AMC
AIME
ratio
AMC 12
AMC 10
geometry
Ross Mathematics Program

what do you guys take with you during the test?

compass, protractor, and the whole deal? i've never really had a need to use them

I've always found a compass to be very useful for making good diagrams.  Graph paper is also usually helpful.

[quote="mathfanatic"][quote="tokenadult"][quote="mathfanatic"]Are scissors allowed?[/quote]

Scissors aren't specifically mentioned, but anyone ought to be able to tear a piece of paper with a ruler (mentioned, and legal).[/quote]

In that case, I'll make sure to bring my razor-blade-edged ruler.  :twisted:[/quote]

that's a good idea...i need to get one...

Your best bet: bring everything they let u bring :-) who knows what could come in handy

Scissors...interesting. There was a question on the AIME...maybe two years ago # 10 i believe that could be solved easily with protractor and ruler.

I'm reasonably sure that you can't bring anything that isn't specifically mentioned in the rules.  However, you can also use the tip of a compass to help cut out diagrams.

I don't know if you guys take this with a bunch of other people, and I realize Stuy was somewhat unusual in having 200+ people qualify, but if you're in a room with other students and you're tearing paper, it will probably really get on their nerves.

unless they all leave after an hour....lol at my school

Gah, Chinaboy writes so freakin loud.  He murders the table.

heh, ya... same here, by the end half the ppl were gone :P

(this doesnt count as discussing the aime right?)

[quote="paladin8"](this doesnt count as discussing the aime right?)[/quote]

No, Jeff, I'm banning you for life from the AMC forum.

:D jkjkjkjk

(I don't think I could do that even if I wanted to.)

My schedule for the day followed as:

8-1140: AIME
1140-1210: Lunch
1210-130: Assembly
135-230: Science

So I went to one clats; AIME needs to be held more times during the year.

I thought people weren't allowed to leave the AIME early ... well, whatever.

I have to take AIME II after school for 3 hours until six.  Then I'll have to ride the train home for about another hour or two.  This blows...:evil:.  And my teachers had to be (whatever I would say that's too bad for this site) and not let me take it in the morning when I'd be more alert.

[quote="cats..."]200 people at Stuy for the AIME?  I assume you give the AIME to everyone, then?[/quote]

Stuy has more than 200 students.  I presume he means that on average, 200 students from Stuy qualify.

[quote="JBL"]Stuy has approximately 3,000 students.  In the time I was there, approximately 1,400 to 1,600 students took the AMC every year with roughly 200-300 advancing to the AIME.  Certainly during all 4 years I was there we had the highest AMC and AIME numbers in the country, and I don't know how far back into the past that extends.[/quote]

dear god...  200-300?  that's an amazingly large number.  but then again, at our school around 200-300 people take the [i]amc[/i]...  so i guess out success rate (1/6) is a bit higher than yours...

At my school, only 10 people are taking the AIME, and I'm the only kid in the ninth or tenth grades taking it.

It looks like about the same [url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2004-12a/04amc12statstats/MN.HTML]number of people in my whole state qualify for the AIME[/url] as qualify at Stuy. Of course, New York City has a population almost twice that of my home state, so that simply means that Stuy gets a good sampling of the more math-able people in a large population area. Competition culture also has a much longer history in New York City than in almost any other part of the United States, so the coaching resources are more lavish, and the coaches more experienced.

The testing group my son was part of yesterday, forty miles from our home, consisted of three boys. All stayed in the testing room for the whole three hours  :)  My eleven-year-old son was actually the state champion on the 10B, so he was the "school winner" in our homeschool group. I think the other two guys taking the AIME yesterday in our group got in on the 12A and the 12B.

Hey, how about my sons school, 17/(30 >), thats a better ratio than Stuy and Z9R4C3s SF school!

But then of course I conveniently concealed the fact that 3000+ students who do not even know about AMC/AIME! And then there are many schools that will have ratios of 1/1, 2/2, etc. Well, those ratios are certainly not better than 1/6 or 200/1400. The bottom line: kudos to Stuy and Z9R4C3S SFBA school, you have fantastic math culture!

We had something like 25/100 (100=40 AMC12+60 AMC10 (or was it the other way?)) last year, with all but four AIME qualifiers coming from the AMC12.

z9r4c3's school will always have more AIME qualifiers than mine. Those silly smart rich people...

[quote="ComplexZeta"]We had something like 25/100 ... [/quote]

Simon, just curious on my part. Was your school (Homestead I believe) a regular school or a home school?

From what I remember, perhaps more regular than most.  But you at least had one outstanding teacher, eh?

[quote="ComplexZeta"]What do you mean more regular than most? We had a teacher who really cared about math and did a lot to help us do well, but unfortunately, he doesn't know how to do a lot of math problems. But  he had a lot of books and resources for us, and they really helped.[/quote]

really?  i thought headley (or however you spell his name) was (short of zuming and titu and the imsa and tj teachers) was the ultimate math coach...  oh well.  we are not smart rich people.  that would be saratoga.  we are just people who happen to do decently on the amc.

Haha...at my school, we're all rich ppl.  Just that the majority of them aren't good at math competitions.

Well, he's definitely very dedicated to helping students learn math, at least if they want to. The great thing is that just about everyone who doesn't like math or isn't good at it is absolutely terrified of him. He used to teach the best class in the Bay Area (if you exclude Harker at any rate; I'm not sure about that), and that was analytic geometry and calculus. For the first quarter (analytic geometry), he would assign some problems, and then the three or four people at each table would work on figuring out the material needed to solve them and then would solve them. Of course, tons of people got horrible grades in his class because they were never actually forced to learn anything before. But the ones who were actually interested really thrived on it. Apparently that was the way schools were before this ridiculous ``no student left behind'' policy came along and made almost all students be left behind.

ohhh, that sounds so very nice.  as opposed to my school, where it's "go study.  and yeah, good luck."  bah

[quote="tokenadult"]The rich people in this neighborhood, who are plentiful, have so many hobbies (sailing, horseback riding, skiiing on mountains far away) that they don't have as much time for studying math as middle-class people like my family. We have a roof over our heads, eat adequately, and spend most of our spare money on books. Evidently that is something that has been a family characteristic for a long time, as I own books that have been owned by members of my mother's family that were printed in the 1600s. Some people seek money as the main thing, and some people spend money to seek knowledge, and in a free country we can all live together peacefully.[/quote]

lol.  mine too.  i have at least a thousand dollars worth of math books sitting in my room.  what sucks is that the amc charges so much for their books (ross honsberger and ivan niven write the nicest books, but only the maa has them, i think) and there's only so much you can learn from the cheap dover paperbacks.

Before I got to college, I used to think that MAA and Springer books at $30 or $40 each were expensive. Then some of my classes had me buy textbooks for over $100 apiece, and I learned that those problem books are actually very cheap. Also, some Dover books are very good, but there are also plenty of bad ones.


Let $ a_1,a_2,...,a_n$ be posive real numbers, and let $ x_i \equal{} \frac {2a_i}{a_{i \minus{} 1} \plus{} a_{i \plus{} 1}}$, where $ a_0 \equal{} a_n$ and $ a_{n \plus{} 1} \equal{} a_1$.
Find the maximum value of $ n$ such that

(a) $ \frac 1{1 \plus{} x_1} \plus{} \frac 1{1 \plus{} x_2} \plus{} ... \plus{} \frac 1 {1 \plus{} x_n} \ge \frac {n}{2}$;

(b) $ \frac 1{1 \plus{} 2x_1} \plus{} \frac 1{1 \plus{} 2x_2} \plus{} ... \plus{} \frac 1 {1 \plus{} 2x_n} \ge \frac {n}{3}$.

A hint: The both inequalities are true for $ n\le 100$. :D

These inequalities are similar to Ando's Conjecture, but easier:

$ \frac {a_1}{(n\minus{}2)a_1\plus{}a_2\plus{}a_3}\plus{}\frac {a_2}{(n\minus{}2)a_2\plus{}a_3\plus{}a_4}\plus{}...\plus{}\frac {a_n}{(n\minus{}2)a_n\plus{}a_1\plus{}a_2} \le 1$.

Both of the above inequalities are true for any $ n$, of course. :wink:

The below nice inequality is also true for any positive numbers $ a_1, a_2,...,a_n$ such that $ a_1a_2...a_n\equal{}1$:

$ \frac {1}{n\minus{}2 \plus{} a_1 \plus{} a_2} \plus{} \frac {1}{n \minus{} 2\plus{} a_2 \plus{} a_3} \plus{} ... \plus{} \frac {1}{n \minus{} 2\plus{} a_n \plus{} a_1} \le 1$.


[quote="10maths_tp0609"]Let a,b,c be positive real numbers. Prove or disprove that:
$\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}\ge \frac{8(a^{2}+b^{2}+c^{2})}{ab+bc+ca}-5$[/quote]
I think, it's true. :D

The post was wrong. Thank you, 10maths_tp0609.

[quote="arqady"][quote="ductrung"]What about this one:
\[6+\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}\ge\frac{9(a^{2}+b^{2}+c^{2})}{ab+bc+ca}\]
I think it's true too. [/quote]
Yes, it's true:
let's assume $c=\max\{a,b,c\}.$ Hence,
$6+\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}\ge\frac{9(a^{2}+b^{2}+c^{2})}{ab+bc+ca}\Leftrightarrow$
$\Leftrightarrow\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\right)\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3\right)\geq\frac{9\left((a-b)^{2}+(c-a)(c-b)\right)}{ab+ac+bc}.$
[/quote]

I don't understand that, I think
$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\ge \frac{9(a^{2}+b^{2}+c^{2})}{ab+bc+ca}$
$\Leftrightarrow\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\right)\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3\right)\geq\frac{9\left((a-b)^{2}+(c-a)(c-b)\right)}{ab+ac+bc}.$

And
$6+\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}\ge\frac{9(a^{2}+b^{2}+c^{2})}{ab+bc+ca}$
$\leftrightarrow \left(\frac{(a+b)^{2}}{a^{2}b^{2}}-\frac{9}{ab+bc+ca}\right)(a-b)^{2}+\left(\frac{(a+c)(b+c)}{a^{2}c^{2}}-\frac{9}{ab+bc+ca}\right)(a-c)(b-c) \ge 0$

How about this one:
$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\ge \frac{14(a^{2}+b^{2}+c^{2})}{ab+bc+ca}-5$

[quote="10maths_tp0609"]Let a,b,c be positive real numbers. Prove or disprove that:
$ \frac {a^{2}}{b^{2}} \plus{} \frac {b^{2}}{c^{2}} \plus{} \frac {c^{2}}{a^{2}}\ge \frac {8(a^{2} \plus{} b^{2} \plus{} c^{2})}{ab \plus{} bc \plus{} ca} \minus{} 5$[/quote]
<=>
$ (a\minus{}b)^2[\frac{(a\plus{}b)^2}{a^2b^2}\minus{}\frac{8}{ab\plus{}bc\plus{}ca}]\plus{}[\frac{(b\plus{}c)(c\plus{}a)}{a^2c^2}\minus{}\frac{8}{ab\plus{}bc\plus{}ca}](a\minus{}c)(b\minus{}c) \geq 0$
But I can't continue! :blush:

[quote="10maths_tp0609"] How about this one:
$ \left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\right)^{2}\ge \frac {14(a^{2} \plus{} b^{2} \plus{} c^{2})}{ab \plus{} bc \plus{} ca} \minus{} 5$[/quote]
<=>$ (a\minus{}b)^2[\frac{\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}3}{ab}\minus{}\frac{14}{ab\plus{}bc\plus{}ca}]\plus{}(a\minus{}c)(b\minus{}c)[\frac{\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}3}{ac}\minus{}\frac{14}{ab\plus{}bc\plus{}ca}] \geq 0$


In circle O, diameter AC is extended half its length to D. Lines EC and DB are tangent to O at C and B respectively. If the radius of the circle is r, then find the area of triangle BFE in terms of r.

Note: ABF are collinear even if it may not look it.

[quote="davidlizeng"]since when is the point F mentioned in the problem??
you can't really do this without knowing where F is[/quote]

F is on lines AB and CE, that should be enough information.

[hide]Draw OB. The triangle OBD is a right triangle with angle BOD is 60 and angle BDO is 30. => BD is (3)^(1/2)r. Triangle ECD is also right triangle with angle CED = 60 and angle EDC = 30. => ED = 2/3^(1/2)r. => Subtract ED from BD and we get BE = 1/3^(1/2)r. Angle chasing shows that triangle BFE is equilateral (spelled correctly?). We know all of its sides and angles. The area is 1/2*(1/3^(1/2)r)^2*sin60 = 3^(1/2)/12r^2.[/hide]


Source: USC Math Contest 1991

Circle $C$ is inscribed in square $PQRS$, and the length of segment $PQ$ is $1$. Inside the square $PQRS$, Circle $D$ is tangent to Circle $C$, segment $PQ$ and segment $QR$, find the radius of Circle $D$.

Fun geometry problem isn't it? :lol:

[quote="shobber"][hide="reminds me of Chemistry"]

It seems that $r=\frac{3-2\sqrt{2}}{2}$.[/hide][/quote]

Yup, show your work though, and how does it remind u of chemistry? :?

Let the distance from $D$ to $Q$ be represented as $r+x$, then we have the following system of equations:

$2r+x = \frac{\sqrt{2}-1}{2} \text{  and  }  (r+x)^2 = 2r^2 \Rightarrow r = \frac{3-2\sqrt{2}}{2}$

[hide]
I used the power of point.
The diagnol of the square if $\sqrt{2}$ and the diameter of the large circle is $1$.
Hence: $r^2= (\frac{\sqrt{2}-1}{2}-2r) \cdot \frac{\sqrt{2}-1}{2}$

The picture remind me of the structure of some substance......[/hide]

[quote="pkothari13"]lol which substance did you think it looked like shobber?[/quote]
Just similar but not all the same.
My poor sketch...This is $\text{NaCl}$. It is supposed to be a square...

[quote="DPopov"]body centered cubic?[/quote]
Yep!
Finding the radius of the atoms is finding the radius of the circle in this problem.

