[hide]Using finite differences, you find the constant at cubic ($18$).

$a+b+c+d=1$
$8a+4b+2c+d=17$
$27a+9b+3c+d=66$
$64a+16b+4c+d=166$

Solve for a,b,c,d.
$a=3$
$b=-1.5$
$c=-.5$
$d=0$

Factored, your equation is: $\frac{1}{2}(6x^3-3x^2-x)$

Evaluate with $x=34$ (34 groups until 100)

$\frac{1}{2}[(6)(34)^3-3(34)^2-34]=\boxed{116161}$

I hope I didn't make any algebraic errors (in solving the equations).[/hide]

[quote=surge]Using finite differences, you find the constant at cubic ($18$).[/quote]

What does this mean?

[quote="DanK"][quote="surge"]Using finite differences, you find the constant at cubic ($18$).[/quote]

What does this mean?[/quote]

It's this ugly process that you learn in Algebra II/Trig that I promptly forgot.

it's some really ugly stuff, but is actually quite useful when determining which degree the polynomial is.
find the difference between two consecutive numbers
if not all of the differences are the same, repeat procedure for the first differences
repeat until all the numbers in the row are constant. that is your finite difference

[hide="Rather screwy method"]
$\sum_{k=0}^{n} (1+3k)^2 = \sum_{k=0}^{n} (1 + 6k + 9k^2) = \left( \sum_{k=0}^{n} 1 \right) + 6 \left( \sum_{k=0}^{n} k \right) + 9 \left( \sum_{k=0}^{n} k^2 \right)$

We know how to evaluate all of these.  We find that the expression becomes

$(n+1) + 6 \frac{n(n+1)}{2} + 9 \frac{n(n+1)(2n+1)}{6}$

For $n = 33$, this becomes

$\boxed{ 116161 }$.
[/hide]


......*. .   *
*. .  *. .  * . . *
*. .  *. .  * . . *
......*. .  *

Picture the 12 asterisks above as dots making a square-shaped
cross, and the periods are to be ignored.

If you wanted to place them as points in the xy-plane,
they could be listed (from lowest row to highest row) as:
A(1,0), B(2,0), C(0,1), D(1,1), E(2,1), F(3,1),
G(0,2), H(1,2), I(2,2), J(3,2), K(1,3), L(2,3)

Two or more dots (all to act only as vertices of squares)
on part of the 12 dots (all of which at a time are to be
connected to form squares are to form different squares
for a total count.  If a set of dots can make more than
one square, then you count all of them.

Summary
------------

You are choosing all ways of 2, 3, and 4 dots at a time in the
12-dot diagram to be vertices only of squares and add
the total.

Note:   If one choice of dots gets the same square (same size
and same location) as another, then it does NOT add to
the total number.

Example) # of squares picking dots B,E,F --> 1
Example) # of squares picking dots B,I --> 2 (one on each side)

Is it 75?

Breakdown:
For choosing 4 dots: 11 ways
For choosing 3 dots: 8 ways
For choosing 2 dots: 56 ways:
At least one of those is probably wrong though :maybe:

There will be a delay in getting back with specifics about this.
I expect about one day.

It is good that you showed some partial work (breakdown).
That is, it makes it more helpful about checking parts of
where subtotals came from.

(Others might post in the meantime too.)

cyberspace,

I got the same subtotals for 3 dots and for 4 dots,

as you, respectively.

However, I got a larger number for 2 dots than you

did, but I want to study the case for 2 dots again.


integration
advanced fields
advanced fields unsolved

$(\Omega, \mu)$ a mesurable space, $\mu( \Omega )=1$. For $1 \leq p < \infty$, suppose that $S$ is a closed subspace of $L^p(\Omega, \mathbb{C})$, included in $L^{\infty}(\Omega, \mathbb{C})$. Show that $S$ is a finite dimensional subspace.

I don't have the time to write down a full proof now but the way to proceed seems clear: take any bounded sequence $(f_{n})$ in $L_{p}$, and extract a subsequence $g_{n}$ such that $\int {g_{n}}$ converges on every interval $I_{k,n}=[k.2^{-n},(k+1).2^{-n}]$, and such that $|\int {(g_{n},I_{k,n})}-lim \ \int {(g_{n},I_{k,n})}| \leq 2^{-n!}$ for every $k$. Now I guess we can prove that $(g_{n})$ is Cauchy for the $L_{p}$ norm, and hence convergent to an element of $S$ (since $S$ is closed). But this proves that $S$ is finite dimensional (by Riesz's theorem).


Find the sum of the digits of the number: $ 10^{2008}\minus{}2008$

I'll post the answer on thursday.

I think I know the answer but I don't know how to "hidden content" something. Can you help???!!!

This is my first post! YAY :D

It ends in 045 right?

To post a hidden answer just type "[;hide=Whatevertext]what's hidden[/hide]" But with out the quote marks and take out the semi colon.

Ok thanks

[hide="Answer"]You do 10000-2008=7992 7+9+9+2=27
Now all the other digits besides the last 4 digits are 9s and there are 2003 other digits
2003*9=18027
18027+27=18054   ->[b]18054[/b] [/hide]

I meant 054 in the first one :P

Yeah, I agree. I got 18063, but I'm pretty sure I'm wrong, since subtracting 2008 takes off one digit.

well I guess ill post the answer now.

ok micrim you almost had it, you were missing one 9.
darkmattter got it right.

Answer:
$ 10^5 \minus{} 2008 \equal{} 97992$
$ 10^6 \minus{} 2008 \equal{} 997992$

notice that the last four digits are always 7992 wich is 27,
and that the number of nines is always the power -4

so here is a equation for the sum of the digits of any 10^x -2008; where x>4:
$ 27 \plus{} 9(x \minus{} 4)$

so just plug in 2008 for x and you get:

$ 10^{2008} \minus{} 2008$:
$ 27 \plus{} 9(2008 \minus{} 4)$
$ \equal{} \boxed{18063}$

Oh I know what I did wrong. 10^2008 has 2008 0's not digits......
 Righttttttttttt... :wallbash_red:

Sweet! :lol:  I made the same mistake as micrim, but a typing error on the calculator accidentally gave me the answer!  :rotfl:

@elio(n)

Yep that works,Although you missed a $ \plus{}9$ when you wrote $ (2008\minus{}4)*9\plus{}9\plus{}9$

[hide] first I did 10 to a low power (to the fifth) minus 2008

I did 100000-2008

I got 97992

so I realized that it always ended in 7992 and the rest were 9s.

I knew that when 10 is to the x power, it has one more digit than x

so 10^2008 has 2009 digits. Taking away 2008 from that number, we get 2008 digits. Then adding the 7 and the 2 together, it becomes 9, so I did 9 x 2007

I got 18063[/hide]

i calculated the answer multiple times and i have concluded that by cheating off of ya'll's answer that the real answer is 10863. yeah :w00t:


let $a,b \in \mathbb{N}$. Prove that $(a,b)=(3,5)$ is only solution for equation $a^3=b^2+2$

Factor it as $a^3 = (b+ \sqrt{-2})(b-\sqrt{-2})$.
Now $\gcd(b+ \sqrt{-2} , b- \sqrt{-2}) = \gcd(b+ \sqrt{-2}, 2 \sqrt{-2} ) =1$ since $\sqrt{2} \nmid b$ because $2 \nmid b$ (the last one can be seen $\mod 8$).
Thus by unique factorisation in $\mathbb{Z}[\sqrt{-2}]$ we get that $b+\sqrt{-2} = (x+y\sqrt{-2})^3$ is a perfect cube.
This gives us:
$x^3-6xy^2=b$
$y(3x^2-2y^2)=3x^2y-2y^3=1$
thus $y = \pm 1$, so $3x^2-2 = y= \pm 1$ giving finally $y=1 , x= \pm 1$.
So $b=x(x^2-6y^2) = \pm 5$, done.

i know solution using ring $\mathbb{Z}[\sqrt{2}]$. I'm intersted in lementary solution if it's possible...

[mod.: never, really never, take anything like numbers into \mathbb{...}]

[quote="bomb"]There is no elementary method.

Bomb[/quote]
I would formulate it in another way: there seems to be no method not being the same as the above one up to elementarisation.


let $ k>0$ be an integer and define $ f(x)\equal{}\sum_{n\equal{}0}^\infty c_n x^n,$ where $ c_{n\plus{}k}\equal{}c_n$ for each $ n\ge0.$ Find the interval of convergence of the series and compute $ f(x)$ explicitly.

don't know how to find $ f(x)$ explicitly.

If not all terms of $ \{ c_n \}$ vanish, then root test is applicable, yielding 1 as the radius of convergence for this series. Also,

\begin{eqnarray*}
f(x) & = & \sum_{n=0}^{\infty} c_n x^n \\
& = & \sum_{q=0}^{\infty} \sum_{r=0}^{k-1} c_{qk+r} x^{qk+r} \qquad (n = qk+r)\\
& = & \sum_{q=0}^{\infty} \sum_{r=0}^{k-1} c_{r} x^r \cdot x^{qk} \\
& = & \left( \sum_{r=0}^{k-1} c_{r} x^r \right) \left( \sum_{q=0}^{\infty} x^{qk} \right) \\
& = & \frac{1}{1-x^q} \sum_{r=0}^{k-1} c_{r} x^r
\end{eqnarray*}


You are given the sequence $a_1 = 1, a_2 = 0, a_3 = 1, a_4 = 0, a_5 = 1, a_6 = 0$ and a prime number $p$.

The rest of the sequence is defined as $a_{n+6} = \sum_{i=0}^{5}a_{n+i} \mod p$

Find all primes $p$ for which the sequence contains $0,1,0,1,0,1$ as some six consecutive terms.

I have been able to show that p must be one of 2,3,7,11 for that to happen. I was able to verify, using a computer that only for 2,3 and 11 does the sequence contain 0,1,0,1,0,1.

So, try the modified problem (which does not involve a computer):

Show that the set of primes p for which the sequence contains 0,1,0,1,0,1 is finite.

In fact, it is a subset of {2,3,7,11}.


Aν $ a,b,c$ είναι θετικοί πραγματικοί αριθμοί τέτοιοι ώστε $ ab\plus{}bc\plus{}ca\plus{}2abc\equal{}1$ να αποδείξετε ότι $ 2(a\plus{}b\plus{}c)\plus{}1\geq 32abc$

Profanos $ abc$ diaforo tu midenos,opote $ ab\plus{}bc\plus{}ca\equal{}1\minus{}2abc$
dld
$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{}\frac{1\minus{}2abc}{abc}$

omos
$ (a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})>\equal{}9$
opote
$ a\plus{}b\plus{}c>\equal{}\frac{9abc}{1\minus{}2abc}$

$ 2(a\plus{}b\plus{}c)>\equal{}\frac{18abc}{1\minus{}2abc}$

diladi  $ 2(a\plus{}b\plus{}c)\plus{}1>\equal{}\frac{16abc\plus{}1}{1\minus{}2abc}$
opote arki ndo
$ \frac{16abc\plus{}1}{1\minus{}2abc}>\equal{}32abc$

isodinama afu $ 2abc<1$ ($ >\equal{}1$ apokliete afu $ a,b,c$ 8etiki pragmatiki..)

$ 16abc\plus{}1>\equal{}32abc\minus{}64(abc)^2$

dld $ (8abc\minus{}1)^2>\equal{}0$ pu isxii kai i isotita isxii profanos otan $ a\equal{}b\equal{}c\equal{}\frac{1}{2}$

Δείτε και εδώ . ¨εχω ποστάρει μια αρκετά καλή λύση only Cauchy  :)
http://www.mathlinks.ro/viewtopic.php?search_id=1564802248&t=80027

Αρκετά δύσκολη ανισότητα. :huh:

Χάρηκα πάντως επειδή μετά από πολύ ώρα βρήκα μια πολύ original λύση. :idea:

Θέτουμε $ a\equal{}(1\minus{}x)/x$ κτλ. οπότε η δοσμένη παίρνει τη μορφή $ x\plus{}y\plus{}z\equal{}2$ (!) ενώ η ανισότητα προς απόδειξη τη μορφή:

$ x/(4\minus{}3x)\plus{}y/(4\minus{}3y)\plus{}z/(4\minus{}3z)>\equal{}1$

(Αν κάτι χρειάζεται περαιτέρω εξήγηση πείτε μου και θα το βάλω αύριο γιατί νυστάζω λίγο)

Πολλαπλασιάζουμε τώρα κάθε κλάσμα πάνω και κάτω με τον αριθμητή του και Andreescu και τέλος.

Νομίζω δικαιούμαι ένα Q.E.D. :D

Χρησιμοποιώντας την ανισότητα $ AM \minus{} HM$ παίρνουμε
$ (a \plus{} b \plus{} c \plus{} 1/2)/4 > \equal{} 4/(1/a \plus{} 1/b \plus{} 1/c \plus{} 2)$
και μετά από πράξεις παίρνουμε
$ (a \plus{} b \plus{} c \plus{} 1/2) > \equal{} 16abc$
Πολλαπλασιάζοντας με 2 παίρνουμε την ζητούμενη


calculus
integration
algebra
partial fractions
calculus computations

A sequence $\{a_n\}$ is defined by $a_n=\int_0^1 x^3(1-x)^n dx\ (n=1,2,3.\cdots)$

Find the constant number $c$ such that $\sum_{n=1}^{\infty} (n+c)(a_n-a_{n+1})=\frac{1}{3}$

This is not so hard by conventional methods, although there may be a cleverer way.  I'm leaving out most of the calculations because I am lazy.

We can simply calculate $a_n - a_{n-1}$ very straight-forwardly.  Adding the integrals adds the integrands, giving us $a_n - a_{n-1} = \int_0^1x^4(1-x)^n dx$.  Doing the RHS by parts and combining like terms, $a_{n+1} = \frac{1+n}{5+n}a_n$ from which it follows that $a_n = \frac{5!n!}{(n+4)!}a_1$.  Direct calculation gives $a_1 = \frac{1}{20}$, so $a_n = \frac{6}{(n+1)(n+2)(n+3)(n+4)}$.  (We also could have done integration by parts 3 times to get this result.)  Now, if we take this and stick it back into the original summation, we find $\sum_{n=1}^{\infty} (n+c)(a_n-a_{n+1})=\frac{1}{3}$,
and we can decompose the expression in that sum in order to write it as
$(c - 1)\sum_{n=1}^{\infty}\frac{24}{(n+1)(n+2)(n+3)(n+4)(n+5)} + \sum_{n=1}^{\infty}\frac{24}{(n+2)(n+3)(n+4)(n+5)} = \frac{1}{3}$

Each of those sums can be gotten by standard partial fractions, although they're a bit messy (especially the first one), and then we have a linear system whose solution is $c = 5$.

Thank your for your reply, JBL.

