limit
integration
calculus
logarithms
trigonometry
real analysis
function

\[\lim_{x\to\infty}\frac{1}{x-a}\int_a^x\frac{1}{t-a}dt\]

anyone help me???  I use Maple I got 0, but I want the detail

Are you sure that $\lim_{x\to\infty}\frac{1}{x-a}\int_a^x\frac{1}{t-a}dt$?

I think that $\lim_{x\rightarrow a}\frac{1}{x-a}\int_a^x\frac{1}{t-a}dt$.

Let $f(x)=\int_a^x \frac{1}{t-a}dt$, since $f(a)=0$, we have

$\lim_{x\rightarrow a} \frac{1}{x-a}\int_a^x \frac{1}{t-a}dt$

$=\lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a}=f'(a)=\left[\frac{1}{x-a}\right]_{x=a}\ [\text{by Fundamental theorem of calculus}]$

$=???$ :?

kunny

[quote="blahblahblah"]Does that integral exist in any reasonable sense?  It's not in $L^1$.

For example, we could say that

$\lim_{x\to\infty}\left[\ln x-\ln\left(\frac {1}{x}\right)\right]=\lim_{x\to\infty}\ln x^2=\int^{\infty}_{0} \frac {1}{t}dt$

But

$\lim_{x\to\infty} \frac{\ln x^2}{x}=0$[/quote]

what do u mean what sense?  and how does it not in $L^1$ norm space?

so are you saying that $\ln(a-a) = \lim_{x\to\infty}\ln{\frac{1}{x}}$ ??

actually I am doing something about cesaro integral, and stuck with some nasty integrals and limits

what about this one?

\[
\lim_{x\to\infty}\frac{1}{x-a}\int_a^x\frac{\cos t}{t-a}\,dt
\]

it should be zero but I dont see how

[quote="Kent Merryfield"]That has the same problem as the first example.

Since for all $x>a,\,\int_a^x\frac{\cos t}{t-a}\,dt=\infty,$ the limit cannot exist.

What did blahblahblah mean by "it's not in $L^1$?

(Setting aside questions of measuability, which are not an issue here) we say that $f\in L^1(S)$ if and only if $\int_S|f(x)|\,dx<\infty.$

So ${\frac1{x^2}\in L^1([1,\infty})$ but ${\frac1{x^2} \not\in L^1((0,\infty}).$

And ${\frac{\sin x}{x}\not\in L^1([0,\infty})$ even though the improper integral converges.[/quote]

So you are saying that the limit is $\infty$??  But in maple I get 0.

In the previous example, $\forall\,x>a$, we also have $\int_a^x\frac{1}{t-a}\,dt = \infty$ but $\lim_{x\to\infty}\frac{\int_a^x\frac{1}{t-a}\,dt}{x-a} = 0$

maybe the same here the limit exists and equal 0

[quote]So you are saying that the limit is $\infty$?? But in maple I get 0. [/quote]
I am saying the limit is $\infty.$ If that's what Maple says, why should we trust it?

However: $\lim_{\delta\to0}\left[\lim_{x\to\infty}\frac1{x-a}\int_{a+\delta}^x \frac{dt}{t-a}\right]=0.$

He did mention something about the Cesaro integral.  Now, I don't know anything about it, and none of the analysis texts I have mention anything about it, but maybe it allows for some sort of weird manipulation here.  (mind you, as far as I know, most of the integrals that are even more 'advanced' than the Lebesgue integral only permit some sort of cancellation between positive and negative parts of functions that are not in $L^1$ to obtain a principal value, which doesn't really apply here)

I know $\lim_{x\to\infty}\int_a^x\frac{\cos t}{t-a}\,dt=\infty$ but also $\lim_{x\to\infty}(x-a)=\infty$ maybe L'Hopital's Rule applies here?

but I dont know how to do this, can someone give me idea, I have a feeling that the limit is 0

L'Hôpital's rule applies to quotients of functions that [b]tend[b] to infinity, divided by functions that tend to infinity. But it isn't just that $\lim_{x\to\infty}\int_a^x\frac{\cos t}{t-a}\,dt=\infty.$ It's that $\int_a^x\frac{\cos t}{t-a}\,dt=\infty$ (or $-\infty$) for all $x.$ And if the numerator or denominator in L'Hôpital's rule is always $\infty$ (or always $0$), then there's nothing you can do.

UNLESS ... If $\cos(a)=0,$ the answer would be different.


These graphs show the four least and the four most populous states in the U.S. in 1996. What is the greatest positive difference in population between any two U.S. states in 1996?

I believe a pic and/or Wikipedia would be nice.

I saw a picture, but then i refreshed and then it dissapeared.

That means it just caught the flu.  :wink:

[quote="isabella2296"]There's supposed to be a picture, but currently the flu is going around AoPS, and symptoms include:

drowsiness
pink eye
rash
sneezing/coughing
lack of diagrams/graphs/pictures[/quote]

I believe the flu is only going around FTW.  Hopefully, the rest of AoPS is safe.

Hopefully. I'd hate to get the flu!!!

*cough cough* *achoo!!!* *ow a rash!* *i have pink eye!* *hey, where'd my pic go?*

[quote="isabella2296"]There's supposed to be a picture, but currently the flu is going around AoPS, and symptoms include:

drowsiness
pink eye
rash
sneezing/coughing
lack of diagrams/graphs/pictures[/quote]

You forgot vomiting.

That isn't one of the symptoms though. I'm a doctor. I should know.  :D

[quote="isabella2296"]... *hey, where'd my pic go?*[/quote]

lol.  :rotfl:  Isn't Lack of diagrams/graphs/pics one of the most annoying diseases?
In your examples for what people say when they catch the flu, you forgot drowsiness. *Yawn* :sleeping:

Oops, lol.

Well let's let this topic die now.

Someone can answer the question by researching the most populous states and least populous states in 1996, and post it, but I'm too lazy to.

Wouldn't this problem be kind of easy if there were a graph?

There's supposed to be a graph. Now please let the topic die.

Good-bye no-pic syndrome! (at least for now.)

[quote]Oops, lol.

[b]Well, let's let this topic die. [/b]There's nothing we can do for those unfortunate patients who have caught the flu.[/quote]

[quote]There's supposed to be a graph. [b]Now please let the topic die.[/b][/quote]

A-[i]hem[/i]. *cough cough*[size=59]ernie!![/size]*cough cough*

I am very, truly sorry for not letting this thread die.

If you didn't reply, then I wouldn't have replied.  :P

[quote="isabella2296"]Please...stop...now...do...not...reply....to...this...let...this...die...now....thank...you...[/quote]

I like your avatar, izzy!

Thanks.  :wink:

But seriously, let this topic die.

But this is a really cool chat room.  And someone tell me why the heck my rating is 3.4.

If you want to chat, go to Games and Fun Factory. And I have a sadly low rating too...it dropped from 4 to 3, and it's currently 3.3. Someone rated one of my posts as spam and I've no idea why. It was an ordinary post that explained something...not like this one. So please, PLEASE just stop it!!!


Prove that the sequence $\left\{\frac{1}{1+nz}\right\}$ is uniform convergent to zero for all z such that $\left|z\right|\geq2.$
Solution from book:
We have $\left|\frac{1}{1+nz}-0\right|<\epsilon$ when $\frac{1}{\left|1+nz\right|}<\epsilon$ or $\left|1+nz\right|>1/\epsilon$. Now $\left|1+nz\right|\leq|1|+|nz|=1+n|z|$ and $1+n|z|\geq|1+nz|>1/\epsilon$ for $n>\frac{1/\epsilon-1}{|z|}$. Thus the sequence converges to zero for $|z|>2$. [snipped rest of solution]

Seems wrong to me. Wouldn't we rather use sth like $|nz|\leq|nz+1|+|-1|$, hence $|nz+1|\geq|nz|-1>1/\epsilon$ when $n>\frac{1/\epsilon+1}{|z|}$?


The cost $ C$ of sending a parcel post package weighing $ P$ pounds, $ P$ and integer, is $ 10$ cents for the first pound and $ 3$ cents for each additional pound. The formula for the cost is:

$ \textbf{(A)}\ C \equal{} 10 \plus{} 3P \qquad\textbf{(B)}\ C \equal{} 10P \plus{} 3 \qquad\textbf{(C)}\ C \equal{} 10 \plus{} 3(P \minus{} 1)$
$ \textbf{(D)}\ C \equal{} 9 \plus{} 3P \qquad\textbf{(E)}\ C \equal{} 10P \minus{} 7$

[hide="Click for solution"]
For $ P\minus{}1$ pounds the charge is $ 3$ cents per pound.  For the first pound the charge is $ 10$ cents.  Thus, $ C\equal{}10\plus{}3(P\minus{}1)$ or $ \boxed{\textbf{(C)}}$.
[/hide]


If $[x]$ denotes the integer part of $x$ then prove that

if   $[x]+[-x]=0$  then  $x \in Z $

 :)

suppose $X$ not be an integer.So it is $x.a$ that $x$ is $[X]$ and $a$ is $\{x\}$ and $a>0$ so $[X]=x$ and $[-X] =x-1$ so $[X]+[-X]= -1$.
so if $[X]+[X-1]=0$ then $X$ is an integer.
I think my proof is ok. Please tell me if there is wrong with it.


Hi Folks,

Is there any good ways of teaching kids how to calculate expressions like 1+2*3 correctly.  Observing that the * must be done before the +.

Does it present much of a problem, or is it just a very minor area of the schools year.

I've developed a computer program that might be useful when teaching this part of mathematics.  I'm wondering if I it would be worth setting up a website to distribute it.  (I've put many months of evenings into developing it and its quite good)

thanks,
Martin

My math teacher used to help us remember the order of operations by saying Please Excuse My Dear Aunt Sally (Parenthases, exponents, multiply, divide, add, subtract.)  These helped me, so hope they will help you too.

People also use PEMDAS (Parentheses, Exponents, Multiplication, Division, Addition, Subtraction).

And it is important to remember that if there are a lot of parenthesis, that you have to work from the inside out.

On the topic of order of operations: do you do logarithms with the same priority as exponentation?


trigonometry
trig identities
Law of Cosines
geometry
angle bisector

I think it will be solved with trigo.It is asking m(CBA)
[hide][img]http://www.sinavbankasi.com/forum/files/soruu_512.jpg[/img][/hide]

The problem says $AD=2BD$, so we can just assign values to $AD$ and $BD$ and solve for $\angle DAB=\angle DCB$.

[color=darkblue][b][u]An equivalent enunciation.[/u][/b] Let $ABC$ be a right triangle ($AB\perp AC$).
Denote the midpoints $D$, $M$ of the segments $[BC]$, $[AD]$ respectively.
Suppose $m(\widehat{BED})=80^{\circ}$. Ascertain $m(\widehat{BEC})$.[/color]

erm, how do you know $AC$ is perpendicular to $AB$?

[hide="Well, if you could use a caculator, then you could do it as follows:"]
For simplicity, say that BD=1, CD=2, and AD=2.  Also, say  that $AB=x$.  Applying the Law of Cosines for Triangles ADB and ACB, we obtain that \[\frac{2^{2}+x^{2}-1^{2}}{2(2)(x)}=\cos A=\frac{4^{2}+x^{2}-BC^{2}}{2(4)(x)}\Rightarrow \frac{x^{2}+3}{4x}=\frac{16+x^{2}-BC^{2}}{8x}\Rightarrow 2x^{2}+6=16+x^{2}-BC^{2}\Rightarrow BC^{2}=10-x^{2}\] Now, we also must find what $x$ is.  So, we have from The Law of Cosines of Triangle ABC that $\cos 80=\frac{x^{2}+1^{2}-2^{2}}{2(x)(1)}=\frac{x^{2}-3}{2x}$.  So, we have that \[2x\cos 80=x^{2}-3\Rightarrow x^{2}-2x\cos 80-3=0\Rightarrow x=\frac{2\cos 80+\sqrt{12+4\cos^{2}80}}{2}=\cos 80+\sqrt{3+\cos^{2}80}\] Note that we ignored the minus part of the plus-minus because the minus would make x, a length, negative, which is impossible.
Now, we can apply the Law of Cosines on angle ABC to obtain that \[\cos \angle ABC=\frac{x^{2}+(\sqrt{10-x^{2}})^{2}-4^{2}}{2(x)(\sqrt{10-x^{2}})}=\] \[\frac{-6}{2x\sqrt{10-x^{2}}}=\frac{-3}{x\sqrt{10-x^{2}}}=\] \[\frac{-3}{(\cos 80+\sqrt{\cos^{2}80+3})(10-(\cos 80+\sqrt{\cos^{2}80+3})^{2}})=\] \[\frac{-3}{[\cos 80+\sqrt{\cos^{2}80+3}][10-(2\cos^{2}80+3+2\cos 80\sqrt{\cos^{2}80+3})]}=\] \[\frac{-3}{(\cos 80+\sqrt{\cos^{2}80+3})(7-2\cos^{2}80-2\cos 80\sqrt{\cos^{2}80+3})}\] Now, we can use a calculator to say that $\cos 80\approx.17364$  When we plug this in, we get that Angle ABC is about 104 degrees, so we round it down to $\boxed{100}$.  I think I made a mistake.
[/hide]

[quote="Shahzada"]The answer is 160.[/quote]
Wait, I don't think that's true. If $\angle CBA=160^\circ$, then $\angle CBD=\angle DBA=80^\circ$, so $BD$ is the angle bisector of $\angle CBA$. Since $BD$ is both the angle bisector and the median to $AC$, we must have $AB=BC$. So $\angle ACB = \angle BAC=10^\circ$, and $BD\perp AC$.

But then, $\tan C=\frac{BD}{DC}=\frac{1}{2}$, but $\tan 10^\circ \ne \frac{1}{2}$... :huh:

Am I missing something?

[hide]
Can't we just let $AD=1$, $BD=2$? Using the sine law we can solve for $\angle DAB$. Then $180-2\angle DAB\doteq 120=\angle CBA$.
[/hide]

[quote="Zellex"]Any hints?

Where do you get these problems Shahzada?[/quote]
This question has been asked by a student.I couldn't solve it.The others from some edition and some of them was created by teachers that their name writing under question.


linear algebra
matrix
ratio
abstract algebra
linear algebra unsolved

I have a 100 X 100 matrix and I want to modify it (by adding any number of additional rows) to obtain almost all the eigenvalues to be quite similar (i.e. the ratio of any 2 eigenvalues should have an upper bound limit of 3 e.g if the max. eigenvalue is 10 the others should be at least 3.33). The main condition on the matrix is that no entries in the matrix can have values of 0 (all entries are from 1 to 15). I think for a 2 X 2 matrix the two eigenvalues can be made equal if the matrix is diagonal (i.e. all entries apart from the main diagonals are zeros). This is not valid because of the restriction of zeros. Then, also, I will have to extend it to the larger matrix.
I have found how to test if the eigenvalues are equal on the following websites:
http://www.fon.hum.uva.nl/praat/manu...values___.html
http://www.terrapub.co.jp/journals/j...2/37020299.pdf
Unfortunately, I don’t have a clue how to make the above square matrix have almost similar eigenvalues. Any help would be much appreciated. Thank you.

Do you want to have $ 100$ different eigenvalues?