YES! I remembered something from Chem 1  :lol:  :lol:

After reading all of y'alls solutions, I see that none of y'all did it the way I did, set up proportions!

You know that the the diagonal of the square $SQ=\sqrt{2}$, then let's call the diameter of the larger circle $AB$ such that it overlaps and is a part of the diagonal. Then you also know that $BQ=\frac{\sqrt{2}-1}{2}$ and $AQ=\frac{\sqrt{2}+1}{2}$. Then let $BC$ be the diameter of the smaller circle, which also overlaps and is part of the diagonal of the square.

Thus:

$\frac{AQ}{AB}=\frac{BQ}{BC}-------->\frac{\frac{\sqrt{2}+1}{2}}{1}=\frac{\frac{\sqrt{2}-1}{2}}{BC}$
We find out that $BC=3-2\sqrt{2}$ thus the radius is $\frac{3-2\sqrt{2}}{2}$.


For all positive integers $ x$, let
\[ f(x) \equal{} \begin{cases}1 & \text{if }x \equal{} 1 \\
\frac x{10} & \text{if }x\text{ is divisible by 10} \\
x \plus{} 1 & \text{otherwise}\end{cases}\]and define a sequence as follows: $ x_1 \equal{} x$ and $ x_{n \plus{} 1} \equal{} f(x_n)$ for all positive integers $ n$. Let $ d(x)$ be the smallest $ n$ such that $ x_n \equal{} 1$. (For example, $ d(100) \equal{} 3$ and $ d(87) \equal{} 7$.) Let $ m$ be the number of positive integers $ x$ such that $ d(x) \equal{} 20$. Find the sum of the distinct prime factors of $ m$.

Hm...it appears that nobody has replied, so I will...I just took this AIME as a practice and for fun, fully enforcing rules, and if i get this one, i get 14...now if i could just pull that off on the real thing...

[hide]We can reverse engineer this process.  First, we begin with 1.  Then, we must take 19 steps, at each of which we either multiply by 10 or subtract 1.  At the first step, we must multiply by 10, giving us $10$.  Within the next 9 steps we must multiply by 10 at least once: otherwise, the number would reach $1$.  By a complement argument, the number of ways to do this is $2^9-1$.  Note that we can only decrease the number by subtracting 1.  Thus, after these next 10 steps, because our number will be at least $20$, we cannot reach 1 again after that, and we need not set any more restrictions.  The number of ways to place operations after that is thus $2^9$, yielding a total of $2^9*(2^9-1)$.  However, we forgot something.  We cannot have a string of subtracting 1's 10 times or more.  If we did, that would imply that the number reached a multiple of 10 and went on adding.  So we must pick out those possibilities from our set.  We can first take out the number of ways to have a string of 10 (-1)'s in a string of 18 operations, then add back in the number of ways to do so if the first 9 of those 18 operations are (-1)'s.  This, again, is by a complement argument.  The number of ways to have a string of 10 A's in a string of 18 A's and B's given that the first 9 are A's is $2^8$, and the number of ways to have a string of 10 A's in a string of 18 A's and B's is $[7(2^6)+6(2^5)+5(2^4)+4(2^3)+3(2^2)+2(2)+1(1)]+2[2^8-1]+1=1280$.  So the answer is then $2^{18}-2^9+2^8-2^8(5)=2^18-2^9-2^{10}=2^9(509)$.  So our final answer is $\boxed{511}$[/hide]

can you explain your solution a little better? I am having difficulty just understanding the problem...
thanks!

If $x_{20}=1$, then $f^{19}(x)=1$.  Given some integer, if we continually perform the function on it, we get 1.  For example, take $354$.  Then we have \[ 354 \implies 355 \implies 356 \implies 357 \implies 358 \] \[ \implies 359 \implies 360 \implies 36 \implies 37 \] \[ \implies 38 \implies 39 \implies 40 \implies 4 \implies 5 \implies 6 \] \[ \implies 7 \implies 8 \implies 9 \implies 10 \implies 1 \]
Here, we had to perform the function $19$ times to first get 1, so $d(354)=19+1=20$.
Now, we know that, in the same way, we can backtrack the function.  Suppose we take the number $1$.  We can either multiply by 10 or subtract 1 at any point (the inverse of the function).  If $d(a)=20$, then $a_{20}=1, a_{19} \neq 1$.  So given 1, if we want to get a possible $a_20$, we have to perform the inverse of the function $19$ times.  At each step, we either add 1 or multiply by 10.
However, we can't add 1 and get a multiple of 10.  Think about it by going forwards.  If you went the other way, and you got to a multiple of 10, would you subtract 1?  You can't: the rules of the function don't allow it.  Anyways, you can count the number of possible strings of subtracting 1 and multiplying by 10.

Just as a note, this is AIME [b]I[/b] 2004 #15. Also, a solution should be added to the resources section.


Given the equation $ \frac {1}{m} \plus{} \frac {1}{n} \equal{} \frac {2}{15}$, where m and n are positive integers, find the value of $ m \plus{} n$ such that m>n and m is not a multiple of n.

[hide="hint"]
[url=http://www.artofproblemsolving.com/Wiki/index.php/Sfft]Simon's Favorite Factoring Trick[/url]
[/hide]

[quote="Ihatepie"]hint
[hide]

1/m+1/n=m+n/mn=2/15
15m+15n=2mn
15(m+n)=2mn
15/2(m+n)=mn

mn must be a multiple of 15.

[/hide][/quote]

Yeah, I got that, but for some reason I got m+n=2 which doesn't really work out...

[hide]
Listing possibilities by factoring $ (2m \minus{} 15)(2n \minus{} 15) \equal{} 225$, we find that $ (m, n)$ could be:

$ (120, 8) \\
(45, 9) \\
(30, 10) \\
(20, 12) \\
(15, 15)$

Clearly, $ (15, 15)$ does not satisfy the requirement that $ m > n$ and only $ (20, 12)$ satifies the requirement that $ m$ is not a multiple of $ n$. So [b]part of[/b] the answer is $ (20, 12)$. [b]You figure out the rest by reading the question[/b] :wink: [/hide]

[b]Edit:[/b] In bold

the answer is not $ (20,12)$, go back and read the question :wink:

I'm posting a hint, duh! The answer is really
[hide="Click here if you really want to know"]$ 32$[/hide]

meh, you did specifically say "The answer is (20,12)"

[hide="getting to the factorization"]
$ \frac{1}{m}\plus{}\frac{1}{n}\equal{}\frac{2}{15}$
$ 15n\plus{}15m\equal{}2mn$
$ 2mn\minus{}15n\minus{}15m\equal{}0$
$ 4mn\minus{}30n\minus{}30m\equal{}0$
$ 4mn\minus{}30n\minus{}30m\plus{}225\equal{}225$
$ (2m\minus{}15)(2n\minus{}15)\equal{}225$
[/hide]


[quote="SomKim"]If $ a,b,c$ are positive real numbers and $ r \ge 0$. Let
$ S_r \equal{} a^{r}(a \minus{} b)(a \minus{} c) \plus{} b^{r}(b \minus{} c)(c \minus{} a) \plus{} c^{r}(c \minus{} a)(c \minus{} b)$
 Then,
    $ S^{2}_2(a^2 \plus{} b^2 \plus{} c^2) \plus{} (ab \plus{} bc \plus{} ca)(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \ge 2abc S_1 S_2$[/quote]

    $ S^{2}_2(a^2 \plus{} b^2 \plus{} c^2) \plus{} (ab \plus{} bc \plus{} ca)(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\geq S^{2}_2(a^2 \plus{} b^2 \plus{} c^2)$

then we just need to prove

$ S^{2}_2(a^2 \plus{} b^2 \plus{} c^2) \geq 2abc S_1 S_2$

or

$ S_2(a^2 \plus{} b^2 \plus{} c^2) \geq 2abc S_1$

 it is equivalent to

$ \sum(a^4(a\minus{}b)(a\minus{}c))\plus{}(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2\geq 0$

easy to prove $ \sum(a^4(a\minus{}b)(a\minus{}c))\ge 0$   by  schur

[quote="hedeng123"]
 it is equivalent to

$ \sum(a^4(a \minus{} b)(a \minus{} c)) \plus{} (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\geq 0$

easy to prove $ \sum(a^4(a \minus{} b)(a \minus{} c))\ge 0$   by  schur[/quote]
You can prove the following ine(hold):

If $ a,b,c$ are positive real numbers and $ r \ge 0$. Let
$ S_r \equal{} a^{r}(a \minus{} b)(a \minus{} c) \plus{} b^{r}(b \minus{} c)(c \minus{} a) \plus{} c^{r}(c \minus{} a)(c \minus{} b)$
 Then,
    $ S^{2}_2(a^2 \plus{} b^2 \plus{} c^2) \plus{} (ab \plus{} bc \plus{} ca)(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \ge 2abc S_1 S_2\plus{} S_2 \cdot S_4$


Let $f \in \mathbb{Z} [X]$, $f=n(X^{2n}-1)+X^2-X+1$.
Find $n$ s.t. $f$ has rational root(s).

To find all such $n$ (or, I suppose, to prove that for many such $n$ it fails) does not seem like an intermediate level question to me ... am I missing something obvious?  (To find a single such n is easy but not very enlightening.)

By the way, for those who don't know the notation, $f \in \mathbb{Z} [X]$ means that $f$ is a polynomial with integer coefficients.

Joel, which forum do you think this topic should go?

I have no idea, but I haven't been able to do anything decent with it.  It seems like an olympiad algebra problem, but I'm honestly not sure because I don't actually know how to solve it (although I suspect I know the answer).  Maybe just pre-olympiad, since that's where everything seems to go?

Shouldn't it be just about any n, since f(0) seems to usually give you numbers opposite to the direction f(x) goes as x approaches either infinity..

On the contrary, I think for just about all n it has a rational root.  Just think about this.  Say n is non-zero.

Then f(0) = 1 - n.  If n is positive, the polynomial tends toward infinity at either extreme, while f(0) is 0 or negative.  This implies the existence of a root.  If n is negative, well I don't think that's considered a polynomial...

[quote="JBL"][quote="perfect_radio"]I don't know how to solve it.
I posted it here because it was in my school book (and that should intermediate, right?).

BTW, it has a solution at the end of the book (it says that only for $n=1$ there is a rational root), but I didn't understand anything from it[/quote]

Well, that depends what level of school you are in.  If you are using the notation $\mathbb{Z} [X]$, I would guess we are not talking American high school level, yes?  The fact that you can't understand the solution makes it seem even less likely that this has an elementary solution.  I'd be interested in seeing the proof, and perhaps we can then figure out how it works :)[/quote]
I'm from Romania, I don't know what the American High School level is  :roll:

Here is the solution:
[quote="10th grade school book"]
$f(1)$ not equal to $0$.
$f(-1)$ not equal to $0$.
Let $\tau = \frac{p}{q}$, $(p,q)=1$ be a rational root.
So $p| \left ( n-1 \right )$, $q|n$.
(I couldn't understand anything beyond this line)
Therefore $\tau = \frac{1}{q}$ (any explanation?????).
Since $\left ( q-p \right ) | f(1)=1$, it follows that $q-1 = \pm 1$, and from this we get $q=2$.
Also, $f(\frac {1}{2})=1$, which is equivalent to $\frac{n}{4^n}+\frac{3}{4}=n$.
For $n=1$, $f=X \cdot (2X-1)$.
For $n \geq 2$ we get the contradiction $n< \frac{1}{4}+\frac{3}{4}=1$
[/quote]

If you understand how it was done, pls give more details

PS: what's the latex for "not equal"???

I agree.  That solution needs more explanation.

LaTeX: \neq: $\neq$

Wow! The standard of Mathematics in Romania is very high as such a problem could appear in textbooks of high school.
May I know which book do you use?
I want to take a look on this book. ( Is it in English? I wonder.... )

the problems in this book are usually easy. and of course it is only in romanian

Let $\tau = \frac{p}{q}$, $(p,q)=1$ be a rational root. WLOG, we may assume that $p>0$. Then $(X-\frac{p}{q})$ divides $f(X)$ in ${\mathbb Q}[X]$. So $(qX-p)$ divides $f(X)$ in ${\mathbb Z}[X]$. Therefore, $q-p | f(1) =1$ and $q+p | f(-1)=3$. It follows that $p=1$ and $q=2$ with a little case work.

hhmmmm, i have another soln:

[quote="My teacher"]
$f=nX^{2n}+X^2-X+(1-n)$
let $x=\frac{p}{q}$ be such a root.
then $f(\frac{p}{q})=n \cdot \frac{p^2}{q^2}+\frac{p^2}{q^2}-\frac{p}{q}+(1-n)$
Upon multiplication by $q^2$ we get:
$q^2f(\frac{p}{q})=n\frac{p^{2n}}{q^{2n-2}}+p^2-pq+q^2(1-n)=0$, so $\frac{np^{2n}}{q^{2n-2}}$ is an integer
For $|q| \geq 2$ we have $|q|^{2n-2} \geq 2^{2n-2} >n$, for $n \geq 2$ (by induction), so the fraction above can't an integer (since $\gcd (p,q) =1$).
So $q=1$ (if $n \leq 2$). $|x|>1|$ implies $f(x)>0$, and $x=-1,0,1$ don't work also.
Therefore $n=1$, which works :)

[/quote]

[quote="fuzzylogic"]Therefore, $q-p | f(1) =1$ and $q+p | f(-1)=3$.[/quote]

I don't understand this part too well :( could you explain?

[quote="fuzzylogic"][quote="perfect_radio"]
[quote="fuzzylogic"]Therefore, $q-p | f(1) =1$ and $q+p | f(-1)=3$.[/quote]

I don't understand this part too well :( could you explain?[/quote]

$f(X)=(qX-p)g(x)$ for some $g\in{\mathbb Z}[X]$, then let $X=1$ and $X=-1$. Hope this clears your doubt :-)[/quote]

this certainly cleared my doubt  :blush:


abstract algebra
group theory
inequalities
superior algebra
superior algebra theorems

So far I know that the finite abelian groups can all be modeled by the direct product of cyclic groups. But what kind of groups can we create with a 'reverse homomorphism'?

