calculus
integration
function
derivative
real analysis
complex numbers
calculus computations

Hi I was wondering, what is the difference between definite and indefinite integral in terms of their definition.

I have a vague idea that definite integrals are definied in terms of riemann sums whereas indefinite integrals are somewhat "directly" related to derivates.

A definite integral (some professors like to call this simply integral) uses boundaries :

$\int_{a}^{b}f\,dx$ is a definite integral : it is an element of$\mathbb{R}$ (or possibly the complex numbers :$\mathbb{C}$, that depends on the scope of your syllabus)

an indefinite integral of a function $f$ would be another function $F$ :

$F(x)=\int_{a}^{x}f\,dx$

the derivative of $F$ is $f$ almost always, but it depends on continuity, I can provide annoying functions

That's all good stuff, but what about Riemann sums? Can those be only applied to definite integrals.

Also, we have to worry about continuity in definite integrals too right?

I'm confused now, you can do Riemann integrals with Riemann sums, or you can have Lebesgue integrals in another way
I don't see how in this case this has to do with the difference between definite and indefinite  :maybe:

Edit : yes, you should worry about things like continuity always  :wink:

It's fairly common to use the phrase "indefinite integral" for such expressions as this:

$\int x^{4}\,dx=\frac15x^{5}+C$

I don't really like that phrase, and usually say "antiderivative" myself. I've noticed that most of the Europeans on this board usually say "primitive."

Keeping fredbel6's "annoying examples" (yes, I know them too) out of this, this is simply a statment that the set of all functions whose derivatives are $x^{4}$ is what you see on the right, for an arbitrary constant $C.$

---

The limit of a collection of Riemann sums is a definite integral.


Let p1, p2, . . . , p25 be primes smaller than 2004. Find the largest T ∈ N
such that every positive integer not exceeding T can be expressed as a
sum of distinct divisors of (p1p2 . . . p25)^2004.

we suppose that $ p_1 < p_2 < ... < p_{25}$
if $ p_1 > 2$ then $ T \equal{} 1$
if $ p_1 \equal{} 2$ then $ \forall k\le 2^{2005}\minus{}1$ $ k$ in base $ 2$ gives $ k\equal{}\sum_{k\equal{}1}^{m}2^{a_k}$ somme some $ a_1 < a_2 < a_3 < ... < a_m\le 2004$ so $ T\ge 2^{2005}\minus{}1$

is there some condition for $ p_i$ ???? because for exemple if $ p_n\equal{}n^{th}$ prime ($ p_1\equal{}2,p_2\equal{}3,p_3\equal{}5...$) $ T\ge \sum_{k\equal{}1}^{2005}(p_{k}^{2005}\minus{}1)$


I was presented with a problem which involved finding the lowest numbers (besides 3 and 4) such that the sum of the squares of two consecutive positive integers is a perfet square.  After attempting to find it algebraically and failing, I resorted to scanning a list of Pythagorean Triples and soon found it: 20,21,29.

But I wanted to know how one could go about finding this mathematically, so I located the two numbers that generate the triple (using $m^2-n^2,2mn,m^2+n^2$) and found them to be 2 and 5.

Of course, there is the 3-4-5 which is generated with 1 and 2.

Continuing through the list of triples, I found 119-120-169 (from 5 and 12) and 696-697-985 (from 12 and 29).

$m,n \Rightarrow (a,b,c)$
$1,2 \Rightarrow (3,4,5)$
$2,5 \Rightarrow (20,21,29)$
$5,12 \Rightarrow (119,120,169)$
$12,29 \Rightarrow (696,697,985)$

After looking at these numbers, I found a pattern.  Take the second number (the "n"), multiply it by 2, and add the m to get a new "n".  Use the old "n" as the new "m" and you get the next pair of generators.  Testing it out, it sure enough worked.

29*2+12=70
$29,70 \Rightarrow (4059,4060,5741)$
70*2+29=169
$70,169 \Rightarrow (23660,23661,33461)$
169*2+70=408
$169,408 \Rightarrow (137903,137904,195025)$

But why is this so?  How would I prove it?  I guess it might become obvious if you find how to algebraically find these answers ($m^2-n^2+2mn = k^2$) but I'm not very good at Number Theory, or Diophantine Equations, or whatever they're called  :?

Well, necessary and sufficient is that $m^2-n^2 \pm 1 =2mn \iff n^2 + 2mn -m^2 = \pm 1 \iff (m-n)^2-2n^2= \mp 1$.
But setting $a:=2m+n , b:=m$ you directly get that $(a-b)^2-2b^2 = (2m+n-m)^2 - 2m^2 = (m+n)^2 - 2 m^2 = n^2 + 2mn - m^2 = \pm 1$, so your formula works.
By doing some simple arguments using Pell's equation (if it's not known to you look for it on the internet), you can also proof that this are all solutions (neglecting negative ones).

Ah, so that shows that it works.  Does anyone know a way to arrive at that number other than by observation?

Yes, there is a way: the already mentioned 'Pell-equations'


[hide="Solution"]

Let $ x\equal{}\sqrt{d^2\plus{}2d}$

Then $ 2d^2\plus{}4d\minus{}2\sqrt{d^2\plus{}2d}\equal{}2x^2\minus{}2x\ge \minus{}0.5$ by AM-GM

So if $ y\equal{}a^2(\phi\minus{}\frac12)\plus{}b^2(\frac1{\phi\minus{}1}\plus{}2)\plus{}2\minus{}4c\ge 0.5$, we have proved the inequality.

Expand and rearrange

$ a^2(2\phi\minus{}1)\plus{}b^2(\frac2{\phi\minus{}1}\plus{}4)\ge 8c\minus{}3$

We know that $ \phi^2\equal{}\phi\plus{}1$, and $ \frac1{\phi}\equal{}\phi\minus{}1$

Now $ 2\phi\minus{}1\equal{}1\plus{}\frac{2}{\phi}$, and $ \frac2{\phi\minus{}1}\plus{}4\equal{}2\phi\plus{}4$

We need to prove $ (a^2\plus{}4b^2)\plus{}(\frac{2a^2}{\phi}\plus{}2b^2\phi)\ge 8c\minus{}3$

$ a^2\plus{}4b^2\equal{}(a\plus{}2b)^2\minus{}4ab\equal{}4c^2\plus{}1\minus{}4ab$

$ \frac{2a^2}{\phi}\plus{}2b^2\phi\minus{}4ab\ge \minus{}4(c\minus{}1)^2$

LHS is non-negative by AM-GM, and inequality is proved

Equality case: $ d\equal{}1\minus{}d$, so $ d\equal{}0.5$

$ c\minus{}1\equal{}0$, so $ c\equal{}1$

$ a\equal{}b\phi,(a\plus{}2b)^2\equal{}5$, so $ a\equal{}\frac{\phi}{\sqrt{\phi\plus{}1}},b\equal{}\frac1{\sqrt{\phi\plus{}1}}$

[/hide]

Woah that was intense.

Just as a note, you could simply them further to get $ a \equal{} 1$ and $ b \equal{} \frac{\sqrt{5} \minus{} 1}{2}$.

Also, I got $ d \equal{} \frac{\sqrt{5} \minus{} 2}{2}$ for the equality case for $ d$...?

Oh fail. But what about d? Is there a mistake in my solution? Can you show yours?

[quote="bboypa"]Define $ \phi$ the positive real root of $ x^2 \minus{} x \minus{} 1$ and let $ a,b,c,d$ be positive real numbers such that $ (a \plus{} 2b)^2 \equal{} 4c^2 \plus{} 1$.
Show that $ \displaystyle 2d^2 \plus{} a^2\left(\phi \minus{} \frac {1}{2}\right) \plus{} b^2\left(\frac {1}{\phi \minus{} 1} \plus{} 2\right) \plus{} 2 \ge 4(c \minus{} d) \plus{} 2\sqrt {d^2 \plus{} 2d}$ and find all cases of equality.
[i](A.Naskov)[/i][/quote]
Notice first that $ \phi >1.$ Applying the Cauchy Schwarz Inequality, we have
\[ \left[ a^{2}\left( \phi \minus{}\frac{1}{2}\right) \plus{}b^{2}\left( \frac{1}{\phi \minus{}1}\plus{}2\right) \right] \left( \dfrac{1}{\phi \minus{}\dfrac{1}{2}}\plus{}\dfrac{4}{\dfrac{1}{\phi \minus{}1}\plus{}2}\right) \geq (a\plus{}2b)^{2}.\]
On the other hand, it is easy to verify that
\[ \dfrac{1}{\phi \minus{}\dfrac{1}{2}}\plus{}\dfrac{4}{\dfrac{1}{\phi \minus{}1}\plus{}2}\equal{}2.\]
Combining this with the above inequality, we get
\[ a^{2}\left( \phi \minus{}\frac{1}{2}\right) \plus{}b^{2}\left( \frac{1}{\phi \minus{}1}\plus{}2\right)
\geq \frac{(a\plus{}2b)^{2}}{2}\equal{}\frac{4c^{2}\plus{}1}{2}.\]
Thus, it is enough to prove that
\[ 2d^{2}\plus{}2c^{2}\plus{}\frac{5}{2}\geq 4(c\minus{}d)\plus{}2\sqrt{d^{2}\plus{}2d}.\]
This inequality is equivalent to
\[ 2(c\minus{}1)^{2}\plus{}\frac{1}{2}\left( 2\sqrt{d^{2}\plus{}2d}\minus{}1\right) ^{2}\geq 0,\]
which is obviously true.

Zhero, you are right, I made a mistake for d. It's in fact $ x\equal{}1\minus{}x$, so $ x\equal{}0.5$, but then $ d$ is what you said...

[quote="bboypa"]Define $ \phi$ the positive real root of $ x^2 \minus{} x \minus{} 1$ and let $ a,b,c,d$ be positive real numbers such that $ (a \plus{} 2b)^2 \equal{} 4c^2 \plus{} 1$.
Show that $ \displaystyle 2d^2 \plus{} a^2\left(\phi \minus{} \frac {1}{2}\right) \plus{} b^2\left(\frac {1}{\phi \minus{} 1} \plus{} 2\right) \plus{} 2 \ge 4(c \minus{} d) \plus{} 2\sqrt {d^2 \plus{} 2d}$ and find all cases of equality.
[i](A.Naskov)[/i][/quote]
Define $ e: \equal{}d\minus{}1,f: \equal{}2b$, then only by am-gm: $ \displaystyle 2e^2\plus{}a^2\left(\phi \minus{} \frac {1}{2}\right) \plus{} f^2\left(\frac {1}{4(\phi \minus{} 1)} \plus{} \frac{1}{2}\right)\minus{}4c\equal{}$ $ \displaystyle \frac{\sqrt{5}(\sqrt{5}\minus{}1)}{2}\left(\frac{4e^2}{5}\plus{}\frac{f^2}{(\sqrt{5}\minus{}1)^2} \right) \plus{} \frac{a^2\sqrt{5}}{2} \plus{} \frac{2e^2}{\sqrt{5}} \minus{}4c$ $ \displaystyle \ge 2ae\plus{}2fe\minus{}4c\equal{}$ $ (e\plus{}1)(a\plus{}f\minus{}2c)\plus{}(e\minus{}1)(a\plus{}f\plus{}2c) \ge 2\sqrt{(e^2\minus{}1)((a\plus{}f)^2\minus{}4c^2)}$ $ \equal{}2\sqrt{e^2\minus{}1}$, that is our aim.[]


function
real analysis
algebra
domain
calculus
derivative
real analysis unsolved

Show that function $ f: [a,b] \to \mathbb{R}$ is Riemann integrable $ \iff f$ is bounded and $ Z$ is a measure zero set, where $ Z \equal{} \{x \in [a,b]:$ f is NOT continuous at $ x \}$

It is one of the many Lebesgue's theorems. You can find it on any advanced book of real analysis.

Let me pose a question:

Is it true that a subset of the real line of Lebesgue measure equal to zero,  is the set of discontinuities of a Riemann integrable function?

The answer is no. The set of discontinuities of [b]any[/b] function is an $ F_{\sigma}$-set; that is, it can be written as the union of countably many closed sets. (Proof is a $ \delta\minus{}\epsilon$ exercise).

A set of Lebesgue measure zero does not have to be Borel, let alone $ F_{\sigma}$.

[quote="mlok"]The answer is no. The set of discontinuities of [b]any[/b] function is an $ F_{\sigma}$-set; that is, it can be written as the union of countably many closed sets. (Proof is a $ \delta \minus{} \epsilon$ exercise).

A set of Lebesgue measure zero does not have to be Borel, let alone $ F_{\sigma}$.[/quote]

Would you give a proof of the problem? Thanks.

