trigonometry
LaTeX
geometry solved
geometry

Well, I have been working on this proof for a while but no matter what way I try it, it doesn't seem to work for me?  I know it will probably very easy for most of you guys/gals so here goes:

\[\displaystyle \frac{1 - \cos x}{1 + \cos x} = (\cot x - \csc x)^2\]

Thanks in advance for your help.

Sincerely,

CSW

*edited to convert to Latex

Here is was I got:

\[ \displaystyle LHS = \frac{1 - \cos x}{1 + \cos x} \times \frac{1 - \cos x}{1 - \cos x} = \frac{(1 - \cos x)^2}{\sin^2 x} = \left( \frac{1 - \cos x}{\sin x}\right)^2 =\]

\[\displaystyle = (\csc x - \cot x)^2 = \csc^2 x - 2\csc x \cot x + \cot^2 x = RHS\]

*edited to convert to Latex

Your calculation is fine, what was the problem?

Yetti

It just didn't seem right for some reason??  His is correct also though right?


The sum of a set of five consecutive integers is $ \minus{}5$. What is the least of these integers?

Let $ x$ be the middle integer. Then, $ (x\minus{}2)\plus{}(x\minus{}1)\plus{}x\plus{}(x\plus{}1)\plus{}(x\plus{}2)\equal{}5x\equal{}\minus{}5 \implies x\equal{}\minus{}1$. Thus, we have $ \minus{}1\minus{}2\equal{}\boxed{\minus{}3}$.

ok!!!! n is the base integer, or the answere, whichever. So, n+n+1+n+2+n+3+n+4=-5. This equals, 5n+10=-5. 5n=-15. n=-3.


algebra
function
domain
polynomial
superior algebra
superior algebra unsolved

[quote="wiosna"]If exist a noncomutative domain whith uniform dimension is infinite ? Give me an example.[/quote]

You should know that a ring has infinite uniform dimension iff it contains an infinite direct sum of its left ideals. So, an example for your question would be $ R\equal{}F \{X_j, \ j \in \mathbb{N} \},$

the ring of polynomials in noncommuting indeterminates $ X_1, X_2, \cdots ,$ over a field $ F$. Because, for example, in $ R$ we have this infinite direct sum $ RX_1 \oplus RX_2 \oplus \cdots.$


geometry
circumcircle
analytic geometry
inradius
incenter
trigonometry
Pythagorean Theorem

A triangle with sides of $ 5$, $ 12$, and $ 13$ has both an inscibed and a circumscribed circle. What is the distance between the centers of those circles?

$ \textbf{(A)}\ \frac{3\sqrt{5}}{2}\qquad
\textbf{(B)}\ \frac{7}{2}\qquad
\textbf{(C)}\ \sqrt{15}\qquad
\textbf{(D)}\ \frac{\sqrt{65}}{2}\qquad
\textbf{(E)}\ \frac{9}{2}$

Since the given triangle is a right triangle, we can plot the coordinates of the center of the circumscribed circle at $ (\frac{5}{2}, 6)$. Using $ A\equal{}rs$, we find that the inradius is $ 2$, so the coordinates of the incenter is $ (2,2)$. Using the Pythagorean Theorem, we find that
\[ d\equal{}\sqrt{4^2\plus{}\frac{1}{2^2}}\equal{}\frac{\sqrt{65}}{2} \rightarrow \boxed{\textbf{D}}.\]

*DIFFERENT* solution ( so i wont get yelled at for reviving an old post)

[hide="Solution"]
The circumradius, $ R$, is half of the hypotenuse, or $ \frac{13}{2}$. We can find the inradius, $ r$, by equating the area of the triangle:

\[ \frac{5 \cdot 12}{2} \equal{} rs\]

where $ s$ is the semiperimeter, which is $ \frac{5 \plus{} 12 \plus{} 13}{2} \equal{} 15$. Solving, we get $ r \equal{} 2$. Euler's Theorem says that the distance between the inradius and circumradius, $ d$, is $ \sqrt{R^2 \minus{} 2Rr} \equal{} \boxed{\frac{\sqrt{65}}{2}}$.
[/hide]

Is there any non-coordinate method? (and without Euler's Theorem).

EDIT: I revived the thread because I did not know how to do the problem.  Obviously if i attempted it and was successful
I wouldn't be posting.  Please don't be so rude.

If you are looking for a proof of Euler's triangle formula, a version can be found [url=http://www.proofwiki.org/wiki/Euler_Triangle_Formula]here[/url].

Hmm... I think the coordinate solution is the easiest here (or if you don't prefer coordinates, dropping perpendiculars and the like to pretend that there are coordinates).  But I do have another solution!

[hide="Solution"]
Note that the circumcenter of any right triangle is just the midpoint of the hypotenuse.

Label the points of the triangle $A,B,C$ such that $AB=5$, $AC=12$, and let $I$ be the incenter.  Consider $\triangle BIC$.  Note that by simple calculations we can determine the inradius of $\triangle ABC$ to be equal to $2$ units.  Suppose this incircle is tangent to $AB$ at $D$.  Then $BD=5-2=3$ and $DI=2$, so $BI=\sqrt{13}$.  Similarly, one may deduce that $CI=\sqrt{104}$.

Denote $M$ the midpoint of $BC$.  Applying LoC to $\triangle BMI, \triangle CMI$ gives \[\begin{cases}BI^2&=BM^2+MI^2-2(BM)(MI)\cos\angle BMI,\\CI^2&=CM^2+MI^2-2(CM)(MI)\cos\angle CMI.\end{cases}\]  Since $\angle BMI$ and $\angle CMI$ are supplementary, their cosines are negatives of each other.  Therefore $2(BM)(MI)\cos\angle BMI=-2(CM)(MI)\cos\angle CMI$.  Thus, adding the two equations together will cancel out these terms and leave us with $BI^2+CI^2=BM^2+CM^2+2MI^2$, which implies that \[2MI^2=BI^2+CI^2-2BM^2=13+104-2\left(\dfrac{169}4\right)=\dfrac{65}2.\]  Hence $MI=\sqrt{\dfrac{65}4}=\dfrac{\sqrt{65}}2$, which is answer choice $\boxed{\textbf{(D)}}$. [/hide]

EDIT: Also, it's okay and generally preferred to revive contest threads since, well, that's what the thread's for in the first place.  bobthesmartypants, you've been getting really sassy lately....

Nice solution, djmathman. I guess once you apply LoC, one can notice that the angles are supplementary and thus can make the manipulation in substituting/adding the equations. However, if I may ask, where did you get the idea to use LoC in the first place considering that many of the variables were unknown? I guess it might just come with experience, but I was wondering.

And, well, bobthesmartypants kind of already warned us of his cheekiness. After all, his username *is* bob[b]thesmartypants[/b].

Well, it's just the derivation for the median length formula in disguise.  I didn't want to pull that result out of a hat, so I rederived it.

(And actually after I went to bed last night I realized that there is an even easier solution: drop the perpendicular from $I$ to $BC$ and call it $F$.  Then use the Pythagorean Theorem on $\triangle IFM$.)

[quote="bobthesmartypants"]Thank you for reviving this thread. *sarcasm*

Why don't you try it yourself? Look at their solutions. Does it necessarily have to be on the coordinate plane? Can you simply find the values of sides?[/quote]
The technical ( ;) ) term for this is "back-seat moderating." It's obnoxious, don't do it! (It should be distinguished from helpful comments to newcomers, which this is not.)

[quote="djmathman"]
EDIT: Also, it's okay and generally preferred to revive contest threads since, well, that's what the thread's for in the first place.  bobthesmartypants, you've been getting really sassy lately....[/quote]

I was having a bad day :(

Apologies to anyone who was offended.

By euler
$OI^2=R(R–2r)$ ,$R\ge 2r$
And $4R[ABC]=abc$,$sr =[ABC)]$
Now we are done by calculate


factorial
abstract algebra
Ring Theory
superior algebra
superior algebra solved

Let $R$ be a factorial ring and $p$ - a prime element of $R$.Prove that the localization ring $A_{(p)}$ is a principal ring.

[quote="sissa"]For $R$ noetherian: $R_{(p)}$ has dimension = height (p) = 1 and its maximal ideal is principal, so $R$ is a PID.[/quote]

I must be missing something obvious, but why does it follow that the ring is PID from that it's maximal ideal is principal ?

This is a well-known characterization of Noetherian local rings of dimension $1$: DVR <=> normal <=> maximal ideal principal <=> dim m/m^2 = 1 <=> PID


If $x,y>0$, $log_yx + log_xy=\frac{10}{3}$ and $xy=144$, then $\frac{x+y}{2}=$

(A) $12\sqrt{2}$
(B) $13\sqrt{3}$
(C) $24$
(D) $30$
(E) $36$

This problem has been bothering me for a while, so thanks for helping out  :)

p.s: If this problem has already been posted somewhere else, or if I've violated a rule, please tell me

Hint, substitute in for the first equation.

Make it in the form $\frac{1}{a}+a=144$ Now solve for $a$, and then plug back in $\log_{x}{y}$....

That should make it MUCH simpler :)

if you need more help, just ask, although i'm sure you can solve it from there


3L reactant vessel contain 0.25 mol. NO and 0.12 mol $ O_{2}$. At equilibrium NO=0.05. Calculate the K vale

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

Initial [ ] $ SO_{2}$ and $ O_{2}$ = 0.06 and 0.05 respectively. At equilibrium $ SO_{3}$ is 0.04 mol/l
Vol. of container is 2L

calculate K

[hide="Answer"]The equilibrium in question is $ 2NO(g)\plus{}O_{2}(g)\rightleftharpoons 2NO_{2}(g)$. The initial concentrations are

$ [NO]_{i}\equal{}\frac{0.25\, mol}{3\, L}\equal{} 0.083\, M$, $ [O_{2}]_{i}\equal{}\frac{0.12\, mol}{3\, L}\equal{} 0.040\, M$, $ [NO_{2}]_{i}\equal{} 0$.

In the equilibrium the concentrations will be

$ [NO]_{eq}\equal{} 0.083\minus{}2x$, $ [O_{2}]_{eq}\equal{} 0.040\minus{}x$, $ [NO_{2}]_{eq}\equal{} 2x$.

But $ [NO]_{eq}\equal{}\frac{0.050\, mol}{3\, L}\equal{} 0.083\minus{}2x\Rightarrow  x \equal{} 0.033\, M$

and so $ [NO]_{eq}\equal{} 0.017\, M$, $ [O_{2}]_{eq}\equal{} 0.0070\, M$, $ [NO_{2}]_{eq}\equal{} 0.066\, M$.

The equilibrium constant will then be $ K \equal{}\frac{[NO_{2}]^{2}_{eq}}{[NO]^{2}_{eq}[O_{2}]_{eq}}\equal{} 2.2\times 10^{3}$.

For the second problem the reasoning is similar:

$ 2SO_{2}(g)\plus{}O_{2}(g)\rightleftharpoons 2SO_{3}(g)$,

and you should get K = 133.[/hide]


calculus
integration
function
complex analysis

Find $ \oint_{|z|\equal{}2}\frac{z^3e^{\frac{1}{z}}}{1\plus{}z}\, {\rm d}z$. Any hints?

Residue theorem. You can do either $ \minus{}1$ and $ 0$, in which case you get a series whose sum is not too hard to evaluate, or, which is better, you can easily find the residue at $ \infty$. ;)

I split it into 3 parts, the residue at 0 I mean:  First write the power series for $ \frac{1}{1\plus{}z}$.  That's easy.  Then multiply that by $ z^3$.  Still ok.  Now multiply that series by the power series for $ e^{1/z}$.  Maybe a little tricky but just look at the first few terms:

$ (z^3\minus{}z^4\plus{}z^5\plus{}...)(1\plus{}\frac{1}{z}\plus{}\frac{1}{2z^2}\plus{}\frac{1}{3!z^3}\plus{}...)$

Now, start picking out the terms with $ 1/z$ right?  The first is $ \frac{1}{4!z}$, second is $ \frac{\minus{}1}{5!z}$ and so on.  So end up with an infinite sum for the residue.  But that sum is not too hard to evaluate considering what the power series is for $ e^x$ when x is 1 and when it's -1 right?

Oh yea, I don't know about you, but I usually check it numerically just to make sure I got the right answer.  Here's the Mathematica code:

[code]
In[8]:= f[z_] := (z^3*Exp[1/z])/(1 + z);
NIntegrate[f[z]*2*I*Exp[I*t] /.
     z -> 2*Exp[I*t], {t, 0, 2*Pi}]

Out[9]= 1.7763568394002505*^-15 - 2.0943951023934675*I
[/code]

Very Thanks fedja and shawtoos! This was the second question of my last exam, the only one i didn't solve. We haven't learned the residue theorem so i mailed the teacher asking the grade and how to solve it... Hopefully i got all the other questions solved and earned 8 points from 10 :D . He said we can do the substitution $ w\equal{}1/z$ and solve this by Cauchy Integral formula. Other option is to use Cauchy Goursat theorem to isolate the singularities, expand in powers of $ z$ and multiply to get the result.

I think gonna be easy learn the residue theorem after shawtoos's explanation, complex analysis rocks!