(focusing on finite groups)
I am not sure if this is a formally defined concept, but what I mean is that if we start with a group $ H$ and we take its elements and 'split' or expand them into a set which is a group under a new binary operation and that satisfies the homomorphism equation $ G\rightarrow H$ (surjective). If we do this we get of course that the identity of $ H$ expands to a normal subgroup of $ G$ and the restrictions on the binary operation that preserve coset multiplication, But the binary table still has some room for freedom after we chose our subgroup for the reverse of identity of $ H$ (any subgroup is ok I think?). How much room do we have? what are the minimal number of equations $ ab\equal{}c$ ($ a,b,c \in G$) such that the whole group is defined? and what kind of groups can we now create with this construction?

The way you describe the problem is very interesting, but unfortunately it has a dull solution.  I'll describe some similar problems that have more interesting answers.

The G that work are exactly the groups G such that there is some surjective homomorphism from G to H.

Take H=Z/2Z, then G=Z/3Z x Z/2Z works, and so does the symmetric group G on n≥2 points.  Its crazy how many groups work.  For this H you just need a normal subgroup of index 2.  In other words, your initial interesting question turns out to have a very dull answer: tons of completely unrelated groups.

However, there is a tiny variation on your question that does have an interesting question:  While G=Z/2Z x Z/2Z does have a surjective homomorphism onto H=Z/2Z, H only needs one generator, while G needs 2.  Suppose instead of all G, I only wanted those G such that every generating set of H when pulled back into G was a generating set of G (just one element of G for each element H, not the whole block).

Such G are called "Frattini Extensions" of H and there are largest such G (at least under some mild hypotheses on H).

A more extreme variation results in a very classical problem:  What if instead of just allowing any old G and surjective homomorphism onto H, we required that the kernel was isomorphic to a specific group N.  Then the set of all such G are called the "extensions" of H by the normal subgroup N, and finding all such G is called the "extension problem".  In terms of your multiplication table picture, this means you decide beforehand how many elements of G each element of H splits into.

You can also be less strict about N: if you require that G have the generator property and that N be contained in the center, then there is a largest such N and some largest such G.  N is called the "Schur multiplier", and when H is a perfect group, there is a unique largest G called the Darstellungsgruppe or universal central extension.

Thank you, JackSchmidt that was very informative. So let me summarize and pose clearer questions.

the construction (consider finite group and finite expansion sets $ [x]$)
We take an arbitrary group $ H$ and for each $ h \in H$ we create a nonempty set $ [h]$ now we define an incomplete/variable binary operation on members of these sets; For all $ a,b \in H$ we have that $ \forall \: a' \in [a]\:$ and $ \: \forall \: b' \in [b]\:$  $ a'b' \in [ab]$.Now we want the set $ G$={$ x | x \in [x]$} to be a group so we attempt to apply the constraints of the group axioms to the set and see what constraints we get in return. take $ e' \in [e]$ ($ e$ id of $ H$) and let $ b_1,...,b_n$ be the members of $ [b]$, then $ e'b_1,...,e'b_n$ are inside $ [eb] \equal{} [b]$ so let us make this $ e'$ our identity in G. so we have filled the binary table entries for all multiplication from left/right by $ e'$. if $ x \in [a]$ and $ y \in [c]$ and $ xy \equal{} e'$ then we have that $ [a][c] \equal{} [ac] \equal{} [e]$ but now $ c$ must equal $ a^{ \minus{} 1}$ So inverses of $ [a]$ are all in $ [a^{ \minus{} 1}]$ this show we must have the size of each of our sets equal. Now since $ [e][e] \equal{} [e]$ we have that $ [e]$ is a subgroup of $ G$. Now multiply each member of $ [b]$ on the left with $ a' \in [e]$ we have that $ a'b_1,....,a'b_n \subseteq [b]$ and in fact equals $ [b]$ (using the inverse of $ a'$ established) thus computing $ (b^{ \minus{} 1}a)b$, $ b \in [b], \: a \in [e]$ we get that this is a member of $ [b^{ \minus{} 1}e][b] \equal{} [e]$ so indeed $ [e]$ is normal in $ G$.

So here are some questions to go along with this:
1. Given an $ H$ will any finite group $ [e]$ be able to make a new group following this construction?
2. Given an $ H$ and an $ [e]$ how many groups can we generate this way? how diverse are they? (this will clearly depend on $ H, [e]$ so maybe look at some specifics, could it be that this purely depends on the sizes of these two?)
3. Applying this generator of groups (call it $ T$) to some set of groups $ A$ (ie creating the set $ T(A)$ of all possible group from every member of $ A$) how do the various sets $ (T^k(A))$ compare. (if 2 can be answered..)
4. What happens when we take an infinite $ H$ or $ [e]$ or both?

[quote="question?"]
So here are some questions to go along with this:
1. Given an $ H$ will any finite group $ [e]$ be able to make a new group following this construction?
[/quote]

Yes, take G to be the direct product of H and [e].

[quote="question?"]
2. Given an $ H$ and an $ [e]$ how many groups can we generate this way? how diverse are they? (this will clearly depend on $ H, [e]$ so maybe look at some specifics, could it be that this purely depends on the sizes of these two?)
[/quote]

This is a question about the theory of extensions.  Look for "Schreier's theory of factor sets" in most group theory texts.  When [e] is an abelian group, then you look for homomorphisms from H into Aut([e]) up to conjugacy, and for each such you consider [e] as an H-module, and then calculate the second cohomology group of H with coefficients in [e].  Then you consider the action of Out(H) on the cohomology, and take representatives.  There is now an approximately one-to-one map between the pairs of conjugacy classes of homomorphisms and cohomology classes and the isomorphism classes of G (sometimes G can still be counted more than once, but it is a little rarer).

It depends on more than the sizes.  Taking H to be cyclic of order 4 and [e] to be cyclic of order 2, there are exactly two such groups G: G cyclic of order 8, or G the direct product of a cyclic group of order 4 and a cyclic group of order 2.  Taking H to be noncyclic of order 4 and [e] to be cyclic of order 2, there are exactly four such groups: the direct product of three copies of a cyclic group of order 4, the direct product of a cyclic group of order 4 with a cyclic group of order 2, the dihedral group of order 8, and the quaternion group of order 8.

[quote="question?"]
3. Applying this generator of groups (call it $ T$) to some set of groups $ A$ (ie creating the set $ T(A)$ of all possible group from every member of $ A$) how do the various sets $ (T^k(A))$ compare. (if 2 can be answered..)
[/quote]

Well it depends on which T, the one that takes both H and [e] or the one that takes just H. If T just takes H, then T=T^2.  Otherwise T(H,[e]) consists entirely of groups of order |H| * |[e]|, so presumably the T^k are disjoint.

[quote="question?"]
4. What happens when we take an infinite $ H$ or $ [e]$ or both?[/quote]

Nothing too special happens other than the calculations are harder.

Thanks for your reply JackSchmidt. Your answer to my question 2 is a bit advanced for me at the moment, hopefully i'll get to it one day. I have some other questions on the topic.

-What kind of groups are those that can be partitioned into disjoint cyclic groups?
-Suppose we have a group $ G$. Take a generating set of minimal size $ A$ and form the set of cyclic subgroups generated by the elements of $ A$ call this $ <A>$. In the case of abelian groups, all such $ <A>$ are the same up to isomorphic groups and multiplicity, a consequence of the fundamental theorem of finite abelian groups. What do we know about such $ <A>$ when $ G$ is not abelian. Do they give us any useful information on the structure of $ G$. The various $ <A>$ are probably not the same, otherwise I think this would suggest a construction of $ G$ from the groups $ <A>$ ? or am I leaping to conclusions here.


20. Let $ a$, $ b$, and $ c$ be real numbers such that $ a\minus{}7b \plus{} 8c \equal{} 4$ and $ 8a\plus{}4b\minus{}c \equal{} 7$.
Then $ a^2\minus{}b^2\plus{}c^2$ is
(A) $ 0$ (B) $ 1$ (C) $ 4$ (D) $ 7$ (E) $ 8$

Could someone explain how we would start this problem? I saw the solution after trying the problem, and I understood the solution, but I wasn't sure how we would would have thought of it.
[hide=" Solution"]
20. (B) We have $ a \plus{} 8c \equal{} 4 \plus{} 7b$ and $ 8a\minus{}c \equal{} 7\minus{}4b$. Squaring both equations and
adding the results yields
$ (a \plus{} 8c)^2 \plus{} (8a\minus{}c)^2 \equal{} (4\plus{}7b)^2 \plus{} (7\minus{}4b)^2.$
Expanding gives $ 65(a^2 \plus{} c^2) \equal{} 65(1 \plus{} b^2)$. So $ a^2 \plus{} c^2 \equal{} 1 \plus{} b^2$, and $ a^2\minus{}b^2 \plus{} c^2 \equal{} 1$.
[/hide]

You have long dashes in place of minus signs, which aren't rendered properly by $ \text{\LaTeX}$.

The solution is somewhat contrived.  There are symmetries in the given equations (the coefficients $ 1, 7, 8, 4$ appearing in a different order and everything) that might lead you to the intended solution if you mess around with a few other things that probably won't work.

A general approach (which I gave in another thread on this problem in the AMC forum):  the given equations are planes in three dimensions, and their intersection is a line (in three dimensions).  Parameterize this line, and plug the parameterization into the expression $ a^2 - b^2 + c^2$.  The parameter disappears.  This is because the level surfaces of $ a^2 - b^2 + c^2$ are [url=http://en.wikipedia.org/wiki/Ruled_surface]ruled surfaces[/url].

Edit:  A good thread where this problem was discussed is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170124&search_id=105981621]here[/url].


geometry
3D geometry
tetrahedron
geometry proposed

How many regular tetrahedrons with side $ \sqrt {2}$ can be fitted into a cuboid box of size 1 by 2 by 2 ?


[color=darkblue]O.K. ! In the future, please don't post here same problems.[/color]

$E\equiv 2\sin 2x+2\sin x-(1-\cos 2x)=4\sin x\cos x+2\sin x-2\sin^{2}x=$ $2\sin x(2\cos x+1-\sin x)=$ $2\sin x\left[2\left(\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}\right)+\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^{2}\right]=$ $2\sin x\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)\left(3\cos\frac{x}{2}+\sin\frac{x}{2}\right)\ .$
$E=0\Longleftrightarrow \sin x=0\ \ \vee\ \ \sin\frac{x}{2}-\cos\frac{x}{2}=0\ \ \vee\ \ 3\cos\frac{x}{2}+\sin\frac{x}{2}=0$ $\Longrightarrow$ $x\in (\pi Z)\cup \left(2\pi Z+\frac{\pi}{2}\right)\cup \left(2\pi Z-2\arctan 3\right)\ .$


Prove that any set of n red points and n blue points in the plane can be grouped in n pairs (each pair consisting of a blue point and a red point) such that no two of the segments joining each pair interesect.

Ah the usefulness of the extremal principle. :lol:

[quote="arccosinus"]Famous problem... I guess it is originally from somewhere in Russia (maybe Kvant, or some olympiad).

Consider all possible configurations of lines between the $n$ pairs. Choose the configuration with the shortest total distance of the lines. Then, if two lines intersect you can (according to the triangle inequality) make the total distance shorter by switching them. Thus, a contradiction which means that the shortest configuration cannot contain intersecting lines.[/quote]

wow.  I have never seen any Olympiad problem with such a short solution-yet incredibly elegant.  I can't see myself coming up with it, though.

[quote="mkkool64"]
wow.  I have never seen any Olympiad problem with such a short solution-yet incredibly elegant.  I can't see myself coming up with it, though.[/quote]
You haven't seen? Check the [u]proposed[/u] solution (not the official one) for problem 5 from ISL 1997. It is one sentence (literary)!


Let $ \{a,b,c\}\subset[ \minus{} 1, \plus{} \infty).$  Prove that:
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \plus{} 2 \ge ab \plus{} ac \plus{} bc \plus{} a \plus{} b \plus{} c\]

[hide]$ a^2\plus{}b^2\plus{}c^2\plus{}abc\plus{}2\ge ab\plus{}bc\plus{}ca\plus{}a\plus{}b\plus{}c$

$ \Leftrightarrow a^2\minus{}2a\plus{}b^2\minus{}2b\plus{}c^2\minus{}2c\plus{}3\ge \minus{}abc\plus{}ab\plus{}bc\plus{}ca\minus{}a\minus{}b\minus{}c\plus{}1$

$ \Leftrightarrow (1\minus{}a)^2\plus{}(1\minus{}b)^2\plus{}(1\minus{}c)^2\ge (1\minus{}a)(1\minus{}b)(1\minus{}c)$

Now we have to prove the inequality

$ x^2\plus{}y^2\plus{}z^2\ge xyz$

where $ x,y,z\in (\minus{}\infty,2]$.  If the right hand side is negative then we're done.  If not, then either all of $ x,y,z$ are positive or only one is.  So, we must have at least one positive variable, without loss of generality $ z$.  We must have $ z\le2$ so we have the inequality

$ 2xy\ge xyz$

$ \Rightarrow 2\sqrt{x^2y^2}\ge xyz$

$ \Rightarrow x^2\plus{}y^2\ge xyz$

$ \Rightarrow x^2\plus{}y^2\plus{}z^2\ge xyz$[/hide]

Yes, cosinator!  :lol:
My proof:
let $ a\equal{}1\plus{}x,$ $ b\equal{}1\plus{}y$ and $ c\equal{}1\plus{}z...$ It's the same!


abc is a three digit number. If acb + bca + bac + cab + cba = 3194, find abc.


For a certain real number $x$, the differences between $x^{1919}$, $x^{1960}$ and $x^{2001}$ are all integers. Prove that $x$ is an integer.

More generally if :

$x^n = x^{(n+p)/2} + N = x^p + P$ with $N, P$ integers, $gcd(n,p) = 1$, $n < p$

Then easy algebra gives $x^n = \dfrac {N^2}{2N-P}$ (case P = 2N treated separatedly) i.e $x^n$ is rationnal and hence $x^p$ also.
But since $gcd(n,p) = 1$, there exists u and v such as $u n + v p = 1$ and so x is rationnal also.