Let $ D$ be the domain of $ f$ and $ Disc(f)$ the set of its discontinuities. The shortest definition of $ Disc(f)$ is probably:

$ Disc(f) \equal{} \cup_{k \equal{} 1}^\infty \cap_{x\in D} \overline{\{y\in D\colon\, \vert f(x) \minus{} f(y)\vert \ge 1/k\}}$ tat is the countably union of closed sets.

There are other definitions. In the book of Oxtoby "measure and category" the author  makes use of the definition
of oscillation of $ f$ at a point.

One more question: Can we characterize the discontinuities of a derivative that is the discontinuities of $ f'$

The set of discontinuities of such a function is an $ F_{\sigma}$ set of the first category, and one can construct a derivative whose set of discontinuities is any given 1st category, $ F_{\sigma}$ set.


hi guys!!

i'm a new entry in the world of mathematics...
help me get a formal proof...

if n is odd and a1,a2,a3....an are the numbers 1,2,3...n arranged is some order prove that the product of (a1-1)(a2-2)....(an-n) is always even......

i feel the thing which can be said is that regardless what the order is there always at least one term such that (ax-x)is even......i.e both ax and x are both either odd or even..this makes the product always even.... :maybe:  :maybe:
 :maybe:

You can take $ a_1, a_2, ... a_n$ to be a collection of ones and zeroes, with one more one than a zero.  On the other hand, the given expression can be written $ (a_1 \minus{} 1)a_2(a_3 \minus{} 1)a_4...(a_n \minus{} 1)$ where there is again one more one than a zero.  In order for this product to be odd, every other term $ a_{2k}$ must be a one (that is, odd).  But...?

Sorry, let me use "odd" and "even" instead of "even" and "odd."  Since $ n$ is odd, among the numbers $ \{ 1, 2, 3, ... n \}$ exactly $ \frac{n\plus{}1}{2}$ are odd and $ \frac{n\minus{}1}{2}$ are even; in other words, there is one more odd number than there are even numbers.

Now look at the even terms $ a_2 \minus{} 2, a_4 \minus{} 4, ... a_{n\minus{}1} \minus{} (n\minus{}1)$.  If any of $ a_2, a_4, ... a_{n\minus{}1}$ are even, then the product will be even, so if we want to make the product odd we must have $ a_2, a_4, ... a_{n\minus{}1}$ be odd.  But since there are $ \frac{n\minus{}1}{2}$ such numbers, there is exactly one odd number $ a_j$ left over, and this odd number must be paired with some [i]odd[/i] term - and this makes the product even!


m, n are relatively prime positive integers. The coefficients of x 2  and x^3 in the expansion of (mx + b)^2000 are equal. Find m + n.

I think I'm close to the answer, here's what I've done:

[hide]

The terms that contain an x 2  is

C(2000, 2) m 2 x 2 b^1998.

Term that contains x^3 is

C(2000, 3) m^3 x^3 b^1997

C(2000, 2) is 1,999,000 and C(2000, 3) is 1,331,334,000.  We can divide the second number by the first and get 666.  Also, since we can't determine what x is, I think we can just take it out, leaving

m^2 b^1998

666m^3 b^1997

666 factors into 2*3^2 * 37

So

m^2 b^1998

2*3^2*37 m^3 b^1997

Now I need to find m + b, but I'm stuck.  I know m needs to be pretty big or at least larger than m since these two equations need to be equal and the second equation has m^3.  Maybe the fact that m and n are relatively prime can help me here...

[/hide]

I hope I was on the right track in my explanation...

Thinking algebraically, one of our most powerful problem solving tactics is simply to look for two ways to express the same quantity and equate them explicitly.

You are pretty close....
Remember coefficient of x^2=coefficient of x^3

So you got

666m^3b^1997=m^2b^1998

Now divide m^2b^1997 on both sides
yields
666m=b

Trick is........ m and b are relatively prime... the only way that's possible is
m=1, b=666.  Remember 1 is relatively prime to all integers.

oh I get it now
thanks!
also, to simplify C(2000, 2) and C(2000, 3), you could do:

(2000 * 1999)/2
(2000 * 1999 * 1998)/6

they both have 2000 and 1999 in common, so taking that out gives 1/2 and 1998/6
we want to divide 1998/6 by 1/2, so (1998*2)/6 gives 666


Find all functions defined on reals such that for every pair of real numbers (x,y)
f(x^2-y^2)=(x-y)(f(x)+f(y)) holds

[b]Please see :[/b]

    http://www.mathlinks.ro/Forum/viewtopic.php?highlight=f%28x%5E2-y%5E2%29%3D%28x-y%29%28f%28x%29%2Bf%28y%29%29&t=8462

 [color=red]posts 7,8[/color]

 [u]babis[/u]


[quote="randomdragoon"]Time yourself. Type for 1 minute without stopping. If you get writers' block just let gibberish come out. Once one minute is over immediately stop typing and post what you have.

----

Once upon a time there was a man named Gibberelin who was a great priest of the fairyland that was in the sky of the utopian paradise lost. Because he was actually an actor in a tv show,P = NP and therefore he learned a great deal of C++ from his nemesis, yu-gi-oh who was not actually in that show but whatever, he was. Gibberelin also decided to coat his pans with teflon. Because teflon was a natural[/quote]

Once upon a time there was a man named Gibberelin who was a great priest of the fairyland that was in the sky of the utopian paradise lost. Because he was actually an actor in a tv show,P= NP and therefore he learned a great deal of C++ from his nemesis

i screwed up!

-jorian

The point of the game is to grab a watch, a clock, or some other device that measures time and then start it, and then click "add reply" and then start typing in the box that is labeld "message body" until the device reaches 1:00 or the equal amount of time. WHile this is happening, you should continuously type until you are done because otherwise it will be bad, and if you run out of things to say just say random stuff like rofl

I think the point asn't to copy what he typed but to just type gibberish for a minute. Fore example, there's a water bottle on my desk and as my desk is shaking ever so slightly due to my typin, the water is going from one side to the other a little bit and the 'image' i get of the object behind my water bottle changes slightly every secon

There once was a guy who liiked to eat pie, but then he had a hat which ate him and the caramel apple. But then the apple started doing integral calculus and proved the Reimann hypothesis but there was a flaw and then OH NO the monkey ate himself. poor monkey.  Until the llamas explode, there will be no pudding. I like pudding. The pudding pro

Well if you are an extremely good spontaneous thinker then the stuff that you write might not be gibberish. But chances are, it will be gibberish. IN unrelated news, did anyone see the new Bleach episode? because it isn't out yet, but I wonder what I will write given enough caffine and an adequate supply of popsicle sticks. Great opportunities to write things that sound like Bushisms? You bet. I bet that if you roll two coins

Everyehere I turn I hurt someone,
But there's nothing I could do to change the things I have done,
I'd do anything within my power, I'd give everything I've got,
But the path I seek is hidden from me now
Brother Bear, I have let you down.

Oooh poetry, nice twist on the idea!
But I am at a loss to write poetry simultaneously. I mean spontaneously.
That's like when I screw up similar sounding words whenever I try to talk spontaneously. Like in NSB, I said Biology was the most "promiscuous" topic. Which is actually sorta true.

But then I said something even more unintelligent. It's a good thing I'm not running for president, or else

Normally Starcraft CD-Keys can be decoded into 12-digit numbers, sometimes 13. there is a 2-3 digit product value, a 7-digit public value, and a 3 digit private value. since there are 3 digits in the private value, there are 1000 cdkeys per public value. (after decoding) For each public value, there is only one working cd key. therefore thare are only 20 mil cdkeys

One time my computer exploded. It really stunk. Unfortunately I made too many typos which caused it to get angyr at me and it exploded. Gr. It liked bananas. Unfortunattely it didn' thave any so now i"m making evne more thysos cause i'm typing too fast to get evertyhing i wanna say in fast enough, but the llamas were too eveil for me to type fast enough th

the product values of the sc keys can only be 01 or 02 if it wants to get past bnet's 0x51 SID_AUTH_CHECK. normally, any key that can be decoded can be installed onto starcraft, but only keys with product values 01 or 02 and private values that match the public values on bnet's database can get past the auth check.

after the keys are decoded, studied, and generated, they are enc

Judas i burhy you in your own blood
aman
a whhole minute>
this is seven seconds
i should get back to doing homework tf we don't have any homework i guess i should go back to doing the AMY packet and wow i'm just koojkijng at my watch not typing backspace not looking up just looking at my watch and it's 333 seconds already, i might've typed 33 but i meant 33 and hey now it's been around 45 sefonds almost over hey hey.
so i'll talk about poop

yeah i'm bored, this blows mons

end

Hi I like cheese. Do you like cheese? I think I type relatively slow. My writing smellls like cheese. Sorry for the many typos that smell like cheese. Am I saying cheese too often? Can you spell motorcycle without any k's? Of course not! What did you think? Did you think that you like cheese? My name is not motorcycle. Can you spell potato? How abou

Hi. I wanna type for 1 minute striahgt, Wthat am I thpying athis reaylry?? I can't berlileve that I'm typing gibberish!

There It stpooed. Do I need good grammer in this game?

I think I'm not doing very good at this game/ I haven't had a full typing class. This game is done.

My typing sucks.

aliens are aliens. did u know that? i didnt think so.
did u know that i like cheese? well i guessed u might from my above posts... etc. hi. ahem. hielllwo what am i talking about? how much potatoes does it cost to eat a burritwo? sorry about the spelling eroors and all the other grammar mistkes, cause im being kinda careless whatever do you like cheese? how many i

LIFE OF ME, BY ME, AND ABOUT ME.
The end. *Applause*

OK, seeing as I still have time... (This is another minute)
goldfish is made of cheese and looks like cheesy potatoe fries hi my name is pio what are we talking about? this is really random. do u like goldfish? why do i have to talk about cheese; it seems like a cheese addict's cheesy language. i really dont eat cheese all that much, u know what im talking about without any k's? no of course you cant, cause im lying.

YAY.
The end. *Cheers and hearty clapping*

Speaking of 0x51, it is the packet id that bugs most bot developpers. Since it involves Checkrevision, bnet is free to change the mpq files and update it without updating the game. Since the mpq files check memory hash, programmers cant jus use them. they have to try to figure out what they do, and try to re-create the checkrevisions. then th

ZOMG GOM GOM GOM GOMG BLAH blah blah I am MR BLAH whahahahahahahahahahahahahahaha moooooooo i am the great chicken who likes pie. I like pie but do not like pie who gets pie but then the why woogle is about to get destroyed by something that isn't destroyable that is THE BLAH withouth the blahness of a blah that has blah. U


vector
AMC
AIME
USA(J)MO
USAMO
geometry
analytic geometry

are vectors and matrices important for preparing AIME and USAMO?

Thanks

I personally have never seen an AIME or USAMO problem that required vectors or matrices.  I have seen a USAMO geometry proof that was slightly simplified with vectors.  But that was not the only solution.

Yeah, though of course everyone loves vectors and matrices, for preparation for these competitions, they are not worth spending much time on. You may still want to go through the basics of them, though.

However, they can still give fairly clean bashes on some problems, at least compared to using coordinates directly. :wink:

ya, they are not worth your time. I haven't seen even one single those problems in AIME yet.

[quote="Poincare"]they won't be necessary but they can be helpful. coordinate bash especially. if you have triangles, and you can move one to make the problem easier, use matrices.[/quote]
Or you could use complex numbers?

Complex numbers typically are better than vectors for bashing, yes (unless its 3d).

There are also some combo problems where knowing linear algebra is pretty useful. (Example: USAMO 2008 #6) and IMO even if it doesn't help you should learn linear algebra anyways + I think linear algebra is actually pretty useful for intuition even though it doesn't always apply...

So learn linear algebra anyways, though it won't help too much for usamo. (Though I haven't seen a solution to USAMO 2008 #6 that didn't use linear algebra...)


[quote="GameBot"]How many seven-cent stamps should be put on a package requiring exactly 47 cents postage if only seven-cent and nineteen-cent stamps are available?[/quote]

$ 7n\plus{}19m\equal{}47$.

Cases are $ m\equal{}0,1,2$.

Thus, $ n\equal{}\frac{47}{7}, 4, \frac{9}{7}$. So 4, since this is the only integer.


calculus
integration
trigonometry
function
real analysis
real analysis solved

Let $l,m,n$ be integers such that $lmn\neq0$.
Evaluate

\[\int^{2\pi}_0 \cos lx\cos mx\cos nx dx\]

$\cos(lx)\cos(mx)\cos(nx)=$
$\frac14(\cos((l+m+n)x)+\cos((l+m-n)x)+\cos((l-m+n)x)+\cos((l-m-n)x)))$

This can be seen by expressing the cosine in terms of complex exponentials and multiplying the expression out.

