\left\lceil \frac{x-100}{10}\right \rceil$ on the AIME to make USAMO.

[quote="SplashD"]what's this index thing, can you explain it?[/quote]
Your USAMO index is equal to your AMC score + 10 x your AIME score. This way, the highest index possible is 300, but the qualifying index is usually in the lower 200s (for example, this year it was 217.5).

ok. then what's the point of qualifying for AIME or USAMO via USAMTS..

I mean if anybody is good enough to qualify for AIME through usamts they would probably get a score of AMC anyways...and if you want to qualify for aime you need to get aroud 100 anyways..

I find this sort of ironical...

because on the usamts homepage Mr.ruscyzk(spelling)
said that it is more benifitial to qualify bothways...

Not all people go to schools were AMC contests are offered.

Actually, to qualify for AIME, you need at least 100 on AMC 12 [i]or[/i] you need to be in the top 5%, whichever is lower.  Say, for example, that the 2007 amc is mad hard and getting a 90 puts you easily in the top 5%.  Then since your USAMTS score is automatically 100, that is beneficial if the AMC is harder that year.  In contrast, if the AMC is easy, and you get a 120 effortlessly, then your AMC score will obviously be better.

The point is that you never know ahead of time how hard the AMC will be that year, so you might as well do both.

so if the top 5 is higher than 100/120 (AMC 12 and 10, respectively) and you get a 100/120 you'd still qualify?

How do you qualify for USAMO? I've heard it's a combination of your AIME and AMC score.

well basically it's by your index which is calculated by AMC+10*AIME score...

This year I think for 10th grade or under you could just qualify for it if you just had a AIME score of 8 and and another AMc score...

[quote="climskywell"]i couldn't find the usamts discussion board.......[/quote]

Er, it should be [i]right below[/i] this forum (unless you have it hidden).

[quote="Xantos C. Guin"]AMC 10 scores are used in when too many 9th and 10th graders qualify by the floor. That almost happened this year, but they let all 8+ AIME scores qualify.[/quote]

what if you didn't take AMC10

by the way, when some people quote they use those bubbly things, how do you do that?

Then your AMC 12 score is used for your index. Or the USAMTS default 100.


trigonometry
inequalities proposed
inequalities

hello, see here
http://mathworld.wolfram.com/TrigonometryAnglesPi11.html
Sonnhard.

you can find more here:
http://www.mathlinks.ro/viewtopic.php?t=46202


Let $ n\geq2,n \in N$ .Find the minimum value of $ |X|$: for every $ n$ diffrent subsets $ B_{1},B_{2},\dots,B_{n}\subset X, |B_{i}|=2$ for every $ i=1,2,\dots,n$ there exist set $ Y$ such that:
1) $ Y \subset X$
2)$ |Y|=n$
3)$ |Y \cap B_{i}|\leq1$ for every $ i=1,2,\dots,n$

let $ S=\{|X|: \ X\ check\ the\ conditions\ 1,2\ and\ 3 \}$
we can suppose that $ X=\{1,2,3,...,k\}$ (we must have $ 2k\ge n(n-1)\ge n$ because we have to choose $ n$ diffrent 2-elements subsets from $ X$).
if we take $ B_{i}=\{i,i+1\}$ for $ i=1,2,...,n$: $ \exists Y\subset X: \ |Y|=n\ and\ |Y \cap \{1,i+1\}|\leq 1$
then $ Y$ if $ Y=\{a_{1},a_{2},...,a_{n}\}$ and $ a_{1}<a_{2}<...<a_{n}$, we must have $ a_{i+1}\ge a_{i}+2$ so $ |X|=k\ge a_{n}\ge 2(n-1)+a_{1}\ge 2n-1$.
then $ min(S) \ge 2n-1$
NOW we take $ X=\{1,2,3,...,2n-1\}$ and $ B_{1}=\{2n-1,2n-2\},B_{2}=\{2n-1,2n-3\}$ and for $ i=3,4,...,n$ $ B_{i}=\{2(i-2),2(i-2)-1\}$ so $ |(B_{1}\cup B_{2})\cap Y|=|\{2n-1,2n-2,2n-3\}\ cap Y|\le 1$ and $ \forall i\in\{3,4,..,n\}: \|Y\cap B_{i}|\le 1$
and it's clair that $ Y\cap B_{i}$ for $ i=3,4,..,n$ and $ B_{1}\cap B_{2}\ cap Y$ are two by two disjoined
then $ |Y|=|Y\cap(\cup_{i=1}^{n}B_{i})|=|((B_{1}\cup B_{2})\cap Y)\cup \cup_{i=3}^{n}(B_{i}\cap Y)|\le n-1$ (impossible)
then $ 2n-1\not\in S$ and $ min(S)\ge 2n$
NOW i let you rpove that $ X=\{1,2,3,...,2n\}$ check the conditions  :roll: and we will have $ min(S)=2n$


Find all functions $f: \mathbb R \to \mathbb R$ such that for any real $x$ and $y$,
\[ f(x\plus{}y)\equal{}f(x)\plus{}f(y)\]
and
\[f(x)\equal{}x^2f\left( \frac 1x \right).\]

[quote="MJ GEO"]$ f$ from $ R$ to $ R$,$ f(x \plus{} y) \equal{} f(x) \plus{} f(y),f(x) \equal{} x^2f(\frac 1x)$[/quote]

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

So $ f(1)(x\plus{}1)^2\equal{}x^2f(1\plus{}\frac 1x)\plus{}f(x\plus{}1)$ $ \equal{}x^2f(1)\plus{}x^2f(\frac 1x)\plus{}f(x)\plus{}f(1)$ $ \equal{}x^2f(1)\plus{}2f(x)\plus{}f(1)$

So $ f(x)\equal{}f(1)x$ $ \forall x$ and it's immediate to verify back that this necessary form is sufficient.

Hence the solution : $ \boxed{f(x)\equal{}ax}$ $ \forall x$


geometry
3D geometry
tetrahedron
AoPS Books

I was recently doing a thing from South Africa when I encountered a problem I had seen before, the exact same thing, numbers and everything, in an AHSME. Mistake, or intentional? Ever seen this before?

oh, it was in one of the AoPS books, something about a tetrahedron, and given the lengths of the sides (but not which numbers go to which sides), plus one match of side-length, figure out one of the sides.


I seen severals bots talking about how being a bot what is a bot and how do you become one

Try searching. They archive things for search engines pretty much. Here are some more links:

http://www.mathlinks.ro/viewtopic.php?search_id=1205226179&t=99267
http://www.mathlinks.ro/search.php
http://www.mathlinks.ro/viewtopic.php?search_id=443874477&t=232151
http://www.mathlinks.ro/viewtopic.php?search_id=1499926424&t=194205


function
calculus
derivative
integration
real analysis
real analysis solved

I don't see the point of this problem. if f'=0 then there must be a f', which means that f is derivable, and thus always positive, since it's increasing ... maybe you wanted to ask something else alekk :?

No Valentin, f can be strictly (I will omit that from now on since in english 'increasing' means strictly increasing, which is not the case in France) increasing and its derivative has not to be strictly positive everywhere.

To the initial question : We construct a function F which is derivable on |R and such that 0 is not an isolated point in the set N = {x/ F'(x) = 0}.

Let f(x) = |x*sin(1/x)| for x =/= 0 and f(0) = 0.
Since 0  \leq  f(x)  \leq  |x| we deduce that f is continuous at x = 0.
Let F(x) =  \int {0,x} f(t)dt.
Then F is derivable on |R and F' = f  \geq  0.

Moreover, for each non-zero integer k, let x_k = 1/(k \pi ). Clearly F'(x_k) = 0, and F'(0) = f(0) = 0.
Since the sequence (x_k) tends to 0, we deduce that 0 is not an isolated point in N.
Note that N = {0}u{x_k / k in Z}, thus N is infinite.

From above, we know that F is nondecreasing on |R.
Now let 0 \leq a <b. Let n be the least subscript such that x_n < b (such a subscript does exist since (x_k) tends to 0 and b > 0).
- If a < x_n then F is increasing on ]x_n,b[ then F(a) \leq  F(x_n) < F(b).
- If x_n < a < b then F is increasing on ]a,b[, thus F(a) < F(b).
It follows that F is increasing on [0,+oo[.
Since F is odd, F is then increasing on |R.

Pierre.

Thanks a lot Pierre ;-)
Here is another question much more difficult : does there exist a strictly increasing continuous function on [a;b] so that the set
E={x | f'(x)=0} is dense on [a;b] ...

any idea ????