Very nice.  Didn't see that.  Letting $ w\equal{}1/z$ over the contour $ |z|\equal{}2$ means that $ w$ traverses the contour $ w\equal{}1/2 e^{\minus{}it}$ which means it's going in negative orientation so need to include a minus sign.  However, when we pull out $ dz\equal{}\minus{}1/w^2 dw$ we can drop them both and I end up with:

$ \mathop\oint\limits_{|z|\equal{}2}\frac{z^3e^{1/z}}{1\plus{}z}dz\equal{}
\mathop\oint\limits_{|w|\equal{}1/2}\frac{e^{w}}{w^4(w\plus{}1)}dw$

Putting in now in the form of Cauchy's Integral Formula:

$ \mathop\oint\limits_{|w|\equal{}1/2}\frac{\frac{e^w}{w\plus{}1}}{w^4}\equal{}\frac{2\pi i f^{'''}(0)}{3!}\equal{}\minus{}4\pi i/6$

Neat!

That's equivalent to what I meant by computing the residue at $ \infty$ :)

as fedja said we only need to find residue at point $ \infty$ it's same that $ \minus{}c_{\minus{}1}$ in laurent expansion of our function or
$ \frac{z^{3}e^{1/z}}{1\plus{}z}\equal{}z^{2}(1\minus{}1/z\plus{}1/z^{2}\minus{}1/z^{3}\plus{}...)(1\plus{}1/z\plus{}1/2z^{2}\plus{}1/3!z^{3}\plus{}...)\equal{}...\plus{}(\minus{}\frac{1}{2}\plus{}1\plus{}\frac{1}{3!}\minus{}1)\frac{1}{z}\plus{}..$


number theory
least common multiple
function

find the remainder when 2004^2003^2002^2001 is divided by 1000.

[hide="hint"]If brute force with a calculator is allowed, then it is best trying to take the $2004^{2003^{2002^{2001}}}(\mod 10000)$ b/c taking it with 1000 will get you the wrong answer, i.e. $4^{9}$.

Or, you can split the mod into 8 and 125 and after calculations, use the chinese remainder theorem[/hide]

I'm pretty sure he did it with *gasp* a calculator  :o.

[quote="krsattack"][hide]Consider 2004^x modulo 1000.  It follows the pattern:
4, 16, 64, 256, 24, 96, 384, 536, 144, 576, 304, 216, 864, 456, 824, 296, 184, 736, 944, 776, 104, 416, 664, 656, 624, 496, 984, 936, 744, 976, 904, 616, 464, 856, 424, 696, 784, 136, 544, 176, 704, 816, 264, 56, 224, 896, 584, 336, 344, 376, 504, 16

Counting carefully, we see the 16 repeats every 50 times.  So now we have to find 2003^2002^2001 modulo 50...
2003^x follows this pattern modulo 50:
3, 9, 27, 31, 43, 29, 37, 11, 33, -1, -3, -9, ...until -33, 1, 3
This gives a pattern of 20.

Now 2002^2001 modulo 20...2002^x has this pattern:
2, 4, 8, 16, 12, 4
This repeats every 4.  Since 2001 is congruent to 1 mod 4, 2002^2001=12(mod 20)
2003^2002^2001=-9=41(mod 50)
2004^2003^2002^2001=[b]816[/b](mod 1000)

There's probably an easier way...[/hide][/quote]

There's definitely an easier way.  For one thing, using my amazing magical abilities I can tell you that powers $\bmod 1000$ have a period dividing $100$ no matter what the base is.

 I can also tell you that powers $\bmod 50$ have a period dividing $20$ no matter what the base is.

Finally, I can definitely tell you that powers $\bmod 20$ have a period dividing $4$ no matter what the base is - and all without doing a single calculation of powers.

:D

I seem to recall posting a solution to this [b]very[/b] recently in a thread a day or two ago.  Let me try to find it.

Edit:  Right, I typed it up and decided not to post it.  So...

[hide="Solution"] $2004^{...}\equiv 4^{...}\bmod 1000$.  Let me define a function.

[b]Definition:[/b]  Let $e(m)$ be the smallest integer $k$ such that $n^{k}\equiv 1 \bmod m$ for all $gcd(n, m) = 1$.

Some of you might think that $e(m) = \phi(m)$, but this is actually not true!  Because of CRT, we actually find that, letting $m = p_{1}^{e_{1}}p_{2}^{e_{2}}...$, we have

$e(m) = lcm \left( \phi(p_{1}^{e_{1}}), \phi(p_{2}^{e_{2}}), ... \right)$

Obviously $4^{...}\equiv 0 \bmod 8$.  Let's take $\bmod 125$ and use CRT later.  $e(125) = \phi(125) = 100$, so

$4^{2003^{...}}\equiv 4^{3^{...}}\bmod 125$

$e(100) = lcm(\phi(4), \phi(25)) = 20$, so

$4^{3^{2002^{2001}}}\equiv 4^{3^{2^{2001}}}\bmod 125$

Finally, $e(20) = lcm(\phi(4), \phi(5)) = 4$, so

$4^{3^{2^{2001}}}\equiv 4^{3^{2^{1}}}\equiv 4^{9}\equiv 1024 \cdot 256 \equiv 24 \cdot 6 \equiv 144 \equiv 19 \bmod 125$

Luckily, $19+125 = 144 = 8 \cdot 18$, so

$2004^{...}\equiv \boxed{ 144 }\bmod 1000$ [/hide]

[quote="t0rajir0u"]$\ldots \equiv 144 \equiv 19 \bmod 125$

Luckily, $19+125 = 144 = 8 \cdot 18$, so

$2004^{...}\equiv \boxed{ 144 }\bmod 1000$ [/quote]

Wait so are you just supposed to find the smallest k where $8|19+125k$? and then that $19+125k$ is $2004^{\ldots}\mod 1000$?...could that possibly be why you said "luckily"?  :fool:

The powers seemed to just reduce $\mod 1000$ in this case too...but it just reduces in smaller values as factors of 1000...so it all works out?  :ewpu:

But the answer is actually 704 and it is a question in one of Mildorf's Mock AIMEs.  :D

From what I see, it's just applications of Euler's totient theorem on the powers of 2004.

[quote="t0rajir0u"]Finally, $e(20) = lcm(\phi(4), \phi(5)) = 4$, so

$4^{3^{2^{2001}}}\equiv 4^{3^{2^{1}}}\equiv 4^{9}\equiv 1024 \cdot 256 \equiv 24 \cdot 6 \equiv 144 \equiv 19 \bmod 125$[/quote]

Ah, I see my silly mistake now.  [hide="Revised"] Of course $gcd(2, 20) \neq 1$.  Instead we must reapply CRT here.  Now, $2^{2001}\equiv 0 \bmod 4$, of course.  Also,

$2^{2001}\equiv 2^{1}\equiv 1 \bmod 5$

By FLT.  Hence $2^{2001}\equiv 16 \bmod 20$ and so

$4^{3^{2^{2001}}}\equiv 4^{3^{16}}\bmod 125$

Curiously, this still does not produce an answer of $704$.  Hmm. [/hide]

I think the problem is that we can change the congruency of the original equation to

$4^{2003^{2002^{2001}}}$ but it doesn't necessarily mean that we can turn this into $4^{3^{2^{1}}}$.

t0rajir0u, I believe that is Carmichael's theorem.

Carmichael?  Hmm, I wasn't aware that the function I defined already had a name.  I'll remember about that $\lambda(n)$ from now on :)

(I was slightly incorrect about its definition, since it turns out that $\lambda(2^{k+2}) = 2^{k}$.)

[quote="deej21"][quote="t0rajir0u"]
$2^{2001}\equiv 2^{1}\equiv 1 \bmod 5$[/quote]

Perhaps you can find your mistake here :)[/quote]

:oops:  Yes, I realized this right after I turned off the computer.  I'm making a lot of mistakes these days...

[hide="One more time"] So $2^{2001}\equiv 2 \bmod 5$ now, and this time it's true!  All in all,

$2^{2001}\equiv 12 \bmod 20$

So that

$4^{3^{2^{2001}}}\equiv 4^{3^{12}}\bmod 125$

We want to find $3^{12}\bmod 100$.  We can write this as

$3^{12}\equiv 27^{4}\equiv 729^{2}\equiv 29^{2}\equiv 841 \equiv 41 \bmod 100$

So that

$4^{3^{12}}\equiv 4^{41}\bmod 125$

Let's write this as

$2^{82}\equiv 128^{11}\cdot 2^{5}\equiv 3^{11}\cdot 32 \equiv 96 \cdot 243^{2}\equiv (-29) \cdot (-7)^{2}\equiv-29 \cdot 49 \equiv \boxed{ 79 }\bmod 125$

Whew.  So we want $a$ such that

$a \equiv 0 \bmod 8$
$a \equiv 79 \bmod 125$

A bit of CRT reveals $a \equiv \boxed{ 704 }\bmod 1000$ to be the correct answer. [/hide]

Knowing that great problem solvers like you make mistakes at times eases me when I make mistakes, like even though I made a mistake I can still be a good solver, kind of reminds me that in the end we're all humans, and we all make mistakes.  :lol:


Due to the nature of the commended solutions I personally doubt that it would be a valid reason to post.

If you think that your solution deserved a commendation, just be happy with it as it stands, and marvel at it's beauty... or something along those lines.

Commendations don't affect your ratings, and, they are meant to be given out for what others feel are "particularly excellent/beautiful" solutions.  A commended solution should, simply put, make a problem look easy.  The type of solution that you don't have to put any thought into to understand.  If you have to argue about weather your solution is commended, then it probably didn't give an instant impression of awesomeness, which is what a commended solution should do.

Cheer up, I expected a commendation as well, but none came.  Getting a solution "officially" commended doesn't make the solution any more/less beautiful, and even commended solutions aren't usually posted (only two or three out of however many there are get posted).  I'll stop ranting now... just take pride in your own solutions, and see how they differ from the commended ones once they are posted.

[quote="banblink182"]If I felt that my solution deserved a commendation, but did not receive one. Is that a valid reason for posting in the "Grading" forum?[/quote]

Not really. I am afraid that giving commendations is very subjective. Sometimes we graders have objective elements that deserve commendations, such as proving uniqueness in a problem that did not require such a proof, but mostly we commend because we feel the solution is better than the rest of them that earned 5 points. And that depends more on other people's solutions than on your solution.

Thus, someone who solved the problem by an especially elegant method could find himself or herself without a commendation because half the correct solutions used that method. In contrast, someone who wrote a plain, bland solution could get a commendation just because most of the other correct solutions were badly written.

The standard is to try to commend one tenth of the complete and correct solutions, but that standard is not rigorously followed by us graders. Some problems just do not lend themselves to commendations, because the answer is so straightforward that there is no way to make a solution stand out. On other problems, I am so relieved that someone has finally solved the problem correctly that I commend one third of the correct solutions.

Erin Schram
USAMTS grader


I never noticed that. Also, it says they are sold out.

[quote="westiepaw"]Sorry if this is off-topic, but I wasn't sure where to put this.  Does anybody know why, after I click on the Alcumus tab, then touch (but don't click) the Community tab, there's an option for T-Shirts?  Whenever I just mouse over the Community tab from the Forum that option doesn't show up....[/quote]Thanks for pointing this out, we'll get that fixed.


If we know that $c(x)$ is a generating function for a sequence $c_{0},c_{1},c_{2},...$ of numbers and \[c(x) = \frac{1-\sqrt{1-4x}}{2x},\] how can we find an explicit formula for $c_{n}$?

I'm not very good with generating functions, so I'm not sure if I've provided enough information.

If it helps, the sequence in question is the Catalan Numbers.

To get you started :
$\sqrt{1-4x}= \sum_{n=1}^{\infty}{{n-3/2}\choose n}4^{n}x^{n}$.

How'd you get that?

Heh. Sorry.

We couldn't just use the binomial theorem?

Well, we have the formula (true for all $a \in \mathbb{R}$) : \[(1-x)^{a}= \sum_{k=0}^{\infty}{{k-a-1}\choose k}x^{k}\] So now, just apply this with $a =-1/2$ and $x : = 4x$.


:rotfl:  :rotfl:  meaningfull problem! I think a have a solution, but I have to take into account some lemmas which I think the proposer isn`t aware of. But hey, we`re here to learn :wink:

Lemma 1.
If $a\geq 1$, then $a+1\geq 2$.
Lemma 2. If $x_{1}\geq 2, x_{2}\geq 2\dots x_{n}\geq 2$, then $x_{1}x_{2}\cdots x_{n}\geq 2^{n}$
I leave the solution to the proposer.
Now the tricky part comes: how do we use these lemmas to prove This Year`s Problem?
Well by applying lemma $1$, we obtain $a_{1}+1\geq 2, a_{2}+1\geq 2\dots a_{2007}+1\geq 2$, and thus by lemma 2 we obtain $(a_{1}+1)(a_{2}+1)\cdots (a_{2007}+1)\geq 2^{2007}.$
Finally, we solve This Year`s Problem by a contradiction argument:

Assume that between the 2007 numbers there is one number $a_{i}$ such that $a_{i}$ is diffferent from $1$. Now, because $a_{i}\geq 1$, we obtain $a_{i}+1>2$. But the other 2006 $a_{j}+1$ numbers are all $\geq 2$, and thus, by lemma 2, we obtain $(a_{1}+1)(a_{2}+1)\cdots (a_{2007}+1)>2^{2007}$. But from the problem condition $(a_{1}+1)(a_{2}+1)\cdots (a_{2007}+1)=2^{2007}$, we obtain a contradiction. Thus, there is no number different from $1$, and thus all the numbers are equal to $1$. The conclusion in the end is thus rewritten as \[ \frac{2007}{2^{2007}}\leq 2007^{2}\cdot 2006+\frac{1}{2^{2007}}\Longleftrightarrow 2007\leq 2007^{2}\cdot 2006\cdot 2^{2007}+1,\]which is indeed tricky, but once you see that $2007^{2}\cdot 2006\cdot 2^{2007}> 2007$, the problem is clear.