Thus the definite integral is zero unless one of $l+m+n,$ $l+m-n,$ $l-m+n,$ or $l-m-n$ is equal to zero. If any of those expressions is zero, then only one of them is zero and the integral is $\frac{\pi}2.$

Thank you for your reply,Kent Merryfield.

kunny

More over here the primitives  of similar functions
with appropriate conditions on $a,b,c$ reals

$\int \sin(ax)\sin(bx)\sin(cx)dx=\frac{-1}{4}%Error. "Letf" is a bad command.
( \frac{\cos{(a-b+c)x}}{a-b+c}+\frac{\cos{(-a+b+c)x}}{-a+b+c}+\frac{\cos{(a+b-c)x}}{a+b-c}-\frac{\cos{(a+b+c)x}}{a+b+c} )$

$\int \sin(ax)\cos(bx)\cos(cx)dx=\frac{-1}{4}%Error. "Letf" is a bad command.
( \frac{\cos{(a+b+c)x}}{a+b+c}-\frac{\cos{(-a+b+c)x}}{-a+b+c}+\frac{\cos{(a+b-c)x}}{a+b-c}+\frac{\cos{(a-b+c)x}}{a-b+c} )$

$\int \cos(ax)\sin(bx)\sin(cx)dx=\frac{1}{4}%Error. "Letf" is a bad command.
( \frac{\sin{(a+b-c)x}}{a+b-c}+\frac{\sin{(a-b+c)x}}{a-b+c}-\frac{\sin{(a+b+c)x}}{a+b+c}-\frac{\sin{(-a+b+c)x}}{-a+b+c} )$

$\int \cos(ax)\cos(bx)\cos(cx)dx=\frac{1}{4}%Error. "Letf" is a bad command.
( \frac{\sin{(a+b+c)x}}{a+b+c}+\frac{\sin{(-a+b+c)x}}{-a+b+c}+\frac{\sin{(a-b+c)x}}{a-b+c}+\frac{\sin{(a+b-c)x}}{a+b-c} )$

Moubinool, you are implicitly assuming that all four of your denominators ($a+b+c$, etc.) are nonzero. In that case, if the question was the definite integral from $0$ to $2\pi$, the integral would be zero. You should consider the case of the definite integral in the case where one of those sums is zero.

Product of three sin or cos in sum see

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=147858#147858


The increasing sequence $ 1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of [i]at most 3[/i] distinct powers of 3.

1. Find the $ 100^{\text{th}}$ term of this sequence.
2. Find the $ n^{\text{th}}$ term of this sequence.

[hide="hint?"] Examine in base 3. It's not too hard to do the first part of the problem then.[/hide]

[hide="1"]$ \binom83\plus{}\binom82\plus{}\binom81\plus{}\binom80\equal{}93$ so the 93rd number on the list is $ 10^8_3$. From that you can find that the 100th number on the list is $ (10^8\plus{}10^3)_3\equal{}\boxed{6588}$[/hide]


topology geometry algebra polynomial calculus derivative function

Does there exist a smooth map $ f : S^2 \rightarrow \mathbb{R}$ having exactly three critical points?

at first I thought, $ f$ couldn't exist but the professor of the course said it does exist by finding a polynomial which it's derivative has exactly 3 zeros and then by stereoghraphy, we can project it to $ S^2 \minus{} \{e_2\}$ and like the proof of 'the fundamental theorem of algebra' from the book "topology from differential view point" written by milnor, we can extend it to $ S^2$ and then we can get the result by a bit modifications. But I don't know more details and also I think by generalizing the process we can find a smooth function on $ S^2$ having $ n$ critical points for any $ n \in \mathbb{N}$, which is quite odd and strange for me!  :maybe:  :(

This is interesting. Your function has to have some bad properties. It cannot be a Morse function for instance, so the Hessian has to be singular somewhere. Or we need to double one of the critical points.

We can construct such functions (not really bad) by projecting a polynomial on $ S^2$ with stereographic mapping, then composing it with the height function in order to change the codomain to $ \mathbb{R}$. Also we can suppose that each pair of roots of derivation of our polynomial has enough distance, so if we change perturb the polynomial a little, the roots of its derivation change also a bit, therefore we will find another polynomial satisfying the same condition and it is a contraction to the fact that the Morse functions are dense in the space of smooth functions.  :huh:

[quote]it is a contraction to the fact that the Morse functions are dense in the space of smooth functions. [/quote]
Not a contradiction. A dense subset may well be tiny (think of polynomials among all continuous functions). So it's not a surprise that you can move within the set of of non-Morse functions. Density says that if you try [b]all possible[/b] small perturbations of a given height function, you'll hit a Morse function. But you only considered very special perturbation: replacing a polynomial by another one.

The result about having any number $ \ge 2$ of critical points isn't geometrically counterintuitive. Think of $ S^2$ as an elastic sphere in $ \mathbb R^3$ and let the height function be just the ordinary height ($ z$-coordinate). Gently poke the sphere somewhere in the northern hemisphere (not on the pole). As you gradually increase the pressure, a critical point appears (if you keep pushing, it will split into two).

[quote="mlok"]
Not a contradiction. A dense subset may well be tiny (think of polynomials among all continuous functions). So it's not a surprise that you can move within the set of of non-Morse functions. Density says that if you try [b]all possible[/b] small perturbations of a given height function, you'll hit a Morse function. But you only considered very special perturbation: replacing a polynomial by another one. [/quote]

I don't understand, by what I said, we can find an open set of non-Morse functions, so you meant this is not an open set, why?

Being open is a relative property: if a topological space $ X$ is a subspace of $ Y$, then a set $ A\subset X$ may be open as a subset of $ X$, but not as a subset of $ Y$.

In our situation $ X$ is the set of functions of the form (polynomial composed with stereographic projection), and $ Y$ is the set of all height functions. What you proved that there is an open subset of $ X$ that consists of non-Morse functions. It does not have to be open in $ Y$, and it's not.

dear mlok, I have got your point, thanks very much.  :)


function
induction
algebra unsolved
algebra

[quote="PWojtaszewski"]Find all functions $ f: N \rightarrow N$ for which $ f(n)$ is a square of an integer for all $ n \in N$, and that satisfy
\[ f(m \plus{} n) \equal{} f(m) \plus{} f(n) \plus{} 2mn
\]
for all $ m,n \in N$.[/quote]
Let $ f(1) \equal{} a \in \mathbb{N}$. Then we have $ f(n) \equal{} an \plus{} n(n\minus{}1)$ for $ n\equal{}1$. We proceed by induction: $ f(n\plus{}1) \equal{} f(n) \plus{} f(1) \plus{}2n \equal{} an\plus{}n(n\minus{}1)\plus{}a\plus{}2n \equal{} a(n\plus{}1) \plus{} (n\plus{}1)n$ And so we have $ f(n) \equal{} an \plus{} n(n\minus{}1)$ for all $ n$.
Inserting gives:
$ a(m\plus{}n) \plus{} (m\plus{}n)(m\plus{}n\minus{}1) \equal{} a(m\plus{}n) \plus{} m(m\minus{}1) \plus{} n(n\minus{}1) \plus{} 2mn$ $ \iff$
$ a(m\plus{}n) \plus{} m^2 \plus{} n^2 \plus{} 2mn \minus{} m \minus{} n \equal{} a(m\plus{}n) \plus{} m^2 \plus{} n^2 \plus{} 2mn \minus{} m \minus{} n$ which is true.

So $ f(n) \equal{} an \plus{} n(n\minus{}1)$ the only solutions for $ a \in \mathbb{N}$

Another solution. Let
$ g(n)\equal{}f(n)\minus{}n(n\minus{}1)$ -> we can prove that g(m+n)=g(m)+g(n) but this is Cauchy functional equation. Hence g(n)=an QED

[quote="Mathias_DK"][quote="PWojtaszewski"]$ f(n)$ is a square of an integer for all $ n \in N$[/quote]

So $ f(n) \equal{} an \plus{} n(n \minus{} 1)$ the only solutions for $ a \in \mathbb{N}$[/quote]

but for what $ a$, $ f(n)$ is a square of an integer for all $ n \in N$ ?

I think the only solution is a=1. Assume there is another solution -> an+n(n-1) is a square for all n and $ a\neq 0,1$. Let m=a-1. We see that $ mn\plus{}n^2$ is always a square, but setting n=|m| we get $ 2m^2\equal{}k^2$ - contradiction.

[quote="PWojtaszewski"]Find all functions $ f: \mathbb{N} \rightarrow \mathbb{N}$ for which $ f(n)$ is a square of an integer for all $ n \in \mathbb{N}$, and that satisfy
\[ f(m \plus{} n) \equal{} f(m) \plus{} f(n) \plus{} 2mn
\]
for all $ m,n \in \mathbb{N}$.[/quote]
i take $ \mathbb{N}\equal{}\{1,2,3,....\}$

$ f(n\plus{}1)\minus{}f(n)\equal{}f(1)\plus{}2n\ge f(n)\plus{}2n\plus{}1$ gives $ f(n)\equal{}f(1)\plus{}\sum_{k\equal{}1}^{n\minus{}1}f(k\plus{}1)\minus{}f(k)\ge 1\plus{}\sum_{k\equal{}1}^{n\minus{}1}2k\plus{}1\equal{}n^2$
then $ \forall n: g(n)\equal{}f(n)\minus{}n^2\ge 0$ and $ g(n\plus{}m)\equal{}g(n)\plus{}g(m)$ gives $ g(n)\equal{}an$ for some $ a\in\mathbb{N}\cup\{0\}$ then $ f(n)\equal{}n(n\plus{}a)$ and $ f(a)\equal{}2a^2$ is squar gives $ a\equal{}0$
finally $ \boxed{f(n)\equal{}n^2}$


Sorry if this is not the right place, but I've forgotten how
the "contradiction-symbol" (which occasionally appears after a proof by
contradiction) looks like. Could someone post some comments and some
pictures, please?

I have seen that in most of the proofs they use to the symbol $!$ like contradiction, but with exactitude I wouldn't know which is the “original one”

Normally, when one presents a proof by contradiction in mathematics
he or she would follow the Latin origin: [i]Reductio Ad Absurdum[/i].

Thus, when a proof begins, one would write down ([i]RAA Proof[/i]) and possibly write down [i]RAA[/i] when contradiction occurs.
However, there are many other ways to show that a contradiction occurred both by words or symbols.

isnt there a symbol which looks a bit like a zikzak arrow or lightning flash?

[url=http://zimmer.csufresno.edu/~larryc/proofs/proofs.contradict.html]How to write a proof by contradiction[/url].

I prefer English to Latin or any kind of symbolic language. Once I saw a textbook in which every single theorem was stated as a boxed displayed formula:
\[\boxed{\exists j_{0}\ \forall \rho>0\ \exists \epsilon>0\ \forall j\ge j_{0}\ \forall x\in F_{j-1}\ 0<r<\epsilon\,\&\, x\notin A_{\rho}\implies B(f(x),r)\subset f(B(x,r^{j}))}\] (Thus example is made up since I don't have that book anymore... thankfully.)

Do you know anything about the sign I mentioned above.

I know that some amateurs use it. I never saw zigzag arrows (or the skull and cross bones for that matter) in a serious mathematical paper.


If ax^2+bx+6=0 does not have two distinct real roots , then the least value of  3a+b is..??

Since the eqn.

$ ax^2 \plus{} bx \plus{} 6 \equal{} 0$ does not have 2 distinct real roots,

$ Discriminant \le 0$

i.e.,
$ b^2 \minus{} 24a \le 0$
$ b^2 \le 24a
 \Rightarrow 3a \ge \dfrac{b^2}{8}
 \Rightarrow 3a \plus{} b \ge \dfrac{b^2}{8} \plus{} b
 \Rightarrow 3a \plus{} b \ge \dfrac{b^2\plus{} 8b}{8}$

Now this is a quadratic in $ b$

The minimum value of the quadratic is $ \dfrac{\minus{}Discriminant}{4} \equal{} \minus{}16$
Thus, the minimum value of $ 3a \plus{} b \equal{} \minus{}2$


If I took the AMC A at one site and the AMC B at a different site, then at which site will I be taking the AIME?  How does all of that work?

Thanks

See the AMC Director's comments in this thread:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135067[/url]

You'll be able to make arrangements to take the AIME at your own school.


calculus Support MIT college geometry email Stanford

Hello

I'm a 13 year old homeschooler finishing 8th grade. My parents are letting me design my own "school" plan for next year, so I was wondering what you think the ideal plan would be.

What I'm doing currently is studying calculus I at the University of Minnesota, taking an honors biology class in a lab space with other homeschoolers, learning Java at a commercial college, taking a CTD distance learning class in Logo, and continuing the EPGY C11C programming class. I also took an AP Chemistry class earlier this year with my homeschooler support group.

