This year's USAMO has 20 qualifiers from the 5% club.

Something about this strikes me as very odd.  Looking at the Summary of High School Results and Awards 2006, I found that only one member of the 5% club scored over 120 on the AMC 10 or the AMC 12 last year and 12 members scored over 100 on the AMC 10 or 12.

However, none of these members are the ones listed in the 2007 USAMO qualifier list.

Secondly, no student from the 5% club last year scored over 5 points on the AIME.

I hold nothing against the 5% Club, but is there some explanation for this? (Perhaps massive recruiting or rigorous problem solving courses?) Otherwise, I find the fact that 20 students qualified to be very unlikely.

Does this mean that the 5% Club is using the fact that they're based in Los Angeles (even though they took the test in Korea) to qualify for the USAMO?

Unless the qualifiers are all US Citizens, shouldn't these spaces be given to US qualifiers?

[quote="Kent Merryfield"]It's not a school, and it appears that the students in question probably do not reside in the U.S. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142291&start=120]this post[/url] by kohjhsd explains at least some of it.[/quote]
[quote="AIME Teacher's Manual Page 5"]US and Canadian Citizens and students residing in the United States (with qualifying scores) are eligible to take the USAMO.[/quote]
[quote="fourierseries"]Unless the qualifiers are all US Citizens, shouldn't these spaces be given to US qualifiers?[/quote]
On the face, your conclusion would appear to be valid.  But we don't know all the facts.  Maybe the CAMC didn't ask where the students reside.  Maybe there is a special agreement between CAMC and this group?  Maybe these students intend to move to the USA before April 24?

The 5% Club is an academic preparation and enrichment program in Los Angeles CA.    The 5% Club registered for the 2006 AMC 10/AMC 12, with about 150 participants in each level contest.   The 5% Club registered again for the 2007 AMC 10/AMC 12 contest with about 10 times as many participants for each contest.

From conversations with the Contest Manager, I know that the 5% Club was providing 2007 AMC Contest administration both in the United States and in Korea.  It appears that in 2007 the 5% Club adminstered the contests in Los Angeles, Irvine, Victorville, and San Diego, and possibly Chicago as well as Korea.

Registration from within the US for a school outside the US is a acceptable practice in registration for the AMC contests and has been done for years.  In past years, registration for schools outside the US was  usually with organizations representing embassy schools, "American Schools" and some schools with APO/FPO addresses.  The reason is that mailing the contests is faster, cheaper and more reliable than mailing directly.  The registration form the 5% Club was much larger than we have seen in the past from these "embassy and American school" organizations, but it is not different in principle or custom.

All the registration information about the students indicates that the  registration was from California.
Our records indicate that the students from the 5% Club did not provide addresses on their answer forms.  The address on the answer form is  the address we use for direct correspondence with the students.  It also provides a cross-check on residence within the US for puposes of USAMO invitation.  Therefore, we do not know where these students came from.  I am attempting to determine that now.

Please note that not providing the address on the answer form is common.  Many of the correspondents on this forum have contacted me personally to inform me that they forgot to include some piece of information. So not providing an address is not an automatic  reason for disqualification .  Until we have information on the residence of all students, we will use the best available information.

Steve Dunbar
Director, American Mathematics Competitions

If they reside in Korea, wouldn't it be possible that they took the AIME after the answers were already posted on AoPS? I know AMCDirector mentioned that certain schools in Asia were taking the AIME later...

If, hypothetically, something did disqualify one or more of them from taking the USAMO officially, would another person take their place or would their be fewer qualifiers?

Unless they received the AIME packets later, Korea should have taken AIME before the US because they are across the international date line, putting them over ten hours ahead of New York.

I doubt that they can change the index so that only 1 or 2 more people qualify to take the USAMO to replace those disqualified because i bet there are quite a few people who missed the index by a little bit.

Apparently there are only 12 of those people.  See Mr. Dunbar's post, "Revised USAMO list is being prepared".  Well, 12 who barely missed the floor tiebreaker, anyway.


each got 10 birds or total they got that much???

I haven't tried the question yet but that's the way the question is....I thought it was wrong at first sight :D

shadysaysurspammed wrote:Two hunters A and B got 10 birds for shooting.Sum of the square of their shots fired is 2880 and the product of the number of shots fired by each was 48 times the product of the number of birds killed by each.If A had fired as often as B and B would have fired as often as A,then B would have shot five more birds than A.Find the number of birds killed by each.

a = # of birds from hunter A

b = # of birds from hunter B

x = # of shots from hunter A

y = # of shots from hunter B

[hide]

a = 10 - b

x^2 + y^2 = 2880

xy = 48ab

ay/x = bx/y + 5

x and y have to be whole numbers

so either (24, 48) or (48, 24), but they're interchangeable (you just have to adjust a and b accordingly), so we'll say x = 24 and y = 48

x = 24

y = 48

2a = b/2 + 5

2a = 20 - 2b

15 = 5b/2

b = 6

a = 4

Hunter A killed four birds and fired 24 times

Hunter B killed six birds and fired 48 times

[/hide]

[quote="riddler"]sorry if this is dumb but if [quote="shady"]B would have shot five more birds than A"[/quote]
then number of birds shot by A=$a$ so $2a+5=10$ this is $a=2.5$????[/quote]

Exactly same doubt here too :oops:

[quote="riddler"]sorry if this is dumb but if [quote="shady"]B would have shot five more birds than A"[/quote]
then number of birds shot by A=$a$ so $2a+5=10$ this is $a=2.5$????[/quote]

YEahh....I don't really get the use of as often as..its bugging me.

so is treething right,shady??? tell the answer


Given the quadratic equation $ ax^2\plus{}bx\plus{}c \equal{} 0$ which has $ r$ and $ s$ as its roots. I know that we can find the quadratic equation with $ (r\plus{}1)$ and $ (s\plus{}1)$ as its roots and even $ r^3$ and $ s^3$ as its roots. But, is there a way to find a quadratic equation with $ 2^r$ ans $ 2^s$ as its roots?

A quadratic always has 2 roots (whether or not they are real, rational, equal,... is another thing).  Therefore, we can always write
$ (x-2^{r})(x-2\&{s})$.  Carry out that multiplication and you have your equation.

I understand that but is it possible to find it in terms of a, b and c?
$ r\plus{}s\equal{}\minus{}\frac{b}{a}$ and $ rs\equal{}\frac{c}{a}$.
What about for $ 2^r\plus{}2^s$ and $ (2^r)(2^s)$?

If you want to find your coefficients $ a$,$ b$ and $ c$, expand JRav's factorization-and voila, you have your form $ ax^2 \plus{} bx \plus{} c$.

On the other hand, do you know Vieta's formulae? It looks like you know it (from your last post, which is a bit confusing). I assumed you wanted to show that you can find the coefficients if solutions are $ r$ and $ s$, and you're asking if the principle is same for this particular problem ($ 2^r$,$ 2^s$). Answer to that is yes.

$ \frac \minus{}{b}{a} \equal{} x_1 \plus{} x_2$

$ \frac {c}{a} \equal{} x_1x_2$

whatever the form of $ x_1$ and $ x_2$ is.


In $ \triangle ABC$, $ \angle C\equal{} 90 \deg$ and $ CD$ is the altitude to $ AB$.

$ r_1$,  $ r_2$ and $ r$ are the in-radii of $ \triangle CAD$, $ \triangle CBD$ and $ \triangle ABC$.

Prove that $ r_1 \plus{} r_2 \le r \times \sqrt(2)$.   When does the equality occur?

[hide="Hint"]
In right-angled triangle with sides of lenghth $ a$, $ b$ and $ c$ and inradius $ r$ holds:

$ r\equal{}\frac {a\plus{}b\minus{}c}{2}$

[/hide]


geometry
perimeter
SFFT
special factorizations

What is "Simon's favorite factoring trick"? How is it used?

I believe its $a+ab+b+1=a(b+1)+(b+1)=(a+1)(b+1)$, or something like that.

nevermind, i found it in another post. in case you're wondering:

\[(x-k)(y-k)=xy-kx-ky+k^2\], where $k$ is a constant and $x$ and $y$ are variables. That's cool!

\begin{eqnarray*}
3m+4n&=&mn-5\\
5&=&mn - 3m - 4n
\end{eqnarray*}

Now what? How do you do it if the coefficients of $m$ and $n$ are different?

[hide]\begin{eqnarray*}
3m+4n&=&mn-5\\
5&=&mn - 3m - 3n - n\\
5+n&=&mn - 3m - 3n\\
5+n&=&(m-3)(n-3) - 9\\
14+n&=&(m-3)(n-3)
\end{eqnarray*}

What do I do with the extra $n$ on the LHS?[/hide]

[quote="Phelpedo"][hide] mn - 3m - 4n +12 = 5+12

(m-4) (n-3) = 17. Is this right?[/hide][/quote]

I think it is. Since 17 is a prime number, $m-4=1$ and $n-3=17$ or $m-4=17$ and $n-3=1$. In  the first case, $m=5$ and $n=20$. Then $3m+4n=mn-5$, so $15+80=95$, which is true! In the second case, $m=21$ and $n=4$. Then $3m+4n=mn-5$, so $63+16=79$, which is also true! Good job Phelpedo.

Here's another one:

What is the minimum value of $a+b+c$ where a,b, and c, are distinct positive integers, if $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{9}{10}$.

[hide="Solution"]
Because the final value of the fraction is $\frac{9}{10}$, the common denomenator has a factor of 2 and a factor of 5. One case is $a=5$. Then $\frac{1}{b}+\frac{1}{c}=\frac{7}{10}$.

\begin{eqnarray*}
\frac{1}{b}+\frac{1}{c}=\frac{7}{10}\\
\frac{b+c}{bc}=\frac{7}{10}\\
10b+10c=7bc\\
0=7bc-10b-10c\\
0=bc-\frac{10}{7}b-\frac{10}{7}c\\
\frac{100}{49}=bc-\frac{10}{7}b-\frac{10}{7}c+\frac{100}{49}\\
0=(b-\frac{10}{7})(c-\frac{10}{7})-\frac{100}{49}\\
\frac{100}{49}=(\frac{7b-10}{7})(\frac{7c-10}{7})\\
100=7(7b-10)(7c-10)\\
\end{eqnarray*}

None of these are integers. Let's try the second case. If $a=2$:
\begin{eqnarray*}
\frac{1}{b}+\frac{1}{c}=\frac{2}{5}\\
\frac{b+c}{bc}=\frac{2}{5}\\
5b+5c=2bc\\
0=2bc-5b-5c\\
0=bc-2.5b-2.5c\\
6.25=bc-2.5b-2.5c+6.25\\
0=(b-2.5)(c-2.5)-6.25\\
6.25=(b-2.5)(c-2.5)\\
625=100(b-2.5)(c-2.5)\\
25=(4b-10)(4c-10)
\end{eqnarray*}

Now we need to find pairs of numbers that multiply to give 25: (1,25) and (5,5). Only (1,25) will work because (5,5) will not give us distinct numbers. So $b=\frac{11}{4}$ and $c=\frac{35}{4}$, but none are integers.

Heh well, looks like this problem I just made up has no distinct integer solutions! (I think)
[/hide]

[quote="Rep123max"]Here is a less obvious use of Simon's Favorite Factoring Trick:

Three non-overlapping regular plane polygons, at least two of which are congruent, all have sides of length 1. The polygons meet at a point A in such a way that the sum of the three interior angles at A is $360^\circ$. Thus the three polygons form a new polygon with A as an interior point. What is the largest possible perimeter that this polygon can have?[/quote]

Err I didn't really use Simon's Favorite Factoring Trick but I have a solution.
[hide]The sum of one of the interior angles of the three polygons have to be equal to 360, and two of them must be congruent.

$\frac {360n - 720}{n} + \frac {180p - 360}{p} = 360$

$\frac {2n - 4}{n} + \frac {p - 2}{p} = 2 \Rightarrow 3np - 4p - 2n = 2np$

Solving for $p$ in terms of $n$ we get

$p = \frac {2n}{n-4} \Rightarrow p = 2 + \frac 8{n-4}$

The perimeter of the new polygon would be equal to $2n + p - 6$ because 6 of the sides are in the interior of the polygon, and the rest will count. The largest value of $n$ is 12 which gives us the maximum perimeter of $21$.[/hide]

chess64, you forgot that $a$ could equal 10, but i tried it out and there aren't any solutions to it anyway  :P nice problem though!

what is the factoring way to figure out Rep123Max's problem?

Check this out: [url]http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=1069.com/Forum/weblog_entry.php?e=10[/url]

It uses the factoring trick to determine how you can tesselate certain polygons. kinda cool, eh?

This is how you do Rep123max's problem with the factoring trick:

[hide] Allow n to be the # of sides of the figure which is duplicated and x to be # of sides of the figure which is not. Three angles, two coming from the duplicated figures and one from the non-duplicated one, meet at one point. This gives us:

(360(n-2))/n+(180(x-2))/x=360

Multiplying out and simplifying gives us nx-4x-2n=0. This is where the factoring trick comes in play.

Factoring gives:

(n-4)(x-2)=8

For the perimeter to be as large as possible, the values n and x have to be the farthest apart, because each time we increase n by 1, we add 2 to the perimeter, but then subract 1 because this requires us to decrease x by 1. Thus, we add 1 to the perimeter each time we increase n and decrease x by 1 each. This proves that n and x have to be as far apart as possible. Allowing n-4=8 and x-2=1, we get n=12 and x=3, thus giving us a perimeter of:

[b]21[/b][/hide]


If ACED and CBFG(F lies on EC) are squares, H is the midpoint ot AC and HE = HG, prove that F is the midpoint of EC.