Your answer is correct.

kunny


Given that BDEF is a square and AB = BC = 1, find the number of square units in the area of the regular octagon.

[asy]pair A,B,C,D,E,F;
A = (-3,1); B = (-3,3); C = (-1,3); D = (3,3); E = (3,-3); F = (-3,-3);
for(int i = 0; i < 4; ++i){
draw(rotate(90*i)*(B--D));
draw(rotate(90*i)*(A--C));
}
label("A",A,W); label("B",B,NW);label("C",C,N);
label("D",D,NE);label("E",E,SE);label("F",F,SW);[/asy]

The side length of the octagon is $ \sqrt{2}$.  The side length of the square, therefore, is $ 2 \plus{} \sqrt{2}$.  Simplify $ (2 \plus{} \sqrt{2})^2$ for the area of the square.  This is equal to $ 4 \plus{} 4\sqrt{2} \plus{}2$, or $ 6 \plus{} 4\sqrt{2}$.  Then, subtract the areas of the $ 4$ triangles.  Each triangle's area is $ \frac{1}{2} \times 1 \times 1$, or $ \frac{1}{2}$.  Subtract $ 2$ from $ 6 \plus{} 4\sqrt{2}$, to receive the final answer of $ \boxed{4 \plus{} 4\sqrt2}$.


calculus
integration
trigonometry
calculus computations

Evaluate $ \int_0^{2\pi } {\frac{{\cos 2x}}{{5 \minus{} 4\sin x}}\,dx} .$

[hide]If $ I$ is the integral, then substituting for $ 2\pi \minus{}x$ gives
\[ I\equal{}\int_0^{2\pi} \frac{\cos 2x}{5\plus{}4\sin x}\ dx\]
Add to the original integral and divide by 2 to get
\[ I\equal{}5\int_0^{2\pi} \frac{\cos 2x}{25\minus{}16\sin^2 x}\ dx\equal{}5\int_0^{2\pi} \frac{\cos 2x}{17\plus{}8\cos 2x}\ dx\]
Note that this can be expressed as
\[ 10\int_0^\pi \frac{\cos 2x}{17\plus{}8\cos 2x}\ dx\equal{}10\int_{\minus{}\frac{\pi}{2}}^\frac{\pi}{2} \frac{\cos 2x}{8\cos 2x \minus{}17}\ dx\]
Let $ u\equal{}\tan x$, so that $ \ du\equal{}(1\plus{}u^2)\ dx$, giving the improper integral
\[ 10\int_{\minus{}\infty}^\infty \frac{u^2\minus{}1}{(1\plus{}u^2)(9\plus{}25u^2)}\ du\equal{}10\int_{\minus{}\infty}^\infty \frac{1}{8(1\plus{}u^2)}\minus{}\frac{17}{8(9\plus{}25u^2)}\ du\]
\[ \equal{}10\left[ \frac{1}{8}\arctan u\minus{}\frac{17}{120}\arctan \frac{5}{3}u\right]_{\minus{}\infty}^\infty \equal{}\boxed{\minus{}\frac{\pi}{6}}\][/hide]

Generally, $ \int_{0}^{2\pi} \frac{e^{i n x}}{5\minus{}4\sin x} \, dx \equal{} \frac{2 \pi}{3} \left( \frac{i}{2} \right)^n$.


"The integer $ 999,919$ is a product of a three digit prime and a four digit prime.  Let A be the sum of the digits of the three digit prime, and B be the sum of the digits of the four digit prime.  Compute the sum $ A \plus{} B$."

Since $ 999,919$ is a product of two primes, then that must mean that it's only factors are $ 1$, $ 999919$, three-digit prime, four-digit prime.  That's as far as I have gotten.  Calculators are allowed, but not TI-89 or higher (or the HP equivalents).  Note that I'm only supposed to spend about 3 minutes on this problem, so sitting around and brute forcing isn't a good idea.

It's one of those notice random things about the number problems:
[hide]
$ 999919 \equal{} 1000000 \minus{} 81 \equal{} 1000^2 \minus{} 9^2 \equal{} (1000 \plus{} 9)(1000 \minus{} 9) \equal{} 1009*991$
I'm going to take the problem's word that those numbers are prime.
Anyway, the desired answer should be $ 1 \plus{} 9 \plus{} 9 \plus{} 9 \plus{} 1 \equal{} \boxed{29}$
[/hide]


integration
inequalities
function
logarithms
real analysis
complex analysis
real analysis unsolved

I understand how to use them in proofs... but I really wish I knew where the equation:

$ \frac{1}{p}\plus{} \frac{1}{q}\equal{}1$ came from!

For example, I can prove that for a bounded linear functional one $ L^p$, $ F(f)\equal{} \int fg$ we have $ \parallel{}F\parallel{} \equal{} \parallel{}g\parallel{}_q$ and the proof is all algebra. But, it just seems like magic. I don't know WHY it works. Can someone give me a hint as to the bigger picture here?

I'm not sure it is the ultimate reason, but Young's inequality $ \frac {x^p}p\plus{}\frac {y^{q}}q\ge xy$, which is behind most of the "magic" concerning the $ L^p$ spaces is just a partial case of $ \Phi(x)\plus{}\Psi(y)\ge xy$ where $ \Phi$ and $ \Psi$ are any convex conjugate functions vanishing at $ 0$ (i.e., for some positive increasing continuous function $ f: [0,\plus{}\infty)\to[0,\plus{}\infty)$ with $ f(0)\equal{}0$, we have $ \Phi(x)\equal{}\int_0^x f$, $ \Psi(x)\equal{}\int_0^x f^{\minus{}1}$ where $ f^{\minus{}1}$ is the inverse function to $ f$. The proof is the same picture with the graph of $ f$ and two curvilinear triangles covering a rectangle. Now, given any $ \Phi$, you can compute $ \Psi$ and it just happens that if $ \Phi(x)\equal{}\frac{x^p}{p}$, then $ \Psi(y)\equal{}\frac{y^q}q$ with $ q\equal{}\frac{p}{p\minus{}1}$.

Another good reason is that if you want to have the inequality $ \int_{\mathbb R} fg\le  \|f\|_p\|g\|_q$, both sides should scale the same way when you change $ f(x)$ and $ g(x)$ to $ f(x/t)$ and $ g(x/t)$. Now, the left hand side scales as $ t$ and the right hand side as $ t^{\frac 1p\plus{}\frac 1q}$. This doesn't explain why everything works when $ \frac 1p\plus{}\frac 1q\equal{}1$ but, at least, explains why it doesn't work otherwise.

The answer in my opinion is just one word: Convexity. That is certainly what I think of when I see the equation $ 1/p\plus{}1/q\equal{}1$. If you look at Rudin's Real and Complex Analysis, you'll find his chapter on $ L^p$ spaces starts with a section on convex functions.

Fedja mentioned the inequality $ xy\leq \frac{x^p}{p}\plus{}\frac{y^q}{q}$. That inequality, and some clever manipulations, are really all that is behind the basic results for $ L^p$ spaces, conjugate indices, duality, etc. And that inequality comes directly from the convexity of the exponential function: $ e^{(1/p)\log{x^p}\plus{}(1/q)\log{y^q}}\leq (1/p)e^{\log{x^p}}\plus{} (1/q)e^{\log{y^q}}$.

Folland remarks in his Real Analysis:
[quote] In some respects it is unfortunate that $ L^p$ spaces were not named $ L^{1/p}$ spaces[/quote]
because of the role of linear relations in $ 1/p$.

[quote="mlok"]because of the role of linear relations in $ 1/p$.[/quote]
For instance, Young's inequality for convolutions: $ \|f*g\|_r\le\|f\|_p\|g\|_q,$ whenever $ \frac1r\equal{}\frac1p\plus{}\frac1q\minus{}1$ and $ p,q,r\in[1,\infty].$

Thanks-- I'll look at Rudin, we skipped convexity... so maybe that's why I'm confused.


calculus
integration
algebra proposed
algebra

Assume that the cubiic equation $4x^3-(a-2)x-(a+4)=0\ (a\in{\mathbb{Z}})$ has a positive rational root not integer.
Find the root satisfying this property.

You said quartic and gave a cubic.  The roots are stated to all be positive rationals but sum to zero (no $x^2$ term).  Some mistake?

He didn't say that [i]all[/i] roots are positive rationals.
I think what he meant was that the equation has at least one positive rational root :roll:

let the non-integral rational root be p/q. then q must divide 4 => q is 2 or 4.

divide the above integral by (qx-p) for each case of p.  factorise and then u'll get an eqn in a and p.  but values of p are limited.

To everyone:Sorry for that I made a typo.I have just edited.

kunny


function
trigonometry
calculus
derivative
inequalities proposed
inequalities

Let $\alpha ,\ \beta ,\ \gamma$ be real numbers. Find the maximum value of $k$ satisying $\alpha^{2}+\beta^{2}+\gamma^{2}+\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma\geq k$.

For any real $x$, we have

$\left|x\right|\geq\left|\sin x\right|\ \Longleftrightarrow\ x^{2}\geq\sin^{2}x\ \Longleftrightarrow x^{2}+\cos^{2}x\geq \sin^{2}x+\cos^{2}x=1$.

Needless to say, equality holds if and only if $x=0$.

[i]Very[/i] meaningful problem...

  darij

Thank you! :) I have no comment more than this. That's very background of this problem.

Just minimize the function x^2+(cosx)^2
As the function is continuous and bounded at the bottom (the function is always grater than 0), the minimal point will be achieved when derivative is 0.
The derivative is (x^2+(cosx)^2)'=2x-2cosxsenx, and is equal with 0 when x=cosxsenx, but this can only happen into the [-1,1] bound. If we plug x=0 into the formula, then we have 0=cos0sen0 which is true. Function f=cosxsenx has derivative (cosx)^2-(senx)^2. As f passes through (0,0), the derivative for this point is 1, so, the y=x line is tangent to senxcosx on (0,0), (cosx)^2-(senx)^2 decreases when x goes from 0 to 1 or -1, so the function cosxsenx is between the x axis and the funcition y=x, this means the only x for which x=cosxsenx, is 0. So, the minimum of x^2+(cosx)^2 is 1 (x=0). Then k=3.


analytic geometry
geometry
geometric transformation
reflection
inequalities

In Cartesian coordinate, let be given two points $A(0,6),B(2,5)$ and line $(d): x-2y+2=0$. Find point $M \in (d)$ such that $|MA-MB|$ takes the greatest value.

Reflect A through d to A', then we get A': (4,-2). Hence by triangular inequality, $MA-MB =MA'-MB \leq AB = \sqrt{17}$, thus $M= A'B \cap d$. So $M: \left( \frac{38}{7}, \frac{26}{7}\right)$.


calculus function integration geometry trapezoid algebra polynomial

HI

I need help with this.
Work out the extended newton cotes formula (with error term) for the case d = 3.

thanks

[quote="Cezar Lupu"]Do you mean Newton-Cotes quadrature formulae? :)[/quote]

yup
thanks

numero1: Numerical analysis is not number theory; your post does not belong here. Numerical analysis is also not complex analysis. It might be calculus - real analysis, but I think the best place to collect together numerical analysis questions would be in College Playground - Advanced Fields.

A moderator should move this topic.

[quote="Cezar Lupu"]I'll post it later with the proof. Now I have to go! ;)[/quote]

Bro, can you post it. I need as soon as possible. :lol:
Thanks a lot :)

I think googling might help :) :
http://www.google.com/search?q=newton+cotes+quadrature+formula

[quote="Cezar Lupu"]Well here it is in six knots. If $f: [a,b]\to\mathbb{R}$ is a $C^{6}[a, b]$ function then there is $c\in (a, b)$ such that

    \[ \int_{a}^{b} f(x)dx=\frac{(b-a))}{90}[7f(a)+32f(\frac{3a+b}{4})+12f(\frac{a+b}{2})+32f(\frac{a+3b}{4})+7f(b)]-\frac{(b-a)^{7}}{1935360}\cdot f^{(6)}(c).  \]

   This is a quite strong quadrature formulae. :)[/quote]

Was that for the case d = 3?
Could you explain what all that means...I am still very confused.
thanks

A Newton-Cotes formula involves equally spaced points within an interval. A "closed" Newton-Cotes formula would use the endpoints; an "open" formula would not.

The closed Newton-Cotes formula using two points is the trapezoid rule, and is a second order method - it integrates first degree polynomials exactly and has an error that depends on the second derivative. The 3-point closed Newton-Cotes formula is what is usually taught in calculus textbooks as Simpson's rule. It's a 4th order method. THe 4-point closed Newton-Cotes forumla is also 4th order. Cezar Lupu has given the formula for the 5-point closed Newton-Cotes formula, which is (as you can see from the error formula provided), 6th order.

One can devise these formulas for all orders, although there are some diminishing returns aspects of pushing them too far.


calculus
integration
real analysis
real analysis unsolved

Does anyone know how to solve $\int_{-\infty}^\infty e^{-x^{2}}dx$?

Yes. It's [i]very[/i] well-known. Here's the most recent [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=150137]appearance[/url].

You may also want to look [url=http://en.wikipedia.org/wiki/Gaussian_integral]here[/url]

Just as a side note, when I read this post it came to my mind that, according to a book I read some time ago, this was Lord Kelvin's favourite integral. It is said, that, during one of his lectures to engineering students he uses the word "mathematician". He then stops, turns to his students and ask them "Do you know what a mathematician is?". Then Lord Kelvin goes to the blackboard and write in it the above integral. Then he says to the class: "A mathematician is someone for whom this integral is as obvious as that twice two makes four is to you."


If f:R→R, solve this functional equation.

f(x-f(y))=f(f(y))+xf(y)+f(x)-1 for all x, y∈R.