Hi Tyl
The number of the eigenvalues will depend on the matrix. The matrix might have 100 eigenvalues and the minimum could be only one e.g if the 100X100 matrix was like this:
1 1 1 1 .....   100 times
2 2 2 2....    100 times
3 3 3 3....    100 times
4 ...
.
.
100 times
In this case, my guess is that the matrix will have only one large eigenvalue because the variance across the columns do not change.

To sum up, I have an arbitrary 100 X 100 matrix which most certainly will have more than one eigenvalues. If this is the case, how many rows will I have to add to the matrix to make the eigenvalues of the newly made matrix quite similar? The entries in the newly added rows should range between 1,2,3... ,14,15.

1) Your example
"100X100 matrix was like this:
1 1 1 1 ..... 100 times
2 2 2 2.... 100 times
3 3 3 3.... 100 times
4 ...
.
.
100 times "
is uncorrect because the eigenvalues are: $ 0,\cdots,0,1\plus{}2\plus{}\cdots\plus{}100$.
2) If you want to obtain such eigenvalues the simplest way is: charge the diagonal, using rows s.t. the row $ i$ is in the form $ [0,\cdots,0,\pm{15},0,\cdots,0]$. This method works only if the elements of the square matrix are small in front of $ 15$.
Beware : the method does not ensure the eigenvalues will be real; the condition about the ratios relates to the modules of the eigenvalues.
Example: $ A\equal{}\begin{pmatrix}12&\minus{}9&10\\7&\minus{}2&14\\9&\minus{}14&8\end{pmatrix}$ is transformed into: $ B\equal{}\begin{pmatrix}27&\minus{}9&10\\7&13&14\\9&\minus{}14&23\end{pmatrix}$. The maximum ratio is $ \approx1.25$ but there exist $ 2$ complex eigenvalues.
If you want only real eigenvalues it is much more difficult because (for example) a matrix in the form $ \begin{pmatrix}a&b\\c&a\end{pmatrix}$ with $ bc<0$ has $ 2$ complex eigenvalues.

Thank you!
Unfortunately, I don't think I made the question clear.


Εστω $P(n!)$ το πληθος των ψηφιων του $n!$.Βρειτε την συναρτηση $f(n)$ που υπολογιζει ακριβως η προσεγγιζει την συναρτηση $P(n!)$ για καθε $n$ φυσικο.

[quote="silouan"]Λοιπόν από τον τύπο του Ramanajuan έχουμε
$ln\ (n!)=nln\ (n)-n+\frac{ln\ (n(1+4n(1+2n)))}{6}+\frac{ln\ (\pi)}{2}$[/quote]

Silouan o tipos opws ton exeis den isxiei (an kai fainetai ekpliktikos, se kati pou mas exei sinithisei o Ramanujan me tis perierges kai asinithistes tautotites tou...)

Sinithws (opws p.x. simvainei gia ton tipo tou Stirling gia to $n!$ eite einai tipos o opoios isxiei proseggistika gia megales times tou $n$ eite einai kapoios tipos pou afora ta akeraia meri twn dio arithmwn. Mipws tha mporouses na elenkseis kai pali ton tipo tou Ramanujan ?

Alexandros

Μπορει τοτε καποιος να μου βρει το πληθος των ψηφιων του $2^{2007}$ ?

Σωστος μαγκα!Εγω προσωπικα δεν ετυχε να το δω την λυση-μεθοδο σε καποιο βιβλιο(το σκεφτηκα μονος) και ειναι ενα παρα πολυ χρησιμο λημμα για τον καθενα μας.Σκεψου π.χ οτι ειναι ευκολο να υπολογισεις το πληθος των ψηφιων οποιουδηποτε πρωτου του Mersene!Aληθεια ετυχε να το δεις σε καποιο βιβλιο οχι ιδιαιτερα υψηλου επιπεδου?Μολις απαντησει καποιος θα ηθελα πολυ να κοιταξεται και να προτινεται ιδεες και για το αρχικο προβλημα που εθεσα στο τοπικ. :)

Όχι δεν το είδα σε κάποιο βιβλίο. Kι εγώ μόνος μου το έβγαλα. Αρκεί να παρατηρήσεις ότι αν π.χ. ο $a$ έχει $n$ ψηφία τότε ισχύει $10^{n-1}\leq a<10^{n}\Leftrightarrow log(a)<n\leq log(a)+1 \Rightarrow n=\left[log(a)+1 \right]$.  :wink:

Oποτε για το πληθος των ψηφιων του $n!$ θα εχουμε $[logn!]+1=[log1+log2+....logn]+1$ oμως δεν ειναι και τοσο πρακτικο να υπολογιζουμε αυτο το αθροισμα ακομα και με επιστημονικο κομπιουτερακι και ιδιως αν ο $n$ ειναι μεγαλος(απο προγραμματα δεν τα παω πολυ καλα.... :lol: )οποτε θα πρεπει να υπολογισουμε αυτο το αθροισμα συναρτηση του $logn$.Eχει κανεις καμια ιδεα? :)

[quote]Oποτε για το πληθος των ψηφιων του n! θα εχουμε [logn!]+1=[log1+log2+....logn]+1 oμως δεν ειναι και τοσο πρακτικο να υπολογιζουμε αυτο το αθροισμα ακομα και με επιστημονικο κομπιουτερακι και ιδιως αν ο n ειναι μεγαλος[/quote]
Gt esi 8a to ipologisis?? :P  To mathematica kseri tin dulia tu... :wink:

 :D

Υπολόγισε πρώτα το $ n!$ και μετά τον $ \log n!$ επίση μπορείς να υπολογίσεις το $ n!$ με εναν πολύ ωραίο τρόπο . Από τους κανόνες των λογαρίθμων προκύπτει ότι $ \ln n!\equal{}\ln 1\plus{}...\plus{}\ln n$ άρα $ n!\equal{}e^{\sum\ln n}$

[color=red]moderetor edit : @mathteen: Καλώς ήρθες . Διάβασε το τόπικ LATEX οδηγίες και προσπάθησε να μην γράφεις σε κεφαλαία .[/color]

[quote="cretanman"][quote="silouan"]Λοιπόν από τον τύπο του Ramanajuan έχουμε
$ ln\ (n!) \equal{} nln\ (n) \minus{} n \plus{} \frac {ln\ (n(1 \plus{} 4n(1 \plus{} 2n)))}{6} \plus{} \frac {ln\ (\pi)}{2}$[/quote]

Silouan o tipos opws ton exeis den isxiei (an kai fainetai ekpliktikos, se kati pou mas exei sinithisei o Ramanujan me tis perierges kai asinithistes tautotites tou...)
[/quote]

Ναι βρε, προφανώς ο Sil τον εννοεί ασυμπτωτικά (με $ \sim$ αντί για $ \equal{}$) :) Κατα τ'άλλα, ο τύπος σωστός είναι...

Cheerio,

Durandal 1707

Μα καλα! :o Ευχαριστω για το ενδιαφερον σας!!!(που το ξετρυπωσατε!! :) )


It's a cross between a camel and a llama. They are very cool, however they enjoy spitting a lot. They like to express themselves through the art of dance. Llamels can be found roaming the mirc channels.

MUAHAHAHAHAHAHAHAHAHA :rotfl:  :rotfl:  :rotfl:

Cross between camel and llama? We already guessed that. Super said we were wrong.

[quote="superllamel"]Why does everyone care so much? Does it really matter?

And math, you are wrong.  :wink:[/quote]

uh I kinda saw after mewto's post...

Hmm. Some posts have vanished. What happened? :!:

Dan is right. That is weird.

Can moderators delete posts?

Maybe some were very irrelevant, although i thought mine was on topic....

Of course they can...if it's inappropriate...obviously the posts need to be deleted by someone...

i should've asked superllamel at mc nats, wat state r u from again?

What state am I from?

I'm from Colorado if I'm who you're asking.

OOOOOOOOH!  i'll see you at mathcounts nats next year than super..

oh well, i guess thats B-A-D!

I'm not going to nats next year. I'm going into 9th grade.  :(

http://www.urbandictionary.com/define.php?term=llamel

lol  :rotfl:  :rotfl:  :rotfl:

[quote="isabella2296"][quote="dynamo729"][quote="guineapiggies"]i still think it's 'lame' with fancy l/absolute values around it :D[/quote]

I agree... but i didnt see this post (promise) and decided that myself.[/quote]

Super-the-absolute-value-of-lame...

I like it!  :D[/quote]
sounds like he's...lame?!?!?!?!? thats not right....

You said that it had two meanings right?

[quote="superllamel"]Why do so many people care?

I don't think it really matters.  :huh:[/quote]

everyone cares because u made everyone interested
anyway, what is a llamel??


linear algebra
matrix
linear algebra unsolved

Let $ A,B \in M_n{\mathbb{(C)}}$  such that $ AB \equal{} BA$. Prove that or disprove :
1)  $ \det \left(A\overline{A} \plus{} \overline{B}B\right) \in \mathbb{R}$
2)  $ \det \left(A\overline{A} \plus{} \overline{B}B\right) \geq 0$
  where $ \overline{X}$ is the matrix formed by conjugating the elements of $ X$
........... :play_ball:  ....

Sorry I post it  :blush:  It is false !
case1)  For $ A \equal{} B \in M_n{\mathbb{(C)}}$ then
 $ A \equal{} X \plus{} iY$ then $ \overline{A} \equal{} X \minus{} iY,X,Y \in M_n{\mathbb{(R)}}$ then $ A\overline{A} \equal{} X^2 \plus{} Y^2 \plus{} i(YX \minus{} XY)$ and  $ \overline{A}A \equal{} X^2 \plus{} Y^2 \minus{} i(YX \minus{} XY)$
 Then  $ A\overline{A} \plus{} \overline{A}A \equal{} X^2 \plus{} Y^2$
 Now for $ A\overline{A} \in M_n{\mathbb{(C)}}$ then $ XY \neq YX$ it is easy for example that $ \det(A\overline{A} \plus{} \overline{A}A) \equal{} \det(X^2 \plus{} Y^2) < 0$
 I hope that you check example  that it is false !
 Example of me $ X \equal{} \begin{bmatrix}1 & \minus{} 2 \\
2 & 0\end{bmatrix},Y \equal{} \begin{bmatrix}1 & 0 \\
0 & 3\end{bmatrix}$ then $ X^2 \plus{} Y^2 \equal{} \begin{bmatrix} \minus{} 2 & \minus{} 2 \\
2 & 5\end{bmatrix}$ then $ \det(X^2 \plus{} Y^2) \equal{} \minus{} 6 < 0$
case2) $ \det\left(A\overline{A} \plus{} \overline{B}B\right)\notin\mathbb{R}$
 Begin $ A\overline{A} \plus{} \overline{B}B \neq \overline{A}A \plus{} B\overline{B}$

Sorry  :blush:  $ A\overline{A} \plus{} \overline{A}A \equal{} 2(X^{2} \plus{} Y^{2})$ and $ \det(A\overline{A} \plus{} \overline{A}A) \equal{} 2^n\det(X^{2} \plus{} Y^{2})$


[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \sum \frac {a^3}{a^3 \plus{} (b \plus{} c)^3} \plus{} 1 \ge 2\sum \frac {a^2}{a^2 \plus{} (b \plus{} c)^2}
\]
Equality holds if and only if $ abc \equal{} 0$. :)[/quote]
It is a problem at a contest of my school!
<=>$ \sum_{cyc} \frac{1}{1\plus{}(\frac{b\plus{}c}{a})^3}\plus{}1 \geq \sum_{cyc} \frac{1}{1\plus{}(\frac{b\plus{}c}{a})^2}$
Put $ \frac{b\plus{}c}{a}\equal{}x;\frac{c\plus{}a}{b}\equal{}y;\frac{a\plus{}b}{c}\equal{}z$
We need prove that:
$ \sum_{cyc} \frac{1}{1\plus{}x^3}\plus{}1 \geq 2\sum_{cyc} \frac{1}{1\plus{}x^2}$
Note that
$ \frac{1}{1\plus{}x}\plus{}\frac{1}{1\plus{}y}\plus{}\frac{1}{1\plus{}z}\equal{}1$
and $ \frac{1}{1\plus{}x^3}\plus{}\frac{1}{1\plus{}x} \geq \frac{2}{1\plus{}x^2}$
Done!

It is equivalent to:
$ (x \minus{} 1)^{3}(x \plus{} 1)x \geq 0$ which holds if $ x \geq 1$

([b]chien than[/b]'s proof works if $ a$, $ b$ and $ c$ are triangle sidelengths.)

[quote="delegat"]

([b]chien than[/b]'s proof works if $ a$, $ b$ and $ c$ are triangle sidelengths.)[/quote]
No! It works for all positive $ a,$ $ b$ and $ c,$ but the chien than's proof is not full.

OK Most probably proposed inequality holds for all nonnegative reals but my comment is related to:
$ \frac {1}{1 \plus{} x^3} \plus{} \frac {1}{1 \plus{} x} \geq \frac {2}{1 \plus{} x^2}$

[quote="arqady"][quote="chien than"]
and $ \frac {1}{1 \plus{} x^3} \plus{} \frac {1}{1 \plus{} x} \geq \frac {2}{1 \plus{} x^2}$
Done![/quote]
Why done?  Explain please. Thank you!
I think, the Chebyshov's inequality ends this problem.[/quote]

Arqady, how do you use Chebychev ?

Thank you very much.

manlio, I have found a mistake in my Chebyshov.  :(

[quote=can_hang2007]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \sum \frac{a^3}{a^3\plus{}(b\plus{}c)^3} \plus{}1 \ge 2\sum \frac{a^2}{a^2\plus{}(b\plus{}c)^2}\]
Equality holds if and only if $ abc\equal{}0$. :)[/quote]

[img]http://s11.sinaimg.cn/middle/006ptkjAzy75Sx4ye9Qaa&690[/img]


LaTeX
email
probability
number theory
divisibility tests

OK, here's the thread for all the problems and answers.  I have the PDF & .tex for the individual problems, and I'm about to get the Team latex from Shobhit.  After I get it all corrected and all, I will TRY to insert images into the PDF.  Also, post solutions for the problems here.  As soon as we get them all, I'll try to make a solution document.
Note: Shobhit is out of town.

EDIT: Attached is the corrected version of the problems.  PM or email me any more corrections.  I also attached the images.  Thanks Joe!

:rotfl:  :rotfl: Those are some sad pictures, maybe someone wants to use Geometer's Sketchpad for the pictures?

And I'm supposed to email people the solutions to those team problems too i guess ;)

The Team Problems will be sent via email. For now, we can discuss the other problems here

Feel free to discuss ALL the problems here, not just the Team questions.  In fact, we might as well discuss the Counting & Number Theory & Proof problems here as well.  So post any questions!

On day 3, we went over divisibility briefly.  A while back I posted some divisibility rules [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=432208#p432208]here[/url].  If you do not know these, I suggest you learn them!  Ideally, you should also learn how to prove them.  To prove them, generally the technique is to view a number, $a_na_{n-1}\cdots a_1a_1$ (the $a_i$ are digits), as $10^na_n+10^{n-1}a_{n-1}+\cdots+10a_1+a_0$ and then use modular arithmetic.

I'll show how one might do this for 7.

$\text{Divisibility by 7: }$ A number, $x$, is divisible by 7 if the alternating sum of each block of 3 digits is divisible by 7.