Again I say, VERY meaningfull problem :wink:

Needless to say, I have got your solution, too.
How about $a_{i}\geq 0\ (i=1,\ 2,\ \cdots 2007)$?
This was my first problem.


I just got back from Penn State ARML, where I was introduced to ultimate frisbee. Despite being sure that I wouldn't like it, I really enjoyed it; it's a terrific game. Anyway, my question was, why is frisbeeing such a large part of the high school competitive math culture in the US?

Ultimate Frisbee = Bad

 :furious:  :furious:  :furious:  :furious:  :furious:  :furious:

I don't know, but Ultimate Frisbee =  :thumbup:, probably the coolest game ever. Can't wait to play at Mu Alpha Theta Nationals...  :D

Psh. Speak for yourself.

Anyways, in my personal unbiased and completely unprofessional opinion, frisbee and math are completely unrelated. Mathematically gifted students love other sports. Those who do not like math enjoy frisbee. The reason it is many times the sport of choice at competitions is that it does not require a complicated field or equipment. Also note that many math students are not athletically talented and frisbee happens to be one of those sports that can be enjoyed by people with very little experience in athletics - even when playing with those who are somewhat more accomplished.

[quote="MithsApprentice"] Also note that many math students are not athletically talented[/quote]

That is not true. There is no relation between them. Maybe you havnt realized but many people are not athletic.

in addition to being lots of fun and not too violent, frisbee is very easy to learn and teach others. i first learned frisbee at arml and still enjoy it every year :)

[quote="MithsApprentice"][quote="GoBraves"]
That is not true. There is no relation between them.[/quote]

Actually it is true. I didn't claim a certain proportion or percentage (or even a majority), only that there were "many" such people. I didn't claim a "relation" between "them" either.

If you can prove to me that every single person interested in math is also athletically gifted (or that there was less than, say, 10 exceptions) I'll assume I'm still correct.[/quote]

It depends on what you call "athletically gifted". In general i dont think i could prove that, but from what i've seen on this forum, there are a large amount of people who are interested in math who also participate in some sort of sport.

Yes....people that enjoy ultimate. It is a very fun sport, and I personally find the ultimate subculture to be very diverse. There are very smart people, and not so smart people. In competitions (not math competitions, but ultimate frisbee tournaments), it is not the case that the players are all smart. I think that frisbee has been adopted as a sport of choice for many smart (and nerdy) people because it is seen as an alternative sport; a sport different from the mainstream of soccer, football, basketball, etc.

I could go on about how addicting ultimate is, but I need to sleep.

i was introduced to ultimate frisbee at ARML as well. it is pretty fun.  i have realized that i suck at throwing the frisbee but i am pretty nasty when it comes it defense.

I think ultimate frisbee is just a fun game to play for any group.. it's easy to learn, doesn't require that many materials (just a frisbee), any amount of people can play it, and it's fun. So it's not just math people who like it. Most people like it.

Yeah, it's fun for everyone, but we like to play during math class :D and my Calc BC class has surprisingly good frisbee players... we're training for MAO Nationals ;) Good stuff.

paladin8 are you training for MAO nationals ultimate frisbee or math? :P  Anyhow, I had already known of the sport but for some reason mostly play it during summer camps (and the ones I attend are like CTY).  I think it might be a little more than a coincidence, although I do believe some of it has to do with what MithsApprentice said.

[quote="probability1.01"]paladin8 are you training for MAO nationals ultimate frisbee or math? :P [/quote]

 :roll: :agent: What do you think? It's in Hawaii, after all. Math is of secondary importance; besides, I don't need training for math :D

Oooh...since we're on the topic, anyone have tips for a good air bounce?

not many people at my school play frisbee
maybe that's why our math team sucks :(
tennis, soccer, and basketball is the best

[quote="xxreddevilzxx"]I am the only frisbee player in my math team. Think about how sad that is.

P.S. paladin, the match shall be in MAO Nationals. You+justin vs. Me+Qiaochu.[/quote]

2v2? What is that supposed to mean? We have a 7-player frisbee team with substitutes, too. We are intense like that. :D

[quote="mcalderbank"]Wait, is your entire team going to MAO?[/quote]

A lot of people. I think we're sending 12 students, of which 9 or 10 can play frisbee :P

[quote="xxreddevilzxx"]Dont worry. WE are sending 70 students... I am sure there willl be more than enough to outhustle and outscore your team in the ultimate frisbee, palandin... :P[/quote]

Yeah. I knew that. But frisbee is 7v7 officially so we won't have to worry about the other 63 people :D Besides, if they can't catch they are pretty much useless.

[quote="xxreddevilzxx"]we will never lose as long as:

1. double-guard Justin
2. dont ever get out of defense
3. whenever you drop, we just try to throw it as far from our goal as possible :D  :D[/quote]

Oh, believe me, Justin is not the best... we have much better players... well at least one, whose name I will not reveal :P and we are practicing like at least once a week (I can throw an awesome forehand now, yeah!).


Hey guys, do you know when there is going to be a math jam for mathcounts?

Probably not for a long while.

I believe Math Jams are targeted to AMC/AIME-type level questions rather than MATHCOUNTS. There might be the occasional MC one (I believe a MATHCOUNTS Math Jam was done Jan, 18, 2006) but I wouldn't get too hopeful.

yeah, no math jam for mathcounts, unless youre talkign about the mathcounts problem series class.


Let ABC be a triangle. Angles A, B and C are acute. On sides BC, CA, AB we have points $A_1, B_2, C_3$. $AA_1$ ia a diameter of circle $T_1$, $BB_1$ - $T_2$, $CC_1$ - $T_3$. Prove that common chords of each two circles goes through the orthocentre of triangle ABC.

I can prove that this all 3 chords have common point, but I don't know how to prove that it is orthocentre..

if its too hard for this section, plz admins move it to another.

Let $B'$ is the second point of intersection of $T_2$ and $AC$, $C'$ is the second point of intersection of $T_3$ and $AB$. Then $BB'\perp AC$ and $CC'\perp AB$, i.e. $BB'$ and $CC'$ are altitudes. We need to prove that $XY$, $BB'$ and $CC'$ are concurrent. Indeed, quadrilateral $BC'B'C$ is inscribed, therefore $BB'$, $CC'$ and $XY$ are three radical axes of three cicles, hence result.

That's the usual case when a problem is posted in two different fores... http://www.mathlinks.ro/Forum/viewtopic.php?t=22558

  darij

Yes, it's also different from Grobber's one... I just wanted to say that posting problems in different threads causes more work to people...

  darij

[quote="Myth"]But my solution has a nice figure![/quote]

I really appreciate that Myth attaches images to his solutions. This is really an impressive and important feature, especially when dealing with geometry problems. Well done, Myth.  :)


Find all pairs $(a, b)$  of positive integers  such that \[(\sqrt[3]{a}+\sqrt[3]{b}-1 )^{2}= 49+20 \sqrt[3]{6}.\]

You also have to prove there are no others... and a sketch how you found them would be nice. :)

I try to find $ m,n\in Z^ \plus{}$ so that $ 49 \plus{} 20\sqrt [3]{6} \equal{} (m\sqrt [3]{6} \plus{} n\sqrt [3]{36} \minus{} 1)^2$ and I have already found them!
Now $ \sqrt [3]{48} \plus{} \sqrt [3]{288} \minus{} 1 \equal{} \sqrt [3]{a} \plus{} \sqrt [3]{b} \minus{} 1\Rightarrow a \equal{} 48,b \equal{} 288$ or $ a \equal{} 288,b \equal{} 48$
(Because $ \sqrt [3]{48} \plus{} \sqrt [3]{288} \minus{} 1\neq 1 \minus{} \sqrt [3]{a} \minus{} \sqrt [3]{b}$)

In fact the original question (I think which this question maybe should have been) is to find [b]a[/b] pair $(a,b)$, rather than find [b]all[/b] pairs $(a,b)$.

The trick is to set $a=6x^3$ and $b=36y^3$. After expanding we have $x^2\sqrt[3]{36}+6y^2\sqrt[3]{6}+12xy=2y\sqrt[3]{36}+(2x+20)\sqrt[3]{6}+48$. So to find a solution, we look to equate the coefficients: $x^2=2y,6y^2=2x+20,12xy=48$. The pair $(2,2)$ can be shown to satisfy these, hence we obtain tdl's solution $(a,b)=(48,288)$ and it's permutation.

If any other solution $(c,d)\not= (a,b)$ exists then it satisfies $\sqrt[3]{c}+\sqrt[3]{d}=\sqrt[3]{a}+\sqrt[3]{b}=2\sqrt[3]{6}+2\sqrt[3]{36}$. This is equivalent to finding all $1\le c\le 277$ for which $6\sqrt[3]{36c^2}+6\sqrt[3]{6c^2}-12\sqrt[3]{36c}-72\sqrt[3]{6c}-144\sqrt[3]{c}+144\sqrt[3]{6}+144\sqrt[3]{6}$ is an integer, I have an inkling it is only $c=48$. Perhaps there exists a nicer method.


MIT
college
blogs
AMC
USA(J)MO
USAMO
summer program

Hi,

I am just curious if any AOPS member has attended RSI. How is it like? I am sure it is an awesome program and I just want to know more about it.

Thanks for any reply.

RSI is an annual summer research program held at MIT and Caltech for rising high school seniors. i applied but didn't get in :(, though the people i know who went had lots of fun there, and many of the RSI alumns were semifinalists at Intel (not sure about finalists though, probably some also went there!)
also, going to RSI pretty much ensures your getting into MIT.

Research Science Institute.... probably THE most prestigious research program for High School students in U.S. It is held by MIT, and I think Cal Tech, starting next yr....(I am not sure though). It is free :D

i've heard its really hard to get in..
i've won two times in intel isef in math (with a team partner) and my team partner applied and didn't get in. and he had a OK sat score, winning at isef, and a couple of IMO's.
I'll apply next year though...

Most Intel Winners research their project at RSI though.
And do you know if there is nay science background required, like previous science researches at other insitutes or some sort?

don't know... but it will help for sure.

Manuel, is that correct?  Your friend had a couple IMO's and didn't get into RSI?  How do you not get into RSI with a couple IMO's?

i don't know... that was really unfair...
His GPA was 3.8
could that be the reason..

A couple of IMO's from what country?  (Remember that many programs will accept very few international students and that making the IMO is relatively easier from some countries than from others.  Also, don't fall into the "I have credential X and therefore am invincible" trap.  As I've said before, given any reasonable set of characteristics, I can design a student around them who will be rejected from a given top-tier university, or in this case from RSI.)

3.8 GPA isn't bad either I dont forsee that as a plausible reason.

RSI pays a lot more attention to the free-response questions (the "essays") on the application than you might think.  They are looking for people who show both an interest in and an aptitude for research as opposed to merely academics.

I say this both as an RSI alumnus and as a former member of the teaching staff.

[quote="JBL"]Remember that many programs will accept very few international students [/quote]

RSI is sort of an outlier in this respect.  While international students do not make up the majority of the body, international students do make up around 1/4 to 1/3 of RSI.  Plus, factor in the fact that the number of international applicants is much much less than the number of domestic applicants.  RSI may be the one case where international attendees are not much more qualified than their domestic counterparts.  Granted, one would be hard-pressed to declare any "Rickoid" underqualified.  Anyway, take what I say with a grain of salt, since everything I know about RSI comes from another source, albeit that source is RSI alumni.

[quote="Mariag"]many of the RSI alumns were semifinalists at Intel (not sure about finalists though, probably some also went there!)[/quote]

[quote="Matt McGann"]Six of the forty finalists were RSI. Of the top 10 winners, RSI took places 3, 4, and 10.[/quote]

Source: [url=http://blogs.mit.edu/madmatt/posts/9137.aspx]Matt's Blog[/url].

[quote="sonar"][quote="JBL"]Remember that many programs will accept very few international students [/quote]

RSI is sort of an outlier in this respect.  While international students do not make up the majority of the body, international students do make up around 1/4 to 1/3 of RSI. [/quote]

Really? Wow, that's a lot of competition to get there... now I'm really impressed at my friend that went there in 2003. I wonder if I have any chance :?

My friend is going this year. I might as well try next year, since all of his scores from Standardized Test, SChool grades arent better than mine. Plus, he doesnt have much accomplishments, whereas I have USAMO/ARML etc. By the way, does USAMO qual count a lot???

It suddenly occured to me that this thread doesn't really have anything to do with college, so I've moved it here.

As long as I'm posting...

Keep in mind that RSI is a program for all sciences, not just math.  They only take about 15-20 math students from the U.S. every summer.  So it is extremely competitive.

Sonar,

While what you assert is true to some extent, you can never refer to any international participant as underqualified, because simply put, you don't honestly know. Granted, CEE receives much less applications from international students, however, you can't deny the fact that this would make the international pool much more self-selecting and hence, they would have more qualified international applicants. I am going to RSI as an international partcipant this summer, and while there are few domestic admits of whom I admire, I do not consider myself academically inferior to any other domestic admit.

I attended RSI 2004 at MIT, and it was AMAZING, probably the best summer of my life. RSI admissions are a mystery; they even joke about it being like throwing darts. Either way, it seems to be working =) My best advice for those of you applying is don't [i]try[/i] (too hard) to get in because they can totally see through that.