If $x = A/B$ then ${A^n = A^{(n+p)/2} B^{(n-p)/2} + N B^n }$ and $B = \pm 1$ since $n-q < 0$


induction
number theory unsolved
number theory

1.4

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

Induction.

From the induction hypothesis, assume the result is true for k, and the two groups are {x_1, x_2, ... x_2^k} and {y_1, y_2, ... , y_2^k}. Now for k+1, divide the 2^(k+1) numbers into {x_1, x_2, ... x_2^k, y_1 + 2^k, y_2 + 2^k, ... , y_2^k + 2^k} and {y_1, y_2, ... y_2^k, x_1 + 2^k, x_2 + 2^k, ... x_2^k + 2^k}.

Observe that all pairs of numbers in the form (Z, Z+2^k) are in different groups, where Z=1,2,..., 2^k. Hence expand (Z+2^k)^m using binomial expansion for all m=1,2,...,k+2, then the problem is reduced to the combination of equalities given from the induction hypothesis. (We effectively bring the exponent down 1 because Z^m cancels out.) :D

Well, this is really strange! This problem is more than well known, it has been discussed on the forum, it is an absolute classic and appears in the China TST?

I am not sure if these are really the Chinese TST for [b]IMO[/b] ... maybe they are TST for something else?!  :?  :?

As I understand, the Chinese team selection occurs in a 2~3 week camp. There, 6 to 8 TST quizzes (composed of 3 problems each) are given, and their total weight is 50%. In the end, a final TST (6 problems) is given, with a total weight of 50%. The combined score is used to select the team.

[quote="Valentin Vornicu"]All 6 problems at once? :)[/quote]

I doubt it. I think they're given 3 problems at a time (for 4 to 4.5 hours).

Then again, I've never been to one of those camps.


this is actually a half life question, but it includes math ( and i need a formula for it) so i figured that i would ask...

Batteries used in heart pacemakers contain plutonim-238.  if your original sample contains 2.57x10^9 atms of plutonium, how much time will pass before the amount is reduced to 5.02x10^6 atoms? the half life is 27.1 years.

is there an easy formula to do this?

[hide]
So, you should be multiplying by $ \frac {1}{2}$ every 27.1 years. This can be represented in a formula by:

$ f(t) \equal{} 2.57 \cdot 10^9 (\frac {1}{2})^{\dfrac{t}{27.1}}$

Plugging in $ 5.02 \cdot 10^6$ for $ f(t)$ and solving for t gives the answer to be about 244 years.

edit:
Uhh... that $ \dfrac{t}{27.1}$ is supposed to be an exponent, though it doesn't look much like it... Any advice on LaTeX?
[/hide]

how do u solve for T?  (considering its an exponent)

First, you wat to get as many terms as you can to one side. So, divide by $ 2.57 \cdot 10^9$:

$ \dfrac{5.02 \cdot 10^6}{2.57 \cdot 10^9} \approx .0019533074 = .5^\dfrac{t}{27.1}$

To get the $ t$ out of being an exponent, you take a log base .5 on both sides.

$ \log _{.5} .0019533074 = \log _{.5} .5^\dfrac{t}{27.1}$

Using the change of base formula, this simplifies to:

$ \dfrac{\log .0019533074}{\log .5} = \dfrac{t}{27.1}$

Now you can just multiply by 27.1 one to solve for t.
It helps to know logs. :)


1. For how many nonnegative integers n does $ x^3 \plus{} (n \minus{} 1)x^2 \plus{} (n \minus{} n^2)x \minus{} n^3$ have all integer roots?

2. Consider the set of all equations $ x^3 \plus{} a_2x^2 \plus{} a_1x \plus{} a_0 \equal{} 0$, where $ a_2, a_1, a_0$ are real constants and $ |a_i| \le 2$ for $ i \equal{} 0, 1, 2$. Let r be the largest positive real number which satisfies at least one of these equations. Find r.

[hide="1"]We expand and factor:
$ x^3\plus{}(n\minus{}1)x^2\plus{}(n\minus{}n^2)x\minus{}n^3\equal{}x^3\plus{}nx^2\minus{}x^2\plus{}nx\minus{}n^2x\minus{}n^3$
$ \equal{}(x\minus{}n)(x^2\plus{}2xn\plus{}n^2\minus{}x)\equal{}(x\minus{}n)(x^2\plus{}(2n\minus{}1)x\plus{}n^2)$.

For this to factor, the discriminant of the quadratic must be a perfect square.  Then since the coefficient of $ x^2$ is 1 and the other coefficients are integers, the roots will be integers.

The discriminant is $ (2n\minus{}1)^2\minus{}4(n^2)(1)\equal{}4n^2\minus{}4n\plus{}1\minus{}4n^2\equal{}1\minus{}4n$.  This is negative unless $ n\equal{}0$, and in this case it is in fact a square.  Thus there is 1 nonnegative integer $ n$ so that that equation has all integer roots.[/hide]

[hide="2"]The greatest $ r$ is the root of $ x^3\minus{}2x^2\minus{}2x\minus{}2$.  I couldn't find any useful closed form of it, and now we show that this is the greatest.

Let $ s>r$ be the root of $ x^3\plus{}ax^2\plus{}bx\plus{}c$ where $ |a|, |b|, |c|\leq 2$.  Then
$ 0\equal{}s^3\plus{}as^2\plus{}bs\plus{}c\equal{}s^3\minus{}2s^2\minus{}2s\minus{}2\plus{}(2\plus{}a)s^2\plus{}(2\plus{}b)s\plus{}(2\plus{}c)$
$ >(2\plus{}a)s^2\plus{}(2\plus{}b)s\plus{}(2\plus{}c)>0$, impossible.

We know that $ s^3\minus{}2s^2\minus{}2s\minus{}2$ is positive, because after the root, it only increases. (shown by taking derivatives...) and we know that $ (2\plus{}a)s^2\plus{}(2\plus{}b)s\plus{}(2\plus{}c)>0$ because $ s$ and $ s^2$ are positive, and so are all of $ 2\plus{}a, 2\plus{}b, 2\plus{}c$.  Hence the root of $ x^3\minus{}2x^2\minus{}2x\minus{}2$ is the greatest $ r$ possible.[/hide]


Find the maximum number of planes in the space, such there are $ 6$ points, that satisfy to the following conditions:
[b]1.[/b]Each plane contains at least $ 4$ of them
[b]2.[/b]No four points are collinear.

We have at max 6 such planes.

A possible construction is $ (A_1,A_2,A_3)$ on a line and $ (B_1,B_2,B_3)$ on another skew line.

Given that we reached six we are not interested in configuration were there are:
1) 5 coplanar points (otherwise we can't find more than 1 plane)
2) no three collinear points (otherwise we can't find more than 3 planes)

So let $ (A_1,A_2,A_3)$ be collinear on the line $ l$. We need to show that there can not be more than 6 planes.
Clearly there are 3 planes containing $ l$. And any other plane can contain at most 1 point from $ l$. but then it would have to contain all other three points. so there are at most three other planes.
Clearly the construction above is the only case when this equality is reached.


inequalities
function
geometry
inequalities unsolved

Hi people!!!

I invite you to visit the [url=http://www.artofproblemsolving.com/Forum/index.php?f=255]Spanish Communities[/url]......

We try to post questions in 2-language versions... (spanish and english)
in order to practice your spanish...
checking and matching the most useful expressions and new words in spanish....

Second Post

 [color=blue][size=150]Inequalities[/size][/color]

Link: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59578

Carlos Bravo :)
Lima -PERU

The second one is in sholtlish, It's created by my teacher. There's $2$ ways to prove this.

The first one is very easy, but I think the following is better:

$aY+bZ+cX \le k^2$

(All of them is easy, too)

Yes you're right!!!

However ....
could you put your solutions??

Thanks in advance!!

Carlos Bravo :)
LIma-PERU

$a(k-b)+b(k-c)+c(k-a) \le k^2$
$\Leftrightarrow k(a+b+c-k) \le ab+bc+ca=a(b+c)+bc$
$\Leftrightarrow (k-a)(k-b-c)+bc \ge 0$
If $a+b$ or $b+c$ or $c+a \le k$, we are done.
If not, $bc \ge \frac{(b+c-a)^2}{4} \ge (b+c-k)(k-a)$
($4bc-(b+c-a)^2 =2ab+2ac+2bc-c^2-b^2-a^2 \ge 0$ because $a,b,c$ are 3 lengths of triangle.)

The second one is quite long! Ok, You should find it in this forum. I'm sure that It's posted before.

Proposed solution:

the function a-> aB+bC+cA is linear [b]"a"[/b] as variable in the intervall of definition [0,k], so the maximum is attained for [b]"a"[/b] belonging to {0,k}. Repeating the same argument for b and c variables, it follows that the maximum is attained for a,b,c belonging all to {0,k}. cheking the eight possible situations we obtain that the maximum is really [b]k²[/b].
 ;)

i must say i've seen this very very nice proof in some book i don't remember... anyway there's the solution

consider an equilateral triangle $ABC$ with $AB=AC=BC=k$ and define the points $M,N,P$ in $BC,AC,AB$ respectively such that $AP=a, PB=X, BM=c, MC=Z, CN=b, AN=Y$. after multiplying the original inequality by $\frac{\sqrt3}{4}$ we find it is equivalent to $(APN)+(CMN)+(PBM)\leq (ABC) \Leftrightarrow 0\leq(MNP)$ which is obviously true.

Ist one can be solved by using expansion of (1-a)(1-b)(1-c) ... and it comes out.
Or a geometry solution exists, was that from the camos guy.

Bomb

If we wanted to read the Spanish communities, we already would be doing so.  Post the problems in this thread, without the useless linkage / advertisement for the other forums.


$ \bigstar$ Problem 10:
Let $ a,b,c,d$ be positive real numbers. Find the min value of the following expression:
\[ \sqrt {\frac {a^2}{a^2 \plus{} \frac {1}{4}ab \plus{} b^2}} \plus{} \sqrt {\frac {b^2}{b^2 \plus{} \frac {1}{4}bc \plus{} c^2}} \plus{} \sqrt {\frac {c^2}{c^2 \plus{} \frac {1}{4}cd \plus{} d^2}} \plus{} \sqrt {\frac {d^2}{d^2 \plus{} \frac {1}{4}da \plus{} a^2}}
\]

you can get arbitarily close to 2 but letting a=c and b=d=a very small value almost equal to zero. :maybe:
doing it another way i got the solution of $ [4*\sqrt2]/3$ but i don't think it can be reached   :(


function
integration
real analysis
real analysis unsolved

Let $f,g: [0,1]\to R$ continuous functions such that $\int_{0}^{1}f(x)dx=\int_{0}^{1}g(x)dx=1$. Prove that we can find an interval $[a,b]$ included in $[0,1]$ such that:
$\int_{a}^{b}f(x)dx=\int_{a}^{b}g(x)dx=\frac{1}{2}$

I hope I stated well the problem:)

I hoped it was not posted :(  However I look for  an elementary approach :P ...I know (maybe I am wrong) that one exists ( related to the famous King's problem)

Can somebody help me and give some hints about that elementary approach...I know that the square [0,1]*[0,1] was divided diagonally in many parts then these parts were coloured with white and black somehow related to the functions f and g. Then was applied in a tricky way the King's problem and maybe other results...But I am totally lost and I can't put together all these :(  Can u help me please?

Please read this topic (king's theorem) http://www.mathlinks.ro/Forum/viewtopic.php?p=785007#785007 and tell me if u can find a solutin using that fact...I know for sure that one ( a   proof from the Book :P ) exists  :)

According to http://en.wikipedia.org/wiki/Proofs_from_THE_BOOK, "The Book" is something Erdos made up and that caught on.

Of course, the proof I am looking for is not in "The Book" (maybe in The Book Erdos has now we can find it at some page or whatever:)...but I can't ask him ) It was just a way of saying things :P
My question is still unsolved :(  Please help


How many square units are in the area of a semicircle with radius 2 inches? Express in terms of $ \pi$.

Area of circle: 2^2pi. Half of that is 2pi


[color=darkblue]Let $ \begin {cases} a.b.c \neq 0\\a\plus{}b\plus{}c\equal{}2\\a.x\plus{}b.y\plus{}c.z\equal{}6 \end {cases}$. Prove that $ \sqrt{16a^2\plus{}a^2x^2}\plus{}\sqrt{16b^2\plus{}b^2y^2}\plus{}\sqrt{16c^2\plus{}c^2z^2}\ge 10$[/color]

Follow CBS.

       $ ((4a)^2\plus{}(ax)^2))((\frac{1}{3})^2\plus{}(\frac{1}{4})^2)\geq(\frac{4a}{3}\plus{}\frac{ax}{4})^2$

    and  similarly,

    $ \sum \sqrt{16a^2\plus{}a^2x^2}\cdot \frac{25}{9\cdot16}\geq\sum{\frac{4a}{3}\plus{}\frac{ax}{4}}$

  and inequality is prove.

from minkowsky $ LHS\geq \sqrt{16(a\plus{}b\plus{}c)^2 \plus{} (ax\plus{}by\plus{}cz)^2}\equal{}\sqrt{64\plus{}36}\equal{}10$
P.S minkowsky is really useful in inequalities like these.

[color=darkblue]Yes. Thank you very much. I can done it by your way. And I can done it by Vector.[/color]


linear algebra
matrix
linear algebra unsolved

Let $ A$ be $ n$x$ n$ matrix. Show that there exsists such $ n$x$ n$ matrix $ B$ that holds $ ABA\equal{}A$.

We assume $ A\in\mathcal{M}_n(k)$ where $ k$ is a field.
If $ A$ is invertible then take $ B\equal{}A^{\minus{}1}$,
else $ A$ is similar, over $ k$, to $ diag(U,N)$ where $ U$ is invertible and $ N$ is nilpotent.
Thus we may assume $ A\equal{}J_r$ where $ J_r$ is the Jordan nilpotent matrix of dimension $ r$. Then take $ B\equal{}{J_r}^T$.