$\text{Proof: }$  We start with $x=a_na_{n-1}\cdot a_1a_0=10^na_n+10^{n-1}a_{n-1}+\cdots+10a_1+a_0.$  Since our divisibilty rule tells us to break the number up into blocks of 3, we should also do that in our proof.  So let's start at the right (i.e. next to the decimal place).  Our first block is $10^2a_2+10a_1+a_0$.  The second block is $10^5a_5+10^4a_4+10^3a_3$.  And so forth on up to $10^{3k+2}a_{3k+2}+10^{3k+1}a_{3k+1}+10^{3k}a_{3k}.$  Now we can factor out a 1000 from each of this (except for the first).  But then $1000\equiv -1\mod 7.$  So this is where we get the alternating sum.  To recap this:

\begin{eqnarray*} x &=& a_0+10a_1+10^2a_2+\cdots+10^na_n\\ &=& (a_0+10a_1+10^2a_2)+(10^3a_3+10^4a_4+10^5a_5)+\cdots+(10^{3k+2}a_{3k+2}+10^{3k+1}a_{3k+1}+10^{3k}a_{3k})\\ &=& (a_0+10a_1+10^2a_2)+10^3(a_3+10a_4+10^2a_5)+\cdots+1000^k(10^2a_{3k+2}+10a_{3k+1}+a_{3k})+\cdots\\ &=& (a_0+10a_1+10^2a_2)-1(a_3+10a_4+10^2a_5)+\cdots+(-1)^k(10^2a_{3k+2}+10a_{3k+1}+a_{3k})+\cdots \end{eqnarray*}

lol, no solutions have been posted yet for anything :rotfl:

$\text{Set 2 Solutions}$

[hide="2"]$\frac{30^{39}}{30^{29}}=30^{10}$.

$30^{10}=5^{10} \cdot 3^{10} \cdot 2^{10}$ and contains $(10+1)(10+1)(10+1)=11^3$ divisors.

$30^{39}=5^{39} \cdot 3^{39} \cdot 2^{39}$ and contains $(39+1)(39+1)(39+1)=40^3$ divisors.

The probability of picking the "right" divisor is therefore $\boxed{(\frac{11}{40})^3}$[/hide]
[hide="5"]
Out of the first $500$ positive integers, $\left floor \frac{500}{7} \right floor=71$ are multiples of $7$ and $\left floor \frac{500}{11} \right floor=45$ are multiples of $11$. However, we have overcounted multiples of $77$, and there are $\left floor \frac{500}{77} \right floor=6$ multipes of those.

So our total probability is $\frac{71+45-6}{500}=\boxed{\frac{11}{50}}$[/hide]
[hide="6"]
Three points on a circle split the circle into three arcs. Let us denote the largest arc formed by these three points $S$. Since $S$ is the largest arc, $360^\circ\geq S\geq 120^\circ$. However, we want the three points to lie on a semicircle, therefore, $360^\circ\geq S\geq 180^\circ$. The probability therefore is $\frac{180}{240}=\boxed{\frac{3}{4}}$[/hide]
[hide="7"]Essentially, we want to find the number of ways $n$ can be partitioned into positive integers denoted as $x_{i}$:

$x_{1}=n$
$x_{1}+x_{2}=n$
$x_{1}+x_{2}+x_{3}=n$
$......$
$......$
$x_{1}+x_{2}+x_{3}+....+x_{n}=n$

For each number of $x_{i}$ terms, $n$ has $\binom{n-1}{i-1}$ possible ordered pairs. Therefore, the number of partitions is $1+\binom{n-1}{1}+\binom{n-1}{2}+\binom{n-1}{3}+....+\binom{n-1}{n-2}+1$[/hide]

ahhh, i hope the latex gets back to normal soon....

Ok, for some reason there are none of the pdf download things here, but w/e.

I remember there was one problem like:

$x_{1}+x_{2}+x_{3}+x_{4}=2006$, find all positive odd ordered pairs or w/e.

[hide="solution"]All the numbers will be written in the form:
$2x_{1}-1+2x_{2}-1+2x_{3}-1+2x_{4}-1=2006$

Thus, it's equivalent to finding the number of positive solutions to:

$x_{1}+x_{2}+x_{3}+x_{4}=1005$

The answer is thus $\binom{1004}{3}$[/hide]


modular arithmetic
number theory unsolved
number theory

[quote="Allnames"]Find all natural solutions of equation: $ x^{2009} \plus{} y^{2009} \equal{} 7^z$[/quote]

Since $ x\plus{}y|x^{2009} \plus{} y^{2009}$, we get $ x\plus{}y\equal{}7^a$ with $ a>0$

Since $ x^7\plus{}y^7|x^{2009} \plus{} y^{2009}$, we get $ x^7\plus{}y^7\equal{}7^b$

Using "Lifting the exponent" lemma, we get :
$ x\plus{}y\equal{}0\pmod 7$ $ \implies$ $ v_7(x^7\plus{}y^7)\equal{}v_7(x\plus{}y)\plus{}v_7(7)$ and so $ b\equal{}a\plus{}1$ and so $ x^7\plus{}y^7\equal{}7(x\plus{}y)$, which is obviously impossible for $ x,y\in\mathbb N$ (if $ x>1$ and $ y>1$ $ x^7>7x$ and $ y^7>7y$ and no solution with $ y\equal{}1$ or $ x\equal{}1$).

So no solution.

Aside from the trivial solutions $ (x,y,z) \equal{} (0, 7^{\alpha}, 2009\alpha), (7^{\alpha}, 0, 2009\alpha)$ with $ \alpha$ non-negative integer, we may assume $ 7$ not to divide $ x,y$ (else we reduce to lower solution). Since $ x^7\plus{}y^7$ and $ x\plus{}y$ are factors of $ x^{2009} \plus{} y^{2009}$ ($ 2009 \equal{} 7\cdot 7\cdot 41$), we have that $ x^7\plus{}y^7\equal{}7^{\alpha}$ and $ x\plus{}y \equal{} 7^{\beta}$, with $ \beta \geq 1$. Then $ y \equal{} 7^{\beta} \minus{} x$, so $ 7^{\alpha} \equal{} x^7\plus{}y^7 \equal{} 7^{\beta\plus{}1}\left (7^{\beta}M \plus{} x^6 \right )$ (for some integer $ M$), hence $ 7^{\beta}M \plus{} x^6 \equal{} 7^{\alpha \minus{} \beta \minus{} 1}$. This implies $ \alpha \equal{} \beta \plus{} 1$.
But $ 7^{\beta \plus{} 1} \equal{} x^7\plus{}y^7 \geq \frac {1} {2^6} (x\plus{}y)^7 \equal{} \frac {1} {2^6} 7^{7\beta}$, therefore $ 7^{6\beta \minus{} 1} \leq 64 < 7^3$, so $ 6\beta \minus{}1 < 3$, in contradiction with $ \beta \geq 1$. Therefore no more solutions are possible.

[quote="mavropnevma"]Aside from the trivial solutions $ (x,y,z) \equal{} (0, 7^{\alpha}, 2009\alpha), (7^{\alpha}, 0, 2009\alpha)$ with $ \alpha$ non-negative integer, ...[/quote]

... which are not solutions since Allnames asked for $ x,y\in\mathbb N$, so different from zero.
 :wink:

Especially you, dear pco, as residing in France, should remember that $ \mathbb{N} \equal{} \{0, 1, \ldots \}$ in many countries, including Romania, France, Italy  :) .

There are 2 notations:

1) Romania, France, Italy? $ \{0,1,\ldots\}\equal{}\mathbb{N},\{1,\ldots\}\equal{}\mathbb{N}^*$

2) other countries? $ \{0,1,\ldots\}\equal{}\mathbb{N}_0,\{1,\ldots\}\equal{}\mathbb{N}$

In Austria $ \mathbb{N}\equal{}\{0$, $ 1$, $ \dots \}$, but I don't understand why we don't just write $ \mathbb{Z_\plus{}}$ or non negative numbers... It's always the same with $ \mathbb{N}$.^^

Sure, and I had a lot of difficulties about this when I entered this forum.

It seems this subject have been discussed many times and that the default situation on these fora is to consider $ 0\notin\mathbb N$  :(

The best thing to do is to avoid confusion in the problem statements (using "integer", "positive integer", "non-negative integers", ...)

By the way, this does not matter. It was just to explain why my previous post stated "no solution"  :wink:


Let $ A\equal{}(a_{ij})_{1 \leq i,j \leq n}$ Such that : $ a_{ij}\equal{}\sum_{k\equal{}1}^{n}{k^{i\plus{}j}}$ for all $ 1 \leq i,j \leq n$
Calculate :$ \det(A)$
 :wink:


calculus
derivative
function
ratio
real analysis
real analysis solved

if u have a parametric function defined by $x=x(t)$ and $y=y(t)$.

and u want to find the second derivative $\dfrac{d^2y}{dx^2}$ what do u do?
the first is "quite" simple $\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$, but for the second i am not sure whether it is just $\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d^2y}{dt}}{\left(\dfrac{dx}{dt}\right)^2}$ or is that complicated that u have to use the division-rule:
$\dfrac{d^2y}{dx^2}=\dfrac{d\left(\dfrac{dy}{dx}\right)}{dx}=\dfrac{dx\dfrac{d^2y}{dx}-dy\dfrac{d^2x}{dx}}{(dx)^2} =\dfrac{\dfrac{dx}{dt}\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}\dfrac{d^2x}{dt^2}}{\left(\dfrac{dx}{dt}\right)^3}$

well which one is true or are they the same? and the most important question is why?

thx

It's the more complicated version.

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right) =\dfrac{d}{dx}\left(\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\right) =\dfrac1{\dfrac{dx}{dt}}\cdot\dfrac{d}{dt}\left(\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\right) =\dfrac1{\dfrac{dx}{dt}}\cdot\left(\dfrac{\dfrac{d^2y}{dt^2}}{\dfrac{dx}{dt}}- \dfrac{\dfrac{dy}{dt}\cdot\dfrac{d^2x}{dt^2}}{\left(\dfrac{dx}{dt}\right)^2}\right)$
$=\dfrac{\dfrac{d^2y}{dt^2}}{\left(\dfrac{dx}{dt}\right)^2}- \dfrac{\dfrac{dy}{dt}\cdot\dfrac{d^2x}{dt^2}}{\left(\dfrac{dx}{dt}\right)^3}$

At least now you've got the details right.

A simple example, showing that the naive version must be wrong: $y=t,x=\sqrt{t}$ for $t>0$. The second derivative of $y$ with respect to $t$ is zero, but the path is a parabola.

but what is wrong with the logic of the first case?

i mean only expand with $dt^2$ when i say $\dfrac{d^2}{dx^2}=\dfrac{\dfrac{d^2y}{dt^2}}{\dfrac{dx^2}{dt^2}}= \dfrac{\dfrac{d^2y}{dt^2}}{\left(\dfrac{dx}{dt}\right)^2}$ and actually thats what we both do in our other proofs

therefore  :?:

Well, the problem is that in $\dfrac{d^2y}{dx^2}$ the misterious expression $d^2y$ stands for the (infinitesimal) second difference with equally spaced values of $x$ while in $\dfrac{d^2y}{dt^2}$ it stands for that with equally spaced values of $t$, which is not the same.

but $\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$, so i dont quite understand that argument. why is that possible for the first but not for the second derivative?

OK, let me try to explain it in another way. Suppose that you are given a table of $t$, $x(t)$ and $y(x(t))$ with equally spaced $t$  where the step $dt$ in $t$ is very small. Let $t_j$; $x_j$ and $y_j$ be the values in your table. You can create the table of the first and the second differences $dx_j=x_{j+1}-x_j$, $d^2x_j=dx_{j+1}-dx_j=x_{j+2}+x_j-2x_{j+1}$ and similarly for $y$.

How do you approximate the first derivative from this table? You just take $\dfrac{dy_j}{dx_j}$. Clearly, here it makes sense to divide by $dt$ and to say that the (approximate) first derivative of $y$ with respect to $x$ equals to the ratio of the approximate derivatives of $y$ and $x$ with respect to $t$. Now, let us look at the second derivatives.
Since the values of $t$ are equally spaced, we, indeed, have the approximation $\frac 1{dt_j}\left(\dfrac{dx_{j+1}}{dt_{j+1}}-\dfrac{dx_{j}}{dt_j}\right)=\dfrac{d^2x_j}{dt^2}$ for the second derivative of $x$ with respect to $t$ and similarly for $y$. But when we try to approximate the second derivative of $y$ with respect to $x$, we get $\frac 1{dx_j}\left(\dfrac{dy_{j+1}}{dx_{j+1}}-\dfrac{dy_{j}}{dx_j}\right)$ and we see that the denominators are different now, so we cannot extract $d^2y_j$ immediately. Instead, we have to write
$\dfrac{1}{dx_j^2dx_{j+1}}(dy_{j+1}dx_j-dy_j dx_{j+1})=\dfrac{1}{dx_j^2dx_{j+1}}(d^2y_j dx_j-dy_j d^2x_j)$, so the approximation for the second derivative we get is $\dfrac{d^2y_j}{dx_j^2}- \dfrac{dy_jd^2x_j}{dx_j^2dx_{j+1}}$. Now, you are welcome to divide by $dt^2$ in the first fraction and by $dt^3$ in the second one.

This is, of course, just one possible meaning you can assign to $d^2y$. There are other possibilities, but the point is that you cannot make the expression $d^2y$ meaningful by itself and have the formulas $y''_t=\dfrac{d^2y}{dt^2}$ and $y''_x=\dfrac{d^2y}{dx^2}$ hold simultaneously: they require two different definitions of $d^2y$, say, the second differences with equally spaced $t$ and $x$ correspondingly. Meanwhile, you can do it with $dy$ perfectly well because the first derivative does not require any equal spacing to be approximated by the difference ratio.

Hope, this helps. :)


where can I find almost all the proofs that were found?  here there are some http://www.cut-the-knot.org/pythagoras/index.shtml but there are over 300

Quote from Wikipedia:

"This theorem may have more known proofs than any other (the law of quadratic reciprocity being also a contender for that distinction); the book [u]Pythagorean Proposition[/u], by Elisha Scott Loomis, contains 367 proofs."


i dont have it, but i doubt anyone will post it. people trade to get mathcounts tests (especially old ones.)

My coach has all of them from 1985. I don't know where she got them. However, she is not my coach anymore  :( .

[quote="AIME15"]I don't think it's legal to post a test online without permission, but if you PM somebody with it, it should be fine.[/quote]
I thought maybe you can only trade tests through EMAIL... :wink:  :maybe:

would anyone like to email me it then =D?


Ever heard of Rooster Teeth? Check the following links:

http://panics.roosterteeth.com/archive/

http://rvb.roosterteeth.com/archive/

I especially liked PANICS. (It's supposed to be [b]mature[/b], but I don't really see the reason for that, except, well, randomly killing people) Please tell me if you enjoyed the "movies" or not.

Also try RvB. For example, the Winter Olympics.

I was expecting some enthusiasm, but really, guys, you're overwhelming me with this unusual amount of comments.

RvB is awesome! I'm saddened that their 100th episode came out and the Blood Gulch Chronicles ended...are they ending the entire series or just the season?


calculus
integration
trigonometry
complex numbers
complex analysis
real analysis
real analysis unsolved

Calculate the integral
$ \int^{2\pi}_0\sin{(2009x \plus{} \sin{x})}dx$
No idea how to start...