I call CH=x,CB=y
 $ \rightarrow x^2\plus{}(2x)^2\equal{}(x\plus{}y)^2\plus{}y^2$
  $ \rightarrow 2x^2\minus{}y^2\minus{}xy\equal{}0$
  $ \rightarrow (x\minus{}y)(2x\plus{}y)\equal{}0$
 $ \rightarrow x\equal{}y$


search
function
algebra
polynomial
quadratics
geometry
geometric transformation

for a sequence like $ 1,3,5,7,9\cdots$  the odd numbers
$ U_n\equal{}2n\minus{}1$ which comes from the arithmetic sequence formula $ U_n\equal{}a\plus{}(n\minus{}1)d$

for this sequence $ 2,3,6,11,18,27\cdots$ where does $ U_n\equal{}n^2\minus{}2n\plus{}3$ comes from ? can we get this from the formula $ U_n\equal{}a\plus{}(n\minus{}1)d$

[quote="binomial_4eva"]for this sequence $ 2,3,6,11,18,27\cdots$ where does $ U_n \equal{} n^2 \minus{} 2n \plus{} 3$ comes from ? can we get this from the formula $ U_n \equal{} a \plus{} (n \minus{} 1)d$[/quote]

This is neither an arithmetic nor a geometric sequence.  However, its [b]first difference[/b] $ 3 \minus{} 2 \equal{} 1, 6 \minus{} 3 \equal{} 3, 11 \minus{} 6 \equal{} 5, 18 \minus{} 11 \equal{} 7, 27 \minus{} 18 \equal{} 9$ is an arithmetic (that is, linear) sequence, so we can write an equation for $ U_n$ that is quadratic.

This is a standard lemma.  Look up the [url=http://www.google.com/search?hl=en&q=method+of+finite+differences&btnG=Google+Search]method of finite differences[/url].

Alternately, because

$ 3 \equal{} 2 \plus{} 1$
$ 6 \equal{} 2 \plus{} 1 \plus{} 3$
$ 11 \equal{} 2 \plus{} 1 \plus{} 3 \plus{} 5$
$ 18 \equal{} 2 \plus{} 1 \plus{} 3 \plus{} 5 \plus{} 7$
$ 27 \equal{} 2 \plus{} 1 \plus{} 3 \plus{} 5 \plus{} 7 \plus{} 9$

we can consider this to be a sequence of [b]partial sums[/b] of the arithmetic sequence $ 1, 3, 5, 7, 9, ...$ (ignoring the initial $ 2$), and work from there using what we know about how to evaluate arithmetic [b]series[/b].  In particular, we know that

$ 1^2 \equal{} 1$
$ 2^2 \equal{} 1 \plus{} 3$
$ 3^2 \equal{} 1 \plus{} 3 \plus{} 5$
...
$ n^2 \equal{} 1 \plus{} 3 \plus{} 5 \plus{} ... \plus{} (2n \minus{} 1)$

so we can easily find an equation for $ U_n$ from there.  I say "an" equation because, as we have discussed, such equations are not unique to a given sequence.

how many kinds of sequences are there ? like these 2, arithmetic and geometric.

There are a lot, but arithmetic and geometric are the two most important ones. Another important one is the harmonic series, but that's mostly in the context of calculus.

There really isn't a specific definition for what constitutes a special sequence, and here's an example why. I could make up one that starts with $ 1$ and the $ n$th term ($ n > 1$) is the $ n\minus{}1$th term with the digital representation of $ n!$ on it, so the sequence would go like $ 1, 21, 621, 24621$, etc. Then I can be all smug and be like "Check out this new sequence I discovered that will become cutting edge mathematics" and preemptively call it the Melon Sequence. But most people including me would agree that the above is silly and insignificant.

To start learning about more advanced sequences:
[url=http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29]Harmonic Series[/url]
[url=http://en.wikipedia.org/wiki/Riemann_zeta]Riemann Zeta Function[/url]
[url=http://en.wikipedia.org/wiki/Fourier_series]Fourier Series[/url]
This is by no means a comprehensive list.

[quote="binomial_4eva"]how many kinds of sequences are there ?[/quote]

There are more kinds of sequences than you can possibly imagine names for.

$ (1)$
does this sequence $ 2,3,6,11,18,27\cdots$ have a name ?like this sequence , what are those sequences called that the second last difference is always an arithmetical sequence,or geometrical sequence.

$ (2)$
how do you get the last differences (the constant) of a sequence without writing all the differences of that sequence down?

$ (3)$
if a sequence has a polynomial formula, what does the root of that polynomial tell us about that sequence? example $ 1,4,9,16,25,36\cdots$ $ U_n\equal{}n^2$ what does the root $ (x\equal{}0)$ tell us about the sequence ?.

$ (4)$
can you get the coefficient of the polynomial formula for a sequence if you dont know the elements in the sequence but you only know that the fourth difference is constant

(1) You might as well call it a "quadratic sequence" or (less specifically) a "polynomial sequence," since it's generated by a quadratic polynomial.  (Polynomial sequences are the only ones whose difference tables eventually reach a constant row.)

(2) You can't unless you already know the formula used to generated the sequence, in which case writing it in terms of the basis of falling powers* will help.

(3) Not very much.  If we're only talking about polynomials over the integers and $ n \in \bf Z$ is a root then $ x \minus{} n$ is a factor so you know that the $ m$th term of the sequence is divisible by $ m \minus{} n$.  More generally, I don't think you really learn much of anything.  (Divisibility means nothing over $ \bf R$ or $ \bf C$.)

(4) Of course not: every fourth-degree polynomial has constant fourth difference and not all fourth-degree polynomials have the same coefficients.

* Every polynomial (say in the variable $ x$) can be written as a unique linear combination (that is, a sum of multiples) of the elements of $ \{1, x, x^2, x^3, \ldots\}$ -- this is what we're normally used to seeing when we think of a polynomial in expanded form.  It's also possible to write every polynomial (in the variable $ x$) as a unique linear combination of the elements of $ \{1, x, x(x \minus{} 1), x(x \minus{} 1)(x \minus{} 2), \ldots\}$.  These are called the [i]falling powers[/i] of $ x$.  For example, we can write $ x^3 \plus{} 3x \plus{} 2 \equal{} x(x \minus{} 1)(x \minus{} 2) \plus{} 3x(x \minus{} 1) \plus{} x \plus{} 2$.  Look at the difference table for the sequence generated by $ x^3 \plus{} 3x \plus{} 2$ to see what my comment (2) means.

apart from question (2) i get everything else you said, thank u very very much.

the sequence of $ x^3\plus{}3x\plus{}2$ is

the seq $ 6,16,38,78,142$
1th diff $ 10,22,40,64$
2th diff $ 12,18,24$
3th diff $ 6,6$

i couldnt figure out how to get the constant (6) from the formula.

Okay, I actually made two mistakes there:

Mistake 1: $ x^3 \plus{} 3x \plus{} 2 \equal{} x(x\minus{}1)(x\minus{}2) \plus{} 3x(x\minus{}1) \plus{} 4x \plus{} 2$.

Mistake 2: I shouldn't have used falling powers; instead, I should have use binomial coefficients.  There's an easy translation between falling powers and binomial coefficients: $ \binom{n}{k}$ is the $ k$th falling power of $ n$ divided by $ k!$.  Then we get $ x^3 \plus{} 3x \plus{}2 \equal{} 6\binom{x}{3} \plus{} 6\binom{x}{2} \plus{} 4\binom{x}{1} \plus{} 2\binom{x}{0}$.

Now, even with that information, you probably still don't see what this has to do with your sequence, and reasonably so.  That's because I start my sequences at $ 0$ instead of at $ 1$, so one or the other of us has to change.  I'd rather make you change ;) -- in that case, your difference table gets an extra first column, making it look like this:
the seq $ 2, 6,16,38,78,142$
1th diff $ 4, 10,22,40,64$
2th diff $ 6, 12,18,24$
3th diff $ 6,6,6$
Okay, now you should be able to see the relation, yes?  (Obviously, this isn't a [i]proof[/i] of the relation -- it's a demonstration of a particular case.  The proof isn't hard, exactly, but it's a decent amount of work to write it all down properly.)

If you can figure out how to correctly formulate the relation my example has suggested, you should be able to state for yourself the relationship between the polynomial that generates a sequence and the eventual row of constants in its difference table.  Once you think you've formulated the general result, test your claim with a few different polynomials.  For example, I recommend:
$ f_1(x) \equal{} ax \plus{} b$
$ f_2(x) \equal{} x^3 \plus{} x$
$ f_3(x) \equal{} 7x^3$
$ f_4(x) \equal{} \minus{}x^3 \minus{} x$
$ f_5(x) \equal{} 3x^4 \plus{} 2x^2 \plus{} 1$
but I'm sure you can come up with a bunch of polynomials yourself ;)

[quote="not_trig"][quote="JBL"]in that case, your difference table gets an extra first column, making it look like this:
the seq $ 2, 6,16,38,78,142$
1th diff $ 4, 10,22,40,64$
2th diff $ 6, 12,18,24$
3th diff $ 6,6,6$[/quote]

This reminds me of differentiation :wink:[/quote]

Indeed.  This is the [url=http://en.wikipedia.org/wiki/Difference_operator]forward difference operator[/url], which shares properties in common with differentiation.

In addition, with the forward difference operator, the falling power takes the place of normal powers. That is, if $ x^^{\underline{k}}$ is the falling power $ x$ with $ k$ terms, that is $ x(x-1)(x-2)(\cdots)(x-k+1)$, then $ \Delta x^{^{\underline{k}}} = kx^{^{\underline{k-1}}}$. Also, $ 2$ take the place of $ e$. $ \Delta 2^x = 2^x$.

i am stuck in this part
[quote]Then we get $ x^3 \plus{} 3x \plus{} 2 \equal{} 6\binom{x}{3} \plus{} 6\binom{x}{2} \plus{} 4\binom{x}{1} \plus{} 2\binom{x}{0}$.[/quote]

how did you get that ? can we write any polynomial in that form (i think you said we can) ? how ?
sorry .i dont get it. can you write a few more poly....s in that form.

now i undrestand where the constant comes from, the $ 6,6,4,2$ in the equation are the first elements of the sequence and differences $ (2,4,6,6)$

So, first, I hope you've understood how to write a polynomial in terms of the falling powers.  It's really a pretty straightforward process: the $ k$th falling power (what Hamster wrote $ x^{\underline k}$, which is the product $ x(x-1)\cdots(x -k + 1)$) is a polynomial of degree $ k$ and leading coefficient 1.  If you have any polynomial (Let's take $ 4x^3 - x^2$ as an example), first you choose the highest-degree term (in this case $ x^3$), and you need to match its coefficient with some falling power.  So, in our example, we need to start with $ 4x^{\underline{3}} = 4x(x - 1)(x - 2) = 4x^3 -12x^2 + 8x$.  Now, move on to the next highest power: we're supposed to have $ x^2$ with a coefficient of -1, but instead we have it with a coefficient of -12, so we need to add $ 11 x^{\underline{2}}$.  This gives us
\begin{align*}
4x^{\underline{3}} + 11 x^{\underline{2}} & = (4x^3 -12x^2 + 8x) + 11(x(x - 1)) \\
& = (4x^3 -12x^2 + 8x) + 11x^2 - 11x \\
& = 4x^3 - x^2 - 3x.
\end{align*}
Now we move on to the next highest power: we're supposed to have $ x$ with a coefficient of 0, but instead we have it with a coefficient of -3.  So, we need to add a $ 3x^{\underline{1}}$ term.  But luckily $ x^{\underline{1}} = x$ so we don't need to do any arithmetic for this one.  Then we have to move on to the next highest power, which in this case is the constant term -- we already have the right constant (0), so we don't have to add anything here, either.  This gives us the final expression
\[ 4x^3 - x^2 = 4x^{\underline{3}} + 11 x^{\underline{2}} + 3x^{\underline{1}}.
\]

Now, what does this have to do with binomial coefficients?  A lot, as it turns out: when $ x$ is an integer and $ k \leq x$,
\begin{align*}
x^{\underline k} & = x(x-1)(x-2)\cdots(x-k+2)(x-k+1) \\
& = \frac{x\cdots(x-k+1)(x-k)(x-k-1)\cdots(2)(1)}{(x-k)(x-k-1)\cdots(2)(1)} \\
& = \frac{x!}{(x-k)!}
\end{align*}
and so $ \binom{x}{k} = \frac{x!}{(x-k)!k!} = \frac{x^{\underline{k}}}{k!}$, or equivalently $ x^{\underline{k}} = \binom{x}{k}\cdot k!$.  Thus, if you can write something in terms of falling powers, you can translate it immediately into binomial coefficients by using this substitution.  Thus is our example we have immediately that
\begin{align*}
4x^3 - x^2 & = 4x^{\underline{3}} + 11 x^{\underline{2}} + 3x^{\underline{1}} \\
& = 4\cdot 3!\binom{x}{3} + 11\cdot 2! \binom{x}{2} + 3\cdot 1!\binom{x}{1} \\
& = 24\binom{x}{3} + 22\binom{x}{2} + 3\binom{x}{1}.
\end{align*}

Now, here's an interesting thing to notice: when we go from a "regular" polynomial to an expression in terms of falling powers, we never change the leading coefficient.  And when we go from the falling powers to the binomial coefficients, we change it in a predictable way.  In other words, if we have a polynomial that looks like
$ P(x) = ax^n + \ldots$ for some $ a \neq 0$, where $ n$ is the highest-degree term, then if we write $ P$ in terms of falling powers we get $ P(x) = ax^{\underline{n}} + \ldots$ (where the thing in the $ \ldots$ has changed from one step to the next).  If we then write this in terms of binomial coefficients, we get $ P(x) = a\cdot n! \binom{x}{n} + \ldots$ (where again something has happened to the stuff in the $ \ldots$).  Now, as you've already noted (although, again, this is something no one has proven on this thread), the final row of constants is just equal to the leading coefficient in this expression, which is $ an!$.  So if [i]all[/i] you're interested in is what that bottom constant row in the difference table is, you can get it just by looking at the largest-degree term in your polynomial.

dunno how to thank you ,really nice explanation, thanks JBL. i learned loads of things

Glad to be of service!  While I'm here, I thought I'd add a little more of interest (like, say, some actual proofs).