Now I'm looking at different possibilities for my next school year. It might be that I could enroll full time at a community college, I can do full time distance learning, I can do full time at a strong high school. I could also do a mix of two of the last three. My main interests are math, computer science, and computer graphics. I do a lot of math competitions and have scored the highest score in my AMC region on the AMC 10, but I haven't done any computer competitions yet.

I spend most of my free time working on a computer role playing game, which my friends and I work on over a Google Group. I do most of the programming and graphics design in addition to beginning the script; at our current rate of progress this will probably take two years :). I also play travelling soccer.

I'm very open minded about where and when I will enroll in college.

What kind of plan for next year would you recommend? What kind of learning experiences do you think I should be sure not to miss? I would appreciate any advice you have for me.

I would suggest that you spend some time getting involved in various contests, like USACO and USAPhO, and obviously the AMC contests. For your other subjects, I'm not really sure.

[quote="probability1.01"]I would suggest that you spend some time getting involved in various contests, like USACO and USAPhO.[/quote]

Do you do these? And if so are they fun? I'm pretty interested in starting USACO.

Go to high school. I'm serious. Go to the strong high school. No, it's not going to be "the best years of your life" of any of that bs, but believe me, when you are older, you will regret not having a high school experience of some sort. It's great and all that you are so interested in computers and math and stuff but you need other skills besides that. You need english and history and [i]social skills[/i]. When discussing college with a lot of my teachers, the main thing that's said to me is that I can make something of myself, wherever I go. You are going to have a lot more opportunities going to a strong high school than going to a community college. It may not be as "cool-sounding" as going to college at 12 or 14 or whatever but remember, you have the rest of your life. Don't fast foward through it.

[quote="bleumoose"]Go to high school. . . . You need english and history and [i]social skills[/i].[/quote]

What would prevent the development of social skills in any place other than high school for someone of the same age? Particularly, what would prevent a student at a community college from developing social skills?

Just wondering . . .

Frankly, I think social bonding is a lot stronger in a high school environment than in a community college one. For one thing, you have social time in between classes with your age peers in HS. Not only are you less likely to be considered an outsider, but you're also more likely to form deeper relationships with the people around you. Besides, it's important to be exposed to a variety of social situations and relationships, something for which high school offers an unique opportunity. Not that homeschool can't possibly provide for these things, but if you can find a good high school near you, I would recommend attending it for a year and then weighing the social experience with the amount of time wasted (which can be quite small if you can work something out with the administration, or there are good programs to begin with).

Even if you go to a strong high school, there's a danger that nobody there (teachers or students) will be as smart or as enthusiastic about learning as you are.  Usually high school is an intellectual desert.  High school can waste an enormous amount of a student's time.

If you can find a high school that has some other hardcore math/computer science students, that could be much better.  But that can be hard to find.

I think sending a smart kid who's enthusiastic about learning to a typical high school can actually be bad for his social skills.  Due to his intelligence and interests alone, he is in danger of not fitting in.  Dumb kids may pick on him all the time.  Even if nobody makes fun of him for being smart, he may still feel pressured to suppress his intelligence and his interests just because nobody else gets it, so what's the point of trying to talk about anything smart or interesting.  I think the social environment in a lot of high schools can take a smart kid and make him shrink.  That's not the way to develop social skills.  High school could conceivably be a good choice, but I wouldn't assume that going to high school is the best way (or even a good way) to develop social skills.

I'm not saying high school is necessarily a bad choice.  If you can find a good high school, it could be a good choice.  But you can develop social skills without going to high school, so I wouldn't worry about that too much.

On the subject of community colleges...I've heard some community colleges are very good, but I wonder if math is really taught that well in most community colleges.  It may not be.  That's something to worry about.  The teachers should know their stuff, hopefully, but the other students may not be that great and the classes may be dumbed down too much as a result.  (However, there are some excellent students who start out in community college then transfer to a highly-ranked research university.)

Then again, math is probably taught well in community college more often than it's taught well in high school.

[quote="Aunt Sally"]Even if you go to a strong high school, there's a danger that nobody there (teachers or students) will be as smart or as enthusiastic about learning as you are.  Usually high school is an intellectual desert.  High school can waste an enormous amount of a student's time.[/quote]

I concur. Right now, there are two people in my school that want a math team, me, and a senior. What made me work hard and try hard in middle school was the local honor of doing well. Now that there are no competitions except the AMC series, I find myself slacking. If I had to learn the material from school on my time, I'd spend maybe two hours a day. Honestly, stay with homeschooling. Your soccer team is plenty for social skills.

[quote="bleumoose"]Go to high school. I'm serious... You need english and history and [i]social skills[/i]. [/quote]

Homeschooling offers me plenty of social time right now. I go to a co-op and take classes there, and I always socialize when I'm there. Also there are monthly "teen events" there which offer even more socializing time. I'm not saying that homeschooling all the way is the best solution (I probably won't), just that homeschooling doesn't cause you to lose any social skills.

Here's my two cents.

Don't focus on your "development" (ie social skills, all that crap).  Do what you enjoy and don't worry about anything else.  We spend too much time defining ourselves and not enough realizing our passions and desires.

Ah, I knew I was opening up Pandora's box with this one, but I really do agree with probability. You need to be around people your age. Now, remember, the person said "strong high school" which is why I immediately gravitated towards that decision. At a strong high school, there's going to be a lot of people who are strong academically and probably people that are strong in other ways. Also, community colleges? Come on, you guys! Community colleges are where people go to get 101 courses out of the way and/or to raise their GPA. It's not the place for an intellegent young person to be. An intellegent young person should be around other intellegent young people.
Mostly, I really believe that you should be around other people your age. I know the arguments: I'm more mature than the people my age. I'm not interested in the things people my age are interested in. I have abilities that greatly surpass those of people my age. Yes, of course, and absolutely; but I still don't think it matters. I think it's really important for everyone to learn to live with people that they may not have much in common with. Now, of course you have to take advantage of your situation. I did. I live in one of the most educationally-backwards states in the country. However, I worked hard to get into a strong public school. Even at the school I attend, I work hard to make challenges for myself. Even here, I could breeze through with Bs and complain about how bored I am, but I don't. I take challenging courses and I work to find other courses that interest me. I have extracurricular activities. I work. I volunteer.

As a side note, I noticed from what you said you are doing right now that you are taking no courses in writing, literature, history, art, or language. You MUST take these courses. I don't care how interested you are in computers or math, you MUST take them. If you listen to none of my other advice, please listen to this!

Last bit: those who say "don't worry about social skills," please remember that, regardless of how intellegent you are, if you show up to a public speaking event dressed in jeans and a ratty t-shirt, people are going to notice. You think I'm making this up, but I'm not.

[quote]As a side note, I noticed from what you said you are doing right now that you are taking no courses in writing, literature, history, art, or language. You MUST take these courses. I don't care how interested you are in computers or math, you MUST take them. If you listen to none of my other advice, please listen to this![/quote]

He did say he's helping to write the script of a videogame.

I don't mean to be overly argumentative, but I don't think smart young students should worry much about becoming one-sided.  If you have a subject that you love and you study it intensely, then you are doing very well.  You can round yourself out later if that becomes interesting.  Just pursue your interests and with time they will cobweb out and branch out more and more...

However, literature, history and art are fascinating subjects...so somebody would be missing out if he never became interested in them.  (To me it doesn't seem that important to study languages.)

Sorry, but
[code]if (writingskills) {
cout << "Ha! I can write a complete sentence.";
}[/code]
hardly constitutes writing skills.

Anyway, I just want to add that if your life had an arbitrary structure, you could do as you pleased. Alas, that is not the case--you have to deal with everyday things once you're older than 18, and you're going to have to write essays all over the place for various programs and college.

I'm not saying that you can't pursue one interest deeply, but ignoring all other subjects at this point is out of the question. Not only would that be ignoring important skills, but it would also be taking a huge amount of perspective out of life. However, considering this is tokenadult, I'm guessing other subjects got incorporated into the homeschool curriculum somehow.

I thought A+ meant he was working on the story and dialogue for the game, in addition to the programming.  That could be some practice in creative writing, conceivably.

Judging by A+'s little writing sample at the beginning of this thread, I feel pretty confident in his writing ability.

But you are right, writing skills are extremely valuable in life.  Very important to develop them.  Maybe it sounded like I was dismissing them too much.  Good post.

I can see the arguments for both sides very strongly.

I'd personally go to high school for a year and try it out.  Agreed, you obviously will be ahead of a lot of the students in most subjects, however, its easy not to get bored.  For instance, I find it easy to keep my marks high, but by being involved in so many things at once, school is challenging, juggling time!

You meet some amazing people that you don't always meet from a soccer club.  And a huge variety of people too.  Also, there may be a particular teacher who has a positive impact on you.  Not just in teaching, but in being a person really.

Another reason I'd push this route is I've known a few people who went to college/university a few years ahead of time.  I don't know...but they just don't seem to fit in with people their own age now.  It also seems like a few vital years of social development was missing.  It really depends on the person.

If school on the other hand is too boring, or you're not sure, look at taking some extra courses at the college too.  I actually did that for a couple years...but I stopped, again,personal opinion.  Its actually a nice mix, you get the challenging courses, but you also get the benefit of a high school experience...which is definitely unique.   :)

If you can find a [i]very[/i] good high school (maybe not overall, but with a strong culture of top students) I would recommend going there. After I finished my 8th grade year, I was planning on going to the high school where everyone else was going (in the end, almost all my friends ended up going to other schools, but yeah) but I moved between districts and happened to get into one of the best public schools in the state/country. Unfortunately, any public environment will be plagued by those who just don't seem to get school, but that's how the real world is, too. I met a suprisingly large number of mathematically talented peers and working with them the last three years has definitely been the highlight of high school. As an added bonus, my school offers around 20 AP classes, so I can always find a reasonable challenge (more so getting a 5 on an AP test than getting an A in the class, but nonetheless, a challenge) outside of mathematics. Basically, going to high school will open you up to a lot more things in life, perhaps not all good, but all real.

Yeah, I'd have to agree with the other people who posted before -- neither community college nor distance learning can offer you the same level of social interaction as going to a real high school. The problem with Community Colleges is that people will be quite a bit older than you and most likely a lot less driven. Furthermore, neither of those two options offer the wealth of extracurricular activities as a good high school might. (I am not sure where you live in MN, but I know this is true for the suburbs around me - Edina, Wayzata, Eden Prairie, etc). While its true that High School can't offer you the same level of mathematical advancement, you don't need to take math or computer science at High School.

If you're worried about your peers not being up to par -- well, at least a few should be; I go to a public high school and a lot of people are quite driven(maybe not in math...). BTW, are you considering joining the local math league or are you in it? That's always pretty fun.

[quote="bleumoose"]Last bit: those who say "don't worry about social skills," please remember that, regardless of how intellegent you are, if you show up to a public speaking event dressed in jeans and a ratty t-shirt, people are going to notice. You think I'm making this up, but I'm not.[/quote]

Seriously, just because your home-schooled doesn't mean you're going to go to a public speaking event in a ratty t-shirt and jeans.  I think he'd have more sense than that.  And I don't see how this has anything to do with social skills.

I think people need to realize though that just because you go to a public school doesn't mean you're going to have social skills, and just because you're home-schooled doesn't mean you don't have social skills.  As far as I know, there is no class in school called "Social Skills."  If you wanted to, you could go throughout your whole school career, only talking to people if it was completley nessesary.  And there are ways to get social skills while being home-schooled.  Soccer teams, church groups, and loads of other stuff.

[quote="h_s_potter2002"]As far as I know, there is no class in school called "Social Skills." [/quote]As far as I'm concerned, the whole outside world is one big class called "Social Skills."   ;)

My point is, it is easier to learn to interact with people when you have more outside contact.  I feel one important attribute one should have as an adult is to be able to get along with all types of people.  As an example, I know too many very "intelligent" people who refuse to talk others they perceive as "unintelligent."  I'm not saying A+ Math is like this, but learning to deal with people of all backgrounds is one of the most important skills one can have in the real world, and as a teenager the best place to gain that type of experience is at a place with people of all backgrounds - the makeup of most public high schools.  It is not 100% guaranteed (nothing is, after all), that going to high school will produce socially ept kids, but for an intelligent kid like A+ Math, I believe high school is the best route.

[quote="h_s_potter2002"] As far as I know, there is no class in school called "Social Skills."[/quote]

MIT has a [url=http://web.mit.edu/slp/charm/]Charm School[/url] -- does that count?  ;-)

amen to the "different people" thing. that was the next point i was going to make. luckily for me, i was able to go to the most different-people school i know of. a WONDERFUL experience, i must say.

I think highschool would be counter productive for A+ for math and science academics.  he's already taken calc I in the time of half a school year, also he's doing programming graphics which is beyond the scope of high school programming from my experience.  So academically i would say highschool is out of the question.