This is IMO 99/6th problem.
[url]http://www.mathlinks.ro/viewtopic.php?p=131856#131856[/url]


trigonometry
factorial
real analysis
real analysis unsolved

First, define:

$ f(x) \equal{} \sum ^\infty_{n \equal{} 0} \frac {x^{4n \plus{} 2}}{(4n \plus{} 2)!}$

Successive differentiating gives:

$ f'(x) \equal{} \sum ^\infty_{n \equal{} 0} \frac {(4n \plus{} 2) x^{4n \plus{} 1}}{(4n \plus{} 2)!} \equal{} \sum ^\infty_{n \equal{} 0} \frac {x^{4n \plus{} 1}}{(4n \plus{} 1)!}$

$ f''(x) \equal{} \sum ^\infty_{n \equal{} 0} \frac {(4n \plus{} 1) x^{4n}}{(4n \plus{} 1)!} \equal{} \sum ^\infty_{n \equal{} 0} \frac {x^{4n}}{(4n)!}$

$ f'''(x) \equal{} \sum ^\infty_{n \equal{} 0} \frac {(4n) x^{4n \minus{} 1}}{(4n)!} \equal{} \sum ^\infty_{n \equal{} 1} \frac {x^{4n \minus{} 1}}{(4n \minus{} 1)!}$

$ f^{(4)}(x) \equal{} \sum ^\infty_{n \equal{} 0} \frac {(4n \minus{} 1) x^{4n \minus{} 2}}{(4n \minus{} 1)!} \equal{} \sum ^\infty_{n \equal{} 1} \frac {x^{4n \minus{} 2}}{(4n \minus{} 2)!} \equal{} \sum ^\infty_{n \equal{} 0} \frac {x^{4n \plus{} 2}}{(4n \plus{} 2)!} \equal{} f(x)$

From this we can find $ f$:

$ f(0) \equal{} 0$

$ f'(0) \equal{} 0$

$ f''(0) \equal{} 1$

$ f'''(0) \equal{} 0$

$ f^{(4)}(x) \equal{} f(x)$

This is a linear differential equation so we can observe the characteristic equation (define $ f(x) \equal{} e^{rx}$):

$ r^4 \equal{} 1$

$ r_1 \equal{} 1\quad r_2 \equal{} \minus{} 1\quad r_3 \equal{} i \quad r_4 \equal{} \minus{} i$

And this boils down to:

$ f(x) \equal{} C_1 e^x \plus{} C_2 e^{ \minus{} x} \plus{} C_3 \sin x \plus{} C_4 \cos x$

Now, from the initial conditions we have:

$ (1) \qquad f(x) \equal{} C_1 e^x \plus{} C_2 e^{ \minus{} x} \plus{} C_3 \sin x \plus{} C_4 \cos x \Rightarrow f(0) \equal{} C_1 \plus{} C_2 \plus{} C_4 \equal{} 0$

$ (2) \qquad f'(x) \equal{} C_1 e^x \minus{} C_2 e^{ \minus{} x} \plus{} C_3 \cos x \minus{} C_4 \sin x \Rightarrow f'(0) \equal{} C_1 \minus{} C_2 \plus{} C_3 \equal{} 0$

$ (3) \qquad f''(x) \equal{} C_1 e^x \plus{} C_2 e^{ \minus{} x} \minus{} C_3 \sin x \minus{} C_4 \cos x \Rightarrow f''(0) \equal{} C_1 \plus{} C_2 \minus{} C_4 \equal{} 1$

$ (4) \qquad f'''(x) \equal{} C_1 e^x \minus{} C_2 e^{ \minus{} x} \minus{} C_3 \cos x \plus{} C_4 \sin x \Rightarrow f'''(0) \equal{} C_1 \minus{} C_2 \minus{} C_3 \equal{} 0$

Summing up $ (2)$ and $ (4)$ we get $ C_1 \equal{} C_2$. Also, $ (1)$ and $ (3)$ give us $ C_1 \plus{} C_2 \equal{} \frac {1}{2} \Rightarrow C_1 \equal{} C_2 \equal{} \frac {1}{4}$. The rest is easy - $ C_3 \equal{} 0$ and $ C_4 \equal{} \minus{} \frac {1}{2}$.

This gives us:

$ f(x) \equal{} \frac {1}{4}\left(e^x \plus{} e^{ \minus{} x}\right) \minus{} \frac {1}{2} \cos x \equal{} \frac {1}{2}\left(\cosh x \minus{} \cos x\right)$

Now, all that we have to see is that your quotient can be written as:

$ Q \equal{} \frac {\sum ^\infty_{n \equal{} 0} \frac {1}{(4n)!} \left(\frac {\pi}{2}\right)^{4n}}{\sum ^\infty_{n \equal{} 0} \frac {1}{(4n \plus{} 2)!} \left(\frac {\pi}{2}\right)^{4n}} \equal{} \frac {f''\left(\frac {\pi}{2}\right)}{\left(\frac {\pi}{2}\right)^{ \minus{} 2} f\left(\frac {\pi}{2}\right)} \equal{} \left(\frac {\pi}{2}\right)^2 \frac {f''\left(\frac {\pi}{2}\right)}{f\left(\frac {\pi}{2}\right)}$

To proceed, we have to calculate $ f''(x)$:

$ f'(x) \equal{} \frac {1}{2}\left(\sinh x \plus{} \sin x\right)$

$ f''(x) \equal{} \frac {1}{2}\left(\cosh x \plus{} \cos x\right)$

And the result is:

$ Q \equal{} \left(\frac {\pi}{2}\right)^2 \frac {\cosh \frac {\pi}{2} \minus{} \cos \frac {\pi}{2}}{\cosh \frac {\pi}{2} \plus{} \cos \frac {\pi}{2}} \equal{} \left(\frac {\pi}{2}\right)^2$

To sum up:

$ \boxed{\boxed{\frac {1 \plus{} \frac {{\pi}^{4}}{2^{4}\cdot 4!} \plus{} \frac {{\pi}^{8}}{2^{8}\cdot 8!} \plus{} \frac {{\pi}^{12}}{2^{12}\cdot 12!} \plus{} \cdots}{\frac {1}{2!} \plus{} \frac {{\pi}^{4}}{2^{4}\cdot 6!} \plus{} \frac {{\pi}^{8}}{2^{8}\cdot 10!} \plus{} \frac {{\pi}^{12}}{2^{12}\cdot 14!} \plus{} \cdots} \equal{} \frac {\pi^2}{4}}}$

Though, in retrospect I suppose, one might have also immediately written the numerator as
\[ \sum_{k = 0}^\infty \left(\frac {\pi}{2}\right)^{4k}\frac {1}{(4k)!} = \frac {1}{2}\left(\cosh \frac {\pi}{2} + \cos \frac {\pi}{2}\right)\]
and the denominator as
\[{ \sum_{k = 0}^\infty \left(\frac {\pi}{2}\right)^{4k}\frac {1}{(4k + 2)!} = \left(\frac {2}{\pi}\right)^2\sum_{k = 0}^\infty \left(\frac {\pi}{2}\right)^{4k + 2}\frac {1}{(4k + 2)!} = \left(\frac {2}{\pi}\right)^2\left(\frac {1}{2}\right)\left(\cosh \frac {\pi}{2} - \cos \frac {\pi}{2}\right)}\]
This is of course in retrospect after glancing at milin's solution, but it's not too unnatural an approach; any time you see a sum of powers of a constant over factorials it's worth eyeballing it to see if you can manipulate it into some combination of trigonometric and exponential series by canceling out unwanted terms.


MATHCOUNTS
AMC
AIME
USA(J)MO
USAMO
AMC 8

im in 5th grade, so how do i join the actual competition? :o  :?:

You'll have to talk to your principal about participating.

When does the actual competition take place? Is there even a winner?

No, just a plaque or something, I believe.

Oh, well that is still rewarding.... ( :first: )

Is MOEMS even that hard, or is it really easy?

Presuming that this test includes 4th graders, its pretty easy.

There are different tests for different grade levels.

In comparison to MathCounts, it's fairly easy. Probably around the same level as the AMC 8.

[quote="AIME15"]I've heard that the AMC 8 is around the same level as MC...

I'd say that Chapter MC Sprint is about as hard as AMC 8, but when you get to the Target and Team Rounds, and higher levels, it becomes harder.

The time restriction on MOEMS is a lot more generous than those on AMC 8 and MathCounts, yet the problems aren't much harder, if at all.[/quote]

1.) AMC 8 is extremely easy. The only problems that are MC Chapter level are like the last 7 of them. I was actually just working on a copy of AMC 8 from 2001. It wasn't as hard as MC. It was about the same level as school.

2.) I think that you get a pretty good amount of time for AMC 8.

You get 40 minutes for 25 problems, which is much less than 3-5 minutes per problem on MOEMS.

Oh!!! Well I didn't expect MOEMS to have so much time. Usually the AMC competitions aren't based on time. At the more difficult rounds, like AIME, USAMO, and IMO you get a lot of time.


If $ a, b$ and $ c$ are positive integers, $ 0 < a < b < c < 10$ and
$ \frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c} \equal{} 1$, what is the value of $ c$?

If $ a \geq 3$ clearly the maximum sum is $ \frac13\plus{}\frac14\plus{}\frac15<1$

Therefore a=2.

WE now have $ 2<b<c<10$ and $ \frac1b\plus{}\frac1c\equal{}\frac12$

If $ b\geq4$ then the maximum sum is $ \frac14\plus{}\frac15<1$

Therefore b=3 and then we have $ c\equal{}\boxed{6}$


Encuentre todos los pares $(a,b)$ de números naturales distintos tales que al invertir el orden de los dígitos de $a^{b}$ se obtiene $b^{a}$.

[color=blue]¿Inviertes el orden de los dígitos de cada variable o del resultado de la operación entre ellas?[/color]

Inviertes el orden de los dígitos de la expresión decimal  del número natural $a^{b}$ y obtienes la expresión decimal del número natural $b^{a}$.

[color=blue]¿Es este un problema de polinomios?[/color]

Creo que no. Al menos mi solucion no va por ahi.


limit
logarithms
geometry
geometric transformation
reflection
algebra
polynomial

My friends (in high school) asked me if there were any formulas to calculate the sum:
$P=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{n}$
I believe that It's impossible but I cannot give them a solution to claim...
Can anybody help me?
Thanks.

Well, it is asymptotic to $\log(n)$ in the sense that
\[ \lim_{n\to\infty}\frac{1}{\log n}\cdot \sum^n_{i=1} \frac{1}{i} = 1  \]
so any closed form needs to reflect that in the long term.  This rules out polynomials and rational functions and whatnot.

Another question:
I've been starting at calculus, can anybody tell me which is the best way to know if [u]a function[/u] has [u]limit[/u] or no?
Thanks.


Demonstratzi ca MP + MH >= MG + MA pt n variabile pozitive, unde

MP=media patratica
MH=media armonica
MG=media geometrica
MA=media aritmetica

Ar fi utila, daca ar fi adevarata.
[u]Contraexemplu[/u]: $n=4$, $a_{1}=1$, $a_{2}=a_{3}=a_{4}=2$ (se verifica).
Deci nu mai e nevoie sa incerci sa o demonstrezi  :)

pt n=4 , a1 = 1, a2=a3=a4=2 inegalitatea este adevarata...

si este adevarata in general...verifica-ti calculele.

Tu sustii ca $\frac{8}{5}+\frac{\sqrt{13}}2 \geq \frac74+\sqrt[4] 8$? Ce e in stanga e $\approx 3.40$, iar ce e in dreapta e $\approx 3.43$...

...:| am calculat eu prost inseamna...eu am gasit problema pt 2 variabile a si b ...si era data si generalizarea....asta in mai multe cartzi...deci nu e inventata sau ceva:)

Cred ca ce ai gåsit tu era [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=sierpinski&t=52180]Inegalitatea lui Sierpinski[/url]. O inegalitate e falså atunci cand nu e adevaratå în fiecare caz, deci nu avea cum så fie asta. :!: Când postezi o ipotezå, poti face si câteva verificåri.


geometry circumcircle inradius trigonometry inequalities perimeter algebra

[quote="Stun"]Prove that
$h_a+h_b+h_c\le 3(R+r)$
with $h_a,h_b,h_c$ are the altitudes, $R$ is the circumradius and $r$ is the inradius[/quote]

One solution I immediately have is the following one:

It is well-known that $h_a=2R\sin B\sin C$, $h_b=2R\sin C\sin A$, $h_c=2R\sin A\sin B$ and $R+r=R\left(\cos A+\cos B+\cos C\right)$, so that your inequality

$h_a+h_b+h_c\leq 3\left(R+r\right)$

can be rewritten as

$2R\sin B\sin C+2R\sin C\sin A+2R\sin A\sin B\leq 3R\left(\cos A+\cos B+\cos C\right)$,

what simplifies to

$2\left(\sin B\sin C+\sin C\sin A+\sin A\sin B\right)\leq 3\left(\cos A+\cos B+\cos C\right)$,

or, equivalently,

$\sin B\sin C+\sin C\sin A+\sin A\sin B\leq \frac32\left(\cos A+\cos B+\cos C\right)$.

But this is easy: Since we have

$\sin B\sin C+\sin C\sin A+\sin A\sin B\leq \left(\cos A+\cos B+\cos C\right)^2$

(see http://www.mathlinks.ro/Forum/viewtopic.php?t=5716 ) and

$\cos A+\cos B+\cos C\leq \frac32$,

we have

$\sin B\sin C+\sin C\sin A+\sin A\sin B\leq \left(\cos A+\cos B+\cos C\right)^2$
$=\left(\cos A+\cos B+\cos C\right)\cdot\left(\cos A+\cos B+\cos C\right)\leq \frac32\cdot\left(\cos A+\cos B+\cos C\right)$,

and we are done.

But I think there is also a more elegant solution: Consider the case when the triangle ABC is acute-angled. Then, a [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24870]well-known formula[/url] states that AH + BH + CH = 2 (R + r), where H is the orthocenter of triangle ABC; thus, $R+r=\frac{AH+BH+CH}{2}$, and your inequality $h_a+h_b+h_c\leq 3\left(R+r\right)$ rewrites as $h_a+h_b+h_c\leq 3\frac{AH+BH+CH}{2}$, or, equivalently, $2\left(h_a+h_b+h_c\right) \leq 3\left(AH+BH+CH\right)$. But if A', B', C' are the feet of the altitudes of triangle ABC, then we have $h_a=AA^{\prime}=AH+HA^{\prime}$, and similarly $h_b=BH+HB^{\prime}$ and $h_c=CH+HC^{\prime}$, so that the inequality $2\left(h_a+h_b+h_c\right) \leq 3\left(AH+BH+CH\right)$ rewrites as

$2\left(\left(AH+HA^{\prime}\right)+\left(BH+HB^{\prime}\right)+\left(CH+HC^{\prime}\right)\right) \leq 3\left(AH+BH+CH\right)$.

This simplifies to $2\left(HA^{\prime}+HB^{\prime}+HC^{\prime}\right)\leq AH+BH+CH$, what immediately follows from the Erdös-Mordell inequality for the triangle ABC and its interior point H (with HA', HB', HC' being the distances from this point H to the sides BC, CA, AB of triangle ABC). But I don't succeed to extend this solution to the case of obtuse-angled triangles ABC. Can someone help?