I suppose it's 0. This is a problem under contour integrals.

multiply your integrand by i and add cos(2009x + sinx) to it.  The integrand becomes exp i(2009x + sinx) / ix, integrated along the unit circle. You need the imaginary part of it. Now use residue theorem. You get 2pi as the answer. Its imaginary part is zero.

We've never learned the residue theorem so I suppose you dont have to use it..

Complex analysis is overkill.  Let $ y \equal{} x \minus{} \pi$, then the integral becomes $ \int_{ \minus{} \pi}^\pi \minus{}\sin \left (2009y \minus{} \sin y \right)dy$.  The integrand is odd, so the integral is $ 0$.

Doh can't believe I couldnt see that :) very nice, thanks!


In Texniccenter does anyone know the command for a line segment bar?

I've tried \bar, but the bar only covers 1 of the  letters.

Look, I never said \overbar.

I said \overline.  ;)  Please pay attention.  :)


Find the larger root of $ x^2 \minus{} 4x \plus{} 4  \equal{} 25$.

hello, rewriting the equation in the form $ x^2\minus{}4x\minus{}21\equal{}0$ and solving the quadratic equation we get $ x_1\equal{}\minus{}3$ or $ x_2\equal{}7$. So the larger root of both is $ x_2\equal{}7$.
Sonnhard.

[b]OR[/b]

We notice that both sides of the given equation are perfect squares, which means we can factor both sides and solve for $x$:

$x^2 - 4x + 4  = 25$

$(x-2)^2=5^2$

$\sqrt{(x-2)^2}=\sqrt{5^2}$

$x-2=\pm5$

Solving for $x-2=5$, we get $x=7$.
Solving for $x-2=-5$, we get $x=-3$.

Clearly, $7>-3$, so the larger root of $ x^{2}-4x+4 = 25 $ is $\boxed{7}$.


real analysis
geometry
3D geometry
advanced fields
advanced fields unsolved

Hello,

Suppose we have $m<n$, and $f:\mathbb{R}^m\rightarrow\mathbb{R}^n$ continuous, and assume that $\mu$ is the Lebesgue measure in $\mathbb{R}^n$. Under what conditions do we have that $\mu(\mbox{Img}(f))=0$. The answer is not just continuity, because the Cardano curve from $\mathbb{R}$ to $\mathbb{R}^2$ (suppose it extended as f(x)=(x,x) out of the interval [0,1]) is a counterexample, because the mesure of the image is 1.

Best,

[quote="fedja"]Suppose that $f$ satisfies the Holder condition $|f(x)-f(y)|\leq C|x-y|^d$ for some $d>0$. Consider the image of the unit cube. Split the cube into $N^m$ cubes of diameter $1/N$. Then the image will be covered by $N^m$ balls of diameter $N^{-d}$ and $n$-dimensional volume $N^{-nd}$. So, if $d>m/n$, the image has measure $0$. For $d=m/n$, you may have a space-filling map (the classical Peano curve is Holder with $d=1/2$). Hope this helps... :)[/quote]

Thank you very much. It seems that Holder's Condition was enough, and in the case I am researching, it is obvius that it holds, and actually, it is obvius that $d>m/n$.

Best,


I was wondering if anyone had the formula or knew about what you needed on amc and aime to qualify for usamo, so if anyone knows that would be really helpful

this has been discussed many many many many many times on this forum

the USAMO index is: your amc score + 10 times your AIME score.

if you are a 11/12 grader you have to meet the index, and if you are 9/10 or below, you can qualify for USAMO, if you score the floor value on the AIME, regardless of you AMC score.

Roughly what index value is needed to qualify?

changes every year, based on the level of difficulty of AIME, but last year the official website says the cutoff was 233 for AIME I, but i know somepoeple who qualified with a 220...so it depends...

[quote="eryaman"][quote="amirhtlusa"]but i know somepoeple who qualified with a 220...so it depends...[/quote]According to the AMC website, the cutoff for those that took the AIME II was a 220.5.[/quote]
talking about those who took the AIME I, and qualified with a <220 :)

I think its because they were the only person in the state.


Find all $(m, n) \in \mathbb{N}^{2}$ that satisfy

$n^2 = 10m^2 + 1$

[quote="Gyan"][hide]Let  $m = \frac{(3+\sqrt{10})^{2k} - (3-\sqrt{10})^{2k} } {2 \sqrt{10}}$
and
$n = \frac{(3+\sqrt{10})^{2k} + (3-\sqrt{10})^{2k} } 2$

Obviously $m$ is an  integer, $n$ is an integer and easy to see that
  $10m^2+1=n^2$

All you have to do is to plug in the value of k=0,1,2,3...
And you get m=0,6, 228 ...:) [/hide][/quote]
lol, where this numbers came from?  :?

From the Pell's equation $n^2 - 10m^2 = 1$.


calculus
integration
trigonometry
real analysis
real analysis unsolved

I have troubles with this integral:

\[ \int \sin (t/2)\tan (t/2)dt\]
Could you solve it?
thanks.

It is very easy integral. Move sint/2 under the differential and then express tant/2 in terms of cost/2. Then it may be solved by substitution maybe...

$ P \equal{} \int \sin(t/2) \tan(t/2)\,dt \equal{} \int \frac {\sin^2(t/2)}{\cos(t/2)}\,dt \equal{}$

$ \equal{} \int \frac {1 \minus{} \cos^2(t/2)}{\cos(t/2)}\,dt \equal{} \int \sec(t/2)\,dt \minus{} \int \cos(t/2)\,dt$.


I think it should be the moudlus of the value  :)

they just depuct it by the modulus value....
DO u not know that combustion is  aexothermic process....

yes i knwo it is an exothermic process but wanted to double check the def. :)


calculus
integration
limit
calculus computations

to determine all values of p for which improper integral converges
$ \int^{1}_{0}\frac {dx}{x^p}$

hello, i can't see no improper integral and we get
$ \int\frac{1}{x^p}\,dx\equal{}\minus{}\frac{x}{(p\minus{}1)x^p}\plus{}C$.
Sonnhard.

[quote="jmerry"]This question makes no sense as written. Integral from where to where?[/quote]
sorry, I forget the limitings.
but now it should be fine

Yes.. for $ p>1$ the indefinite integral exists. But your question was on improper right?? Where are the limits?? :?:

Why doesn't  the improper integral converge for $ P\le 1$?

liguangming's answer is wrong. It converges for $ p<1$, and diverges for $ p\ge 1$.

For $ \int_{0}^{1}\frac {dx}{x^p}$ the integral diverges for $ p\equal{}1$ the $ log(0)$=$ \minus{}\infty$.

For any $ p<1$ Considering $ p \in \mathbb{Z}$ the integral converges..Infact for $ p\equal{} \frac {1}{2}$ the integral doesn't exists in Riemann sense and is understood as

$ lim_{a \to 0^{\plus{}}} \int _{a}^{1} \frac {1}{\sqrt x}.dx\equal{}lim_{a \to 0^{\plus{}}} \left[ 2 \sqrt 1 \minus{}2 \sqrt a \right ]\equal{}2$ :)

[quote="jmerry"]liguangming's answer is wrong. It converges for $ p < 1$, and diverges for $ p\ge 1$.[/quote]
jmerry you are right, the answer should be $ p < 1$

[quote="DineshManohar"]Why so $ p < 15 ?$[/quote]
sorry, I always make some sort of mistake like this. It sure be $ p<1$.

hello, we get
$ \int_{0}^{1}\frac{1}{x^p}\,dx\equal{}\lim_{x\to 0^{\plus{}}}\frac{x^{\minus{}p\plus{}1}\minus{}1}{p\minus{}1}$
Sonnhard.


geometry
3D geometry
algebra
polynomial
number theory
modular arithmetic
superior algebra

I need help proving why the sum of 3 consecutive cubes is always divisible by 9? I've done the arithmetic but and see that it does happen but I don't know why?! [/hide]

Compute $ (n\minus{}1)^3 \plus{} n^3 \plus{} (n\plus{}1)^3$.

I did but I just get 3a^3+6a..am i doing something crazy wrong retarded??

What you've done so far is right but you're not done yet. Try to prove that $ a^3\plus{}2a$ is divisible by $ 3$.

ugh! i don't know how to prove it! i see it works but im lost!!

Polynomials $ f(a)$ have the property that $ f(a \plus{} 3)$ leaves the same remainder upon division by $ 3$ as $ f(a)$ does (why?).  It therefore suffices to test the values $ a \equal{} 0, 1, 2$.

Alternately, write $ 3a \plus{} (a^3 \minus{} a)$ and cite [url=http://en.wikipedia.org/wiki/Fermat%27s_little_theorem]Fermat's little theorem[/url] for $ p \equal{} 3$.  This is a little stronger than one needs; along the same lines, you could instead write

$ a^3 \plus{} 2a \equal{} 3 \left( a \plus{} {a \plus{} 1 \choose 3} \right)$,

where the term in parentheses is clearly an integer.

I'm stupid! That doesn't make any sense...sorry...

Never mind the other two comments; focus on the first one.  If $ f(a)$ is a polynomial with integer coefficients, then $ \frac{f(a \plus{} 3) \minus{} f(a)}{3}$ is an integer, so $ f(a \plus{} 3) \equiv f(a) \bmod 3$.  Does this part make sense?

Observe that:
$ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc \plus{} (a\plus{}b\plus{}c)[a(a\minus{}b) \plus{} b(b\minus{}c) \plus{} c(c\minus{}a)]$

when $ a \equal{} n, b \equal{} n \minus{}1, c \equal{} n\minus{}2$
We have:
1: $ [a(a\minus{}b) \plus{} b(b\minus{}c) \plus{} c(c\minus{}a)] \equal{} 3$ (double check it I may be wrong).

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

3: $ abc \equal{} n(n\minus{}1)(n\minus{}2) \equal{} 3k \plus{} r$, where $ r \equal{} 0, 1, 2$. Clearly the equation does hold only for $ r \equal{} 0$.

Putting it all together, you get the result you wanted.

In my opinion, this is the simplest way to see that $ a^3 \plus{} 2a$ is divisible by 3. (if the more abstract arguments are scary, anyway)

Divide the problem into three cases.  Case I: $ a$ is divisible by $ 3$.  Then obviously $ a^3 \plus{} 2a \equal{} a(a^2 \plus{} 2)$ is divisible by $ 3$.

Case II: $ a \equal{} 3k \plus{} 1$ for some integer $ k$.  Then $ a^2 \plus{} 2 \equal{} (3k \plus{} 1)^2 \plus{} 2 \equal{} 9k^2 \plus{} 6k \plus{} 3$ is divisible by $ 3$.

Case III: $ a \equal{} 3k \plus{} 2$ for some integer $ k$.  I'm sure that you can do this one just like case II. :)

Someone comfortable with modular arithmetic can skip all of this, of course, and just observe that the cubes of $ \{0,1,2\}$ (mod 3) are just $ \{0,1,8\}$ (mod 9).


integration
calculus
calculus computations

evaluate $ \sum_{m\equal{}1}^\infty\sum_{n\equal{}1}^\infty\;\dfrac{(m\minus{}1)!\cdot(n\minus{}1)!}{(m\plus{}n)!}$ :D


logarithms
integration
trigonometry
calculus
calculus computations

here is part $ \text{(i)}$ of a question i asked for HW  yesterday  :D

given

$ \psi_0\;(x) \equal{} \frac {d}{dx} \;\ln{\Gamma(x)}\; \equal{} \; \frac {d}{dx} \;\ln\left(\int_0^\infty t^{x \minus{} 1} e^{ \minus{} t}\;dt \right)$

show

$ \psi_0\;\left(\frac {1}{8}\right) \equal{} \minus{} \left( \gamma \plus{} 4\ln\; 2 \plus{} \pi\left( \frac {1}{2} \plus{} \frac {1}{\sqrt {2}}\right) \plus{} \frac {1}{\sqrt {2}} \;\left( \ln\;\frac { \sqrt {2} \plus{} 1} { \sqrt {2} \minus{} 1} \right) \right)$

$ m, n \in \mathbb{N} \ (m<n)$
$ \displaystyle{\psi_0\left( \frac{m}{n}\right) \equal{} \minus{} \gamma \minus{} \ln n \minus{} \frac{\pi}{2} \cot \left( \frac{m\pi}{n} \right) \plus{} \sum_{r\equal{}1}^{n\minus{}1} \cos \left( \frac{2rm \pi}{n} \right) \ln \left( 2 \sin \frac{r \pi}{n} \right) }$ :D


1. What are diastereomers? Select from the following the correct statement(s):

(a) Structural isomers.
(b) Nonsuperimposable non-mirror images.
(c) Geometrical isomers.
(d) Superimposable non-mirror images.
(e) Stereoisomers.
(f) Nonsuperimposable mirror images.

2. If two molecules are diastereomers then they will have different melting and boiling points, different NMR spectra, and differents heats of combustion, but they will have identical solubilities in water, densities, and acid strenghts. Is this true or false?

because there is no reason(may be   :maybe: ) for them to be so similr in properties

That's right: all those properties are, in principle, different. Just compare, for example, fumaric and maleic acids: the solubility in water of the second is more than 100 times greater than the first one.


Cindy was asked by her teacher to subtract $ 3$ from a certain number and then divide the result by $ 9$. Instead, she subtracted $ 9$ and then divided the result by $ 3$, giving an answer of $ 43$. What would her answer have been had she worked the problem correctly?

$ \textbf{(A)}\ 15 \qquad \textbf{(B)}\ 34 \qquad \textbf{(C)}\ 43 \qquad \textbf{(D)}\ 51 \qquad \textbf{(E)}\ 138$

$ 43\times 3 \plus{}9\equal{}138$ is her original number.  Doing it correctly would yield $ \frac{138\minus{}3}{9}\equal{}15$.

Let the original number be $x$.
$\dfrac {x -9}{3} = 43 \Rightarrow x = 43 \times 3 + 9 = 138$.
So the answer is $\dfrac {138 - 3}{9} = \boxed {15} \Rightarrow \boxed {A}$.

Let the original number be X
(X-9)/3=43
X=138

So the answer is (138-3)/9=15


calculus
integration
calculus computations

Evaluate $ \int_{ \minus{} 1}^1 \frac {|x|}{1 \plus{} e^x}\ dx$.

\[
\begin{array}{l}
 I = \int\limits_{ - 1}^1 {\frac{{\left| x \right|}}{{1 + e^x }}dx}  = \int\limits_{ - 1}^1 {\frac{{\left| x \right|}}{{1 + e^{ - x} }}dx}  = \int\limits_{ - 1}^1 {\frac{{\left| x \right|e^x }}{{1 + e^x }}dx}  \\
  \Rightarrow 2I = \int\limits_{ - 1}^1 {\frac{{\left| x \right|}}{{1 + e^x }}dx}  + \int\limits_{ - 1}^1 {\frac{{\left| x \right|e^x }}{{1 + e^x }}dx}  = \int\limits_{ - 1}^1 {\left| x \right|dx}  = 1 \\
  \Rightarrow I = \frac{1}{2} \\
 \end{array}
\]

Those solutions are the most typical solutions that Japanese High School students can easily understand, two guys. :lol:

Here is my solution:

Let $f(x)=\int_{-x}^{x} \frac{|t|}{1+e^t}dt$,

$f'(x)=\frac{|x|}{1+e^{x}}-\frac{|x|}{1+e^{-x}}\cdot (-1)=\left(\frac{1}{1+e^{x}}+\frac{1}{1+e^{-x}}\right)|x|$

$=\left(\frac{1}{1+e^{x}}+\frac{e^{x}}{e^{x}+1}\right)|x|=|x|$

$\therefore \int_0^1 f'(x)dx=\int_0^1 |x|dx$

$\Longleftrightarrow f(1)-f(0)=\frac 12\cdot 1\cdot 1$, thus $f(0)=0$, yielding

$\int_{-1}^1 \frac{|x|}{1+e^{x}}dx=f(1)=\frac 12.$


Prove that there exist infinitely many quadruples of positive integers $ (x,y,u,v)$ such that $ xy+1, xu+1, xv+1, yu+1, yv+1, uv+1$ are all perfect squares.