[quote="onEnterFrame"]you can never refer to any international participant as underqualified[/quote]
As for the international students, I personally found many of them to be much more qualified than some of the domestic participants, and they certainly made RSI much more colorful and enjoyable.

Hello,
I have not been to RSI yet but am going this summer. I am pretty sure we are only at MIT; I think they did CalTech and MIT last year, and for some reason decided not to do CalTech this year. I don't know about future years, though.
I agree that the emphasis on essays must be very strong. I have never made USAMO, etc. (though my research field is Planetary Science, not math), nor done ISEF, etc (Though I have won other national competitions), but my essays were built around showing them how much of a passion for my field I have, and how that passion led me to research. That, and my recs probably got me in.
All the 2005 Rickoids-to-be have been in communication, and I must say that all of them are fantastically qualified -- regardless of nationality.
In response to the original question, yes, I think it is going to be fun; this is the only program in which all the kids started communicating with everyone else the day they got contact info (my mailbox had ~50+ unread messages after ARML.

[quote="manuel"]Is the SAT score important??
 i mean i am an international student and I did not do good in the english.[/quote]

Was that done intentionally the irony with the "I did not do good in the english"?  Either way it's pretty funny.

[quote="onEnterFrame"] you can never refer to any international participant as underqualified, because simply put, you don't honestly know.[/quote]

You missed my point entirely  :P .  I was simply saying that attendees are more or less on level ground--usually international attendees of selective programs/colleges are much more qualified than the average domestic, as JBL inferred, which I was under the impression was not the case at RSI, but olgaman has informed us otherwise.

[quote="DPatrick"]It suddenly occured to me that this thread doesn't really have anything to do with college, so I've moved it here.[/quote]

Yes, and if one does relate it with college, he or she has missed the point.

[quote="paladin8"]Really? Wow, that's a lot of competition to get there... now I'm really impressed at my friend that went there in 2003. I wonder if I have any chance :?[/quote]

Yes, around 75 particpants and around 50 of those are domestic.

Hey people . I wanna ask a favour. Please visit [url=http://www.iisc.ernet.in/kvpy]KVPY[/url] and post how you compare this with the RSI and other US Research summer programs.
I was one of just 40 selected all over India and attended this camp. I wanna know how this compare with the situatution in US regarding RSI. This KVPY is actually a scholarship and the summer camps and later research projects are just a part. Just remember that India has a much larger population than US(just a statistic ) and there are the famous IITs here. I'm not bragging. Just bringing things into context that just 40 are selected in my (SA) Stream.
So please compare this with the RSI and also post what you think about its importance for admission to MIT, Caltech, 0princeton etc. Will it be really valued??? If you want, I can post the subject matter they taught us.

The whole point of RSI is that they actually do research there, as opposed to other summer camps, where the focus is more on learning than original discovery.  RSI is more in the vein of the American REUs or the Canadian NSERC USRAs, except that the participants are high schoolers rather than undergrads.  RSI is also partly famous because of its alumni - although some of the participants are rather 'normal', they get their share of Gabriel Carrolls and other brilliant young minds.

As for your assertion about India having a much larger population and a (limited) number of schools of some renown, it is pretty meaningless - India doesn't produce many international olympiad medalists, nor does your program draw many, if any, students from abroad (for example, Vess on this forum went to RSI from Bulgaria, and he is an unbelievably bright young mathematician).  I suspect that your acceptance will be (and should be) treated on a par with attending one of the prestigious, but not VERY prestigious summer programs like PROMYS or MathCamp.  It's not, however, going to be put on a par with attending RSI or MOP or anything like that.

Regarding research, We actually did research at IISC where, we interacted with Profs. is specific fields. I chose Mathematics. Also, we have research projects continuing with univ. profs. close to our homes.

Regarding popularity, I believe the KVPY began in 1999 and it will take some time for International students to know about it. And please realise that I am not bragging and just asking for info. Your post has a bit of sarcasm Please don't mind!

I also suggest visiting the website I have given!

[quote="Rushil"]Your post has a bit of sarcasm[/quote]

No, it doesn't.

I skimmed the website you provided, but really have no desire to go through it in much detail.

Any ways, thanks for your comments. If you want to discuss more on KVPY or hea others etc, plz nisit the KVPY forum in College Section. I think discussing more about it here would take the forum away from the original topic RSI


You write a list containing all of the whole numbers between 200 and 499. How many times does the digit 5 appear?

Since $ 5$ will not appear in the hundreds place at all in the range of numbers,
we can simply find the number of fives between $ 200$ and $ 299$ and multiply
that amount by $ 3$...

First let's look at range of $ 250$ to $ 259$...

There are $ 10$ between $ 250$ and $ 259$, but in the number $ 255$ five appears
twice.

So there are $ 11$ fives between $ 250$ and $ 259$

From $ 200$ to $ 299$, there is one five for every ten numbers except for the range
$ 250$ to $ 259$....

So there are $ 9$ fives in the range $ 200$ to $ 299$ excluding the $ 11$ fives already found.

So there are $ 20$ fives between $ 200$ and $ 299$...

$ 20\times3$

$ \boxed{60}$


does it mean that the Bay Area represents Northern California?

Because, when I checked the list of states, there was Bay Area and Southern California.

And I figured out that Bay Area is Northern California, but is that true?

Geographically, no, but I'd say about 95-99% of all NorCal AoPSers are in Bay Area.

yeah, the bay area has pretty much all the AoPSers in NorCal.


Leaderboard:

theone853-5
MysticTerminator-3
MathFiend-2
Leeyuhang-1

1.If n is a positive integer and n(n + 1) is divided by 3, the remainder can be:
a.) 0 only
b.) 2 only
c.) 0 or 1 only
d.) 0 or 2 only
e.) 0,1, or 2

2. A team's record is 20 wins and 25 losses.  To qualify for the playoffs a team must win 60% of its games played.  What is the number of wins of the remaining 15 games necessary for the team to qualify for playoffs?
a.)4
b.)9
c.)12
d.)15
e.) impossible to achieve

3. A tank is 1/6 full of gasoline.  If 2 gallons are added, then the tank is 1/4 full.  What is the total capcity of the tank in gallons?

4. The legs of a right-angled triangle are 5 and 10 while the hypotenuse of a similar triangle is 15.  What is the area of the larger triangle?

5. If x^0.3 = 10, then what does x^0.4 equal?

MysitcTerminator gets #3.
I editted #2, my apologies.  MathFiend's answer is incorrect

Mystic gets #2 & 4.

Do you want me to give possible answers for number 5?

huh...I though MT and I posted at the same time for # 2, but oh well...

sure, I'd like to see answers for #5...

wait can x be complex? in that case

5.[hide]10^(4/3)e^(i*:pm:2:pi:/3) in addition to my earlier answer[/hide]

a bunch of these are from joel's contest.

yeah...I said that, but they seem to be ignoring me...

MysticTerminator gets #5.  His original answer 10^4/3 was correct.

For #2, my computer lists Mystic's answer coming in a minute earlier than MathFiend's answer.  Do other's show it coming in at the same time?

Joel's questions come from the source I gave him.  There are two types of tests from the source: Probe I questions and Probe II questions.  I told him I use questions from Probe I for my contests and suggested he use questions from Probe II.  It's not really that big a deal.  It's not like these contests mean anything, plus Joel's contest requires solutions, mine doesn't.

he must've come in only a few seconds before me then, cuz his message didn't show when I answered and I knew it before I tried to type...but oh well...

Updated leaderboard after 3 rounds:

MysticTerminator-8
Theone-5
MathFiend-3
Leeyuhang-1

Dad needs computer for awhile...


calculus
integration
function
real analysis
geometric series
real analysis unsolved

Hi all

I am working on a problem involving finding the integral from 0 to 1 of the product of the cantor staircase function with x^n for some arbitrary integer n.

Now I know what I want to do, namely if we say phi is the staircase function, then let phi_m be the staircase function restricted to the first m generations of removed intervals ( ie. (1/3, 2/3) + (1/9,2/9) + (7/9,8/9) +... up to the intervals with denominator 3^m), and then apply limits.

Now the part I'm stuck on is very annoying because it seems so trivial. I looked at these integrals and I can see how to find their limit, namely what happens is as m increases, you get an a sequence of geometric series, that should each be summable for any k. It is kind of difficult to explain but hopefully you know what I mean,
For example: if we assume k = 1, we get for the first few iterations:
(1) int(phi_1*x) = 1/2 *(2/3-1/3)
(2) int(phi_2*x) = "(1)" + 1/2*1/3*"(1)" + 3/4*(8/9-7/9)
(3) int(phi_3*x) = "(2)" + 1/2*1/3*"(2)" + 5/8*(20/27-19/27) + 7/8*(26/27-25/27)
(4) int(phi_4*x) = "(3)" + 1/2*1/3*"(3)" + etc...

So hopefully you can see what I mean, where we will be able to find the limit by writing
int(phi_n*x) = (phi_1*x) + [(phi_2*x) - (phi_1*x)] + [(phi_3*x) - (phi_2*x)] +... and then summing up infinitely many geometric series (In the limiting case).

So that is what I [i]want[/i] to do, but my problem is to do this I need to find a general closed formula for the endpoints of the m-th generation intervals because in the general case when k is arbitrary, I will have to sum up a geometric series with factor (b^k - a^k) where (a,b) is some interval that is removed. (For the example above it didn't matter because when k=1, this b-a will always be one over some power of 3, and we know precisely how many such intervals there are, but in general I think I need to actually know what the endpoints are!)

So anyways that is where I am stuck and it seems very annoying because I know how I want to attack this problem but I can't think of a non-recursive formula to express the endpoints of the intervals. If anyone can help me out or suggest a different way to attack the problem that would circumvent this issue then please let me know.

Thanks very much.

You might want to consider doing an integration by parts:

$ \int_0^1x^nF(x)\,dx\equal{} \left.\frac1{n\plus{}1}x^{n\plus{}1}F(x)\right|_0^1\minus{} \frac1{n\plus{}1}\int_0^1x^{n\plus{}1}\,dF(x)$

The boundary terms evaluate to $ \frac1{n\plus{}1},$ leaving us with that $ dF(x)$ integral. Perhaps you can control the singular measure given by $ dF$ well enough to compute that Stieltjes integral.

The problem actually started as integrating with respect to the measure dF(x). I am not really familiar with lebesgue-stieltjes integrals, other then knowing you can do integration by parts, so I immediately did that to get into the form I said.

I think what you are suggesting then is to try and treat it like a riemann integral, and get a sequence of partitions of [0,1] such that the "F"-measures of each sub-intervals of each partition will somehow be easy to compute, and then calculate upper/lower sums and take the limit? Is that correct? Sorry for such a basic question but like I said I'm just not very familiar with these integrals.

I would just use self-similarity:
\[ I_n\equal{}\int_0^1 x^n\,dF(x)\equal{}\int_0^{1/3}\plus{}\int_{2/3}^1\equal{}\frac1{2\cdot 3^n}[I_n\plus{}\int_0^1(2\plus{}x)^n\,dF]\]
If you open the parentheses, you'll get a recurrence relation that will allow you to compute each particular value. I do not know if there is a neat closed form expression for this integral though :(


If $47^{100}$ has 468 digits, then how many digits does $47^{17}$ have?

I think $47^{100}$ has 168 digits? Typo?

Is there any other way to do this beside using log? :?:

Not that I know of. Why does this bother you?

i dont really know.. but log is the basic way to figure out how many digit a number has...  :maybe:

There are two basic ways

1) Use logs

2) Use computers


floor function
function
inequalities
algebra
AMC

Suppose $ a, b,$ and $ c$ are positive integers with $ a \plus{} b \plus{} c \equal{} 2006$, and $ a!b!c! \equal{} m\cdot10^n$, where $ m$ and $ n$ are integers and $ m$ is not divisible by 10. What is the smallest possible value of $ n$?

$ \textbf{(A) } 489 \qquad \textbf{(B) } 492 \qquad \textbf{(C) } 495 \qquad \textbf{(D) } 498 \qquad \textbf{(E) } 501$

[hide="Answer"]We are looking for powers of $10$, so we need $5$'s, as $2$'s appear much more commonly than $5$'s. The way to partition with fewest factors of $5$ relative to the actual number is to have one less than a power of $5$. The largest power of $5$ which we can use is $625$, so we use $624$ twice, and $758$ for the last one. This gives us $152+152+188=492$ powers of $10$. $\boxed{B}$[/hide]

[hide]
Since $2$'s appear more frequently than $5$'s, $n$ is the number of factors of $5$ in $a!b!c!$.

The number of factors of $5$ in $a!$ is $\lfloor \frac{a}{5}\rfloor+\lfloor\frac{a}{5^2}\rfloor +\lfloor\frac{a}{5^3}\rfloor +\lfloor\frac{a}{5^4}\rfloor =\sum_{i=1}^{4} \lfloor \frac{a}{5^i}\rfloor$
So $n=\sum_{i=1}^{4} (\lfloor\frac{a}{5^i}\rfloor +\lfloor\frac{b}{5^i}\rfloor +\lfloor\frac{c}{5^i}\rfloor)$

The floor function has the following property:
$\lfloor x\rfloor +\lfloor y\rfloor\ge \lfloor x+y\rfloor -1$
Similarly, for three numbers, we have
$\lfloor x\rfloor +\lfloor y\rfloor +\lfloor z\rfloor\ge \lfloor x+y+z\rfloor -2$

Therefore, we must have
$n\ge \sum_{i=1}^{4} \left(\lfloor\frac{a+b+c}{5^i}\rfloor -2\right)$
$\Leftrightarrow n\ge \sum_{i=1}^{4} \left(\lfloor\frac{2006}{5^i}\rfloor -2\right) =401+80+16+3-8=\boxed{492}$

As JesusFreak197 mentioned, $492$ can be achieved when $(a,b,c)=(624,624,758)$.[/hide]

Hi.