  Darij

Here  is a  stronger  inequality: Let  [tex] w_a , w_b , w_c [/tex]  are  bisectors  of  the  angles  A , B , C . Then  [tex] w_a+w_c+w_c\leq3(R+r) [/tex]

Prove :
[tex] (\Sigma w_a)^2=(\Sigma\frac{2bc}{b+c}cos\frac{A}{2})^2\leq(\Sigma\sqrt{bc}cos\frac{A}{2})^2\leq\Sigma{ab}\Sigma{cos^2\frac{A}{2}\leq4(R+r)^2\frac{9}{4}[/tex] , so  we  have  the  result.
I hope  you  know  how  to prove  the  well  know  ineqs:
1) [tex] \Sigma ab\leq4(R+r)^2[/tex]
2)[tex]\Sigma cos^2\frac{A}{2}\leq\frac{9}{4}[/tex]

[quote="Stun"]If $ h_a,h_b,h_c$ are the altitudes of a triangle $ ABC$, $ R$ is the circumradius and $ r$ is the inradius, prove that

$ h_a \plus{} h_b \plus{} h_c\le 3(R \plus{} r)$.[/quote]Actually, the following stronger inequality holds

$ h_a \plus{} h_b \plus{} h_c\le 2R \plus{} 5r$,

with equality if and only if $ \triangle ABC$ is equilateral.

[b]Proof[/b] A formula states that

$ h_a \plus{} h_b \plus{} h_c \equal{} \frac {s^2 \plus{} r^2 \plus{} 4Rr}{2R}$,

where $ s$ is the semi-perimeter of $ \triangle ABC$, and the inequality rewrites as

$ s^2\leq 4R^2 \plus{} 6Rr \minus{} r^2$,

which immediately follows from the well-known inequalities $ s^2 \leq 4R^2 \plus{} 4Rr \plus{} 3r^2$ and $ R\geq 2r$.

On the other hand, the Horizon Mood's inequality holds

$ h_a \plus{} w_b \plus{} m_c\le 3(R \plus{} r)$,

where $ w_b$ is the angle-bisector and $ m_c$ is the median.

[b]Proof[/b] It is well-known that

$ h_a \equal{} \frac {2\sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}}{a}, w_b \equal{} \frac {2\sqrt {cas(s \minus{} b)}}{c \plus{} a},m_c \equal{} \frac {\sqrt {2a^2 \plus{} 2b^2 \minus{} c^2}}{2}$,

$ R \equal{} \frac {4abc}{\sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}},r \equal{} \sqrt {\frac {(s \minus{} a)(s \minus{} b)(s \minus{} c)}{s}}$,

where $ a,b,c$ are sides of $ \triangle ABC$.

Substitute $ a \equal{} y \plus{} z,b \equal{} z \plus{} x,c \equal{} x \plus{} y$, the geometric inequality is equivalent to the following algebraic inequality:

$ 8xyz(x \plus{} y \plus{} z)(2y \plus{} z \plus{} x) \plus{} 8y(y \plus{} z)(x \plus{} y \plus{} z)\sqrt {zx(x \plus{} y)(y \plus{} z)}$

$ \plus{} 2(y \plus{} z)(2y \plus{} z \plus{} x)\sqrt {xyz(x \plus{} y \plus{} z)[4z(x \plus{} y \plus{} z) \plus{} (x \minus{} y)^2]}$

$ \leq 3(y \plus{} z)^2(z \plus{} x)(x \plus{} y)(2y \plus{} z \plus{} x) \plus{} 12xyz(y \plus{} z)(2y \plus{} z \plus{} x)$

for positive real numbers $ x,y,z$.

Since $ [y(z \plus{} x) \plus{} 2zx]^2 \minus{} 4zx(x \plus{} y)(y \plus{} z) \equal{} y^2(z \minus{} x)^2\geq 0$, we have to prove

$ 2(y \plus{} z)(2y \plus{} z \plus{} x)\sqrt {xyz(x \plus{} y \plus{} z)[4z(x \plus{} y \plus{} z) \plus{} (x \minus{} y)^2]}$

$ \leq 3(y \plus{} z)^2(z \plus{} x)(x \plus{} y)(2y \plus{} z \plus{} x) \plus{} 12xyz(y \plus{} z)(2y \plus{} z \plus{} x)$

$ \minus{} 8xyz(x \plus{} y \plus{} z)(2y \plus{} z \plus{} x) \minus{} 4y(y \plus{} z)(x \plus{} y \plus{} z)[y(z \plus{} x) \plus{} 2zx]$

$ \equal{} (3y^2 \minus{} 2yz \plus{} 3z^2)x^3 \plus{} (5y^3 \plus{} 9yz^2 \plus{} 6z^3)x^2 \plus{} (y \plus{} z)^2(2y^2 \plus{} 8yz \plus{} 3z^2)x$

$ \plus{} yz(y \plus{} z)^2(2y \plus{} 3z)$,

or, equivalently,

$ F(x,y,z)\equiv [(3y^2 \minus{} 2yz \plus{} 3z^2)x^3 \plus{} (5y^3 \plus{} 9yz^2 \plus{} 6z^3)x^2 \plus{} (y \plus{} z)^2(2y^2 \plus{} 8yz \plus{} 3z^2)x$

$ \plus{} yz(y \plus{} z)^2(2y \plus{} 3z)]^2 \minus{} 4xyz(y \plus{} z)^2(x \plus{} y \plus{} z)(2y \plus{} z \plus{} x)^2[4z(x \plus{} y \plus{} z) \plus{} (x \minus{} y)^2]$

$ \geq 0$,

with equality if and only if $ x \equal{} y \equal{} z$.

[i]Case 1[/i]: $ x \equal{} \min\{x,y,z\}.F(x,x \plus{} s,x \plus{} t)$

$ \equal{} 768(4s^2 \minus{} 8st \plus{} 5t^2)x^8 \plus{} 64(112s^3 \minus{} 100s^2t \minus{} 84st^2 \plus{} 161t^3)x^7$

$ \plus{} 16(420s^4 \plus{} 152s^3t \minus{} 664s^2t^2 \plus{} 546st^3 \plus{} 699t^4)x^6$

$ \plus{} 16(196s^5 \plus{} 382s^4t \minus{} 100s^3t^2 \plus{} 299s^2t^3 \plus{} 963st^4 \plus{} 390t^5)x^5$

$ \plus{} 4(180s^6 \plus{} 796s^5t \plus{} 973s^4t^2 \plus{} 1806s^3t^3 \plus{} 3305s^2t^4 \plus{} 2284st^5 \plus{} 476t^6)x^4$

$ \plus{} 16(s \plus{} t)(4s^6 \plus{} 38s^5t \plus{} 108s^4t^2 \plus{} 264s^3t^3 \plus{} 326s^2t^4 \plus{} 145st^5 \plus{} 19t^6)x^3$

$ \plus{} 4t(s \plus{} t)^2(12s^5 \plus{} 112s^4t \plus{} 313s^3t^2 \plus{} 307s^2t^3 \plus{} 82st^4 \plus{} 5t^5)x^2$

$ \plus{} 4st^2(s \plus{} t)^3(16s^3 \plus{} 48s^2t \plus{} 40st^2 \plus{} 5t^3)x \plus{} s^2t^2(2s \plus{} 3t)^2(s \plus{} t)^4\geq 0$,

which is clearly true for

$ 112s^3 \minus{} 100s^2t \minus{} 84st^2 \plus{} 161t^3 \equal{} 4(28s \plus{} 31t)(s \minus{} t)^2 \plus{} t^2(52s \plus{} 37t)$,

$ 420s^4 \plus{} 152s^3t \minus{} 664s^2t^2 \plus{} 546st^3 \equal{} 2s(210s \plus{} 271t)(s \minus{} t)^2 \plus{} 2st(225s^2 \plus{} 2t^2)$.

[i]Case 2[/i]: $ y \equal{} \min\{x,y,z\}. F(y \plus{} t,y,y \plus{} s)$

$ \equal{} 768(5s^2 \minus{} 2st \plus{} t^2)y^8 \plus{} 64(161s^3 \plus{} 81s^2t \plus{} 23st^2 \plus{} 7t^3)y^7$

$ \plus{} 16(699s^4 \plus{} 1166s^3t \plus{} 716s^2t^2 \plus{} 74st^3 \plus{} 5t^4)y^6$

$ \plus{} 16s(390s^4 \plus{} 1281s^3t \plus{} 1523s^2t^2 \plus{} 457st^3 \plus{} 17t^4)y^5$

$ \plus{} 4s(476s^5 \plus{} 2660s^4t \plus{} 5545s^3t^2 \plus{} 3274s^2t^3 \plus{} 541st^4 \plus{} 4t^5)y^4$

$ \plus{} 16s^2(s \plus{} t)(19s^4 \plus{} 160s^3t \plus{} 454s^2t^2 \plus{} 191st^3 \plus{} 20t^4)y^3$

$ \plus{} 4s^2(s \plus{} t)^2(5s^4 \plus{} 86s^3t \plus{} 375s^2t^2 \plus{} 95st^3 \plus{} 5t^4)y^2$

$ \plus{} 4s^3t(s \plus{} t)^3(5s^2 \plus{} 44st \plus{} 5t^2)y \plus{} 9s^4t^2(s \plus{} t)^4\geq0$.

[i]Case 3[/i]: $ z \equal{} \min\{x,y,z\}.F(z \plus{} s,z \plus{} t,z)$

$ \equal{} 768(s^2 \plus{} 4t^2)z^8 \plus{} 64(7s^3 \plus{} 52s^2t \plus{} 148st^2 \plus{} 112t^3)z^7$

$ \plus{} 16(5s^4 \plus{} 102s^3t \plus{} 1080s^2t^2 \plus{} 1304st^3 \plus{} 420t^4)z^6$

$ \plus{} 16t(13s^4 \plus{} 549s^3t \plus{} 1804s^2t^2 \plus{} 1158st^3 \plus{} 196t^4)z^5$

$ \plus{} 4t( \minus{} 4s^5 \plus{} 521s^4t \plus{} 3302s^3t^2 \plus{} 5613s^2t^3 \plus{} 2044st^4 \plus{} 180t^5)z^4$

$ \plus{} 16t^2(16s^5 \plus{} 180s^4t \plus{} 561s^3t^2 \plus{} 566s^2t^3 \plus{} 110st^4 \plus{} 4t^5)z^3$

$ \plus{} 4st^2(5s^5 \plus{} 81s^4t \plus{} 424s^3t^2 \plus{} 751s^2t^3 \plus{} 460st^4 \plus{} 36t^5)z^2$

$ \plus{} 4s^2t^3(s \plus{} t)(5s^3 \plus{} 38s^2t \plus{} 72st^2 \plus{} 40t^3)z \plus{} s^2t^4(s \plus{} t)^2(3s \plus{} 2t)^2$

$ \equal{} 768(s^2 \plus{} 4t^2)z^8 \plus{} 64(7s^3 \plus{} 52s^2t \plus{} 148st^2 \plus{} 112t^3)z^7$

$ \plus{} 8(9s^4 \plus{} 204s^3t \plus{} 2160s^2t^2 \plus{} 2608st^3 \plus{} 840t^4)z^6$

$ \plus{} 16t(13s^4 \plus{} 549s^3t \plus{} 1804s^2t^2 \plus{} 1158st^3 \plus{} 196t^4)z^5$

$ \plus{} 4t^2(521s^4 \plus{} 3302s^3t \plus{} 5613s^2t^2 \plus{} 2044st^3 \plus{} 180t^4)z^4$

$ \plus{} 16t^2(16s^5 \plus{} 180s^4t \plus{} 561s^3t^2 \plus{} 566s^2t^3 \plus{} 110st^4 \plus{} 4t^5)z^3$

$ \plus{} 4st^2(3s^5 \plus{} 81s^4t \plus{} 424s^3t^2 \plus{} 751s^2t^3 \plus{} 460st^4 \plus{} 36t^5)z^2$

$ \plus{} 4s^2t^3(s \plus{} t)(5s^3 \plus{} 38s^2t \plus{} 72st^2 \plus{} 40t^3)z \plus{} s^2t^4(s \plus{} t)^2(3s \plus{} 2t)^2$

$ \plus{} 8z^2s^4(z^2 \minus{} st)^2\geq0$.
[quote="nickolas"]Let  $ w_a , w_b , w_c$  are  bisectors  of  the  angles  $ A , B , C$, then  $ w_a \plus{} w_c \plus{} w_c\leq3(R \plus{} r)$.[/quote]Here  is a  stronger  inequality:

$ w_a \plus{} w_c \plus{} w_c\leq 9r \plus{} k(R \minus{} 2r)$,

where $ k \equal{} 2.1707\cdots$ be a root of the following irreducible polynomial

$ 72k^5 \minus{} 1497k^4 \plus{} 12450k^3 \minus{} 51109k^2 \plus{} 102227k \minus{} 78656$;

with equality if $ b \equal{} c \equal{} 1,a \equal{} 0.62892\cdots$ be a root of the following irreducible polynomial

$ a^5 \plus{} 12a^4 \plus{} 39a^3 \plus{} 17a^2 \minus{} 7a \minus{} 14$.


Whats $ \frac{f_{1}}{5^{1}}+\frac{f_{2}}{5^{2}}+...+\frac{f_{n}}{5^{n}}+.......$ where $ f_{1}=1$,$ f_{2}=2$, and $ f_{n}=f_{n-1}+f_{n-2}$

Theres a simple common fractional answer to this.  Its just a simple trick..

call that sum S.  You can find an equation soley for S if you multiply that by 5 and subtract the originial S from it.


1. Prove that there exists an infinite sequence of pairwise co-prime positive $>1$, none of which divides a number of the form $2^{m}+2^{n}+1$, where $m,n \in \mathbb{N}$.

2*. [Open to me. ] Are there infinitely many primes which do not divide any number of the form $2^{m}+2^{n}+1$?

1. Consider $a=2^{p}-1$. If $ak=2^{m}+2^{n}+1,m\ge n$, then $k=-1+2^{l}a$ a - odd and $a2^{l+p}-2^{l}a-2^{p}=2^{m}+2^{n}.$
It give $l=p$ and $a2^{2p}+(a+1)=2^{m-p}+2^{n-p}$ contradition for prime p$\ge 3$.
Obviosly $(2^{p}-1,2^{q}-1)=1$ for different primes p and q.
2. Let $P|2^{p}-1$ prime divisor of Mersen's number (p prime). Then $r_{m}=2^{m}(mod \ P), r_{n}=2^{n}.$
$P|2^{m}+2^{n}+1\to r_{m}+r_{n}=P-1$.
Obviosly if P is mersen's prime it give contradition. If there are infinetely many Mersenn's primes, then 2. is true.
But it is true for many (I think for infinetely many) prime divisisors of Mersen's numbers.
For example $2^{11}-1=23*89$ and $2^{n}(mod \ 89)=1,2,4,8,16,32,64,39,78,67,45.$ Obviosly $r_{n}+r_{m}\not = 88$.


calculus
integration
calculus computations

If $ f(0) \equal{} g(0) \equal{} 0$ and  $ g'$ and $ f''$  are continuous, show that:
$ \int_{0}^{a} f(x)g''(x) dx \equal{} f(a)g'(a)\minus{}f'(a)g(a)\plus{}\int_{0}^{a} f''(x)g(x) dx$

That is straightfor using successively integration by parts.