Hint: (But that may give the whole solution away) so peek at one hint at a time:
[hide="hint0"]
One such set is $ 1,3,8,120$ [/hide]

[hide="hint1"]Any two numbers whose difference is 2 will form a pair where ab+1 would be perfect square...
 Or IOW one can take two numbers $ n \pm 1$ to start ....[/hide]
[hide="hint2"]  add ... $ 4n$   and now  $ (n \pm 1 )$ also form such a pair (or three pairs :) [/hide]
[hide]Product of four consecutive integers +1 is a perfect square...

Next Hint gives away the answer, so do not peek unless you all ready have guessed the answer..

[hide]So if  the next number is $ 2.(2n)(2n-1)(2n+1)$ you are set...[/hide][/hide]

Now for a little more challenge:

Prove 1 -  Find   sets  of  positive integers $ a,b,c$ such that they form arithmetic   progression  and $ ab+1$, $ bc+1$ and $ ca+1$ all are perfect squares... prove there are infinite such  sets..

[hide](Example one such set is (1,8,15)) [/hide]

Prove 2 -  If $ (ab+1)(bc+1)(ca+1)$ is a perfect square, all three factors (that is $ (ab+1),(bc+1),(ca+1)$  are perfect square too.


invariant
vector
linear algebra
linear algebra unsolved

Let T be a linear operator on a vector space V, and let W be a T-invariant subspace of V. Define $ \overline{T}: V/W\rightarrow V/W$ by $ \overline{T}(v\plus{}W)\equal{}T(v)\plus{}W$ for any $ v\plus{}W\in V/W$.

Let $ \eta: V\rightarrow V/W$ be the linear transformation defined by $ \eta(v)\equal{}v\plus{}W$. Show that $ \eta T\equal{}\overline{T}\eta$.

Directly from definitions:
$ (\eta T) (v) \equal{} \eta (T(v)) \equal{} T(v) \plus{} W \equal{} \overline{T}(v \plus{} W) \equal{}\overline T(\eta(v)) \equal{} (\overline{T}\eta) (v)$ for every $ v\in V$.


In trapezoid $ ABCD$ we have $ \overline{AB}$ parallel to $ \overline{DC}$, $ E$ as the midpoint of $ \overline{BC}$, and $ F$ as the midpoint of $ \overline{DA}$. The area of $ ABEF$ is twice the area of $ FECD$. What is $ AB/DC$?

$ \textbf{(A)}\ 2\qquad
\textbf{(B)}\ 3\qquad
\textbf{(C)}\ 5\qquad
\textbf{(D)}\ 6\qquad
\textbf{(E)}\ 8$

[hide="Solution"]
Let $ a = DC$ and $ b = AB$ such that $ FE = \frac {a + b}2$. Let half the height of trapezoid $ ABCD$ be $ h$ such that we must have, by the area condition,

$ 2(\frac12h(a + \frac {a + b}2)) = \frac12h(b + \frac {a + b}2)\implies2a + (a + b) = \frac {a + b}2 + b$

$ \implies6a + 2b = (a + b) + 2b\implies5a = b\implies\frac {AB}{DC} = \frac ba = \boxed{\textbf{(C)}\ 5}$.
[/hide]

Edit:  2 stupid mistakes in one day.   :(

No, they're not similar. You're saying $ \frac aa\equal{}\frac{\frac{a\plus{}b}2}b$?

Let DC=x, EY=y, and AB=z. Both ABEF and DCEF have the same height, so we have $ y\plus{}z\equal{}2(x\plus{}y)$, or
$ z\equal{}2x\plus{}y$.
Also, since $ DC \parallel EF \parallel AB$ and EF is midway between AB and CD,
$ y\equal{}\frac{x\plus{}z}{2}$.
Substitution yields z=5x, so the answer is 5.
When I work on these problems, it helps me to have a lot of variables.

Sorry about the necro, but why can't we use similarity?

1. If $ \overline{AB}$ is parallel to $ \overline{DC}$, $ E$ is the midpoint of $ \overline{BC}$, and $ F$ is the midpoint of $ \overline{DA}$, then $ \overline{FE}$ is parallel to $ \overline{AB}$ and $ \overline{DC}$ (using theorems).

2. If $ \overline{FE}$ is parallel to $ \overline{AB}$ and $ \overline{DC}$, then $ ABCD$ and $ FECD$ are similar (all angles of both figures are congruent).

3. If $ ABEF$ is twice the area of $ FECD$, $ ABCD$ is three times the area of $ FECD$.

4. If $ ABCD$ is three times the area of $ FECD$, then the ratio of $ \overline{FE}$ to $ \overline{AB}$ is $ \frac {1}{\sqrt {3}}$.

5. $ \frac {AB \plus{} DC}{2} \equal{} FE$. That means that the length of $ \overline{DC}$ is $ 2 \minus{} \sqrt {3}$ if $ FE \equal{} 1$.

6. Therefore $ AB/DC \equal{} \frac {\sqrt {3}}{2 \minus{} \sqrt {3}} \equal{} \frac {\sqrt {3}(2 \plus{} \sqrt {3})}{(2 \minus{} \sqrt {3})(2 \plus{} \sqrt {3})} \equal{} 3 \plus{} 2\sqrt {3}$.

Now this seems correct according to all the theorems in my book, but obviously $ 3 \plus{} 2\sqrt {3}$ is not an answer choice. Where is the error?

[quote="12markkram34"]Sorry about the necro, but why can't we use similarity?
2. If $ \overline{FE}$ is parallel to $ \overline{AB}$ and $ \overline{DC}$, then $ ABCD$ and $ FECD$ are similar (all angles of both figures are congruent).[/quote]
Not only should the angles be congruent, the corresponding side lengths must be proportional, too.

Yes that's what I'm talking about. If the ratio of side lengths is $ a: b$, the ratio of the areas is $ a^{2}: b^{2}$.

Err, one side is the [b]same[/b] in both trapezoids, even if the others are different and of equal proportion.

That's hax :O! Angle congruence similarity only works in triangles?

Are all rectangles similar? Clearly not. But they have the same angles.

[hide="Wat lol"]
I called the area of ABFE $2x$ and the area of CEFD $x$. Then I just called CD=3 and tried random answer choices assuming that the altitude of ABFE is 2 and that altitude is equal to the altitude of CEFD. Is that a safe assumption? It got me to the right answer C...[/hide]

From NICE's Areas document.

[hide=Solution]Let $AB=a$ and $CD=b$. Then since $EF$ is the midsegment of the trapezoid, we have $EF=\frac{a+b}{2}$. The condition that $[ABEF]=2[FECD]$ implies $AB+EF=2(EF+CD)$, or $a+\frac{a+b}{2}=2\left(\frac{a+b}{2}+b\right)$. Expanding and simplifying gives $\frac{3a+b}{2}=a+3b$, or $3a+b=2a+6b$. This is equivalent to $a=5b$, for an answer of $5 \implies \boxed{\text{(C)}}$.[/hide]

I got this right in alcumus, but it is simple though. You just use variables


Given 40 feet of fencing, what is the greatest possible number of square feet in the area of a rectangular pen enclosed by the fencing?

It's $ 100$ because for the area to be the greatest, the side lengths have to be the closest to each other. So we divide 40 by 4, and we get 10, and it's a square so $ 10^2\equal{}100$

To prove that the greatest area possible is a square, we use AM-GM. Suppose the side lengths are $ a$ and $ b$. Given $ 2(a \plus{} b) \equal{} 40$, we have $ \frac{a \plus{} b}{2} \equal{} 10$. Using AM-GM, we have $ \frac{a \plus{} b}{2} \ge \sqrt{ab}$. Since $ \frac{a\plus{}b}{2} \equal{} 10$, we can make a substitution. $ 10 \ge \sqrt{ab}$. Squaring both sides gives $ 100 \ge ab$ and $ \boxed{100}$ is the maximum area. To show that it can be achieved, let $ a, b \equal{} 10$.

[hide="this is my answer"]1.when there is a problem like this we always want the #'s that are the closest to each other

2.and we are told the perimeter is 40

3.we want the #'s that adds up to this periemeter as close as possible so the answer is $\boxed{10}$[/hide]

 :spam:   :spam:   :spam:   :spam:


Find all values of $ x$ such that $ \frac{x\minus{}a}{b} \plus{} \frac{x\minus{}b}{a} \equal{} \frac{b}{x\minus{}a} \plus{} \frac{a}{x\minus{}b}$, where $ a$ and $ b$ are constants.

Let $ X \equal{} \frac {b}{x \minus{} a},\ Y \equal{} \frac {a}{x \minus{} b},$ the given equation can be written as $ X \plus{} Y \equal{} \frac {1}{X} \plus{} \frac {1}{Y}$

$ \therefore XY(X \plus{} Y) \equal{} X \plus{} Y\Longleftrightarrow (X \plus{} Y)(XY \minus{} 1) \equal{} 0.$

Case 1: $ X \plus{} Y \equal{} 0\Longleftrightarrow \frac {x \minus{} a}{b} \plus{} \frac {x \minus{} b}{a} \equal{} 0$

$ \Longleftrightarrow (a \plus{} b)x \equal{} a^2 \plus{} b^2$

If $ a \plus{} b\neq 0, \text{then}\ x \equal{} \frac {a^2 \plus{} b^2}{a \plus{} b}$

If $ a \plus{} b \equal{} 0$ and $ a^2 \plus{} b^2 \equal{} 0\Longleftrightarrow a \equal{} b \equal{} 0$, since $ ab\neq 0$ No Solution.

If $ a \plus{} b \equal{} 0$ and $ a^2 \plus{} b^2\neq 0$, then No solution.

Case 2: $ XY \equal{} 1\Longleftrightarrow \frac {x \minus{} a}{b}\cdot \frac {x \minus{} b}{a} \equal{} 1$, Solving this gives $ x \equal{} 0,\ a \plus{} b.$

My solution(easier than kunny :P ):
$ \frac {x \minus{} a}{b} \minus{} \frac {b}{x \minus{} a} \equal{} \frac {a}{x \minus{} b} \minus{} \frac {x \minus{} b}{a} \Leftrightarrow \frac {(x \minus{} a)^2 \minus{} b^2}{b(x \minus{} a)} \equal{} \frac {a^2 \minus{} (x \minus{} b)^2}{a(x \minus{} b)}\Longleftrightarrow \frac {(x \minus{} a \minus{} b)(x \minus{} a \plus{} b)}{b(x \minus{} a)} \minus{} \frac {(x \minus{} a \minus{} b)(x \minus{} b \plus{} a)}{a(x \minus{} b)} \equal{} 0$
So $ x \equal{} a \plus{} b$,or:
$ \frac {(x \minus{} a \plus{} b)}{b(x \minus{} a)} \minus{} \frac {(x \minus{} b \plus{} a)}{a(x \minus{} b)} \equal{} 0$
Which after expanding gives:
$ ax^2 \minus{} a^2x \plus{} bx^2 \minus{} b^2x \equal{} 0\Longleftrightarrow x[x(a \plus{} b) \minus{} (a^2 \plus{} b^2)] \equal{} 0$
So $ x \equal{} 0$ or
$ x(a \plus{} b) \equal{} a^2 \plus{} b^2$
See kunnys solution now.....

Another Solution: $ y \equal{} x \minus{} \frac {1}{x}\ (x\neq 0)$ is just an increasing function. :lol:

$ \frac{1}{2}\left(X \minus{} \frac {1}{X} \plus{} Y \minus{} \frac {1}{Y}\right)\equal{}0$, or mid pont.

Thanks kunny, that was a good way to do it. Now if only I can understand how the solution manual did it.  :maybe:

[quote="kunny"]Another Solution: $ y \equal{} x \minus{} \frac {1}{x}\ (x\neq 0)$ is just an increasing function. :lol:

$ \frac {1}{2}\left(X \minus{} \frac {1}{X} \plus{} Y \minus{} \frac {1}{Y}\right) \equal{} 0$, or mid pont.[/quote]

But $ f(x)\equal{}x \minus{} \frac {1}{x}$ has a discontinuity at $ x\equal{}0$ so it is increasing on $ (\minus{}\infty,0)$ and $ (0,\infty)$ but not on $ \mathbb{R}$.

Furthermore, how do you get the $ XY\equal{}1$ solution then?

Yet another solution

$ \frac {x \minus{} a}{b} \plus{} \frac {x \minus{} b}{a} \equal{} \frac {b}{x \minus{} a} \plus{} \frac {a}{x \minus{} b}$

$ \frac {(a)(x \minus{} a) \plus{} (b)(x \minus{} b)}{ab} \equal{} \frac {(a)(x \minus{} a) \plus{} (b)(x \minus{} b)}{(x \minus{} a)(x \minus{} b)}$

Note that the numerators are the same, therefore
$ ab \equal{} (x \minus{} a)(x \minus{} b)$

$ ab \equal{} x^2 \minus{} ax \minus{} bx \plus{} ab$

$ 0 \equal{} x^2 \minus{} x(a \plus{} b)$

$ 0 \equal{} x(x \minus{} (a \plus{} b))$

$ x \equal{} 0, a \plus{} b$


Hi im new to this AoPS forum community. I guess i label myself as a sufficient math student...wanting to get better at it. I wanted to know if you guys know any good books(textbooks, problem books etc) that will help on the AMC, AIME, and those competitions. I have lots of problems in probability and i was wondering if y'all had any suggestions...

Thanks,

lifeisgood =)

Go to resources on the left hand side of this website. IT should give you a lot of information.

My favorite way to get better is to do old tests and look at the solutions if I'm stuck.  You might also consider taking an AoPS class (also can be found on the left navigation bar thingy).  Mainly, do lots of problems.

Just wondering, does a question like this belong in this part of the AoPS forum? I suspect it doesn't.

Personally, I would have put it in the AMC forum (since his questino was directed toward AMC competitions) or the other problem solving topics.


Here's a simple proposition about an equilic quadrilateral $ ABCD$ where $ AD=BC$ and $ \angle ADC+\angle BCD = 120^\circ$.

$ P$, $ M$ and $ L$ are midpoints of $ AC$, $ CD$ & $ DB$ respectively. Show that $ \triangle PML$ is equilateral.

Let $ DA\cap BC=E$,since $ \angle ADC+\angle BCD=120\Rightarrow\angle CED=60$.Since $ AD=BC$,$ PM=\frac{AD}{2},PM\parallel AD$,and$ ML=\frac{BC}{2},ML\parallel BC\Rightarrow ML=MP,\angle PML=\angle CED=60$.


um....this topic is kinda pointless....you could have just pmed him.

Yes. In the future if you want to suggest something to someone personally, please PM them. Or else actually find a place where he did not use the hide tag and post in that thread. Instead of posting it in a forum for puzzles and brainteasers...