Consider the difference table for the polynomial $ f(x) \equal{} \binom{x}{n}$.  (Here $ n$ is just some constant.)  We know that $ f(0) \equal{} f(1) \equal{} f(2) \equal{} \ldots \equal{} f(n \minus{} 1) \equal{} 0$ and that $ f(n) \equal{} 1$.  So, if we start the difference table, it looks like this:
The seq: $ 0, 0, 0, \ldots, 0, 0, 1, \ldots$  (and there are $ n$ 0s there before the 1)
1st dif: $ 0, 0, 0, \ldots, 0, 1, \ldots$  (now there are only $ n \minus{}1$ 0s)
2nd dif: $ 0, 0, \ldots, 1, \ldots$ (now $ n \minus{} 2$)
$ \vdots$
$ n\minus{}1$ dif: $ 0, 1, \ldots$
$ n$th dif: $ 1, \ldots$

But here's the thing: our sequence in generated by a degree-$ n$ polynomial, so the $ n$th differences are just a constant!  So, $ f(x) \equal{} \binom{x}{n}$ gives us a difference table so that the first diagonal ($ f(0), \Delta f(0), \ldots$) is all-0, except that the $ n$th difference at 0 is equal to 1.  So what?  Well, here's the nice part: if you have two sequences and you add them, their difference tables add.  If you have a sequence and you multiply it by a constant, the entire difference table gets multiplied by the same constant.  And, of course, when our sequence is generated by a polynomial, all you need is that first diagonal to be able to reconstruct the entire sequence.  So, suppose you have a polynomial sequence with a difference table that looks like this:

The seq: $ a_0(0), \ldots$
1st dif: $ a_1(0), \ldots$
2nd dif: $ a_2(0), \ldots$
$ \vdots$
$ k$th dif: $ a_k(0), \ldots$ (and this last row is known to be constant, i.e. we're dealing with a degree-$ k$ polynomial to generate our sequence),

I can say to you, "Hey, I know a polynomial that generates exactly this difference table: to get the $ a_0(0)$ at the beginning of the first row, I need to take $ a_0\binom{x}{0}$; to get the $ a_1(0)$ at the beginning of the second row, I need to take $ a_1\binom{x}{1}$; and so on."  The key idea here is that the difference tables of the $ \binom{x}n$ are so nice that they don't interact with each other -- when I add in a multiple of $ \binom{x}{m}$ for some $ m$, I know it will [i]only[/i] change the value in the $ m$th row along that first diagonal.  Then, since two sequences with the same difference table are the same, we have exactly that my polynomial (written in terms of the $ \binom{x}{n}$) and whatever polynomial we started with must be the same.  Going the other direction, writing a polynomial in terms of these binomial coefficients gives me exactly the difference table I've claimed in my previous post.

Now, a quick aside on something Hamster noted, using some calculus:  the function $ \exp(x) \equal{} e^x$ has the nice property that $ \frac{d}{dx}\exp(x) \equal{} \exp(x)$.  Why exactly does it have this property?  Well, $ e^x \equal{} 1 \plus{} x \plus{} \frac{x^2}{2!} \plus{} \frac{x^3}{3!} \plus{} \ldots$ is a sum of polynomials of the form $ p_n(x) \equal{} x^n/n!$.  What's special about these polynomials?  They are the unique polynomials such that if we take the $ k$th derivative $ p_n^{(k)}(x)$ and look at the value at $ 0$, we get that $ p_n^{(k)}(0) \equal{} \delta_{k, n}$.  This is fancy-talk for saying that the value is 0 unless $ k \equal{} n$, in which case the value is 1.  Now, I just described some polynomials that $ f_n(x)$ that have the property that the $ k$th difference $ \Delta^kf_n(x)$ taken at $ x \equal{} 0$ gives $ \Delta^kf_n(x) \equal{} \delta_{k, n}$, i.e. the $ k$th difference at 0 is 0 unless $ k \equal{} n$, in which case it's 1.  So, this suggests that the natural analogue of $ \exp(x)$ should be the sum $ E(x) \equal{} \sum_{n \geq 0} f_n(x) \equal{} \sum_{n \geq 0}\binom{x}{n}$.  What is this sum?  When $ x$ is an integer (the only case we really deal with for finite difference tables), $ \binom{x}{n} \equal{} 0$ whenever $ n > x$.  So, we can actually rewrite $ E(x) \equal{} \sum_{n\equal{} 0}^x \binom{x}{n}$.  But a well-known combinatorial identity tells us that this is just $ E(x) \equal{} 2^x$.  So, this (I hope) gives some intuition on why it should be that $ 2^x$ is the analogue of $ e^x$, i.e. it's more than just a freak accident but actually has a lot to do with the underlying frameworks.

how do we find the constant term of this $ U_n\equal{}2^{n\minus{}1}$ or more generally $ U_n\equal{}a(r)^{n\minus{}1}$

Polynomial sequences are the only sequences that have an eventual constant row.  (Neither direction of the equivalence is very hard to prove.)  You could easily verify for yourself that the first differences of the sequence $ V_n \equal{} 2^n$ are $ 2^{n \plus{} 1} \minus{} 2^n \equal{} 2^n$, i.e. its first differences (and so all further differences) just spit the same sequence back to you.  Your $ U_n$ is just my $ V_n$ shifted, so to get one from the other you shift the entire difference table.

For $ r^n$, you can do the same calculation to see that the first differences are $ r^{n \plus{} 1} \minus{} r^n \equal{} (r \minus{} 1)r^n$, i.e. the difference operator is the "multiply by $ r \minus{} 1$" operator for this sequence.  (Multiplying by a constant changes nothing, of course.)  Extending the analogy from the end of my previous post, this says that in some sense $ r \minus{} 1$ is the difference table equivalent of $ \ln r$ (since $ \frac {d}{dx}(r^x) \equal{} (\ln r)r^x$).

sorry i couldnt explain the problem completly in my previous post, i was trying to ask these questions ,

$ (1)$how can we find the constant term of a sequence that ends in a geometric sequence. a sequence like this :

The seq:$ 3,3,6,24,192,3072,\dots$
1st diff: $ 1,2,4,8,16,\ldots$
2nd diff:$ 2,2,\ldots$

$ (2)$ is there a general formula for the $ n_{th}$ term of all sequences that ends in geometric sequence like this one
The seq:$ 3,3,6,24,192,3072,\dots$
1st diff: $ 1,2,4,8,16,\ldots$
2nd diff:$ 2,2,\ldots$
that ends in a geometric sequence.

like this one that i found for the sequences that ends in an arithmetics sequence
if we expand the general arithmetic formula $ U_n \equal{} a \plus{} (n \minus{} 1)d$ it becomes
$ U_n \equal{} a \plus{} \frac {1}{1!}(n \minus{} 1)d_1 \plus{} \frac {1}{2!}(n \minus{} 1)(n \minus{} 2)d_2 \plus{} \cdots \plus{} \frac {1}{t!}(n \minus{} 1)\cdots(n \minus{} t)d_t \equal{} a \plus{} {{n \minus{} 1}\choose1}d_1 \plus{} {{n \minus{} 1}\choose2}d_2 \plus{} \cdots \plus{} {{n \minus{} 1}\choose t}d_t \equal{} a \plus{} \sum^{t}_{i \equal{} 1}\binom{n \minus{} 1}{i}d_{i}$ so $ U_n \equal{} a \plus{} \sum^{t}_{i \equal{} 1}\binom{n \minus{} 1}{i}d_{i}$
i love this kind of general formulas

so for question $ (2)$ what i am asking is if we can expand the geometric formula $ G_n \equal{} a(r)^{n \minus{} 1}$ like the one above.

$ (3)$ As you said the constant term of sequences that ends in arithmetic sequence is $ C \equal{} an!$
what would the formula for the constant term of a sequence that ends in geometric sequence be ?

According to the posts above, there will be no constant difference row in geometric sequence:
$ f(x)\equal{}a\cdot r^x$
$ \Delta[f](x)\equal{}a\cdot r^x(r\minus{}1)$
$ \Delta^n[f](x)\equal{}a\cdot r^x(r\minus{}1)^n$
When $ r\equal{}2$, for all natural number $ n$ there is $ \Delta^n[f](x)\equal{}a\cdot 2^x$, as @JBL said.

[quote="timwu"]According to the posts above, there will be no constant difference row in geometric sequence:
$ f(x) \equal{} a\cdot r^x$
$ \Delta[f](x) \equal{} a\cdot r^x(r \minus{} 1)$
$ \Delta^n[f](x) \equal{} a\cdot r^x(r \minus{} 1)^n$
When $ r \equal{} 2$, for all natural number $ n$ there is $ \Delta^n[f](x) \equal{} a\cdot 2^x$, as @JBL said.[/quote]
i am confused, of course not constant [size=150]difference[/size], but has a constant [size=150]term[/size], its $ 2$ isnt it ?
The seq:$ 3,3,6,24,192,3072,\dots$
1st diff: $ 1,2,4,8,16,\ldots$
2nd diff:$ 2,2,\ldots$

What do you mean by "constant term"?  You're being very unclear.

And your example doesn't help at all.  You're not taking the differences between consecutive terms in the sequence; you're taking the [i]quotients[/i], which isn't what we're talking about.  (If you take the base-$ 2$ logarithm of the sequence, you'll get JBL's result for polynomial sequences, which is also not what we're talking about.)

If a sequence is polynomial, then its "constant term" is the coefficient of the $ x^0$ term.  But if a sequence is not polynomial there is not necessarily an unambiguous choice of "constant term."  Consider, for example, the sequence

$ a_n \equal{} 2 \cos^2 \frac{\pi n}{6} \equal{} \cos \frac{\pi n}{3} \plus{} 1$


According to my Algebra 2/Trig teacher, many lines are tried and each time the calculator calculates the distance from each point to the line and squares this distance.  It then adds up all these squares and records the values.  After it tries enough lines it finds the line with the least sum of all the squares of the distances and gives you this line.  Does anybody know how the coefficient of variation, r is calculated?

From my statistics book:

\[r=\frac{1}{n-1}\sum\left(\frac{x_i-\overline{x}}{s_x}\right)\left(\frac{y_i-\overline{y}}{s_y}\right)\]

n is the number of values in the list

$\overline{x}$ is the mean of the x's(same with the y's)

$s_x$ is the standard deviation of the x's(ditto with y)


if i hav a math i dea, how does it become an "official" conjecture?

you write it up and ask some scientists or mathematician ppls and see if ur anywhere near possibly right.. and then.. u try to prove it, but if you can't then it becomes one of those famous unprovable things that take centuries for someone to prove... =) Good luck if ur planning on doing something like that! sounds cool.. hehe

For making up a question you can't solve?

theres alot of mathematicians that made up unsolvable problems like fermat and goldbach and there famous now.  and what about those unsolvable millenium problems?


function
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I have a question on statistics, and I don't know which forum to post it on.

I have two large sets of numbers, each with the same size, say 1000 elements.

But to simplify let's say A{1,5,7,4} and B{1.5,6,14,3.5}

I want to find some kind of formula for "how close" the two sets are to each other.

These are the characteristics I would prefer:
(1) the differences |A_i - B_i| are important, not the ratios. (In other words, if A_i=4 and B_i=5, that would give the same result as A_i=0.01, B_i=1.01)
(2) the value returned is large when the sets are close and small when the sets are very different
(3) if two points are far apart, even if all the others are really close, this has a very dramatic effect on the function - it pushes it down a lot. Therefore averaging the differences will not be a good idea.
(4) it is not a computational nightmare. For instance, if I make a Lagrange polynomial for one, then it will have a degree of about 1000 and that would not be good.

Thank you for anyone who can help me find such a function. (There may be some other restrictions that I haven't thought of yet, but I think I found all of them.)

My first reaction would be to do a $\chi^2$ test of homogeneity with the two sets as two rows of a matrix (you'd first order the sets in increasing or decreasing value).  Then either the resulting $\chi^2$ test statistic or the p-value associated with it could be used as a measure of the "difference" between the two sets (with the p-value more precisely indicating the probability that if the two sets were samples from propulations with identical relative distributions, that the resultings samples would be at least as different as the sets you come up with).  The p-value would satisfy condition 3, since it will be large if the sets are similar to each other and small if they are very different.

Do you mean for us to consider the sets as ordered?  One thing you might consider is taking a large power mean of the differences, rather than the arithmetic mean.  This will be more sensitive to a single large difference.  How large a power to choose depends how sensitive you want it to be.

Power mean:  for positive reals $x_1, x_2, \ldots, x_n$, the $q$th power mean
$f_q = (\frac{x_1^q + x_2^q + \ldots + x_n^q}{n})^{\frac 1q}$ is a continuous, increasing function for $q$ on the positive or negative reals.  It has the properties that as $q$ goes to infinity, $f_q \to \max{x_i}$, as $q$ goes to negative infinity, $f_q \to \min{x_i}$ and as $q\to 0$,  $f_q$ approaches the geometric mean of the $x_i$.


How many numbers between $ 1$ and $ 200$ inclusive are divisible by the square of a prime number?

[hide="Solution"]Just so I can be lazy, I'm not going to add floor symbols for any division, so pretend that any divisions I take are done by a computer that does integer division. :lol:

We use inclusion-exclusion:

The prime squares less than $ 200$ are $ 4$, $ 9$, $ 25$, $ 49$, $ 121$, and $ 169$.