But, all work and no play makes A+ a dull boy, we being social creates [u]need[/u] social skills/developement, like the other guy said
[code]cout << salutation.waitingforbus();
cin >> "This bus is 5 minutes late";
cout << joke.busalwayslate();
while (posvibe >= 0)     {
     converse();
     if(posvibe == 10)   {
         question.gethernumber();
         break;
     }
}[/code]
I hope i wasnt being offensive, if it was i apoligize, but stuff like getting a girls number is hard and everything has to go perfectly, it takes real good social skills.
I go to a community college and basically the only social areas are the smoking area and the cafeteria, other than that people are either walking to class or walking to their cars.  I would reccomend getting involving in a lot of clubs and going to a community college.

[quote]but stuff like getting a girls number is hard and everything has to go perfectly, it takes real good social skills.[/quote]

Getting a girl's number is easy.  If a girl's next to you, just say, "How's it going?"  If she responds nicely, talk to her a little bit.  If she still seems interested after a few minutes of talking, say, "maybe I could get your phone number" or "let me get your phone number" or "what's your email?"  (I recommend going for email.)  Then leave.  Easy.

[quote="oooooh ya"][code]cout << salutation.waitingforbus();
cin >> "This bus is 5 minutes late";
cout << joke.busalwayslate();
while (posvibe >= 0)     {
     converse();
     if(posvibe == 10)   {
         question.gethernumber();
         break;
     }
}[/code][/quote]That's absolutely hilarious :) You should consider the posibility of a non-ending loop :D thus the correct code is

[code]tryouts = 0;
cout << salutation.waitingforbus();
cin >> "This bus is 5 minutes late";
cout << joke.busalwayslate();
while (posvibe >= 0 && tryouts++ < tryout_limit )
{
     converse();
     if(posvibe == 10)   {
         question.gethernumber();
         break;
     }
}[/code]

tokenadult made a post saying that Stanford has created an online high school...Sounds perfect for students like A+Math.  Here's a link to tokenadult's post:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=83497&highlight=

Calculus in 8th grade??

If you can attend a magnet school (i.e. TJ, Exeter, ect.) then go there, otherwise stay with homeschooling.

[quote]They (tokenadult and A+ Math) are related (e.g, father/son).[/quote]

lol.  I thought I was being so helpful.

A+MATH, here's my opinion: I love this sentence, is really true for me:"hich-shools do not give you intelligence". What I'm trying to say is that it doesn't matter so much in what high school you study, if you want to learn and improve your knowledge, you'll do it. But I think that is REALLY IMPORTANT (maybe one of the most important things in your life) to go to high school. Not only becuase of knowledge, because of social skills (friends, girlfriends, etc), and because there you'll find what things do you like, and what to study or work, in order to be happy. Although I love math, I know that being very good at math is not the only important thing in the world...

Let me give you an example: Here in Argentina, there is a very good high school student at math. He always gets good results in our National Math Olympiad. He also went to same Olympiads, held in another countries. But let me tell you something: he's not happy (you will be surely thinking: why?). Well, he's in love with a girl of his age (both are 17). And she isn't in love with him. He is very smart, and he gets prizes in all the olympiads he participate (all). However, he's not happy.

With this I'm trying to tell you that math (or another subjets' knowledge) is not the only thing that matters in the world. Maybe you're still young and you don't believe me (as I don't believe it when I was 13) but when you are older and grow up, you'll give me the reason (I repeat, surely now you're thinking that I'm saying stupid, not true and non-sense things).

Just go to high school, there you'll find yourself, and you'll find what to do in your life......

Be wise about [i]who[/i] you socialize with. That might seem stupid but it's not, when I tell you this.

A boy came from China, at around 11 years of age. His grandfather was a well-known chemist and his dad was a professor. He was an excellent student in China, even with the competition found there.
So he went to school here, and hung out with bad kids. In two years, he could not sit down and study anymore. He ended up not being able to go to a university. His parents paid for him to go to college, but he failed every class. Why? He never attended class. He always hung out with his girlfriend during class.

Now his life is doomed. He has no job; he just lives in his parent's house.

So go to high school, but be careful.


algebra
polynomial
linear algebra
linear algebra solved

Let $A,B \in M_3(C)$ such that  $A^2B^2=I_3$.

Prove that $det(AB-BA)=0$

Posted by Silviu on romanian forum

You are right harazi.
Having: $AB^2A=I_n$,we have $det(AB-BA)^3=3Tr[AB^2A(AB-BA)]=Tr(AB-BA)=0.$
So $det(AB-BA)=0.$

Here another solution with Cayley-Hamilton

$M=AB-BA$, $tr(M)=0$

polynomial characteristic of $M$ is $p(x)=x^3-tr(M)x^2+tr(M^{*})x-det(M)$

Cayley-Hamilton theorem $p(M)=0$

(1) $M^3-tr(M)M^2+tr(M^{*})M-det(M)I_3=0$ take the trace

$tr(M^3)-3det(M)=0$

But $tr(M^3)=tr((AB-BA)^3)=0$ see harazi post

you get  $3det(AB-BA)=0$

Remark: (1) reduce $M^3-tr(M^{*})M=0$  we know $M=AB-BA$ is not inversible

If $rank(AB-BA)\leq 1$ then $M^{*}=0$ and $M^3=0$, $M$ is nilpotent

If $rank(AB-BA) = 2$ then $rank(M^{*})=1$ then $M^{*}^2=tr(M^{*})M^{*}$

here is another approach:
It easy to get from the condition that the matrices $AB$and $BA$ commute, and that $ABBA=I_3$, so if we denote the eigenvalues of $AB$ by $\lambda_1,\lambda_2,\lambda_3$, then the eigenvalues of $BA$ must be $\frac{1}{\lambda_1},\frac{1}{\lambda_2},\frac{1}{\lambda_3}$ (as follows from $ABBA=I_3$), but it is known that matrices $AB$and $AB$ have the same characterisctic polynomials, thus they have the same eigenvalues.
Thus supposing contrary $det(AB-BA)\neq 0$ wlog it must be $\lambda_1=\frac{1}{\lambda_2},\lambda_2=\frac{1}{\lambda_3},\lambda_3=\frac{1}{\lambda_1}$ but then $\lambda_1=\frac{1}{\lambda_2}=\lambda_3=\frac{1}{\lambda_1}$, so  one of the eigenvalues of $AB-BA$ is $\lambda_1-\frac{1}{\lambda_1}=0$, which contradicts with our supposi

[quote="harazi"]If I did not loose my minds, then from the condition we have $ (AB-BA)^3= (AB)^3-(BA)^3+3(BA-AB)$ and thus the trace of $(AB-BA)^3$ is 0 and since the trace of $AB-BA$ is also 0, it follows that the determinant is also 0. Sorry if I made stupid mistakes.[/quote]

Trace (A) =0  ==>  det(A) = 0

This is false in general, can you explain me why it is true in this particular case?

I suppose it is since you started saying that this determinant is a trace of something. I can't follow that step though :) <no idea of an alike theorem>

Very nice problem btw!

Peter, I just use that a,b,c the eigenvalues of AB-BA must verify $ a+b+c=0, a^3+b^3+c^3=0$ by the conditions on the trace. Now, it is not difficult to see that one of a,b,c is 0, thus the determinant, equal to abc is 0.


Here is a problem that I like to propose to my 12th grade students:

Suppose you are given an iron nail. The nail is not galvanized, so there is no cathodic protection of the iron. Explain how it is possible to keep this nail stored, submerged in the water of a glass, for several weeks. You are not supposed to give some kind of treatment to the nail - for example, by painting it -, the nail is to remain "untouched".

I am not sure if this is correct but I will try anyway.

In order for the iron nail to corrode we would need both oxygen and water. Now since you want the nail submerged in water so we must attempt to discard the oxygen from the mixture.

You may be able to boil the water in order to get rid of the oxygen present and then put a layer of oil on top of water surface to prevent further oxygen from dissolving into the water.

Very nice, that is the correct answer. As you can see, it mixes together corrosion, temperature dependence of gas solubility, and density, that's why I think this is an instructive problem.
The only problem is that oxygen is more soluble in oil than in water (and also diffuse more quickly in oil), so some oxygen can still reach the water surface. To minimize the effect of some oxygen that eventually reach the water we can add to the water some hydrazine($ NH_{2}NH_{2}$), a very soluble liquid that readily reduces oxygen to water,

$ NH_{2}NH_{2}(aq)+O_{2}(g) \longrightarrow N_{2}(g)+2H_{2}O(l)$.

nice solution bandishedtraittor (i really liked writin this  :D )
and nice question carcul  :)


What is the correct pronunciation of the Russian letter "ы"? I remember reading some years ago the following "recipe" to pronounce it: open your lips as if you were saying a "oo" sound (as in b[b]oo[/b]k), but say an "ee" sound instead. Is this correct?

Not quite though what you get this way may be not very far from the correct sound; at least when I tried your recipe, I got something slightly resembling "ы" despite the way I normally pronounce it is completely different. The lips go pretty much the same way as with English "e" but the throat seems to be completely relaxed (this is, probably, what the "oo" lip position was assumed to achieve) and the tongue is somewhere in the middle of the mouth rather then in the forward position it takes for "e". The online guide to russian pronunciation says something like
[quote]
no exact and even similar counterpart in English, this sound is very hard to describe, you should pronounce [i:] as in "mean" then lower the middle part of your tongue a bit and bring your lips to neutral position, then you will hear somewhat similar to Russian sound. But better listen to the examples: мыло (soap), дышать (to breathe), забыть (forget), крыса (rat)
[/quote]
which I find almost totally useless as a description. I could easily teach you the sound if you were sitting next to me but to explain how to get it in written is far above my abilities :(.

According to Wikipedia (which I consider a bad source, but don't have anything better in my own reference library) the letter ы corresponds to the unrounded high central vowel written in IPA as [ɨ], "barred i."

Some English speakers do have this vowel, but only in unstressed syllables, like schwa.  It's the vowel in the last syllable of the word "ros[b]es[/b]"  Other people pronounce that as the lax high front unrounded vowel [ɪ], like in "it."

So, maybe, see if you pronounce the "e" in "roses" differently from the "i" in "it," and if you do, pronounce ы as that sound?

Does that sound about right, fedja?  I feel bad dispensing advice when I have no experience with Russian whatsoever...

Here's a link to an IPA chart which will pronounce the sounds for you when you click on them, if that's any help.

[url]http://hctv.humnet.ucla.edu/departments/linguistics/VowelsandConsonants/course/chapter1/chapter1.html[/url]

Ah, if I only knew how to pronounce "e" in "roses" correctly! :roll: At least, when I listen to native English speakers it is usually not quite the same as "i" in "it" but definitely not "ы" either. The problem seems to be that they usually keep the tongue near the teeth which doesn't make the right sound (at least, my  attempts to make the "ы" sound keeping the tongue in the front position were completely unsuccessful). I tried to pronounce "roses" holding the tongue back with my finger; "s" became really terrible but "e" came a bit closer to "ы". I'm afraid this vowel gives English speaking people as much headache as "th" gave me when I was learning English :) The best I can offer is to retract your tongue from the teeth to the middle of the mouth so that it slightly touches the bottom, relax your throat and other facial muscles completely and try to pronounce "it" to the best of your abilities under these conditions. It will, probably, still not give you exactly the right sound but, at least, you'll get some idea of what it is like. Just don't consider it an expert advice: I have never tried to teach somebody to say "ы" online before and all I posted here is just the result of a few amateurish experiments in front of a mirror :P.

Ah, I completely understand!    :blush:

One reason that the "e" in "roses" may not sound the same as "ы" is that the English version is always unstressed and very short-- bad conditions for consciously trying to hear vowel quality.  Or, it could indeed be a different vowel.  It does sound like you're describing a central vowel, though, but it's hard to tell how high or low in the mouth.

I wish we could figure this out with sound spectrograms!  It's relatively easy to tell some consonants apart, but with vowels, the variations in voice quality between individual speakers swamp the fine distinctions between vowels.  With a given set of vowels, you could tell the difference between, say, [i] and [u] but not [i] and [ɪ].  The human mind can still adapt to speaker differences better than computers   :)

[quote]all I posted here is just the result of a few amateurish experiments in front of a mirror Razz.[/quote]

That was the first homework assignment in my phonetics class!  I think real phoneticians have to resign themselves to just looking and sounding really goofy on occasion.  The less amateurish experiments involve getting a test subject to agree to be x-rayed producing while speaking sounds-- after drinking a barium enema or dangling a chain of metallic beads down their tongue and throat so the soft tissue will show up in the x-rays!  This was the golden age of phonetics research... :P

Ok, thanks for the explanations.

Another question: what is exactly the pronunciation expected when a consonant is followed by "ь" at the end of a word?