[quote="pbornsztein"]No Valentin, f can be strictly (I will omit that from now on since in english 'increasing' means strictly increasing, which is not the case in France) increasing and its derivative has not to be strictly positive everywhere.
[...]
Pierre.[/quote]
ah yes, I rushed, you are right Pierre :)

[quote="alekk"]does there exist a strictly increasing continuous function on [a;b] so that f'(x) has infinitely many zeroes?[/quote]
we can construct an easier example... (i mean it is easy to think out this example).
consider sin x on [-\pi/2,\pi/2].
its derivative is 0 at -\pi/2 and \pi/2.
now we join the part of curves. it looks like this
[code]
         __
      __|
   __|
__|[/code]
every
[code]
  _
_|
[/code]
is sin x on [-\pi/2, \pi/2].
now we get an increasing continuous function (for example, defined on [1,\infty) whose derivative is 0 at infinite many points.
we need an upbound. we need only to make the height of each part of the curve smaller and smaller. the original sin x from -\pi/2 to \pi/2 has a height of 2, so next we join a curve stretched to height 1, (like 1/2 sin x), the third has a height of 1/2 (1/4 sin x), ....
so f(x) tends to 2+1+1/2+1/4+...=4 as x tends to \infty.

let g(x)=f(1/x), for x \in (0,1]. and g(0) = 4.
we get a decreasing continuous function on a closed interval ([0,1] here) with 0 derivatives at infinite many points.

and i like to say, again, this problem is exactly an exercise in an analysis textbook....  :D

But it's very difficult to find an example such that {x | f'(x)=0} is dense on [a;b]... i wonder if there exists such a function.. :D

I think I also asked this question and noone answered

I think you asked for a function which isn't strictly increasing, but for which the set of x s.t. f'(x)>0 is dense. Am I right? Are the 2 problems equivalent?

The derivative of f, f', should be primitivable (of course!), but not Riemann integrable. I know such functions exist, but I don't know any examples.

There exist functions f : |R -> |R for which f' exists everywehre and such that, for each a< b, f is not monotonous on [a,b]. For such a function, it is clear that each of the sets P = {x / f'(x) > 0}, N = {x / f'(x) < 0} and Z = {x / f'(x) = 0} is dense on |R (the third one because, f' has the intermediate values property).
Such functions have been constructed by Denjoy (first half of 20th century), and there are others papers on them since then (but, this is not so easy...).

Thus, the two problems are not equivalent, and Alekk's and Harazi's question remain open since the functions above do not answer their demand. Have to look for something else....

Pierre.

Can someone please show me an example of a non-integrable derivative? (When I say "integrable" I mean "Riemann-integrable"; it should be defined on a compact interval) I looked on the Internet and I found some non-integrable differentiable functions, but the derivatives weren't bounded, so of course the functions weren't integrable.

I'd like to see a "clean" examle of a function f:[a,b]->R, bounded, which has a primitive, but isn't Riemann-integrable.

Maybe I'm too demanding :D.

Yes, there exists an increasing function f on [0,1], which derivative vanishes on the dense subset (but this subset must have cardinality of continuum). Let {r_n} (n=1,2,...) be enumeration of rationals from [0,1]. Take g=sum(n=1..infinity) {1/n^2}*(x-r_n)^{1/3}. Then clearly g increases on [0,1] and is continious. The formal derivative
sum({1/3n^2}*(x-r_n)^{-2/3}) is finite on some set E. We claim that g' exists on E and g'=(infinity) on [0,1]/E. We omit the proof, it is not hard to verify by definition. Then the function g^{-1} strictly increases and its derivative vanishes in each r_n.


Let be a function $ f: \mathbb{R}\to\mathbb{R}$ .
Prove that :
$ a)$ If $ g: \mathbb{R}\to\mathbb{R}$ is strictly increasing $ \Rightarrow$ $ f\plus{}g$ is increasing , then $ f$ is increasing.
$ b)$ If $ g: \mathbb{R}\to\mathbb{R}$ is strictly increasing $ \Rightarrow$ $ f\cdot g$(their product) is increasing , then $ f$ is increasing.

Suppose $ a < b$, $ f(a) > f(b)$.

For part a), define $ g$ so that $ f(a) < g(a) < \frac{f(a)\plus{}f(b)}{2} < g(b) < f(b)$. This gives a contradiction of $ f\plus{}g$ increasing.

For part b), suppose $ f(b) < 0$. Then $ g(b) < 0$ and $ g(a) < 0$ gives a contradiction of $ f \cdot g$ increasing. So suppose $ f(a), f(b)$ are both positive. Pick $ g$ to have all positive values. Now there exists a function $ g^\prime$ such that $ e^{g^\prime(x)} \equal{} g(x)$, and a function $ f^\prime$ such that at $ x \equal{} a,b$, $ e^{f^\prime(x)} \equal{} f(x)$. Proceed with the proof of part a) for $ f^\prime, g^\prime$ to get the contradiction. (Actually, I'm kind of suspicious of this argument...)

I think you misunderstood the text of the problem . Maybe I was a little bit ambiguous . I will reword it :

$ \text{For part a)}$ .We know that the functions $ f,g: \mathbb{R}\to\mathbb{R}$ . We know also that $ g$ is strictly $ \text{increasing}$ and $ f\plus{}g$ is $ \text{increasing}$ . [b]We have to prove that[/b]  $ f$ is $ \text{increasing}$

$ \text{For partb)}$. We know that $ f,g: \mathbb{R}\to\mathbb{R}$ , $ g$ is strictly $ \text{increasing}$ and $ fg$ is $ \text{increasing}$ . [b]We have to prove that[/b] $ f$ is $ \text{increasing}$.

The problem seems false as stated.

For a), take $ f(x) \equal{} \minus{}x, g(x) \equal{} 2x$.  Similarly, for b) take $ f(x) \equal{} e^{\minus{}x}, g(x) \equal{} e^{2x}$.

Er, the correction you gave is false... and the one that you have in the original post is a decent problem. :maybe: Actually the correction is just a stronger version of the one I solved, too strong apparently.

Where did the problem come from? It sounds like you might have misinterpreted it.

I only considered a), and I think that the poblem might be correct.
If ($ g$ is strictly increasing function) is suficient condition for $ f\plus{}g$ to be increasing then $ f$ has to be increasing, becouse, if $ f$ is strictly decreasing then for $ f$ we can choose function which decreases faster then $ g$ increases and it would be counter example for:
$ g$ is strictly increasing function $ \Rightarrow$ $ f\plus{}g$ is increasing function.

I will correct the problem now . There should be "for every $ g: \mathbb{R}\to\mathbb{R}$ " in the text .

$ \text{For part a)}$ .We know that the functions $ f,g: \mathbb{R}\to\mathbb{R}$ . We know also that  $ \forall g$ is strictly $ \text{increasing}$ and $ f \plus{} g$ is $ \text{increasing}$ . [b]We have to prove that[/b]  $ f$ is $ \text{increasing}$

$ \text{For partb)}$. We know that $ f,g: \mathbb{R}\to\mathbb{R}$ , $ \forall g$ is strictly $ \text{increasing}$ and $ fg$ is $ \text{increasing}$ . . [b]We have to prove that[/b] $ f$ is $ \text{increasing}$.


Find the functions $f$ defined on $\mathbb{R}_{+}$ , such that $f^{2}(x+y)-f^{2}(x-y)=4f(x)f(y)$

$g(x)=f(x)^{2}$ for all $x\in\mathbb{R}_{+}$, and you are [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=131875]here[/url].

[hide]Suppose we let $y=0$, then we have

$0=4f(x)f(0)$
$0=f(x)f(0)$

So either $f(x)=0$ for all $x$, or $f(0)=0$. If the second case is true, let $x=y$

$f^{2}(2x)=4f(x)f(x)$
$f(2x)=2f(x)$
Thus $f(x)=2x$

So either $f(x)=0$ or $f(x)=2x$[/hide]

[quote="gopherhole112"]Suppose we let $y=0$[/quote]
No, $f$ is defined on $\mathbb{R}_{+}$.


if x is odd, which implies $x \equiv 1 mod 2$

Then, why is $x^2 \equiv 1 mod8$ ??
I read this from a book, I just need an explaination plz.

because if we consider $x \equiv 1,3,-3,-1 \mod 8$ and $1^2 \equiv 1 \mod 8, 3^2 \equiv 1 \mod 8$.
so It become more obvious. ;)

Pretty much, consider all cases of odd numbers (mod 8), and they have residue 1.

Let $x=2k+1$ for $k\in Z$ then

$x^2\equiv (2k+1)^2\equiv 4k(k+1)+1 \mod 8$

But $k(k+1)$ is always even . So $4k(k+1)+1\equiv 0+1\equiv 1 \mod 8$


[quote="dgreenb801"]Show that if $ a,b,c > 0$ and $ abc \equal{} 1$, then
$ \sum \frac {2}{(a \plus{} 1)^2 \plus{} b^2 \plus{} 1} \le 1$
(This is not my own by the way, the one I posted earlier was wrong, this is a save).[/quote]
$ \sum \frac {2}{(a \plus{} 1)^2 \plus{} b^2 \plus{} 1} \equal{} \sum \frac {2}{a^2 \plus{} 2a \plus{} b^2 \plus{} 2} \le \sum \frac {1}{ab\plus{}a\plus{}1} \equal{} 1$


How do you center the contents of a single page in a $\LaTeX$ document?