It is a particular case of pseudoinverse matrix theory (Moore-Penrose):
if $ A$ is a $ (m,n)$ matrix then there exists an unique $ (n,m)$ matrix $ B$ s.t.:
$ ABA\equal{}A,BAB\equal{}B,AB$ and $ BA$ are hermitian matrices.
cf: wikipedia Moore-Penrose.


Um, I really wanted to go to MathPath, but not being a National MC winner type person, I am going to have to pay \$4000 or more.

I've heard this is really fun, but you know, my parents are not very eager to spend \$4000 on this.

Your input on the valubility of Mathpath please?

_________________________________________________________________
This is not my signature. If you cannot figure out why, \pm me.

[quote="pinkmuskrat"]First of all, the scholarships for AMC and MC are not [i]huge[/i], in terms of percent of total tuition. http://www.mathpath.org/fees/schol.htm

It is really fun, and definitely a very worthwhile experience. You are exposed to so many new ideas and concepts. You learn approaches and unique perspectives, and also a dash of theorems. The month is filled with laughter, excitement, and passion with kids and professors who love math as much as we do. For myself, and for many others, it was the greatest month ever.

You should contact Mr. M and MathPath to discuss your specific concerns with schedule, finances, parents, etc. Also on the site you may want to check out the Mission, FAQ, etc.

(I'm not sure if I should be posting Mr. M's email.  :P ) so here's this: http://www.mathpath.org/contact.htm[/quote]

So what I mean is really, is it worth \$4000 (or possibly somewhat less depending on application date)?

Please, I'd really appreciate answer. Especially directed to past MathPathers.

[quote="laughinghead505"]Huh, since when are there scholarships?
I don't think anyone can give you a definite answer. It was worthwhile for me, but the experience is different for each person. Look at the website, talk to past MathPathians, and decide for yourself.
And as far as I know, application date has no effect on price.[/quote]

You are given a 1% reduction for every month earlier than the application deadline that you submit.

:maybe: Wow, they've changed a lot since I went (06-07).

If you like math, you'll really, really enjoy mathpath. You'll have a lot of fun learning all about different mathematical topics (many times topics you didn't know even existed) and math history, and will enjoy getting to know lots of people who share your interest in math. For me, mathpath was a great experience. Though it's not cheap (I didn't qualify for any scholarships), mathpath is [b]definately worth it[/b]. I would highly recommend it to anyone who enjoys math.

Bottom line:[u]MathPath is a great camp. You should convince your parents to let you go.[/u]

I will answer your question: Yes, it is definitely worth $ \$$5000.

I know I know. Still convincing is much better said than done. P.s. Just in case you guys are wondering, I am not in any low income/financial situation. :)

Hi I was at MathPath 2009. I applied for financial assitance(whatever it is, it should be in the application form) and I only had to pay $ 2000. Anyway, I'd say it's worth$2500. Also, if you are applying for financial aid, you must be really good in math which is shown through your answers to the question.

I was at MP08 and 09 and was one of the best experiences I've ever had. Although I already enjoyed math before I came here, going here really solidified my love and interest in mathematics - not just competitions, but topics that go beyond the regular scope of competition based programs, and more into research of mathematics. Also, meeting and making friends with other like-minded friends really made me feel part of a community and happy with the knowledge that I wasn't alone and that there were actual people out there instead of just avatars on a computer screen. I strongly recommend MathPath.

MathPath was one of the best experiences that I have had with math. The culture of MathPath not only excepts nerds but actually embraces them. At MathPath I felt the most comfortable being myself that I ever have been. I would say the MathPath is definitely worth the cost.


you don't pay tax in the tip [hide="easy"]  $15*1.10$(the tax) $+5 = 1.5+15+5= 16.5+5= 21.5$[/hide] :showoff:  :rotfl:

[hide]
The meal costs:
15
The tax is costs:
1.5
The tip costs:
5
The total is
21.5[/hide]

[quote="pascal_1623"]Bobby is going out to dinner.
He pays 15 dollars for his meal.
He gives a five dollar tip and pays 10% tax.
How much did Bobby spend?[/quote]

[hide]
15 dollars from the meal
Then add tax (which I notice when at resturants, tip is after tax) so it is 1.5 dollars
5 dollar tip

15+1.5+5=21.50

[/hide]

[hide]
The tax is $.1\cdot15=1.50$ The tip is 5
Add together and get 21.50[/hide]

[hide]Meal=15.Tax=1.5.Tip=5. The sum of all these is 21.5[/hide]
Whatever happened to using Billy Bob?  :P

Easy as 3.14159
Dinner cost  15 dollars; 10% of 15 is 1.5; Tip was 5 dollars
Add together to get a total sum of...

[hide]2150 cents[/hide]

IF you had to pay tax on the tip :) it would be $22 if not, it would be$21.50
 :rotfl:

[hide] The meal is \[15$, the tax is \]: 0136112cbe1.50$, and the tip is \[5.00$. Adding these, we get \]: 0136112cbe21.50$[/hide]

Bobby is cool. So is ice cream. :icecream:

Sorry about my earlier answer. I messed up.  :blush:  :oops:  :oops_sign:

[hide]Anway, the meal is 15 dollars, the tax is one dollar and fifty cents, and the tip is five dollars. The total is 21 dollars and 50 cents.[/hide]

There's tons of answers already, do we really need to bump up a 6 month old topic?


A relief airplane is delivering a food package to a group of people stranded on a very small island. The island is too small for the plane to land on, and the only way to deliver the package is by dropping it. The airplane flies horizontally with constant speed of 290 mph at an altitude of 600m . The positive x and y directions are defined as usual. For all parts, assume that the "island" refers to the point at a distance D from the point at which the package is released.

a) After a package is ejected from the plane, how long will it take for it to reach sea level from the time it is ejected? Assume that the package, like the plane, has an initial velocity of 290 mph in the horizontal direction.

b) If the package is to land right on the island, at what horizontal distance  from the plane to the island should the package be released?

c) What is the speed v_f of the package when it hits the ground?

[hide]Let $ x,h,v$ respectively be the horizontal distance, height and the horizontal velocity of the airplane at the moment it releases the reliefs and let $ t$ be the time it takes for the reliefs to reach the ground.

The vertical component of the velocity is $ v\cos 90^{\circ}=0$. Hence $ h=\frac 12gt^2$ which gives $ t=\sqrt{\frac{2h}{g}}$. This can be used to answer (a).

The horizontal component of the acceleration of the relief is $ g\cos 90^{\circ}=0$. Hence the velocity is uniform and thus $ x=vt=v\sqrt{\frac{2h}{g}}$ thus we may find the answer to (b).

The moment the reliefs hit the ground, it's velocity becomes vertical and so $ v_f^2=0+2gh\implies v_f=\sqrt{2gh}$ from which we can find the answer to (c).[/hide]

a) Apply: $ y\left(t\right) \equal{} y_0 \plus{} v_y\left(0\right) t \minus{}\frac{1}{2}g t^2$

Solve for $ t$.

b) Apply: $ x\left(t\right) \equal{} v_x\left(0\right) t$.

Insert $ t$ from part (a).

c) Apply: $ v_{yf}^{2} \equal{} v_{y}^{2}\left(0\right) \plus{} 2gh$

and $ v_{xf} \equal{} v_{x}\left(0\right)$ to,

$ v_{f}^{2} \equal{} v_{yf}^{2} \plus{} v_{xf}^2$.


[quote="mostel"]Σας παραθέτω την άσκηση που είχε το τεστ στην έλλειψη... όχι ότι είναι κάτι το ιδιαίτερο, απλώς έχει μπόλικες πράξεις.

[color=red][b]ΑΣΚΗΣΗ[/b][/color]

Να βρεθεί η εξίσωση της έλλειψης $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ με  $(a>b)$που εφάπτεται στις:

$\varepsilon_{1}: \quad x-4y+10=0$
$\varepsilon_{2}: \quad x+y-5=0$

Have fun !  :lol:[/quote]

 Γενικά , αν η έλλειψη που έγραψες εφάπτεται στην ευθεία $y = lx+k$ , τότε $a^{2}l^{2}+b^{2}= k^{2}$. Ίσως αυτό βοηθάει λίγο στο να γίνουν οι πράξεις ανθρώπινες.

Ακριβώς κύριε Στεργίου !

Στο τεστ το πήρα ως δεδομένο ότι είναι έτσι και δεν έβαλα απόδειξη ! (είναι προφανής άλλωστε)

Ελπίζω να μη μου κόψει μονάδες :huh: :P

Πως μπορούμε όμως να το αποδείξουμε αυτο? :idea:

Πάρε την εξίσωση της έλλειψης, βάλε όπου y αυτό που έχει παραπάνω και κάνε πράξεις


function
integration
calculus
derivative
calculus computations

Prove that if $ f(x)$ is an odd function, then $ \int_{\minus{}u}^u f(x)\,dx\equal{}0.$

Prove that if $ f(x)$ is an even function, then $ \int_{\minus{}u}^u f(x)\,dx\equal{}2\int_0^u f(x)\,dx.$

This could be quite trivial, but, how do you prove that an odd function satisfies $ f(\minus{}x)\equal{}\minus{}f(x)\,?$ And for an even function $ f(\minus{}x)\equal{}f(x)\,?$

I actually found the proofs. Anyway, if someone want to explain the last part of the post.

That odd and even functions satisfy $ f(\minus{}x)\equal{}\minus{}f(x)$ and $ f(\minus{}x)\equal{}f(x)$, respectfully is a matter of definition in most treatments. Furthermore, the derivative of an odd (even) function is an even (odd) function.

Yeah, thanks.

When I posted this, I realized that they were a definition.

Advice to my calculus students, repeated in class dozens of times: if you see an integral over an interval symmetric about the origin, [b]always[/b] check whether the function is odd or even (or even whether it can be easily split into an even part plus an odd part), and [b]always[/b] take advantage of these identities. Even the "fold it over and double it" identity for even functions. The latter is something that might not make much difference for a machine, but for a human doing the problem by hand, it greatly improves the odds of getting the answer without an arithmetic mistake.

$ \int_{\minus{}3}^3x^4\minus{}6x^3\minus{}2x^2\plus{}7x\,dx\equal{}
\left.\frac{x^5}5\minus{}\frac{6x^4}4\minus{}\frac{2x^3}3\plus{}\frac{7x^2}2\right|_{\minus{}3}^3$

as a hand calculation is a sign error (or several sign errors) waiting to happen.

$ \int_{\minus{}3}^3x^4\minus{}6x^3\minus{}2x^2\plus{}7x\,dx\equal{}
\left.2\left(\frac{x^5}5\minus{}\frac{2x^3}3\right)\right|_{0}^3$

is something that probably will be calculated correctly.


trigonometry Gauss function LaTeX factorial algebra polynomial

[hide]
Use complex numbers
[hide]DeMoivre's Theorem[/hide]
[/hide]

Yes, that is a good hint.  Another hint would be: [hide="hint"]Gauss Summation FOrmula Variations[/hide].

Masoud Zargar

[quote="boxedexe"]Simplify:

$S=\sin \alpha+ \sin 2\alpha + \sin 3\alpha +...+ \sin 1000\alpha$

Masoud Zargar[/quote]

So we want: $\Im{\sum_{i = 1}^{1000}e^{i\alpha} }$, right?
But I can't simplify that, though.  :(

Can I give the answer or should I let people try more?

Masoud Zargar

How about a [size=200]big [/size]hint? :)

[quote="boxedexe"]Okay, here's a [size=150]big[/size] hint:

[hide="hint"]
Figure out a formula, using Gaussian variations for the sum of sine functions.
[/hide]

Masoud Zargar[/quote]

That's the problem; I don't know what Gaussian variations are. Was I on the right track before?

[quote="boxedexe"]Well, Gaussian variations are basically formulas deduced from Gauss' Summation formula $\frac{n(n+1)}{2}$.  If you have msn, I will be glad to fully explain them to you, including some techniques.

Masoud Zargar[/quote]

I know what you are tralking about, but I don't have the answer to this still. How about you just run through the first few steps of the problem? Its so frustrating...... But I like this problem though. :lol:

Lol, okay...I am not going to go straight to the formula, but I'll pave the path for you.  First, we will try to find some similarities between $\frac{n(n+1)}{2}$ and teh unknown formula of the summation of sine functions.  Since we have $\alpha$ in all of the terms in $S=\sin\alpha+\sin 2\alpha+\sin 3\alpha+...+\sin 1000\alpha$, we will multiply $\frac{n(n+1)}{2}$ by $\alpha$, which become $\frac{n(n+1)\alpha}{2}=\alpha + 2\alpha+ 3\alpha +...+n\alpha$.  If we consider $\sin$ as an ordinary algebraic function for teh moment, we'll have $\sin(\alpha+ 2\alpha + 3\alpha +...+ n\alpha)=\sin \frac{n(n+1)\alpha}{2}$... I will add to this if you still need some hints.  This is fun. :)

Masoud Zargar

Masoud, your explanation to me was in vain :D . I found an alternative solution. Sorry!