I found one source on the web that says that that letter indicates palatalization (pronouncing [j] as an offglide, basically the same as saying -y- at the same time as you say a consonant) but that the situation is really more complicated than that.  But this is really out of my league.  Why don't you try the SEELANGS (Slavic & East European Languages and Literatures) discussion list?  They would probably be able to give you more specific help.


Rewrite the following inequality as
$\sum_{cyc}\frac{a^{2}}{b+c+d}\ge\frac{4}{3}\sqrt{a^{2}+b^{2}+c^{2}+d^{2}}$
Now, I think the rest is easy :P

[quote="chien than"]Prove that:$\frac{a^{2}}{b+c+d}+\frac{b^{2}}{c+d+a}+\frac{c^{2}}{d+a+b}+\frac{d^{2}}{a+b+c}\geq \frac{4}{3}$
where $a;b;c;d$ are positive real numbers satisfying $a^{2}+b^{2}+c^{2}+d^{2}=1$[/quote]
I think, it's wrong: $a=b=c=d=\frac{1}{2}.$ :wink:

[quote="chien than"]Prove that:$\frac{a^{2}}{b+c+d}+\frac{b^{2}}{c+d+a}+\frac{c^{2}}{d+a+b}+\frac{d^{2}}{a+b+c}\geq \frac{4}{3}$
where $a;b;c;d$ are positive real numbers satisfying $a^{2}+b^{2}+c^{2}+d^{2}=1$[/quote]

PLEASE STOP SPAMS.

XIN DUNG LAM O Nhiem dien dan nua. Cac bai post cua ban deu da co tren dien dan nay, hoac trong sach. Phai co dong gop thuc su chu.


In Huffman coding, 0 and 1 are assigned to the branches of the binary tree. Is there any rule to assign 0 to left and 1 to right branch? Or we can assign sometime 0 to left 1 to right and sometime 1 to left and 0 to right branch.

Uhm, the point is that a code like 01011 should point to a specific node in the tree, and how could it if there's not a fixed rule that 1 means right and 0 means left?

Actually, I'd say that it doesn't really matter ultimately.  As long as you end up with an encoding table where you specify what code encodes what symbol, you should be fine.  However, I would definitely suggest sticking with the common convention for multiple reasons.


linear algebra
matrix
vector
integration
calculus
linear algebra unsolved

Let $A=\displaystyle
\left[
			\begin{array}{cccc}1&\frac{1}{2}&\dots&\frac{1}{n}
			\\\frac{1}{2}&\frac{1}{3}&\dots&\frac{1}{n+1}
			\\\vdots&\vdots&\ddots&\vdots
			\\\frac{1}{n}&\frac{1}{n+1}&\dots&\frac{1}{2n-1}
			\end{array}\right]$

prove that A is invertible, and that its inverse has integer entries.

This well-known matrix goes by the name "Hilbert matrix."

It is notoriously ill-conditioned; that is, if $x$ is a solution of $Ax=B$ and if $y$ is a vector chosen so that the distance from $Ay$ to $b$ is very small, the distance from $y$ to $x$ may be large.

Besides the algebraic problems zscool has posed, can you prove this more or less analytic fact?

The spectral radius of $A$ remains bounded as $n\to\infty.$

and how can one prove that the spectral radius of $H_n$ converges toward Pi ?

$H_n$ is invertible because it is the matrix of the following
scalar product of $R_{n-1} [X]$ if we work in the
base $(1, x, x^2, .. ,x^{n-1})$:
$(f,g)=\int_{0}^{1} f(t)g(t)dt$

Here is a proof  that $H={H_{n}}^{-1}$ has integral entries:
notice that $H_{i,j}=\frac{1}{a_i+b_j}$ where $a_i=b_i=i - \frac {1} {2}$.
consider the equation $HX=C$. It is equivalent to
$\sum_{i=1}^{n} \frac{x_i}{a_j+b_i} = c_j$
Let $F$ be:
$F = \sum_{i=1}^{n} \frac{x_i}{X+b_i} = \frac{Q}{P}$
where $P=(X+b_1)(X+b_2)..(X+b_n)$, $Deg(Q) \leq (n-1)$
So the equation is equivalent to:$Q(a_j)=P(a_j)c_j$.
Hence you can find $Q$ with Lagrange interpolation and a few
computation shows that the coefficient of ${{H_{n}}^{-1}}$ are integral.

this was proposed in the RomSh last year at the 11-th class, as being part a) of a problem. The second part was to prove that the sum of the entries of the inverse is $n^2$.

Another Solution Suppose $x_1,\dots,x_n,y_1,\dots,y_n$ are variable in $\mathbb R$ and we want to find the $det(A)$ which $A$ is an $n\times n$ matrix that the element $(i,j)$ of $A$ is $\frac 1{x_i+y_j}$. You see $det (A)=\frac{\prod_{i>j}(x_i-x_j)(y_i-y_j)}{\prod_{i,j}(x_i+y_j)}$. So if you set $x_i=y_i=i-\frac 12$ then $det(A)\neq0$ and also you can easily prove that all coefficients of $A{-1}$ are integers.


The land of $ \pi$ plays a game called "Pie." In this game, two teams each choose a integer between 1 and 10 inclusive. The number of positive integers less than 100 that are divisible by both numbers are added together. A square is then created with that area. A circle is created entirely inside the square with a radius of $ 1$. Both teams them attempt to shoot a dart into the circle. What is the expected value of the probability that both teams hit the circle?

Really nasty casework, I know, but it's $ \pi$ day, so bear with me.

Happy Pi day. I would try to work that out and may do later but I don't have my calculator with me and I don't want to go and get it at the moment. I will probably try the problem later.  :maybe:


MATHCOUNTS
number theory
modular arithmetic

[quote="spiders_hate_chestnut"]martin, could you post (or pm) some problems on mods?[/quote]
Do you mind if I work on them, too?

lol, I came up with some mod problems, but I'm not Martin

lol go ahead, cause i have none at the moment.

ok, I'll post a few, but I didn't feel like making these up

1) For what integer $ n$ from 0-7 satisfies $ 2x\equiv 3\mbox{ (mod 7)}$? (problem is somewhat trivial, I know.  It comes from a math meet)

2) Consider a set $ S$ with the following properties.

a) 2 is in $ S$
b) for every integer $ x$ in $ S$, $ 3x$ and $ x\plus{}7$ are also in $ S$
c) the only numbers in $ S$ are defined by the above 2 properties

What is the smallest integer greater than 2004 that is not in $ S$?
(MATHCOUNTS 2004)

And this is one that I did when I got bored in math class

3) Find a divisibility rule for $ 19$.  (hint: $ 2\times 10\equiv 1 \mbox{ (mod 19)}$)  Ok, this one is harder than the previous 2, and the rule is probably pretty useless.  Also, there are actually 2 rules for 19.

Have you completed the ones I posted on the materials page?  (i think I gave them out in class as well)

When I get back I can loan you the AoPS book on Number Theory.  It has a section on mod arithmetic.

[quote="alanchou"]ok, I'll post a few, but I didn't feel like making these up

1) For what integer $ n$ from 0-7 satisfies $ 2x\equiv 3\mbox{ (mod 7)}$? (problem is somewhat trivial, I know.  It comes from a math meet)

2) Consider a set $ S$ with the following properties.

a) 2 is in $ S$
b) for every integer $ x$ in $ S$, $ 3x$ and $ x \plus{} 7$ are also in $ S$
c) the only numbers in $ S$ are defined by the above 2 properties

What is the smallest integer greater than 2004 that is not in $ S$?
(MATHCOUNTS 2004)

And this is one that I did when I got bored in math class

3) Find a divisibility rule for $ 19$.  (hint: $ 2\times 10\equiv 1 \mbox{ (mod 19)}$)  Ok, this one is harder than the previous 2, and the rule is probably pretty useless.  Also, there are actually 2 rules for 19.[/quote]
I am not very familiar with mods..
But #1 was pretty easy.
[hide]$ 5$[/hide]

Rest, I do not know..

[quote="alanchou"]I'll give a hint for #2

[hide="hint"]
Consider all of the elements of $ S$ in mod 7
[/hide][/quote]

Hmm.. 2004 = 2(mod 7)... That's all I know, but my guess in on 2006, because I saw $ 3x$ as 3(mod 7), which I don't think is right, but I think it's better than nothing..

[hide="continuing on the hint"]
If we start at 2 and repeatedly add 7, all numbers $ y$ so that $ y\equiv 2\mbox{ (mod 7)}$ and $ y\geq 2$ are in set $ S$.  Now multiply 2 by 3 and you get 6.  Then, through repeatedly adding 7, all numbers $ z$ so that $ z\equiv 2\mbox{ (mod 9)}$ and $ z\geq 6$ are generated.  Continue on in this manner, finding which numbers mod 7 are not in $ S$.
[/hide]

Rule: Never post a question from Nat. Target 2004.

Bad memories.  :mad:

lol, just venting...

[hide="Further hint"]To change the value mod 7, you clearly need to multiply by 3. Checking the first few powers of 3 mod 7 gives us something.[/hide]

i dont think im even understanding the question correctly.

could 13 be in the set?
(2x3) + 7?

[quote="xpmath"]Yes.
Since 2 is in the set, 6 is also in there. Hence, 13 is there too.
THe problem's reaaaaally clear if you look at values of powers of 3s taken mod 7. Here's a list.
[hide]3=3 (mod 7)
9=2 (mod 7)
27=6 (mod 7)
81=4 (mod 7)
243=5 (mod 7)
729=1 (mod 7).
Hey, look at that. All of the values mod 7 besides 0 are in there. What does that tell us?[/hide][/quote]

The answer should be divisible by 7.

$ 13$ is there, because $ 13 \equal{} 6 (\text{mod}7)$, right?

So the answer would be $ 2009$?

Hmm... I am not really comfortable with this.. I don't think I completely understood it..

That's what I get. The only numbers not in the set are those divisible by 7.

[quote="xpmath"]THe problem's reaaaaally clear if you look at values of powers of 3s taken mod 7. [/quote]

Sorry, I am kind of slow at this.. How could you tell if the problem wanted you to look for the values of powers of 3s taken by mod 7?

First of all, $ x\plus{}7$ determines that it's mod 7, right?

And $ 3x$ tells us to find the values of powers of 3s, right?

Well, you can only change the value it is mod 7 by multiplying by 3, which tells us to look at the powers mod 7. Don't worry if you don't understand it, it was like #7 or #8 on Nationals.

[quote="alanchou"]
And this is one that I did when I got bored in math class

3) Find a divisibility rule for $ 19$.  (hint: $ 2\times 10\equiv 1 \mbox{ (mod 19)}$)  Ok, this one is harder than the previous 2, and the rule is probably pretty useless.  Also, there are actually 2 rules for 19.[/quote]

I approched it different way, but it's kind of weird.

$ 10\equiv \minus{}9 (\text{mod} 19)$

And I picked a number, for example $ 2128$ which ,by the way, is divisible by $ 19 (0\equiv (\text{mod}19)$ and expressed it with 10s, like this:

$ 2(10)^3 \plus{}1(10)^2 \plus{}2(10)^1 \plus{}8 \rightarrow 2(\minus{}9)^3 \plus{}1(\minus{}9)^2 \plus{}2(\minus{}9)^1 \plus{}8$

$ 2(\minus{}9)^3 \plus{}1(\minus{}9)^2 \plus{}2(\minus{}9)^1 \plus{}8 \equal{} \minus{}1458\plus{}81\minus{}18\plus{}8\equal{} \minus{}1387$

Then I figured that $ \minus{}1387$ is divisible by $ 19$.

I played around with other numbers and figured that a number to be divisible by $ 19$, the number achieved by modding out should be divisible by $ 19$

I derived this from divisibillity rule for $ 11$... Sorry if this makes no sense..

Well, there is an easier divisibility rule for 19, and it's very similar to the divisibility rule for 7.

[hide="7"]
Let $ a\equal{}10t\plus{}d$.  In mod $ 7$, this is $ a\equiv 3t\plus{}d$.  Then, $ 3t\equiv a\minus{}d$.  Multiplying both sides by 5, we get $ t\equiv 5a\minus{}5d$.  Subtracting $ 2d$ from both sides yields $ t\minus{}2d\equiv 5a$.  Therefore, if $ t\minus{}2d$ is divisible by 7, then 5a, and thus $ a$ is divisible by seven.  If we added $ 5d$ instead of subtracting $ 2d$, we would get $ t\plus{}5d\equiv 5a$, so if $ t\plus{}5d$ is divisible by seven, then $ a$ is divisible by 7.
[/hide]
Use a similar method for 19.

where did "t" and "d" and those variables came from??
i kinda get it

We have numbers $ t$ and $ d$ so that $ t$ and $ d$ are integers and $ 0\leq d\leq 9$.  $ a$ is defined to be $ 10t\plus{}d$.

uh.... what

are these the problems that would be on MC competition?
if it is, its kinda hard.
it could be just me, though.