I just got this puzzule from my friend. I want to share it with everyone:

  To walk one mile in desert, an elephant must eat one pound of bananas. The maximum load of bananas the elephant can walk with is 1000lb. A desert is 2000 miles wide and there is no banana in it. On both side of the desert an unlimited supply of bananas is stored.
1. Can the elephant cross the desert? [Hint: the elephant can make cashes.]
2. If the answer is possible, what is the shortest distance it should cover?
3. Can the elephant cross the desert if it is N miles wide? [N is a finite but maybe is very big number]

  Have fun! :D

i guess that the elephant could take trips back and forth and put a supply of bananas in the middle of the desert, and go back and repeat, i.e. go 500 mi, deposit 500 bananas, and do that 2 times, then go 500 mi past, and make sure to have enough bananas at each place, and in this way, the elephant could go infinite distance, since it could make check points with infinite bananas, and repeat

you mean, caches, like it can drop bananas in a part of the desert and return to them later(assuming it can find the bananas and they don't dessicate and all that jazz)

Mine is probably redudant, but:

1. Go 250 mi., drop 500, go back. (500 mi.)
2. Go 250 mi., pick up 250, go 250 more, drop 500, go back 250, pick up 250, go back. (1000 mi.)
3. Same as 1. (500 mi.)
4. Same as 2, but go out 250 more miles and drop 500 at the 750 mi. point. (1500 mi.)
5. Same as 1. (500 mi.)
6. Same as 2. (1000 mi.)
7. Same as 1. (500 mi.)
8. Same as 4, but go to the 1000 mi. point. (2000 mi.)
9. Repeat steps 1 - 8. (7500 mi.)
10. Go all the way across (2000 mi.)

Total: 17000 mi.

As I said, probably redundant.

Yeah, it could do something like
Take 1000, go some distance, drop some, use the rest of the bananas to come back.  Repeat.  Essentially that point in the desert now has an infinite number of bananas.  Hence the elephant has moved the starting point forward, and shortened the distance.  So the elephant can travel any distance required.


function
topology
real analysis
real analysis unsolved

Suppose that $: \mathbb R^{m}\to \mathbb R$ satisfies two conditions i)For each compact set $K,f(K)$ is compact. ii)For any nested decreasing sequence of compacts$(K_{n}),\ f(\bigcap K_{n})=\bigcap f(K_{n})$.
Prove that $f$ is continuous.

You have a confusing typo. State the range of f & where is K contained, just in case. It may be useful to know whether the range of f is open, closed, or whatnot.

Consider a sequence $x_{n}\to x$ in $\mathbb R^{n}$. We shall prove that $f(x_{n}) \to f(x)$.

Let $K_{n}= \{x_{i}: i\geq n\}\cup \{x\}$. Each set $K_{n}$ is compact. So, by the condition, each set $f(K_{n})$ is compact. Let $y$ be a limit point of the sequence $f(x_{i})$. Since all except a finite number of $f(x_{i})$ belong to $f(K_{n})$ and $f(K_{n})$ is closed, $y\in f(K_{n})$. We have
\[y\in \cap_{n=1}^\infty f(X_{n}) = f(\cap_{n=1}^\infty X_{n}) = f(\{x\}). \]
We proved that $y= f(x)$. So $f(x_{i})$ is a sequence in a compact set, whose only limit point is $f(x)$. Therefore, $f(x_{n}) \to f(x)$. This concludes the proof.

Here's my solution. Let $R^{p}$be the range of f and take a nonempty closed set $C \subset R^{p}$and let b be a limit point of $f^{-1}(C)$. Take a decreasing sequence of closed balls $(B(b;{\textstyle{1 \over n}})) = (K_{n})$ so that $\cap K_{n}= b$.

$\cap (f(K_{n}) \cap C) = ( \cap f(K_{n})) \cap C = \{ f(b)\}= f( \cap K_{n}) \cap C)$

Therefore $b \in f^{-1}(C)$. $f^{-1}(C)$ is closed.

I hope this is right. I did not use the first property so it might not be.  :oops:


In a coding system, a single bit is encoded as a triple, consisting of a short binary string S and 2 integers, K and N. A single "processing step" takes a binary string and replaces each 1 by 0, and each 0 by the input string S. The bit that was encoded can be recovered by applying this "processing step" successively K times, beginning with the string 1, and then extracting the N_th bit from the resulting string. Your job is to decode, that is, given S, K and N, extract the said bit.

INPUT
The input file will contain several test cases. The first line contains an integer t, the number of test-cases. The succeeding lines contain the test-cases.

Input Format:
Each test-case comprises the following:

Line 1: K
Line 2: N
Line 3: S

OUTPUT
         The output should comprise the solutions to all the test cases, in the order of the test cases in the input. There should NOT be any blank line in the output.

Output Format:
Line 1: 1 or 0, the N_th bit that is extracted after K processing steps.

POINTS and CONSTRAINTS
Total points: 200

Easy Set (50 points):   1<=t<=20, 0<=K<=2000, 1<=N<=1000, 1<=Length of S<=10.
Hard set:   1<=t<=20, 0<=K<=1000000, 1<=N<=100000000000000, 1<=Length of S<=10.

Time Limit: 5 seconds

SAMPLE INPUT
2
3
5
001
2
1
101

SAMPLE OUTPUT
0
1

Explanation: In the first test-case, the 3 processing steps are: 1 -> 0 -> 001 -> 0010010. The 5th bit of 0010010 is 0.

can sumone explain the solution for this

OK, here's a hint: suppose you start with "1" and do $n$ steps. Can you calculate the length of the resulting string? Can you calculate the length of the resulting string if you start with "0"? Can you calculate this for every possible $n$ from 1 to 1000000 and for both starting strings ("0" and "1") in linear time?


function
analytic geometry
graphing lines
slope
linear algebra

[quote="GameBot"]According to the linear function represented in this table, what is the value of $ x$ when $ y \equal{} 8$?
\[ \begin{tabular}{ c | c } x & y \\
\hline \minus{} 4 & 23 \\
1 & 20 \\
6 & 17 \\
\end{tabular}
\]
[/quote]

$ \Delta x \equal{} 5 \to \Delta y \equal{} \minus{} 3$.
So,
$ 8 \minus{} 17 \equal{} \minus{} 9 \equal{} 3\times( \minus{} 3) \equal{} 3 \times \Delta y$.
Therefore, $ 3\times\Delta x \equal{} 3 \times 5 \equal{} 15$.
Thus, $ 6 \plus{} 15 \equal{} \boxed{21}$.

[hide="Solution"]Since the function is linear, we can find the slope, then the equation, then substitute in $y=8$ to find $x$.
First, let's find the slope.
These are the given points:
$(-4, 23)$
$(1, 20)$
$(6, 17)$

Let's take the bottom two points and use the slope formula:
$\frac{17-20}{6-1}
=-\frac{3}{5}$

Therefore the slope is $-\frac{3}{5}$.

We use the point-slope formula to find the equation using the point $(1, 20)$.
$y-20 = -\frac{3}{5}(x-1)$

$y-20 =-\frac{3}{5}x + \frac{3}{5}$

$y = -\frac{3}{5}x + 20\frac{3}{5}$

Now we plug in $y=8$ into the equation:
$8 = -\frac{3}{5}x + 20\frac{3}{5}$

$\frac{3}{5}x = \frac{63}{5}$

$x = \boxed{21}$[/hide]

[quote="minimario"][quote="GameBot"]According to the linear function represented in this table, what is the value of $ x$ when
$ y \equal{} 8$?
\begin{tabular}{ c | c }
  x&y\\ \hline
   -4&23\\
   1&20\\
   6&17\\
\end{tabular}[/quote]

What is up with this graph[/quote]
Maybe he didn't put the $$.

[quote="GameBot"]
\begin{tabular}{ c | c }
  x&y\\ \hline
   -4&23\\
   1&20\\
   6&17\\
\end{tabular}[/quote]

[quote="Kouichi Nakagawa"]\[ \begin{tabular}{ c | c } x & y \\
\hline \minus{} 4 & 23 \\
1 & 20 \\
6 & 17 \\
\end{tabular}
\][/quote]

Notice the differences. Simply a coding error.


Can someone explain to me where my reasoning went wrong, becuase obviously the proof is wrong.

---------------------------------------------------------------------------------------------------
1*1 = 12
12 = 1
02 = 0

x2 = x
x = 1, 0

infinity2 = infinty
so if we substituted infinity into the equation x2 = x, we could say that infinity = 1 or infinty = 0

But obviously that's not true.

You can't treat infinity like a real number.  If you do a lot of things will be screwed up.  Infinity means growing without bound.  Since you can't define infinity as one exact number you can't use it in equations like it is one exact number.  The main point is that you have to be careful when working with infinity.

You really can't do infinity2 = infinity. It doesn't make sense. The closest you can get is

[tex]\displaystyle \lim_{x\rightarrow\infty}x^2 = \infty[/tex].

I get it now, infinity isn't a real number like 23 or something.

But what is [tex]\displaystyle \lim_{x\rightarrow\infty}x^2 = \infty[/tex]?

That is read the limit as x grows w/out bound of x squared goes to inifinity.  It just means that if you picked values of x to put into x^2 and each time you increased the value you chose that the result would keep getting bigger and bigger but it would not approach any value.  Limits aren't getting started but I'll leave this topic here since it was informational.

Does $\displaystyle \lim_{x\rightarrow\infty}x^2 = \infty$ mean that $x^{2}$ approaches $\infty$?

Yes as x approach infinity so does x^2.  This should be pretty easy to see.  If you look at a graph of it you will see a parabola.  The parabola keeps increasing without bound as the x value increases.

Right so as the value of x gets bigger and bigger (approaches infinity), the value of x^2 gets bigger and bigger (infinity), so the limit (what happens to f(x) as x approaches infinity) is equal to infinity.


The expressions $ a \plus{} b \plus{} c, ab \plus{} ac \plus{} bc,$ and $ abc$ are called the elementary symmetric expressions on the three letters $ a, b, c;$ symmetric because if we interchange any two letters, say $ a$ and $ c,$ the expressions remain algebraically the same. The common degree of its terms is called the order of the expression. Let $ S_k(n)$ denote the elementary expression on $ k$ different letters of order $ n;$ for example $ S_4(3) \equal{} abc \plus{} abd \plus{} acd \plus{} bcd.$ There are four terms in $ S_4(3).$ How many terms are there in $ S_{9891}(1989)?$ (Assume that we have $ 9891$ different letters.)

Am I missing something or is it:

$ \displaystyle {9891 \choose 1989}$

[quote="SimonM"]Am I missing something or is it:

$ \displaystyle 9891 \choose 1989$[/quote]

You are missing the fact that the \displaystyle command, the way you use it, only applies to the $ 9891$ and not the whole binomial coefficient. Brackets help:

$ \displaystyle {9891 \choose 1989}$.

Otherwise, the answer is right. Orl, what have I told you about the quality of ILL problems? ;)

  darij


The complex number $ w$ equals $ \minus{} 128 \minus{} 128\cdot\sqrt {3}\cdot i$, where $ i^2 \equal{} \minus{} 1$, and $ a$ and $ b$ are real numbers. There are four distinct complex numbers $ v$ such that $ v^4 \equal{} w$. Find these four numbers, and express them in $ a \plus{} bi$ form.

This is from the CJML website, right?

[hide="Solution"]
Let $ \text{\cis}\ \theta = \cos \theta + i \sin \theta$. Then $ -128-128\cdot 3 \cdot i = 256\ \text{cis}\ 240^{\circ}$, and so by De Moivre's, $ v = (256\ \text{cis}\ 240^{\circ})^{1/4} = 4 \text{cis}\ 60+90k$ where $ k \in \{0,1,2,3\}$. The answers are
\begin{align*}
4\ \text{cis}\ 60  &= 2 + 2\sqrt{3} i\\
4\ \text{cis}\ 150 &= -2\sqrt{3} + 2i\\
4\ \text{cis}\ 240 &= -2 - 2\sqrt{3} i\\
4\ \text{cis}\ 330 &= 2\sqrt{3} - 2i
\end{align*}
[/hide]


linear algebra
matrix
vector
advanced fields
advanced fields solved

What is a spectrum??????????? (Not sure if it belongs to this forum, if not please move it to the correct one).......... thanks.....

[quote="amfulger"]Spectrum of what?[/quote]

I don't know, any example.... I just heard this name and I wanted to know something about it.....

The set of the eigenvalues of a matrix  (in linear algebra).

[quote="eugene"]For example if it is spectrum of a linear operator $A$ in a vector space $V$, then the spectrum is the set of the eigenvalues of $A$, i.e. the set of $\lambda$ for which there exists $x \neq 0$ such that $Ax= \lambda x$.[/quote]

Thanks eugene. I find this: http://mathworld.wolfram.com/Spectrum.html


geometry
3D geometry
quadratics
trigonometry
algebra
quadratic formula

[quote="The Will"]How do I solve for x

x^3=216

I know its 6, but what are the steps to get the answer?

Thanks for any help.  :)[/quote]

You just cube root it to get the answer...

But actually techinically, there are 3 answers, except the other 2 are imaginary.
You can use DeMoivre's Therom and find the answers are actually

6(cos0+i*sin0)
6(cos 120+i*sin120)
6(cos 240+i*sin240) where i=sqrt{-1}

and the things are in degrees..

cube root of $216$ is it's only real answer.but to find the conjugate none-real roots you have to use the cubic formula.
it has 3 answer which are.
$x=6$
$x=-3-5.19615i$
$x=-3+5.19615i$

[quote="The Will"]How do I solve for x

x^3=216

I know its 6, but what are the steps to get the answer?

Thanks for any help.  :)[/quote]
[hide="All solutions"]$x^{3}= 216$
$x^{3}-216 = 0$
$(x-6)(x^{2}+6x+36) = 0$

Hence, $x=6$ or a root of $x^{2}+6x+36$. By the quadratic formula,
$x = 6,-3\pm 3i\sqrt{3}$[/hide]

EDIT: Grr.... I always forget the negative sign...

how did u change the roots from decimal to $x=3\pm3i\sqrt{3}$ ? isnt it $x=-3\pm3i\sqrt{3}$?
can you explain please how u changed from decimal to that form? is there a general rule?

[quote="The Will"]How do I solve for x

x^3=216

I know its 6, but what are the steps to get the answer?[/quote]

The first thing you might try is assuming $x$ is an integer.  That assumption works out nicely in this case, since you don't have to test many values of $x$ until that works, but if that assumption doesn't pan out, then $x$ is not going to be very nice at all, and all you can do is take a cube root with a calculator or what have you.