...Moderator?

Well, I requested to be a moderator here...

so did I :lol:
lets just not post  in this topic again.


inequalities
trigonometry
inequalities unsolved

In an acute angled triangle ABC proof that
     SinA + SinB + SinC >2

besides jordan's inequality (which states that $ \sin \alpha\geq \dfrac{2\alpha}{\pi}$ for $ 0 < \alpha < \dfrac{\pi}{2}$) you can try also this:

$ \sum \sin A>\sum \sin^2 A \equal{} 2 \plus{} 2\cos A\cos B\cos C > 2$  :D


How would I prove that pi is irrational?


trigonometry
calculus
calculus computations

In general, $ e^{z} \equal{} \sum_{k \equal{} 0}^{\infty}\frac {z^{k}}{k!}$ or $ e^{r(cos\theta \plus{} i\sin\theta)} \equal{} \sum_{k \equal{} 0}^{\infty}r^{k}\frac {(\cos k\theta \plus{} i\sin k\theta)}{k!}$

Hence, $ e^{r\cos\theta}\cos(r\sin\theta) \equal{} \sum_{k \equal{} 0}^{\infty}r^{k}\frac {\cos k\theta}{k!}$ and $ e^{r\cos\theta}\sin(r\sin\theta) \equal{} \sum_{k \equal{} 0}^{\infty}r^{k}\frac {\sin k\theta}{k!}$

$ \therefore\sum_{n \equal{} 1}^{\infty}\frac {\sin(\sqrt {2}n)}{n!} \equal{} \sum_{n \equal{} 0}^{\infty}\frac {\sin(\sqrt {2}n)}{n!} \equal{} e^{\cos\sqrt {2}}\sin(\sin\sqrt {2})$ :lol:

You are right, I thought it takes a little more time of mathlinkers to solve it :D :(

You forgot $ r$:

$ e^{r\cos\theta}\sin(r\sin\theta) \equal{} \sum_{k \equal{} 0}^{\infty}r^{k}\frac {\sin k\theta}{k!}$ ;) the same in $ \cos$.


probability
geometry
geometric transformation
AMC
AIME

Two three-letter strings, $aaa$ and $bbb$, are transmitted electronically. Each string is sent letter by letter. Due to faulty equipment, each of the six letters has a 1/3 chance of being received incorrectly, as an $a$ when it should have been a $b$, or as a $b$ when it should be an $a$. However, whether a given letter is received correctly or incorrectly is independent of the reception of any other letter. Let $S_a$ be the three-letter string received when $aaa$ is transmitted and let $S_b$ be the three-letter string received when $bbb$ is transmitted. Let $p$ be the probability that $S_a$ comes before $S_b$ in alphabetical order. When $p$ is written as a fraction in lowest terms, what is its numerator?

[hide]If the string bbb becomes aaa in translation, there's no way the aaa string can be before the bbb string in alphabetical order.  Therefore, it's chances won't be considered.
If the string bbb becomes any sequence with 2 As in it, the chances that it happens is $(1/3)^2(2/3)=2/27$.  If bbb turns into aab, only aaa can be before it.  If bbb turns into aba, only aaa and aab are before it.  If bbb turns into baa, then aaa, aab, aba, and abb are before it.  So $(2/27)(8/27+8/27+4/27+8/27+4/27+4/27+2/27)=2/27*38/27=76/729$.
If the string bbb becomes any sequence with 1 A in it, the chances that it happens is $(2/3)^2(1/3)=4/27$.  If bbb turns into abb, then aaa, aab, and aba come before it.  If bbb turns into bab, then aaa, aab, aba, abb, baa all come before it.  If bbb turns into bba, then aaa, aab, aba, abb, baa, and bab all come before it.  $(4/27)(8/27+2[4/27]+8/27+3[4/27]+2/27+8/27+3[4/27]+2[2/27])=4/27*62/27=248/729$
If string bbb becomes any sequence with 0 As in it, the chances that it happens is $(2/3)^3=8/27$.  The chances that aaa's transformations will be ahead in ABC order than bbb is $1-(1/3)^3$.  That's $26/27*8/27=208/729$.
Adding the three, $532/729$; so the answer is 532.[/hide]

I don't think you need all that casework (I know I would forget something)

[hide]Let S(a,n) mean the nth letter of string S(a)
Compare the first letter of the string S(a) to the first letter of the string S(b)
There is a (2/3)^2=4/9 chance that S(a,1) comes before S(b,1)
There is a (1/3)(2/3)2=4/9 that S(a,1) is the same as S(b,1)

If S(a,1)=S(b,1), then you do the same for the second letters of the strings.  But you have to multiply the 4/9 chance that S(a,2) comes before S(b,2) as there is a 4/9 chance we will get to this step.

Similarly, if S(a,2)=S(b,2), then there is a (4/9)^3 chance that we will get to comparing the third letters and that S(a) comes before S(b).

So we have (4/9)+(4/9)^2+(4/9)^3=4/9+16/81+64/729=532/729[/hide]

It is an AIME problem I believe

[url]http://www.artofproblemsolving.com/Wiki/index.php/1991_AIME_Problems/Problem_10[/url]

can someone elaborate on solution 1 of the problem?

The first solution is just casework; they just didn't show all of their calculations. Each entry of the box counted the number of ways a certain string could occur.

Also, you should be more specific with your question. I don't know how many times AwesomeToad has told you this but it kinda gets on my nerves.

Sorry,

Also I was referring to solution 1 on the page of the actual problem, not this one.

http://www.artofproblemsolving.com/Wiki/index.php/1991_AIME_Problems/Problem_10

[hide="Slick Solution I Found"]

Let $P_i$ denote the probability that when we have $i$ more moves to be made that $S_a$ will be before $S_b$ in the alphabet.  We then have a probability of $\frac{4}{9}$ that $a\to a, b\to b$.  We also have a $\frac{4}{9}$ probability that $a\to b, b\to b$ or $a\to a, b\to a$ indicating we would go on to the next state.  Therefore \[S_i=\frac{4}{9}+\frac{4}{9}S_{i-1}\]
Also, $S_1=\frac{4}{9}$ (since $S_0=0$ therefore we have
$\begin{cases}  S_3=\frac{4}{9}+\frac{4}{9}S_2 \\ S_2=\frac{4}{9}+\frac{4}{9}S_1 \\ S_1=\frac{4}{9} \end{cases}$
Solving gives $S_2=\frac{52}{81}, S_3=\frac{532}{729}$.  My method also helps to generalize for if there are $4$ letters or more (but then the computations would be terrible). [/hide]


Hey guys,

I needed some help with my homework and the first place i thought was this.

Please dont give me the answer, but give me help that brings me to the answer.

Horizon offers 700 minutes for $ 45.99 a month, additional minutes are 6 cents each.
Stingular offers 700 minutes for$29.99 a month, additional minutes are 35 cents each.

1.For each plan write an expression for the total cost per month in terms of the number of additional numbers used.

2. For how many total minutes of calls per month is Horizon's plan a better deal?

please help me with this question dont solve it.

Hey,

A good place to start would be to do some examples.

Working out the following should start you on the road to the answer:

How much does 100 minutes in a month cost for the Horizon plan?  How much does 100 minutes in a month cost for the Stingular plan?

How much does 1000 minutes in a month cost in each plan?

If you write out your work (rather than do it in your head), you should see the pattern arise.

[quote="uanaka"]H

Horizon offers 700 minutes for $ 45.99$ a month, additional minutes are 6 cents each. Stingular offers 700 minutes for$ 29.99$ a month, additional minutes are 35 cents each.

1.For each plan write an expression for the total cost per month in terms of the number of additional numbers used.

2. For how many total minutes of calls per month is Horizon's plan a better deal?[/quote]

Hints:

For Horizon, 45.99 is a fixed price.  .06 will change depending on the number of minutes, which you can call $ x$.  Same for Stingular.  Set equations equal and that should help.

[hide="Answers:"]

[hide="1"]

Horizon has a fixed constant of 45.99, and each extra minute is 6 cents.  Our equation is:

$ 49.99 \plus{} .06(x \minus{} 700)$ where $ x$ is the number of minutes.  Do you see why I had $ x \minus{} 100$?  It is becaue the problem asks for each [b]additional[/b] minute.

Stingular's equation is then:

$ 29.99 \plus{} .35(x \minus{} 700)$[/hide]

[hide="2"]

Set them equal and you'll get the price where they are the same for $ x$ number of minutes:

$ 49.99 \plus{} .06(x \minus{} 700) \equal{} 29.99 \plus{} 0.35(x \minus{} 700) \implies 49.99 \plus{} .06x \minus{} 42 \equal{} 29.99 .35x \minus{} 245 \implies 7.99 \plus{} .06x \equal{} .35x \minus{} 215.01 \implies 223 \equal{} .29x \implies x \equal{} 768.97$

Thus Horizon is better for $ > 768$ minutes of calls per month.

[/hide]
[/hide]

Could someone post a hint on #2? Also how would finding how much 100 minutes for each plan help?

[quote="uanaka"]If i write the expression as an inequality, would it be like this

Horizon

45.99+.06x(>or equal to) 700 minutes
29.99+.35x(>or equal to) 700 minutes[/quote]

that is wrong.  Remember the price that you already pay has 700 minutes already included, so you want to know how many extra minutes you have.  that would be simply $ x \minus{} 700$.  See my other post for the correct equations.

#2 Hint:

Use [i]the right[/i] equations and set them equal.  OR graph them.[/i]

[hide="#2 Hint"]Notice how at the end of one month, Singular's plan costs a lot less than Horizon's as long as you stay under or equal to $ 700$ minutes worth of minutes.  However, as you keep on using more and more minutes, Singular's $ 35$ cents is slowly reaching Horizon's total, which only adds $ 6$ cents to each extra minute.  So, try to make an inequality that allows you to represent this.  Ask yourself: At what point does this extra $ 29$ cents per minute end up making Horizon a better plan?[/hide]

So would the inequality be like

25.99+x(> or equal to) 700??

Your questions asked for the EXPRESSIONS not inequality.  IN the expression, it won't be equal to anything (well, you could say $ y$)

So Horizon's equation is x= Additional minute

x+45.99??

Remember each additional minute costs $ 0.06$ dollars.  Also, remember that you already have $ 700$ minutes, so you want to know how many minutes over $ 700$, which is $ x \minus{} 700$.  The equation is then $ 49.99 \plus{} 0.06(x \minus{} 700)$


Why is it impossible to divide by 0? If any number divided by itself is 1, why not zero? :huuh:

I don't know if this question is mathcounts worthy, but I'll say something anyway. Let's set up this equation. $\frac{x}{0}=y$ where x does not equal 0 and y can be anything. If we multiply both sides by 0, we get $x=(0)(y)=0$ so the only thing x could be is 0. But if we had $\frac{0}{0}$, then could it be 1 since we are dividing the same number by itself? Or would it be 0 since 0 divided by anything is 0? Therefore, we proved first of all that all numbers (except 0) divided by 0 is impossible because then we would have something like 5=0. :?: Then in the 2nd part I proved that 0/0 is also undefined because we have 2 possible answers. As you would have it, any number divided by 0 is [b]undefined[/b].

my teacher explained this

cross products!

notice how $\frac{0}{2}=\frac{2}{x}$ is simplified to
$2x=0$, in cross products, so $x$ must be $0$

now for dividing by $0$

look at $\frac{2}{0}=\frac{x}{4}$ is the same as
$8=0x$ using cross products, which is impossible, thus it is called undefined

-jorian

Hm well either your teacher is confused or you are confused. Take your first example
\[\frac{0}{2}= \frac{2}{x}\]
In fact this is the same thing as $0*x=4$ which has no solutions either.

The reason that $0$ is undefined is that it cannot be defined without contradicting many rules of arithmetic, as kstan013 points out. However,
\[\lim_{x \rightarrow 0}\frac{1}{x}= \infty \]
\[\lim_{x \rightarrow \infty}\frac{1}{x}= 0 \]
So there are ways of interpreting $\frac{k}{0}$. What's more tricky is $\frac{0}{0}$.

There's already a rather animated debate going on in the Classroom Math forum. Topic name: "Infinete" [sic].

Locked until further notice, this has been discussed MANY times.


Find $ x,y$ in terms of $ a,b$ from the equation
$ \frac{a^{2}}{x^{2}}\minus{}\frac{b^{2}}{y^{2}}\equal{}8(y^{4}\minus{}x^{4})$
$ ax\minus{}by\equal{}x^{4}\minus{}y^{4}$.

After substituting $ y\equal{}tx$ we get the algebraic equation of 6 order. I'm not sure there is a nice way to solve it..

[hide]in first equation join the fractions. Then replace with the second ecuation ax-by. And exclude $ x^{4}\minus{}y^{4}$. Then add two resulting equations.[/hide]

[quote="borislav_mirchev"]in first equation join the fractions. Then replace with the second ecuation ax-by. And exclude $ x^{4} \minus{} y^{4}$. Then add two resulting equations.[/quote]
Have you done it accurately? May be i misunderstood, but it does not give any simplification.

My idea was to regard this as a system in $ a,b$, but nothing remotely useful came out of that.

First equation is: $ \frac {(ax \minus{} by)(ax \plus{} by)}{x^{2}y^{2}} \equal{} 8(x^{4} \minus{} y^{4})$
Substitute in first equation ax-by from the second. $ (x^{4} \minus{} y^{4})$ may be eliminated if you are careful enough to examine $ |x| \equal{} |y|$.
$ ax \plus{} by \equal{} \frac {8}{x^{2}y^{2}}$
$ ax \minus{} by \equal{} x^{4} \minus{} y^{4}$
Then multiply these equations and took as homogeneous equation. Is it OK?

Of course it is not ok. The 1st equation is $ \frac {(ay \minus{} bx)(ay \plus{} bx)}{x^2y^2} \equal{} 8(x^4 \minus{} y^4)$.
And even if we change the 1st equation to be $ \frac{a^2}{y^2}\minus{}\frac{b^2}{x^2}\equal{}8(y^4\minus{}x^4)$, i dont see how your method works.


How many three-digit numbers are there such that no two digits differ by more than 3.

[hide]262[/hide]
I think. I didn't check it.

Dingzhou, why do you have to torture us with counting questions? It makes me feel like a preschooler. :(

counting is a very important part of life....why else would mathcounts give you dull pointless problems of counting?

who invented counting???
and to keep this "on topic"...

Is my answer correct or not?

...lol--so close to being able to rate others' posts...
EDIT: Finally!!!

lol why are you trying to get higher post count so that u can rate polls? no stealing ideas(i thought of that first) :P

Ok then. [b][i][u]Proooooooooooove it.[/u][/i][/b]jkjk lol

@math154
stop trying to raise your post count!!!!!!!!!!!
feel the wrath of the "emoticons"!!!!!

 :spam:  :bomb:  :spam:

I'm not trying to get my post count up. I'm trying to get it down, with no success.

你是个...(____)

Wow. I can't believe the Chinese characters worked.

[hide]
Okay, if the first digit is 9 (0 not possible), then we have 4*4 possibilities.
If the first digit is 1 or 8, then we have 5*5 possibilities.
If the first digit is 2 or 7, then we have 6*6 possibilities.
And if the first digit is 3, 4, 5, or 6, we have 7*7 possibilities.