In [url]https://www.artofproblemsolving.com/Wiki/index.php/2006_AMC_12B_Problems/Problem_22[/url], can you guys explain where the solution is going from the phrase "More precisely" till the end? Thanks. :)

Which part of that isn't clear? I can rephrase things but I'm not sure how much sense it will make.

The first step is to establish the inequality \[ \left\lfloor \frac ak \right\rfloor + \left\lfloor \frac bk  \right\rfloor + \left\lfloor \frac ck \right\rfloor \ge  \left\lfloor \frac{a+b+c}{k} \right\rfloor -2 \] for all positive integers $a,b,c,k$.  Applying this inequality for $k=5,25,125,625$ to the long expression for $n$ gives us that \[ n \ge \left\lfloor \frac{2006}{5} \right\rfloor + \dots + \left\lfloor \frac{2006}{625} \right\rfloor = 492 \] because $a+b+c=2006$. (I guess the solution used $n$ by mistake here for the numerators.) Anyways this means $n \ge 492$ is a lower bound. Finally, the solution chooses $a=b=624$ to show that equality can actually occur, and hence $492$ is actually a minimum.


floor function
ratio
arithmetic series
geometric series

[quote="nemomath"]I don't understand how the equations are suppose to work. I seem to be struggling with the problems.
Find the sum of each series:
a.)
7,3,1,-1,...,-189=__________
b.)
1/5+2/5+4/5+8/5+... for 15 terms =__________
c.)
The sum of the multiples of 3 from 1 to 500 =__________

How exactly do I solve these problems?[/quote]

For part (a), it is not possible to see what the rule is.. maybe you missed a 5 there... I am sure it is an arithmetic series or something.

[hide="(b)"]Part (b) is found by applying the geometric series sum formula... take out 1/5 first and get
\[\frac15 \sum_{k=0}^{14}2^{k}= \frac{2^{15}-1}{5}. \]
[/hide]
[hide="(c)"] There are $\left\lfloor \frac{500}{3}\right\rfloor = 166$ multiples of $3$. Therefore this is
\[\sum_{k=1}^{166}3k = 3 \sum_{k=1}^{166}k = 3 \cdot \frac{166 \cdot 167}{2}= \boxed{41583}. \]
[/hide]

[hide]b)
$\frac{2^{0}}{5}+\frac{2^{1}}{5}+\frac{2^{2}}{5}\cdots \frac{2^{14}}{5}=?$

This is a finite geometric series. The sum of a finite geometric series is calculated by:
$\frac{a\cdot(1-r^{n})}{1-r}$ where:
$a=$first term
$r=$common ratio
$n=$number of terms
so...
$\frac{1\cdot(1-2^{15})}{1-2}=32767$
Now just divide that by 5:
$32767/5=\boxed{6553 \frac{2}{5}}$[/hide]

Oops here is what I was trying to say:
a.)
7,3,-1,...,-189=__________

Can you please check my work for this:
Q1.)
Find term 4 of the sequence: a_n=4(3)^(n-1)
A1.)
a_4=4*3^(4-1)
a_4=108

[quote="nemomath"]Oops here is what I was trying to say:
a.)
7,3,-1,...,-189=__________[/quote]

[hide]The general formula is $a_{n}=-4n+11$ (common difference $d$ is $3-7=(-1)-3=...=-4$), so there are $50$ terms since $a_{50}=-189$. Then the sum is
\[\text{(average of the terms)}\cdot \text{(number of terms)}= \frac{(7-189) \cdot 50}{2}= \boxed{-4550}. \]
[/hide]

601st post :(

Can you please check my work for this:
Q1.)
Find term 4 of the sequence: a_n=4(3)^(n-1)
A1.)
a_4=4*3^(4-1)
a_4=108
Is there a way to enter subsets on a graphing calculator I have a TI-89.

[quote="nemomath"]Can you please check my work for this:
Q1.)
Find term 4 of the sequence: $a_{n}=4 \cdot 3^{n-1}$
A1.)
$a_{4}=4 \cdot 3^{4-1}$
$a_{4}=108$[/quote]
That is correct.
[quote="nemomath"]
Is there a way to enter subsets on a graphing calculator I have a TI-89.[/quote]
For what subsets exactly?

For what subsets exactly?
Like for a_n and a_1.

Also, how do I find the sum of things like this:
sum_3(2){n-1}=?


Let n be a natural number. Prove
\[ {1+2^{n}+3^{n}+4^{n}\,mod\, 10=\begin{cases}4 & n \bmod 4=0 \\ 0 & n \bmod 4 \ne 0 \end{cases}}\]

[b]Fact:[/b] if $ n, p \in \mathbb{N}_{0}$ and $ p$ is prime, then $ \sum_{a=1}^{p-1}a^{n}\equiv \left\{\begin{array}{ll}0 & \mbox{if }(p-1) \nmid n \\-1 & \mbox{if }(p-1) \mid n\end{array}\right. \bmod p$ (here as usual, $ \mathbb{N}= \{1, 2, \ldots\}$ and $ \mathbb{N}_{0}= \mathbb{N}\cup \{0\}$). For a proof, follow [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=40171]the link[/url].

Now fix $ n \in \mathbb{N}_{0}$ and let $ s_{n}= 1^{n}+2^{n}+3^{n}+4^{n}$. Then obviously $ s_{n}\equiv 0 \bmod 2$ and $ s_{n}\equiv \left\{\begin{array}{ll}0 & \mbox{if }4 \nmid n \\ 4 & \mbox{if }4 \mid n\end{array}\right. \bmod 5$ by the fact. So trivially $ s_{n}\equiv 0 \bmod 10$, if $ 4 \nmid n$, and $ s_{n}\equiv 4 \bmod 10$, if $ 4 \mid n$.


function
inequalities
inequalities unsolved

Let $\ a_{i}\geq 0$ such that $\sum_{i=1}^{n}a_{i}=1$
 Prove that $\sum_{i=1}^{n}\frac{a_{i}}{2-a_{i}}\geq\frac{n}{2n-1}$
         good luck  :D

The function $f(x)=\frac{x}{2-x}$ is convex for $x < 2$.
So using Jensen inequality we can prove it easily. :D

I don't know Jensen . Can you post other case? Thankyou !  :roll:


geometry
geometric transformation
national olympiad

ok, as Valentin says, a post for each bmo :)

[b]BMO 1984:[/b]

1: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822800#822800[/url]

2: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822803#822803[/url]

3: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822804#822804[/url]

4: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822805#822805[/url]

Don't edit, post a new reply for each please.

[b]BMO 1985:[/b]

1: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822819#822819[/url]

2: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822824#822824[/url]

3: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822826#822826[/url]

4: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822828#822828[/url]

[b]BMO 1986:[/b]

1: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822841#822841[/url]

2: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822847#822847[/url]

3: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822851#822851[/url]

4: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=345517#345517[/url]

PS: if the single observation that Cezar in his post mentioned that problem 4, is from 1985, but it's from 1986. :)

[b]BMO 1987:[/b]

1: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822867#822867[/url]

2: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822870#822870[/url]

3: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822872#822872[/url]

4: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822878#822878[/url]

[b]BMO 1988:[/b]

1: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822891#822891[/url]

2: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822893#822893[/url]

3: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822895#822895[/url]

4: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822897#822897[/url]

[b]BMO 1989:[/b]

1: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822904#822904[/url]

2: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822909#822909[/url]

3: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822912#822912[/url]

4: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822915#822915[/url]

[b]BMO 1990:[/b]

1: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822924#822924[/url]

2: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822931#822931[/url]

3: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822933#822933[/url]

4: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822937#822937[/url]

[b]BMO 1991: [/b]

1: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822946#822946[/url]

2: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822954#822954[/url]

3: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822959#822959[/url]

4: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=822973#822973[/url]

game over :)

[quote="edriv"]I think we should say to pohoatza at least (1991 - 1984) times "thank you"  :)

Maybe when I come back (from Balkan  :P ) I'll continue this work. Maybe also with a translation in Italian, when Valentin finishes the new version of the Resources section![/quote]It's in beta right now, we will make it available in whatever languages you guys want to contribute to :)

good luck at the BMO, edriv, let's hear a gold from you :)


modular arithmetic
algebra
polynomial
arithmetic sequence
number theory solved
number theory

Show that the equation $x^2 + y^2 + z^2 = ( x-y)(y-z)(z-x)$ has infintely many solutions in integers $x,y,z$.

First One-liner Solution:
Let $x= 6m(54m^2+1), y=(6m+1)(54m^2+1), z=(6m-1)(54m^2+1)$, where $m \in \mathbb{Z}$.

Second One-liner Solution:
Let $x = 2m(6m^2+1), y = (2m+1)(6m^2 + 1), z = (2m-1)(6m^2+1)$, where $m \in \mathbb{Z}$.

My Motivation (how i came up with these 2 solutions):
We think that 3 variables are too many variables to play with, so the trick is to let $y = x+k, z=x-k$. This reduces the equation to $3x^2 + 2k^2 = 2k^3$, which is equivalent to $x^2 = \frac{2(k-1)k^2}{3}$. Since $x \in \mathbb{Z}$, we must have $k \equiv 1 \pmod{3}$. We can easily check that $k \equiv 4 \pmod{9}$ results in a contradiction.

For the case of $k \equiv 1 \pmod{9}$, letting $k = 9k_0+1$, we have $x^2 = (6k_0){(9k_0+1)}^2$. Now, $(6k_0)$ must be a perfect square, so letting $6k_0 = 6m^2$ for some $m \in \mathbb{Z}$, we get our first one-liner solution.

For the case of $k \equiv 7 \pmod{9}$, letting $k = 9k_0+7$, we have $x^2 = 2(3k_0+2){(9k_0+7)}^2$. Now, $2(3k_0+2)$ must be a perfect square, so letting $(3k_0+2) = 2m^2$ for some $m \in \mathbb{Z}$, we get our second one-liner solution.

Here is a polynomial solution that is not in arithmetic progression

 $ x \equal{} (6m \minus{} 1)(18m^2 \plus{} 2m \plus{} 1)$, $ y \equal{} (6m \plus{} 1)(18m^2 \plus{} 2m \plus{} 1)$, $ z \equal{} (6m \plus{} 2)(18m^2 \plus{} 2m \plus{} 1)$

[quote="Agr_94_Math"]Actually, Mr. CR Pranesachar, leader IMO team India many times, in a camp, told us that infinite solutions existing for this equation are all  3 element Arithmetic Progressions.
Even in dblue's solution, he set $ y\equal{}x\plus{}k$ and $ z\equal{}x\minus{}k$ to get $ (y,x,z)$ in AP.
Can some try to disprove or prove it?[/quote]

Hmm, may be i can proof this one.

WLOG, $x \le y \le z$. So, assume that, $x = a - k$, $y = a$, $z = a + k$. So, subtitute to the problem. So,

$(a-k)^2 + a^2 + (a+k)^2 = (-k)(-k)(2k)$. So, $3a^2 + 2k^2 = 2k^3 \implies 3a^2 = 2k^2(k-1)$. So, we can assume that $a = kb$. So,

$3k^2b^2 = 2k^2(k-1)$. So, $3b^2 = 2(k-1) \implies k = \frac{3b^2+2}{2}$. And, since $k$ must be integer, So, assume that $b = 2c$. So, $k = 6c^2 + 1$.

And, we obtain $y = a = kb = 2kc = 2c(6c^2 + 1) = 12c^3 + 2c$.

$x = y -  k = 12c^3 + 2c - 6c^2 - 1$
$z = y + k = 12c^3 + 2c + 6c^2  + 1$.

So, one of the solution is $(x, y, z) = (12c^3 + 2c - 6c^2 - 1, 12c^3 + 2c, 12c^3 + 2c + 6c^2  + 1)$. CMIIW

This problem was in the Hungarian National Olympiad, 2nd category, 2nd round,  problem 1. 2018/19.

I solved it back then, I tried picking three $x,y,z$ and manipulating them to create a solution, and showed that there exists always an arbitrarily big solution, here is my proof

Say, you have three $x,y,z$ numbers. If you take a number $k$, and multiply all three numbers, then the square sum's value grows exactly $k^2$ times bigger, because it's previous value was $x^2+y^2+z^2$, and now it's $(kx)^2+(ky)^2+(kz)^2$. However, the right side's new value is $k^3$ times it's previous value. Thus, we have to find some $x,y,z$ numbers, for which, the right side divides the left side. Why? Because if $x^2+y^2+z^2=M(x-y)(y-z)(z-x)$, then using $k=M$, $(Mx,My,Mz)$ will be a solution in our original equation.
But that's not hard, if you take $x=n-1, y=n, z=n+1$ then the right side value is 2, and the left side, if n is odd, is a (big) even number.