I dont know how to specify a certain page, but you can use \begin{center} and \end{center} to signify what you want centered.

Sorry, I wasn't very specific.  I meant how do you center it [i]vertically[/i].

Try adding \vspace*{\fill} to the beginning and end of the page.

Awesome!...That's exactly what I was trying to do.

Thanks.

Or, if you want to type less, I think you could also use \vfill instead...

Yes, that might work too.  The LaTeX manual by Lamport says that spaces produced by \vfill disappear at the beginning or end of a page.  \vfill is equivalent to a blank line followed by \vspace{\fill}.  To be sure, I recommended \vspace*{\fill} instead.

Yeah, Mr. Patrick suggested \vfill but it didn't work at all for me.  I might have done it wrong or something.


A ball is shot from the lower left part of the
table along a path of 45 degrees, as shown.
After contact with a side, it continues along a
path that is a reflection of the path prior to
contact. The line of reflection is the line
perpendicular to the side of the table the ball
hit, at the point of contact. The first point of
contact is labeled $ A$. After initially shot, how
many times will the ball touch a side of the
table before it reaches a corner of the table?

[asy]unitsize(0.5inch);
draw((0,0)--(6,0)--(6,8)--(0,8)--cycle);
draw((0,1)--(6,1));
draw((0,2)--(6,2));
draw((0,3)--(6,3));
draw((0,4)--(6,4));
draw((0,5)--(6,5));
draw((0,6)--(6,6));
draw((0,7)--(6,7));
draw((1,0)--(1,8));
draw((2,0)--(2,8));
draw((3,0)--(3,8));
draw((4,0)--(4,8));
draw((5,0)--(5,8));
dot((1,1));
draw((1,1)--(6,6),Arrow);
label("$A$",(6,6),E);[/asy]


xy + yz + xz - 2(xyz) <= 7/27

whr x + y + z = 1 and x,y,z>0

Ok, let us have $ (x,y,z)\equal{}(\frac{1}{3}\plus{}a,\frac{1}{3}\plus{}b,\frac{1}{3}\plus{}c)$, so we have $ a\plus{}b\plus{}c\equal{}0$
Now after some computations:
$ xy\plus{}yz\plus{}zx\minus{}2xyz\equal{}\frac{1}{3}\plus{}\frac{2}{3}(a\plus{}b\plus{}c)\plus{}(ab\plus{}bc\plus{}ca)\minus{}2*(\frac{1}{27}\plus{}\frac{1}{9}(a\plus{}b\plus{}c)\plus{}\frac{1}{3}(ab\plus{}bc\plus{}ca))\equal{}\frac{7}{27}\plus{}\frac{1}{3}(ab\plus{}bc\plus{}ca)\leq \frac{7}{27}\plus{}\frac{1}{3}*(\frac{1}{3}(x\plus{}y\plus{}z)^{2})\equal{}\frac{7}{27}$
Q.E.D. :wink:


limit
function
integration
calculus
derivative
calculus computations

If the mean value theorem is expressed in the form

$F(x)=F(0)+xF'(\theta x)$

where $\theta$ depends on $x$ , $0<\theta<1$ , find $\lim_{x\rightarrow 0}\theta$ ,assuming $F''(0)\neq 0$ .

We have that $F(x)=a_{0}+a_{1}x+a_{2}x^{2}+o(x^{2}),$ with the extra provision that $a_{2}\ne0.$ We can also express $F'(x)=a_{1}+2a_{2}x+o(x).$

The equation $F(x)=F(0)+xF'(\theta x)$ becomes

$a_{0}+a_{1}x+a_{2}x^{2}+o(x^{2})=a_{0}+a_{1}x+2a_{2}\theta x+o(x^{2}).$

We can then rearrange that to

$a_{2}(1-2\theta)x^{2}=o(x^{2}).$ Now divide by $x^{2}:$

$a_{2}(1-2\theta)=o(1).$

(Note that $\theta=\theta(x)$ is a function of $x.$)

Take the limit as $x\to 0.$ We have that

$a_{2}(1-2\lim_{x\to0}\theta(x))=0$

Since $a_{2}\ne 0,$ we have $\lim_{x\to0}\theta(x)=\frac12,$

This does critically depend on $F''(0)\ne 0.$

[quote="Kent Merryfield"]We have that $F(x)=a_{0}+a_{1}x+a_{2}x^{2}+o(x^{2})$[/quote]

This is true if F is analytic at 0, but does the same result hold if F is once or twice differentiable but not analytic?

$g$ is differentiable at zero iff $g(x)=g(0)+xg'(0)+o(x).$ All that is is a rearrangement of $\lim_{x\to0}\frac{g(x)-g(0)}x=g'(0).$

So that gives us that $f'(x)=a_{1}+2a_{2}x+o(x).$

(There is a tacit assumption here that $f'(x)$ exists on an interval. It looks like I also need to assume that $f'(x)$ is integrable on an interval to do what I'm going to do next.)

To get the expansion we want for $f(x)$ we integrate this:

$f(x)=a_{0}+\int_{0}^{x}a_{1}+2a_{2}t+o(t)\,dt$

Now, can we claim that $\int_{0}^{x}o(t)\,dt=o(x^{2})?$

Consider $h(t)$ with the property that for any $\epsilon>0,$ there exists $\delta>0$ such that is $|t|<\delta$ then $|h(t)|<\epsilon |t|.$ That's what we mean by $h(t)=o(t).$

Suppose $|x|<\delta.$ Then,

$\left|\int_{0}^{x}h(t)\,dt\right|\le\epsilon\int_{0}^{|x|}\epsilon|t|\,dt<\epsilon x^{2}.$

So we have the expansion I used, based only on the function being continuously differentiable and having a second derivative at $0.$


Let $ a,b,c\geq 0$, no two of which are zero.Prove that:
$ \frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\geq\frac{a}{a^2\plus{}bc}\plus{}\frac{b}{b^2\plus{}ca}\plus{}\frac{c}{c^2\plus{}ab}$.

No one can solve?   :(  :?:

[hide="to arqady"]arqady, please post your's or Vasc's solution.  [/hide]

Immediately follows from [b]Vornicu-Schur[/b] inequality,as it can be rewrited as:
$ \sum_{a,b,c}\frac {(a \minus{} b)(a \minus{} c)}{(b \plus{} c)(a^2 \plus{} bc)}\geq 0$

Let a=0 , b>0, c>0  .Please try the inequality by this case .1/(b+c)>= 1/b,is it possible, e.g . when b=c = 1/2 ?


geometry
3D geometry
prism
USAMTS
modular arithmetic

$Question 1$
Take a triangular prism. Its cross-section could be a triangle of any shape. The prism is of infinite length.

Can you cut this prism With a plane such that the resulting intersection is an Equilateral Triangle?

$Question 2$
There is an integer whose first digit (the most significant digit) is $3$. If you take this $3$ from the beginning of the number and put it at the end, you get another integer that is $50\%$ more than the original number.

What are these numbers?

EDIT::: Please do not answer $\#2$ as I am told that this is an open question. Apologies for the same.

Number two is really similar to #1 of the USAMTS, could you wait till round 1 is over to post this?

[hide="2"]
The initial number is $3\cdot10^{k}+a$, where $a<10^{k}$

$1.5 \cdot (3\cdot10^{k}+a) = 10a+3 \Leftrightarrow$
$3\cdot (3\cdot10^{k}+a) = 2\cdot (10a+3) \Leftrightarrow$
$3\cdot 3\cdot10^{k}+3\cdot a = 20a+3\cdot 2 \Leftrightarrow$
$\boxed{17a = 3\cdot (3\cdot10^{k}-2)}$

So $17 \mid 3\cdot10^{k}-2 \Leftrightarrow$
$3\cdot10^{k}\equiv 2 \pmod{17}\ \ (1)$

At this point we notice that 10 is a primitive root modulo 17, and also $3 \equiv 2 \cdot 10 \pmod{17}$

So we have $(1) \Leftrightarrow$
$2 \cdot 10 \cdot 10^{k}\equiv 2 \pmod{17}\Leftrightarrow$
$10^{k+1}\equiv 1 \equiv 10^{16}\pmod{17}\Leftrightarrow$
$16 \mid k+1 \Leftrightarrow$

$k \equiv 15 \pmod{16}$

So for every $k \equiv 15 \pmod{16}$ the number $a=\frac{3\cdot (3\cdot10^{k}-2)}{17}= \frac{9\cdot10^{k}-6}{17}$ is an integer less than $10^{k}$, so we have a solution[/hide]


$ (\forall)n\in \mathbb{N}\ ,\ a,b\in \mathbb{R}_\plus{}$ prove that :

$ (a\plus{}b)^n(a^n\plus{}b^n)\le 2^n(a^{2n}\plus{}b^{2n})$

Idea.
[hide]
Suppose that $ a$ and $ b$ be positive reals, and let $ p$ and $ q$ be both positive or both negative reals.