[hide="Took me 2 hours to get this....."]$S=\sin\alpha+\sin 2\alpha+\sin 3\alpha+...+\sin 1000\alpha$

Multiply both sides by $2\sin\alpha$, and used the identity $\cos{(a-b)}- \cos{(a+b)} = 2\sin{a}\sin{b}$ and watch the terms cancel. You get

$2S\sin\alpha = \cos\0 + \cos\alpha - \cos 1000\alpha - \cos 1001\alpha$

$S= \frac{ 1 + \cos\alpha - \cos 1000\alpha - \cos 1001\alpha}{2\sin\alpha}$

Hopefully, I am done.[/hide]

I'd still like to see your solution, though.
Edit: Is this simplified? :(

Nice job!  Hoever, my solution is a little smaller than that. try simplifying that.  Even though I consider mathematics, physics, chemistry, and basically knowledge my life, I am tired and need to sleep.  I'll write down my entire solution and proofs in several hours (maybe in 7 hours).:)

Masoud Zargar

Well, I just woke up, so I'm ready for some mathematics.  Here's my solution:

[hide]
First, we will try to find some similarities between $\frac{n(n+1)}{2}$ and the unknown formula of the summation of sine functions.  Since we have $\alpha$ in all of the terms in
\[S=\sin\alpha+\sin 2\alpha+\sin 3\alpha+...+\sin 1000\alpha,\]
we will multiply $\frac{n(n+1)}{2}$ by $\alpha$, which becomes
\[\frac{n(n+1)\alpha}{2}=\alpha + 2\alpha+ 3\alpha +...+n\alpha\].
If we consider $\sin$ as an ordinary algebraic function for the moment, we'll have
\[\sin(\alpha+ 2\alpha + 3\alpha +...+ n\alpha)=\sin \frac{n(n+1)\alpha}{2}\].
We then use a variation of Gauss' formula $\frac{n(n+1)}{2}$ to deduce the following.  Let
\[S=\sin\alpha + \sin 2\alpha +...+ \sin(n-1)\alpha + \sin n\alpha\]
\[S=\sin n\alpha + \sin(n-1)\alpha +...+\sin 2/alpha + \sin\alpha.\]
Since we know that Gauss used 2 vertical columns to figure out a formula for $\sum_{x=1}^{n}x$, we will do the same thing with sine functions and sum-to-product formula, which is
\[\sin\alpha + \sin\beta = 2\sin (\alpha + \beta )/2 \cdot  \cos (\alpha - \beta )/2\],
we get
\[2S=2[\sin(1+n)\alpha/2 \cdot \cos(1-n)\alpha/2 + \sin(1+n)\alpha/2\]
\[\cdot \cos(3-n)\alpha +...+\sin(n+1)\alpha/2\]
\[\cdot \cos(n-3)\alpha/2 +\sin(n+1)\alpha/2 \cdot \cos(n-1)\alpha/2]\]
\[=2\sin(n+1)\alpha/2 \cdot[\cos(n-1)\alpha/2+\cos(3-n)\alpha/2+...\]
\[+\cos(n-3)\alpha/2+\cos(n-1)\alpha/2].\]
Now, we have a very tedious equation to deal with; one way would be to use the product-to-sum formula \[\sin\alpha\cdot\cos\beta=\frac{\sin(\alpha-\beta)+\sin(\alpha+\beta)}{2}.\]
This product-to-sum formula leads us to
\[2S\sin\alpha/2=\sin(n+1)\alpha/2\cdot[\sin n\alpha/2\] \[+\sin(1-n/2)\alpha+\sin(-1+n/2)\alpha+\sin(2-n/2)\alpha+...\]
\[+\sin(2-n/2)\alpha+\sin(-1+n/2)\alpha+\sin(1-n/2)\alpha+\sin n\alpha/2.\]
We know that $\sin(-1+n/2)\alpha=-\sin(1-n/2)\alpha$, which also applies for many other terms in the equation.  As a result, the terms will cancel each other out, and we'll end up with
\[2S\sin\alpha/2=2\sin(n+1)\alpha/2\cdot\sin n\alpha/2\]
\[S=\frac{\sin(n+1)\alpha/2\cdot\sin n\alpha/2}{\sin\alpha/2}.\]
We can then use this to simplify $S=\sin\alpha+\sin 2\alpha+\sin 3\alpha+...+\sin 1000\alpha$:
\[S=\frac{\sin 500.5\alpha\cdot\sin 500\alpha}{\sin\alpha/2}.\]
[/hide]

I hope I didn't make any mistakes with all of this $\LaTeX$. :)

Masoud Zargar

[quote="kunny"][quote="boxedexe"]Simplify:

$S=\sin \alpha+ \sin 2\alpha + \sin 3\alpha +...+ \sin 1000\alpha$

Masoud Zargar[/quote]

We are to calculate $S\sin \frac{\alpha}{2}$ or $S\cos \frac{\alpha}{2}$[/quote]

The Kunny's solution is the most simple solution. I supose w.l.o.g$.\ \alpha =1$. Therefore,

$2S_1 \sin \frac 12 =\sum\limits_{k=1}^n 2\sin k\cdot \sin \frac 12 =\sum\limits_{k=1}^n \left( \cos \frac {2k-1}{2} -\cos \frac {2k+1}{2} \right)=\cos \frac 12 -\cos \frac {2n+1}{2}=$

$2\sin \frac{n+1}{2}\cdot \sin \frac n2\Longrightarrow S_1=\frac {\sin \frac {n+1}{2}\cdot \sin \frac n2 }{\sin \frac 12 }.$ In conclusion, $S_{\alpha}=\frac {\sin \left( \frac {n+1}{2}\cdot \alpha \right)\cdot \sin \left( \frac n2 \cdot \alpha \right) }{\sin \left ( \frac 12 \cdot \alpha \right)}.$

.

I believe this is an interesting problem. For teh question I pruposed, my solution is the longest. :(

[quote="blahblahblah"]$123....$, that is a very nice solution, nicer, in fact, than mine (even though mine requires no ingenuity).  To do it the 'other' way, the following manipulations work:

$\sum_k \sin (k\alpha)=\text{Im}\left(\sum_k e^{ik\alpha}\right)=\text{Im}\left(\sum_k (e^{i\alpha})^k\right)$

That is just a geometric series; now hack it out.[/quote]

Sorry, I am not very familliar with some things.  Could you please tell me what $\text{Im}$ mean? PLease give an example if possible.

Masoud Zargar

$\text{Im}$ just means the imaginary part of a complex number ($\text{Re}$ is the real).  For example, $\text{Im}(e^{\frac{\pi}{6}i}) = 0.5i$ and $\text{Re}(e^{\frac{\pi}{6}i}) = \displaystyle\frac{\sqrt{3}}{2}$

This is an idea, but I haven't actually cranked it out: could you use the alternate definition of $\sin x = \displaystyle\frac{e^{ix}-e^{-ix}}{2i}$ and sum up two geometric series?  I'm not sure if this is what blahblahblah was saying also.

[quote="Gyan"]Okay, this is also an old  classic problem,  if you have not seen it before, see if you can do it"
what is the value of  this (product .).
$sin 1^o \cdot  sin 2^o \cdot sin 3^o \dots  sin 177^o \cdot sin 178^o \sin 179^o$[/quote]

I think I have an idea how to solve this problem...I just need to figure out a formula for tehj product of sine functions, which may be found using factorial so on and so forth.

Masoud Zargar

The conjecture that you can sum trigonometric functions in a manner analogous to polynomials is a bit of a lucky guess that works out in the earlier case, but it doesn't necessarily work if you try and extend it to multiplication.  You should always be a bit careful of methods that lack rigor.

[quote="Gyan"]Okay, this is also an old  classic problem,  if you have not seen it before, see if you can do it"
what is the value of  this (product .).
$sin 1^o \cdot  sin 2^o \cdot sin 3^o \dots  sin 177^o \cdot sin 178^o \sin 179^o$[/quote]

I am stumped....... perhaps if I multiplied this by something??????

Sorry to double post, but I can't figure this out.
Help would be appreciated.

The discussion  for the product part can./is continued in
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53316]www.artofproblemsolving.com/Forum/viewtopic.php?t=53316[/url]

you could try and pairing terms...i don't know what else would work

if you pair up $\sin{x}*\sin{180-x}$, then you get $\sin^2{x}$ for x 1 to 89
and then if you do $\sin^2{x}*\sin^2{90-x}$ is $\dfrac{\sin^2{2x}}{4}$ for x 1 to 44 times sin45
i am not sure what to do next, but i think that i have seen this before


Two standard 4-sided dice are tossed. What is the probability
that the sum of the numbers shown on the dice is a multiple of 3?
Express your answer as a common fraction.

Sorry.
I misunderstood only one dice.

1-234: 2+3+4=9=3\times3
2-134: 1+3+4=8=3
3-124: 1+2+4=7=3
4-123: 1+2+3=6=3\times2
So,
$ \begin{array}{|c\parallel{}c|c|c|c|} \hline & 1 & 2 & 3 & 4 \\
\hline \hline 1 & \color{red}18\color{black} & 17 & 16 & \color{red}15\color{black} \\
\hline 2 & 17 & 16 & \color{red}15\color{black} & 14 \\
\hline 3 & 16 & \color{red}15\color{black} & 14 & 13 \\
\hline 4 & \color{red}15\color{black} & 14 & 13 & \color{red}12\color{black} \\
\hline \end{array}$

Thus, $ \frac {6}{4^2} \equal{} \boxed{\frac {3}{8}}$.


Given a rational number in lowest terms, Alvin calculates the product of the numerator and denominator. For how many rational numbers in lowest terms between 0 and 1 will 10! be the resulting product?

Let $ \frac {p}{q}$ be the rational number, where $ p$ and $ q$ are coprime integers. If the number is less than 1, we must have that $ p < q$. So we have also that $ pq \equal{} 10!$.

The problem is now a lot easier: Find the number of pairs of factors of $ 10!$.

$ 10! \equal{} 2^8 \cdot 3^4 \cdot 5^2 \cdot 7$

There are $ 9 \cdot 5 \cdot 3 \cdot 2 \equal{} 270$ factors, so there are $ \frac {270}{2} \equal{} \boxed{135}$ pairs.

EDIT: oh so stupid.. Didn't see that  :blush:

[quote="modularmarc101"]Let $ \frac {p}{q}$ be the rational number, where $ p$ and $ q$ are coprime integers. If the number is less than 1, we must have that $ p < q$. So we have also that $ pq \equal{} 10!$.

The problem is now a lot easier: Find the number of pairs of factors of $ 10!$.

$ 10! \equal{} 2^8 \cdot 3^4 \cdot 5^2 \cdot 7$

There are $ 9 \cdot 5 \cdot 3 \cdot 2 \equal{} 270$ factors, so there are $ \frac {270}{2} \equal{} \boxed{135}$ pairs.[/quote]How are these pairs coprime?

But, 6! and 7! are not coprime and yet the product is 10!. This is another pair counted in his solution.

[quote="Complex_Ninja"]Basically u need either $ 2^8$, $ 3^4$, $ 5^2$, $ 7$ or $ 1$ either in the denominator or the numerator. Then subtract the pairs where the numerator is $ >$ denominator. So its 31 - something[/quote]

The values of the following fractions are greater than 0 and less than 1:

$ \frac {1}{10!}$

$ \frac {2^8}{3^4\cdot5^2\cdot7}$

$ \frac {2^8\cdot7}{3^4\cdot5^2}$

$ \frac {3^4}{2^8\cdot5^2\cdot7}$

$ \frac {3^4\cdot7}{2^8\cdot5^2}$

$ \frac {5^2}{2^8\cdot3^4\cdot7}$

$ \frac {5^2\cdot7}{2^8\cdot3^4}$

$ \frac {7}{2^8\cdot3^4\cdot5^2}$

8 total rational numbers in lowest terms. . . Answer


limit
induction
algebra
polynomial
calculus
calculus computations

Let $ a_n$ be a sequence defined by $ a_1\equal{}1,\ a_2\equal{}3,\ a_{n\plus{}2}\equal{}6a_{n\plus{}1}\minus{}5a_n\ (n\geq 1)$.
Find $ \lim_{n\to\infty} \frac{a_n}{a_{n\plus{}1}}$.

[hide="Solution"]Rearranging the terms, we get this: $ a_{n\plus{}2}\minus{}a_{n\plus{}1}\equal{}5(a_{n\plus{}1}\minus{}a_n)\implies \left(1\minus{}\frac{a_{n\plus{}1}}{a_{n\plus{}2}}\right)\cdot a_{n\plus{}2}\equal{}5\left(1\minus{}\frac{a_n}{a_{n\plus{}1}}\right)\cdot a_{n\plus{}1}$

$ 1\minus{}\frac{a_{n\plus{}1}}{a_{n\plus{}2}}\equal{}5\left(1\minus{}\frac{a_n}{a_{n\plus{}1}}\right)\cdot \frac{a_{n\plus{}1}}{a_{n\plus{}2}}$

Let $ L\equal{}\lim_{n\to\infty}\frac{a_n}{a_{n\plus{}1}}$.

$ 1\minus{}L\equal{}(5\minus{}5L)L\implies L\equal{}1,\frac 15$.

It's clear that the sequence is monotonically increasing, however, so we expect $ L<1$, and conclude that $ L\equal{}\boxed{\frac 15}$.[/hide]

[hide="Cool observation/question"]
After finding $ a_{n\plus{}2}\minus{}a_{n\plus{}1}\equal{}5(a_{n\plus{}1}\minus{}a_n)$, we can write this as $ \Delta a_{n\plus{}2}\equal{}5\Delta a_{n\plus{}1}$.

Now noting that $ \lim_{n\to\infty}a_n\to\infty$, we see that the limit is of the indeterminate form $ \frac\infty\infty$.

And $ \lim_{n\to\infty}\frac{a_n}{a_{n\plus{}1}}\equal{}\lim_{n\to\infty}\frac{\Delta a_n}{\Delta a_{n\plus{}1}}\equal{}\boxed{\frac 15}$

This is like a discrete analog of L'Hopital's rule. Does this generalize somehow, or is it coincidental?
[/hide]

Your answer is correct.

Here is the solution for my students.

Rewrite the given reccursion in two ways,

$ a_{n \plus{} 2} \minus{} a_{n \plus{} 1} \equal{} 5(a_{n \plus{} 1} \minus{} a_n),\ a_{n \plus{} 2} \minus{} 5a_{n \plus{} 1} \equal{} a_{n \plus{} 1} \minus{} 5a_n$

$ \therefore a_{n \plus{} 1} \minus{} a_n \equal{} 5^{n \minus{} 1}(a_2 \minus{} a_1) \equal{} 2\cdot 5^{n \minus{} 1}$
$ a_{n \plus{} 1} \minus{} 5a_n \equal{} a_2 \minus{} 5a_1 \equal{} \minus{} 2$
Subtracting both sides gives $ 4a_n \equal{} 2\cdot 5^{n \minus{} 1} \plus{} 2$, yielding $ a_n \equal{} \frac {5^{n \minus{} 1} \plus{} 1}{2}\ (n\geq 1)$, thus $ \lim_{n\to\infty} \frac {a_n}{a_{n \plus{} 1}} \equal{} \lim_{n\to\infty} \frac {5^{n \minus{} 1} \plus{} 1}{5^n \plus{} 1} \equal{} \frac 15$

hello, alternatively you can prove by induction that $ a_n\equal{}\frac{1}{10}5^n\plus{}\frac{1}{2}$.
Sonnhard.