Assume there are finitely many solutions. Then there exists a "biggest" solution, say $x \geq y \geq z$, assume here $x=N$. But take numbers $x=2N, y=2N+1, z=2N+2$, then the product equals 2, whereas the square sum is a large even number, say 2*S, then $2SN, S(2N+1), S(2N+2)$ is a "bigger" solution for the equation.


geometry
geometric transformation
number theory proposed
number theory

For $A\subset\mathbb Z$ and $a,b\in\mathbb Z$. We define $aA+b: =\{ax+b|x\in A\}$. If $a\neq0$ then we calll $aA+b$ and $A$ to similar sets. In this question the Cantor set $C$ is the number of non-negative integers that in their base-3 representation there is no $1$ digit. You see \[C=(3C)\dot\cup(3C+2)\ \ \ \ \ \ (1)\] (i.e. $C$ is partitioned to sets $3C$ and $3C+2$). We give another example $C=(3C)\dot\cup(9C+6)\dot\cup(3C+2)$.

A representation of $C$ is a partition of $C$ to some similiar sets. i.e. \[C=\bigcup_{i=1}^{n}C_{i}\ \ \ \ \ \ (2)\] and $C_{i}=a_{i}C+b_{i}$ are similar to $C$.
We call a representation of $C$ a primitive representation iff union of some of $C_{i}$ is not a set similar and not equal to $C$.
Consider a primitive representation of Cantor set. Prove that
a) $a_{i}>1$.
b) $a_{i}$ are powers of 3.
c) $a_{i}>b_{i}$
d) (1) is the only primitive representation of $C$.

(a), (b) and (c) do not depend on the fact that the partition is primitive. Unless we're dealing with the trivial partition $C=C$ (so let's assume that $n\ge 2$), if for some $i$ one of (a), (b), (c) doesn't hold, then $a_{i}C+b_{i}$ isn't even contained in $C$:

(a)

If some $a_{i}=1$ then for some non-zero $b$ we must have $C+b\subset C$. This means that $b\in C$, so if $2\cdot 3^{k}$ is the largest summand in the decomposition of $b$ in base $3$, then $2\cdot 3^{k}+b\not\in C$.

(b)

Let $a$ be a positive integer which is not a power of $3$, and $b$ a non-negative integer. We want to prove that $aC+b\not\subset C$.

First notice that $aC+b\subset C$ is in fact equivalent to $aC\subset C$, so we can forget about the $b$. Also, we can assume that $a$ is not divisible by $3$. Finally, notice that the digits of $a$ in base $3$ must be $0,1$, for otherwise $2a$ would not belong to $C$. Since we assumed $a$ is not equal to $1$, it must be of the form $1+3^{k}+t\cdot 3^{k+1}$. This, however, means that $a\cdot(2+2\cdot 3^{k})$ does not belong to $C$, so we have our contradiction.

(c)

Let $a=3^{k}$, and $b\ge a$. As above, we want to show that $aC+b\not\subset C$.

If $aC+b\subset C$, then $b\in C$. If $2\cdot 3^{u},\ u\ge k$ is the largest summand in the decomposition of $b$ in base $3$, then $2\cdot 3^{u}+2\cdot 3^{u}\not\in C$. This contradicts the fact that $aC+b\subset C$ because $2\cdot u\in aC$.

Due to (a), (b), (c), we can interpret the problem purely combinatorially. What we want is to partition the set of sequences with terms $0,2$ (with only finitely many $2$'s) into "translations" of this set, i.e. sets of the form $S_{k}(u)$ where $k$ is a positive integer and $u$ is a $0,2$-sequence of length $k$, consisting of those $0,2$-sequences whose first $k$ digits are given by $u$. Of course, we can replace $0,2$ by $0,1$ respectively. When translated back into numbers, this time using base $2$, (d) becomes:

Suppose we partition the set $\mathbb N$ of non-negative integers into progressions of the form $A_{i}=\{a_{i}m+b_{i}\ |\ m\ge 0\},\ i=\overline{1,n},\ n\ge 2$, with $1<a_{i}=2^{k_{i}}$  and $b_{i}<a_{i}$. If this partiton is primitive (in the sense described above by Omid), then it must be the partition into even and odd integers.

Assume first that $k_{1}=\ldots=k_{t}<k_{t+1}\le\ldots\le k_{n}$. The progressions $A_{i}, i\ge t+1$ must cover a certain fixed residue class $r<2^{k_{1}}$ modulo $2^{k_{1}}$ which does not appear among the $A_{i}, i\le t$. Since each residue class modulo $2^{k_{i}}, i\ge t+1$ is contained in a single residue class modulo $2^{k_{1}}$, we find that the union of several of the progressions $A_{i}, i\ge t+1$ is precisely the progression $\{2^{k_{1}}m+r\ |\ m\ge 0\}$. This contradicts the primitivity of the pertition, and shows that in fact all $k_{i}$ must be equal. But now we split the progressions $\{2^{k_{i}}m+b_{i}\}$ into those which have even $b_{i}$ and those which have odd $b_{i}$. The union of the first group is the set of even integers, and the union of the second group is the set of odd integers. Since the partition is primitive, it must coincide with this one, and we're done.


Does anyone know how to prove the following using the definition of big-O?

If $f$ in big-O $g$, then $|f - g|$ in big-O $g$.

$|f - g|$ is a function that maps n to $|f(n) - g(n)|$


a, b, c be real number. Prove

$ a^{2}+b^{2}+c^{2}\ge 2(a-\frac1a+1+b-\frac1b+1+c-\frac1c)+a+b+c$

doesn't work
set them all $ 1.1$
or very small and negative
there is a correct version
$ a,b,c > 0$ $ a+b+c\leq 3$.then prove the inequality at hand.


function
inequalities
algebra unsolved
algebra

find all fuctions $\mathbb{R} \rightarrow \mathbb{R}$ such that
                                     \[ f(x-y)-xf(y)\leq 1-x  \]

$\forall x,y \ \in \mathbb{R}$

 :)

Let $g : = f -1$ and substitute $x$ for $x-y$. It thus suffices to find all functions $g: \mathbb{R} \to \mathbb{R}$
such that for all real $x$ and $y$ we have $g(x) \le (x+y) g(y)$.
Let $g$ such a function. By interchanging variables we have $g(x) \le (x+y) g(y)$ and $g(y) \le (x+y) g(x)$.
For $x+y >0$ this implies $g(x) \le (x+y) g(y) \le (x+y)^2 g(x)$.
For given $x$ we may choose $y_1$ such that $x+y_1 = 1/2$ and $y_2$ such that $x+y_2 = 2$.
For these values of $y$ the above inequality gives $g(x) \le \frac 1 4 g(x)$ and $g(x) \le 4 g(x)$,
i.e.  $g(x) \le 0$ and $g(x) \ge 0$.
So $g = 0$, and hence $f = 1$.
It is easy to check that this really is a solution to the given functional inequality.

[quote=socrates]Find all fuctions $\mathbb{R} \rightarrow \mathbb{R}$ such that
                                     \[ f(x-y)-xf(y)\leq 1-x  \]
for all $ x,y \ \in \mathbb{R}.$ [/quote]

[b]Here is my solution.[/b] Let $f$ be a functions satisfy the conditions of the problem. Put $y=x-1$ into the functional inequality, we obtain $f(1)-xf(x-1)\le 1-x$ for all real number $x.$ Put $y=1$ into the functional inequality, we obtain $f(x-1)-xf(1)\le 1-x.$ We deduce that \[(1-x)\left[f(x-1)+f(1)\right] \le 2(1-x) \]
for all real number $x.$ Put $x=0$ into the functional inequality and change $y$ with $-y,$ yields $f(y)\le 1,$ for all real number $y.$ Now, we see that for all real number $x> 1,$ we have \[(1-x)\left[f(x-1)+f(1)\right]\ge (1-x)(1+1)=2(1-x).\]
Thus, for all real number $x>1,$ we have $f(x-1)+f(1)=2,$ but $f(1),f(x-1)\le 1,$ hence $f(x-1)=f(1)=1.$ We deduce that $f(x)=1$ for all real number $x>0.$ Put $x>0\ge y$ into the functional inequality, we obtain $x\left[f(y)-1\right]\ge 0,$ which means that $f(y)\ge 1,$ hence $f(y)=1$ for all $y\le 0.$  We conclude that $f(x)=1,$ for all real number $x,$ which indeed is a solution.


In a school of 320 students, 85 students are in the band, 200 students are on sports teams, and 60 students participate in both activities.  How many students are involved in either band or sports?

[hide="solution"]85+200-60=225 that participate in at least 1.
[/hide]

[quote="jli"][quote="bigboy82892"][hide="Or."]You could argue that the question asked for only band OR sports, not both. So the answer would be $\boxed{165}$.[/hide][/quote]

How did you get [color=white]165[/color]?[/quote]

He isn't counting the 60 that do both band and sports.

[quote="bigboy82892"][hide="Or."]You could argue that the question asked for only band OR sports, not both. So the answer would be $\boxed{165}$.[/hide][/quote]
The question is "...involved in either band [b]or[/b] sports?" What you're saying would be xor. However, there is the either. How should this problem be interpreted?


vector
analytic geometry
induction
linear algebra
linear algebra unsolved

What if the maximal number of vectors in $R^n$ such that the scalar product of any two of them is at most 0. Care must be paid, since it seems the same as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=351&t=31714]a problem posted by Alekk[/url]. But is it?

Oh, it's certainly not the same, since in that other problem we had to have negativ scalar products (not merely non-positive :)), and the answer for it was $n+1$, while the answer for this one is $2n$ (I assume the vectors are non-null; if we also add $0$, we can have $2n+1$ such vectors, of course), but it can be solved in a very similar way.

Suppose WLOG that the vectors have norm $1$ and that one of the vectors is $e_1=(1,0,\ldots,0)$. The first coordinates of all the other vectors are $\le 0$, so the vectors we obtain by projecting our system (without $e_1$) on the orthogonal complement of $e_1$ also satisfies the condition. This means that this projection must have (induction hypothesis; I forgot to mention, we perform induction on $n$) at most $2(n-1)+1$ vectors, with equality only if one of the vectors is $0$, i.e. only if one of the initial vectors was $-e_1$. It means that, except, eventually, for $e_1,-e_1$, we have at most $2(n-1)$ other vectors (it's easy to see that unless a vector has all its last $n-1$ coordinates equal to $0$, it's the only one in our system projected on its image in $e_1^{\perp}$, sdo there is an injection from the set of vectors in our initial system different from $\pm e_1$ and the projections on $e_1^{\perp}$).

Can someone post a link to the solutions if the scalar product must be < 0?
Thanks

Here's an interesting extension of the result, which can be obtained through basically the same inductive process:

If we have $n+k,\ k\ge 1$ such vectors (non-zero vectors in $\mathbb R^{n}$, each two of which have non-positive scalar product), then we can partition the set of vectors into $k$ mutually orthogonal subsets.


Putnam algebra polynomial Stanford college function geometry

Let $p(z)$ be a polynomial of degree $n,$ all of whose zeros have absolute value $1$ in the complex plane. Put $g(z)=\frac{p(z)}{z^{n/2}}.$ Show that all zeros of $g'(z)=0$ have absolute value $1.$

I'd really like to see a solution to this one.  I don't think anyone at Stanford solved it--we all hit brick walls somewhere in the course of the solution.

The semi-official solutions are up at http://www.unl.edu/amc/a-activities/a7-problems/putnam/ if you would like to see multiple solutions.  The only person who solved it at my university was one of the profs invigilating the Putnam.

This is what I did. Well, the solution is incomplete and I would be glad if anyone could point out errors if any. Let $p(z) = C(z - z_1)(z - z_2)...(z - z_n)$, where $|z_1| = |z_2| = ... = |z_n| = 1$. Note that $\frac{p'(z)}{p(z)} = \frac 1{z - z_1} + \frac 1{z - z_2} + ... \frac 1{z - z_n}$ Now, after differentiating $g(z)$, we obtain $g'(z) = \frac{p(z)}{z^{\frac n2}} (\frac 1{z - z_1} + \frac 1{z - z_2} + ... \frac 1{z - z_n} - \frac n{2z})$. Therefore, one set of zeroes of $g'(z)$ occur exactly where $p(z) = 0$ and we know that the zeroes of $p(z)$ all have magnitude 1.

The remaining part of the solution seems to be tricky. I did try assuming $|z| > 1$ and  $|z| < 1$ when $\frac 1{z - z_1} + \frac 1{z - z_2} + ... \frac 1{z - z_n} - \frac n{2z}) = 0$ and getting some contradiction. It didn't work though :(

This was a pretty hard problem for A3 in my opinion :(

Here's my solution, just found today.

Let $h(z)=np(z)-2zp'(z)$; this is a polynomial of degree $n$ with the same zeros as $g'$. Also, $h(z)$ has a zero of multiplicity $k-1$ at $r$ if $p$ has a zero of multiplicity $k$ at $r$.

Now, if $p(z)=a(z-r_1)(z-r_2)\cdots (z-r_n)$, $p'(z)=\sum_{j=1}^n\frac{p(z)}{z-r_j}$, so
$h(z)=\sum_{j=1}^n p(z)-2z\frac{p(z)}{z-r_j}=p(z)\cdot\sum_{j=1}^n\frac{-z-r_j}{z-r_j}$. Define
$f(z)=i\frac{h(z)}{p(z)}=\sum_{j=1}^n-i\frac{z+r_j}{z-r_j}$. Since each fraction $-i\frac{z+r_j}{z-r_j}$ takes on only real values on the unit circle, the restriction of $f$ to the unit circle is real-valued.

Now, we use the intermediate value theorem; since the fraction $-i\frac{z+r_j}{z-r_j}$ is large positive for $z$ clockwise from $r_j$ and large negative for $z$ counterclockwise from $r_j$, $f$ has a zero between each pair of roots of $p$. Adding in the extra copies of the roots of $p$, $g$ has roots of total multiplicity at least $n$ on the unit circle. Since $g$ is a polynomial of degree at most $n$, these are all of the roots of $g$, and we are done.