Then $ (a^p \minus{} b^p)(a^q \minus{} b^q)\ge 0$ (because the functions $ x^p$ and $ x^q$ are both increasing or both decreasing in $ \mathbb{R}^ \plus{}$).
Then $ a^{p \plus{} q} \plus{} b^{p \plus{} q}\ge a^pb^q \plus{} a^qb^p$, then
\[ a^{p \plus{} q} \plus{} b^{p \plus{} q}\ge\dfrac{(a^p \plus{} b^p)(a^q \plus{} b^q)}{2}
\]
Appliying this inequality many times one arrives at the desired inequality.

Note: I know this is Chebyshev at the begining and a sort of power mean at the end when $ p \equal{} q$, but i like the facts that these ideas are so elemental that one can mention them just like that.
[/hide]

:)

Cauchy-Schwarz,
$ a^{2n} \plus{} b^{2n} \geq \frac {1}{2}(a^n \plus{} b^n)^2$
Holder
$ (a^n \plus{} b^n)(1 \plus{} 1)^{n \minus{} 1}\geq(a \plus{} b)^n$
then,
$ 2^n(a^{2n} \plus{} b^{2n}) \geq 2^{n \minus{} 1}(a^n \plus{} b^n)(a^n \plus{} b^n)\geq(a \plus{} b)^n(a^n \plus{} b^n)$


calculus
integration
real analysis
advanced fields
advanced fields unsolved

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

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

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

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

Thank you very much!!!

Hi mathmean,

I just read this through quite quickly but to me seems to boil down to the integral
$ I\equal{}\int_\mathbb{R} \frac{1}{|x|^r}dx$
which we know diverges for all $ r$, but for a neighborhood around zero,
$ I$ converges if $ r<1$ which is satisfied if $ 0<\beta<\alpha$ (do a
Maclaurin expansion of $ S$ to see this). At infinity though something
else is needed (a property of $ f$ maybe?) and probably a statement about the integral
that describes $ S$. Perhaps you can get a minus sign or something from
it and continue.

Just some thoughts.

Hi, Hormander,
thank you very much!
You are totally right, this problem is solved in that way.
I have posted a new question in the forum:  Calculus - Real Analysis » Unsolved and Proposed Problems.
Under this title: A technical question, about integrals.
Could you read it for me, thanks for your help :)


Χαιρετώ ολους τους mathlinkers.Σκέφτηκα μια ασκηση και θα ηθελα να σας την προτεινω ;)  .Η ασκηση ειναι η παρακατώ:         Aν x,y,k φυσικοι αριθμοι να βρεθουν οι λυσεις ωστε το κλασμα (x^k+y^k)/(x^k-y^k) να ειναι φυσικος.Περιμενω την λυση σας.

Να βοηθήσω λιγο.Αν θεσουμε οπου x+y=a και x-y=b τοτε για κ=2 δεν εχει λυσεις.

Έστω (x^k+y^k)/(x^k-y^k)=n και gcd(x,y)=d, x=da, y=db, gcd(a,b)=1.  Τότε (a^k)/(b^k)=(n+1)/(n-1) και έχουμε τις εξής 2 περιπτώσεις:

i) n=2m, m>=1. Τότε gcd(n+1,n-1)=1 οπότε a^k=2m+1 και b^k=2m-1 και τελικά k=1, a=2m+1, b=2m-1.
ii) n=2m+1, m>=0. Tότε (a^k)/(b^k)=(m+1)/m άρα a^k=m+1 και b^k=m, οπότε k=1, a=m+1, b=m.

Οι λύσεις λοιπόν είναι (x,y,k)=(d(2m+1), d(2m-1), 1) και (x,y,k)=(d(m+1), dm, 1)    ;)


analytic geometry
geometry
perimeter
circumcircle
cyclic quadrilateral

$\text{Overview of Circles}$

Everyone knows what a circle is, it's a perfectly round shape in which every point on the circle is equidistant from the center of the circle.

The general equation of a circle is $(x-h)^2+(y-p)^2=r^2$, in which $r$ is the radius and the point $(h,p)$ is the center of the circle on a coordinate plane. If you see an equation in this form, you can automatically assume it's a circle. Also, to get to this form, you can simply complete the square. (note that you must be given $x^2$ and $y^2$ in the equation)

A circle has exactlly $360$ degrees, or in other words $2\pi$ radians. You can convert between degrees and radians through the fact that $\pi$ radians$=180$ degrees.

The general formula for the area of a circle is $\pi \cdot r^2$ and the circumference (perimeter) of the circle is $2 \pi r$. However, the area of the sectors of a circle can also be determined.

The general formula of the sector is $\frac{\theta}{360}\pi \cdot r^2$, which makes sense since we're only looking for the area of the part of the circle which is cut off by the angle $\theta$. Simlarily, the arc length of the circle can also be determined by $\frac{\theta}{360} 2\pi \cdot r$

$\text{More advanced Theorems}$

The first thing we must note about a circle is that it is perfectly cyclic. The power of point theorem shows this. If we have chords $AB$ and $CD$ that intersect each other at point $P$, we know that $AP \cdot PB=CP \cdot PD$. Which is the first rule of the power of point theorem.

Another thing we should know are the line segments that have to do with circles. A tangent line is a line which touches the circle at only 1 point, and a secant is a segment that touches the circle at two points. The power of point theorem states two things regarding these segments.

1) If two tangent lines $AB$ and $AC$ begin from the same point $A$, then $AB=AC$
2) If the tangent line $AB$ and the secant $AE$ begin from the same point $A$ and $AE$ intersects the outside of the circle at point $D$, then $AB^2=AE \cdot AD$

All three of these rules are called the power of point theorem.

$\text{Circles and 2-D Geometry Shapes}$

Circles, triangles, and polygons are all connected in various ways. Firstly, we will talk about circles and triangles.

Every triangle can be inscribed inside a circle (the circumcircle), which associates with the fact that a triangle has $180$ degrees. From this, we can also assume that a $90$ degree triangle/right triangle inscribed inside a circle has its hypotenuse as the diameter of the circle. How do we know this? Well, there are many ways we can find the degree measurements of certain arcs. If two radii of the circle are apart at $\angle{\theta}$, then the measure of the arc degree is $\theta$. If two line segments begin at the same point on the circle itself and are apart at angle $\theta$, then the arc degree is $2 \cdot \theta$. If two chords intersect each other and create the two arc degree measurements of $\theta$ and $\alpha$, their verticle angle measurement is $\frac{\theta+\alpha}{2}$. Thus, from these degree values, we are able to see how and why a right triangle has its hypotenuse as the diameter of the circle.

Now going along with circumcircles, we can see how the circumradii can be formed from the vertices of the triangle and touch at the circumcenter. The circumradius can be found from $R=\frac{abc}{4A}$, where $A$ is the area of the circle and $a,b,c$ are the sidelengths. Manipulation of the formula can also give you the area of the triangle itself.

Now just as circles can be formed outside of a triangle, they can also be formed inside them as well, which is called the incircle. The incircle's formula can be given as $rs=A$, where $s$ is the semiperimeter of the circumtriangle and $A$ is its area.

Excircles are the three circles which are formed by the tangent lines of the extentions of the vertices of the triangle. The sum of the exradii is given by $4R-r$

Note that from circles, many great formulas of triangles are derived... (Ceva, Stewart's,etc)

Some quadrilaterals can also be inscribed in quadrilaterals. These are called cyclic quadrilaterals. We are not going to go in depth with cyclic quadrilaterals because our general focus is on circles. But cyclic quadrilaterals are those that can be inscribed in a circle, thus having the sum of the angles of two adjacent vertices equal to $180$ degrees. The diagonals of the quadrilateral must obey the power of point theorem also in order for it to be considered cyclic. For a cyclic quadrilateral, the area is $A=\sqrt{(s-a)(s-b)(s-c)(s-d)}$, where $a,b,c,d$ are the sidelengths. Ptolemy's theorem is also very useful for cyclic quadrilaterals. In a cyclic quadrilateral $ABCD$, $AC \cdot BD=AB \cdot CD+BD \cdot AC$

More pictures

Excircles from mathworld

[img]http://mathworld.wolfram.com/images/eps-gif/FeuerbachTriangle_650.gif[/img]


Triangle $ABC$ is isosceles with base $AC$. Points $P$ and $Q$ are respectively on $CB$ and $AB$ such that $AC=AP=PQ=QB$. Find the measure of angle $B$.