[quote="binomial_4eva"]is there a general rule?[/quote]

Yes.  We will be making use of Euler's formula,

$e^{ix}= \cos x+i \sin x$

And the corollary, De Moivre's Formula,

$(\cos x+i \sin x)^{n}= \cos nx+i \sin nx$

First we will find the cube roots of $1$.  To get the cube roots of $216$ we merely multiply by $6$.  Now, when we have

$(\cos x+i \sin x)^{3}= \cos 3x+i \sin 3x = 1$

We know that $3x = 0, 2\pi, 4\pi \implies x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}$.  Hence the three cube roots of $1$ are given by

$e^{0}= 1$
$e^{i \frac{2\pi}{3}}= \cos \frac{2\pi}{3}+i \sin \frac{2\pi}{3}=-\frac{1}{2}+i \frac{ \sqrt{3}}{2}$
$e^{i \frac{4\pi}{3}}= \cos \frac{4\pi}{3}+i \sin \frac{4\pi}{3}=-\frac{1}{2}-i \frac{\sqrt{3}}{2}$

Thank you all very much for your answers.  :lol:

Now I need to go figure out how to get the cube root "without a calculator"

How do you know that its x^3 is the cube root of 216? It is, but what there tells you that x is the cube root of 216?  :huh:

[quote="t0rajir0u"][quote="The Will"]but what there tells you that x is the cube root of 216?  :huh:[/quote]

That's the definition of a cube root.  If $x^{3}= 216$, then we say that $x$ is a cube root of $216$, just the same way if $x^{2}= 36$, then we say that $x$ is a square root of $36$.[/quote]

lol ok

Thanks makes sense.

why are people still posting roots of unity when like 50 people have already posted the same exact solution and also it doesnt belong here


algorithm
number theory
greatest common divisor
calculus
integration
induction
modular arithmetic

My friend showed me this problem. What does $ (a_1,...a_n)|c$ mean?

Show that given $ a_1,...a_n$, the equation $ a_1x_1\plus{}..a_nx_n\equal{}c$ has integer solutions iff $ (a_1,...a_n)|c$.
How can solutions be found using the Euclidean Algorithm and substitution?

$ (a_1, \ldots, a_n)$ is the greatest common divisor of the numbers $ a_1$, $ a_2$, ..., $ a_n$.

The vertical bar $ |$ denotes "divides". Thus the condition you quote says that $ c$ must be divisible by the greatest common divisor of the $ a_i$ in order for the linear equation to have a solution.

The key to this is to see how the Euclidean algorithm can be used to write the gcd of a set of numbers as a linear combination in those numbers. Start with two numbers, say 30 and 105. Apply the Euclidean algorithm to find their gcd. Now trace back and try to use the equations you got along the way to write the gcd $ d$ as a combination $ 30a \plus{} 105b \equal{} d$. Generalize.

- R

[quote="randomgraph"]

Now trace back and try to use the equations you got along the way to write the gcd $ d$ as a combination $ 30a \plus{} 105b \equal{} d$. Generalize.

- R[/quote]

What equations did we get along the way? How can the gcd of two numbers be a combination of the two numbers? Like 3 is the gcd of 3 and 15... What would be the point of writing $ 3\equal{}3*1\plus{}0*15$?

[hide="If GCD does not divide c then no solution"]If the GCD of the set of $ a_i$ does not divide $ c$, then note that any integral linear combination of all $ a_i$ will contain a factor that does not divide $ c$, and we have no solution.[/hide]

[hide="Simplification of other direction"]If the GCD of the set $ a_i$ does divide $ c$, then let $ a_i/GCD[\{ a_i\}] \equal{} b_i$, and let $ c/GCD[\{ a_i \}] \equal{} d$. Then, $ GCD[\{ b_i\}]\equal{}1$ and $ a_1 x_1 \plus{} \ldots \plus{} a_n x_n \equal{} c \iff b_1 x_1 \plus{} \ldots \plus{} b_n x_n \equal{} d$. It suffices to prove the latter, in which the GCD is 1.[/hide]

[hide="Induction"]
[b]Base case (n=2):[/b]

See [url]http://en.wikipedia.org/wiki/Euclidean_algorithm#Proof[/url]

[b]Inductive Step:[/b] Suppose it works for the case of $ n\equal{}m$. We want to show there is a solution to the case of $ n\equal{}m\plus{}1$, which takes the form:

$ (b_1 x_1 \plus{} \ldots \plus{} b_m x_m) \plus{} b_{m\plus{}1} x_{m\plus{}1} \equal{} d$

Let $ k$ be the GCD of the first $ m$ values of $ b_i$. Let $ f_i \equal{} b_i / k, \; \forall 1 \leq i \leq m$. We then have:

$ (f_1 x_1 \plus{} \ldots f_m x_m)k \plus{} b_{m\plus{}1} x_{m\plus{}1} \equal{} d$

We know that $ GCD[b_1 , \ldots b_{m\plus{}1}]\equal{}1 \implies GCD[k,b_{m\plus{}1}]\equal{}1$
We also know that $ GCD[f_1 \ldots f_m]\equal{}1$.

From the case of $ n\equal{}2$, we know there is a solution $ (x_*,x_{m\plus{}1})\equal{}(u,v)$ to $ kx_*\plus{}b_{m\plus{}1} x_{m\plus{}1} \equal{} d$.
From the case of $ n\equal{}m$, we know there is a solution to $ f_1 x_1 \plus{} \ldots f_m x_m\equal{}u$. Combining the two, we find that there is a solution to $ (b_1 x_1 \plus{} \ldots \plus{} b_m x_m) \plus{} b_{m\plus{}1} x_{m\plus{}1} \equal{} d$; that is, the case of $ n\equal{}m\plus{}1$

The induction is complete.[/hide]

Let's take those one by one...

[quote="aufha"]
What equations did we get along the way?
[/quote]
Let's say you're calculating the gcd of 63 and 13 using the Euclidean Algorithm. It looks something like this:

$ 63 \equal{} 4*13 \plus{} 11$
$ 13 \equal{} 1*11 \plus{} 2$
$ 11 \equal{} 5*2 \plus{} 1$

so the gcd is 1, and tracing back we see that

$ 1 \equal{} 11 \minus{} 5*2 \equal{} 11 \minus{} 5*(13 \minus{} 1*11) \equal{} \minus{} 5*13 \plus{} 6*11 \equal{} \minus{} 5*13 \plus{} 6*(63 \minus{} 4*13) \equal{} \minus{} 29*13 \plus{} 6*63$

and here you go: the gcd is a linear combination with integer coefficients of the original numbers.

[quote="aufha"]
How can the gcd of two numbers be a combination of the two numbers?
[/quote]
I'm not sure I understand what you're asking exactly. How is it possible? Well, we just saw an example, and you mentioned one yourself.

[quote="aufha"]
Like 3 is the gcd of 3 and 15... What would be the point of writing $ 3 \equal{} 3*1 \plus{} 0*15$?
[/quote]

I would write this as $ 3 \equal{} 1*3 \plus{} 0*15$ just to be consistent, but that is correct. What's the point? Well, the gcd of two integers is:

1. The largest number which divides them both, and
2. The smallest positive number which can be written as a linear combination (with integer coefficients) of the two of them.

This is actually a very important observation, useful in many number-theoretic and ring-theoretic problems and proofs. It is true that if one of the two numbers happens to [b]be[/b] the gcd, this expression doesn't seem very illuminating. That doesn't mean it's not helpful or useful in other cases.

The point of my previous post was to try and show the way for you to see how to solve the problem. If that wasn't clear enough, I hope that now it is - if not, please ask again.

- R

This is also known as the [url=http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity#Generalizations]generalized form[/url] of [url=http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity]Bezout's Identity[/url].

This not only holds for the integers $ \mathbb{Z}$, but also for any [url=http://mathworld.wolfram.com/EuclideanRing.html]Euclidean ring[/url].

[hide="An example of how to solve such a system"]
Take a look at the equation $ 7x \plus{} 5y \equal{} 11$. Remember that if $ (a,b) \nmid c$ (does not divide c), there would be no solutions. However, since $ (7,5) \equal{} 1 \mid 11$, we can proceed. To solve it, first eliminate the $ y$ term by writing the equation in $ \mod 5$. We have $ 7x \equal{} 11 \equiv 1 \pmod 5$. We add multiples of 5 until we can divide out the $ 7$: $ 7x \equiv 21 \pmod 5$, so $ x \equiv 3 \pmod 5$. If we let $ x \equal{} 5k \plus{} 3$ for $ k \in \mathbb{Z}$, we can solve for $ y$. A simple solution would be $ (3, \minus{} 2)$.
In the more general case, solutions to $ ax \plus{} by \equal{} c, (a,b) \equal{} 1$ are of the form $ \left(k,\displaystyle\frac {c \minus{} ak}{b}\right)$, with $ k$ being any solution to $ ak \equiv c \pmod b$.
[/hide]


Find $ f(10)$, if $ f(n) \equal{} n \minus{} 10$ when $ n > 100$ and $ f(n) \equal{} f(f(n \plus{} 11))$ in all other cases.

Please show me the smart way to do this problem? I think mine is just dumb, and probably wrong too.

[hide]We see that $ f(10) \equal{} f(f(f(f(f(f(f(f(f(f(109))))))))))$.

Simplyfing the $ f(f(f(109)))$, we see that it becomes $ f(f(99))$, becoming $ f(f(f(110))) \equal{} f(f(100)) \equal{} f(f(f(111))) \equal{} f(f(101)) \equal{} f(91) \equal{} f(f(102)) \equal{} f(92) \equal{} f(f(103)) \equal{} \ldots \equal{} f(98) \equal{} f(f(109))$. As we keep going, we see there is a pattern. Every pair of operations on the numbers increases the number by one, until attains a value above a $ 100$, at which time the number dips below $ 100$. Each time this cycle operation gets rid of one left parentheses.

I believe the answer is $ 91$? Please correct me if I'm wrong.[/hide]

No my friend, the answer is indeed $ 91$  :)  The official solution is also along these lines, although a bit more rigorous.

can someone guide me to where i can find about these kinda maths i mean the topic is it polynomials or something else,, a link or a site, or a book. 123456789 which theorem or what you used to solve these with?.please

@binomial_4eva, I didn't use a "theorem." I used logic.

binomial_4eva, if you notice that $ f(f(f(109))) \equal{} f(f(99)) \equal{} f(f(f(110))) \equal{} f(f(100)) \equal{} f(f(f(111))) \equal{} f(f(101)) \equal{} f(91) \equal{} f(f(102)) \equal{} f(92) \equal{} f(f(103)) \equal{} f(93) \equal{} f(f(104)) \equal{} f(94) \equal{} f(f(105)) \equal{} f(95) \equal{} f(f(106)) \equal{} f(96) \equal{} f(f(107)) \equal{} f(f(97)) \equal{} f(f(108)) \equal{} f(98) \equal{} f(f(109))$, (as 123456789 pointed out) then we basically have $ f(f(109)) \equal{} f(f(f(109)))$ and repeatedly applying this to $ f(10) \equal{} f(f(f(f(f(f(f(f(f(f(109))))))))))$ we have $ f(10) \equal{} f(f(109))$.  Similar to what we have above, $ f(f(109))) \equal{} f(99) \equal{} f(f(110)) \equal{} f(100) \equal{} f(f(111)) \equal{} f(101) \equal{} 91$, so our answer is indeed $ 91$.

No theorem needed, just lots of algebra.


geometry
trigonometry
rectangle
number theory
relatively prime
trig identities
Law of Cosines

I just drew pentagon $LINGO$, where $\overline{LI} \cong \overline{IN} \cong \overline{NG} \cong \overline{GO} \cong \overline{OL}$.  Also, $\overline{LI}\parallel\overline{NG}$ and $\overline{LI}\perp\overline{IN}$.  If $LN=\sqrt2$:

A) Compute the product of all the possible areas of pentagon $LINGO$.
B) Assume that $LINGO$ is a convex polygon.  If I drew $\overline{IO}$, its length could be expressed in the form $\frac{\sqrt{a+b\sqrt{c}}}d$.  Compute the ordered quadruple $(a,b,c,d)$ if $a,b,c,d\in\mathbb{N}$, $a$ and $b$ are relatively prime, and $c$ is prime.

[hide="A"]So basically you have square $LING$ and equilateral triangle $LGO$. $LN=\sqrt{2} \implies LI=IN=NG=GO=OL=LG=1$. Therefore the area is $1^2+\frac{1^2 \sqrt{3}}{4}= \boxed{1+\frac{\sqrt{3}}{4}}$.[/hide]

[hide="B"] By Law of cosines $IO^2 = 2-2\cos(\frac{5\pi}{6})=2+\sqrt{3}$
$IO=\sqrt{2+\sqrt{3}} \therefore (a,b,c,d) = \boxed{(2,1,3,1)}$[/hide]

Cincodemayo:
[hide]
Part A wants the product of all possible areas... There is one more :)
[/hide]

Hmm... Law of Cosines... didn't think of that.  I guess that works too.  What I had in mind was this:
[hide]Extend $\overline{LI}$ to point $X$ so that $\overline{LX}\perp\overline{OX}$.  $LX=1+\frac{\sqrt3}2$ and $OX=\frac12$, so apply the Pythagorean Theorem to get $\sqrt{2+\sqrt3}$.  So the answer would be $\boxed{2,1,3,1}$.[/hide]

And A is wrong:
[hide]Point $O$ could be also positioned inside square $LING$, making the area [hide]$1-\frac{\sqrt{3}}4$.  So the answer is [hide]$\frac{13}{16}$.[/hide][/hide][/hide]

I'll give out more parts to do.  If $LINGO$ is convex:
C) Extend $\overline{GO}$ to point $A$ so that $L$, $I$, and $A$ are collinear.  Draw point $B$ so that $N$, $G$, and $B$ are collinear.  Now draw $C$, the midpoint of $\overline{AB}$.  If the length of $LC$ can be expressed as $\frac{a+b\sqrt{c}}d$, compute the ordered quadruple $(a,b,c,d)$ if $a,b,c,d\in\mathbb{N}$, $a$ and $b$ share no common factor, and $c$ is prime.
D) Prove that $\bigtriangleup LGO \cong \bigtriangleup ABO$.
E) Name all line segments with a length of $\sqrt3$.
F) If you drew lines connecting $A$ to $I$ and $B$ to $N$, you'd form a quadrilateral, $ABNI$.  What is this quadrilateral's area?

These build upon each other.

[hide="Hint for C"]Pythagorean Theorem[/hide]
[hide="Second hint for C"]$\bigtriangleup LAC$[/hide]
[hide="Another method for C"]Law of Cosines[/hide]
[hide="Hint for E"]There are two.[/hide]

Assuming $\overline{LO}$ is extended to $B$...
[hide="C"]
If you draw it out, you have that $\triangle OAB$ is an equilateral triangle w/sides 1.  $OC$ would then be $\frac{\sqrt3}{2}$.  Since $m\angle LOC=150$.  Then by law of cosines, $LC^2=1+\frac34-\frac{\sqrt3}{2}\cos150=\frac{13}{4}$ so $LC=\frac{\sqrt{13}}{2}$ so $(a,b,c,d)=(0,1,13,2)$?
[/hide]

An error by saying $\frac{\sqrt3}2$ in your calculations, but you got the right answer in the end.