MC never asks to derive something, I just felt like using that one...

#2 is one where you should definitely learn the method for, because problems don't come up that say "USE MODULAR ARITHMETIC FOR THIS PROBLEM"


parameterization
search
algebra unsolved
algebra

[quote="zzzzzzz"]Solve the system
(x+y)(x+y+z) = 18
(y+z)(x+y+z) = 30
(z+x)(x+y+z) = 2A
in terms of the parameter A.

please give a solution or a hint.[/quote]

Pleeeaaaaase, search before posting ! A simple search on "rmo 1992" gives the result : http://www.mathlinks.ro/Forum/viewtopic.php?t=56246


Show that if $3 \leq d \leq 2^{n+1}$, so $d$ does not divide $a^{2^{n}}+1$  with $a$ intenger.

Suppose on the contrary, and let $k$ be the order of $a \mod{d}$. Then $k|2^{n+1}$ but $k \not |2^{n}$, so $k=2^{n+1}$. Because of that: $2^{n+1}=k|\phi(d)$, so $2^{n+1}\leq \phi(d) < d$, contradiction.

All prime divisors are $=2$ or $\equiv 1 \mod 2^{n+1}$, and $2$ can only occure once.
To see this, look for the order of $a \mod q$ for $q$ being a prime divisor of $a^{2^{n}}-1$ and look $\mod 4$ to exclude factor $4$.

I know if $gcd(x,y)=1$ then $x^{2^{n}}+y^{2^{n}}$ has only prime factor form $2^{n+1}m+1$


superior algebra
superior algebra unsolved

Let F be the field F5 (i.e. Fp where p=5). How many subspaces of each dimension does the space F^3 have?
How would I proceed about this?

cheers,

Count linearly independent sequences of size $ k$, and divide by the number of possible ordered bases for any particular subspace. It's analogous to a binomial coefficient, with the same symmetry.


[quote="minhtuan_lequydon_dn"]Vision for ABCD and 1 straight road around the back of D, find the position d to the total square distance from A, B, C and D reach the largest and smallest   :oops:[/quote]
It's don't hard  :blush:


calculus
integration
algebra
polynomial
absolute value
inequalities unsolved
inequalities

I think the general problem is unsolved, but it would be nice to see some results about it. The purpose is to give some estimations of the minimal positive non-zero value that must take $\sqrt{a_1}+...+\sqrt{a_n}-\sqrt{b_1}-...-\sqrt{b_n}$ when $a_i, b_i$ are natural numbers. Some estimations, even rude?

What does modal verb "must" mean?
Infimum equals 0: take $a_i=b_i+1$ and $b_i$ large.

Oh, sorry, I didn't express in a right way. In fact, the statement of the problem is fairly stupid. I will write it again:
  For a fixed $n$ and variable $N$ find a good constant $ k(N)>0$ with the property that if $a_1,...,a_n, b_1,...,b_n$ are natural numbers between $1$ and $N$ and $\sum_{i=1}^{n}{\sqrt{a_i}-\sqrt{b_i}}>0$ then $ \sum_{i=1}^{n}{\sqrt{a_i}-\sqrt{b_i}}> k(N)$. I apologise again for the previous statement.

Since the product of all $\le 4^n$ not-vanishing expressions of form $\pm a_\pm a_2\dots \pm b_1\dots\pm b_n$ is integral (they are roots of a unitary polynomial over ${\mathbb Z}$), absolute value of each such expression is at least $(2n\sqrt{N})^{-4^n}$, because each other expression has absolute value at most $2n\sqrt{N}$. So, if $n$ is fixed, we get $k(N)=N^{-c(n)}$. Are you searching for something better?

Yes, I also found this bound, but couldn't find anything better.


Suppose we form all k-digit palindromes that consist only of 8's an 9's such that

there is atleast one of each digit. What is the smallest value of k such that

 there is at leat 2004 numbers in the list?

The answer that i got was 21 but the book says its "23" :o

I think they forgot to consider the part that states "atleast one of each"(or maybe im wrong)

Maybe some1 an help me decide if im wrong or if the book has a typo?

Ups srry the book says its 21 and im the one that got 23 :oops:

This is a problem that can be handled well with a recursion.

Yes IK, but do u get 23 or 21 for yor answer?

[hide="Well, I got.."]
21. Let $ p_n$ be the number of palindromes of length $ n$, ignoring the at least one 8/9 condition. By recursion, to get a $ n\plus{}2$-length palindrome, we have to append one number to the front and end of a $ n$-length palindrome, eg $ p_{n\plus{}2} \equal{} 2p_n$. The answer we want is $ p(n) \minus{} 2 \equal{} 2^{\lceil n/2 \rceil} \minus{} 2$, and $ 2^{\lceil n/2 \rceil} \minus{} 2 > 2004 \Longrightarrow \left\lceil \frac n2 \right\rceil \ge 11 \Longrightarrow n \equal{} 21$.

Remember that the at least one of each digit condition does not disinclude a particular string from being included in the string of length $ n\plus{}2$.
[/hide]

You can get the identical function to what azjps posted without a recursion, too.

[hide="Answer"]Starting from the left, the first $ \lceil n/2 \rceil$ digits define the number, since the remaining digits must mirror these. Each has two possibilities, thus $ 2^{\lceil n/2 \rceil}$. However, we must discount the two cases that consist of only one distinct digit (all 8's or all 9's). Thus, $ p_n \equal{} 2^{\lceil n/2 \rceil}\minus{}2$

We want $ 2^{\lceil n/2 \rceil}\minus{}2 \geq 2004 \implies 2^{\lceil n/2 \rceil} \geq 2006$. Taking the next power of $ 2$, we have $ 2^{\lceil n/2 \rceil} \equal{} 2048 \equal{} 2^{11} \implies \lceil n/2 \rceil \equal{} 11 \implies n \geq 21$[/hide]

O_o I must have missunderstood what the term "one of each" meant, so in this

list of 2048 palindromes would one of them would one of them be 999999.....(21

 9's)? I tought they were asking for numbers such that they contain both digits

 like 9889 but not 8888.

 :wallbash_red:


Η παρακατω ανισοτητα ειναι απο το βιβλιο της Μαθηματικης Αναλυσης 1 για τους φοιτητες του ΕΜΠ, με παραγωγηση βγαινει αμεσος αλλα επηδη στο βιβλιο αναφερεται σαν εφαρμογη στις ανισοτητες Cauchy, Bernouli, Weighted Means, Minkowski και Holder, ψαχνω για μια λυση που να βασιζεται μονο σε αυτες τις ανισοτητες και σε απλες μαθηματικες πραξεις, χωρις παραγωγους και χωρις μαθηματικη επαγωγη.

Αν $ n \in N^{*}$, νδο:
$ \sqrt[n]{n} \leq 1 \minus{} \frac{2}{n} \plus{} \frac{2}{\sqrt{n}}$

ΥΓ: Ισος να ειναι κατι απλο που απλα εγω δεν το εχω δει ακομα...
ΥΓ2: Δεν εχω δοκιμασει ακομα να χρησημοποιησω το δυωνυμικο θεωρημα, αλλα δεν νομιζω να βοηθαει καπου...

[quote="NickNafplio"]ΥΓ2: Δεν εχω δοκιμασει ακομα να χρησημοποιησω το δυωνυμικο θεωρημα, αλλα δεν νομιζω να βοηθαει καπου...[/quote]

Kι όμως.

Αρκεί να δειχθει ότι:

$ 1 \minus{} \frac {2}{n} \plus{} \frac {2}{\sqrt {n}}\geq \sqrt [n]{n}$

$ n \minus{} 2 \plus{} 2\sqrt {n}\geq n\sqrt [n]{n}$

$ \underbrace{1 \plus{} 1 \plus{} \ldots \plus{} 1}_{n \minus{} 2} \plus{} \sqrt {n} \plus{} \sqrt {n}\geq n\sqrt [n]{1\cdot1\cdot\ldots\cdot1\cdot\sqrt {n}\cdot\sqrt {n}}$

Καλά Χριστούγεννα! :)

Στέλιος

Ευχαριστω. Τελικα ηταν πολυ πιο απλο απο οσο περιμενα :| ...

cmad το εψαξα λιγο και με το δυωνυμικο θεωρημα και παλι δεν βρηκα κατι... μπορεις να γραψεις και τη δικια σου λυση καποια στιγμη?

Καλα χριστουγεννα σε ολους :)

Πωπω, μετά την elegant λύση του mostel εγώ ντρέπομαι να πω τη χαζή μου λύση :oops:

[hide]
Αλλα for the sake of completeness...

$ \frac{2}{\sqrt{n}} \minus{} \frac{2}{n} \geq 0 \Rightarrow \left( 1 \plus{} \frac{2}{\sqrt{n}} \minus{} \frac{2}{n} \right)^n \geq 1 \plus{} n \left( \frac{2}{\sqrt{n}} \minus{} \frac{2}{n} \right) \plus{} \binom{n}{2} \left( \frac{2}{\sqrt{n}} \minus{} \frac{2}{n} \right)^2 \equal{} 2n \plus{} \frac{4}{\sqrt{n}} \minus{} \frac{2}{n} \minus{} 2 \sqrt{n} \minus{} 1$.

Παρατηρώντας ότι $ n \plus{} \frac{4}{\sqrt{n}} \geq 1 \plus{} 2 \sqrt{n} \plus{} \frac{2}{n}$ (γιατι; ), παίρνεις αυτό που θέλεις.
[/hide]

[quote="cmad"]Πωπω, μετά την elegant λύση του mostel εγώ ντρέπομαι να πω τη χαζή μου λύση :oops:

[hide]
Αλλα for the sake of completeness...

$ \frac {2}{\sqrt {n}} \minus{} \frac {2}{n} \geq 0 \Rightarrow \left( 1 \plus{} \frac {2}{\sqrt {n}} \minus{} \frac {2}{n} \right)^n \geq 1 \plus{} n \left( \frac {2}{\sqrt {n}} \minus{} \frac {2}{n} \right) \plus{} \binom{n}{2} \left( \frac {2}{\sqrt {n}} \minus{} \frac {2}{n} \right)^2 \equal{} 2n \plus{} \frac {4}{\sqrt {n}} \minus{} \frac {2}{n} \minus{} 2 \sqrt {n} \minus{} 1$.[/quote]

Ωραια... εμενα δεν μου εβγαινε γιατι απο χαζομαρα προσπαθουσα να αποφυγω τους δυωνυμικους συντελεστες που ειναι μεγαλυτεροι απο $ n$

[quote="cmad"]Παρατηρώντας ότι $ n \plus{} \frac {4}{\sqrt {n}} \geq 1 \plus{} 2 \sqrt {n} \plus{} \frac {2}{n}$ (γιατι; ), παίρνεις αυτό που θέλεις.
[/hide][/quote]

$ \frac{(n \minus{} 2)(\sqrt{n} \minus{} 1)^2}{n} \geq 0$ :wink:

Προφανος ομως η λυση που ζηταει ο καθηγητης της Αναλυσης στο βιβλιο του ειναι αυτη του κυριου Στελιου, η οποια βασιζεται σε ενα αρκετα γνωστο τεχνασμα αλλα για καποιο λογο δεν μου εκοψε να την βγαλω ετσι. :| [/hide]

[quote="mostel"][quote="NickNafplio"]
Προφανος ομως η λυση που ζηταει ο καθηγητης της Αναλυσης στο βιβλιο του ειναι αυτη του κυριου Στελιου, ...[/quote]

Με έκανες και γέλασα  :D

Δεν είμαι κύριος.. Κι εγώ πρωτοετής είμαι :)[/quote]

Α οκ, δεν το ηξερα  :P

Βαζω αλλη μια παρομοια ανισοτητα απο το ιδιο βιβλιο με την οποια εχω κολλησει εδω και 3 ωρες... φαινεται πολυ απλη αλλα με δυσκολευει ακομα περισσοτερο απο την προηγουμενη γιατι νομιζω πως τα εχω δοκιμασει ολα...
Χωρις να γινει χρηση παραγωγων και αλλων ανισοτητων εκτος των Cauchy, Bernouli, Holder, Minkowski και Weighted Means, νδο:
$ \left(1 \plus{} \frac {1}{n}\right)^n < 4$, οπου $ n:$ θετικος ακεραιος.