[hide]
in degrees:

(using equivalent angles from the isosceles triangles, working from the base angles up to the vertex angle)
$\angle{PAC}= 180-(2\angle{C})$
$\angle{BAP}= \angle{C}-(\angle{PAC}) = 3\angle{C}-180$
$\angle{BQP}= 180-\angle{BAP}= 360-3\angle{C}$

$\angle{B}= \frac{180-\angle{BQP}}{2}= \frac{3\angle{C}-180}{2}$

in the large isosceles triangle
$\angle{B}+2\angle{C}= 180$

now we have a system that we can solve for $\angle{B}$

$\angle{B}= \frac{3\frac{180-\angle{B}}{2}-180}{2}$

$4\angle{B}= 3(180-\angle{B})-360$

$7\angle{B}= 180$

$\angle{B}= \frac{180}{7}$
[/hide]


Hi guys, it's my first time come to this forum...
I am looking for somebody to help me to solve this question:

let A be an m x n matrix, where m < n. Then， (the transpose of A)  X  A is an n X n matrix. Show that det((transpose of A) X A)= 0  m is row and n is column

please help me ,cheers...

I think this works: We add m-n rows at he bottom of A and call the new matrix B. Then
(trans A) x A=(trans B) x B
det ((trans. A) x A)=det ((trans. B) x B)=det(trans. B) x det(B)=0 since determinant is multiplicative.
QED

http://www.mathlinks.ro/Forum/viewtopic.php?t=84676 why do you post the same problem twice?


S29

An invertible number is one which can be read upside down on a standard calculator. For example, 6592 can be read upside down as 2659 on a calculator. The first invertible number is $ 1$, the 10th invertible number is $ 15$ and the 15th invertible number is $ 21$. What is the 2007th invertible number?

The usable digits are 0, 1, 2, 5, 6, 8, 9, so this maps into base seven: 0, 1, 2, 3, 4, 5, 6.

The 2007th invertable number reads as 2007 in base 7 (5565) mapped into our usable digits (8898).

So, the 2007th inveratble number is 8898 (in base 10).

Similarly, if you want to exclude numbers that, when inverted, begin with a 0 when flipped upside down, you have to remove multiples of 7, which can be done by calculating 2007 + floor(2007/7) and writing it in base 7 (6454), and mapping it into our usable digits (9686).

Nice observation.. One thing, isn't it 2007 - floor (2007/7) and not plus?


$ a*b \equal{} 4*c$
$ a*c \equal{} 6*b$
$ c*b \equal{} 8*a$
$ a^2 \plus{} b^2 \plus{} c^2 \equal{} ?$

$ a\plus{}b\equal{}2*c\minus{}1$
$ a\plus{}c\equal{}15$
$ c\plus{}b\equal{}a\plus{}14$
$ a\plus{}b\minus{}c\equal{}?$

[hide="Hint"]
Solve with substitution but stop simplifying when you see a term you recognize.
[/hide]

[hide]By rearranging the terms in the first equation, we get that $ c\equal{}ab/4$
By plugging this in, we get that $ a(ab/4)\equal{}6b$
$ a^2b/4\equal{}6b$
$ a^2b\equal{}24b$
$ a^2\equal{}24$

By using the same concept, we find that $ b^2\equal{}32$ and $ c^2\equal{}48$
So now we see that $ a^2\plus{}b^2\plus{}c^2\equal{}24\plus{}32\plus{}48\equal{}104$[/hide]

[hide]In the second problem, we can also rearrange the terms in the actual question to find that
$ a\plus{}b\minus{}c\equal{}(2c\minus{}1)\minus{}c$
Thus, this reduces to $ c\minus{}1$
By simplifying the problem, you allow yourself to only have to isolate one variable.
By rearranging to find the variable $ a$ in terms of $ c$, you can find the numeric value of $ c$
$ a\plus{}c\equal{}15$
$ c\equal{}15\minus{}a$

$ c\plus{}b\equal{}a\plus{}14$
$ (15\minus{}a)\plus{}b\equal{}a\plus{}14$
$ 15\plus{}b\equal{}2a\plus{}14$
$ b\equal{}2a\minus{}1$

$ a\plus{}b\equal{}2c\minus{}1$
$ a\plus{}(2a\minus{}1)\equal{}2c\minus{}1$
$ 3a\minus{}1\equal{}2c\minus{}1$
$ 3a\equal{}2c$
$ a\equal{}2c/3$

$ a\plus{}c\equal{}15$
$ (2c/3)\plus{}c\equal{}15$
$ 5c/3\equal{}15$
$ c\equal{}9$

So the value of $ c\minus{}1\equal{}9\minus{}1\equal{}8$[/hide]

[b]1)[/b] You should state whether $ a,b,c$ can be $ 0$.
For $ a,b,c \neq 0$, multiply three equations we get $ abc\equal{}4 \times 6 \times 8$
$ 4 \times 6 \times 8\equal{}abc\equal{}4c^2 \Leftrightarrow c^2\equal{}48$, similarly $ b^2\equal{}32$, $ a^2\equal{}24$,
So $ a^2\plus{}b^2\plus{}c^2\equal{}48\plus{}32\plus{}24\equal{}104$

[b]2)[/b] $ (a\plus{}b)\plus{}(a\plus{}c)\minus{}(c\plus{}b)\equal{}(2c\minus{}1)\plus{}15\minus{}(a\plus{}14) \Leftrightarrow 3a\equal{}2c$
So $ a\plus{}b\equal{}(3a)\minus{}1$ gives $ b\equal{}2a\minus{}1$, $ a\plus{}c\equal{}15 \Leftrightarrow \frac{5a}{2}\equal{}15 \Leftrightarrow a\equal{}6$
, $ c\equal{}\frac{3 \times 6}{2}\equal{}9$, $ b\equal{}2 \times 6\minus{}1\equal{}11$

So $ a\plus{}b\minus{}c\equal{}6\plus{}11\minus{}9\equal{}8$


[u][b]The author of this posting is : iura[/b][/u]
____________________________________________________________________

Very easy problem:

Given a unit circle, and a set $S$ consisting of $2004$ disjoint arcs of this circle. Assume that the set $S$ has the following property: If three points $A, B, M$ on our circle belong to the set $S$, and the point $M$ is the midpoint of one of the two arcs $AB$ (the smaller arc or the larger arc), then there is one of the arcs from the set $S$ which contains all three points $A, B, M$.

Find the biggest possible sum of lengths of arcs in $S$.

As I remember, Iura is the author of this problem.
If it so easy, why did noone post solution? ;)

i think that if 2 arcs of measures $A$ and $B$ belong to the set, then an arc of measure $\frac{A+B}{2}$ doesnt. i thnk in this way one can prove a bound. am I right Iura?

[quote="iura"]
Given a unit circle, and a set $S$ consisting of $2004$ disjoint arcs of this circle. Assume that the set $S$ has the following property: If three points $A, B, M$ on our circle belong to the set $S$, and the point $M$ is the midpoint of one of the two arcs $AB$ (the smaller arc or the larger arc), then there is one of the arcs from the set $S$ which contains all three points $A, B, M$.

Find the biggest possible sum of lengths of arcs in $S$.[/quote]

 :D


geometry
3D geometry
algebra
polynomial
function
calculus
geometric transformation

Is there a sequence for cubes, such as the one for squares? (addition of odd numbers)

For squares

1+3+5+7.....

I've noticed that in the sequence 1, 7, 19, 37, 61, 91, 127...... if you subtract each term from the term before it, you get 6, 12, 18, 24, 30, 36......

Why is this?

How are the functions related though? Can you take the sequence 6, 12, 18..... and use it to make an equation defining $x^{3}$?

Look at wikipedia under calculus of finite differences.

Each case is kind of trivial, but there is a sequence for cubes, fourth powers, fifth powers, etc.

What are the sequences? Is it too advanced for intermediate topics?

[quote="paladin8"]It's finite differences. If $f$ is a polynomial of degree $n$, then

$f(x+1)-f(x)$

is a polynomial of degree $n-1$.[/quote]

This is kind of off-topic, but is that related to the fact that for a polynomial $f(x)$ of degree $n$, $f'(x)$ is a polynomial of degree $n-1$?

Yes it is, in fact they are near to be the same.
Both are so-called $\delta$-operators $\delta: K[x] \to K[x]$, here with $K=\mathbb{R}$, which:

- are linear (thus $\delta(ap+bq)=a\delta(p)+b\delta(q)$ for all $a,b \in K, p,q \in K[x]$).
- commute with translations (thus $\delta(p(x+a)) = (\delta p)(x+a)$ for all $a \in K, p \in K[x]$).
- $\delta(x)=c \in K^{*}$.