Again, assume that $LINGO$ is convex.
G) The length of $\overline{NC}$ can be expressed in the format $\frac{\sqrt{a+b\sqrt{c}}}d$, where $a,b,c,d\in\mathbb{N}$, $a$ and $b$ are relatively prime, and $c$ is prime.  Compute the ordered quadruple $(a,b,c,d)$.

Hints for D:
[hide]$\overline{LI}\parallel\overline{NG}$.  $L$, $I$, and $A$ are collinear, and $B$, $G$, and $N$ are collinear.  What can you conclude about $\overline{ILA}$ and $\overline{NGB}$?[/hide]
[hide]Alternate Interior Angles[/hide]
[hide]$\overline{LG}\parallel\overline{IN}$[/hide]

[hide="D"]
ASA:
both triangles are equilateral with side length 1: GBA and LAB are right. GAB and LBA are $60^\circ$ b/c LAG and LBG are $30^\circ$. $AB \cong IN =1$, and the same with $\triangle LOG$. I don't feel like typing out the formalities but you get the gist of it. [/hide]

[hide="E"] $\boxed{AL \text{ and }BG}$.$\angle ALB$ and $\angle BGA$ are $30^\circ$ and $AB$ is $1$ so $LB=AG=\sqrt{3}$.[/hide]

[hide="F"] $AINB$ is a rectangle (right angles), and $BN=BG-NG=\sqrt{3}-1$. So the area is $1(\sqrt{3}-1)=\boxed{\sqrt{3}-1}$[/hide]

[hide="G"] $NC^2 = NB^2 + BC^2 \implies NC=\sqrt{(\sqrt{3}-1)^2+(\frac{1}{2})^2}=\frac{\sqrt{17-8\sqrt{3}}}{2}$. So $(a,b,c,d)=\boxed{17,-8,3,4}$[/hide]


geometry
circumcircle
geometric transformation
homothety
ratio
power of a point
radical axis

Given a triangle $ ABC$. The angle bisectors of the angles $ ABC$ and $ BCA$ intersect the sides $ CA$ and $ AB$ at the points $ B_1$ and $ C_1$, and intersect each other at the point $ I$. The line $ B_1C_1$ intersects the circumcircle of triangle $ ABC$ at the points $ M$ and $ N$. Prove that the circumradius of triangle $ MIN$ is twice as long as the circumradius of triangle $ ABC$.

[color=darkblue][b]Remark[/b] ([u]Bulgaria, 1997[/u]). If the side $AB$ separates the points $M$ and $C$ then $\frac{1}{BM}=\frac{1}{AM}+\frac{1}{CM}$ and $\frac{1}{CN}{=\frac{1}{AN}+\frac{1}{BN}\ .}$[/color]

[color=darkred][b]Generalization[/b] ([u]own[/u]). Let $ABC$ be a triangle inscribed in the circle $w$ and for the points $X\in (AC)\ ,$ $Y\in (AB)$ define the points $\{M,N\}=XY\cap w\ .$ Then for any point $D\in \{M,N\}$ there is the following relation : \[ \boxed {\left|b\cdot \frac{YB}{YA}\cdot \frac{1}{DB}-c\cdot \frac{XC}{XA}\cdot \frac{1}{DC}\right|=a\cdot \frac{1}{AD}}\ . \] [b]Remark.[/b] If $I\in BX\cap CY$ then we obtain the above problem from Bulgaria, 1997.[/color]

Here is my solution to the problem.

I base my proof on the following lemma (I predicted this lemma, and fortunately it is true)

[i]Lemma[/i] Suppose $(C,h)$ and $(O, r)$ are two given distinct circles. A point $P$ moves in the circle $(C)$ such that the tangent at $P$ intersects $(O)$ at two points $A$ and $B$. Then the circumcenter $J$ of $ABC$ moves on a circle with center $O$.

[hide="Proof"] Up till now, I just have a brute force proof. Maybe I 'm not good enough today to find a synthetic one. But anyway, let $OC = m, AB =c, CA=b, BC = a, S_{ABC} = S, R_{ABC} = R$ and $M$ be the midpoint of $AB$. Then by Leibnitz theorem, we obtain
$a^2(b^2+c^2-a^2)OA^2 + b^2(c^2+a^2-b^2)OB^2 + c^2(a^2+b^2-c^2)OC^2$

$= \sum (2b^2c^2 - a^4) KJ^2+a^2(b^2+c^2-a^2)JA^2 + b^2(c^2+a^2-b^2)JB^2 + c^2(a^2+b^2-c^2)JC^2$

Thus $OJ^2 = r^2 - R^2 + \frac{(m^2-r^2)c^2(a^2+b^2-c^2)}{16S^2}$

$= r^2 - R^2 + \frac{(m^2-r^2)(2h^2+PA^2+PB^2-AB^2)}{4h^2}$

$= r^2 - R^2 + \frac{PA^2.PB^2}{4h^2} + \frac{(m^2-r^2)(2h^2+PA^2+PB^2-AB^2)}{4h^2}$

$= \frac{2r^2+2m^2-h^2}{4} + \frac{2(m^2-r^2)PA.PB - PA^2.PB^2 - h^2(PA^2+PB^2)}{4h^2}$

$= \frac{2r^2+2m^2-h^2}{4} + \frac{2(m^2-r^2)(PO^2- r^2) - (PO^2-r^2)^2 - h^2(2PO^2 +2r^2 - 4MO^2)}{4h^2}$

$= u + \frac{2m^2PO^2 - PO^4 - 2h^2PO^2 + 4h^2MO^2}{4h^2}$, here $u$ is a constant in term of $m, h, r$.

It is now sufficient to show that the numerator is a constant in term of $m, h, r$. We have $PO^2 = MO^2 + m^2 - (h-MO)^2 = m^2-h^2 + 2hMO$. Replacing this into the numerator we have

$4h^2MO^2 + 2m^2(m^2+2hMO - h^2) - (m^2+2hMO - h^2)^2 - 2h^2(m^2+2hMO - h^2)$

$= (m^2-h^2)^2$.

Hence the lemma is proved. [/hide]

Consider the $A$-excircle is $(C)$. You may notice that if we let the chord $AB$ in the lemma be $BC, AC, AB$ respectively, then the circumcenter $J$ will be the midpoints of $I_aI, I_aI_c, I_aI_b$ which lie on a circle center $O$ (however, we know that this circle is in fact the circumcircle $(O)$). Let the chord $AB$ in the lemma be the two chords of circumcircle $(O)$ that are tangent with $(I_a)$ at the two circle's intersections. Then without difficulty, you can see that the midpoints of $I_aM$ and $I_aN$ are the circumcenter $J$ in this time which lie on the circumcircle $(O)$ too. Hence, the homothety center $I_a$ with ratio $2$ maps these midpoints to $I, I_b, I_c, M, N$. Therefore they both lie on a circle the circumradius of which is twice as long as the circumradius of $ABC$.

@ Virgil: I was aware of the Bulgarian problem; it was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=3807 . The generalization can basicly be proven in the same way.

@ Grobber: It is not that hard to come up with the idea; actually, what can a circle passing through I and having its radius equal to twice the circumradius of triangle ABC be? The circumcircle of triangle $II_bI_c$ is the only such circle known, so one tries to show that it coincides with the circumcircle of triangle MIN, and this is not particularly hard.

@ Treegoner: Your Lemma was [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19000]a Tournament of Towns 2004 problem[/url] ;) .

  Darij

Thank you, Darij! Your proof to the lemma is wonderful.  :)

[quote="marko avila"]wait i  dont understand. as i  know  I_aI_bI_c has ABC  as its nine point circle. why is ABC  the nine point circle of II_bI_c?[/quote]

Because $A$ is the foot of altitude from $I$ to $I_{b}I_{c}$
$C$ is the foot of altitude from $I_{b}$ to $II_{c}$
$B$ is the foot of altitude from $I_{c}$ to $II_{b}$

The conclusion follows  :)

[quote=grobber]Man.. You have to wonder how these people manage to solve the problems in contest :). I have a pretty short and simple proof, but it took me a while to see what was going on, and I used dynamic geometry software.

Let $I_b,I_c$ denote the $B$ and, respectively, $C$-excenters of $ABC$. If we prove that the circle $(MIN)$ passes through these two points we're done, because then $(MIN)$ would be the circumcircle of $II_bI_c$, while the circumcircle of $ABC$ is its nine-point circle. $ABC$ is the orthic triangle of $II_bI_c$, so $B_1C_1$ is the orthic axis of $II_bI_c$, which coincides with the radical axis of its circumcircle and nine-point circle. These two circles do intersect, because $\angle I_bII_c>\frac\pi 2$, so the two points of intersection must be precisely the points of intersection between $B_1C_1$ and $(ABC)$, i.e. $M$ and $N$.[/quote]

Sweet!!

Call the circumcircle $\omega$ and call the circle $I I_b I_c$ as $\Gamma$. it’s quite well known that $AIBI_c$ and $AICI_b$ are cyclic. Call them $\omega_1$ and $\omega_2$. Now we get that $C_1$ is the radical center of $\omega\Gamma\omega_1$ and $B_1$ is the radical center of $\omega\Gamma\omega_2$ so $B_1,C_1$ are on the radical axis of $\omega$ and $\Gamma$. Thus $M,N$ are the intersections of $\omega$ and $\Gamma$ so $IMNI_b I_c$ is cyclic. Now because of the homothety taking $\omega$ to $\Gamma$ we conclude the problem.

Let $P,Q$ be the midpoints of minor arcs $AC$ and $AB$ and let $X,Y$ be the $B,C$ excenters. By PoP, we have that $(NC_1)(MC_1) = (AC_1)(C_1B) = (IC_1)(C_1Y)$ so $Y$ lies on $(MIN)$ and similarly so does $X$. Since $P,Q$ are midpoints of $IX, IY$, $(MIN)$ has double the radius of $(PIQ)$.

But since $QA = QI$ and $PI =PA$, we have $\triangle PAQ$ and $\triangle PIQ$ are congruent so the circumradii are equal too, so $(MIN)$ has double the radius of $(ABC)$, as desired.

Denote by $I_A,I_B,I_C$ respective excenters of $ABC.$ Since $AICI_B$ is cyclic, point $B_1$ (and analogously $C_1$) lies on radical axis of $\odot (ABC),\odot (II_BI_C),$ therefore $M,N\in \odot (II_BI_C).$ We are done, because in triangle $I_AI_BI_C$

$\bullet \text{ } I$ is the orthocenter;
$\bullet \text{ } \odot (ABC)$ is the nine-point circle.

Well, I have pretty complicated proof.. Also special thanks to L567!
[asy]/* File unicodetex not found. */

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import graph; size(14.2cm);
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pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -37.7, xmax = 12.5, ymin = -8.9, ymax = 21.5;  /* image dimensions */
pen wrwrwr = rgb(0.4,0.4,0.4);

draw((-7.2,6)--(-7.7,-2.2)--(2.5,-2.2)--cycle, linewidth(0.4) + wrwrwr);
 /* draw figures */
draw((-7.2,6)--(-7.7,-2.2), linewidth(0.4) + wrwrwr);
draw((-7.7,-2.2)--(2.5,-2.2), linewidth(0.4) + wrwrwr);
draw((2.5,-2.2)--(-7.2,6), linewidth(0.4) + wrwrwr);
draw(circle((-2.5,1.7), 6.4), linewidth(0.4) + wrwrwr);
draw(circle((-4.8,13.3), 12.7), linewidth(0.4) + wrwrwr);
draw((-8.9,1.2)--(-4.8,0.5), linewidth(0.4) + wrwrwr);
draw((-4.8,0.5)--(3.5,3.6), linewidth(0.4) + wrwrwr);
draw((-8.9,2)--(1.6,6.5), linewidth(0.4) + wrwrwr);
draw((-13,3.5)--(2.5,-2.2), linewidth(0.4) + wrwrwr);
draw((-7.7,-2.2)--(1.6,6.5), linewidth(0.4) + wrwrwr);
draw((1.6,6.5)--(8,12.5), linewidth(0.4) + wrwrwr);
draw((-8.9,1.2)--(3.5,3.6), linewidth(0.4) + wrwrwr);
draw((-13,3.5)--(8,12.5), linewidth(0.4) + wrwrwr);
draw((-2.6,-4.7)--(-7.2,6), linewidth(0.4) + wrwrwr);
draw((-26.2,-2.1)--(-7.7,-2.2), linewidth(0.4) + wrwrwr);
draw((-26.2,-2.1)--(-13,3.5), linewidth(0.4) + wrwrwr);
draw((-26.2,-2.1)--(-8.9,1.2), linewidth(0.4) + wrwrwr);
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dot((-7.2,6),linewidth(5pt) + dotstyle);
label("$A$", (-7,6.4), NE * labelscalefactor);
dot((-7.7,-2.2),linewidth(5pt) + dotstyle);
label("$B$", (-7.5,-1.8), NE * labelscalefactor);
dot((2.5,-2.2),linewidth(5pt) + dotstyle);
label("$C$", (2.7,-1.8), NE * labelscalefactor);
dot((-2.8,2.4),linewidth(4pt) + dotstyle);
label("$B_1$", (-2.7,2.6), NE * labelscalefactor);
dot((-7.4,1.5),linewidth(4pt) + dotstyle);
label("$C_1$", (-7.3,1.8), NE * labelscalefactor);
dot((-4.8,0.5),linewidth(4pt) + dotstyle);
label("$I$", (-4.7,0.8), NE * labelscalefactor);
dot((-8.9,1.2),linewidth(4pt) + dotstyle);
label("$M$", (-8.8,1.5), NE * labelscalefactor);
dot((3.5,3.6),linewidth(4pt) + dotstyle);
label("$N$", (3.6,3.9), NE * labelscalefactor);
dot((-8.9,2),linewidth(4pt) + dotstyle);
label("$M_C$", (-8.8,2.3), NE * labelscalefactor);
dot((1.6,6.5),linewidth(4pt) + dotstyle);
label("$M_B$", (1.7,6.8), NE * labelscalefactor);
dot((-13,3.5),linewidth(4pt) + dotstyle);
label("$I'$", (-12.9,3.8), NE * labelscalefactor);
dot((8,12.5),linewidth(4pt) + dotstyle);
label("$I'_{1}$", (8.1,12.8), NE * labelscalefactor);
dot((-2.6,-4.7),linewidth(4pt) + dotstyle);
label("$K$", (-2.4,-4.4), NE * labelscalefactor);
dot((-26.2,-2.1),linewidth(4pt) + dotstyle);
label("$T$", (-26,-1.8), NE * labelscalefactor);
dot((-3.7,-2.2),linewidth(4pt) + dotstyle);
label("$A_1$", (-3.5,-1.9), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */[/asy]
Introduce the following points

    $\bullet$ $I_C,I_B$ as the $C,B$ excenters
    $\bullet$ $A_1$ as $AI\cap BC$
    $\bullet$ $M_C, M_B$ as midpoint of arc $AB,AC$ respectively.