[quote]Χωρις να γινει χρηση παραγωγων και αλλων ανισοτητων εκτος των Cauchy, Bernouli, Holder, Minkowski και Weighted Means, νδο:
$ \left(1 \plus{} \frac {1}{n}\right)^n < 4$, οπου $ n:$ θετικος ακεραιος.[/quote]

Το ότι $ e < 4$ θεωρείται γνωστό ή όχι? :P

Durandal 1707

[quote="cmad"]Αυτή η συζήτηση ανήκει στο φοιτιτικό, by the way, γιατί όλες οι ερωτήσεις αυτές είναι κλασικές ερωτήσεις πρώτου κεφαλαίου ανάλυσης (και όχι τόσο δύσκολες ώστε να είναι για ολυμπιάδες...) Απόδειξε ότι: αν $ a_n \equal{} \left( 1 \plus{} \frac {1}{n} \right)^n$ και $ b_n \equal{} \left( 1 \plus{} \frac {1}{n} \right)^{n \plus{} 1}$, τότε $ a_1 < a_2 < a_3 < \ldots < b_3 < b_2 < b_1 \equal{} 4$.[/quote]

Οκ την καταφερα επιτελους... Ευχαριστω για τη βοηθεια
Η ασκηση παντος ειναι μεν ευκολη για καποιον που γνωριζει οτι η $ b_n$ ειναι φθινουσα, για καποιον που δεν το γνωριζει ομως δεν θα ελεγα το ιδιο, το θεωρω αρκετα δυσκολο να παει το μυαλο καποιου στο να αποδειξει πρωτα οτι η $ b_n$ ειναι φθινουσα για να το χρησημοποιησει μετα στη λυση της ασκησης, ασχετα αν η αποδειξη της μονοτονιας της $ b_n$ δεν ειναι και τοσο δυσκολη...

Αν θεωρειτε οτι το θεμα δεν ανηκει στις ολυμπιαδες, ας μεταφερθει στα φοιτιτικα...


At ddtoanhoc in vietnam,
one member wrote   [tex]a,b,c,d\ge0[/tex].
[tex](a^2+3)(b^2+3)(c^2+3)(d^2+3)\ge\frac{128}{3}(ab+ac+ad+bc+bd+cd)[/tex].

It was posted by spdf.
http://www.mathlinks.ro/Forum/viewtopic.php?t=36878
A general case was proposed by vasc eleven years ago, an elegant proof can be found here
http://www.mathlinks.ro/Forum/viewtopic.php?p=134537#p134537


[size=150][b]Još samo mjesec dana!
[/b][/size]
 :pilot:

[img]http://upload.wikimedia.org/wikipedia/commons/f/fb/PlazaCastillaMadrid.JPG[/img]
[color=green][b]
Vidimo se u Madridu![/b][/color]

[size=75]Ovaj post nema nikakvog smisla osim izražavanja mog oduševljenja. Slobodno se pridružite u iskazivanju istoga. Sad idem probati riješiti teoriju brojeva koju nisam na pripremama. Nije kao da ću uspjeti.[/size]

to te ja pitam morniče, carski post!

Inače mi svaki dan na pripremama imamo po jedan [b]tulum[/b] od 4 sata. I to ubitačan. Najozbiljnije. Tulumarimo ko ludi.

Je li tulumarite sa proslogodisnjim shortlistom, ili sa nekim starim olimpijskim zadacima?  :lol:

Nažalost, moram priznati da ne znam... u svakom slučaju ima više izvora s obzirom da sam neke zadatke već vidio, ali ih većinu nisam. Vjerojatno su neki od onih težih sa prošlogodišnjeg shortlista.

No, ionako do njih najčešće ni ne stignemo (jer se mučimo već na onima za koje mislimo da su "lakši"), pa zapravo ni ne mogu točno procijeniti težinu.

[quote="Goblin"][quote="djuro"]Inače mi svaki dan na pripremama imamo po jedan [b]tulum[/b] od 4 sata. I to ubitačan. Najozbiljnije. Tulumarimo ko ludi.[/quote]

Ako nije problem posalji neke interesantne zadatke, da se i mi malo zabavimo.[/quote]

Pa evo jednog koji danas nitko nije riješio: $ a,b,c,d > 0$ i $ a\plus{}b\plus{}c\plus{}d\equal{}1$, naći maksimum od $ a^2b\plus{}b^2c\plus{}c^2d\plus{}d^2a$.

@Djuro: Skužio sam zakaj je TB s 1. dana MEMOa trivijalna - čim sam malo porazmislio, postalo je očito.


Let $ a,b,c>0$ such that $ ab\plus{}ac\plus{}bc \ge 4 \ge a\plus{}b\plus{}c$.  Prove that at least two of $ |a\minus{}b| \le 2$, $ |b\minus{}c| \le 2$, $ |c\minus{}a| \le 2$ are true.

[quote="archimedes1"]Let $ a,b,c > 0$ such that $ ab \plus{} ac \plus{} bc \ge 4 \ge a \plus{} b \plus{} c$.  Prove that at least two of $ |a \minus{} b| \le 2$, $ |b \minus{} c| \le 2$, $ |c \minus{} a| \le 2$ are true.[/quote]

We suppose to at least two of $ |a \minus{} b| \le 2$, $ |b \minus{} c| \le 2$, $ |c \minus{} a| \le 2$ are false,
Without loss of generality $ |a\minus{}b|>2,|b\minus{}c|>2$
$ \Rightarrow \frac{a^2\plus{}b^2}{6}\minus{}\frac{1}{3}>ab, \frac{b^2\plus{}c^2}{6}\minus{}\frac{1}{3}>bc.$
We have $ 4\le ab\plus{}bc\plus{}ca<\frac{a^2\plus{}b^2}{6} \minus{} \frac{1}{3}\plus{}\frac{b^2\plus{}c^2}{6}\minus{}\frac{1}{3}\plus{}\frac{c^2\plus{}a^2}{6}\plus{}\frac{2(ab\plus{}bc\plus{}ca)}{3}$
$ \equal{}\frac{(a\plus{}b\plus{}c)^2\minus{}4}{3}$
$ \Rightarrow 16<(a\plus{}b\plus{}c)^2$, It is conflict with $ a\plus{}b\plus{}c\le 4$.

Wow, it is going to be a very nice proof!

[quote="NextPeace"]
$ \Rightarrow \frac {a^2 \plus{} b^2}{6} \minus{} \frac {1}{3} > ab, \frac {b^2 \plus{} c^2}{6} \minus{} \frac {1}{3} > bc.$
[/quote]
It can be $ \Rightarrow \frac {a^2 \plus{} b^2}{6} \minus{} \frac {1}{3} > \frac{1}{3}ab, \frac {b^2 \plus{} c^2}{6} \minus{} \frac {1}{3} > \frac{1}{3}bc,$ but changes nothing.
Nice proof, NextPeace!


hello people!
i have this exercise.

[img]http://img337.imageshack.us/img337/7394/75664022iq2.jpg[/img]

thank you

Wrong forum, [b]kingabcd[/b]. This should be in College Playground, perhaps under Advanced Fields subforum.

#1 How do you prove tautologies? Truth tables certainly are the key to this.

#2 [u]Theory of types[/u] is not in any high school curriculum, not even math competitions. Try combinations of deduction calculus like modus ponens, rule T, etc...

Depends on what your logical axioms are. Which set are you using?

good morning

seamusoboyle what do you mean?

For example, 10000th User says to use truth tables in the first one. There are systems where you cant use truth tables, and the only way to proceed forward is through use of the logical axioms given

Good morning

In the exercise does not  say something more.
It's exactly what you see.
So you can use anything you want.

Thank you


calculus
integration
logarithms
trigonometry
calculus computations

Your answer is correct, Solar Plexsus. :)

Simplify $\sqrt{1-\sin \theta}+\sqrt{1+\sin \theta}\ \left(0\leq\theta \leq \frac{\pi}{2}\right).$


Given x>=y>=z>=0 such that x+y+z=90.Find the maximum and minimum of
 f(x,y,z)=cosx*siny*cosz

I think I'm missing something.

[hide]
Obviously, $ 0 \le x,y,z \le 90^\circ$.

Thus, $ 0 \le \cos(x), \sin(y), \cos(z) \le 1$.

Therefore, $ 0 \le \cos(x)\sin(y)\cos(z) \le 1$.

These minimum and maximum values can be achieved:
$ f(90^\circ,0^\circ,0^\circ) \equal{} \cos(90^\circ)\sin(0^\circ)\cos(0^\circ) \equal{} 0$
$ f(0^\circ,90^\circ,0^\circ) \equal{} \cos(0^\circ)\sin(90^\circ)\cos(0^\circ) \equal{} 1$

So, the minimum and maximum values are $ 0$ and $ 1$ respectively.
[/hide]

it seems I have the wrong notes, I will check
Pleas wait.

The problem has conditions x>=y.=z
your solution is wrong


I need help with these trig problems. I'm not that great at trig, so dont skip steps when explaining the solution. Also, If you're using any formula, include it in your solution.

1. If the equation $ x^2 \plus{} 4x \sin y \plus{} \tan y \equal{} 0$, $ 0 < y < \pi/2$ has repeated roots, then what does $ y$ equal?

2.The equation $ x^2 \minus{}(2k \plus{} 1) x \plus{} k^2 \minus{} 1 \equal{} 0$ has two roots $ \sin y$ and  $ \cos y$, where  $ y$ and $ k$ are both real constants. What is $ k$?

3.Given that 0 is less than or equal to y which is less than or equal to 2 $ \pi$ , the number of possible solutions to $ 8 cos^6(2y)\minus{}7 cos^3(2y)\minus{}1 \equal{} 0$ is what?

4.How many solutions does $ \cos(x\plus{}\pi/6)\equal{}0.5$, where x is between 0 and 1000 inclusive?

5. If $ 2^x\equal{} \cos(y/2)$ and $ a^x \equal{} \sin y$, then $ \sin(y/2)$ is equal to what?

I wonder how many people you PMed about this...
[hide="1"]
We want $ (x \minus{} r)(x \minus{} r) \equal{} x^2 \plus{} 4x\sin{y} \plus{} \tan{y}$ (because a quadratic can have at most 2 roots, which means exactly one root is repeated), so $ 4x\sin{y} \equal{} \minus{} 2rx$ and $ \tan{y} \equal{} r^2$. First equation implies $ 2\sin{y} \equal{} \minus{} r$, so we want $ (2\sin{y})^2 \equal{} \tan{y} \implies 4\sin{y}\cos{y} \equal{} 1$ (we can divide both sides by sin(y) here because it is non-zero). Use the double angle formula here and you are done.
[/hide]

[hide="2"]
By vieta's, $ \sin{y} \plus{} \cos{y} \equal{} 2k \plus{} 1$ and $ \sin{y}\cos{y} \equal{} k^2 \minus{} 1$, and $ (\sin{y} \plus{} \cos{y})^2 \minus{} 2\sin{y}\cos{y} \equal{} (2k \plus{} 1)^2 \minus{} 2(k^2 \minus{} 1)$. However, this is also equal to sin^2(y)+cos^2(y) which is equal to 1. It should be easy to solve for k now.[/hide]

[hide="3"]
Let $ u \equal{} cos^3(2y)$. Note that the roots of the resulting quadratic are 1 and -1/8. Then you can set each of these equal to cos^3(2y) and find how many values of y work.[/hide]

[hide="4"]
We know $ \cos{(\frac {\pi}{3} \plus{} 2k\pi)} \equal{} 0.5$, and $ \cos{(\frac {4\pi}{3} \plus{} 2k\pi)} \equal{} 0.5$, for all integers k. Then you can set x equal to both of them and find how many x work.[/hide]

[hide="5"]
$ \sin{y} \equal{} 2\sin{y/2}\cos{y/2}$[/hide]


function
limit
real analysis
real analysis unsolved

Suppose $ (f_n)$ is a sequence of continuous functions on an interval $ I$ that converges uniformly on $ I$ to a function $ f$. If $ (x_n) \subseteq I$ converges to $ x_o \in I$ , show that $ lim (f_n(x_n)) \equal{} f(x_o)$.

This follows from the fact that the uniform limit of a sequence of continuous functions is continuous.  If $ x_{n}\to x$, then $ f(x)\equal{}\lim_{n\to\infty}f_{n}(x)\equal{}\lim_{m\to\infty}\lim_{n\to\infty}f_{n}(x_{m})$.  The fact that $ f$ is continuous gives $ f(x)\equal{}\lim_{m\to\infty}f(x_{m})\equal{}\lim_{n\to\infty}\lim_{m\to\infty}f_{n}(x_{m})$.  The two limits are equal so they can be interchanged.


The rectangular plot with largest possible area has a perimeter of 86 feet. The plot has whole number dimensions and is enclosed with fencing. How many square feet are in the area of the plot?

Since we want the greatest area, we want the dimensions which are the closest together. So just let $ a, b$ = the two dimensions. We have $ 2(a \plus{} b) \equal{} 86 \Rightarrow a \plus{} b \equal{} 43$. The two integers that are the closest together that satisfy this are $ a \equal{} 22, b \equal{} 21$. Now we have the dimensions, just multiply. $ 22 \cdot 21 \equal{} \boxed{462}$