These have a lot of similar properties:

- $\delta(a)=0$ for $a \in K$.
- They lower the degree of a polynomial by $1$, thus map those with degree $\leq n$ surjectively to those with degree $\leq n-1$.

There is a unique "basic system" $p_{0}, p_{1},...$ of polynomials, meaning that:
- $p_{0}= 1$
- $p_{n}(0)=0$
- $\mathrm{deg}(p_{n}) =n$
- $\delta p_{n}= n p_{n-1}$

Try to find them for $\delta_{0}$ = differential operator and $\delta_{1}=\Delta_{1}$ = finite difference operator.

And with those one has:

[u]General Taylor formula:[/u] \[f(x+y) = \sum_{k}\frac{\delta^{k}f(x)}{k!}p_{k}(y)\] with $\delta^{k}$ being $\delta$ iterated $k$ times.

[u]General binomial identity:[/u] \[p_{n}(x+y) = \sum_{k=0}^{n}\binom nk p_{k}(x) p_{n-k}(y)\] If $\delta$ is the differential operator, these two get the well known formulas one is used to (also note that the second is pure polynomial identity without any $\delta$!).

One can also describe one $\delta$-operator as power series of any other, and a lot more such stuff. Don't know, but looking for "Umbral Calculus" may bring something thats worth reading.

[quote="mustafa"]I think that I found a function for cubes

n
_
\
/   $3n(n+1)+1$
_
x=0

That is supposed to be the E thingy that sums all of the stuff :rotfl:[/quote]

[quote="mna851"]Well that sequence works because $\ (x+1)^{2}-x^{2}= 2x+1$.  So the sequence for cubes would be $\ (x+1)^{3}-x^{3}= 3x^{2}+3x+1$.  So the sequence would be $\ 1+7+19+...$[/quote]

Which is what I wrote.

Yeah :rotfl:

Sorry about that. I just didn't quite comprehend that I guess.... I got it a different way, and I didn't quite realize that it was the same way as yours, or something like that....

Good job :coolspeak:

[quote="mustafa"]I think that I found a function for cubes

n
_
\
/   $3n(n+1)+1$
_
x=0

That is supposed to be the E thingy that sums all of the stuff :rotfl:[/quote]
$\sum^{n}_{x=0}3n(n+1)+1$

[quote="mna851"]Well that sequence works because $\ (x+1)^{2}-x^{2}= 2x+1$.  So the sequence for cubes would be $\ (x+1)^{3}-x^{3}= 3x^{2}+3x+1$.  So the sequence would be $\ 1+7+19+...$[/quote]

Did you notice that these will always just be binomial expansions without the first term?  For example, $(x+1)^{4}-x^{4}=4x^{3}+6x^{2}+4x+1$.  So we can just use the Binomial Theorem and be like $(x+1)^{n}-x^{n}=nx^{n-1}+\binom{n}{2}x^{n}-2+...+\binom{n}{n-2}x^{2}+nx+1$
Now we have a formula for all of them!

Really, though, it'd probably be easier to just sub your values into the unexpanded form to get your sequence...


hey people i couldnt understand this question !
it may be very silly but..... :blush:  :blush:
$ wat will be the slope of a line perpendicular to another whose slope is infinite$
$ i used the formula m1 *m2 \equal{}\minus{}1$

Correct your syntax.  It's hard to read.

hey i am sorry the question is wat will be the slope of a line perpendicular to another line whose slope is infinite ?
$ (m1  *  m2 \equal{} \minus{}1)$

line whose slope is undefined(infinite)=a line parallel to $ Y\minus{}axis$

line perpendicular to that,a line parallel to $ X\minus{}axis$
So,$ slope\equal{}0$(any line parallel to X-axis has slope 0) :)


geometry
circumcircle
conics
hyperbola
geometry unsolved

Assume we have a general point $ P$ in the plane. Find all $ P$ that have the circumcircles of cevian triangles
passing through $ F$ - the Feuerbach point.

Is it a well known hyperbola ?
I know Feuerbach hyperbola is for pedal circumcircles and probably there is one for my question.

Actually the locus you ask for is the same Feuerbach hyperbola. This is a generalization of [url=http://forumgeom.fau.edu/FG2001volume1/FG200117index.html]Emelyanov's theorem[/url], which was initially discussed in the Hyacinthos forum, and subsequently studied by me and Paul Yiu in a relatively recent FG paper.
See [url=http://forumgeom.fau.edu/FG2007volume7/FG200723index.html]Cosmin Pohoata and Paul Yiu, On a product of two points induced by their cevian triangles, [i]Forum Geometricorum[/i], 7 (2007) 169-180[/url] for more details. A generalization due to Brianchon and Poncelet is also available on page 177.


linear algebra
matrix
linear algebra unsolved

How many elements does the following set have ? $ \{ A \in \mathcal{M}_n(\mathbb{N})\ | \det(A)\neq 0,A^{\minus{}1} \in \mathcal{M}_n(\mathbb{N})\}$

[quote="loup blanc"]It's the set of permutation matrices.[/quote]

Can you be more specific ? I don't quite understand .thanks ;)

1) If $ A$ and $ A^{\minus{}1}$ are in $ \mathcal{M}_n(\mathbb{N})$ then on each $ A$-column and on each $ A$-row there exists exactly one non zero element. (write with the $ A_{i,j}$ the equalities $ AA^{\minus{}1}\equal{}A^{\minus{}1}A\equal{}I$). Let $ \alpha_i$ be the element of the column $ i$.
2) $ det(A)\equal{}\pm\alpha_1\cdots\alpha_n$. Necessarily $ det(A)\equal{}\pm{1}$; then $ \alpha_i\equal{}1$ and $ A$ is a permutation matrix $ P_{\sigma}$.
Conversely $ A^{\minus{}1}\equal{}P_{\sigma^{\minus{}1}}\in\mathcal{M}_n(\mathbb{N})$.

[quote="olorin"]Is $ 0\in\mathbb{N}$ here?[/quote]
Yes it is .


calculus
integration
real analysis
geometric series
real analysis solved

I don't know if this is a "computation" or an actual problem. If this forum is not appropriate plase move it.

First show that
\[ (1+t)^{-1/2} = 1 + \frac12 t + \frac38 t^2 - R(t) \]
where $0 < R(t) \le \frac{5t^3}{16}$ when $t > 0$.

Then show that
\[ \sum^\infty_{k=0} \frac{2^{-k}}{\sqrt{100+2^{-k}}} = 2\cdot10^{-1} - \frac23 \cdot 10^{-3} + \frac37 \cdot 10^{-5} - F \]
where $0 < F < \frac13 \cdot 10^{-7}$

The first bit is easy to show, but the second isn't. Can someone point me in the right direction... what does the first part even have to do with the second? It's not a Riemann sum or anything :(

A corrcection: that should be

$(1+t)^{-1/2}=1-\frac 12t+\frac38t^2-R(t).$

Hint: "Factor out what's big."

$\frac{2^{-k}}{\sqrt{100+2^{-k}}}=\frac{2^{-k}}{10}\left(1+\frac{2^{-k}} {100} \right)^{-1/2}$

$=\frac{2^{-k}}{10}\left(1-\frac{2^{-k}}{200}+\frac{3(2^{-k})^2}{8(100)^2} -R\right)= \frac{2^{-k}}{10}-\frac{2^{-2k}}{2000}+\frac{3\cdot2^{-3k}}{8\times 10^5}-R^*$

Now add this up over $k.$ You have some geometric series to sum. I've left some of the heavy lifting for you to do - figuring out in detail what I mean by $R$ and $R^*$, above.

Alright. A little messy but not too hard. I didn't think of using maclaurin expansions for the discrete case. I obviously should have given the first part of the exercise, but I was stuck on trying to approximate the series with an integral first. Thanks. This also helped me solve a whole slew of other problems. I don't know why I didn't think of taylor  :blush:


A = {(x;y) in RxR / x^2-y^2 = 3x - 4y }

B = {(x;y) in RxR / 2xy = 4x+ 3y}

C = { x^2+y^2 /(x;y) is in the intersection of A and B}

find  all (x;y) in C

It is quite easy:
from (2) we have
$ x\equal{}\frac{y}{2(y\minus{}2)}$
Replace to the (1)

Bad title http://www.mathlinks.ro/Forum/viewtopic.php?t=144403 . Locked!


Really, are you satisfied with such a solution? I really hope there is a smart solution.

[quote="romano"]Prove that : $ a,b,c \geq 0 $
 $ 2.\sum_{sym}a^6 + \sum_{sym}a^4bc + 2\sum_{sym}a^3b^2c + \sum_{sym}a^3b^3 \geq \sum_{sym}a^2b^2c^2 + 5.\sum_{sym}a^4b^2 $[/quote]
sorry can you tell me exactly which inequality , vasc has lots of inequalities

maybe this is  http://www.mathlinks.ro/Forum/topic-26104.html

see my proof


What determines the layout of ligands as they approach the metal complex ion?