Answer $ \fbox{334}$
[/hide]

@math154: post full solutions instead of simply answers.

@math154 and mathking123: don't make spammy comments.

i thought of the method that distracted used, but i dont think that it works completely.

the problem is that you are counting numbers such as 537 as "good", where in fact it is "bad" in the sense that the 3 and the 7 differ by more than 3. the main problem is the third number is dependent on not only the first digit, as in your "solution", but it is also dependent on the second digit.

that's the beauty of counting.... :rotfl:  :rotfl:  :rotfl:

Meh, I was sleepy.
Okay...revised version.
[hide]
If the first digit is 9 (0 not possible), then we have 4 possibilities for the second. (6, 7, 8, 9)
16 possiblities

If 1 or 8, then we have 5. (0, 1, 2, 3, 4) or (5, 6, 7, 8, 9)
25-2=23 possibities each

If 2 or 7, then 6. (0, 1, 2, 3, 4, 5) or (4, 5, 6, 7, 8, 9)
36-6=30 possibilities each.

If 3, 4, 5, or 6, then 7. (0 to 6) , (1 to 7), (2 to 8), or (3 to 9)

49-12=35 possibilities each.

Answer $ \fbox{262}$

[/hide]

OMG!! That means I actually got it right the first time I did it!


I just randomly picked 40 users. You can still play, even if you aren't on this list.

Wickedestjr: 31
alingaling: 29
Yongyi781: 30
hello.world: 30
12markkram34: 30
Canton: 30
lotsofmath: 30
asethi01: 30
Havingfitz: 30
isabella2296: 30
CountdownKing: 30
hurdler: 30
ernie: 30
chenhsi: 30
budi713: 30
dragon96: 30
james41: 30
pythag011: 30
eshen: 30
erabel: 30
Brut3Forc3: 30
stevenmeow: 30
ProtestanT: 30
TeddyRocks: 30
eggylv999: 30
Bacteria: 30
dnkywin: 30
Xantos C. Guin: 30
Ketvoman: 30
tinytim: 30
InhibitorOfNostalgia: 30
JBL: 30
rrusczyk: 30
BOGTRO: 30
mewto55555: 30
Urc: 30
superllamel: 30
MCrawford: 30
1=2: 30
#H34N1: 30
michaelwong256: 30

Wickedestjr: 29
alingaling: 29
Yongyi781: 30
hello.world: 30
12markkram34: 30
Canton: 30
lotsofmath: 30
asethi01: 30
Havingfitz: 30
isabella2296: 30
CountdownKing: 30
hurdler: 30
ernie: 30
chenhsi: 30
budi713: 30
dragon96: 30
james41: 30
pythag011: 30
eshen: 30
erabel: 30
Brut3Forc3: 30
stevenmeow: 30
ProtestanT: 30
TeddyRocks: 30
eggylv999: 30
Bacteria: 30
dnkywin: 30
Xantos C. Guin: 30
Ketvoman: 30
tinytim: 30
InhibitorOfNostalgia: 30
JBL: 30
rrusczyk: 30
BOGTRO: 30
mewto55555: 30
Urc: 30
superllamel: 30
MCrawford: 30
1=2: 30
#H34N1: 30
michaelwong256: 30

Wickedestjr: 30
alingaling: 29
Yongyi781: 30
hello.world: 30
12markkram34: 30
Canton: 30
lotsofmath: 30
asethi01: 30
Havingfitz: 30
isabella2296: 30
CountdownKing: 30
hurdler: 30
ernie: 30
chenhsi: 30
budi713: 30
dragon96: 30
james41: 30
pythag011: 30
eshen: 30
erabel: 30
Brut3Forc3: 30
stevenmeow: 30
ProtestanT: 30
TeddyRocks: 30
eggylv999: 28
Bacteria: 30
dnkywin: 30
Xantos C. Guin: 30
Ketvoman: 30
tinytim: 30
InhibitorOfNostalgia: 30
JBL: 30
rrusczyk: 30
BOGTRO: 30
mewto55555: 30
Urc: 30
superllamel: 30
MCrawford: 30
1=2: 30
#H34N1: 30
michaelwong256: 30

take away from others, other than me alingaling.

Wickedestjr: 30
alingaling: 30
Yongyi781: 30
hello.world: 30
12markkram34: 30
Canton: 30
lotsofmath: 30
asethi01: 30
Havingfitz: 30
isabella2296: 30
CountdownKing: 30
hurdler: 30
ernie: 30
chenhsi: 30
budi713: 30
dragon96: 30
james41: 30
pythag011: 30
eshen: 30
erabel: 30
Brut3Forc3: 30
stevenmeow: 30
ProtestanT: 30
TeddyRocks: 30
eggylv999: 28
Bacteria: 30
dnkywin: 30
Xantos C. Guin: 30
Ketvoman: 30
tinytim: 30
InhibitorOfNostalgia: 30
JBL: 30
rrusczyk: 30
BOGTRO: 30
mewto55555: 30
Urc: 30
superllamel: 30
MCrawford: 30
1=2: 30
#H34N1: 30
michaelwong256: 28

I took away from michaelwong256 because he was the last name on this list.

Wickedestjr: 31
alingaling: 30
Yongyi781: 30
hello.world: 30
12markkram34: 30
Canton: 30
lotsofmath: 30
asethi01: 30
Havingfitz: 30
isabella2296: 30
CountdownKing: 30
hurdler: 30
ernie: 30
chenhsi: 30
budi713: 30
dragon96: 30
james41: 30
pythag011: 30
eshen: 30
erabel: 30
Brut3Forc3: 30
stevenmeow: 30
ProtestanT: 30
TeddyRocks: 30
eggylv999: 28
Bacteria: 30
dnkywin: 30
Xantos C. Guin: 30
Ketvoman: 28
tinytim: 30
InhibitorOfNostalgia: 30
JBL: 30
rrusczyk: 30
BOGTRO: 30
mewto55555: 30
Urc: 30
superllamel: 30
MCrawford: 30
1=2: 30
#H34N1: 30
michaelwong256: 28

took away from Ketvoman, because I barely know him.

Wickedestjr: 31
alingaling: 30
Yongyi781: 30
hello.world: 30
12markkram34: 30
Canton: 30
lotsofmath: 30
asethi01: 30
Havingfitz: 30
isabella2296: 30
CountdownKing: 30
hurdler: 30
ernie: 30
chenhsi: 30
budi713: 30
dragon96: 30
james41: 30
pythag011: 31
eshen: 30
erabel: 30
Brut3Forc3: 30
stevenmeow: 30
ProtestanT: 30
TeddyRocks: 30
eggylv999: 28
Bacteria: 30
dnkywin: 30
Xantos C. Guin: 30
Ketvoman: 28
tinytim: 30
InhibitorOfNostalgia: 28
JBL: 30
rrusczyk: 30
BOGTRO: 30
mewto55555: 30
Urc: 30
superllamel: 30
MCrawford: 30
1=2: 30
#H34N1: 30
michaelwong256: 28

took away from IhibtorOfNOstalgia and increased myself.

Wickedestjr: 31
alingaling: 30
Yongyi781: 30
hello.world: 30
12markkram34: 30
Canton: 30
lotsofmath: 30
asethi01: 30
Havingfitz: 30
isabella2296: 30
CountdownKing: 30
hurdler: 30
ernie: 30
chenhsi: 31 GO MAFIASCUM!!!
budi713: 30
dragon96: 30
james41: 30
pythag011: 31
eshen: 30
erabel: 30
Brut3Forc3: 30
stevenmeow: 30
ProtestanT: 30
TeddyRocks: 30
eggylv999: 28
Bacteria: 29 infectious
dnkywin: 30
Xantos C. Guin: 30
Ketvoman: 28
tinytim: 30
InhibitorOfNostalgia: 28
JBL: 30
rrusczyk: 30
BOGTRO: 30
mewto55555: 30
Urc: 30
superllamel: 30
MCrawford: 30
1=2: 30
#H34N1: 30
michaelwong256: 28

[size=200]guys for now on hurt=-2 and heal=+1[/size]

Wickedestjr: 32
alingaling: 30
Yongyi781: 30
hello.world: 30
12markkram34: 30
Canton: 30
lotsofmath: 30
asethi01: 30
Havingfitz: 30
isabella2296: 30
CountdownKing: 30
hurdler: 30
ernie: 30
chenhsi: 29
budi713: 30
dragon96: 30
james41: 30
pythag011: 31
eshen: 30
erabel: 30
Brut3Forc3: 30
stevenmeow: 30
ProtestanT: 30
TeddyRocks: 30
eggylv999: 28
Bacteria: 29 infectious
dnkywin: 30
Xantos C. Guin: 30
Ketvoman: 28
tinytim: 30
InhibitorOfNostalgia: 28
JBL: 30
rrusczyk: 30
BOGTRO: 30
mewto55555: 30
Urc: 30
superllamel: 30
MCrawford: 30
1=2: 30
#H34N1: 30
michaelwong256: 28

This topic's locked because Wickedestjr and alingaling are the same account.

Wickedestjr, stop creating new accounts and impersonating them or else I'll take this to the admins and have your accounts banned.

Plus, this thread is really annoying.


We call a finite string of the symbols X and O balanced
iff every substring of consecutive symbols has a difference of at
most 2 between the number of Xs and the number of Os
 Find the number of balanced strings of length n.

The solution at Kalva website is not nice. I want to find the better
solution.

herer's my solution
 Let b_n denote the number of balanced strings of length n. We
establish the recursion b_(n+2) = 2b_n + 2 for n  \geq  1. This recursion and the initial
conditions b_1 = 2, b_2 = 4 yield the  result by induction.(try it!!!)
Let x_n, y_n, z_n denote the number of balanced strings of length n ending with XX,the number that end in XO and whose last doubled letter is X, and the number that end in XO and whose last doubled letter is O, respectively. (To keep things consistent , we count a purely alternating string as if its last doubled letter were both X and O.) The counts remain the same if X and O are interchanged. Thus bn = 2x_n+2y_n+2z_n−2
for n  \geq  1 (we subtract 2 to avoid double-counting the purely alternating strings).
If a balanced string of length n+2 ends in XX, the remainder must end in OO, or end in XO and have last doubled letter O. Thus
 x_(n+2) = x_n +z_n. If a balanced string of length n + 2 ends in XO and has last doubled letter X, the remainder (the ﬁrst n symbols) must end in XO and have last doubled letter X, or end in OX and have last doubled letter O; conversely, appending XX to the end of such a balanced string
results in a balanced string of length n + 2. Thus y_(n+2) = 2y_n. Finally, if a balanced string of length n + 2 ends in XO and has last doubled letter O, the remainder must end in OO, or end in XO and have last doubled letter O. Thus z_(n+2) = x_n + z_n as
well. Putting this together,
b_(n+2) = 2x_(n+2) + 2y_(n+2) + 2z_(n+2) − 2
= 2(x_n + z_n) + 2(2y_n) + 2(x_n + z_n) − 2
= 2(b_n + 2) − 2,
the desired recursion.
now solve it and get the answers
i hope i've made things pretty clear (atleast i've tried hard to do that :D )
best regards

nice sol!!
But you have a mistake
I think you must consider n is odd and n is even.
When n is even then bn=2x_n+2y_n+2z_n-2
When n is odd  then bn=2x_n+2y_n+2z_n

Ex : n=3 then bn=6 , xn=1,yn=1,zn=1.

when n is old , y_(n+2)=2yn+1.
Is it right?


calculus integration logarithms trigonometry function limit algebra

So you could just put $ k \equal{} \cos(\theta)$, and solve it that way?

or maybe its a hint to rewrite the bottom as

$ x^2\plus{}2x\cos \theta \plus{}1 \equal{} x^2 \plus{}2x\cos \theta \plus{} \cos^2 \theta \plus{}\sin^2 \theta\equal{}(x\plus{} \cos \theta )^2 \plus{} \sin^2 \theta$ !!

but i dont see why that would be of any use.

Fairly late, but now I think it's time to give an answer.

$ \int_{0}^{\infty} \frac {\log(1 + x^2)}{x^2 + 2x\cos\theta + 1} \, dx = \frac {1}{\sin \theta} [ \frac {\pi}{4} \log ( \frac {1 + \sin \theta}{1 - \sin \theta} ) + \theta \log |2 \cos\theta| + \frac {\mathrm{Li}_2 ( - e^{2i\theta}) - \mathrm{Li}_2 ( - e^{ - 2i\theta})}{4i} ]$

[hide="Solution"]Let $ I(\alpha) = \int_{0}^{\infty} \frac {\log(\alpha^2 + x^2)}{x^2 + 2x\cos\theta + 1} \, dx$.

Then it is easy to show that $ I(0) = 0$, and what we want to calculate is now the value of $ I(1)$.

Differentiating $ I(\alpha)$, we have $ I'(\alpha) = 2\alpha \int_{0}^{\infty} \frac {1}{(x^2 + 2x\cos\theta + 1)(\alpha^2 + x^2)} \, dx$.

Here, integrand is rational function, so we can evaluate (though somewhat cumbersome) the value of this integral, giving

$ I'(\alpha) = \frac {2\alpha}{\alpha^4 + 2\alpha^2 \cos 2\theta + 1} [ \frac {\theta}{\sin\theta} (\alpha^2 + \cos 2\theta) + \cos \theta \log \alpha^2 + \frac {\pi}{2} ( \frac {1}{\alpha} - \alpha ) ]$.

Thus integrating both side w.r.t. $ \alpha$ from 0 to 1 and partially applying the substitution $ x = \alpha^2$, we have

$ (1) \quad I(1) = \int_{0}^{1} \frac {1}{x^2 + 2x \cos 2\theta + 1} [ \frac {\theta}{\sin\theta} (x + \cos 2\theta)$ $ + \cos \theta \log x ] \, dx + \int_{0}^{1} \frac {\pi (1 - \alpha^2)}{\alpha^4 + 2\alpha^2 \cos 2\theta + 1} \, d\alpha$

Here, the last term of RHS is again integral of a rational function. So we can evaluate it. Or you can use the taylor expansion

$ (2) \quad \frac {\cos \theta}{x^2 + 2x \cos 2\theta + 1} = \frac {1}{2\sin\theta} \sum_{n = 1}^{\infty} ( - 1)^{n - 1}x^{n - 1} \sin (2n \theta)$

to obtain

$ (3) \quad \int_{0}^{1} \frac {\pi (1 - \alpha^2)}{\alpha^4 + 2\alpha^2 \cos 2\theta + 1} \, d\alpha = \frac {\pi}{2} \frac {\tanh^{ - 1} (\sin \theta)}{\sin \theta}$

Now, separating the leading integral of RHS into two integrals, we immediately find that the former one is easily calculated. That is,

$ (4) \quad \frac {\theta}{\sin\theta} \int_{0}^{1} \frac {x + \cos 2\theta}{x^2 + 2x \cos 2\theta + 1} \, dx = \frac {\theta}{\sin\theta} (\log |\cos\theta| + \log 2)$

Thus combining (1), (3) and (4), we have

$ \int_{0}^{\infty} \frac {\log(1 + x^2)}{x^2 + 2x\cos\theta + 1} \, dx = \frac {1}{\sin \theta} [ \frac {\pi}{2} \tanh^{ - 1} (\sin \theta) + \theta \log |2 \cos\theta| ]$ $ + \int_{0}^{1} \frac {\cos \theta \log x}{x^2 + 2x \cos 2\theta + 1} \, dx$

Therefore it remains to evaluate $ \int_{0}^{1} \frac {\cos \theta \log x}{x^2 + 2x \cos 2\theta + 1} \, dx$. But using (2) again, we have

\begin{eqnarray*} \int_{0}^{1} \frac {\cos \theta \log x}{x^2 + 2x \cos 2\theta + 1} \, dx & = & \frac {1}{2\sin\theta} \sum_{n = 1}^{\infty} ( - 1)^{n - 1} \sin (2n\theta) \int_{0}^{1} x^{n - 1} \log x \, dx \\
& = & \frac {1}{2\sin\theta} \sum_{n = 1}^{\infty} \frac {( - 1)^{n}}{n^2} \sin (2n\theta) \\
& = & \frac {1}{4 i \sin\theta} [ \mathrm{Li}_2 ( - e^{2i\theta}) - \mathrm{Li}_2 ( - e^{ - 2i\theta}) ] \end{eqnarray*},

completing the proof.[/hide]

Using this formula, we can obtain some special values for this integral. For example,

$ \int_{0}^{\infty} \frac {\log(1 + x^2)}{1 + x + x^2} \, dx = \frac {\pi}{\sqrt {3}} \log(2 + \sqrt {3}) + \frac {1}{12}[ \psi_1 ( \frac {2}{3} ) - \psi_1 ( \frac {1}{3} ) ]$.