Case 1: Squares comprising one prime. $ 200/4 \equal{} 50$, $ 200/9 \equal{} 22$, $ 200/25 \equal{} 8$, $ 200/49 \equal{} 4$, $ 200/121 \equal{} 1$, $ 200/169 \equal{} 1$.

Case 2: Squares comprising two primes.$ 200/36 \equal{} 5$, $ 200/100 \equal{} 2$, $ 200/196 \equal{} 1$

$ 4\times 9 \times 25 > 200$, so no squares comprising three primes exist.

$ 50\plus{}22\plus{}8\plus{}4\plus{}1\plus{}1\minus{}5\minus{}2\minus{}1 \equal{} 78$

Therefore, there are exactly $ \boxed{78}$ numbers between $ 1$ and $ 200$ divisible by a prime square.

[/hide]


logarithms
integration
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this is the last question of the day from me, i promise.  :D   :P

(since i asked this in class, thought i would also ask here, LOL!)...... have fun....  :)

show

$ \dfrac{1}{2}\cdot \sum_{k\equal{}1}^\infty\; \dfrac{(\minus{}1)^{k\plus{}1}}{ \dbinom{2k}{k}\cdot k^2} \;\;\equal{}\;\; \boxed{ \left(\, \sinh^{\minus{}1}\,\left(\dfrac{1}{2}\right)\, \right)^2}$

I think you love gamma and beta function's. :lol:
another way:
more generally $ your sum ... \equal{} \frac {1}{2}\sum ( \minus{} 1)^{k \plus{} 1}\frac {(k!)^{2}}{(2k)!k^{2}}$
we already know that :
$ f(x)\equal{} \sum_{n \equal{} 0}^{\infty}\frac {(n!)^{2}}{(2n)!}x^{n} \equal{} \frac {4}{4 \minus{} x} \minus{} \frac {4\sqrt {|x|}}{(4 \minus{} x)^{3/2}}\ln \frac {\sqrt {|x|} \minus{} \sqrt {4 \minus{} x}}{2}$ for $ x < 0$
answer will be :
$ \minus{} \int\frac {\int f(x)/x}{x}|_{x \equal{} \minus{} 1}$ or something like this  :)
few calcualtion and we are done

[quote="Extremal"]I think you love [b][size=150][color=blue]gamma[/color] [/size][/b]and beta function's. :lol:  [/quote]

yes sir !!  :D that i most certainly do. :thumbup:

i have never seen anything more elegant and beautiful ! (after all i worship euler and ramanujan... (not literally of-course) :) )

i have this huge poster of ramanujan in my studyroom (a colleague sent it to me from cambridge), it says

$ 2\,\sqrt [4]{\; \dfrac{3^2}{3^2\minus{}1} \cdot \dfrac{5^2\minus{}1}{5^2} \cdot \dfrac{7^2}{7^2\minus{}1} \cdot \dfrac{9^2\minus{}1}{9^2}  \cdot \dfrac{11^2}{11^2\minus{}1}\cdot \quad  \ldots\;}  \;\;\equal{}\;\; \boxed { \dfrac{ \Gamma\left( \dfrac{1}{4}\right) }{\sqrt\pi} }$


function
topology
advanced fields
advanced fields unsolved

Let $(X,d)$ be a compact metric space and $f: X\rightarrow X$ be a function such that $\forall x, y \in X$ we have:
$d(f(x),f(y)) \leq d(x,y)$ ,where equation holds only when $x=y$.
Prove that the function $f$ has a unique fixed point.

already posted many times .. $x\to d(x,f(x))$ is continuous and achieves its minimum somewhere in the compact. From here it is easy.

it is Banach fixed point theorem.
f is a contraction here.


probability
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probability and stats

If $ P$ and $ Q$ are two probability measures on $ \mathbb{N}_0\equal{}\{0,1,2, \, . \, . \, .\}$ then $ \parallel{} P\minus{}Q\parallel{} : \equal{}\sup_{B\subset \mathbb{N}}|P(B)\minus{}Q(B)|$
Can someone show that how is the above equal to $ \frac{1}{2} \sum_{k\ge 0}|P(k)\minus{}Q(k)|$?
Thanks.

hint: introduce the set $ A\equal{}\{i: P(i) > Q(i) \}$

Hi all,

I must have missed something because I can't find the 1/2 factor.
Intuitively we have that the "sup" is reached for $ B \equal{} \mathbb{N}$ otherwise you could add some i not in B and have a bigger "sup" if P(i) is not equal to Q(i).

So I get $ \parallel{} P \minus{} Q\parallel{}: \equal{} \sup_{B\subset\mathbb{N}}|P(B) \minus{} Q(B)| \equal{} \sum_{i\ge 0}|P(i) \minus{} Q(i)|$

Could someone show me where I am wrong

thx

[quote="TheBridgeTwo"]Intuitively we have that the "sup" is reached for $ B \equal{} \mathbb{N}$ otherwise you could add some i not in B and have a bigger "sup" if P(i) is not equal to Q(i).[/quote]
I think, this is where you may be wrong!

First, we observe that  $ \sum_{i: P(i) > Q(i)}\left(P(i) \minus{} Q(i)\right)\equal{}\sum_{i: P(i) < Q(i)}\left(Q(i) \minus{} P(i)\right)$

Therefore, $ \sum_{k\ge 0}|P(k)\minus{}Q(k)|\equal{}\sum_{i: P(i) > Q(i)}\left(P(i) \minus{} Q(i)\right)\plus{}\sum_{i: P(i) < Q(i)}\left(Q(i) \minus{} P(i)\right)$

$ \equal{}2. \sum_{i: P(i) > Q(i)}\left(P(i) \minus{} Q(i)\right)$

$ \equal{}2.\sup_{B\subset\mathbb{N}_{0}}|P(B) \minus{} Q(B)|$

$ \equal{}2.\parallel{} P \minus{} Q\parallel{}$

Am I right alekk?

yes, this is the idea. But you might want to justify a little bit more carefully this step:
\[ \ldots \equal{} \sum_{i: P(i) > Q(i)}\left(P(i) \minus{} Q(i)\right)  \equal{} \sup_{B\subset\mathbb{N}_{0}}|P(B) \minus{} Q(B)| \]
which is essentially the most important one  :)

[quote="TheBridgeTwo"]Could someone show me where I am wrong[/quote]  More specifically, $ |P({\bf N}) \minus{} Q({\bf N})| \equal{} |1 \minus{} 1| \equal{} 0$ is certainly not the supremum of a sometimes-positive function :).  (You are treating $ |P({\bf N}) \minus{} Q({\bf N})|$ as if it were a sum of absolute values, but in fact it is the absolute value of a sum.)

thx JBL,Alekk, and limsup

For the point underlined by alekk

let B an event and A the set defined in alekk's post

then we have

$ |P(B) \minus{} Q(B)| \equal{} |\sum_{i\in B}P(i) \minus{} Q(i)|$

$ \equal{} Max(\sum_{i\in B\cap A}P(i) \minus{} Q(i) \minus{} \sum_{i\in B\cap A^c}P(i) \minus{} Q(i), \minus{} [\sum_{i\in B\cap A}P(i) \minus{} Q(i) \minus{} \sum_{i\in B\cap A^c}P(i) \minus{} Q(i)])$

$ \le Max(\sum_{i\in  A}P(i) \minus{} Q(i),\sum_{i\in A^c}Q(i) \minus{} P(i)) \equal{} \sum_{i \in A}P(i) \minus{} Q(i) \equal{} |P(A) \minus{} Q(A)|$

because we have $ \sum_{i\in B\cap A}P(i) \minus{} Q(i) \equal{} \sum_{i\in B\cap A^c}Q(i) \minus{} P(i)$ which can be seen from the fact that

$ 1 \equal{} \sum_{i\ge 0}P(i) \equal{} \sum_{i\in B\cap A}P(i) \plus{} \sum_{i\in B\cap A^c}P(i) \equal{} \sum_{i\in B\cap A}Q(i) \plus{} \sum_{i\in B\cap A^c}Q(i) \equal{} \sum_{i\ge 0}Q(i) \equal{} 1$

So the sup is indeed reached with $ B \equal{} A$.

I think your arguments are true only when $ B \equal{} \mathbb{N}_{0}$.

Would you please elaborate limsup I don't see what's wrong with my proof

thx

[quote="TheBridgeTwo"]$ 1 \equal{} \sum_{i\ge 0}P(i) \equal{} \sum_{i\in B\cap A}P(i) \plus{} \sum_{i\in B\cap A^c}P(i) \equal{} \sum_{i\in B\cap A}Q(i) \plus{} \sum_{i\in B\cap A^c}Q(i) \equal{} \sum_{i\ge 0}Q(i) \equal{} 1$[/quote]
How is this true unless $ B\equal{}\mathbb{N}_{0}$?

I cannot edit the post so here is the corection :

[quote]
because we have

$ \sum_{i\in A}P(i) \minus{} Q(i) \equal{} \sum_{i\in A^{c}}Q(i) \minus{} P(i)$ which can be seen from the fact that

$ 1 \equal{} \sum_{i\ge 0}P(i) \equal{} \sum_{i\in A}P(i) \plus{} \sum_{i\in A^{c}}P(i) \equal{} \sum_{i\in A}Q(i) \plus{} \sum_{i\in A^{c}}Q(i) \equal{} \sum_{i\ge 0}Q(i) \equal{} 1$

[/quote]

This was used in the equality :

$ Max(\sum_{i\in A}P(i) \minus{} Q(i),\sum_{i\in A^{c}}Q(i) \minus{} P(i)) \equal{} \sum_{i\in A}P(i) \minus{} Q(i)$


analytic geometry
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real analysis
real analysis unsolved

Decide whether the following subset of $ \ell^2$ is compact:
$ A: \equal{}\{a\in\ell^2|\ \ |a_k|\leq\frac{1}{k}\mbox{ for all }k\}$

Direct from the definition (every open covering has a finite subcovering), or using completeness+precompactness seems not convenient, so I expect I'll need to use "every sequence in A has a converging subsequence". After some playing with it, I suspect it is compact (i.e. I didn't manage to come up with a sequence without converging subsequence).

Take a sequence $ (a_n)_n$ in A, i.e. the n-th term is a sequence $ (a_n^k)_k$ with $ |a_n^k|\leq\frac{1}{k}$.  I need to construct a converging subsequence... This is so general, that I don't know how to begin.
Any help? Thanks in advance!

Use the metric properties to your advantage and construct a Cauchy sequence. Since the product of finitely many intervals $ [\minus{}\frac1k,\frac1k]$ is compact, we can find a subsequence for which the first $ n$ coordinates converge. Then the remaining coordinates contribute at most $ \frac1n$ to the squared distance between points, and we can go far enough out so that the squared distances are at most $ \frac2n$. Pick some elements, then apply this same idea to that bounded subsequence with a larger $ n$. Repeating infinitely, this constructs a Cauchy subsequence, which must converge.

Alternately, this product is compact in the product topology, where convergent sequences are those that converge pointwise. Show that a sequence that converges pointwise which is dominated by an element of $ \ell^2$ converges in $ \ell^2$.

Thanks! That's very helpful :)
I guess $ A': \equal{} \{a\in\ell^{2}|\ \ |a_{k}| < \frac {1}{k}\mbox{ for all }k\}$ (with strict inequality) is not compact then?

It's not. A compact set does need to be closed.


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Let $ A$ and $ B$ be two Abelian groups, and define the sum of two homomorphisms $ \eta$ and $ \chi$ from $ A$ to $ B$ by \[ a( \eta\plus{}\chi)\equal{}a\eta\plus{}a\chi \;\textrm{for all}\ \;a \in A\ .\] With this addition, the set of homomorphisms from $ A$ to $ B$ forms an Abelian group $ H$. Suppose now that $ A$ is a $ p$-group ( $ p$ a prime number). Prove that in this case $ H$ becomes a topological group under the topology defined by taking the subgroups $ p^kH \;(k\equal{}1,2,...)$ as a neighborhood base of $ 0$. Prove that $ H$ is complete in this topology and that every connected component of $ H$ consists of a single element. When is $ H$ compact in this topology? [L. Fuchs]

every descending chain $ U_k$ of subgroups of a group $ G$ defines the structure of a topological group with these subgroups as a neighborhood basis of $ 0$. there is only one way to define the topology and this works: a subset $ S$ is open in $ G$ if for every $ s \in S$ there is some $ k$ such that $ s \plus{} U_k \subseteq S$. in particular, $ H$ is a topological group with the $ p^k H$ as a neighborhood basis of $ 0$.

the completeness states that the canonical homomorphism $ H \to \lim_n H / p^k H$ is an isomorphism. if $ h$ is in the kernel, and $ a \in A$, we may choose $ k$ such that $ p^k a \equal{} 0$. since $ h \in p^k H$, this shows $ h(a) \equal{} 0$. now let $ (\overline{h_k})_k$ be an element in the projective limit, i.e. $ h_k \in H$ such that $ h_i \equiv h_k mod p^k H$ for all $ i \geq k$. if $ a \in A$, choose again $ k$ such that $ p^k a \equal{} 0$ and define $ h(a) : \equal{} h_k(a)$. this is well-defined since we are free to make $ k$ larger without changing $ h_k(a)$. this yields our desired preimage $ h \in H$.

[b]- Every connected component of $ H$ is a singleton: [/b]

The connected component $ C$ of the zero element is a closed subgroup of $ H$. If we can show that $ C$ is contained in every $ p^nH$, then we are done, since then $ C\subseteq \cap_n p^nH\equal{}\{0\}$ and the connected component of $ h\in H$ is $ h\plus{}C$.