I know there are the five d orbitals, $ d_{x^2 \minus{} y^2}$, $ d_{z^2}$, $ d_{xy}$, $ d_{yz}$, and $ d_{xz}$, and as the ligands approach the ion, the energy levels of the orbitals split according to where the ligands are placed.

What I don't get is, how can you determine if a complex ion will form bonds with its ligands in a tetrahedral, octahedral, or square planar structure?

it depends on no. of empty orbitals  in the d-orbital

and also the no of ligands that can make the the metal ion most stable :)


Let $x,y,z$ be distinct complex numbers. Prove that:
\[ \frac{x^3}{(x-y)(x-z)} + \frac{y^3}{(y-z)(y-x)} + \frac{z^3}{(z-x)(z-y)} = x+y+z.  \]

Not a very nice solution:
Let $f(x,y,z)=(x+y+z)(x-y)(y-z)(z-x)+x^3(y-z)+y^3(z-x)+z^3(x-y)$.
[b]Regard f as a polynomial in x.[/b]
Coefficient of $x^3$ = -(y-z)+(y-z)=0
Coefficient of $x^2$=(y-z)[-(y+z)+y+z]=0
[b]Degree of f is at most 1.[/b]
$f(y,y,z)=0+y^3(y-z)+y^3(z-y)+0=0$
$f(z,y,z)=0+z^3(y-z)+y^3(z-y)+0=0$
Therefore, y and z are two distinct roots of f.
This implies that f is a zero polynomial in x.
$(x+y+z)(x-y)(y-z)(z-x)+x^3(z-y)+y^3(z-x)+z^3(x-y)=0$
Re-arrange terms the result follows.

A generalization is the following:

Let $x_1,\ldots,x_n$ be $n$ distinct complex numbers, then:
\[ \sum_{i=1}^n\frac{x_i^n}{\prod^n_{j=1,\\ j\not=i}(x_i-x_j)} = \sum_{i=1}^n x_i.  \]
[hide="hint"]
Use Lagrange Interpolation
[/hide]

[hide="further hint"]
Use two different ways to find the unique polynomial of degree <= n-1 such that $f(x_i)=x_i^n$.
[/hide]


geometry
3D geometry
sphere
tetrahedron
incenter
linear algebra
matrix

What is the formula to get radius of sphere inscribed in irregular tetrahedron, when all edges are given??

When tetrahedron is regular we have
r=1/3R=1/3*sqrt(6)/4*a
where a is an edge of tetrahedron, R -  radius of circumscribed sphere, r -  radius of inscribed sphere.

But how to find r when we are given an irregular tetrahedron with edges WXYZ (we know edges: WX, WY, WZ, XY, XZ, YZ)??

It might help finding the incenter of the tetrahedron, then working from there...

I can calculate volume from the proper matrix determinant and the surface area from Heron's formula, but could you please tell me what to do then??


inequalities
trigonometry
geometry
geometry proposed

Show that in any $\triangle ABC$ there are the following chain of inequalities :

$\left(\frac{2r}{R}\right)^{2}\le\prod(\cos B+\cos C)\le \frac{2r}{R}\le\prod\sin\left(A+\frac{B}{2}\right)\le\left(\frac{2p}{3\sqrt 3R}\right)^{2}\le 1$.

We will use $r=4R\prod\sin{\frac{A}{2}}$.

1st inequality: By $\cos A+\cos B=2\sin\frac{C}{2}\cos\frac{A-B}{2}$, e.t.c. and above equality this inequality is equivalent to $8\prod \sin\frac{A}{2}\leq \prod\cos\frac{A-B}{2}$ or $8\prod\sin A\leq \prod (\sin A+\sin B)$, it is true by AM-GM.

2nd inequality: By above equality and $\cos A+\cos B=2\sin\frac{C}{2}\cos\frac{A-B}{2}\leq 2\sin{\frac{C}{2}}$, e.t.c. we're done.

We have $\prod\sin(A+\frac{B}{2})=\prod\cos\frac{B-C}{2}$ therefore
The third is $8\prod\sin\frac{A}{2}\le\prod\cos\frac{B-C}{2}$ as above

The fourth $\Leftrightarrow \prod\cos\frac{B-C}{2}\le \frac{4}{27}(\sum\sin A)^{2}\Leftrightarrow \prod\cos\frac{B-C}{2}\cos\frac{A}{2}\le \frac{4}{27}(\sum\sin A)^{2}\prod\cos\frac{A}{2}\Leftrightarrow \frac{1}{8}\prod(\sin B+\sin C)\le \frac{1}{27}(\sum\sin A)^{3}$(Note $4\prod\cos\frac{A}{2}=\sum\sin A$) which is trivial by AM-GM

The fifth is trivial

[quote="gemath"]We have $\prod\sin(A+\frac{B}{2})=\prod\cos\frac{B-C}{2}$ [/quote]
New for me, please post its prove, thanks!  :wink:

Small trick  :D
$A+\frac{B}{2}=\frac{2A+B}{2}=\frac{180-C+A}{2}=90-\frac{C-A}{2}\Rightarrow\sin(A+\frac{B}{2})=\cos\frac{C-A}{2}$
Is it hard with a genius about higher math N.T.TUAN?

[quote="gemath"]Small trick  :D
$A+\frac{B}{2}=\frac{2A+B}{2}=\frac{180-C+A}{2}=90-\frac{C-A}{2}\Rightarrow\sin(A+\frac{B}{2})=\cos\frac{C-A}{2}$
[/quote]
Thanks, But
[quote]
Is it hard with a genius about higher math N.T.TUAN?[/quote]
Please don't joke on this forum!  :mad: If you want you can PM to me!

[quote="N.T.TUAN"]
[quote]
Is it hard with a genius about higher math N.T.TUAN?[/quote]
Please don't joke on this forum!  :mad: If you want you can PM to me![/quote]
Sorry for this spam but really you didn't have any joke in this forum my friend ?  :D
[quote="N.T.TUAN"]
It is not easy for me. I think that i must choose : these inequalities or B&S-D  :rotfl:[/quote]
Really when anybody worked with higher math, they always discount the elementary math and elementary problems, they forgot when they began to do math then what problem they learn, follow me all higher or elementary is only relative, don't discount any problem of mathematics, when you study Riemann geometry don't forget Euclide.

Following text is TRUE!
[quote="gemath"]
Really when anybody worked with higher math, they always discount the elementary math and elementary problems, they forgot when they began to do math then what problem they learn, follow me all higher or elementary is only relative, don't discount any problem of mathematics, when you study Riemann geometry don't forget Euclide.[/quote]
Hey, my friend! Just two point:
1) If you quote my post , please quote all http://www.mathlinks.ro/Forum/viewtopic.php?t=148721&sid=e7d3faa48eb2602930923a6c49bff5d7
2) In that topic I mean that: I thanks he for problems, but I don't have much time to slove it.
That is all!


A man was three times as old as his daughter, and 10 years ago he was 4 times as old as his son. His son is two years older then his daughter. His son is 12 years old. How old is he?

I don't get this. If 10 years ago he was 4 times as old as his son, that means he was 8 and his son was 2? :huh:

Nope. I'm evil. Almost as evil as the guy with "some easy problems". I will reveal the answer if no one can get it.

Think, people. 4 equations. 3 unknowns. Internally contradictory, in fact.

No matter how you handle it this guy had really early or really late children. I guess I should think up the numbers before I think up the equations.

[hide="hint!"]He has two sons! Otherwise it's contradictory, unless he has two daughters, but that is not true in this situation.[/hide]

[hide]Ok, if we discount the middle part of the question (about the son who was 4 times younger 10 years ago) then maybe this man is 30?  :maybe: [/hide]


The set of points $ (x,y)$ such that $ |x\minus{}3|\leq y\leq4\minus{}|x\minus{}1|$ defines a region in the $ xy$-plane. Compute the area of this region.

[hide="Answer"]
This is a rectangle with endpoints (0,3), (1,4), (3,0), (4,1).

We can easily find that the dimensions in the $ 45^{\circ}$ and $ \minus{}45^{\circ}$ directions are $ \sqrt{2}$ and $ 3 \sqrt{2}$, respectively, so the area is $ 6$[/hide]


Let $k$ be a positive integer. Prove that one can partition the set $\{ 0,1,2,3, \cdots ,2^{k+1}-1 \}$ into two disdinct subsets $\{ x_1,x_2, \cdots, x_{2k} \}$ and $\{ y_1, y_2, \cdots, y_{2k} \}$ such that $\sum_{i=1}^{2^k} x_i^m =\sum_{i=1}^{2^k} y_i^m$ for all $m \in \{ 1,2, \cdots, k \}$.