I dont understand this:

$ {{I}^{'}}\left( \alpha  \right)\equal{}2\alpha \int\limits_{0}^{\plus{}\infty }{\frac{dx}{\left( {{x}^{2}}\plus{}2x\cos \theta \plus{}1 \right)\left( {{x}^{2}}\plus{}{{\alpha }^{2}} \right)}}$

$ \text{Put }\frac{2\alpha }{\left( {{x}^{2}}\plus{}2x\cos \theta \plus{}1 \right)\left( {{x}^{2}}\plus{}{{\alpha }^{2}} \right)}\equal{}\frac{A}{{{x}^{2}}\plus{}2x\cos \theta \plus{}1}\plus{}\frac{B}{{{x}^{2}}\plus{}{{\alpha }^{2}}}$

$ \Rightarrow A\left( {{x}^{2}}\plus{}{{\alpha }^{2}} \right)\plus{}B\left( {{x}^{2}}\plus{}2x\cos \theta \plus{}1 \right)\equal{}2\alpha$

$ \Rightarrow \underset{x\to i.\alpha }{\mathop{\lim }}\,\left[ A\left( {{x}^{2}}\plus{}{{\alpha }^{2}} \right)\plus{}B\left( {{x}^{2}}\plus{}2x\cos \theta \plus{}1 \right) \right]\equal{}2\alpha$

$ \Rightarrow B\left( \minus{}{{\alpha }^{2}}\plus{}2i.\alpha .\cos \theta \plus{}1 \right)\equal{}1\Rightarrow B\equal{}\frac{2\alpha }{\left( \minus{}{{\alpha }^{2}}\plus{}2i.\alpha .\cos \theta \plus{}1 \right)}\equal{}\frac{\minus{}2\alpha }{{{\alpha }^{2}}\minus{}2i.\alpha .\cos \theta \minus{}1}$

$ {{x}^{2}}\plus{}2x\cos \theta \plus{}1\equal{}{{\left( x\plus{}\cos \theta  \right)}^{2}}\plus{}{{\left( \sin \theta  \right)}^{2}}\equal{}0\Rightarrow \left( x\plus{}\cos \theta  \right)\equal{}i.\sin \theta$

$ \Rightarrow x\equal{}\minus{}\cos \theta \plus{}i.\sin \theta \equal{}\cos \left( \pi \minus{}\theta  \right)\plus{}i.\sin \left( \pi \minus{}\theta  \right)\equal{}{{e}^{i\left( \pi \minus{}\theta  \right)}}$

$ \Rightarrow \underset{x\to {{e}^{i\left( \pi \minus{}\theta  \right)}}}{\mathop{\lim }}\,\left[ A\left( {{x}^{2}}\plus{}{{\alpha }^{2}} \right)\plus{}B\left( {{x}^{2}}\plus{}2x\cos \theta \plus{}1 \right) \right]\equal{}2\alpha$

$ \Rightarrow A\left( {{e}^{2i\left( \pi \minus{}\theta  \right)}}\plus{}{{\alpha }^{2}} \right)\equal{}1\Leftrightarrow A\equal{}2\alpha .{{\left( {{e}^{2i\left( \pi \minus{}\theta  \right)}}\plus{}{{\alpha }^{2}} \right)}^{\minus{}1}}$

$ \Rightarrow {{I}^{'}}\left( \alpha  \right)\equal{}2\alpha \int\limits_{0}^{\plus{}\infty }{\frac{dx}{\left( {{x}^{2}}\plus{}2x\cos \theta \plus{}1 \right)\left( {{x}^{2}}\plus{}{{\alpha }^{2}} \right)}}\equal{}\int\limits_{0}^{\plus{}\infty }{\frac{A.dx}{\left( {{\left( x\plus{}\cos \theta  \right)}^{2}}\plus{}{{\left( \sin \theta  \right)}^{2}} \right)}}\plus{}\int\limits_{0}^{\plus{}\infty }{\frac{B.dx}{\left( {{x}^{2}}\plus{}{{\alpha }^{2}} \right)}}$

$ \equal{}A.\frac{1}{\sin \theta }.\left[ \arctan \left( \frac{x\plus{}\cos \theta }{\sin \theta } \right) \right]_{x\equal{}0}^{x\equal{}\plus{}\infty }\plus{}B.\frac{1}{\alpha }.\left[ \arctan \frac{x}{\alpha } \right]_{x\equal{}0}^{x\equal{}\plus{}\infty }$

$ \equal{}\frac{A}{\sin \theta }.\left( \frac{\pi }{2}\minus{}\arctan \left( \cot \theta  \right) \right)\plus{}\frac{B}{\alpha }.\frac{\pi }{2}$

$ \equal{}2\alpha .{{\left( {{e}^{2i\left( \pi \minus{}\theta  \right)}}\plus{}{{\alpha }^{2}} \right)}^{\minus{}1}}.\theta \minus{}\frac{2\alpha }{{{\alpha }^{2}}\minus{}2i.\alpha .\cos \theta \minus{}1}.\frac{\pi }{2\alpha }$
I dont understand why you get:

$ {I}'\left( \alpha  \right)\equal{}\frac{2\alpha }{{{\alpha }^{4}}\plus{}2{{\alpha }^{2}}\cos 2\theta \plus{}1}\left[ \frac{\theta }{\sin \theta }\left( {{\alpha }^{2}}\plus{}\cos 2\theta  \right)\plus{}\cos \theta .\log \left( {{\alpha }^{2}} \right)\plus{}\frac{\pi }{2}\left( \frac{1}{\alpha }\minus{}\alpha  \right) \right]$
Thus integrating both side w.r.t. $ \alpha$ from 0 to 1 and partially applying the substitution    $ x\equal{}{{\alpha }^{2}}$, we have

$ (1)\quad I(1)\equal{}\int\limits_{0}^{1}{\frac{dx}{{{x}^{2}}\plus{}2x\cos 2\theta \plus{}1}\left( \frac{\theta }{\sin \theta }\left( x\plus{}\cos 2\theta  \right)\plus{}\cos \theta .\log x \right)}\plus{}\int\limits_{0}^{1}{\frac{\pi \left( 1\minus{}{{\alpha }^{2}} \right).d\alpha }{{{\alpha }^{4}}\plus{}2{{\alpha }^{2}}.\cos 2\theta \plus{}1}}$
and I dont understand this:

$ (2)\quad \frac{\cos \theta }{{{x}^{2}}\plus{}2x.\cos 2\theta \plus{}1}\equal{}\frac{1}{2\sin \theta }\sum\limits_{n\equal{}1}^{\infty }{{{\left( \minus{}1 \right)}^{n\minus{}1}}.{{x}^{n\minus{}1}}.\sin \left( 2n\theta  \right)}$

$ \cos x\equal{}\sum\limits_{k\equal{}0}^{2m}{\frac{{{\left( \minus{}1 \right)}^{k}}}{\left( 2k \right)!}}{{x}^{2k}}$

Please help me to explain this:

$ \int\limits_{0}^{1}{\frac{\cos \theta .\log x.dx}{{{x}^{2}}\plus{}2x\cos 2\theta \plus{}1}}\equal{}\frac{1}{2\sin \theta }\sum\limits_{n\equal{}1}^{\infty }{{{\left( \minus{}1 \right)}^{n\minus{}1}}}\sin \left( 2n\theta  \right)\int\limits_{0}^{1}{{{x}^{n\minus{}1}}.\log x\,.dx}$


linear algebra
matrix
vector
linear algebra unsolved

Any integer $2x2$ matrix with integer entries and trace 0 can be written as $AB-BA$ where $A,B$ also have integer entries.

Let's go for brute force.

Let $A=\left(\begin{array}{cc} x&y\\z& 0\end{array}\right)$ , $B=\left(\begin{array}{cc} u&v\\w& 0\end{array}\right)$

We have to solve $AB-BA=\left(\begin{array}{cc} a&b\\c& -a\end{array}\right)$
This is equivalent to $\left\{\begin{array}{ccc} wy-vz=a\\ vx-yu=b \\ zu-wx=c\end{array}\right.$
This is easily seen to be equivalent to $(x,y,z)\times (u,v,w)=(a,c,b)$ [b](*)[/b], where " $\times$ " denotes the vector product of two vectors in $\mathbb{Z}^3$.

If $a=b=c=0$, then we have trivial solutions $A=B=0_2$, so suppose they are not all $0$.
Let $d=gcd(a,b,c)$. Then $a=a'd,b=b'd,c=c'd$ for some integers $a',b',c'$ and $gcd(a',b',c')=1$. Suppose we solved the problem for these numbers. Then multiply $x,y,z$ by $d$ to obtain a solution for $a,b,c$. This just goes to prove that it suffices to solve (*) for $gcd(a,b,c)=1$.

So let $gcd(a,b,c)=1$.
Let $f:\mathbb{Z}^3\to \mathbb{Z}$, $f(m,n,p)=am+cn+bp$.
Because $gcd(a,b,c)=1$, $f$ is surjective. Let $(a_1,c_1,b_1),(a_2,c_2,b_2)$ be a basis in $\ker f$. Then $(a_1,c_1,b_1)\times (a_2,c_2,b_2)$ is collinear to $(a,c,b)$.

What remains to prove is that this product is $\pm (a,c,b)$

[quote="amfulger"]Let's go for brute force.

Let $A=\left(\begin{array}{cc} x&y\\z& 0\end{array}\right)$ , $B=\left(\begin{array}{cc} u&v\\w& 0\end{array}\right)$

We have to solve $AB-BA=\left(\begin{array}{cc} a&b\\c& -a\end{array}\right)$
This is equivalent to $\left\{\begin{array}{ccc} wy-vz=a\\ vx-yu=b \\ zu-wx=c\end{array}\right.$
[/quote]

Let $y=0$.
We have to solve
$\left\{\begin{array}{ccc} vz=-a\\ vx=b \\ zu-wx=c\end{array}\right.$

If $a=b=0$, then let $v=0,z=c,u=1,x=0,w=0$. This is a solution.

Else, let $v=gcd(a,b)$ and let $z=-\frac av$, $x=\frac bv$. $gcd(z,x)=1$ so there exist $w,u\in \mathbb{Z}$ such that $xw-uz=c$.
This is kind of simpler than my previous incomplete solution isn't it?

Hey, hey, hey, what is with this hurry to throw it on solved problems?  :D . This was the warm up!! Of course, the case of $2x2$ matrices works fine, but... the theorem is true in higher dimensions too...  ;) . Should we just forget such a nice problem? I guess not... ;)

[quote="harazi"]the theorem is true in higher dimensions too...  ;) . Should we just forget such a nice problem? I guess not... ;)[/quote]

As Harazi said it is not finish, you must find a proof for higher dimension

Sorry. I didn't realize you use [b]2[/b] to denote any integer. I must have skipped that reading class.

Nice try, amfulger! I will repeat what I said: the first problem, with 2 was just to warm ourselves. But anyway, if you really insist, you can move it to solved section and forget about it.  ;)

Does the proof for the $\mathbb Z$ case have anything in common with the (long...) proof for the $\mathbb C$ case, which I can barely understand?


Two numbers, one of which is 3 more than twice the other, have a sum of 66. Find the larger of the two numbers.

let the smaller number be x.

then the other number is 2x+3.

so 3x+3=66, 3x=63.

x=21.

the larger number = 66-21=45

answer ; 45


Let $a_1,a_2,....a_{1999}$ be a sequence of nonegative integers such that for any integers $i,j$, with $i+j<=1999$
we have :

                $a_i+a_j<=a_{i+j}<=a_i+a_j+1$
Prove that there are infinetely many real numbers $x$ such that $a_n=[nx]$ for each $n=1,2,....1999$.

Just a word of advice: Instead of <=,

Use $\leq$s (\leq, or mouseover)

For $k$ denote $I_k$ the interval of numbers $x$ for which $[kx]=a_k$.
Thus $I_k=[\frac {a_k}k;\frac {a_k+1}k )$
We prove that $I_k$ and $I_l$ intersect for any $k,l$. Let $k<l$.
Indeed this is equivalent to $\frac{a_k+1}k >\frac {a_l}l$(*) and
$\frac {a_l+1}l >\frac {a_k}k$(**).
This is equivalent to
$la_k+l>ka_l$ and $ka_l+k>la_k$.
Now let $l=kj+r$
Thus we get $a_l\leq ja_k+a_r+j$ and
$ka_l \leq kja_k+ka_r+kj=la_k+l+r(a_r-a_k-1)\leq la_k+l$
From the other side we get $a_l \geq ja_k+a_r$ and so
$ka_l+k \geq kja_k+ ka_r+k=la_k-ra_k+ka_r+k\geq la_k$ because $ka_r+k\geq ra_k$
(look at first result (*) proved, replacing $l$ by $k$ and $k$ by $r$)
Thus every two of the interrvals intersect and so all intervals have intersection, which is the desired $x$
QED

I kind of got the same result but couldnt put it on paper. NICE!


The sum of seven integers is $ \minus{}1$. What is the maximum number of the seven integers that can be larger than $ 13$?

$ \textbf{(A)}\ 1\qquad
\textbf{(B)}\ 4\qquad
\textbf{(C)}\ 5\qquad
\textbf{(D)}\ 6\qquad
\textbf{(E)}\ 7$

I'm not sure if if that problem is entirely correct because for any six integers larger than 13, there will be one negative number that will make the sum $ \minus{}1$.

It almost seems too easy.

[hide="Click for solution"]
Obviously, not all seven integers can be larger than $ 13$.  If six of them were each $ 14$, then the seventh could be $ \minus{}(6 \times 14)\minus{}1$, so the sum would be $ \minus{}1$.  Thus, the answer is $ \boxed{\textbf{(D)}}$.
[/hide]

You can have any six positive real integers which can then be offset by one negative integer to give -1 as the sum.

[quote="mathwizarddude"]Shouldn't the answer be "infinitely many" then?[/quote]

Except that there are only seven integers  :P


That's right. I made this problem directly from an ordinary inequality $x^{2}+1\geqq 2x$.