My idea during the couple of minutes where I worked on it was to take a mobius transformation sending $S^1$ to $\mathbb{R}$, e.g, a Blaschke factor or something like that, and then see if things worked out better.  Unfortunately, I realized that I don't really know how to make this work, or even if it can be made to work.

FieryHydra, your solution doesn't work: Where $z=z_i$, the equation you give for $g(x)$ goes to 0/0, so you can't claim it's a root.  I tried something not unlike that, but the fact that the nice formula for $p'(z)$ fails when $z=z_i$ makes things a pain.

I actually proved in the course of working on it $z_i$ is a root of $g(z)$ if and only if it's a multiple root.

Sly Si, thanks for pointing out the flaw in my solution. You are absolutely right. Since I couldn't come up with a reasonable solution during the examination, I was just being obtuse in writing that solution! :) Good job! ;)

[quote="blahblahblah"]My idea during the couple of minutes where I worked on it was to take a mobius transformation sending $S^1$ to $\mathbb{R}$, [/quote]
This is a good idea. Good knowledge of fractional linear transformations could really help you here. Indeed, we are to show that all solutions of $\sum_{j=1}^n\frac z{z-z_j}=\frac n2$ lie on the unit circumference. But, for every $z_j$ with $|z_j|=1$, the mapping $z\mapsto \frac z{z-z_j}$ maps the unit disk to the half-plane $\Re z<\frac 12$ and the outside of the unit disk to the half-plane $\Re z>\frac 12$. The rest should be clear.

This is a pretty darn hard problem for an A3.  However, if you know Rouche's Theorem (which is mentioned in an aside after like the third "official" solution provided  :P ), the problem becomes more or less trivial (I think).  Rouche was one of the first things I thought of, because it's a great tool for showing that the zeroes of a rational function have to lie in certain parts of the complex plane.

Downsides:  I've forgotten how to prove Rouche (i.e. I have no idea off the top of my head why it's true without looking in Ahlfors' complex analysis text), and I didn't learn Rouche until I was a graduate student, hence ineligible for Putnam.  Ouch!

Major props to anyone who actually managed to solve this beast during the contest.

Care to say why it becomes more or less trivial?  I know Rouche's theorem, and I don't think it gives an immediate answer - there's still a fair bit of work to be done.

Ahhh!  I'm sorry, blah^3 - it turns out Rouche's theorem is definitely not the theorem I was thinking of.  I was thinking of something different, which I'm not even sure is true anymore.  I'm gonna have to get back to you after I consult with Ahlfors.

The LFT proof, though, is quite nice, and a lot more instructive than anything I'm going to be able to produce using this theorem.

[quote]Since each fraction -i\frac{z+r_j}{z-r_j} takes on only real values on the unit circle, the restriction of f to the unit circle is real-valued. [/quote]

I don't understand this part.. Can someone elaborate?

Look at the geometry; $z$ is the vertex of a right angle between $r_{j}$ and $-r_{j}$, so $\frac{z-r_{j}}{z-(-r_{j})}$ is pure imaginary.

Here is a solution that should not be too difficult to follow.

[quote="fedja"]Indeed, we are to show that all solutions of $ \sum_{j \equal{} 1}^n\frac z{z \minus{} z_j} \equal{} \frac n2$ lie on the unit circumference. [/quote]
This is the approach I took during training this year, but I can't for the life of me remember what I did after this step.  I do remember that the solution didn't take too long, so perhaps my solution was incorrect.

Wow this is hard for an A3.  Thanks to Po-Shen Loh for giving me the hint $(*)$.

Let $r_1,r_2,\ldots, r_n$ denote the roots of $p(z)$.  Differentiating the expression for $g$ yields \[g'(z) = \frac{d}{dz}\left[\frac{p(z)}{z^{n/2}}\right] = \frac{p'(z)}{z^{n/2}} - \frac n2\cdot\frac{p(z)}{z^{n/2+1}}.\] Let $z$ now be a zero of $g'(z)$, and assume that $z$ is not a zero of $p$ (else we're done).  Then the given equality rearranges to $p'(z) = \tfrac n{2z}p(z)$, or
\[
\frac{n}{2z} = \frac{p'(z)}{p(z)} = \sum_{i=1}^n\frac{1}{z-r_i}.
\]
Now note that subtracting $\tfrac{1}{2z}$ from each term on the RHS yields
\[
0 = \sum_{i=1}^n\frac{z+r_i}{z-r_i} \stackrel{(*)}=\sum_{i=1}^n\frac{(z+r_i)(\overline{z-r_i})}{|z-r_i|^2} = \sum_{i=1}^n\frac{|z|^2 - 1 + (\overline z r_i - z\overline{r_i})}{|z-r_i|^2}.
\]
Since $\overline z r_i - z\overline{r_i}$ is purely imaginary, taking the real part of both sides yields
\[
(|z|^2 - 1)\left(\sum_{i=1}^n\frac{1}{|z-r_i|^2}\right) = 0.
\]
As the second term is always positive, we must conclude that $|z| = 1$. $\blacksquare$

My solution is a bit different from any of those presented in the Putnam archive or here. As such, I wanted to share it.

Put $g(z) = p(z)/z^{n/2}$, where p is a polynomial of degree n with all roots on the unit circle.
\[ g'(z) = p'(z)z^{-n/2} - \frac{n}{2}p(z)z^{-\frac{n}{2}-1} = 0 \]
\[ \frac{zp'(z) - \frac{n}{2}p(z)}{z^{\frac{n}{2}+1}} = 0 \]
Claim: the above statement is true if and only if the numerator is zero. One direction is clear. For the other direction, if the numerator is zero and the denominator is non-zero, then the above is certainly true. If the denominator is zero, then $z=0$. For the numerator to also be zero, then $p(0)=0$. This cannot occur though because all zeroes of p lie on the unit circle.
\[ zp'(z) - \frac{n}{2}p(z) = 0 \]
\[ \frac{zp'(z)}{p(z)} = \frac{n}{2} \]
\[ \sum_{i=1}^n\frac{z}{z-z_i} = \frac{n}{2} \ \ \ \ \ \ \ \ \      (*) \]
where the z_i are the zeroes of p(z). Below, we start by taking the conjugate of the LHS above, which will retain equality with the RHS, as it is real. Because $|z_i|^2=1$, we also know that $\bar{z_i}=\frac{1}{z_i}$:
\[ \sum\frac{\bar{z}}{\bar{z}-\bar{z_i}} = \sum \frac{\bar{z}}{\bar{z}-\frac{1}{z_i}} = \sum \frac{\bar{z}z_i}{\bar{z}z_i - 1} = \sum (1 + \frac{1}{\bar{z}z_i-1}) = n + \sum \frac{1}{\bar{z}z_i-1} = \frac{n}{2} \]
\[ \sum \frac{1}{1 - \bar{z}z_i} = \frac{n}{2}    \ \ \ \ \ \ \ \ \ \           (**) \]
As all statements are equivalences, $\bar{z}$ satisfies (**)  iff $g'(z)=0$.
Returning to equation (*), we now multiply numerator and denominator by $\bar{z}$, and we set $z\bar{z}=|z|^2=k$, which must be real and positive, as we've shown that z is nonzero.
\[ \sum \frac{z\bar{z}}{z\bar{z}-\bar{z}z_i} = \frac{k}{k-\bar{z}z_i} = \frac{1}{1-\frac{\bar{z}}{k}z_i} = \frac{n}{2}  \]
Now, note that $\frac{\bar{z}}{k}$ satisfies (**) iff $\bar{z}$ satisfies (**) iff z is a root of g' iff $\frac{z}{k}$ is a root of g'.
If $k \neq 1$, then any root z of g' implies that g' has infinite roots, as $z, z/k, z/k^2, ... , z/k^n, ...$ must also be roots of g'.
We know from above that this cannot be the case as p is a polynomial. Therefore, there are no roots z of g' such that $|z|^2=k \neq 1.$
Then $|z|^2 = 1$ for all roots of g', and we are done.

Write $p(x)=(x-r_1)\cdots(x-r_n)$ where $|r_i|=1$. Let $k\le n$ be the number of distinct $r_i$.

Let $z$ be an arbitrary complex number on the unit circle. It is easy to verify that $g'(z)=0$ if and only if
\[\sum_{i=1}^n\frac{1}{1-r_i/z} = \frac{n}{2}.\]
Note that $\frac{1}{1+x}$ always has real part $\frac{1}{2}$ for $|x|=1$. We now find $k$ points on the unit circle such that
\[k(z):=\sum_{i=1}^n\Im\frac{1}{1-r_i/z} = 0,\]
none of which are equal to $r_i$ for any $i$. For any $\theta$, we have
\[k(r_ie^{i\theta}) = \frac{\sin\theta}{|1-e^{i\theta}|^2} + \sum_{\substack{1\le j\le n \\ j\ne i}}\Im\frac{1}{1-\frac{r_j}{r_ie^{i\theta}}}.\]
As $\theta\to 0^+$, this is manifestly some huge positive number. Similarly, as $\theta\to 0^-$, this is manifestly some huge negative number. Note that $k$ is continuous in the interval between two of the $r_i$ on the unit circle, as long as that interval has no other roots. Therefore, by IVT, there is some root of $k$ in each of those intervals. There are $k$ such intervals, so our claim is proven.

Now, we have to find an additional $n-k$ roots on the unit circle. Indeed, if $r_i$ has multiplicity $m_i$, then it is easy to see that $g'(z)$ has root $r_i$ with multiplicity $m_i-1$, so we in fact have $n-k$ roots that are the $r_i$ (with multiplicity), so we have accounted for all $n$ roots of $g'(z)$, all of which are on the unit circle. This proves that all of its $n$ roots are on the unit circle, as desired.


Ring Theory
number theory
relatively prime
number theory unsolved

x, y, z = all the integer and relatively prime
$ x^2 \plus{} 2y^2 \equal{} 3z^2$
catch the all x, y, z

actually above equation is equivalent to following
$ ab \equal{} 2cd$
$ a \minus{} b \equal{} c \plus{} d$
a, b=relatively prime and different parity each other
c, d=relatively prime and different parity each other

ex)(some solutions)
$ x,y,z\equal{}e^2$ for interger e
x=23, y=13, z=17
...

There exist integers $ a, b$ such that $ x \equal{} a \minus{} 2b, y \equal{} a \plus{} b$.  (It's enough that $ x \equiv y \bmod 3$, which one gets by inspection $ \bmod 9$.)  This gives

$ a^2 \plus{} 2b^2 \equal{} z^2$

and now for relatively prime solutions we observe that $ \mathbb{Z}[ \sqrt { \minus{} 2} ]$ is a UFD and set $ a \plus{} b \sqrt { \minus{} 2} \equal{} (c \plus{} d \sqrt { \minus{} 2})^2 \equal{} (c^2 \minus{} 2d^2) \plus{} 2cd \sqrt { \minus{} 2}$, which gives the general solution

$ (x, y, z) \equal{} (c^2 \minus{} 4cd \minus{} 2d^2, c^2 \plus{} 2cd \minus{} 2d^2, c^2 \plus{} 2d^2)$.

Note the special case $ (c, d) \equal{} (e, 0)$ that you observed.  See also the geometric approach [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?p=1084535#1084535]here[/url].


inequalities
combinatorics proposed
combinatorics

Suppose that $r_{1},r_{2},...,r_{n}$ are real numbers.  Prove that there exists a set $S$ such that $S \subseteq \{1,2,...,n\}$ and
\[1\leq |S\cap \{i,i+1,i+2\}| \leq 2 \]
for every $1\leq i \leq n-2$, and
\[\left| \sum_{i\in S}r_{i}\right| \geq \frac{1}{6}\sum_{i=1}^{n}\left|r_{i}\right|. \]

I am very interested. It is in my short list. :)

Let $A_1$ be positive $r_i$'s such that $i=3k+1$, $B_1$ negative $r_i$'s such that $i=3k+1$

Define in a similar way $A_2$, $B_2$, $A_3$, $B_3$

Now we have that if we choose $S$ to be composed with $A_iUB_iUA_jUB_j$ or $A_iUB_iUA_j$ or $A_iUB_iUB_j$ we get satisfactory $S$. Now we only have to deal with the sums:

Let $A_i$, and $B_i$ denote the sum of the absolut value of the elements of that set.

We can suppose $A_1+A_2+A_3>B_1+B_2+B_3$

and we get:

$|A_1+A_2-B_1|+|A_1+A_2-B_2|+|A_1+A_3-B_1|+|A_1+A_3-B_3|+|A_2+A_3-B_2|+|A_2+A_3-B_3|>
|A_1+A_2+A_3+B_1+B_2+B_3|$

so one of the sums is at elast $1/6$ of the last sum, the set obtained as the intersection of the summed objects works then!.


modular arithmetic
number theory
relatively prime
number theory unsolved

Employ the Fermat's little theorem to prove that

$1^{p-1}+2^{p-1}+3^{p-1}+\ldots+(p-1)^{p-1}\equiv-1 \pmod{p}$

I tried but I didn't get anything  :(

Since $p$ is a prime, all bases are relatively prime to $p$.  So, $a^{p-1}\equiv 1\pmod p$.  As a result, the sum is $\equiv p-1\equiv-1\pmod p$.