Proof of $ C\subseteq p^nH$: Write $ C$ as the disjoint union of cosets of the subgroup $ C_n: \equal{}C\cap p^nH$ of $ C$. $ p^nH$ is an open subgroup of $ H$ such that, with $ C_n$, every coset $ h\plus{}C_n$ ($ h\in C$) is an open subset of $ C$ (equipped with the topology inherited from $ H$). By connectedness of $ C$, just one coset is sufficient to cover $ C$. But $ C_n\equal{}C$ is equivalent to $ C\subseteq p^nH$.

[b]- $ H$ is compact iff $ H/p^nH$ is finite for every $ n$:[/b]

Let $ H$ be compact. The disjoint union of cosets of the open subgroup $ p^nH$ is an open cover of $ H$. Since this covering is disjoint, it cannot be refined such that there are only finitely many cosets, hence the index $ [H: p^nH]$ is finite.
If all quotients $ H/p^nH$ are finite, they are compact when equipped with the discrete topology. By Tychonov/Axiom of Choice, the product space $ \prod_nH/p^nH$ is compact. $ H$ is compact as a closed subspace of the product space.

[quote="Third Edition"]Since this covering is disjoint, it cannot be refined such that there are only finitely many cosets, hence the index $ [H: p^nH]$ is finite. [/quote]
could you explain this argument?

You could use the following lemma:

Let $ H$ be a compact topological group and $ H'$ an open subgroup of H. Then $ [H: H']$ is finite.
[hide="Proof:"] Let $ (h_i)_{i\in I}$ be a set of representatives of $ H/H'$. Then $ h_iH'$ is an open subset of $ H$ (since, for fixed $ h_0\in H$, the translation $ h\mapsto h_0h$ is a homeomorphism of $ H$) and we have the disjoint union \[ H\equal{}\stackrel{\cdot}{\bigcup}_{i\in I}h_iH'\]Because  this is an open covering of the compact space $ H$, there is a finite subcovering, i.e. there is a finite subset $ I'\subset I$ such that \[ H\equal{}\stackrel{\cdot}{\bigcup}_{i\in I'}h_iH'\]The initial covering is disjoint, hence we must have $ I\equal{}I'$, thus $ [H: H']\equal{}|I|\equal{}|I'|$ is finite.[/hide]

ok thx (I knew the lemma but forgot the compactness...).


calculus
integration
trigonometry
function
vector
Support
real analysis

Let $ |\ |\equal{}|\ |_{C^0}$ be the supremum norm on $ C^0$.Define an integral transformation $ T: C^0\longrightarrow C^0$ by $ T(f)(X)\equal{}\int_0^xf(t)dt$
$ i)$ show that $ T$ is continuous and find its norm.
$ ii)$ find $ T(\cos (nt)),\ n\equal{}1,2,...$
$ iii)$is the set of functions $ K\equal{}\{T(\cos (nt)\equal{}n\in \mathbb{N}$ closed?bounded?compact?
$ iiii)$ is $ T(K)$ compact?

[hide="Hint for (i)"]Let $ (V,\left\|\cdot\right\|_{V})$ and $ (W,\left\|\cdot\right\|_{W})$ be normed vector spaces, and let $ T: V\rightarrow W$ be a linear map.  Then the following are equivalent:
1) T is Lipschitz
2) T is uniformly continuous
3) T is continuous
4) T is continuous at $ 0\in V$
5) there is a constant $ C<\infty$ such that $ \left\|T(x)\right\|_{W}\leq C\left\|x\right\|_{V}$ for all $ x\in V$.

Show one of these is true and you're done.[/hide]

[quote="fedja"]Normally $ C^0$ means either the space of continuous functions with compact support on the entire real line or the space of the functions that tend to $ 0$ at $ \infty$. In both cases $ T$ doesn't even map $ C^0$-functions to $ C^0$-functions, which means that your definition of $ C^0$ can be neither of those. So, what is $ C^0$ in your problem?[/quote]
$ |f|_{c^0}\equal{}\max\{|f(t)|: t\in [0,1]\}$defined on the infinite-dimensional vector space $ C^0$ of continuous functions $ f: [0,1]\rightarrow \mathbb{R}$


symmetry
combinatorics unsolved
combinatorics

24 points are equally spaced on a circle so that the distance between any two adjacent points is
1. How many different ways can we choose 8 points so that the arc length between any two chosen
points is not 3 or 8.

I get 258.

$\begin{array}{llllllll}1&4&7&10&13&16&19&22\\9&12&15&18&21&24&3&6\\17&20&23&2&5&8&11&14\end{array}$

If we write the numbers 1 to 24 in an array where adjacent terms differ by three or eight then the problem is equivalent to finding a 'path' through the array, containing one number from each column and no two adjacent numbers from the same row (where the last and first columns are considered adjacent).

If we start with $1$, we notice that the number of partial paths ending at any given element is the sum of the number of partial paths ending at each of the two elements in the previous column but not in the same row.  This allows us to create an array containing the number of paths ending at each point:

$\begin{array}{llllllll}1&0&2&2&6&10&22&42\\0&1&1&3&5&11&21&43\\0&1&1&3&5&11&21&43\end{array}$

The $86$ paths which do not end on $22$ are all valid, and by symmetry it is easy to see that there are the same number of paths starting with $9$ and $17$, for a total of $256$.

Incidentally, it is possible to derive general formulae for the entries in the second array, which would hold for larger values, and could be extended to similar problems.  In particular, the entries in the top row are $\frac{2^{n-1}-2}{3}$ for $n$ even, and $\frac{2^{n-1}+2}{3}$ for $n$ odd, whilst the corresponding entries in the other two rows are $\frac{2^{n-1}+1}{3}$ and $\frac{2^{n-1}-1}{3}$ respectively.  However, with such a small array in this particular problem, it is easier to compute by hand.

I have an idea a little different.
First, draw the first table of Diarmuid. The number of paths are equal to number of that:
we have a circle which is cut into 8 sectors. We have 3 colours for them. Now the adjacent sectors must be coloured in different colours.
And this is a classical problem of number sequence.


group theory
abstract algebra
inequalities
superior algebra
superior algebra unsolved

Let c(G) be the number of conjugate classes for finite group G and H be a subgroup of G. Whether or not c(H)<2c(G)?

Not necessary. Take $ H \equal{} G$.

(I read the inequality too fast. I'm blind.)

Well, if H=G we do have c(H) < 2c(H), right?

But in fact, the answer to the original question is "no". There are finite groups G with proper subgroups H such that H has more than twice as many conjugacy classes as G. I believe the smallest counterexample is a group of size 57. In GAP notation it is SmallGroup(57, 1). It has 9 conjugacy classes but it has a subgroup with 19 conjugacy classes (EDIT: of course, this subgroup is a cyclic group of order 19).

Other examples are SmallGroup(68,3), SmallGroup(75, 2) and there are others. SmallGroup(93, 1) has 13 conj classes but has an element of order 31 (and therefore, a cyclic subgroup with 31 conjugacy classes).

- R


Hey! i found this weird pattern but i dont know what good it is! so if someone else has ever seen this pattern EXPLAIN to me why it works! or i will [b]hunt you down and annoy you until you do.[/b]

well look at this:

0^2 + 0^3=0
0^2+1^3=1
1^2+2^3=9=3^2
3^2+3^3=6^3
.......
.......
........

hey notice how the the first and last numbers are triangle numbers? well yea the pattern continues but why does this work?

1) You've posted this in a wrong section of the forum.  (I've now moved it.)

2) This appears to be a consequence of the fact that $ 1 \plus{} 2 \plus{} \ldots \plus{} n \equal{} \frac {n(n \plus{} 1)}{2}$ and $ 1^3 \plus{} 2^3 \plus{} \ldots \plus{} n^3 \equal{} \left(\frac {n(n \plus{} 1)}{2}\right)^2$, but you haven't given enough information about what this pattern is for me to be sure.  The two results I've quoted can be proven inductively or with a variety of cute tricks (try searching the forum for more info) -- I don't think there's any "deep" reason why the two formulas are so closely related.

That looks like the right result.  The last line should be $ 6^2$ (clearly $ 6^3$ is far too large).  See [url=http://mathworld.wolfram.com/NicomachussTheorem.html]Nichomachus' theorem[/url].  As far as I know, this result is unique among results about power sums.

look if we put letters in this we have this pattern:

1^2+2^2=3^2 than
3^2+3^3=6^2

if you see:

1 2 3
3 3 6
6 4 10
10 5 15
15 6 21

triangle numbers

That pattern is explained by exactly the identity given by JBL.  Plug in a few values of $ n$ and see what happens.


When I go to look at a transcript of a class, sometimes the background is dark blue instead of the creamy white color.  This happens to about 1 out of 3 transcripts.  Then after a while, it changes to light then to dark etc.
I am using Mac OS 10.3.9 , Firefox 2.0.0.11 , I use ML skin on AoPS, but that shouldn't affect this.
Is this a CSS problem?
I attached a PDF that has the menu and copyright as an example.

If you refresh, do the colors correct themselves?  By about once in every three times, do you mean on the same transcript or that 1 in every 3 unique transcripts have this problem?

Possibly same problem as http://www.mathlinks.ro/viewtopic.php?t=195102
Your browser might be doing this to you, in which case we can't help you (we can recommend you upgrade your browser to the latest version though, or use another one instead like Firefox or Opera).

This is probably related:

Part of the 2008 AIME I Math Jam transcript's background is dark. Half of the transcript has a white background (note: this is directly in the middle in terms of height), while the rest (above and below the white part) is dark.

This might involve some kind of limit on the length of the white part. It gets displayed as large as possible, and centered in terms of height.

Browser/OS: Firefox 2.0.0.13 on Ubuntu, also confirmed on Windows XP.


If $f(\dfrac{x}{x^2+x+1})=\dfrac{x^2}{x^4+x^2+1}$ then
find  $f(x)$.

Saman Gharib

thank you arne for your nice solution :)


[quote="led14pa2"]i dont know which forum this belongs on, it might be too easy for this one but how many symmetry planes does a football have if u ignore the laces and writing. also, how many symmetry planes does an oblique cylinder have? thanks.[/quote]

I don't know if I am understanding your question, but I would say that the answer is infinity in both cases.

In the case of a football (assuming that you are talking about american football, and not soccer), then you can take any plane that contains the line from extreme to extreme (infinity of them). Also you can consider the plane containing the equator of the ball.

In the case of an oblique cylinder, (assuming that the perpendicular projection of the upper circle in the plane containing the lower circle gives you two concentric circles) then, you can consider any plane that passes for the segment containing those two circles.

[quote="RC-7th"]I kinda doubt you made that yourself.[/quote]

Because of the "Totem Graphic International" written there?

Actually, because of the "Totem Graphics Incorporated" written there.  :D

Psst. It's called Google Images. I was usin' this as an example.

maybe he meant a soccer ball, the one made of 12 pentagons and 20 exagons (am i right with the number of faces?), the same one as fullerene :)
this could be more interesting, maybe!

sorry for the confusion, i meant an american football, but u guys think it is infinitely many for both the american football and the oblique cylinder

led14 said ignore the writing and the laces, but I don't think there's anything there about ignoring the seams. . .

oops sorry, i meant to say to ignore the seams as well


[hide="da answer"]
$\frac{100}{7}=14 R2$

That means 14 weeks and 2 days will have passed.
14 weeks from Tuesday is Tuesday.
2 days after tuesday is [b]Thursday[/b][/hide]

this is problem 1 in elementary school matholympiads.

[hide]The answer is thursday because 100/7 is 14 r2, so tuesday+2 is thursday.[/hide]

[hide]Thursday because 100/7 = 14 Remainder 2 14 weeks from tuesday would be tuesday and 2 days after tuesday is thursday.[/hide]

[quote="MCrawford"]Suppose today is Tuesday.  What day of the week will it be 100 days from now?[/quote]

[hide]$\frac{100}{7}$ has a remainder of 2

so it is Thursday[/hide]

I'm sorry if this sounds really stupid, but what's the point of posting if so many people before you have posted the answer with a relatively same explanation?  Unless you have a very different solution, or if you have a different answer...?


articles
analytic geometry
graphing lines
slope
LaTeX

Hi,
When I use the minipage environment (for side-by-side figures or side-by-side text and figure), the whole matter gets shifted into the left margin. Is this because I am specifying too much width in the minipage option (I did try reducing this but got the same result) or is there something else I need to fix? Any help is appreciated.
SSN.

Sorry for bumping this post, but I thought that if I post some code it might help you understand/diagnose the problem better.

[code]
\documentclass[12 pt]{article}

\usepackage{graphicx}
\usepackage{enumerate}
\usepackage{indentfirst}
\usepackage[fleqn,intlimits]{amsmath}
\usepackage{setspace}
\usepackage{hhline}
\usepackage[headings]{fullpage}

\begin{document}

\begin{enumerate}
\item For the three fracture surfaces in Figures \ref{2}, \ref{3} and
\ref{4}
\begin {enumerate}
\item 1, 2, 3 are all ductile fractures \item 1, 2, 3 are all brittle
fractures \item 2 \& 3 are intergranular fractures \item 2 \& 3 are
transgranular fractures \item can't tell
\end {enumerate}

\begin{figure}[!h]
\begin{minipage}{3.5 in}
\item Which is false
\begin {enumerate}
\item if material is loaded up to a point on the curve between points 2 and
3, necking will be observed. \item material is not deformed
permanently upon loading up to point 1. \item plastic deformation
starts at point 2. \item if the slope of the linear region is
decreased, the resistance to elastic deformation is decreased.
\end{enumerate}
\end{minipage}
\hfill
\begin{minipage}{2.5 in}
\begin{center}
\includegraphics[angle=270,width=2.5 in]{Figure5.eps}
\caption{Question 6} \label{5}
\end{center}
\end{minipage}
\end{figure}

\end{enumerate}
\end{document}

 [/code]

The problem I have is that the minipage environment is shifted to the left by one tab space (with respect to the regular document matter), no matter how much width I specify in the environment argument. If one of you could try the code (I have posted the figure as an attachment) and see if you experience the same problem or, even better, have a solution it'd be a great help. Thanks.
SSN.