Hi;

[hide]
Using the auxiliary polynomial solve the recurrence to get:

$ a_n \equal{} \frac {5^n \plus{} 5}{10}$

$ \\
\frac {a_n}{a_{n \plus{} 1}} \equal{} \cfrac {\cfrac {5^n \plus{} 5}{10} }{\cfrac {5^{n \plus{} 1} \plus{} 5}{10} } \equal{} \frac {5^n \plus{} 5}{5^{n \plus{} 1} \plus{} 5}, \\
 \\
 \\
\lim_{n\to\infty}\frac {5^n \plus{} 5}{5^{n \plus{} 1} \plus{} 5} \equal{} \frac {1}{5}$[/hide]


inequalities trigonometry calculus logarithms function algebra domain

How do you show that
\[\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}\leq\frac{1}{8}\]
where $\alpha ,\beta ,\gamma$ are  angles of a triangle.
[color=red]with use of multivariable calculus, considering it as a extremevalue problem?[/color]
I assume we should use that $\alpha+\beta+\gamma =\pi$ as an constraint.

[color=green][EDIT: methods like lagrange multipliers or something.][/color]

[hide="Hint"]We should prove the following inequality.
\[\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}\leq \left( \frac{3}{\pi}\right)^{3}\frac{\alpha}{2}\cdot \frac{\beta}{2}\cdot \frac{\gamma}{2}\]
$\log\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)$ is concave.
Please try to use Jensen　ineq for this.[/hide]

Thanks a lot but the aim was to
solve this by calculus methods, like lagrange multipliers or
othe extremevalue problem methods.

We're trying to maximize the function $f(\alpha,\beta,\gamma)=\sin\frac{\alpha}2\sin\frac{\beta}2 \sin\frac{\gamma}2$ for $\alpha,\beta,\gamma\in[0,\pi]$ subject to the constraint $\alpha+\beta+\gamma=\pi.$

Lagrange multipliers leads to

$\frac12\cos\frac{\alpha}2\sin\frac{\beta}2 \sin\frac{\gamma}2=\lambda$
$\frac12\sin\frac{\alpha}2\cos\frac{\beta}2 \sin\frac{\gamma}2=\lambda$
$\frac12\sin\frac{\alpha}2\sin\frac{\beta}2 \cos\frac{\gamma}2=\lambda$

If any of $\alpha,\beta,$ or $\gamma$ equals $0,$ then $f(\alpha,\beta,\gamma)=0.$ Since this function is positive otherwise, such points are not the maximum we seek. Hence we assume that $\alpha,\beta,$ and $\gamma$ are all nonzero.

Eliminating $\lambda$ between the first two equations and dividing by $\sin\frac{\gamma}2,$ we get

$\cos\frac{\alpha}2\sin\frac{\beta}2=\sin\frac{\alpha}2\cos\frac{\beta}2$ which becomes

$\sin\frac{\alpha}2\cos\frac{\beta}2-\cos\frac{\alpha}2\sin\frac{\beta}2=0$ or $\sin\left(\frac{\alpha}2-\frac\beta{2}\right)=0.$

This leads us to $\alpha=\beta.$ The alternatives would be things like $\alpha-\beta=2\pi,$ which lie outside our region.

A very similar computation with the second and third equations yields $\beta=\gamma.$

Hence our only nontrivial critical point is $\alpha=\beta=\gamma=\frac{\pi}3.$ With our function being zero on the boundaries of its domain and positive in the interior, this critical point must be a maximum. Plugging in these values gives us our desired inequality.

This was outstanding! Thanks a lot!
 :lol:


Let be $ x,y\in \mathbb{R}$ . Show that :

i)$ \sin^2\ \frac{x\minus{}y}{2}\plus{}\cos \ x\cdot \cos \ y\ge 0$

ii)$ \sin^2\ \frac{x\plus{}y}{2}\minus{}\sin \ x\cdot \sin \ y\ge 0$

Let's apply Werner:

\[ \cos\alpha \cdot \cos\beta \equal{} \frac{1}{2}(\cos(\alpha \plus{} \beta) \plus{} \cos(\alpha \minus{} \beta))\]
\[ \sin\alpha \cdot \sin\beta \equal{} \frac{1}{2}(\cos(\alpha \minus{} \beta) \minus{} \cos(\alpha \plus{} \beta))\]

So

i) $ \sin^2\ \frac{x\minus{}y}{2}\plus{}\cos \ x\cdot \cos \ y\ge 0 \Rightarrow \frac{1}{2}(1 \minus{} \cos(x\minus{}y)) \plus{} \frac{1}{2}(\cos(x\plus{}y) \plus{} \cos(x\minus{}y) \ge 0$
$ \Rightarrow \cos(x\plus{}y) \ge \minus{}1$ which is true

ii)

$ \sin^2\ \frac{x\plus{}y}{2}\minus{}\sin \ x\cdot \sin \ y\ge 0 \Rightarrow \frac{1}{2}(1 \minus{} \cos(x\plus{}y)) \minus{} \frac{1}{2}(\cos(x\minus{}y) \plus{} \cos(x\plus{}y) \ge 0$
$ \Rightarrow \cos (x\minus{}y) \le 1$ which is true

hello, to i) applying the addition formulas we get $ \sin^2\left(\frac{x\minus{}y}{2}\right)\plus{}\cos(x)\cos(y)\equal{}\frac{1}{2}\left(1\plus{}\cos(x\plus{}y)\right)\geq0$
Sonnhard.

hello, by the same method we get for ii)
$ \sin^2\left(\frac{x\plus{}y}{2}\right)\minus{}\sin(x)\sin(y)\equal{}\frac{1}{2}\left(1\minus{}\cos(x\minus{}y)\right)\geq0$.
Sonnhard.


geometry
3D geometry
algebra
factorization
sum of cubes

[quote="sweepsfairie"]I have a few problems that I am stuck on I am studying for a test and its driving me crazy!!!
I will write it out because I don't know how to get signs on here!!

x to the fifth power + 27x squared   I need to factor this completely so far I have

x squared (xcubed + 27)  but I can't figure out how to factor it any further than that.

Then I have
8x cubed + 24x squared -2x-6  I have so far
8x squared (x+3) - 2(x+3) so then I combine it to
(8x squared -2) (x+3)  what do I do after that?  Do I make the first one 4x squared -1?

Thanks!![/quote]

[hide="1"]$a^3+b^3=(a+b)(a^2-ab+b^2)$ :)  [/hide]
[hide="2"]For $(8x^2-2)$, you make the first one $2(4x^2-1)$ and then for $4x^2-1$, $a^2-b^2=(a-b)(a+b)$  :) [/hide]

Then for the second problem, what happens to the 2 in 2(4x squared -1)??

So for my answer I would have (2x-1)(2x+1)(x+3)?

And for the first problem, I still have no idea what you are telling me!

For the second one, just factor out the two. It is 2(2x-1)(2x+1)(x+3).

For the first one, it is a sum of two cubes. Find what the two cubes are after factoring out the x and apply it.

Your teacher may or may not want you to do sum of cubes, it's not really middle school level and what you did may be enough, you should ask him/her


Two non-zero real numbers, $ a$ and $ b$, satisfy $ ab \equal{} a \minus{} b$. Which of the following is a possible value of $ \frac {a}{b} \plus{} \frac {b}{a} \minus{} ab$?

(A)$ \minus{} 2$

(B)$ \minus{} \frac {1}{2}$

(C)$ \frac {1}{3}$

(D)$ \frac {1}{2}$

(E)$ 2$

[hide="solution"]

$ \frac {a}{b} \plus{} \frac {b}{a} \minus{} ab$

$ \equal{} \frac {a^2 \plus{} b^2 \minus{} a^2 b^2}{ab}$

$ \equal{} \frac {a^2 \plus{} b^2 \minus{} (ab)^2}{ab}$

$ \equal{} \frac {a^2 \plus{} b^2 \minus{} (a \minus{} b)^2}{ab}$

$ \equal{} \frac {a^2 \plus{} b^2 \minus{} (a^2 \plus{} b^2 \minus{} 2ab)}{ab}$

$ \equal{} \frac {2ab}{ab}$

$ \equal{} \boxed{2}$

The very fact that I did the problem may mean that this belongs in the HS Basics forum...

[/hide]

Oh haha thanks for your help.  I was thinking I might have put this in the wrong forum as I was skimming through the other Int. threads :D

[quote="leoxnlin"]this belongs in the HS Basics forum...[/quote]  No, this question is of an appropriate difficulty for Intermediate.  It's a common fallacy to think that because you can solve problems in forum X now, any problems you solve in the future belong in forum X.  This falacy is responsible for a great deal of "difficulty creep" that leads to people doing things like posting USAMO questions in Pre-Olympiad :huh:.

So, what is the appropriate level of difficulty for this intermediate forum? (broadly speaking)


geometry
3D geometry
prism
trigonometry
ratio

If an octagonal prism with regular octagonal bases, like the one in the link below, has side lengths of 5 cm and a height of 22 cm, show how you can find the surface area of the prism?

http://mathworld.wolfram.com/images/eps-gif/OctagonalPrism_800.gif

Just in case if you are confused, the side  length is the length of the bases and the height is the length of the lateral edge.

The picture you added shows a pentagonal prism.

In a right prism, with the sides perpendicular to the bases, the side faces are rectangles. You would know how to find their area, so the lateral area is no problem.

The area of the bases is no problem if they are regular polygons (octagons or pentagons, or whatever). There are formulas, based on thinking of the polygon as a collection of triangles. You can consider a regular octagon to be made of 8 triangles. Each triangle would have a side of the octagon for a base, and all the triangles meet at the center of the octagon. To find the area of those triangles, you need their height. Geometry and/or trigonometry tells you the ratio of the side of the octagon to that height.


vector
analytic geometry
real analysis
real analysis unsolved

Suppose $\sup\limits_{n\geq 1} |a_n| < \infty$. We define a linear operator in $l^p (p\geq 1)$
\[y=Tx,\quad \eta_i=a_i\xi_i,\quad i=1,2,\dots\]
where $x=(\xi_1,\xi_2,\dots,\xi_n,\dots), y=(\eta_1,\eta_2,\dots,\eta_n,\dots),$
Prove that $T$ is a bounded operator, and $\|T\| = \sup\limits_{n\geq 1} |a_n|$

Why did you post it in the "unsolved" section? :? You did say it was easy in the title :).

First of all, it's clear that $\|Tx\|_p\le \sup |a_n|\cdot\|x\|_p$, so the operator is bounded and it has norm at most $\sup\|a_n\|$. On the other hand, we can take the sequence of vectors $x_n$ to have the $n$'th coordinate equal to $1$ and all the other coordinates equal to $0$ to see that the norm of the operator is actually equal to $\sup|a_n|$.


[hide]Here's my solution:
Dividing both sides by $ 5^x, we have: (2/5)^x \plus{} (3/5)^x \equal{} 1$.
If $ x > 1, then (2/5)^x < 1,(3/5)^x < 1, so the LHS < 1$(unreasonable)
If $ x \equal{} < \minus{} 1, then (2/5)^x > 1,(3/5)^x > 1, so the LHS > 1$(unreasonable)
Therefore, x is just equal to either $ {0,1}$.
Plug and chug we easily show that x=1 is the unique solution :) [/hide]

[quote="Zhero"]Prove that (1, 1) is the only solution to the Diophantine equation $ 2^x \plus{} 3^x \equal{} 5^y$.[/quote]

The RHS is $ 5^y$, not $ 5^x$...

Using mod 4, we see x is even (or 1).
So $ 4^n\plus{}9^n \equal{} 5^y$
But this is false mod 5, as
$ 2 \cdot (\minus{}1)^n \not \equiv 0 \pmod 5$

dbgreenb801 got it. Have a cookie for you!


inequalities
real analysis
geometry
geometric transformation
dilation
function
integration

Again I need some help with analysis... :wallbash_red:

This is Young's Inequality.

Take $1 \le p, q \le \infty$, where $\frac{1}{p}+\frac{1}{q}-1 = \frac{1}{r}\ge 0$. Suppose $f \in L^{p}(\mathbb{R}^{n})$ and $g \in L^{q}(\mathbb{R}^{n})$. Then we have the convolution $f*g \in L^{r}(\mathbb{R}^{n})$ and indeed $\|f*g\|_{r}\le \|f\|_{p}\cdot \|g\|_{q}$

I think it is a beautiful inequality but as usual I am not satisfied when the author just asserts it! Perhaps it is my own fault for trying to read a text too advanced for my level of education. But I have not yet given up and so I still want to understand these things. Can someone help me with this? Again, I am equally happy if someone will prove this for me or if someone will give me some guidance so I can finish the proof myself. Thanks!

edit: fixed the statement of the inequality.

Just take logs to the both sides and use the convexity of logarithm. That is how I proved one special case of Young's inequality and I believe the same works in the qeneral case too.

Ah, one of my favorites!

That's $\|f*g\|_{r}\le\|f\|_{p}\cdot\|g\|_{q}.$ Not a sum!
And what puuhikki said falls well short of being convincing.

One starting place: Assume that such an inequality holds for some set of indices $p,q,r.$ Assume also that we are talking about $\mathbb{R}$ (or $\mathbb{R}^{n}$) - the dilation argument is no good outside that. Consider fixed (nice) functions $f$ and $g$ Work out what dilating the functions - replacing $f(x)$ by $f(\lambda x)$ and $g(x)$ by $g(\lambda x)$ does to their norms and their convolutions. Since $\lambda$ is free to take on any value in $(0,\infty),$ the exponent of $\lambda$ must be the same on both sides of the inequality.

That gives the condition $\frac1r+1=\frac1p+\frac1q$ as a necessary condition. That's no proof that it is true, of course, but it explains why we would limit ourselves to those exponents.

There are several possible paths for the proof.