[quote="cuopbientcd"]general problem :
 If $p$ is prime then $\sum_{i=1}^{p-1}i^{n}=-1 mod p$ if $n =0 mod p-1$ and $=0 mod p$ if $n\#0 mod p-1$[/quote]
you want say
$\forall (k,s)\in N\times [|0,p-2|]$; $\sum_{i=1}^{p-1}i^{k(p-1)+s}=-E(10^{-s})\ (mod\ p)$

for example:
for $k=1,s=0$
$1^{p-1}+2^{p-1}+...+(p-1)^{p-1}\equiv-E(10^{0})\ (mod\ p)\equiv-1\ (mod\ p)$


I was reading chapter 6 (Polynomials) in AoPS, and I can't seem to understand the reasoning behind page 85-86 ("Find an equation whose roots are 3 greater than those of x^4 - 3x^3 - 3x^2 + 4x - 6." You need to expand all the (x - 3)^... terms using synthetic division.).

Shouldn't you keep dividing by (x - 3), not (x + 3)?

In the answer, it shows repeated synthetic division with -3 (corresponding to x + 3).

Please help! I have come back to this section several times, and I have reread the section carefully but I still do not understand this example.

Thanks!

Yeah, I had the same trouble, and posted here about this a while back. I still dont understand it fully. Sorry to be of no help, but just wanted to show that youre not teh only one.

Perhaps this will help:

What polynomial has roots which are 3 more than those of f(x)?

Suppose the roots of f(x) are a,b,c,d.  Therefore,

f(x) = k(x-a)(x-b)(x-c)(x-d)

f(x-3) = k(x-3-a)(x-3-b)(x-3-c)(x-3-d) = k(x-(a+3))(x-(b+3))(x-(c+3))(x-(d+3))

and you've got your polynomial with roots 3 larger than the original.

The rest of that section is describing a shortcut to get to the polynomial, and is really not nearly as important as understanding the above (though understanding the machinery of that section does give you a great look at the utility of the polynomial 'division' expression f(x) = d(x) q(x) + r(x).)

Thanks-

Also, how does the shortcut work? I don't quite understand why you divide by (x + 3) and not (x - 3).

Doh... that wasnt the thing that i had had trouble with. It was that other part of the chapter where multiply divisions of -3 occured. Dont worry about explaining it though.

no- that's what I'm  referring to (the multiple divisions with -3)

Well I'm not exactly sure of whats going on here since I don't have the books, but I'm guessing that since you want all the roots 3 greater, you do stuff with x-3. Its like you want to shift the graph 3 to the right, which is the same as just considering f(x-3).. then when, say, x=3 in the new equation, x=0 in the first..

But then again, maybe this isn't what you are talking about..


integration
calculus
calculus computations


Factor by grouping: x^2y - 4y + 12 - 3x^2

Let's show some work!

$ x^2y \minus{} 4y \plus{} 12 \minus{} 3x^2$

$ (x^2y \minus{} 3x^2) \plus{} (\minus{}4y \plus{} 12)$

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

$ (y\minus{}3)(x^2\minus{}4)$

$ \boxed{(x\plus{}2)(x\minus{}2)(y\minus{}3)}$


function
search
algebra unsolved
algebra

Search the forum about a theorem of Erdos on increasing multiplicative functions. There it is proved a much more general result (by Pierre, I think). By the way, in this form the problem is very simple: just put $n^{r}$ between powers of $2$ for all $n,r$.


function
limit
inequalities
calculus
real analysis
triangle inequality
calculus computations

I've read a couple of old threads about this, but I'm still confused.

Definition: The function $ f$ approaches the limit $ l$ near $ a$ means: for all $ \epsilon > 0$ there is some $ \delta > 0$ such that, for all $ x$, if $ 0 < |x \minus{} a| < \delta$, then $ |f(x) \minus{} l| < \epsilon$

Okay so lets take an example like:

$ \lim_{x \to\ 1}\frac{ (x \minus{} 1)(x \plus{} 1)}{ (x\minus{}1)}\equal{}L$

Do we have to find the limit first and then test it?  Also this only tells you what could be a limit right?

Looking at the problem we see that $ L \equal{} 2$

We can pick any $ \epsilon > 0$

Say we pick $ \epsilon \equal{} .1$ - than $ x$ will be within $ .1$ of $ L$

Next we have $ 0 < |x \minus{} a| < \delta$

So if we let $ x \equal{} 2.1$ we have $ 1.1 < \delta$ - we can then choose any $ \delta > 1.1$

If we choose $ \delta \equal{} 2$ then $ x$ must be within $ 2$ of $ a$

Then we have $ |f(x) \minus{} L| < \epsilon$

So we can choose any $ x$ within $ 2$ of $ a$ - lets use $ x \equal{} 2$

Which gives us - $ (\frac{(2)(3)}{1} \Rightarrow 6$

So the $ f(x) \equal{} 6$

Thus $ |6 \minus{} 2| <  .1$

Thus any limit $ L$ must satisfy this inequality

Which $ 2$ doesn't.....

Thanks for the help!

Two points:

In the first place, the $ \epsilon$-$ \delta$ formality is primarily used to prove theorems, not deal with specific limits. And as you already noted, the definition doesn't [i]find[/i] the limit for you. You have to do that through other means.

Second: what matters most about the definition are the logical connections - which you don't seem to have internalized. The quantifiers matter. The provision of the definition is that for any $ \epsilon$ (no matter how small) there exists a $ \delta$ (small enough) such that ...

So $ \delta$ depends on $ \epsilon;$ a proof would consist of a procedure for finding a suitable $ \delta$ for any possible $ \epsilon.$

So once you've picked your $ \epsilon\equal{}.1$ you don't have a free hand with $ \delta$ - you have to pick a $ \delta$ that's good enough. $ \delta\equal{}.1$ would work. $ \delta\equal{}10^{\minus{}10}$ would also work - it's overkill, of course, but sometimes you're just in an overkill sort of mood. If you let $ \epsilon\equal{}10^{\minus{}10},$ then you'd have to use $ \delta\equal{}10^{\minus{}10}$ (or smaller). You have to show that you can deal with any possible $ \epsilon.$

Side note: for the purpose of finding a limit as $ x\to1,$ your function need not be defined at $ x\equal{}1,$ and even if it is, you don't care about its value there. Hence, you have no reason for not writing the function you're trying to find the limit of as $ x\plus{}1$ (at least for $ x\ne 1.$)

Your gave an excellent explanation!  I think I actually understand it now!  I'm taking AP Calculus at my school, but it is not rigorous by any means - we never prove anything, and the class just focuses on the stupid AP exam... So I have to learn formal definitions and proofs (calculus!) on my own  :(   However, I often get stuck (I looked over this for a few hours), so its awesome to have you guys for help!

You definitely won't have to worry about $ \epsilon$-$ \delta$ proofs in AP Calculus, and probably not in college either until you take an advanced calculus or real analysis course.

Here's another example, without specific numbers:
[i]Prove that $ \lim_{x\rightarrow a}x^{2}\equal{}a^{2}$.[/i]
[b]Solution:[/b]
First assume that $ \left|x^{2}\minus{}a^{2}\right|<\epsilon$; our goal is to find a $ \delta>0$ that works.  Write $ \left|x^{2}\minus{}a^{2}\right|\equal{}\left|x\minus{}a\right|\left|x\plus{}a\right|<\epsilon$.  Now $ |x\minus{}a|$ is what we want to solve for, but how about $ \left|x\plus{}a\right|$?  The triangle inequlaity will make an appearance or two here.  Let's assume that $ |x\minus{}a|<1$ (we'll correct for this later).  Then $ \left|x\plus{}a\right|\leq|x|\plus{}|a|$.  Now we just need to put a bound on $ |x|$.  By the triangle inequality, $ |x|\minus{}|a|\leq |x\minus{}a|<1$, so $ |x|<\delta\plus{}|a|<1\plus{}|a|$.  So putting it together, we get that $ |x\plus{}a|<1\plus{}|a|\plus{}|a|\equal{}1\plus{}2|a|$.  Now returning to our first equation, $ |x^{2}\minus{}a^{2}|\equal{}|x\minus{}a\parallel{}x\plus{}a|<|x\minus{}a|(1\plus{}2|a|)<\epsilon$.  Hence $ |x\minus{}a|<\frac{\epsilon}{1\plus{}2|a|}$.  In order to correct for the assumption that $ |x\minus{}a|<1$, we simply choose the minimum of 1 and the term above.  That is, set $ \delta\equal{}\min\left(1,\frac{\epsilon}{1\plus{}2|a|}\right)$.  Now we complete the proof: we need to show that the $ \delta$ above works.  We have that $ \left|x^{2}\minus{}a^{2}\right|\equal{}\left|x\minus{}a\right|\left|x\plus{}a\right|<\delta\cdot\left|x\plus{}a\right|<\frac{\epsilon\cdot\left|x\plus{}a\right|}{1\plus{}2\left|a\right|}<\epsilon$, which completes the proof.

[hide="Exercise"]
Prove that $ \lim_{x\rightarrow a}x^{3}\equal{}a^{3}$.[/hide]


Let   $ \frac {p}{q} \equal{} 1 \minus{} \frac {1}{2} \plus{} \frac {1}{3} \minus{} \frac{1}{4} \plus{} ... \minus{} \frac {1}{2k} \plus{} \frac {1}{2k \plus{} 1}$, where $ \gcd (p,q) \equal{} 1$. Is it possible to choose $ k$ such that $ p$ is divisible by $ 1997$.

Take $ k\equal{}665$:

$ S\equal{}\frac1{666}\plus{}\frac1{667}\plus{}\ldots\plus{}\frac1{1331}\equal{}(\frac1{666}\plus{}\frac1{1331})\plus{}(\frac1{667}\plus{}\frac1{1330})\plus{}...\plus{}(\frac1{998}\plus{}\frac1{999})$.

In each term, the numerator must be divisible by 1997. So we must have

$ S\equal{}\frac{1997N}{666\cdot667\cdot\ldots\cdot1331}$.

Since 1997 is prime, and all of the terms in the denominator are less than 1997, it follows that the numerator of $ S$ is divisible by 1997.

Hence, the answer is "Yes".

Maybe you miss the first part. Sorry, I thought it was well-known.

$ S\equal{}\frac11\plus{}\frac12\plus{}...\plus{}\frac1{2k\plus{}1}\minus{}2(\frac12\plus{}\frac14\plus{}...\plus{}\frac1{2k})\equal{}\frac11\plus{}...\plus{}\frac1{2k\plus{}1}\minus{}\frac11\minus{}...\minus{}\frac1k\equal{}\frac1{k\plus{}1}\plus{}...\plus{}\frac1{2k\plus{}1}$.


The postal service will mail a package in the shape of a rectangular
prism only if the sum of the height and girth (the perimeter of the base)
does not exceed 108 inches. Given that the dimensions of the package,
in inches, are integers, how many inches long, to the nearest inch, is the
longest fishing pole that could be shipped in a box with a square base?

You want the height(length of fishing pole) to be as long as possible, so you want the perimeter to be as small as possible, by making the square base side length $ 1$ inch. So the perimeter of the base is $ 4$ inches, and in order to maximize the length, you make the sum equal to $ 108$ inches, which is the maximum amount. Now you get $ 108\minus{}4\equal{}104$ as the length.

[I am going to use spacecadet's info as a basis]

You could but it says to the nearest inch.
So a^2 + b^2 = c^2.
104^2 + 1^2 = c^2
10816 + 1 = c^2
10817 = c^2
104.0048 ≈ c

So rounding it still becomes 104 in.

[quote=flipifiystacks]Why can't the fishing pole be put in diagonally?[/quote]

The fishing pole can be put in diagonally. You want to find the maxima of the height, so the sol is $\sqrt{104^2+1^2+1^2}\implies \boxed{104}.$


[hide][b]2 hours becuase after that the first one is 4 hours 2nd is 2 hours[/b][/hide]

[quote="dragonx"][quote="buzzer11"][quote="zoa"]hi

the ques is :-

Q)  one candle was guaranteed to burn for 6 hours and the other for 4 hours. they were lit at the same       time.after some time one was twice as long as the other. for how long had they been burning???

plz reply as soon as possible.....

or mail at zoa_mast@yahoo.com[/quote]

[hide]
Let h be the number of hours.
Each hour, one candle burns 1/6 of its height, and the other burns 1/4 of its height.

$1-\frac{1}{6}h = 2-2\frac{1}{4}h$
$\frac{h}{6}= 2\frac{h}{4}-1$
$h = 3h-6$
$2h = 6$
$h =3$

Thus, it has been 3 hours since the candles started to burn.[/hide][/quote]

well if it was like that then 6-3=3 and 4-3=1 and 1*2 is not equal to 3[/quote]

I think you're misunderstanding the problem. You're calculating time. We need to calculate the length of the candles.

The question is a bit ambiguous, because we need the height of the candles at the beginning. We could have both candles the same, or we could have one 1 ft. tall and the other 100 ft tall. If they are the same, Buzzer11 is correct.

I assumed they were the same height.

Which is probably correct.

Well, if they didn't start at the same height, it would be a strange problem, so you should assume they are the same height.. Oh and I agree that buzzer11 is correct


A circle passes through the vertex $C$ of a rectangle $ABCD$ and touches its sides $AB$ and $AD$ at points $M$ and $N$ respectively.
Suppose the distance from $C$ to $MN$ is $2$

Find the area of $ABCD$

Wow, this one is really hard  :huh: I have no idea, how would you do it?

It's a bit vague. Do you mean the perpendicular distance from C to MN?

Exactly, the distance here is perpendicular distance