The source of the problem appears to be the figure environment. If you remove it or start it just after \begin{document} you don't get the same effect. OK, so you get "\caption outside float" error but that [i]might[/i] be solved using the ccaption package.


Given any four distinct positive real numbers, show that one can choose three numbers $A,B,C$ from among them, such that all three quadratic equations \begin{eqnarray*} Bx^2 + x + C &=& 0\\ Cx^2 + x + A &=& 0 \\ Ax^2 + x +B &=& 0 \end{eqnarray*} have only real roots, or all three equations have only imaginary roots.

Let the four numbers be $a, b, c, d$ with $0<a\leq b \leq c \leq d$.  If $bc\leq\dfrac{1}{4}$ then $ab$ and $ac$ are also no more than $\dfrac{1}{4}$.  Otherwise $bc>\dfrac{1}{4}$ and so $bd$ and $cd$ are also greater than $\dfrac{1}{4}$.  The former implies that $1-4ab$, $1-4ac$, $1-4bc$ are all nonnegative and so setting $A=a$, $B=b$, $C=c$ gives all equations real roots.  The latter implies that $1-4bc$, $1-4bd$, $1-4cd$ are all negative and so  setting $A=b$, $B=c$, $C=d$ gives all equations negative roots.

I may be wrong....but from my sources the INMO 1999 Question 3 reads the same as Rushil's except that the fact that the reals are positive is not given.....can this be right??

@Rijul saini:In the official question paper,it does say that the reals are positive!

INMO 1999 Question 3 was wrong since it did not mention that A,B,C were positive. it is possible only when A,B,C are positive. a counter example is -2,-1,1,2

Indeed positivity is needed. A simple approach runs like that, for the numbers being $0<a\leq b\leq c\leq d$. Consider the sets $(0,1/2]$ and $(1/2,+\infty)$. If any three numbers belong to a same set, we're done, so assume  $0<a\leq b\leq 1/2 < c\leq d$. If $ac > 1/4$ then $ad>1/4$ and we can choose $a,c,d$. If $bc > 1/4$ then $bd>1/4$ and we can choose $b,c,d$. Otherwise $ac\leq 1/4$ and $bc \leq 1/4$, and we can choose $a,b,c$.

The problem is a simple application of PHP
Now,  let the four numbers be $a,b,c,d$
Now,  look at the numbers of the form $1-4xy$that can be formed by the numbers a, b,c,d.
Now,  the to tall possible such numbers is $4C2=6$
now,  these numbers can be either positive or negative.
So we need to divide these $6$ numbers in two intervals.Say one of theses numbers is already settled now a apply PHP on these $5$numbers.Hence, by PHP $3$ numbers fall in the same interval.Now,select any three of the numbers that fall in the same interval.
Now,  note that if three numbers of the form.     $1-4ab,1-4ac,1-4ad$ cannot be selected.Now, if three of the numbers fall in the same interval then select three numbers from the other three interval,while if this interval has any other number apart from these three numbers then select that number and two of these numbers.And, we are done.:)


abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a group,
if $ \forall a,b \in G (ab)^{x\plus{}1}\equal{}a^{x\plus{}1}b^{x\plus{}1}$ then $ \forall a,b \in G a^xb \equal{} ba^x$

[quote="coma"]Let $ G$ be a group,
if $ \forall a,b \in G (ab)^{x \plus{} 1} \equal{} a^{x \plus{} 1}b^{x \plus{} 1}$ then $ \forall a,b \in G a^xb \equal{} ba^x$[/quote]
for which $ x$? if for all then by using fact $ (ab)^{ \minus{} 1} \equal{} b^{ \minus{} 1}a^{ \minus{} 1}$
and let $ x \equal{} \minus{} 2$ then we have
$ (ab)^{ \minus{} 1} \equal{} a^{ \minus{} 1}b^{ \minus{} 1}$
and in finnaly $ b^{ \minus{} 1}a^{ \minus{} 1} \equal{} a^{ \minus{} 1}b^{ \minus{} 1}$ and $ ab \equal{} ba$ which means that your group is comutative. then I can say that
$ a^{x}b \equal{} ba^{x}$

What if $ G$ is the group of octonions?
Then $ G$ has order 8 and its center is $ {\minus{}1,\plus{}1}$. If we choose $ x\equal{}7$ it follows that $ a^7$ is in the center for all $ a\in G$, which is not true.

That's the quaternion group- octonions aren't associative.

The example works for any non-abelian finite group; if $ n$ is the order, $ (ab)^n\equal{}a^nb^n\equal{}e$ but $ a^{n\minus{}1}b\equal{}a^{\minus{}1}b$ and $ ba^{n\minus{}1}\equal{}ba^{\minus{}1}$ are not always equal.

yes, of course jmerry.
Thanks for the correction.

That's a completely different problem. Unlike this one, it's correct.


modular arithmetic
number theory unsolved
number theory

Find all triples $ (p,q,n)$ such that
$ q^{n\plus{}2} \equiv 3^{n\plus{}2} \pmod{p^n}$
$ p^{n\plus{}2} \equiv 3^{n\plus{}2} \pmod {q^n}$
where $ p,q>0$ are odd primes and $ n>1$ is an integer.

[quote="praneeth"]It had been already submitted in Own & Proposed Problems.[/quote]
Oh :blush: ... I am really sorry, I assume problems from national olympiad will post in unsolved problems section, so I think nobody had post it before.Please delete this post together with  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=184715]this[/url] if necessary.


Can anyone kindly go through once of the hypersurface section in the lecture notes ?

http://www.math.ust.hk/%7Emamyan/ma204/lecture.pdf
p.260-262

Thank you.

Can you be more specific in your request?

Very good lecture notes by the way! The teacher sounds very good.

I'm very confused about the definition of tangent space here. In particular, I don't understand why for the case the preimage represents the hypersurface, the tangent space is like that.

By the way, his lecture notes is worse than his lectures.  :lol:


geometry
logarithms
function
limit
perimeter
pigeonhole principle
calculus

Hi.
I tried to do some problems on an introduction to proofs packet given at my high school math team. When I asked the co-captain when and where I could ask about the questions if I was stumped, he said "you can always just post aops"... so here:
Could you point me in the right direction or give me suggestions with these problems? All of my work on these problems are trivial... I'm stumped on all of them.

[b]1. Prove that an irrational raised to an irrational power can be rational.[/b]

I'm clueless. What does it actually mean to raise something to an irrational power? You can't express such a number with exponents and nth roots... The statement isn't even intuitive... or is it?

[b]2. Prove that $ \sqrt {2}$ is irrational. [/b]

I think the question wants a more rigorous proof. All my life I've thought of an irrational number as a number that can't be expressed as the quotient of two integers.
$ \frac {a}{b} \equal{} \sqrt {2}$ where a and b are integers.
I hoped I could find some contradiction with that statement.

[b]3. Show that you cannot cover a circular disk with two circular disks of smaller diameter.[/b]

Since this has to do with area, I thought of pigeon hole like that box problem in aops volume 1. I don't know how to break up a circular disk in a clever way, though.

[b]4. Show that any set of n integers has some subset whose sum is divisible by n.[/b]

It works for one in the set. It works for two...

let a/b=$ \sqrt{2}$ by definition, a and b are relatively prime integers if we assume that sqrt 2 is rational.

this means that $ a^2\equal{}2b^2$. so a must be a multipl of 2. let a=2m.

then $ 4m^2\equal{}2b^2\rightarrow2m^2\equal{}b^2$ However, this means that b must also be a multiple of two. but this is impossible because we assumed a and b were relatively prime.

1. $ \sqrt {2}^{\log_{\sqrt {2}} 10} \equal{} 10$. I'll leave it to you to prove that $ \log_{\sqrt {2}} 10$ is irrational.

And these problems probably would belong in Intermediate, if not HSB.

The definition of raising something to an irrational power comes from the fact that we want exponential functions to be continuous. So, what you do to evaluate $ a^b$ is to take a sequence of rational numbers $ b_1, b_2, b_3, \ldots$ which converge to $ b$. Then, $ a^b \equal{} \lim_{n\to\infty}a^{b_n}$. So, for example, $ \sqrt {2}^{\sqrt {2}}$ is what the sequence $ \sqrt {2}^{1}, \sqrt {2}^{1.4}, \sqrt {2}^{1.41}, \sqrt {2}^{1.414}, \ldots$ converges to.

For number 3, my favorite way is to use perimeter. A smaller disk cannot cover half or more of the perimeter of the larger disk, so therefore some of the perimeter must be left uncovered by two disks. You can also consider lines perpendicular to the line connecting the centers of the smaller disks.

Number four is pigeonhole. Choose a nice collection of $ n$ sets, then try to create a set from them whose sum is divisible by $ n$.

Thank you all for responding. I'm still very confused, though.

First, aside from the way GoColts0626 showed, are there methods to prove a number irrational or rational?

What does it mean for $ b_1,b_2,b_3...$ to converge to b?
I mean, $ b_n$ doesn't equal b las n approaches the limit if the the terms from $ b_1$ to $ b_n$ add to b. How do you know that $ \sqrt {2}^{1}, \sqrt {2}^{1.4}, \sqrt {2}^{1.41}, \sqrt {2}^{1.414}, \ldots$ converges to $ \sqrt {2}^{\sqrt {2}}$?

If the perimeter of the larger circle cannot be covered, does it imply that neither can the area? Why or why not?

Finally, I'm confused about number four. Why do we want to choose a collection of n sets? Aren't we looking for numbers within a single set with n integers which add to a multiple of n? How do we apply pigeonhole?

$ b_1,b_2,b_3,\ldots$ converging to $ b$ means that for every $ \epsilon > 0$, there exists an $ N$ such that for every $ m>N$, $ |b\minus{}b_m|<\epsilon$. Basically the terms of the sequence approach $ b$ and would equal $ b$ if we continued it infinitely far.
$ \sqrt{2}^1, \sqrt{2}^{1.4}, \sqrt{2}^{1.41},\sqrt{2}^{1.414},\ldots$ converges to $ \sqrt{2}^{\sqrt{2}}$ because $ 1,1.4,1.41,1.414,\ldots$ is a sequence of rational numbers converging to $ \sqrt{2}$, so our definition kicks in.

If the perimeter can't be covered, the thing can't be covered because the perimeter is part of the area.

There are $ n$ different remainders upon division by $ n$. If all of the remainders of the $ n$ sets are different, then one is $ 0$. Otherwise two are the same by pigeonhole. See if you can finish it from there.

An alternative way for 3:

It suffices to show that for any two points in a circle, there is a third point in the circle closer (or equally close to) to the center than to either of the two original points.  This holds because you can always draw a diameter such that both points are on the same side (or one or both are on the diameter itself).

For #4...well this is a well-known trick, but it's nice when you see it.  I'll leave you to look at Hamster's hint, because it's gratifying when you figure it out.

[quote="Hamster1800"]The definition of raising something to an irrational power comes from the fact that we want exponential functions to be continuous. So, what you do to evaluate $ a^b$ is to take a sequence of rational numbers $ b_1, b_2, b_3, \ldots$ which converge to $ b$. Then, $ a^b \equal{} \lim_{n\to\infty}a^{b_n}$. So, for example, $ \sqrt {2}^{\sqrt {2}}$ is what the sequence $ \sqrt {2}^{1}, \sqrt {2}^{1.4}, \sqrt {2}^{1.41}, \sqrt {2}^{1.414}, \ldots$ converges to.[/quote]

I don't like this way of defining general exponentiation (in particular, it's not at all clear how to show that a sequence of algebraic numbers converges to a number that may or may not be transcendental).  It's much more useful to define

$ a^b \equal{} e^{b \ln a}$

where $ e^x, \ln x$ can be defined in other ways that require calculus and are totally independent of the intuitive concept of exponentiation.  In fact, this is how a calculator evaluates square roots.

For the irrationals, we can just use $ e^{\ln N} \equal{} N$.

[hide="Circles"]Construct a rectangle such that the longer sides are both tangent to both circles (which may or may not overlap), and the shorter sides are each tangent to one circle. The circles are fully enclosed in this rectangle, but the shorter dimension of the rectangle is shorter than the diameter of the original circle, so this rectangle, and thus the circles inside it, can never fully cover the square.[/hide]

[hide="Subsets"]Let $ b_i \equal{} \sum _{k\equal{}1}^i a_k$. There are $ n$ different such sums. If, for $ j > i$,  $ b_i \equiv b_j \pmod n$, then $ \sum _{k\equal{}i\plus{}1} ^ \equal{} b_j \minus{} b_i  \equiv 0 \pmod{n}$. If this is not the case, all of the sums have different remainders in $ \pmod{n}$, and each of $ 0, 1, 2, \ldots n\minus{}1$ must be represented, so one of these sums is itself divisible by $ n$.[/hide]


1. How do you name an alkene with more than one double bonds present?

e.g. $ (CH_3)_2C\equal{}CHCH_2CH_2C(CH_3)\equal{}CHCH_2CH\equal{}C(CH_3)CH\equal{}CH_2$

2. It is known that the stability of $ \text{trans}$ isomers are greater than that of their $ \text{cis}$ isomers counterpart due to steric strain. Is there an analogous rule with $ E,Z$ configuration?

3. What's the differentce between $ \text{steric strain}$ and $ \text{steric crowding}$?

4. What is $ \text{hyperconjugation}$ and how does it explain the fact that more highly substituted alkenes are stabler (and also for carbocations).