In his [i]Trigonometric Series[/i], in a very early chapter, Zygmund gives a very nice proof of this based. This is the most direct use of real analysis to prove this, and I may come back to this later.

Another approach depends on first proving the Riesz-Thorin interpolation theorem.

Here's a version of the Riesz-Thorin theorem.

Suppose $T$ is a linear operator defined on the simple functions in a metric space mapping into measurable functions on the metric space.

Suppose that $\exists\,M_{0}$ and $M_{1}$ such that :

$\|Tf\|_{q_{0}}\le M_{0}\|f\|_{p_{0}}$ and

$\|Tf\|_{q_{1}}\le M_{1}\|f\|_{p_{1}}$

Suppose $\frac1{p_{t}}=\frac{1-t}{p_{0}}+\frac{t}{p_{1}}$ and $\frac1{q_{t}}=\frac{1-t}{q_{0}}+\frac{t}{q_{1}}$ for some fixed $0\le t\le 1.$ Then

$\|Tf\|_{q_{t}}\le M_{1}\|f\|_{p_{t}}$

Now, fix one of the functions in convolution and consider it a linear operator on the other function.

We have the $p=1,q=1,r=1$ case as a standard exercise in Fubini's theorem.

We have the $p=1,q=\infty,r=\infty$ case by a direct argument.

Interpolate between those and we have the inequality true for the $p=1,q$ arbitrary, $r=q$ case.

Now consider the $1<p<\infty,q=p'$ (the dual exponent), $r=\infty$ case. This is true by Hölder's inequality. Interpolate between this and our previous case, and you have the full theorem.

--

Also of interest: Beckner's theorem. In the specific case of $\mathbb{R}^{n},$ and in the away-from-the-boundary case (none of $p,q,$ or $r$ is one or infinity), then

$\|f*g\|_{r}\le C_{p,q}\|f\|_{p}\cdot\|g\|_{q}$ where $C_{p,q}<1.$

The exact constant can be computed from the knowledge that Gaussians are extremal functions.

[quote="Kent Merryfield"]Zygmund gives a very nice proof of this based.[/quote]
On what?

Based on an extension of Hölder's inequality. It's Theorem 1.15 on page 37 of the 1968 edition, and Zygmund doesn't name the theorem for Young or anyone else.

First, WLOG, he assumes that $f\ge0,g\ge0.$ After all, $|f*g|\le |f|*|g|$ on a pointwise basis.

Let $\lambda,\mu,\nu$ be positive numbers satisfying

$\frac1{\lambda}+\frac1{\mu}+\frac1{\nu}=1.$

Write $f(x-t)g(t)=f^{p/\lambda}g^{q/\lambda}\cdot f^{p(1/p-1/\lambda)}\cdot g^{q(1/q-1/\lambda)}.$

We use the (recursively generated) extension of Hölder's inequality to claim that

$f*g(x)\le \left[\int f^{p}(x-t)g^{q}(t)\,dt\right]^{1/\lambda}\left[\int f^{p\mu(1/p-1/\lambda)}(x-t)\,dt\right]^{1/\mu}\left[\int g^{q\nu(1/q-1/\lambda)}(t)\,dt\right]^{1/nu}$

Now let $\lambda=r,\ \frac1p-\frac1{\lambda}=\frac1{\mu}, \ \frac1q-\frac1{\lambda}=\frac1{\nu}.$

The last two terms in the inequality are then $\|f\|_{p}^{p/\mu}$ and $\|g\|_{q}^{q/\nu}$

$f*g(x)\le \|f\|_{p}^{p/\mu}\|g\|_{q}^{q/\nu}\left[\int f^{p}(x-t)g^{q}(t) \,dt\right]^{1/r}$

$(f*g(x))^{r}\le \|f\|_{p}^{pr/\mu}\|g\|_{q}^{qr/\nu}\int f^{p}(x-t)g^{q}(t) \,dt$

Integrate that, and use the already known $L^{1}*L^{1}$ result and we get that

$\|f*g\|_{r}^{r}\le \|f\|_{p}^{pr/\mu}\|g\|_{q}^{qr/\nu}\|f\|_{p}^{p}\|g\|_{q}^{q}$

Now sort out the exponent arithmetic.


This is a question from the MOEMS 2008: Person A can cross the river on his boat in 5 minutes. Person B can do it in 7 minutes. If Person A & B go back and forth from one side of the river to the other, and if they simultaneously cross the river, how many times would they meet in 35 minutes? how do you do this? thanks!

[hide]
Every time the faster boat crosses the river, the slower boat will meet him exactly once. Therefore, the faster boat will cross the river 7 times (35 / 5 = 7), so they must meet exactly $ \boxed 7$ times.[/hide]

AAAAA!!!! I meant faster boat. :ninja:  :arrow:

ah right! 7 times...havent thought of that, thanks!:)


$ \log p \plus{} \log q \equal{} \log (p \plus{} q)$ only if:

$ \textbf{(A)}\ p \equal{} q \equal{} 0 \qquad\textbf{(B)}\ p \equal{} \frac {q^2}{1 \minus{} q} \qquad\textbf{(C)}\ p \equal{} q \equal{} 1$
$ \textbf{(D)}\ p \equal{} \frac {q}{q \minus{} 1} \qquad\textbf{(E)}\ p \equal{} \frac {q}{q \plus{} 1}$

[hide="Click for solution"]
$ \log p \plus{} \log q \equal{} \log (pq)$.  Thus, $ pq\equal{}p\plus{}q \implies p\equal{}\frac{q}{q\minus{}1}$ or $ \boxed{\textbf{(D)}}$.
[/hide]


It is from [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=18745]http://www.mathlinks.ro/Forum/viewtopic.php?t=18745[/url].
For $ a,b,c,d \ge 0$ Prove:
\[ \sum a^4 \plus{} 2abcd\geq \sum ab(a^2 \plus{} b^2)/2\]

I am sorry,I find it is a sepcial case of suranyi's inequality .See [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Suranyi&t=14008]www.mathlinks.ro/Forum/viewtopic.php?highlight=Suranyi&t=14008[/url]

There is a pretty short proof of the particular n = 4 case. In fact, we have to prove the inequality

$2\left(a^4+b^4+c^4+d^4+2abcd\right)\geq ab\left(a^2+b^2\right)+ac\left(a^2+c^2\right)$
$+ad\left(a^2+d^2\right)+bc\left(b^2+c^2\right)+bd\left(b^2+d^2\right)+cd\left(c^2+d^2\right)$

for any four nonnegative reals a, b, c, d. In fact, consider the inequality

$a\left(2a+d\right)\left(a-b\right)\left(a-c\right)+b\left(2b+d\right)\left(b-c\right)\left(b-a\right)+c\left(2c+d\right)\left(c-a\right)\left(c-b\right)$
$\geq 0$;

this inequality is obviously true by [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=387639#p387639]the Vornicu-Schur inequality[/url], applied to the numbers a, b, c (in fact, if we assume WLOG that $a\geq b\geq c$, then $a\left(2a+d\right)\geq b\left(2b+d\right)$). Similarly, the following inequalities are also true:

$b\left(2b+a\right)\left(b-c\right)\left(b-d\right)+c\left(2c+a\right)\left(c-d\right)\left(c-b\right)+d\left(2d+a\right)\left(d-b\right)\left(d-c\right)$
$\geq 0$;
$c\left(2c+b\right)\left(c-d\right)\left(c-a\right)+d\left(2d+b\right)\left(d-a\right)\left(d-c\right)+a\left(2a+b\right)\left(a-c\right)\left(a-d\right)$
$\geq 0$;
$d\left(2d+c\right)\left(d-a\right)\left(d-b\right)+a\left(2a+c\right)\left(a-b\right)\left(a-d\right)+b\left(2b+c\right)\left(b-d\right)\left(b-a\right)$
$\geq 0$.

Thus,

$2\left(a^4+b^4+c^4+d^4+2abcd\right)-\left(ab\left(a^2+b^2\right)+ac\left(a^2+c^2\right)\right.$
$\left. +ad\left(a^2+d^2\right)+bc\left(b^2+c^2\right)+bd\left(b^2+d^2\right)+cd\left(c^2+d^2\right)\right)$
$=\frac13\left(a\left(2a+d\right)\left(a-b\right)\left(a-c\right)+b\left(2b+d\right)\left(b-c\right)\left(b-a\right)+c\left(2c+d\right)\left(c-a\right)\left(c-b\right)\right)$
$+\frac13\left(b\left(2b+a\right)\left(b-c\right)\left(b-d\right)+c\left(2c+a\right)\left(c-d\right)\left(c-b\right)+d\left(2d+a\right)\left(d-b\right)\left(d-c\right)\right)$
$+\frac13\left(c\left(2c+b\right)\left(c-d\right)\left(c-a\right)+d\left(2d+b\right)\left(d-a\right)\left(d-c\right)+a\left(2a+b\right)\left(a-c\right)\left(a-d\right)\right)$
$+\frac13\left(d\left(2d+c\right)\left(d-a\right)\left(d-b\right)+a\left(2a+c\right)\left(a-b\right)\left(a-d\right)+b\left(2b+c\right)\left(b-d\right)\left(b-a\right)\right)$
$\geq 0$,

so that

$2\left(a^4+b^4+c^4+d^4+2abcd\right)\geq ab\left(a^2+b^2\right)+ac\left(a^2+c^2\right)$
$+ad\left(a^2+d^2\right)+bc\left(b^2+c^2\right)+bd\left(b^2+d^2\right)+cd\left(c^2+d^2\right)$

is proven.

I wonder whether this idea of proof could be generalized for n > 4. This would require generalizing Vornicu-Schur to several variables...

  Darij

[quote="darij grinberg"]

$2\left(a^4+b^4+c^4+d^4+2abcd\right)-\left(ab\left(a^2+b^2\right)+ac\left(a^2+c^2\right)\right.$
$\left. +ad\left(a^2+d^2\right)+bc\left(b^2+c^2\right)+bd\left(b^2+d^2\right)+cd\left(c^2+d^2\right)\right)$
$=\frac13\left(a\left(2a+d\right)\left(a-b\right)\left(a-c\right)+b\left(2b+d\right)\left(b-c\right)\left(b-a\right)+c\left(2c+d\right)\left(c-a\right)\left(c-b\right)\right)$
$+\frac13\left(b\left(2b+a\right)\left(b-c\right)\left(b-d\right)+c\left(2c+a\right)\left(c-d\right)\left(c-b\right)+d\left(2d+a\right)\left(d-b\right)\left(d-c\right)\right)$
$+\frac13\left(c\left(2c+b\right)\left(c-d\right)\left(c-a\right)+d\left(2d+b\right)\left(d-a\right)\left(d-c\right)+a\left(2a+b\right)\left(a-c\right)\left(a-d\right)\right)$
$+\frac13\left(d\left(2d+c\right)\left(d-a\right)\left(d-b\right)+a\left(2a+c\right)\left(a-b\right)\left(a-d\right)+b\left(2b+c\right)\left(b-d\right)\left(b-a\right)\right)$
$\geq 0$,

  Darij[/quote]
Very very nice.I only wonder how do you come out this one?

:!:  :!:  Darij, how do you come out the LHS is equal to your discovering inequality?


$ x$ is a root of the equation $ \frac 1{\sqrt{x}}\plus{}\frac 1{\root 3\of{x}}\equal{}\frac 1{\root 6\of{x}}$. What's the value of $ 1\plus{}\frac 1x$?

Multiply both sides by $ \sqrt {x}$ to get $ 1 \plus{} {\root 6\of{x}}\equal{} {\root 3\of{x}}$. Let $ {\root 6\of{x}} \equal{} n$. Then we have $ 1 \plus{} n \equal{} n^2$, so $ n \equal{} (1 \plus{} \sqrt {5})/2$. Raise that to the sixth to get x, take the recipriocal, and add 1 to get the answer. I don't know how to simplify the decimal.

After simplifying all the radicals, I got $ 10\minus{}4\sqrt{5}$ as the answer.

Unfortunately the answer is $ 18$. :(

Using Wolfram Mathematica...:

$ \text{Solve}\left[\frac{1}{\sqrt{x}}\plus{}\frac{1}{\sqrt[3]{x}}\text{\equal{}\equal{}}\frac{1}{\sqrt[6]{x}}, x\right]$

$ \left\{\left\{x\to 9\plus{}4 \sqrt{5}\right\}\right\}$

$ \text{Simplify}\left[9\plus{}4 \sqrt{5}\plus{}\frac{1}{9\plus{}4 \sqrt{5}}\right]$

$ 18$

 :lol:

[quote="dogcat"]Unfortunately the answer is $ 18$. :(

Using Wolfram Mathematica...:

$ \text{Solve}\left[\frac {1}{\sqrt {x}} \plus{} \frac {1}{\sqrt [3]{x}}\text{ \equal{} \equal{} }\frac {1}{\sqrt [6]{x}}, x\right]$

$ \left\{\left\{x\to 9 \plus{} 4 \sqrt {5}\right\}\right\}$

$ \text{Simplify}\left[9 \plus{} 4 \sqrt {5} \plus{} \frac {1}{9 \plus{} 4 \sqrt {5}}\right]$

$ 18$

 :lol:[/quote]

You asked for $ 1 \plus{} \frac {1}{x}$.  You computed $ x \plus{} \frac {1}{x}$.  The answer is actually...

[hide]$ 10 \minus{} 4\sqrt {5}$[/hide]

$ x \equal{} 9 \plus{} 4\sqrt {5}$, as he posted. So $ 1/(9 \plus{} 4\sqrt {5}) \equal{} (9 \minus{} 4\sqrt {5})/(9 \plus{} 4\sqrt {5})(9 \minus{} 4\sqrt {5}) \equal{} 9 \minus{} 4\sqrt {5}$, so adding 1 should be $ 10 \minus{} 4\sqrt {5}$, as I posted before, not +. You computed x+1, not (1/x)+1.

I'm really sorry. It's my fault. I mistyped it. :blush: (the question was $ x\plus{}\frac 1x$...)


Calculate $ \displaystyle\sum_{n\equal{}1,\ n\ \text{odd}}^{\plus{}\infty}\frac{n^2}{(n^2\minus{}4)^2}$