[quote="N.T.TUAN"]I know one solution of this problem,

[hide]use Morse's sequence
see http://mathworld.wolfram.com/Thue-MorseSequence.html

or http://en.wikipedia.org/wiki/Thue-Morse_sequence[/hide][/quote]
[b]Lemma[/b]
Set{$n+1;n+2;...;n+2^{k}$}can be patitioned into two sets A,B satisfying the question.
[b]proof[/b]
Induction by k
(Sorry-I'm very busy ) :)

I have a question which doesn't seem to be trivial: How can we prove that the sets $S=\{1,2,...,2^{n}\},$ $S_{0}=\{n|n\in S, t_{n}=0\}$ and $S_{1}=\{n|n\in S, t_{n}=1\},$ where $\{t_{n}\}_{n=0}^{\infty}$ is the Thue-Morse sequence, have the property
\[\sum_{x\in S_{0}}x^{s}=\sum_{y\in S_{1}}y^{s}\]
for $s=0,1,...,n-1$  :?:

Thanks

Split all the numbers in groups based on the parity of the sum of it's digits in binary.From [url=https://artofproblemsolving.com/community/q1h1524734p9148805]this[/url] (that is any of the generalizations listed there) we're done.


How many positive integers less than $ 1000$ are $ 6$ times the sum of their digits?

$ \textbf{(A)}\ 0 \qquad
\textbf{(B)}\ 1 \qquad
\textbf{(C)}\ 2 \qquad
\textbf{(D)}\ 4 \qquad
\textbf{(E)}\ 12$

[hide="solution"]It must be a multiple of 3, so the sum of it's digits is a multiple of 3. Therefore the number is a multiple of 9, and so is the sum of it's digits. Therefore the number must be a multiple of $ 6*9\equal{}54$. 54 works, but all the three-digit multiples don't work, as can easily be seen by checking them. B.[/hide]

eh I did this problem the cheap way. :)

[hide="slightly faster"]Clearly the number has to be a multiple of 6. The maximum sum of digits of a 3 digit number is 27, when the number is 999. Thus, the number is at most 27*6=162. The hundreds digit must thus be 1. This makes the maximum sum of digits 19, when the number is 199. Thus, the number is at most 19*6=114. Now there are only three numbers to check: 102, 108, 114, clearly these all fail.[/hide]

[hide="Solution"]Obviously, no one-digit number works. We try two digit numbers; $ 10a \plus{} b \equal{} 6a \plus{} 6b$ or $ 4a \equal{} 5b$. The only solution with $ a$ and $ b$ digits is $ 54$. For three digit numbers $ \overline{abc}$, note that the maximum is six times the maximum digital sum, or $ 6\cdot27 \equal{} 162$. The first digit is therefore one. We now have $ 100\plus{}10b\plus{}c\equal{}6\plus{}6b\plus{}6c$, or $ 94 \equal{} 5c \minus{} 4b$. However, we must have $ 5c\minus{}4b\le5c\le45$ as $ c$ is a digit, a contradiction. So our only solution is $ 54\rightarrow\boxed{\textbf{(B)}}$.[/hide]

What i did

[hide]
Lets say the number is in the form 100a + 10b + c
a,b,c can be anything from the set {0,1,2..9} which means we have included one digit and two digit numbers
From the question we get
100a + 10b + c = 6(a + b + c)
But 100a has to be < 6(a + b + c) --> 100a < 27 ---> a = 0

now solving we get
4a = 5c
From the set {0,1,2,3..9} The only possible pair for (a,c) is (5,4)
Therefore the only solution is
                                        [img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/c/d/8/cd86286fa615eee10d1daf97202c736adf83d050.gif[/img]
[/hide]

[quote="Complex_Ninja"]
From the question we get
100a + 10b + c = 6(a + b + c)
But 100a has to be < 6(a + b + c) --> 100a < 27 ---> a = 0
[/quote]

Shouldn't this be 100a < 162?

[hide="@Truffles"]I think it should be:

$ 100a > 6(a \plus{} b \plus{} c)$, when $ a \ge 2$ because the most $ b$ and $ c$ can be is $ 9$.

So $ a$ can either be $ 0$ or $ 1$.

To me, I think worthawholebean placed the most straightforward solution.[/hide]

But if $ 100a > 6(a\plus{}b\plus{}c)$ then obviously $ 100a \plus{} 10b \plus{} c \neq  6(a\plus{}b\plus{}c)$.  So I don't understand how you got to this.

Yes, that's the case when a is 2 or greater. They were using that to show why a could only be 1 or 0.

I accidentally reversed one of the signs.  :oops: Sorry that it didn't make any sense. It has been corrected now. Hope you can understand it better.

sorry for revival :P

$100a+10b+c=6(a+b+c)$ mod 9
a+b+c=6(a+b+c) mod 9
5(a+b+c) = 0 mod 9
a+b+c must be a multiple of 9

9,18...
therefore, 6(a+b+c) = 54, 108, 162...

however, since 9 divides a+b+c, then 9 divides 6(a+b+c) and therefore divides our number, so the digits must add up to 9. Hence, the only number is 6*9 = 54

[hide="My Solution"]
We know that no $ 1 $-digit number works. We have only the $ 2 $-digit numbers left.
Let's say the number is in the form $ ab $ where $ a $ and $ b $ are digits.

The value of the number is $ 10a + b $. This is $ 6 $ times the sum of the digits ($ a + b $). Therefore:

$ 10a + b = 6(a+b) $

$ 10a + b = 6a+6b $

$ 4a = 5b $

$ a = 5 \text { and } b = 4 $

The only number that works is $ 54 $. The answer is:

$ \boxed {\textbf {(B)}} $
[/hide]

Case 1. This positive integer is a $3$ digit number. Let $\overline{abc}$ be this positive integer.
$$\implies 100a + 10b + c = 6(a + b + c)$$$$\implies 100a + 10b + c = 6a + 6b + 6c$$$$\implies 94a + 4b - 5c$$We clearly see that there are no such 3-digit positive integers that satisfy this.

Case 2. This positive integer is a $2$ digit number. Let $\overline{ab}$ be this positive integer.
$$\implies 10a + b = 6(a + b)$$$$\implies 10a + b = 6a + 6b$$$$\implies 4a - 5b = 0$$$$\implies 4a = 5b$$$$\implies 4a = 5b = 20$$This is true because it needs to be a multiple of $4$, as well as a multiple of 5, and it can't be any other multiple of $20$, because then $a > 10$.
Anyways, back to the question.
$$\implies a = 5$$$$\implies b = 4$$Hence there is only one $2$ digit integer that satisfies this which is $54$.

Case 3. This positive integer is a one digit number. Let $a$ be this number.
$$\implies a = 6a$$$$\implies 5a = 0$$We clearly see that there are no such one digit positive integers that satisfy this.

We see that only $\boxed{1}$ positive integer satisfies this condition, which is $54$.


function
algebra
polynomial
combinatorics unsolved
combinatorics

Its not a hard problem.

We know $(1+x)^n = nC_0 + nC_{1}x ... nC_{n}x^n$

Similarly  $(x + 1)^n = nC_0x^n + nC_{1}x^{n-1} ... nC_{n}$

Multiply Both

then you will get

              $(1+x)^{2n} = (nC_0 + nC_{1}x ... nC_{n}x^n)$ * $(nC_0x^n + nC_{1}x^{n-1} ... nC_{n})$

Now compare the coefi. of $x^n$ in Left hand and right hand
the equality is proved.

Sorry but I don't see how the equality has been proven...

Its a classic generating functions problem: If you can show that two polynomials are equal, then they must be equal term-by-term (or coefficient-by-coefficient).

My favorite proof of this identity goes as follows:

Flip $n$ pennies and $n$ nickels.  How many ways can the coins come up so that the number of heads on the pennies is the same as the number of tails on the nickels?

Method 1: Pick some $k$.  Pick $k$ of the pennies to come up heads.  Pick $k$ of the nickels to come up tails.  Thus there are $\sum_{k=0}^n \binom{n}{k}^2$ ways.

Method 2: The numbers are the same only when exactly $n$ coins come up heads.  Thus there are $\binom{2n}{n}$ ways.

Its simple.
The coefi. of $x^n$ in the left hand side is $2nC_n$.
The R.H.S is
${nC_0}^2*x^n + {nC_1}^2*x^n + {nC_2}^2*x^n + {nC_3}^2 *x^n+ .... {nC_n}^2*x^n +$ Other terms

Hence the Coe. of $x^n$  in the R.H.S is
${nC_0}^2 + {nC_1}^2 + {nC_2}^2 + {nC_3}^2 + .... {nC_n}^2$

Hence coe. in R.H.S and L.H.S are equal hence it is solved.

Now i am sure you will get it. if you have other problem then mail to me.