[color=#6FA8DC][b]Claim[/b][/color]: The circumradius of $II_CI_B$ is twice the circumradius of $ABC$.

[b]Proof[/b]: By incenter-excenter lemma, note that $M_C$ is the midpoint of $II_C$ and $M_B$ is the midpoint of $II_B$.
So the circumradius of $II_CI_B$ is twice the circumradius of $IM_CM_B$.

So we will show that the circumradius of $IM_CM_B=$ the circumradius of $ABC$. For that, note that by sine law,
$$ \frac{\sin \angle IM_CM_B}{ IM_B}=\frac{\sin \angle IM_CM_B}{M_BC}=\frac{\sin\angle M_BBC}{M_BC}.$$

[color=#6FA8DC][b]Claim[/b][/color]: $ I_CI_B,C_1B_1,BC$ are concurrent

[b]Proof[/b]: Let $C_1B_1\cap BC=T.$ Note that $(T,A_1;B,C)=-1.$ But note that $\angle I_CAA_1=90, \angle BAA_1=\angle CAA_1$. So $T-I_C-A-I_B$.

Now, it is well known that $I_CI_BCB$ is cyclic. So by POP, we have $$TM\cdot TN=TB\cdot TC=TI_C\cdot TI_B\implies II_CI_BMN\text{ is cyclic }.$$
And we are done as The circumradius of $II_CI_BMN$ is twice the circumradius of $ABC$.

Quite nice problem.

Let $I_B$ and $I_C$ be the $B$ and $C$-excenters respectively. Notice that $B_1$ is the radical center of $(IACI_B)$, $(ABC)$, and $(I_BII_C)$, and similar for $C_1$, which implies $M, N$ both lie on $(I_BII_C)$. On the other hand by incenter-excenter lemma the radius of $(I_BII_C)$ is equal to the radius of $(I_CI_AI_B)$, which is twice the circumradius of $(ABC)$, as needed.

Absolutely brutal
[hide = solution]
Let $I_A$ denote the $A$-excentre of $ABC$. Define $I_B, I_C$ analogously. Let line $BB_1$ meet $\odot(MIN)$ at point $X$. Observe that $B_1I \cdot B_1X = B_1M \cdot B_1N = B_1A \cdot B_1C = B_1A \cdot B_1I_B$ where the last equality follows due to the existence of $\odot(AIBI_B)$. It follows that $X = I_B$, in particular $I_B \in \odot(MIN)$. Similarily one deduces $I_C \in \odot(MIN)$. As $I$ is the orthocentre of $\triangle I_AI_BI_C$ it follows that $\odot(MIN)$ and $\odot(I_AI_BI_C)$ have equal radius. Finally, remark that $\odot(ABC)$ is the nine point circle of $\triangle I_AI_BI_C$ and so it has half the radius as $\odot(I_AI_BI_C)$, as required. $\square$
[/hide]


linear algebra
matrix
linear algebra unsolved

$A$ is n by n, invertible matrix on field $K$

$AB$=$BA$ for every n by n invertible matrix $B$ in $K$

Find $A$

I would say if and only if $A = \lambda I_n$ ?

Taking $B = I_n + \Delta_i^j$ with $\Delta_i^j$ is the matrix with all null entries except for $(i,j)=1$
$B$ is invertible, $B^{-1}$ is easy to compute, it is $I_n - \Delta_i^j$
$AB = BA$ for all $B$ gives that $A$ is diagonal.


ok as MOST of you know, my first problem i mean I'M BACK!!!!!!!!!!!!!!!!! (the title of the problem)  will not be so ... difficult :twisted:  ok lets start:

(mods...don't delete this please!)

there's a person named... bob...he wants to destroy the world.  there are 7 continents in the world(everybody knows that) and lets just say that there are 4 on one side and 3 on the other.  for one continent, it took him 23 nuclear bombs and it took 14 hours.  for another it took 35 nuclear bombs and that took 21 hours.  for another it took 53 nuclear bombs and that one took 1 day and 8 hours and so on (look for a pattern).  that is the side with 4 continents so you need to figure out how many nuclear bombs and how many hours it took to destroy the last side. oh yeah the ones in italics are the questions.  also figure out how many nuclear bombs and how many hours in total it took to destroy the first half.  now for the other side.  sadly, it was easier to destroy it so...it took him 76 nuclear bombs to destroy one continent and 59 hours.  then another continent took 61 nuclear bombs and it took 47 hrs.  (look for a different pattern)  find the number of bombs and hrs it took to destroy the last continent (round to the nearest one).  second to last but not least, find the number of bombs and hrs that destroyed the side with 3 continents. last but not least, add up all of the total # of bombs and hrs it took to destroy earth.  and PLEASE DON'T DELETE THIS, IT TOOK ME 20 MINS TO CREATE! i triple checked my work so i know my answer is right.  i'll post my answer on a different topic but don't look at it unless you know your answer. [hide]1st half 80 bombs, 48.5 hrs and the second half 40 bombs 37.4 hrs 1st half 191 bombs in all, 115.5 hrs 2nd half 177 bombs and 143.4 hrs[/hide]

oh yeah i forgot to tell you guys... [size=200][u][i][b]don't look at the spoilers!!!!!!!!!!![/b][/i][/u][/size]

[quote="gameworld7"][i][size=100]figure out how many nuclear bombs and how many hours it took to destroy the last side.[/size][/i][/quote]

I think that should be continent instead of side for the last word.

i never even knew there were spoilers! THANK YOU for telling me!

HOW could ANYONE not tell that you are the same, geeky, author of I'm Back! (j/k)

[quote="yif man12"]Don't post the answer in another topic. That just clogs up the forum.

Also, am I the only one who sees no pattern whatsoever there? How can you see a pattern with only 2 terms? It could be a million things.

Example- Find the next term in this pattern.

3,8

It could be 13,24,3/8,63, or a ton of stuff.[/quote]

Since you asked, no, you're not.

replying to the person that said that 2 examples is not enough, i agree but lets just say that the pattern is the same example:
2,5,? well yeah it could be *5/2 or +3 but lets also say in my problem it is dividing and another operation

That's definitely not specific enough. It could be a million different things. That would make the question totally invalid. I'll give you another example.

3,8

The pattern is dividing and then another operation. Is it divide by 3 and times 8? Or divide by 6 and plus 7.5? There's an infinite number of possible combinations.

Also, stop being so paranoid about people looking at your spoiler. It's getting to be really pathetic.

[quote="gameworld7"]do i really really really really really need to make myself clear ok i'll say it 1 more time.  no 5 more times. no 7 more times is good.
[size=200][u][i][b]don't look at the spoilers!!!!!!!!!!!
don't look at the spoilers!!!!!!!!!!!
don't look at the spoilers!!!!!!!!!!!
don't look at the spoilers!!!!!!!!!!!
don't look at the spoilers!!!!!!!!!!!
don't look at the spoilers!!!!!!!!!!!
don't look at the spoilers!!!!!!!!!!![/b][/i][/u][/size] :evil:  :evil:  :evil: :evil: :evil: :evil:[/quote]

Holy crackers. We have cracker alert. And no I don't see the pattern anyways. These problems are so vague. Since you asked a "nice" question I'll give you a "nice" answer. Read the Bible. I think it predicts the Armageddon in there, though I may be wrong.

Hmm, I'm surprised no one has inserted a "certain president's" name in this topic yet.

That "certain president" - was that Carter or Clinton.

Besides, it is silly to put the answer right afterwards and expect no one to look. I looked first thing :P , as soon as I was told not to.

Billy

yif man12 wrote:A+MATH wrote:[hide]I looked at the spoiler[/hide] Don't look at the spoiler! Don't look at the spoiler!!!

OMG! You DID NOT just write that!

[hide]LOL[/hide]DON'T LOOK AT THE SPOILER!!!!!!!!!!!

[u][i][b][size=200] :twisted:  :twisted:  :twisted:  :twisted:  :twisted:  :twisted:  :twisted:  :twisted:  :twisted:  :twisted:  :twisted:  :twisted:  :twisted:  :twisted:  :twisted:  :twisted:  :twisted:  :twisted:  :twisted:  :twisted:  :twisted:  :twisted: [/size][/b][/i][/u] don't look at the spoilers! and you make me so sad :(  :(  :(

beta wrote:that gameworld7 or whoever you are,

please stop posting your tedious problems. They don't make sense and are poorly worded. and everyone will just get annoyed.....  and NO ONE cares you are back.....

THANK YOU!

yif wrote:Why don't you stop whining about the spoilers and address the incompetency of your problem? If you didn't want people to know the answers, DONT POST THEM!

thank you again!

[hide](not that im any better)[/hide]

[quote="gameworld7"][b]PLEASE DON'T DELETE THIS, IT TOOK ME 20 MINS TO CREATE![/b][/quote]

Then I kindly advise you to spend your valuable time working on something more worthwhile (and coherent).

yif man12 wrote:[hide]you will die in seven days... (wonder why no one has done this yet. oh well.)[/hide]OMGOMGOMG DON'T LOOK AT THE SPOILER OR ELSE!!!

uh.... WHAT was THAT?... ur as disturbing as ken...

[quote="Athena"]but the spoilers.....ITS SO TEMPTING!!! i cant live without the spoilers!! dont tell me not 2 look at them i NEED them!!! WOE IS ME!!!  um...i didnt read the question...but i saw something about nuclear bombs.....and i just read the replies.

[quote]uh.... WHAT was THAT?... ur as disturbing as ken...[/quote]

well gee, SUNNYRAY. maybe that was SARCASM or AN IMITATION of some randomly disturbing person that is randomly called gameworld7 who does not exist because the topic here is just our imaginations....ooooooh aahhhhh.[/quote]

Gee, how'd you know?

I'm starting to feel sorry for theone, and anyone else who goes to Miller.

I think we went a little out of control there. I don't think we should be posting things as blunt as some of the previous comments on here. But hey I'm not a moderater so there is nothing I can do.  ;)

[quote] I think we went a little out of control there. I don't think we should be posting things as blunt as some of the previous comments on here.[/quote]

Yeah, I agree, that was kind of mean to world7. I'm sure that he/she tried to make a good problem and just worded it a little wrong. I think the problem is funny, I mean who would have that many nuclear bombs? :D  :D  :D

oh yeah i think i'm a he and i think that i don't like to be called "world7" and theone doesn't go to miller anymore and lets say that the fed gov. gave all the nukes for hmmm... 5 trillion bucks?

JUST edit your post and delete the spoilers lol

Ken- helpful hint- double check ur problem with a partner before u post it, quantity= spam, please also stop spamming- post more thought out problems

Other people- please stop spamming this topic!

NoTe: I am also back... start looking for problems...

first i'm too lazy to edit my post and second, umm, is there a second? um i DON'T want to double check or whatever


so i decided to start a book club to discuss books. ive been reading the heir books by cinda williams chima from where my battle of the roses game came from. they are pretty good books with all the suspense and turns of events

First rule of Book Club: Don't talk about Book Club

[quote="evilhamster"][quote="guineapiggies"][quote="mathguy999"]girls in my school say Twilight is good.[/quote]

haha it is really good. but not just girls say that i know a couple of guys that like that series =][/quote]
That doesn't sound right...
At my school just about every girl loves the series. No boys are talking about it.[/quote]

... me too... (twilight is about girls and vampires falling in love)
We have a twilight lover who might go to watch the twilight movie on the first day.

guys like me in my school don't talk about twilight, but some of the you-know-what guys do. it sounds kind of a weird book to me, but they say its good.

In my school, the girls are crazy over twilight( I despise twilight).

So do I. At a recent math competition, one of the rounds' problems were all about Twilight. *groan*

hm i just finished reading The Death Collector...

i luv twilight!! the movie is coming out this friday!!! and yes. i do know about 2 or 3 guys that are going to see it. and no they are not you-know-what. they are just awesome like that  :D you should read it!! oh another good book is... umm... let's see... kite runner i guess. (random book i randomly saw on my shelf just now that happens to be good) read it for a class over the summer... i'd say for 8th grade and above...but it's good =]

I've read Kite Runner before.  It's really sad.

At my school some of the girls are like "Have you read Twilight?" and if you say no then their eyes bug out of their sockets and they practically scream "YOU HAVEN'T READ TWILIGHT?" At least not all the girls in my school do that. But it's annoying.

Exactly why aren't people talking about Brisingr and the Iliad?

hm

i think i'll read brisingr soon

i've read eragon and eldest and they are pretty  good

I put a hold on Brisingr at our library system, but I am about hold #200, so it probably won't arrive for another month or so.
Anybody read Pendragon? I haven't read it, but a lot of my classmates like the book, boys and girls alike. I'm wondering if it's really that good.

I've read Pendragon. I definately recommend it.

MacHale's last book in this series should be coming out. I'm really excited for it since the last one had hell breaking loose - I'm not sure if the Travelers can actually score a win at the end.  :|

Well, I mean, being "children's books" (I loosely apply that term) the good side will win.  Even in adult books the good side frequently wins.
Some of my fav books/authors:
Footprints of God (All other books by Greg Iles r crazy)
Blasphemy (similar to above, about AI)
Joseph Finder (workplace crime)
Michael Crichton (Fiction about controversial topics)
Lincoln Child and Douglas Preston (crime with a twist?)
etc

The Physics of Superheroes by James Kakalios is hilarious and very entertaining not only to the geek side but to everyone :)


Να προρδιοριστούν όλες οι γεωμετρικές σειρές με θετικούς όρους που έχουν την εξής ιδιότητα:  αν $ S$ είναι το άθροισμα της σειράς τότε για κάθε αριθμό  $ a$ με  $ 0<a<S$,  υπάρχει υποσειρά (πεπερασμένη ή άπειρη) της δοσμένης σειράς που έχει άθροισμα $ a$.

Έχει μείνει καιρό αναπάντητο,οπότε μπορεί κανείς μήπως να βάλει τη λύση?


For some reason or other, I'm having one heck of a time with this simple problem: write ${ \sin^{-1}(\tan(x)})$ purely in terms of $ x$ (no trig functions present).

I'm not sure, but I don't think there's a simple answer .. for one thing, the range of $ \tan x$ and the domain of $ \sin ^{\minus{}1} x$ don't match up.

If it was the other way around ($ \sin (\tan ^{\minus{}1} x)$) then the answer would be pretty simple. Perhaps it was a typo.

Okay, I thought the same thing, but in my cal class the teacher was adamant about simplifying statements of this nature.