[hide="Some tips"]1. The purpose is to number the longest carbon chain (the one that includes the biggest number of double bonds) in order to give the lowest numbers to the double bonds. In your example the name would be

[hide="Name of 1"]3,7,11-trimethyldodeca-1,3,6,10-tetraene.[/hide]

2. E,Z-configuration and cis-trans configurations are the same thing. However, the "E,Z" designation is more recent and the one in common use nowadays.

3. They are the same concept.

4. Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=161878]here[/url] (post number 4).[/hide]

Hyperconjugation: The stabilizing effect due to orbital overlap between the carbon-carbon $ \pi$ bond and a correctly oriented carbon-hydrogen $ \sigma$ bond

Is that a good enough definition?

Also, how does this stabilizing effect actually arise? I mean most of the sources I read simply states that it exists.

I don't think $ \text{E,Z}$ and $ \text{trans \minus{} cis}$ are the same convention because how would you name:

$ \text{(Z) \minus{} 1 \minus{} bromo \minus{} 1 \minus{} chloro \minus{} 2 \minus{} iodo \minus{} 1 \minus{} propene}$ using $ \text{cis \minus{} trans}$?

[quote="BanishedTraitor"]Is that a good enough definition?[/quote]

It doesn't have to be with a carbon-hydrogen sigma bond, it can also be with a carbon-carbon sigma bond. And the stabilizing effect refers to empty or half-filled p orbitals, not pi orbitals.

[quote="BanishedTraitor"]Also, how does this stabilizing effect actually arise?[/quote]

Most probably, it is simply an inductive effect: alkyl groups are electron donating.

[quote="BanishedTraitor"]I don't think E,Z and trans-cis are the same convention[/quote]

Z = cis and E = trans: what is the meaning of each? They mean that the groups with highest priority (according to Cahn-Ingold-Prelog rules) are in opposite sides of the double bond (trans or E) or on the same side (Z or cis).


For $ a>0,\ b>0,\ c>0,$ prove :

$ (a\plus{}b\plus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)\plus{}(a\plus{}b\plus{}c)(c\plus{}a\minus{}b)(a\plus{}b\minus{}c)$
$ \plus{}(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)\minus{}(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)\leqq (a\plus{}b)(b\plus{}c)(c\plus{}a)$

[quote="kunny"]For $ a > 0,\ b > 0,\ c > 0,$ prove :

$ (a \plus{} b \plus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b) \plus{} (a \plus{} b \plus{} c)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)$
$ \plus{} (a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a) \minus{} (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)\leqq (a \plus{} b)(b \plus{} c)(c \plus{} a)$[/quote]

$ (a \plus{} b \plus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b) \plus{} (a \plus{} b \plus{} c)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)$
$ \plus{} (a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a) \minus{} (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)\equal{}8abc\equal{}2\sqrt{ab}2\sqrt{ac}2\sqrt{bc}\leq(a\plus{}b)(a\plus{}c)(b\plus{}c)$

$ a\equal{}b\equal{}c$

I think it will work although I haven't tried yet!
Using homogenous. we can assume that a+b+c=1
Then we have:(1-2a)(1-2b)+(1-2c)(1-2a)+(1-2b)(1-2c)<=(1-c)(1-b)(1-a)+(1-2a)(1-2b)(1-2c)
After expansion, we seem to get the solution(maybe using AM-GM 1 or 2 times more). Another way is let c=min(a,b,c) and a=c+x,b=c+y and proving the coefficient>0!In general, it's ugly inequality :lol:

[quote="kunny"][quote="zaya_yc"]
$ (a \plus{} b \plus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b) \plus{} (a \plus{} b \plus{} c)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)$
$ \plus{} (a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a) \minus{} (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b) \equal{} 8abc$[/quote]

How did you calulate that?[/quote]

$ P\equal{}\frac{a\plus{}b\plus{}c}{2}$ --->

 $ (a \plus{} b \plus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b) \plus{} (a \plus{} b \plus{} c)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)\plus{}(a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a) \minus{} (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b) \equal{}8p(p\minus{}a)(p\minus{}b)\plus{}8p(p\minus{}b)(p\minus{}c)\plus{}8p(p\minus{}c)(p\minus{}a)\minus{}8(p\minus{}a)(p\minus{}b)(p\minus{}c)\equal{}$
$ \equal{}8p(p^{2}\minus{}(a\plus{}b)p\plus{}ab)\plus{}8p(p^{2}\minus{}(b\plus{}c)p\plus{}bc)\plus{}8p(p^{2}\minus{}(a\plus{}c)p\plus{}ac)\minus{}8(p^{3}\minus{}p^{2}(a\plus{}b\plus{}c)\plus{}p(ab\plus{}ac\plus{}bc)\minus{}abc)\equal{}8p(3p^{2}\minus{}2(a\plus{}b\plus{}c)p\plus{}ab\plus{}ac\plus{}bc)\minus{}8(p^{3}\minus{}2p^{3}\plus{}p(ac\plus{}ab\plus{}bc)\minus{}abc)\equal{}8p(3p^{2}\minus{}2p^{2}\plus{}ab\plus{}ac\plus{}bc)\minus{}8p^{3}\minus{}8p(ab\plus{}ac\plus{}bc)\plus{}8abc\equal{}8p^{3}\plus{}8p(ab\plus{}ac\plus{}bc)\minus{}8p^{3}\minus{}8p(ab\plus{}ac\plus{}bc)\plus{}8abc\equal{}8abc$

Thanks. Here is my solution.

Let $ a\plus{}b\minus{}c\equal{}x,\ b\plus{}c\minus{}a\equal{}y,\ c\plus{}a\minus{}b\equal{}z$, we have $ a\plus{}b\plus{}c\equal{}x\plus{}y\plus{}z.$ Thus $ L.H.S\equal{}(x\plus{}y)(y\plus{}z)(z\plus{}x)\equal{}8abc$, which follows your solution, zaya_yc. :lol:


function
LaTeX
algebra
functional equation
algebra solved

Find all funtions $ f: R_{+} \rightarrow R_{+} $ such that

$f(x+yf(x))=f(x)f(y) \quad \forall x,y \in R_{+}$

and there exist only finitely many $x \in R_{+}$ such that $f(x)=1$

[quote="grobber"]Does $R_+$ contain $0$? (It does in Romanian notation; the one without $0$ is $R_+^*$ here :))[/quote]

no, it unfortunately doesn't :)

the one including $0$ is often denoted by $R_{0}^{+}$ in italy

Assume $x_0$ is a solution to the equation $f(x)=1$. Then make $x=x_0$ in the functional equation, and get $f(x+y)=f(y),\ \forall y\in\mathbb R_+^*$ (I'll use the Romanian notation :)). This means that $f$ is a periodic function of period $x$, so $f(nx)=1,\ \forall n\in \mathbb N^*$, so there are infinitely many $x$ for which $f(x)=1$, a contradiction, so there is no solution to the eqn $f(x)=1\ (*)$.

Now assume there are $a,x>0$ s.t. $f(x+a)=f(x)\Rightarrow f(\frac a{f(x)})=1$, which contradicts $(*)$. This means that $f$ is injective.

$f(x)f(y)=f(x+yf(x))=f(y+xf(y))$ because $f(x)f(y)$ is symmetric in $x,y$. This means that $x+yf(x)=y+xf(y)\Rightarrow f(x)=kx+1$, and the conditions show that $k>0$. Conversely, each function of the form $f(x)=kx+1,\  k>0$ satisfies the conditions, so this is the answer.

Bonjour (:-)) 14 07 0409h21 Saint Jean de Trzy France

Vous trouverez votre quation fonctionnelle dans l'Enseignement mathmatique
(Revue Suisse internationale Case postale 240 CH-1211 Genve 24 Suisse)
Janvier 1964 page 147

Moubi said to me that the langage of this forum is english :
The roumanian do not understand french ?
Why I have not french flag ?

Grobber wrote : javascript:emoticon(':(')
"
Now assume there are  s.t.  f(x+a)=f(x) ====> f(a/f(x))=1"
I do not understand if f(x)=0 ?

Moubi said to me that the langage of this forum is english :
The roumanian do not understand french ?
Why I have not french flag ? How copy in text the smileys in mathlinks

You will find your functional equation in the mathematical Education (Seen again(Revised) international Switzerland POB 240 CH-1211 Geneva 24 Switzerland(Swiss)) in January, 1964 p 147

[quote="VIDIANI"]
Now assume there are  s.t.  f(x+a)=f(x) ====> f(a/f(x))=1"
I do not understand if f(x)=0 ?[/quote]

As talpuz said, $R_+$ doesn't contain $0$, and $f$ takes values in $R_+$, so..

wow!

thanks grobber for your answer

i noticed that if the function was injective the solution would be trivial, but i couldn't prove that f was injective :)

???? 'What's the sound of one hand clapping ?'
I do not understand ?

In french $R+=[0,+_infty[ $ and $R+^*=]0,+\infty[$
T+ roumanian is R+^* french

That is the international notation for $[0,+\infty[$ and $]0,+\infty[$ ????

Come on Vidiani! Go to the topic 'training in Latex' and the 'Training in English' one too  :D  :D
You may preview you post before posting by using the appropriate button (near the one to send the post). Thus, you may avoid some latex problems.

Pierre.

i think that $R_+$ is the international symbol for the set of POSITIVE reals, and $R_0^+$ denotes the set of NON-NEGATIVE reals (including $0$)

(at least, Arthur Engel agrees with me ;) )


floor function
induction
number theory unsolved
number theory

I hope you can help me with this:

Prove that for every positive integer $m$ there exists a positive integer $a$ such that $\lfloor (1+\sqrt{3})^{2m+1}\rfloor= 2^{m+1}a$, where $\lfloor . \rfloor$ is the integer part of a number.

Adriana

In other words, we want to prove that $2^{m+1}|[(1+\sqrt{3})^{2m+1}]$. Consider the sequence $(a_{n})$ given by:
$a_{0}=2, a_{1}=2$ and $a_{n+2}= 2a_{n+1}+2a_{n}$. Standard method for solving recurrence relations gives us $a_{n}=(1+\sqrt{3})^{n}+(1-\sqrt{3})^{n}$. If $n$ is odd then $-1 < (1-\sqrt{3})^{n}<0$ so we must have $[(1+\sqrt{3})^{2m+1}]=(1+\sqrt{3})^{2m+1}+(1-\sqrt{3})^{2m+1}$ (since $a_{n}$ is an integer, for every $n$). So, we are left with proving that $2^{m+1}|a_{2m+1}$. But that's easy. Using induction and the recurrence relation we can prove that $2^{[\frac{n}{2}]+1}|a_{n}$ for every $n \geq 0$.

[hide]
Consider the sequence
\[\frac{\left( 1+\sqrt{3}\right)^{2m+1}+\left( 1-\sqrt{3}\right)^{2m+1}}{2^{m+1}}\]
\[=\left( \frac{1+\sqrt{3}}{2}\right) \left( 2+\sqrt{3}\right)^{m}+\left( \frac{1-\sqrt{3}}{2}\right) \left( 2-\sqrt{3}\right)^{m}\]
(This is so because $\left( 1\pm \sqrt{3}\right)^{2}=2\left( 2\pm \sqrt{3}\right)$.) Let this sequence be $a_{m}$. Then we have
\[a_{0}=1,\ a_{1}=5,\ a_{m+2}=4a_{m+1}-a_{m}\]
(the recurrence is straightforward induction and follows from the fact that $x=2+\sqrt{3},\ y=2-\sqrt{3}$ implies $x+y=4,\ xy=1$). In particular, $a_{m}$ is integer for all nonnegative $m$, so
\[\left( 1+\sqrt{3}\right)^{2m+1}+\left( 1-\sqrt{3}\right)^{2m+1}\]
is an integer multiple of $2^{m+1}$. But also $\left( 1-\sqrt{3}\right)^{2m+1}$ lies strictly between -1 and 0, so we have
\[\left\lfloor \left( 1+\sqrt{3}\right)^{2m+1}\right\rfloor =\left( 1+\sqrt{3}\right)^{2m+1}+\left( 1-\sqrt{3}\right)^{2m+1}\]
which completes the proof. (We can also show that the integer is [i]not[/i] divisible by $2^{m+2}$ since $a_{m}$ is always odd.)
[/hide]


Find all the function $ f: \mathbb{R}\rightarrow \mathbb{R}$ such that $ f(xf(y)\plus{}f(x))\equal{}2f(x)\plus{}xy$

hi!
for $ x\equal{}1$ we have $ f(f(y)\plus{}f(1))\equal{}2f(1)\plus{}y$ ==> f is bijective.
now let $ f(0)\equal{}a$ and $ f(\minus{}1)\equal{}b$
for $ x\equal{}y$ we have: $ f(f(x)(x\plus{}1))\equal{}2f(x)\plus{}x^{2}$ $ (1)$
and for $ x\equal{}\minus{}1$ ==> $ a\equal{}2b\plus{}1$ $ (2)$

since f is surjective,so there exist $ c$ such that: $ f(c)\equal{}0$
so for $ x\equal{}c$ and $ y\equal{}0$ we have: $ f(ac)\equal{}0\equal{}f(c)$
f is injective so $ ac\equal{}c$ ==> $ a\equal{}1$ and  from $ (2)$ $ b\equal{}0$
so $ f(\minus{}1)\equal{}0$ and $ f(0)\equal{}1$.
for $ y\equal{}\minus{}1$ we have $ f(f(x))\equal{}2f(x)\minus{}x$
now if we consider a sequence $ (x_{n})$ defined by: $ x_{0}\equal{}x$ and $ f(x_{n})\equal{}x_{n\plus{}1}$
we have : $ x_{n\plus{}2}\equal{}2x_{n\plus{}1}\minus{}x_{n}$,then we solve the equation $ r^{2}\minus{}2r\plus{}1\equal{}0$.
this equation has one solution $ r\equal{}1$ so : $ x_{n}\equal{}\alpha\plus{}\beta*n$ and for $ n\equal{}1$ we have :
$ x_{1}\equal{}f(x)\equal{}x\plus{}\beta$ (because $ x_{0}\equal{}x\equal{}\alpha$).
then we put our function in the initial f.e we conclude that $ \beta\equal{}1$
so $ f(x)\equal{}x\plus{}1$.

[i]How you can get $ c \ \neq \ 0$ :maybe: [/i]

if $ f(0) \equal{} 0$ we get: $ f(f(x)) \equal{} 2f(x)$ ==> $ f(x) \equal{} 2x$ and that is wrong, so $ c\neq0$ ;)

f is bijective===> f is injective and surjective  :maybe:

Okay, I get how $ f(f(y)\plus{}f(1))\equal{}2f(1)\plus{}y$ proves its injective but not how it proves it's surjective

[quote="dgreenb801"]Okay, I get how $ f(f(y) \plus{} f(1)) \equal{} 2f(1) \plus{} y$ proves its injective but not how it proves it's surjective[/quote]
Plugging in $ y\minus{}2f(1)$ for $ y$ gives $ f(f(y\minus{}2f(1))\plus{}f(1))\equal{}y$ for all $ y$; thus, the function can take on any value.

an other proof :
we have $ f(f(y)\plus{}f(1))\equal{}2f(1)\plus{}y$
since $ h(y)\equal{}2f(1)\plus{}y$ is bijective so $ f(f(y)\plus{}f(1))$ is bijective too==> $ f$ is surjective and $ f(y)\plus{}f(1)$ is injective.
then,since $ g(y)\equal{}f(y)\plus{}f(1)$ is injective so $ f(y)$ is injective too.

conclusion: f is bijective.


[quote="lomos_lupin"]Well.This problem is [i]very[/i] well known and its more like an exercise but its good to have a rigorous proof for it.

Problem.Let $A$ and $B$ be two points. Let $P$ be a continous path between them with lenght $\ell$.
Show that
\[\ell \geq AB \]
Where AB is the lenght of the segment.

PS.Problem is easy but its not  obvious.[/quote]

Wow...i didn't even know that this could be a [i]problem[/i]! Hmmm....rigourous.....well....

Assume wlog, that $P$ is [i]"piecewise linear". [/i]Prove induction on the number of pieces $n$ :lol:

Induction between $n$ and $n+1$: use triangle inequality...and base case=1=when $P=AB$

Yeah this problem is obvious. You just have to take into account that $P$ can be a curve. Then you can look at a curve as an infinite number of line segments and then we proceed to Chen's solution. :)

Suppose we have determined $M$ to be the shortest continuous curve with endpoints $A$ and $B$, for some unequal points $A$ and $B$.  Consider the curve $M'$ with endpoints $A$ and $C$ where $C$ is any point on $M$.  By hypothesis, $M'$ is homothetic to $M$.  So $A, B, C$ are collinear.  It follows $M$ must be a line segment.

[quote] Assume wlog, that  is "piecewise linea",
[/quote]
[quote="rem"]that $P$ can be a curve. Then you can look at a curve as an infinite number of line segments [/quote]
This is not rigorous by any means.

But yet There exists a rigorous calculus base solution
But its brute force.
This result was known [i]long [/i] before calculus was invented.

Another approach, Similar to kileel's idea, is that
 you can  [i] transform [/i] your curve into a part of a cricle.


a, b, c and d  are in N.  let:

(P) :  a+b < c+d
(Q) : (a+b) (c+d) < ab + cd
(R) : (a+b) cd < ab (c+d)

prove that at least one of (P) or (Q) or (R) is false.

Bad title http://www.mathlinks.ro/Forum/viewtopic.php?t=144403 . Locked!


inequalities algebra polynomial function

Maverick's first step follows from Cauchy !
The second step follows from maclaurin's inequalities.

Do you know symmetric polynomials ? If not, let me know.

For a certain positive integer n, the symmetric polynomials of n positive numbers a(1), a(2), ..., a(n) are denoted s(1), s(2), ..., s(n).
For example, if n = 3 then we have
a(1) = (a + b + c)/3
a(2) = (ab + bc + ca)/3
a(3) = abc
Now for any n we have :
a(1) >= a(2)^(1/2) >= a(3)^(1/3) >= ... >= a(n)^(1/n).
This is Maclaurins inequality.

Try to apply this to the above posts !
If you don't know how to do this, let me know ...

[color=cyan]Maverick, it really isn't helpful at all if you don't write out full solutions.  Putting in the important steps does nothing for many of us who have less familiarity with these types of problems.[/color]

Ah, yes, I have indeed seen what you call maclaurin's inequalities before, just under a different name and have not yet had a chance to apply them.

Thanks!

I simply had to make this post!!

You know.. this is a very clear example of Jensen.
Consider WLOG that a+b+c+d=1 and apply Jensen to the convex function f: (0,+oo)-->R, f(x)=1/x with a,b,c and d as weights.

cheers! :D :D

I'll work out Lagrangia's idea ...

Since the inequality is homogenous in a, b, c and d we can assume WLOG that a + b + c + d = 1. Now notice that the function f(x) = 1/x is convex and increasing on the interval ]0,+:inf:[ so we have by Jensen's inequality :

a f(b + 2c + 3d) + b f(c + 2d + 3a) + ... + d f(a + 2b + 3c)
:ge: f(4(ab + bc + cd + da + ac + bd))
:ge: f(24(a + b + c + d)/16)
= 2/3

where 4(ab + bc + ... + bd) :ge: 24(a + b + c + d)/16 follows immediately from Maclaurin's inequalities. :)


Find all possible integral solutions $ (x,y,z)$ such that

$ x^2\plus{}y^2\plus{}z^2\plus{}2xy\plus{}2x(z\minus{}1) \plus{}2y(z\plus{}1)$ is a perfect square.

Note that $ x^2\plus{}y^2\plus{}z^2\plus{}2xy\plus{}2x(z\minus{}1)\plus{}2y(z\plus{}1)\equal{}(x\plus{}y\plus{}z)^2\plus{}2(y\minus{}x)$. If $ y>x$, $ x^2\plus{}y^2\plus{}z^2\plus{}2xy\plus{}2x(z\minus{}1)\plus{}2y(z\plus{}1)\ge(x\plus{}y\plus{}z\plus{}1)^2$, we easily get contradiction. Similarly if $ y<x$, $ x^2\plus{}y^2\plus{}z^2\plus{}2xy\plus{}2x(z\minus{}1)\plus{}2y(z\plus{}1)\le(x\plus{}y\plus{}z\minus{}1)^2$, another contradiction follows after a few manipulation. So $ x\equal{}y$ which is indeed a solution, so $ (x,y,z)\equal{}(a,a,b)$.


Note: these are by no means supposed to be standard MathCounts problems, and some of them will take well over however much time you get on a MathCounts test to solve.

[quote="happyme"]As many of you may know, CatalystofNostalgia (I hope I spelled his name correctly) had a Problem of the Day thread that ended at problem number 95. I wanted to make an archive of all the problems (some of them were really good and intuitive) and so I did. I also realize that its States/Nationals time coming up so many people are desperate for problems. I believe that if you can solve all of these problems, then you have a pretty good understanding of math. If you want, I can make another .pdf with all the answers but that might take more time.

Enjoy the problems![/quote]
Have a lot of free time eh?

Actually, it was more of an experiment for me to get used to typesetting in $ \text{\LaTeX}$ since I'll need it for USAMTS next year. Also, like Cat said, these were Problems of the DAY, so its okay if you spend a while on them.

Also, avoid using a calculator unless absolutely necessary. Also, using programs to do the work will not help you get the answer in a contest where a proctor prevents you from doing so.

err.... where can I check my answers? :P:):* :roll:

Thank you :) That was  really good idea  put here all that problems :)

I'm almost done with the Answer Key and solutions will follow...

Answer Key is attached, solutions will be posted later...

Sweet! :D Thanks!
Your hard work (both of you) is much appreciated! :)

I have updated the answer key and the problems (with Asymptote thanks to i_like_pie).


Let [tex]a,b,c[/tex] be positive number .Prove that:
[tex]6(b+c-a)(a^2+b^2+c^2)+27abc\leq 10(a^2+b^2+c^2)^{\frac32}[/tex]

[quote="ondrob"]It reminds me inequality which was not very long ago in one Slovak seminar:
For reals $a,b,c$ prove: $6(a+b+c)(a^2+b^2+c^2)\leq 27abc+10(a^2+b^2+c^2)^{\frac{3}{2}}$[/quote]
It's Vietnam 2000 and it's easy!I think my problem is harder than yours a lot!

[quote="Singular"]substitute  x = b+c-a etc.

$3x(\sum x^2 +  xy) + \frac{27}{8}(2xyz + \sum_{sym} x^2y) \leq 5(\sum x^2 + xy)^{3/2}$

x maximal wlog

maybe this help?[/quote]
I don't think so.It's hard to prove it with new variables!

I have a solution.
First, denote $f(a,b,c)=10(a^2+b^2+c^2)^{\frac{3}{2}}-6(b+c-a)(a^2+b^2+c^2)-27abc$.
WLOG we can assume a to be the smallest. (Otherwise R.H.S. remains unchanged and L.H.S. is smaller)
Now, it can be simply shown that
$f(a,b,c) \ge f(a,t,t)$ where $t=\sqrt{\frac{b^2+c^2}{2}}$.
Proof:
\[
\begin{array}{l}
 f(a,b,c) = \left( {a^2  + b^2  + c^2 } \right)\left( {10\left( {a^2  + b^2  + c^2 } \right)^{\frac{1}{2}}  - 6(b + c - a)} \right) - 27abc \\
  = \left( {a^2  + 2t^2 } \right)\left( {10\left( {a^2  + 2t^2 } \right)^{\frac{1}{2}}  - 6(b + c - a)} \right) - 27abc \\
  \ge \left( {a^2  + 2t^2 } \right)\left( {10\left( {a^2  + 2t^2 } \right)^{\frac{1}{2}}  - 6(2t - a)} \right) - 27at^2  = f(a,t,t) \\
 \end{array}
\]
Now, it remains a two-variable inequality with $t \ge a$ (because a is the smallest)
\[
(a^2  + 2t^2 )(10\sqrt {a^2  + 2t^2 }  - 6(2t - a)) \ge 27at^2
\]
Substitute $x=\frac{t}{a}$ (so $x \ge 1$), then
\[
(1 + 2x^2 )(10\sqrt {1 + 2x^2 }  - 6(2x - 1)) \ge 27x^2
\]
To prove this, first we use:
\[
9(1 + 2x^2 ) = 9(1 + 8\left( {\frac{x}{2}} \right)^2 ) \ge \left( {1 + 8\left( {\frac{x}{2}} \right)} \right)^2  = (1 + 4x)^2
\]
Then we have to prove
\[
(1 + 2x^2 )(\frac{{10}}{3}(1 + 4x) - 6(2x - 1)) \ge 27x^2
\]
After expanding, it is equal to $8x^3-25x^2+4x+28 \ge 0$ for $x \ge 1$.
Put $x=2y$ and we have to prove $64y^3+8y+28 \ge 100y^2$. But this directly follows from weighted amgm: $64y^3+8y+28 \ge 100 y^{\frac{3(64)+8}{100}}=100y^2$.
This finishes our proof, and equal holds when x=2 or t=2a or b=c=2a.

Oh,Really,my problem is weaker than Ondrob's one :Let a'=-a,b'=b,c'=c(because a,b,c are real numbers) and we have done.

No,my problem for a,b,c be positive numbers .So mine is weaker than ondrob's one
Here is my solution for ondrob's one:
Assume [tex]a^2+b^2+c^2=9[/tex](because it's homogeneos) and /c/=max {/a/,/b/,/c/} then [tex]c^2\geq 3[/tex].
Given ineq becomes:
[tex]2(a+b+c)\leq abc+10 [/tex]
[tex]<->(2(a+b+c)-abc)^2\leq 100[/tex]
By Cauchy schwart we have:
[tex](2(a+b)+(2-ab)c)^2\leq (8+(ab)^2-4ab)(9+2ab)=100+(2ab-7)(ab+2)^2[/tex](check it)
But[tex] 2ab\leq a^2+b^2=9-c^2\leq 6<7[/tex]
And we have done

[quote="keira_khtn"][quote="zhaobin"]yeah,you are right.
but your inequality can be real numbers.
BTW,nice solution.[/quote]
I don't think so...a,b,c have to be positive[/quote]
why?
your solution is okay for reals

Here is my solution for the first one:The given inequal is equivalent to:
[tex](a^2+b^2+c^2)(10(a^2+b^2+c^2)^{\frac12}-6(b+c-a))\geq 27abc[/tex].Notice:the equal occurs iff [tex]b=c=2a[/tex] then we wrote:[tex]\frac{10}{3}\sqrt{(a^2+b^2+c^2)(1+2^2+2^2)}-6(b+c-a)\geq \frac{10}{3}(a+2b+2c)-6(b+c-a)=\frac{28}{3}a+\frac{2}{3}b+\frac23 c[/tex]
Applying AM-Gm we obtain:[tex]a^2+4.\frac{b^2}{4}+4.\frac{c^2}{4}\geq \frac{9}{4^{\frac89}}a^{\frac29}b^{\frac89}c^{\frac89}(1)[/tex]
[tex]\frac{28}{3}a+\frac{2}{3}b+\frac23c=7.\frac43a+\frac{2}{3}b+\frac23c\geq 3.4^{\frac89}a^{\frac79}b^{\frac19}c^{\frac19}(2)[/tex]
Multiping (1) and (2) it follows!
And we have to have a,b,c>0 to use Am-Gm